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https://plainmath.net/secondary/geometry | High school geometry questions and answers
Recent questions in Geometry
Minkowski difference of two convex polygonsI just want to make sure that the following algorithm is correct for computing the Minkowski difference of two shapes A,B:Where CH(S) is the convex hull of the set S and a,b are the vertices of the two polygons.
Sara Fleming 2022-09-27
How to recognize what type of probability distribution to use in solving probability problems?Bernoulli's, Binomial, Geometric, Hypergeometric, Negative binomial, Poisson's, Uniform, Exponential, Normal, Gamma, Beta, Chi square, Student's distribution.I would like to know how and when to use each of these distributions when solving problems in probability. If possible, make analogy with combinatorics (when we use permutations, variations and combinations).
mydaruma25 2022-09-27
A certain discrete random variable has probability generating function: ${\pi }_{x}\left(q\right)=\frac{1}{3}\frac{2+q}{2-q}$Compute . (Hint: the formula for summing a geometric series will help you expand the denominator)."
Wevybrearttexcl 2022-09-26
Distance between (-2,1,3) and (-1,4,-2)?
Medenovgj 2022-09-26
What is the distance between (3,-14,15) and (12,-21,16)?
koraby2bc 2022-09-26
Distance between (11,-13,-5) and (9,-14,4)?
Kody Whitaker 2022-09-26
What is the distance between (-11,-11) and (21,-22)?
Averi Fields 2022-09-26
Let ABC be a triangle and $\mathrm{\Omega }$ be its circumcircle, the internal bisectors of angles A, B, C intersect $\mathrm{\Omega }$ at ${A}_{1},{B}_{1},{C}_{1}$. The internal bisectors of ${A}_{1},{B}_{1},{C}_{1}$ intersect Omega ${A}_{2},{B}_{2},{C}_{2}$. If the smallest angle of $\mathrm{△}ABC$ is 40 degrees, find the smallest angle of $\mathrm{△}{A}_{2}{B}_{2}{C}_{2}$.
Marcus Bass 2022-09-26
Geometric probabilityInside a square of side 2 units , five points are marked at random. What is the probability that there are at least two points such that the distance between them is at most $\sqrt{2}$ units?
Medenovgj 2022-09-26
Let $Q\left(z\right)=\left(z-{\alpha }_{1}\right)\cdots \left(z-{\alpha }_{n}\right)$ be a polynomial of degree $>1$ with distinct roots outside the real line.We have$\sum _{j=1}^{n}\frac{1}{{Q}^{\prime }\left({\alpha }_{j}\right)}=0.$Do we have a proof relying on rudimentary techniques?
gaby131o 2022-09-25
Probability of Observing N particles in a given volume?I'm having an issue with a probability problem concerning solutions.Assume there is an "observational region" in a dilute solution with a volume V, and as solutes move across its boundary, the number N of solute molecules inside the observation region fluctuates.Divide V into M regions of volume v each with n particles. The solution is dilute enough that $n=0$ or 1 (there is no v with more than one particle of solute), and each cell is occupied ($n=1$) with probability $p=\left({\rho }_{0}\right)v$.If W(N) is the number of configurations of the observation volume when N solutes are present, what is the probability P(N) of observing a given value of N, in terms of p,W(N),M, and N.I know the probability $P\left({n}_{1},{n}_{2},\dots ,{n}_{M}\right)$ of finding the system in a particular configuration in the observation volume is $p\left(N\right)={p}^{N}\left(1-p{\right)}^{M-N},$, (Bernoulli Distribution), and since there are N particles in M spaces then the maximum number of configurations is $\frac{M!}{\left(N!\left(M-N\right)!\right)}$I'm not sure where to go from here.
Harrison Mills 2022-09-25
Distance between (3,-1,1) and (4,1,-3)?
tarjetaroja2t 2022-09-25
Gram Determinant equals volume?I have been trying to solve this problem of finding the 'n-volume' of a paralleletope spanned by m vectors, where clearly $m\le n$. In general, for computational purposes, what I have managed to do is define volume as the product of absolute values of vectors obtained by gram-schmidt orthogonalizationn. (Makes sense right? That's the natural interpretation when we say volume)I had to do two things, firstly to show that this definition of volume is a well defined one (i.e. any set of orthogonal vectors obtained by the process will give the same volume), and secondly to find a quick way to do this. I managed to prove the first one by induction, but the second part is a little bit of a problem. I managed to obtain formulae for small dimensions as 2,3 or even 4 but this process is impractical for any bigger dimensions as the substitutions for smaller dimensions into the formula for the next dimension becomes exponentially complicatedHow does one prove that the gram determinant is equal to the volume of a paralleletope spanned by a set of vectors?
Melina Barber 2022-09-25
Suppose we have the following:Can this be proven without making assumptions for conditional or indirect proofs?
deiluefniwf 2022-09-25
You have a quadrilateral ABCD. I want to find all the points x inside ABCD such that$angle\left(A,x,B\right)=angle\left(C,x,D\right)$Is there a known formula that gives these points ?
basaltico00 2022-09-25 | 2022-09-29 13:42:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 30, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8082137703895569, "perplexity": 447.5247461322236}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335355.2/warc/CC-MAIN-20220929131813-20220929161813-00267.warc.gz"} |
https://www.commonlounge.com/javascript-loops-break-and-continue-997305fa58194089b4a04b8c35e96c2c | # JavaScript: Loops, break and continue
September 26, 2018
In this tutorial you’ll learn about loops in JavaScript, one of the most important concepts in programming. In plain English, loop-statements correspond to “Do this action X times.” or “Keep doing that action as long as this condition is true.”
You’ll also learn about two important keywords break and continue. They help in controlling the flow of the program by jumping to statements, skipping certain statements etc.
# for loops
Programmers don’t like to repeat themselves. Programming is all about automating things. Let’s say we have an array of names of people we wish to greet. We can write something like this:
var names = ['Alan', 'Bob', 'Alice'];
console.log("Hello, " + names[0]);
console.log("Hello, " + names[1]);
console.log("Hello, " + names[2]);
Output:
Hello, Alan
Hello, Bob
Hello, Alice
But we don’t want to greet every person by their name manually, right? That’s where loops come in handy.
for (var i = 0; i < names.length; i++) {}
This is a for loop. A for statement behaves the same way as the if statement with a difference that it keeps executing its following code block until the condition remains true. The above statement evaluates to
for (var i = 0; i < 3; i++) {}
because, names.length is equal to 3.
A for loop has three parts separated by semicolons.
1. Initialization — In our example, the variable i is initialized with value 0.
2. Checking a condition — Then we check if i < 3 is true.
3. Update — If it is true, we execute the following code block and increment i by 1. i++ increments i by 1 and is shorthand for i = i + 1 or i += 1.
The letters i, j, k are generally used for loop variables (it’s kind of a convention).
Let’s see what all values can i take when this loop is executed.
• i = 0; i is less than 3; i is incremented by 1.
• i = 1; i is still less than 3; i is incremented by 1.
• i = 2; i is still less than 3; i is incremented by 1.
• i = 3; i is equal to 3; the condition is false; the loop ends.
So i can take the values 0, 1, 2. We can traverse through our names array using i!
var names = ['Alan', 'Bob', 'Alice'];
for (var i = 0; i < names.length; i++) {
console.log("Hello, " + names[i]);
}
Output:
Hello, Alan
Hello, Bob
Hello, Alice
You can also fill an array with elements using a loop:
var numbers = new Array(100); // creates an empty array with 100 elements
for (var i = 0; i < 100; i++) {
numbers[i] = i + 1; // this will put 1 at index 0, 2 at 1, 3 at 2 and so on.
}
console.log(numbers);
Output:
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100
This populates the numbers array with natural numbers from 1 to 100.
How to traverse objects?
var obj = {"name": "Commonlounge", "type": "website"};
for (var key in obj) {
console.log(key + ": " + obj[key]);
}
Here, the loop variable is key, and it takes the values "name" and "type".
Output:
name: Commonlounge
type: website
# while loops
In for loops we are certain about the number of times we wish to run the loop. But, what if we aren’t?
Consider, this example. Suppose we want to know how many times a given number can be divided by 2 before it is less than or equal to 1. Yes, we can do the math to figure out the answer, but we are programmers and programmers are lazy. So, we use loops. In this section, we will look at while loops. Before we continue loops, a quick note:
Note: To make things more interesting, from this point on, we will incorporate our Javascript examples in HTML. Don’t worry if you don’t know anything about HTML yet. There are just a couple things to keep in mind:
1. We will use document.write() function, instead of console.log to print out the values we care about. document.write is a special function that lets us write to the HTML document.
2. We will use prompt() function to capture the input from the user.
For example, to take the number as input from the user. Remember the prompt function?
var i = prompt("Enter a number");
That’s it.
Coming back to loops. For a while loop, we initialize the loop variable outside the loop, unlike the for loop. Here, our loop variable is the number itself, because we are going to divide it by 2 in every loop iteration.
As we say above, the structure of a while loop is simple:
while (<condition is true>) {
// codeblock
}
We want to know the number of times the given number can be divided by 2, so we make another variable called counter, which stores the answer.
Are you excited to try your first while loop? Hit the Try it now button below and see what happens!
<html>
<script>
var i = prompt("Enter a number");
var counter = 0;
while (i > 1) {
i /= 2; // same as i = i / 2
counter++;
document.write("Number is " + i + " after " + counter + " divisions", "<br>");
}
document.write("Answer = " + counter, "<br>");
</script>
</html>
If you give input 10, you will see the following output:
Number is 5 after 1 divisions
Number is 2.5 after 2 divisions
Number is 1.25 after 3 divisions
Number is 0.625 after 4 divisions
Answer = 4
Did you notice the operator /= in line 7 of the code? This is called an augmented assignment. It is equivalent to
i = i / 2;
Similar to this, we have
i += 2; // same as i = i + 2
i -= 2; // same as i = i - 2
i *= 2; // same as i = i * 2
i %= 2; // same as i = i % 2
i **= 2; // same as i = i ** 2
etc., every operator can be used as described above.
# Writing a while loop as a for loop
You can always convert a while loop to for loop and vice-versa. For example, the while loop example we saw can be converted to a for loop as follows:
<html>
<script>
var i = prompt("Enter a number");
var counter = 0;
for (; i > 1; i /= 2) {
counter++;
}
document.write("Answer = " + counter, "<br>");
</script>
</html>
Notice, that since we have already initialized the loop variable outside of the loop, we do not need to do it again inside the loop (if we do, it will change the value of i to the value we set inside the loop).
We can also write this as:
<html>
<script>
var counter = 0;
for (var i = prompt("Enter a number"); i > 1; i /= 2) {
counter++;
}
document.write("Answer = " + counter, "<br>");
</script>
</html>
This resembles more with the structure we discussed first.
# Exercise: Writing a for loop as a while loop
Convert the for-loop in the following code into a while-loop:
<html>
<script>
var numbers = [1, 2, 3, 4, 5, 6];
for (var i = 0; i < numbers.length; i++) {
var square = numbers[i] * numbers[i];
document.write("The square of " + numbers[i] + " is " + square, "<br>");
}
</script>
</html>
Please try the exercise above yourself, before looking at the solution below!
<html>
<script>
var numbers = [1, 2, 3, 4, 5, 6];
var i = 0
while (i < numbers.length) {
var square = numbers[i] * numbers[i];
document.write("The square of " + numbers[i] + " is " + square, "<br>");
i++;
}
</script>
</html>
# Break and continue
break and continue give us some more control over what code is executed inside a loop.
If a break is encountered, JavaScript immediately stops iterating over the loop (it also ignores all the code after the break within the block, if any). Here’s an example; save the following code in "loops.html":
<html>
<script>
var a = [1, 5, 3, 2, 4, 6];
for (var i = 0; i < a.length; i++) {
document.write(a[i], "<br>");
if (a[i] === 2)
break; // quit the loop when a[i] === 2
document.write(-a[i], "<br>"); // when element === 2, this line too won't execute
}
</script>
</html>
Here’s the code output:
1
-1
5
-5
3
-3
2
Note: 4 and 6 don’t get printed at all. 2 gets printed, but -2 does not, since break happens before that.
If a continue is encountered, JavaScript ignores the code after the continue within the block, but “continues” iterating over the rest of the elements. Here’s an example:
<html>
<script>
var a = [1, 5, 3, 2, 4, 6];
for (var i = 0; i < a.length; i++) {
document.write(a[i], "<br>");
if (a[i] <= 2)
continue; // don't execute the rest if element <= 2
document.write(-a[i], "<br>"); // when element <= 2, this line too won't execute
}
</script>
</html>
Here’s the code output:
1
5
-5
3
-3
2
4
-4
6
-6
Note: All numbers get printed. However, for 1 and 2, the negative values don’t get printed because of the continue.
Lastly, break and continue can be used inside while-loops as well.
# Summary
In the last few exercises you learned about:
• loops – you learnt two types of loops, the for-loop and the while-loop
• break and continuebreak can be used to exit the loop, and continue can be used to skip some parts of the loop body
Quiz time! | 2022-08-12 17:40:22 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2982959747314453, "perplexity": 1491.5785714499357}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571745.28/warc/CC-MAIN-20220812170436-20220812200436-00422.warc.gz"} |
https://solvedlib.com/n/12-8-pts-calculate-ag-for-the-reaction-at-25-0-c-2-fe-s,1827289 | # 12. (8 pts) Calculate AG? for the reaction at 25.0 %C 2 Fe(s) 3Clz(g) v> 2FeClz(s)Species Fels) Clzlg) FeClzAH?f (kJ/mol)s?
###### Question:
12. (8 pts) Calculate AG? for the reaction at 25.0 %C 2 Fe(s) 3Clz(g) v> 2FeClz(s) Species Fels) Clzlg) FeClz AH?f (kJ/mol) s? (Umolk) 27.78 223.08 142.3 3344.0
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https://blog.rexdf.org/2013/03/codeforces-175-a/ | # Codeforces 175 A
A. Slightly Decreasing Permutations
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Permutation p is an ordered set of integers p1, p2, …, pn, consisting of n distinct positive integers, each of them doesn’t exceed n. We’ll denote the i-th element of permutation p as pi. We’ll call number n the size or the length of permutation p1, p2, …, pn.
The decreasing coefficient of permutation p1, p2, …, pn is the number of such i (1 ≤ i < n), that pi > pi + 1.
You have numbers n and k. Your task is to print the permutation of length n with decreasing coefficient k.
Input
The single line contains two space-separated integers: n, k (1 ≤ n ≤ 105, 0 ≤ k < n) — the permutation length and the decreasing coefficient.
Output
In a single line print n space-separated integers: p1, p2, …, pn — the permutation of length n with decreasing coefficient k.
If there are several permutations that meet this condition, print any of them. It is guaranteed that the permutation with the sought parameters exists.
Sample test(s)
input
5 2
output
1 5 2 4 3
input
3 0
output
1 2 3
input
3 2
output
3 2 1
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int main()
{
int n,k;
cin>>n>>k;
cout<<(k+1);
for(int i=k;i>0;i--)
cout<<" "<<i;
for(int i=k+2;i<=n;i++)
cout<<" "<<i;
cout<<endl;
return 0;
}
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 2018-06-22 03:37:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4421428442001343, "perplexity": 1206.8991266891085}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267864343.37/warc/CC-MAIN-20180622030142-20180622050142-00377.warc.gz"} |
https://www.physicsforums.com/threads/urgent-help-in-friction.53637/ | # Urgent Help in Friction
1. Nov 22, 2004
### primarygun
Imagine there is an object on the surface.
The surface is uniform smooth(with same friction over the all surface and friction is greater than 0).
Let's define the friction be f.
Act a f force to horizontally to the object continuously, it is continuous added.
It is obviously stationary.
I'm quite confused of here then. If there is 1N force pushed to it in the same direction to the f force added by hand.
What will be observed?
a.Move with acceleration first, then remains the velocity later on.
b.1.If f>kinetic friction, then the object accelerates first then decelerates to stop.
b.2. If f> kinetic friction, then the object will accelerates continuously forward.
Which one is correct?
2. Nov 22, 2004
### maverick280857
Okay I think I see what you mean but you've not worded it correctly. First off, the friction coefficient is defined differently for zero and nonzero relative motion. In the former case, it is called the coefficient of static friction $$\mu_{s}$$ and in the latter, the coefficient of kinetic friction $$\mu_{k}$$. Usually $$\mu_{k} < \mu_{s}$$.
For the simple case described by you, the actual static frictional force f and the maximum possible static frictional force are related by the inequality,
$$f \leq f_{s,max} = \mu_{s}N$$
where N (=mg for zero vertical motion of the block) is the normal reaction on the block.
You need to know that for as long as f is less than fsmax, friction is a readjusting force..it equals itself to the applied force so that the net force on the body is zero and there is zero relative motion. Now if F is the applied force which is gradually increased from 0 to f, no motion will occur. At F = f, motion is "just" about to occur. As soon as F > f however, motion occurs but now f equals $$f_{k} = \mu_{k}N$$ which is less than its original value (and therefore also less than F). Hence there is a net force on the body.
This explanation can be made clearer if you draw a graph of the friction force versus applied force. At F = fsmax, there is a kink in the graph (which consists of a straight line of slope = 1 for F < fsmax and a horizontal line for F > fsmax) which corresponds to this dynamic switch from static to kinetic friction (in mathematical terminology the frictional force is not a continuous function of applied force). This graph of course can be obtained through experiment but since it holds for most simple pairs of surfaces you can find it in your physics textbook.
Now attempt your question again. Hope that helps...
Cheers
Vivek
3. Nov 22, 2004
### primarygun
b2? right?
4. Nov 22, 2004
### maverick280857
Yup. Good luck! :-)
5. Nov 23, 2004
### primarygun
How can we prove the presence of kinetic friction?
6. Nov 27, 2004
### maverick280857
Well its not something that can be proved mathematically but it can be proved logically using an argument such as the one that follows:
Suppose we exert a force F on an object that is resting on a surface whose coefficient of static friction is known (perhaps through tables or a measurement carried out using more advanced methods than the simple theoretical ones). As mentioned earlier the body will not move for as long as F is less than the maximum static frictional force. Now if F is gradually increased to a point where it "just" exceeds the maximum static frictional force, motion "just" begins and if the force F is kept constant in magnitude and direction, the acceleration vector of the body is constant (assuming no nonlinear behavior of the friction force). The net acceleration of the body is given by
$$a_{net} = \frac{F_{net}}{M}$$
and the acceleration that the applied force F would alone impart the body is
$$a_{F} = \frac{F}{M}$$
If $$a_{F} < a_{net}$$ (as will be the case) there must be some force which reduces the effect of F and that force can be safely thought of as the kinetic frictional force. Upon subsequent measurements you will find that it is less (usually) than the static frictional force.
Again, this "argument" depends on a LOT of assumptions some of which are implicit and so cannot be taken as an experimental or theoretical proof of friction, which is already known to exist. What you can prove however, through this approach is that kinetic friction exists after motion starts and not before it. And taking this a bit further you can plot the graph I had mentioned earlier.
Hope that helps...
Cheers
Vivek | 2018-03-18 08:32:36 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6975638270378113, "perplexity": 439.4914589637537}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257645550.13/warc/CC-MAIN-20180318071715-20180318091715-00303.warc.gz"} |
https://encyclopediaofmath.org/wiki/Nottingham_group | # Nottingham group
The subgroup of automorphisms of the field of formal power series $\mathbf{F_p}[[t]]$ consisting of those automorphisms of the form $f(t) \mapsto f(g(t))$ where $g(t) \in t + O(t^2)$.
It is a finitely generated, just infinite pro-$p$ group of finite width. However, every separable pro-$p$ group can be embedded in it as a closed subgroup. | 2023-02-05 05:05:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9831767678260803, "perplexity": 157.9296785264544}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500215.91/warc/CC-MAIN-20230205032040-20230205062040-00600.warc.gz"} |
https://www.qpute.com/2019/08/09/focus-entangling-photon-sources-on-a-tiny-bridge-via-qpute-com/ | /Focus: Entangling Photon Sources on a Tiny Bridge (via Qpute.com)
Focus: Entangling Photon Sources on a Tiny Bridge (via Qpute.com)
Physics 12, 90
Researchers entangled a pair of atomic-scale light emitters in a micrometer-scale device, which could potentially be useful for quantum communication and cryptography.
For quantum information technologies such as quantum cryptography to become practical, they will need to be implemented with miniaturized devices. Quantum entanglement is a key ingredient for these technologies, and a team of researchers has now demonstrated a chip-based structure that can reliably produce entanglement between a pair of photon-emitting sites in a crystal. Such a device might find uses in long-distance optical quantum-communication networks and quantum computing circuits.
Quantum entanglement, where two or more objects have interdependent quantum states, is central to quantum information processing. A pair of quantum-scale objects can, in principle, be entangled if they can emit identical photons. If it’s impossible to tell which of the pair is the source of a detected photon, then the emitters are entangled. But creating quantum emitters that produce identical photons is difficult with miniaturized devices, because small, uncontrollable variations in the microscopic structure can shift the emission frequencies. Quantum emitters often need to be individually “tuned” to compensate for such shifts.
One promising type of emitter can be created within a diamond crystal when two neighboring carbon atoms in the crystal lattice are replaced by a single silicon atom (called a silicon vacancy, or SiV, center). If excited by the absorption of a photon, the SiV center will emit a red photon. Typically, however, emission from multiple SiV centers in a diamond crystal spans a range of wavelengths because each silicon atom’s environment is slightly different from the others.
Marko Lončar of Harvard University and his co-workers have shown previously that the emission wavelength can be adjusted for an individual SiV center by deforming the diamond lattice in its vicinity (1, 2). The Harvard team has now applied this “strain tuning” principle to two separate SiV centers within the same diamond microstructure. The technique brings the two emission wavelengths into concordance so that photons emitted by the two sources are indistinguishable. In the new device, the SiV centers sit within a horizontal bar, or beam, of diamond several tens of micrometers (
$𝜇\text{m}$
) long and about 1
$𝜇\text{m}$
wide, suspended from supports at both ends. Gold electrodes deposited on one end of the beam and on the substrate below it can be charged by applying voltage, so that they attract one another, bending the beam slightly.
The researchers implanted silicon atoms, forming SiV centers at particular locations in the diamond beams using microlithographic methods. They then excited two of these, about 30
$𝜇\text{m}$
apart, using laser light delivered from above the device. The diamond beam trapped the photons emitted by the excited SiV centers and guided them into an optical fiber for detection.
As the researchers adjusted the voltage between the electrodes, the emission wavelength of the closest SiV center changed smoothly. But the second center experienced little bending, and so its emission wavelength remained constant. At a certain voltage, the two wavelengths were essentially identical, so that the emitting SiV centers could become entangled.
To detect that entanglement, “we exploit the fact that the entangled state produces a photon faster than you would expect from a system with two independently excited atoms, a phenomenon called superradiance,” explains Bart Machielse, a graduate student on the Harvard team. “This means we can indirectly probe the creation of the entangled state by looking at the rate at which we get a second photon after we get the first.”
The researchers say that their structure could act as a component of a quantum repeater, a system that allows photonic signals to be transmitted over long distances. Repeaters use a series of intermediate entangled pairs to allow entanglement of widely separated quantum objects.
“This is a very important achievement, as it allows one to bring and keep in resonance two SiV centers,” which are good candidates for making quantum bits (qubits), says quantum optics specialist Christoph Becher of the Saarland University in Germany. Although strain tuning of solid-state emitters is not a new idea, this work shows the first application to entanglement, he says.
In principle, says Lončar, the same approach could work with many more than two photons, although Becher cautions that this vision for a quantum optical technology will require much more engineering work. Still, Lončar likes to think big. “My dream is to have an array of, say, 100 qubits on the same chip, each on a beam with separate control,” he says. “Then we push a button, and each beam deflects just enough to bring them into resonance. That would be amazing.”
This research is published in Physical Review X.
–Philip Ball
Philip Ball is a freelance science writer in London. His latest book is How To Grow a Human (University of Chicago Press, 2019).
References
1. Y. Sohn et al., “Controlling the coherence of a diamond spin qubit through its strain environment,” Nat. Commun. 9, 2012 (2018).
2. S. Meesala et al., “Strain engineering of the silicon-vacancy center in diamond,” Phys. Rev. B 97, 205444 (2018).
Related Articles
This is a syndicated post. Read the original post at Source link . | 2019-08-19 16:23:30 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.39749762415885925, "perplexity": 1863.7971226898094}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027314852.37/warc/CC-MAIN-20190819160107-20190819182107-00213.warc.gz"} |
https://www.iacr.org/cryptodb/data/paper.php?pubkey=29970 | ## CryptoDB
### Paper: Synchronous Consensus with Optimal Asynchronous Fallback Guarantees
Authors: Erica Blum Jonathan Katz Julian Loss DOI: 10.1007/978-3-030-36030-6_6 Search ePrint Search Google Typically, protocols for Byzantine agreement (BA) are designed to run in either a synchronous network (where all messages are guaranteed to be delivered within some known time $\varDelta$ from when they are sent) or an asynchronous network (where messages may be arbitrarily delayed). Protocols designed for synchronous networks are generally insecure if the network in which they run does not ensure synchrony; protocols designed for asynchronous networks are (of course) secure in a synchronous setting as well, but in that case tolerate a lower fraction of faults than would have been possible if synchrony had been assumed from the start.Fix some number of parties n, and $0< t_a< n/3 \le t_s < n/2$. We ask whether it is possible (given a public-key infrastructure) to design a BA protocol that is resilient to (1) $t_s$ corruptions when run in a synchronous network and (2) $t_a$ faults even if the network happens to be asynchronous. We show matching feasibility and infeasibility results demonstrating that this is possible if and only if $t_a + 2\cdot t_s < n$.
##### BibTeX
@article{tcc-2019-29970,
title={Synchronous Consensus with Optimal Asynchronous Fallback Guarantees},
booktitle={Theory of Cryptography},
series={Lecture Notes in Computer Science},
publisher={Springer},
volume={11891},
pages={131-150},
doi={10.1007/978-3-030-36030-6_6},
author={Erica Blum and Jonathan Katz and Julian Loss},
year=2019
} | 2022-11-28 20:57:03 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5517450571060181, "perplexity": 2912.401027956431}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710662.60/warc/CC-MAIN-20221128203656-20221128233656-00281.warc.gz"} |
https://www.maplesoft.com/support/help/maple/view.aspx?path=OrthogonalSeries/Multiply&L=E | Multiply - Maple Help
OrthogonalSeries
Multiply
multiply two series
Calling Sequence Multiply(S1, S2)
Parameters
S1 - finite orthogonal series S2 - orthogonal series
Description
• The Multiply(S1, S2) function multiplies the series S2 by the finite series S1. The routine uses an adaptation of the Horner scheme.
• The result is expanded in terms of the polynomials of S2.
• This command is part of the OrthogonalSeries package, so it can be used in the form Multiply(..) only after executing the command with(OrthogonalSeries). However, it can always be accessed through the long form of the command by using OrthogonalSeries[Multiply](..).
Examples
> $\mathrm{with}\left(\mathrm{OrthogonalSeries}\right):$
> $\mathrm{S1}≔\mathrm{ChangeBasis}\left(1+{x}^{2},\mathrm{LaguerreL}\left(n,\frac{1}{3},x\right)\right)$
${\mathrm{S1}}{≔}\frac{{37}{}{\mathrm{LaguerreL}}{}\left({0}{,}\frac{{1}}{{3}}{,}{x}\right)}{{9}}{-}\frac{{14}{}{\mathrm{LaguerreL}}{}\left({1}{,}\frac{{1}}{{3}}{,}{x}\right)}{{3}}{+}{2}{}{\mathrm{LaguerreL}}{}\left({2}{,}\frac{{1}}{{3}}{,}{x}\right)$ (1)
> $\mathrm{S2}≔\mathrm{Create}\left(u\left(m\right),\mathrm{Charlier}\left(m,\frac{1}{7},x\right)\right)$
${\mathrm{S2}}{≔}{\sum }_{{m}{=}{0}}^{{\mathrm{\infty }}}{}{u}{}\left({m}\right){}{\mathrm{Charlier}}{}\left({m}{,}\frac{{1}}{{7}}{,}{x}\right)$ (2)
> $\mathrm{Multiply}\left(\mathrm{S1},\mathrm{S1}\right)$
$\frac{{4225}{}{\mathrm{LaguerreL}}{}\left({0}{,}\frac{{1}}{{3}}{,}{x}\right)}{{81}}{-}\frac{{3892}{}{\mathrm{LaguerreL}}{}\left({1}{,}\frac{{1}}{{3}}{,}{x}\right)}{{27}}{+}\frac{{532}{}{\mathrm{LaguerreL}}{}\left({2}{,}\frac{{1}}{{3}}{,}{x}\right)}{{3}}{-}{104}{}{\mathrm{LaguerreL}}{}\left({3}{,}\frac{{1}}{{3}}{,}{x}\right){+}{24}{}{\mathrm{LaguerreL}}{}\left({4}{,}\frac{{1}}{{3}}{,}{x}\right)$ (3)
> $\mathrm{SimplifyCoefficients}\left(\mathrm{Multiply}\left(\mathrm{S1},\mathrm{S2}\right),\mathrm{collect},u\right)$
$\left(\frac{{57}{}{u}{}\left({0}\right)}{{49}}{-}\frac{{9}{}{u}{}\left({1}\right)}{{7}}{+}{2}{}{u}{}\left({2}\right)\right){}{\mathrm{Charlier}}{}\left({0}{,}\frac{{1}}{{7}}{,}{x}\right){+}\left(\frac{{134}{}{u}{}\left({1}\right)}{{49}}{-}\frac{{9}{}{u}{}\left({0}\right)}{{49}}{-}\frac{{46}{}{u}{}\left({2}\right)}{{7}}{+}{6}{}{u}{}\left({3}\right)\right){}{\mathrm{Charlier}}{}\left({1}{,}\frac{{1}}{{7}}{,}{x}\right){+}\left({\sum }_{{m}{=}{2}}^{{\mathrm{\infty }}}{}\left(\left(\frac{{57}}{{49}}{+}\frac{{4}}{{7}}{}{m}{+}{{m}}^{{2}}\right){}{u}{}\left({m}\right){+}\frac{{u}{}\left({m}{-}{2}\right)}{{49}}{+}\left(\frac{{5}}{{49}}{-}\frac{{2}{}{m}}{{7}}\right){}{u}{}\left({m}{-}{1}\right){+}\left({-}\frac{{23}}{{7}}{}{m}{-}\frac{{9}}{{7}}{-}{2}{}{{m}}^{{2}}\right){}{u}{}\left({m}{+}{1}\right){+}\left({{m}}^{{2}}{+}{3}{}{m}{+}{2}\right){}{u}{}\left({m}{+}{2}\right)\right){}{\mathrm{Charlier}}{}\left({m}{,}\frac{{1}}{{7}}{,}{x}\right)\right)$ (4)
> $\mathrm{Multiply}\left(\mathrm{S2},\mathrm{S2}\right)$ | 2022-05-28 18:08:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 10, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9686163067817688, "perplexity": 1567.2976898877996}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652663016949.77/warc/CC-MAIN-20220528154416-20220528184416-00473.warc.gz"} |
https://dynasor.materialsmodeling.org/tutorials/time_convergence.html | # Time convergence¶
Here, a short demonstration on how time-correlation typically converges with total simulation time is provided for FCC Al at $$T =$$ 300 K. To get a converged correlation function, e.g. in order to extract a phonon lifetime, one require on the order 100-1000 times longer simulation than the lifetime itself. Rather than running a single long MD simulation one can also many shorter simulations and average their correlation functions.
In the figure below, the decay time of the correlation function is about 1-2 ps. The $$C_L(q, t)$$ is pretty well-converged after 200 ps and very well converged after 2000 ps.
Current correlation $$C_L(q, t)$$ for simulations of varying lengths. | 2023-02-06 23:27:54 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8808660507202148, "perplexity": 1275.6256844504214}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500365.52/warc/CC-MAIN-20230206212647-20230207002647-00828.warc.gz"} |
https://en.x-mol.com/paper/article/1356347058825838592 | Find Paper, Faster
Example：10.1021/acsami.1c06204 or Chem. Rev., 2007, 107, 2411-2502
On topological Hochschild homology of the K(1)‐local sphere
Journal of Topology (IF1.582), Pub Date : 2021-02-01, DOI: 10.1112/topo.12182
Gabriel Angelini‐Knoll
We compute mod $( p , v 1 )$ topological Hochschild homology of the connective cover of the $K ( 1 )$‐local sphere spectrum for all primes $p ⩾ 3$. This is accomplished using a May‐type spectral sequence in topological Hochschild homology constructed from a filtration of a commutative ring spectrum. | 2022-01-21 23:11:04 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9488117694854736, "perplexity": 2779.3996588087302}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320303717.35/warc/CC-MAIN-20220121222643-20220122012643-00164.warc.gz"} |
https://www.physicsforums.com/threads/divergence-what-am-i-doing-wrong.115853/ | # Homework Help: Divergence what am I doing wrong
1. Mar 29, 2006
I don't understand what I am doing wrong here.
I'm supposed to show that this function is divergence free.
$$\vec v = \left( \frac{x}{x^2+y^2}, \frac{y}{x^2+y^2} \right)$$
I ran the divergence through with my TI-89 at it equals 0. But, I want to calculate it by hand, so it would be easier to do this in polar coordinates (and this is my problem).
$$\vec v(r,\theta) = \left ( \frac{\cos \theta}{r}, \frac{\sin \theta}{r} \left)$$
Standard divergence in cylindrical coordinates (dropping the z component)
$$div\,\, \vec u = \frac{1}{r} \left( \frac{\partial}{\partial r} (rF_r) +\frac{\partial}{\partial \theta} (F_\theta) \right)$$
$$div\,\, \vec v = \frac{1}{r}\left( \frac{\partial}{\partial r} (\cos \theta ) + \frac{\partial}{\partial \theta}\left (\frac{\sin \theta}{r} \right) \right)$$
Now this is obviously not 0, since the $\sin \theta$ is not going anywhere with the partials. So what am I doing wrong?
Thanks,
2. Mar 29, 2006
### Tom Mattson
Staff Emeritus
You made a subtle mistake here. Your vector above is still just $\vec{v}=(v_x,v_y)$. You just changed the expressions for the components. You do not have $\vec{v}=(v_r,v_{\theta})$ yet. To get that you have to transform the basis vectors from $\hat{e}_x$ and $\hat{e}_y$ to $\hat{e}_r$ and $\hat{e}_{\theta}$.
3. Mar 29, 2006
I'm pretty sure I follow what you are saying. Well, I at least understand it. I am unfortunatly too tired to tackle it, so I will do it in the morning... But I should be good with what you said :)
thanks man
4. Mar 29, 2006
### J77
...but I think he's pretty much there.
Since the $$\cos$$ and $$\sin$$ are bounded, and $$1/r\rightarrow 0$$ as $$r\rightarrow\infty$$
(with the exception at $$r=0$$)
5. Mar 29, 2006
### Tom Mattson
Staff Emeritus
No, he is not even close. This has nothing to do with taking a limit. The divergence should vanish identically in polar coordinates, just as it does in rectangular coordinates. His mistake is exactly what I said it was: He is plugging the rectangular components of $\vec{v}$ into the polar form of $\vec{\nabla}$.
$v_x=\frac{x}{x^2+y^2}=\frac{\cos(\theta)}{r}$
Note that this is still just $v_x$
$v_y=\frac{y}{x^2+y^2}=\frac{\sin(\theta)}{r}$
Note that this is still just $v_y$
He needs to find the components $v_r$ and $v_{\theta}$, which both contain contributions from $v_x$ and $v_y$. This is done by transforming the basis vectors, just like I said.
6. Mar 29, 2006
### J77
I'm having a very bad brain day | 2018-06-20 14:06:46 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8222988843917847, "perplexity": 439.9513782855161}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267863519.49/warc/CC-MAIN-20180620124346-20180620144346-00242.warc.gz"} |
https://math.stackexchange.com/questions/683294/determining-linear-independence-dependence-non-trivial-solution | # Determining Linear Independence/Dependence & non-trivial solution.
I put the vectors in a matrix and reduced it, solved the determinant and got 0. This tells me that the vectors are linearly dependent. I am not sure how to figure out the non-trivial relation. This is my reduced matrix.
$$\begin{pmatrix} 1& 0&-1/4 \\ 0& 1& 1\\ 0& 0& 0 \end{pmatrix}$$
$$A = \left\{\begin{matrix} -60\\ -4\\ -72 \end{matrix}\right. : B = \left\{\begin{matrix} -5\\ -1\\ -5 \end{matrix}\right. : C = \left\{\begin{matrix} 10\\ 0\\ 13 \end{matrix}\right.$$
__A +__B +___C = 0
Find coefficients.
• Your reduced matrix corresponds to the system of equations, $a+b-(1/4)c=0,b+c=0$. Find a non-zero solution to that system. That solution will be the coefficients in the relation. – Gerry Myerson Feb 20 '14 at 8:42
• Since they're dependent, can I set A = 1? – KnowledgeGeek Feb 20 '14 at 8:47
• Also, shouldn't your equation just read a - (1/4)c? – KnowledgeGeek Feb 20 '14 at 8:48
• How can you set $A=1$, when $A=\pmatrix{-60\cr-4\cr-72\cr}$? Or do you mean $a$ when you write $A$? In that case, set $a=1$, and see what happens. – Gerry Myerson Feb 20 '14 at 8:49
• My 1st equation comes from the first row. The first row is $(1,1,-1/4)$. So, my 1st equation is $a+b-(1/4)c=0$. – Gerry Myerson Feb 20 '14 at 8:50 | 2020-07-07 15:33:33 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8229999542236328, "perplexity": 469.4956201569432}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655893487.8/warc/CC-MAIN-20200707142557-20200707172557-00483.warc.gz"} |
https://global.ihs.com/doc_detail.cfm?document_name=ISO%206932&item_s_key=00042045 | # Cold-reduced carbon steel strip with a maximum carbon content of 0,25 %
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This International Standard describes cold-reduced carbon steel strip with a maximum mass fraction of carbon of 0,25 %, furnished to two levels of closer tolerances than cold-reduced carbon steel sheet, with specific quality, specific hardness requirements or mechanical properties, specific edge, and specific finish.
NOTE This International Standard does not apply to the product in narrow widths known as cold-reduced carbon steel sheet slit from wider widths (see ISO 3574), nor does it include cold-reduced carbon steel strip with a mass fraction of carbon over 0,25 % (see ISO 4960).
Cold-reduced carbon steel strip is produced with a maximum mass fraction of the specified carbon not exceeding:
- 0,15 % for material specified to mechanical properties;
- 0,25 % for material specified to temper (hardness) requirements | 2018-02-22 09:01:14 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9503539204597473, "perplexity": 11738.233607123657}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891814079.59/warc/CC-MAIN-20180222081525-20180222101525-00585.warc.gz"} |
https://enwiki.academic.ru/dic.nsf/enwiki/183464 | # Multiset
Multiset
In mathematics, the notion of multiset (or bag) is a generalization of the notion of set in which members are allowed to appear more than once. For example, there is a unique set that contains the elements a and b and no others, but there are many multisets with this property, such as the multiset that contains two copies of a and one of b or the multiset that contains three copies of both a and b. The term "multiset" was coined by Nicolaas Govert de Bruijn in the 1970s.[1] The use of multisets in mathematics predates the name "multiset" by nearly 90 years: Richard Dedekind used multisets in a paper published in 1888.[2]
## Overview
The number of times an element belongs to the multiset is the multiplicity of that member. The total number of elements in a multiset, including repeated memberships, is the cardinality of the multiset. For example, in the multiset {a, a, b, b, b, c} the multiplicities of the members a, b, and c are respectively 2, 3, and 1, and the cardinality of the multiset is 6. To distinguish between sets and multisets, a notation that incorporates brackets is sometimes used: the multiset {2,2,3} can be represented as [2,2,3].[3] In multisets, as in sets and in contrast to tuples, the order of elements is irrelevant: The multisets {a, b} and {b, a} are equal.
## Formal definition
Within set theory, a multiset may be formally defined as a 2-tuple(A, m) where A is some set and m : A → N≥1 is a function from A to the set N≥1 = {1, 2, 3, ...} of positive natural numbers. The set A is called the underlying set of elements. For each a in A the multiplicity (that is, number of occurrences) of a is the number m(a). If a universe U in which the elements of A must live is specified, the definition can be simplified to just a multiplicity function mU : U → N from U to the set N = {0, 1, 2, 3, ...} of natural numbers, obtained by extending m to U with values 0 outside A. This extended multiplicity function is the multiplicity function called 1A below. Like any function, the function m may be defined as its graph: the set of ordered pairs { (am(a)) : a in A }. With these definitions the multiset written as { aab } is defined as ({ ab },{ (a, 2), (b, 1) }), and the multiset { ab } is defined as ({ ab },{ (a, 1), (b, 1) }).
The concept of a multiset is a generalization of the concept of a set. A multiset corresponds to an ordinary set if the multiplicity of every element is one (as opposed to some larger natural number). However to replace set theory by "multiset theory" so as to have multisets directly into the foundations is not easy: a privileged role would still have to be given to (ordinary) sets when defining maps, as there is no clear notion of maps (functions) between multisets. It can be done[not specific enough to verify], with the result that classical theorems such as the Cantor–Bernstein–Schroeder theorem or Cantor's theorem, when generalized to multisets, are false[examples needed]; they remain true only in the case of finite multisets[citation needed]. In addition, the notion of a set as a "class of items satisfying a certain property" – i.e. the extension of a predicate – is used throughout mathematics, and this notion lacks a sensible generalization to multisets with multiple memberships.
An indexed family, ( ai ), where i is in some index-set, may define a multiset, sometimes written { ai }, in which the multiplicity of any element x is the number of indices i such that ai = x. The condition for this to be possible is that no element occurs infinitely many times in the family: even in an infinite multiset, the multiplicities must be finite numbers.
## Multiplicity function
The set indicator function of a normal set is generalized to the multiplicity function for multisets. The set indicator function of a subset A of a set X is the function
$\mathbf{1}_A : X \to \lbrace 0,1 \rbrace \,$
defined by
$\mathbf{1}_A(x) = \begin{cases} 1 &\text{if }x \in A, \\ 0 &\text{if }x \notin A. \end{cases}$
The set indicator function of the intersection of sets is the minimum function of the indicator functions
$\mathbf{1}_{A\cap B}(x) = \min\{\mathbf{1}_A(x),\mathbf{1}_B(x)\}.$
The set indicator function of the union of sets is the maximum function of the indicator functions
$\mathbf{1}_{A\cup B}(x) = \max\{{\mathbf{1}_A(x),\mathbf{1}_B(x)}\}.$
The set indicator function of a subset is smaller than or equal to that of the superset
$A\subseteq B \Leftrightarrow \forall x \mathbf{1}_{A}(x) \le \mathbf{1}_{B}(x).$
The set indicator function of a cartesian product is the product of the indicator functions of the cartesian factors
$\mathbf{1}_{A \times B}(x,y) = \mathbf{1}_A(x) \cdot\mathbf{1}_B(y).$
The cardinality of a (finite) set is the sum of the indicator function values
$|A|=\sum_{x\in X} \mathbf{1}_{A}(x).$
Now generalize the concept of set indicator function by releasing the constraint that the values are 0 and 1 only and allow the values 2, 3, 4 and so on. The resulting function is called a multiplicity function and such a function defines a multiset. The concepts of intersection, union, subset, cartesian product and cardinality of multisets are defined by the above formulas.
The multiplicity function of a multiset sum, is the sum of the multiplicity functions
$\mathbf{1}_{A \uplus B}(x) = \mathbf{1}_A(x) + \mathbf{1}_B(x).$
The multiplicity function of a multiset difference is the zero-truncated subtraction of the multiplicity functions
$\mathbf{1}_{A \setminus B}(x) = \max(0, \mathbf{1}_A(x) - \mathbf{1}_B(x)).$
The scalar multiplication of a multiset by a natural number n may be defined as:
$\mathbf{1}_{n \otimes A}(x) = n \times \mathbf{1}_A(x). \,$
A small finite multiset, A, is represented by a list where each element, x, occurs as many times as the multiplicity, 1A(x), indicates.
$\{1,1,1,3\} \cap \{1,1,2\} = \{1,1\} \,$
$\{1,1\} \cup \{1,2\} = \{1,1,2\}\,$
$\{1,1\} \subseteq \{1,1,1,2\}\,$
$\left|\{1,1\}\right|=2 \,$
$\{1,1\} \times \{1,2\} = \{(1,1), (1,1), (1,2), (1,2)\} \,$
$\{1,1\} \uplus \{1,2\} = \{1,1,1,2\} \,$
## Examples
One of the simplest and most natural examples is the multiset of prime factors of a number n. Here the underlying set of elements is the set of prime divisors of n. For example the number 120 has the prime factorization
$120 = 2^3 3^1 5^1\,$
which gives the multiset {2, 2, 2, 3, 5}.
A related example is the multiset of solutions of an algebraic equation. A quadratic equation, for example, has two solutions. However, in some cases they are both the same number. Thus the multiset of solutions of the equation could be { 3, 5 }, or it could be { 4, 4 }. In the latter case it has a solution of multiplicity 2.
## Free commutative monoids
The free commutative monoid on a set X (see free object) can be taken to be the set of finite multisets with elements drawn from X, with the monoid operation being multiset sum and the empty multiset as identity element. Such monoids are also known as (finite) formal sums of elements of X with natural coefficents. The free commutative semigroup is the subset of the free commutative monoid which contains all multisets with elements drawn from X except the empty multiset.
Free abelian groups are formal sums (i.e. linear combinations) of elements of X with integer coefficients. Equivalently, they may be seen as signed finite multisets with elements drawn from X.
## Counting multisets
The number of multisets of cardinality k, with elements taken from a finite set of cardinality n, is called the multiset coefficient or multiset number. This number is written by some authors as $\textstyle\left(\!\!{n\choose k}\!\!\right)$, a notation that is meant to resemble that of binomial coefficients; it is used for instance in (Stanley, 1997), and could be pronounced "n multichoose k" to resemble "n choose k" for $\tbinom nk$. Unlike for binomial coefficients, there is no "multiset theorem" in which multiset coefficients would occur, and they should not be confused with the unrelated multinomial coefficients that occur in the multinomial theorem. The value of multiset coefficients can be given explicitly as
$\left(\!\!{n\choose k}\!\!\right) = {n(n+1)(n+2)\cdots(n+k-1)\over k!} = {n + k - 1 \choose k},$
where the last expression expresses it as binomial coefficient; many authors in fact avoid separate notation and just write binomial coefficients. So, the number of such multisets is the same as the number of subsets of cardinality k in a set of cardinality n + k − 1. The analogy with binomial coefficients can be stressed by writing the numerator in the above expression as a rising factorial power
$\left(\!\!{n\choose k}\!\!\right) = {n^{\overline{k}}\over k!},$
to match the expression of binomial coefficients using a falling factorial power:
${n\choose k} = {n^{\underline{k}}\over k!}.$
There are for example 4 multisets of cardinality 3 with elements taken from the set {1,2} of cardinality 2 (n=2, k=3), namely : {1,1,1}, {1,1,2}, {1,2,2}, {2,2,2}. And there are also 4 subsets of cardinality 3 in the set {1,2,3,4} of cardinality 4 (n+k-1 = 4), namely : {1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}.
One simple way to prove the equality of multiset coefficients and binomial coefficients given above, involves representing multisets in the following way. First, consider the notation for multisets that would represent { a, a, a, a, a, a, b, b, c, c, c, d, d, d, d, d, d, d } (6 as, 2 bs, 3 cs, 7 ds) in this form:
$\bullet \bullet \bullet \bullet \bullet \bullet \mid \bullet \bullet \mid \bullet \bullet \bullet \mid \bullet \bullet \bullet \bullet \bullet \bullet \bullet$
This is a multiset of cardinality 18 made of elements of a set of cardinality 4. The number of characters including both dots and vertical lines used in this notation is 18 + 4 − 1. The number of vertical lines is 4 − 1. The number of multisets of cardinality 18 is then the number of ways to arrange the 4 − 1 vertical lines among the 18 + 4 − 1 characters, and is thus the number of subsets of cardinality 4 − 1 in a set of cardinality 18 + 4 − 1. Equivalently, it is the number of ways to arrange the 18 dots among the 18 + 4 − 1 characters, which is the number of subsets of cardinality 18 of a set of cardinality 18 + 4 − 1. This is
${18+4-1 \choose 4-1}={18+4-1 \choose 18} = 1330,$
so that is the value of the multiset coefficient
$\left(\!\!{4\choose18}\!\!\right).$
One may define a generalized binomial coefficient
${n \choose k}={n(n-1)(n-2)\cdots(n-k+1) \over k!}$
in which n is not required to be a nonnegative integer, but may be negative or a non-integer, or a non-real complex number. (If k = 0, then the value of this coefficient is 1 because it is the empty product.) Then the number of multisets of cardinality k in a set of cardinality n is
$\left(\!\!{n\choose k}\!\!\right)=(-1)^k{-n \choose k}.$
This fact led Gian-Carlo Rota to ask "Why are negative sets multisets?".[citation needed] He considered that question worthy of the attention of philosophers of mathematics.
### Recurrence relation
A recurrence relation for multiset coefficients may be given as
$\left(\!\!{n\choose k}\!\!\right) = \left(\!\!{n\choose k - 1}\!\!\right) + \left(\!\!{n-1\choose k}\!\!\right) \quad \mbox{for } n,k>0$
with
$\left(\!\!{n \choose 0}\!\!\right) = 1,\quad n\in\N, \quad\mbox{and}\quad \left(\!\!{0 \choose k}\!\!\right) = 0,\quad k>0.$
The above recurrence may be interpreted as follows. Let [n] := {1, ..., n} be the source set. There is always exactly one (empty) multiset of size 0, and if n = 0 there are no larger multisets, which gives the initial conditions.
Now, consider the case in which n,k > 0. A multiset of cardinality k with elements from [n] might or might not contain any instance of the final element n. If it does appear, then by removing n once, one is left with a multiset of cardinality k − 1 of elements from [n], and every such multiset can arise, which gives a total of
$\left(\!\!{n\choose k - 1}\!\!\right)$ possibilities.
If n does not appear, then our original multiset is equal to a multiset of cardinality k with elements from [n − 1], of which there are
$\left(\!\!{n-1\choose k}\!\!\right).$
Thus,
$\left(\!\!{n\choose k}\!\!\right) = \left(\!\!{n\choose k - 1}\!\!\right) + \left(\!\!{n-1\choose k}\!\!\right).$
## Polynomial notation
The set {x} may be represented by the monomial x. The set of subsets, { {}, {x} } , is represented by the binomial 1 + x.
The set {x,y} may be represented by the monomial x·y. The set of subsets, { {}, {x}, {y}, {x,y} } , is represented by the polynomial
(1 + x)·(1 + y) = 1 + x + y + x·y.
The multiset {x,x} may be represented by the monomial x·x = x2. The multiset of submultisets, { {}, {x}, {x}, {x,x} }, is represented by the polynomial
(1 + x)2 = 1 + x + x + x·x = 1 + 2·x + x2.
The multiset of submultisets of the multiset xn is
$(1+x)^n=\sum_{k=0}^n{n \choose k}\cdot x^k.$
That is why the binomial coefficient counts the number of k-combinations of an n-set.
The multiset xK·yN−K , containing N elements, K of which are hits, is called a statistical population, and a submultiset is called a statistical sample. The set of samples is
(1 + x)K·(1 + y)N−K
which by the binomial theorem equals
$\sum_{n=0}^N\sum_{k=0}^K{K \choose k}\cdot{N-K \choose n-k}\cdot x^k\cdot y^{n-k}.$
So the number of n-samples with k hits is
${K \choose k}\cdot{N-K \choose n-k}.$
See hypergeometric distribution and inferential statistics for further on the distribution of hits.
The infinite set of finite multisets of elements taken from the set {x}:
{ {}, {x}, {x,x}, {x,x,x}, ... }
may be represented by the formal power series
S = 1 + x + x2 + x3 + ... = 1 + xS .
The formal solution,
S = (1 − x)−1,
makes sense as a set of multisets, but the intermediate result, 1−x, does not make sense as a set of multisets.
The infinite set of finite multisets of elements taken from the set x·y is
(1 − x)−1·(1 − y)−1 = 1 + (x + y) + (x2 + x·y + y2) + ...
The special case y=x : The infinite multiset of finite multisets of elements taken from the multiset x2 is
(1 − x)−2 = 1 + 2·x + 3·x2 + ...
The general case: The infinite multiset of finite multisets of elements taken from the multiset xn is
$(1-x)^{-n}=\sum_{k=0}^\infty{-n \choose k} \cdot(-x)^k$ .
This explains why "multisets are negative sets". The negative binomial coefficients count the number of k-multisets of elements taken from an n-set.
## Cumulant generating function
A non-negative integer, n, can be represented by the monomial xn . A finite multiset of non-negative integers, say {2, 2, 2, 3, 5}, can likewise be represented by a polynomial f(x), say f(x) = 3·x2 + x3 + x5 .
It is convenient to consider the cumulant generating function g(t) = log(f(et)), say g(t) = log(3·et + et + et) .
• The cardinality of the multiset is eg(0) = f(1), say 3 + 1 + 1 = 5.
• The derivative g is g '(t) = f(et)−1·f '(et)·et, say g '(t) = (3·et + et + et)−1·(6·et + 3·et + 5·et) .
• The mean value of the multiset is μ = g '(0) = f(1)−1·f '(1), say μ = (3+1+1)−1·(6+3+5) = 2.8 .
• The variance of the multiset is σ2 = g ' '(0) .
The numbers ( μ, σ2, ··· ) = ( g '(0), g ' '(0), ··· ) are called cumulants.
The infinite set of non-negative integers {0, 1, 2, ···} is represented by the formal power series 1 + x + x2 + ··· = (1 − x)−1. The mean value and standard deviation are undefined. Nevertheless it has a cumulant-generating function, g(t) = −log(1−et). The derivative of this cumulant-generating function is g '(t) = (et−1)−1.
A finite multiset of real numbers , A = { Ai }, is represented by the cumulant generating function
$g_A(t) = \log \left(\sum_i e^{t \cdot A_i}\right) .$
This representation is unique: different multisets have different cumulant generating functions. See partition function (statistical mechanics) for the case where the numbers in question are the energy levels of a physical system.
The cumulant-generating function of a multiset of n real numbers having mean μ and standard deviation σ is: g(t) = log(n) + μ·t + 2−1·(σ·t)2 + ··· , and the derivative is simply: g '(t) = μ + σ2·t + ···
The cumulant-generating function of set, {k}, consisting of a single real number, k, is g(t) = k·t , and the derivative is the number itself: g '(t) = k . So the concept of the derivative of the cumulant generating function of a multiset of real numbers is a generalization of the concept of a real number.
The cumulant-generating function of a constant multiset, {k, k, k, k, ··· , k} of n elements all equal to the same real number k, is g(t) = log(n)+k·t , and the derivative is the number itself: g '(t) = k , irrespective of n.
The cumulant-generating function of the multiset of sums of elements of two multisets of numbers is the sum of the two cumulant-generating functions:
\begin{align} g_{A+B}(t) & = \log \left(\sum_i \sum_j e^{t\cdot(A_i+B_j)}\right) = \log \left(\sum_i\sum_j e^{t\cdot A_i}\cdot e^{t\cdot B_j}\right) \\ & = \log \left(\sum_i e^{t\cdot A_i}\cdot \sum_j e^{t\cdot B_j}\right) = \log \left(\sum_i e^{t\cdot A_i}\right)+ \log\left(\sum_j e^{t\cdot B_j}\right) \\ & = g_A(t) + g_B(t). \end{align}
There is unfortunately no general formula for computing the cumulant generating function of a product
$g_{A\cdot B}(t) = \log \left(\sum_i\sum_j e^{t\cdot A_i\cdot B_j}\right) \,$
but the special case of a constant times a multiset of numbers is:
$g_{k\cdot A}(t) = \log \left(\sum_i e^{t\cdot (k\cdot A_i)}\right) = \log \left(\sum_i e^{(t\cdot k)\cdot A_i}\right) = g_A\left(k\cdot t\right).$
The multiset 2·A = {2·Ai} is not the same multiset as 2×A =A+A = {Ai+Aj}. For example, 2·{+1,−1} = {+2,−2} while 2×{+1,−1} = {+1,−1} + {+1,−1} = {+1+1,+1−1,−1+1,−1−1} = {+2,0,0,−2}.
$g_{k\times A}(t) = k\cdot g_{A}(t).$
The standard normal distribution is like a limit of big multisets of numbers.
$\lim_{k \rarr \infty}k^{-1}\cdot (k^2\times \{+1,-1\}).$
This limit does not make sense as a multiset of numbers, but the derivative of the cumulant generating functions of the multisets in question makes sense, and the limit is well defined.
\begin{align} \lim_{k \rarr \infty} g'_{k^{-1}\cdot (k^2\times \{+1,-1\})}(t) & = \lim_{k \rarr \infty} \frac{d(k^2\cdot \log(e^{+t\cdot k^{-1}}+e^{-t\cdot k^{-1}}))}{dt} \\ & = \lim_{k \rarr \infty} \frac{d(k^2\cdot \log(2)+2^{-1}\cdot t^2+\cdots)}{dt}=t. \end{align}
The constant term k2·log(2) vanishes by differentiation. The terms ··· vanish in the limit. So for the standard normal distribution, having mean 0 and standard deviation 1, the derivative of the cumulant generating function is simply g '(t) = t . For the normal distribution having mean μ and standard deviation σ, the derivative of the cumulant generating function is g '(t) = μ + σ2·t .
## Notes
1. ^ Knuth, Donald E. (1998). The Art of Computer Programming – Vol. 2: Seminumerical Algorithms (3rd edition ed.). Addison Wesley. pp. 694. ISBN 0201896842. Knuth also lists other names that were proposed for multisets, such as list, bunch, bag, heap, sample, weighted set, collection, and suite.
2. ^ Syropoulos, Apostolos (2001), p. 347
3. ^ Hein, James L. (2003). Discrete mathematics. Jones & Bartlett Publishers. pp. 29–30. ISBN 0763722103.
## References
• Blizard, Wayne D. (1989) "Multiset theory," Notre Dame Journal of Formal Logic, Volume 30, Number 1, Winter: pp. 36–66. doi:10.1305/ndjfl/1093634995 http://projecteuclid.org/euclid.ndjfl/1093634995 MR990203 0668.03027
• Bogart, Kenneth P. (2000). Introductory combinatorics, 3rd. ed. San Diego CA: Harcourt.
• Gessel, Ira M., and Stanley, Richard P. (1995) "Algebraic enumeration" in Graham, R. L., Grötschel, M., & Lovász, L., eds., Handbook of combinatorics, Vol. 2. Elsevier: 1021–1061. ISBN 0-444-82351-4, 0-444-88002-X, 0-262-07171-1, 0-262-07169-X.
Multisets are discussed on pp. 1036–1039.
• Hickman, J. L. (1980) "A note on the concept of multiset," Bulletin of the Australian Mathematical Society 22: 211–17.
• Stanley, Richard P. (1997, 1999) Enumerative Combinatorics, Vols. 1 and 2. Cambridge University Press. ISBN 0-521-55309-1, 0-521-56069-1.
• Syropoulos, Apostolos (2001) "Mathematics of Multisets" in C. S. Calude et al., eds., Multiset processing: Mathematical, computer science, and molecular computing points of view, LNCS 2235. Springer-Verlag: 347–358.
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• Amenable number — An amenable number is an integer for which there exists a multiset of as many integers as the original number that can be either added up or multiplied together to give the original number. To put it algebraically, for an integer n , there is a… … Wikipedia | 2020-01-23 12:06:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 43, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8656003475189209, "perplexity": 504.2918089207899}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250610004.56/warc/CC-MAIN-20200123101110-20200123130110-00420.warc.gz"} |
https://planetmath.org/DualOfACoalgebraIsAnAlgebraThe | # dual of a coalgebra is an algebra, the
Let $R$ be a commutative ring with unity. Suppose we have a coassociative coalgebra $(C,\Delta)$ and an associative algebra $A$, both over $R$. Since $C$ and $A$ are both $R$-modules, it follows that $\mathrm{Hom}_{R}({C},{A})$ is also an $R$-module. But in fact we can give it the structure of an associative $R$-algebra. To do this, we use the convolution product. Namely, given morphisms $f$ and $g$ in $\mathrm{Hom}_{R}({C},{A})$, we define their product $fg$ by
$(fg)(x)=\sum_{x}f(x_{(1)})\cdot g(x_{(2)}),$
where we use the Sweedler notation
$\Delta(x)=\sum_{x}x_{(1)}\otimes x_{(2)}$
for the comultiplication $\Delta$. To see that the convolution product is associative, suppose $f$, $g$, and $h$ are in $\mathrm{Hom}_{R}({C},{A})$. By applying the coassociativity of $\Delta$, we may write
$((fg)h)(x)=\sum_{x}(f(x_{(1)})\cdot g(x_{(2)}))\cdot h(x_{(3)})$
and
$(f(gh))(x)=\sum_{x}f(x_{(1)})\cdot(g(x_{(2)}))\cdot h(x_{(3)}).$
Since $A$ has an associative product, it follows that $(fg)h=f(gh)$.
In the foregoing, we have not assumed that $C$ is counitary or that $A$ is unitary. If $C$ is counitary with counit $\varepsilon\colon C\to R$ and $A$ is unitary with identity $1\colon R\to A$, then their composition $1\circ\varepsilon\colon C\to A$ is the identity for the convolution product.
###### Example.
Let $C$ be a coassociative coalgebra over $R$. Then $R$ itself is an associative $R$-algebra. The algebra $\mathrm{Hom}_{R}({C},{R})$ is called the algebra dual to the coalgebra $C$.
We have seen that any coalgebra dualizes to give an algebra. One might expect that a similar construction could be performed on $\mathrm{Hom}_{R}({A},{R})$ to give a coalgebra dual to $A$. However, this is not the case. Thus coalgebras (based on “factoring”) are more fundamental than algebras (based on “multiplying”).
(The proof will be provided at a later stage).
Remark on Al/gebraic DualityMirror or tangled ‘duality’ of algebras and ‘gebras’:
An interesting twist to duality was provided in Fauser’s publications on al/gebras where mirror or tangled ‘duality’ has been defined for Grassman-Hopf al/gebras. Thus, an algebra not only has the usual reversed arrow dual coalgebra but a mirror (or tangled) gebra which is quite distinct from the coalgebra.
Note: The dual of a quantum group is a Hopf algebra.
## References
• 1 W. Nichols and M. Sweedler, Hopf algebras and combinatorics, in Proceedings of the conference on umbral calculus and Hopf algebras, ed. R. Morris, AMS, 1982.
• 2 B. Fauser: A treatise on quantum Clifford Algebras. Konstanz, Habilitationsschrift.
arXiv.math.QA/0202059 (2002).
• 3 B. Fauser: Grade Free product Formulae from Grassman–Hopf Gebras. Ch. 18 in R. Ablamowicz, Ed., Clifford Algebras: Applications to Mathematics, Physics and Engineering, Birkhäuser: Boston, Basel and Berlin, (2004).
• 4 J. M. G. Fell.: The Dual Spaces of C*–Algebras., Transactions of the American Mathematical Society, 94: 365–403 (1960).
Title dual of a coalgebra is an algebra, the DualOfACoalgebraIsAnAlgebraThe 2013-03-22 16:34:20 2013-03-22 16:34:20 mps (409) mps (409) 8 mps (409) Derivation msc 16W30 GrassmanHopfAlgebrasAndTheirDualCoAlgebras DualityInMathematics QuantumGroups dualities of algebraic structures | 2021-01-26 13:14:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 41, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.924993634223938, "perplexity": 675.0376932721609}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610704799741.85/warc/CC-MAIN-20210126104721-20210126134721-00119.warc.gz"} |
https://www.clutchprep.com/physics/practice-problems/137145/consider-the-wire-loop-you-used-in-part-a-for-the-lab-you-placed-the-bottom-sect | Force and Torque on Current Loops Video Lessons
Concept
# Problem: Consider the wire loop you used in part A. For the lab, you placed the bottom section of the loop in the magnetic field (between two magnets in a box with opening in between for loop to go in) and measured the force upon it when current was flowing through the wire. Imagine instead that both the bottom and one side of the loop were placed in the magnetic field (but not the top or other side). What would be the net direction of the force upon this loop? Explain carefully and refer back to equations if needed. Please specify the NET direction!
###### FREE Expert Solution
Force on a current-carrying conductor:
$\overline{){\mathbf{F}}{\mathbf{=}}{\mathbf{B}}{\mathbf{i}}{\mathbf{L}}{\mathbf{s}}{\mathbf{i}}{\mathbf{n}}{\mathbf{\theta }}}$
The angle between the current and the magnetic field is 90°. Thus:
$\overline{){\mathbf{F}}{\mathbf{=}}{\mathbf{B}}{\mathbf{i}}{\mathbf{L}}}$
Right-hand rule for currents in B-fields: Fingers→ field direction, Thumb→ current direction, and palm→ force direction.
FAD is directed in the positive x-axis.
FAB is directed in the positive y-axis.
83% (435 ratings)
###### Problem Details
Consider the wire loop you used in part A. For the lab, you placed the bottom section of the loop in the magnetic field (between two magnets in a box with opening in between for loop to go in) and measured the force upon it when current was flowing through the wire. Imagine instead that both the bottom and one side of the loop were placed in the magnetic field (but not the top or other side). What would be the net direction of the force upon this loop? Explain carefully and refer back to equations if needed. Please specify the NET direction! | 2021-09-20 10:58:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5998536944389343, "perplexity": 635.3769099868879}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057036.89/warc/CC-MAIN-20210920101029-20210920131029-00220.warc.gz"} |
https://electronics.stackexchange.com/questions/182592/vhdl-testbench-variable-clock-wave-generation | # VHDL testbench variable clock/wave generation
A testbench for one sub-entity in my system currently defines a helper process to generate a clock-like waveform at the command of the main stimulus process. A simplified version of it is:
shared variable gen_period : time := 10 us;
signal gen_all : boolean := false; -- wavegen enabled
signal gen_A : boolean := false; -- run this signal
signal A_in : std_logic; -- manual input (wavegen disabled)
signal A_gen : std_logic; -- generated output (internal)
signal A_out : std_logic; -- final output
...
A_generate: process
if gen_A then
A_gen <= '1';
wait for gen_period/2;
A_gen <= '0';
wait for gen_period/2;
else
A_gen <= '0';
wait for 1 ns;
end if;
end process;
A_out <= A_gen when gen_all else A_in;
The full system has a few more signals, but this is the basic idea. gen_all applies to all signals and allows a manual signal pattern to be applied when disabled, while gen_A is basically a clock enable for that specific signal. The other important feature is that the period of the clock is variable.
Putting this all together it means that the main testbench stimulus process can either manually generate edges or can just request a pulse train at a specific interval and then wait for a longer time (typically gen_period * N) to generate multiple pulses. It works quite well.
Now though I'm interested in generalising this a bit, in particular to instantiate several such independent generators in the testbench for a higher-level design that contains multiple of the original component. However I'm having trouble finding the right way to do so.
My first attempt was to wrap the code above into its own entity, declaring all but A_gen as ports of the entity. Doing this required changing gen_period from a variable to a signal, since AFAIK ports must be signals.
Unfortunately, back in the original testbench stimulus process, whenever it tried to assign a new period it had to do this as a signal assignment, which didn't take effect until later. Additionally the wait for gen_period * 5 calls appeared to be using the original value for gen_period rather than one just assigned. (This is not surprising behaviour for signals, but in this case it's undesired.)
Is there a better way to encapsulate this functionality, and retain instant behaviour of the variable? The code is only going to be used in a testbench, so it does not need to be synthesisable. I'm using Xilinx ISim.
(I realise that I can do a for .. generate to instantiate multiple instances in the higher level testbench, but this doesn't allow me to share the code between separate testbenches, which is also desirable to avoid duplication.)
I can live with having only one instance of the gen_period variable shared between all generators in the high level testbench, but I would prefer to have one per generator instance.
Since the answers seem to be focusing on "clock" rather than "waveform", it seems that I need to clarify the usage a bit more. This is inside a testbench:
stim_proc : process
begin
-- (reset and other setup instructions here)
-- generate "slow" pulse train
gen_all <= true;
gen_period := 250 ns;
gen_A <= true;
wait for gen_period * 20;
gen_A <= false;
wait for gen_period;
-- (perform tests on logic for slow pulses here)
-- generate "fast" pulse train
gen_period := 10 ns;
gen_A <= true;
wait for gen_period * 50;
gen_A <= false;
wait for gen_period;
-- (perform tests on logic for fast pulses here)
-- generate asymmetric pulses
gen_all <= false;
A_in <= '1';
wait for 100 ns;
A_in <= '0';
wait for 200 ns;
A_in <= '1';
wait for 150 ns;
A_in <= '0';
wait for 50 ns;
-- (perform tests on logic for asymmetric pulses here)
-- etc
end process;
These are the simple ones; there are a few more complex ones involving multiple signals happening simultaneously, but that's not really important for this question. The important point is that this is all sequential code and both the pulse generation and the waiting need to act on the most recently set period, not any delayed value.
You could use a procedure with an out parameter. The procedure can be declared in a package and called with different parameters for different 'clock' signals.
Here is an example of a clock_gen procedure from VHDL-extras:
subtype duty_cycle is real range 0.0 to 1.0;
procedure clock_gen( signal Clock : out std_ulogic; signal Stop_clock : in boolean;
constant Clock_period : in delay_length; constant Duty : duty_cycle := 0.5 ) is
constant HIGH_TIME : delay_length := Clock_period * Duty;
constant LOW_TIME : delay_length := Clock_period - HIGH_TIME;
begin
Clock <= '0';
while not Stop_clock loop
wait for LOW_TIME;
Clock <= '1';
wait for HIGH_TIME;
Clock <= '0';
end loop;
end procedure;
The constant input period can be exchanged with a variable for simulation environments.
Usage example:
architecture rtl of my_entity is
signal SimStop : boolean := false;
signal my_clock1 : std_ulogic;
signal my_clock2 : std_ulogic;
begin
clock_gen(my_clock1, SimStop, 10.000 ns);
clock_gen(my_clock2, SimStop, 6.666 ns);
-- ...
process
begin
--
-- some stimuli
--
wait for 10 ns;
SimStop := true;
wait;
end process;
end architecture;
Example to generate a waveform with an generic waveform:
type T_TIME_VECTOR is array(NATURAL range <>) of TIME;
constant myWaveform : T_TIME_VECTOR := (
-- generate "slow" pulse train
20 * 250 ns,
250 ns,
-- generate "fast" pulse train
50 * 10 ns,
10 ns
);
procedure generateWaveform(signal Wave : out BOOLEAN; Waveform: T_TIME_VECTOR; InitialValue : BOOLEAN) is
variable State : BOOLEAN := InitialValue;
begin
Wave <= State;
for i in Waveform'range loop
wait for Waveform(i);
State := not State;
Wave <= State;
end loop;
end procedure;
signal mySignal : BOOLEAN;
-- begin of architecture
generateWaveform(mySignal, myWaveform);
• This looks more promising but it still seems too concurrent for the required usage. It's fine for a clock, but I need a controlled waveform. Aug 3 '15 at 0:29
• @Miral You can produce any signal with this method, it's just a question on how complex your procedure is. A procedure is as powerful as a process - both are sequential. You process example is again evaluated from the beginning, if one run is complete. You can imitate this in a procedure, while framing your code with a endless-loop. See my example: while not Stop_clock loop ... end loop;. Aug 6 '15 at 8:44
• But I don't see how this improves anything. I would still have to have uniquely-named signals for each separate waveform, and the interval can't be a parameter because it has to be changed inside the stimulus process. See the updated question. Aug 9 '15 at 23:53
• @Miral As I said, it's just a question of the procedure's complexity. I added an example, that generates a waveform for type boolean. As an input it takes a VCD-like list of delays between toggle events. This example can be extended to every type. You could hand-over a second list of std_logic_vectors, which are assigned to the wave signal at every listed event. Aug 10 '15 at 8:50
• Ok, I admit that's possible, but it still seems like a considerable departure from the current code, and brittle to changes in the test ordering. I was hoping to keep the tests immediately adjacent to the generation, as in the question. (Also BTW note that your code doesn't do the same thing as mine -- during that 20 * 250 ns period mine is generating multiple pulses.) Aug 11 '15 at 7:45
You can define the clock generator as an entity also. Instead of trying to assign gen_period as a signal, try defining it as a generic in the entity definition as shown below:
entity clock_gen is
generic (
gen_period : time := 10 us
);
port (
signal gen_all : boolean := false; -- wavegen enabled
signal gen_A : boolean := false; -- run this signal
signal A_in : std_logic; -- manual input (wavegen disabled)
signal A_gen : std_logic; -- generated output (internal)
signal A_out : std_logic -- final output
);
end entity clock_gen;
Now, when you try to instantiate the entity in your higher-level module, it will allow you to define the gen_period parameter separately for each instance of clock_gen. The default value of the parameter will be considered as 10 us if it is not mentioned separately during instantiation. Two examples of usage in architecture are shown below:
gen1: clock_gen
generic map (gen_period => 100 us)
port map (
gen_all => <some_port1>;
gen_A => <some_port2>;
A_in => <some_port3>;
A_gen => <some_port4>;
A_out => <some_port5>
);
gen2: clock_gen
generic map (gen_period => 1 us)
port map (
gen_all => <some_port1>;
gen_A => <some_port2>;
A_in => <some_port3>;
A_gen => <some_port4>;
A_out => <some_port5>
);
If you have multiple instances of clock_gen with the same gen_period, you can even call the instantiation inside a for ... generate procedure in your code.
• Generics are not variable at runtime so this does not answer the question. Jul 31 '15 at 13:42
• @BrianDrummond Sure, however, simply making gen_period a port solves the problem. In addition, an entity allows more efficient forms of clk to be used than a procedure. IE: Clk <= not Clk after tperiod_Clk/2 ; Aug 1 '15 at 6:35
• As specified, generics are not sufficient because they're not variable, and making it a port didn't work because it introduces delayed-write semantics that are undesirable. And the Clk <= not Clk after` trick is only suitable for infinite run clocks, which this is definitely not. Aug 3 '15 at 0:14 | 2021-10-26 09:08:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.36599454283714294, "perplexity": 5244.919114354197}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323587854.13/warc/CC-MAIN-20211026072759-20211026102759-00460.warc.gz"} |
https://quantumcomputing.stackexchange.com/questions/16208/how-can-we-construct-a-control-control-y-rotation-ccry-gate-in-qiskit | # How can we construct a control-control y-rotation (CCRy) gate in Qiskit?
Qiskit has a CRy gate, however I couldn't find a CCRy (double control Ry) gate implementation. How can we construct the CCRy circuit given below in Qiskit without any ancillary qubits?
Edit: A quick note on another solution, when using the mcry gate to perform control-control y-rotation. I had been initializing quantum circuits via:
from qiskit.circuit import QuantumCircuit
circ = QuantumCircuit(4, 4)
circ.mcry(q_controls=[0, 1], q_target=2, q_ancillae=None) # Gives error saying q_target needs to be a qubit
As shown in the linked solution, one needs to define quantum registers and pass those objects as parameters to the quantum circuit to avoid this issue. In general, this makes me wonder if the latter is a better practice than what I had been doing previously (shown above).
## 1 Answer
There is no CCRy gate in the library but there is the multi-controlled multi-target gate (MCMT) class that you can use. Or, you could also do it as follow:
from qiskit import QuantumCircuit,QuantumRegister
from qiskit.circuit.library.standard_gates import RYGate
from qiskit.circuit import Parameter
import matplotlib.pyplot as plt
qr=QuantumRegister(3)
circ=QuantumCircuit(qr)
a=Parameter('a') # You can replace a with your choice of angle here
CCRY=RYGate(a).control(2)
circ.append(CCRY,qr)
print(circ)
q0: ────■────
│
q1: ────■────
┌───┴───┐
q2: ┤ RY(a) ├
└───────┘
• Thank you very much! That's extremely helpful. – Faiyaz Hasan Feb 25 at 2:55
• Nice. I suppose CCRY may be a little better name for CCCRY. – Adam Zalcman Feb 25 at 2:57
• @AdamZalcman Ha! I was using it for something else where I had 3 control qubits.... forgot to change the name back. :) – KAJ226 Feb 25 at 3:10
• @FaiyazHasan No problem!! Glad I can help. Welcome to QCSE. – KAJ226 Feb 25 at 3:11 | 2021-04-20 06:09:41 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6461508274078369, "perplexity": 5152.77797971089}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618039379601.74/warc/CC-MAIN-20210420060507-20210420090507-00001.warc.gz"} |
https://fykos.org/problems/start?tasktag=elProud | # Search
## electric current
### (10 points)5. Series 33. Year - P. there will be light
Estimate the time that passes between the flip of a light switch and the turning on of the light source. Make independent estimates for a light bulb, fluorescent lamp, LED light bulb and Neon tube light. Discuss as many factors influencing the time as you can.
Dodo throws the circuit breaker.
### (10 points)5. Series 33. Year - S. min and max
We are sorry. This type of task is not translated to English.
They had to wait a lot for Karel.
### (10 points)4. Series 33. Year - S.
We are sorry. This type of task is not translated to English.
### (3 points)1. Series 33. Year - 2. battery issue on holidays
How long does it take for a fully charged car battery ($12 \mathrm{V}$, $60 \mathrm{Ah}$) to run out, when someone forgets to turn off the daytime running lights, locks the car and walks away? Specifically we are interested in a situation with two head lights H4 (each running with $55 \mathrm{W}$) and two rear lights P21/5W (each running with $5 \mathrm{W}$). For simplicity, assume no transport losses between the battery and the lights, that there is no other significant consumption of power and that the voltage on the battery stays constant.
### (12 points)3. Series 32. Year - E. indexed capacitor
Find an electrolytic capacitor and a resistor and measure their capacity and resistance, respectively. You cannot measure these quantities directly. We recommend a choice of parameters, such that $RC\approx 20 \mathrm{s}$.
Be aware of maximum allowed voltage on the capacitor and the capacitor's polarity.
Dodo was measuring resonance in labs.
### (10 points)3. Series 32. Year - P. personal power bank
Last battery percentages in your mobile phone are almost gone, your power bank is dead, or you left it at home and 230 is also not in the sight. Wouldn't it be awesome if you could have your own source of electrical energy with you all the time?
• Suggest several different tools, which would be able to produce electrical energy just from your body resources.
• Discuss their maximum power and efficiency. What devices could you supply with electricity using this method?
• Discuss its effects on your health and physical condition. Which body organs would fail first?
As a possible solution, consider a system of small turbines located in your bloodstream. Support all arguments with accurate calculations.
Jachym had a feeling that he is missing some energy.
### (8 points)1. Series 32. Year - 5. damned circuit
a) Determine the resistance between points A and B of the infinite grid in the picture. The point A is directly connected to two resistors with resistances $R_a$ and $R_b$. Each of these resistors is connected to two more resistors with $R_a$ and $R_b$ etc.
b) Replace all the resistors with capacitors of capacitances $C_a$ and $C_b$. What is the total capacitance of the circuit?
Yet again, Karel wanted something unendingly infinite.
### (3 points)6. Series 31. Year - 2. hot wire
Calculate the current, that needs to pass through a metal wire of a diameter $d = 0{,}10 \mathrm{mm}$ located in a vacuum bulb, so that its temperature stays at $T = 2 600 K$. Assume the surface of the wire radiates like an ideal black body and neglect any losses by heat conduction. The resistivity of the material of the wire at the given temperature is $\rho = 2{,}5 \cdot 10^{-4} \mathrm{\Ohm \cdot cm}$. \taskhint {Hint}{Use the Stefan-Boltzmann's law.}
Danka was contemplating the light bulb efficiency
### (7 points)4. Series 31. Year - 4. solve it yourself
We have a black box with three outputs (A, B, and C). We know that it consists of $n$ resistors with the same resistance but we don't know the circuit diagram. So we measure the resistance between each pair of outputs $R\_{AB} = 3 \mathrm{\Omega }$, $R\_{BC} = 5 \mathrm{\Omega }$ a $R\_{CA} = 6 \mathrm{\Omega }$. Your task is to find the minimum possible $n$ and calculate the corresponding resistance of one resistor.
Matěj solved it quickly.
### (6 points)0. Series 31. Year - 3.
We are sorry. This type of task is not translated to English. | 2020-04-08 04:46:11 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4877694845199585, "perplexity": 1356.123242005786}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585371810617.95/warc/CC-MAIN-20200408041431-20200408071931-00386.warc.gz"} |
https://tldr.dendron.so/notes/common.moro.html | # moro
• Invoke moro without parameters, to set the current time as the start of the working day:
moro
• Specify a custom time for the start of the working day:
moro hi {{09:30}}
• Invoke moro without parameters a second time, to set the current time at the end of the working day:
moro
• Specify a custom time for the end of the working day:
moro bye {{17:30}}
• Add a note on the current working day:
moro note {{3 hours on project Foo}}
• Show a report of time logs and notes for the current working day:
moro report
• Show a report of time logs and notes for all working days on record:
moro report --all | 2021-06-22 14:37:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7643957734107971, "perplexity": 3521.619069241706}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623488517820.68/warc/CC-MAIN-20210622124548-20210622154548-00044.warc.gz"} |
https://tex.stackexchange.com/questions/439130/international-rules-how-to-layout-a-vector-correctly | # International rules how to layout a vector correctly
Here in Germany it is common to write a vector as a one column matrix.
The vector character itself is setup as e.g.: \vv{a}
I have however seen vectors setup as one row matrix where the components are separated with commas, and the vector character is setup in BOLDMATH without any arrow above it.
Any expert who knows the international rules how to layout a vector properly so it is internationally accepted?
Are there any such rules or are they country-depending?
And as I could see that my question was a bit misleading, I shall be more precise. I am setting up a sty file containing some vector calculations and I am not knowing which conventions for vector notation are present in other countries than Germany ...
The idea is to setup a global command for typesetting a vector to quickly change layout country-depending.
So my question is: What notation of vectors are common in various countries, e.g. one column matrix, one row matrix, one row matrix transposed, etc. and maybe a clever command change them globally and to know if there is a international rule how to typeset vectors.
Any hints appreciated which notation in what country, like
Germany: one column matrix UK: ... USA: ... France: ... Spain: ...
Hope to have been more understandful now :-)
• Pick up a style and stick to it. I somewhat adopted the boldmath notation over arrows in research papers (CS) for brevity and wide understanding. As for row vs. column – might it be that they save space? Because, you can write (1,2,3)^T instead of 1-2-3 column. – Oleg Lobachev Jul 3 '18 at 22:05
• @Oleg Lobachev ... so no rules -- just personal preferences? – user151328 Jul 3 '18 at 22:11
• I'd look at papers in the target area (a proper subfield of CS, math, etc.) and mimic them. – Oleg Lobachev Jul 3 '18 at 22:35
• So, I suggest two different things here, as I just noticed. The first comment is about writing a thesis or something, where you are completely in control. The second is about a scientific paper, where you'd better fulfil the expectations of the community. But actually, these things are so minor that you can use your own preferences and then just wait for reviewer's feedback. I don't think a paper would be rejected because it notes vectors down in a different established manner than most in the field. – Oleg Lobachev Jul 3 '18 at 22:37
• you are asking a mathematical rather than tex question so it's off topic here, it depends what you are doing but sometimes the choice of row or column is just a typographic choice but sometimes the row and column vector denote different objects so it is not a stylistic choice especially if anything is being represented by matrix multiplication. – David Carlisle Jul 4 '18 at 8:00
So, more related to TeX: define a \newcommand for vectors, say \vec and for values in a vector, say, \vecvals. Assign them to \vv and column matrix pmatrix magic, as you want, e.g.
\newcommand{\vec}[1]{\vv{#1}}
Then, you'd be able to change them fast to \boldmath and row matrix (with an optional ^T) if requested:
\newcommand{\vec}[1]{\boldmath{#1}}
In your actual text you use only your \vec commands, so, a quick global style change is a matter of minutes.
• I think \v exists. Why not use \vec and redefine in case a change is wanted? – Manuel Jul 3 '18 at 22:47
• Point taken. I've updated the answer. – Oleg Lobachev Jul 3 '18 at 23:13 | 2019-06-26 12:34:20 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8063482642173767, "perplexity": 1637.5773988108137}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560628000306.84/warc/CC-MAIN-20190626114215-20190626140215-00077.warc.gz"} |
http://mathhelpforum.com/math-topics/187256-deflection-beam-efect-errors-span-load-data.html | # Math Help - Deflection of beam. Efect of % errors in span and load data.
1. ## Deflection of beam. Efect of % errors in span and load data.
OK so I'm practicsing some of the questions out of my text book for when I start my course in 2 weeks and finding this one pretty difficult, but here it goes.
Bestend Properties is refurbishing an old hospital that was built in the 1950's. The building's main structure is a steel frame. The company's structural engineers are trying to determine the strength of the existing floor beams by loading them up and measuring the deflection.
The Formula that relates to deflection to the strength and loadings on the beam is:
E=5wL^4/384ly
Where:
E - Is Young's Modulus (strength of material)
w - Is the measured uniformly distributed load on the beam
l - Is the moment of inertia of the beam (a constant due to the cross sectional shape)
L - Is the measured span of the beam.
y - Is the deflection of the loaded beam
After the on-site testing was done, it was discovered that some of the measuring equipment was wrongly calibrated. The errors that were discovered were as follows:
The span L was 5% too long.
The load w was 3% too low.
Using you knowledge of binomial theory, produce a formula to show the net percentage effect of these calibration errors on the value of E (Young's Modulus).
I'm not even sure where to start, if anyone could help me I would really appreciate it.
Thanks
2. ## Re: Can someone help me with this, I have no clue?
Hi Chip6891,
I think you would do this beam to beam and calculate wrong values then right values. Find % difference. You need size and weight of beam, span , uniform load in kips and midpoint deflection in inches.Also the moment of inertia of cross section
3. ## Re: Can someone help me with this, I have no clue?
There are several methods for dealing with this sort of problem, but since the binomial theorem is mentioned I suppose that it has to be this one ...
Replace w with w + $\delta$w and L with L + $\delta$L. Those are the 'original' values together with there respective (actual) errors.
As a result of these changes, E will change to E + $\delta$E.
Now multiply out the RHS (using the binomial theorem for the L to the power four term), retaining the term for E and first order terms only. (i.e ignore any terms involving the products of two or more $\delta$'s.)
The E's on both sides will cancel.
Now divide both sides of the resulting equation by 100E (replacing E with its original expression on the RHS) and you should arrive at a result that says that the percentage error in E is approximately equal to the sum of the percentage error in w plus 4 times the percentage error in L.
4. ## Re: Deflection of beam. Efect of % errors in span and load data.
Hi, thanks for getting back.
So when you say "Replace w with w + w and L with L + L."
Should step one in solving the equation now look like this?
E=5(w + w)(L + L^4)/384ly
or is this wrong? It's ok maths isn't my strongest point
5. ## Re: Deflection of beam. Efect of % errors in span and load data.
Hi Chips891,
I thought about my first answer and I see a simplification.
E= 5 wl* l^3/ 384 I y where wl is uniform load in kips ( lb per inch) y is deflection in inches
E1/E2 = wl1 (l1)^3/wl2 (l2)^3
wl1/wl2 =1.05 (l1)^3 /(l2)^3= (.95)^3
E2/E1= 1.05* (.95)^3 = .88 so the first E would be lowered by 12%
bjh
6. ## Re: Deflection of beam. Efect of % errors in span and load data.
The first line should look like this ...
E + $\delta$E = 5(w + $\delta$w)(L + $\delta$L $)^4$/384ly
The changes in w and L bring about a change $\delta$E in E.
7. ## Re: Deflection of beam. Efect of % errors in span and load data.
Hi BobP,
The formula for the deflection of a beam is as given by theOP and is used to calculate this amount using 29* 10^6 psi as E. It can be used to calculate E given a deflection by inserting the other values.In addition to the simplification I show I applied this to a real beam with load and span.If you have an interest I would post it
bjh
8. ## Re: Deflection of beam. Efect of % errors in span and load data.
OK, so I have done that.
So for the next step do I have to use Binomial Theorem and expand (L+&L)^4 ?
9. ## Re: Deflection of beam. Efect of % errors in span and load data.
Hi Chip6891,
What does your book ask you to do? Are you asked for a statistical analysis after testing all beams in the structure? What course are you going to take?
bjh
10. ## Re: Deflection of beam. Efect of % errors in span and load data.
Hi bjh
I'm starting the BTEC National in Construction in about 2 weeks time. I'm doing a practice question out of the BTEC National in Construction and Civil Engineering text book, its a distinction question.
The question is asking me to produce a formula using Binomial Theorem to show the net percentage of the errors to the origional question asked above. I've never had to learn maths on this scale before, but I've got to learn it for this course if I want to get the distinction questions.
11. ## Re: Deflection of beam. Efect of % errors in span and load data.
Hello again Chip6891,
I read what I could find on line for BTECand see that it is a2year course in Construction and Civil Engineering.I don't understand what the Binomial Theorem would have to do with this.If it said use basic algebra I have done that.I did make a mistake in post 5.E1 in this question has a higher modulus than E2(post revised)where E2 is the one with corrections.
It is very important in similar problems that all units are consistant.
bjh
12. ## Re: Deflection of beam. Efect of % errors in span and load data.
Hi bjh
Yea to be honest I'm a little stumped to how to apply binomial theorem to this equation.
I'm only doing one year of the BTEC Natational as a bridge to HNC in construction. This is because I am already a qualified tradesmen (Carpenter) to level 3. Also because of this I don't think I have to do the maths unit, but I thought it would help as this leads to more professional positions in construction.
It does show me a worked example in the textbook which I do not fully understand.
Find the approximate percentage error in the calculated volume of a right circular cone if the radius is taken as 2% too small and the hight as 3% too large.
Volume (V) of a right cone = 1/3(Pir^2h)
The error in height ($\delta$h) = h3/100
The error in radius ($\delta$r) = r2/100
V+$\delta$V = 1/3(Pi(r-r*2/100)^2(h+h*3/100)
V+$\delta$V = 1/3(Pir^2h(1-2/100)(1-3/100))
Since the errors in height and its radius are small when compared to the origional lengths, we can approximate using binomial theory that:
V+$\delta$V = 1/3Pir^2h(1-2*2/100)(1+3/100)
Multiplying out the brackets:
V+$\delta$V = 1/3Pir^2h(1-1/100) approximate, ignoring the last small term.
V+$\delta$V = V(1-1/100)
$\delta$V = -V(1/100)
Therefore. for a right circular cone with a radius taken as 2% too small and a height taken as 3% too large, the calculated volume will be 1% too small.
13. ## Re: Deflection of beam. Efect of % errors in span and load data.
We seem to have two threads running wrt this problem, I confess to knowing very little about the 'practical' side of the problem, I'm simply looking at it from a maths point of view.... given percentage changes in the values of w and L, derive a formula for the resulting percentage change in E.
The binomial expansion isn't difficult when you see the pattern, look at the expansions for 2, 3 and 4.
$(a+b)^2 = a^2 + 2ab + b^2,$
$(a+b)^3 = a^3 + 3a^2b+3ab^2+b^3,$
$(a+b)^4 = a^4 + 4a^3b+6a^2b^2+4ab^3+b^4.$
You get the coefficients from Pascal's triangle.
If you see the pattern and you have Pascal's triangle it's easy to write out the expansions for 5, 6, ...
Only the first two terms in the expansion of (L + $\delta$L $)^4$ are required, because we are going to ignore terms involving the products of two $\delta$'s. (We assume that one $\delta$ is small so that the product of any two of them will be small enough to be negligible.)
The next step is to multiply out the brackets on the RHS.
14. ## Re: Deflection of beam. Efect of % errors in span and load data.
Hi Chip,
I agree with your answer which I calculated using basic algebra.Good luck with your course and future in construction
15. ## Re: Deflection of beam. Efect of % errors in span and load data.
Hi guys
Well anyways I had a little time the weekend to play around with what you have told me, and this is what I have got so far to the original question, I'm just wondering if it is right?
E+&E = (w+&w)(L+&L)^4/384ly
E+&E = (w+&w)(L^4+4L^3&L)/384ly
E+&E = (w+&w*3/100)(L^4+4L^3*5/100)
E+&E = [5wL^4(&w*3/100)(4L^3*5/100)]
E+&E = [5wL^4(1*3/100)(12*5/100)]
I was just wondering if this was right so far, and if so what would I have to do next? I'm guessing I have to multiply out the brackets, but would I have to ignore any terms?
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http://www.citizendia.org/Elementary_particle | In particle physics, an elementary particle or fundamental particle is a particle not known to have substructure; that is, it is not known to be made up of smaller particles. Particle physics is a branch of Physics that studies the elementary constituents of Matter and Radiation, and the interactions between them If an elementary particle truly has no substructure, then it is one of the basic building blocks of the universe from which all other particles are made. The Universe is defined as everything that Physically Exists: the entirety of Space and Time, all forms of Matter, Energy In the Standard Model, the quarks, leptons, and gauge bosons are elementary particles. The Standard Model of Particle physics is a theory that describes three of the four known Fundamental interactions together with the Elementary particles In Physics, a quark (kwɔrk kwɑːk or kwɑːrk is a type of Subatomic particle. Leptons are a family of fundamental Subatomic particles comprising the Electron, the Muon, and the Tauon (or tau particle as well as their In Particle physics, gauge bosons are Bosonic particles that act as carriers of the fundamental forces of nature [1][2]
## Overview
All elementary particles are either bosons or fermions (depending on their spin). In Particle physics, bosons are particles which obey Bose-Einstein statistics; they are named after Satyendra Nath Bose and Albert Einstein In Particle physics, fermions are particles which obey Fermi-Dirac statistics; they are named after Enrico Fermi. In Quantum mechanics, spin is a fundamental property of atomic nuclei, Hadrons and Elementary particles For particles with non-zero spin The spin-statistics theorem identifies the resulting quantum statistics that differentiates fermions from bosons. The spin-statistics theorem in Quantum mechanics relates the spin of a particle to the statistics obeyed by that particle Particle statistics refers to the particular description of particles in Statistical mechanics. According to this methodology: particles normally associated with matter are fermions, having half-integer spin; they are divided into twelve flavours. Matter is commonly defined as being anything that has mass and that takes up space. In Particle physics, fermions are particles which obey Fermi-Dirac statistics; they are named after Enrico Fermi. In Mathematics, a half-integer is a Number of the form n + 1/2 where n is an Integer. In Particle physics, flavour or flavor (see spelling differences) is a Quantum number of Elementary particles related to their Particles associated with fundamental forces are bosons, having integer spin. In Physics, a fundamental interaction or fundamental force is a mechanism by which particles interact with each other and which cannot be explained in terms In Particle physics, bosons are particles which obey Bose-Einstein statistics; they are named after Satyendra Nath Bose and Albert Einstein The integers (from the Latin integer, literally "untouched" hence "whole" the word entire comes from the same origin but via French [3]
Quarksup, down, charm, strange, top, bottom
Leptonselectron neutrino, electron, muon neutrino, muon, tau neutrino, tau
Gauge bosonsgluon, W and Z bosons, photon
Other bosons — Higgs boson, graviton
## Standard Model
Main article: Standard Model
The Standard Model of particle physics contains 12 flavours of elementary fermions, plus their corresponding antiparticles, as well as elementary bosons that mediate the forces and the still undiscovered Higgs boson. In Particle physics, fermions are particles which obey Fermi-Dirac statistics; they are named after Enrico Fermi. In Physics, a quark (kwɔrk kwɑːk or kwɑːrk is a type of Subatomic particle. The up quark is a particle described by the Standard Model theory of Physics. The down quark is a first-generation Quark with a charge of -(1/3 e. The charm Quark is a second-generation quark with an electric charge of +(2/3 e. The strange quark is a second- generation Quark with a charge of &minus(1/3 e and a strangeness of &minus1 The top quark is the third- generation up-type Quark with a charge of +(2/3 e. The bottom quark is a third-generation Quark with a charge of − e. Leptons are a family of fundamental Subatomic particles comprising the Electron, the Muon, and the Tauon (or tau particle as well as their Neutrinos are Elementary particles that travel close to the Speed of light, lack an Electric charge, are able to pass through ordinary matter almost The electron is a fundamental Subatomic particle that was identified and assigned the negative charge in 1897 by J Neutrinos are Elementary particles that travel close to the Speed of light, lack an Electric charge, are able to pass through ordinary matter almost The muon (from the letter mu (μ--used to represent it is an Elementary particle with negative Electric charge and a spin of 1/2 Neutrinos are Elementary particles that travel close to the Speed of light, lack an Electric charge, are able to pass through ordinary matter almost The tau lepton (often called the tau, tau particle, or occasionally the tauon; symbol) is a negatively charged Elementary particle with In Particle physics, bosons are particles which obey Bose-Einstein statistics; they are named after Satyendra Nath Bose and Albert Einstein In Particle physics, gauge bosons are Bosonic particles that act as carriers of the fundamental forces of nature Gluons ( Glue and the suffix -on) are Elementary particles that cause Quarks to interact and are indirectly responsible for the The W and Z bosons are the Elementary particles that mediate the Weak force. In Physics, the photon is the Elementary particle responsible for electromagnetic phenomena The Higgs Boson is a hypothetical massive scalar Elementary particle predicted to exist by the Standard Model of Particle physics In Physics, the graviton is a hypothetical Elementary particle, a Boson to be exact that mediates the force of Gravity in the framework The Standard Model of Particle physics is a theory that describes three of the four known Fundamental interactions together with the Elementary particles In Particle physics, fermions are particles which obey Fermi-Dirac statistics; they are named after Enrico Fermi. to most kinds of particles, there is an associated antiparticle with the same Mass and opposite Electric charge. In Particle physics, bosons are particles which obey Bose-Einstein statistics; they are named after Satyendra Nath Bose and Albert Einstein The Higgs Boson is a hypothetical massive scalar Elementary particle predicted to exist by the Standard Model of Particle physics However, the Standard Model is widely considered to be a provisional theory rather than a truly fundamental one, since it is fundamentally incompatible with Einstein's general relativity. Albert Einstein ( German: ˈalbɐt ˈaɪ̯nʃtaɪ̯n; English: ˈælbɝt ˈaɪnstaɪn (14 March 1879 – 18 April 1955 was a German -born theoretical General relativity or the general theory of relativity is the geometric theory of Gravitation published by Albert Einstein in 1916 There are likely to be hypothetical elementary particles not described by the Standard Model, such as the graviton, the particle that would carry the gravitational force or the sparticles, supersymmetric partners of the ordinary particles. In Physics, the graviton is a hypothetical Elementary particle, a Boson to be exact that mediates the force of Gravity in the framework Gravitation is a natural Phenomenon by which objects with Mass attract one another " Sparticle " is a merging of the words supersymmetric and particle. In Particle physics, supersymmetry (often abbreviated SUSY) is a Symmetry that relates elementary particles of one spin to another particle that
### Fundamental fermions
Main article: fermion
The 12 fundamental fermionic flavours are divided into three generations of four particles each. In Particle physics, fermions are particles which obey Fermi-Dirac statistics; they are named after Enrico Fermi. In Particle physics, a generation is a division of the Elementary particles Between generations particles differ only by their Mass. Six of the particles are quarks. In Physics, a quark (kwɔrk kwɑːk or kwɑːrk is a type of Subatomic particle. The remaining six are leptons, three of which are neutrinos, and the remaining three of which have an electric charge of −1: the electron and its two cousins, the muon and the tau lepton. Leptons are a family of fundamental Subatomic particles comprising the Electron, the Muon, and the Tauon (or tau particle as well as their Neutrinos are Elementary particles that travel close to the Speed of light, lack an Electric charge, are able to pass through ordinary matter almost The muon (from the letter mu (μ--used to represent it is an Elementary particle with negative Electric charge and a spin of 1/2 The tau lepton (often called the tau, tau particle, or occasionally the tauon; symbol) is a negatively charged Elementary particle with
First generationelectron: e−electron-neutrino: νeup quark: udown quark: d Second generationmuon: μ−muon-neutrino: νμcharm quark: cstrange quark: s Third generationtau: τ−tau-neutrino: ντtop quark: tbottom quark: b
#### Antiparticles
Main article: antimatter
There are also 12 fundamental fermionic antiparticles which correspond to these 12 particles. The electron is a fundamental Subatomic particle that was identified and assigned the negative charge in 1897 by J Neutrinos are Elementary particles that travel close to the Speed of light, lack an Electric charge, are able to pass through ordinary matter almost The up quark is a particle described by the Standard Model theory of Physics. The down quark is a first-generation Quark with a charge of -(1/3 e. The muon (from the letter mu (μ--used to represent it is an Elementary particle with negative Electric charge and a spin of 1/2 Neutrinos are Elementary particles that travel close to the Speed of light, lack an Electric charge, are able to pass through ordinary matter almost The charm Quark is a second-generation quark with an electric charge of +(2/3 e. The strange quark is a second- generation Quark with a charge of &minus(1/3 e and a strangeness of &minus1 The tau lepton (often called the tau, tau particle, or occasionally the tauon; symbol) is a negatively charged Elementary particle with Neutrinos are Elementary particles that travel close to the Speed of light, lack an Electric charge, are able to pass through ordinary matter almost The top quark is the third- generation up-type Quark with a charge of +(2/3 e. The bottom quark is a third-generation Quark with a charge of − e. In Particle physics and Quantum chemistry, antimatter is the extension of the concept of the Antiparticle to Matter, where antimatter is composed The positron e+ corresponds to the electron and has an electric charge of +1 and so on:
First generationpositron: e+electron-antineutrino: $\bar{\nu}_e$up antiquark: $\bar{u}$down antiquark: $\bar{d}$ Second generationpositive muon: μ+muon-antineutrino: $\bar{\nu}_\mu$charm antiquark: $\bar{c}$strange antiquark: $\bar{s}$ Third generationpositive tau: τ+tau-antineutrino: $\bar{\nu}_\tau$top antiquark: $\bar{t}$bottom antiquark: $\bar{b}$
#### Quarks
Main article: quark
Quarks and antiquarks have never been detected to be isolated, a fact explained by confinement. The positrons or antielectron is the Antiparticle or the Antimatter counterpart of the Electron. The positrons or antielectron is the Antiparticle or the Antimatter counterpart of the Electron. In Physics, a quark (kwɔrk kwɑːk or kwɑːrk is a type of Subatomic particle. Color confinement, often called just confinement, is the Physics phenomenon that Color charged particles (such as Quarks cannot be isolated singularly Every quark carries one of three color charges of the strong interaction; antiquarks similarly carry anticolor. In Particle physics, color charge is a property of Quarks and Gluons which are related to their Strong interactions in the context of Quantum In particle physics the strong interaction, or strong force, or color force, holds Quarks and Gluons together to form Protons and Color charged particles interact via gluon exchange in the same way that charged particles interact via photon exchange. Gluons ( Glue and the suffix -on) are Elementary particles that cause Quarks to interact and are indirectly responsible for the In Physics, the photon is the Elementary particle responsible for electromagnetic phenomena However, gluons are themselves color charged, resulting in an amplification of the strong force as color charged particles are separated. Unlike the electromagnetic force which diminishes as charged particles separate, color charged particles feel increasing force. Electromagnetism is the Physics of the Electromagnetic field: a field which exerts a Force on particles that possess the property of
However, color charged particles may combine to form color neutral composite particles called hadrons. In Physics, a bound state is a composite of two or more building blocks ( particles or bodies) that behaves as a single object In Particle physics, a hadron ( from the ἁδρός hadrós, " stout, thick " ( A quark may pair up to an antiquark: the quark has a color and the antiquark has the corresponding anticolor. The color and anticolor cancel out, forming a color neutral meson. In Particle physics, a meson is a strongly interacting Boson &mdashthat is a Hadron with integer spin. Alternatively, three quarks can exist together, one quark being "red", another "blue", another "green". These three colored quarks together form a color-neutral baryon. Baryons are the family of Subatomic particles with a Baryon number of 1 Symmetrically, three antiquarks with the colors "antired", "antiblue" and "antigreen" can form a color-neutral antibaryon. Baryons are the family of Subatomic particles with a Baryon number of 1
Quarks also carry fractional electric charges, but since they are confined within hadrons whose charges are all integral, fractional charges have never been isolated. Electric charge is a fundamental conserved property of some Subatomic particles which determines their Electromagnetic interaction. Note that quarks have electric charges of either +2/3 or −1/3, whereas antiquarks have corresponding electric charges of either −2/3 or +1/3.
Evidence for the existence of quarks comes from deep inelastic scattering: firing electrons at nuclei to determine the distribution of charge within nucleons (which are baryons). Deep inelastic scattering is the name given to a process used to probe the insides of Hadrons (particularly the Baryons, such as Protons and Neutrons The electron is a fundamental Subatomic particle that was identified and assigned the negative charge in 1897 by J The nucleus of an Atom is the very dense region consisting of Nucleons ( Protons and Neutrons, at the center of an atom In Physics a nucleon is a collective name for two Baryons the Neutron and the Proton. If the charge is uniform, the electric field around the proton should be uniform and the electron should scatter elastically. In Physics, the space surrounding an Electric charge or in the presence of a time-varying Magnetic field has a property called an electric field (that can Low-energy electrons do scatter in this way, but above a particular energy, the protons deflect some electrons through large angles. The recoiling electron has much less energy and a jet of particles is emitted. A jet is a narrow cone of Hadrons and other particles produced by the Hadronization of a Quark or Gluon in a Particle physics This inelastic scattering suggests that the charge in the proton is not uniform but split among smaller charged particles: quarks.
### Fundamental bosons
Main article: boson
In the Standard Model, vector (spin-1) bosons (gluons, photons, and the W and Z bosons) mediate forces, while the Higgs boson (spin-0) is responsible for particles having intrinsic mass. In Particle physics, bosons are particles which obey Bose-Einstein statistics; they are named after Satyendra Nath Bose and Albert Einstein In Quantum mechanics, spin is a fundamental property of atomic nuclei, Hadrons and Elementary particles For particles with non-zero spin Gluons ( Glue and the suffix -on) are Elementary particles that cause Quarks to interact and are indirectly responsible for the In Physics, the photon is the Elementary particle responsible for electromagnetic phenomena The W and Z bosons are the Elementary particles that mediate the Weak force. The Higgs Boson is a hypothetical massive scalar Elementary particle predicted to exist by the Standard Model of Particle physics Mass is a fundamental concept in Physics, roughly corresponding to the Intuitive idea of how much Matter there is in an object
#### Gluons
Main article: gluon
Gluons are the mediators of the strong interaction and carry both colour and anticolour. Gluons ( Glue and the suffix -on) are Elementary particles that cause Quarks to interact and are indirectly responsible for the In particle physics the strong interaction, or strong force, or color force, holds Quarks and Gluons together to form Protons and In Particle physics, color charge is a property of Quarks and Gluons which are related to their Strong interactions in the context of Quantum Although gluons are massless, they are never observed in detectors due to colour confinement; rather, they produce jets of hadrons, similar to single quarks. In experimental and applied Particle physics and Nuclear engineering, a particle detector, also known as a radiation detector, is a device used to Color confinement, often called just confinement, is the Physics phenomenon that Color charged particles (such as Quarks cannot be isolated singularly A jet is a narrow cone of Hadrons and other particles produced by the Hadronization of a Quark or Gluon in a Particle physics In Particle physics, a hadron ( from the ἁδρός hadrós, " stout, thick " ( In Physics, a quark (kwɔrk kwɑːk or kwɑːrk is a type of Subatomic particle. The first evidence for gluons came from annihilations of electrons and positrons at high energies which sometimes produced three jets — a quark, an antiquark, and a gluon. In Particle physics, a three-jet event is an event with many particles in final state that appear to be clustered in three jets A single jet consists
#### Electroweak bosons
Main article: W and Z bosons
There are three weak gauge bosons: W+, W, and Z0; these mediate the weak interaction. The W and Z bosons are the Elementary particles that mediate the Weak force. The W and Z bosons are the Elementary particles that mediate the Weak force. The weak interaction (often called the weak force or sometimes the weak nuclear force) is one of the four Fundamental interactions of nature The massless photon mediates the electromagnetic interaction. In Physics, the photon is the Elementary particle responsible for electromagnetic phenomena Electromagnetism is the Physics of the Electromagnetic field: a field which exerts a Force on particles that possess the property of
#### Higgs boson
Main article: Higgs boson
Although the weak and electromagnetic forces appear quite different to us at everyday energies, the two forces are theorized to unify as a single electroweak force at high energies. The Higgs Boson is a hypothetical massive scalar Elementary particle predicted to exist by the Standard Model of Particle physics In Particle physics, the electroweak interaction is the unified description of two of the four Fundamental interactions of nature Electromagnetism and the This prediction was clearly confirmed by measurements of cross-sections for high-energy electron-proton scattering at the HERA collider at DESY. HERA ( Hadron-Elektron-Ringanlage, or Hadron-Electron Ring Accelerator was a Particle accelerator at DESY in Hamburg. The DESY ( D eutsches E lektronen Sy nchrotron "German Electron Synchrotron" is the biggest German research center for Particle physics The differences at low energies is a consequence of the high masses of the W and Z bosons, which in turn are a consequence of the Higgs mechanism. The Higgs mechanism is Spontaneous symmetry breaking in a Gauge theory. Through the process of spontaneous symmetry breaking, the Higgs selects a special direction in electroweak space that causes three electroweak particles to become very heavy (the weak bosons) and one to remain massless (the photon). In Physics, spontaneous symmetry breaking occurs when a system that is symmetric with respect to some Symmetry group goes into a Vacuum state Although the Higgs mechanism has become an accepted part of the Standard Model, the Higgs boson itself has not yet been observed in detectors. The Higgs Boson is a hypothetical massive scalar Elementary particle predicted to exist by the Standard Model of Particle physics Indirect evidence for the Higgs boson suggests its mass lies below 200-250 GeV. [4] In this case, the LHC experiments may be able to discover this last missing piece of the Standard Model.
## Beyond the Standard Model
Although all experimental evidence confirms the predictions of the Standard Model, many physicists find this model to be unsatisfactory due to its many undetermined parameters, many fundamental particles, the non-observation of the Higgs boson and other more theoretical considerations such as the hierarchy problem. The Standard Model of Particle physics is a theory that describes three of the four known Fundamental interactions together with the Elementary particles The Higgs Boson is a hypothetical massive scalar Elementary particle predicted to exist by the Standard Model of Particle physics In Theoretical physics, a hierarchy problem occurs when the fundamental parameters ( couplings or masses of some Lagrangian are vastly different (usually There are many speculative theories beyond the Standard Model which attempt to rectify these deficiencies.
### Grand unification
One extension of the Standard Model attempts to combine the electroweak interaction with the strong interaction into a single 'grand unified theory' (GUT). Grand Unification, grand unified theory, or GUT refers to any of several very similar unified field theories or models in Physics that In Particle physics, the electroweak interaction is the unified description of two of the four Fundamental interactions of nature Electromagnetism and the In particle physics the strong interaction, or strong force, or color force, holds Quarks and Gluons together to form Protons and Such a force would be spontaneously broken into the three forces by a Higgs-like mechanism. In Physics, spontaneous symmetry breaking occurs when a system that is symmetric with respect to some Symmetry group goes into a Vacuum state The Higgs mechanism is Spontaneous symmetry breaking in a Gauge theory. The most dramatic prediction of grand unification is the existence of X and Y bosons, which cause proton decay. In Particle physics, the X and Y bosons are hypothetical Elementary particles analogous to the W and Z bosons, but corresponding to a new type of force In Particle physics, proton decay is a hypothetical form of Radioactive decay in which the Proton decays into lighter Subatomic particles However, the non-observation of proton decay at Super-Kamiokande rules out the simplest GUTs, including SU(5) and SO(10). Super-Kamiokande, or Super-K for short is a neutrino observatory in the city of Hida, Gifu Prefecture, Japan.
### Supersymmetry
Main article: supersymmetry
Supersymmetry extends the Standard Model by adding an additional class of symmetries to the Lagrangian. In Particle physics, supersymmetry (often abbreviated SUSY) is a Symmetry that relates elementary particles of one spin to another particle that The Lagrangian, L of a Dynamical system is a function that summarizes the dynamics of the system These symmetries exchange fermionic particles with bosonic ones. In Particle physics, fermions are particles which obey Fermi-Dirac statistics; they are named after Enrico Fermi. In Particle physics, bosons are particles which obey Bose-Einstein statistics; they are named after Satyendra Nath Bose and Albert Einstein Such a symmetry predicts the existence of supersymmetric particles, abbreviated as sparticles, which include the sleptons, squarks, neutralinos and charginos. " Sparticle " is a merging of the words supersymmetric and particle. In Particle physics, a sfermion is any of the class of spin -0 Superpartners of ordinary Fermions appearing in supersymmetric extensions In Particle physics, a sfermion is any of the class of spin -0 Superpartners of ordinary Fermions appearing in supersymmetric extensions In Particle physics, the neutralino is a hypothetical particle part of the doubling of the menagerie of particles predicted by supersymmetric theories The chargino is a hypothetical Supersymmetric particle. It refers to the mass Eigenstates of a charged Superpartner, i Each particle in the Standard Model would have a superpartner whose spin differs by 1/2 from the ordinary particle. In Quantum mechanics, spin is a fundamental property of atomic nuclei, Hadrons and Elementary particles For particles with non-zero spin Due to the breaking of supersymmetry, the sparticles are much heavier than their ordinary counterparts; they are so heavy that existing particle colliders would not be powerful enough to produce them. In Particle physics, supersymmetry breaking is the process to obtain a seemingly non- Supersymmetric physics from a supersymmetric theory which is a necessary step A collider is a type of a Particle accelerator involving directed beams of particles. However, some physicists believe that sparticles will be detected when the Large Hadron Collider at CERN begins running. The European Organization for Nuclear Research (Organisation Européenne pour la Recherche Nucléaire known as CERN
### String theory
Main article: string theory
String Theory is a theory of physics where all "particles" that make up matter and energy are comprised of strings (measuring at the Planck length) that exist in an 11-dimensional (according to M-theory, the leading version) universe. String theory is a still-developing scientific approach to Theoretical physics, whose original building blocks are one-dimensional extended objects called strings In Theoretical physics, M-theory is a new limit of String theory in which 11 dimensions of Spacetime may be identified These strings vibrate at different frequencies which determine mass, electric charge, color charge, and spin. A string can be open (a line) or closed in a loop (a one-dimensional sphere, like a circle). As a string moves through space it sweeps out something called a world sheet. String theory predicts 1- to 10-branes (a 1-brane being a string and a 10-brane being a 10-dimensional object) which prevent tears in the "fabric" of space using the uncertainty principle (e. In Theoretical physics, a membrane, brane, or p -brane is a spatially extended mathematical concept that appears in String theory In Quantum physics, the Heisenberg uncertainty principle states that locating a particle in a small region of space makes the Momentum of the particle uncertain g. the electron orbiting a hydrogen atom has the probability, albeit small, that it could be anywhere else in the universe at any given moment).
String theory posits that our universe is merely a 4-brane, inside which exist the 3 space dimensions and the 1 time dimension that we observe. The remaining 6 theoretical dimensions are either very tiny and curled up (and too small to affect our universe in any way) or simply do not/cannot exist in our universe (because they exist in a grander scheme called the "multiverse" outside our known universe).
One particularly interesting prediction of string theory is the existence of extremely massive counterparts of ordinary particles due to vibrational excitations of the fundamental string. Another important prediction is the existence of a massless spin-2 particle behaving like the graviton. In Physics, the graviton is a hypothetical Elementary particle, a Boson to be exact that mediates the force of Gravity in the framework
### Preon theory
Main article: preon
According to preon theory there are one or more orders of particles more fundamental than those (or most of those) found in the Standard Model. In Particle physics, preons are postulated "point-like" particles conceived to be subcomponents of Quarks and Leptons The word was coined The Standard Model of Particle physics is a theory that describes three of the four known Fundamental interactions together with the Elementary particles The most fundamental of these are normally called preons, which is derived from "pre-quarks". In essence, preon theory tries to do for the Standard Model what the Standard Model did for the particle zoo that came before it. The Standard Model of Particle physics is a theory that describes three of the four known Fundamental interactions together with the Elementary particles In Particle physics, the term particle zoo is used colloquially to describe a relatively extensive list of the known elementary particles that almost look like hundreds of species Most models assume that almost everything in the Standard Model can be explained in terms of three to half a dozen more fundamental particles and the rules that govern their interactions. Interest in preons has waned since the simplest models were experimentally ruled out in the 1980s. | 2013-06-19 08:12:13 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 9, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6559720039367676, "perplexity": 612.6517266520344}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368708144156/warc/CC-MAIN-20130516124224-00092-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://www.gradesaver.com/textbooks/science/physics/conceptual-physics-12th-edition/chapter-7-think-and-explain-page-129-130/85 | ## Conceptual Physics (12th Edition)
Published by Addison-Wesley
# Chapter 7 - Think and Explain: 85
#### Answer
If mass is doubled while speed stays the same, the momentum mv turns into (2m)v, and increases by a factor of 2. The KE, $\frac{1}{2}mv^{2}$, becomes $\frac{1}{2}(2m)v^{2}$ and also increases by a factor of 2.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 2018-05-26 17:52:13 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7283110022544861, "perplexity": 1855.449252286551}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794867841.63/warc/CC-MAIN-20180526170654-20180526190654-00010.warc.gz"} |
https://math.stackexchange.com/questions/2156093/exponential-equation-6x8x15x-9x10x12x/2156104 | # exponential equation: $6^x+8^x+15^x=9^x+10^x+12^x$
What are the solutions of this equation? Or at least in which interval are they? $$6^x+8^x+15^x=9^x+10^x+12^x$$ I tried to find an increasing function, or use some inequalities but I got nothing out of it...
• Where does this equation come from? – Henning Makholm Feb 22 '17 at 11:15
• $x=0, x=2$ tis a solution. – S.C.B. Feb 22 '17 at 11:17
• have you tried plotting it – mercio Feb 22 '17 at 11:17
• $0$ is obvious and WA got $2$ too. – Zubzub Feb 22 '17 at 11:17
Let $a=3^x$ and $b=2^x$ and $c=5^x$.
Then we have that $$ab+b^3+ac=a^2+bc+ab^2$$ $$ab+b^3+ac-a^2-bc-ab^2=0$$ $$a(b-a)+b^2(b-a)-c(b-a)=0$$ $$(b-a)(a+b^2-c)=0$$
Now $a=b \implies x=0$ and $a+b^2-c=0 \implies 3^x+4^x=5^x$ which gives solution for $x=2$ only and not for any higher integer $x$ by Fermat's Last Theorem.
So these are the $2$ solutions.
• Love this use of FLT ! – Zubzub Feb 22 '17 at 11:26
• Can we use FLT here when there is no guarantee of $x$ being an integer? – didgogns Feb 22 '17 at 11:35
• No, but there's a simple way around this: Dividing both sides by $5^x$ gives $\left(\frac{3}{5}\right)^x + \left(\frac{4}{5}\right)^x = 1$, but the quantity on the l.h.s. is a strictly decreasing function of $x$, so the equation has at most one solution. – Travis Feb 22 '17 at 11:59
Hint: your equation can be factorized as $$\left(2^x-3^x\right) \left(2^{2 x}+3^x-5^x\right)=0$$
• How did you get this? I mean, it doesn't seem obvious. – S.C.B. Feb 22 '17 at 11:19
• I know this equation from Mathlinks – Dr. Sonnhard Graubner Feb 22 '17 at 11:20
• @S.C.B We can write the equation as: $$2^{3x} + 2^x3^x +3^x5^x = 2^{2x}3^x + 3^{2x} + 2^x5^x$$ $$\Rightarrow 2^x2^{2x} +2^x3^x-2^x5^x = 3^x2^{2x} +3^x3^x-3^x5^x$$ – Rohan Feb 22 '17 at 11:21
• i think this doesn't help at all – Dr. Sonnhard Graubner Feb 22 '17 at 11:22
• @Dr.SonnhardGraubner What is Mathlinks? – S.C.B. Feb 22 '17 at 11:22 | 2019-09-17 23:14:22 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5094958543777466, "perplexity": 527.2910069575439}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514573124.40/warc/CC-MAIN-20190917223332-20190918005332-00201.warc.gz"} |
http://community.boredofstudies.org/13/mathematics-extension-1/348462/cambridge-hsc-mx1-textbook-marathon-q-3.html | # Thread: Cambridge HSC MX1 Textbook Marathon/Q&A
1. ## Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread
Originally Posted by davidgoes4wce
$Cambridge Exercise 3E Q4 (a)$
$A stone is dropped from a lookout 500 metres above the valley floor. Take g =10 m/s^{2}, ignore air resistance, take downwards as positive, and use the lookout as the origin of displacement- the equation of motion is then \ddot{x}=10$
$(a) Replace \ \ddot{x} \ by \frac{d}{dx} (\frac{1}{2} v^2) and show that v^{2}=20x. Hence find the impact speed.$
$(b) \ Explain why, during the fall, v=\sqrt{20x} rather than v=-\sqrt{20x}$
$(c) Integrate to find the displacement-time function, and find how long it takes to fall.$
$Ans a) 100 m/s b) Downwards is positive, so while the stone is falling, its velocity is positive c) x=5t^{2}, 10 seconds$
Just had a bit of trouble with part (a)
I have nothing else to do on a Monday night and have no social life, so I decided to answer my own question.
2. ## Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread
The amplitude of a particle moving in simple harmonic motion is 5 metres, and its acceleration when 2 metres from its mean position is 4 m/s^2. Find the speed of the particle at the mean position and when it is 4 m from the mean position.
do i form an equation of: x double dot = -n^2 (x - 2) ??
which becomes 4 = - n^2 (x-2) but then i dont know how to solve for n
or is the equation just 4 = -n^2 x 2
but then how do i get rid of the negative to get a value for n.
yeah i know really easy question, i just havent grasped the concept of shifting away from the centre. any reply will help heaps thanks!
3. ## Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread
Originally Posted by hedgehog_7
The amplitude of a particle moving in simple harmonic motion is 5 metres, and its acceleration when 2 metres from its mean position is 4 m/s^2. Find the speed of the particle at the mean position and when it is 4 m from the mean position.
do i form an equation of: x double dot = -n^2 (x - 2) ??
which becomes 4 = - n^2 (x-2) but then i dont know how to solve for n
or is the equation just 4 = -n^2 x 2
but then how do i get rid of the negative to get a value for n.
yeah i know really easy question, i just havent grasped the concept of shifting away from the centre. any reply will help heaps thanks!
We can set the mean position to be x = 0.
So we can say v^2 = n^2 (A^2 - x^2), where A is the amplitude, n is the usual n (angular frequency is what n is).
(If we have centre of motion (mean position) x0 instead, the formula becomes v^2 = n^2 (A^2 - (x - x0)^2).)
We already know A = 5 (given), so v^2 = n^2 (25 - x^2).
Using SHM differential equation, we have x-double-dot = -n^2 x. We know x-double-dot = -4 when x = 4. (Remember in SHM, acceleration and x - x0 have opposite signs at all times.)
So -4 = -n^2 * 4, so n = 1.
So v^2 = 25 - x^2.
Now you can find v^2 when |x| = 4; at this moment, v^2 = 25 - 16 = 9, so the speed is 3 m/s.
4. ## Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread
Originally Posted by InteGrand
We can set the mean position to be x = 0.
So we can say v^2 = n^2 (A^2 - x^2), where A is the amplitude, n is the usual n (angular frequency is what n is).
(If we have centre of motion (mean position) x0 instead, the formula becomes v^2 = n^2 (A^2 - (x - x0)^2).)
We already know A = 5 (given), so v^2 = n^2 (25 - x^2).
Using SHM differential equation, we have x-double-dot = -n^2 x. We know x-double-dot = -4 when x = 4. (Remember in SHM, acceleration and x - x0 have opposite signs at all times.)
So -4 = -n^2 * 4, so n = 1.
So v^2 = 25 - x^2.
Now you can find v^2 when |x| = 4; at this moment, v^2 = 25 - 16 = 9, so the speed is 3 m/s.
question, why is x double dot equal to -4 and not just 4? because isnt the acceleration just 4m/s^2. the problem i had with that was then since
x double dot = - n^2 x
is the x (the distance for the mean position) can it be equal to 2 or -2 ? since it can be above or below the mean position?? can i take it as x = -2, so when i sub it into x double dot, i can actually solve for n?
5. ## Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread
Originally Posted by hedgehog_7
question, why is x double dot equal to -4 and not just 4?
As I said, in simple harmonic motion, the acceleration and displacement from centre of motion are always opposite signs (this is because the acceleration is always trying to push the object back to its centre of motion, as you can visualise by imagining a pendulum for instance). So when x = 4 (displacement from centre of motion), the acceleration must be negative.
6. ## Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread
Originally Posted by hedgehog_7
question, why is x double dot equal to -4 and not just 4? because isnt the acceleration just 4m/s^2. the problem i had with that was then since
x double dot = - n^2 x
is the x (the distance for the mean position) can it be equal to 2 or -2 ? since it can be above or below the mean position?? can i take it as x = -2, so when i sub it into x double dot, i can actually solve for n?
If you take x to be negative, the acceleration you plug in needs to be positive.
And when they said the acceleration is 4 m s-2, that is a bit misleading, since what they really were referring to was the magnitude of the acceleration. But as I said, the acceleration and displacement have opposite signs, so you can deduce everything from that to solve for n.
7. ## Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread
Originally Posted by InteGrand
If you take x to be negative, the acceleration you plug in needs to be positive.
And when they said the acceleration is 4 m s-2, that is a bit misleading, since what they really were referring to was the magnitude of the acceleration. But as I said, the acceleration and displacement have opposite signs, so you can deduce everything from that to solve for n.
oh ok thanks for clarifying, so when it states that the displacement from the mean position is 2, i CAN take it as x = 2 or x = -2? considering above and below respecitvely?
8. ## Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread
Originally Posted by hedgehog_7
oh ok thanks for clarifying, so when it states that the displacement from the mean position is 2, i CAN take it as x = 2 or x = -2? considering above and below respecitvely?
Yeah. You only need to take one of them to find n.
9. ## Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread
How common is Volumes in the HSC Exams? I rarely see them o.O
http://i.imgur.com/kzIHB2w.jpg
ew imperial
yuck
11. ## Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread
Originally Posted by si2136
How common is Volumes in the HSC Exams? I rarely see them o.O
http://i.imgur.com/kzIHB2w.jpg
Which part do you need help with? Just finding the volume? Once we've found that, multiplying it by 0.25 will give us the mass in tons.
12. ## Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread
Originally Posted by InteGrand
Which part do you need help with? Just finding the volume? Once we've found that, multiplying it by 0.25 will give us the mass in tons.
Yes, the volume. And are feet and foot the same (dumb question)?
13. ## Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread
Originally Posted by si2136
Yes, the volume. And are feet and foot the same (dumb question)?
Yeah they're the same. I think you need to use the fact that AP is one inch and convert it to feet. From Google, 1 inch = 0.0833333 foot = 5/60 foot = 1/12 foot.
This means viewing everything with origin A, the equation of the inner parabola is y2 = 160(x – (1/12)) (since it's shifted right by 1/12 compared to the outer parabola).
We can now find the volume as follows. Find the volume obtained by rotating the outer parabola about the x-axis from A to X (i.e. from x = 0 to x = 10). Then subtract off the volume obtained by rotating the inner parabola about the x-axis from P to X (i.e. from x = 1/12 to x = 10).
14. ## Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread
Originally Posted by InteGrand
Yeah they're the same. I think you need to use the fact that AP is one inch and convert it to feet. From Google, 1 inch = 0.0833333 foot = 5/60 foot = 1/12 foot.
This means viewing everything with origin A, the equation of the inner parabola is y2 = 160(x – (1/12)) (since it's shifted right by 1/12 compared to the outer parabola).
We can now find the volume as follows. Find the volume obtained by rotating the outer parabola about the x-axis from A to X (i.e. from x = 0 to x = 10). Then subtract off the volume obtained by rotating the inner parabola about the x-axis from P to X (i.e. from x = 1/12 to x = 10).
Yes, you do need to convert. But that's just blasphemy! How is someone expected to do this question without that knowledge
15. ## Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread
Originally Posted by si2136
Yes, you do need to convert. But that's just blasphemy! How is someone expected to do this question without that knowledge
I doubt they'd ask it in the HSC without providing a conversion rate though, so don't worry.
16. ## Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread
Originally Posted by InteGrand
We can set the mean position to be x = 0.
So we can say v^2 = n^2 (A^2 - x^2), where A is the amplitude, n is the usual n (angular frequency is what n is).
(If we have centre of motion (mean position) x0 instead, the formula becomes v^2 = n^2 (A^2 - (x - x0)^2).)
We already know A = 5 (given), so v^2 = n^2 (25 - x^2).
Using SHM differential equation, we have x-double-dot = -n^2 x. We know x-double-dot = -4 when x = 4. (Remember in SHM, acceleration and x - x0 have opposite signs at all times.)
So -4 = -n^2 * 4, so n = 1.
So v^2 = 25 - x^2.
Now you can find v^2 when |x| = 4; at this moment, v^2 = 25 - 16 = 9, so the speed is 3 m/s.
Is there something wrong with the solution for that question because it shows the speed to be $v=5 \sqrt{2} \ and \ 3 \sqrt{2}$
17. ## Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread
Originally Posted by davidgoes4wce
Is there something wrong with the solution for that question because it shows the speed to be $v=5 \sqrt{2} \ and \ 3 \sqrt{2}$
Oops, no, it's correct, I realise I misread the Q., the acceleration was 4 at x = 2; I accidentally did it at x = 4. So I got a different value for n. The actual value for n would be sqrt(2).
18. ## Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread
Originally Posted by InteGrand
Oops, no, it's correct, I realise I misread the Q., the acceleration was 4 at x = 2; I accidentally did it at x = 4. So I got a different value for n. The actual value for n would be sqrt(2).
I fixed up your mistake and redid the solution and was able to get the speeds.
19. ## Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread
Hey guys,
Sorry I'm a bit new to using this. :P
Could I get some help with Exercise 10I question 10???
There are two round tables, one oak and one mahogany, each with five seats. In how many
ways may a group of ten people be seated?
Thanks
20. ## Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread
There are (5-1)! ways to arrange the 5 people in the first table then another (5-1)! ways to arrange in the other table. But for the first table, any 5 of the 10 people can be chosen and the remaining 5 chosen for the 2nd table.
So the answer is 10C5*4!*4! = 145152
21. ## Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread
Cambridge EX 10 D Q 19
$Two dice are rolled. A three appears on at least one of the dice. Find the probability that the sum of the uppermost faces is greater than seven.$
Thought the possible sample outcomes was : (3,5), (3,6), (5,3) and (5,6).
Am I missing something?
22. ## Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread
Originally Posted by davidgoes4wce
Cambridge EX 10 D Q 19
$\\ Two dice are rolled. A three appears on at least one of the dice. Find the probability that the sum of the uppermost faces is greater than seven.$
Thought the possible sample outcomes was : (3,5), (3,6), (5,3) and (5,6).
Am I missing something?
$The favourable outcomes are as follows: (3,5) , (3,6), (5,3), (6,3) (Thus there are four favourable outcomes.) The number of possible outcomes (where a three appears on at least one of the dice) is 6^2 - 5^2 = 11.$
$\\ Hence the probability is \frac{4}{11}.$
23. ## Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread
Can understand it a bit better now, my thinking initially was 4/36 but can see why its 4/11, since the total possible combinations with a 3 are: (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (1,3), (2,3), (4,3), (5,3) and (6,3).
Only 4 of those have combined value above 8 or more.
24. ## Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread
Ex 3A Q 9 (h)
$A particle moves according to x=10 \ cos \ \frac{\pi \ t}{12}, in units of metres and seconds$
$Use the graph and table of values to find when the particle is more than 15 metres from its initial position$
$I got the following graph :$
Should the solution be between the range?
8 < t < 16, 32 < t < 40, 54 < t< 60 ?
8 < t < 16
25. ## Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread
Originally Posted by davidgoes4wce
Ex 3A Q 9 (h)
$A particle moves according to x=10 \ cos \ \frac{\pi \ t}{12}, in units of metres and seconds$
$Use the graph and table of values to find when the particle is more than 15 metres from its initial position$
$I got the following graph :$
Should the solution be between the range?
8 < t < 16, 32 < t < 40, 54 < t< 60 ?
8 < t < 16
Since they only wanted you to consider when you could use the graph AND the table of values...
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• | 2018-10-22 20:25:03 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 18, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7833376526832581, "perplexity": 1114.1033001384017}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583515539.93/warc/CC-MAIN-20181022201445-20181022222945-00224.warc.gz"} |
https://math.stackexchange.com/questions/2594895/how-to-evaluate-sums-in-the-form-sum-k-infty-infty-e-pi-n-k2 | # How to evaluate sums in the form $\sum_{k=-\infty}^\infty e^{-\pi n k^2}$
Online, one may find the values of the following sums: $$\sum_{k=-\infty}^\infty e^{-\pi k^2}=\frac{\pi^{1/4}}{\Gamma(3/4)}$$ $$\sum_{k=-\infty}^\infty e^{-2\pi k^2}=\frac{\pi^{1/4}(6+4\sqrt 2)^{1/4}}{2\Gamma(3/4)}$$ $$\sum_{k=-\infty}^\infty e^{-3\pi k^2}=\frac{\pi^{1/4}(27+18\sqrt 3)^{1/4}}{3\Gamma(3/4)}$$ Can someone show me how to prove at least one of these? I've already tried using the reside theorem but had no luck with that.
## 2 Answers
The sums in question are nothing but values of Jacobi theta function defined by $$\vartheta_{3}(q)=\sum_{n\in\mathbb{Z}}q^{n^2}\tag{1}$$ Evaluation of these functions for certain specific values of $$q$$ is done via the help of their friends called elliptic integrals. Before discussing the problem of evaluation of theta functions, it is best to give preliminary information about elliptic integrals.
We start with a number $$k\in(0,1)$$ called elliptic modulus and define another number $$k'=\sqrt{1-k^2}$$ called complementary (to $$k$$) modulus. The following equation $$K(k) =\int_{0}^{\pi/2}\frac{dx}{\sqrt{1-k^2\sin^{2}x}}\tag{2}$$ defines complete elliptic integral of first kind $$K(k)$$ for modulus $$k$$. The expressions $$K(k), K(k')$$ are usually denoted by $$K, K'$$ respectively if $$k$$ is known from context. It is then a wonderful surprise that if the values $$K, K'$$ are known the value of modulus $$k$$ can be obtained via Jacobi theta functions with argument $$q=e^{-\pi K'/K}$$ (also called nome) $$k=\frac{\vartheta_{2}^{2}(q)}{\vartheta_{3}^{2}(q)},\,\vartheta_{2}(q)=\sum_{n\in\mathbb{Z}}q^{(n+(1/2))^2}\tag{3}$$ Also under these circumstances we have $$\vartheta_{3}^{2}(q)=\frac{2K}{\pi}\tag{4}$$ The most interesting aspect of these functions and integrals was understood by Ramanujan and he championed the idea of a modular equation to which we turn next.
The function $$f(k) =K(k') /K(k)$$ is strictly decreasing and maps interval $$(0,1)$$ to $$(0,\infty)$$ and hence if $$p$$ is a positive real number then there exists a unique number $$l\in(0,1)$$ such that $$\frac{K(l')} {K(l)} =p\frac{K(k')} {K(k)} \tag{5}$$ (here $$l'=\sqrt{1-l^2}$$ is complementary to $$l$$, also $$K(l), K(l')$$ are usually denoted by $$L, L'$$). Thus given $$k\in(0,1),p\in(0,\infty)$$ we have a new modulus $$l$$ such that the above equation holds and if $$p$$ is fixed in our discussion then $$l$$ is a function of $$k$$. Jacobi proved in his Fundamenta Nova that if $$p$$ is a positive rational number then the relationship between $$k, l$$ is algebraic and this relationship between $$k, l$$ in form of an algebraic equation is called a modular equation of degree $$p$$. It is a computational challenge to find such equations for large values of $$p$$ and Ramanujan was an expert in finding such modular equations.
Let $$P(k, l) =0$$ be the modular equation of degree $$p$$ where $$P$$ is a polynomial with rational coefficients. Ramanujan added another constraint in this equation namely $$l=k'$$ so that $$k=l'$$ then the equation $$P(k, k') =0$$ shows that both $$k, l=k'$$ are algebraic numbers. And in that case equation $$(5)$$ leads us to $$\frac{K(l')} {K(l)} =\sqrt{p}, \frac{K(k')} {K(k)} =\frac{1}{\sqrt{p}}\tag{6}$$ and thus we have the following theorem
Theorem: If $$p$$ is a positive rational number and $$K(k') /K(k) =\sqrt{p}$$ then $$k$$ is an algebraic number and such values of $$k$$ are called singular moduli.
From now on $$p$$ will denote a positive rational number unless otherwise stated. Note that if $$Q=\exp(-\pi K(l') /K(l))$$ then equation $$(5)$$ shows that $$Q=q^{p}$$. From equation $$(3)$$ it follows that a modular equation can also be thought as an algebraic relation between theta functions of arguments $$q$$ and $$q^{p}$$. It can also be proved by differentiating the equation $$(5)$$ that the ratio $$K(k) /K(l)$$ can be expressed as an algebraic expression in $$k, l$$. Ramanujan expressed many of his modular equations as algebraic expressions for $$K/L$$ and that's what we need here.
We start with $$q=e^{-\pi}$$ so that $$K=K', k=k'=1/\sqrt{2}$$ and the value of integral $$K$$ is easily evaluated for this value of $$k$$ which gives the desired value of the first sum in question. From equation $$(4)$$ it follows that $$\frac{\vartheta_{3}(q)}{\vartheta_{3}(q^p)}=\sqrt{\frac{K}{L}}$$ And as discussed earlier the ratio $$K/L$$ can be expressed as an algebraic function of $$k, l$$ therefore the evaluation of $$\vartheta_{3}(q^p)$$ can be performed if the value of $$l$$ as well expression for $$K/L$$ in terms of $$k, l$$ is known. The value of $$l$$ can be obtained by solving the modular equation $$P(k, l) =0$$ as $$k=1/\sqrt{2}$$ is known.
To sum up we need the modular equation connecting $$k, l$$ as well as expression for $$K/L$$ in terms of $$k, l$$. These are well known and have a simple form if $$p=2$$ and are famously known as Landen transformation $$l=\frac{1-k'}{1+k'},\frac{K}{L}=\frac{2}{1+k'}\tag{6}$$ Thus putting $$k'=1/\sqrt{2}$$ we get $$K/L=4/(2+\sqrt{2})$$ and the second sum in question $$\vartheta_{3}(e^{-2\pi})=\sqrt{\frac{L}{K}}\vartheta_{3}(e^{-\pi})=\frac{\pi^{1/4}\sqrt{2+\sqrt{2}}}{2\Gamma(3/4)}$$ has the value as mentioned in your post.
For $$p=3$$ we have the following modular equation $$\sqrt{kl} +\sqrt{k'l'} =1\tag{7}$$ Putting $$k=k'=2^{-1/2}$$ we get $$\sqrt{l} +\sqrt{l'} =2^{1/4}$$. With some effort the value of $$l$$ can be obtained. To get the value of $$K/L$$ we differentiate equation $$(5)$$ and get $$\frac{dl} {dk} =p\frac{ll'^{2}L^{2}}{kk'^{2}K^{2}}$$ Thus $$\left(\frac{L} {K} \right) ^{2}=\frac{kk'^{2}}{3ll'^{2}}\frac{dl}{dk}\tag{8}$$ Differentiating equation $$(7)$$ with respect to $$k$$ we get $$\sqrt{\frac{l} {k}}+\sqrt{\frac{k} {l}} \frac{dl} {dk} - \frac{k}{k'} \sqrt{\frac{l'} {k'}} - \frac{l} {l'}\sqrt{\frac{k'}{l'}} \frac{dl} {dk} =0$$ ie $$\frac{dl}{dk} =\dfrac{\sqrt{\dfrac{l}{k}}-\dfrac{k}{k'}\sqrt{\dfrac{l'}{k'}}}{\dfrac{l}{l'}\sqrt{\dfrac{k'}{l'}}-\sqrt{\dfrac{k}{l}}}$$ and putting this value of $$dl/dk$$ in equation $$(8)$$ we get the value of $$L^2/K^2$$ in terms of $$l, l'$$ (the value $$k=k'=2^{-1/2}$$ being used in the process) as $$\left(\frac{L} {K} \right) ^{2}=\frac{1}{6\sqrt{ll'}}\cdot\frac{1}{l+l'+\sqrt{ll'}}\tag{9}$$ Since the values of $$l, l'$$ and their square roots are known we can get the value of ratio $$L^2/K^2$$ in form of a radical expression.
The algebraic calculations are formidable and I managed to get $$\sqrt{l} =\frac{\sqrt{2}-\sqrt[4]{3} (\sqrt{3}-1)} {2^{5/4}}, \sqrt{l'} =\frac{\sqrt{2}+\sqrt[4]{3} (\sqrt{3}-1)} {2^{5/4}}, \sqrt{ll'} =\frac{(\sqrt{3}-1)^{2}}{2\sqrt{2}}$$ and $$l=\frac{(\sqrt{3}-1)(\sqrt{2}-\sqrt[4]{3})}{2}, l'=\frac{(\sqrt{3}-1)(\sqrt{2}+\sqrt[4]{3})}{2}$$ Using these values in equation $$(9)$$ we can show that $$\left(\frac {L} {K} \right) ^{2}=\frac {3+2\sqrt{3}}{9}$$ and thus the desired value of the third sum is obtained.
For $$p=5$$ Ramanujan gives the modular equation $$\frac{5L}{K}=\frac{1+(\alpha^5/\beta)^{1/8}} {1+(\alpha\beta^3)^{1/8}}, \frac{K} {L} =\frac{1+((1-\beta)^5/(1-\alpha))^{1/8}}{1+((1-\alpha)^3(1-\beta))^{1/8}}$$ where $$\alpha=k^2,\beta=l^2$$. Using $$\alpha=\frac{1}{2},\beta=\frac {1}{2}-6\sqrt{161\sqrt{5}-360}=\frac{1-\sqrt{1-\phi^{-24}}}{2}$$ (value of $$\beta$$ is obtained using value of class invariant $$G_{25}=(1+\sqrt {5})/2=\phi$$) corresponding to nomes $$e^{-\pi}, e^{-5\pi}$$ we can get the value of $$L/K$$ in algebraic form. It is hard to believe, but this thread tell us that the final result has a very simple algebraic form given by $$\frac{L} {K} =\frac{\sqrt{5}+2}{5}$$ The algebra involved can be simplified a lot if we use a denesting by Ramanujan for $$\beta^{1/8},(1-\beta)^{1/8}$$ given here (see equation $$(7)$$).
• Wow, thanks for the detailed explanation! It will probably take me a while to go through and understand it all. – Franklin Pezzuti Dyer Jan 8 '18 at 20:28
• @Nilknarf: All the facts stated in this answer are proved in my blog posts. Search for "modular equation" in this archive page and you will find detailed treatment of these topics. I have covered approaches given by both Ramanujan and Jacobi. And yes its a difficult topic which will require some time. – Paramanand Singh Jan 9 '18 at 0:00
• Okay, awesome! Thanks for all the help! – Franklin Pezzuti Dyer Jan 9 '18 at 0:01
• @ParamanandSingh Your blog posts are beautiful fair play!! – Colm Bhandal Jan 9 '18 at 14:19
See: Ramanujan's Notebooks Volume 3, Chapter 17, Example(i). pp 103.
See also: Ramanujan's Notebook Volume 5 chapter 35. Values of Theta-Functions P. 325.
• Are these online by any chance? – Colm Bhandal Jan 6 '18 at 22:20
• @ColmBhandal: they should be available online if you search enough. But those are unofficial copies not like the legal ones available for some old classic texts. – Paramanand Singh Jan 9 '18 at 13:21 | 2021-01-18 04:20:48 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 119, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9064989686012268, "perplexity": 145.57886352904745}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703514121.8/warc/CC-MAIN-20210118030549-20210118060549-00215.warc.gz"} |
https://www.snapxam.com/problems/103093959/limit-as-x-approaches-0-of-3x-1 | # Step-by-step Solution
Go!
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## Step-by-step explanation
Problem to solve:
$\lim_{x\to0}\left(3x+1\right)$
Learn how to solve limits by direct substitution problems step by step online.
$\lim_{x\to0}\left(3x\right)+\lim_{x\to0}\left(1\right)$
Learn how to solve limits by direct substitution problems step by step online. Evaluate the limit of 3x+1 as x approaches 0. The limit of a sum of two functions is equal to the sum of the limits of each function: \displaystyle\lim_{x\to c}(f(x)\pm g(x))=\lim_{x\to c}(f(x))\pm\lim_{x\to c}(g(x)). The limit of a constant is just the constant. The limit of the product of a function and a constant is equal to the limit of the function, times the constant: \displaystyle \lim_{t\to 0}{\left(at\right)}=a\cdot\lim_{t\to 0}{\left(t\right)}. Evaluate the limit by replacing all occurrences of x by 0.
$1$
$\lim_{x\to0}\left(3x+1\right)$
### Main topic:
Limits by direct substitution
### Time to solve it:
~ 0.02 s (SnapXam) | 2020-10-31 18:34:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9914540648460388, "perplexity": 798.1919874174488}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107922411.94/warc/CC-MAIN-20201031181658-20201031211658-00415.warc.gz"} |
https://mail-archives.us.apache.org/mod_mbox/spark-dev/201507.mbox/%3CCA+B-+fysUo6Ga8rigdDeKOoaiP6TW6Ah32CT6A_Oy9gWujbUMw@mail.gmail.com%3E | # spark-dev mailing list archives
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From Debasish Das <debasish.da...@gmail.com>
Subject Re: Confidence in implicit factorization
Date Sun, 26 Jul 2015 08:19:03 GMT
Yeah, I think the idea of confidence is a bit different than what I am
looking for using implicit factorization to do document clustering.
I basically need (r_ij - w_ih_j)^2 for all observed ratings and (0 -
w_ih_j)^2 for all the unobserved ratings...Think about the document x word
matrix where r_ij is the count that's observed, 0 are the word counts that
are not in particular document.
The broadcasted value of gram matrix w_i'wi or h_j'h_j will also count the
r_ij those are observed...So I might be fine using the broadcasted gram
matrix and use the linear term as \sum (-r_ijw_i) or \sum (-rijh_j)...
I will think further but in the current implicit formulation with
confidence, looks like I am really factorizing a 0/1 matrix with weights 1
+ alpha*rating for . It's a bit different from LSA model.
On Sun, Jul 26, 2015 at 12:34 AM, Sean Owen <sowen@cloudera.com> wrote:
> confidence = 1 + alpha * |rating| here (so, c1 means confidence - 1),
> so alpha = 1 doesn't specially mean high confidence. The loss function
> is computed over the whole input matrix, including all missing "0"
> entries. These have a minimal confidence of 1 according to this
> formula. alpha controls how much more confident you are in what the
> entries that do exist in the input mean. So alpha = 1 is low-ish and
> means you don't think the existence of ratings means a lot more than
> their absence.
>
> I think the explicit case is similar, but not identical -- here. The
> cost function for the explicit case is not the same, which is the more
> substantial difference between the two. There, ratings aren't inputs
> to a confidence value that becomes a weight in the loss function,
> during this factorization of a 0/1 matrix. Instead the rating matrix
> is the thing being factorized directly.
>
> On Sun, Jul 26, 2015 at 6:45 AM, Debasish Das <debasish.das83@gmail.com>
> wrote:
> > Hi,
> >
> > Implicit factorization is important for us since it drives recommendation
> > when modeling user click/no-click and also topic modeling to handle 0
> counts
> > in document x word matrices through NMF and Sparse Coding.
> >
> > I am a bit confused on this code:
> >
> > val c1 = alpha * math.abs(rating)
> > if (rating > 0) ls.add(srcFactor, (c1 + 1.0)/c1, c1)
> >
> > When the alpha = 1.0 (high confidence) and rating is > 0 (true for word
> > counts), why this formula does not become same as explicit formula:
> >
> >
> > For modeling document, I believe implicit Y'Y needs to stay but we need
> > explicit ls.add(srcFactor, rating, 1.0)
> >
> > I am understanding confidence code further. Please let me know if the
> idea
> > of mapping implicit to handle 0 counts in document word matrix makes
> sense.
> >
> > Thanks.
> > Deb
> >
>
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View raw message | 2021-07-31 00:37:28 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.715388834476471, "perplexity": 5927.214615327896}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154032.75/warc/CC-MAIN-20210730220317-20210731010317-00434.warc.gz"} |
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Get Full Access to Discrete Mathematics And Its Applications - 7 Edition - Chapter 2.3 - Problem 79e
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# a) Show that if a set S has cardinality m, where m is a
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## Solution for problem 79E Chapter 2.3
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Problem 79E
Problem 79E
a) Show that if a set S has cardinality m, where m is a positive integer, then there is a one-to-one correspondence between S and the set {1. 2,...,m).
b) Show that if S and T are two sets each with m elements. where m is a positive integer, then there is a one-to-one correspondence between S and T.
Step-by-Step Solution:
Step 1 of 3
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##### ISBN: 9780073383095
Discrete Mathematics and Its Applications was written by and is associated to the ISBN: 9780073383095. Since the solution to 79E from 2.3 chapter was answered, more than 390 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: Discrete Mathematics and Its Applications, edition: 7. The answer to “a) Show that if a set S has cardinality m, where m is a positive integer, then there is a one-to-one correspondence between S and the set {1. 2,...,m).________________b) Show that if S and T are two sets each with m elements. where m is a positive integer, then there is a one-to-one correspondence between S and T.” is broken down into a number of easy to follow steps, and 58 words. The full step-by-step solution to problem: 79E from chapter: 2.3 was answered by , our top Math solution expert on 06/21/17, 07:45AM. This full solution covers the following key subjects: Where, correspondence, show, Integer, set. This expansive textbook survival guide covers 101 chapters, and 4221 solutions.
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Unlock Textbook Solution | 2022-08-18 11:42:03 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4926638603210449, "perplexity": 6083.363416673679}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882573193.35/warc/CC-MAIN-20220818094131-20220818124131-00173.warc.gz"} |
http://tavodraugas.lt/n1uqh3n/65d1b6-5-htp-for-ssri-withdrawal-reddit | East of this agonic line, the magnetic North Pole is to the west of the geographic North Pole and a correction must be applied to a compass indication to get a true direction. A dip circle is taken to geomagnetic equator. Then the angle of dip is (a) 0° (b) 30° (c) 45° (d) 90° A magnetic field B is applied perpendicular to the plane of the sample. The same needle is then allowed to vibrate in the horizontal plane and the time period is again found to be 2 sec. The same needle is then allowed to vibrate in the horizontal plane and the time period is again found to be 2 seconds. The time period of vibration is found to be 2 seconds. An electron with charge -e and mass m travels at a speed v in a plane perpendicular to a magnetic field of magnitude B. For a vertical plane other than magnetic meridian $\alpha>0$ or $\cos \alpha<1$ so $\theta^{\prime}>\theta$ In a vertical plane other than magnetic meridian angle of dip is more than in magnetic meridian. When the plane of the coil is normal to the magnetic meridian, a neutral point is observed at the centre of the coil. A voltage measured perpendicular to the plane is indicative of the CME of hot electrons. Real dip theta=40.9^@ Let the vertical component of earth's magnetic field be V and its horizontal component at magnetic meridian be H. Then the angle of dip theta at magnetic meridian will be given by tantheta=V/H.....[1] When the dip needle is suspended at an angle of 30^@ to the earth magnetic meridian then it makes an angle 45^@ with the horizontal. The electron follows a circular path of radius R. In a time, t, the electron travels halfway around the circle. Illustration of a thin-film CME measurement geometry in the presence of a strong in-plane electric field (E). Click hereto get an answer to your question ️ A vertical circular coil of radius 0.1 m and having 10 turns carries a steady current. The needle is allowed to move in a vertical plane perpendicular to the magnetic meridian. A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22° with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at … Textbook Solutions 6926. Maharashtra State Board HSC Science (Electronics) 11th. Here the needle and vertical scale are in a plane perpendicular to the magnetic meridian. Important Solutions 17 ... perpendicular to the magnetic axis of Earth. This sets the needle and vertical scale exactly in magnetic meridian. The time period of vibration is found to be 2 sec. Magnetic meridian is the plane . The needle will stay (a) in horizontal direction only (b) in vertical direction only (c) in any direction except vertical and horizontal (d) in the direction it is released To get the true heading, you need to first read the magnetic compass, then either add an Easterly, or subtract a Westerly, magnetic variation; based on the isogonic lines. If BH = 0.314 × 10^-4 T , the current in the coil is : Reuse & Permissions × Another magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at with the horizontal . Then the angle of dip is The box is then rotated through 90 o position on the horizontal circular scale. A magnetic field is directed perpendicular to the plane of a circular coil of area 0.2 m² and 250 turns. The magnitude of the horizontal component of the earth ' magnetic field at the place is known to be . A dip needle vibrates in the vertical plane perpendicular to the magnetic meridian. 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Touchstone Electric Fireplace, X2 Power Lithium Battery Motorcycle, Johnson's Blue Geranium Companion Plants, Duster 2012 Model Second Hand Price, Williamson County, Tn Zoning, Trader Joe's Orange Chicken, Pleurozium Schreberi Characteristics, Associate Degree In Nursing Schools Near Me, Where To Buy Eurmax Canopy, K Balaji Ias, River Rats Canoe, Restaurant Frozen Pizza, | 2021-09-19 19:19:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5000232458114624, "perplexity": 764.9986790122599}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780056900.32/warc/CC-MAIN-20210919190128-20210919220128-00126.warc.gz"} |
https://socratic.org/questions/how-do-you-solve-7a-3-5b-and-2a-b-0 | # How do you solve 7a=3-5b and 2a+b=0?
May 9, 2016
The solution for the system of equations is:
color(green)(a = -1
color(green)(b = 2
#### Explanation:
$7 a = 3 - 5 b$..........equation $\left(1\right)$
$2 a + b = 0$
color(green)(b = -2a..............equation $\left(2\right)$
Solving by substitution:
Substituting equation $2$ in $1$:
$7 a = 3 - 5 b$
$7 a = 3 + \left(- 5\right) \cdot \textcolor{g r e e n}{\left(- 2 a\right)}$
$7 a = 3 + 10 a$
$7 a - 10 a = 3$
$- 3 a = 3$
$a = \frac{3}{- 3}$
color(green)(a = -1
Finding $b$ from equation $2$ :
$b = - 2 a$
$b = - 2 \cdot \left(- 1\right)$
color(green)(b = 2 | 2019-09-18 03:01:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 21, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2534559667110443, "perplexity": 10444.350766528974}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514573176.51/warc/CC-MAIN-20190918024332-20190918050332-00030.warc.gz"} |
https://r.789695.n4.nabble.com/Re-stopifnot-td4756281.html | # Re: stopifnot
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## Re: stopifnot
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## Re: stopifnot -- eval(*) inside for()
>>>>> Suharto Anggono Suharto Anggono via R-devel >>>>> on Sun, 31 Mar 2019 15:26:13 +0000 writes: > Ah, with R 3.5.0 or R 3.4.2, but not with R 3.3.1, 'eval' > inside 'for' makes compiled version behave like > non-compiled version. Ah.. ... thank you for detecting that " eval() inside for()" behaves specially in how error message get a call or not. Let's focus only on this issue here. I'm adding a 0-th case to make even clearer what you are saying: > options(error = expression(NULL)) > library(compiler) > enableJIT(0) > f0 <- function(x) { x ; x^2 } ; f0(is.numeric(y)) Error in f0(is.numeric(y)) (from #1) : object 'y' not found > (function(x) { x ; x^2 })(is.numeric(y)) Error in (function(x) { (from #1) : object 'y' not found > f0c <- cmpfun(f0) ; f0c(is.numeric(y)) so by default, not only the error message but the originating call is shown as well. However, here's your revealing examples: > f <- function(x) for (i in 1) {x; eval(expression(i))} > f(is.numeric(y)) > # Error: object 'y' not found > fc <- cmpfun(f) > fc(is.numeric(y)) > # Error: object 'y' not found I've tried more examples and did not find any difference between simple interpreted and bytecompiled code {apart from "keep.source=TRUE" keeping source, sometimes visible}. So I don't understand yet why you think the byte compiler plays a role. Rather the crucial difference seems the error happens inside a loop which contains an explicit eval(.), and that eval() may even be entirely unrelated to the statement in which the error happens [above: The error happens when the promise 'x' is evaluated, *before* eval() is called at all]. > Is this accidental feature going to be relied upon? [i.e. *in stopifnot() R code (which in R-devel and R 3.5.x has had an eval() inside the for()-loop)] That is a good question. What I really like about the R-devel case: We do get errors signalled that do *not* contain the full stopifnot() call. With the newish introduction of the exprs = { ... ... } variant, it is even more natural to have large exprs in a stopifnot() call, and when there's one accidental error in there, it's quite unhelpful to see the full stopifnot(..........) call {many lines of R code} obfuscating the one statement which produced the error. So it seems I am asking for a new feature in R, namely to temporarily say: Set the call to errors to NULL "in the following". In R 3.5.x, I had used withCallingHandlers(...) to achieve that and do even similar for warnings... but needed to that for every expression and hence inside the for loop and the consequence was a relatively large slowdown of stopifnot().. which triggered all the changes since. Whereas what we see here ["eval() inside for()"] is a cheap automatic suppression of 'call' for the "internal errors", i.e., those we don't trigger ourselves via stop(simplError(...)). ______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-devel
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## Re: [External] Re: stopifnot -- eval(*) inside for()
On Mon, 1 Apr 2019, Martin Maechler wrote: >>>>>> Suharto Anggono Suharto Anggono via R-devel >>>>>> on Sun, 31 Mar 2019 15:26:13 +0000 writes: > > > Ah, with R 3.5.0 or R 3.4.2, but not with R 3.3.1, 'eval' > > inside 'for' makes compiled version behave like > > non-compiled version. > > Ah.. ... thank you for detecting that " eval() inside for()" behaves > specially in how error message get a call or not. Don't count on that remaining true indefinitely. The standard behavior is better and we'll eventually get the case where 'eval' and a few others are called to behave the same. Best, luke > Let's focus only on this issue here. > > I'm adding a 0-th case to make even clearer what you are saying: > > > options(error = expression(NULL)) > > library(compiler) > > enableJIT(0) > > > f0 <- function(x) { x ; x^2 } ; f0(is.numeric(y)) > Error in f0(is.numeric(y)) (from #1) : object 'y' not found > > (function(x) { x ; x^2 })(is.numeric(y)) > Error in (function(x) { (from #1) : object 'y' not found > > f0c <- cmpfun(f0) ; f0c(is.numeric(y)) > > so by default, not only the error message but the originating > call is shown as well. > > However, here's your revealing examples: > > > f <- function(x) for (i in 1) {x; eval(expression(i))} > > f(is.numeric(y)) > > # Error: object 'y' not found > > fc <- cmpfun(f) > > fc(is.numeric(y)) > > # Error: object 'y' not found > > I've tried more examples and did not find any difference > between simple interpreted and bytecompiled code {apart > from "keep.source=TRUE" keeping source, sometimes visible}. > So I don't understand yet why you think the byte compiler plays > a role. > > Rather the crucial difference seems the error happens inside a > loop which contains an explicit eval(.), and that eval() may > even be entirely unrelated to the statement in which the error > happens [above: The error happens when the promise 'x' is > evaluated, *before* eval() is called at all]. > > > > Is this accidental feature going to be relied upon? > > [i.e. *in stopifnot() R code (which in R-devel and R 3.5.x has > had an eval() inside the for()-loop)] > > That is a good question. > What I really like about the R-devel case: We do get errors > signalled that do *not* contain the full stopifnot() call. > > With the newish introduction of the exprs = { ... ... } variant, > it is even more natural to have large exprs in a stopifnot() call, > and when there's one accidental error in there, it's quite > unhelpful to see the full stopifnot(..........) call {many lines > of R code} obfuscating the one statement which produced the > error. > > So it seems I am asking for a new feature in R, > namely to temporarily say: Set the call to errors to NULL "in > the following". > In R 3.5.x, I had used withCallingHandlers(...) to achieve that > and do even similar for warnings... but needed to that for every > expression and hence inside the for loop and the consequence > was a relatively large slowdown of stopifnot().. which > triggered all the changes since. > > Whereas what we see here ["eval() inside for()"] is a cheap > automatic suppression of 'call' for the "internal errors", i.e., > those we don't trigger ourselves via stop(simplError(...)). > > ______________________________________________ > [hidden email] mailing list > https://stat.ethz.ch/mailman/listinfo/r-devel> -- Luke Tierney Ralph E. Wareham Professor of Mathematical Sciences University of Iowa Phone: 319-335-3386 Department of Statistics and Fax: 319-335-3017 Actuarial Science 241 Schaeffer Hall email: [hidden email] Iowa City, IA 52242 WWW: http://www.stat.uiowa.edu______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-devel
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## Re: stopifnot
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## Re: [External] Re: stopifnot -- eval(*) inside for()
In reply to this post by Tierney, Luke >>>>> Tierney, Luke >>>>> on Mon, 1 Apr 2019 12:41:08 +0000 writes: > On Mon, 1 Apr 2019, Martin Maechler wrote: >>>>>>> Suharto Anggono Suharto Anggono via R-devel >>>>>>> on Sun, 31 Mar 2019 15:26:13 +0000 writes: >> >> > Ah, with R 3.5.0 or R 3.4.2, but not with R 3.3.1, 'eval' >> > inside 'for' makes compiled version behave like >> > non-compiled version. >> >> Ah.. ... thank you for detecting that " eval() inside for()" behaves >> specially in how error message get a call or not. > Don't count on that remaining true indefinitely. The standard behavior > is better and we'll eventually get the case where 'eval' and a few > others are called to behave the same. > Best, > luke Yes, Suharto did indeed mention that it may not be a good idea to rely on that behavior, and I did agree ... though my agreement was a bit buried in other stuff. Note that I have been asking if this "accidental" but internally consistent behavior for the current situation, could not be made a feature that the user can ask for ... without having to use a handler (which had been a real slowdown when used inside stopifnot() in R 3.5.3) : [................] [................] >> So it seems I am asking for a new feature in R, >> namely to temporarily say: Set the call to errors to NULL "in >> the following". >> In R 3.5.x, I had used withCallingHandlers(...) to achieve that >> and do even similar for warnings... but needed to that for every >> expression and hence inside the for loop and the consequence >> was a relatively large slowdown of stopifnot().. which >> triggered all the changes since. >> >> Whereas what we see here ["eval() inside for()"] is a cheap >> automatic suppression of 'call' for the "internal errors", i.e., >> those we don't trigger ourselves via stop(simpleError(...)). So, for me as programmeR, it would be nice to be able to ask for "this" behavior explicitly in a special case as here, where "no-call" error messages are preferable .. because the call can be really large and is known to be "almost surely" distracting rather than helpful. Martin ______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-devel
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## Re: stopifnot -- eval(*) inside for()
In reply to this post by Martin Maechler With f <- function(x) for (i in 1) x fc <- cmpfun(f) (my previous example), error message of fc(is.numeric(y)) shows the originating call as well, while error message of f(is.numeric(y)) doesn't. Compiled version behaves differently. Even with f <- function(x) for (i in 1) {x; eval(expression(i))} fc <- cmpfun(f) , error message of fc(is.numeric(y)) shows the originating call in R 3.3.1. As I see, error message only has one line of call. If the deparsed call spans more than one line, the rest is not shown. In 'stopifnot' in R 3.5.x, each is wrapped in 'tryCatch' which is wrapped again in 'withCallingHandlers'. Just one wrapping may be enough. The 'withCallingHandlers' construct in 'stopifnot' in R 3.5.x has no effect anyway, as I said before (https://stat.ethz.ch/pipermail/r-devel/2019-February/077386.html). Also, 'tryCatch' (or 'withCallingHandlers' ...) can wrap the entire 'for' loop. The slowdown can be less than in R 3.5.x. -------------------------------------------- On Mon, 1/4/19, Martin Maechler <[hidden email]> wrote: Subject: Re: [Rd] stopifnot -- eval(*) inside for() Cc: [hidden email] Date: Monday, 1 April, 2019, 5:00 PM >>>>> Suharto Anggono Suharto Anggono via R-devel >>>>> on Sun, 31 Mar 2019 15:26:13 +0000 writes: > Ah, with R 3.5.0 or R 3.4.2, but not with R 3.3.1, 'eval' > inside 'for' makes compiled version behave like > non-compiled version. Ah.. ... thank you for detecting that " eval() inside for()" behaves specially in how error message get a call or not. Let's focus only on this issue here. I'm adding a 0-th case to make even clearer what you are saying: > options(error = expression(NULL)) > library(compiler) > enableJIT(0) > f0 <- function(x) { x ; x^2 } ; f0(is.numeric(y)) Error in f0(is.numeric(y)) (from #1) : object 'y' not found > (function(x) { x ; x^2 })(is.numeric(y)) Error in (function(x) { (from #1) : object 'y' not found > f0c <- cmpfun(f0) ; f0c(is.numeric(y)) so by default, not only the error message but the originating call is shown as well. However, here's your revealing examples: > f <- function(x) for (i in 1) {x; eval(expression(i))} > f(is.numeric(y)) > # Error: object 'y' not found > fc <- cmpfun(f) > fc(is.numeric(y)) > # Error: object 'y' not found I've tried more examples and did not find any difference between simple interpreted and bytecompiled code {apart from "keep.source=TRUE" keeping source, sometimes visible}. So I don't understand yet why you think the byte compiler plays a role. Rather the crucial difference seems the error happens inside a loop which contains an explicit eval(.), and that eval() may even be entirely unrelated to the statement in which the error happens [above: The error happens when the promise 'x' is evaluated, *before* eval() is called at all]. > Is this accidental feature going to be relied upon? [i.e. *in stopifnot() R code (which in R-devel and R 3.5.x has had an eval() inside the for()-loop)] That is a good question. What I really like about the R-devel case: We do get errors signalled that do *not* contain the full stopifnot() call. With the newish introduction of the exprs = { ... ... } variant, it is even more natural to have large exprs in a stopifnot() call, and when there's one accidental error in there, it's quite unhelpful to see the full stopifnot(..........) call {many lines of R code} obfuscating the one statement which produced the error. So it seems I am asking for a new feature in R, namely to temporarily say: Set the call to errors to NULL "in the following". In R 3.5.x, I had used withCallingHandlers(...) to achieve that and do even similar for warnings... but needed to that for every expression and hence inside the for loop and the consequence was a relatively large slowdown of stopifnot().. which triggered all the changes since. Whereas what we see here ["eval() inside for()"] is a cheap automatic suppression of 'call' for the "internal errors", i.e., those we don't trigger ourselves via stop(simplError(...)). ______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-devel | 2019-10-21 08:21:12 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5382020473480225, "perplexity": 13596.252601537948}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570987763641.74/warc/CC-MAIN-20191021070341-20191021093841-00405.warc.gz"} |
https://www.openmodelica.org/doc/OpenModelicaUsersGuide/latest/omedit.html | # OMEdit – OpenModelica Connection Editor¶
OMEdit – OpenModelica Connection Editor is the new Graphical User Interface for graphical model editing in OpenModelica. It is implemented in C++ using the Qt graphical user interface library and supports the Modelica Standard Library that is included in the latest OpenModelica installation. This chapter gives a brief introduction to OMEdit and also demonstrates how to create a DCMotor model using the editor.
OMEdit provides several user friendly features for creating, browsing, editing, and simulating models:
• Modeling – Easy model creation for Modelica models.
• Pre-defined models – Browsing the Modelica Standard library to
access the provided models.
• User defined models – Users can create their own models for
immediate usage and later reuse.
• Component interfaces – Smart connection editing for drawing and
editing connections between model interfaces.
• Simulation – Subsystem for running simulations and specifying
simulation parameters start and stop time, etc.
• Plotting – Interface to plot variables from simulated models.
## Starting OMEdit¶
A splash screen similar to the one shown in Figure 5 will appear indicating that it is starting OMEdit. The executable is found in different places depending on the platform (see below).
### Microsoft Windows¶
OMEdit can be launched using the executable placed in OpenModelicaInstallationDirectory/bin/OMEdit/OMEdit.exe. Alternately, choose OpenModelica > OpenModelica Connection Editor from the start menu in Windows.
### Linux¶
Start OMEdit by either selecting the corresponding menu application item or typing “OMEdit” at the shell or command prompt.
### Mac OS X¶
The default installation is /Application/MacPorts/OMEdit.app.
## MainWindow & Browsers¶
The MainWindow contains several dockable browsers,
• Libraries Browser
• Documentation Browser
• Variables Browser
• Messages Browser
Figure 6 shows the MainWindow and browsers.
The default location of the browsers are shown in Figure 6. All browsers except for Message Browser can be docked into left or right column. The Messages Browser can be docked into top or bottom areas. If you want OMEdit to remember the new docked position of the browsers then you must enable Preserve User's GUI Customizations option, see section General.
### Filter Classes¶
To filter a class click Edit > Filter Classes or press keyboard shortcut Ctrl+Shift+F. The loaded Modelica classes can be filtered by typing any part of the class name.
### Libraries Browser¶
To view the Libraries Browser click View > Windows > Libraries Browser. Shows the list of loaded Modelica classes. Each item of the Libraries Browser has right click menu for easy manipulation and usage of the class. The classes are shown in a tree structure with name and icon. The protected classes are not shown by default. If you want to see the protected classes then you must enable the Show Protected Classes option, see section General.
### Documentation Browser¶
Displays the HTML documentation of Modelica classes. It contains the navigation buttons for moving forward and backward. It also contains a WYSIWYG editor which allows writing class documentation in HTML format. To see documentation of any class, right click the Modelica class in Libraries Browser and choose View Documentation.
### Variables Browser¶
The class variables are structured in the form of the tree and are displayed in the Variables Browser. Each variable has a checkbox. Ticking the checkbox will plot the variable values. There is a find box on the top for filtering the variable in the tree. The filtering can be done using Regular Expression, Wildcard and Fixed String. The complete Variables Browser can be collapsed and expanded using the Collapse All and Expand All buttons.
The browser allows manipulation of changeable parameters for Re-simulating a Model. It also displays the unit and description of the variable.
### Messages Browser¶
Shows the list of errors. Following kinds of error can occur,
• Syntax
• Grammar
• Translation
• Symbolic
• Simulation
• Scripting
See section Messages for Messages Browser options.
## Perspectives¶
The perspective tabs are loacted at the bottom right of the MainWindow:
• Welcome Perspective
• Modeling Perspective
• Plotting Perspective
• Debugging Perspective
### Welcome Perspective¶
The Welcome Perspective shows the list of recent files and the list of latest news from https://www.openmodelica.org/. See Figure 10. The orientation of recent files and latest news can be horizontal or vertical. User is allowed to show/hide the latest news. See section General.
### Modeling Perspective¶
The Modeling Perpective provides the interface where user can create and design their models. See Figure 11.
The Modeling Perspective interface can be viewed in two different modes, the tabbed view and subwindow view, see section General.
### Plotting Perspective¶
The Plotting Perspective shows the simulation results of the models. Plotting Perspective will automatically become active when the simulation of the model is finished successfully. It will also become active when user opens any of the OpenModelica’s supported result file. Similar to Modeling Perspective this perspective can also be viewed in two different modes, the tabbed view and subwindow view, see section General.
### Debugging Perspective¶
The application automatically switches to Debugging Perpective when user simulates the class with algorithmic debugger. The prespective shows the list of stack frames, breakpoints and variables.
## Modeling a Model¶
### Creating a New Modelica Class¶
Creating a new Modelica class in OMEdit is rather straightforward. Choose any of the following methods,
• Select File > New Modelica Class from the menu.
• Click on New Modelica Class toolbar button.
• Click on the Create New Modelica Class button available at the left bottom of Welcome Perspective.
• Press Ctrl+N.
### Opening a Modelica File¶
Choose any of the following methods to open a Modelica file,
• Select File > Open Model/Library File(s) from the menu.
• Click on Open Model/Library File(s) toolbar button.
• Click on the Open Model/Library File(s) button available at the right bottom of Welcome Perspective.
• Press Ctrl+O.
(Note, for editing Modelica system files like MSL (not recommended), see Editing Modelica Standard Library)
### Opening a Modelica File with Encoding¶
Select File > Open/Convert Modelica File(s) With Encoding from the menu. It is also possible to convert files to UTF-8.
### Model Widget¶
For each Modelica class one Model Widget is created. It has a statusbar and a view area. The statusbar contains buttons for navigation between the views and labels for information. The view area is used to display the icon, diagram and text layers of Modelica class. See Figure 14.
Drag the models from the Libraries Browser and drop them on either Diagram or Icon View of Model Widget.
### Making Connections¶
In order to connect one component model to another the user first needs to enable the connect mode () from the toolbar.
Move the mouse over the connector. The mouse cursor will change from arrow cursor to cross cursor. To start the connection press left button and move while keeping the button pressed. Now release the left button. Move towards the end connector and click when cursor changes to cross cursor.
## Simulating a Model¶
The simulation options for each model are stored inside the OMEdit data structure. They have the following sequence,
• Each model has its own simulation options.
• If the model is opened for the first time then the simulation options are set to default.
• experiment and __OpenModelica_simulationFlags annotations are applied if the model contains them.
• After that all the changes done via Simulation Setup window are preserved for the whole session. If you want to use the same settings in the future sessions then you should store them inside experiment and __OpenModelica_simulationFlags.
The OMEdit Simulation Setup can be launched by,
• Selecting Simulation > Simulation Setup from the menu. (requires a model to be active in ModelWidget)
• Clicking on the Simulation Setup toolbar button. (requires a model to be active in ModelWidget)
• Right clicking the model from the Libraries Browser and choosing Simulation Setup.
### General Tab¶
• Simulation Interval
• Start Time – the simulation start time.
• Stop Time – the simulation stop time.
• Number of Intervals – the simulation number of intervals.
• Interval – the length of one interval (i.e., stepsize)
• Simulate with steps (makes the interactive simulation synchronous; plots nicer curves at the expense of performance)
• Simulation server port
• Integration
• Method – the simulation solver. See section Integration Methods for solver details.
• Tolerance – the simulation tolerance.
• Jacobian - the jacobain method to use.
• DASSL/IDA Options
• Root Finding - Activates the internal root finding procedure of dassl.
• Restart After Event - Activates the restart of dassl after an event is performed.
• Initial Step Size
• Maximum Step Size
• Maximum Integration Order
• C/C++ Compiler Flags (Optional) – the optional C/C++ compiler flags.
• Number of Processors – the number of processors used to build the simulation.
• Build Only – only builds the class.
• Launch Transformational Debugger – launches the transformational debugger.
• Launch Algorithmic Debugger – launches the algorithmic debugger.
• Launch Animation – launches the 3d animation window.
### Output Tab¶
• Output Format – the simulation result file output format.
• Single Precision - Output results in single precision (only for mat output format).
• File Name Prefix (Optional) – the name is used as a prefix for the output files.
• Result File (Optional) - the simulation result file name.
• Variable Filter (Optional)
• Protected Variables – adds the protected variables in result file.
• Equidistant Time Grid – output the internal steps given by dassl instead of interpolating results into an equidistant time grid as given by stepSize or numberOfIntervals
• Store Variables at Events – adds the variables at time events.
• Show Generated File – displays the generated files in a dialog box.
### Simulation Flags Tab¶
• Model Setup File (Optional) – specifies a new setup XML file to the generated simulation code.
• Initialization Method (Optional) – specifies the initialization method.
• Equation System Initialization File (Optional) – specifies an external file for the initialization of the model.
• Equation System Initialization Time (Optional) – specifies a time for the initialization of the model.
• Clock (Optional) – the type of clock to use.
• Linear Solver (Optional) – specifies the linear solver method.
• Non Linear Solver (Optional) – specifies the nonlinear solver.
• Linearization Time (Optional) – specifies a time where the linearization of the model should be performed.
• Output Variables (Optional) – outputs the variables a, b and c at the end of the simulation to the standard output.
• Profiling – creates a profiling HTML file.
• CPU Time – dumps the cpu-time into the result file.
• Enable All Warnings – outputs all warnings.
• Logging (Optional)
• stdout - standard output stream. This stream is always active, can be disabled with -lv=-stdout
• assert - This stream is always active, can be disabled with -lv=-assert
• LOG_DASSL_STATES - outputs the states at every dassl call.
• LOG_DEBUG - additional debug information.
• LOG_DSS - outputs information about dynamic state selection.
• LOG_DSS_JAC - outputs jacobian of the dynamic state selection.
• LOG_DT_CONS - additional information about dynamic tearing (local and global constraints).
• LOG_EVENTS - additional information during event iteration.
• LOG_EVENTS_V - verbose logging of event system.
• LOG_INIT - additional information during initialization.
• LOG_IPOPT - information from Ipopt.
• LOG_IPOPT_JAC - check jacobian matrix with Ipopt.
• LOG_IPOPT_HESSE - check hessian matrix with Ipopt.
• LOG_IPOPT_ERROR - print max error in the optimization.
• LOG_JAC - outputs the jacobian matrix used by dassl.
• LOG_LS - logging for linear systems.
• LOG_LS_V - verbose logging of linear systems.
• LOG_NLS - logging for nonlinear systems.
• LOG_NLS_V - verbose logging of nonlinear systems.
• LOG_NLS_HOMOTOPY - logging of homotopy solver for nonlinear systems.
• LOG_NLS_JAC - outputs the jacobian of nonlinear systems.
• LOG_NLS_JAC_TEST - tests the analytical jacobian of nonlinear systems.
• LOG_NLS_RES - outputs every evaluation of the residual function.
• LOG_NLS_EXTRAPOLATE - outputs debug information about extrapolate process.
• LOG_RES_INIT - outputs residuals of the initialization.
• LOG_RT - additional information regarding real-time processes.
• LOG_SOLVER_V - verbose information about the integration process.
• LOG_SOLVER_CONTEXT - context information during the solver process.
• LOG_SOTI - final solution of the initialization.
• LOG_STATS_V - additional statistics for LOG_STATS.
• LOG_SUCCESS - This stream is always active, can be disabled with -lv=-LOG_SUCCESS.
• LOG_UTIL.
• Additional Simulation Flags (Optional) – specify any other simulation flag.
### Archived Simulations Tab¶
Shows the list of simulations already finished or running. Double clicking on any of them opens the simulation output window.
## Plotting the Simulation Results¶
Successful simulation of model produces the result file which contains the instance variables that are candidate for plotting. Variables Browser will show the list of such instance variables. Each variable has a checkbox, checking it will plot the variable. See Figure 12.
### Types of Plotting¶
The plotting type depends on the active Plot Window. By default the plotting type is Time Plot.
#### Time Plot¶
Plots the variable over the simulation time. You can have multiple Time Plot windows by clicking on New Plot Window toolbar button ().
#### Plot Parametric¶
Draws a two-dimensional parametric diagram, between variables x and y, with y as a function of x. You can have multiple Plot Parametric windows by clicking on the New Plot Parametric toolbar button ().
#### Array Plot¶
Plots an array variable so that the array elements' indexes are on the x-axis and corresponding elements' values are on the y-axis. The time is controlled by the slider above the variable tree. When an array is present in the model, it has a principal array node in the variable tree. To plot this array as an Array Plot, match the principal node. The principal node may be expanded into particular array elements. To plot a single element in the Time Plot, match the element. A new Array Plot window is opened using the New Array Plot Window toolbar button ().
#### Array Parametric Plot¶
Plots the first array elements' values on the x-axis versus the second array elements' values on the y-axis. The time is controlled by the slider above the variable tree. To create a new Array Parametric Plot, press the New Array Parametric Plot Window toolbar button (), then match the principle array node in the variable tree view to be plotted on the x-axis and match the principle array node to be plotted on the y-axis.
## Re-simulating a Model¶
The Variables Browser allows manipulation of changeable parameters for re-simulation. After changing the parameter values user can click on the re-simulate toolbar button (), or right click the model in Variables Browser and choose re-simulate from the menu.
## 3D Visualization¶
Since OpenModelica 1.11 , OMEdit has built-in 3D visualization, which replaces third-party libraries (such as Modelica3D) for 3D visualization.
### Running a Visualization¶
The 3d visualization is based on OpenSceneGraph. In order to run the visualization simply right click the class in Libraries Browser an choose “Simulate with Animation” as shown in Figure 15.
One can also run the visualization via Simulation > Simulate with Animation from the menu.
When simulating a model in animation mode, the flag +d=visxml is set. Hence, the compiler will generate a scene description file _visual.xml which stores all information on the multibody shapes. This scene description references all variables which are needed for the animation of the multibody system. When simulating with +d=visxml, the compiler will always generate results for these variables.
### Viewing a Visualization¶
After the successful simulation of the model, the visualization window will show up automatically as shown in Figure 16.
The animation starts with pushing the play button. The animation is played until stopTime or until the pause button is pushed. By pushing the previous button, the animation jumps to the initial point of time. Points of time can be selected by moving the time slider or by inserting a simulation time in the Time-box. The speed factor of animation in relation to realtime can be set in the Speed-dialog. Other animations can be openend by using the open file button and selecting a result file with a corresping scene description file.
The 3D camera view can be manipulated as follows:
Operation Key Mouse Action
Move Closer/Further none Wheel
Move Closer/Further Right Mouse Hold Up/Down
Move Up/Down/Left/Right Middle Mouse Hold Move Mouse
Move Up/Down/Left/Right Left and Right Mouse Hold Move Mouse
Rotate Left Mouse Hold Move Mouse
Shape context menu Right Mouse + Shift
Predefined views (Isometric, Side, Front, Top) can be selected and the scene can be tilted by 90° either clock or anticlockwise with the rotation buttons.
The shapes that are displayed in the viewer can be selected with shift + right click. If a shape is selected, a context menu pops up that offers additional visualization features
The following features can be selected:
Change Transparency The shape becomes either transparent or intransparent.
Make Shape Invisible The shape becomes invisible.
Change Color A color dialog pops up and the color of the shape can be set.
Apply Check Texture A checked texture is applied to the shape.
Apply Custom Texture A file selection dialog pops up and an image file can be selected as a texture.
Remove Texture Removes the current texture of the shape.
## Interactive Simulation¶
Warning
Interactive simulation is an experimental feature.
Interactive simulation is enabled by selecting interactive simulation in the General tab of the simulation settings.
There are two main modes of execution: asynchronous and synchronous (simulate with steps). The difference is that in synchronous (step mode), OMEdit sends a command to the simulation for each step that the simulation should take. The asynchronous mode simply tells the simulation to run and samples variables values in real-time; if the simulation runs very fast, fewer values will be sampled.
When running in asynchronous mode, it is possible to simulate the model in real-time (with a scaling factor just like simulation flag -rt, but with the ability to change the scaling factor during the interactive simulation). In the synchronous mode, the speed of the simulation does not directly correspond to real-time.
## How to Create User Defined Shapes – Icons¶
Users can create shapes of their own by using the shape creation tools available in OMEdit.
• Line Tool – Draws a line. A line is created with a minimum of two
points. In order to create a line, the user first selects the line tool from the toolbar and then click on the Icon/Diagram View; this will start creating a line. If a user clicks again on the Icon/Diagram View a new line point is created. In order to finish the line creation, user has to double click on the Icon/Diagram View.
• Polygon Tool – Draws a polygon. A polygon is created in a similar
fashion as a line is created. The only difference between a line and a polygon is that, if a polygon contains two points it will look like a line and if a polygon contains more than two points it will become a closed polygon shape.
• Rectangle Tool – Draws a rectangle. The rectangle only contains two
points where first point indicates the starting point and the second point indicates the ending the point. In order to create rectangle, the user has to select the rectangle tool from the toolbar and then click on the Icon/Diagram View, this click will become the first point of rectangle. In order to finish the rectangle creation, the user has to click again on the Icon/Diagram View where he/she wants to finish the rectangle. The second click will become the second point of rectangle.
• Ellipse Tool – Draws an ellipse. The ellipse is created in a
similar way as a rectangle is created.
• Text Tool – Draws a text label.
• Bitmap Tool – Draws a bitmap container.
The shape tools are located in the toolbar. See Figure 19.
The user can select any of the shape tools and start drawing on the Icon/Diagram View. The shapes created on the Diagram View of Model Widget are part of the diagram and the shapes created on the Icon View will become the icon representation of the model.
For example, if a user creates a model with name testModel and add a rectangle using the rectangle tool and a polygon using the polygon tool, in the Icon View of the model. The model’s Modelica Text will appear as follows:
model testModel
annotation(Icon(graphics = {Rectangle(rotation = 0, lineColor = {0,0,255}, fillColor = {0,0,255}, pattern = LinePattern.Solid, fillPattern = FillPattern.None, lineThickness = 0.25, extent = {{ -64.5,88},{63, -22.5}}),Polygon(points = {{ -47.5, -29.5},{52.5, -29.5},{4.5, -86},{ -47.5, -29.5}}, rotation = 0, lineColor = {0,0,255}, fillColor = {0,0,255}, pattern = LinePattern.Solid, fillPattern = FillPattern.None, lineThickness = 0.25)}));
end testModel;
In the above code snippet of testModel, the rectangle and a polygon are added to the icon annotation of the model. Similarly, any user defined shape drawn on a Diagram View of the model will be added to the diagram annotation of the model.
## Global head section in documentation¶
If you want to use same styles or same JavaScript for the classes contained inside a package then you can define __OpenModelica_infoHeader annotation inside the Documentation annotation of a package. For example,
package P
model M
annotation(Documentation(info="<html>
</html>"));
end M;
<script type=\"text/javascript\">
function HelloWorld() {
}
</script>"));
end P;
In the above example model M does not need to define the javascript function HelloWorld. It is only defined once at the package level using the __OpenModelica_infoHeader and then all classes contained in the package can use it.
In addition styles and JavaScript can be added from file locations using Modelica URIs. Example:
package P
model M
annotation(Documentation(info="<html>
</html>"));
end M;
<script type=\"text/javascript\">
src=\"modelica://P/Resources/hello.js\">
}
</script>"));
end P;
Where the file Resources/hello.js then contains:
function HelloWorld() {
}
## Settings¶
OMEdit allows users to save several settings which will be remembered across different sessions of OMEdit. The Options Dialog can be used for reading and writing the settings.
### General¶
• General
• Language – Sets the application language.
• Working Directory – Sets the application working directory. All files are generated in this directory.
• Toolbar Icon Size – Sets the size for toolbar icons.
• Preserve User’s GUI Customizations – If true then OMEdit will remember its windows and toolbars positions and sizes.
• Terminal Command – Sets the terminal command. When user clicks on Tools > Open Terminal then this command is executed.
• Terminal Command Arguments – Sets the terminal command arguments.
• Hide Variables Browser – Hides the variable browser when switching away from plotting perspective.
• Activate Access Annotations – Activates the access annotations for the non-encrypted libraries. Access annotations are always active for encrypted libraries.
• Libraries Browser
• Library Icon Size – Sets the size for library icons.
• Show Protected Classes – If enabled then Libraries Browser will also list the protected classes.
• Modeling View Mode
• Tabbed View/SubWindow View – Sets the view mode for modeling.
• Default View
• Icon View/DiagramView/Modelica Text View/Documentation View – If no preferredView annotation is defined then this setting is used to show the respective view when user double clicks on the class in the Libraries Browser.
• Enable Auto Save
• Auto Save interval – Sets the auto save interval value. The minimum possible interval value is 60 seconds.
• Enable Auto Save for single classes – Enables the auto save for one class saved in one file.
• Enable Auto Save for one file packages – Enables the auto save for packages saved in one file.
• Welcome Page
• Horizontal View/Vertical View – Sets the view mode for welcome page.
• Show Latest News – if true then displays the latest news.
### Libraries¶
• System Libraries – The list of system libraries that should be loaded every time OMEdit starts.
• Force loading of Modelica Standard Library – If true then Modelica and ModelicaReference will always load even if user has removed them from the list of system libraries.
• Load OpenModelica library on startup – If true then OpenModelica package will be loaded when OMEdit is started.
• User Libraries – The list of user libraries/files that should be loaded every time OMEdit starts.
### Text Editor¶
• Format
• Line Ending - Sets the file line ending.
• Byte Order Mark (BOM) - Sets the file BOM.
• Tabs and Indentation
• Tab Policy – Sets the tab policy to either spaces or tabs only.
• Tab Size – Sets the tab size.
• Indent Size – Sets the indent size.
• Syntax Highlight and Text Wrapping
• Enable Syntax Highlighting – Enable/Disable the syntax highlighting.
• Enable Code Folding - Enable/Disable the code folding. When code folding is enabled multi-line annotations are collapsed into a compact icon (a rectangle containing "...)"). A marker containing a "+" sign becomes available at the left-side of the involved line, allowing the code to be expanded/re-collapsed at will.
• Match Parentheses within Comments and Quotes – Enable/Disable the matching of parentheses within comments and quotes.
• Enable Line Wrapping – Enable/Disable the line wrapping.
• Autocomplete
• Enable Autocomplete – Enable/Disable the autocomplete.
• Font
• Font Family – Shows the names list of available fonts. Sets the font for the editor.
• Font Size – Sets the font size for the editor.
### Modelica Editor¶
• Preserve Text Indentation – If true then uses diffModelicaFileListings API call otherwise uses the OMC pretty-printing.
• Colors
• Items – List of categories used of syntax highlighting the code.
• Item Color – Sets the color for the selected item.
• Preview – Shows the demo of the syntax highlighting.
### MetaModelica Editor¶
• Colors
• Items – List of categories used of syntax highlighting the code.
• Item Color – Sets the color for the selected item.
• Preview – Shows the demo of the syntax highlighting.
### CompositeModel Editor¶
• Colors
• Items – List of categories used of syntax highlighting the code.
• Item Color – Sets the color for the selected item.
• Preview – Shows the demo of the syntax highlighting.
### C/C++ Editor¶
• Colors
• Items – List of categories used of syntax highlighting the code.
• Item Color – Sets the color for the selected item.
• Preview – Shows the demo of the syntax highlighting.
### Graphical Views¶
• Extent
• Left – Defines the left extent point for the view.
• Bottom – Defines the bottom extent point for the view.
• Right – Defines the right extent point for the view.
• Top – Defines the top extent point for the view.
• Grid
• Horizontal – Defines the horizontal size of the view grid.
• Vertical – Defines the vertical size of the view grid.
• Component
• Scale factor – Defines the initial scale factor for the component dragged on the view.
• Preserve aspect ratio – If true then the component’s aspect ratio is preserved while scaling.
### Simulation¶
• Simulation
• Matching Algorithm – sets the matching algorithm for simulation.
• Index Reduction Method – sets the index reduction method for simulation.
• Target Language – sets the target language in which the code is generated.
• Target Compiler – sets the target compiler for compiling the generated code.
• OMC Command Line Options – sets the OMC command line options for the simulation.
• Ignore __OpenModelica_commandLineOptions annotation – if true then ignores the __OpenModelica_commandLineOptions annotation while running the simulation.
• Ignore __OpenModelica_simulationFlags annotation – if true then ignores the __OpenModelica_simulationFlags annotation while running the simulation.
• Save class before simulation – if true then always saves the class before running the simulation.
• Switch to plotting perspective after simulation – if true then GUI always switches to plotting perspective after the simulation.
• Close completed simulation output windows before simulation – if true then the completed simulation output windows are closed before starting a new simulation.
• Delete intermediate compilation files – if true then the files generated during the compilation are deleted automatically.
• Delete entire simulation directory of the model when OMEdit is closed – if true then the entire simulation directory is deleted on quit.
• Output
• Structured – Shows the simulation output in the form of tree structure.
• Formatted Text – Shows the simulation output in the form of formatted text.
### Messages¶
• General
• Output Size - Specifies the maximum number of rows the Messages Browser may have. If there are more rows then the rows are removed from the beginning.
• Reset messages number before simulation – Resets the messages counter before starting the simulation.
• Font and Colors
• Font Family – Sets the font for the messages.
• Font Size – Sets the font size for the messages.
• Warning Color – Sets the text color for warning messages.
• Error Color – Sets the text color for error messages.
• Always quit without prompt – If true then OMEdit will quit without prompting the user.
• Show item dropped on itself message – If true then a message will pop-up when a class is dragged and dropped on itself.
• Show model is defined as partial and component will be added as replaceable message – If true then a message will pop-up when a partial class is added to another class.
• Show component is declared as inner message – If true then a message will pop-up when an inner component is added to another class.
• Show save model for bitmap insertion message – If true then a message will pop-up when user tries to insert a bitmap from a local directory to an unsaved class.
• Always ask for the dragged component name – If true then a message will pop-up when user drag & drop the component on the graphical view.
• Always ask for what to do with the text editor error – If true then a message will always pop-up when there is an error in the text editor.
### Line Style¶
• Line Style
• Color – Sets the line color.
• Pattern – Sets the line pattern.
• Thickness – Sets the line thickness.
• Start Arrow – Sets the line start arrow.
• End Arrow – Sets the line end arrow.
• Arrow Size – Sets the start and end arrow size.
• Smooth – If true then the line is drawn as a Bezier curve.
### Fill Style¶
• Fill Style
• Color – Sets the fill color.
• Pattern – Sets the fill pattern.
### Plotting¶
• General
• Auto Scale – sets whether to auto scale the plots or not.
• Plotting View Mode
• Tabbed View/SubWindow View – Sets the view mode for plotting.
• Curve Style
• Pattern – Sets the curve pattern.
• Thickness – Sets the curve thickness.
### Figaro¶
• Figaro
• Figaro Library – the Figaro library file path.
• Tree generation options – the Figaro tree generation options file path.
• Figaro Processor – the Figaro processor location.
### Debugger¶
• Algorithmic Debugger
• GDB Path – the gnu debugger path
• GDB Command Timeout – timeout for gdb commands.
• GDB Output Limit – limits the GDB output to N characters.
• Display C frames – if true then shows the C stack frames.
• Display unknown frames – if true then shows the unknown stack frames. Unknown stack frames means frames whose file path is unknown.
• Clear old output on a new run – if true then clears the output window on new run.
• Clear old log on new run – if true then clears the log window on new run.
• Transformational Debugger
• Always show Transformational Debugger after compilation – if true then always open the Transformational Debugger window after model compilation.
• Generate operations in the info xml – if true then adds the operations information in the info xml file.
### FMI¶
• Export
• Version
• 1.0 – Sets the FMI export version to 1.0
• 2.0 – Sets the FMI export version to 2.0
• Type
• Model Exchange – Sets the FMI export type to Model Exchange.
• Co-Simulation – Sets the FMI export type to Co-Simulation.
• Model Exchange and Co-Simulation – Sets the FMI export type to Model Exchange and Co-Simulation.
• FMU Name – Sets a prefix for generated FMU file.
• Platforms - list of platforms to generate FMU binaries.
• Import
• Delete FMU directory and generated model when OMEdit is closed - If true then the temporary FMU directory that is created for importing the FMU will be deleted.
### OMTLMSimulator¶
• General
• Path - path to OMTLMSimulator bin directory.
• Manager Process - path to OMTLMSimulator managar process.
• Monitor Process - path to OMTLMSimulator monitor process.
### OMSimulator¶
• General
• Working Directory - working directory for OMSimulator files.
• Logging Level - OMSimulator logging level.
## __OpenModelica_commandLineOptions Annotation¶
OpenModelica specific annotation to define the command line options needed to simulate the model. For example if you always want to simulate the model with a specific matching algorithm and index reduction method instead of the default ones then you can write the following code,
model Test
annotation(__OpenModelica_commandLineOptions = "--matchingAlgorithm=BFSB --indexReductionMethod=dynamicStateSelection");
end Test;
The annotation is a space separated list of options where each option is either just a command line flag or a flag with a value.
In OMEdit right click inside the icon/diagram view of the model and choose Properties. Then OMC Command Line Options and in the text field write --matchingAlgorithm=BFSB --indexReductionMethod=dynamicStateSelection.
If you want to ignore this annotation then use setCommandLineOptions("--ignoreCommandLineOptionsAnnotation=true"). In OMEdit Tools > Options > Simulation check Ignore __OpenModelica_commandLineOptions annotation.
## __OpenModelica_simulationFlags Annotation¶
OpenModelica specific annotation to define the simulation options needed to simulate the model. For example if you always want to simulate the model with a specific solver instead of the default DASSL and would also like to see the cpu time then you can write the following code,
model Test
annotation(__OpenModelica_simulationFlags(s = "heun", cpu = "()"));
end Test;
The annotation is a comma separated list of options where each option is a simulation flag with a value. For flags that doesn't have any value use () (See the above code example).
In OMEdit open the Simulation Setup and set the Simulation Flags then in the bottom check Save __OpenModelica_simulationFlags annotation inside model and click on OK.
If you want to ignore this annotation then use setCommandLineOptions("--ignoreSimulationFlagsAnnotation=true"). In OMEdit Tools > Options > Simulation check Ignore __OpenModelica_simulationFlags annotation.
## Debugger¶
For debugging capability, see Debugging.
## Editing Modelica Standard Library¶
By default OMEdit loads the Modelica Standard Library (MSL) as a system library. System libraries are read-only. If you want to edit MSL you need to load it as user library instead of system library. We don't recommend editing MSL but if you really need to and understand the consequences then follow these steps,
• Go to Tools > Options > Libraries.
• Remove Modelica & ModelicaReference from list of system libraries.
• Add \$OPENMODELICAHOME/lib/omlibrary/Modelica X.X/package.mo under user libraries.
• Restart OMEdit.
## State Machines¶
### Creating a New Modelica State Class¶
Follow the same steps as defined in Creating a New Modelica Class. Additionally make sure you check the State checkbox.
### Making Transitions¶
In order to make a transition from one state to another the user first needs to enable the transition mode () from the toolbar.
Move the mouse over the state. The mouse cursor will change from arrow cursor to cross cursor. To start the transition press left button and move while keeping the button pressed. Now release the left button. Move towards the end state and click when cursor changes to cross cursor.
A Create Transition dialog box will appear which allows you to set the transition attributes. Cancelling the dialog will cancel the transition.
Double click the transition or right click and choose Edit Transition to modify the transition attributes.
### State Machine Simulation¶
Support for Modelica state machines was added in the Modelica Language Specification v3.3. A subtle problem can occur if Modelica v3.2 libraries are loaded, e.g., the Modelica Standard Library v3.2.2, because in this case OMC automatically switches into Modelica v3.2 compatibility mode. Trying to simulate a state machine in Modelica v3.2 compatibility mode results in an error. It is possible to use the OMC flag --std=latest in order to ensure (at least) Modelica v3.3 support. In OMEdit this can be achieved by setting that flag in the Tools > Options > Simulation dialog. | 2018-09-19 13:10:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.26641061902046204, "perplexity": 5937.7259794491565}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267156224.9/warc/CC-MAIN-20180919122227-20180919142227-00043.warc.gz"} |
http://mathhelpforum.com/statistics/124100-marbles-bags.html | 1. ## Marbles in Bags
Bag A contains 3 black marbles and 1 white marble and Bag B contains 1 black marble and 3 white marbles. If two marbles are randomly taken from Bag A and placed in Bag B, then a marble is randomly selected from Bag B, what is the probability that the marble selected from Bag B is black?
I've tried to draw a tree, but I must be doing something wrong cause I can't get the correct answer. Could someone give me some guidance?
2. Hello mneox
Originally Posted by mneox
Bag A contains 3 black marbles and 1 white marble and Bag B contains 1 black marble and 3 white marbles. If two marbles are randomly taken from Bag A and placed in Bag B, then a marble is randomly selected from Bag B, what is the probability that the marble selected from Bag B is black?
I've tried to draw a tree, but I must be doing something wrong cause I can't get the correct answer. Could someone give me some guidance?
The number of ways of selecting $2$ marbles from the four in bag A is $\binom42=6$.
The number of ways of selecting the white marble and one of the black marbles from bag A $= 3$.
Therefore the probability that one of the marbles chosen from bag A is white is $\tfrac36 = \tfrac12$; and the probability that both are black is also $\tfrac12$.
So, when a marble is drawn from bag B, half the time it will be chosen from a total of $2$ black and $4$ white, and half the time from $3$ black and $3$ white. So the probability that it is black is
$\frac12\times \frac26 + \frac12\times \frac36 = \frac{5}{12}$ | 2013-05-19 15:11:24 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9202381372451782, "perplexity": 107.84652900784478}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368697745221/warc/CC-MAIN-20130516094905-00043-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://mathematicalypse.wordpress.com/2016/04/06/linear-assignment/ | Linear Assignment
Problem 1. For this problem use the $4\times 4$ matrix
$A = \left[ \begin{array}{cccc} 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 6 \\ 1 & x & 7 & y \\ 1 & 0 & 0 & 1 \end{array}\right]$
(a) Are there values that we can assign $x$ and $y$ so that $A$ is not an isomorphism? Explain your answer, and, if your answer is yes, provide two such values.
(b) Are there values that we can assign $x$ and $y$ so that $A$ is an isomorphism? Explain your answer, and if your answer is yes, provide two such values.
Answer 1. (a) Note that if we choose $x = 4$ and $y = 10$, then row 3 equals the sum of row 1 and row 2 which implies $A$ is not an isomorphism.
(b) In general, we can compute that the determinant of $A$ is $44 - 6x - 2y$, so provided this expression is never zero, then $\det A \neq 0 \iff A$ is an isomorphism. In particular, we can choose $x = 0 = y$ so that $\det A = 44.$
Problem 2. For this problem let $A$ be a square, $n \times n$ matrix.
(a) Explain why $\det A = 0 \iff$ there exists a non-zero vector $\vec{v} \in \mathbb{R}^n$ such that $A\vec{v} = \vec{0}$.
(b) Explain why a non-zero vector $\vec{v} \in \mathbb{R}^n$ is an eigenvector for $A$ (with eigenvalue $\lambda$) $\iff \det\left(A-\lambda\,I_n\right) = 0$.
Answer 2. (a) We have already shown that there exists a non-zero vector $\vec{v}$ such that $A\vec{v} = \vec{0} \iff A$ is not one-to-one. (By the Rank Nullity Theorem, this also implies that $A$ is also not onto.)
We have also discussed that $\det A = 0 \iff A$ is singular, i.e. $A$ is an isomorphism (from $\mathbb{R}^n$ to $\mathbb{R}^n$). Being an isomorphism requires that $A$ be linear, one-to-one, and onto. Since matrix multiplication is always linear, the determinant equals zero if and only if $A$ is either not one-to-one or not onto. Again, by the Rank Nullity Theorem, $A$ is not one-to-one $\iff \, A$ is not onto.
Hence, $\det A = 0 \iff A$ is not one-to-one $\iff \, \exists \, \vec{v} \neq \vec{0}, A\vec{v} = \vec{0}$.
(b) By definition, an eigenvector must satisfy $\vec{v} \neq \vec{0}$. Therefore, $\vec{v}$ is an eigenvector for $A$ (with eigenvalue $\lambda$) if and only if $A\vec{v} = \lambda\vec{v} \iff A\vec{v} = \lambda\,I\vec{v} \iff A\vec{v} - \lambda I\vec{v} = \vec{0} \iff (A-\lambda I)\vec{v} = \vec{0} \iff \det(A-\lambda I) = 0$ since, by our work in part (a), the matrix $A-\lambda I$ “kills” a non-zero vector if and only if its determinant is zero.
Problem 3. Let $h \in \mathcal{L}(\mathbb{R}^2, \mathbb{R}^2)$ be the linear transformation
$h(x,y) = (x + 2y, 2x + y)^T$
(a) Compute the matrix representation $\text{Rep}_{\mathcal{E}_2, \mathcal{E}_2} (h) = M$.
(b) Use your work in part (b) of Problem 2 to compute all of the possible eigenvalues of the matrix $M$ from part (a) of this problem. (Note: in this case you should find two distinct eigenvalues $\lambda_1, \lambda_2 \in \mathbb{R}$.)
(c ) Find a change-of-basis matrix $P$ so that
$P^{-1}\, M \, P = \left[ \begin{array}{cc} \lambda_1 & 0 \\ 0 & \lambda_2 \end{array}\right]$
(d) Take a moment to draw yourself a congratulatory picture (say, of a rainbow or a flower or something else nice), since you just diagoanlized your first matrix! Yay you!
(a) This matrix representation is given by
$\left( \begin{array}{cc} 1 & 2 \\ 2 & 1 \end{array} \right)$
(b) An eigenvalue will necessarily make the polynomial $p(\lambda) = \det(A-\lambda I)$ equal zero. Therefore we can compute these eigenvalues by solving
$\displaystyle p(\lambda) = \det \left( \begin{array}{cc} 1-\lambda & 2 \\ 2 & 1-\lambda \end{array}\right) = (1-\lambda)^2 - 4 = \lambda^2 - 2\lambda -3 = 0$
Factoring and/or the quadratic formula tells us that this polynomial has two roots, namely $\lambda_1 = 3$ and $\lambda_2 = -1$.
(c ) As discussed in class, we will have
$P^{-1} = \left[ \begin{array}{ccc} | & & | \\ \vec{v}_1 & \cdots & \vec{v}_n \\ | & & | \end{array}\right]$
where the vectors $\mathcal{B} = \{\vec{v}_1, \cdots, \vec{v}_n\}$ are an eigenbasis (with respect to the given matrix). Therefore, to construct this matrix we need to find the associated eigenbasis.
This can be accomplished by first finding a basis for the eigenspace $E_{\lambda_1} = \mathcal{N}(A-\lambda_1\,I)$ and then finding a basis for the eigenspace $E_{\lambda_2} = \mathcal{N}(A-\lambda_2\,I)$. Finding these bases corresponds to solving two homogeneous systems:
$\displaystyle \left( \begin{array}{cc|c} -2 & 2 & 0 \\ 2 & -2 & 0 \end{array}\right) \Rightarrow E_{\lambda_1} = \left\{ y\left(\begin{array}{c} 1 \\ 1 \end{array}\right) : y \in \mathbb{R} \right\}$
$\displaystyle \left(\begin{array}{cc|c} 2 & 2 & 0 \\ 2 & 2 & 0 \end{array}\right) \Rightarrow E_{\lambda_2} = \left\{ y\left(\begin{array}{c} 1 \\ -1 \end{array}\right) : y \in \mathbb{R}$
Each eigenspace is one-dimensional, and so we may select a single basis vector from each. Our eigenbasis is, for instance,
$\mathcal{B} = \left\{ \left(\begin{array}{c} 1 \\ 1 \end{array}\right), \, \left\{ y\left(\begin{array}{c} 1 \\ -1 \end{array}\right)$
As a result we have
$\left( \begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array}\right)\,\left(\begin{array}{cc} 1 & 2 \\ 2 & 1 \end{array}\right)\,\left( \begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array}\right)^{-1} = \left(\begin{array}{cc} 3 & 0 \\ 0 & -1 \end{array}\right)$
Problem 4. Compute the area of the parallelogram shown in the picture below.
(Note: unit lengths are marked on the axes above.)
Answer 4. This can be computed by writing down the two black vectors, $\vec{v}_1$ and $\vec{v}_2$, as columns in a matrix
$A = \left[ \begin{array}{cc} | & | \\ \vec{v}_1 & \vec{v}_2 \\ | & | \end{array}\right ]$.
Then the desired area is $|\det A|$.
Problem 5. (a) Find $A^{-1}$ given
$A = \left[ \begin{array}{cc} \pi & 1 \\ 2 & 4 \end{array}\right]$.
(b) Find $A^{-1}$ given
$A = \left[ \begin{array}{ccc} 1 & 2 & 3 \\ 1 & 1 & 1 \\ 3 & 5 & 8 \end{array}\right]$.
If we want to interpret $A$ as a change-of-basis matrix, $A = \text{Rep}\left(\text{id}_{\mathbb{R}^3}\right)_{\mathcal{E}_3, \mathcal{B}}$, what are the basis vectors $\mathcal{B} = \{ \vec{v}_1, \vec{v}_2, \vec{v}_3 \}$?
Answer 5. (a) The matrix $A^{-1}$ is given by
$A^{-1}= \frac{1}{4\pi-2}\left[\begin{array}{cc}4 & -1 \\ -2 & \pi \end{array}\right]$
(b) The matrix $A^{-1}$ is given by
$A^{-1} = \left[\begin{array}{ccc} -3 & 1 & 1 \\ 5 & 1 & -2 \\ -2 & -1 & 1 \end{array}\right]$
The basis $\mathcal{B}$ contains the vectors
$\mathcal{B} = \left\{ \left(\begin{array}{c} -3 \\ 5 \\ -2 \end{array}\right), \left(\begin{array}{c} 1 \\ 1 \\ -1 \end{array}\right), \left(\begin{array}{c} 1 \\ -2 \\ 1 \end{array}\right)\right\}$
Problem 6. Let $D$ denote an $n \times n$ diagonal matrix:
$D = \left[ \begin{array}{cccc} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_n \end{array} \right]$.
Under what conditions is $D$ invertible? Write down the inverse matrix $D^{-1}$ (assuming the conditions you specified are met).
Answer 6. $D$ is invertible provided each diagonal entry $\lambda_i \neq 0$. When this condition is met, the inverse matrix is a diagonal matrix with entries
$\displaystyle (D)_{i,i} = \frac{1}{\lambda_i}$
Problem 7. The trace of a square matrix $A \in \mathcal{M}_{n\times n}$ is the sum of the diagonal entries of $A$:
$\displaystyle \text{tr}A = \sum_{i=1}^n \! a_{ii} = a_{11} + a_{22} + \cdots + a_{nn}$.
(a) Prove that the trace, regarded as a function $\text{tr}:\mathcal{M}_n \to \mathbb{R}$ is a homomorphism.
(b) A proof is written below. Write down the theorem this proof demonstrates.
Theorem: Given two matrices, $\text{tr}(AB) = \text{tr}(BA)$.
proof. Let $A, B \in \mathcal{M}_{n\times n}$.
Let us notate the $(i, j)$-entry of $A$ by $a_{ij}$, the $(i, j)$-entry of $B$ by $b_{ij}$, the $(i, j)$-entry of the product $AB$ by $c_{ij}$, and the $(i, j)$-entry of the product $BA$ by $d_{ij}$.
Then the traces of these matrices is given by the sum of their respective diagonal entries so that
$\text{tr}A = a_{11} + a_{22} + \cdots + a_{nn}$
$\text{tr}B = b_{11} + b_{22} + \cdots b_{nn}$
$\text{tr}AB = c_{11} + c_{22} + \cdots c_{nn}$
$\text{tr}BA = d_{11} + d_{22} + \cdots d_{nn}$
We want to prove that $\text{tr}(AB) = \text{tr}(BA)$, i.e. that
$c_{11} + c_{22} + \cdots + c_{nn} = d_{11} + d_{22} + \cdots d_{nn}$.
To prove this, we first note that
$c_{ij} = \text{(row i)}_A \cdot \text{(col j)}_B \text{ and } d_{ij} = \text{(row i)}_B \cdot \text{(col j)}_A$
This lets us replace each $c_{ii}$ expression appearing in the trace of $AB$ in terms of the entries of $A$ and $B$. In particular, we find
$\text{tr}(AB) = \text{(row 1)}_A\cdot\text{(col 1)}_B + \text{(row 2)}_A\cdot\text{(col 2)}_B + \cdots + \text{(row n)}_A\cdot \text{(col n)}_B$
$= \left(a_{11}b_{11} + a_{12}b_{21} + \cdots + a_{1n}b_{n1}\right) + \left(a_{21}b_{12} + a_{22}b_{22} + \cdots a_{2n}b_{n2}\right) + \cdots + \left(a_{n1}b_{1n} + a_{n2}b_{2n} + \cdots a_{nn}b_{nn}\right)$
$\displaystyle = \left( \sum_{i=1}^n a_{1i}b_{i1} \right) + \left( \sum_{i=1}^n a_{2i}b_{i2}\right) + \cdots + \left(\sum_{i=1}^n a_{ni}b_{in}\right) = \sum_{i,j = 1}^n a_{ji}b_{ij}$
Similarly, the trace of the product $BA$ works out as follows:
$\text{tr}(BA) = \text{(row 1)}_B\cdot\text{(col 1)}_A + \text{(row 2)}_B\cdot\text{(col 2)}_A + \cdots + \text{(row n)}_B\cdot \text{(col n)}_A$
$= \left(b_{11}a_{11} + b_{12}a_{21} + \cdots + b_{1n}a_{n1}\right) + \left(b_{21}a_{12} + b_{22}a_{22} + \cdots b_{2n}a_{n2}\right) + \cdots + \left(b_{n1}a_{1n} + b_{n2}a_{2n} + \cdots b_{nn}a_{nn}\right)$
$\displaystyle = \left( \sum_{i=1}^n b_{1i}a_{i1} \right) + \left( \sum_{i=1}^n b_{2i}a_{i2}\right) + \cdots + \left(\sum_{i=1}^n b_{ni}a_{in}\right) = \sum_{i,j = 1}^n b_{ji}a_{ij} = \sum_{i, j = 1}^n a_{ij}b_{ji}$
Note that in this last sum if we interchange the names of the indices $i$ and $j$, we obtain the exact same expression for the $\text{tr}(AB)$. This implies that these two sums are equal, completing the proof. $\square$
(c ) Suppose that a square matrix $A$ is diagonalizable. Explain why $\det A =$ the product of $A$‘s eigenvalues. Using part (b) from this problem, explain why $\text{tr}A =$ the sum of $A$‘s eigenvalues.
Answer 7. (a) Proving the trace is linear is straightforward.
(b) The Theorem claims that $\text{tr}(AB) = \text{tr}(BA)$.
(c ) First, we can use the fact that for determinants, $\det(AB) = \det(A)\det(B)$. Suppose $A$ diagonalizes to $D$ so that $P^{-1}AP = D$. We then have
$\det D = \det(P^{-1}AP) = \det(P^{-1})\det(A)\det(P) = \det A$
and since $D$ is a diagonal matrix, by our (many) formula(s) for computing the determinant we find that
$\lambda_1\lambda_2\cdots\lambda_n = \det D = \det A$
where $\lambda_i$ is an eigenvalue for $A$.
For the trace claim, we can perform a similar bit of magic. Note that
$\text{tr}(D) = \text{tr}(P^{-1}AP) = \text{tr}(P^{-1}PA) = \text{tr}(A)$
and since the trace of $D$ is $\text{tr}D = \lambda_1 + \cdots \lambda_n$ we find that the trace of $A$ is the sum of its eigenvalues.
Problem 8. (a) Given two square matrices, $A, B \in \mathcal{M}_{n\times n}$, prove that $(AB)^T = B^T\,A^T$.
(b) Suppose $A$ is a non-singular matrix. Prove that $\left(A^{-1}\right)^T = \left(A^T\right)^{-1}$.
Answer 8. (a) Let $C = AB$, and now that the $(i, j)$ entry of $C$ is the number
$c_{ij} = (\text{row}_i A)\cdot (\text{col}_j B)$
so that the $(i, j)$ entry of $C^T$ is
$c_{ji} = (\text{row}_j A)\cdot(\text{col}_i B)$
We now need to argue that the $(i, j)$ entry of the product $B^TA^T$ equals $c_{ji}$. Of course
$\left(B^TA^T\right)_{i,j} = (\text{row}_i B^T)\cdot(\text{col}_j A^T) = (\text{col}_i B)\cdot (\text{row}_j A) = c_{ji}$.
(b) To prove the claim we simply use part (a) and multiply:
$AA^{-1} = I \Rightarrow \left(AA^{-1}\right)^T = I^T = I$
and by part (a) we have
$\left(A^{-1}\right)^T\,A^T = I \Rightarrow \left(A^{-1}\right)^T = \left(A^T\right)^{-1}$
Problem 9. (Bonus) For this problem, get help from a math student who has taken Algebra I.
Prove that the set $\text{GL}_n(\mathbb{R}) = \{ \text{ all invertible, } n \times n \text{ matrices with real entries } \}$ is a group under matrix multiplication.
Answer 9. (definition) A group is a set, $G$, with a binary operation $\star : G\times G \to G$ satisfying three conditions.
(1) $\exists \, e \in G, g\star e = g = e \star g \, \, \forall \, g \in G$
(2) $\forall \, g \in G, \exists \, h \in G, g\star h = e = h \star g$
(3) $\star$ is associative; i.e. $\forall \, a, b, c \in G, a\star(b\star c) = (a\star b)\star c)$.
To prove that $\text{GL}_n(\mathbb{R})$ is a group under matrix multiplication, we must show these three conditions hold (where $A\star B$ means $AB$).
First, the $n \times n$ identity matrix, $I_n \in \text{GL}_n(\mathbb{R})$, satisfies property (1) since for all square matrices $AI = A = IA$.
Second, given an invertible matrix $A \in \text{GL}_n(\mathbb{R})$, we have that $A^{-1} \in \text{GL}_n(\mathbb{R}$ with $AA^{-1} = I = A^{-1}A$.
Third, by the definition of matrix multiplication it follows that $A(BC) = (AB)C$ — in fact, the easiest way to argue this is to view matrix multiplication not as “rows times columns,” but as representing the composition of homomorphisms.
fdfd | 2017-07-23 22:46:41 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 182, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9793462157249451, "perplexity": 354.2258450310743}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549424623.68/warc/CC-MAIN-20170723222438-20170724002438-00628.warc.gz"} |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/07._Periodic_Properties_of_the_Elements/7.8%3A_Group_Trends_for_Selected_Nonmetals | 7.8: Group Trends for Selected Nonmetals
Learning Objectives
• To gain a descriptive understanding of the chemical properties of Hydrogen, the group 16, 17 and 18 elements.
Non-metallic character is the ability to be reduced (be an oxidizing agent), form acidic hydroxides, form covalent compounds with non-metals. These characteristics increase with a larger nuclear charge and smaller radius, with no increase in shielding. The most active non-metal would be the one farthest up and to the right -- not including the noble gases (non-reactive.)
Hydrogen
Hydrogen has a 1s1 electron configuration and is placed above the alkali metal group. Hydrogen is a non-metal, which occurs as a gas (H2) under normal conditions.
• Its ionization energy is considerably higher (due to lack of shielding, and thus higher $$Z_{eff}$$) than the rest of the Group 1 metals and is more like the nonmetals
• Hydrogen generally reacts with other nonmetals to form molecular compounds (typically highly exothermic)
• Hydrogen reacts with active metals to form metal hydrides which contain the H- hydride ion:
$2Na_{(s)} + H_{2(g)} \rightarrow 2NaH_{(s)} \label{7.8.1}$
• Hydrogen can also lose an electron to yield the aqueous $$H^+_{(aq)}$$ hydronium ion.
Group 16: The Oxygen Family
As we proceed down group 16 the elements become more metallic in nature:
• Oxygen is a gas, the rest are solids
• Oxygen, sulfur and selenium are nonmetals
• Tellurium is a metalloid with some metal properties
• Polonium is a metal
Oxygen can be found in two molecular forms, O2 and O3 (ozone). These two forms of oxygen are called allotropes (different forms of the same element in the same state)
$3O_{2(g)} \rightarrow 2O_{3(g)}\;\;\; \Delta H = 284.6\; kJ / mol \label{7.8.2}$
the reaction is endothermic, thus ozone is less stable that O2
Oxygen has a great tendency to attract electrons from other elements (i.e. to "oxidize" them)
• Oxygen in combination with metals is almost always present as the O2- ion (which has noble gas electronic configuration and is particularly stable)
• Two other oxygen anions are observed: peroxide (O22-) and superoxide (O2-)
Sulfur
Sulfur also exists in several allotropic forms, the most common stable allotrope is the yellow solid S8 (an 8 member ring of sulfur atoms). Like oxygen, sulfur has a tendency to gain electrons from other elements, and to form sulfides (which contain the S2- ion). This is particular true for the active metals:
$16Na_{(s)} + S_{8(s)} \rightarrow 8Na_2S_{(s)}\label{7.8.3}$
Note: most sulfur in nature is present as a metal-sulfur compound. Sulfur chemistry is more complex than that of oxygen.
Group 17: The Halogens
"Halogen" is derived from the Greek meaning "salt formers"
• Astatine is radioactive and rare, and some of its properties are unknown
• All the halogens are nonmetals
• Each element consists of diatomic molecules under standard conditions
Colors of diatomic halogens: (not flame colors)
• Fluorine: pale yellow
• Chlorine: yellow green
• Bromine: reddish brown
• Iodine: violet vapor
The halogens have some of the most negative electron affinities (i.e. large exothermic process in gaining an electron from another element)
$X_2 + 2e^- \rightarrow 2X^-\label{7.8.4}$
• Fluorine and chlorine are the most reactive halogens (highest electron affinities). Fluorine will remove electrons from almost any substance (including several of the noble gases from Group 18).
Note
The chemistry of the halogens is dominated by their tendency to gain electrons from other elements (forming a halide ion)
In 1992, 22.3 billion pounds of chlorine was produced. Both chlorine and sodium can be produced by electrolysis of molten sodium chloride (table salt). The electricity is used to strip electrons from chloride ions and transfer them to sodium ions to produce chlorine gas and solid sodium metal
Chlorine reacts slowly with water to produce hydrochloric acid and hypochlorous acid:
$Cl_{2(g)} + H_2O_{(l)} \rightarrow HCl_{(aq)} + HOCl_{(aq)}\label{7.8}.5$
Hypochlorous acid is a disinfectant, thus chlorine is a useful addition to swimming pool water
The halogens react with most metals to form ionic halides:
$Cl_{2(g)} + 2Na_{(s)} \rightarrow 2NaCl_{(s)}\label{7.8.6}$
Group 18: The Noble Gases
• Nonmetals
• Gases at room temperature
• monoatomic
• completely filled 's' and 'p' subshells
• large first ionization energy, but this decreases somewhat as we move down the group
Rn is highly radioactive and some of its properties are unknown
They are exceptionally unreactive. It was reasoned that if any of these were reactive, they would most likely be Rn, Xe or Kr where the first ionization energies were lower.
Note
In order to react, they would have to be combined with an element which had a high tendency to remove electrons from other atoms. Such as fluorine.
Compounds of noble gases to date:
$$XeF_2$$ $$XeF_4$$ $$XeF_6$$
only one compound with Kr has been made
$$KrF_2$$
No compounds observed with He, Ne, or Ar; they are truly inert gases. | 2020-10-30 10:47:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6265087723731995, "perplexity": 4270.075211433466}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107910204.90/warc/CC-MAIN-20201030093118-20201030123118-00215.warc.gz"} |
https://www.nature.com/articles/s41598-018-34085-4?error=cookies_not_supported&code=a84aaaf4-ca87-4880-b88e-261ed67a09de | # Acceleration-induced pressure gradients and cavitation in soft biomaterials
## Abstract
The transient, dynamic response of soft materials to mechanical impact has become increasingly relevant due to the emergence of numerous biomedical applications, e.g., accurate assessment of blunt injuries to the human body. Despite these important implications, acceleration-induced pressure gradients in soft materials during impact and the corresponding material response, from small deformations to sudden bubble bursts, are not fully understood. Both through experiments and theoretical analyses, we empirically show, using collagen and agarose model systems, that the local pressure in a soft sample is proportional to the square of the sample depth in the impact direction. The critical acceleration that corresponds to bubble bursts increases with increasing gel stiffness. Bubble bursts are also highly sensitive to the initial bubble size, e.g., bubble bursts can occur only when the initial bubble diameter is smaller than a critical size (≈10 μm). Our study gives fundamental insight into the physics of injury mechanisms, from blunt trauma to cavitation-induced brain injury.
## Introduction
The transient dynamic response of biological materials during rapid mechanical loading1,2,3 is becoming increasingly important due to emerging medical implications, i.e., injury mechanisms under blast, ballistic, or impact exposures4,5,6. When exposed to these threats, a biological system, e.g., human brain or skin, is rapidly accelerated, which results in the acceleration-induced pressure gradient. Depending on the amplitude and duration of such mechanical inputs, the biological system may be subjected to everything from small deformations to considerably larger damage associated with sudden bubble bursts. For the accurate prediction and assessment of potential injuries, a fundamental understanding of how soft materials respond under rapid loading conditions is critical.
Despite this persistent need, the underlying mechanisms that govern biomaterial deformation are not fully understood, mainly due to the complex coupling of fluid- and solid-like characteristics. In addition, accurate characterization of soft biomaterial properties, including biological hydrogels or tissues, remains challenging due to their soft, labile nature7,8,9,10, nonlinear material response11,12,13,14,15, and strain-rate-dependent material properties16,17.
As an example, while it is well known that material properties of soft materials greatly depend on their concentration as well as loading conditions, experimental measurements on the key properties such as surface tension and elasticity, in particular, at fast loading conditions are still very limited. Established experimental techniques18 to quantify surface tension are mostly for liquid-phase solutions, and as a result, the induced surface tension for gel-phase samples (i.e., fully polymerized hydrogels) is not well characterized. It is worth noting that there are several emerging technologies specifically designed for measuring the surface tension of soft materials19,20,21. For elastic modulus measurements, many characterization techniques8,9,22, including high loading conditions17, are readily available for soft material samples, but the quantitative values for soft biological tissue samples (e.g. elastic modulus) can significantly vary depending on the characterization methods23.
One notable advance is in the characterization of cavitation properties for soft materials under an impulsive force15,24 where the critical acceleration that corresponds to the onset of the bubble formation and bursts is quantitatively measured. While the acceleration-induced pressure is the primary driving force for this rapid material deformation associated with bubble dynamics in pure water and gelatin samples, the spatio-temporal dynamics in soft material samples during mechanical impact remains largely unknown.
Here we experimentally and theoretically consider transient bubble dynamics in biologically-relevant soft materials, collagen and agarose, that are commonly used as an extracellular matrix (ECM) for 3-dimensional (3D) cell culture. For quantitative characterization of soft materials, we utilize a recently developed experimental setup that is based on a drop tower system10,15. First, we directly visualize the acceleration-induced pressure gradient during impact utilizing 1% collagen samples with optically visible, macro air bubbles. By analyzing high speed camera images, the maximum bubble size is correlated to the mechanical input and sample height while monitoring key bubble behaviors, including burst and collapse. We further quantify the critical mechanical inputs that trigger cavitation nucleation and bursting in agarose samples as a function of the sample stiffness. Finally, we develop a theoretical model for predicting transient bubble dynamics.
## Results
We experimentally considered transient bubble dynamics in soft material samples during mechanical impact. For this study, a series of drop tower experiments were performed in which the amplitude of each impact was controlled by changing vertical height (hdrop) of a movable mass (impactor) with respect to the customized sample holder (Fig. 1a). The holder was designed to accommodate a transparent cuvette containing the soft material sample. Once released, the impactor was accelerated toward the holder by gravity (g) until collision. During impact, high speed cameras and an accelerometer were utilized for concurrent visual observation and impulse force quantification. Detailed sample preparation and experimental procedures are discussed in the Methods section.
It is important to note that a soft foam sheet is placed between (1) the sample holder and rigid anvil as well as (2) the impactor and sample holder, respectively (see Fig. 1a). Because these foams are readily deformable upon mechanical impact, the sample holder is free to move vertically against the foam stiffness. In addition, these foams prevent a direct collision between the rigid parts including the holder, anvil, and impactor, which would result in generating shock waves and propagating waves to the sample via surrounding solid structures, i.e., the sample holder and cuvette. By utilizing the soft foam sheets, we achieve the smooth acceleration profiles, which are critical to study acceleration-induced cavitation, as shown in Fig. 1c–g without any pulse-like signals. In-depth dynamics analysis of the experimental setup can be found elsewhere10.
### Acceleration-induced pressure gradient and transient bubble dynamics
Here we focus on the characterization of acceleration-induced pressure gradients in soft materials during impact. For direct visualization of the pressure gradient we utilized collagen, the most abundant extracellular protein in vivo and one of the most commonly used proteins for 3D cell culture25,26,27, prepared in a transparent cuvette with randomly distributed, macro air bubbles (Fig. 1b). To quantify the correlation between bubble size and input acceleration, bubble radii (r), indicated by the arrows in Fig. 1b, were measured by performing image analysis of each high speed image frame (see the Methods section). Next, the bubble size was correlated with the acceleration measurements as shown in Fig. 1c–g.
Figure 1c–g show the normalized bubble radius (λ = r/r0, vertical axis on the left) utilizing 5 different drop heights, i.e., hdrop = 4, 6, 8, 10, and 12 cm, respectively. The corresponding acceleration signals induced by these mechanical impacts at t = 0 are also shown (vertical axis on the right) where the amplitudes of the acceleration, |aamp|, were 228.52, 300.78, 376.95, 448.24, and 516.60 g, respectively, and t is time. To measure the initial radius (r0), the mean average of each target bubble was obtained over 49 high speed image frames from t = −2 ms to = 0 ms (see Table S1 for the initial radius (r0) and vertical location (hb) of each bubble with respect to the top surface of the collagen sample (Fig. 1a) where the average bubble radius of all 8 bubbles was 0.52 mm). As expected, before the impact (t < 0), r/r0 = 1 for all bubbles. Supplementary Movies 15 show the acceleration measurement and corresponding high speed camera images for Fig. 1c–g, respectively.
The results in Fig. 1 indicate that an increase in bubble size depends on not only the drop height (hdrop), but the vertical location (hb) of each bubble. For example, with regards to drop height dependence, the maximum normalized radius (λmax = rmax/r0) of Bubble 7 increased by 40% when drop height increased from 4 cm to 12 cm (Fig. 1c,g, respectively). With regards to vertical location, bubbles of the same initial radii, for example Bubble 1 at the top of the field of view and Bubble 8 at the bottom, exhibited radically different deformations with λmax = 1.08 for Bubble 1 and λmax = 1.31 for Bubble 8 (4 cm drop height). This effect only became more pronounced as the drop height increased.
To quantitatively characterize the correlation between λmax, |aamp|, and hb, we plotted λmax for each bubble as a function of |aamp| in Fig. 2a, which shows that the slopes of the λmax and |aamp| curves strongly depend on hb. Furthermore, we found that the slope is proportional to $${h}_{b}^{2}$$ utilizing the linear least-squares fitting in Fig. 2b. Based on these experimental observations, we concluded that λmax is linearly proportional to (|aamp|/g)(hb/H)2 as summarized in the inset of Fig. 2b, i.e., λmax = 0.0023(|aamp/g|)(hb/H)2 + 1.0153 with R2 = 0.9911 where H, the total height of the collagen sample (Fig. 1a), was 36 mm. It is worth noting that dynamic size characterization of individual bubbles became increasingly difficult with increasingly higher drops because bubble became large enough to directly interact with either other bubbles or cuvette walls. As an example, the radius of Bubble 8 is not available Fig. 1g because it merged with a neighboring bubble (see Supplementary Movie 5).
Another important observation from Fig. 2a is that bubble bursts, a sharp increase in bubble radius (λmax 1) with a small increase in a mechanical input, were not observed even for a considerably large input acceleration (=516.60 g). This is interesting because it was reported that input accelerations of ~500 g triggers bubble bursts even in much stiffer gelatin samples without macro bubbles15, e.g., the nominal shear moduli for 7.5% gelatin28 and 1% collagen9 are approximately 20 kPa and 10 Pa, respectively, at relatively low strain rates. In general, the higher stiffness results in increasing resistance against bubble expansion and, as a result, requires higher accelerations to trigger bubble bursts. Because our observations are contradictory to this statement, we hypothesized that, in addition to material stiffness, bubble burst in soft material samples is also dependent upon initial bubble size, as is further discussed below.
For validation of our experimental approach, we also measured the size of the reference (rref), a non-circular solid, plastic inclusion at the bottom of the cuvette. Unlike bubbles, the reference was rigid and, as a result, rref/r0 ≈ 1 regardless of drop height. The small fluctuation in rref/r0 observed for t > 0 was likely due to illumination variations as the cuvette holder moved vertically with respect to the lights during impact. We estimated standard error in size of the reference when |aamp| > 0 (Fig. 2a) and found that the mean and 95% confidence interval were rref/r0 = 1.03 ± 0.02 (indicated by the gray area). This confirmed a vibration induced measurement error which was significantly smaller than λmax values for the bubbles and, therefore, we concluded that our observations/analysis on the bubble size are not due to measurement artifacts.
### Bubble burst in soft materials
For the burst of bubbles in soft materials, we utilized agarose samples, another common material for 2D/3D cell culture29,30, for two main reasons: (1) agarose is considerably stiffer than collagen and hence we can explore bubble dynamics for a wide range of material stiffnesses and (2) unlike collagen, agarose samples can be readily prepared without pre-existing macro bubbles. Mechanical stiffness of biological samples typically covers a broad range from 0.1~10 kPa for brain tissues31,32,33 to 100 kPa for human aorta tissue34. Agarose with its concentration dependent stiffness can be easily tuned to mimic various target tissues, offering an attractive advantage as a tissue simulant. Second, optically visible bubbles (>1 μm) generally do not exist in biological systems such as tissues and organs. Therefore, characterizing gels without macro bubbles during impact is important for understanding potential bubble formation, also known as cavitation nucleation, in these systems and the possible damage mechanisms associated with cavitation nucleation, bubble burst, and collapse.
For this study, we first characterized the critical acceleration (acr) that corresponds to cavitation nucleation followed by bubble burst in the agarose samples. To do so we prepared agarose samples without optically visible bubbles and detected bubbles that nucleated directly as a result of impact. Eighteen agarose samples at each concentration (0.3%, 0.58%, 0.9%, and 1.5% agarose) were prepared in individual cuvettes following the protocol discussed in the Methods section. The onset of cavitation was detected by visual observation of the high speed camera video of the impacted sample. Note that the smallest bubble radius that was optically detectable was on the order of ~100 μm due to the limited spatial resolution of the high speed camera images.
Figure 3a,b show acceleration measurement and the corresponding material response for a 0.3% agarose sample, respectively, during impact (see Supplementary Movie 5). In Fig. 3b, no bubble was optically detected at t = 0 ms and then a cavitation bubble (see the zoom-in view) was first observed at tcr = 0.44 ms. Finally tcr in Fig. 3a was directly correlated to the acceleration signal to determine the corresponding acceleration (acr). The same experimental procedure above was repeated for all 72 agarose samples to characterize cavitation properties (see Tables S2–5 for the detailed data). Supplementary Movies 69 show representative experimental results for 0.3%, 0.58%, 0.9%, and 1.5% agarose, respectively. To minimize the possible effect of accumulated damage to samples on the measured critical acceleration values, we replaced a sample by a new one whenever cavitation nucleation was detected.
Figure 3c shows the critical acceleration as a function of elastic modulus of agarose samples where $${E}_{a}=85{c}_{a}^{1.8}\,{\rm{kPa}}$$16 was used to convert agarose concentration (ca) to elastic modulus (Ea). Similarly, cavitation properties of gelatin samples15 were also converted using Eg = 8cg kPa28 for direct comparison. In using these relationships, we note the that static elastic modulus of soft gels as a function of gel concentration is well established in the literature16,28. Also, it is known that elastic properties within small material deformation of gels are relatively insensitive to strain rates at low strain rates (~10/min)35,36 as well as high strain rates (~1000/s)37,38. In addition, previous studies on laser- as well as acoustically-induced cavitation bubbles in gels showed that use of strain-rate independent nominal elastic modulus captures bubble dynamics reasonably well14,39.
Both agarose and gelatin samples quantitatively show a similar trend in increasing the critical acceleration with increasing elastic modulus, i.e., a sharp increase from pure water to ~10 kPa followed by a more modest slope for higher elastic modulus. This trend can be explained by (1) concentration-dependent material properties of soft gels and (2) interplay between pre-existing material defects, e.g., micro/nano-bubbles, and gel molecules. As briefly discussed earlier, gel stiffness including collagen, agarose, and gelatin increases with increasing concentration16,22,28. For stiffer gels, more energy is needed for local deformation of a gel associated with bubble expansion and, as a result, larger mechanical inputs, i.e., higher accelerations, is expected to trigger cavitation nucleation and bubble bursts.
Regarding the bubble-gel interplay in gels without macro bubbles, small traces of nanoscale air bubbles (~hundreds nanometers in size) in ultrapure water have been detected utilizing dynamic light scattering40. Also it is known that dissolved gases in water tend to accumulate at the interfaces between solids and water to form nanoscale bubbles41,42. Because all samples, including agarose, gelatin, and pure water, in Fig. 3c were prepared using ultrapure water and had either a cuvette-gel or -water interface, submicron-scale bubbles, although optically invisible, likely existed in the samples.
For pure water, the shape of a bubble is mainly determined by surface energy minimization, i.e., a sphere, because water is a rigidity-free liquid medium. Unlike pure water, a bubble (or material defect) in gels would directly interact with the neighboring gel structures. Due to gel stiffness, a bubble cannot change its shape without deforming the gel structures and, as a result, the bubble shape at equilibrium is determined by surface energy of a bubble as well as stored-strain energy in gel. As a result, a bubble can be non-spherical, which increases its area-to-volume ratio. It was theoretically shown that a bubble with a larger area-to-volume ratio requires higher mechanical inputs for cavitation nucleation15, explaining the sharp increase in acr from pure water to the lower concentrations of gels in Fig. 3c.
Compared to collagen with macro bubbles, we noticed distinctively different dynamic responses in agarose, gelatin, and pure water without macro bubbles. While the maximum normalized radius of each bubble is linearly proportional to the amplitude of acceleration for collagen (λmax |aamp|), the bubble dynamics in the agarose, gelatin, and pure water samples was highly non-linear and stochastic. As an example, no bubbles were optically observed when t < tcr and then millimeter scale bubbles suddenly appeared for t > tcr. Also, the maximum bubble radius in the agarose sample in Fig. 3b was rmax = 1.13 mm at t = 0.58 ms while the smallest bubble radius that was optically detectable was r = 180 μm at tcr = 0.44 ms (Fig. 3b). This result indicates that λmax values for the micro/nano-scale bubbles are expected to be much larger than macro bubbles, e.g., λmax for agarose (>6.28) is considerably larger than collagen (<1.8, Fig. 2a). Finally, all bubbles nucleated in agarose, gelatin, and pure water were fully collapsed, e.g., r0 = 0, after completion of each impact.
So far, we have experimentally shown that (1) $$({\lambda }_{max}-1)\propto {h}_{b}^{2}|{a}_{amp}|$$ and under these conditions that (2) bubble bursts do not occur for relatively large bubbles (~0.5 mm) and (3) the critical acceleration increases with increasing concentration (or equivalently stiffness) of gels. The first conclusion, i.e., mechanical impacts result in pressure gradients in a soft gel, has a key application toward characterization of acceleration-induced pressure (p) in soft material samples. Currently, pah (see Eq. S3 in the supplementary document) is often used to estimate p where a and h are acceleration and a coordinate of the sample in the impact direction because directly measuring p in pure water24 and soft gels15 during impact is still experimentally challenging. In this regard, we theoretically consider below how p, λ and h are linked to each other and show pah is not a good estimation for the acceleration-induced pressure gradient for soft gels. It is worth noting that the Navier-Stokes derivations shown in the Supplementary Information are simplified, 1D, and just for water not a gel.
The other conclusions have important implications for biological studies investigating injury mechanisms due to rapid mechanical loading, e.g., possible traumatic brain injury mechanisms associated with impact or blast. As an example, air bubbles with a wide range of radii from several hundreds of micrometers to a few millimeters are often introduced into biosamples such as soft gels or brain tissues to mimic bubble-brain interplay in vitro43. However, because key characteristics of bubble dynamics in such soft materials are sensitive to the initial bubble size, use of relatively large bubbles may not realistically capture the bubble-brain interplay when a human head is exposed to impact or blast. In addition, because target organs, e.g., brain or skin, could have significantly different material properties, quantitative characterization of cavitation properties for different soft biosamples becomes directly relevant for developing reliable cavitation criteria in the scope of cavitation-induced injuries. Next we theoretically consider the important role of initial bubble size and gel stiffness on bubble dynamics in a soft gel and discuss possible mechanisms that capture the key experimental observations above.
### Theoretical Framework
Consider a bubble in a soft material sample in Fig. 4a with an emphasis on the pressure and bubble size relation in a soft material sample. Upon mechanical impact, the sample is rapidly accelerated by ain that results in acceleration-induced pressure (pa) in the sample. Based on the experimental results above, pa is a function of time (t), the amplitude of input acceleration (aamp), and the bubble location (hb) in a gel. In the following analysis, we assume that bubble radius (r) is relatively small such there is no pressure gradient from the top of the bubble to its bottom, i.e., pa(t, aamp, hb) ≈ pa(t, aamp, hb ± r). We also assume constant temperature and an incompressible material model, i.e., viscoelastic neo-Hookean material. These are reasonable assumptions for bubble expansion, our main focus, because a rate of change of bubble radius during bubble expansion is typically much slower than bubble collapse.
In fluid mechanics, the dynamics of a spherical bubble in Newtonian liquids has been well understood utilizing the Rayleigh–Plesset equation44. Unlike Newtonian liquids, soft gels have material stiffness and generally relax to an original configuration after loading, i.e., a Kelvin-Voigt viscoelastic model13. The Rayleigh–Plesset equation has been modified by incorporating the Kelvin-Voigt and neo-Hookean models13,39 to theoretically capture bubble dynamics in a viscoelastic medium as follows. The governing equation for the dynamics of a spherical bubble at hb in Fig. 4a can be written as13,44
$$\frac{{p}_{v}-{p}_{a}(t,{a}_{amp},{h}_{b})}{\rho }+\frac{{p}_{G0}}{\rho }{(\frac{{r}_{0}}{r})}^{3k}=r\ddot{r}+\frac{3}{2}{\dot{r}}^{2}+\frac{4\nu \dot{r}}{r}+\frac{2\gamma }{\rho r}+\frac{\,\mu }{2\rho }[5-4\frac{{r}_{0}}{r}-{(\frac{{r}_{0}}{r})}^{4}]$$
(1)
where pG0 (=p(0) − pv(T) + 2γ/r0) is the initial partial pressure of the contaminant gas in a bubble, pv is vapor pressure, k is the polytropic index (k = 1 for constant temperature), T is the ambient temperature, ρ, μ, ν and γ are the density, shear modulus, viscosity, and surface tension of the sample, respectively, and r0 and r(t) are initial and current radius of the bubble, respectively, and the overdot denotes the derivative with respect to time. Recent studies that compare various Rayleigh-Plasset-like questions show that the Kelvin-Voigt and neo-Hookean models match well with experimental measurements on cavitation bubbles in soft gels14.
Equation 1 can be readily solved by using numerical solvers (see the Methods section) for the given initial conditions, $$r(0)={r}_{0}\,{\rm{and}}\,\dot{r}(0)=0.$$ For use of reasonable mechanical inputs that are directly relevant to actual impact, pa(t, aamp, hB) = pamp(aamp, hb)an(t) is used as a driving force where an(t) is normalized acceleration signal measured during an impact experiment and pamp is the maximum amplitude of pa at $$t={t}_{max}^{in}$$ (Fig. 4b, vertical axis on the left). The nominal material properties used in the numerical studies are summarized in Table S6 in Supplementary document.
One representative result in Fig. 4b (yellow line, vertical axis on the right) shows the transient dynamics of an air bubble (r0 = 50 μm) due to pa(t) (red line, vertical axis on the left) where pamp = 18 kPa at $$\,{t}_{max}^{in}=0.547\,{\rm{ms}}$$. The bubble radius gradually increases in $$0\le t\le {t}_{max}^{out}$$where the “x” marker indicates the maximum radius rmax at $$t={t}_{max}^{out}$$. Then, although it is beyond our main focus, for $$t > {t}_{max}^{out}$$ the bubble rapidly collapses, i.e., dr/dt → − ∞ as r/r0 → 0, followed by bubble radius oscillation and, as a result, theoretical analysis of bubble collapse must appropriately account for change in temperature and compressibility of material.
To study the effect of initial bubble size on the bubble dynamics, we consider various r0 ranging from 0.5 to 500 μm. The left column of Fig. 5 shows the time evolution of the normalized bubble size (λ) with increasing input pressure (pamp). The initial bubble radii (r0) in a, c, e, and g are r0 = 500, 50, 5, and 0.5 μm, respectively. For each r0, the normalized bubble radius (λ(t) = r(t)/r0) is shown in time for a given pamp, which increases linearly. Note that “x” markers indicate the normalized maximum radius (λmax = rmax/r0) at $$t={t}_{max}^{out}$$ for a given pamp. The results in the left column are summarized in b, d, f, and h to show how $${t}_{max}^{out}$$ and λmax (vertical axis on the left) as well as pamp and λmax (vertical axis on the right) are correlated.
In general, λmax and $${t}_{max}^{out}$$ in the left column of Fig. 5 increase with increasing pamp; however, their specific trends strongly depend on r0. For relatively large initial radius (r0 = 500 μm, Fig. 5a,b), the effect of surface tension, which is inversely proportional to bubble size (see Eq. 1), becomes less important and, as a result, the maximum bubble radius (rmax/r0) as well as the time ($${t}_{max}^{out}$$) corresponding to rmax smoothly increase as a function of pamp. As an example, the maximum radius vs pamp plot for r0 ≥ 500 μm (Fig. S1 in the supplementary document) can be reasonably well captured by a linear approximation. Because the maximum bubble size for r0 = 0.52 mm is linearly proportional to amplitude of input acceleration as well as input pressure, we conclude that $${p}_{amp}\propto {h}_{b}^{2}|{a}_{amp}|$$, in contrast to the commonly assumed pha (see the supplementary document for details).
On the other hand, the slope of the $${t}_{max}^{out}$$and λmax curves is about zero within 1 ≤ r/r0 < 1.1 for r0 = 50,5, and 0.5 μm in Fig. 5d,f,h, respectively. This result indicates that input acceleration and bubble response are in phase, i.e., $${t}_{max}^{out}\approx {t}_{max}^{in}$$, likely due to smaller inertial effects. Then a sharp increase in λmax is observed in the pamp − λmax curves for r0 = 0.5 and 5 μm (Fig. 5f,h, respectively) near λmax ≈ 1.1 even with a small increase in pamp, which is the key signature of the bubble burst44,45. For example, λmax for r0 = 0.5 μm (see Fig. 5h) increases by 3 orders of magnitude from $${r}_{cn}^{-}/{r}_{0}$$ to $${r}_{cn}^{+}/{r}_{0}$$ at $${p}_{amp}^{cn}$$ where $${p}_{amp}^{cn},$$ the critical amplitude of pressure for cavitation nucleation, can be obtained by dpamp/max = 0 at $$r={r}_{cn}^{-}$$ and $$={r}_{cn}^{+}$$. This rapid expansion of a bubble is observed only for small r0 because of the transition in the dominant mechanism of material response from surface tension to inertia with increasing r.
The results of the numerical simulations match well with the experimentally observed, dynamic response of bubbles in a soft material. First, bubble bursts are indeed sensitive to initial bubble size. Experimentally we have observed bubble bursts in agarose, gelatin, and pure water likely with submicron-scale bubbles40,41,42, but no bubble bursts in collagen with macro bubbles (r0 ~ 0.5 mm), as shown above. Also it is confirmed that the use of macro bubbles for quantitative characterization of acceleration-induced pressure gradients in the sample offers a unique advantage because of a simple relation between pressure and bubble size without rapid bubble bursts.
On the contrary, the size of the small-scale bubbles becomes increasingly unstable when approaching the critical pressure ($${p}_{amp}^{cn}$$). As a result, experimental establishment of the pa and r relation near $${p}_{amp}^{cn}$$ becomes impractical. In addition, the simulation predicts that small bubble bursts are accompanied by a sharp increase in radius, which we also experimentally observed. As an example, assuming the initial bubble size is in the range of hundreds nanometer40 as discussed earlier, λmax for the agarose sample in Fig. 3a becomes 103~104 due to the millimeter-scale rmax, consistent with the numerical prediction in Fig. 5h.
It is worth noting that unlike rapid bubble expansion, the numerical studies predict that bubble collapse occurs regardless of the initial radius (r0) when the maximum bubble radius is sufficiently larger than r0 (see the left column of Fig. 5). This prediction matches with experimental observations (see Fig. S2 and Supplementary Movies 69). When impact occurs, bubbles expand during which significant energy is stored in the gel, i.e., surface energy and elastic energy associated with the material deformation. Upon the completion of the impact, input acceleration rapidly decreases and, as a result, the relaxation of the stored energy becomes dominant force that drives bubble collapse.
As discussed earlier, experimental measurements to quantify surface tension and elasticity of soft material samples, in particular, at fast loading conditions are still very limited. In this regard, although our numerical studies above using nominal soft material properties have revealed important characteristics of bubble dynamics in the sample, the current lack of reliable material property data makes quantitative predictions difficult. Therefore, in the following studies we perform non-dimensional analysis to further investigate bubble dynamics in the dimensionless parameter space of material properties.
By introducing $$\lambda (\tilde{t})=r(\tilde{t})/{r}_{0},$$ $$\tilde{r}={r}_{0}/{r}_{cr,w}$$, and $$\tilde{t}=t/{t}_{max}^{in}$$ Eq. 1 can be nondimensionalized as follows
$${\tilde{r}}^{3}{\mathscr{A}}(\lambda \ddot{\lambda }+\frac{3}{2}{\dot{\lambda }}^{2})+\tilde{r}[2 {\mathcal B} \frac{\dot{\lambda }}{\lambda }+\frac{1}{2}(5-\frac{4}{\lambda }-\frac{1}{{\lambda }^{4}})-{\tilde{p}}_{v}(1-\frac{1}{{\lambda }^{3}})+{\tilde{p}}_{a}]+{\mathscr{C}}(\frac{1}{\lambda }-\frac{1}{{\lambda }^{3}})=0$$
(2)
where $${\mathscr{A}}=\rho {({r}_{cr}/{t}_{max}^{in})}^{2}/\mu$$, $${\mathcal B} =2(\nu \rho )/\mu {t}_{max}^{in}$$, $${\mathscr{C}}=2\gamma /{r}_{cr}\mu$$, $${\mathop{p}\limits^{ \sim }}_{v}(\mathop{t}\limits^{ \sim })={p}_{v}(\mathop{t}\limits^{ \sim })/\mu ,{\mathop{p}\limits^{ \sim }}_{a}(\mathop{t}\limits^{ \sim })={p}_{a}(\mathop{t}\limits^{ \sim })/\mu$$, and rcr, w is the critical radius of pure water at the onset of rapid bubble expansion or equivalently at bubble burst. Note that $${\mathscr{A}}, {\mathcal B} ,\,{\rm{and}}\,{\mathscr{C}}$$, which are normalized by elastic modulus, are associated with inertia (Cauchy number), viscosity (Deborah number), and surface tension (Weber number) terms, respectively. For perturbation analysis of Eq. 2, we apply small input pressures $$\tilde{p}$$a($$\tilde{t}$$) = $$\tilde{p}$$q($$\tilde{t}$$) to the bubble where $$\tilde{p}$$ is the amplitude of the non-dimensionalized pressure and 0 < |q($$\tilde{t}$$)| 1. Upon the application of the input pressure, the bubble oscillates in time about the initial radius r0, which can be written as λ($$\tilde{t}$$) = λ0(1 + f($$\tilde{t}$$)) where 0 < |f($$\tilde{t}$$)| 1. By substituting the small input pressure and corresponding radius to Eq. 2, we obtain
$${\mathscr{A}}{\tilde{r}}^{3}\ddot{f}+2 {\mathcal B} \tilde{r}\dot{f}+(2{\mathscr{C}}+\tilde{r}(4-3{\tilde{p}}_{v}))f=-\,\tilde{p}q$$
(3)
The non-dimensional frequency of damped vibration ($${\tilde{\omega }}_{d}$$) for the bubble can be written as
$${\tilde{\omega }}_{d}=\frac{{\omega }_{d}}{{t}_{max}^{in}}=\frac{1}{\tilde{r}}{((4-\frac{3{p}_{v}}{\mu })\frac{\mu }{\rho {r}_{cr}^{2}}+\frac{4\gamma }{\rho {r}_{cr}^{3}}(\frac{1}{\tilde{r}})-\frac{4{\nu }^{2}}{{r}_{cr}^{4}}{(\frac{1}{\tilde{r}})}^{2})}^{\frac{1}{2}}\,$$
(4)
Note that ωd increases with increasing elastic modulus and surface tension while decreasing with increasing viscosity. It is worth mentioning that when the size of pre-existing bubbles or defects at cavitation nucleation sites is much smaller than the critical size of pure water, i.e., $$\tilde{r}$$ 1, the last term with ν in Eq. 4 becomes insignificant due to the $$\tilde{r}$$−2 term and, as a result, ωd ≈ ωN where ωN is the natural frequency of the bubble. This result indicates that the effect of viscosity is not critical for dynamics of submicron-scale bubbles. Furthermore when ν ≈ 0, Eq. 2 can be rewritten as13,44
$$\dot{\lambda }={(\frac{\mu }{\rho })}^{\frac{1}{2}}\frac{{t}_{max}^{in}}{\tilde{r}{r}_{cr,w}}{(\frac{2{\rm{\Delta }}p}{3\mu }(1-\frac{1}{{\lambda }^{3}})+\frac{2{p}_{Go}}{\mu {\lambda }^{3}}\mathrm{log}\lambda -\frac{2\gamma }{\mu {r}_{cr,w}}\frac{1}{\tilde{r}\lambda }(1-\frac{1}{{\lambda }^{2}})+\frac{1}{3}{{\mathbb{E}}}^{\ast })}^{\frac{1}{2}}\,$$
(5)
where Δp = pv − pa and $${\mathbb{E}}$$* = −5 + 6λ−1 + 2λ−3 − 3λ−4. When a sufficiently large pa, i.e., $${p}_{amp} > {p}_{amp}^{cr}$$, is applied to the small bubble or defect, we have λ(t) = λmax 1 due to bubble bursts and $$\dot{\lambda }=0$$ at $$t={t}_{max}^{out}$$. Using these known conditions at $$t={t}_{max}^{out}$$, the first order approximation of Eq. 5 becomes
$${\lambda }_{max}=\frac{3}{2}(\frac{2\gamma }{{r}_{0}\mu }-2)/(\frac{{\rm{\Delta }}p}{\mu }-\frac{5}{2})\,$$
(6)
In Fig. 6a, the result of Eq. 6 (dash-dot lines) for the nominal material sample in Table S6 is directly compared with the numerical simulation (solid and dashed lines) for 20 ≥ λ ≥ 1 and 20 kPa ≥ pamp ≥ 1 kPa. For relatively large r0 (≥9 μm), the first order approximation matches with the direct simulation reasonable well. In addition, Equation 6 correctly predicts the critical r0 and pamp that bound the two distinctive regimes with and without bubbles burst. First, from the denominator of Eq. 6, λmax → ∞, the key signature of bubble burst, as Δp/μ → 5/2, i.e., Δpcr = 20 kPa. Second, λmax < 0 in Eq. 6 when r0 < γ/μ and, as a result, the sum of all terms within the square root in Eq. 5 becomes negative. Therefore, $${r}_{0}^{cr}={r}_{0}={\gamma }/{\mu }$$ is a bifurcation point from real to imaginary number solutions for $$\dot{\lambda }$$, i.e., $${r}_{0}^{cr}=9\,{\rm{\mu }}{\rm{m}}.$$
Note that Eq. 6 fails to capture the pamp and λmax relation for $${r}_{0}^{cr} < 9\,{\rm{\mu }}{\rm{m}}$$ because rapid changes in bubble size due to bubble bursts cannot be appropriately captured by the first order approximation. In other words, the effect of the viscosity on bubble expansion must be appropriately considered after bubble burst. The same conclusion can be made from Eq. 4 because $${\tilde{\omega }}_{d}$$ is very sensitive to bubble size for relatively large bubbles due to the $$\tilde{r}$$−2 term. Equation 3 could be utilized, e.g., Fig. 6, for analytical prediction of the critical acceleration. However, Eq. 3 still requires a reasonable r0 estimate while direct measurement of the submicron-scale bubbles or material flaw remains very challenging experimentally.
While we have used a deterministic model to study cavitation behavior in soft gel, experimental measurements on the critical acceleration for cavitation nucleation are distributed over a broad range for a given stiffness of soft gel samples. This suggests that cavitation nucleation is stochastic in nature and, as a result, developing a theoretical frame work that capture stochastic distribution of the critical mechanical input at the onset of cavitation nucleation would be interesting future work and we believe can be guided by the distributions of data obtained using our technique.
## Conclusion and Discussion
Soft gels that mimic properties of biological samples are increasingly utilized as tissue simulants for the assessment of potential damage, e.g., blunt injuries, to a human body against rapid mechanical inputs. As a result, capabilities to accurately characterize and predict transient, dynamic response of soft materials under fast loading rate conditions have become critical for such biomedical applications. In this regard, we have performed experimental and theoretical investigations on (1) acceleration-induced pressure gradients in biologically-relevant soft gels during impact and (2) the corresponding critical transition in the material response from small deformation to a sudden bubble burst.
We have characterized collagen samples with macro air bubbles (about 0.5 mm in radius) and monitored the dynamic response of individual bubbles during impact. This study reveals that the maximum bubble radius (rmax) monotonically increases as a function of amplitude of input acceleration (ain) and is also proportional to the square of the depth of individual bubbles (hb) in the sample, i.e., rmaxah2. This simple size-pressure relation offers an attractive advantage in utilizing large bubbles for experimental, quantitative characterization of acceleration-induced pressure gradients. We have also considered agarose samples without the macro air bubbles and quantified the critical acceleration that triggers the onset of bubble bursts, which increases with increasing gel stiffness. Finally, we have found that bubble bursts are sensitive to the initial bubble size by comparing experimental data for different types of gels with and without macro bubbles.
To explain these key experimental observations, we have developed theoretical framework. The model predicts the onset of bubble bursts can be triggered only when the initial bubble size is smaller than the critical size (about 10 micron). For micro/nanoscale bubbles (<10 μm), their size becomes unstable as it approaches a critical radius due to the transition from surface tension to inertia dominant material deformation. The model also shows that the maximum radius and amplitude of input pressure (pamp) are linearly correlated for larger bubbles (>500 μm). Based on $${r}_{max}\propto {a}_{in}{h}_{b}^{2}$$ and pamp relations, we empirically determine that the pressure gradient in a soft material sample during impact is simply $${p}_{amp}\propto {a}_{in}{h}_{b}^{2}$$.
Our study on how biological soft materials respond to rapid mechanical inputs is an important step to understanding possible underlying injury mechanism(s) in blunt trauma and cavitation-induced brain injury. As an example, cavitation is being conjectured as a possible mechanism for mild to moderate traumatic brain injury (mTBI)46,47,48. However, the current understanding of dynamic cavitation in biological materials, e.g., brain tissues, is still very limited and, as a result, a range of mechanical impact that is relevant to cavitation-induced brain injury is still unknown. To address the current limitation, we have used collagen, agarose, and gelatin as a tissue simulant and have experimentally considered the acceleration-induced pressure gradient in the simulant and quantified the critical acceleration corresponding to cavitation nucleation. This quantitative measurements on the cavitation properties can be utilized for developing injury criteria for cavitation-induced TBI.
In addition, our experimental approaches have the potential for clinically- and biologically-relevant studies that require use of 3-dimentional (3D) extracellular matrix (ECM) such as collagen and agarose. As an example, a reasonable 3D in vitro model that biologically and mechanically represents target organs could be developed by culturing fibroblast or neurons in collagen/agarose at a specific concentration to match known mechanical stiffnesses of skin or brain, respectively. Biological studies that utilize well-designed 3D cell-ECM systems and mechanical inputs directly relevant to blunt injuries would provide unique opportunities to probe the specific cell-ECM interplay during impact and shed light on key mechanism(s) in various forms of traumatic injuries.
## Methods
### Materials
For 1% [g/ml] collagen samples (Collagen Type I, Rat Tail, Corning, Bedford, MA), we prepare 24 ml of ice cold collagen solution by mixing 2.4 ml of Sterile 10X phosphate buffered saline, 8.1 μl of 1 N NaOH, and 18 ml of sterile deionized water with 7 ml of collagen solution at 3.42 mg/ml. Then the solution is thoroughly mixed using a 10-ml pipette and vortex mixer to have randomly distributed bubbles in the sample. Immediately after the mixing, 4 ml of 1% collagen solution is added to six individual transparent cuvettes using a 10 ml pipette tip and cured in an incubator overnight.
For 0.3%, 0.58%, 0.9%, and 1.5% [g/ml] agarose samples, we add 80 ml of room temperature water into a Pyrex beaker with a magnetic stir bar. Then put the beaker on a magnetic stirrer and slowly add the required amount of agarose powder (0.24, 0.46, 0.72, and 1.2 g for 0.3%, 0.58%, 0.9%, and 1.5% agarose, respectively) into the beaker to avoid the formation of clusters. Measure the total weight of the beaker and solution before heating and boil the solution for 10 mins with a Pyrex beaker cover while stirring. To compensate evaporation, measure the weight of the beaker and add water to bring it back to its initial weight. After cooling the beaker for about 10–20 minutes, insert 4 ml of agarose solution into the individual eighteen cuvettes using a 10 ml pipette and cure them at room temperature overnight.
### Experimental setup
The integrated drop tower system consists of a conventional drop tower impact system (Dynatup 9210, Instron, Norwood, MA), two high speed cameras (Fastcam SA-X2RV, Photron, San Diego, CA), a cuvette and holder, accelerometers, and a data acquisition system. Soft gel samples are prepared inside standard plastic cuvettes for ease of handling. Each cuvette is sealed with a cap and then glued onto the cuvette holder which consists of two horizontal plates connected by four vertical columns.
For optical observation of samples, the two independent high speed cameras are mounted on two camera stands such that the cameras can concurrently capture the front and side views of the cuvette, typically at 25 k or 50 k frames per second (fps). Data acquisition is performed using a system consisting of an NI PXIe-8135 embedded controller and an NI PXI-6115 multifunction I/O module using SignalExpress 2014 data acquisition software (all from National Instruments Corp., Austin, TX). A three-axis ICP-based accelerometer (model #356A01, PCB Piezotronics, Depew, NY) is connected to three channels of the data acquisition system through an ICP signal conditioner (model #480B21, PCB Piezotronics). Data are acquired at a rate of 1 MHz and triggered off the vertical axis of the accelerometer. More details on the experimental setup and procedure can be found elsewhere10,15.
### Image analysis for bubble dynamics
For experimental measurement of bubble size during impact, high-speed images are analyzed at each frame (25,000 or 50,000 frames/sec) using ImageJ. To differentiate bubbles from the background during the image analysis, the same threshold setting is used for all image frames. All pixels that are above the threshold due to bubbles are counted and converted to the corresponding area. Then, radius of each bubble is calculated using r = (A/π)1/2 where is the projected area of the bubbles. Following the same procedure, we also monitor the effective radius of a solid structure, reference, at the bottom of a cuvette.
### Statistical analysis for experimental data
For the statistical analysis, mean and std functions of Matlab are utilized to compute the mean and standard deviation of experimental data.
### Numerical simulation and optimization
ODE45 function of Matlab are used to simulate dynamic response of bubbles in a soft gel (Eq. 1).
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## Acknowledgements
This work has been funded by the Naval Research Laboratory’s Institute for Nanoscience. W.K. thanks for generous funding from the Computational Cellular Biology of Blast (C2B2) program (Dr. Timothy Bentley) through the Office of Naval Research. The authors thank Drs J. Thomas and A. Geltmacher for kindly allowing use of experimental instruments in their laboratories and Dr. T. O’Shaughnessy for his technical support with the data acquisition system.
## Author information
W.K. conceived the idea, designed the research, carried out the experiments, and performed the theoretical analyses. W.K. and M.R. analyzed and interpreted the results. W.K. drafted the manuscript and all authors contributed to the writing of the manuscript.
Correspondence to Wonmo Kang.
## Ethics declarations
### Competing Interests
The authors declare no competing interests.
Publisher’s note: Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
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Kang, W., Raphael, M. Acceleration-induced pressure gradients and cavitation in soft biomaterials. Sci Rep 8, 15840 (2018). https://doi.org/10.1038/s41598-018-34085-4 | 2020-04-10 10:14:48 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7409236431121826, "perplexity": 2918.8832547875145}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585371893683.94/warc/CC-MAIN-20200410075105-20200410105605-00508.warc.gz"} |
http://mathhelpforum.com/calculus/52023-optimum-turning-point.html | # Thread: Optimum Turning Point
1. ## Optimum Turning Point
Time to pull in the big guns. This is a frustrating twist on a classic problem:
A cop is chasing a burglar, both running east along an east-west shore line. (The water is to the north; land is to the south.) When the cop is 50m behind him, the burglar jumps into the water and begins swimming in a straight line 30 degrees north of the shore (a bearing of 60 degrees from North). The burglar swims at a rate of 2m/s. The cop runs along the shore for a short time, then jumps in the water and swims in a straight line on a course that will intercept the burglar. The cop can run 5m/s along the shore, and can swim at a rate of 3m/s. How many meters should the cop run before jumping in the water in order to apprehend the burglar in the shortest amount of time? At what angle from the shore should he turn?
I can get as far as drawing a picture, and coming up with times for three specific points along the shore. Creating a general equation to optimize, though, has me stumped. Any help would be appreciated.
2. I began with the assumption that the shortest time would be obtained in such a way that the cop passes the point at which the burglar leaves the shore, and then some time after that, begins to swim in order to intercept the burglar. So I stated:
$t_1$ is the time between the cop reaching the point where the burglar left the shore (B) and the point at which he begins to swim (A).
$t_2$ is the time the cop takes after he begins to swim before he catches the burglar.
It takes the cop 10 seconds to reach B. Using the law of cosines,
$(3t_2)^2 = [20 + 2(t_1 + t_2)]^2 + (5t_1)^2 - 2[20 + 2(t_1 + t_2)](5t_1)\cdot \frac{\sqrt{3}}{2}$.
Does that make sense or did I lose you?
3. Your equation makes sense- over the weekend I came up with a similar equation, though I had assumed the turning point was before the 50m, and used the cosine of 150.
Now the hard part: how do I find the optimum point off of that? I know I want to minimize the quantity (t1+t2), but...
I guess I'm not sure how to take this derivative, and what I should take it with respect to. | 2017-12-13 15:12:41 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6567341685295105, "perplexity": 598.5177561365563}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948527279.33/warc/CC-MAIN-20171213143307-20171213163307-00184.warc.gz"} |
https://imadali.net/posts/embeddings/ | You can think of a neural network embedding is another form of dimensionality reduction. You’re taking a bunch of tokens (words, movies, games, etc) and instead of one-hot encoding them you want to map them down to a lower dimensional space.
For example, suppose you have a collection of 1,000 tokens. To one-hot encode them means having a very sparse vectors of length 1,000 for each word (where a 1 exists at the index that represents the token, 0 otherwise).
So basically you’re taking very high dimensional vectors and mapping them down to a low dimensional space. Preferably you don’t want the low dimensional space to be random. You want it to have some structure. For example, vectors that represent similar tokens are close to one another, adding vectors together gets you close to a vector that represents a good mix of the two tokens, etc).
I like using generated (synthetic) data. It gives you a good understanding of how the data you’re modeling was generated. Below I show how to discover an embedding space from some generated data and and simple predictive visualization.
# Generating Data
Using numpy we generate some data that we can create embeddings with. Suppose we have 5 tokens that have associated effects (i.e. parameters that we can use to generate the data from). Using the tokens and the token effects we generate 10,000 observations from the normal distribution.
N = 10_000
n_tokens = 5
token_effect = np.linspace(start=-20, stop=20, num=5)
# reorder the tokens (not necessary but more realistic)
token_effect = np.random.choice(token_effect, n_tokens, replace=False)
# generate tokens for each observation
x = np.random.choice(n_tokens, N)
# simulate an outcome for each observation's token
y = np.random.normal(token_effect[x], scale=2, size=N)
The tokens x[:5] might look something like array([4, 4, 3, 2, 2, 0]) and the outcome y[:5] might look something like array([ 18.64902242, 19.22041727, -20.25015504, 11.30246504, 11.15165945, -10.05891201]).
# Creating Embeddings
Knowing how many tokens we have and what we want our embedding space to look like we can create our embedding model with keras. We define the dimensionality of the embedding layer with the number of unique tokens and the output dimension (which should be substantially less than the number of tokens).
model = keras.models.Sequential()
model.add(layers.Embedding(input_dim = n_tokens, output_dim = 4, input_length=1))
model.add(layers.Dense(units=10, activation='relu'))
model.add(layers.Dense(units=8, activation='linear'))
model.add(layers.Dense(units=6, activation='linear'))
model.add(layers.Dense(units=4, activation='linear'))
model.add(layers.Dense(units=1, activation='linear'))
model.summary()
Now we compile/train the model and plot the loss to confirm that our model configuration (choice of layers, activation functions, loss function, etc) is appropriate.
model.compile(optimizer='rmsprop', loss='mse')
history = model.fit(x=x, y=y, epochs=100, batch_size=1_000, validation_split=0, verbose=0)
# Embedding Similarity
Our token effect might look something like array([-10., 0., 10., 20., -20.]). In this case we would expect token 3 (which maps to a token effect of 20) to be closest to token 2 (which maps to a token effect of 10) and farthest from token 4 (which maps to a token effect of -20).
The function below computes the similarity (Euclidean distance) for an indexed embedding vector against all the other embedding vectors. np.argmax(token_effect) gives us the token value that maps to the largest token effect.
def similarities(target_index, embedding_matrix):
result = []
for i in range(0,embedding_matrix.shape[0]):
s = np.linalg.norm(embedding_matrix[target_index,] - embedding_matrix[i,])
result.append(s)
return(np.array(result))
similarities(np.argmax(token_effect), embedding)
The above might return similarity values like array([0.7304573 , 0.5255725 , 0.32273838, 0. , 1.0848211 ] which confirm that the embedding corresponding to token 3 is closest to token 2 and farthest from token 4.
This is a convenient result. For illustrative purposes we used a small number of tokens. But you can see how calculating similarities between vectors of an embedding matrix would be preferred to the one-hot encoded counterpart in situations where you might have thousands of tokens.
# Visualizing Predictions
Here’s what the outcome variables looks like, color coded by token.
And here are the predictions using the model along with the outcome variable. This plot doesn’t really provide any new information. It just reiterates the fact that the loss from training is sufficiently low. | 2021-03-06 04:58:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6486057043075562, "perplexity": 1425.5161704314423}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178374391.90/warc/CC-MAIN-20210306035529-20210306065529-00382.warc.gz"} |
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https://enacademic.com/dic.nsf/enwiki/3348639 | # Gaussian measure
In mathematics, Gaussian measure is a Borel measure on finite-dimensional Euclidean space R"n", closely related to the normal distribution in statistics. There is also a generalization to infinite-dimensional spaces. Gaussian measures are named after the German mathematician Carl Friedrich Gauss.
Definitions
Let "n" ∈ N and let "B"0(R"n") denote the completion of the Borel "σ"-algebra on R"n". Let "λ""n" : "B"0(R"n") → [0, +∞] denote the usual "n"-dimensional Lebesgue measure. Then the standard Gaussian measure "γ""n" : "B"0(R"n") → [0, +∞] is defined by
:$gamma^\left\{n\right\} \left(A\right) = frac\left\{1\right\}\left\{sqrt\left\{2 pi\right\}^\left\{n int_\left\{A\right\} exp left\left( - frac\left\{1\right\}\left\{2\right\} | x |_\left\{mathbb\left\{R\right\}^\left\{n^\left\{2\right\} ight\right) , mathrm\left\{d\right\} lambda^\left\{n\right\} \left(x\right)$
for any measurable set "A" ∈ "B"0(R"n"). In terms of the Radon-Nikodym derivative,
:$frac\left\{mathrm\left\{d\right\} gamma^\left\{n\left\{mathrm\left\{d\right\} lambda^\left\{n \left(x\right) = frac\left\{1\right\}\left\{sqrt\left\{2 pi\right\}^\left\{n exp left\left( - frac\left\{1\right\}\left\{2\right\} | x |_\left\{mathbb\left\{R\right\}^\left\{n^\left\{2\right\} ight\right).$
More generally, the Gaussian measure with mean "μ" ∈ R"n" and variance "σ"2 > 0 is given by
:$gamma_\left\{mu, sigma^\left\{2^\left\{n\right\} \left(A\right) := frac\left\{1\right\}\left\{sqrt\left\{2 pi sigma^\left\{2^\left\{n int_\left\{A\right\} exp left\left( - frac\left\{1\right\}\left\{2 sigma^\left\{2 | x - mu |_\left\{mathbb\left\{R\right\}^\left\{n^\left\{2\right\} ight\right) , mathrm\left\{d\right\} lambda^\left\{n\right\} \left(x\right).$
Gaussian measures with mean "μ" = 0 are known as centred Gaussian measures.
The Dirac measure "δ""μ" is the weak limit of $gamma_\left\{mu, sigma^\left\{2^\left\{n\right\}$ as "σ" → 0, and is considered to be a degenerate Gaussian measure; in contrast, Gaussian measures with finite, non-zero variance are called non-degenerate Gaussian measures.
Properties of Gaussian measure
The standard Gaussian measure "γ""n" on R"n"
* is a Borel measure (in fact, as remarked above, it is defined on the completion of the Borel sigma algebra, which is a finer structure);
* is equivalent to Lebesgue measure: $lambda^\left\{n\right\} ll gamma^\left\{n\right\} ll lambda^\left\{n\right\}$, where $ll$ stands for absolute continuity of measures;
* is supported on all of Euclidean space: supp("γ""n") = R"n";
* is a probability measure ("γ""n"(R"n") = 1), and so it is locally finite;
* is strictly positive: every non-empty open set has positive measure;
* is inner regular: for all Borel sets "A",
:$gamma^\left\{n\right\} \left(A\right) = sup \left\{ gamma^\left\{n\right\} \left(K\right) | K subseteq A, K mbox\left\{ is compact\right\} \right\},$
so Gaussian measure is a Radon measure;
* is not translation-invariant, but does satisfy the relation
:$frac\left\{mathrm\left\{d\right\} \left(T_\left\{h\right\}\right)_\left\{*\right\} \left(gamma^\left\{n\right\}\right)\right\}\left\{mathrm\left\{d\right\} gamma^\left\{n \left(x\right) = exp left\left( langle h, x angle_\left\{mathbb\left\{R\right\}^\left\{n - frac\left\{1\right\}\left\{2\right\} | h |_\left\{mathbb\left\{R\right\}^\left\{2^\left\{2\right\} ight\right),$
:where the derivative on the left-hand side is the Radon-Nikodym derivative, and ("T""h")∗("γ""n") is the push forward of standard Gaussian measure by the translation map "T""h" : R"n" → R"n", "T""h"("x") = "x" + "h";
* is the probability measure associated to a normal probability distribution:
:$Z sim mathrm\left\{Normal\right\} \left(mu, sigma^\left\{2\right\}\right) implies mathbb\left\{P\right\} \left(Z in A\right) = gamma_\left\{mu, sigma^\left\{2^\left\{n\right\} \left(A\right).$
Gaussian measures on infinite-dimensional spaces
It can be shown that there is no analogue of Lebesgue measure on an infinite-dimensional vector space. Even so, it is possible to define Gaussian measures on infinte-dimensional spaces, the main example being the abstract Wiener space construction. A Borel measure "γ" on a separable Banach space "E" is said to be a non-degenerate (centred) Gaussian measure if, for every linear functional "L" ∈ "E"∗ except "L" = 0, the push-forward measure "L"∗("γ") is a non-degenerate (centred) Gaussian measure on R in the sense defined above.
For example, classical Wiener measure on the space of continuous paths is a Gaussian measure.
ee also
* Cameron-Martin theorem
Wikimedia Foundation. 2010.
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• Pushforward measure — In mathematics, a pushforward measure (also push forward or push forward) is obtained by transferring ( pushing forward ) a measure from one measurable space to another using a measurable function.DefinitionGiven measurable spaces ( X 1, Sigma;1) … Wikipedia | 2021-01-26 12:41:11 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 9, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9327868819236755, "perplexity": 1268.228110730901}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610704799741.85/warc/CC-MAIN-20210126104721-20210126134721-00604.warc.gz"} |
http://math.stackexchange.com/questions/239965/solving-problem-3-29-in-spivak%c2%b4s-calculus-on-manifolds-without-using-change-of-v?answertab=oldest | # Solving problem 3-29 in Spivak´s Calculus on Manifolds without using change of variables
Problem 3-29 (p. 61) in the section treating Fubini´s theorem reads:
Use Fubini´s theorem to derive an expression for the volume of a set of $\mathbb{R}^{3}$ obtained by revolving a Jordan-measurable set in the $yz$-plane about the $z$-axis.
By making some simplying assumptions about the plane region and changing variables to cylindrical coordinates we can obtain an expression for the volume. However, the material on the change of variables is treated two sections later (p. 66). Thus, is there is a way to solve the problem without using change of variables?
The definition of Jordan-measurable can be found on p. 56 (See also Theorem 3-9, p.55), but I sumarize it here: Spivak defines a bounded subset $C$ of $\mathbb{R}^{n}$ to be Jordan measurable if the topological boundary of $C$ has measure $0$, i.e. if for any $\varepsilon>0$ there is a cover $\{U_{i}\}$ of $\mathrm{Bd}(C)$ by closed $n$-cubes such that $\sum_{i}v(U_{i})<\varepsilon$ (where $v(U_{i})$ denotes the $n$-dimensional volume of the rectangle $U_{i}$). If $C$ is Jordan measurable it is contained inside some closed $n$-cube $A$ and the characteristic function $\chi_{C}$ is integrable on $A$. The integral $\int_{A}\chi_{C}$ is called the $n$-dimensional volume of $C$.
-
The following argument in fact uses Fubini's theorem (twice!), but makes simplifying assumptions about the given domain $D$ in the meridian half-plane. I use $(r,z)$ instead of $(y,z)$ as coordinates in this "abstract" plane.
Assume that $D$ is given in the form $$D:=\{(r,z)\ |\ a\leq z\leq b,\ 0<c(z)\leq r\leq d(z)\}\ ,$$ with $a<b$ and $c(\cdot)$, $d(\cdot)$ integrable. Then the three-dimensional body $B$ whose volume we have to compute is given by $$B=\{(x,y,z)\ |\ a\leq z\leq b,\ c(z)\leq \sqrt{x^2+y^2}\leq d(z)\}\ .$$ For given $z\in[a,b]$ let
$$B_z:=\{(x,y)\ |\ c(z)\leq \sqrt{x^2+y^2}\leq d(z)\}$$ denote the intersection of $B$ with the horizontal plane at level $z$. The set $B_z$ is an annulus with inner radius $c(z)$ and outer radius $d(z)$. Therefore its area is given by $${\rm area}(B_z)=\pi\bigl(d^2(z)-c^2(z)\bigr)=2\pi\int_{c(z)}^{d(z)}r \ dr$$ (writing this area as an integral is a trick).
Now comes Fubini: \eqalign{{\rm vol}(B)&=\int_B 1\ {\rm d}(x,y,z)=\int_a^b \int_{B_z} 1\ {\rm d}(x,y)\ dz\cr &=\int_a^b {\rm area}(B_z)\ dz =2\pi\int_a^b \int_{c(z)}^{d(z)} r\ dr \ dz \cr &=2\pi \int_D r\ {\rm d}(r,z)\ .\cr}
-
I may be mistaken, but isn´t this solution basically a change of variables to cylindrical coordinates? – John Nov 19 '12 at 19:07
@John: Heavens; what have you expected, given that in the final formula there is a $2\pi$ in front and an extra factor $r$ in the integral? Using some elementary geometry about areas of circles it was possible to avoid Jacobians, etc. $--$ This comment is my last word on this matter. – Christian Blatter Nov 19 '12 at 19:28
Remark: I don't have a copy of Spivak's book at hand.
You don't need the full strength of the change of variables formula.
Assume that you are given a domain $D$ in the $(y,z)$-plane such that $y>0$ for all $(y,z)\in D$. Then a tiny rectangle $Q:=[y-u, y+u]\times[z-v,z+v]\subset D$ produces under rotation a cylindrical ring $R$ of exact volume $${\rm vol}(R)=2\pi y\cdot {\rm area}(Q)$$ (this is one of Guldin's rules).
Consider now approximations of $D$ by suitable unions of "almost disjoint" rectangles $Q_i$ $\ (1\leq i\leq N)$, and let $R_i$ be the rings generated by the $Q_i$. Then the volume of the considered body $B$ is approximatively given by $${\rm vol}(B)\doteq\sum_{i=1}^N{\rm vol}(R_i)=\sum_{i=1}^N 2\pi y_i\ {\rm area}(R_i)\doteq 2\pi \int_D y \ {\rm d}(y,z)\ ;\qquad(1)$$ whereby the $\doteq$'s can be made as exact as desired. As the outer terms in $(1)$ each have a fixed value these two values have to be equal, i.e., we obtain $${\rm vol}(B)= 2\pi \int_D y \ {\rm d}(y,z)\ .$$ (Depending on the analytical apparatus developed for the Jordan measure and integral you can replace $(1)$ by a squeeze argument or similar.)
-
Could you clarify the point where Fubini´s theorem is used in your solution? – John Nov 19 '12 at 12:25
@John: ${\rm d}(y,z)$ denotes the "area element" in ${\mathbb R}^2$. Spivak's integral notation seems to omit it. "Fubini" as a naked theorem has not been used. You have to bring in some elementary geometry, or the factor $2\pi$ wouldn't make its appearance in the final formula. – Christian Blatter Nov 19 '12 at 13:18 | 2015-05-25 03:38:02 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9699496030807495, "perplexity": 196.94380666887017}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1432207928350.51/warc/CC-MAIN-20150521113208-00155-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://kx.lumerical.com/t/plane-wave-source-power-of-order-10e-14-watts/5471 | # Plane wave source power of order 10e-14 Watts
#1
Dear Numerical Team,
as I have written in my previous post, I am using FDTD solutions to irradiate the surface to get Electric field, that would be used for further calculations. Could you explain me, please, how can I set an input electric field and/or the source power and intensity? For example, I would like to use a cw laser, power of 200mW.
I know about the possibility to check the source power using following commands:
?"Source power (Watts): "+num2str(sp);
?"Source intensity (Watts/um^2): " + num2str(I1e-12);
?"Ensure Intensity
Area=Power: " + num2str(I*area/sp);
However I do not understand, what they mean. The result power I get is of order 10e-14 Watts . Could you explain their meaning or send a link to the explanation, please?
Best regards,
Diana
How to pump a ring resonator
FDTD source amplitude unit
Scale bar in surface plot units
#2
In order to interpret the power results from a simulation, there are a couple of things we should discuss.
First, the actual power injected in the FDTD simulation by a plane wave source depends on the amplitude setting, the area of the source and the shape of the injected pulse.
• Amplitude: For beam sources, such as plane waves, this is the peak electric field amplitude in units of V/m.
• Area of the source: Wavefronts from plane wave sources are infinite. Therefore, the amount of power carried by a plane wave is not a well-defined quantity: it depends on the area of the plane wave source. As my colleague @bkhanaliloo mentioned in his reply to your earlier post, this is the reason why we use amplitude to define the source. The fact that the wavefronts of a plane wave extend to infinity also means that you should use Bloch/periodic boundaries and the source area will extend all the way to these boundaries. Therefore, the source area will be the same as the area of the simulation region in the plane perpendicular to the injection axis.
• Pulse shape: In FDTD simulations, a pulse is injected in time-domain (you can see the pulse shape and spectrum in the source settings). This means that not all frequencies have the same amplitude. Therefore, in principle the power will depend on frequency as well.
The frequency dependence due to the pulse shape can be easily removed using the CW normalization. This is the normalization state used by default and it is the easiest to interpret the results for linear simulations because it allows you to find the CW (or impulse) response of the system, i.e. the response as if all frequencies were injected with the same amplitude, independent of the source pulse used to excite the system.
If you want to find out how much power was injected in the simulation (with and without CW normalization) you can use the sourcepower script function. In the CW norm state this will return the power in Watts, which will be essentially flat because of the CW normalization (for more details visit this KB page). This power will still depend on the area of the plane wave source and the amplitude. If you have a periodic structure you will normally simulate one unit cell only, so this will restrict the source area to be the same as the area of the unit cell. You can still change the amplitude setting to modify the sourcepower result; however, in linear simulations it is not necessary to do this as explained next.
You mentioned that you have a CW laser with power of 200mW. Presumably, this is the total power injected by the laser over the laser spot. To relate this power to the results from a simulation with a plane wave source, it is more meaningful to use the source intensity (power per unit area) instead of the net power. Assuming the power is uniform over the laser spot, the source intensity would be the power divided by the spot area.
The transmission results from DFT monitors are normalized to sourcepower (in both CW norm and no norm states); in other words, the transmission is a fraction of the power injected by the source. Thus, if you want to find the transmitted intensity in the experiment you simply need to multiply the normalized transmission by the laser intensity. Then you can multiply by area to get power; for example, if you multiply by the spot are you get the total transmitted power, or if you multiply by the area of the unit cell of a periodic structure you get the power transmitted per unit cell.
As you can see, you normally don’t need to tune the amplitude setting. This discussion applies to linear simulations only. In non-linear simulations the amplitude setting is usually very important.
How can I get mode source poynting vector?
realtion between amplitude and intensity of electric field
Why is Vector plot scale different from E-field profile scale?
#3
Thank you very much for the answer! | 2018-09-20 21:03:16 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8312090039253235, "perplexity": 437.3582228595078}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267156613.38/warc/CC-MAIN-20180920195131-20180920215531-00294.warc.gz"} |
https://theopenacademy.com/content/forces-density-and-pressure | < Classical Mechanics
Lecture 4 Part 1 Part 2 Part 3 Part 4
Forces, Density and Pressure
Classical Mechanics
o The same volume of two different fluids can have very different masses because they have different densities.
- Density is the mass (in kilograms) of an object per volume (in metres cubed), denoted by $\rho=\frac{m}{V}$ where $\rho$ = density $(kg/m^3)$, m = mass (kg), V = volume $(m^3)$
o Archimedes was asked to find a way to measure the volume of the king's crown without damaging it in any way (King Hierro II suspected that the crown made by his goldsmith was not purely gold).
- The solution came to him as he was taking a bath. He noticed that as he got in to the tub, the water level would rise. He figured out that the amount of volume of his body submerged displaced an equal volume of water.
- All he needed to do was submerge the crown in water, and measure the volume of water it pushed out of the way. This would be equal to the volume of the crown.
- As the story goes, he was so excited by his discovery that he jumped out of the tub and ran naked through the streets yelling "Eureka!" which translates from Greek as "I have found it."
o This is known as the Archimedes' Principle, which says that the force of buoyancy that an object will feel when it is immersed in a fluid is equal to the weight of the of the fluid displaced by the object.
- This just means that if you place an object with a volume of (for example) 3 $m^3$ fully into some water, it will displace 3 $m^3$ of water. The buoyant force the object will feel is equal to the weight of that 3 $m^3$ of water.
- It is important to note that only the volume of the object that is submerged into the fluid can displace any volume, so that is the only amount considered when you figure out the buoyant force.
We can figure out a formula for buoyancy based on the formula we have for weight.
- Keep in mind that the buoyancy depends on the mass of the fluid displaced.
- Density $\rho\ = \frac{m}{V} (\frac{mass}{volume}) \quad in \quad (kg/m^3) \rightarrow m = \rho\ V$. This is handy, since we often talk about fluids in terms of density and volume, instead of mass.
$F_g = mg \rightarrow F_B = g m_{fluid} \rightarrow F_B = g \rho_{fluid}\ V_{sub}$
A motor boat is being launched at a lake. When it is placed in the water, it sinks into the water enough to displace 4300000 $cm^3$ of water. Assuming that this is enough for the boat to float, determine the mass of the motor boat.
1. 430000 $g$
2. 4300 $kg$
3. 43000 $kg$
o Imagine that you have a container of fluid.
- From what we've learned so far, we know that the pressure the fluid exerts on the sides of the container is the same everywhere. If they were not, the fluid would no longer be static.
- Now we exert a force somewhere on the outside of the container. The pressure inside the container will increase.
- The important part is that, according to Pascal's Principle, the pressure will increase everywhere in the fluid, not just where you are applying the force.
o Assume a hydraulic lift that lifts a large mass as an example of Pascal's principle application.
- On one side we have a small cylinder (filled with an incompressible liquid) with a piston. This is where we will exert a force downwards $(\Delta\ P = \frac{F_1}{A_1})$.
- This is connected by a pipe to another, larger cylinder, with a large piston, where the large mass will be lifted by a force upwards $(\Delta\ P = \frac{F_2}{A_2})$.
- Based on Pascal's principle the change in pressure must be the same at both sides i.e $\frac{F_1}{A_1} = \frac{F_2}{A_2}$ and since $A_2 > A_1$ then $F_2 > F_1$ for the sides to be equal. This just means that we can exert a small force on the small piston and get a bigger force on the big piston.
- The volume of the fluid being moved is $V = A_1 d_1$ and $V = A_2 d_2$ but $d_1 > d_2$ and substituting this in the above equality: $\frac{F_1}{A_1} (A_1 d_1) = \frac{F_2}{A_2} (A_2 d_2) \rightarrow F_1 d_1 = F_2 d_2$
I want to build a hydraulic press to be able to squeeze all my gold bars down to thin gold disks. The small piston has an radius of 1.0 $cm$ and I will be able to exert a force of 230 $N$ on it. If the large piston has a radius of 30 $cm$ determine the force against the gold bars.
1. 0.256 $N$
2. 155,000 $N$
3. 207,000 $N$
Part 1 Part 2 Part 3 Part 4 Lecture 5 | 2019-03-24 11:25:23 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.652734100818634, "perplexity": 335.94678806000377}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912203438.69/warc/CC-MAIN-20190324103739-20190324125739-00516.warc.gz"} |
https://www.experts-exchange.com/questions/21336015/Sed-Command.html | Want to protect your cyber security and still get fast solutions? Ask a secure question today.Go Premium
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# Sed Command
Posted on 2005-03-03
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Hi All,
How to replace a "\" character to "\\" characters with the sed command?
0
Question by:ee_lcpaa
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LVL 46
Expert Comment
ID: 13448569
I assume that you want to run sed from the shell, possibly a script, without the editor actually opening in a "window".
Try this:
sed -e 's/\\/\\\\/g' < SourceFile > DestinationFile
Kent
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Expert Comment
ID: 13452888
$echo "\foo\bar" | sed 's!\\!\\\\!g' \\foo\\bar 0 Author Comment ID: 13455544 Dear Kent & Tintin, I tried the two methods proposed by you, but I go the following error: sed function 's/\\/\\\\/g' cannot be parsed or sed function 's!\\!\\\\!g' cannot be parsed Below lists my script FILEONE=/tmp/pwd.txt ACCTINFO=cat$FILEONE | sed 's/\\/\\\\/g'
The content of FILEONE is mydomain\myaccount mypassword
I hope ACCTINFO will be euqal to "mydomain\\myaccount my paasword" after calling the sed function.
0
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Expert Comment
ID: 13455627
What Unix flavour are you running?
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Author Comment
ID: 13456039
Hi Tintin,
I am using HP Tru64 Unix Version 5.1B.
0
LVL 46
Accepted Solution
Kent Olsen earned 500 total points
ID: 13463780
Hi ee_lcpaa,
This is going to look pretty weird, but try this:
ACCTINFO=cat $FILEONE | sed 's/\\\\/\\\\\\\\/g' The backslashes may be being interpretted by both the shell and sed. If so when the shell gets done with: > ACCTINFO=cat$FILEONE | sed 's/\\/\\\\/g'
what gets passed to sed is:
sed 's/\/\\/g'
The first slashed is now escaped because of the backslash, so a parse error occurs.
You can also turn on statement trace with stty to see what the sed statement looks like.
Kent
0
LVL 51
Expert Comment
ID: 13474335
ACCTINFO="sed 's/\\/\\\\/g' \$FILEONE`"
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Course of the Month13 days, 7 hours left to enroll | 2018-01-19 00:14:18 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.35214751958847046, "perplexity": 14813.30380562664}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084887660.30/warc/CC-MAIN-20180118230513-20180119010513-00395.warc.gz"} |
https://mathemerize.com/the-distribution-below-gives-the-weights-of-30-students-of-a-class-find-the-median-weight-of-the-students/ | # The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
## Solution :
We prepare the following table to compute the median :
We have : n = 30, So, $$n\over 2$$ = 15
The cumulative frequency just greater than $$n\over 2$$ is 19 and the corresponding the class is (55 – 60).
Thus, (55 – 60) is the median class such that $$n\over 2$$ = 15, l = 55, f = 6, cf = 13 and h = 5.
Substituting these values in the formula,
Median = l + ($${n\over 2} – cf\over f$$)(h)
= 55 + ($$15 – 13\over 6$$)(5) = 55 + $$2\over 6$$(5) = 55 + 1.67 = 56.67
Hence, the median weight is 56.67 kg | 2023-01-29 21:26:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5897144675254822, "perplexity": 609.4785709838025}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499768.15/warc/CC-MAIN-20230129211612-20230130001612-00830.warc.gz"} |
https://www.scielo.br/j/prod/a/J6QH4TYKqHSvPB9xK8YBvJm/?lang=en | # Abstract
## Paper aims
To study fall-accident cases in order to analyze the commonly missing or not adequately applied risk management measures (RMM) and its consequences depending on falling height.
## Originality
First study to analyze failed RMM for preventing falls from height.
## Research method
The study reviewed court cases published by the journal “Safety & Health Practitioner”. NIOSH recommendations were used to define RMM to apply to this study.
## Main findings
Finally, in 98% of analyzed cases, the fall from height was a result of several non-adequate or missing RMM: in 81.6% procedures of work, 65.8% guardrails and edge protection, 60.5% risk assessment, and 60.5% platforms or scaffolds. It can be concluded that falls from height pose a significant risk for workers, which could be prevented by adequately apply RMM.
## Implications for theory and practice
The focus in the prevention of falls should be given on most common RMM.
Keywords
Injury; Fall accidents; Risk control; Workplace fatalities; Safety in construction
# 1. Introduction
Every day, people die as a result of occupational accidents or work-related diseases. In total, it reaches more than 2.78 million deaths and some 374 million non-fatal work-related injuries and illnesses each year (International Labour Organization, 2017International Labour Organization – ILO. (2017). Safety and health at work. Retrieved in 2017, November 30, from http://www.ilo.org/global/topics/safety-and-health-at-work/lang--en/index.htm
http://www.ilo.org/global/topics/safety-...
). The human cost of this daily adversity is vast, and the economic burden of poor occupational safety and health practices is estimated at 3.94% of global Gross Domestic Product each year (International Labour Organization, 2017International Labour Organization – ILO. (2017). Safety and health at work. Retrieved in 2017, November 30, from http://www.ilo.org/global/topics/safety-and-health-at-work/lang--en/index.htm
http://www.ilo.org/global/topics/safety-...
). Globally, among all, unintentional injuries represent a major public health problem and a leading cause of deaths (Centers for Disease Control and Prevention, 2017Centers for Disease Control and Prevention – CDC. (2017). Ten leading causes of death and injury. Retrieved in 2017, November 30, from https://www.cdc.gov/injury/wisqars/LeadingCauses.html
). After road traffic injuries, falls represent the second leading cause of unintentional injury deaths worldwide. An estimation is a number of 646 000 fatal falls and some 37.3 million non-fatal falls each year, severe enough to require medical attention (World Health Organization, 2017World Health Organization – WHO. (2017). Falls. Retrieved in 2017, November 30, from http://www.who.int/mediacentre/factsheets/fs344/en/
http://www.who.int/mediacentre/factsheet...
). The construction industry represents the most influential group in these numbers, with around 21.4% of USA’s workers fatalities, where the leading causes were falls (38.8%) (Occupational Safety and Health Administration, 2017Occupational Safety and Health Administration – OSHA. (2017). Construction’s “fatal four”. Retrieved in 2017, November 9, from https://www.osha.gov/oshstats/commonstats.html
https://www.osha.gov/oshstats/commonstat...
) and around 31% of UK’s workers fatalities, where the primary cause of falls from height (20%) (Bomel, 2003Bomel. (2003). Falls from height: prevention and risk control effectiveness (Research Report, 428). Sudbury: HSE Books.). The severity of fall-risk was investigated in many studies, analyzing the risk depending on occupation, age and location (Beavers et al., 2006Beavers, J. E., Moore, J. R., Rinehart, R., & Schriver, W. R. (2006). Crane-related fatalities in the construction industry. Journal of Construction Engineering and Management, 132, 901-910. http://dx.doi.org/10.1061/(ASCE)0733-9364(2006)132:9(901).
http://dx.doi.org/10.1061/(ASCE)0733-936...
; Dong et al., 2009Dong, X. S., Fujimoto, A., Ringen, K., & Men, Y. (2009). Fatal falls among Hispanic construction workers. Accident Analysis & Prevention, 41(5), 1047-1052. http://dx.doi.org/10.1016/j.aap.2009.06.012. PMid:19664444.
http://dx.doi.org/10.1016/j.aap.2009.06....
; Johnson et al., 1999Johnson, H. M., Singh, A., & Young, R. H. F. (1999). Fall protection analysis for workers on residential roofs. Journal of Construction Engineering and Management, 124(5), 418-428. http://dx.doi.org/10.1061/(ASCE)0733-9364(1998)124:5(418).
http://dx.doi.org/10.1061/(ASCE)0733-936...
). Some went further, analyzing heights from which people mostly fell, the type and value of projects where fall-accidents mostly occurred (Huang et al., 2003Huang, X., Hinze, J., & Asce, M. (2003). Analysis of construction worker fall accidents. Journal of Construction Engineering and Management, 129, 262-271. http://dx.doi.org/10.1061/(ASCE)0733-9364(2003)129:3(262).
http://dx.doi.org/10.1061/(ASCE)0733-936...
; Kang et al., 2017Kang, Y., Siddiqui, S., Suk, S. J., Chi, S., & Kim, C. (2017). Trends of fall accidents in the U. S. construction industry. Journal of Construction Engineering and Management, 143(8), 1-7. http://dx.doi.org/10.1061/(ASCE)CO.1943-7862.0001332.
http://dx.doi.org/10.1061/(ASCE)CO.1943-...
).
Despite all of these studies and the risk of falling from height is clearly identified as a challenge to be solved. Even after several studies have investigated the reasons why they continue to occur and solutions to minimize hazards or eliminate their risk, the number of accidents due to falls from height continues to grow.
The objective of this study was to analyze the consequences depending on falling height and to investigate the risk management measures that were commonly missing or not adequately applied in preventing and controlling at the time when falls from height occurred.
# 2. Methodology
The methodology of this review was based on the PRISMA Statement for Reporting Systematic Reviews and Meta-Analyses (Liberati et al., 2009Liberati, A., Altman, D. G., Tetzlaff, J., Mulrow, C., Gøtzsche, P. C., Ioannidis, J. P., Clarke, M., Devereaux, P. J., Kleijnen, J., & Moher, D. (2009). The PRISMA statement for reporting systematic reviews and meta-analyses of studies that evaluate health care interventions: explanation and elaboration. Annals Internal Medicine, 151(4), W65-94. http://dx.doi.org/10.7326/0003-4819-151-4-200908180-00136. PMid:19622512.
http://dx.doi.org/10.7326/0003-4819-151-...
). The searching process was conducted by using the Brazilian CAPES searching tool (Coordenação de Aperfeiçoamento de Pessoal de Nível Superior, 2017Coordenação de Aperfeiçoamento de Pessoal de Nível Superior – CAPES. (2017). Portal Periódicos CAPES. Retrieved in 2017, August 11, from http://www.periodicos.capes.gov.br/
http://www.periodicos.capes.gov.br/...
), by using the institutional IP address of the University of Pernambuco federate credentials. The following two keywords were defined: “fall” AND “height”. The selection process included first applying the exclusion, and afterward inclusion criteria.
## 2.1. Exclusion and inclusion criteria
The review only included court cases, as studies, published in the English language by the journal “Safety & Health Practitioner” (Institution of Occupational Safety & Health, 2017Institution of Occupational Safety & Health. (2017). The safety & health practitioner. Retrieved in 2017, October 20, from http://www.shponline.co.uk/
http://www.shponline.co.uk/...
), as this journal made all the analyzed information open access. Afterward, the studies were excluded if repeated, then screened and excluded by title, considering only those related to falls from height, excluding if the fall height was unknown, if falls were from a standing height, if the person fell on material which absorbed the impact, or if suffered multiple falls. As inclusion criteria, only accidents were considered, while suicidal and homicidal events were excluded.
Additionally, this study included a previously conducted systematic review on falls from height (Zlatar & Barkokébas, 2018Zlatar, T., & Barkokébas, B. J. (2018). Building information modelling as a safety management tool for preventing falls from height (1st ed., pp. 15-21). Mauritius: LAP Lambert Academic Publishing.). This article serves as state of the art on the topic of falls from height, give indicators for the data analysis (fall accidents by height and by location) and develop the discussion part by comparing the results from this study with the results from previously conducted studies.
## 2.2. Data analysis
Statistical analysis was done by using excel statistical toolbox. The data were analyzed in accordance with rules specified in the following sections:
1. A
Fall height and place
In order to be able to compare data with previous studies (Huang et al., 2003Huang, X., Hinze, J., & Asce, M. (2003). Analysis of construction worker fall accidents. Journal of Construction Engineering and Management, 129, 262-271. http://dx.doi.org/10.1061/(ASCE)0733-9364(2003)129:3(262).
http://dx.doi.org/10.1061/(ASCE)0733-936...
; Kang et al., 2017Kang, Y., Siddiqui, S., Suk, S. J., Chi, S., & Kim, C. (2017). Trends of fall accidents in the U. S. construction industry. Journal of Construction Engineering and Management, 143(8), 1-7. http://dx.doi.org/10.1061/(ASCE)CO.1943-7862.0001332.
http://dx.doi.org/10.1061/(ASCE)CO.1943-...
), the cases were distributed in four height-groups as given by previous studies:
• Falls from a height between 0 to 3 m;
• Falls from a height between 3 to 6.1 m;
• Falls from a height between 6.1 to 9.0 m;
• Falls from a height of more than 9.1 m.
The analyzed results include activities which were conducted before falling, the fall height and where it occurred.
1. B
Consequence analysis
In order to better analyze the consequences of falls from height, the cases were divided into four groups according to the consequences:
• Nothing injured (bruising, minor burns, and blisters, minor cuts on the head);
• Temporary disability (fractured leg, ankle, ribs);
• Permanent disability (serious spinal injuries or paralyzed from the waist down);
• Death (including instant death and death which occurred after some time, but which was related to injuries suffered by the fall).
The consequences of the falls were then related and grouped according to the height of fall, determined by studies previously mentioned section in the “A) Determination of fall-height”.
1. C
Risk management analysis
For risk management analysis, five categories were selected in order to evaluate which measures were applied to prevent and control workplace hazards, and therefore minimize or eliminate safety hazards. For this study, the NIOSH recommendations (National Institute for Occupational Safety and Health, 2018National Institute for Occupational Safety and Health – NIOSH. (2018). Hierarchy of controls. Washington.) on the hierarchy of controls were reflected, considering the following categories and measures:
• Risk Assessment (including identification and evaluation of the risk);
• Elimination (to physically remove the hazard) or Substitution (to replace the hazard);
• Engineering Controls (to isolate people from the hazard, including the use of work platforms, scaffolds, ladders, stepladders, guardrails, handrails, barriers, edge protection, and nets);
• Administrative Controls (to change the way people work, including procedure, method, and plan of work, training certification, signs, lighting, warning labels, and supervision);
• Personal Protective Equipment – PPE (to protect the worker).
This recommendation is commonly accepted by Safety at work engineers and practitioners to always start with the most effective possible measure (elimination), and when not feasible to apply it, go to the next measure of the hierarchy.
# 3. Results
The identification process resulted in 386 studies. All were screened thoroughly in order to exclude those that were not in accordance with the exclusion and inclusion criteria. Finally, 114 cases were included in this analysis (illustrated in Table 1A of the Appendix A).
## 3.1. Fall height and place
In the included studies, falling height ranged between 1.2 to 42 meters, where the numbers were: 19 cases between 0 to 3 m; 52 cases between 3 to 6.1 m; 21 cases between 6.1 to 9.0 m; and 22 cases of more than 9.1 m. The distribution of cases per group is illustrated in Figure 1.
Figure 1
Distribution of encountered cases by fall height.
The location of all included cases was in the United Kingdom, ranging from the year 2003 to 2014. Nevertheless, this review did not analyze fall cases fluctuation during the years on one specific territory, but was primarily focused on consequences depending on fall height, among other analyzed questions. The building height and type was not specified by included articles. The type of working activity was mostly (in 65 cases, 57%) related to construction working activities (building, reforming or demolishing buildings), in three cases it was related to leisure time, while other (in 46 cases) were related to other working activities, such as sewage maintenance, vehicle repairing or boat building.
Figure 2 illustrates the most common places where falls from height occurred: on scaffolds/platforms (26-22.8%); roofs (30-26.3%); collapses, including collapses of floors, walls and staircases (4-3.5%); through opening, including falls through stairwells, trapdoors, lift wells or the glass panels in construction (15-13.2%); ladders and stepladders (10-8.8%); lifting, including lifting’s with forklifts (10-8.8%), and other (19-16.7%).
Figure 2
Most common places for falls from heights.
## 3.2. Consequence analysis
The consequences depending on fall height were illustrated in Table 1, showing the number of cases and percentages for each of the four analyzed consequences.
As it could be seen from the Table 1, the consequence of not having anything injured was present only in fall heights below 6.1 meters.
Table 1
Fall consequences per height groups.
## 3.3. Risk management analysis
Figure 3 illustrates missing or non-adequate safety procedures. In total, 5 main categories with 11 measures were illustrated: category 1 – identification, evaluation and risk control (measure 1); category 2 – risk elimination/prevention (measure 2); category 3 – engineering controls and measures (measures 3, 4, 5 and 6); category 4 – administrative controls and measures (measures 7, 8, 9 and 10); and category 5 – using of PPE. The data for each analyzed measure was divided into: missing (if the measure was not applied); not adequate (if the measure was not appropriate); additionally (if the measure should be revised if appropriate); and total (the total number the three mentioned situations).
Figure 3
Measures failed while working at heights. Measures: (1) Risk Assessment; (2) Risk Elimination (Prevention); (3) Work platform, Scaffold; (4) Ladder/Stepladder; (5) Guardrails, Handrails, Bariers, Edge Protection; (6) Nets; (7) Procedure of work (method, plan); (8) Training and Certification; (9) Signs, Lighting, Warning labels; (10) Supervision; (11) Personal Protective Equipment.
# 4. Discussion
## 4.1. Fall height and place
Table 2 illustrates groups depending on falling height and compares the results from this study with results from two other studies. It is important to notice that percentages were a product of the analyzed cases and that in reality, it is probable to expect a much higher number of falls from lower heights, where falls are probably passing not recorded.
Table 2
Fall accidents by height.
As it could be concluded from Table 2, this study found a lower number of cases with falling heights between 0 to 3.0 meters. The results from falling heights between 3.0 to 6.1 meters are in accordance with the findings from one study (Kang et al., 2017Kang, Y., Siddiqui, S., Suk, S. J., Chi, S., & Kim, C. (2017). Trends of fall accidents in the U. S. construction industry. Journal of Construction Engineering and Management, 143(8), 1-7. http://dx.doi.org/10.1061/(ASCE)CO.1943-7862.0001332.
http://dx.doi.org/10.1061/(ASCE)CO.1943-...
). The percentage of falls between 6.1 to 9.0 meters was higher, while the percentage of falls from heights ≥9.1 meters was in between both previously conducted studies.
In Table 3, the fall accidents by location were compared with findings from other studies.
Table 3
Fall accidents by location.
Data analyzed through this review show that falls from height occur mostly when working on roofs, scaffolds, and platforms, representing almost 50% of all analyzed cases. Therefore, workers working at these positions are most endangered, were all mentioned risk management measures and procedures should be applied and revised on a regular basis. In accordance with the Table 3, some studies concluded that scaffolders and roofers are among the most exposed working activities, which is understandable as they spent more time working on heights (Bobick, 2005Bobick, T. G. (2005). Falls through roof and floor openings and surfaces, including skylights: 1992-2000. Journal of Construction Engineering and Management, 130(6), 895-907. http://dx.doi.org/10.1061/(ASCE)0733-9364(2004)130:6(895).
http://dx.doi.org/10.1061/(ASCE)0733-936...
; Wong et al., 2016Wong, L., Wang, Y., Law, T., & Lo, C. T. (2016). Association of root causes in fatal fall-from-height construction accidents in Hong Kong. Journal of Construction Engineering and Management, 142(7), 1-12. http://dx.doi.org/10.1061/(ASCE)CO.1943-7862.0001098.
http://dx.doi.org/10.1061/(ASCE)CO.1943-...
), and as they typically carry heavy and bulky materials on slippery and inclined walking/working surfaces (Wiersma & Charles, 2006Wiersma, M., & Charles, M. (2006). Occupational injuries and fatalities in the roofing contracting industry. Journal of Construction Engineering and Management, 131(11), 1233-1240.). Further-on, innovative safety solutions should be considered, because as compared with one study (Cheung & Chan, 2012Cheung, E., & Chan, A. P. C. (2012). Rapid demountable platform (RDP): a device for preventing fall from height accidents. Accident Analysis & Prevention, 48, 235-245. http://dx.doi.org/10.1016/j.aap.2011.05.037. PMid:22664686.
http://dx.doi.org/10.1016/j.aap.2011.05....
) comparing scaffolds, it could benefit to the safety of workers, reduce the cost of the equipment in use, increase durability and speed of setting the equipment, among other advantages. Most of the challenges about falls from height might be solved through tools (Ezisi & Issa, 2018Ezisi, U., & Issa, M. H. (2018). Case study application of prevention through design to enhance workplace safety and health in Manitoba heavy construction projects. Canadian Journal of Civil Engineering, (204), 1-36.) for implementing the approach of Prevention through Design.
## 4.2. Consequence of falls from height
Other studies did not analyze the consequence of falls from height; therefore it was not possible to compare the results. By comparing consequences among analyzed studies, the number of cases which resulted in no injury was very low (5; ≈4% of all analyzed cases). With only 5 cases it could be assumed that this is probably the most biased category, as it is reasonable to assume that many low-altitude fall cases happen on a daily basis, but most of them end with no or light injuries, therefore ending up unreported.
The number of cases which resulted in a temporary disability was the highest (51; ≈45% of all analyzed cases). Although workers did not suffer the more severe consequence, it can be seen that falls from height temporarily disabled further working activities, where it is probable to expect rehabilitation costs and loss of production.
Serious consequences were represented in a high number of cases, the permanent disability was encountered in 17 (≈15%), while deaths in 41 (≈36% of all analyzed cases). The fatal falls from a height above 9.1 m were responsible for 33.9% of fatal falls, which is in accordance with the findings from another study where falls above 9.1 m (30 feet in the article) were accounted for more than one-third of fatal falls (Dong et al., 2017Dong, X. S., Largay, J. A., Choi, S. D., Wang, X., Cain, C. T., & Romano, N. (2017). Fatal falls and PFAS use in the construction industry: findings from the NIOSH FACE reports. Accident Analysis & Prevention, 102, 136-143. http://dx.doi.org/10.1016/j.aap.2017.02.028. PMid:28292698.
http://dx.doi.org/10.1016/j.aap.2017.02....
).
Figure 4 illustrates the severity of the consequence depending on fall height (distance) and the percentage of occurrence of each consequence. It also illustrates the logarithmic tendency lines (chosen because they minimize the overall R2 value) with their equations for each consequence. The severity of injuries varied according to the falling height. Although falling from any altitude may result in any of considered consequences, the results show that falls from heights above 20m should result in death consequence, while other consequences could happen only by chance, therefore set up to height until 20m. Some cases were removed for the construction of the interpolations as have been considered as cases by chance and therefore withdrawn from Figure 4 (For example, the percentage of death consequences gradually increased as falling height increased, reaching 75% of death cases on height of 10m, and then being 100% on heights from 12 to 42m. From analyzed data, on falling height of 16m, there was a death consequence of 50%, not following the logical trend, and therefore considered as cases by chance and withdrawn from the figure).
Figure 4
Consequences depending on falling height.
The Figure 4 illustrating the tendencies of consequences depending of falling height, show that as the fall height increase there was a tendency of:
• Decrease for consequence “nothing injured” $y=−3.112lnx+13.03$;
• Decrease for consequence “temporary disability” $y=−19.84lnx+83.599$;
• Increase for consequence “permanent disability” $y=15.437lnx+1.0332$;
• Increase for consequence “death” $y=40.243lnx–25.992$.
It is also important to notice that in some cases a consequence resulted in temporary disability, while it could as easily result with death. For example, in one case fall resulted in a person being on life support machines for 10 days (The Safety & Health Practitioner, 2006bThe Safety & Health Practitioner. (2006b). Fall from height: three firms fined over bus garage plunge. Retrieved in 2017, October 15, from http://link.galegroup.com/apps/doc/A141175247/AONE?u=capes&sid=AONE&xid=a3f90bf6
) or in another case, being unable to return to work for 2 years (The Safety & Health Practitioner, 2013bThe Safety & Health Practitioner. (2013b). Miscommunication led to worker’s stairwell plunge. Retrieved in 2017, October 15, from http://link.galegroup.com/apps/doc/A341129969/AONE?u=capes&sid=AONE&xid=e790bfb1
).
The lowest altitude from which the person died was 1.8 m. By analyzing death cases from low altitudes, it was noticed that all died due to falling headfirst, received severe head injuries, fractured skulls or hit their head on the kerb (The Safety & Health Practitioner, 2005The Safety & Health Practitioner. (2005). Fall from height: unsecured ladder implicated in worker’s fatal fall. Retrieved in 2017, October 15, from http://link.galegroup.com/apps/doc/A132848205/AONE?u=capes&sid=AONE&xid=21be3201
, 2006aThe Safety & Health Practitioner. (2006a). Fall from height: engineer fell from forklift truck while repairing door. Retrieved in 2017, October 15, from http://link.galegroup.com/apps/doc/A143775620/AONE?u=capes&sid=AONE&xid=711faf9d
, 2009The Safety & Health Practitioner. (2009). Company director “wholly culpable”. Retrieved in 2017, October 15, from http://link.galegroup.com/apps/doc/A204090998/AONE?u=capes&sid=AONE&xid=e226e1db
, 2010aThe Safety & Health Practitioner. (2010a). Construction firm fined over death at premier-league club. Retrieved in 2017, October 15, from http://link.galegroup.com/apps/doc/A243044576/AONE?u=capes&sid=AONE&xid=cb36c9a5
, bThe Safety & Health Practitioner. (2010b). Death of Polish worker a wake-up call to construction bosses. Retrieved in 2017, October 15, from http://link.galegroup.com/apps/doc/A218591885/AONE?u=capes&sid=AONE&xid=e744f410
, 2013aThe Safety & Health Practitioner. (2013a). Lack of work-at-height checks contributed to ladder death. Retrieved in 2017, October 15, from http://link.galegroup.com/apps/doc/A318915529/AONE?u=capes&sid=AONE&xid=1607eff8
). These findings are in accordance with a study (Türk & Tsokos, 2004Türk, E. E., & Tsokos, M. (2004). Pathologic features of fatal falls from height. The American Journal of Forensic Medicine and Pathology, 25(3), 194-199. http://dx.doi.org/10.1097/01.paf.0000136441.53868.a4. PMid:15322459.
http://dx.doi.org/10.1097/01.paf.0000136...
) which found that head trauma was the cause of death in 11 of the 19 cases that were from 9m or less (58%). Therefore, as head injuries were found to be responsible for deaths on lower heights, it can be concluded that helmets would be an effective life-protection equipment for lower heights. On the other hand, analyzed deaths from heights over 10m (Türk & Tsokos, 2004Türk, E. E., & Tsokos, M. (2004). Pathologic features of fatal falls from height. The American Journal of Forensic Medicine and Pathology, 25(3), 194-199. http://dx.doi.org/10.1097/01.paf.0000136441.53868.a4. PMid:15322459.
http://dx.doi.org/10.1097/01.paf.0000136...
) were caused mainly due to polytrauma (72%), and in only ≈24% cases (8/33) by head trauma.
In practice, falls from height typically occur when carrying heavy and bulky materials on slippery and inclined walking/working surfaces (Wiersma & Charles, 2006Wiersma, M., & Charles, M. (2006). Occupational injuries and fatalities in the roofing contracting industry. Journal of Construction Engineering and Management, 131(11), 1233-1240.). Therefore, for working activities when this is the case, wearing helmets could be considered for activities on the same level, while for activities on height, special attention should be taken in applying risk management measures.
## 4.3. Risk management analysis
Figure 3 illustrates a total percentage of 11 failed risk management measures for analyzed cases. The administrative measure - the procedure of work (method and plan) was found to be the most common safety measure noted as “not adequate” or as “should be revised”, within 81.6% of analyzed cases. The engineering measure - guardrails, handrails, barriers and edge protection were found to be the second most failed safety measure with 65.8% (where it was missing in 33.3% of cases). Further two most commonly failed measures were risk assessment (60.5%) and the engineering measure - work platform/scaffold (60.5%). Inadequate PPE or missing PPE was noticed in 56.1% of the cases. By comparison, one previously conducted study found that in 48% of the cases workers fell due to their loss in balance while not wearing adequate fall protection devices (Wong et al., 2016Wong, L., Wang, Y., Law, T., & Lo, C. T. (2016). Association of root causes in fatal fall-from-height construction accidents in Hong Kong. Journal of Construction Engineering and Management, 142(7), 1-12. http://dx.doi.org/10.1061/(ASCE)CO.1943-7862.0001098.
http://dx.doi.org/10.1061/(ASCE)CO.1943-...
).
It is also interesting to notice that training and certification were missing in 19.3% of the workers. This is important because training increases workers' perception and reaction to risk and, when conducted regularly, can improve safety performance and therefore the worker is more likely to identify, evaluate and control risks (Chan et al., 2008Chan, A. P. C., Wong, F. K. W., Chan, D. W. M., Yam, M. C. H., Kwok, A. W. K., Lam, E. W. M., & Cheung, E. (2008). Work at height fatalities in the repair, maintenance, alteration, and addition works. Journal of Construction Engineering and Management, 134, 527-535. http://dx.doi.org/10.1061/(ASCE)0733-9364(2008)134:7(527).
http://dx.doi.org/10.1061/(ASCE)0733-936...
; Hinze & Gambatese, 2003Hinze, J., & Gambatese, J. (2003). Factors that influence safety performance of specialty contractors. Journal of Construction Engineering and Management, 129, 159-164. http://dx.doi.org/10.1061/(ASCE)0733-9364(2003)129:2(159).
http://dx.doi.org/10.1061/(ASCE)0733-936...
; Rodríguez-Garz et al., 2015Rodríguez-Garz, I., Lucas-Ruiz, V., Martínez-Fiestas, M., & Delgado-Padial, A. (2015). Association between perceived risk and training in the construction industry. Journal of Construction Engineering and Management, 141(5), 1-9. http://dx.doi.org/10.1061/(ASCE)CO.1943-7862.0000960.
http://dx.doi.org/10.1061/(ASCE)CO.1943-...
). In addition, it is essential to consider that training should be conducted in accordance with the individual characteristics of workers as age, position, trade, number of years of work, past experience with accidents, and personality, which was all found to contribute on how effective would be the training (Kim et al., 2011Kim, E., Yu, I., Kim, K., & Kim, K. (2011). Optimal set of safety education considering individual characteristics of construction workers. Canadian Journal of Civil Engineering, 38(5), 506-518. http://dx.doi.org/10.1139/l11-024.
http://dx.doi.org/10.1139/l11-024...
). Kang found that workers were not equipped with fall protection in 70.7% of cases, and were equipped incorrectly in 17.9% of cases (Kang et al., 2017Kang, Y., Siddiqui, S., Suk, S. J., Chi, S., & Kim, C. (2017). Trends of fall accidents in the U. S. construction industry. Journal of Construction Engineering and Management, 143(8), 1-7. http://dx.doi.org/10.1061/(ASCE)CO.1943-7862.0001332.
http://dx.doi.org/10.1061/(ASCE)CO.1943-...
). Although this could not be directly compared with results from this study, the same conclusions could be adopted – there is an urgent need to improve working safety culture and adopt adequate occupational risk management measures.
Missing or not adequate supervision was found in 22.8% of the analyzed cases. One study found that supervision is important as scaffolders failed to anchor their harness, not due to poor safety attitude, but due to a subjective norm (perceived social pressure) (Goh & Binte Sa’adon, 2015Goh, Y. M., & Binte Sa’adon, N. F. (2015). Cognitive factors influencing safety behavior at height: a multimethod exploratory study. Journal of Construction Engineering and Management, 141(6), 1-8. http://dx.doi.org/10.1061/(ASCE)CO.1943-7862.0000972.
http://dx.doi.org/10.1061/(ASCE)CO.1943-...
).
The 11 risk management measures illustrated in Figure 3 were further analyzed by each case separately. It was noted that most of the cases had failed several risk management measures. In the following Figure 5 were illustrated all 114 cases (100%) by the number of failed risk management measures (both missing and not adequate risk management measures) by each case, where 1 failed measure was only present in 2% of analyzed cases, 2 failed measures in 15%, 3 in 19%, 4 in 20%, 5 in 15%, 6 in 19% and 7 in 10% of analyzed cases.
Figure 5
Number of failed risk management measure from the analyzed cases.
As it is shown in Figure 5, only 2% of the analyzed cases could be associated with one failed risk management measure, while in other 98%, the fall from height was a result of several non-adequate or missing risk management measures. Therefore, it is possible to conclude that in the majority of cases, falls from height were not a coincidence or an unlucky event due to only one fail, but that it could be promptly easily noted due to various failures, and prevented with daily safety screening of the working process.
## 4.4. Future studies
In order to be able to analyze fall consequences further and understand better how some factors benefit to the survival of falling from greater heights, there is a need to include more data on persons which fell and explain how it occurred. For example, fall (impact) energy could be calculated through data on fall height and human body mass: E=mgh (J).
Results on calculations regarding fall energy for four different persons (body mass of 60, 75, 90 and 105kg) were illustrated in Figure 6.
Figure 6
Human body mass an its relation to fall energy and fall height.
As it is illustrated in Figure 6, the fall energy of 10,500J correspond to fall of a human with a body mass of 105kg from a height of 10m, 90kg from 12m, 75kg from 14m and 60kg from 18m. Therefore, a fall impact from the same height would be much lower for those humans having lower body mass compared with those with higher, representing a heavy person less chance of surviving a fall.
Future studies should also consider exploring with more details the employment conditions of workers which suffered falls from height, including the type of employment contract, the age, and experience of the worker.
# 5. Limitations
The limitations of this study lay in analyzing cases which were reported and recorded by the reviewed source. A bias probably lays in the number of no reported cases of falls from height, especially when the fall resulted in minor or no injuries which could be expected in falls from lower heights. Therefore, the percentages on injuries and death occurrences might not correspond to actual values, especially for lower heights. Finally, the analyzed data do not contain information on workers body mass, which would be interesting to analyze as it might have influenced the energy of fall impact, explaining why some persons survived falls from greater heights. Analyzing more cases would help in more consistent results and therefore understanding better the consequences of falls from heights, and possibly result with more or different consequence-based groups.
# 6. Conclusions
The falls from height represent one of the leading risks, causing more than 2.78 million deaths and some 374 million non-fatal work injuries each year. Through the analysis of included studies, it was found that a typical accident of falls from height would be in 45.6% from heights between 3 to 6.1 meters and in 49.1% occurring from scaffolds or roofs. The consequences this fall would result in death if the person fell on head and suffered head trauma, while if not, the percentage representing survival would be ≈55%, depending on the persons mass and also material on which he would fall. As the data show, there would be a percentage of ≈98% that several risk assessment measures were not applied. Among those not applied (failed) measures the reason would be: in 81.6% the procedures of work (administrative measure); in 65.8% the guardrails, handrails, barriers and edge protection (engineering measure); in 60.5% risk assessment; and in 60.5% work platform/scaffold (engineering measure). Therefore, it can be concluded that falls from height pose a great risk for workers, which could be prevented by adequately apply management measures.
Future studies should include more cases with data on body mass of persons which fell from heights, and evaluate how falling height affect each body part.
# Appendix A Falls from height: analysis of 114 cases.
This appendix file contains 4 tables, which illustrate all included and analysed cases within the article “Falls from Height: Analysis of 114 Cases”:
- Table 1A: Included articles, illustration of the article title, reference, year, type of industry and age of the injured worker;
- Table 2A: Included articles, illustration of the falling height by articles, consequence, injured body parts and recovery period;
- Table 3A: Included articles, illustration of the measures which were Not Appropriate (NA), were missing (0) or should be Additionally (A) considered among each one of included cases;
- Table 4A: Included articles, illustration of accidents which were related to most common falling places.
Table 1A
Included articles, illustration of the article title, reference, year, type of industry and age of the injured worker.
Table 2A
Included articles, illustration of the falling height by articles, consequence, injured body parts and recovery period.
Table 3A
Included articles, illustration of the measures which were Not Appropriate (NA), were missing (0) or should be Additionally (A) considered among each one of included cases.
Table 4A
Included articles, illustration of accidents which were related to most common falling places.
# Acknowledgements
This project was financially supported by the Brazilian Ministry of Education through the Program for Coordination and Improvement of Higher Level Personnel (PNPD/CAPES). Many thanks for all the support from the Faculty of Engineering, University of Porto (FEUP), Federal University of Pernambuco (UFPE) and to the University of Pernambuco (UPE).
• How to cite this article: Zlatar, T., Lago, E. M. G., Soares, W. A., Baptista, J. S., & Barkokébas Junior, B. (2019). Falls from height: analysis of 114 cases. Production, 29, e20180091. https://doi.org/10.1590/0103-6513.20180091.
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» http://dx.doi.org/10.1097/01.paf.0000136441.53868.a4
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# Publication Dates
• Publication in this collection
13 May 2019
• Date of issue
2019 | 2022-06-29 23:41:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 4, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.47865045070648193, "perplexity": 7889.861448402027}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103645173.39/warc/CC-MAIN-20220629211420-20220630001420-00415.warc.gz"} |
https://socratic.org/questions/562f9459581e2a4299123973 | # Question 23973
Nov 17, 2015
"Be: " ["He"] 2s^2
#### Explanation:
The first thing to do when trying to write an atom's electron configuration is to figure you exactly how many electrons must be accounted for.
To do that, grab a periodic table and look for that atom's atomic number.
In your case, beryllium, $\text{Be}$, is located in period 2, group 2 of the periodic table, and has an atomic number equal to $4$. This means that a neutral beryllium atom will contain $4$ protons in its nucleus and $4$ electrons surrounding its nucleus.
Therefore, your electron configuration must account for $4$ electrons. The complete electron configuration for beryllium will be
$\text{Be: } 1 {s}^{2} 2 {s}^{2}$
Now, in order to use the noble gas shorthand notation, you must first identify which noble gas comes immediately before beryllium in the periodic table.
In this case, the only option available is helium, $\text{He}$. Helium has $2$ electrons surrounding its nucleus, so its electron configuration will look like this
$\text{He: } \textcolor{b l u e}{1 {s}^{2}}$
Notice that the configuration that accounts for the first two electrons is identical for both atoms. This means that you can replace it in beryllium's configuration to get
$\text{Be: } \textcolor{b l u e}{1 {s}^{2}} 2 {s}^{2}$
The electron configuration of helium is written like this, $\left[\text{He}\right]$, which means that the noble gas shorthand for beryllium will be
"Be: " ["He"] 2s^2# | 2020-08-11 07:40:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 13, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6347415447235107, "perplexity": 892.0297114555267}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439738735.44/warc/CC-MAIN-20200811055449-20200811085449-00294.warc.gz"} |
https://physics.stackexchange.com/questions/122235/critical-temperature-difference-between-ising-and-xy-model | # Critical temperature difference between Ising and XY model
The following formula gives the critical coupling (more precisely the ratio of the spin-spin coupling over the temperature) for $O(n)$ models on a triangular lattice:
$$\text{e}^{-2K}=\frac{1}{\sqrt{2+\sqrt{2-n}}}$$
with $K=\beta J$
Numerically, it says that:
Ising model (n = 1) has $K \approx 0.27$
XY model (n=2) has $K \approx 0.17$
Thus, the critical temperature for the XY model is higher than the Ising model. I've been thinking about it but I can't come out with a reason of why allowing the order parameter to take continuous values means that we need to go higher in temperature to destroy order. Is there a (semi) intuitive reason for that?
• Your intuition is correct. It is not difficult to prove that the 2-point function of the Ising model is always an upper bound on the corresponding quantity for the XY model. In particular, if $\beta_c^{XY}$ denotes the inverse temperature at which the Kosterlitz-Thouless phase transition occurs, then $\beta_c^{XY}\geq 2\beta_c^{I}$, where $\beta_c^I$ is the inverse critical temperature for the Ising model. You can find the proof here. Note that the formula you give above is not for the $O(n)$ spin model, but for the $O(n)$ loop model. – Yvan Velenik Jun 28 '14 at 17:37
• No, the loop model is an approximation of the O(n) spin model. You can look, for example, at the explanations in Nienhuis' Les Houches lectures on Loop models, which can be downloaded from his homepage. – Yvan Velenik Jul 1 '14 at 8:30
• ps: In the Ising case, I managed to very precisely check the critical coupling value given by the formula above. That's why it's even more puzzling for me. – Learning is a mess Jul 1 '14 at 15:41
• The value of the critical points of the spin and loop models should not be related at all. However, the critical behaviors are expected to be the same (they should belong to the same universality class). – Yvan Velenik Jul 1 '14 at 16:53
• I have no idea what the value of the coupling constant of the spin O(2) model on the triangular lattice is. There are very efficient cluster methods for this model (as far as I know, I am not a specialist in the numerical aspects) which should provide precise estimates. – Yvan Velenik Jul 1 '14 at 16:56 | 2020-02-21 16:30:08 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7687211036682129, "perplexity": 300.6286811762453}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875145533.1/warc/CC-MAIN-20200221142006-20200221172006-00376.warc.gz"} |
http://www.solutioninn.com/cond-nast-traveler-conducts-an-annual-survey-in-which-readers | # Question
Condé Nast Traveler conducts an annual survey in which readers rate their favorite cruise ship. All ships are rated on a 100-point scale, with higher values indicating better service.
A sample of 37 ships that carry fewer than 500 passengers resulted in an average rating of 85.36, and a sample of 44 ships that carry 500 or more passengers provided an average rating of 81.40 (Condé Nast Traveler, February 2008). Assume that the population standard deviation is 4.55 for ships that carry fewer than 500 passengers and 3.97 for ships that carry 500 or more passengers.
a. What is the point estimate of the difference between the population mean rating for ships that carry fewer than 500 passengers and the population mean rating for ships that carry 500 or more passengers?
b. At 95% confidence, what is the margin of error?
c. What is a 95% confidence interval estimate of the difference between the population mean ratings for the two sizes of ships?
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https://www.physicsforums.com/threads/wind-power-generator.40383/ | # Wind Power generator
1. Aug 24, 2004
### darkar
The power generated from the aerogenerator is from the kinetic energy of the wind which pass throught it. So, after the wind passed through the aerogenerator, it's kinetic energy decreases. So, will the temperature of the wind drop after passing through the generator since its kinetic energy is lost?
2. Aug 24, 2004
### Integral
Staff Emeritus
I would say no to this. The average velocity with respect to the ground will change. The question is how is the microscopic motions which are used to find an average KE of a molecule and therefore the air temperature related to the air velocity wrt the ground. I believe you find that the velocity for the temperature is relative to the neighboring air molecules, so it is velocity that is superimposed on the velocity of air wrt the ground. Thus we can can have a high velocity COLD wind. (If you have ever experienced a winter in Chicago you will now that wind can be cold!) This means that the air molecules are moving slowly among themselves but a large mass of air is moving at a high speed. It is the energy (velocity) wrt the ground which drives an air turbine, not the temperature of the air. So no the energy removed to drive a wind turbine does not NECESSARILY reduce the temperature of the air.
3. Aug 26, 2004
### darkar
There is friction at the moving parts of the generator, so there should be heat build up. So, will the temperature of the wind will eventually increase?
4. Aug 26, 2004
### Locrian
From my understanding, most wind generators are not designed to turn rapidly, so there is very little build up of heat.
5. Aug 26, 2004
### Clausius2
Not quite Mr. PF Mentor. The wind generator acts like a turbine rotor. There is an exchange of kinetic energy between fluid and mechanical environment.
In particular, the flow is being accelerated relative to blades. Then pressure decreases as relative velocity to the blade increases. It is possible that such decreasing will be small, because of the constant pressure of the surroundings. The generalized Bernoulli equation for total enthalpy flow shows that total temperature and static temperature will decrease as the flow passes trough blades.
End of first round.
6. Aug 26, 2004
### LURCH
I don't think so.
The wind will lose velocity, and this kinetic energy will go to mean places. First, some of the energy (hopefully most of the energy) will be converted to electricity by the generator. Secondly, since no method of power generation is 100% efficient, some energy will be lost to entropy. Most of this energy will be converted into heat due to friction. Friction between the blades and the pair will keep the air directly, while friction within the mechanism of the generator will cause the moving parts to heat up, then radiate this heat out into the wind.
7. Aug 26, 2004
### Clausius2
If you think so, the same reasoning would be employed with turbines. Do you think air behind a blade turbine is hotter due to entropy generation? I hope you don't think that.
8. Aug 26, 2004
### Staff: Mentor
Lurch didn't say the air would heat up, just that there would be heat generated by the interaction of the wind and the turbine - or if you prefer, heat transferred from the wind to the turbine - and then some of it back to the wind.
My take on this is that the turbine does remove some heat from the air (it has to via conservation of energy), but the amount is insignificant.
9. Aug 26, 2004
### Phobos
Staff Emeritus
IIRC, wind power generators have a max efficiency around 35%. I'd have to dig back into my college textbooks to give you the details.
10. Aug 26, 2004
### Clausius2
The increasing of entropy causes that behind the turbine blades the air is not as cool as it would be if there was not entropy increasing at all (isentropic flow). But, as I stated before, and with generalized (compressible) Bernoulli equation as background of my words, the air is cooled by the turbine blades.
You have said something that has taken me aback. The amount of heat removed by a conventional turbine of air is not insignificant at all.!. In particular the decreasing of total enthalpy is the work per unit mass obtained in the rotating shaft. Do you think this amount is insignificant in a turbine of a nuclear plant?. Surely I have interpreted wrong your words. But in some way you have reason, keep on reading please.
Let's see. Name 1 at blade inlet, and 2 at blade oulet:
$$\tau=h_{t1}-h_{t2}$$; here $$\tau$$ is the work obtained per unit mass and ht is the total enthalpy. This is one of the so-called Euler equations of Turbomachinery. It is similar to energy N-S equation viewed from an inertial frame.
In order to obtain work, ht1>ht2. Thus, the outlet total temperature must be lower than the inlet one. What happens with the static temperature?.
At the inlet: $$\frac{T_{t1}}{T_{1}}=1+\frac{\gamma-1}{2}M_{1}^2$$ where M is #Mach. Moreover: $$\frac{T_{t2}}{T_{2}}=1+\frac{\gamma-1}{2}M_{2}^2$$ at the outlet.
All this stated above is true for isentropic and steady flow. Lately, we will include non-isentropic effects. If Mach# is M<<1, then kinetic energy variations are much smaller than thermal energy itself. So that, $$\tau$$ will be of the order of $$c_{p}(T_{1}-T_{2})$$. Here you can see the reason for my surprising. With isentropic flow, all the work obtained in the shaft is due to static thermal variations at M<<1.
If the flow is not isentropic, the work obtained would be shorten via the isentropic efficiency. So that, the real temperature decreasing must be smaller than that forecasted by isentropic equations. The effect of viscosity on internal dissipation, wall friction, and heat losses is contained in this experimental factor.
I have not said nothing about M>>1. This have not sense, bacause wind generators never have peripheral velocities larger than sound speed. It would cause shock waves at the blade leading edge, and structural bendings. Only high powered rotors, like vapor turbines, have local #Mach supersonic in concrete points.
What happens at M<<<<<1 , I mean, at quasi-incompressible flow?. The Euler equation for isentropic flow is transformed into:
$$\tau=\frac{P_{1}-P_{2}}{\rho}$$. Why? This is because for an incompressible fluid: $$Tds=cdT$$ and $$dh=\frac{dP}{\rho}$$. This is, there is no variation of internal energy at all in an isentropic flowby.
In particular, non isentropic effects would cause a temperature increment. But, as you have said, this increment is very very small. In fact we can obtain an order of this increment:
$$\frac{\delta T}{T}=\frac{\delta P}{P}=\frac{\rho U^2}{P}=\gamma M^2$$
What does it mean?. At very low #Mach, increments of temperature are of the order of the #Mach powered to two.
To sum up, the temperature will remain roughly constant in an hydraulic turbine (hydroelectric power plants), a conventional house fan, compressors at M<<1, and maybe a wind generator at low rotaing speeds.
By contrast, high speed turbines, like nuclear power plant generators, car's turbocompressor, or aircraft's turbines, would cause a severe temperature decreasing.
But in all cases, temperature must be suffer a decreasing.
An example: A 900 KW electric generator connected to some vapor turbine, would need a mass flow of vapor of the order of 10 Kg/s. So that, the decreasing of temperatures will be of the order of:
$$\deltaT=\frac{900.000}{1000*10}=90ºC$$
Best regards for everybody.
Last edited: Aug 26, 2004
11. Aug 26, 2004
### Integral
Staff Emeritus
If what you say is true then there should be a easily measurable temperature drop through a large wind turbine. Is this the case? So why not set up a large wind turbine up wind of your house to not only to power it, but to cool it on a hot day? Sounds like with a large enough wind turbine we should be able to freeze water in the back draft, why has no one done this?
Last edited: Aug 26, 2004
12. Aug 26, 2004
### pervect
Staff Emeritus
Maybe I'm missing something, but wouldn't the mach number M of a wind turbine necessarily be very low?
The speed of sound at sea level is 761 miles/hour, and if the wind speed driving the turbine is a brisk 20 miles/hour wind, the mach number should be 20/761 = .02
So it seems to me that your own arguments suggest that a wind turbine shouldn't cause much temperature decrease.
Let's look at this another way. 20 mph is about 9 m/s, and the density of air is very roughly around 1kg/m^3 (it depends on temperature). So one cubic meter of air moving at 9 m/s would mass around 1kg and have an energy of 40 joules. Regardless of whether you use a Cp of 1 kilojoule/kg, or a Cv of .7 kilojoule/kg, the maximum possible temperature increase is much less than a degree, somewhere around .05 degrees.
Figures from
Last edited by a moderator: May 1, 2017
13. Aug 26, 2004
### krab
I'm with Integral. Clausius2, I'm not impressed with your arguments. Integral says temp. won't NECESSARILY decrease, meaning he can imagine mechanical techniques that turn the convective motion of wind into mechanical energy, without changing the air's temperature. Then you say "not quite", and mention a PARTICULAR method in which temperature does decrease. This is obviously not a valid refutation.
14. Aug 26, 2004
### Integral
Staff Emeritus
This is the sort of temperature change that I would be ok with. Due to mixing behind the blades it would not be measurable.
Does the water temperature change after a water wheel? I think not, if anything it will be increased due to a greater surface area being exposed to the (possiliby) warmer air. This is much more analogous to a wind turbine the any thermal engine.
Last edited: Aug 26, 2004
15. Aug 26, 2004
### Staff: Mentor
Well, I don't know what about what I said would have led you to start talking about nuclear reactors. A wind turbine is a wind turbine. Its only a heat engine in the very loosest sense of the term.
16. Aug 27, 2004
### Clausius2
Yes! It is just what I meant.
All the mathematical arguments posted before states that under the hypothesisof incompressible and isentropic flow, no temperatures variations are produced at all. Under the hypothesis ofincompressible and non-isentropic flow, there will be an small increasing of the temperature. Such increasing will be valued by the isentropic efficiency. Under the hyphotesis of compressible and non isentropic flow, there is an important temperature decreasing at #Mach of order 1, and less important decreasing for #Mach <<1.
Surely, the hypothesis nearest wind generators behaviour is the average between the first and the last ones.
Last edited: Aug 27, 2004
17. Aug 27, 2004
### Sariaht
18. Aug 27, 2004
### Clausius2
Yes it would be the case, but you will never be able of design a large wind turbine with a high speed rotor, because usual air streams don't have the neccesary velocity to enhance a #Mach important.
I was not talking about nuclear reactors itself, but about nuclear power plants. All of them are provided with a Rankine thermodynamic cicle. All of them have a turbine where vapor is despressurized (therefore cooled) enough to convert it in liquid. I think you know the temperature jump between core reactor exit and condenser entrance is not insignificant, and such jump is provided by a vapor turbine. It has nothing to do with wind generator (see the post just above this), because in wind gen. #Mach is usually much smaller than 1. Their behaviour is nearer incompressible flow. But it is an interesting question to discuss what happens when #Mach is much smaller than 1, then the flow is roughly incompressible. Such approaching states that in an ideal case there won't be a T decreasing at all (incompressible case), but if we allow a little compressible behaviour then we will see little T decreasings surely being counteracted with heat generation by means of friction. The total result will be a short variation of temperature.
Are you agree?.
Last edited: Aug 27, 2004
19. Aug 27, 2004
### Clausius2
What is not valid is your null capacity of reading what I've posted above. I have mentioned a PARTICULAR method, such method is extracted by flow equations themselves. THIS IS OBVIOUSLY A VALID REFUTATION. You should read all what I have posted. Surely I have difficults with the language, but science people are all of us understable. It would be a misconception not to mention about why temperature does or does not decrease. I would have a million of non-structured ideas of why this could happen. But the difference is I have showed you physically and mathematically where, when and how the temperature will decrease and will increase. Read carefully.
Last edited: Aug 27, 2004
20. Aug 27, 2004
### Integral
Staff Emeritus
Clausius2,
I must apologies for not reading your post closely enough myself. I missed the lines:
$$\frac{\delta T}{T}=\frac{\delta P}{P}=\frac{\rho U^2}{P}=\gamma M^2$$
and
Just doing some order of magnitude calculations I get the speed of sound in air to be ~1 km/s and the speed of wind to on the order of .01 km/s so am right in saying that the Mach number of wind is on the order of 10-5 ?
If so then by your approximations the temperature change for a wind turbine is on the order of 10-10 C. This is entirely unmeasurable therefore essentially none existent. Looks to me like the main error is yours when you said:
To my first post. Which was right on the money. | 2018-08-20 17:15:55 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6903505325317383, "perplexity": 1029.7189965085004}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221216718.53/warc/CC-MAIN-20180820160510-20180820180510-00533.warc.gz"} |
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04/30/17
#### Exponential and logarithmic models
The half-life of Palladium-100 is 4 days. After 24 days a sample of Palladium-100 has been reduced to a mass of 1 mg.
College Algebra Word Problems Algebra Word Problems
04/24/17
#### Rebecca picked 5/8 of a bushel of blueberries yesterday. Today, she picked 5/7 of a bushel of blueberries. How much more did she pick today than yesterday?
I need to know how many more she picked yesterday than today?
College Algebra Word Problems
03/14/17
#### A grocery store bought milk for $2.60 per half gallon and stored it in two refrigerators. A grocery store bought milk for$2.60 per half gallon and stored it in two refrigerators. During the night, one refrigerator malfunctioned and ruined 15 half gallons. If the remaining milk is sold... more
College Algebra Word Problems
02/17/17
#### How many gallons of a 90% antifreeze solution must be mixed with 60 gallons of 25% antifreeze to get a mixture that is 80% antifreeze?
How many gallons of a 90% antifreeze solution must be mixed with 60 gallons of 25% antifreeze to get a mixture that is 80% antifreeze?
College Algebra Word Problems College Algebra
02/08/17
#### Solve the equation
A man has 40 ft. of fencing to put around a rectangular garden. If the length is 4 times the width, find the dimensions of his garden. length ftwidth ft
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Choose an expert and meet online. No packages or subscriptions, pay only for the time you need. | 2022-06-29 15:14:36 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2960459589958191, "perplexity": 2096.5747270757197}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103640328.37/warc/CC-MAIN-20220629150145-20220629180145-00123.warc.gz"} |
http://dharmath.blogspot.com/2014/09/inspired-by-problem-from-brilliant.html | ## Friday, September 5, 2014
### Rotating projected ball
Inspired by a problem from brilliant.org A solid spherical ball with radius $R$ is projected from a rough ground with a velocity $v$ at angle $\theta$ with the horizontal, traveling to the right. At the same time it is given an angular velocity of $\omega$ clockwise such that the axis of rotation is perpendicular to the plane of projection.
At some point, the ball bounced on the ground, with coefficient of restitution $\frac{1}{2}$ and coefficient of friction is enough so that the ball did not slip while touching the ground. Find the initial angle such that the horizontal distance traveled by the ball to the point of second bounce be maximized.
### Solution
The initial velocity components are $v_x = v\cos\theta,v_y = v \sin \theta$. When the ball first touches the ground again, it will do so after a $t = 2v_y/g$ time, covering a distance of $d_1=tv_x = 2v_xv_y/g$.
At that point, the new velocity will have components $v'_x,v'_y$ respectively. We will calculate each component separately.
The vertical velocity is affected by the coeff. of restitution. Since the ground is assumed to be stationary (mass of the ball << mass of the Earth), then $v'_y = v_y / 2$
The horizontal velocity is affected by the friction. At the point of contact, the ball tends to go to the left, so the friction tends to go to the right. Suppose that the impulse caused by friction is $I$, then we have the following:
Linearly, the friction helps the linear motion: $$I = M(v'_x - v_x)$$ Rotationally, the torque caused by the friction slows down the angular motion: $$IR = \frac{2}{5}MR^2 (\omega - \frac{v'_x}{R})$$ (the last term is because we're assuming that the ball doesn't slip when it touches the ground). Solving for $v'_x$ we have: $$MR(v'_x - v_x) = \frac{2}{5}MR^2 (\omega - \frac{v'_x}{R})$$ $$v'_x - v_x = \frac{2}{5}R (\omega - \frac{v'_x}{R})$$ $$v'_x - v_x = \frac{2}{5}R\omega - \frac{2v'_x}{5}$$ $$v'_x = \frac{5}{7}( v_x+\frac{2}{5}R\omega ) = \frac{5v_x+2R\omega}{7}$$ Similar to the distance to the first bounce, the distance between first and second bounce is $$d_2 = 2v'_x v'_y/g = \frac{v_y(5v_x+2R\omega)}{7g}$$ Total distance is therefore: $$d_1+d_2 = \frac{2v_xv_y}{g} + \frac{v_y(5v_x+2R\omega)}{7g} = \frac{19v_xv_y + 2Rv_y\omega}{7g}$$ $$= \frac{v}{7g}(19v\sin\theta\cos\theta + 2R\sin\theta\omega)$$ Now we wish to find $\theta$ that maximizes: $$19v\sin\theta\cos\theta + 2R\omega\sin\theta$$ which can be done in various ways (calculus, double-angle formula etc). | 2018-03-22 15:53:45 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8952020406723022, "perplexity": 210.74392948372832}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257647892.89/warc/CC-MAIN-20180322151300-20180322171300-00564.warc.gz"} |
https://chemistry.stackexchange.com/questions/75396/why-add-acid-to-rid-of-soluble-carbonate-or-hydroxide-impurities-will-it-not-al | # Why add acid to rid of soluble carbonate or hydroxide impurities? Will it not also react with halide I'm testing for
When identifying metal halides with silver ions, the positive ion $\ce{Ag+}$ will form $\ce{AgX}$ with halide in a solution. However, $\ce{H+}$ is added (typically in the form of an acid like $\ce{HNO3}$) to solution to get rid of $\ce{OH-}$ and $\ce{CO3^{2-}}$ ions first. Will the $\ce{H+}$ not also react with the halide $\ce{X-}$?
Hydrogen halides are strong acids, meaning that their conjugate bases $\ce{X-}$ are extremely weak. Thus, adding $\ce{H+}$ will neutralize stronger bases such as $\ce{OH-}$ first before $\ce{X-}$. | 2019-07-21 06:56:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6663104891777039, "perplexity": 2399.689378819202}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195526931.25/warc/CC-MAIN-20190721061720-20190721083720-00521.warc.gz"} |
https://cforall.uwaterloo.ca/trac/changeset/34a6b2e659212c174177b49d4c952d1b2fe26cf0/doc | # Changeset 34a6b2e for doc
Ignore:
Timestamp:
Sep 25, 2018, 5:33:02 PM (4 years ago)
Branches:
aaron-thesis, arm-eh, cleanup-dtors, deferred_resn, enum, forall-pointer-decay, jacob/cs343-translation, jenkins-sandbox, master, new-ast, new-ast-unique-expr, no_list, persistent-indexer, pthread-emulation, qualifiedEnum
Children:
Parents:
461eed2 (diff), a32346b (diff)
Note: this is a merge changeset, the changes displayed below correspond to the merge itself.
Use the (diff) links above to see all the changes relative to each parent.
Message:
Merge branch 'master' of plg2:software/cfa/cfa-cc
Location:
doc
Files:
6 edited
Unmodified
Removed
• ## doc/proposals/virtual.txt
r461eed2 first polymorphic parameter). Once a function in a trait has been marked as virtual it defines a new function that takes in that trait's reference and then dynamically calls the underlying type implementation. Hence a trait reference becomes a kind of abstract type, cannot be directly instantiated but can still be used. Instances of a trait are created by wrapping an existing instance of a type that implements that trait. This wrapper includes all the function pointers and other values required to preform the dynamic look-up. These are chosen by the normal look-up rules at the point of abstraction. One of the limitations of this design is that it does not support double is also restricted, initially forbidden, see extension. Ownership of the underlying structure is also a bit of a trick. Considering the use cases for trait object, it is probably best to have the underlying object be heap allocated and owned by the trait object. Extension: Multi-parameter Virtual Traits: context, for instance if the cast occurs on the right hand side of an assignment. Function look-up follows the same rules as relaxed (behavioural) inheritance. Traits can be upcast and down cast without losing information unless the trait is cast down to a structure. Here there are two options. Abstraction Time Binding: The more efficient and consistant with other parts of CFA. Only the trait types use dynamic look-up, if converveted back into a structure the normal static look-up rules find the function at compile time. Casting down to a structure type can then result in the loss of a set of bindings. Construction Time Binding: For more consistant handling of the virtual structs, they are always considered wrapped. Functions are bound to the instance the moment it is constructed and remain unchanged throughout its lifetime, so down casting does not lose information. (We will have to decide between one of these two.) Extension: Multiple Parents We have so far been silent on how the vtable is created, stored and accessed. Creation happens at compile time. Function pointers are found by using the same best match rules as elsewhere (additional rules for defaults from the parent may or may not be required). For strict virtual this must happen at the global scope and forbidding static functions, to ensure that a single unique vtable is created. Similarly, there may have to be stricter matching rules for the functions that go into the vtable, possibly requiring an exact match. Relaxed virtual could relax both restrictions, if we allow different vtable at different conversion (struct to trait reference) sites. If it is allowed local functions being bound to a vtable could cause issues when they go out of scope, however this should follow the lifetime rules most C programs already follow implicitly. Most vtables should be stored statically, the only exception being some of the relaxed vtables that could have local function pointers. These may be able to be stack allocated. All vtables should be immutable and require no manual cleanup. The vtables for the two types might be handled slightly differently and then there is also the hierarchy data for virtual casts. The hierarchy data is simple conceptually. A single (exactly one copy) pointer for each type can act as the identity for it. The value of the pointer is its parent type, with the root pointer being NULL. Additional meta-data can accompany the parent pointer, such as a string name or the vtable fields. They types of each vtable can be constructed from the definitions of the traits (or internal nodes). The stand alone/base vtable is the same for both kinds of inheritance. It may be argumented differently however (include parent /this pointer in hierachal inheritance). Creation of the actual vtable is tricky. For classical single implementation semantics we would assemble the functions and create one vtable at compile time. However, not only does this not give CFA-like behaviour, it is impossible generally because types can satify assertions in different ways at different times and stop satifying them. A special set of harder rules could be used, instead we have decided to try creating multiple vtables for each type. The different vtables will all implement the same type but not always in the same way. Storage has some issues from creation. If the contents of every vtable could be determained at compile time they could all be created and stored statically. However since thunks can be constructed on the stack and become the best match, that isn't always possible. Those will have to be stored in dynamic memory. Which means that all vtables must be stored dynamically or there must be a way to determain which ones to free when the trait object is destroyed. Access has two main options:
• ## doc/theses/aaron_moss_PhD/phd/.gitignore
r461eed2 templates/ code/a.out thesis.pdf thesis.aux
• ## doc/theses/aaron_moss_PhD/phd/Makefile
r461eed2 introduction \ background \ generic-types \ type-environment \ resolution-heuristics \
r461eed2 tabsize=5, % N space tabbing xleftmargin=\parindentlnth, % indent code to paragraph indentation %mathescape=true, % LaTeX math escape in CFA code $...$ escapechar=\$, % LaTeX escape in CFA code keepspaces=true, % • ## doc/theses/aaron_moss_PhD/phd/frontpgs.tex r461eed2 % L I S T O F F I G U R E S % ----------------------------- % \addcontentsline{toc}{chapter}{List of Figures} % \listoffigures % \cleardoublepage % \phantomsection % allows hyperref to link to the correct page \addcontentsline{toc}{chapter}{List of Figures} \listoffigures \cleardoublepage \phantomsection % allows hyperref to link to the correct page % GLOSSARIES (Lists of definitions, abbreviations, symbols, etc. provided by the glossaries-extra package) • ## doc/theses/aaron_moss_PhD/phd/generic-types.tex r461eed2 \label{generic-chap} Talk about generic types. Pull from Moss~\etal\cite{Moss18}. A significant shortcoming in standard C is the lack of reusable type-safe abstractions for generic data structures and algorithms. Broadly speaking, there are three approaches to implement abstract data structures in C. One approach is to write bespoke data structures for each context in which they are needed. While this approach is flexible and supports integration with the C type checker and tooling, it is also tedious and error prone, especially for more complex data structures. A second approach is to use !void*!-based polymorphism, \eg{} the C standard library functions !bsearch! and !qsort!, which allow for the reuse of common functionality. However, basing all polymorphism on !void*! eliminates the type checker's ability to ensure that argument types are properly matched, often requiring a number of extra function parameters, pointer indirection, and dynamic allocation that is otherwise not needed. A third approach to generic code is to use preprocessor macros, which does allow the generated code to be both generic and type checked, but errors in such code may be difficult to locate and debug. Furthermore, writing and using preprocessor macros is unnatural and inflexible. Figure~\ref{bespoke-generic-fig} demonstrates the bespoke approach for a simple linked list with !insert! and !head! operations, while Figure~\ref{void-generic-fig} and Figure~\ref{macro-generic-fig} show the same example using !void*!- and !#define!-based polymorphism, respectively. \begin{figure} \begin{cfa} struct int_list { int value; struct int_list* next; }; void int_list_insert( struct int_list** ls, int x ) { struct int_list* node = malloc(sizeof(struct int_list)); node->value = x; node->next = *ls; *ls = node; } int int_list_head( const struct int_list* ls ) { return ls->value; }$\C[\textwidth]{// all code must be duplicated for every generic instantiation}$struct string_list { const char* value; struct string_list* next; }; void string_list_insert( struct string_list** ls, const char* x ) { struct string_list* node = malloc(sizeof(struct string_list)); node->value = x; node->next = *ls; *ls = node; } const char* string_list_head( const struct string_list* ls ) { return ls->value; }$\C[\textwidth]{// use is efficient and idiomatic}$int main() { struct int_list* il = NULL; int_list_insert( &il, 42 ); printf("%d\n", int_list_head(il)); struct string_list* sl = NULL; string_list_insert( &sl, "hello" ); printf("%s\n", string_list_head(sl)); } \end{cfa} \caption{Bespoke code for linked list implementation.} \label{bespoke-generic-fig} \end{figure} \begin{figure} \begin{cfa} // single code implementation struct list { void* value; struct list* next; };$\C[\textwidth]{// internal memory management requires helper functions}$void list_insert( struct list** ls, void* x, void* (*copy)(void*) ) { struct list* node = malloc(sizeof(struct list)); node->value = copy(x); node->next = *ls; *ls = node; } void* list_head( const struct list* ls ) { return ls->value; }$\C[\textwidth]{// helpers duplicated per type}$void* int_copy(void* x) { int* n = malloc(sizeof(int)); *n = *(int*)x; return n; } void* string_copy(void* x) { return strdup((const char*)x); } int main() { struct list* il = NULL; int i = 42; list_insert( &il, &i, int_copy ); printf("%d\n", *(int*)list_head(il));$\C[2in]{// unsafe type cast}$struct list* sl = NULL; list_insert( &sl, "hello", string_copy ); printf("%s\n", (char*)list_head(sl));$\C[2in]{// unsafe type cast}$} \end{cfa} \caption{\lstinline{void*}-polymorphic code for linked list implementation.} \label{void-generic-fig} \end{figure} \begin{figure} \begin{cfa}$\C[\textwidth]{// code is nested in macros}$#define list(N) N ## _list #define list_insert(N) N ## _list_insert #define list_head(N) N ## _list_head #define define_list(N, T)$\C[0.25in]{ \textbackslash }$struct list(N) { T value; struct list(N)* next; };$\C[0.25in]{ \textbackslash }\C[0.25in]{ \textbackslash }$void list_insert(N)( struct list(N)** ls, T x ) {$\C[0.25in]{ \textbackslash }$struct list(N)* node = malloc(sizeof(struct list(N)));$\C[0.25in]{ \textbackslash }$node->value = x; node->next = *ls;$\C[0.25in]{ \textbackslash }$*ls = node;$\C[0.25in]{ \textbackslash }$}$\C[0.25in]{ \textbackslash }\C[0.25in]{ \textbackslash }$T list_head(N)( const struct list(N)* ls ) { return ls->value; } define_list(int, int);$\C[3in]{// defines int\_list}$define_list(string, const char*);$\C[3in]{// defines string\_list}\C[\textwidth]{// use is efficient, but syntactically idiosyncratic}$int main() { struct list(int)* il = NULL;$\C[3in]{// does not match compiler-visible name}$list_insert(int)( &il, 42 ); printf("%d\n", list_head(int)(il)); struct list(string)* sl = NULL; list_insert(string)( &sl, "hello" ); printf("%s\n", list_head(string)(sl)); } \end{cfa} \caption{Macros for generic linked list implementation.} \label{macro-generic-fig} \end{figure} \CC{}, Java, and other languages use \emph{generic types} to produce type-safe abstract data types. Design and implementation of generic types for \CFA{} is the first major contribution of this thesis, a summary of which is published in \cite{Moss18}. \CFA{} generic types integrate efficiently and naturally with the existing polymorphic functions in \CFA{}, while retaining backward compatibility with C in layout and support for separate compilation. A generic type can be declared in \CFA{} by placing a !forall! specifier on a !struct! or !union! declaration, and instantiated using a parenthesized list of types after the generic name. An example comparable to the C polymorphism examples in Figures~\ref{bespoke-generic-fig}, \ref{void-generic-fig}, and \ref{macro-generic-fig} can be seen in Figure~\ref{cfa-generic-fig} \TODO{test this code}. \begin{figure} \begin{cfa} forall(otype T) struct list { T value; list(T)* next; };$\C[\textwidth]{// single polymorphic implementation of each function}\C[\textwidth]{// overloading reduces need for namespace prefixes}$forall(otype T) void insert( list(T)** ls, T x ) { list(T)* node = alloc();$\C{// type-inferring alloc}$(*node){ x, *ls };$\C{// concise constructor syntax}$*ls = node; } forall(otype T) T head( const list(T)* ls ) { return ls->value; }$\C[\textwidth]{// use is clear and efficient}$int main() { list(int)* il = 0; insert( &il, 42 );$\C{// inferred polymorphic T}\$ printf("%d\n", head(il)); list(const char*)* sl = 0; insert( &sl, "hello" ); printf("%s\n", head(sl)); } \end{cfa} \caption{\CFA{} generic linked list implementation.} \label{cfa-generic-fig} \end{figure} \section{Design} Though a number of languages have some implementation of generic types, backward compatibility with both C and existing \CFA{} polymorphism presented some unique design constraints for this project. The guiding principle was to maintain an unsurprising language model for C programmers without compromising runtime efficiency. A key insight for this design was that C already possesses a handful of built-in generic types (\emph{compound types} in the language of the standard\cit{}), notably pointer (!T*!) and array (!T[]!), and that user-definable generics should act similarly. \subsection{Related Work} One approach to the design of generic types is that taken by \CC{} templates\cit{}. The template approach is closely related to the macro-expansion approach to C polymorphism demonstrated in Figure~\ref{macro-generic-fig}, but where the macro-expansion syntax has been given first-class language support. Template expansion has the benefit of generating code with near-optimal runtime efficiency, as distinct optimizations can be applied for each instantiation of the template. On the other hand, template expansion can also lead to significant code bloat, exponential in the worst case\cit{}, and the costs of increased instruction cache pressure at runtime and wasted developer time when compiling cannot be discounted. The most significant restriction of the \CC{} template model is that it breaks separate compilation and C's translation-unit-based encapsulation mechanisms. Because a \CC{} template is not actually code, but rather a sort of recipe'' to generate code, template code must be visible at its call site to be used. C code, by contrast, only needs a !struct! or function declaration to call that function or use (by-pointer) values of that type, a desirable property to maintain for \CFA{}. Java\cit{} has another prominent implementation for generic types, based on a significantly different approach than \CC{}. The Java approach has much more in common with the !void*!-polymorphism shown in Figure~\ref{void-generic-fig}; since in Java nearly all data is stored by reference, the Java approach to polymorphic data is to store pointers to arbitrary data and insert type-checked implicit casts at compile-time. This process of \emph{type erasure} has the benefit of allowing a single instantiation of polymorphic code, but relies heavily on Java's object model and garbage collector. To use this model, a more C-like language such as \CFA{} would be required to dynamically allocate internal storage for variables, track their lifetime, and properly clean them up afterward. \TODO{Talk about Go, maybe Rust, Swift, etc. as well; specifically mention fat pointer'' polymorphism} \TODO{Talk about Cyclone as well, and why my generics are more powerful} \subsection{\CFA{} Generics} The generic types design in \CFA{} draws inspiration from both \CC{} and Java generics, capturing the better aspects of each. Like \CC{} template types, generic !struct!s and !union!s in \CFA{} have macro-expanded storage layouts, but, like Java generics, \CFA{} generic types can be used with separately-compiled polymorphic functions without requiring either the type or function definition to be visible to the other. The fact that the storage layout of any instantiation of a \CFA{} generic type is identical to that of the monomorphic type produced by simple macro replacement of the generic type parameters is important to provide consistent and predictable runtime performance, and to not impose any undue abstraction penalty on generic code. As an example, consider the following generic type and function \TODO{test this}: \begin{cfa} forall( otype R, otype S ) struct pair { R first; S second; }; pair(const char*, int) with_len( const char* s ) { return (pair(const char*), int){ s, strlen(s) }; } \end{cfa} In this example, !with_len! is defined at the same scope as !pair!, but it could be called from any context that can see the definition of !pair! and a declaration of !with_len!. If its return type was !pair(const char*, int)*!, callers of !with_len! would only need the declaration !forall(otype R, otype S) struct pair;! to call it, in accordance with the usual C rules for opaque types. !with_len! is itself a monomorphic function, returning a type that is structurally identical to !struct { const char* first; int second; }!, and as such could be called from C given an appropriate redeclaration and demangling flags. However, the definition of !with_len! depends on a polymorphic function call to the !pair! constructor, which only needs to be written once (in this case, implicitly by the compiler according to the usual \CFA{} constructor generation\cite{Moss18}) and can be re-used for a wide variety of !pair! instantiations. Since the parameters to this polymorphic constructor call are all statically known, compiler inlining can eliminate any runtime overhead of this polymorphic call. \CFA{} deliberately does not support \CC{}-style partial specializations of generic types. A particularly infamous example in the \CC{} standard library is !vector!, which is represented as a bitstring rather than the array representation of the other !vector! instantiations. Complications from this inconsistency (chiefly the fact that a single bit is not addressable, unlike an array element) make the \CC{} !vector! unpleasant to use in generic contexts due to the break in its public interface. Rather than attempting to plug leaks in the template specialization abstraction with a detailed method interface, \CFA{} takes the more principled position that two types with an unrelated data layout are in fact unrelated types, and should be handled with different code. Of course, to the degree that distinct types are similar enough to share an interface, the \CFA{} !trait! system allows one to be defined, and objects of types implementing that !trait! to be operated on in the same polymorphic functions. Since \CFA{} polymorphic functions can operate over polymorphic generic types, functions over such types can be partially or completely specialized using the usual overload selection rules. As an example, the !with_len! function above could be an optimization of the following more general function: \begin{cfa} forall(otype T, otype I | { I len(T); }) pair(T, I) with_len( T s ) { return (pair(T,I)){ s, len(s) }; } \end{cfa} \section{Implementation} % forall constraints on struct/union constrain default constructor (TODO check with Rob) % TODO discuss layout function algorithm, application to separate compilation % TODO put a static const field in for _n_fields for each generic, describe utility for separate compilation % TODO mention impetus for zero_t design % TODO mention use in tuple-type implementation % TODO put a static const field in for _n_fields for each generic, describe utility for separate compilation (actually no, you need to be able to see the type for it to be sized) % mention that tuples are implemented on top of a per-arity generic type \section{Performance Experiments} % TODO pull benchmarks from Moss et al. \section{Future Work} % mention future work adding non-type generic parameters, like ints % taking advantage of generic layout functions to provide field assertions in forall qualifiers % mention packed generic layouts (significantly more complex layout function, but possible)
Note: See TracChangeset for help on using the changeset viewer. | 2023-01-30 20:37:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.27663716673851013, "perplexity": 7310.88789208155}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499829.29/warc/CC-MAIN-20230130201044-20230130231044-00252.warc.gz"} |
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# Computational Learning Theory
## Presentation on theme: "Computational Learning Theory"— Presentation transcript:
Computational Learning Theory
Content Introduction Probably Learning an Approximately Correct Hypothesis Sample Complexity for Finite Hypothesis Spaces Sample Complexity for the Infinite Hypothesis Space The Mistake Bound Model of Learning Summary
Introduction Goal: Theoretical characterisation of the difficulty of several types of ML problems Capabilities of several types of ML algorithms Answer to the questions: Under what condition is successful learning possible and impossible? Under what condition is a particular ML algorithm assured to learn successfully? PAC: Identify classes of hypotheses that can or cannot be learned given a polynomial number of training examples Define a natural complexity measure for hypothesis space that allows to limit the number of training examples required for inductive learning
Introduction 2 Task Questions Given: Goal: Inductive learning of
Some training example , Space of candidate hypotheses H Goal: Inductive learning of Questions Sample complexity: How many training examples are needed for a learner to converge (with high probability) to a successful hypothesis? Computational complexity: How much computational effort is needed for a learner to converge (with high probability) to a successful hypothesis? Mistake bound: How many training examples will the learner misclassify before converging to a successful hypothesis?
Introduction 3 Possibility to set quantitative bounds on these measures, depending on attributes of the learning problem such as: The size or complexity of the hypothesis space considered by the learner The accuracy to which the target concept must be approximated The probability that the learner will output a successful hypothesis The manner in which training examples are represented to the learner
Content Introduction Probably Learning an Approximately Correct Hypothesis The Problem Setting Error of the Hypothesis PAC Learnability Sample Complexity for Finite Hypothesis Spaces Sample Complexity for the Infinite Hypothesis Space The Mistake Bound Model of Learning Summary
Probably Learning an Approximately Correct Hypothesis
PAC (Probably approximately correct) Probably learning a approximately correct solution Restriction: We only consider the case of learning boolean valued concepts from noise free training data Result can be extended to the more general scenario of learning real-valued target functions (Natarajan 1991) Result can be extended learning from certain types of noisy data (Laird 1988, Kearns and Vazirani 1994)
The Problem Setting Names
X - set of all possible instances over which target functions may be defined C - set of target concepts that our learner might be called upon to learn D - probability distribution which is generally not known to the learner as stationary: distribution does not change over time T - set of training examples H - space of candidate hypotheses Each target concept c in C corresponds to some subset of X or equivalent to some boolean-valued function Searched: After observing a sequence of training examples of c, L must output some h from H, which estimates c. Evaluation of success of L: Performance of h over new instances drawn randomly from X according to D
Error of the Hypothesis
True error: error of h with respect to c observable L can only observe the performance of h over a training example Training error: Fraction of training examples misclassified by h Analysis: how probable is it that the observed training error for h gives a misleading estimate of the true
PAC Learnability Goal: characterise classes of target concepts that can be reliably learned from a reasonable number of randomly drawn training examples and a reasonable amount of computation Possible definition of the success of the training: for search : Problems Multiple hypotheses consistent with the training examples Non-representative training set Definition of PAC-Learning: Consider a concept class called C defined over a set of instances X of length n and a learner L using a hypothesis space H. C is PAC-learnable by L using h if for all , the distribution D over X, an such that , and such that , the learner L will with a probability of at least output a hypothesis such that , in a time that is polynomial in , , n and size(c)
Content Introduction Probably Learning an Approximately Correct Hypothesis Sample Complexity for Finite Hypothesis Spaces Agnostic learning and Inconsistent Hypotheses Conjunctions of Boolean Literals Are PAC-Learnable Sample Complexity for the Infinite Hypothesis Space The Mistake Bound Model of Learning Summary
Sample Complexity for Finite Hypothesis Spaces
Definition: Sample complexity of the learning problem is the required number of training examples which are necessary for successful learning Depending on the constraints of the learning problem Consistent Learner: It outputs a hypothesis that perfectly fits the training data whenever possible Question: can a bound be derived for the number of training examples required by any consistent learner, independent of the specific alg. it uses to derive a consistent hypothesis? -> YES Significance of the version space : every consistent learner outputs a hypothesis belonging to the version space Therefore to limit the number of examples needed by any consistent learner we need only to limit the number of examples needed to assure that the version space contains no unacceptable hypotheses
Sample Complexity for Finite Hypothesis Spaces 2
Definition of -exhausted (Haussler 1988): Consider a hypothesis space H, target concept c, instance distribution D and a set of training examples T of c. The version space is said to be -exhausted with respect to c and D, if every hypothesis h in has an error less than with respect to c and D Picture: is 0.3 exhausted but not 0.1-exhausted
Sample Complexity for Finite Hypothesis Spaces 3
Theorem -exhausting the version space (Haussler 1988) If and D is a sequence of independent randomly drawn examples of some c then for any the probability that is not -exhausting (with respect to c) is less than or equal Important information: given the upper limit of the misclassification, using choose Hint 1: m grows linearly in , logarithmically in , and logarithmically in the size of H Hint 2: bound can be substantially overestimated:
Agnostic learning and Inconsistent Hypotheses
Problem: Consistent hypotheses are not always possible (H does not contain c) Agnostic learning: choose hypothesis where for example Searched , so that if => with high possibility
Agnostic learning and Inconsistent Hypotheses 2
Analogous: m independent coin flips showing “head” with some probability (m distinct trials of a Bernoulli experiment) Hoeffding boundary: characterise the deviation between the true probability of some event and its observed frequency over m independent trials => Requirement: The error of must be limited => Interpretation: Given choose: m depends logarithmically on H and on but m now grows as
Conjunction of Boolean Literals are PAC-Learnable
Example: C is the class where the target concept is described by a conjunction of boolean literals (a literal is any boolean variable or its negation) Is C PAC-learnable ->YES Any consistent learner will require only a polynomial number of training examples to learn any c in C Suggesting a specific algorithm that uses polynomial time per training example: Assumption H=C from the Theorem of Haussler follows: M grows linearly in the number of literals n, linearly in and logarithmically in
Conjunction of Boolean Literals are PAC-Learnable 2
Example with numbers: 10 boolean variables: Wanted: Safety 95% that the error of the hypothesis => algorithm with polynomial computing time Find-S Algorithm computes for each new positive training example the intersection of the literals shared by the current hypothesis and the new training example using time linear in n
Find-S: Finding a Maximally Specific Hypothesis
Use the more_general_than partial ordering: Begin with the most specific possible hypothesis in H Generalise this hypothesis each time it fails to cover an observed positive example 1. Initialise h to the most specific hypothesis in H 2. For each positive training instance x For each attribute constraint in h If the constrain is satisfied by x Then do nothing Else replace in h by the next more general constraint that is satisfied by x 3. Output hypothesis h
Find-S: Finding a Maximally Specific Hypothesis (Example)
1. Step: 2. Step: 1.Example + 1 Step: 3. Step: substituting a '?' in place of any attribute value in h that is not satisfied by new example 3.negative Example: FIND-S algorithm simply ignores every negative example 4.Step:
Content Introduction Probably Learning an Approximately Correct Hypothesis Sample Complexity for Finite Hypothesis Spaces Sample Complexity for the Infinite Hypothesis Space Sample Complexity and the VC Dimension The Vapnik-Chervonenkis Dimension The Mistake Bound Model of Learning Summary
Sample Complexity for Infinite Hypothesis Spaces
Disadvantage of the estimation before: Weak boundary In the case of an infinite hypothesis space it cannot be used Def: Shattering a Set of Instances A set of instances S is shattered by a hypothesis space H if and only if for every dichotomy of S there exists some hypothesis in H consistent with this dichotomy here the measuring is not based on the number of distinct hypotheses in |H| but on the number of distinct instances form X that can be completely discriminated using H
Shattering a Set of Instance
Follows from the definition: is not shattered by h <=> : from the aspect of all hypotheses
The Vapnik-Chervonenkis Dimension
S instance => different dichotomy Definition Vapnik-Chervonenkis Dimension: The Vapnik-Chervonenkis Dimension, VC(H), of hypothesis space H defined over the instance space X is the size of the largest finite subset of X shattered by H. If arbitrarily large finite sets of X can be shattered by H then Example: Let H the set of intervals on real numbers VC(H) =?
The Vapnik-Chervonenkis Dimension 2
Example: Let , H the set of linear decision surface in the x, y plane; VC(H) = shattering is obviously general case irregular special case no shattering possibility
Sample Complexity and the VC Dimension
Earlier: the number of randomly drawn examples suffice to probably approximately learn any c in C Theorem: Lower bound on sample complexity Consider any concept class C such that , any learner L, and any and , then there exists a distribution D and a target concept in C such that: if L observes fewer examples than then with probability at least , L outputs a hypothesis h having Hint: Both boundaries are logarithmic in and linear in VC(H)
Content Introduction Probably Learning an Approximately Correct Hypothesis Sample Complexity for Finite Hypothesis Spaces Sample Complexity for the Infinite Hypothesis Space The Mistake Bound Model of Learning The Mistake Bound for the FIND-S Algorithm The Mistake Bound for the HALVING Algorithm Optimal Mistake Bounds WEIGHTED-MAJORITY Algorithm Summary
The Mistake Bound Model of Learning
Mistake bound model: the learner is evaluated by the total number of mistakes it makes before it converges to the correct hypothesis. Problem Inductive learning It receives a set of training examples but after each x, the learner must predict the target value c(x) before it is shown the correct target value by the trainer Success: exact/PAC-learning How many mistakes will the learner make in its predictions before it learns the target concept. It is significant in practical application when the learning must be done while the system is in actual use Exact learning:
The Mistake Bound for the Find-S Algorithm
Assumption: , H: conjunction of up to n boolean literals and their negations Learning without noisy Find-S algorithm: Initialise h as the most specific hypothesis For each positive training instance Remove from h any literal that is not satisfied by x Output hypothesis h Can we prove the total number of mistakes that Find-S will make before exactly learning C ->YES Note: No error on negative instances Step 1: any additional error => maximal n+1 errors (case )
The Mistake Bound for the HALVING Algorithm
Refine the version space Maintaining the version space through majority vote decision Halving algorithm = + Every error => maximal Note: reduction of the version space also in the case of correct prediction Extension: WEIGHTED-MAJORITY Algorithm ( weighted vote)
Optimal Mistake Bounds
Question: What is the optimal mistake bound for an arbitrary concept class C – the lowest worst case mistake bound in respect to all possible learning algorithms Let H=C for algorithm A: For example: Littlestone (1987)
WEIGHTED-MAJORITY Algorithm
Generalisation of the Halving Algorithm Weighted vote among the pool of prediction algorithms Learns by altering the weight associated with each prediction algorithm Advantage: Accommodate inconsistent training data Note: => Halving algorithm Theorem: Relative mistake bound for WEIGHTED-MAJORITY Let T be any sequence of training examples, let A be any set of n prediction algorithms, and let k be the minimum number of mistakes made by any algorithm in A for the training sequence T. Then the number of mistakes over T made by the WEIGHTED-MAJORITY algorithm using is at most
WEIGHTED-MAJORITY Algorithm 2
denotes the prediction algorithm in the pool A of algorithms denotes the weight associated with For each i initialise For each training example Initialise and to 0 For each prediction algorithm If then If then predict If then predict If then predict 0 or 1 at random for c(x) For each prediction algorithm in A do If then
Content Introduction Probably Learning an Approximately Correct Hypothesis Sample Complexity for Finite Hypothesis Spaces Sample Complexity for the Infinite Hypothesis Space The Mistake Bound Model of Learning Summary
Summary PAC learning versus exact learning
Consistent and inconsistent hypothesis, agnostic learning VC-Dimension: complexity of hypothesis space - largest subset of instances that can be shattered Bound on the number of training examples sufficient for successful learning under the PAC model Mistake bound model: Analyse the number of training examples a learner will misclassify before it exactly learns the target concept WEIGHTED-MAJORITY Algorithm: combines the weighted votes of multiple prediction algorithms to classify new instances
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Ads by Google | 2018-02-20 12:03:55 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8572356700897217, "perplexity": 1474.1832780853138}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891812938.85/warc/CC-MAIN-20180220110011-20180220130011-00061.warc.gz"} |
https://stats.stackexchange.com/questions/144818/does-breimans-random-forest-use-information-gain-or-gini-index/167153 | # Does Breiman's random forest use information gain or Gini index?
I would like to know if Breiman's random forest (random forest in R randomForest package) uses as a splitting criterion (criterion for attribute selection) information gain or Gini index? I tried to find it out on http://www.stat.berkeley.edu/~breiman/RandomForests/cc_home.htm and in documentation for randomForest package in R. But the only thing I found is that Gini index can be used for variable importance computing.
• I also wonder if trees of random forest in randomForest package are binary or not. – somebody Apr 4 '15 at 16:51
The randomForest package in R by A. Liaw is a port of the original code being a mix of c-code(translated) some remaining fortran code and R wrapper code. To decide the overall best split across break points and across mtry variables, the code uses a scoring function similar to gini-gain:
$GiniGain(N,X)=Gini(N)-\frac{\lvert N_{1} \rvert }{\lvert N \rvert }Gini(N_{1})-\frac{\lvert N_{2} \rvert }{\lvert N \rvert }Gini(N_{2})$
Where $X$ is a given feature, $N$ is the node on which the split is to be made, and $N_{1}$ and $N_{2}$ are the two child nodes created by splitting $N$. $\lvert . \rvert$ is the number of elements in a node.
And $Gini(N)=1-\sum_{k=1}^{K}p_{k}^2$, where $K$ is the number of categories in the node
But the applied scoring function is not the exactly same, but instead a equivalent more computational efficient version. $Gini(N)$ and |N| are constant for all compared splits and thus omitted.
Also lets inspect the part if the sum of squared prevalence in a node(1) is computed as $\frac{\lvert N_{2} \rvert }{\lvert N \rvert }Gini(N_{2}) \propto |N_2| Gini(N_{2}) = |N_2| (1-\sum_{k=1}^{K}p_{k}^2 ) = |N_2| \sum \frac{nclass_{2,k}^2}{|N_2|^2}$
where $nclass_{1,k}$ is the class count of target-class k in daughter node 1. Notice $|N_2|$ is placed both in nominator and denominator.
removing the trivial constant $1-$ from equation such that best split decision is to maximize nodes size weighted sum of squared class prevalence...
score= $|N_1| \sum_{k=1}^{K}p_{1,k}^2 + |N_2| \sum_{k=1}^{K}p_{2,k}^2 = |N_1|\sum_{k=1}^{K}\frac{nclass_{1,k}^2}{|N_1|^2} + |N_2|\sum_{k=1}^{K}\frac{nclass_{2,k}^2}{|N_2|^2}$ $= \sum_{k=1}^{K}\frac{nclass_{2,k}^2}{1} |N_1|^{-1} + \sum_{k=1}^{K}\frac{nclass_{2,k}^2}{1} |N_1|^{-2}$ $= nominator_1/denominator_1 + nominator_2/denominator_2$
The implementation also allows for classwise up/down weighting of samples. Also very important when the implementation update this modified gini-gain, moving a single sample from one node to the other is very efficient. The sample can be substracted from nominators/denominators of one node and added to the others. I wrote a prototype-RF some months ago, ignorantly recomputing from scratch gini-gain for every break-point and that was slower :)
If several splits scores are best, a random winner is picked.
This answer was based on inspecting source file "randomForest.x.x.tar.gz/src/classTree.c" line 209-250 | 2019-09-15 20:52:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5583598613739014, "perplexity": 1463.3775215034352}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514572289.5/warc/CC-MAIN-20190915195146-20190915221146-00358.warc.gz"} |
https://solvedlib.com/novak-corp-was-experiencing-cash-flow-problems,14939 | # Novak Corp. was experiencing cash flow problems and was unable to pay its $115,000 account payable... ###### Question: Novak Corp. was experiencing cash flow problems and was unable to pay its$115,000 account payable to Pina Corp. when it fell due on September 30, 2020. Pina agreed to substitute a one-year note for the open account. The following two options were presented to Novak by Pina Corp.: Option 1: A one-year note for $115,000 due September 30, 2021. Interest at a rate of 10% would be payable at maturity. Option 2: A one-year non-interest-bearing note for$126,500. The implied rate of interest is 10%. Assume that Pina Corp. has a December 31 year end. Assuming Novak Corp. chooses Option 1, prepare the entries required on Pina Corp.'s books on September 30, 2020, December 31, 2020, and September 30, 2021. (Credit account titles are automatically indented when the amount is entered. Do not indent manually. If no entry is required, select "No Entry" for the account titles and enter o for the amounts. Round answers to 0 decimal places, e.g. 5,275. Record journal entries in the order presented in the problem.) Date Account Titles and Explanation Debit Credit September 30, 2020 Notes Receivable Accounts Receivable December 31, 2020 Interest Receivable Interest Income September 30, 2021 Cash Interest Receivable Interest Income Notes Receivable
Assuming Novak Corp. chooses Option 2, prepare the entries required on Pina Corp.'s books on September 30, 2020, December 31, 2020, and September 30, 2021. (Credit account titles are automatically indented when the amount is entered. Do not indent manually. If no entry is required, select "No Entry" for the account titles and enter o for the amounts. Round answers to O decimal places, e.g. 5,275. Record journal entries in the order presented in the problem.) Date Account Titles and Explanation Debit Credit September 30, 2020 - Notes Receivable Accounts Receivable December 31, 2020 - Notes Receivable Interest Income September 30, 2021 Notes Receivable Interest Income (To record interest income) September 30, 2021 Cash Notes Receivable (To record the collection of the note receivable)
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##### Question 13 of 15Vlev Pollcles Current Attempt in ProgressAt New York City; thc carth' 5 magnetic feldhas Vertlcr component0n5 0 $Tthat points donnnrd Mecrpendiculi-totho ground) and a horizontal component = 180 * 10$ Tthat points toward gcographic north (parallel to thc ground} What is the magnitude ofthe magnetic forceona 7.40-mlong straight wire that carrics curtent 0f 46.5 A perpendicularly Into Ihet grourd?NumberUnits
Question 13 of 15 Vlev Pollcles Current Attempt in Progress At New York City; thc carth' 5 magnetic feldhas Vertlcr component0n5 0 $Tthat points donnnrd Mecrpendiculi-totho ground) and a horizontal component = 180 * 10$ Tthat points toward gcographic north (parallel to thc ground} What is the... | 2023-03-21 23:43:51 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.385913223028183, "perplexity": 6291.748340733282}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296943747.51/warc/CC-MAIN-20230321225117-20230322015117-00610.warc.gz"} |
https://gamedev.stackexchange.com/questions/124273/object-smooth-orbital-limit-around-another-object-depending-on-mouse-drag | # object smooth orbital limit around another object depending on mouse drag
I am working in Unity.
I have special objects that can each be dragged freely around in 2d.
The objects cannot get too close to other objects (minimal Vector3.distance allowed is 1.5f).
I also have special connectors, which I can apply freely between 2 of any of these special objects. once 2 objects are connected, dragging them is further restricted.
The 2 objects cannot get too far between each other (max Vector3.distance allowed is 5f).
Both of these restrictions work, my issue is that once these restrictions occur, the objects are pretty much just stuck in place until dragged away from the restriction they tried to pass.
I want the object being dragged to orbit smoothly around the object it's connected to, once they reach any of the 2 restrictions.
The best example I can think of is the angry birds slingshot,
that once it reaches it's maximum pullback, you can just orbit along this max distance smoothly up or down depending on where you are trying to drag, this same smooth effect needs to occur both when too close and too far.
Hope I explained myself thoroughly I tried to find answers and try myself but could not reach any working solution.
The closest I think I got to a solution is with rays and the on point of ray function like in the video I provided below but I have no idea how could that be implemented in my sort of game.
thanks,
Remember the BBQ.
link to the sort of angry birds pulling mechanics I am talking about: https://unity3d.com/learn/tutorials/topics/physics/making-angry-birds-style-game-part-1
Image provided is very inaccurate to the boundaries of each restriction, just to make it seem more clear.
If I understand your question correctly this can be done using a bit of vector math. I've created an example in Javascript so please keep in mind that many of the mathematical operations I'm doing here can be done in Unity really easily using the built-in classes and methods.
The ball will follow your mouse when you hover it over the window. I recommend you run it in Chrome if possible. Also once you click "Run Snippet" I recommed you click the "Full page" link, the reason for this is because the mouse coordinates don't account for the scrollbar, so if you scroll down the mouse coordinates will no longer be accurate.
The function called "mousemove" is where the magic happens.
Ask away if you need any clarification.
var OUTER_BOUNDS = 160;
var INNER_BOUNDS = 40;
var canvas = document.getElementById("mainCanvas");
var ctx = canvas.getContext("2d");
var centerPos = {
x: (canvas.width / 2),
y: (canvas.height / 2)
};
var ballPos = {
x: -1000,
y: -1000
};
function mousemove(event) {
//This is where the good stuff happens.
//Set the ball to be at the position of the mouse.
ballPos.x = event.clientX;
ballPos.y = event.clientY;
//Calculate the distance from the center to the ball. This is simple linear algebra.
var dX = ballPos.x - centerPos.x;
var dY = ballPos.y - centerPos.y;
var distance = Math.sqrt(Math.pow(dX, 2) + Math.pow(dY, 2));
if (distance > 0) {
//Calculate a unit vector that points from the center towards the ball. This is the same as normalizing the vector.
var direction = {
x: dX / distance,
y: dY / distance
};
//If you square your inner and outer bounds you can perform this same comparison we're about to make without using the square root operation.
if (distance > (OUTER_BOUNDS - BALL_RADIUS)) {
//The ball is outside the outer bounds. We will calculate a new position for it using the unit vector above.
ballPos.x = centerPos.x + (direction.x * (OUTER_BOUNDS - BALL_RADIUS));
ballPos.y = centerPos.y + (direction.y * (OUTER_BOUNDS - BALL_RADIUS));
} else if (distance < (INNER_BOUNDS + BALL_RADIUS)) {
//The ball is too close to the center. We use the same method as above to calculate a new position for it.
ballPos.x = centerPos.x + (direction.x * (INNER_BOUNDS + BALL_RADIUS));
ballPos.y = centerPos.y + (direction.y * (INNER_BOUNDS + BALL_RADIUS));
}
}
};
function run() {
window.requestAnimationFrame(run);
draw();
}
function draw() {
//Draw background.
ctx.fillStyle = "#444444";
ctx.fillRect(0, 0, canvas.width, canvas.height);
//Draw the center ball.
strokeCircle(centerPos.x, centerPos.y, OUTER_BOUNDS, "#FF0000");
strokeCircle(centerPos.x, centerPos.y, INNER_BOUNDS, "#FF0000");
//Draw the moveable ball.
}
function fillCircle(x, y, radius, color) {
ctx.fillStyle = color;
ctx.beginPath();
ctx.arc(x, y, radius, 0, Math.PI * 2);
ctx.fill();
}
function strokeCircle(x, y, radius, color) {
ctx.strokeStyle = color;
ctx.lineWidth = 2;
ctx.beginPath();
ctx.arc(x, y, radius, 0, Math.PI * 2);
ctx.stroke();
}
run();
html,
body {
margin: 0;
}
<canvas id="mainCanvas" width="400" height="400"></canvas>
• This looks great!! thank you so much for the feedback and that code snippet is above and beyond! – Remember The BBQ Jun 21 '16 at 21:39
• I'll try implementing it to my game when I have the time and see if I'll need any clarification. Thanks again!! :) – Remember The BBQ Jun 21 '16 at 21:40
Instead of hard restrictions you can use opposing forces.
You want to simulate a rubber band between the two objects. That sort of force can be simulated by applying a push every frame that equals "distance to object" to the power of "your rubber band's springiness". If you want to increase the effective distance that object A can travel away from object B without being pulled back then you subtract a "distance constant" from that value and do a check at the end which sets any negative values for force to zero. At some other distance threshold you can set the rubber band to break.
• thanks for the feedback I will try to implement this idea when I have the time :) – Remember The BBQ Jun 21 '16 at 16:13
• pushing the object should be done with addforce? Because I don't have any physics in the game I am wondering if this would work if this is what you meant. – Remember The BBQ Jun 21 '16 at 16:13
• It would be ineffective now that I think about it since the object re positions itself on every LateUpdate() call based on the position of the mouse, wouldn't that just make the whole adding force meaningless? – Remember The BBQ Jun 21 '16 at 16:29
• unless by rubber band you mean using springjoint2d? – Remember The BBQ Jun 21 '16 at 16:51
• Yes, pushing should be done via addforce. It sounds like you'll have to change your mouse code as well so that it doesn't reposition itself, but rather is attracted to the point of the mouse click (perhaps raycasted out to some distance). – Tealr Jun 29 '16 at 6:06 | 2020-01-19 22:38:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3977533280849457, "perplexity": 1676.7543917131172}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250595282.35/warc/CC-MAIN-20200119205448-20200119233448-00382.warc.gz"} |
https://mathoverflow.net/questions/214830/extension-by-zero-in-sobolev-spaces/214833 | # Extension by zero in Sobolev spaces
Let $\Omega$ be an open bounded set of $R^n$, and let $\omega$ be an open subset of $\Omega$ s.t $\overline{\omega} \subset \Omega.$
For $f\in H_0^1(\omega)$, it is known that the extension of $f$ to $\Omega$ by $0$ is an element of $H_0^1(\Omega).$
I wonder if the result remains true when we replace $H_0^1$ with $H_0^1\cap H^2$.
It does not remain true. If $\omega=B(0,1)$ and $\Omega=B(0,2)$ and $f(x)=1-|x|^2$, then $f\in H^1_0(\omega)\cap H^2(\omega)$ but the extension by zero is not in $H^2(\Omega)$. | 2019-04-23 12:51:58 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9133818745613098, "perplexity": 31.438010304811165}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578602767.67/warc/CC-MAIN-20190423114901-20190423140901-00312.warc.gz"} |
https://www.primerpy.com/post/leetcoding/content/167-twosum/ | # 167. Two Sum
## Problem
• Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number.
• The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2.
• Note:
• You may assume that each input would have exactly one solution and you may not use the same element twice.
Example:
Input: numbers = [2,7,11,15], target = 9
Output: [1,2]
Explanation: The sum of 2 and 7 is 9. Therefore index1 = 1, index2 = 2.
## Thought Process
• create two pointers: head=0 and tail=n-1
• if sum == target, return head+1, tail+1 (since not zero based)
• if sum < target, move head up, head += 1, new sum = nums[head] + nums[tail]
• if sum > target, move tail down, tail -= 1, new sum = nums[head] + nums[tail]
• since we cannot use the same element twice, if head == tail, break loop and return false if the target is not found
## White Board
Below is the white board:
## Code
class Solution:
def two_sum(self, nums, target):
i, j = 0, len(nums) - 1
while i < j:
sum = nums[i] + nums[j]
if sum == target:
return (i+1, j+1)
elif sum < target:
i += 1
else:
j -= 1
return False
def test_two_sum():
import random
random.seed(42)
nums = random.sample(range(1, 100), 20)
print("Original Array: ")
print(nums)
nums = sorted(nums)
print("Sorted Array: ")
print(nums)
target = random.sample(range(1,100),1)[0]
print("Target: ", target)
ts = Solution().two_sum(nums, target)
print("Solution: ")
print(ts)
test_two_sum()
Original Array:
[82, 15, 4, 95, 36, 32, 29, 18, 14, 87, 70, 12, 76, 55, 5, 28, 30, 65, 78, 72]
Sorted Array:
[4, 5, 12, 14, 15, 18, 28, 29, 30, 32, 36, 55, 65, 70, 72, 76, 78, 82, 87, 95]
Target: 26
Solution:
(3, 4)
## Complexity
• Time complexity is O(n), two pointers loop from head and tail at the same time
• Space complexity is O(1), as I didn’t create new spaces
## Optimization
I think what I provided is the optimal solution. I took advantage of the property that the array is already sorted. No need to optimize for the sake of optimizing. | 2019-10-17 14:03:17 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.17956310510635376, "perplexity": 3047.1550844321114}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986675316.51/warc/CC-MAIN-20191017122657-20191017150157-00456.warc.gz"} |
https://pos.sissa.it/396/293/ | Volume 396 - The 38th International Symposium on Lattice Field Theory (LATTICE2021) - Oral presentation
An update on QCD+QED simulations with C* boundary conditions
J. Luecke*, L. Bushnaq, I. Campos, M. Catillo, A. Cotellucci, M.E.B. Dale, P. Fritzsch, M.K. Marinkovic, A. Patella and N. Tantalo
Full text: pdf
Pre-published on: May 16, 2022
Published on:
Abstract
We present two novelties in our analysis of fully dynamical QCD+QED ensembles with C* boundary conditions. The first one is the explicit computation of the sign of the Pfaffian. We present an algorithm that provides a significant speedup compared to traditional methods. The second one is a reweighting of the mass in the context of the RHMC. We have tested the techniques on both pure QCD and QCD+QED ensembles with pions at $m_{\pi^\pm}\approx400$ MeV, a lattice spacing of $a\approx0.05$ fm, a fine-structure constant of $\alpha_{\mathrm{R}}=0$ and $0.04$.
DOI: https://doi.org/10.22323/1.396.0293
How to cite
Metadata are provided both in "article" format (very similar to INSPIRE) as this helps creating very compact bibliographies which can be beneficial to authors and readers, and in "proceeding" format which is more detailed and complete.
Open Access
Copyright owned by the author(s) under the term of the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License. | 2022-06-28 05:37:53 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.334401398897171, "perplexity": 1830.6569644033973}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103355949.26/warc/CC-MAIN-20220628050721-20220628080721-00089.warc.gz"} |
https://lifemath.wordpress.com/2007/05/12/on-the-size-of-shoe-racks/ | # On the size of shoe racks
On a web discussion board, I conjectured the following:
The diversity of the size of women’s shoe-racks can be expressed in mathematical fashion as a distribution of a particular form, called a “power law“, meaning that the probability of a woman’s shoe-rack attaining a certain size $x$ is proportional to $(1/x)^y$, where $y \ge 1$.
When a distribution of some property has a power law form, the system looks the same at all length scales. Therefore, if one were to look at the distribution of rack-sizes for one arbitrary range, say, just racks with 100 to 1000 shoes, it would look the same as for a different range, say, 1 to 10 shoes. In other words, “zooming” in or out in the distribution, one keeps obtaining the same result. It also means that if one can determine the distribution of shoes per rack for a range of shoes, one can then predict the distribution for many other ranges.
Equally interesting, power law distributions have very long tails, meaning there is a non-zero probability of finding racks extremely large compared to the average. This finite probability of finding large racks is quite striking and can be illustrated by the example of the heights of individuals following the familiar normal distribution. It would be very surprising to find someone measuring two or three times the average human height of 5’10”. On the other hand, a power law distribution makes it possible to find a rack many times larger than average. Power laws also imply that the system’s average behavior is not typical. A typical size is one that is encountered most frequently; the average is the sum of all the sizes divided by the number of women. If one were to select a group of shoe-racks at random and count the number of shoes in each of them, the majority would be smaller than average.
A similar analysis can be carried out for women’s cloth cupboards, with $y \in [3, \infty)$.
Men don’t have shoe-racks! So an analogy between the show-rack size and the size of the cupboard would not be possible. But here is one fact that gives you the basic idea: I’ve two pairs of jeans, one is torn at several places. I have a few Ts, a few bought and a few won in different competitions held at my place. Apart from these prized possessions, I also have a couple of shorts and a track suite, a couple of towels, a few pairs of rotten socks and a few undies. | 2017-06-28 07:17:15 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 4, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7185444831848145, "perplexity": 460.5978236388218}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128322873.10/warc/CC-MAIN-20170628065139-20170628085139-00098.warc.gz"} |
https://warwick.ac.uk/fac/sci/physics/research/epp/people/alex_ward/ | Skip to main content Skip to navigation
# Alex Ward
### Ph.D. Research
2nd year Ph.D. candidate with the Monash-Warwick Alliance in Particle PhysicsLink opens in a new window. My research is conducted at the Large Hadron Collider beauty (LHCbLink opens in a new window) experiment based at CERNLink opens in a new window, Geneva. I am focusing primarily on rare electroweak penguin decays, specifically the angular analysis of the $B0 (Bs0) -> pi+ pi- mu+ mu-$ decay modes.
My supervisors are Dr Michal KrepsLink opens in a new window and Dr. Tom BlakeLink opens in a new window at the University of Warwick and Professor Ulrik EgedeLink opens in a new window at Monash University.
### Research Topic
The study of Rare Decays within particle physics has produced some interesting results in recent years which have produced tension with the Standard Model and promising signs of New Physics. The particle decays pertinent to my research project involve same-charge, namely $b -> s$ and $b -> d$ quark transitions. These FCNC (Flavour Changing Neutral Currents) decays are rare and occur only at loop level [1], shown below in examples of the penguin and box Feynman diagrams.
My research involves the analysis of the angular distribution of the $B0 (Bs0) -> pi+ pi- mu+ mu-$ decay modes. The measurement of multiple observables related to the angles between the decay components, as shown below, will provide a comparison with the predictions of the Standard Model [2]. The nature of the 4-body decay gives sensitivity to potential New Physics and is a promising test of the current predictions [3].
### Previous Education
I graduated in 2017 from Swansea University having conducted a final year project entitled 'Detecting Radiation from a CERN Detector' in which I used a portable Medipix pixel detector to measure radiation attenuation.
In 2019 I embarked upon a Masters by Research degree at the University of Glasgow. Here I gained my first experience of working within the LHCb experiment where I researched the lifetime of charm baryons, culminating in the writing of my thesis entitled 'Measurement of the $Omega_c0$ baryon lifetime with LHCb'.
This experience led me to having a vested interest in the research of the LHCb and in pursuing a Ph.D. within the collaboration.
### References
[1] T. Blake et al., Rare b-hadron decays at the lhc, Annu. Rev. Nucl. Part. Sci. 65 (2015) 113. [arxiv.1501.03309Link opens in a new window]
[2] C. Bobeth, G. Hiller, and G. Piranishvili, Cp asymmetries in $B -> Kbar* (-> Kbar pi) l+ l-$ and Untagged $B_s (Bbar_s) -> phi (-> K+ K-) l+ l-$ decays at nlo, JHEP 0807 (2008) 106. [arxiv.0805.2525]Link opens in a new window
[3] L. Cappiello, O. Cata, and G. DAmbrosio, Standard model prediction and new physics tests for $D0 -> hhll$ JHEP 2013 (2013) . [arxiv.9808289Link opens in a new window]
### Contact Details:
Email: alex dot ward at warwick dot ac dot uk
Office: P449
Address: Department of Physics, University of Warwick, Coventry, CV4 7AL
Roles:
• LHCb UK Student Meeting Organiser
• LHCb UK Social Media Coordinator | 2022-08-13 14:41:13 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 8, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6505332589149475, "perplexity": 3822.8436237576284}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571959.66/warc/CC-MAIN-20220813142020-20220813172020-00578.warc.gz"} |
https://web2.0calc.com/questions/what-is-x-if-the-equation-is-2-tan-1-x-160-tan-1-x-60 | +0
# what is x if the equation is 2(tan-1(x/160)) = tan-1(x/60)?
0
436
3
what is x if the equation is 2(tan-1(x/160)) = tan-1(x/60)?
Guest Jul 9, 2015
### Best Answer
#1
+19206
+10
what is x if the equation is 2(tan-1(x/160)) = tan-1(x/60) ?
$$\small{\text{ \begin{array}{rclrcl} 2 \cdot \left[ \arctan{ \left( \dfrac{ x } { 160 } \right) } \right] &=& \arctan{ \left( \dfrac{ x } { 60 } \right) } \qquad | \qquad \varphi = \arctan{ \left( \dfrac{ x } { 160 } \right) } \\\\ 2\cdot \varphi &=& \arctan{ \left( \dfrac{ x } { 60 } \right) } \qquad |\qquad \tan{()}\\\\ \tan{ (2\cdot \varphi) } &=& \dfrac{x}{60} \\\\ && &\tan{ (2\cdot \varphi) } &=& \dfrac{ 2\cdot\tan{(\varphi)} } { 1-[\tan{(\varphi)]^2} } \\\\\ && &\tan{ (2\cdot \varphi) } &=& \dfrac{ 2\cdot\tan{( \arctan{ \left( \dfrac{ x } { 160 } \right) } )} } { 1-[\tan{( \arctan{ \left( \dfrac{ x } { 160 } \right) } )]^2} } \\\\\ && &\tan{ (2\cdot \varphi) } &=& \dfrac{ 2\cdot \dfrac{ x } { 160 } } { 1- \left( \dfrac{ x } { 160 } \right)^2 } \\\\\ \dfrac{ 2\cdot \dfrac{ x } { 160 } } { 1- \left( \dfrac{ x } { 160 } \right)^2 } &=& \dfrac{x}{60} \\\\ \dfrac{ 1 } { 1- \left( \dfrac{ x } { 160 } \right)^2 } &=& \dfrac{4}{3} \\\\ 1- \left( \dfrac{ x } { 160 } \right)^2 &=& \dfrac{3}{4} \\\\ \left( \dfrac{ x } { 160 } \right)^2 &=& \dfrac{1}{4} \qquad | \qquad \pm\sqrt{}\\\\ \dfrac{ x } { 160 } &=& \pm0.5 \\\\ x &=& \pm0.5 \cdot 160 \\\\ \mathbf{x_1} & \mathbf{=} & \mathbf{80} \\\\ \mathbf{x_2} & \mathbf{=} & \mathbf{-80} \\\\ \end{array} }}$$
heureka Jul 10, 2015
Sort:
### 3+0 Answers
#1
+19206
+10
Best Answer
what is x if the equation is 2(tan-1(x/160)) = tan-1(x/60) ?
$$\small{\text{ \begin{array}{rclrcl} 2 \cdot \left[ \arctan{ \left( \dfrac{ x } { 160 } \right) } \right] &=& \arctan{ \left( \dfrac{ x } { 60 } \right) } \qquad | \qquad \varphi = \arctan{ \left( \dfrac{ x } { 160 } \right) } \\\\ 2\cdot \varphi &=& \arctan{ \left( \dfrac{ x } { 60 } \right) } \qquad |\qquad \tan{()}\\\\ \tan{ (2\cdot \varphi) } &=& \dfrac{x}{60} \\\\ && &\tan{ (2\cdot \varphi) } &=& \dfrac{ 2\cdot\tan{(\varphi)} } { 1-[\tan{(\varphi)]^2} } \\\\\ && &\tan{ (2\cdot \varphi) } &=& \dfrac{ 2\cdot\tan{( \arctan{ \left( \dfrac{ x } { 160 } \right) } )} } { 1-[\tan{( \arctan{ \left( \dfrac{ x } { 160 } \right) } )]^2} } \\\\\ && &\tan{ (2\cdot \varphi) } &=& \dfrac{ 2\cdot \dfrac{ x } { 160 } } { 1- \left( \dfrac{ x } { 160 } \right)^2 } \\\\\ \dfrac{ 2\cdot \dfrac{ x } { 160 } } { 1- \left( \dfrac{ x } { 160 } \right)^2 } &=& \dfrac{x}{60} \\\\ \dfrac{ 1 } { 1- \left( \dfrac{ x } { 160 } \right)^2 } &=& \dfrac{4}{3} \\\\ 1- \left( \dfrac{ x } { 160 } \right)^2 &=& \dfrac{3}{4} \\\\ \left( \dfrac{ x } { 160 } \right)^2 &=& \dfrac{1}{4} \qquad | \qquad \pm\sqrt{}\\\\ \dfrac{ x } { 160 } &=& \pm0.5 \\\\ x &=& \pm0.5 \cdot 160 \\\\ \mathbf{x_1} & \mathbf{=} & \mathbf{80} \\\\ \mathbf{x_2} & \mathbf{=} & \mathbf{-80} \\\\ \end{array} }}$$
heureka Jul 10, 2015
#2
+26625
+5
There is also the trivial solution x= 0
.
Alan Jul 10, 2015
#3
+92194
+5
I have only just now had a chance to look at your answer Heureka.
I really like it.
I have added this thread address to our "Great Answers to Learn From" sticky thread. :)
Melody Jul 16, 2015
### 16 Online Users
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. See details | 2018-04-22 00:58:22 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9271973967552185, "perplexity": 10200.426737773658}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125945484.57/warc/CC-MAIN-20180422002521-20180422022521-00151.warc.gz"} |
https://pressbooks.online.ucf.edu/phy2048tjb/chapter/3-3-first-law-of-thermodynamics/ | Chapter 3 The First Law of Thermodynamics
# 3.3 First Law of Thermodynamics
OpenStax and Paula Herrera-Siklody
Learning Objectives
By the end of this section, you will be able to:
• State the first law of thermodynamics and explain how it is applied
• Explain how heat transfer, work done, and internal energy change are related in any thermodynamic process
Now that we have seen how to calculate internal energy, heat, and work done for a thermodynamic system undergoing change during some process, we can see how these quantities interact to affect the amount of change that can occur. This interaction is given by the first law of thermodynamics. British scientist and novelist C. P. Snow (1905–1980) is credited with a joke about the four laws of thermodynamics. His humorous statement of the first law of thermodynamics is stated “you can’t win,” or in other words, you cannot get more energy out of a system than you put into it. We will see in this chapter how internal energy, heat, and work all play a role in the first law of thermodynamics.
Suppose Q represents the heat exchanged between a system and the environment, and W is the work done by or on the system. The first law states that the change in internal energy of that system is given by $Q-W$. Since added heat increases the internal energy of a system, Q is positive when it is added to the system and negative when it is removed from the system.
When a gas expands, it does work and its internal energy decreases. Thus, W is positive when work is done by the system and negative when work is done on the system. This sign convention is summarized in Table 3.1. The first law of thermodynamics is stated as follows:
First Law of Thermodynamics
Associated with every equilibrium state of a system is its internal energy $E_\text{int}$. The change in $E_\text{int}$ for any transition between two equilibrium states is
$$$\tag{3.7} \Delta E_\text{int} = Q-W$$$
where Q and W represent, respectively, the heat exchanged by the system and the work done by or on the system.
Process Convention Heat added to system $Q > 0$ Heat removed from system $Q < 0$ Work done by system $W > 0$ Work done on system $W < 0$
The first law is a statement of energy conservation. It tells us that a system can exchange energy with its surroundings by the transmission of heat and by the performance of work. The net energy exchanged is then equal to the change in the total mechanical energy of the molecules of the system (i.e., the system’s internal energy). Thus, if a system is isolated, its internal energy must remain constant.
Although Q and W both depend on the thermodynamic path taken between two equilibrium states, their difference $Q-W$ does not. Figure 3.7 shows the pV diagram of a system that is making the transition from A to B repeatedly along different thermodynamic paths. Along path 1, the system absorbs heat Q1 and does work W1; along path 2, it absorbs heat Q2 and does work W2, and so on. The values of Qi and Wi may vary from path to path, but we have
$Q_1 - W_1 = Q_2 - W_2 = \cdots = Q_i - W_i = \cdots,$
or
$\Delta E_\text{ent1} = \Delta E_\text{int2} = \cdots = \Delta E_{\text{int}i} = \cdots.$
That is, the change in the internal energy of the system between A and B is path independent. In the chapter on potential energy and the conservation of energy, we encountered another path-independent quantity: the change in potential energy between two arbitrary points in space. This change represents the negative of the work done by a conservative force between the two points. The potential energy is a function of spatial coordinates, whereas the internal energy is a function of thermodynamic variables. For example, we might write $E_\text{int}(T,p)$ for the internal energy. Functions such as internal energy and potential energy are known as state functions because their values depend solely on the state of the system.
Often the first law is used in its differential form, which is
$$$\tag{3.8} dE_\text{int} = dQ - dW.$$$
Here $dE_\text{int}$ is an infinitesimal change in internal energy when an infinitesimal amount of heat dQ is exchanged with the system and an infinitesimal amount of work dW is done by (positive in sign) or on (negative in sign) the system.
Example 3.2
Changes of State and the First Law
During a thermodynamic process, a system moves from state A to state B, it is supplied with 400 J of heat and does 100 J of work.
(a) For this transition, what is the system’s change in internal energy?
(b) If the system then moves from state B back to state A, what is its change in internal energy?
(c) If in moving from A to B along a different path, ${W^\prime}_{AB} = 400J$ of work is done on the system, how much heat does it absorb?
Strategy
The first law of thermodynamics relates the internal energy change, work done by the system, and the heat transferred to the system in a simple equation. The internal energy is a function of state and is therefore fixed at any given point regardless of how the system reaches the state.
Solution
a. From the first law, the change in the system’s internal energy is
$\Delta E_{\text{int}AB} = Q_{AB} - W_{AB} = 400\text{J} - 100\text{J} =300\text{J}.$
b. Consider a closed path that passes through the states A and B. Internal energy is a state function, so $\Delta E_\text{int}$ is zero for a closed path. Thus
$\Delta E_\text{int} = \Delta E_{\text{int}AB} + \Delta E_{\text{int}BA} = 0,$
and
$\Delta E_{\text{int}AB} = -\Delta_{\text{int}BA}.$
This yields
$\Delta E_{\text{int}BA} = -300\text{J}.$
c. The change in internal energy is the same for any path, so
$\begin{array}{rcl}\Delta E_{intAB} & = & \Delta {E^\prime}_{intAB} = {Q^\prime}_{AB} - {W^\prime}_{AB} ; \\ 300J & = & {Q^\prime}_{AB} - (-400J),\end{array}$
and the heat exchanged is
$Q\prime _{AB} = -100\text{J}.$
The negative sign indicates that the system loses heat in this transition.
Significance
When a closed cycle is considered for the first law of thermodynamics, the change in internal energy around the whole path is equal to zero. If friction were to play a role in this example, less work would result from this heat added. Example 3.3 takes into consideration what happens if friction plays a role.
Notice that in Example 3.2, we did not assume that the transitions were quasi-static. This is because the first law is not subject to such a restriction. It describes transitions between equilibrium states but is not concerned with the intermediate states. The system does not have to pass through only equilibrium states. For example, if a gas in a steel container at a well-defined temperature and pressure is made to explode by means of a spark, some of the gas may condense, different gas molecules may combine to form new compounds, and there may be all sorts of turbulence in the container—but eventually, the system will settle down to a new equilibrium state. This system is clearly not in equilibrium during its transition; however, its behavior is still governed by the first law because the process starts and ends with the system in equilibrium states.
Example 3.3
Polishing a Fitting
A machinist polishes a 0.50-kg copper fitting with a piece of emery cloth for 2.0 min. He moves the cloth across the fitting at a constant speed of 1.0 m/s by applying a force of 20 N, tangent to the surface of the fitting.
(a) What is the total work done on the fitting by the machinist?
(b) What is the increase in the internal energy of the fitting? Assume that the change in the internal energy of the cloth is negligible and that no heat is exchanged between the fitting and its environment.
(c) What is the increase in the temperature of the fitting?
Strategy
The machinist’s force over a distance that can be calculated from the speed and time given is the work done on the system. The work, in turn, increases the internal energy of the system. This energy can be interpreted as the heat that raises the temperature of the system via its heat capacity. Be careful with the sign of each quantity.
Solution
a. The power created by a force on an object or the rate at which the machinist does frictional work on the fitting is $\vec{{F}} \cdot \vec{{v}} = -Fv$. Thus, in an elapsed time $\Delta t$ (2.0 min), the work done on the fitting is
$\begin{array}{rcl}W & = & -Fv\Delta t = -(20N)(0.1m/s)(1.2 \times {10}^2 s) \\ & = & -2.4 \times {10}^3 J.\end{array}$
b. By assumption, no heat is exchanged between the fitting and its environment, so the first law gives for the change in the internal energy of the fitting:
$\Delta E_\text{int} = -W = 2.4 \times 10^3\text{J}.$
c. Since $\Delta E_\text{int}$ is path independent, the effect of the $2.4 \times 10^3\text{J}$ of work is the same as if it were supplied at atmospheric pressure by a transfer of heat. Thus,
$2.4 \times 10^3\text{J} = mc\Delta T = (0.50\text{kg})(3.9 \times 102\text{J/kg}\cdot ^\circ \text{C})\Delta T,$
and the increase in the temperature of the fitting is
$\Delta T = 12 ^\circ \text{C}$
where we have used the value for the specific heat of copper, $c = 3.9 \times 10^2\text{J/kg} ^\circ \text{C}$.
Significance
If heat were released, the change in internal energy would be less and cause less of a temperature change than what was calculated in the problem.
The quantities below represent four different transitions between the same initial and final state. Fill in the blanks.
Q (J) W (J) $\Delta E_\text{int}(\text{J})$ -80 -120 90 40 -40
Example 3.4
An Ideal Gas Making Transitions between Two States
Consider the quasi-static expansions of an ideal gas between the equilibrium states A and C of Figure 3.6. If 515 J of heat are added to the gas as it traverses the path ABC, how much heat is required for the transition along ADC? Assume that $p_{1}=2.10×10^5\text{N/m}^2$, $p_2 = 1.05 \times 10^5 \text{N/m}^2$, $V_1 = 2.25 \times 10^ {-3}m^3$, and $V_2 = 4.50 \times 10^{-3}m^3$.
Strategy
The difference in work done between process ABC and process ADC is the area enclosed by ABCD. Because the change of the internal energy (a function of state) is the same for both processes, the difference in work is thus the same as the difference in heat transferred to the system.
Solution
For path ABC, the heat added is $Q_{ABC} = 515\text{J}$ and the work done by the gas is the area under the path on the pV diagram, which is
$W_{ABC} = p_1(V_2 - V_1) = 473 \text{J}.$
Along ADC, the work done by the gas is again the area under the path:
$W_{ADC} = p_2(V_2 - V_1)=236 \text{J}.$
Then using the strategy we just described, we have
$Q_{ADC} - Q_{ABC} = W_{ADC} - W_{ABC},$
$Q_{ADC} = Q_{ABC} + W_{ADC} - W_{ABC} = (515 + 236 −473)\text{J} = 278\text{J}.$
Significance
The work calculations in this problem are made simple since no work is done along AD and BC and along AB and DC; the pressure is constant over the volume change, so the work done is simply $p\Delta V$. An isothermal line could also have been used, as we have derived the work for an isothermal process as $W = nRT\text{ln}\frac{V_2}{V_1}$.
Example 3.5
Isothermal Expansion of an Ideal Gas
Heat is added to 1 mol of an ideal monatomic gas confined to a cylinder with a movable piston at one end. The gas expands quasi-statically at a constant temperature of 300 K until its volume increases from V to 3V.
(a) What is the change in internal energy of the gas?
(b) How much work does the gas do?
(c) How much heat is added to the gas?
Strategy
(a) Because the system is an ideal gas, the internal energy only changes when the temperature changes.
(b) The heat added to the system is therefore purely used to do work that has been calculated in Work, Heat, and Internal Energy.
(c) Lastly, the first law of thermodynamics can be used to calculate the heat added to the gas.
Solution
a. We saw in the preceding section that the internal energy of an ideal monatomic gas is a function only of temperature. Since $\Delta T = 0$, for this process, $\Delta E_\text{int} = 0$.
b. The quasi-static isothermal expansion of an ideal gas was considered in the preceding section and was found to be
$\begin{array}{rl}W & = nRT\ln\frac{V_2}{V_1} = nRT\ln\frac{3V}{V} \\ & = (1.00\text{mol})(8.314\text{J/K} \cdot \text{mol})(300\text{K})(\ln3) = 2.74 \times {10}^3 \text{J}.\end{array}$
c. With the results of parts (a) and (b), we can use the first law to determine the heat added:
$\Delta E_\text{int} = Q - W = 0,$
$Q = W = 2.74 \times 10^{3}\text{J}.$
Significance
An isothermal process has no change in the internal energy. Based on that, the first law of thermodynamics reduces to $Q = W$.
Why was it necessary to state that the process of Example 3.5 is quasi-static?
Example 3.6
Vaporizing Water
When 1.00 g of water at 100°C changes from the liquid to the gas phase at atmospheric pressure, its change in volume is $1.67 \times 10^{-3}\text{m}^3$.
(a) How much heat must be added to vaporize the water?
(b) How much work is done by the water against the atmosphere in its expansion?
(c) What is the change in the internal energy of the water?
Strategy
We can first figure out how much heat is needed from the latent heat of vaporization of the water. From the volume change, we can calculate the work done from $W = p\Delta V$ because the pressure is constant. Then, the first law of thermodynamics provides us with the change in the internal energy.
Solution
a. With Lv representing the latent heat of vaporization, the heat required to vaporize the water is
$Q = mL_v = (1.00\text{g})(2.26 \times 10^3\text{J/g}) = 2.26 \times 10^3\text{J}.$
b. Since the pressure on the system is constant at $1.00 \text{atm} = 1.01 \times 10^5\text{N/m}^2$, the work done by the water as it is vaporized is
$W = p\Delta V = (1.01 \times 10^5\text{N/m}^2)(1.67 \times 10^{-3}m^3) = 169\text{J}.$
c. From the first law, the thermal energy of the water during its vaporization changes by
$\Delta E_\text{int} = Q - W = 2.26 \times 10^3\text{J} - 169\text{J} = 2.09 \times 10^3\text{J}.$
Significance
We note that in part (c), we see a change in internal energy, yet there is no change in temperature. Ideal gases that are not undergoing phase changes have the internal energy proportional to temperature. Internal energy in general is the sum of all energy in the system.
When 1.00 g of ammonia boils at atmospheric pressure and $-33.0^\circ\text{C}$, its volume changes from 1.47 to $1130 \text{cm}^3$. Its heat of vaporization at this pressure is $1.37 \times 10^6\text{J/kg}$. What is the change in the internal energy of the ammonia when it vaporizes?
Interactive
View this site to learn about how the first law of thermodynamics. First, pump some heavy species molecules into the chamber. Then, play around by doing work (pushing the wall to the right where the person is located) to see how the internal energy changes (as seen by temperature). Then, look at how heat added changes the internal energy. Finally, you can set a parameter constant such as temperature and see what happens when you do work to keep the temperature constant (Note: You might see a change in these variables initially if you are moving around quickly in the simulation, but ultimately, this value will return to its equilibrium value). | 2021-12-04 11:32:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.9996767044067383, "perplexity": 1408.0874127229024}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964362969.51/warc/CC-MAIN-20211204094103-20211204124103-00563.warc.gz"} |
https://kerodon.net/tag/00MB | # Kerodon
$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$
## 6.1 Expanded Tags
The tags in this section were retired because they contained an explanation which was later expanded upon.
Example 6.1.0.1 (Simplicial Sets of Dimension $\leq 0$). The contents of this tag have been expanded to subsection §1.1.4.
Let $S_{\bullet }$ be a simplicial set. Then the $0$-skeleton $\operatorname{sk}_0( S_{\bullet } )$ can be identified with the coproduct $\coprod _{v \in S_0} \{ v\}$, indexed by the collection of all vertices of $S_{\bullet }$. In particular, the simplicial set $S_{\bullet }$ has dimension $\leq 0$ if and only if it is isomorphic to a coproduct of copies of $\Delta ^0$. We therefore obtain an equivalence of categories
$\xymatrix { \{ \text{Simplicial Sets of Dimension \leq 0} \} \simeq \{ \text{Sets} \} . }$
Example 6.1.0.2. The contents of this example have been expanded to §1.2.6.
Let $G$ be a directed graph (Definition 1.1.5.1) and let $S_{\bullet }$ denote the associated simplicial set of dimension $\leq 1$ (Proposition 1.1.5.9). Then the homotopy category $\mathrm{h} \mathit{S}_{\bullet }$ can be described explicitly as follows:
• The objects of $\mathrm{h} \mathit{S}_{\bullet }$ are the vertices of the graph $G$.
• Given a pair of vertices $v,w \in \operatorname{Vert}(G)$, a morphism from $v$ to $w$ in $\mathrm{h} \mathit{S}_{\bullet }$ is given by a path from $v$ to $w$ in the directed graph $G$: that is, an ordered sequence of edges $(e_1, e_2, \ldots , e_ n)$ satisfying $s( e_1 ) = v$, $t(e_ n) = w$, and $t( e_ i ) = s( e_{i+1} )$ for $0 < i < n$. Here $s,t: \operatorname{Vert}(G) \rightarrow \operatorname{Edge}(G)$ denote the source and target maps. Moreover, we allow $n=0$ in the case $v = w$ (the empty sequence is regarded as the identity morphism from the vertex $v$ to itself).
• Composition of morphisms in $\mathrm{h} \mathit{S}_{\bullet }$ is given by concatenation of paths. More precisely, given morphisms $f = (e_1, e_2, \ldots , e_ m)$ from $u$ to $v$ and $g = (e'_1, e'_2, \ldots , e'_ n)$ from $v$ to $w$, the composition $g \circ f$ is given by the sequence $(e_1, e_2, \ldots , e_ m, e'_1, \ldots , e'_ n)$. | 2020-05-25 20:06:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9913297295570374, "perplexity": 97.2873507805782}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347389355.2/warc/CC-MAIN-20200525192537-20200525222537-00286.warc.gz"} |
https://brilliant.org/practice/line-exploration-2/?chapter=introduction-18 | # Line Exploration
In the following problems, we'll use visualizations with sliders (like the one below) to reason about lines and the equations that describe them.
In the visual above, can you imagine what would happen if you were able to keep moving the slider further and further to the right?
# Line Exploration
The slider below modifies one of the parameters of a linear equation. What could we reasonably expect to happen to the line if the slider were to move to a negative value?
# Line Exploration
These steepness (called the slope) of the two lines are controlled by the sliders below. What would happen to the lines if the sliders had the same value?
Hint: pay attention to the value labels on the sliders; even if the sliders are in the same position, they may not correspond to the same value.
# Line Exploration
In the equation $$y=kx$$, any change to the value of $$x$$ results in a change to the value of $$y$$.
If adding $$3$$ to $$x$$ results in $$y$$ growing by $$12$$, what is the value of $$k$$?
# Line Exploration
The slider below controls the slope of all the lines. If the lines below are described by equations in the form $$y=mx+b$$, which parameter is the slider controlling?
# Line Exploration
The equation that powers the visualization below is $$y=mx + b$$. $$m$$ controls the slope of the line and $$b$$ controls its position. Is it possible to make every line shown above in the format $$y = mx + b ?$$
# Line Exploration
We want the line shown to pass through the marked red point at $$(2,2) .$$ Which values of $$m$$ will work? Mark all that apply and note that $$b$$ can vary. $y = mx + b$
## Line Exploration
### Introduction
Select one or more
# Line Exploration
In these last few questions, we've seen how the equation $$y=mx+b$$ relates to the graph of a line and particularly how the parameter $$m$$ describes its slope.
In the course ahead, we'll further investigate this equation and others, developing a deeper intuition about how changing one part of an equation affects the shape of its graph.
× | 2019-02-18 13:32:55 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5672472715377808, "perplexity": 594.7988559808903}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247486480.6/warc/CC-MAIN-20190218114622-20190218140622-00201.warc.gz"} |
https://test.biostars.org/p/166896/ | Biostar Beta. Not for public use.
Alternative for BGI WEGO
0
Entering edit mode
4.6 years ago
Amk • 120
@Amk21971
Hi Friends,
Do any one have a suggestion on the alternative for the BGI WEGO? As it is not updated since 2009. My goal is to compare GO terms of two species.
Thank you.
Amk
WEGO GO Comparison • 2.3k views
0
Entering edit mode
This is an update that I wanted to share. It seems that the wego wesite got a makeover recently and appears quite better
Just have a look http://wego.genomics.org.cn/
Though I am still struggling to get a good quality bar plot in png format.
1
Entering edit mode
5.2 years ago
Ying W ♦ 3.9k
@Ying W595
you could try using DAVID but i am not too sure how common web based GO enrichment tools are nowadays. There was a post made a couple years ago but some of those links are dead: https://www.biostars.org/p/4302/
Gene Ontology website used to curate a list of tools but they removed it after warning that it was out of date for a while. An archived list of resources can be found here: https://web.archive.org/web/20140625053522/http://geneontology.org/GO.tools.shtml
Here is another archived list of tools: https://web.archive.org/web/20141002022923/http://neurolex.org/wiki/Category:Resource:Gene_Ontology_Tools
0
Entering edit mode
Hi Ying,
Thank you for the suggestion. Links found to be useful. As you said we can't rely on any of the tool as the updation is quite far away for those tools.
Thank you.
Amk
0
Entering edit mode
4.6 years ago
liketree36 • 0
@liketree3610967
If you want an alternative for WEGO, PloGO is deserved to look into, I think. :) http://onlinelibrary.wiley.com/doi/10.1002/pmic.201100445/abstract
It's an R package tool. Here's an introduction of the paper.
This R package contains tools for plotting gene ontology information in a similar fashion to the WEGO web gene ontology plotting tool Ye et al. (2006). However it was designed to incorporate information about abundance in addition to the gene ontology annotation, as well as handle multiple les and allow for a small selection of gene ontology categories of interest.
0
Entering edit mode
I am not able to search it in bioconductor package repo
Neither, I could install it.
> biocLite("plogo")
BioC_mirror: https://bioconductor.org
Using Bioconductor 3.6 (BiocInstaller 1.28.0), R 3.4.4 (2018-03-15).
Installing package(s) 'plogo'
Warning: unable to access index for repository https://www.stats.ox.ac.uk/pub/RWin/src/contrib:
Could this be a proxy issue?
> sessionInfo()
R version 3.4.4 (2018-03-15)
Platform: i386-w64-mingw32/i386 (32-bit)
Running under: Windows 7 x64 (build 7600)
Matrix products: default
locale:
[1] LC_COLLATE=English_India.1252 LC_CTYPE=English_India.1252 LC_MONETARY=English_India.1252 LC_NUMERIC=C
[5] LC_TIME=English_India.1252
attached base packages:
[1] stats graphics grDevices utils datasets methods base | 2021-01-17 04:16:39 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.17033988237380981, "perplexity": 7069.869919113555}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703509104.12/warc/CC-MAIN-20210117020341-20210117050341-00024.warc.gz"} |
https://cracku.in/9-which-number-will-complete-the-series-6-8-17-19-28-x-ssc-cgl-08-sep-evening-shift | Question 9
# Which number will complete the series?6 , 8 , 17 , 19 , 28 , 30 , ?
• Free SSC Study Material - 18000 Questions
• 230+ SSC previous papers with solutions PDF
OR | 2020-01-18 08:27:51 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8739288449287415, "perplexity": 2172.1882360297504}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250592394.9/warc/CC-MAIN-20200118081234-20200118105234-00402.warc.gz"} |
https://events.mpifr-bonn.mpg.de/indico/event/43/contributions | Bonn-Dwingeloo Neighbourhood VLBI Meeting
4 July 2017
MPIfR
Europe/Berlin timezone
Home > Contribution List
Contribution List
Displaying 24 contributions out of 24
Session: Session I
We present total and linearly polarised 3 mm GMVA images of a
sample of blazars and radio galaxies from the VLBA-BU-BLAZAR 7 mm monitoring
program aimed to probe the innermost regions of AGN jets and locate the sites
of gamma-ray emission observed by Fermi-LAT. The reduced opacity at 3 mm and improved angular resolution, of the order of 50 $\mu$arcseconds, allow us to estimate the angu ... More
Presented by Dr. Carolina CASADIO on 4 Jul 2017 at 11:23
Session: Session I
Very Long Baseline Interferometry (VLBI) Observations at 86$\,$GHz reach a resolution of about 50 $\mu$as and sample the scales as small as 10$^3$--10$^4$ Schwarzschild radii of the central black hole in Active Galactic Nuclei (AGN), and uncover the jet regions where acceleration and collimation of the relativistic flow takes place. Synchrotron radiation becomes optically thin at millimetre wavel ... More
Presented by Ms. Dhanya G. NAIR on 4 Jul 2017 at 10:59
Session: Session III
The high resolution and high sensitivity to diffuse emission at low radio frequencies make LOFAR a unique discovery instrument. I will outline how I use its potential to time the activity of AGN hosted by massive, multi-core galaxies found in nearby galaxy clusters.
Presented by Dr. Aleksandar SHULEVSKI on 4 Jul 2017 at 15:42
Session: Session I
Using Very Long Base Interferometry (VLBI), the Bar and Spiral Legacy (BeSSeL) survey has provided distances and proper motions for maser-bearing young massive stars (Reid et al. 2009,2014), allowing an accurate measure of the spiral Galactic structure and kinematics. By the same technique, we are planning to map the inner Galaxy and bulge using positions and velocities of SiO masers stars (Bulge ... More
Presented by Prof. Huib VAN LANGEVELDE on 4 Jul 2017 at 11:35
Session: Session II
Fast radio bursts (FRBs) are a new radio transient phenomena that puzzles scientists over their nature. These millisecond, Janksky bright single pulses are seen with dispersion measures (DMs) several times larger then those caused by the electron density in our own Milky Way and therefore thought to be of extra-galactic origin. Their non-repetitive behaviour (only one FRB has seen to repeat thus f ... More
Presented by Leonard HOUBEN on 4 Jul 2017 at 13:30
Session: Session II
I will present new results on spectral index imaging of several AGN using multi-frequency space VLBI observations with RadioAstron, in combination with ground array images. I will introduce a GUI-based Python tool for spectral analysis of VLBI data, that was used to produce the results.
Presented by Ms. Laura VEGA GARCÍA on 4 Jul 2017 at 14:42
Session: Session III
Radio observations play an important role to understand the processes involved in the formation and evolution of stellar and substellar objects. In this context, we have made several contributions to binary stars belonging to the AB Doradus moving group, namely, AB Dor A/C (Guirado et al. 2006; 2011), AB Dor Ba/Bb (Azulay et al. 2015), and HD 160934 A/c (Azulay et al. 2014; 2017). In these cases, ... More
Presented by Dr. Rebecca S. AZULAY on 4 Jul 2017 at 15:54
Session: Session II
As a participant in the Black Hole Cam project, JIVE undertook to develop a fringe-fitting tool in the Casa dataprocessing package closely modeled on the AIPS FRING task. A Python prototype has been completed and is now in preliminary use by colleagues at Radboud University Nijmegen; the C++ code to apply the calibrations is now mature, and a first draft of a C++ implementation is now complete alt ... More
Presented by Dr. Des SMALL on 4 Jul 2017 at 14:18
Session: Session III
Many active galactic nuclei (AGN) show a spectral peak in their radio spectra. Peaked-spectrum radio sources (including gigahertz-peaked spectrum and compact steep spectrum sources) are radio galaxies that often display small angular extents, suggesting that they are either very young AGN or are confined by a dense surrounding medium. We here present a spectacular sample of 1484 low-frequency peak ... More
Presented by Dr. Joseph CALLINGHAM on 4 Jul 2017 at 16:06
Session: Session I
We performed full polarimetric VLBA observations of water masers towards the Turner-Welch Object in the W3(OH) high-mass star forming complex. This object drives a synchrotron jet, which is quite exceptional for a high-mass protostar, and is associated with a strongly polarized water maser source, W3(H$_2$O), making it an optimal target to investigate the role of magnetic fields on the innermo ... More
Presented by Dr. Ciriaco GODDI
Session: Session I
Understanding how mass accretion and jet formation occurs near the central engine of AGN has been one of major challenges in modern astrophyiscs. The apparent size of the jet forming region is desperately small even for an extraordinary massive BH system (10Rs=0.001pc for a BH mass of 10^9Msun), making a direct comparison of theories and observations difficult. Therefore, it is indispensable to im ... More
Presented by Mr. Jae-Young KIM on 4 Jul 2017 at 11:47
Session: Session II
SKA phase 2 sensitivity could be achieved at relatively low cost if we can piggyback a radio astronomy receiver on solar concentrators, of which four square kilometres are operating in the form of solar power towers. However the mirror arrays form a speckle pattern rather than a point focus, but we can restore full sensitivity in principle using a focal-plane array to collect the power from ... More
Presented by Dr. Alan ROY on 4 Jul 2017 at 13:54
Session: Session III
After a review of the robopol project and its main findings in the angle domain, we will present average R-band optopolarimetric data, as well as variability parameters, from the first and second RoboPol observing season. We investigate whether gamma-ray-loud and gamma-ray-quiet blazars exhibit systematic differences in their optical polarization proper- ties. We find that gamma-ray-loud blazars h ... More
Presented by Dr. Emmanouil ANGELAKIS on 4 Jul 2017 at 16:18
Session: Session II
The Event Horizon Telescope (EHT) project aims to image the supermassive black holes in the center of our Galaxy (Sgr A*) and M87 at event horizon scales using very long baseline interferometry (VLBI) at (sub-)millimetre wavelengths. The EHT currently consists of a number of telescopes around the globe. A telescope in Africa will greatly improve the uv-coverage and the image quality. We prop ... More
Presented by Dr. Cornelia MUELLER on 4 Jul 2017 at 14:30
Session: Session II
The powerful radio galaxy Cygnus A is one of the very few targets where the jet acceleration and collimation region can be probed directly through VLBI imaging. Through a rich, multi-wavelength VLBI data set, we were able to reconstruct the two-sided collimation profile of the outflow, and to follow its evolution from scales of a few hundreds to millions Schwarzschild radii. Preliminary results su ... More
Presented by Dr. Biagina BOCCARDI on 4 Jul 2017 at 14:06
Session: Session I
The jets of powerful radio galaxies are known to play a vital role in regulating the gas distribution of the host galaxy. Evidence for this includes observations of fast outflows of neutral hydrogen gas detected in absorption in a number of radio galaxies, though these observations mostly lacked the resolution to pinpoint the location of the outflow with respect to the jet system. However, this ca ... More
Presented by Dr. Robert SCHULZ on 4 Jul 2017 at 10:35
Session: Session III
I will briefly discuss the interpretation of the peculiar light curve of J1415+1320, that shows time-symmetric and recurring U-shaped features across the cm-wave and mm-wave bands, which we call Symmetric Achromatic Variability (SAV). Although a common proposal to explain simmilar features in blazar radio light curves are Extreme Scattering Events (ESEs), this is shown to be not viable for this pa ... More
Presented by Dr. Walter MAX-MOERBECK on 4 Jul 2017 at 15:30
Session: Session II
Blazars, a subclass of AGN jets, show extreme flux variability across the electromagnetic spectrum from radio to gamma-rays. A challenge to theoretical interpretation is the rapid flux variability at GeV energies, which implies an origin from ultra-compact emission regions (< sub-pc). The exact location of the gamma-ray emitting region within the AGN is also controversially discussed. The pri ... More
Presented by Ms. Efthalia TRAIANOU on 4 Jul 2017 at 14:54
Session: Session I
The $\gamma$-ray sky is strongly dominated by blazars, i.e. AGN with relativistic jets oriented closely with our line of sight. Radio galaxies are their misaligned counterpart, and make up about $\sim$1-2% of all AGN observed by *Fermi*-LAT. Nonetheless, they provide us with a view of AGN jets which is less biased by Doppler boosting effects, and allow us to test jet production and emission models ... More
Presented by Mr. Roberto ANGIONI on 4 Jul 2017 at 10:47
Session: Session III
We use the VLBI technique to obtain precise astrometric measurements to several young stars in nearby star-forming regions belonging to the galactic structure known as the Gould's Belt. These measurements will allow us to investigate the structure of the clouds and their internal kinematics and eventually to characterize the overall dynamics of the Belt. I will present the current status of the pr ... More
Presented by Dr. Gisela Noemí ORTIZ-LEÓN on 4 Jul 2017 at 16:42
Session: Session III
Stars similar to our Sun undergo radical changes as their lives draw to an end. After swelling up to huge red giants, they start to pulsate, blowing off large clouds of gas and dust, to create intricately shaped and mesmerizing planetary nebulae. The remaining part of the star collapses under gravity, creating a compact white dwarf to slowly cool and fade away. One of the pivotal questions in this ... More
Presented by Dr. Gabor OROSZ on 4 Jul 2017 at 16:30
Session: Session I | 2021-01-23 09:54:52 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4892706573009491, "perplexity": 4904.598148373215}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703537796.45/warc/CC-MAIN-20210123094754-20210123124754-00798.warc.gz"} |
http://ompf2.com/viewtopic.php?t=789 | ## PMC-ERPT
Practical and theoretical implementation discussion.
Zelcious
Posts: 42
Joined: Mon Jul 23, 2012 11:05 am
### PMC-ERPT
Is it the mutated path that goes back into the re-sampling step or the original. If it's the mutation then is pmc-erpt unbiased? Should violate detailed balance right, since the number of mutation is dependent of path intensity.
Trivia: I did a little calculation last where it tried to come up with the simplest case where basing the mutation length on intensity violates detailed balance.
Two states with intensity 1,2 respectively does the trick if you let mutation length be equal to intensity.
Btw, are there any known methods dealing with the more relaxed general balance?
Dietger
Posts: 50
Joined: Tue Nov 29, 2011 10:33 am
### Re: PMC-ERPT
The paper couldn't convince me that PMC-ERPT is unbiased either (or even consistent for that matter). I recently reread the PMC sources referenced in the paper and I still don't see how the authors got from there to the presented PMC-ERPT algorithm. Aside from the lack of proofs, the authors skip over some details which makes it impossible for me to figure out how their method is supposed to work. Interesting observation: the papers Cornell box comparison of ERPT and PMC-ERPT seems to converge to different images...
PS. I initially tried to write down in more detail why the paper does not make sense to me, but I got too frustrated and gave up If anyone really wants to know I could give it another go. Or if someone can explain to me why I am an idiot and the paper makes perfect sense, please do! | 2020-11-30 13:31:39 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8164144158363342, "perplexity": 787.4176068362541}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141216175.53/warc/CC-MAIN-20201130130840-20201130160840-00517.warc.gz"} |
https://en.wikipedia.org/wiki/Anti-reflection_coating | # Anti-reflective coating
(Redirected from Anti-reflection coating)
Uncoated glasses lens (top) versus lens with antireflective coating. Note the tinted reflection from the coated lens.
An antireflective or anti-reflection (AR) coating is a type of optical coating applied to the surface of lenses and other optical elements to reduce reflection. In typical imaging systems, this improves the efficiency since less light is lost due to reflection. In complex systems such as telescopes and microscopes the reduction in reflections also improves the contrast of the image by elimination of stray light. This is especially important in planetary astronomy. In other applications, the primary benefit is the elimination of the reflection itself, such as a coating on eyeglass lenses that makes the eyes of the wearer more visible to others, or a coating to reduce the glint from a covert viewer's binoculars or telescopic sight.
Many coatings consist of transparent thin film structures with alternating layers of contrasting refractive index. Layer thicknesses are chosen to produce destructive interference in the beams reflected from the interfaces, and constructive interference in the corresponding transmitted beams. This makes the structure's performance change with wavelength and incident angle, so that color effects often appear at oblique angles. A wavelength range must be specified when designing or ordering such coatings, but good performance can often be achieved for a relatively wide range of frequencies: usually a choice of IR, visible, or UV is offered.
## Applications
Anti-reflective coatings are often used in camera lenses, giving lens elements distinctive colours.
Anti-reflective coatings are used in a wide variety of applications where light passes through an optical surface, and low loss or low reflection is desired. Examples include anti-glare coatings on corrective lenses and camera lens elements, and antireflective coatings on solar cells.[1]
### Corrective lenses
Opticians may recommend "anti-reflection lenses" because the decreased reflection enhances the cosmetic appearance of the lenses. Such lenses are often said to reduce glare, but the reduction is very slight.[2] Eliminating reflections allows slightly more light to pass through, producing a slight increase in contrast and visual acuity.
Antireflective ophthalmic lenses should not be confused with polarized lenses, which decrease (by absorption) the visible glare of sun reflected off surfaces such as sand, water, and roads. The term "antireflective" relates to the reflection from the surface of the lens itself, not the origin of the light that reaches the lens.
Many anti-reflection lenses include an additional coating that repels water and grease, making them easier to keep clean. Anti-reflection coatings are particularly suited to high-index lenses, as these reflect more light without the coating than a lower-index lens (a consequence of the Fresnel equations). It is also generally easier and cheaper to coat high index lenses.
### Photolithography
Antireflective coatings are often used in microelectronic photolithography to help reduce image distortions associated with reflections off the surface of the substrate. Different types of antireflective coatings are applied either before or after the photoresist, and help reduce standing waves, thin-film interference, and specular reflections.[3][4]
## Types
### Index-matching
The simplest form of anti-reflective coating was discovered by Lord Rayleigh in 1886. The optical glass available at the time tended to develop a tarnish on its surface with age, due to chemical reactions with the environment. Rayleigh tested some old, slightly tarnished pieces of glass, and found to his surprise that they transmitted more light than new, clean pieces. The tarnish replaces the air-glass interface with two interfaces: an air-tarnish interface and a tarnish-glass interface. Because the tarnish has a refractive index between those of glass and air, each of these interfaces exhibits less reflection than the air-glass interface did. In fact, the total of the two reflections is less than that of the "naked" air-glass interface, as can be calculated from the Fresnel equations.
One approach is to use graded-index (GRIN) anti-reflective coatings, that is, ones with nearly continuously varying index of refraction.[5] With these, it is possible to curtail reflection for a broad band of frequencies and incidence angles.
### Single-layer interference
The simplest interference anti-reflective coating consists of a single thin layer of transparent material with refractive index equal to the square root of the substrate's refractive index. In air, such a coating theoretically gives zero reflectance for light with wavelength (in the coating) equal to four times the coating's thickness. Reflectance is also decreased for wavelengths in a broad band around the center. A layer of thickness equal to a quarter of some design wavelength is called a "quarter-wave layer".
The most common type of optical glass is crown glass, which has an index of refraction of about 1.52. An optimal single-layer coating would have to be made of a material with an index of about 1.23. Unfortunately, there are no solid materials with such a low refractive index. The closest materials with good physical properties for a coating are magnesium fluoride, MgF2 (with an index of 1.38), and fluoropolymers (which can have indices as low as 1.30, but are more difficult to apply).[6] MgF2 on a crown glass surface gives a reflectance of about 1%, compared to 4% for bare glass. MgF2 coatings perform much better on higher-index glasses, especially those with index of refraction close to 1.9. MgF2 coatings are commonly used because they are cheap, and when they are designed for a wavelength in the middle of the visible band, they give reasonably good anti-reflection over the entire band. Researchers have produced films of mesoporous silica nanoparticles with refractive indices as low as 1.12, which function as antireflection coatings.[7]
### Multi-layer interference
By using alternating layers of a low-index material like silica and a higher-index material, it is possible to obtain reflectivities as low as 0.1% at a single wavelength. Coatings that give very low reflectivity over a broad band of frequencies can also be made, although these are complex and relatively expensive. Optical coatings can also be made with special characteristics, such as near-zero reflectance at multiple wavelengths, or optimal performance at angles of incidence other than 0°.
### Absorbing
An additional category of anti-reflection coatings is the so-called "absorbing ARC". These coatings are useful in situations where high transmission through a surface is unimportant or undesirable, but low reflectivity is required. They can produce very low reflectance with few layers, and can often be produced more cheaply, or at greater scale, than standard non-absorbing AR coatings. (See, for example, US Patent 5,091,244.) Absorbing ARCs often make use of unusual optical properties exhibited in compound thin films produced by sputter deposition. For example, titanium nitride and niobium nitride are used in absorbing ARCs. These can be useful in applications requiring contrast enhancement or as a replacement for tinted glass (for example, in a CRT display).
### Moth eye
Moths' eyes have an unusual property: their surfaces are covered with a natural nanostructured film, which eliminates reflections. This allows the moth to see well in the dark, without reflections to give its location away to predators.[8] The structure consists of a hexagonal pattern of bumps, each roughly 200 nm high and spaced on 300 nm centers.[9] This kind of antireflective coating works because the bumps are smaller than the wavelength of visible light, so the light sees the surface as having a continuous refractive index gradient between the air and the medium, which decreases reflection by effectively removing the air-lens interface. Practical anti-reflective films have been made by humans using this effect;[10] this is a form of biomimicry.
Such structures are also used in photonic devices, for example, moth-eye structures grown from tungsten oxide and iron oxide can be used as photoelectrodes for splitting water to produce hydrogen.[11] The structure consists of tungsten oxide spheroids of several 100 micrometer size coated with a few nanometers thin iron-oxide layer.[12][13]
### Circular polarizer
Reflections are blocked by a circular polarizer
A circular polarizer laminated to a surface can be used to eliminate reflections.[14][15] The polarizer transmits light with one chirality ("handedness") of circular polarization. Light reflected from the surface after the polarizer is transformed into the opposite "handedness". This light cannot pass back through the circular polarizer because its chirality has changed (e.g. from right circular polarized to left circularly polarized). A disadvantage of this method is that if the input light is unpolarized, the transmission through the assembly will be less than 50%.
## Theory
An anti-reflection coated window, shown at a 45° and a 0° angle of incidence
There are two separate causes of optical effects due to coatings, often called thick-film and thin-film effects. Thick-film effects arise because of the difference in the index of refraction between the layers above and below the coating (or film); in the simplest case, these three layers are the air, the coating, and the glass. Thick-film coatings do not depend on how thick the coating is, so long as the coating is much thicker than a wavelength of light. Thin-film effects arise when the thickness of the coating is approximately the same as a quarter or a half a wavelength of light. In this case, the reflections of a steady source of light can be made to add destructively and hence reduce reflections by a separate mechanism. In addition to depending very much on the thickness of the film and the wavelength of light, thin-film coatings depend on the angle at which the light strikes the coated surface.
### Reflection
Whenever a ray of light moves from one medium to another (for example, when light enters a sheet of glass after travelling through air), some portion of the light is reflected from the surface (known as the interface) between the two media. This can be observed when looking through a window, for instance, where a (weak) reflection from the front and back surfaces of the window glass can be seen. The strength of the reflection depends on the ratio of the refractive indices of the two media, as well as the angle of the surface to the beam of light. The exact value can be calculated using the Fresnel equations.
When the light meets the interface at normal incidence (perpendicularly to the surface), the intensity of light reflected is given by the reflection coefficient, or reflectance, R:
${\displaystyle R=\left({\frac {n_{0}-n_{S}}{n_{0}+n_{S}}}\right)^{2},}$
where n0 and nS are the refractive indices of the first and second media respectively. The value of R varies from 0 (no reflection) to 1 (all light reflected) and is usually quoted as a percentage. Complementary to R is the transmission coefficient, or transmittance, T. If absorption and scattering are neglected, then the value T is always 1 − R. Thus if a beam of light with intensity I is incident on the surface, a beam of intensity RI is reflected, and a beam with intensity TI is transmitted into the medium.
For the simplified scenario of visible light travelling from air (n0 ≈ 1.0) into common glass (nS ≈ 1.5), value of R is 0.04, or 4%, on a single reflection. So at most 96% of the light (T = 1 − R = 0.96) actually enters the glass, and the rest is reflected from the surface. The amount of light reflected is known as the reflection loss.
In the more complicated scenario of multiple reflections, say with light travelling through a window, light is reflected both when going from air to glass and at the other side of the window when going from glass back to air. The size of the loss is the same in both cases. Light also may bounce from one surface to another multiple times, being partially reflected and partially transmitted each time it does so. In all, the combined reflection coefficient is given by 2R/(1 + R). For glass in air, this is about 7.7%.
### Rayleigh's film
As observed by Lord Rayleigh, a thin film (such as tarnish) on the surface of glass can reduce the reflectivity. This effect can be explained by envisioning a thin layer of material with refractive index n1 between the air (index n0) and the glass (index nS). The light ray now reflects twice: once from the surface between air and the thin layer, and once from the layer-to-glass interface.
From the equation above and the known refractive indices, reflectivities for both interfaces can be calculated, denoted R01 and R1S respectively. The transmission at each interface is therefore T01 = 1 − R01 and T1S = 1 − R1S. The total transmittance into the glass is thus T1ST01. Calculating this value for various values of n1, it can be found that at one particular value of optimal refractive index of the layer, the transmittance of both interfaces is equal, and this corresponds to the maximal total transmittance into the glass.
This optimal value is given by the geometric mean of the two surrounding indices:
${\displaystyle n_{1}={\sqrt {n_{0}n_{S}}}.}$
For the example of glass (nS ≈ 1.5) in air (n0 ≈ 1.0), this optimal refractive index is n1 ≈ 1.225.[16][17]
The reflection loss of each interface is approximately 1.0% (with a combined loss of 2.0%), and an overall transmission T1ST01 of approximately 98%. Therefore, an intermediate coating between the air and glass can halve the reflection loss.
### Interference coatings
The use of an intermediate layer to form an anti-reflection coating can be thought of as analoguous to the technique of impedance matching of electrical signals. (A similar method is used in fibre optic research, where an index-matching oil is sometimes used to temporarily defeat total internal reflection so that light may be coupled into or out of a fiber.) Further reduced reflection could in theory be made by extending the process to several layers of material, gradually blending the refractive index of each layer between the index of the air and the index of the substrate.
Practical anti-reflection coatings, however, rely on an intermediate layer not only for its direct reduction of reflection coefficient, but also use the interference effect of a thin layer. Assume the layer's thickness is controlled precisely, such that it is exactly one quarter of the wavelength of light in the layer (λ/4 = λ0/(4n1), where λ0 is the vacuum wavelength). The layer is then called a quarter-wave coating. For this type of coating a normally incident beam I, when reflected from the second interface, will travel exactly half its own wavelength further than the beam reflected from the first surface, leading to destructive interference. This is also true for thicker coating layers (3λ/4, 5λ/4, etc.), however the anti-reflective performance is worse in this case due to the stronger dependence of the reflectance on wavelength and the angle of incidence.
If the intensities of the two beams R1 and R2 are exactly equal, they will destructively interfere and cancel each other, since they are exactly out of phase. Therefore, there is no reflection from the surface, and all the energy of the beam must be in the transmitted ray, T. In the calculation of the reflection from a stack of layers, the transfer-matrix method can be used.
Interference in a quarter-wave anti-reflection coating
Real coatings do not reach perfect performance, though they are capable of reducing a surface reflection coefficient to less than 0.1%. Also, the layer will have the ideal thickness for only one distinct wavelength of light. Other difficulties include finding suitable materials for use on ordinary glass, since few useful substances have the required refractive index (n ≈ 1.23) that will make both reflected rays exactly equal in intensity. Magnesium fluoride (MgF2) is often used, since this is hard-wearing and can be easily applied to substrates using physical vapor deposition, even though its index is higher than desirable (n = 1.38).
Further reduction is possible by using multiple coating layers, designed such that reflections from the surfaces undergo maximal destructive interference. One way to do this is to add a second quarter-wave thick higher-index layer between the low-index layer and the substrate. The reflection from all three interfaces produces destructive interference and anti-reflection. Other techniques use varying thicknesses of the coatings. By using two or more layers, each of a material chosen to give the best possible match of the desired refractive index and dispersion, broadband anti-reflection coatings covering the visible range (400–700 nm) with maximal reflectivities of less than 0.5% are commonly achievable.
The exact nature of the coating determines the appearance of the coated optic; common AR coatings on eyeglasses and photographic lenses often look somewhat bluish (since they reflect slightly more blue light than other visible wavelengths), though green and pink-tinged coatings are also used.
If the coated optic is used at non-normal incidence (that is, with light rays not perpendicular to the surface), the anti-reflection capabilities are degraded somewhat. This occurs because the phase accumulated in the layer relative to the phase of the light immediately reflected decreases as the angle increases from normal. This is counterintuitive, since the ray experiences a greater total phase shift in the layer than for normal incidence. This paradox is resolved by noting that the ray will exit the layer spatially offset from where it entered and will interfere with reflections from incoming rays that had to travel further (thus accumulating more phase of their own) to arrive at the interface. The net effect is that the relative phase is actually reduced, shifting the coating, such that the anti-reflection band of the coating tends to move to shorter wavelengths as the optic is tilted. Non-normal incidence angles also usually cause the reflection to be polarization-dependent.
### Textured coatings
Reflection can be reduced by texturing the surface with 3D pyramids or 2D grooves (gratings). These kind of textured coating can be created using for example the Langmuir-Blodgett method.[18]
If wavelength is greater than the texture size, the texture behaves like a gradient-index film with reduced reflection. To calculate reflection in this case, effective medium approximations can be used. To minimize reflection, various profiles of pyramids have been proposed, such as cubic, quintic or integral exponential profiles.
If wavelength is smaller than the textured size, the reflection reduction can be explained with the help of the geometric optics approximation: rays should be reflected many times before they are sent back toward the source. In this case the reflection can be calculated using ray tracing.
Using texture reduces reflection for wavelengths comparable with the feature size as well. In this case no approximation is valid, and reflection can be calculated by solving Maxwell equations numerically.
Antireflective properties of textured surfaces are well discussed in literature for a wide range of size-to-wavelength ratios (including long- and short-wave limits) to find the optimal texture size.[19]
## History
As mentioned above, natural index-matching "coatings" were discovered by Lord Rayleigh in 1886. Harold Dennis Taylor of Cooke company developed a chemical method for producing such coatings in 1904.[20][21]
Interference-based coatings were invented and developed in 1935 by Alexander Smakula, who was working for the Carl Zeiss optics company.[22][23][24] Anti-reflection coatings were a German military secret until the early stages of World War II.[25] Katharine Burr Blodgett and Irving Langmuir developed organic anti-reflection coatings known as Langmuir–Blodgett films in the late 1930s.
## References
1. ^ Hemant Kumar Raut; V. Anand Ganesh; A. Sreekumaran Nairb; Seeram Ramakrishna (2011). "Anti-reflective coatings: A critical, in-depth review". Energy & Environmental Science. 4: 3779–3804. doi:10.1039/c1ee01297e.
2. ^ Duffner, Lee R (Feb 27, 2015). "Anti-reflective Coating - American Academy of Ophthalmology". Anti-reflective Coating - American Academy of Ophthalmology. American Academy of Ophthalmology. Retrieved Jan 22, 2016.
3. ^ Understanding bottom antireflective coatings
4. ^ Yet, Siew Ing (2004). Investigation of UFO defect on DUV CAR and BARC process. 5375. SPIE. pp. 940–948. Bibcode:2004SPIE.5375..940Y. doi:10.1117/12.535034.
5. ^ Zhang, Jun-Chao; Xiong, Li-Min; Fang, Ming; He, Hong-Bo (2013). "Wide-angle and broadband graded-refractive-index antireflection coatings" (PDF). Chinese Physics B. 22 (4): 044201. Bibcode:2013ChPhB..22d4201Z. doi:10.1088/1674-1056/22/4/044201. Retrieved 13 May 2016.
6. ^ "Opstar AR fluoride coatings and application methods". Archived from the original on 29 January 2011.
7. ^
8. ^ "Nanostructured Surfaces" (PDF). Fraunhofer Magazine (2): 10. 2005. Retrieved 2009-06-17.
9. ^ "Moth-eye Antireflective Microstructures" (PDF). Reflexite Corporation. 2006. Retrieved 2009-06-17.
10. ^ "Novel film inspired by moths" (Press release). Pro-talk. 3 December 2003. Archived from the original on 2014-12-13. Retrieved 2009-06-17.
11. ^ Boudoire, Florent; Toth, Rita; Heier, Jakob; Braun, Artur; Constable, Edwin C. (2014). "Photonic light trapping in self-organized all-oxide microspheroids impacts photoelectrochemical water splitting". Energy Environ Sci. 7: 2680–2688. doi:10.1039/C4EE00380B.
12. ^ "Photoelectrochemical Water Splitting Can Be Achieved with Self-Organized, All-Oxide Electrodes". Materials Research Society. 2014. Retrieved 2014-07-24.
13. ^ "Photonic light trapping in self-organized all-oxide microspheroids impacts photoelectrochemical water splitting". Authors. 2014. Retrieved 2014-05-01.
14. ^
15. ^ Information Display. Society for Information Display. 2006.
16. ^ Krepelka, J. (1992). "Maximally flat antireflection coatings" (PDF). Jemná Mechanika A Optika (3–5): 53. Archived from the original (PDF) on 12 January 2011. Retrieved 2009-06-17.
17. ^ Moreno, I.; Araiza, J.; Avendano-Alejo, M. (2005). "Thin-film spatial filters" (PDF). Optics Letters. 30 (8): 914–916. Bibcode:2005OptL...30..914M. doi:10.1364/OL.30.000914. PMID 15865397.
18. ^ Ching-Mei Hsu, Stephen T. Connor, Mary X. Tang, and Yi Cui. "Wafer-scale silicon nanopillars and nanocones by Langmuir–Blodgett assembly and etching". Applied Physics Letters. 93.
19. ^ A. Deinega; et al. (2011). "Minimizing light reflection from dielectric textured surfaces". JOSA A. 28: 770. Bibcode:2011JOSAA..28..770D. doi:10.1364/josaa.28.000770.
20. ^ MacLeod, H A (2001). Thin Film Optical Filters (3rd ed.). CRC. p. 4. ISBN 9780750306881.
21. ^ British Patent 29561, 31 December 1904
22. ^ "History of Camera Lenses from Carl Zeiss - 1935 - Alexander Smakula develops anti-reflection coating". Zeiss.com. Retrieved 15 June 2013.
23. ^ "Lens coating". Zeiss.com. Archived from the original on 1 January 2013. Retrieved 15 June 2013.
24. ^ Patent DE 685767, "Verfahren zur Erhoehung der Lichtdurchlaessigkeit optischer Teile durch Erniedrigungdes Brechungsexponenten an den Grenzflaechen dieser optischen Teile", published 1935-11-01, assigned to Zeiss Carl FA
25. ^ | 2018-07-22 15:15:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7344362735748291, "perplexity": 2037.8312359192603}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676593302.74/warc/CC-MAIN-20180722135607-20180722155607-00210.warc.gz"} |
https://www.studyadda.com/ncert-solution/11th-chemistry-some-basic-concepts-of-chemistry_q2/490/31898 | • # question_answer 2) Calculate the amount of water produced by the combustion of 16 g of methane.
$\underset{\underset{{}}{\mathop{1mol}}\,}{\mathop{C{{H}_{4}}+2{{O}_{2}}(g)}}\,\to C{{O}_{2}}(g)+\underset{\underset{{}}{\mathop{1mol}}\,}{\mathop{2{{H}_{2}}O(g)}}\,$ Number of moles of $C{{O}_{2}}=\frac{\text{mass}\,\text{in}\,\text{garm}}{\text{Gram}\,\text{molecular}\,\text{mass}}$ $=\frac{22}{44}=0.5$ Gram molecular mass Number of moles of methane = Number of moles of $C{{O}_{2}}$ formed $=\text{ }0.5\text{ }mol$ | 2020-09-19 06:41:45 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6838805675506592, "perplexity": 735.7697751115667}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400190270.10/warc/CC-MAIN-20200919044311-20200919074311-00269.warc.gz"} |
https://votefidler.org/page/7/ | ## James Holzhauer Broke ‘Jeopardy!,’ But Is Broken So Bad?
Note: This article discusses the results of the June 3, 2019, episode of “Jeopardy!”
James Holzhauer claims not to remember many particulars of how he lost on “Jeopardy!” for the first time, other than that he blanked on a clue about the city of Albany and his opponent quickly took control of the board, landing a game-changing Daily Double. Before long, it was all over. Monday’s episode marked the end of Holzhauer’s two-month reign as one of the winningest, and certainly the most radical, champions in the decades-long history of the trivia game show “Jeopardy!”
Holzhauer finished Monday’s game in second place with $24,799 behind Emma Boettcher’s$46,801. But during his 32-win run, he averaged about $77,000 per game — an average nearly identical to the record for the single richest game ever played before he took the podium in early April. In the process, he laid claim to the entirety of the top 10 highest-scoring games of all time, including one single half-hour haul of$131,127. It was a historic run driven by immaculate trivia knowledge, disciplined strategy and calculated aggression.
But other records will forever remain just out of reach. Holzhauer’s streak ended with total winnings of $2,462,216 — less than$60,000 shy of Ken Jennings’s record $2,520,700 which was amassed over a nearly incomprehensible 74 straight wins in 2004. Holzhauer will sit second on the all-time money list until the arrival of some other great champion. (Or he might sit there forever, which seems more likely.)2 “I played every day exactly according to my game plan, so I have no regrets,” Holzhauer told FiveThirtyEight a few days before the fateful episode aired. Holzhauer rewrote swaths of the show’s record books. But his biggest contribution may be to “Jeopardy!” strategy. Holzhauer exploited the game’s Daily Doubles to their fullest, hunting them down and betting big on them. Over his 32 wins (and one loss), Holzhauer — a professional sports bettor from Vegas — not only got significantly richer but likely changed how the venerable game show will be played. Holzhauer was such an effective and alien force that opponents began to mimic his style out of desperation, like growling at a hungry lion in hopes of scaring it away. They hopped wildly around the game’s board whenever they could, picking big-dollar clues early, searching madly for the Daily Doubles and betting big when they found them — just the sort of unalloyed aggression that had quickly become Holzhauer’s trademark and the fuel for his success. “Many of my opponents played like I do, but I’m not sure they would have done so without provocation,” Holzhauer said. “You don’t want to inadvertently make your opponents play a better strategy. In a sense, I may have helped bring about my own downfall.” Life as a longtime “Jeopardy!” champion is a strange one, chronologically speaking. Holzhauer has been watching the world wonder when his streak will end, all the while knowing exactly when it would happen. On its taping dates, the show records five episodes back-to-back, with just a change of clothes in between. The episodes don’t air until much later. If he could alter time, maybe buy a time machine with that$2,462,216, would he have approached the game any differently?
“The only things I would do differently from the start of my run: never wear a sport coat, which interfered a little with my buzzer form, and use gel insoles in my dress shoes,” Holzhauer said.
“Both were fixed by the second taping day.”
The insoles seem to have worked. Holzhauer has earned a spot in the pantheon of the “Jeopardy!” greats, and he gives himself prominent placement there. “I think there is a nebulous top three of Ken Jennings, Brad Rutter and me,” he said. “Ken’s 74-game streak remains the most impressive achievement in the show’s history.”
Rutter is no slouch, either — he has won more money than any other “Jeopardy!” contestant, and he’s a man who has never lost to a human. Rutter’s initial winning streak was ended by the show’s rules at the time, which limited a defending champion to five appearances. But between those appearances and the show’s Tournament of Champions, Ultimate Tournament of Champions and Million Dollar Masters, Rutter won $4,688,436. (Here’s a free idea for the “Jeopardy!” producers: Holzhauer vs. Jennings vs. Rutter in the Ultimate Tournament of Ultimate Champions.) A couple of days after we first emailed, however, Holzhauer followed up to amend his initial assessment. “Amidst all the people comparing me to Ken and Brad, I totally forgot about the two greatest Jeopardy champions of all time: Cindy Stowell, who won six games while dying of cancer, and Eddie Timanus, who’s … blind and was an undefeated five-time champ in his initial run. It’s impossible for me to compare myself to them, so perhaps they should be in their own category.” Holzhauer’s plan for now is a return to normalcy. “The 19-year-old version of James would be thrilled by the opportunities” that the winning streak has brought his way, “but married parent James is hoping to keep his home life settled.” The game that made him famous, however, has been left unsettled. Lots of esteemed competitive pursuits have been “broken” lately. Baseball. Basketball. Even the spelling bee just last week. An innovative strategy or an outlier talent can deeply alter the games we’ve played for decades. In the process, the cadence or tenor of the game might be rendered unrecognizable for someone who hadn’t seen it in a few years. These innovative strategies are often driven by mathematical analysis, data and statistical rigor — things that a sports bettor from Vegas must embrace in order to eat. I asked Holzhauer if “Jeopardy!” now belonged in this category of sabermetrically altered pursuits. “I can see the parallels, for sure,” he said. “At its heart, all these shifts are just attempts to increase your chances of winning. Why would anyone not want to maximize their chances?” Plenty of outlets have written that, thanks to Holzhauer, “Jeopardy!” is now broken. But there’s art in that. While the game may look a bit different than it did before, it may also be closer to perfection — to an ideal expression of trivia game-show strategy. Broken is beautiful. From ABC News: ## How Tottenham Could Shock Us One More Time This could finally be the year when a true underdog wins the Champions League again. The last time a club not among the world’s very elite hoisted the trophy was in 2012, when Chelsea won it all.1 In the six years since, the ruling class of Bayern Munich, Barcelona and Real Madrid captured Europe’s most prestigious club tournament each year. In this context, if Liverpool triumphs on Saturday, despite being clearly among the best teams in the world according to the FiveThirtyEight Soccer Power Index, the Reds would still be breaking ground. But the real underdog here is Tottenham. Spurs rate as the 12th-best club in the world by SPI. And in many ways, the story of Tottenham is the incredible journey it took just to get close to the world top 10. Spurs languished in the middle of the English Premier League table through most of the 2000s, and unlike Manchester City, which also made the leap to the EPL elite in this millennium, Tottenham didn’t luck into an owner willing to spend at a massive deficit to take the club up the ladder. Instead, Spurs had to ratchet up their spending slowly and play the market carefully. Of the Premier League’s current “big six” clubs,2 only Tottenham has come close to breaking even on its transfer dealings over the past decade — receiving about as much money in transfer fees as it has paid out. In this same time, Tottenham has managed to increase its wage spending from about$85 million in 2009-10 to 187 million reported last season. The money saved on transfers has, to a significant degree, been reinvested into wages. By carefully managing the club’s incoming and outgoing funds on transfers, chairman Daniel Levy has built up the quality of the squad and paid to keep better and better players. And it’s worked. Tottenham has finished at least sixth in every season this decade, including now four consecutive berths in the Champions League.3 In the previous decade, Spurs finished in the top six only twice (back-to-back fifth-place finishes in 2005-06 and 2006-07). Tottenham has become a regular Champions League competitor through a process of careful business management backing up good player development. This same “do more with less” mentality is reflected in the team’s performances this season under the management of Mauricio Pochettino. Faced with injuries up and down the squad, Pochettino has had to improvise. His team has given at least 1,500 minutes to 14 different outfield players,4 most among the big six sides in the Premier League. Tottenham’s 10 most-used outfield players have covered only 66 percent of the team’s total minutes, the smallest share among the big six. The team’s injuries have hit the midfield particularly hard: Eric Dier missed most of the second half of the season after an appendectomy, Harry Winks has missed time with ankle and groin injuries, Victor Wanyama was unavailable for most of the season with knee problems, and Mousa Dembele was injured in November and has not played for Tottenham since.5 Dier, Dembele and Wanyama were three of Spurs’ four leading midfielders in tackles and interceptions won per 90 minutes this season; without them, the club had no true ball-winners in the center of the pitch. The loss of these more defensively sound midfielders changed Tottenham’s approach, forcing Pochettino to dial back his preferred high-pressing, high-possession style. The team’s pressing rate has dropped to its lowest level under Pochettino at 47 percent.6 Instead, Pochettino has had to develop a more counterattacking approach. Tottenham has remained one of the best attacking teams in the league, with about 50 open-play expected goals created — good for fifth overall. But unlike the other clubs that rank high in this metric, Spurs complete surprisingly few passes into or within the penalty area. Arsenal completed 459 open-play passes into or within the penalty area to Tottenham’s 310. And yet Spurs have created only about three fewer expected goals from open play. This speaks to an unusually direct attacking style. Tottenham no longer pins its opponents back with a strangling high press but instead builds its attacks quickly to free individual players to run at the defense, in an attacking style that has been compared to college football’s “air raid.” To make passes inside the penalty area, a team must have multiple players already in the vicinity, and if one of these high-risk passes is left incomplete, fewer players are in position to stop a potential counterattack. Without the ball-winners in midfield to recover possession, such a deliberate approach would put Tottenham under too much pressure in transition. So instead, the team works more directly, bypassing midfield and quickly feeding a forward to face a defender one-on-one. This is the attacking style that Liverpool will have to defend against on Saturday in Madrid. Liverpool has consistently been able to use its midfield press to control matches and hold possession. But against Spurs, Jurgen Klopp’s side will have to be particularly wary of the direct counterattacks that have become the team’s go-to play. If Liverpool takes risks to overload Spurs defensively in possession, looking for some of these killer penalty area passes, it will be the Reds who are at risk of being countered. The table may be set for a more slow-paced and tactical Champions League final than either Klopp’s or Pochettino’s reputations might suggest. In the most recent meeting of the two sides on March 31, there were just 2.6 expected goals created from 25 shots. This is almost exactly average for a Premier League match7 but is far from the high-scoring barn-burner one might hope for from the matchup. If Liverpool won’t risk sending extra runners to support its forwards, while Tottenham invites some pressure and looks to counterattack, the game may slow down as it did at Anfield in March. The final, then, could come down to one team or the other relying on its superstar forward for a moment of magic. For Liverpool, this is not such a complicated question. Mohamed Salah had a great season, has returned to fitness after a head injury and is expected to be ready to go at full strength. Harry Kane is a bigger question for Tottenham. The English striker suffered his fifth ankle injury in the past three seasons against Manchester City in April and has not played since. He has rejoined the team and traveled to Spain, but can Spurs hope to get a vintage Kane performance? Last season when Kane returned from an ankle injury, his performance fell off badly. He had been running hot, averaging 0.98 expected goals and assists per 90 minutes in the 10 matches before he got hurt, and he dropped to about 0.73 in the 10 matches after his return. The same thing happened again this season after Kane suffered an ankle injury against Manchester United: His shot statistics declined from 0.88 xG+xA/90 to 0.52. But his ankle injuries in the fall of 2016 and spring of 2017 didn’t result in such a drastic downturn in performance. In 2017, he seemed to come back stronger than he was before the injury, going on a rampage down the stretch with 1.04 xG+xA/90 compared with 0.74 beforehand. Tottenham’s counterattacking style, and the team’s efficiency in turning passes around the penalty area into scoring chances, run best through an elite center forward. If Kane can return to form within a single game, that would give Spurs the best chance at a Champions League update. Getting an upset win probably requires a little bit of good fortune, and there is no better possible stroke of fortune for Tottenham than a fit Harry Kane. Check out our latest soccer predictions. ## Does Toronto’s Game 1 Win Spell Doom For The Warriors? sara.ziegler (Sara Ziegler, assistant sports editor): One game into the NBA Finals, and #WarriorsIn4 is already over. But what a first game! The Toronto Raptors led for most of the contest but weren’t able to put away the Golden State Warriors until the very end. tchow (Tony Chow, video producer): I’m gonna be honest. I was second screening Game 1 because my eyes were glued to the Scripps National Spelling Bee. I learned some new words that I’m gonna try to sneak in here, so you all better have your dictionaries ready — I’m about to drop some neil (Neil Paine, senior sportswriter): Omg, Tony sara.ziegler: Tony Though, I’m not gonna lie, I turned to that after the game was over. neil: Fortunately, the NBA can’t declare an eight-way tie for the championship. (Sorry, Celtics.) sara.ziegler: Chris, you’re in Toronto right now. What was the game like up close? chris.herring (Chris Herring, senior sportswriter): The atmosphere was incredible, and loud — both during the game and then pretty wild after. The fans here are insane. I think the game was what we hoped it would be, after years of watching relatively uncompetitive series with a team that couldn’t defend Golden State well enough. The Raptors’ defense is no joke, and it challenged the Warriors all game long. Toronto presents real problems for a club missing someone like Kevin Durant. neil: Yeah, Chris, this was the Warriors’ 20th-worst shooting game of the season by effective field-goal percentage. They still managed to get to the line, but they had a lot of turnovers, and Toronto held the non-Steph Curry scorers mostly in check. Fred VanVleet even did an admirable job keeping Curry from truly exploding. chris.herring: The Warriors shot 23 percent on contested shots last night, the worst mark they’ve had in a playoff game in the Steve Kerr era, according to ESPN Stats & Information Group. neil: And you have to think that Durant — one of the best tough-shot makers in history — would have boosted that some. chris.herring: Yeah. I’m really curious as to where Curry is going to have problems with VanVleet — we mentioned in our preview that he’d done very little scoring this season — averaging just 10 points per 100 possessions when VanVleet is the man defending him. That continued last night. sara.ziegler: FiveThirtyEight’s most valuable player (valuable in the most literal sense), Pascal Siakam had an amazing NBA Finals debut, scoring 32 points on 14-of-17 shooting. How surprised were you at how well he played? tchow: You could say Siakam was shining bright like a pendeloque last night. neil: LOL Our model definitely thinks highly of Siakam as a player, but I’m not going to say I saw him scoring 30+ going in. He had 32 points on 82 percent shooting!!! That’s the fourth-best shooting percentage in a 30-plus-point finals game EVER, according to Basketball-Reference.com. chris.herring: The Warriors got a lot of questions here about Siakam after last night’s performance. Draymond Green said it’s clear that Siakam is “a guy” now — meaning that we might not have treated him as a difference-maker before, but we sure as hell will now. sara.ziegler: He’s a now. chris.herring: Golden State basically acknowledged leaving certain guys open to begin the game in hopes of taking away Kawhi Leonard. That process worked, in a way. Leonard wasn’t efficient. But as a result, everyone else — particularly Siakam and Marc Gasol, who played brilliantly — got going. Danny Green was also himself again. And Golden State was never able to turn off that faucet. neil: Siakam might be a problem for the Warriors going forward. They didn’t have many good options to stop him. He scored 16 directly on Draymond. He also showcased his dangerous range as a 3-point shooter when rotations broke down or he trailed the play. chris.herring: I understand why GSW was willing to take that gamble with Siakam. He’s become very good from the corners but is right around 30 percent — if not worse — from above the arc. The real issue was letting him get whatever he wanted in transition. He was 5 of 5 in transition and hit 11 shots in a row at one point — the longest streak in a finals game over the last 20 years. As good as he is, that simply can’t happen in a game like that if you’re the Warriors. Golden State gave credit to Siakam but also largely chalked the game up to them not having seen this Raptors club before. They hadn’t played since early December, and Toronto has added Gasol, while Kawhi obviously took turns in and out of the lineup to rest. sara.ziegler: Yeah, no one was expecting this from Siakam, so game-planning it would have been tricky. chris.herring: I feel like I should get my apology in now. Although I don’t know if I’m apologizing to a person or an algorithm. neil: Or are you apologizing directly to CARMELO Anthony? Lol. chris.herring: Our model narrowly had Toronto winning this series. I ruled that possibility out pretty swiftly last week. But Thursday’s game was enough for me to think that their defense is good enough to win the series — particularly if Durant doesn’t return, and perhaps even if Durant is back but doesn’t jell right away after the long layoff. neil: I wanted to go back to what you said about loading up to stop Kawhi. Klay Thompson and Andre Iguodala did a good job limiting his efficiency, although it seems like that played a little into Toronto’s hands. Jackie MacMullan had a great reaction story about just how many other efficient options the Raptors have now if a team tries to focus too much on Kawhi. Only two of the seven Raptors who played at least 10 minutes averaged fewer than 1.2 points per individual possession, according to Basketball-Reference. (For reference, the Warriors as a team averaged 1.17 points per possession in the game.) sara.ziegler: And even with Kawhi bottled up, he still scored 23. neil: And! I worry about Iguodala’s health after he came up limping late. He did the bulk of the job guarding Leonard. tchow: So far, it looks like he’ll be OK, though. chris.herring: Yeah. That was the one other concern we mentioned in the preview: While the Warriors clearly could use Durant on offense, their defense becomes really, really thin on the wings without him. Especially if Iguodala is hurt or isn’t himself. This is now the second time he’s been banged up — he didn’t play in Game 4 against the Blazers, either. Speaking of Durant: The Raptors’ starting front court outscored Golden State’s 75-18. neil: sara.ziegler: Wow How much of a problem is that for the Warriors? If there’s no scoring help for Steph and Klay? neil: Certainly Draymond wasn’t much of a factor. Yes, he got the rare 10-10-10 triple double, but he also shot 2 of 9 from the floor and was a minus-8. chris.herring: They’re now 29-2 when he records a triple-double. neil: And both losses have come this postseason. chris.herring: I think what we saw yesterday is this: The Warriors, without KD, don’t have anyone who can shoot outside of Curry and Thompson. sara.ziegler: That seems … bad. chris.herring: I think Quinn Cook is probably the most reliable guy outside of those two. neil: How weird is it to think about the Warriors not having enough shooting? chris.herring: That’s where Durant’s ability to get his own shot comes in handy. He forces enough defensive attention to where he can play other guys open. Generally speaking, Steph often commands a second defender’s attention, so that’s enough to get someone else open and get the ball moving. It’s a tougher task when the other team can guard him and everyone else straight up. sara.ziegler: And Klay doesn’t really create his own shots. chris.herring: We haven’t talked much about DeMarcus Cousins’s return, but that’s both the blessing and the curse of having him You hope he can create an occasional double-team. But by the same token, his spot could have been used on a guard — and I think some people were of that opinion when they first got him: that the Warriors might have been better served by having another shooter. neil: Yeah, I thought the Warriors might go smaller and take somebody like Gasol out of the game, but either Kevon Looney or Jordan Bell played most of the game, and Gasol logged nearly 30 minutes. Meanwhile, Cousins played eight minutes and didn’t really do much of note. chris.herring: He looked a little rusty, but he made a few really nice passes. It’s tough to get your first playing time in weeks and weeks at this level, in the finals. Same may be true of Durant, honestly, if and when he comes back. tchow: It feels like it might be too soon to judge Cousins, but this is the problem of reintroducing someone like him back into the lineup during the finals. chris.herring: Exactly. neil: And that might be one of the ways our model was overrating the Warriors. It considered him one of the biggest talents of the series, which is true, but didn’t factor in the injury comeback. chris.herring: Not to mention the fact that Golden State has been better with Cousins off the court this season. Albeit with Durant playing more often than not. tchow: Yeah, Neil, it probably did overrate the Warriors because of his return. He ranked as the fifth most valuable player (behind Curry, Durant, Leonard and Lowry) according to our projections. neil: And at full health, that might be true in terms of skills. But that was a lot to expect with him easing back into playing. sara.ziegler: While Cousins did play a bit, the other injured Warrior was spotted high-fiving teammates behind the scenes. What did you make of Durant traveling with the team? neil: It has to be an encouraging sign for his chances of returning sooner rather than later, right? sara.ziegler: Is there a chance he plays in Game 2? chris.herring: No, it sounds like he won’t. Kerr was pretty firm about him needing to practice before having a chance to play. They’ll have another two practices — today and again on Saturday — before Game 2. But it doesn’t sound like he’ll be ready to practice here in Toronto before they suit up again Sunday. neil: The good thing about the finals is the sheer gap in days between games. Game 1 on a Thursday — Game 2 … all the way on Sunday. sara.ziegler: He has at least resumed “basketball activities,” which is my favorite phrase in all of basketball. neil: That reminds me, I need to go to the gym and “resume basketball activities” as well. sara.ziegler: So what do the Warriors need do to even the series? neil: Well, it seems obvious that Siakam won’t be down for 30+ again, so they have that going for them. chris.herring: Be a little less focused on stopping Kawhi to make sure that the other Raptors don’t overtake Jurassic Park again. And they have to slow Toronto down in transition, where the Raptors can be wildly efficient. It’s more of a question as to what they do differently on offense. But getting more stops and creating more opportunities to get out and run off those misses will ease some of that concern, I’d think. neil: Yeah, and that probably played a part in Toronto’s 24-17 disparity on fast-break points as well. Not enough stops turning into chances the other way. tchow: They have to play with rhathymia. (Am I using that right?) Just be the fun-loving Warriors we know. sara.ziegler: LOL tchow: I also agree with Neil in that the Warriors could afford to play smaller and get Gasol out of the game. He’s been solid all playoffs like an imbirussú for the Raptors. Otherwise, the Raptors could embarrass you again. Calembour intended. (OK, now I’m just forcing it.) neil: Tony, you’re banned from watching the spelling bee at work ever again. chris.herring: It’s a lot tougher for the Warriors to dictate the tempo without Durant. Playing smaller alone doesn’t get it done if you don’t have enough shooting to force the Raptors to come out and guard you on the perimeter. sara.ziegler: It’s interesting to me, too, that Kyle Lowry didn’t add much on offense again. He had as many field goals as charges forced. If he heats up, that’s a different wrinkle for Toronto. neil: Lowry continued his trend of being associated with strong Raptors play (+11) despite garbage individual stats. chris.herring: Frankly, if they’re getting what they got from everyone else — Green, Gasol and Siakam — they don’t need Lowry to do anything but bring energy. He had massive moments in that last series, and he’s always going to give you what he has on defense. It also helps a ton that VanVleet can stay attached to Curry so well in the minutes that Lowry is taking a breather. tchow: VanVleet was draped over Curry like a ferraiolone and actually guarded Curry for more possessions than Lowry in the end (33 possession vs. 16). chris.herring: O_______o neil: I’ve come around on this, Tony, and I applaud your spelling work here. tchow: Can we all pretend to be a marmennill for a minute? What do you think is going to happen now? Do the Warriors still three-peat? Do the Raptors pull this off? sara.ziegler: Our model (which accepts Chris’s apology) now has the Raptors at 63 percent to win it all. That feels right to me. chris.herring: The Raptors are the lone team that the Warriors haven’t beaten this season, and they have now won all three matchups against Golden State. I expect Golden State to respond. But stuff will get SO interesting if Toronto takes Game 2 as well. neil: 63 percent kinda makes more sense than our pre-series projection, to be honest. Home teams that win Game 1 of the NBA Finals win the series 78 percent of the time, historically. So this suggests that Toronto has far less of a talent edge than the typical home team that takes a 1-0 finals lead. Which is definitely true. tchow: This is anecdotal, but I was chatting with my cousin who lives in Toronto during last night’s game, and he said: “There’s just one guy outside our building somewhere screaming at the top of his lungs, ‘Let’s go, Raptors, over and over.” I can’t imagine what that guy will scream if the Raptors pull this off. That city is gonna be WILD. neil: I love seeing how excited Toronto fans are. (Drake aside.) Nav Bhatia was going nuts trying to distract Warrior free-throw shooters. chris.herring: I decided to walk home last night, about 35 minutes to my hotel. These two people were shouting “Let’s go, Raptors!” for entire blocks. I thought it was a crowd of people, and it was actually just those two guys. But between that, and all the car horns going off last night, people are on a noisy cloud here right now. Sort of how Milwaukee was to begin the last series. So we’ll see how it plays out. tchow: The city is gonna be as loud as a large flock of emberizines. From ABC News: Check out our latest NBA predictions. ## Can You Win The Lotería? Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. There are two types: Riddler Express for those of you who want something bite-size and Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,8 and you may get a shoutout in next week’s column. If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter. ## Riddler Express From Taylor Firman, the unluck of the draw: Lotería is a traditional Mexican game of chance, akin to bingo. Each player receives a four-by-four grid of images. Instead of a comically large rotating bin of numbered balls, the caller randomly draws a card from a deck containing all 54 possible images. If a player has that image on their grid, they mark it off. The exact rules can vary, but in this version, the game ends when one of the players fills their entire card (and screams “¡Lotería!”). Each of the 54 possible images can only show up once on each card, but other than that restriction, assume that image selection and placement on each player’s grid is random. One beautiful day, you and your friend Christina decide to face off in a friendly game of Lotería. What is the probability that either of you ends the game with an empty grid, i.e. none of your images was called? How does this probability change if there were more or fewer unique images? Larger or smaller player grids? ## Riddler Classic From Ben Wiles, a mathematical trip across the pond: My favorite game show is “Countdown” on Channel 4 in the UK. I particularly enjoy its Numbers Game. Here is the premise: There are 20 “small” cards, two of each numbered 1 through 10. There are also four “large” cards numbered 25, 50, 75 and 100. The player asks for six cards in total: zero, one, two, three or four “large” numbers, and the rest in “small” numbers. The hostess selects that chosen number of “large” and “small” at random from the deck. A random-number generator then selects a three-digit number, and the players have 30 seconds to use addition, subtraction, multiplication and division to combine the six numbers on their cards into a total as close to the selected three-digit number as they can. There are four basic rules: You can only use a number as many times as it comes up in the six-number set. You can only use the mathematical operations given. At no point in your calculations can you end on something that isn’t a counting number. And you don’t have to use all of the numbers. For example, say you ask for one large and five smalls, and you get 2, 3, 7, 8, 9 and 75. Your target is 657. One way to solve this would be to say 7×8×9 = 504, 75×2 = 150, 504+150 = 654 and 654+3 = 657. You could also say 75+7 = 82, 82×8 = 656, 3-2 = 1 and 656+1 = 657. This riddle is twofold. One: What number of “large” cards is most likely to produce a solvable game and what number of “large” cards is least likely to be solvable? Two: What three-digit numbers are most or least likely to be solvable? ## Solution to last week’s Riddler Express Congratulations to Adam Martin-Schwarze of Sequim, Washington, winner of last week’s Riddler Express! Last week we met a soccer coach who was trying to assemble a team of 11 players in a very specific way. He had an infinite pool of players to choose from, each of whom wore a unique number on their jersey such that there was one player for every number. That number also happened to be the number of games it took on average for that player to score a goal. The coach wanted his team to average precisely two goals per game, and he also wanted his weakest player to be as good as possible. What number does the ideal weakest player wear? What are the numbers of the other 10 players the coach should select? The weakest player selected for the team wears the number 24. The other 10 players wear the numbers 1, 5, 6, 8, 9, 10, 12, 15, 18 and 20. Let’s quickly check that everything adds up correctly. If a player’s number is 5, say, then that player scores an average of 1/5 goals per game. If their number is 6, they average 1/6 goals per game, and so on. So our team as a whole averages 1/1 + 1/5 + 1/6 + 1/8 + 1/9 + 1/10 + 1/12 + 1/15 + 1/18 + 1/20 + 1/24 = exactly 2 goals per game, just like the coach wanted! I’m not aware of a more elegant method of solving this coaching conundrum than basic guess-and-check. There are many possibilities to consider, but one thing we do know is that the fractions we’re adding up to try to get to a sum of 2 are of a specific type: 1 divided by a whole number, which are also known as Egyptian fractions. These made a prominent appearance in the so-called Rhind papyrus, which also happens to be the oldest known collection of math puzzles. We also know that we want the worst player on the team to be as good as possible — that is, to have as big a fraction as possible — and that there are 11 players on the team. So first we might check 1/1 + 1/2 + 1/3 + … + 1/11, i.e., the best possible team, but that sum equals about 3, too big for our coach. We then might check the sums of the possible sets of fractions between 1/1 and 1/12, and then the possible sets between 1/1 and 1/13, and so on, rejigging the sums until we find something that gets us to exactly 2. It turns out that none of these will add up to exactly 2 until we get to testing out those fractions between 1/1 and 1/24, and specifically those fractions listed above. Solver David DeSmet shared a handy computer program he wrote to churn through all these possibilities, and Martin Piotte shared his thorough accounting of the possible teams. ## Solution to last week’s Riddler Classic Congratulations to Curtis Bennett of Long Beach, California, winner of last week’s Riddler Classic! Last week, three astronauts were at the edge of their Mars lander, staring down at the surface of the red planet. Each wanted to be the first human to step foot on the planet, but they wanted to pick who it would be using a fair and efficient method. They could, for example, use a fair coin, assign each astronaut an outcome — heads-heads, heads-tails and tails-heads — and flip the coin twice. If the result was tails-tails, they could simply restart the process. However, that method could take a long time and there was exploration to be done. Another approach, however, would be to use an “unfair coin” — one for which the probabilities of heads and tails are not equal. Is it possible to make a fair choice among three astronauts with a fixed number of flips of an unfair coin? You were able to set the coin’s probability of heads to any number you like between 0 and 1. You could flip the coin as many times as you like, as long as that was some known, fixed number. And, you could assign any combination of possible outcomes to each of the three astronauts. Indeed it was possible, though who knew coin flips could get so complicated. This puzzle’s submitter, Dean Ballard, walks us through his solution: What makes this problem interesting is that at first glance it appears to require searching an overwhelmingly large space of possibilities. The trick to solving this more easily lies in a simplifying assumption. Instead of dealing with three different probability functions for the three different astronauts, we can assign two of them sets of head-tail combinations that will give them the equal chances of winning, independent of the weighting of our coin. This way, we can only worry about two things at once rather than three. We will need at least four coin flips to make this work. Let $$p(H)$$ be the probability that our specially designed coin lands heads. With four flips we have 16 possible outcomes: HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH and TTTT. Note that subsets of these 16, such as {HHHT, HHTH, HTHH, THHH}, all have the same probability — in this specific case, $$p(H)^3 (1 – p(H))$$. So let’s call these “3H1T” — three heads, one tail. Expressing the 16 outcomes this way gives us: one 4H, four 3H1T, six 2H2T, four 1H3T and one 4T, or in another piece of shorthand, [1, 4, 6, 4, 1]. Let’s say we assign Astronaut A just the HHHH and TTTT outcomes, and evenly divide the other 14 between Astronauts B and C. This gives us A = [1, 0, 0, 0, 1], B = [0, 2, 3, 2, 0], and C = [0, 2, 3, 2, 0]. Each of these defines a probability function for each astronaut. (Note that the sum A + B + C = [1, 4, 6, 4, 1], so all outcomes have been assigned.) The probability for Astronaut A equals 1 when $$p(H) = 0$$ or $$p(H) = 1$$, so it is concave up. The function for B and C equals 0 when $$p(H) = 0$$ or $$p(H) = 1$$, so it is concave down. Since both functions are continuous, as long as A’s value is less than B’s and C’s value when $$p(H) = 0.5$$ (which is true in this case), there must be a solution between 0 and 0.5, and another between 0.5 and 1. Here is what those solutions look like graphically. On the x-axis is $$p(H)$$ and on the y-axis is the probability that an astronaut wins the contest. Let’s quickly check the solution at point A, where the $$p(H)$$, the probability of our coin landing heads, equals about 0.24213. As we mentioned above, Astronaut A gets to take the first step if the coin lands with four heads or four tails. This happens with probability $$0.24213^4 + (1-0.24213)^4$$ = 0.3333, or a third, which is exactly what we want. Since the other two astronauts have been assigned probabilistically equivalent outcomes, we know they must have equal chances, which must also be a third. So we’ve successfully devised a fair method that will give us a result in a known and fixed number of unfair coin flips! For extra credit, you faced the same question but with five astronauts. I’ll spare you all the gory details, but suffice it to say that solver Zach Wissner-Gross devised one method that used eight flips of a coin that came up heads about 81.7 percent of the time. The colors correspond to the lucky astronaut who will get to make history based on the flips shown on the axes of the diagram. So, for example, if the coin came up eight straight heads — relatively likely given the weighting of the coin — then Astronaut Yellow gets to take those first steps. ## Want more riddles? Well, aren’t you lucky? There’s a whole book full of the best puzzles from this column and some never-before-seen head-scratchers. It’s called “The Riddler,” and it’s in stores now! ## Want to submit a riddle? Email me at [email protected] ## Hiring An SEO? Hear From Google What SEOs Do This Google video “How to hire an SEO” isn’t new but it’s to the point and vital to setting expectations. I encourage both SEOs and clients to watch this video and learn what Google says you should look for in an SEO. ## How to hire an SEO [embedded content] Transcript hi I’m Maile Ohye and I work with Google search I like to share advice to help you hire a useful SEO and prevent hiring a bad SEO one who you might pay a lot of money without positive results or even worse one who implements shady practices on your website that result in a reduction in search rankings SEO stands for search engine optimization – some SEO seems like black magic having worked with Google search for over a decade what I’ve learned is that first it’s not black magic and second if you want long-term success there aren’t any quick magical tricks that an SEO will provide so that your site ranks number one it’s important to note that an SEO potential is only as high as the quality of your business or website so successful SEO helps your website put your best foot forward so that it ranks appropriately in the spot where an unbiased potential customer would expect your site to be seen a successful SEO also looks to improve the entire searcher experience from search results to clicking on your website and potentially converting a good SEO will recommend best practices for a search friendly site from basic things like descriptive page titles for a blog or small business to more complex things like language markup for a multilingual global site SEO is ensure that you’re serving your online customers a good experience especially those coming from a search engine and that your site is helpful whether they’re using a desktop computer or mobile phone in most cases the SEO will need four months to a year to help your business first implement improvements and then see potential benefit my strongest advice when working with an SEO is to request if they corroborate their recommendation with a documented statement from Google either in a Help Center article video or Google a response in a forum that supports both one the SEO description of the issue that needs to be improved to help with ranking and to the approach they prescribed to accomplishing this tasks requesting these two bits of information will help prevent hiring a poor SEO who might otherwise convince you to do useless things like add more words to the keyword meta tag or by links because if you search for google advice on this topic you’d see blog posts and videos from us that clearly explain that adding keywords to the meta tag wouldn’t help furthermore while google uses links for page rank our documentation highlights that we strongly advise against the approach of buying links for the purpose of increasing page rank one basic rule is that in a majority of cases doing what’s good for SEO is also doing what’s good for your online customers things like having a mobile-friendly website good navigation and building a great brand additionally if you’re a more established brand with complicated legacy systems then good search friendly best practices likely involved paying off some of your site’s technical debt such as updating your infrastructure so that your website is agile and able to implement features faster in the long term if you own a small local business you can probably do the initial work yourself check out our 30-minute video series on how to build an online presence for your local business now if you still believe you want to hire an SEO here’s a general process one conduct a two way interview with your potential SEO check that they seem generally interested in you and your business to check their references three act four and you’ll probably have to pay for a technical and search audit 4 decide if you want to hire let’s break this down and start with step 1 conduct a two-way interview in the interview here are some things to look for a good SEO doesn’t focus only on search engine ranking but how they can help your business so they should ask questions like what makes your business content and/or service unique and therefore valuable to customers they want to know this information to make sure it’s highlighted on your website for your current and potential new audience – what does your common customer look like and how do they currently find your website 3 how does your business make money and how can search help for what other channels are you using offline advertising social networks 5 who are your competitors what do they do well online and potentially offline if the SEO doesn’t seem interested in learning about your business from a holistic standpoint look elsewhere it’s difficult to do good SEO without knowing about a business’s goals their customers and other existing marketing efforts SEO should complement your existing work the second step in hiring an SEO is to check references if your potential SEO provides prior clients be sure to check their references you want to hear from past clients that the SEO was able to provide useful guidance and worked effectively with their developers designers UX researchers and our marketers a good SEO should feel like someone you can work with learn from experiment with and who generally cares about you and your business not just getting your site the highest rank as ultimately those techniques rarely last long if they work at all they’ll want to educate you and your staff on how search engines work so that SEO becomes part of your general business operations step 3 is to request a technical and search audit if you trust your SEO candidate give them restricted view not full or right access to your Google search console data and even your analytics data before they actually modify anything on your website have them conduct a technical and search audit to give you a prioritized list of what they think should be improved for SEO if you’re a larger business you can hire multiple SEO to run audits and prioritize improvements see what each has to say and then determine who you could work with the best in the audit the SEO should prioritize improvements with a structure like one the issue to the suggested improvement 3 an estimate on the overall investment in other words the time energy or money it would take for your developers to implement the improvement and for Google search as well as searchers and customers to recognize the improvement the SEO will need to talk with your developers to better understand what technical constraints may exist for the estimated positive business impact the impact might be a ranking improvement that will lead to more visitors and conversions or perhaps the positive impact comes from a back-end change that cleans up your site and helps your brand be more agile in the future five a plan of how to iterate and improve on the implementation or perhaps how to experiment and fail fast should the results not meet expectations that covers the structure of the technical and search audit now let’s talk about each of these audits individually in the technical audit your SEO should be able to review your site for issues related to internal linking crawl ability URL parameters server connectivity and response codes to name some if they mention that your site has duplicate content problems that need to be corrected make sure they show you the specific URLs that are competing for the same query or that they explained it should be cleaned up for long term site health not initial growth I mention this because lots of duplicate content exists on web sites and often it’s not a pressing problem in this search audit your potential SEO will likely break down your search queries into categories like branded and unbranded terms branded terms are those with your business or website’s name like a search for Gmail is a branded term while the search for email is an unbranded or general keyword an SEO should make sure that for branded queries such as Gmail your website is providing a great experience that allows customers who know your brand or website to easily find exactly what they need and potentially convert they might recommend improvements that help the entire searcher experience from what the searcher sees in search results to when they click on a result and use your website for unbranded queries an SEO can help you better make sense of the online competitive landscape they can tell you things like here are the types of queries it would make sense for your business to rank but here’s what your competition is done and why I think they rank where they do for instance perhaps your competition has great reviews really shareable content or they run a highly reputable site an SEO will provide recommendations for how to improve rankings for these queries and the entire searcher experience they’ll introduce ideas like update obsolete content they might say your site is suffering because some of your well ranking content is obsolete has poor navigation a useless page title or isn’t mobile-friendly let’s improve these pages and see if more website visitors convert and purchase or if they can micro convert meaning that perhaps they subscribe or share content improve internal linking your SEO might say your site is suffering because some of your best articles are too far from the homepage and users would have a hard time finding it we can better internally link to your content to feature it more prominently generate buzz the SEO might say you have great content but not enough people know we can try to get more user interaction and generate buzz perhaps through social media or business relationships this will help us attract more potential customers and perhaps garner natural links to your site learn from the competition your SEO might explain here’s what your competitors do well can you reach parity with this and potentially surpass them in utilities or can you better show customers your business’s unique value again a good SEO will try to prioritize what ideas can bring your business the most improvement for the least investment and what improvements may take more time but help growth in the long term once they talk with you and other members of your team such as developers or marketers they’ll help your business forge a path ahead the last thing I want to mention is that when I talk with SEO s one of the biggest holdups to improving away site isn’t there recommendation but it’s the business making time to implement their ideas if you’re not ready to commit to making SEO improvements while getting an SEO audit may be helpful make sure that your entire organization is on board else your SEO improvements may be non-existent regardless of who you hire so that wraps it up thanks for watching and best of luck to you and your business Hits: 2 Hiring An SEO? Hear From Google What SEOs Do ## The WNBA Is Uniquely Suited To Survive Its Many Star Absences When it was announced this week that Seattle Storm point guard Sue Bird would miss the 2019 WNBA season with a knee injury, it was just the latest in a long line of maladies that have struck the league this year. In addition to Bird, the sport has seen absences take out her teammate Breanna Stewart (reigning WNBA MVP), 2014 MVP Maya Moore of Minnesota,1 Atlanta’s Angel McCoughtry, Dallas’s Skylar Diggins-Smith,2 Indiana’s Victoria Vivians, Las Vegas’s Lindsay Allen and Phoenix’s Diana Taurasi for either all or most of the 2019 season. It’s a shocking and unprecedented spate of missing stars that figures to drastically change the championship picture for multiple teams, including — probably first and foremost — the Storm, who won the 2018 WNBA title based on huge contributions from Stewart and Bird. Together, the players listed above generated 30.5 wins last season according to Wins Created — a combination of Basketball-Reference.com’s Win Shares and the Player Efficiency Rating-based Estimated Wins Added metric.3 If all of them do miss the whole year, it would represent the most wins from any single season that were then completely removed from the WNBA ecosystem in the next: Even if we hedge on Taurasi, who currently plans to come back in the second half of the season after missing 10 to 12 weeks of action, the resulting 24.5-win tally would rank fifth-highest in WNBA history before accounting for any of the additional players who will inevitably join the ranks of the absent as the season goes on. (Diggins also plans to return from her pregnancy leave before season’s end.) The season structure of women’s pro basketball — including the year-round overseas grind that stars must engage in to make money — probably has contributed to the WNBA’s problems this spring. With so much mileage being put on top players’ bodies, a nightmare offseason like this was bound to happen sooner or later. But if there is a silver lining for the WNBA (and there isn’t much of one when considering the prospect of a season spent without players of Moore’s or Stewart’s caliber), it’s that the WNBA draws from an exceptionally deep pool of talent. With only 12 teams in the league and an entire world of basketball players to pull from, it might be the most competitive sports league on the planet — at least in terms of the likelihood that any given youth player actually makes the highest level of the sport. My colleague Ben Morris wrote a few years ago about this depth of talent in women’s basketball. He found that for every player who makes a Division I college roster, there are 87 high school players participating in girl’s basketball across the U.S. The only sport whose rosters were more competitive was men’s basketball, in which 101 high schoolers participate for every player who manages to make a Division I team. And, again, that was framed in terms of college basketball. Relative to the number of roster slots available in the WNBA, the amount of talent swimming in the pool becomes even more staggering. In the 2018 season, 157 players suited up for one of the WNBA’s 12 teams. According to an updated version of the same data set Morris used, there were 412,407 girls basketball players in U.S. high schools during the 2017-18 season — meaning there were about 2,627 high school girls playing basketball for every roster slot available in the (American) pros. By comparison, over the same span, there were 540 NBA players and 551,373 participating boys basketball players — or 1,021 for every top-level pro roster slot. By this accounting, it’s more than two and a half times easier (!) for the typical U.S. boys basketball player to make the NBA than it is for his counterpart in girls basketball to make the WNBA. Even if you grant the presence of extra roster slots available overseas,4 the amount of untapped talent in women’s basketball is mind-boggling. That depth of talent won’t necessarily be able to easily replace all of the stars who’ll be missing this season’s WNBA action. But it does mean there are more opportunities than ever for players in the next tier to showcase their skills. And that could be a good thing — because as the numbers show, the WNBA (perhaps more so than any other sport) has a group of unheralded players waiting in the wings, ready to take advantage of the chance. ## Are The Democratic Debates Already A Mess? Welcome to FiveThirtyEight’s weekly politics chat. The transcript below has been lightly edited. sarahf (Sarah Frostenson, politics editor): Republicans struggled with setting debate criteria during the 2016 presidential election because of their large and unwieldy field, and Democrats seem as though they’ll have their own issues in 2020. We already count 20 candidates who have qualified for the first two debates via one of the two criteria the Democratic National Committee has set up: receiving at least 1 percent in at least three qualifying polls or having 65,000 people donate to their campaign, with at least 200 donors in 20 different states. The DNC has said that it will cap participation at 20 candidates, so the next candidate who qualifies, via one of the two criteria for entry, will trigger the tiebreaker rules. Those get complicated fast, but the topline is: If more than 20 candidates qualify, then meeting both the polling and donor requirements will be paramount for candidates — those who do will get first dibs on debate lecterns. But why is it so hard to figure out a fair metric for inclusion? Is there a better way to determine who makes the debate stage? julia_azari (Julia Azari, political science professor at Marquette University and FiveThirtyEight contributor): It’s difficult to figure out a fair metric for inclusion because the whole process is weird. Ideally, it’s both inclusive and efficient (i.e., it narrows options for a nominee relatively quickly), but it’s not really possible to do both at the same time. geoffrey.skelley (Geoffrey Skelley, elections analyst): Right, and in the aftermath of the 2016 Democratic nomination, when the DNC was criticized for “rigging” the debates for Hillary Clinton, the DNC really wants to seem transparent and inclusive. natesilver (Nate Silver, editor in chief): So, 1) It’s good to have objective criteria, 2) as objective criteria go, fundraising and high-quality polling is perfectly fine, but 3) the DNC set the bar too low. Getting donations from 65,000 people is not that hard. And polling at 1 percent in any of three polls out of the many, many polls out there is even easier, probably. sarahf: Although, to be clear, the DNC is not counting all polls from all pollsters. It has said, however, that it’ll consider both national and early-state polls, and qualifying polls can come from 18 different organizations). geoffrey.skelley: Yeah, it’s still pretty easy to qualify via three polls at 1 percent or more — 19 Democrats have already done that. However, if the DNC had set the threshold at 2 percent or more, just eight candidates would meet that mark. ##### Only 8 candidates are polling at 2 percent or more Democratic presidential candidates by whether they have received at least 1 percent or 2 percent support in at least three polls that would qualify them for the first Democratic presidential debates, as of May 21, 2019 IN at least 3 DEBATE-QUALIFYING POLLS, HAS SUPPORT OF … Candidate 1 percent or more 2 percent or more Joe Biden Cory Booker Pete Buttigieg Kamala Harris Amy Klobuchar Beto O’Rourke Bernie Sanders Elizabeth Warren Steve Bullock Julian Castro Bill de Blasio John Delaney Tulsi Gabbard Kirsten Gillibrand John Hickenlooper Jay Inslee Tim Ryan Eric Swalwell Andrew Yang Michael Bennet Seth Moulton Marianne Williamson For candidates deemed “major” by FiveThirtyEight. Sources: Polls, Media reports natesilver: Yeah, hitting 1 percent is soooooooooo easy. Like people can literally just pick your name at random almost. The DNC is spending too much time trying to avoid mistakes they think were made in the previous Democratic nomination process when there are probably more lessons to be learned from the Republican nomination process. geoffrey.skelley: Well, part of what the DNC wanted to avoid was the mistakes the Republicans made in the 2016 cycle with prime time and undercard debates. nrakich (Nathaniel Rakich, elections analyst): I think the Democrats have already done a better job than Republicans did in 2016. The DNC has said that they’ll randomly distribute candidates across the nights, rather than hold “varsity” and “junior varsity” debates. I think that’s a good move. natesilver: Oh, I’m not sure I agree with that, Nathaniel. nrakich: How is a junior varsity debate better, Nate? My problem with splitting the candidates up by tier is that it requires splitting hairs between a candidate who gets, say, 3 percent in a poll and a candidate who gets 4 percent. (Margins of error are real!) I guess it’s fine to argue that you think the threshold should be higher and there should be only one main debate, but if you are going to split the candidates into two debates, I think randomly doing it is the only good way. natesilver: Well, if you wind up stuck in the JV debate because you poll at 2 percent rather than at 3 percent, I don’t have much sympathy for you, even though that’s a minor difference. nrakich: But the debates are candidates’ chance to raise their polling numbers up from that 2 or 3 percent. Debates should start off inclusive but probably get less inclusive as we get closer to voting. Like, the New Hampshire debate three days before the primary should probably only have the candidates with a serious chance of winning that primary. nrakich: My beef with using polling averages as a debate criterion is that they assume that candidates can be precisely ranked by their standing in the polls. But in reality, polls are imprecise instruments, and you can’t do much more than lump candidates into rough categories (and even those have fuzzy boundaries). For example, all candidates polling between 0 and 5 percent are basically in the same spot. julia_azari: I agree with Nathaniel here. I would also add that these differences don’t, in my mind, clearly differentiate candidates. And does it really matter if it’s 20 or 22 candidates on the stage? Either isolate the top-tier candidates or let everyone in. sarahf: Julia, the number of evenings we have to devote to watching the debates is at stake! julia_azari: If other people haven’t blocked off all of 2019 and 2020 to watch debates, that’s not my problem. People want an open nomination process. This is where that goes. nrakich: Some pollsters have also said that they are uncomfortable with their work influencing elections. Their role is as measurers, not active participants. natesilver: Meh, the pollsters complain too much. If you believe in the quality of your poll, you shouldn’t have any problem with it being used as an objective metric. I think they should literally have tiers on stage based on where you’re polling. nrakich: Nate take natesilver: So like Joe Biden and Bernie Sanders are on the top tier and have big giant podiums. And Swalwell is in the cheap seats in like a broom closet. julia_azari: This chat is a serious warning about overpopulated debates, and there are only five of us. natesilver: I do think for this first debate, they might as well just let everyone in. And then set the criteria a lot higher for future debates. geoffrey.skelley: But the polling average tiebreaker might not even solve things. Say there are a few candidates who have a bunch of polls in which they are hitting only 1 percent. If the polling average can’t settle a tie, it comes down to the number of qualifying polls a candidate has. But what if three or four candidates have the same number of qualifying polls? It’s going to be a mess one way or the other. natesilver: Again, though, I’m realllllllly not sympathetic to the borderline cases. The primary has been underway for a while now, and if you can’t both get 65,000 donors AND poll at 1 percent in three polls, there’s probably something pretty wrong with you. And I’d rather give more time to, say, Cory Booker or Amy Klobuchar to make their cases and less to Eric Swalwell or Bill de Blasio. julia_azari: This is a recurring problem for parties. They try to solve a lot of these problems informally by limiting who runs.But when these conversations break down like they did in 2016, the formal solutions — like trying to come up with a fair threshold for inclusion in a debate with so many candidates — show why those problems were being solved informally: It’s a mess. natesilver: Do we think the debate rules factored into how many candidates have decided to run? Mike Gravel, whom we don’t consider a major candidate yet, explicitly seems to have run based on the possibility that he’d get 65,000 donors and therefore some sort of platform to talk about U.S. imperialism or whatever. nrakich: Good question. Probably not? There are other ways to get media attention aside from the debates — it looks like every candidate is getting a CNN town hall, for example. And a few candidates have jumped in so late that it’s not clear whether they’ll make the debates at all, like Seth Moulton and Michael Bennet. So why are they running? geoffrey.skelley: I don’t know — it could have pushed a few candidates who were on the fence. julia_azari: That’s hard to know, but what’s interesting to me is that not that long ago, debates were mostly about getting the top-tier candidates to show up. Now, even though the evidence that they matter is somewhat mixed, they’ve taken on this whole different significance because of the record number of candidates and the scramble for inclusion. sarahf: So what good are debates, Julia, especially this far out? julia_azari: Well, the default position in political science tends to be that not that many people are watching and that those who are have already made up their minds. But the latter point is a bit different for a primary debate, since partisanship doesn’t shape decisions in the same way. sarahf: Right, here at FiveThirtyEight, we’ve been saying things won’t get interesting until the debates! julia_azari: So on the one hand, there’s not really hard evidence that debates affect who wins the primary. (Studies do suggest that debates might affect citizens’ perceptions of personality and viability to win the nomination.) But usually the primary is … not that competitive. The 2008 Democratic primary really stood out in this regard, because there were two strong contenders through most of the primary season, making the contest a real competition. sarahf: Yeah, I think the debates will stand out this year, too, as they’ll be one of the first opportunities for people to get to hear from the candidates directly (outside of a CNN town hall, which, as FiveThirtyEight’s Clare Malone has noted, can be overly orchestrated to begin with). geoffrey.skelley: And primary debates can certainly make or break a candidate — earlier this year, I examined their effects. Rick Perry in the 2012 GOP primary debates really stands out to me because after he defended Texas’s in-state tuition policy for undocumented immigrants, his standing among Republicans plummeted. It was much worse than when he forgot the name of the third federal agency he wanted to dismantle! nrakich: I feel like the debates are one of the events in the Olympic Games that are the primary season. You have to participate in them and be rated favorably by the judges (the media) in order to win gold. natesilver: But quite a few people watch at least relative to the size of the Democratic electorate, don’t they? Here’s some ratings data on the 2016 Democratic primaries from Wikipedia: By comparison, 31 million people voted in the Democratic primaries in 2016. So having an audience of 16 million for the first debate isn’t bad compared with 31 million! nrakich: It’s interesting how viewership dropped off so starkly after the first debate. natesilver: That may have happened because I don’t think either Sanders or Clinton were particularly interesting debaters. They were perfectly competent, but not interesting. sarahf: Do you think candidates who go the second night will be disadvantaged? I realize Democrats aren’t splitting the debates into a varsity and JV debate, but maybe one debate will be enough for folks? geoffrey.skelley: Depends on who is in each debate. If it’s a random draw but a number of leading candidates end up in one debate, that debate will probably get the most attention. natesilver: There might be a wee bit of fatigue, Sarah, but it probably depends more on the draw. If Biden, Sanders and Warren are all on the second night, that’s the one most people will care about. But if the heavyweights are all on the first night, the second night could feel like more of a JV affair. geoffrey.skelley: Yeah, but if the heavyweights are all grouped together, I think that could still be good for some of the underdog candidates. It could give them an opportunity to stand out without facing the same “main event” vs. “undercard” judgment that was explicit in how the GOP handled things in 2016. julia_azari: I don’t know. I’m going to remain on team skepticism about 2016 Republican type ratings. It’s possible that people will tune into these debates with a genuine eye toward actually deciding between candidates or learning more about some candidates. But I don’t expect that these debates will draw in Trump-level ratings. The Democratic field is crowded, but it doesn’t have an animating rivalry between two candidates and it’s not a clown show. sarahf: … at least not yet! There’s still so much we don’t know. julia_azari: But people weren’t watching in 2016 because they wanted to hear the finer points of Marco Rubio’s tax plan vs. Ted Cruz’s. There was a show-biz factor with Trump, to put it politely. And he delivered consistently enough. nrakich: I dunno, Julia, I’m pretty worked up about the Swalwell vs. Hickenlooper rivalry. sarahf: Nathaniel Is there another debate matchup you all are looking forward to? natesilver: Trump was uniquely unpredictable in the context of the debates, so I’m not sure whether there will be a point of comparison. But you will have the dynamic of other candidates working to take the front-runner down, which has both potential risks and rewards for the front-runner. I think the first debate is probably more likely to hurt Biden than help him, however. geoffrey.skelley: The lack of a Trump-like figure will certainly make a difference. But it could get really interesting if Biden and Sanders are on stage the same night. One could easily imagine Sanders going after Biden straight away, just as he did with Hillary Clinton in 2016. natesilver: I mean, I think debates sometimes tend to cause reversion toward the fundamentals. So if we think Biden’s numbers are a little bit inflated right now by a post-announcement boost, and I think they probably are, he’s more likely to decline than improve. julia_azari: Counterpoint: Biden is actually quite good in these settings. His experience helps as he’ll be less likely to go deer-in-the-headlights on a specific question. And he really knows how to work emotion, if you recall his performance in the 2008 VP debates. natesilver: Who do we expect to be an effective debater? Kamala Harris? Elizabeth Warren? Pete Buttigieg? Although, maybe it’s not a good thing if expectations are high. Everyone’s going to expect Harris to be super incisive with every response and for Buttigieg to speak Norwegian or something. julia_azari: I am OUT if I have to learn Norwegian for these debates. I think people expect Warren to be wonky and unlikable, but my impression is that she’s actually pretty good in front of a crowd, so maybe she’ll do well. natesilver: For Warren, I think you can argue that she is someone for whom the fundamentals are misaligned. She’s an “objectively” strong candidate and “should” be doing better (I know how loaded those terms are — it’s a chat, so give me a break). Maybe the same is true for Harris. So they both stand to gain. Or to put it another way, if Harris and Warren don’t benefit from the debates, then maybe we have to start concluding that they’re products that voters just don’t like very much for whatever reason. sarahf: So who … do we think won’t make the debate stage? Because it does seem as though we’re headed toward some sort of tiebreaker, right? nrakich: Maybe Marianne Williamson? She’s the only candidate currently who’s qualified via the donors criterion but not the polling criterion. sarahf: If Marianne Williamson is the one who’s cut … it’s kind of like what was the point of the DNC introducing the 65,000-unique-donor threshold anyway. geoffrey.skelley: But Williamson only needs to earn 1 percent support in one more survey to qualify via polls. So I actually like her chances if it comes down to a polling-average tiebreaker because she might hit both the polling and donor criteria. And yeah, Sarah, that’s a big question mark: How many of the candidates who have qualified via polls but not via donors will actually get 65,000 donors? It sounds like Inslee is close on the donor count, for instance. But what about John Hickenlooper or Kirsten Gillibrand or John Delaney, etc.? I haven’t found any new information about their donor counts. natesilver: There’s no particular reason to limit it to 20 candidates instead of 21 or whatever. nrakich: We live in a base 10 world, Nate. Get used to it. natesilver: But it just sort of seems to defeat the purpose of being inclusive if you’re excluding just Williamson. Moulton might not make it. geoffrey.skelley: Yeah, Moulton is the one who is really up a creek without a polling paddle — he doesn’t have a single qualifying survey yet. nrakich: The new hot take: I should be considered a serious candidate for president even though I have raised no money whatsoever. natesilver: Sorry, but you’re not a major candidate according to our criteria, Rakich. sarahf: OK, so as we’ve discussed, there are pros and cons to having a debate stage as wide-ranging and inclusive as what the DNC has settled on. But it’s also really hard to do any of this fairly. So to end today’s chat, what would you have liked to see the DNC do differently? julia_azari: I mean, the DNC is in somewhat of a no-win position, but given that I’m not sure they can actually regain (or gain) legitimacy by having 20-candidate debates, it might have made sense to just raise the thresholds to begin with. nrakich: Overall, I think the DNC did well. The criteria are arbitrary, sure, but they’ve turned out to be well-calibrated, at least for someone like me who wants initial debates to include (almost) everyone. geoffrey.skelley: I think 10 Lincoln-Douglas debates between pairs of candidates would be the best approach. Oh sorry, Newt Gingrich took over my Slack account for a second there. But seriously, I think the DNC could’ve made a case for higher thresholds, such as polling at 2 percent instead of 1 percent. nrakich: I think this chat did convince me that stricter thresholds are appropriate for later in the primary season, closer to the actual voting. We’ll see if the DNC agrees. natesilver: I think maybe there should have been both a money qualifier and a donors qualifier for the donor threshold. Like, you have to raise donations from 65,000 people and raise at least5 million, or something.
That’s basically what airlines’ frequent flier programs do now — you have to fly a certain amount of miles and spend a certain amount of money.
nrakich: The DNC should be more like airlines — there’s a winning electoral position!
natesilver: ThE AiRLiNe InDuStRy Is UnFaIrLy MaLiGnEd
julia_azari: This debate has been canceled due to mechanical failure. Tomorrow, we fly you to Poughkeepsie instead of Atlanta.
natesilver: And if I were the DNC, I’d stipulate my criteria for future debates sooner rather than later. Because otherwise it’s going to look like they’re engineering the rules around which candidates they do/don’t like.
geoffrey.skelley: Which would defeat the point of being so inclusive in the first place.
## It Turns Out The Vintage Warriors Are Still Pretty Good At Basketball
sara.ziegler (Sara Ziegler, assistant sports editor): The NBA conference finals are just three games old, but we’ve already seen two of the most entertaining games of the entire playoffs.
After Golden State easily dispatched Portland in Game 1 in the West, Milwaukee needed a furious comeback to take down Toronto in the East’s first game. And then came Thursday night, when the Trail Blazers led the Warriors by as many as 17 points in the third quarter, but Golden State used a 27-8 run to get back into the game. The teams traded leads down the stretch, but the Warriors prevailed.
tchow (Tony Chow, video producer): The “Warriors are better without Kevin Durant” crowd has gotten REALLY loud.
I’m not stupid enough to say they’re better without KD, but I can see the argument being made that they might be more fun to watch?
natesilver (Nate Silver, editor in chief): Tony, that feels like a way to rationalize the idea that KD will feel dejected or something by the Warriors because they can win without him so he’ll have to come to the Knicks.
sara.ziegler: LOL
tchow: I’m still auditioning for my Knicks GM job, Nate.
chris.herring (Chris Herring, senior sportswriter): I think they are more fun to watch this way, for sure. It’s a good reminder of what they were before Durant ever signed with them. The up-tempo, heavy ball-movement, “we can be down by 15, but still come back to beat you” Warriors.
I think Portland losing on Thursday was pretty brutal. It’s sounding more and more like Durant won’t be back in the conference finals, and a win would have gone a long way toward making this a series again. It’s hard to imagine them winning four of the next five.
tchow: You’re not kidding about the heavy ball movement, Chris. Per Second Spectrum, the Warriors have averaged 42 more passes per 100 possessions when KD was not on the floor during these playoffs.
natesilver: I guess the question is whether the Warriors could win grind-it-out, slower-paced, half-court-type games at the same rate without KD.
chris.herring: And that’s the thing. When the Warriors play that way, it’s changing the pace of the game. If you have a game with fewer possessions, I’d venture to guess it leaves things to random chance more often and helps the underdog.
Kind of why Virginia was seen as vulnerable in the NCAA Tournament for so long. (A loss to UMBC helps with that, too.)
natesilver: Beating Portland twice at home is just not all that rigorous a test, however.
tchow: That’s important to keep in mind. All the Warriors did was hold home court.
chris.herring: It may not be. But the Blazers played really well on Thursday, and then that third quarter happened. I just think we’re used to these sorts of onslaughts at this point.
tchow: Yeah, even with that scoreline at halftime, after the first three minutes of the third quarter, I think all of us kinda went, “Oh, the Warriors are winning this.”
natesilver: The Game 6 closeout against Houston, in a game where the Rockets played pretty well, was impressive. But I’m still not sure I really have a great sense for how Golden State is going to match up with Milwaukee or Toronto, with or without KD.
sara.ziegler: A Portland win would have completely changed the tone of this series. And it was close to happening — even after the Warriors stormed back!
natesilver: “Were the Blazers actually close to winning or was it all just an illusion” is a fun epistemological question. I mean, obviously, a win probability model or whatever would have them ahead for a lot of the game. But the Warriors have made SO many third-quarter comebacks over the years that I just don’t really know.
sara.ziegler: When the Blazers were up 8 with 4:28 left, I thought they could really win it.
Silly me.
chris.herring: I grow somewhat tired of the Curry vs. Curry storyline at times. But it was pretty awesome to see Seth play so well last night, and to try to get into his brother’s head at one point.
Crazy to think that, if Pau Gasol were healthy, there would be two sets of brothers playing against each other this round.
tchow: That’s very interesting. I’m kinda loving the Curry vs. Curry storyline. It’s pretty cool IMO to have siblings play against each other at such high stakes.
I found myself pingponging between “Where’s Steph? OK, where’s Seth now?” when they were both on the court.
chris.herring: I like the storyline. I just think it’s being milked pretty heavily in terms of showing their parents in the crowd, that’s all. But Seth was huge last night.
I think the challenge for Portland is that there’s a lot of “your turn, my turn” from Damian Lillard and CJ McCollum. McCollum owned the first half, and then Dame got hot in the second half.
And it kind of feels like they may need more of a balance, or another huge bench performance from someone, to get over this hump.
natesilver: What if Seth Curry woke up one day and had Steph Curry’s skills, and vice versa? That feels like a weird/bad movie plot.
tchow: “Freaky Friday 2”
natesilver: Would the Blazers play McCollum at the 3 or something? It would be a really weird team.
chris.herring: I already feel like it’s a weird team as is.
Credit to them for adjusting heavily after how bad Game 1 was.
tchow: You knew they had to do something about that pick-and-roll defense.
chris.herring: Enes Kanter was back at the free-throw line in Game 1 and then moved much farther up to contain their pick and rolls in Game 2. That made Golden State’s looks far more challenging, which you almost have to do in order to have a chance.
sara.ziegler: The Blazers didn’t get much on offense from Kanter on Thursday, though. What was going on there?
chris.herring: His impact is going to be a bit less on a night where they shoot as well as they did from three. Because he doesn’t get any offensive rebounds that way.
But also, when he’s playing so much higher up on D, it probably wears him down a bit.
Not to mention the fact that he’s fasting during daylight hours, which seems like such a tough thing to do during such a high-stakes series.
sara.ziegler: That does seem brutal.
chris.herring: Now THAT storyline I find fascinating.
sara.ziegler: I can barely edit when I’m hungry. Can’t imagine trying to play basketball at the highest level!
natesilver: If I fasted during daylight hours, I don’t think I could even do a Slack chat, let alone play in an NBA game.
sara.ziegler: Haha
tchow: Muslim soccer players do it all the time! (during Ramadan)
It is pretty cool the Blazers have three Muslim players on the roster (Kanter, Jusuf Nurkic and Al-Farouq Aminu).
chris.herring: Hakeem Olajuwon did it as well, and apparently Kanter reached out to him to figure out what all he did to maintain his game during that stretch of the postseason.
natesilver: I didn’t realize that the dates of Ramadan shift around a lot from year to year. It doesn’t always coincide with the playoffs.
sara.ziegler: What, if anything, can the Blazers do to turn the tide as the series heads back to Portland?
chris.herring: I think it goes without saying that they did enough to win Thursday.
You’d imagine they can control the tempo better at home than they did at Oracle, where the Warriors play extremely fast and in transition during those ridiculous comebacks. I think maybe Terry Stotts would call timeout when he feels one of those runs coming on. And they need to clean up some mistakes, in terms of fouling and taking care of the ball. Andre Iguodala made a great steal on Lillard on the final play, and Lillard had that pretty brutal foul on Steph while he was shooting a three late.
tchow: I’m actually not sure what else they can do. They played well on Thursday and still lost. I feel for Portland fans, I really do. But our predictions give them a 6 percent chance of making it to the finals which seems … high?
chris.herring: Realistically, unless Golden State has another major injury, that was probably it. I don’t see a whole lot of adjustments for a scenario where you were in control most of the game. You just have to finish the game. Period.
natesilver: I guess the one piece of good news for Portland is that it’s not obvious that KD’s going to play any time soon.
tchow: Chris mentioned that they needed another huge bench performance to have a chance, but both Rodney Hood and Seth Curry had pretty decent games. I don’t know where else it could come from. Zach Collins?
sara.ziegler: Meyers Leonard! He had a pretty good game.
chris.herring: Collins had five fouls in eight minutes yesterday, somehow. Leonard was impactful, though.
tchow: Yeah, some of those Collins fouls were bad fouls, too.
chris.herring: That’s why it’s hard to see Portland doing this: Everything seems really scattered right now.
Also, props to Draymond Green for raising his game to a ridiculous level lately. You can’t mention the Warriors looking like the Warriors of old without talking about how incredible he’s been on both ends.
natesilver: Maybe Draymond secretly hates KD and so ups his effort level when KD is out?
sara.ziegler: LOL. I kind of want that to be true. Since the NBA is just a soap opera, at its core.
tchow: “The Plays of Our Lives”
I’m sorry.
sara.ziegler: OMG, yes.
Moving on to the East: Chris, you wrote after Game 1 that the Raptors would likely be kicking themselves for letting that get away from them. How important was that outcome to the series?
chris.herring: Not nearly as much of a killer as Game 2 for Portland. But still potentially big.
There’s that saying that a series hasn’t begun until a road team wins a game. And on some level, that may be true. I just think that if you’re going to beat Milwaukee, it makes sense to grab the winnable game when it’s there. And the Bucks played really poorly in some regards, yet they still won. They are a complete team, whereas the Raptors look very stilted on offense at times.
And it’s part of why I continue to like Milwaukee’s chances of winning this whole thing.
tchow: It’s been really impressive seeing how well the Bucks have continued to play when Giannis Antetokounmpo is not on the floor.
natesilver: The thing I’d hate if I were a Raptors fan is that I felt like my team played pretty well in Game 1, and it still wasn’t enough. Obviously, not everything was perfect — the cold shooting in the fourth quarter — but it felt like a relatively fair contest.
chris.herring: Yeah. I guess there are two ways to view it:
1) Lowry is probably never going to shoot like that again.
2) There’s probably no way they’ll ever get less of a contribution from the rest of the team than they did in Game 1.
tchow: 3) Brook Lopez will not have a game like that again.
sara.ziegler: Lopez was EVERYWHERE.
chris.herring: I’m not completely sure about No. 3! If Toronto doesn’t go smaller, the Raptors are going to have to sacrifice something defensively. I don’t know that he’ll have almost 30 again, but the Raps are going to dare Brook and guys like him to prove they can make that shot as opposed to letting Giannis run wild in the paint.
That’s the risk.
sara.ziegler: To your second point, Chris, you can’t imagine a scenario happening again where no Raptor aside from Lowry makes a single shot in an entire quarter.
chris.herring: Yeah, those stats — 0 for 15 aside from Lowry in the fourth, and 1 for 23 in the second half outside of Lowry and Leonard — were some of the more insane ones I’ve ever seen.
And the one second-half basket that someone else made was a buzzer-beating 3 by Pascal Siakam in the third! One he wouldn’t have even taken if not for how much time was left.
tchow: The last time Lopez had a double-double while scoring more than 20 points was … one second, I’m still scrolling up on Basketball-Reference.
sara.ziegler: LOL
chris.herring: That part is true. But him scoring a bunch wouldn’t shock me based on how they’re defending him. Brook isn’t the biggest rebounder, in part because he’s more concerned with boxing out and making sure a teammate collects the miss. (But also, their minutes are longer in the playoffs, meaning he’ll have more chances.)
tchow: Found it! Nov. 3, 2017, when he was on the Lakers. And it was the Lopez revenge game because they played the Nets.
chris.herring: Remember: Milwaukee was 11 of 44 from three! That’s 25 percent. So the Bucks left a ton of points on the table. And many of them were wide-open shots.
As I was saying, I think Toronto may want to consider playing a little smaller. That would potentially crank up the tempo to a level Lopez isn’t comfortable with, and potentially give him more defensive responsibility, to where he has to come out farther to defend.
natesilver: I dunno, I feel weird about slicing-and-dicing the Raptors’ shooting stats into so many little pieces. Overall, they shot 15 of 42 on threes, which is pretty average/good.
chris.herring: Lowry was 7 of 9 by himself!
natesilver: They didn’t shoot great on twos, but a lot of teams don’t do that well against MIlwaukee. They made 85 percent of their free throws.
chris.herring: The other Raptors will likely shoot better. But Milwaukee did plenty to make Kawhi Leonard get his points. This team is really great at pushing star scorers to drive with their weaker hand.
tchow: Sixers should take note. Too soon?
sara.ziegler: LOL
chris.herring: The statistics illustrated that in Game 1. Leonard drove 15 times, and 11 of them were to his left. During the season, he drove to his right a little more than 57 percent of the time.
sara.ziegler: That seems to be a huge focus for the Bucks — and it looks like it’s paying off. But again, the Raptors almost stole Game 1. It would be huge for them to get Game 2 tonight.
chris.herring: Agreed.
While I still think Milwaukee is clearly the stronger team in this matchup, I wouldn’t be foolish enough to say that Toronto is out of this, regardless of what happens tonight. This is a more evenly matched set of opponents than with Portland and Golden State, clearly.
sara.ziegler: So let’s end on some soft predictions. How long will each series go?
tchow: I’m predicting a gentleman’s sweep for the Western Conference finals.
natesilver: Yeah, five games seems like the smartest bet.
sara.ziegler: It would be only fair to the Curry parents.
tchow: I believe Dame and CJ can do enough to get at least one win in Portland.
chris.herring: Agreed on the West.
In the East, I’ll go six, with the Bucks winning. Though if Milwaukee wins tonight, I wouldn’t be shocked if they closed it in five.
natesilver: I’m going to go seven games for the East. Despite what I said earlier about Game 1 being a bearish indicator for Toronto, I still think they’re a liiiiiiittttle underrated, and Nick Nurse probably has more ways to make adjustments than Mike Budenholzer does.
tchow: I think it’ll be Bucks in six, too.
natesilver: I have a hot take.
sara.ziegler:
WHAT IF DURANT HAS PLAYED HIS LAST GAME FOR THE WARRIORS?!?!?
sara.ziegler: Oooooooh
tchow: * searches in google * Durant Knicks jersey
chris.herring: That doesn’t sound as crazy to me as some people might think.
If it’s a more serious strain, and it’s closer to a month than it is a one-week or two-week injury, then the NBA Finals or the middle of the finals would be more realistic for him.
But if the finals aren’t competitive …
natesilver: So Knicks fans should be rooting for a Warriors sweep?
chris.herring: I don’t know. It would be really interesting. If the Warriors win easily without him, it would be weird for him to stay if he wants validation. If the Warriors LOSE, it gets interesting. Because, obviously, the last time the Warriors lost, he went and signed with them.
tchow: I just really want Curry to win his first finals MVP trophy.
sara.ziegler: Would THAT push KD to the Knicks?
natesilver: I think the BEST-case scenario for the Knicks would be if the Warriors are like up 3-1 over Milwaukee in the finals, and then KD comes back and they LOSE.
tchow: grinchgrin.gif
Check out our latest NBA predictions.
## A Lot Of Americans Say They Don’t Want A President Who Is Over 70. Really?
Welcome to Pollapalooza, our weekly polling roundup.
## Poll of the week
Gallup recently released new data on Americans’ willingness to vote for presidential candidates with certain traits. About 1,000 adults were asked3 whether they’d vote for a well-qualified candidate who was nominated by their party and was black, gay or had one of 10 other characteristics that are rarely or never seen in presidential nominees.
Almost all Americans said they’d be comfortable voting for a woman (94 percent), or a Catholic (95 percent), Hispanic (95 percent) or black (96 percent) candidate. But there are characteristics that big swaths of Americans said would be disqualifying — in particular being older than 70, being an atheist and being a socialist.
##### What types of candidates would Americans NOT vote for?
Share of respondents to an April survey who said they would not vote for a “generally well-qualified” presidential candidate from their own party if the candidate had each of the following characteristics
Democrats Independents Republicans Overall
Socialist 24% 48% 80% 51%
Atheist 28 33 56 39
Older than 70 35 37 37 37
Muslim 14 26 62 33
Younger than 40 21 28 34 28
Gay or lesbian 17 18 39 24
Evangelical Christian 27 20 6 18
Jewish 5 9 5 7
Woman 3 6 9 6
Catholic 4 6 3 5
Hispanic 3 3 8 5
Black 1 4 5 3
Source: Gallup
These results are fairly similar to what Gallup found when it previously asked this question, in 2015. There were a couple of interesting exceptions, however. Americans in 2019 said they were slightly more comfortable with a candidate who is an evangelical Christian (the share who said they’d vote for such a candidate rose from 73 percent in 2015 to 80 percent this year) or a Muslim (from 60 percent to 66 percent). Socialists, meanwhile, remained unpopular (47 percent in both 2015 and 2019).
So with Democrats obsessed with finding an “electable” candidate, does this mean that Bernie Sanders (who’s over 70 and identifies as a democratic socialist) and Joe Biden (who’s over 70) have big problems? Not so fast. So how seriously am I taking these numbers?
For the 2020 presidential election, I’m not taking them too seriously. Thirty-seven percent of Republicans said they would not back a GOP presidential candidate over the age of 70. Well … yep, President Trump was 70 on Election Day in 2016, and he’ll be 74 in 2020. I’ll bet that more than 63 percent of Republicans will vote for him — his job approval rating among GOP voters is currently in the 90s. In short, it’s important to remember that the survey question asks about categories of people, not individuals. The negative feelings that some Americans might have toward the idea of a gay or socialist presidential candidate, for example, might not apply to Pete Buttigieg or Sanders specifically.
On the other hand, these numbers could be understating some Americans’ resistance to certain characteristics. In particular, I’d view the numbers on ethnicity, race and gender skeptically. It could be true that virtually all Americans are comfortable with a black, female or Hispanic president, as the Gallup data implies. But I’d expect Americans who aren’t comfortable to be unlikely to express that view to a pollster. So I wouldn’t use this data to suggest that, say, Julian Castro wouldn’t run into electoral problems caused by racism or Elizabeth Warren because of sexism if either were the Democratic nominee.
In terms of which groups might face overt discrimation in the U.S., I’m taking these numbers more seriously. The results generally lined up with my expectations of which categories of people Americans are both somewhat wary of and willing to say so to another person.
Being a socialist is an expression of left-wing political views, so it’s natural and unsurprising that a lot of Americans, particularly Republicans, would openly oppose a socialist candidate. Similarly, it’s not surprising that some Americans wouldn’t want a president who is in her 70s as president (maybe they suspect that person wouldn’t have the energy for the job) or who is younger than 40 (a lack of experience). This is also a view that is perhaps not particularly controversial to express — columns suggesting that Biden (76) and Sanders (77) are too old to be running for president are published regularly.
What views about candidates are more controversial? Disqualifying people based on gender, race, ethnicity or sexuality. Again, I’d expect some Americans with negative attitudes toward certain religious groups, racial groups and sexual orientations not to admit that to a pollster.
Here’s where it gets interesting, however: The share of Americans who were willing to tell a pollster that they would not back an atheist, evangelical Christian, gay or Muslim presidential candidate was nonetheless fairly high. That lines up with how these four groups are treated in American culture — they face open, direct criticism based on their identities. (I don’t want to cast all parties as equal here — Republicans’ high level of opposition to an atheist or Muslim candidate jumps out.)
In terms of understanding the diversity of the Democratic Party, I’m taking these numbers very seriously. I’ve written that Biden is essentially the candidate of the un-woke Democrat (or maybe “less woke” is more accurate) and that those voters still represent a substantial bloc of the Democratic Party. This data is more evidence of that bloc’s existence. I was surprised that the share of Democrats who are uncomfortable with an evangelical Christian president was matched by about an equal share wary of a president who is an atheist or a socialist, since the Democratic Party is often characterized as becoming less religious and more liberal on economic issues. The share of Democrats who said they would not vote for a gay or Muslim candidate was also larger than I anticipated.
## Other polling bites
• 46 percent of likely Democratic primary voters in South Carolina say they would vote for Biden, according to a new Post and Courier/Change Research poll, with only two of his rivals reaching double digits. Sanders (15 percent) and Kamala Harris (10 percent) are far behind the former vice president, as is the rest of the 2020 Democratic field.
• Biden leads in Pennsylvania too, with 39 percent of the vote, according to a new Quinnipiac University survey. The only other candidate in double digits was Sanders (13 percent).
• The Quinnipiac survey also found Biden leading Trump 53 percent to 42 percent in Pennsylvania in a hypothetical general election matchup. Sanders also bested Trump (50-43).
• In the Republican nomination contest, Trump leads former Massachusetts Gov. William Weld 72 percent to 12 percent in New Hampshire, according to a recent Monmouth University survey.
• 61 percent of Americans support same-sex marriage, and 31 percent oppose it, according to a Pew Research Center survey. Support for same-sex marriage varied by party (75 percent of Democrats and Democratic-leaning independents, compared with 44 percent of Republicans and GOP-leaning independents). It varied by race (62 percent of white Americans, 58 percent of Hispanic Americans, 51 percent of black Americans). And it varied by religion (79 percent of those who are religiously unaffiliated, 66 percent of white mainline Protestants, 61 percent of Catholics, 29 percent of white evangelical Protestants).
• 47 percent of registered voters rated the economy as “excellent” or “good,” according to a new Fox News poll.
• Also from that Fox News poll: The share of voters who said Trump hasn’t been tough enough with North Korea is up to 50 percent; that number was 19 percent in September 2017.
## Trump approval
According to FiveThirtyEight’s presidential approval tracker, 42.0 percent of Americans approve of the job Trump is doing as president, while 53.1 percent disapprove (a net approval rating of -11.1 points). At this time last week, 42.4 percent approved and 52.7 percent disapproved (for a net approval rating of -10.3 points). One month ago, Trump had an approval rating of 42.1 percent and a disapproval rating of 52.3 percent, for a net approval rating of -10.2 points.
Check out all the polls we’ve been collecting ahead of the 2020 elections.
## How Many Soldiers Do You Need To Beat The Night King?
Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. There are two types: Riddler Express for those of you who want something bite-size and Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,4 and you may get a shoutout in next week’s column. If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.
## Riddler Express
From Tom Hanrahan, three colorful journeys:
In grade school, you may have learned about the three primary colors — blue, yellow and red — and the three secondary colors — green (blue + yellow), purple (red + blue) and orange (yellow + red).
And now it’s time to put that knowledge to use. Try to get through the maze below, a nine-by-nine grid of lines, three times: once as blue, once as yellow, and once as red.
If you are blue, you may only travel on lines that include the color blue. So you may travel on lines that are blue, green, purple or white (which contains all colors). You may not travel on orange, yellow, red or black (which contains no colors). The analogous rules hold for your trips as yellow and red.
In all three cases, you are attempting to travel between the same two points on the maze’s edge. Send me links, pictures or descriptions of your strategy!
## Riddler Classic
From Greg Burnham, it had to happen eventually, at long last and not a moment too soon, The Riddler meets “Game of Thrones”:
At a pivotal moment in an epic battle between the living and the dead, the Night King, head of the army of the dead, raises all the fallen (formerly) living soldiers to join his ranks. This ability obviously presents a huge military advantage, but how big an advantage exactly?
Forget the Battle of Winterfell and model our battle as follows. Each army lines up single file, facing the other army. One soldier steps forward from each line and the pair duels — half the time the living soldier wins, half the time the dead soldier wins. If the living soldier wins, he goes to the back of his army’s line, and the dead soldier is out (the living army uses dragonglass weapons, so the dead soldier is dead forever this time). If the dead soldier wins, he goes to the back of their army’s line, but this time the (formerly) living soldier joins him there. (Reanimation is instantaneous for this Night King.) The battle continues until one army is entirely eliminated.
What starting sizes of the armies, living and dead, give each army a 50-50 chance of winning?
## Solution to last week’s Riddler Express
Congratulations to Stuart Tooley of Edinburgh, United Kingdom, winner of last week’s Riddler Express!
Last week I gave you the following sequence of numbers and asked you what number came next?
2
6
10
3
8
9
4
7
?
The correct missing number is 8.
The numerical solution actually had more to do with letters, and with my favorite game, Scrabble. Starting at the top of the list: The number 2 is spelled “two,” which is worth six points if spelled with Scrabble tiles, so 6 becomes the next number in the list. The number 6 is spelled “six” which is worth 10 points in Scrabble, so 10 comes next in the list, and so on. To find the missing number, we write 7 as “seven” and tally that it’s worth eight points, so 8 is our answer.
## Solution to last week’s Riddler Classic
Congratulations to Brian Hare of Raleigh, North Carolina, winner of last week’s Riddler Classic!
Last week we were introduced to five brothers who had joined the Riddler Baseball Independent Society, or RBIs. Each of them enjoyed a career of 20 seasons, with 160 games per season and four plate appearances per game.5 Given that their batting averages were .200, .250, .300, .350 and .400, what were each brother’s chances of beating DiMaggio’s 56-game hitting streak at some point in his career? (Streaks could span across seasons.)
Their chances of besting DiMaggio were, respectively, approximately 0, 0, 0.01, 0.8 and 13.9 percent. Put another way, their chances were roughly 1-in-9,000,000,000, 1-in-3,000,000, 1-in-8,000, 1-in-130 and 1-in-7.
To get there, first we need to compute the chances that each brother get at least one hit in any given game. Suppose a brother’s batting average is A. The chances that that brother does not get a hit in a given game is (1-A)^4, because he gets four at bats. So the chances a brother does get a hit in any given game is 1 minus that. That gives the brothers a chance of 0.5904, 0.683594, 0.7599, 0.821494 and 0.8704.
We then need to turn those individual game chances into streak chances. To put a bit of mathematical structure on this problem, we are looking for the probability $$P$$ that at least $$r$$ consecutive games with a hit appears in a sequence of total length $$n$$ games given a probability $$p$$ of a hit in each game. In our case, $$r=57$$, $$n=3,200$$ (20 seasons of 160 games), and we calculated $$p$$ in the paragraph above.
As solver Michael Branicky explained, this probability, assuming the number of games played is large enough to contain a sufficiently long streak, must satisfy the following recursion:
\begin{equation*} P_{n+1} = P_n + ( 1 – P_{n-r} ) (1-p) p^r \end{equation*}
This is because a streak of length $$r$$ in $$n+1$$ games either occurred in $$n$$ games, or occurs for the first time in game $$n + 1$$. From there, the calculation can be done with a bit of computer code. Michael also illustrated the chances of breaking the streak by a player’s batting average:
As it happened, the brothers also had a cousin with a whopping .500 average, but he would get banned from the league after 10 seasons after testing positive for performance enhancers. What were his chances of beating the streak? Despite the shortened career, they were about 93.3 percent, an answer we can arrive at with the same approach described above.
As solver Chris Jones concluded, “Moral of the story: Doping pays off for record-breaking, bros.”
## Want more riddles?
Well, aren’t you lucky? There’s a whole book full of the best puzzles from this column and some never-before-seen head-scratchers. It’s called “The Riddler,” and it’s in stores now!
## Want to submit a riddle?
Email me at [email protected]
CORRECTION (May 17, 2019, 10:48 a.m.): An earlier version of the Riddler Express displayed the wrong color for one line. The leftmost horizontal bar in the fourth row from the top should be white, not black. The graphic has been updated. | 2019-07-23 05:12:36 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.20716175436973572, "perplexity": 2477.9130478365564}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195528869.90/warc/CC-MAIN-20190723043719-20190723065719-00065.warc.gz"} |
https://guitarknights.com/guitar-chords-to-learn-guitar-karaoke.html | Many influences are cited as antecedents to the modern guitar. Although the development of the earliest "guitars" is lost in the history of medieval Spain, two instruments are commonly cited as their most influential predecessors, the European lute and its cousin, the four-string oud; the latter was brought to Iberia by the Moors in the 8th century.[7]
The loud, amplified sound and sonic power of the electric guitar played through a guitar amp has played a key role in the development of blues and rock music, both as an accompaniment instrument (playing riffs and chords) and performing guitar solos, and in many rock subgenres, notably heavy metal music and punk rock. The electric guitar has had a major influence on popular culture. The guitar is used in a wide variety of musical genres worldwide. It is recognized as a primary instrument in genres such as blues, bluegrass, country, flamenco, folk, jazz, jota, mariachi, metal, punk, reggae, rock, soul, and many forms of pop.
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The ratio of the spacing of two consecutive frets is {\displaystyle {\sqrt[{12}]{2}}} (twelfth root of two). In practice, luthiers determine fret positions using the constant 17.817—an approximation to 1/(1-1/ {\displaystyle {\sqrt[{12}]{2}}} ). If the nth fret is a distance x from the bridge, then the distance from the (n+1)th fret to the bridge is x-(x/17.817).[15] Frets are available in several different gauges and can be fitted according to player preference. Among these are "jumbo" frets, which have much thicker gauge, allowing for use of a slight vibrato technique from pushing the string down harder and softer. "Scalloped" fretboards, where the wood of the fretboard itself is "scooped out" between the frets, allow a dramatic vibrato effect. Fine frets, much flatter, allow a very low string-action, but require that other conditions, such as curvature of the neck, be well-maintained to prevent buzz.
The playing of (3-5 string) guitar chords is simplified by the class of alternative tunings called regular tunings, in which the musical intervals are the same for each pair of consecutive strings. Regular tunings include major-thirds tuning, all-fourths, and all-fifths tunings. For each regular tuning, chord patterns may be diagonally shifted down the fretboard, a property that simplifies beginners' learning of chords and that simplifies advanced players' improvisation. On the other hand, in regular tunings 6-string chords (in the keys of C, G, and D) are more difficult to play.
Open tuning refers to a guitar tuned so that strumming the open strings produces a chord, typically a major chord. The base chord consists of at least 3 notes and may include all the strings or a subset. The tuning is named for the open chord, Open D, open G, and open A are popular tunings. All similar chords in the chromatic scale can then be played by barring a single fret.[16] Open tunings are common in blues and folk music,[17] and they are used in the playing of slide and bottleneck guitars.[16][18] Many musicians use open tunings when playing slide guitar.[17]
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Electric guitars and bass guitars have to be used with a guitar amplifier and loudspeaker or a bass amplifier and speaker, respectively, in order to make enough sound to be heard by the performer and audience. Electric guitars and bass guitars almost always use magnetic pickups, which generate an electric signal when the musician plucks, strums or otherwise plays the instrument. The amplifier and speaker strengthen this signal using a power amplifier and a loudspeaker. Acoustic guitars that are equipped with a piezoelectric pickup or microphone can also be plugged into an instrument amplifier, acoustic guitar amp or PA system to make them louder. With electric guitar and bass, the amplifier and speaker are not just used to make the instrument louder; by adjusting the equalizer controls, the preamplifier, and any onboard effects units (reverb, distortion/overdrive, etc.) the player can also modify the tone (aka timbre or "colour") and sound of the instrument. Acoustic guitar players can also use the amp to change the sound of their instrument, but in general, acoustic guitar amps are used to make the natural acoustic sound of the instrument louder without changing its sound that much.
Anyone playing and/or teaching guitar needs staff paper, blank tab, guitar chord charts, guitar scale charts, and fretboard diagrams to chart their guitar lessons and musical ideas. You can find books with some combination of these blank charts and grids, but you can’t find one with all of them organized in a practical way. That’s why we chose to design our own.
##### Justin is an instructor with that rare combination that encompasses great playing in conjunction with a thoughtful, likable personality. Justin's instruction is extremely intelligent because he's smart enough to know the 'basics' don't have to be served 'raw' - Justin keenly serves the information covered in chocolate. Justin's site is like a free pass in a candy store!
A few years back, I dusted off the ol' Takamine I got in high school to try some 'music therapy' with my disabled son, who was recovering from a massive at-birth stroke. This reignited my long dormant passion to transform myself from a beach strummer to a 'real' musician; however, as a single mom, taking in-person lessons was financially difficult. Then I found Justinguitar! Flash forward to today; my son is almost fully recovered (YAY!), my guitar collection has grown significantly, and I'm starting to play gigs. None of this would have been possible without your guidance and generosity, Justin. Thank you for being part of the journey!
When all is said and done, the only way to know for sure what a certain set of strings will sound like on your guitar is to take them for a spin. For that reason, it's not a bad idea to try out a few different ones that interest you to see which ones you like best - then you can stock up! However you like to approach your string shopping, the amazing selection you'll find here is sure to satisfy.
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Adding a minor seventh to a major triad creates a dominant seventh (denoted V7). In music theory, the "dominant seventh" described here is called a major-minor seventh, emphasizing the chord's construction rather than its usual function.[27] Dominant sevenths are often the dominant chords in three-chord progressions,[18] in which they increase the tension with the tonic "already inherent in the dominant triad".[28]
F major. This is fairly similar to the C, but a little more difficult to play. Press the fourth string down at the third fret with your ring finger, the third string down at the second fret with your middle finger, and the first and second strings down at the first fret with your index. You just flatten your index finger down across the two strings; lower your thumb if you struggle. You don't play the fifth or sixth strings in this chord.
Learn the C chord. The first chord we will cover is a C chord—one of the most basic chords in music. Before we do, let's break down just what that means. A proper chord, whether played on a piano, a guitar, or sung by well-trained mice, is simply three or more notes sounded together. (Two notes is called a "diad," and while musically useful, is not a chord.) Chords can also contain far more than three notes, but that's well beyond the scope of this article. This is what a C chord looks like on the guitar:
In Mexico, the popular mariachi band includes a range of guitars, from the small requinto to the guitarrón, a guitar larger than a cello, which is tuned in the bass register. In Colombia, the traditional quartet includes a range of instruments too, from the small bandola (sometimes known as the Deleuze-Guattari, for use when traveling or in confined rooms or spaces), to the slightly larger tiple, to the full-sized classical guitar. The requinto also appears in other Latin-American countries as a complementary member of the guitar family, with its smaller size and scale, permitting more projection for the playing of single-lined melodies. Modern dimensions of the classical instrument were established by the Spaniard Antonio de Torres Jurado (1817–1892).[12]
Spiral bound guitar book arrived on time as promised. As reference book for guitar chords, it's quite convenient to use for all levels of guitar expertise. It also provides alternatives to play a certain chord. It's easy to follow and to use. Using the tabs near the edge of the page, chords are arranged from A to G & "other chords". Obviously, the guitar greenhorn needs to learn a few basic chords first, and this book builds on those skills. Although the first edition was published in 2006, guitar chords don't really change, unlike other fields of study, so it's relevant today as it was years ago. I deducted 1 star because the back cover arrived crumpled, and I like to keep my books pristine. This book is supposed to be brand new. The person who packed the box was not careful. I still recommend this guitar book as a quick reference. It's faster to use this than look up chords individually on the web.
The headstock is located at the end of the guitar neck farthest from the body. It is fitted with machine heads that adjust the tension of the strings, which in turn affects the pitch. The traditional tuner layout is "3+3", in which each side of the headstock has three tuners (such as on Gibson Les Pauls). In this layout, the headstocks are commonly symmetrical. Many guitars feature other layouts, including six-in-line tuners (featured on Fender Stratocasters) or even "4+2" (e.g. Ernie Ball Music Man). Some guitars (such as Steinbergers) do not have headstocks at all, in which case the tuning machines are located elsewhere, either on the body or the bridge.
With the C major chord, put that shape on the guitar for thirty seconds, take it off, shake it out, and repeat the process a few times. As you’re making the shape, remember to come right behind the frets on the tips of your fingers. When you’re starting out, you may have to place each finger down one at a time, but that’s natural. You’ll get better with time and eventually be able to go right to the chord.
The guitar is a fretted musical instrument that usually has six strings.[1] It is typically played with both hands by strumming or plucking the strings with either a guitar pick or the finger(s)/fingernails of one hand, while simultaneously fretting (pressing the strings against the frets) with the fingers of the other hand. The sound of the vibrating strings is projected either acoustically, by means of the hollow chamber of the guitar (for an acoustic guitar), or through an electrical amplifier and a speaker.
Play the A Major. This is another "big chord," sonically. There are several ways to play this. You can use one finger across the 2nd fret of the B, G, and D strings (playing C#, A, and E, respectively), or any other combination of fingers. For this example, we'll use the 4th finger on the B string, 3rd finger on the G string, and 2nd finger on the D string.
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Learn a D major. This chord only requires the bottom four strings. Place your index finger on the 3rd string, 2nd fret. Your ring finger then goes on the 2nd string, 3rd fret, and your middle finger is the 1st string, second fret. You'll form a little triangle shape. Only strum these three strings and the 4th string -- the open D -- to sound out the chord.
Open tunings improve the intonation of major chords by reducing the error of third intervals in equal temperaments. For example, in the open-G overtones tuning G-G-D-G-B-D, the (G,B) interval is a major third, and of course each successive pair of notes on the G- and B-strings is also a major third; similarly, the open-string minor-third (B,D) induces minor thirds among all the frets of the B-D strings. The thirds of equal temperament have audible deviations from the thirds of just intonation: Equal temperaments is used in modern music because it facilitates music in all keys, while (on a piano and other instruments) just intonation provided better-sounding major-third intervals for only a subset of keys.[65] "Sonny Landreth, Keith Richards and other open-G masters often lower the second string slightly so the major third is in tune with the overtone series. This adjustment dials out the dissonance, and makes those big one-finger major-chords come alive."[66]
Guitar chords are dramatically simplified by the class of alternative tunings called regular tunings. In each regular tuning, the musical intervals are the same for each pair of consecutive strings. Regular tunings include major-thirds (M3), all-fourths, augmented-fourths, and all-fifths tunings. For each regular tuning, chord patterns may be diagonally shifted down the fretboard, a property that simplifies beginners' learning of chords and that simplifies advanced players' improvisation.[70][71][72] The diagonal shifting of a C major chord in M3 tuning appears in a diagram.
With the advent of YouTube tutorials for just about everything, it's only fitting that there are a lot of videos out there claiming to be able to teach you how to play guitar. While you might think this is a quick and easy way to become a pro, you'll want to make sure you have all of the information before diving in headfirst. Online Tutorials When you're surfing the Internet, you'll come across many different websites with prerecorded tutorials to help you learn guitar online. But the keywo
The three notes of a major triad have been introduced as an ordered triplet, namely (root, third, fifth), where the major third is four semitones above the root and where the perfect fifth is seven semitones above the root. This type of triad is in closed position. Triads are quite commonly played in open position: For example, the C-major triad is often played with the third (E) and fifth (G) an octave higher, respectively sixteen and nineteen semitones above the root. Another variation of the major triad changes the order of the notes: For example, the C-major triad is often played as (C,G,E), where (C,G) is a perfect fifth and E is raised an octave above the perfect third (C,E). Alternative orderings of the notes in a triad are discussed below (in the discussions of chord inversions and drop-2 chords).
Stacking the C-major scale with thirds creates a chord progression Play (help·info), traditionally enumerated with the Roman numerals I, ii, iii, IV, V, vi, viio. Its major-key sub-progression C-F-G (I-IV-V) is conventional in popular music. In this progression, the minor triads ii-iii-vi appear in the relative minor key (Am)'s corresponding chord progression.
Of course, there are a few ways to narrow down the string options. For starters, since guitars come in different scales, you need a set that's the right length for your instrument. You also need to match the type of guitar: electric strings for an electric guitar, acoustic strings for acoustic. If you play an acoustic-electric, you'll usually be looking for acoustic strings since those instruments use non-magnetic pickups. For classical and Latin guitar types directly descended from ancient gut-stringed instruments, the right strings are generally going to be nylon.
Getting to grips with how chords are formed gives you a basic introduction to music theory and helps you understand the ways you can alter them to create more interesting sounds. All chords are built from certain notes in scales. The C major scale is the easiest, because it just runs C, D, E, F, G, A and B. These notes are numbered (usually using Roman numerals) in that order, from one (I) to seven (VII).
You need to place one finger on whatever fret you want to bar and hold it there over all of the strings on that fret. The rest of your fingers will act as the next finger down the line (second finger barring, so third finger will be your main finger, and so on). You can also buy a capo, so that you don't have to deal with the pain of the guitar's strings going against your fingers. The capo bars the frets for you. This also works with a ukulele.
## The fingerboard, also called the fretboard, is a piece of wood embedded with metal frets that comprises the top of the neck. It is flat on classical guitars and slightly curved crosswise on acoustic and electric guitars. The curvature of the fretboard is measured by the fretboard radius, which is the radius of a hypothetical circle of which the fretboard's surface constitutes a segment. The smaller the fretboard radius, the more noticeably curved the fretboard is. Most modern guitars feature a 12" neck radius, while older guitars from the 1960s and 1970s usually feature a 6-8" neck radius. Pinching a string against a fret on fretboard effectively shortens the vibrating length of the string, producing a higher pitch.
My personal opinion on the topic, as one musician to another, is that the best thing you can possibly do when trying to figure out which string to go with is to try out as many different brands and types of guitar strings as you can. Strings are cheap enough that most people are going to be able to afford to experiment, and the truth of the matter is that you’re probably not going to really know what works best for you until you have hands on experience.
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The intensive A.A. curriculum prepares musicians to perform in any professional situation, along with learning professional development skills such as basic computer use, EPK creation, resume and bio writing, and social media as a tool for business and networking. The Associate of Arts Degree is intended to equip students with the knowledge and training needed to become professional performers in today’s music industry. | 2019-06-25 21:35:23 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2432003915309906, "perplexity": 2697.750881276067}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627999948.3/warc/CC-MAIN-20190625213113-20190625235113-00032.warc.gz"} |
https://www.expii.com/t/use-of-polynomials-over-zpz-in-number-theory-10926 | Expii
# Use of Polynomials over Z/pZ in Number Theory - Expii
Polynomials over Z/pZ are quite useful for encoding various kinds of information in number theory. Applications include the proofs of Wilson's theorem, Euler's criterion, and the existence of primitive roots mod p; as well as the determination of the sign of the quadratic Gauss sum. | 2021-01-23 17:51:46 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9467962980270386, "perplexity": 321.7914219774014}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703538226.66/warc/CC-MAIN-20210123160717-20210123190717-00055.warc.gz"} |
https://www.physicsforums.com/threads/de-of-the-form-y-by-a.243575/ | # DE of the form y'' + by' = a
1. Jul 5, 2008
### Four
Hi, I'm trying to solve the following equation
y'' + by' = a
But my answer doesn't make sense:
The question:
an object is flying through space, with velocity could be approximated as:
v_next = v_current + a*dt - damp*v*dt
dt - time increment taken repetitively
a - acceleration
damp - a constant
For large dt the approximation is inappropriate, find an equation that will do for large dt.
My go:
it looks like the above equation is "similar" to
x'' = a - damp*x'
x'' + damp*x' = a
part 1: x_c
$$x'' + damp*x = 0 => r*r + damp*r = 0; r = 0, r = \frac{-1}{damp}$$
$$x_c = c_1 + c_2e^{\frac{-t}{damp}}$$
part 2: x_p
x(t) = k*t
x'' + damp*k = a
(k*t)'' = 0
$$k = \frac{a}{damp}$$
$$x_p = \frac{a*t}{damp}$$
part 3:
$$x = x_c + x_p = c_1 + c_2e^{\frac{-t}{damp}} + \frac{a*t}{damp}$$
we want x' approximation so
$$x' = \frac{-c_2}{damp}e^{\frac{-t}{damp}} + \frac{a}{damp}$$
what doesn't make sense is lets say damp -> 0 then x' should be a streight line but it doesn't look like it?
Where may I have gone wrong?
Thank you
Last edited: Jul 5, 2008
2. Jul 5, 2008
### smallphi
Your solution for nonzero damp looks fine to me: the system starts with some initial speed and then approaches asymptotically the so called "terminal speed", v(terminal) = a/damp, at which the pulling force balances the friction exactly: m*a = m*damp*v(terminal).
Your solution doesn't apply to damp=0 case because you assumed you got two distinct roots of the characteristic equation and hence the full general solution of the homogeneous equation in part 1. That assumption breaks down when damp =0.
When damp=0, the homogeneous diff. equation in part 1 is x" = 0.
You get a double zero root of the characteristic equation hence only one exponent which can't capture the full general solution required to depend on two arbitrary constants not one. In such cases the general prescription tells you to look for solution of other types not exp(kt). In this case the solution is a linear function xc = c1 + c2*t. The double zero root in the exponent produces only the c1 term.
Last edited: Jul 5, 2008 | 2017-03-26 17:34:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8622094392776489, "perplexity": 1963.371299652467}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218189244.95/warc/CC-MAIN-20170322212949-00452-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/inverse-matrix.559196/ | # Inverse Matrix
1. Dec 11, 2011
### AlonsoMcLaren
Is it always true that (see the attachment)?
#### Attached Files:
• ###### Untitled.png
File size:
2 KB
Views:
75
2. Dec 11, 2011
### AlonsoMcLaren
I am assuming that phi is a matrix and its elements depend on t, another variable.
3. Dec 12, 2011
### HallsofIvy
In order for this to make sense, $\phi$ would have to be a one-to-one function from a set of matrices onto itself. And it asks if the inverse function, appled to matrix x, is the same as the inverse of $\phi(x)$. | 2018-02-24 16:49:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7652836441993713, "perplexity": 1444.7961800954595}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891815843.84/warc/CC-MAIN-20180224152306-20180224172306-00218.warc.gz"} |
http://www.theinfolist.com/html/ALL/s/logical_disjunction.html | TheInfoList
In
logic Logic is an interdisciplinary field which studies truth and reasoning. Informal logic seeks to characterize Validity (logic), valid arguments informally, for instance by listing varieties of fallacies. Formal logic represents statements and ar ...
, disjunction is a
logical connective In logic Logic is an interdisciplinary field which studies truth and reasoning Reason is the capacity of consciously making sense of things, applying logic Logic (from Ancient Greek, Greek: grc, wikt:λογική, λογική, la ...
typically notated $\lor$ whose meaning either refines or corresponds to that of natural language expressions such as "or". In
classical logic Classical logic (or standard logic) is the intensively studied and most widely used class of deductive logic Deductive reasoning, also deductive logic, is the process of reasoning from one or more statements (premises) to reach a logical conclusio ...
, it is given a
truth function In logic Logic (from Ancient Greek, Greek: grc, wikt:λογική, λογική, label=none, lit=possessed of reason, intellectual, dialectical, argumentative, translit=logikḗ)Also related to (''logos''), "word, thought, idea, argument, ac ...
al
semantics Semantics (from grc, σημαντικός ''sēmantikós'', "significant") is the study of reference Reference is a relationship between objects in which one object designates, or acts as a means by which to connect to or link to, another o ...
on which $\phi \lor \psi$ is true unless both $\phi$ and $\psi$ are false. Because this semantics allows a disjunctive formula to be true when both of its disjuncts are true, it is an ''inclusive'' interpretation of disjunction, in contrast with
exclusive disjunction Exclusive or or exclusive disjunction is a logical operation In logic Logic is an interdisciplinary field which studies truth and reasoning Reason is the capacity of consciously making sense of things, applying logic Logic ...
. Classical proof theoretical treatments are often given in terms of rules such as
disjunction introduction Disjunction introduction or addition (also called or introduction) is a rule of inference In the philosophy of logic, a rule of inference, inference rule or transformation rule is a logical form consisting of a function which takes premises, ana ...
and
disjunction elimination In propositional logic Propositional calculus is a branch of logic Logic (from Ancient Greek, Greek: grc, wikt:λογική, λογική, label=none, lit=possessed of reason, intellectual, dialectical, argumentative, translit=logikḗ)Al ...
. Disjunction has also been given numerous non-classical treatments, motivated by problems including
Aristotle's sea battle argument Future contingent propositions (or simply, future contingents) are statements about states of affairs in the future that are ''contingency (philosophy), contingent:'' neither necessarily true nor necessarily false. The problem of future contingents ...
,
Heisenberg Werner Karl Heisenberg (; ; 5 December 1901 – 1 February 1976) was a German theoretical physicist and one of the key pioneers of quantum mechanics Quantum mechanics is a fundamental Scientific theory, theory in physics that provides a de ...
's
uncertainty principle In quantum mechanics Quantum mechanics is a fundamental Scientific theory, theory in physics that provides a description of the physical properties of nature at the scale of atoms and subatomic particles. It is the foundation of all quant ...
, as well the numerous mismatches between classical disjunction and its nearest equivalents in natural language.
# Notation
In logic and related fields, disjunction is customarily notated with an infix operator $\lor$. Alternative notations include $+$, used mainly in
electronics The field of electronics is a branch of physics and electrical engineering that deals with the emission, behaviour and effects of electrons The electron is a subatomic particle In physical sciences, subatomic particles are smaller than ...
, as well as $\vert$ and $\vert\!\vert$ in many
programming language A programming language is a formal language In logic, mathematics, computer science, and linguistics, a formal language consists of string (computer science), words whose symbol (formal), letters are taken from an alphabet (computer science) ...
s. The English word "or" is sometimes used as well, often in capital letters. In
Jan Łukasiewicz Jan Łukasiewicz (; 21 December 1878 – 13 February 1956) was a Polish logician and philosopher who is best known for Polish notation and Łukasiewicz logic. He was born in Lemberg, a city in the Austrian Galicia, Galician Kingdom of Austria-Hungar ...
's prefix notation for logic, the operator is A, short for Polish ''alternatywa'' (English: alternative).
# Classical disjunction
## Semantics
Classical disjunction is a
truth function In logic Logic (from Ancient Greek, Greek: grc, wikt:λογική, λογική, label=none, lit=possessed of reason, intellectual, dialectical, argumentative, translit=logikḗ)Also related to (''logos''), "word, thought, idea, argument, ac ...
al operation which returns the
truth value In logic Logic is an interdisciplinary field which studies truth and reasoning Reason is the capacity of consciously making sense of things, applying logic Logic (from Ancient Greek, Greek: grc, wikt:λογική, λογική, la ...
"true" unless both of its arguments are "false". Its semantic entry is standardly given as follows: :: $\models \phi \lor \psi$ if $\models \phi$ or $\models \psi$ or both This semantics corresponds to the following
truth table A truth table is a mathematical table Mathematical tables are lists of numbers showing the results of a calculation with varying arguments. Tables of trigonometric functions were used in ancient Greece and India for applications to astronomy ...
:
## Defined by other operators
In systems where logical disjunction is not a primitive, it may be defined as :$A \lor B = \neg A \to B$. This can be checked by the following truth table:
## Properties
The following properties apply to disjunction: *
Associativity In mathematics, the associative property is a property of some binary operations, which means that rearranging the parentheses in an expression will not change the result. In propositional logic, associativity is a Validity (logic), valid rule ...
: $a \lor \left(b \lor c\right) \equiv \left(a \lor b\right) \lor c$ *
Commutativity In mathematics Mathematics (from Greek: ) includes the study of such topics as numbers (arithmetic and number theory), formulas and related structures (algebra), shapes and spaces in which they are contained (geometry), and quantities and ...
: $a \lor b \equiv b \lor a$ *
Distributivity In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). It ...
: $\left(a \land \left(b \lor c\right)\right) \equiv \left(\left(a \land b\right) \lor \left(a \land c\right)\right)$ :::$\left(a \lor \left(b \land c\right)\right) \equiv \left(\left(a \lor b\right) \land \left(a \lor c\right)\right)$ :::$\left(a \lor \left(b \lor c\right)\right) \equiv \left(\left(a \lor b\right) \lor \left(a \lor c\right)\right)$ :::$\left(a \lor \left(b \equiv c\right)\right) \equiv \left(\left(a \lor b\right) \equiv \left(a \lor c\right)\right)$ *
Idempotency Idempotence (, ) is the property of certain operations in mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), an ...
: $a \lor a \equiv a$ *
Monotonicity Figure 3. A function that is not monotonic In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus ...
: $\left(a \rightarrow b\right) \rightarrow \left(\left(c \lor a\right) \rightarrow \left(c \lor b\right)\right)$ :::$\left(a \rightarrow b\right) \rightarrow \left(\left(a \lor c\right) \rightarrow \left(b \lor c\right)\right)$ *Truth-preserving: The interpretation under which all variables are assigned a
truth value In logic Logic is an interdisciplinary field which studies truth and reasoning Reason is the capacity of consciously making sense of things, applying logic Logic (from Ancient Greek, Greek: grc, wikt:λογική, λογική, la ...
of 'true', produces a truth value of 'true' as a result of disjunction. *Falsehood-preserving: The interpretation under which all variables are assigned a
truth value In logic Logic is an interdisciplinary field which studies truth and reasoning Reason is the capacity of consciously making sense of things, applying logic Logic (from Ancient Greek, Greek: grc, wikt:λογική, λογική, la ...
of 'false', produces a truth value of 'false' as a result of disjunction.
# Applications in computer science
Operators Operator may refer to: Mathematics * A symbol indicating a mathematical operation * Logical operator or logical connective in mathematical logic * Operator (mathematics), mapping that acts on elements of a space to produce elements of another sp ...
corresponding to logical disjunction exist in most
programming language A programming language is a formal language In logic, mathematics, computer science, and linguistics, a formal language consists of string (computer science), words whose symbol (formal), letters are taken from an alphabet (computer science) ...
s.
## Bitwise operation
Disjunction is often used for
bitwise operation In computer programming, a bitwise operation operates on a bit string, a bit array or a Binary numeral system, binary numeral (considered as a bit string) at the level of its individual bits. It is a fast and simple action, basic to the higher le ...
s. Examples: * 0 or 0 = 0 * 0 or 1 = 1 * 1 or 0 = 1 * 1 or 1 = 1 * 1010 or 1100 = 1110 The or operator can be used to set bits in a
bit field A bit field is a data structure Image:Hash table 3 1 1 0 1 0 0 SP.svg, 315px, A data structure known as a hash table. In computer science, a data structure is a data organization, management, and storage format that enables efficient access and ...
to 1, by or-ing the field with a constant field with the relevant bits set to 1. For example, x = x , 0b00000001 will force the final bit to 1, while leaving other bits unchanged.
## Logical operation
Many languages distinguish between bitwise and logical disjunction by providing two distinct operators; in languages following C, bitwise disjunction is performed with the single pipe operator (, ), and logical disjunction with the double pipe (, , ) operator. Logical disjunction is usually short-circuited; that is, if the first (left) operand evaluates to true, then the second (right) operand is not evaluated. The logical disjunction operator thus usually constitutes a sequence point. In a parallel (concurrent) language, it is possible to short-circuit both sides: they are evaluated in parallel, and if one terminates with value true, the other is interrupted. This operator is thus called the parallel or. Although the type of a logical disjunction expression is boolean in most languages (and thus can only have the value true or false), in some languages (such as
Python PYTHON was a Cold War contingency plan of the Government of the United Kingdom, British Government for the continuity of government in the event of Nuclear warfare, nuclear war. Background Following the report of the Strath Committee in 1955, the ...
and
JavaScript JavaScript (), often abbreviated JS, is a programming language A programming language is a formal language In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), ma ...
), the logical disjunction operator returns one of its operands: the first operand if it evaluates to a true value, and the second operand otherwise.
## Constructive disjunction
The
Curry–Howard correspondence In programming language theory and proof theory Proof may refer to: * Proof (truth), argument or sufficient evidence for the truth of a proposition * Alcohol proof, a measure of an alcoholic drink's strength Formal sciences * Formal proof, a c ...
relates a constructivist form of disjunction to
tagged unionIn computer science, a tagged union, also called a variant type, variant, variant record, choice type, discriminated union, disjoint union, sum type or coproduct, is a data structure used to hold a value that could take on several different, but fixe ...
types.
# Set theory
The
membership Member may refer to: * Military juryA United States military "jury" (or "Members", in military parlance) serves a function similar to an American civilian jury, but with several notable differences. Only a Courts-martial in the United States, Gene ...
of an element of a union set in
set theory Set theory is the branch of mathematical logic that studies Set (mathematics), sets, which can be informally described as collections of objects. Although objects of any kind can be collected into a set, set theory, as a branch of mathematics, i ...
is defined in terms of a logical disjunction: $x\in A\cup B$ if and only if $\left(x\in A\right)\vee\left(x\in B\right)$. Because of this, logical disjunction satisfies many of the same identities as set-theoretic union, such as
associativity In mathematics, the associative property is a property of some binary operations, which means that rearranging the parentheses in an expression will not change the result. In propositional logic, associativity is a Validity (logic), valid rule ...
,
commutativity In mathematics Mathematics (from Greek: ) includes the study of such topics as numbers (arithmetic and number theory), formulas and related structures (algebra), shapes and spaces in which they are contained (geometry), and quantities and ...
,
distributivity In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). It ...
, and
de Morgan's laws In propositional calculus, propositional logic and Boolean algebra, De Morgan's laws are a pair of transformation rules that are both Validity (logic), valid rule of inference, rules of inference. They are named after Augustus De Morgan, a 19th ...
, identifying
logical conjunction In logic Logic is an interdisciplinary field which studies truth and reasoning Reason is the capacity of consciously making sense of things, applying logic Logic (from Ancient Greek, Greek: grc, wikt:λογική, λογική, ...
with
set intersection In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). I ...
,
logical negation In logic Logic (from Ancient Greek, Greek: grc, wikt:λογική, λογική, label=none, lit=possessed of reason, intellectual, dialectical, argumentative, translit=logikḗ)Also related to (''logos''), "word, thought, idea, argume ...
with
set complement In , the complement of a , often denoted by (or ), are the not in . When all sets under consideration are considered to be s of a given set , the absolute complement of is the set of elements in that are not in . The relative complement of ...
.
# Natural language
The classical denotation for $\lor$ does not precisely match the
denotation The denotation of a word is its central sense A sense is a biological system used by an organism for sensation, the process of gathering information about the world and responding to Stimulus (physiology), stimuli. (For example, in the human bod ...
of disjunctive statements in
natural language In neuropsychology Neuropsychology is a branch of psychology. It is concerned with how a person's cognition and behavior are related to the brain and the rest of the nervous system. Professionals in this branch of psychology often focus on ...
s such as
English English usually refers to: * English language English is a West Germanic languages, West Germanic language first spoken in History of Anglo-Saxon England, early medieval England, which has eventually become the World language, leading lan ...
. Notably, classical disjunction is inclusive while natural language disjunction is often understood exclusively. :1. Mary is patriotic or quixotic. ::$\rightsquigarrow$ Mary is not both patriotic and quixotic. This inference has sometimes been understood as an
entailment Logical consequence (also entailment) is a fundamental concept Concepts are defined as abstract ideas A mental representation (or cognitive representation), in philosophy of mind Philosophy of mind is a branch of philosophy that studies th ...
, for instance by
Alfred Tarski Alfred Tarski (; January 14, 1901 – October 26, 1983), born Alfred Teitelbaum,School of Mathematics and Statistics, University of St Andrews ''School of Mathematics and Statistics, University of St Andrews''. was a Polish-American logician ...
, who suggested that natural language disjunction is
ambiguous Ambiguity is a type of meaning in which a phrase, statement or resolution is not explicitly defined, making several interpretations plausible. A common aspect of ambiguity is uncertainty Uncertainty refers to Epistemology, epistemic sit ...
between a classical and a nonclassical interpretation. More recent work in
pragmatics In linguistics Linguistics is the scientific study of language, meaning that it is a comprehensive, systematic, objective, and precise study of language. Linguistics encompasses the analysis of every aspect of language, as well as the m ...
has shown that this inference can be derived as a
conversational implicature An implicature is something the speaker suggests or implies with an utterance, even though it is not literally expressed. Implicatures can aid in communicating more efficiently than by explicitly saying everything we want to communicate. This phenom ...
on the basis of a
semantic Semantics (from grc, σημαντικός ''sēmantikós'', "significant") is the study of reference Reference is a relationship between objects in which one object designates, or acts as a means by which to connect to or link to, another ...
denotation which behaves classically. However, disjunctive constructions including
HungarianHungarian may refer to: * Hungary, a country in Central Europe * Kingdom of Hungary, state of Hungary, existing between 1000 and 1946 * Hungarians, ethnic groups in Hungary * Hungarian algorithm, a polynomial time algorithm for solving the assignmen ...
''vagy... vagy'' and ''soit... soit'' have been argued to be inherently exclusive, rendering un
grammaticality In linguistics Linguistics is the science, scientific study of language. It encompasses the analysis of every aspect of language, as well as the methods for studying and modeling them. The traditional areas of linguistic analysis include ...
in contexts where an inclusive reading would otherwise be forced. Similar deviations from classical logic have been noted in cases such as free choice disjunction and simplification of disjunctive antecedents, where certain modal operators trigger a
conjunction Conjunction may refer to: * Conjunction (astronomy), in which two astronomical bodies appear close together in the sky * Conjunction (astrology), astrological aspect in horoscopic astrology * Conjunction (grammar), a part of speech * Logical conjun ...
-like interpretation of disjunction. As with exclusivity, these inferences have been analyzed both as implicatures and as entailments arising from a nonclassical interpretation of disjunction. :2. You can have an apple or a pear. ::$\rightsquigarrow$ You can have an apple and you can have a pear (but you can't have both) In many languages, disjunctive expressions play a role in question formation. For instance, while the following English example can be interpreted as a
polar question Polar may refer to: Geography Polar may refer to: * Geographical pole A geographical pole or geographic pole is either of the two points on Earth where its axis of rotation intersects its surface. The North Pole lies in the Arctic Ocean ...
asking whether it's true that Mary is either a philosopher or a linguist, it can also be interpreted as an
alternative question ''AlterNative: An International Journal of Indigenous Peoples'' (formerly ''AlterNative: An International Journal of Indigenous Scholarship'') is a quarterly peer-reviewed academic journal An academic or scholarly journal is a periodical publica ...
asking which of the two professions is hers. The role of disjunction in these cases has be analyzed using nonclassical logics such as
alternative semanticsAlternative semantics (or Hamblin semantics) is a framework in formal semantics and logic Logic (from Ancient Greek, Greek: grc, wikt:λογική, λογική, label=none, lit=possessed of reason, intellectual, dialectical, argumentative, t ...
and
inquisitive semantics Inquisitive semantics is a framework in logic Logic (from Ancient Greek, Greek: grc, wikt:λογική, λογική, label=none, lit=possessed of reason, intellectual, dialectical, argumentative, translit=logikḗ)Also related to (''logos' ...
, which have also been adopted to explain the free choice and simplification inferences. :3. Is Mary a philosopher or a linguist? In English, as in many other languages, disjunction is expressed by a
coordinating conjunction In grammar, a conjunction (list of glossing abbreviations, abbreviated or ) is a part of speech that connects words, phrases, or clauses that are called the conjuncts of the conjunctions. This definition may overlap with that of other parts of s ...
. Other languages express disjunctive meanings in a variety of ways, though it is unknown whether disjunction itself is a
linguistic universal A linguistic universal is a pattern that occurs systematically across natural languages, potentially true for all of them. For example, ''All languages have noun A noun (from Latin ''nōmen'', literally ''name'') is a word that functions as the ...
. In many languages such as Dyirbal and
Maricopa Maricopa can refer to: Places * Maricopa, Arizona, United States, a city ** Maricopa Freeway, a piece of I-10 in Metropolitan Phoenix ** Maricopa station, an Amtrak station in Maricopa, Arizona * Maricopa County, Arizona, United States * Maricopa ...
, disjunction is marked using a verb
suffix In linguistics Linguistics is the scientific study of language A language is a structured system of communication used by humans, including speech (spoken language), gestures (Signed language, sign language) and writing. Most languag ...
. For instance, in the Maricopa example below, disjunction is marked by the suffix ''šaa''.
*
Affirming a disjunct The formal fallacy of affirming a disjunct also known as the fallacy of the alternative disjunct or a false exclusionary disjunct occurs when a deductive logic, deductive argument takes the following logical form: :A or B :A :Therefore, not B ...
*
Bitwise OR In computer programming Computer programming is the process of designing and building an executable computer program to accomplish a specific computing result or to perform a specific task. Programming involves tasks such as: analysis, gener ...
*
Boolean algebra (logic) In mathematics and mathematical logic, Boolean algebra is the branch of algebra in which the values of the variable (mathematics), variables are the truth values ''true'' and ''false'', usually denoted 1 and 0, respectively. Instead of elementary a ...
* Boolean algebra topics *
Boolean domain In mathematics Mathematics (from Greek: ) includes the study of such topics as numbers (arithmetic and number theory), formulas and related structures (algebra), shapes and spaces in which they are contained (geometry), and quantities and t ...
*
Boolean function In mathematics Mathematics (from Greek: ) includes the study of such topics as numbers ( and ), formulas and related structures (), shapes and spaces in which they are contained (), and quantities and their changes ( and ). There is no ge ...
*
Boolean-valued function A Boolean-valued function (sometimes called a predicate or a proposition In linguistics and logic, a proposition is the meaning of a declarative sentence. In philosophy, "Meaning (philosophy), meaning" is understood to be a non-linguistic enti ...
*
Disjunctive syllogism In classical logicClassical logic (or standard logic) is the intensively studied and most widely used class of deductive logic. Classical logic has had much influence on analytic philosophy, the type of philosophy most often found in the English-s ...
*
Disjunction elimination In propositional logic Propositional calculus is a branch of logic Logic (from Ancient Greek, Greek: grc, wikt:λογική, λογική, label=none, lit=possessed of reason, intellectual, dialectical, argumentative, translit=logikḗ)Al ...
*
Disjunction introduction Disjunction introduction or addition (also called or introduction) is a rule of inference In the philosophy of logic, a rule of inference, inference rule or transformation rule is a logical form consisting of a function which takes premises, ana ...
*
First-order logic First-order logic—also known as predicate logic, quantificational logic, and first-order predicate calculus—is a collection of formal system A formal system is an used for inferring theorems from axioms according to a set of rules. These rul ...
*
Fréchet inequalitiesIn probabilistic logicThe aim of a probabilistic logic (also probability logic and probabilistic reasoning) is to combine the capacity of probability theory to handle uncertainty with the capacity of deductive logic to exploit structure of formal pr ...
*
Free choice inferenceFree choice is a phenomenon in natural language where a disjunction In logic Logic (from Ancient Greek, Greek: grc, wikt:λογική, λογική, label=none, lit=possessed of reason, intellectual, dialectical, argumentative, translit=lo ...
* Hurford disjunction * Logical graph *
Logical value In logic Logic is an interdisciplinary field which studies truth and reasoning Reason is the capacity of consciously making sense of things, applying logic Logic (from Ancient Greek, Greek: grc, wikt:λογική, λογική, la ...
* Operation *
Operator (programming) In computer programming Computer programming is the process of designing and building an executable computer program to accomplish a specific computing result or to perform a particular task. Programming involves tasks such as analysis, ge ...
*
OR gate The OR gate is a digital logic gate that implements logical disjunctionit behaves according to the truth table to the right. A HIGH output (1) results if one or both the inputs to the gate are HIGH (1). If neither input is high, a LOW output ( ...
*
Propositional calculus Propositional calculus is a branch of logic Logic is an interdisciplinary field which studies truth and reasoning. Informal logic seeks to characterize Validity (logic), valid arguments informally, for instance by listing varieties of fal ...
* Simplification of disjunctive antecedents
# Notes
*
George Boole George Boole (; 2 November 1815 – 8 December 1864) was a largely self-taught Autodidacticism (also autodidactism) or self-education (also self-learning and self-teaching) is education Education is the process of facil ...
, closely following analogy with ordinary mathematics, premised, as a necessary condition to the definition of "x + y", that x and y were mutually exclusive. , and practically all mathematical logicians after him, advocated, on various grounds, the definition of "logical addition" in a form which does not necessitate mutual exclusiveness. | 2022-08-14 15:28:39 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 26, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6539019346237183, "perplexity": 2947.1770611011366}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882572043.2/warc/CC-MAIN-20220814143522-20220814173522-00180.warc.gz"} |
https://quantumcomputing.stackexchange.com/questions/12966/given-a-channel-phix-sum-k-c-kx-sigma-k-are-there-always-f-k-ge0-such | # Given a channel $\Phi(X)=\sum_k c_k(X)\sigma_k$, are there always $F_k\ge0$ such that $\Phi(X)=\sum_k \operatorname{tr}(F_k X)\sigma_k$?
Fix a finite number of states $$\sigma_k$$, and consider a channel of the form $$\Phi(X)=\sum_k c_{k}(X)\sigma_k.$$
For $$\Phi$$ to be linear and trace-preserving we must have: $$c_k(X+X') = c_k(X) + c_k(X'), \qquad \sum_k c_k(X)=1.$$ In other words, the coefficients must be linear functionals $$c_k\in\mathrm{Lin}(\mathcal X)^*$$ for all $$k$$.
Does this imply that there must be some positive operators $$F_k\ge0$$ such that $$c_k(X)=\operatorname{Tr}(F_k X)$$ for all $$k$$ (which in turn would imply $$\sum_k F_k=I$$ and thus that $$\{F_k\}_k$$ is a POVM)? What's a good way to show this?
To this end, pick a linearly independent set $$\{\sigma_k\}$$ which spans the full matrix space (over $$\mathbb C)$$, that is, a basis. (This is always possible, as the positive operators span the hermitian ones over $$\mathbb R$$.)
Then pick a dual basis $$\sigma'_\ell$$ such that $$\mathrm{tr}[\sigma'_\ell \sigma_k]=\delta_{k\ell}\ .$$
Then, $$\Phi(X) = \sum_k \mathrm{tr}[\sigma'_k X]\,\sigma_k$$ is the identity channel, which cannot be written as a POVM $$F_k\ge0$$ followed by a preparation of $$\sigma_k$$ (as that channel would be entanglement breaking).
(Note that this shows that the dual basis $$\sigma'_\ell$$ has non-positive elements. This is not surprising, since otherwise the scalar product $$\mathrm{tr}[\sigma'_\ell\sigma_k]\ge0$$ for all $$k,\ell$$.)
• by "identity channel" you mean $\Phi(X)=X$? But if there are such states $\sigma_k\ge0$ s.t. $\Phi(X)=X=\sum_k c_k(X)\sigma_k$ for all $X$, then $c_k(X)=\operatorname{Tr}(\sigma_k X)$ and thus $\sum_k\sigma_k=I$ if $\Phi$ is trace-preserving (assuming these are an orthonormal basis... but if they are not, are we ensured that they can be used to decompose any $X$?) – glS Jul 19 '20 at 16:51
• ah, I think I got it. There is a (non-orthogonal) basis of states $\sigma_k$ such that we can write $X=c_k(X)\sigma_k$ for all $X$. However, it is not true that $c_k(X)=\operatorname{Tr}(\sigma_k X)$ because the basis is not made of orthogonal operators (and now I understand why you used the notion of dual basis here). So I guess the hypothesis is true only as long as restrict $\sigma_k$ to be orthogonal operators. – glS Jul 19 '20 at 17:07
• @glS The hypothesis is true if and only if the channel is entanglement breaking. I would have to look up myself whether an entanglement breaking channel can always be written with $\sigma_k$ an orthogonormal basis. On the spot, I don't see why. Just because this is not the case in my example does not mean it is required. – Norbert Schuch Jul 19 '20 at 17:25 | 2021-04-17 21:16:01 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 22, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9322410225868225, "perplexity": 141.75496553395274}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038464045.54/warc/CC-MAIN-20210417192821-20210417222821-00412.warc.gz"} |
https://www.physicsforums.com/threads/verifying-this-trigonometric-identity.457034/ | # Verifying this Trigonometric Identity
Hey guys. How are you all doing? I'm helping my younger brother out with his trigonometry homework. He is dealing with verifying trigonometric identities. However, he has the problem that I am getting nowhere with. Hope you all can help. Thanks in advance. :)
## Homework Statement
Verify (1-cos^2 (a))(1+cos^2(a)) = 2sin^2 (a) -sin^4 (a). I cant simplify the (1+cos^2(a)). Also can not tell if I can simplify the other side as well.
## Homework Equations
sin^2 a + cos^2 a = 1
## The Attempt at a Solution
So using the Pythagorean identity, I have been able to simplify this to:
(sin^2 (a))(1+cos^2) ) = 2sin^2 (a) -sin^4 (a).
I am just stuck in simplifying the part after sin^2 (a). Also cant seem to simplify the other side. Any assistance is awesome. | 2020-08-07 01:57:14 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8796271681785583, "perplexity": 1746.2421378797922}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439737050.56/warc/CC-MAIN-20200807000315-20200807030315-00144.warc.gz"} |
https://docs.openstack.org/tripleo-docs/latest/install/post_deployment/build_single_image.html | # Building a Single Image¶
The openstack overcloud image build --all command builds all the images needed for an overcloud deploy. However, you may need to rebuild a single one of them. Use the following commands if you want to do it:
openstack overcloud image build --type {agent-ramdisk|deploy-ramdisk|fedora-user|overcloud-full}
If the target image exist, this commands ends silently. Make sure to delete a previous version of the image to run the command as you expect.
Moreover, you can build the image with an extra element of your choice using the --builder-extra-args argument:
openstack overcloud image build --type overcloud-full \
--builder-extra-args overcloud-network-midonet
Note
Make sure the element is available in the \$ELEMENTS_PATH environment variable
After the new image is built, it can be uploaded using the same command as before, with the --update-existing flag added:
openstack overcloud image upload --update-existing
Note that if the new image is a ramdisk, the Ironic nodes need to be re-configured to use it. This can be done by re-running:
openstack overcloud node configure --all-manageable
Note
If you want to use custom images for boot configuration, specify their names in --deploy-kernel and --deploy-ramdisk options.
Now the new image should be fully ready for use by new deployments. | 2018-10-23 21:07:22 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5578268766403198, "perplexity": 4691.159018265891}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583517376.96/warc/CC-MAIN-20181023195531-20181023221031-00487.warc.gz"} |
http://codeforces.com/blog/upobir | Codeforces celebrates 10 years! We are pleased to announce the crowdfunding-campaign. Congratulate us by the link https://codeforces.com/10years. ×
### upobir's blog
By upobir, 3 months ago, ,
One of my favorite algorithms out there is FWHT, but sadly there are not many tutorials about it. So here's a simple attempt by me (feedback is appreciate).
Note: I will be explaining FWHT via DFT/FFT and as such good understanding of theory behind DFT/FFT is required. Also, I won't be going into Hadamard matrices, since I don't know much about them myself.
What is FWHT? Suppose you have two sets $(1, 2, 2)$ and $(3, 4, 5)$ and want to perform all possible xor operations between these two sets, the resulting set being $(1, 1, 2, 4, 5, 6, 6, 7, 7)$. This is commonly called xor convolution. Naively you'd do this in $O(n^2)$ time, where $n=2^k$, $k$ = maximum number of bits (you maintain a frequency count for each possible value and loop on all possible pair). FWHT let's you do this in $O(n\log n)$ time via some black magic FFT tricks. We first build the frequency arrays upto $n=2^k$, apply FWHT to turn them into some mysterious point-value form, perform pointwise multiplication and finally perform inverse FWHT on the resulting array to recover resultant frequency array. But how does this work? And why should you know it?
Let's answer the first question. Though we will need to learn two facts about FFT first:
1) When we want to multiply $(1+2x)$ and $(3+4x)$, we first extend their size to 4, then transfer them to point-value form, then perform pointwise multiplication, and then finally perform inverse transform. Notice the first step, we extend the size because their product is of size 3 and so will require at least 3 distinct $(x, y)$ point-value pairs (we use 4 cause it's a power of two). What if we didn't do this size extention? You will see that after the last step you get the polynomial $(11+10x)$ instead of $(3+10x+8x^2)$. The explanation is that we used quadratic roots of unity $1, -1$. They don't have any square, instead for them, $x^2=1$, so the quadratic term "wraps around", contributing to the constant term. The lesson is that if you don't use sufficiently granular roots, the larger powers "wrap around".
2) We use FFT to multiply monovariable polynomials, what about multivariable polynomials? Everything is same as before, we just need a suitable point value form so that we can recover the polynomial back from it. For a monovariable polynomial of size $n$, we need $n$ distinct values (we use roots of unity in FFT for divide and conquer speeed up). In a multivariable polynomial, take $P(x, y)$ $= (1+2x+3y+4xy)$ for example, all we need is distinct value sets for each of the variables. Here, x is of degree 1 and y is of degree 1. So we can just use two distinct value sets for them. For example, we can use $10, 20$ and $30, 40$ for $x$ and $y$ respectively. Then the values $P(10, 30), P(10, 40), P(20, 30), P(20, 40)$ uniquely define the polynomial $P(x, y)$. We could've used $10, 20$ for both variables and that'd also work.
We incorporate this into multivariable FFT by by dividing the polynomials into polynomial of $x$ grouping by $y$'s power, like $P(x, y) = (1+2x) +y(3+4x) = Q_0(x) + yQ_1(x)$. First let's apply the FFT on the $Q_i$, to get $Q_0(1), Q_0(-1), Q_1(1), Q_1(-1)$. Then we just note that applying FFT on $Q_0(1)+yQ_1(1)$ and $Q_0(-1)+yQ_1(-1)$ will give us $P(1,1), P(1, -1)$ and $P(-1, 1), P(-1, -1)$. So the lesson is, multivariable FFT is just normal FFT with some grouping of values.
Now to FWHT
Like it was,previously said, FWHT will convert the frequency arrays to some magical form where performing pointwise multiplication somehow performs xor. Note the similarity with polynomial multiplication here. In polynomial multiplication, $ax^i$ and $bx^j$ multiply to merge into $abx^{i+j}$, we perform this merging for all possible pairs and group them according to $x$'s power. In FWHT, what we want is for $ax^i$ and $bx^j$ to merge into $abx^{i \oplus j}$ (considering the frequency array as a polynomial). We can do an interesting thing here, break variables into more varibles, utilizing a number's binary form! Let me explain. Suppose in a problem xor convolution requires 3 bits maximum, then terms like $ax^3$ and $bx^5$ are brokem into $ax_2^0x_1^1x_0^1$ and $bx_2^1x_1^0x_0^1$ (cause 3 is 011 and 5 is 101).
In this new form, there are variables representing each bit. So if we can somehow make multiplication behave well in single bit powers, then xor convolution will become easy! More formally, if we can somehow make it that $x_i^0 \cdot x_i^0=x_i^1\cdot x_i^1=x_i^0$ and $x_i^0 \cdot x_i^1 = x_i^1$ then $ax_2^0x_1^1x_0^1$ and $bx_2^1x_1^0x_0^1$ will multiply to form $abx_2^1x_1^1x_0^0$ ($3 \oplus 5 = 6$). This is last obstacle. But if you notice the equations, you will see that they are basically $x_i^p \cdot x_i^q= x_i^{(p+q) \pmod 2}$. And now you can see why I had talked so much about FFT with limited domain at the beginning. If we use the numbers $1, -1$ then as $x^2=1$, behavior's $x_i^p \cdot x_i^q= x_i^{(p+q) \pmod 2}$ has been achieved!
Now bringing it all together, we think of the terms $ax^i$ as being broken upto variables that represent the exponent's binary form. We will multiply this new polynomial using multivariable FFT. And to make sure that they behave like xor-ing, we use the domain $1,-1$ for each variable.(this wraps the $x^2$ term back to constants). So FWHT is now just a fancy FFT, huzzah!
Of course we don't need to do actual FFT algorithm for the variables, a polynomial $a+bx$ can be converted to point value form of $(a+b, a-b)$ by simply hardcoding it. And for inverting the transform, observe that if $u=a+b$ and $v=a-b$ then $a=\frac{u+v}{2}$ and $b=\frac{u-v}{2}$. This is why codes for FWHT perform division by 2 in inverting (after doing ordinary FWHT).
I am not providing any code since one can find many good implementations online.
But now to the second question I asked at the beginning, why should you learn it? The answer is simple really, I believe you should understand an algorithm to be able to modify it to specific needs. For example, I set a problem in Codechef's November Long challenge, where you will definitely need to modify the traditional FWHT (along with some more tricks). If you have found more problems requiring modifications on FWHT, please comment so. Also, constructive criticism is appreciated.
EDIT: Forgot to add two things. First of all there are some well known common modifications of FWHT: "Or convolution" and "And convolution". They are the same thing as xor convolution, except you do, 'or' or 'and'. The "or convolution" is almost the same except the behavior we want is $x_i^0 \cdot x_i^0 = x_i^0$ and $x_i^0 \cdot x_i^1 = x_i^1 \cdot x_i^0 = x_i^1 \cdot x_i^1 = x_i^1$. For this, obviously $1$ is a candidate, another illegal cunning candidate is $0$ (we'll use $0^0 = 1$). So we use the domain $1, 0$. "And convolution" is just "Or convolution" on complements, which in terms of implementation is just reversing the logic of "Or convolution".
And finally, I learned this interpretation of FWHT from this helpful blog. But I believe the explanation there is not elaborate. Thus my attempt to write this blog.
• +165
By upobir, history, 3 months ago, ,
What is the best process to see if a problem's idea has already been used somewhere or not? There are so many online judges and contests. Is there any good way to find out? It sucks to propose a problem somewhere and later find out it's an almost copy from some oj.
• +25
By upobir, 10 months ago, ,
Sometimes when you are getting WA on a problem, possibly due to missing some corner case or a bug in your code, all you really need are the perfect test cases. Of course you don't know the perfect test cases, so you have to resort to something else. And that something else is checking your code on a bunch of randomly generated testcases. I recently learnt about it and so, am writing this so other noobs like me can get a little help.
First of all, this isn't an unique idea. I myself learnt about it from errichto's youtube video. You should check it out if you haven't seen it before. So the idea is simple, you will write two new c++ programs, one to generate random test cases. And another to get solution for those test cases, but unlike your actual solution code, this will definitely give the correct answer (perhaps not efficiently/ via brute force). You run the test generator a bunch of times and save the tests in a fine input.in. After each generation, you also solve the problem for the test case in output.out (your original solution) and output2.out (your brute force solution). And then you compare the solutions with a script code of your OS.
The test generation part is easy for a lot of problems, you need to just randomly generate a bunch of data. Sometimes these need to follow a strict restriction but that depends on the problem. For random numbers I use
mt19937 rng(chrono::system_clock::now().time_since_epoch().count());
long long random(long long a, long long b){
return a + rng() % (b - a + 1);
}
Anyways, about the script part, which is the main goal of this blog. For linux script you can just see errichto's video. For windows sctipt, put all three codes TestGenerator.cpp, solution.cpp, brute.cpp in one folder, compile them, and write the following code in checker.bat file.
@echo off
for /l %%x in (1, 1, 100) do (
TestGenerator > input.in
solution < input.in > output.out
brute < input.in > output2.out
fc output.out output2.out > diagnostics || exit /b
echo %%x
)
echo all tests passed
The code basically runs a loop 100 times. Each time it calls TestGenerator.exe and puts the randomly generated output in input.in (the > command redirects stdout to input.in). Then solution.exe and brute.exe are run by redirecting stdin to input.in. The output are put in output.out and output2.out respectively. fc is the command for comparing files, if the files have difference then "exit /b" command stops the batch file. Otherwise if there is no difference, the test case number x will be printed showing you the progress. The advantage of the exit command is that after exiting, you will find the corner test case in input.in and you can debug your code. So all you gotta do is just open cmd in the folder of the files and write checker.bat and run.
I myself don't know that much windows scripting, I learned about it from yeputons's comment in this post. Anyways hope this helps. Forgive any mistakes please.
• +32
By upobir, history, 13 months ago, ,
I have been stuck at this problem for quite a long time now. Can someone please provide a detailed solution. I know the trick to be used is broken profile dp, but after that I am in a dead end. Thanks in advance. | 2020-02-18 08:09:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6209757924079895, "perplexity": 750.3929326139709}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875143635.54/warc/CC-MAIN-20200218055414-20200218085414-00334.warc.gz"} |
https://www.gradesaver.com/textbooks/math/applied-mathematics/elementary-technical-mathematics/chapter-15-section-15-6-other-average-measurements-and-percentiles-exercise-page-518/25 | ## Elementary Technical Mathematics
$0.05\times50=2.5$. 2.5 rounds up to 3, so the 5th percentile is the 3rd piece of data in this ordered set of 50 numbers. The 5th percentile is 23. | 2022-08-07 17:06:58 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6517664790153503, "perplexity": 736.5755169750513}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882570651.49/warc/CC-MAIN-20220807150925-20220807180925-00067.warc.gz"} |
https://www.gamedev.net/forums/topic/680466-the-near-view-is-not-shown-properly/ | • 10
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The Near View Is Not Shown Properly
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Hello!
I have encountered some problems and I don't know what might be causing it. The problem is that I can't properly see what is being draw near to the camera.
This is how I am setting up my frustum but I have played around with it alot and it looks like that it is not the issue. http://pastebin.com/Zq9iZm9Y
This is how it looks right now: http://imgur.com/a/uxhgs
You can see that if I rotate the world around that there is a wall/path behind me as well but as you can see it is being rendered strangely. If anybody knows what might be causing this please feel free to share the information.
Regards!
Edited by Skyzyx
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Hard to really tell without knowing more about how big everything is. Have you tried reducing your near plane value?
Matrix.frustumM(projectionMatrix, 0, -aspectRatio, aspectRatio, -1, 1, 1, 150);
One of those is probably it. I don't know what you are using there so I'm not entirely sure but you could try something like:
Matrix.frustumM(projectionMatrix, 0, -aspectRatio, aspectRatio, -1, 1, 0.1, 150);
Is this the function you are using?
https://developer.android.com/reference/android/opengl/Matrix.html#frustumM(float[], int, float, float, float, float, float, float)
You might be better of using perspectiveM to setup your projection.
Edited by Nanoha
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I tried to set the near plane to 0.1f or even 0.01f but everything gets stretched out. I tried using perspectiveM and setting the near plane to 0.1f seemed to do the trick with a 90 degree angle. Thanks for the help!
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Normally for anything related to the physical world, 1.0 is one meter. So if your near plane needs to be at 0.01 it means the thing is one centimeter away from your eyeball. Unless that thing is a pair of glasses, you don't want it anywhere near that close.
Consider what you are doing with that line:Matrix.frustumM(projectionMatrix, 0, -aspectRatio, aspectRatio, -1, 1, 1, 150);
You are making an orthgraphic view, basically a flat view used for HUD or map views. The view has six walls: the left wall is -aspectRatio meters from your eyeball, the right wall is aspectRatio meters from your eyeball, the ceiling and floor are one meter above and below your eyes. The nearest thing you can see is one meter away (which is probably good) and you are looking at a deep box of 150 meters in depth.
When you change to perspectiveM you are looking out more like a fanned out view, more like what your eyeball actually sees. You can set an angle for your field of view, your aspect ratio, and near/far planes. You will probably want to see starting at about a meter or half-meter out, and quite far into the distance.
To get a real life view of what your field of view should be, hold up a protractor to your eyeball and look at your monitor. Center it on your screen and look at the angle it spans. If you want it to look similar to your monitor being a view of the world it should be near 60 degrees or so. If you're looking for a wider view maybe up to 90, an extended camera view you might want even higher than that. The fov can be changed for artistic reasons, like looking in a sniper's scope which can have a fov under 6 degrees, maybe as low as 2 degrees or so depending on the lens you are simulating.
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I might add that a 55mm lens is your normal human view lens with cameras. The more you go below that the more you will get a "fish eye" perspective that will warp things. The more above that you go the more you will get a telescopic view of things where things appear to be slower together than the really are.
I've been watching a great lecture on perspective in drawing that probably applies here. This is basically mathematics and the physical world we live in.
You're basically creating vanishing points like in drawing. The closer they get, the more the whole view is squished together like with a telephoto camera lens. The further they are moved apart, the more it stretches everything in view and the more you get that "fish eye" effect.
The field of view should be between about 45 and 60 to get a natural view the way humans normally see things. Beyond that in either direction it will be "unnatural" for better or worse.
And looking at the pictures, I can't tell that anything is rendered wrong. I don't know what you intend it to look like. If it's drawing "inside out" that's a totally different problem.
Edited by BBeck | 2018-04-26 18:55:53 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.22041888535022736, "perplexity": 906.88287677783}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125948464.28/warc/CC-MAIN-20180426183626-20180426203626-00454.warc.gz"} |
https://solvedlib.com/n/choice-and-briefly-hh-u-aid-financial-describe-your-design,11726582 | # Choice and briefly HH U aid financial describe your design drawn 1 H This information will be used to set
###### Question:
choice and briefly HH U aid financial describe your design drawn 1 H This information will be used to set the level of financial aid, wants to know how alphabetizcd list variable - but have not yet graduated, it can 1 would you suggest cxpcct studemIS [0 1 U Groontaing 97 5 Jels 'Justify your
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What is the delta of a short position in 600 European call options on silver futures? The options mature in 8 months and the silver futures contract matures in 9 months. The current 9 month futures price is $30.00 per ounce. The exercise price of the option is$31.00 per ounce. The risk-free interes... | 2022-08-09 22:04:29 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4929192066192627, "perplexity": 6058.240136092021}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571090.80/warc/CC-MAIN-20220809215803-20220810005803-00171.warc.gz"} |
http://cvgmt.sns.it/paper/4558/ | # Rectifiability of the jump set of locally integrable functions
created by delnin on 08 Jan 2020
[BibTeX]
Preprint
Inserted: 8 jan 2020
Last Updated: 8 jan 2020
Year: 2020
Abstract:
In this note we show that for every measurable function on $\mathbb{R}^n$ the set of points where the blowup exists and is not constant is $(n-1)$-rectifiable. In particular, for every $u\in L^1_{loc}(\mathbb{R}^n)$ the jump set $J_u$ is $(n-1)$-rectifiable.
Keywords: Rectifiability, bounded variation, jump set, Blowup | 2020-06-04 12:00:04 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9151849746704102, "perplexity": 1255.1234664631847}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347439928.61/warc/CC-MAIN-20200604094848-20200604124848-00230.warc.gz"} |
http://www.lastfm.com.br/user/froggafiend/library/music/Elvis+Presley/_/You+Gave+Me+A+Mountain?setlang=pt | # Biblioteca
Música » Elvis Presley »
## You Gave Me A Mountain
39 execuções | Ir para página da faixa
Faixas (39)
Faixa Álbum Duração Data
You Gave Me A Mountain 3:15 Jan 23 2013, 17h49
You Gave Me A Mountain 3:15 Jan 16 2013, 12h46
You Gave Me A Mountain 3:15 Jan 3 2013, 9h32
You Gave Me A Mountain 3:15 Jan 2 2013, 9h40
You Gave Me A Mountain 3:15 Jan 1 2013, 11h03
You Gave Me A Mountain 3:15 Jan 1 2013, 10h01
You Gave Me A Mountain 3:15 Jan 1 2013, 9h22
You Gave Me A Mountain 3:15 Dez 23 2012, 11h48
You Gave Me A Mountain 3:15 Dez 23 2012, 10h08
You Gave Me A Mountain 3:15 Dez 19 2012, 2h08
You Gave Me A Mountain 3:15 Dez 18 2012, 9h53
You Gave Me A Mountain 3:15 Dez 18 2012, 9h22
You Gave Me A Mountain 3:15 Dez 15 2012, 9h55
You Gave Me A Mountain 3:15 Out 21 2012, 11h42
You Gave Me A Mountain 3:15 Out 19 2012, 22h39
You Gave Me A Mountain 3:15 Out 2 2012, 12h36
You Gave Me A Mountain 3:15 Ago 29 2012, 8h07
You Gave Me A Mountain 3:15 Ago 25 2012, 15h44
You Gave Me A Mountain 3:15 Abr 6 2012, 10h38
You Gave Me A Mountain 3:15 Mar 14 2012, 19h10
You Gave Me A Mountain 3:15 Mar 14 2012, 16h11
You Gave Me A Mountain 3:15 Mar 9 2012, 13h34
You Gave Me A Mountain 3:15 Mar 1 2012, 3h07
You Gave Me A Mountain 3:15 Mar 1 2012, 1h36
You Gave Me A Mountain 3:15 Fev 29 2012, 14h38
You Gave Me A Mountain 3:15 Fev 19 2012, 9h33
You Gave Me A Mountain 3:15 Fev 4 2012, 10h05
You Gave Me A Mountain 3:15 Fev 4 2012, 9h51
You Gave Me A Mountain 3:15 Jan 25 2012, 16h11
You Gave Me A Mountain 3:15 Jan 2 2012, 0h44
You Gave Me A Mountain 3:15 Dez 25 2011, 14h59
You Gave Me A Mountain 3:15 Dez 25 2011, 13h33
You Gave Me A Mountain 3:15 Dez 24 2011, 9h27
You Gave Me A Mountain 3:15 Dez 9 2011, 12h41
You Gave Me A Mountain 3:15 Dez 9 2011, 11h27
You Gave Me A Mountain 3:15 Dez 3 2011, 3h15
You Gave Me A Mountain 3:15 Set 14 2011, 17h11
You Gave Me A Mountain 3:15 Set 12 2011, 9h02
You Gave Me A Mountain 3:15 Set 12 2011, 8h58 | 2014-07-22 13:10:33 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9453175663948059, "perplexity": 14484.703741340441}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1405997858581.26/warc/CC-MAIN-20140722025738-00217-ip-10-33-131-23.ec2.internal.warc.gz"} |
https://bookdown.org/huckley/Physical_Processes_In_Ecosystems/1-11-calcint-problems2.html | 1. One way to use least-squares is to approximate one function by another. Here we approximate the curve $y = 6.1(x+5.0)e^{-0.045x}$ by a straight line $y = Ax + B$
Evaluate both $$\partial (E^2)/\partial A$$ and $$\partial (E^2)/\partial B$$, then set both equal to zero to obtain two equations in A and B. The goal is to minimize: $E^2 = \int^{30}_3[6.1(x+5.0)e^{-0.045x}-Ax-B]^2dx$ (Hint: when evaluating $$\partial E^2/\partial A$$ and $$\partial E^2/\partial B$$, differentiate under the integral sign.)
1. The stomates in leaves are small pores which permit the exchange of gases and water vapor. The stomates change shape from a near circle to a slit in order to adjust this exchange in response to varying environmental conditions. Each stomate is roughly elliptical in shape with constant perimeter throughout the shape change. The area of the opening is then given by $A= \frac{2b}{a}\int^a_{-a}(a^2-x^2)^{1/2}dx$ where $$2a$$, $$2b$$ are the lengths of the major and minor axes, respectively, of the ellipse.
1. First evaluate this integral using either of two trigonometric substitutions: let $$x = a \cos \theta$$ or let $$x = a \sin \theta$$. Be sure to check the limits on $$\theta$$ and the sign of each trig function in the transformed integrand. Your result must be non-negative since the area is non-negative. For a fixed perimeter, say $$35 \mu$$, the area should be maximal when the hole is a circle, i.e. $$a=b$$. Demonstrate this by evaluating the area when $$a=b=5.57$$, and when $$a=5$$, $$b=6.09$$. (The perimeter formula is $$P = 2\pi\sqrt{(a^2+b^2)/2}$$.)
2. Use your area formula (the answer to part a) to investigate the dilution of the sun’s energy flux due to the angle at which the rays strike the earth’s surface. The dilution is caused by a fixed amount of solar energy being spread over a larger area (as the rays become more horizontal) and thus the energy per unit area decreases as compared to the energy density of vertical rays. The situation is pictured below, giving a top and side view for two inclination angles. The sunlight passes through a circular hole in an opaque sheet and strikes the earth at an angle $$R$$ from the perpendicular.
The illuminated area changes from a circle ($$R=0$$) to an elongated ellipse (as $$R$$ increases). For a given diameter $$B$$, write the illuminated area $$A$$ as a function of the angle of inclination $$R$$. Check your result by choosing values for $$B$$ and $$R$$ and calculating the dilution $$D$$, which equals the circle’s area divided by the ellipse’s area. Does $$D = 0.6$$ mean more dilution (less energy density) than $$D = 0.3$$ or vice versa? The inclination angle in Minnesota changes from about 20° in summer to 65° in winter. What is the resulting change in energy density, measured by $$D$$? | 2021-03-01 07:49:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9034894704818726, "perplexity": 271.5188570072309}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178362133.53/warc/CC-MAIN-20210301060310-20210301090310-00565.warc.gz"} |
http://www.solutioninn.com/following-are-descriptions-of-transactions-and-other-financial-events-for | # Question
Following are descriptions of transactions and other financial events for the City of Tetris for the year ending December 2010. Not all transactions have been included here. Only the General Fund formally records a budget. No encumbrances were carried over from 2009.
Paid salary for police officers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. \$ 21,000
Received government grant to pay ambulance drivers . . . . . . . . . . . . . . . . . . 40,000
Estimated revenues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232,000
Received invoices for rent on equipment used by fire department
during last four months of the year . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3,000
Paid for newly constructed city hall . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1,044,000
Made commitment to acquire ambulance . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111,000
Received cash from bonds sold for construction purposes . . . . . . . . . . . . . . 300,000
Placed order for new sanitation truck . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154,000
Paid salary to ambulance drivers—money derived from state
government grant given for that purpose . . . . . . . . . . . . . . . . . . . . . . . . . . . 24,000
Paid for supplies for school system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16,000
Made transfer from General Fund to eventually pay off a long-term debt . . . 33,000
Received but did not pay for new ambulance . . . . . . . . . . . . . . . . . . . . . . . . 120,000
Levied property tax receivables for 2010. City anticipates that
95 percent (\$190,000) will be collected during the year and 5 percent
will be uncollectible . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200,000
Acquired and paid for new school bus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40,000
(not previously accrued) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14,000
Made appropriations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225,000
The following questions are independent although each is based on the preceding information. Assume that the government is preparing information for its fund-based financial statements.
a. What is the balance in the Budgetary Fund Balance account for the budget for the year? Is it a
debit or credit?
b. Assume that 60 percent of the school supplies are used during the year so that 40 percent remain. If the consumption method is being applied, how is this recorded?
c. The sanitation truck that was ordered was not received before the end of the year. The commitment will be honored in the subsequent year when the truck arrives. What reporting is made at the end of 2010?
d. Assume that the ambulance was received on December 31, 2010. Provide all necessary journal entries on that date.
e. Give all journal entries that should have been made when the \$33,000 transfer was made to eventually pay off a long-term debt.
f. What amount of revenue would be recognized for the period? Explain the composition of this total.
g. What are the total expenditures? Explain the makeup of this total. Include (b) here.
h. What journal entry or entries were prepared when the bonds were issued?
Sales4
Views134 | 2016-10-27 15:22:29 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9264052510261536, "perplexity": 67.30448500154164}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988721347.98/warc/CC-MAIN-20161020183841-00058-ip-10-171-6-4.ec2.internal.warc.gz"} |
http://ee.usc.edu/~ashutosn/CommNetS2016/dokuwiki/doku.php?id=a_fast_drift_method_for_convex_programs | # CommNetS
### Site Tools
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a_fast_drift_method_for_convex_programs
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Title: A fast drift method for convex programs
Abstract:
This paper considers convex programs with a general (possibly non-differentiable) convex objective function and Lipschitz continuous convex inequality constraint functions. A simple algorithm is developed and achieves an $O(1/t)$ convergence rate. Similar to the classical dual subgradient algorithm and the ADMM algorithm, the new algorithm has a parallel implementation when the objective and constraint functions are decomposable. However, the new algorithm has faster $O(1/t)$ convergence rate compared with the best known $O(1/\sqrt{t})$ convergence rate for the dual subgradient algorithm with averaged primals. Further, it can solve convex programs with nonlinear constraints, which cannot be handled by the ADMM algorithm. The new algorithm is applied to a multipath network utility maximization problem and yields a decentralized flow control algorithm with fast $O(1/t)$ convergence rate.
Bio: Hao Yu is currently a PhD student in EE department at University of Southern California, advised by Prof. Michael J. Neely. He received the B.Eng. in Electrical Engineering from Xi’an Jiaotong University, China, and the Mphil. degree in Electrical Engineering from the Hong Kong University of Science and Technology, Hong Kong. His research interests are in the areas of design and analysis of optimization algorithms, network optimization and network coding. | 2020-02-19 23:21:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5241367220878601, "perplexity": 1051.1151750272204}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875144429.5/warc/CC-MAIN-20200219214816-20200220004816-00001.warc.gz"} |
https://socratic.org/questions/how-do-you-find-the-sum-of-the-infinite-geometric-series-sigma-2-2-3-n-from-n-0--1 | How do you find the sum of the infinite geometric series Sigma 2(-2/3)^n from n=0 to oo?
Apr 14, 2017
$\frac{6}{5}$
Explanation:
First, take the $2$ out of the sigma so it's easier to deal with:
$2 \sum {\left(- \frac{2}{3}\right)}^{n}$
This is in the form:
$\sum {\left(u\right)}^{n}$ where $| u | < 1$. That means, that $\sum {u}^{n} = \frac{1}{1 - u}$
Our $u$, in this case, is $- \frac{2}{3}$. Simply plug into the formula.
$2 \left(\frac{1}{1 - \left(- \frac{2}{3}\right)}\right) = 2 \left(\frac{1}{\frac{5}{3}}\right) = 2 \left(\frac{3}{5}\right) = \frac{6}{5}$ | 2019-03-23 02:16:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9850311875343323, "perplexity": 261.8952801757185}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912202711.3/warc/CC-MAIN-20190323020538-20190323042538-00251.warc.gz"} |
http://math.stackexchange.com/questions/53161/how-do-i-show-that-e1-x-x2x2-5x-4-x12-2x-1-has-only-one-po | # How do I show that $e^{1/x} = x(2x^2 + 5x + 4)/[(x+1)^2 (2x-1)]$ has only one positive real solution?
I need to show that, for the function $f(x) = x^2 (e^{1/x}-1) - x^2/(x+1)$, there is some $x_0 > 0$ such that $f(x)$ is decreasing on $(0,x_0)$ and increasing on $(x_0,\infty)$.
By setting $f'(x) = 0$ I can see that $f(x)$ is decreasing on $(0,1/2)$, and taking the series expansion of $f(x)$ at $\infty$ tells me that $f(x) \to 3/2$ as $x \to \infty$, but this isn't enough. The problem boils down to the question in the title.
I just can't seem to make it give up the goods.
-
I could simplify the problem quite a bit but left a simpler problem for you to solve. We have
$$f(x) = \left(e^{\frac{1}{x}}-1\right) x^2-\frac{x^2}{x+1}$$
using some basic algebra you can factor its derivate to:
$$f'(x) = \frac{2 e^{\frac{1}{x}} x^3-2 x^3+3 e^{\frac{1}{x}} x^2-5 x^2-4 x-e^{\frac{1}{x}}}{(x+1)^2}$$
So what is left to show is that
$$h(x)= e^{\frac{1}{x}} x^3-2 x^3+3 e^{\frac{1}{x}} x^2-5 x^2-4 x-e^{\frac{1}{x}}$$
has exactly one real root in $(0, \infty)$, but this is not hard because if we look at the graph of h we see that
And therefore assume that $$h'(x)=(x+1) \left(e^{\frac{1}{x}} \left(\frac{1}{x^2}+6 x-\frac{1}{x}-2\right)-6 x-4\right)>0$$
Where you can also use the notation
$$h'(t)=h'(1/x)=\frac{(t+1) \left(-4 t+e^t ((t-2) t (t+1)+6)-6\right)}{t^2}$$
So it remains again to show that
$$(t+1) \left(t^3-t^2-2 t+6\right) e^t\geq 2 \left(2 t^2+5 t+3\right)$$
Then we are done if we apply the mean value theorem. It should be doable to show this.
-
Setting $t= 1/x$ and working with $t$ should simplify things a bit... – Aryabhata Jul 22 '11 at 20:29
@Aryabhata: Thanks! I used your suggestion to reduce it even further. – Listing Jul 22 '11 at 20:41
Thanks Listing, following in this manner did eventually yield the solution. Also, @Aryabhata: yes, I had the same idea in the shower :) – Antonio Vargas Jul 23 '11 at 0:42
No problem, good to see I could help. – Listing Jul 23 '11 at 9:19 | 2015-10-09 12:57:08 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9429117441177368, "perplexity": 188.68395417705216}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-40/segments/1443737929054.69/warc/CC-MAIN-20151001221849-00091-ip-10-137-6-227.ec2.internal.warc.gz"} |
https://wiki.apertium.org/w/index.php?title=Bosnian-Croatian-Montenegrin-Serbian_and_Slovenian&diff=next&oldid=39700 | # Difference between revisions of "Bosnian-Croatian-Montenegrin-Serbian and Slovenian"
## Work plan
This is a draft workplan for development of the Bosnian-Croatian-Montenegrin-Serbian and Slovenian translator.
• Trimmed coverage means the coverage of both of the morphological analysers after being trimmed according to the bilingual dictionary of the pair, that is, only containing stems which are also in the bilingual dictionary.
• Testvoc for a category means that the category is testvoc clean, in both translation directions.
• Evaluation is taking ${\displaystyle n}$ words and performing an evaluation for post-edition word error rate (WER). The output for those ${\displaystyle n}$ words should be clean.
Week Dates Trimmed coverage Reached Testvoc Evaluation Notes
0 23/04—21/05 45% `<abbr>` `<ij>` 500 words Preliminary evaluation. Translate the story in both directions with total coverage and without diagnostics. Get a baseline WER. Work on disambiguation, the morphological ambiguities in the story should be resolved.
1 21/05—27/05 50% `<cnjcoo>` `<cnjadv>` `<cnjsub>` `<pr>` - Conjunctions and prepositions should be clean. Verbs should be tagged for transitivity.
2 28/05—03/06 53% `<adv>` -
3 04/06—10/06 59% `<num>` 200 words Work on disambiguation. Numerals should be clean.
4 11/06—17/06 63% `<prn>` `<det>` -
5 18/06—24/06 68% `<n>` -
6 25/06—01/07 70% `<adj>` 500 words Midterm evaluation. Work on disambiguation.
7 02/07—08/07 73% - -
8 09/07—15/07 75% - -
9 16/07—22/07 77% - 200 words Work on disambiguation.
10 23/07—29/07 80% `<vblex>` -
11 30/07—05/08 - - - Closed categories clean.
12 06/08—12/08 - - - Open categories clean.
13 13/08—19/08 - - - Disambiguation and transfer.
14 20/08—24/08 - - - Final evaluation. Tidying up, releasing
## Improving coverage - A two week report
The work on improving the coverage for the slovenian -> bosnian-croatian-montenegrin-serbian language pair should be done 4 hours per day, for two weeks straight.
Day 0 denotes the current state of the language pair, and each subsequent entry will denote the coverage and testvoc reached for that day.
The corpora used for this task can be found here: http://www.statmt.org/europarl/v7/sl-en.tgz
Coverage was calculated using the following script
```
echo total;
cat europarl-v7.sl-en.sl | apertium -d ~/Apertium/apertium-sh-sl/ sl-sh-morph | sed 's/\\$[^\^]*\^/\$\n^/g' | wc -l
# This expression evaluates to 14340199
echo unknown;
cat europarl-v7.sl-en.sl | apertium -d ~/Apertium/apertium-sh-sl/ sl-sh-morph | sed 's/\\$[^\^]*\^/\$\n^/g' | grep '*' | wc -l
coverage = 1 - unknown / total
```
Dat Dates Coverage reached Coverage reached (2) Target Notes
0 20-02-2013 87.12% - -
1 01-03-2013 88.39% - [87.56, 88.24]
2 02-03-2013 90.05% - [87.78, 88.80]
3 03-03-2013 91.53% - [88.00, 89.36]
4 04-03-2013 92.36% - [88.22, 89.93]
5 05-03-2013 92.95% - [88.44, 90.49]
6 06-03-2013 93.55% - [88.66, 91.05]
7 07-03-2013 94.11% - [88.88, 91.61]
8 08-03-2013 94.54% - [89.10, 92.17]
9 09-03-2013 94.95% 87.52% [89.32, 92.74]
10 10-03-2013 95.11% 91.46% [89.54, 93.30]
11 11-03.2013 95.26% [89.76, 93.86]
12 13-03-2013 95.41% 93.23% [89.98, 94.42]
13 15-03-2013 95.50% 93.57% [90.20, 94.99]
14 [90.00, 95.00]
The work on improving the coverage for the slovenian -> bosnian-croatian-montenegrin-serbian language pair should be done 2 hours per day, for two weeks straight.
The two tasks do not have to be run in parallel.
Day Dates Testvoc reached Notes
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14 | 2022-01-21 12:16:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.17522192001342773, "perplexity": 9572.504120262422}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320303356.40/warc/CC-MAIN-20220121101528-20220121131528-00541.warc.gz"} |
https://www.shaalaa.com/question-bank-solutions/find-the-equations-of-a-line-containing-the-point-a-3-4-and-making-equal-intercepts-on-the-co-ordinate-axes-equations-of-lines-in-different-forms_164771 | # Find the equations of a line containing the point A(3, 4) and making equal intercepts on the co-ordinate axes. - Mathematics and Statistics
Sum
Find the equations of a line containing the point A(3, 4) and making equal intercepts on the co-ordinate axes.
#### Solution
Let the equation of the line be
x/"a" + y/"b" = 1
Since, the required line make equal intercepts on the co-ordinate axes.
∴ a = b
∴ (i) reduces to x + y = a …(ii)
Since the line passes through A(3, 4).
∴ 3 + 4 = a
i.e. a = 7
Substituting a = 7 in (ii) to get
x + y = 7.
Is there an error in this question or solution?
#### APPEARS IN
Balbharati Mathematics and Statistics 1 (Commerce) 11th Standard Maharashtra State Board
Chapter 5 Locus and Straight Line
Exercise 5.3 | Q 8 | Page 73 | 2021-04-11 03:04:55 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5935366749763489, "perplexity": 2048.913711034879}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038060927.2/warc/CC-MAIN-20210411030031-20210411060031-00250.warc.gz"} |
https://www.juanmontesinos.com/posts/2020/06/14/weakly-supervised | # Paper: End-to-end weakly-supervised semantic alignment
Published:
PDF
Year: 2018
Conference: CVPR
Goal:Aligning two mages depicting objects of the same category
Note: Parts of the paper may be quoted without indication.
# Paper Resume: End-to-end weakly-supervised semantic alignment
## Contributions:
• End-to-end CNN
• Weakly supervised (image pairs)
• differentiable soft inlier scoring module, inspired by the RANSAC
## Introduction:
In this work authors study the problem of finding category-level correspondence, or semantic alignment, where the goal is to establish dense correspondence between different objects belonging to the same category. This is also an extremely challenging task because of the large intra-class variation, changes in viewpoint and presence of background clutter.
The work is supposed to address previous limitations:
• the image representation and the geometric alignment model are not trained together in an end-to-end manner.
• Supervised methods require strong supervision in the form of ground truth correspondences.
The outcome is that the image representation can be trained from rich appearance variations present in different but semantically related image pairs.
## Core Idea
Here we can see the core method proposed in the paper:
The network is trained using pairs of images. It estimates a transformation to align them. At the very end, the quality of the alignment is estimated. The network is trained to maxize that quality without using strong supervision.
### Feature Extractor
The input is a set of images $(I^s,I^t)$ which pass through a siamese network. The output shape is a 2D grid of shape HxW with d channels. Each point of the grid is called $f_{ij} \epsilon R^d$
## Pairwise feature matching
Pairwise similarities–> normalized correlation function:
$S:{\Bbb R}^{h\times w\times d}\times{\Bbb R}^{h\times w\times d} \rightarrow {\Bbb R}^{h\times w\times h\times w}$
$s_{ijkl}=S(f^s,f^t)_{ijkl}=\frac{ \langle f^s_{ij:},f^t_{kl:} \rangle }{\sqrt{\sum_{a,b} \langle f^s_{ab:},f^t_{kl:} \rangle^2 }}$
The denominator performs a normalization operation with the effect of down-weighing ambiguous matches, by penalizing features from one image which have multiple highly-rated matches in the other image.
## Geometric transformation estimation
Done by transformation regression CNN: $G: {\Bbb R}^{h\times w\times h\times w}\rightarrow {\Bbb R}^K$
Where $K$ are the degrees of freedom (depending on if affine transformation…).
## Soft-inlier count
Inspired by RANSAC.
It sums matching scores over all possible matches. RANSAC formulates an hypotesis for all the possible cases and takes as valid the most scored one.
The RANSAC algorithm is the following one:
$c_R=\sum m(p_i)$ where $m(p)= 1 \text{ if } d(p,l)\lt \text{t } 0 \text{ otherwise}.$
The problem is this algorithm is not differentiable since you are setting strict values based on a threshold. Thus, they reformulate the algorithm as the sum of the match scores over all possible matches penalized by the RANSAC algorithm. This way differentiation is taken into account by the sum and gradients are adjusted by the coefficient each term is multiplied by.
The new algorithm:
$c=\sum s_{ij}m_{ij}$
$m_{ij}$ are the same as in the RANSAC meanwhile $s_{ij}$ are the sum of the matched scores. | 2021-09-29 02:32:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5829764604568481, "perplexity": 2136.4215931473536}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780061350.42/warc/CC-MAIN-20210929004757-20210929034757-00272.warc.gz"} |
http://zbmath.org/?q=an:0594.34016 | # zbMATH — the first resource for mathematics
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A stability result for differential inclusions in Banach spaces. (English) Zbl 0594.34016
In this paper one existence theorem for the differential inclusion (1) $\stackrel{˙}{x}\in F\left(t,x\right)$, $x\left({t}_{0}\right)={x}_{0}$ in a separable Banach space is proved. The multifunction F has nonempty, compact, convex values and satisfies the Caratheodory-type conditions. The proof uses the Ky Fan fixed point theorem and some properties of integral of multifunctions. Next the continuous dependence of solutions to (1) on the right-hand side is studied where the convergence of ${F}_{n}$ to F is unerstand in Kuratowski-Mosco sense. At the end one theorem on convergence of the sets of fixed points of some sequence of multifunctions, say $\left\{{F}_{n}\right\}$, (with Lipschitz constants smaller than 1) to the set of fixed points of the limit of ${F}_{n}$ in a Banach space with Frechet-differentiable norm is proved. A theorem of the same type is proved for the set of integrable selectors of a sequence of multifunctions.
Reviewer: Z.Wyderka
##### MSC:
34A60 Differential inclusions 28B20 Set-valued set functions and measures; integration of set-valued functions; measurable selections 54C65 Continuous selections 49R50 Variational methods for eigenvalues of operators (MSC2000) 93B05 Controllability | 2014-04-20 13:31:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 5, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8554368615150452, "perplexity": 5076.563382562583}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609538787.31/warc/CC-MAIN-20140416005218-00224-ip-10-147-4-33.ec2.internal.warc.gz"} |