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https://www.physicsforums.com/threads/perturbation-theory.345425/ | # Perturbation Theory
1. Oct 13, 2009
### latentcorpse
a particle moves in one dimension in the potential
$V(x)=\infty \forall |x|>a, V(x)=V_0 \cos{\frac{\pi x}{2a}} \forall |x| \leq a$
now the unperturbed state that i use is just a standard infinite square well.
anyway the solution says that perturbation theory is only valid provided that the energy scale of the "bump" (set by $V_0$) is less than the difference in energy between square well states.
Q1: is this just a fact: perturbation theory is only applicable providing the perturbation is less than the energy difference of the states of teh unperturbed system?
it then explains the above mathematically by saying:
$V_0 << \frac{{\hbar}^2 \pi^2}{8ma^2}(2n-1)$
i can't for the life of me see where the RHS of that comes from. arent the energy levels dependent on $n^2$ and not $n$?
2. Oct 14, 2009
### lanedance
i think if you look at the difference in energy is given by
$$n^2- (n-1)^2 = n^2 - (n^2 - 2n+1) = 2n-1$$
i think less than is probably not strong enough, i think the perturbation has to be "small" relative to the energy difference. Looking in Ballentine, this can be seen if you look at the first order contribution to the eigenvector. It effectively contains the ratio of the perturbation to the energy level difference. Higher order terms carry the ratio at higher powers, so for the perturbation sum to converge (and quickly) the ratio must be small | 2018-01-19 06:01:33 | {"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": 0.7343526482582092, "perplexity": 390.0525571426151}, "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-2018-05/segments/1516084887746.35/warc/CC-MAIN-20180119045937-20180119065937-00793.warc.gz"} |
https://brilliant.org/problems/there-can-be-infinitely-many/ | # There Can Be Infinitely Many
A positive integer $$n$$ is called sacred if it is divisible by all odd integers $$a$$ for which $$n \geq a^2$$. Determine the sum of all sacred numbers.
As an arbitrary example, $$n=15$$ is sacred because it is divisible by $$1$$ and $$3$$.
× | 2019-04-26 06:47:50 | {"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": 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": 0.5438226461410522, "perplexity": 493.349744050318}, "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-2019-18/segments/1555578760477.95/warc/CC-MAIN-20190426053538-20190426075538-00287.warc.gz"} |
https://mathematica.stackexchange.com/questions/95513/what-is-the-definition-of-head-in-mathematica | # What is the definition of head in Mathematica? [closed]
I know (by practice) what a head is, but I am unable to find the appropriate words to actually define it. I would like to see a precise definition of head using the appropriate technical words involved. Hopefully, this should provide me with a better understanding of how Mathematica works.
• Note that as stated in the documentation, "Heads need not be symbols", e.g., FixedPointList[Head, f[x][y][z]] // Most returns {f[x][y][z], f[x][y], f[x], f, Symbol} – Bob Hanlon Sep 25 '15 at 21:39
• An answer may be found in the first tutoria,l Everything Is an Expression, linked in the documentation for Head. – Michael E2 Sep 26 '15 at 1:30 | 2020-03-30 20:37:18 | {"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": 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": 0.26925230026245117, "perplexity": 1069.670268817837}, "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/1585370497301.29/warc/CC-MAIN-20200330181842-20200330211842-00054.warc.gz"} |
https://www.answers.com/Q/How_Many_Square_Feet_are_in_130_square_yards | Math and Arithmetic
Area
# How Many Square Feet are in 130 square yards?
###### Wiki User
130 square yards = 1,170 square feet.
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## Related Questions
### How many square yards in 168 square feet?
168 square feet = about 18.7 square yards. (18.6666667 square yards)
### How many square yards are in 43200 square feet?
4,800 square yards. (divide square feet by 9 to get square yards)
### How many square yards are 324 square feet?
36 square yards. (divide square feet by 9 to get square yards).
### How many square yards are there in 700 square feet?
700 square feet = about 77.8 square yards. (77.7777778 square yards)
### How many feet are equal to 4840 square yards?
You either convert yards to feet, or square yards to square feet. But you can't convert square yards to feet.
### How many square yards are in 655 square feet?
9 square feet = 1 square yard 18 square feet = 2 square yards 27 square feet = 3 square yards . . . 655 square feet = 727/9 square yards
### How many square feet is 21 square yards?
189 square feet. (multiply square yards by 3 to get square feet)
### How many yards are in 360 square feet?
There are 40 square yards in 360 square feet. (just divide square feet by nine to get square yards).
### How many square yards are in 2808 square feet?
There are 312 square yards in 2,808 square feet.
### How many square feet in 5 square yards?
There are 45 square feet in 5 square yards
### How many square yards are in 45 square feet?
There are five square yards in 45 square feet.
## Still have questions?
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Previously Viewed | 2021-03-08 04:10:16 | {"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": 0.8749765753746033, "perplexity": 3806.4807434548948}, "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/1614178381803.98/warc/CC-MAIN-20210308021603-20210308051603-00067.warc.gz"} |
https://www.physicsforums.com/threads/derivation-of-equation-for-sliding-object.288604/ | # Derivation of equation for sliding object
1. Jan 30, 2009
### hitek0007
1. The problem statement, all variables and given/known data
1. The problem:
Derive an expression in its simplest form to show the relationship between the mass, ramp angle, and acceleration of the sliding object. Explain clearly but briefly the effect of each term in the expression on the actual acceleration, as the ramp angle changes.
No known variables. Context of this question: part of a lab, in which we found values of friction and coefficients of friction through measuring acceleration of objects sliding down a ramp.
2. Relevant equations
a=Fnet/m
Fgramp=mgsin(x)
Ffk=ukmgcos(x)
3. The attempt at a solution
a=Fnet/m
a=(Fgramp-Fk)/(m)
a=(mgsin(x)-ukmgcos(x))/(m)
a=gsin(x)-ukgcos(x)
a=g(sin(x)-ukcos(x))
Is this the right equation? If so, mass has no effect on the acceleration. Acceleration increases as angle increases.
However, my teacher told me that the equation is supposed to look like:
a=_________+_________
I suppose it is possible the 2nd term is negative... but I am not sure.
Are there different equations? We also calculated ideal and measured accelerations to find the value of friction. Is this any use?
Thanks!
2. Jan 30, 2009
### americanforest
Looks like you've got it right.
$$a=g(sin\theta-\mu cos\theta)$$ where $$\theta$$ is the angle between the ramp and the horizontal. Note that $$a$$ is the acceleration when the object has been released and slides down the ramp. If the object has been pushed up the ramp and is in the process of sliding up then a slightly different equation governs its acceleration. | 2017-04-28 10:32:26 | {"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": 0.6990988850593567, "perplexity": 527.1436257820825}, "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-2017-17/segments/1492917122933.39/warc/CC-MAIN-20170423031202-00431-ip-10-145-167-34.ec2.internal.warc.gz"} |
https://collegephysicsanswers.com/openstax-solutions/contestant-winter-sporting-event-pushes-450-kg-block-ice-across-frozen-lake | Question
A contestant in a winter sporting event pushes a 45.0-kg block of ice across a frozen lake as shown in Figure 5.21(a). (a) Calculate the minimum force F he must exert to get the block moving. (b) What is the magnitude of its acceleration once it starts to move, if that force is maintained?
Question Image
1. $50\textrm{ N}$
2. $0.7 \textrm{ m/s}^2$
Solution Video
# OpenStax College Physics Solution, Chapter 5, Problem 18 (Problems & Exercises) (6:56)
View sample solution
## Calculator Screenshots
Video Transcript
Submitted by raynellmcclellan on Tue, 05/19/2020 - 11:47
Why do given the solution as 50 if it comes out as 51. Are we suppose to be rounding to the nearest tenth?
Submitted by ShaunDychko on Wed, 05/20/2020 - 15:43
Yes, exactly, we're rounding to one significant figure. This is due to the of the coefficient of static friction, which has only one significant figure. When multiplying by a number with one sig. fig., the answer gets only one sig. fig. also.
All the best,
Shaun | 2020-07-11 19:51:20 | {"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": 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": 0.4030017852783203, "perplexity": 2544.8595848227965}, "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-2020-29/segments/1593655937797.57/warc/CC-MAIN-20200711192914-20200711222914-00177.warc.gz"} |
http://books.duhnnae.com/2017/jun/149631423257-Jet-schemes-of-toric-surfaces-a-short-version-Hussein-Mourtada.php | # Jet schemes of toric surfaces a short version
For $m\in \IN, m\geq 1,$ we determine the irreducible components of the $m-th$ jet scheme of a toric surface $S.$ For $m$ big enough, we connect the number of a class of these irreducible components to the number of exceptional divisors on the minimal resolution of $S.$ | 2017-10-24 11:25:30 | {"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": 0.7533653974533081, "perplexity": 134.62259744526506}, "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-2017-43/segments/1508187828411.81/warc/CC-MAIN-20171024105736-20171024125736-00688.warc.gz"} |
http://talkstats.com/threads/copula-based-var-calculation-in-r.69320/ | copula-based VaR calculation in R
maaarten9
New Member
I'm working on a value-at-risk calculation using copulas on different stock market indices. I know how to fit the copula, but I can't figure out how to apply the VaR approach in the next step. The concept of copulas is relatively new to me and has proven to be very challenging for an average master student..I defined 3 periods in which I want to investigate the evolution of the VaR over time. When running the code, R returns a value for the VaR. But when running the code for another time period, R gives the same value as the previous period..Am I overlooking/forgetting something? The code I provided shows the bivariate example of china and india using the normal copula. I plan to extent it with the t and clayton copula in a further stage.
Code:
library(copula)
cop_model = normalCopula(dim = 2)
m = pobs(as.matrix(cbind(CHINA_INDIA$CHINA.LOG[571:406],CHINA_INDIA$INDIA.LOG[571:406])))
#pseudo-observations
fit = fitCopula(cop_model, m, method = "ml")
coef(fit)
tau(normalCopula(param = coef(fit)))
cor(CHINA_INDIA$CHINA.LOG[571:406],CHINA_INDIA$INDIA.LOG[571:406], method = "kendall") #check whether correlation is more or less preserved
set.seed(1559)
u = rCopula(500, normalCopula(coef(fit), dim = 2)) #simulate some observations from the copula
cdf = pCopula(u, normalCopula(coef(fit), dim = 2) #construct cdf of the copula
library(PerformanceAnalytics)
VaR(cdf, p=0.95)
Last edited by a moderator: | 2020-07-09 02:30:38 | {"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": 0.17821195721626282, "perplexity": 1731.5118074427662}, "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-2020-29/segments/1593655897844.44/warc/CC-MAIN-20200709002952-20200709032952-00466.warc.gz"} |
http://pubman.mpdl.mpg.de/pubman/faces/viewItemOverviewPage.jsp?itemId=escidoc:2110444 | de.mpg.escidoc.pubman.appbase.FacesBean
English
# Item
ITEM ACTIONSEXPORT
Released
Journal Article
#### Constraining Astrophysical Neutrino Flavor Composition from Leptonic Unitarity
##### MPS-Authors
http://pubman.mpdl.mpg.de/cone/persons/resource/persons30951
Rodejohann, Werner
Werner Rodejohann - ERC Starting Grant, Junior Research Groups, MPI for Nuclear Physics, Max Planck Society;
1407.3736.pdf
(Preprint), 2MB
##### Supplementary Material (public)
There is no public supplementary material available
##### Citation
Xu, X., He, H.-J., & Rodejohann, W. (2014). Constraining Astrophysical Neutrino Flavor Composition from Leptonic Unitarity. Journal of Cosmology and Astroparticle Physics, 2014(12): 039. doi:10.1088/1475-7516/2014/12/039.
Cite as: http://hdl.handle.net/11858/00-001M-0000-0025-BE6F-3
##### Abstract
The recent IceCube observation of ultra-high-energy astrophysical neutrinos has begun the era of neutrino astronomy. In this work, using the unitarity of leptonic mixing matrix, we derive nontrivial unitarity constraints on the flavor composition of astrophysical neutrinos detected by IceCube. Applying leptonic unitarity triangles, we deduce these unitarity bounds from geometrical conditions, such as triangular inequalities. These new bounds generally hold for three flavor neutrinos, and are independent of any experimental input or the pattern of leptonic mixing. We apply our unitarity bounds to derive general constraints on the flavor compositions for three types of astrophysical neutrino sources (and their general mixture), and compare them with the IceCube measurements. Furthermore, we prove that for any sources without $\nu_\tau$ neutrinos, a detected $\nu_\mu$ flux ratio $< 1/4$ will require the initial flavor composition with more $\nu_e$ neutrinos than $\nu_\mu$ neutrinos. | 2018-01-24 03:51:10 | {"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": 0.8179743885993958, "perplexity": 6023.586720246819}, "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-2018-05/segments/1516084893300.96/warc/CC-MAIN-20180124030651-20180124050651-00589.warc.gz"} |
https://matholympiad.org.bd/forum/viewtopic.php?f=13&t=6227 | ## BDMO Secondary National 2021 #7
Mehrab4226
Posts:230
Joined:Sat Jan 11, 2020 1:38 pm
BDMO Secondary National 2021 #7
কোনো ধনাত্মক পূর্ণসংখ্যা $$n$$-এর জন্য $$s(n)$$ আর $$c(n)$$ হলো যথাক্রমে $$n$$-এর পূর্ণবর্গ আর পূর্ণঘন উৎপাদকের সংখ্যা। একটা ধনাত্মক পূর্ণসংখ্যা $$n$$-কে নায্য বলা হবে যদি $$s(n)=c(n)>1$$ হয়। $$80$$-এর চেয়ে ছোট কতগুলো নায্য সংখ্যা আছে?
For a positive integer $n$ , let $s(n)$ and $c(n)$ be the number of divisors of $n$ that are perfect squares and perfect cubes respectively. A positive integer $n$ is called fair if $s(n)=c(n)>1$ . Find the number of fair integers less than $80$.
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré
Pro_GRMR
Posts:46
Joined:Wed Feb 03, 2021 1:58 pm
### Re: BDMO Secondary National 2021 #7
Mehrab4226 wrote:
Sat Apr 10, 2021 1:26 pm
For a positive integer $n$ , let $s(n)$ and $c(n)$ be the number of divisors of $n$ that are perfect squares and perfect cubes respectively. A positive integer $n$ is called fair if $s(n)=c(n)>1$ . Find the number of fair integers less than $80$.
Let $n$ be in prime factorized form $n= p_1^{e_1}p_2^{e_2}\dots p_n^{e_n}$
We note that number of divisors of $p_1^{e_1}$ that are perfect square is $\lfloor{\frac{e_1}{2}}\rfloor+1$ similarly $\lfloor{\frac{e_1}{3}}\rfloor+1$ for perfect cubes.
So, $n$ has a total of $(\lfloor{\frac{e_1}{2}}\rfloor+1)(\lfloor{\frac{e_2}{2}}\rfloor+1)\dots(\lfloor{\frac{e_n}{2}}\rfloor+1)$ perfect square divisors and $(\lfloor{\frac{e_1}{3}}\rfloor+1)(\lfloor{\frac{e_2}{3}}\rfloor+1)\dots(\lfloor{\frac{e_n}{3}}\rfloor+1)$ perfect cube divisors.
So, for a number to be fair, $\lfloor \frac{e_i}{2} \rfloor = \lfloor \frac{e_i}{3} \rfloor$, which means that $e_i$ is either $1$ or $3$. So, a positive number $n$ is fair if and only if all its powers in the prime factorization are $1$ or $3$.
Now, we count the number of these numbers below $80$ and get all possible values of $n$ are $8, 24, 27, 40, 54, 56$ and so, $\boxed{6}$ values in total.
"When you change the way you look at things, the things you look at change." - Max Planck | 2021-10-23 20:25:26 | {"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": 0.808952271938324, "perplexity": 799.6525792055301}, "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-2021-43/segments/1634323585768.3/warc/CC-MAIN-20211023193319-20211023223319-00580.warc.gz"} |
https://hsm.stackexchange.com/questions/3734/the-origins-of-complex-differentiation-integration | # The origins of complex differentiation/integration
What questions led to the invention of complex differentiation/integration? How were their definitions agreed upon?
Real differentiation/integration has an obvious meaning. To extend calculus to the complex numbers, why would this be done, is it even meaningful to call it 'differentiation'/'integration'?
There is no difference in differentiation. Derivative of a complex function of a complex variable is defined by the same formula $$f'(z)=\lim_{h\to 0} (f(z+h)-f(z))/h$$ as for the real variable. Concerning integration, one reason was the desire to investigate integrals of real functions. One of the consequences (and motivations) of Cauchy theory is that it evaluates integrals of real functions on real intervals by manipulating with integrals in the complex domain.
But there is a more important and more profound reason. Many equations (differential and functional) can be solved in the form of power series. This was essentially discovered by Newton, and he considered this his main discovery in Calculus. A power series, if it converges anywhere except $x=0$, converges in a disk in the complex plane, and it is impossible to understand its properties while staying in the real domain only. This is the most important motivation of extending Calculus to complex domain.
Here is a simple example: we have a power series expansion $$f(x)=\frac{1}{1+x^2}=1-x^2+x^4-x^6+\ldots.$$ It converges in the real domain for $-1<x<1$ only. But the left hand side $f(x)=1/(1+x^2)$ is a nice function on the whole real line. By looking at this function on the real line only it is impossible to understand what happens at $\pm 1$, why the series suddenly stops to converge. | 2019-08-23 17:50:11 | {"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": 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": 0.769256591796875, "perplexity": 231.12138988814993}, "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-2019-35/segments/1566027318952.90/warc/CC-MAIN-20190823172507-20190823194507-00336.warc.gz"} |
https://dsp.stackexchange.com/questions/15464/wavelet-thresholding/15487 | # Wavelet thresholding
What is the difference between soft thresholding and hard thresholding. Where we use soft and hard thresholding in image for denoising. I understand that in hard thresholding, the coefficients below threshold value are set to zero and the value above the threshold is set to one. Please explain me about soft threshold. Please explain whether the threshold value is the intensity value of the image. For example if the intensity value ranges between 0 to 255. In case of hard thresholding if the threshold value is considered as 100 then the values below 100 is set to 0.The value above 100 are retained. Is this correct? Please correct me.
• Google search finds this paper: fceia.unr.edu.ar/~jcgomez/wavelets/Donoho_1995.pdf You should try reading it and coming back with more specific questions. – MackTuesday Apr 7 '14 at 16:51
• Hard thresholding what you are saying is only correct for the set to zero while the other coefficients are left unajusted. In soft thresholding coefficients are all adjusted based on based on MAD and other elements of the equation. – Barnaby May 23 '15 at 9:25
For a given threshold $\lambda$ (that can be dependent on resolution level), and value of wavelet coefficient $d$, hard thresholding is defined as:
$D^H(d|\lambda)=\begin{cases} 0,& \text{for } |d| \leq \lambda\\ d,& \text{for } |d| > \lambda \end{cases}$
whereas soft thresholding is governed by following equation:
$D^S(d|\lambda)=\begin{cases} 0,& \text{for } |d| \leq \lambda\\ d-\lambda,& \text{for } d > \lambda \\ d+\lambda,& \text{for } d < -\lambda \\ \end{cases}$
Figure below depicts both cases:
The soft thresholding is also called wavelet shrinkage, as values for both positive and negative coefficients are being "shrinked" towards zero, in contrary to hard thresholding which either keeps or removes values of coefficients.
In case of image de-noising, you are not working strictly on "intensity values", but wavelet coefficients. You probably remember that you can decompose your image into wavelet levels, like in case of lovely Lena. Assuming that wavelet transform gives sparse coefficients, mostly close to zero, and noise level is lower than wavelet coefficients, you can simply threshold these. Although if you wish, you can perform hard/soft thresholding on each decomposition level with a different value of $\lambda$. When it is done, then you just have to reconstruct your image from all decomposition levels and voila, noise should be removed!
Below you have two examples of de-noised image via hard and soft thresholding respectively (same $\lambda$). Obviously soft thresholding gives more smooth image- if you can notice that with such a poor resolution ;) Courtesy of MATLAB. | 2021-06-18 18:36:06 | {"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": 0.6543461084365845, "perplexity": 739.6466706118983}, "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-2021-25/segments/1623487640324.35/warc/CC-MAIN-20210618165643-20210618195643-00627.warc.gz"} |
https://chemistry.stackexchange.com/questions/42736/disproportionation-of-hydrogen-peroxide/42740 | # Disproportionation of hydrogen peroxide
Hydrogen peroxide is decomposed as follows: $$\ce{2H2O2 -> 2H2O +O2}$$ This is a disproportionation redox reaction of $\ce{H2O2}$ involving the 2 half reactions $$\ce{H2O2 -> O2 + 2H^+ + 2e-}$$ $$\ce{H2O2 + 2e- + 2H^+ -> 2H2O}$$ But I noticed semantically that it can also be the sum of the two following half reactions: $$\ce{2H2O2 -> 2O2 + 4H^+ + 4e-}$$ $$\ce{O2 + 4H^+ + 4e- -> 2H2O}$$ Both reactions are feasible according to the following:
$$\begin{array}{ccc} \text{Oxidised species} & \text{Reduced species} & E^\circ (\mathrm{V}) \\ \hline \ce{H2O2} & \ce{H2O} & 1.763 \\ \ce{O2} & \ce{H2O} & 1.23 \\ \ce{O2} & \ce{H2O2} & 0.695 \\ \end{array}$$
Can it happen in the second way? If yes, can you explain the mechanism in simple terms?
• If it is what I think it is, and the fourth equation is just a typo (H2O should be H2O2), then yes, it is formally correct. However, if you just keep H2O2 by itself, without any oxygen, it will still decompose. This isn't captured by your proposed pair of half-equations, which essentially says that you need oxygen gas for the total reaction (i.e. sum of 2 half-reactions) to proceed. Dec 25, 2015 at 7:16 | 2022-10-03 04:44:41 | {"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": 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": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8860586881637573, "perplexity": 614.3732545569497}, "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-2022-40/segments/1664030337398.52/warc/CC-MAIN-20221003035124-20221003065124-00481.warc.gz"} |
https://anuragarnab.github.io/instances.html | Abstract: Traditional Scene Understanding problems such as Object Detection and Semantic Segmentation have made breakthroughs in recent years due to the adoption of deep learning. However, the former task is not able to localise objects at a pixel level, and the latter task has no notion of different instances of objects of the same class. We focus on the task of Instance Segmentation which recognises and localises objects down to a pixel level. Our model is based on a deep neural network trained for semantic segmentation. This network incorporates a Conditional Random Field with end-to-end trainable higher order potentials based on object detector outputs. This allows us to reason about instances from an initial, category-level semantic segmentation. Our simple method effectively leverages the great progress recently made in semantic segmentation and object detection. The accurate instance-level segmentations that our network produces is reflected by the considerable improvements obtained over previous work at high $$AP^r$$ IoU thresholds. | 2022-08-19 04:18:39 | {"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": 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": 0.3666735887527466, "perplexity": 853.560118835347}, "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-2022-33/segments/1659882573623.4/warc/CC-MAIN-20220819035957-20220819065957-00009.warc.gz"} |
https://math.stackexchange.com/questions/830516/partial-fractions-decomposition-of-the-gamma-function | # Partial Fractions Decomposition of the Gamma Function
I'm currently dealing with a problem my professor raised (since I just studied the Mittag-Leffler's Partial Fractions Theorem). The problem is to derive a partial fractions decomposition of the Gamma Function that displays its (simple) poles and principal parts.
So in my textbook, the Gamma Function is defined as the following limit:
$$\lim_{n \to \infty}\frac{n!\ n^z}{z(z+1)...(z+n)}$$
where $z$ is an arbitrary complex number that is not $0, -1, -2,...$
With the residue of each pole $z_v=-v,\ v=0,1,2,...$ as
$$a_{-1}=\frac{(-1)^v}{v!}$$
We have the principal parts of the Gamma Function at each pole $z_v$ as
$$h_v(z)=\frac{(-1)^v}{v!\ (z+v)}$$
And so it is sufficient to show that
$$\sum_{v=1}^{\infty} \left[h_v(z)-g_v(z) \right]$$ is uniformly convergent on $|z|\leq R,\ R>0$ , where $g_v(z)$ is defined to be the first few finite terms of the series expansion of $h_v(z)$ around $z=0$.
By considering $v$ large enough so that $|z_v|=v>2R$ and taking $g_v(z)=0$, we have that for $|z|\leq R$
$$\left|\frac{(-1)^v}{v!\ (z+v)} \right|=\frac{1}{v!\ |z+v|}<\frac{2}{v!\ v}$$
So the series above converges uniformly and hence there exists an entire function $G(z)$ such that
$$\Gamma(z) = G(z)+\frac{1}{z} + \sum_{v=1}^{\infty}\frac{(-1)^v}{v!\ (z+v)}$$
And now I am stuck at finding $G(z)$. Anyone can provide some help?
• You can identify your function from dlmf.nist.gov/5.9#E4, but I have no clue how to give a proof with your lim/product definition of $\Gamma$. With the integral definition you can find a proof in Lebedev's Special functions Ch.1.1 Jun 11 '14 at 14:04
• Hmm I just need a clue about the function $G(z)$... Jun 11 '14 at 14:27
• Maybe it is a bit late: In your notation the NIST/Lebedev formula is $$\Gamma(z)=\int_{1}^{\infty}t^{z-1}e^{-t}dt+\sum_{v=0}^{\infty}\frac{(-1)^{v}}{(z+v)v!}$$ and $G(z) = \int_{1}^{\infty}t^{z-1}e^{-t}dt$ is the function you seek. According to Lebedev (who refers to Titchmarsh, The Theory of Functions) the function $G(z)$ is entire. Jun 12 '14 at 6:52 | 2021-12-01 15:51:11 | {"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": 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": 0.9604445099830627, "perplexity": 143.189338233606}, "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-2021-49/segments/1637964360803.6/warc/CC-MAIN-20211201143545-20211201173545-00195.warc.gz"} |
https://keio.pure.elsevier.com/ja/publications/artinmazur-zeta-functions-of-generalized-beta-transformations | Artin–mazur zeta functions of generalized beta-transformations
抄録
In this paper, we study the Artin–Mazur zeta function of a generalization of the well-known β-transformation introduced by Góra [Invariant densities for generalized β-maps. Ergodic Theory Dynam. Systems 27 (2007), 1583–1598].We show that the Artin– Mazur zeta function can be extended to a meromorphic function via an expansion of 1 defined by using the transformation. As an application, we relate its analytic properties to the algebraic properties of β.
本文言語 English 85-103 19 Kyushu Journal of Mathematics 71 1 https://doi.org/10.2206/kyushujm.71.85 Published - 2017 はい
• 数学 (全般)
フィンガープリント
「Artin–mazur zeta functions of generalized beta-transformations」の研究トピックを掘り下げます。これらがまとまってユニークなフィンガープリントを構成します。 | 2021-09-19 21:40:39 | {"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": 0.8610842227935791, "perplexity": 1111.3007822907866}, "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-2021-39/segments/1631780056900.32/warc/CC-MAIN-20210919190128-20210919220128-00104.warc.gz"} |
http://physics.stackexchange.com/questions/69840/closed-form-for-shape-tension-of-an-elastic-cable-slung-between-two-points | # Closed form for shape/tension of an elastic cable slung between two points
Given the 2D coordinates of two points, $a$ and $b$, between which an elastic cable of known length, $l$, mass per unit length, $m$, and the spring constant, $e$, is slung, I need to compute the shape of the cable, and also the horizontal tension, $t$, in the cable.
So far I have the equations for the x and y coordinates of the cable, parameterized by $p$, which is the distance along the unstretched cable: $$f_x(p) = \frac{t}{mg} \sinh^{-1}\left(\frac{mgp}{t}\right) + \frac{tp}{e} + c_x.\\ f_y(p) = \sqrt{\left(\frac{t}{mg}\right)^2 + p^2} + \frac{mgp^2}{2e} + c_y.\\$$ where $c_x$ and $c_y$ are constants, which leads me to the following triplet of simultaneous equations: $$f_x(q) - f_x(r) = a_x - b_x.\\ f_y(q) - f_y(r) = a_y - b_y.\\ |q-r| = l.$$ in three unknowns, $t$, $q$ and $r$ (given that the constants cancel), where $q$ and $r$ are the values of the parameter $p$ at the points $a$ and $b$ respectively.
How would you compute those unknowns, and can it be done in closed form?
-
Does $t$ depend on the position? Is it the tension at some extremity? – fffred Jul 2 at 20:34
– Qmechanic Jul 3 at 0:13
fffred: No, the value $t$ is constant along the cable. It is the $x$ (horizontal) component of the tension at any point in the cable. – user664303 Jul 3 at 8:54
Qmechanic: Thanks. I googled "catenary" before posting, and this lead to what I have already. – user664303 Jul 3 at 8:59 | 2013-12-21 05:44:43 | {"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": 0.9348980188369751, "perplexity": 367.4925371320821}, "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-2013-48/segments/1387345775028/warc/CC-MAIN-20131218054935-00099-ip-10-33-133-15.ec2.internal.warc.gz"} |
http://mathhelpforum.com/discrete-math/147201-how-many-five-digit-integers-10-000-99-999-divisible-5-a-print.html | # How many five digit integers (10,000-99,999) are divisible by 5?
• Jun 1st 2010, 12:31 AM
taurus
How many five digit integers (10,000-99,999) are divisible by 5?
When I did it I got 2000*10-2000=18000. But I was marked wrong but cant see where I went wrong.
• Jun 1st 2010, 12:52 AM
undefined
Quote:
Originally Posted by taurus
When I did it I got 2000*10-2000=18000. But I was marked wrong but cant see where I went wrong.
The answer is indeed 18000. I don't know why it got marked wrong.
• Jun 1st 2010, 02:08 AM
simplependulum
The minimum is $10000$ and the maxumum is $99995$ so the total number is :
$\frac{99995-10000}{5} + 1$
$= 18000$
• Jun 1st 2010, 02:41 AM
Quote:
Originally Posted by taurus
When I did it I got 2000*10-2000=18000. But I was marked wrong but cant see where I went wrong.
Maybe it was not the final answer that was marked wrong.
Possibly a certain method was being examined to arrive at the answer.
You either calculated the amount of non-zero numbers up to 100,000 divisible by 5
and got 20,000, by dividing 100,000 by 5 as every 5th number is a multiple of 5....
then you subtracted the amount of numbers up to 10,000 divisible by 5 and got 2,000.
Subtracting these gives 18,000.
Unfortunately, this is the amount of non-zero numbers from 10,001 to 100,000 inclusive that are divisible by 5.
It is of course good enough, but doesn't answer the question directly.
The examiner may have thought this is what you did.
Then again, you may have done the following...
Your answer is good enough because you can start from 0, thereby finding the amount of numbers from 0 to 99,999 divisible by 5
and the amount of numbers from 0 to 9,999 divisible by 5.
Maybe the examiner didn't think through your method.
From 10,000 to 99,999 the numbers divisible by 5 end in 0 or 5.
The first (most significant) digit can be any of nine from 1 to 9.
The 2nd digit can be any of 10.
The 3rd digit can be any of 10.
The 4th digit can be any of 10.
The 5th digit can be either of 2.
That's 9(10)(10)(10)2=18,000 | 2017-11-18 04:39:15 | {"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": 0, "codecogs_latex": 4, "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": 0.5468441843986511, "perplexity": 751.577272473325}, "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-2017-47/segments/1510934804610.37/warc/CC-MAIN-20171118040756-20171118060756-00462.warc.gz"} |
http://mymathforum.com/algebra/339392-i-don-t-think-there-enough-information-here.html | My Math Forum I don't think there is enough information here...
Algebra Pre-Algebra and Basic Algebra Math Forum
March 5th, 2017, 09:51 PM #1 Senior Member Joined: Jan 2017 From: US Posts: 104 Thanks: 5 I don't think there is enough information here... At a diving competition, divers jump off a 64 foot high cliff. Solve for how long it will take the diver to hit the water. Show your work and explain the steps you used to solve. Correct me if I'm wrong, but I feel like you would need more information (i.e., the velocity of the dive) to solve this. If I'm wrong could someone please help me with this? Thanks in advance!
March 5th, 2017, 09:52 PM #2 Senior Member Joined: Sep 2015 From: USA Posts: 1,653 Thanks: 840 assume $v_0=0$ Thanks from Indigo28
March 11th, 2017, 02:36 PM #3 Newbie Joined: Feb 2017 From: Omaha, Nebraska Posts: 8 Thanks: 0 Math Focus: Algebra and Trigonometry Think this might be what your wanting to do: h(t) = 16t^2 where h is height and t^2 is time squared, so plug in the dive distance of 64 feet and solve for t.
March 11th, 2017, 03:16 PM #4 Math Team Joined: Jul 2011 From: Texas Posts: 2,678 Thanks: 1339 $h(t) = 64-16t^2$
March 12th, 2017, 04:00 AM #5 Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 If the initial velocity has zero vertical component it plays no role.
March 12th, 2017, 10:52 AM #6 Newbie Joined: Feb 2017 From: Omaha, Nebraska Posts: 8 Thanks: 0 Math Focus: Algebra and Trigonometry Here's the answer I get... With initial velocity of zero, assuming zero resistance and seconds for time: (Diver accelerates due to constant force of gravity.) h(t) = 16t^2 64t = 16 t^2 64/16 = t^2 / t 4 = t Last edited by y2kevin; March 12th, 2017 at 10:58 AM. Reason: add note about diver and gravity
March 12th, 2017, 11:01 AM #7 Global Moderator Joined: Dec 2006 Posts: 18,245 Thanks: 1439 No. With h(t) = 64 - 16t², one gets h(t) = 0 when t = 2.
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Contact - Home - Forums - Cryptocurrency Forum - Top | 2017-12-15 02:27:14 | {"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": 0.6892391443252563, "perplexity": 5122.532033040548}, "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-2017-51/segments/1512948563083.64/warc/CC-MAIN-20171215021156-20171215041156-00649.warc.gz"} |
https://revisionworld.com/a2-level-level-revision/maths/statistics/binomial-distribution | ## Title
Binomial Distribution
Quick revise
If a discrete random variable X has the following probability density function (p.d.f.), it is said to have a binomial distribution:
• P(X = x) = nCx q(n-x)px, where q = 1 - p
p can be considered as the probability of a success, and q the probability of a failure.
Note: nCr (“n choose r”) is more commonly written , but I shall use the former because it is easier to write on a computer. It means the number of ways of choosing r objects from a collection of n objects (see permutations and combinations).
If a random variable X has a binomial distribution, we write X ~ B(n, p) (~ means ‘has distribution…’).
n and p are known as the parameters of the distribution (n can be any integer greater than 0 and p can be any number between 0 and 1). All random variables with a binomial distribution have the above p.d.f., but may have different parameters (different values for n and p).
Example
A coin is thrown 10 times. Find the probability density function for X, where X is the random variable representing the number of heads obtained.
The probability of throwing a head is ½ and the probability of throwing a tail is ½. Therefore, the probability of throwing 8 tails is (½)8
If we throw 2 heads and 8 tails, we could have thrown them HTTTTTHTT, or TTHTHTTTTT, or in a number of other ways. In fact, the total number of ways of throwing 2 heads and 8 tails is 10C2 (see the permutations and combinations section).
Hence the probability of throwing 2 heads and 8 tails is 10C2× (½)2× (½)8 . As you can see this has a Binomial distribution, where n = 10, p = ½.
You can see, therefore, that the p.d.f. is going to be: P(X = x) = 10Cx (½)(10-x) (½)x .
From this, we can work out the probability of throwing, for example, 3 heads (put x = 3).
Expectation and Variance
If X ~ B(n,p), then the expectation and variance is given by:
• E(X) = np
• Var(X) = npq
Example
In the above example, what is the expected number of heads thrown?
E(X) = np
Now in the above example, p = probability of throwing a head = ½ . n = number of throws = 10
Hence expected number of heads = 5.
This is what you would expect: if you throw a coin 10 times you would expect 5 heads and 5 tails on average.
Rate: | 2022-01-22 06:13:35 | {"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": 0.8566871881484985, "perplexity": 318.22823240348856}, "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-2022-05/segments/1642320303747.41/warc/CC-MAIN-20220122043216-20220122073216-00549.warc.gz"} |
https://math.stackexchange.com/questions/2721756/how-to-pronounce-this-notation | # How to pronounce this notation
Reading this paper : https://arxiv.org/pdf/1608.04644.pdf the following term : $$L_\infty$$ is referenced in the context of :
How is $L_\infty$ pronounced ? Is it just "L infinity" ? | 2021-12-02 18:14:01 | {"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": 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": 0.6887209415435791, "perplexity": 904.3047294357041}, "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-2021-49/segments/1637964362287.26/warc/CC-MAIN-20211202175510-20211202205510-00476.warc.gz"} |
https://eccc.weizmann.ac.il/keyword/18339/ | Under the auspices of the Computational Complexity Foundation (CCF)
REPORTS > KEYWORD > DE MORGAN FORMULA:
Reports tagged with De Morgan Formula:
TR21-002 | 8th January 2021
Pooya Hatami, William Hoza, Avishay Tal, Roei Tell
#### Fooling Constant-Depth Threshold Circuits
We present new constructions of pseudorandom generators (PRGs) for two of the most widely-studied non-uniform circuit classes in complexity theory. Our main result is a construction of the first non-trivial PRG for linear threshold (LTF) circuits of arbitrary constant depth and super-linear size. This PRG fools circuits with depth $d\in\mathbb{N}$ ... more >>>
ISSN 1433-8092 | Imprint | 2021-01-23 05:18:43 | {"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": 0.30972591042518616, "perplexity": 7343.0404792931895}, "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/1610703533863.67/warc/CC-MAIN-20210123032629-20210123062629-00606.warc.gz"} |
https://ora.ox.ac.uk/objects/uuid:09dad8c8-ab57-4029-9e3f-2cc0acb95402 | Report
### Subsumption of concepts in DL FL0 for (cyclic) terminologies with respect to descriptive semantics is PSPACE−complete
Abstract:
We prove the PSPACE-completeness of the subsumption problem for(cyclic) terminologies with respect to descriptive semantics in a simple Description Logic \\cal FL_0, which allows for conjunctions and universal value restrictions only, thus solving the problem which was open for more than ten years
### Authors
Publisher:
Max−Planck−Institut für Informatik
ISSN:
0946-011X
UUID:
Local pid:
cs:880
Deposit date:
2015-03-31 | 2022-07-01 10:50:00 | {"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": 0.9765961170196533, "perplexity": 9608.820239574461}, "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/1656103940327.51/warc/CC-MAIN-20220701095156-20220701125156-00442.warc.gz"} |
http://stats.stackexchange.com/questions/35038/given-two-sets-of-data-what-could-explain-similar-means-but-different-standard | # Given two sets of data, what could explain similar means but different standard deviations?
Given two sets of data of user activity, both of which appear to be in an exponential distribution, I have calculated the mean and standard deviations using both a mean/deviation and a sample mean/deviation (sample size = 30, number of samples = 10k):
### A (size: 627,000):
• Raw -- μ = 45.947, σ = 114.2, σ/√n = 0.14422
### B (size:3570):
• Raw -- μ = 46.43, σ = 116.1
Using the above data, it seems that the two means differ by a statistically significant amount, and thus allow us to say with confidence that the average for B is greater than the average of A.
### A
• Sampling -- μ = 46.174, σ = 21.256
### B
• Sampling -- μ = 46.786, σ = 21.366
Using the sampling data standard deviations, we see that the difference in means (0.612) is much less than the deviation, it seems that the means do not differ by a statistically significant amount.
So given the above, which is right? Can we say that these data sets differ? If the underlying distribution is exponential, are the above tests even accurate?
- | 2013-12-11 14:36:45 | {"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": 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": 0.9116023778915405, "perplexity": 950.6389003937992}, "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-2013-48/segments/1386164037630/warc/CC-MAIN-20131204133357-00032-ip-10-33-133-15.ec2.internal.warc.gz"} |
https://brilliant.org/problems/combinatorial-straight-lines/ | # Combinatorial Straight Lines
Discrete Mathematics Level 4
If the coefficients $$A$$ and $$B$$ of the equation of a straight line $$Ax + By = 0$$ are two distinct digits from the numbers $$0,1,2,3,6,7$$, then the number of distinct straight lines is
× | 2016-10-28 06:26:26 | {"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": 0.7566916346549988, "perplexity": 271.6493713270305}, "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-2016-44/segments/1476988721558.87/warc/CC-MAIN-20161020183841-00304-ip-10-171-6-4.ec2.internal.warc.gz"} |
http://physics.stackexchange.com/tags/dipole/new | # Tag Info
## New answers tagged dipole
0
If the total charge of the system is zero, the dipole moment does not depend on distance. much same like: if total momentum of a system is zero, the angular momentum does not depend on the origin of reference. Dipole moment is the intrinsic property of a system (subtract total charge to zero first); Angular momentum is the intrinsic property of a system ...
3
The electric dipole moment is defined as $$p = \int r \; dq$$ In the case of a pair of charges for which both charges are of the same magnitude, the choice of the origin turns out to be irrelevant: $$p = \mathbf{r_1} q - \mathbf{r_2} q = q(\mathbf{r_1} - \mathbf{r_2}) = q\mathbf{d}$$ where $\mathbf{d}$ is the distance between the charges. However, when ...
1
If you take a permanent magnet, and place a sheet of paper over it. Now sprinkle iron filings on it, and you pretty much get this diagram. This has been the mainstay of field theory since Faraday's time. A test charge at rest will begin to move in the direction of the field line. Since there is nowhere that it can rest where there is more than one ...
0
At any point the electric field is the vector sum of the fields from the two charges. So while the fields from $A$ and $B$ are indeed in opposite directions at your point $p$ you just add them (well, subtract their magnitudes since they're in opposite directions) and this gives you the net field. I wouldn't take the field lines too seriously. They are not ...
1
Dipole $\def\vp{{\vec p}}\def\ve{{\vec e}}\def\l{\left}\def\r{\right}\def\vr{{\vec r}}\def\ph{\varphi}\def\eps{\varepsilon}\def\grad{\operatorname{grad}}\def\vE{{\vec E}}$ $\vp:=\ve Ql$ constant $l\rightarrow 0$, $Q\rightarrow\infty$. \begin{align} \ph(\vr,\vr') &= \lim_{l\rightarrow0}\frac{Ql\ve\cdot\ve}{4\pi\eps_0 l}\l(\frac{1}{|\vr-\vr'-\ve\frac ...
Top 50 recent answers are included | 2014-04-19 12:12:26 | {"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": 0.929149329662323, "perplexity": 233.00837265986812}, "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-2014-15/segments/1397609537186.46/warc/CC-MAIN-20140416005217-00443-ip-10-147-4-33.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/find-elementary-matrix-e-such-that-b-ea.270363/ | # Homework Help: Find elementary matrix E such that B=EA
1. Nov 8, 2008
### subopolois
1. The problem statement, all variables and given/known data
im having problems with this question, i dont know how they got their answer. the question is: find elementary matrix E such that B=EA
A=-1 2 B= 1 -2 (these are matrices)
0 1 0 1
2. Relevant equations
elementary row operations
3. The attempt at a solution
-1 2|1 0 (row 1x-1) 1 -2|-1 0 (row 1+2 row 2) 1 0|-1 2
0 1|0 1 0 1|0 1 0 1|0 1
the answer in my book says its -1 0 but i dont know how they got that
0 1
2. Nov 8, 2008
### gabbagabbahey
If you have learned about matrix inverses, the solution should be fairly simple...A quick calculation shows that $\text{det}(A) \neq 0$ and so its inverse exists...what do you get when you multiply both sides of the equation $B=EA$ from the right by $A^{-1}$? | 2018-07-17 02:55:41 | {"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": 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": 0.6690493822097778, "perplexity": 985.3166934872756}, "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-2018-30/segments/1531676589537.21/warc/CC-MAIN-20180717012034-20180717032034-00442.warc.gz"} |
http://joshkos.blogspot.com/2007/05/texbook.html | $\newcommand{\defeq}{\mathrel{\mathop:}=}$
## 2007/05/27
### TeXbook
--
Quote from preface of TeXbook:
This manual is inteded for people who have never used TeX before, as well as for experienced TeX hackers. In other words, it's supposed to be a panacea that satisfies everybody, at the risk of satisfying nobody.
Knuth 的書寫真是幽默生動 XD。或許鍛鍊文字最好的方法，就是像 Knuth 一樣，研究一個主題就寫一本書出來 XD。
Labels:
skusi5/27/2007 12:53 pm 說：
yen35/27/2007 3:44 pm 說： | 2018-03-22 00:02:39 | {"extraction_info": {"found_math": true, "script_math_tex": 1, "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": 0.9994189739227295, "perplexity": 11829.73888598108}, "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/1521257647707.33/warc/CC-MAIN-20180321234947-20180322014947-00599.warc.gz"} |
https://www.sarthaks.com/2712833/acceleration-gravity-earth-height-above-surface-earth-distance-location-from-centre-earth | # The acceleration due to gravity of Earth at a height above the surface of Earth is 1 mm/s2. The distance of this location from the centre of Earth is
24 views
in Physics
closed
The acceleration due to gravity of Earth at a height above the surface of Earth is 1 mm/s2. The distance of this location from the centre of Earth is (Assume g = 10 m/s2, radius of earth = 6400 km)
1.
3200 km
2.
7650 km
3.
8640 km
4.
9600 km
by (53.7k points)
selected
Correct Answer - Option 4 :
9600 km
The correct answer is option 4) i.e. 9600 km
CONCEPT:
• Acceleration due to gravity on the surface of Earth of mass M and radius Re is denoted by g.
• It has an approximated uniform value of 9.8 m/s2 on the surface of Earth.
The acceleration due to gravity at a depth 'd' below the surface of Earth is given by
$⇒ g' = g(1- \frac{d}{R_e})$
• The acceleration due to gravity at a height 'h' above the surface of Earth is given by
$⇒ g'' = g(1+ \frac{h}{R_e})^{-2}$
$⇒ g'' = g(1- \frac{2h}{R_e})$ for h << Re
EXPLANATION:
Given that:
g'' = 1 mm/s2 = 10-3 m/s2
$⇒ g'' = g(1- \frac{2h}{R_e})$
$⇒ 10^{-3}= 10(1- \frac{2h}{6400})$
⇒ h = 3200 km
Therefore, distance from the centre of earth = R + h = 6400 + 3200 = 9600 km. | 2022-12-08 02:37:58 | {"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": 0.8908458948135376, "perplexity": 917.2449695781604}, "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/1669446711232.54/warc/CC-MAIN-20221208014204-20221208044204-00016.warc.gz"} |
https://brilliant.org/problems/box-and-box-and-box/ | Box and box and box!!
Algebra Level pending
[x/3] +[x/5] +[x/7] =[x/10]. Find all possible +ve integral solutions of the equation. Here , [m] denotes the greatest integer less than or equal to m.
× | 2017-10-18 04:02:29 | {"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": 0.9711319804191589, "perplexity": 2495.752633707936}, "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-2017-43/segments/1508187822739.16/warc/CC-MAIN-20171018032625-20171018052625-00781.warc.gz"} |
https://encyclopediaofmath.org/wiki/Matrix_of_transition_probabilities | # Matrix of transition probabilities
The matrix $P _ {t} = \| p _ {ij} ( t) \|$ of transition probabilities in time $t$ for a homogeneous Markov chain $\xi ( t)$ with at most a countable set of states $S$:
$$p _ {ij} ( t) = {\mathsf P} \{ \xi ( t) = j \mid \xi ( 0) = i \} ,\ \ i, j \in S.$$
The matrices $\| p _ {ij} ( t) \|$ of a Markov chain with discrete time or a regular Markov chain with continuous time satisfy the following conditions for any $t > 0$ and $i, j \in S$:
$$p _ {ij} ( t) \geq 0,\ \ \sum _ {j \in S } p _ {ij} ( t) = 1,$$
i.e. they are stochastic matrices (cf. Stochastic matrix), while for irregular chains
$$p _ {ij} ( t) \geq 0,\ \ \sum _ {j \in S } p _ {ij} ( t) \leq 1,$$
such matrices are called sub-stochastic.
By virtue of the basic (Chapman–Kolmogorov) property of a homogeneous Markov chain,
$$p _ {ij} ( s+ t) = \sum _ {k \in S } p _ {ik} ( s) p _ {kj} ( t),$$
the family of matrices $\{ {P _ {t} } : {t > 0 } \}$ forms a multiplicative semi-group; if the time is discrete, this semi-group is uniquely determined by $P _ {1}$. | 2021-10-22 11:59:15 | {"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": 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": 0.8379265666007996, "perplexity": 389.0623682764327}, "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/1634323585507.26/warc/CC-MAIN-20211022114748-20211022144748-00408.warc.gz"} |
https://watercharity.com/book/gushie-soak-away-pit-sanitation-project-ghana/ | Select Page
# Gushie Soak-Away Pit Sanitation Project – Ghana
Location
Gushie, Northern Region, Ghana
Community Description
Gushie is a small farming village in the Northern Region of Ghana, approximately 50 km north of Tamale, the regional capital. Gushie has a population of just under 1,000 people, and does not have electricity.
The village is mostly comprised of people from the Dagomba and Fulani tribes, and Islam is practiced as the primary religion. November to May marks the dry season in the region, after which the rainy season brings rains and humid weather from June to October.
Because villagers live in compounds consisting of several family members living in small mud huts connected by a large communal area, the bathing area for an average compound must support at least 10-15 people. These areas are often little more than a half wall of mud with a small hole in the corner for run-off water, and families use these areas both as a place for bucket bathing and as makeshift latrines for liquid waste.
The lack of basic sanitation facilities and proper water management has become one of the most urgent threats to the health of the village, particularly during the rainy months when flooding is a common occurrence. The excess water and waste from bathing areas simply flows outside onto the ground and, collects in stagnant pools of liquid waste that often intersect with footpaths or other living areas. The overall health of the village is compromised throughout the year because these brackish ponds provide an ideal place for the breeding of mosquitos, disease-carrying flies, and waterborne parasites.
Project Description
This project consists of the building of soak-away pits for twenty households in the Gushie community.
The pits will be constructed behind the bathing areas of these compounds using PVC piping that will feed the excess water and liquid waste into deep, covered pits filled with rocks, so that the water is properly disposed of. This will deter contaminated flooding and the accompanying breeding of insects.
The majority of households in the community are participating in the project, which will begin with pits being built in areas that directly intersect often-used walking paths and places where villagers spread their crops for drying.
This project is being led by a group called Fara Je Maahim, which is comprised of young adults from the community who meet regularly to organize small projects such as community clean-up days and to financially support fellow members in the village in times of death or other hardships. The members of this group range from 15 to 25 years of age, and have consistently set an example of leadership in Gushie by initiating educational programs and carrying out small but effective health projects that have addressed issues like trash removal and malaria prevention.
The group has committed to lead the project efforts by identifying which households currently have improper water drainage systems and sponsoring training sessions to educate the community on the gravity of current sanitation deficiencies. Members from this group will also assist and oversee each household in the construction of the pits from start to finish.
Community members will provide all of the labor for each soak-away pit by digging the holes, collecting rocks to fill the pits, constructing the drainage pipes, and covering the pits so they can be used for liquid waste management for many years to come.
Project funds will be used to purchase the materials for the project.
Project Impact
This sanitation project will directly benefit 400 people in Gushie. However, this project will more broadly improve the health and sanitation of the entire community of 1,000 by minimizing the existence of stagnant water and thereby reducing the presence of mosquitos, flies, and waterborne parasites.
Peace Corps Volunteer Directing Project
Katie Kirouac
$300.00 Donations Collected to Date$300.00 | 2022-05-28 17:08:05 | {"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": 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": 0.20090420544147491, "perplexity": 4978.057474823793}, "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-20220528154416-20220528184416-00118.warc.gz"} |
https://www.quantumstudy.com/with-what-velocity-must-an-electron-travel-so-that-its-momentum-is-equal-to-that-of-a-photon-with-a-wavelength-of-5000-a/ | # With what velocity must an electron travel so that its momentum is equal to that of a photon with a wavelength of 5000 A°
Q: With what velocity must an electron travel so that its momentum is equal to that of a photon with a wavelength of 5000 A°
(h = 6.6 × 10-34 Js ,me = 9.1 × 10-31 Kg)
Sol: $\large m v = \frac{h}{\lambda}$
$\large v = \frac{h}{m \lambda}$
$\large v = \frac{6.62 \times 10^{-34}}{9.1 \times 10^{-31} \times 5000 \times 10^{-10}}$
v = 1450 m/s | 2021-09-26 21:11:14 | {"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": 0.3985602557659149, "perplexity": 526.8261086240262}, "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/1631780057973.90/warc/CC-MAIN-20210926205414-20210926235414-00164.warc.gz"} |
http://mathhelpforum.com/new-users/210393-research-scholar-lecturer-uos.html | research scholar and lecturer at UOS
Well i am a research scholar as well as lecturer of mathematics
doing work on algebra, representation theory, group theory and combinatorics | 2014-04-20 00:46:59 | {"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": 0.8054121732711792, "perplexity": 1528.3274940117612}, "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-2014-15/segments/1397609537804.4/warc/CC-MAIN-20140416005217-00152-ip-10-147-4-33.ec2.internal.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/algebra-1/chapter-8-polynomials-and-factoring-cumulative-test-prep-multiple-choice-page-529/12 | ## Algebra 1
Published by Prentice Hall
# Chapter 8 - Polynomials and Factoring - Cumulative Test Prep - Multiple Choice - Page 529: 12
#### Answer
H) $\frac{1}{2}$
#### Work Step by Step
To answer the question, one must realize with the given points that for each time the $x$ value increases by 1, the $y$ value increases by 8. To find the $x$ value you must see what the value is when $y$ is equal to zero, meaning that it would have to move either 4 places up or down depending on which point used for reference. Since you are only moving half the amount necessary to move a full $x$ intercept, you can infer that the $y$ value is $0$ at $x=1/2$.
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. | 2020-02-18 15:56:26 | {"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": 0.5947989225387573, "perplexity": 493.35430306552166}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "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-2020-10/segments/1581875143784.14/warc/CC-MAIN-20200218150621-20200218180621-00383.warc.gz"} |
https://www.bartleby.com/solution-answer/chapter-28-problem-42p-principles-of-economics-2e-2nd-edition/9781947172364/if-gdp-is-and-the-money-supply-is-what-is-velocity/c9e64585-726f-11e9-8385-02ee952b546e | Chapter 28, Problem 42P
### Principles of Economics 2e
2nd Edition
Steven A. Greenlaw; David Shapiro
ISBN: 9781947172364
Chapter
Section
### Principles of Economics 2e
2nd Edition
Steven A. Greenlaw; David Shapiro
ISBN: 9781947172364
Textbook Problem
# If GDP is and the money supply is what is velocity?
To determine
The velocity of money, given the nominal GDP and money supply.
Explanation
The velocity of money is pace of circulation of money in the economy. The formula for calculating velocity of money is given below.
velocityofmoney=Nominal GDPMoneySupplyv=YMS
Given Information:
Money supply, MS= $400 Nominal GDP =$1500
### Still sussing out bartleby?
Check out a sample textbook solution.
See a sample solution
#### The Solution to Your Study Problems
Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!
Get Started | 2019-11-17 02:27:23 | {"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": 0.26920509338378906, "perplexity": 14999.101119460945}, "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-2019-47/segments/1573496668782.15/warc/CC-MAIN-20191117014405-20191117042405-00358.warc.gz"} |
https://brilliant.org/weekly-problems/2018-11-19/intermediate/?problem=fun-time-2 | # Problems of the Week
Contribute a problem
# 2018-11-19 Intermediate
This annulus has been formed from two concentric circles. A chord of length $t$ of the larger circle is then drawn tangent to the inner circle.
If you only know the value of $t,$ can you calculate the area of the annulus?
French mathematician Joseph Bertrand posed the following problem in 1889:
You have 3 identical boxes, each containing 2 coins: the first box 2 gold coins, the second box 2 silver coins, and the third box 1 gold and 1 silver coin.
Your friend shuffles the boxes at random. Then, you choose a box and pull a coin out of it, and it's gold.
What is the probability that the other coin in the same box is also a gold coin?
I have a square puzzle consisting of $10 \times 10 = 100$ pieces.
If I pick two pieces at random, what is the probability that they fit together, i.e. they lie next to each other in the puzzle?
If the probability can be written as $\frac ab$ with coprime positive integers $a$ and $b,$ give your answer as $a+b.$
Details and Assumptions:
• Every piece is equally likely to be picked. I don't look whether the first piece has a straight edge or similar tricks.
• Every piece only fits together with its neighboring pieces, and there is a unique solution to the puzzle.
Four congruent semicircles are packed into a square.
What proportion of the square is filled?
The sum of the ages of my five nieces is 47. Their ages are positive integers, and any two of them have a common divisor greater than 1.
How old is the eldest?
× | 2020-02-27 02:32:16 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 11, "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": 0.6251048445701599, "perplexity": 350.48286098336945}, "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/1581875146643.49/warc/CC-MAIN-20200227002351-20200227032351-00504.warc.gz"} |
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition/chapter-r-review-of-basic-concepts-r-4-factoring-polynomials-r-4-exercises-page-44/58 | ## Precalculus (6th Edition)
$(2p+q-5)^{2}$
Test whether this is a perfect square, $(A-B)^{2}=A^{2}-2AB+B^{2}$ First term: $A^{2}=(2p+q)^{2}\Rightarrow A=2p+q$ Third term: $B^{2}=(5)^{2}\Rightarrow B=5$ Test:$\qquad$ does $-2AB$ equal the middle term? $-2AB=-2(2p+q)(5)=-10(2p+q)\qquad$ ... yes, it does. $\Rightarrow$This is a perfect square, $[(2p+q)-5]^{2}=(2p+q-5)^{2}$ | 2020-05-31 16:07:39 | {"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": 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": 0.23827023804187775, "perplexity": 1470.9634617005318}, "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/1590347413551.52/warc/CC-MAIN-20200531151414-20200531181414-00497.warc.gz"} |
https://scriptinghelpers.org/questions/80468/should-i-use-gameplayers-or-gamegetserviceplayers | New: Nitro Boost our Discord server and receive full donation perks here on the website! Join the Scripting Helpers Discord Server to learn more! You can also Support on Patreon as always.
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0
# Should I use game.Players or game:GetService("Players")?
Edited 3 months ago
I have used Roblox Lua some time, and in every script I ever write, when I want to access, let's say, Players, I do
local players = game.Players
However, I have frequently seen others use
local players = game:GetService("Players")
instead. Is this "better" or more efficient in any way? To me, it just seems a bit more complicated, and I only use :GetService() for Services like TweenService or DebrisService, which aren't found in the Explorer.
1
BashCaster 2370
3 months ago
game.Players indexes a child of game named Players. If, for some reason, you renamed the Players service, game.Players will either index a nil value or, if a different child has the name Players, it will index that child. Thus, GetService() is considered the standard go-to for referencing services. GetService(serviceName) returns the service with the name serviceName. For every service (except the Workspace and any service that can't be retrieved), GetService() is a MUST. (You can do game:GetService("Workspace"), although doing so is mostly irrelevant because of the workspace shared variable.)
0
Thanks, this helps. so, in other words, don't use 'game.Players', and use 'workspace' instead of 'game.Workspace', right? RiskoZoSlovenska 239 — 3mo
1 | 2019-09-16 04:24:56 | {"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": 0.32750949263572693, "perplexity": 5796.719478236693}, "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/1568514572484.20/warc/CC-MAIN-20190916035549-20190916061549-00394.warc.gz"} |
http://clay6.com/qa/49980/the-length-of-string-of-a-simple-pendulum-as-measured-by-a-meter-scale-was- | # The length of string of a simple pendulum as measured by a meter scale was found to be $98.2 \;cm$ The diameter of the bob was measured by a vernier caliper and it was found to be $3.62\;cm$ The time for 10 oscillations was found to be $21.2\; s$ Suggest method to improve the accuracy of the result.
Percentage error introduced by limited accuracy of measurement of l is only $0.1\%$ while due to measurement of time is $0.94\%$ .
As $\delta t$ is fixed equal to least count of stop watch , to improve the situation , t must be increased .
In another words time for more number of oscillation must be recorded. | 2017-10-23 08:13:30 | {"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": 0.7228134274482727, "perplexity": 286.43419863351716}, "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-2017-43/segments/1508187825812.89/warc/CC-MAIN-20171023073607-20171023093607-00190.warc.gz"} |
http://clay6.com/qa/31275/the-correct-order-of-first-ionization-potential-is- | Browse Questions
# The correct order of first ionization potential is :
$(a)\;F > He > Mg > N > O \\(b)\;He > E > N > O > Mg \\ (c)\;He > O > f > Mg \\ (d)\;N > F > He> O > Mg$
The correct order of first ionization potential is $He > E > N > O > Mg$
Hence b is the correct answer. | 2016-12-11 08:10:14 | {"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": 0.9126361012458801, "perplexity": 1325.3260010416025}, "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-2016-50/segments/1480698544358.59/warc/CC-MAIN-20161202170904-00404-ip-10-31-129-80.ec2.internal.warc.gz"} |