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# Problem 2.1 ## Insertion sort on small arrays in merge sort Although merge sort runs in $\Theta(\lg{n})$ worst-case time and insertion sort runs in $\Theta(n^2)$ worst-case time, the constant factors in insertion sort can make it faster in practice for small problem sizes on many machines. Thus, it makes sense to coarsen the leaves of the recursion by using insertion sort within merge sort when subproblems become sufficiently small. Consider a modification to merge sort in which $n/k$ sublists of length $k$ are sorted using insertion sort and then merged using the standard merging mechanism, where $k$ is a value to be determined. 1. Show that insertion sort can sort the $n/k$ sublists, each of length $k$, in $\Theta(nk)$ worst-case time. 2. Show how to merge the sublists in $\Theta(n\lg(n/k))$ worst-case time. 3. Given that the modified algorithm runs in $\Theta(nk + n\lg(n/k))$ worst-case time, what is the largest value of $k$ as a function of $n$ for which the modified algorithm has the same running time as standard merge sort, in terms of $\Theta$-notation? 4. How should we choose $k$ in practice? ### 1. Sorting sublists This is simple enough. We know that sorting each list takes $ak^2 + bk + c$ for some constants $a$, $b$ and $c$. We have $n/k$ of those, thus: $$\frac{n}{k}(ak^2 + bk + c) = ank + bn + \frac{cn}{k} = \Theta(nk)$$ ### 2. Merging sublists This is a bit trickier. Sorting $a$ sublists of length $k$ each takes: $$T(a) = \begin{cases} 0 & \text{if } a = 1, \\ 2T(a/2) + ak & \text{if } a = 2^p, \text{if } p > 0. \end{cases}$$ This makes sense, since merging one sublist is trivial and merging $a$ sublists means splitting dividing them in two groups of $a/2$ lists, merging each group recursively and then combining the results in $ak$ steps, since have two arrays, each of length $\frac{a}{2}k$. I don't know the master theorem yet, but it seems to me that the recurrence is actually $ak\lg{a}$. Let's try to prove this via induction: Base. Simple as ever: $$T(1) = 1k\lg1 = k \cdot 0 = 0$$ Step. We assume that $T(a) = ak\lg{a}$ and we calculate $T(2a)$: \begin{align} T(2a) &= 2T(a) + 2ak = 2(T(a) + ak) = 2(ak\lg{a} + ak) = \\ &= 2ak(\lg{a} + 1) = 2ak(\lg{a} + \lg{2}) = 2ak\lg(2a) \end{align} This proves it. Now if we substitue the number of sublists $n/k$ for $a$: $$T(n/k) = \frac{n}{k}k\lg{\frac{n}{k}} = n\lg(n/k)$$ While this is exact only when $n/k$ is a power of 2, it tells us that the overall time complexity of the merge is $\Theta(n\lg(n/k))$. ### 3. The largest value of k The largest value is $k = \lg{n}$. If we substitute, we get: $$\Theta(n\lg{n} + n\lg{\frac{n}{\lg{n}}}) = \Theta(n\lg{n})$$ If $k = f(n) > \lg{n}$, the complexity will be $\Theta(nf(n))$, which is larger running time than merge sort. ### 4. The value of k in practice It's constant factors, so we just figure out when insertion sort beats merge sort, exactly as we did in exercise 1.2.2, and pick that number for $k$. ### Runtime comparison I'm implemented this in C and in Python. I added selection for completeness sake in the C version. I ran two variants, depending on whether merge() allocates its arrays on the stack or on the heap (stack won't work for huge arrays). Here are the results: STACK ALLOCATION ================ merge-sort = 0.173352 mixed-insertion = 0.150485 mixed-selection = 0.165806 HEAP ALLOCATION =============== merge-sort = 1.731111 mixed-insertion = 0.903480 mixed-selection = 1.017437 Here's the results I got from Python: merge-sort = 2.6207s mixed-sort = 1.4959s I can safely conclude that this approach is faster. ### C runner output merge-sort = 0.153748 merge-insertion = 0.064804 merge-selection = 0.069240 ### Python runner output merge-sort = 0.1067s mixed-sort = 0.0561s ### C code #include <stdlib.h> #include <string.h> #define INSERTION_SORT_TRESHOLD 20 #define SELECTION_SORT_TRESHOLD 15 void merge(int A[], int p, int q, int r) { int i, j, k; int n1 = q - p + 1; int n2 = r - q; #ifdef MERGE_HEAP_ALLOCATION int *L = calloc(n1, sizeof(int)); int *R = calloc(n2, sizeof(int)); #else int L[n1]; int R[n2]; #endif memcpy(L, A + p, n1 * sizeof(int)); memcpy(R, A + q + 1, n2 * sizeof(int)); for(i = 0, j = 0, k = p; k <= r; k++) { if (i == n1) { A[k] = R[j++]; } else if (j == n2) { A[k] = L[i++]; } else if (L[i] <= R[j]) { A[k] = L[i++]; } else { A[k] = R[j++]; } } #ifdef MERGE_HEAP_ALLOCATION free(L); free(R); #endif } void merge_sort(int A[], int p, int r) { if (p < r) { int q = (p + r) / 2; merge_sort(A, p, q); merge_sort(A, q + 1, r); merge(A, p, q, r); } } void insertion_sort(int A[], int p, int r) { int i, j, key; for (j = p + 1; j <= r; j++) { key = A[j]; i = j - 1; while (i >= p && A[i] > key) { A[i + 1] = A[i]; i = i - 1; } A[i + 1] = key; } } void selection_sort(int A[], int p, int r) { int min, temp; for (int i = p; i < r; i++) { min = i; for (int j = i + 1; j <= r; j++) if (A[j] < A[min]) min = j; temp = A[i]; A[i] = A[min]; A[min] = temp; } } void mixed_sort_insertion(int A[], int p, int r) { if (p >= r) return; if (r - p < INSERTION_SORT_TRESHOLD) { insertion_sort(A, p, r); } else { int q = (p + r) / 2; mixed_sort_insertion(A, p, q); mixed_sort_insertion(A, q + 1, r); merge(A, p, q, r); } } void mixed_sort_selection(int A[], int p, int r) { if (p >= r) return; if (r - p < SELECTION_SORT_TRESHOLD) { selection_sort(A, p, r); } else { int q = (p + r) / 2; mixed_sort_selection(A, p, q); mixed_sort_selection(A, q + 1, r); merge(A, p, q, r); } } ### Python code from itertools import repeat def insertion_sort(A, p, r): for j in range(p + 1, r + 1): key = A[j] i = j - 1 while i >= p and A[i] > key: A[i + 1] = A[i] i = i - 1 A[i + 1] = key def merge(A, p, q, r): n1 = q - p + 1 n2 = r - q L = list(repeat(None, n1)) R = list(repeat(None, n2)) for i in range(n1): L[i] = A[p + i] for j in range(n2): R[j] = A[q + j + 1] i = 0 j = 0 for k in range(p, r + 1): if i == n1: A[k] = R[j] j += 1 elif j == n2: A[k] = L[i] i += 1 elif L[i] <= R[j]: A[k] = L[i] i += 1 else: A[k] = R[j] j += 1 def merge_sort(A, p, r): if p < r: q = int((p + r) / 2) merge_sort(A, p, q) merge_sort(A, q + 1, r) merge(A, p, q, r) def mixed_sort(A, p, r): if p >= r: return if r - p < 20: insertion_sort(A, p, r) else: q = int((p + r) / 2) mixed_sort(A, p, q) mixed_sort(A, q + 1, r) merge(A, p, q, r)

http://math.stackexchange.com/questions/732445/my-first-induction-proof-very-simple-number-theory-please-mark-grade

## Theorem The sum of the cubes of three consecutive natural numbers is a multiple of 9. ## Proof First, introducing a predicate $P$ over $\mathbb{N}$, we rephrase the theorem as follows. $$\forall n \in \mathbb{N}, P(n) \quad \text{where} \quad P(n) \, := \, n^3 + (n + 1)^3 + (n + 2)^3 \text{ is a multiple of 9}$$ We prove the theorem by induction on $n$. ### Basis Below, we show that we have $P(n)$ for $n = 0$. $$0^3 + 1^3 + 2^3 = 0 + 1 + 8 = 9 = 9 \cdot 1$$ ### Inductive step Below, we show that for all $n \in \mathbb{N}$, $P(n) \Rightarrow P(n + 1)$. Let $k \in \mathbb{N}$. We assume that $P(k)$ holds. In the following, we use this assumption to show that $P(k + 1)$ holds. By the assumption, there is a $i \in \mathbb{N}$ such that $i \cdot 9 = k^3 + (k + 1)^3 + (k + 2)^3$. We use this fact in the following equivalent transformation. The transformation turns the sum of cubes in the first line, for which we need to show that it is a multiple of 9, into a product of 9 and another natural number. $(k + 1)^3 + (k + 2)^3 + (k + 3)^3 \\ = (k + 1)^3 + (k + 2)^3 + k^3 + 9k^2 + 27k + 27 \\ = k^3 + (k + 1)^3 + (k + 2)^3 + 9k^2 + 27k + 27 \quad | \text{ using the induction hypothesis} \\ = 9i + 9k^2 + 27k + 27 \\ = 9 \cdot i + 9 \cdot k^2 + 9 \cdot 3k + 9 \cdot 3 \\ = 9 \cdot (i + k^2 + 3k + 3)$ We see that the above product has precisely two factors: 9 and another natural number. Thus the product is a multiple of 9. This completes the induction. - Looks fine to me! –  Braindead Mar 30 '14 at 13:02 someone should make something along the lines of markExchange instead of stackExchange. Marking instead of answering sounds fresh and good to me –  Nicholas Kyriakides Mar 31 '14 at 3:59 Looks good. You could maybe add a Halmos to the end, to be extra if not overly formal, $\square$. Check out math.stackexchange.com/questions/56606/… –  guydudebro Mar 31 '14 at 4:42 @NicholasKyriakides There already is codereview.stackexchange.com, which is essentially "marking code". Perhaps a proofreview StackExchange would be useful. –  Ixrec Mar 8 at 21:19 I do not think that this is a real question but if you were my student i would give you an A. It is all fine to me. - It's fine, here's a simpler proof without induction: $n^3\equiv n\ (\text{mod }3)$, because it obviously holds for $n=-1,0,1$. Therefore $3n^3\equiv3n\ (\text{mod 9})$ and $$(n-1)^3+n^3+(n+1)^3\equiv3n^3+6n\equiv0\ (\text{mod }9)$$ - i.e. $\ {\rm mod}\ 9\!:\ 3n^3\!+6n\equiv 3(n^3\!-n)\equiv 0\$ by little Fermat. –  Bill Dubuque Mar 30 '14 at 13:16 Formulation, base case, inductive hypothesis, inductive step, it all looks good. :) One might also conclude with a clarifying statement about what has been done - that the hypothesis is true for all $n \in \Bbb N$. - What do you think about the following statement? "By the basis, the inductive step, and the principle of induction; $P(n)$ is true for all $n \in \mathbb{N}$." –  DracoMalfoy Mar 30 '14 at 13:51

http://math.stackexchange.com/questions/263640/how-can-i-algebraically-prove-that-2n-1-is-not-always-prime

# How can I algebraically prove that $2^n - 1$ is not always prime? This question is from Elementary Number Theory by W. Edwin Clark. Is $2^n - 1$ always prime, or not? Prove. Is this a start? $x^n - 1 = ( x - 1)(1 + x + x^2 \cdots x^{n - 1})$. So, $2^n - 1 = \sum \limits _{i = 0}^{n - 1} 2^i.$ Will I reach a solution through the above, or is there any other way? I know that the property doesn't hold true for $n = 1,4,6$ et al but I want an algebraic proof. - I think the question can be interpreted as "prove that there are infinitely many $n$ such that $2^n-1$ is not prime". Otherwise, as @Hurkyl mentioned, you have already proved your own statement. –  akkkk Dec 22 '12 at 10:40 HINT: If $n=ab$, then $2^a-1$ is a factor of $2^n-1$. - And what would be the factored form of $2^{ab} - 1$? :-) P.S.: Hey Brian! –  Parth Kohli Dec 22 '12 at 10:40 Gee, I wonder: $2^{ab}-1=\left(2^a\right)^b-1^b=\ldots~$ nope, too hard for me. :-) Actually, the really cute way to see it is to write the numbers in binary: $2^{ab}$ is a string of $ab$ $1$’s. –  Brian M. Scott Dec 22 '12 at 10:43 @DumbCow - try using the factorisation you have given in the question with an appropriate choice of $x$ –  Mark Bennet Dec 22 '12 at 10:43 @Markbennet: Oh yes, thanks. –  Parth Kohli Dec 22 '12 at 10:44 I know that the property doesn't hold true for n=1,4,6 et al. I just want to clearly point out that this statement all by itself (or possibly with a calculation demonstrating the truth of the statement) constitutes a proof of the question asked. - Take $n=4$. Then $2^n-1=16-1=15=3\cdot 5$ which is not a prime. The statement is proven, that is $2^n-1$ is not always a prime. EDIT: Why this is a formal proof: We want to prove that $$\neg (\forall n\in \mathbb{N})(2^n-1\in \mathbb{P})$$ or equivalently that $$(\exists n\in \mathbb{N})\neg (2^n-1\in \mathbb{P})$$ or even $$(\exists n\in \mathbb{N})(2^n-1\notin \mathbb{P})$$ Since $\exists 4\in \mathbb{N}$ and $2^4-1\notin \mathbb{P}$, the statement is proven. - It is a proof... –  Parth Kohli Dec 22 '12 at 10:37 @DumbCow This is a proof! –  Nameless Dec 22 '12 at 10:37 It should be a formal proof :) –  Parth Kohli Dec 22 '12 at 10:37 @DumbCow I believe I now fully justified why this is a rigorous proof –  Nameless Dec 22 '12 at 10:46 Looks legit pretty much :) –  Parth Kohli Dec 22 '12 at 10:51 If $n$ is even say $n = 2m$ then $2^n - 1 = 2^{2m } -1 = (2^m + 1)(2^m - 1)$ which is not prime. More generally if $n$ is composite then by the formula for the sum of a geometric series we get that... - I like it :-). So, the main point of our proof is that it is not prime since it has factors. –  Parth Kohli Dec 22 '12 at 10:38 You might like to ponder why $2047 = 23 \times 89$ is a different kind of example from those already given in previous answers. -

http://www.virtualmuseum.finearts.go.th/sudbury-cable-lcpi/page.php?12d573=diagonal-of-parallelogram

Calculations include side lengths, corner angles, diagonals, height, perimeter and area of parallelograms. where is the two-dimensional cross product and is the determinant.. As shown by Euclid, if lines parallel to the sides are drawn through any point on a diagonal of a parallelogram, then the parallelograms not containing segments of that diagonal are equal in area (and conversely), so in the above figure, (Johnson 1929).. Some of the properties of a parallelogram are that its opposite sides are equal, its opposite angles are equal and its diagonals bisect each other. This is the currently selected item. The Diagonals of a Parallelogram Abcd Intersect at O. There are several rules involving: the angles of a parallelogram ; the sides of a parallelogram ; the diagonals of a parallelogram Area = 6 m × 3 m = 18 m 2. It is done with the help of law of cosines . The diagonals of a parallelogram bisect each other. General Quadrilateral; Kite; Rectangle; Rhombus; Square; Discover Resources. In mathematics, the simplest form of the parallelogram law (also called the parallelogram identity) belongs to elementary geometry.It states that the sum of the squares of the lengths of the four sides of a parallelogram equals the sum of the squares of the lengths of the two diagonals. 1 answer. MCQ in Plane Geometry. Therefore, a square has all the properties of a rectangle and a rhombus. Apply the formula from the Theorem. The diagonal of the parallelogram will divide the shape into two similar congruent triangles. Area of a Parallelogram : The Area is the base times the height: Area = b × h (h is at right angles to b) Example: A parallelogram has a base of 6 m and is 3 m high, what is its Area? Next lesson. Try this Drag the orange dots on each vertex to reshape the parallelogram. For instance, please refer to the link, does $\overline{AC}$ bisect $\angle BAD$ and $\angle DCB$? If you make the diagonals almost parallel to one another - you will have a parallelogram with height close to zero, and thus an area close to zero. The diagonals of the Varignon parallelogram are the bimedians of the original quadrilateral. פרבולה וכפל - מה הקשר? Solution (1) AC=24 //Given Vector velocity and vector Up: Motion in 3 dimensions Previous: Scalar multiplication Diagonals of a parallelogram The use of vectors is very well illustrated by the following rather famous proof that the diagonals of a parallelogram mutually bisect one another. The rectangle has the following properties: All the properties of a parallelogram apply (the ones that matter here are parallel sides, opposite sides are congruent, and diagonals bisect each other). P inoyBIX educates thousands of reviewers and students a day in preparation for their board … With that being said, I was wondering if within parallelogram the diagonals bisect the angles which the meet. Because the parallelogram has adjacent angles as acute and obtuse, the diagonals split the figure into 2 pairs of congruent triangles. That is, each diagonal cuts the other into two equal parts. Check the picture. You get the equation = . These parallelograms have different areas. The answer is “maybe.” Diagonals of rhombi, which are parallelograms, do bisect the angles. Proof: Diagonals of a parallelogram. Test the conjecture with the diagonals of a rectangle. Please do Subscribe on YouTube! The two bimedians in a quadrilateral and the line segment joining the midpoints of the diagonals in that quadrilateral are concurrent and are all bisected by their point of intersection. What are the diagonals of a parallelogram? The diagonals of a parallelogram. Proof: The diagonals of a kite are perpendicular. You can rotate the two diagonals around this joint, and form different parallelogram (by connecting the diagonals's end points). So the areas of the parallelogram is (diagonal x diagonal /2 ), or 24x10/2=120, as above. The Perimeter is the distance around the edges. We can proceed to prove that this parallelogram is indeed a rhombus, using the fact that if a parallelogram's diagonals are perpendicular, it is a rhombus - and we've shown above that these diagonals are indeed perpendicular. Make a conjecture about the diagonals of a parallelogram. Opposite sides are congruent. Proof: Rhombus diagonals are perpendicular bisectors. A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel. Answered by | 16th Aug, 2017, 04:15: PM. Proof: Rhombus area. Thus, the diagonals of a parallelogram bisect each other. Show that it is a rhombus. Diagonals of rectangles and general parallelograms, however, do not. You can use the calculator for each formula. Practice: Prove parallelogram properties. In the figure below diagonals AC and BD bisect each other. Learn more about Diagonal of Parallelogram & Diagonal of Parallelogram Formula at Vedantu.com A parallelogram where all angles are right angles is a rectangle! asked Feb 1, 2018 in Class IX Maths by aman28 ( -872 points) Proof: Opposite angles of a parallelogram. A diagonal of a parallelogram bisects one of its angles. 3. Since the angles are acute or obtuse, two of the shorter sides of the triangles, both acute and obtuse are congruent. The diagonals of a parallelogram bisect each other. The length of the shorter diagonal of a parallelogram is 10.73 . Find the area of the parallelogram whose diagonals are represented by the vectors - 4 i +2 j + k & 3 i – 2 j - k. asked Aug 22, 2018 in Mathematics by AnujPatel (53.5k points) vectors; 0 votes. person_outlineTimurschedule 2011-03-28 14:49:28. A parallelogram has two diagonals. Parallelogram definition, a quadrilateral having both pairs of opposite sides parallel to each other. The shape has the rotational symmetry of the order two. The diagonals are perpendicular bisectors of each other. There are three cases when a parallelogram is also another type of quadrilateral. The parallelogram has the following properties: Opposite sides are parallel by definition. Show that it is a rhombus. Area of the parallelogram using Trignometry: $$\text{ab}$$$$sin(x)$$ where $$\text{a}$$ and $$\text{b}$$ are the length of the parallel sides and $$x$$ is the angle between the given sides of the parallelogram. In a parallelogram, the diagonals bisect each other, so you can set the labeled segments equal to one another and then solve for . In a parallelogram, the sides are 8 cm and 6 cm long. A square may be considered as rectangle which has equal adjacent sides, or a rhombus with a right angle. Online Questions and Answers in Plane Geometry. Calculate the angle between diagonals of a parallelogram if given 1.Sides and diagonal 2.Sides and area of a parallelogram. Definition of Quadrilateral & special quadrilaterals: rectangle, square,... All Questions Ask Doubt. A parallelogram is a quadrilateral made from two pairs of intersecting parallel lines. The properties of the parallelogram are simply those things that are true about it. The diagonals of a parallelogram bisect each other. The adjacent angles of the parallelogram are supplementary. If ∠Boc = 90° and ∠Bdc = 50°, Then ∠Oab = - Mathematics If ∠Boc = 90° and ∠Bdc = 50°, Then ∠Oab = - Mathematics Question By … The diagonals bisect each other. So we have a parallelogram right over here. Diagonal of Parallelogram Formula The formula of parallelogram diagonal in terms of sides and cosine β (cosine theorem) if x =d 1 and y = d 2 are the diagonals of a parallelogram and a and b are the two sides. Then, substitute 4.8 for in each labeled segment to get a total of 11.2 for the diagonal … Rectangle: Rectangle is a special case of parallelogram in which measure of each interior angle is $$90^\circ$$. Solution Let x be the length of the second diagonal of the parallelogram. Proofs of general theorems . Special parallelograms. A parallelogram is a quadrilateral with opposite sides parallel. Video transcript. The pair of opposite sides are equal and they are equal in length. The diagonals of a parallelogram bisect each other. Calculator computes the diagonals of a parallelogram and adjancent angles from side lengths and angle. See more. More Questions in: Plane Geometry. Which additional tool will you use? View Solution: Latest Problem Solving in Plane Geometry. The diagonals of a parallelogram are not equal. These properties concern its sides, angles, and diagonals. Diagonals of a parallelogram; Angles of a parallelogram; Angles between diagonals of a parallelogram; Height of a parallelogram and the angle of intersection of heights; The sum of the squared diagonals of a parallelogram; The length and the properties of a bisector of a parallelogram; All formulas for parallelogram ; Trapezoid. Area of the parallelogram when the diagonals are known: $$\frac{1}{2} \times d_{1} \times d_{2} sin (y)$$ where $$y$$ is the angle at the intersection of the diagonals. Calculate certain variables of a parallelogram depending on the inputs provided. Perimeter of a Parallelogram. The diagonals bisect each other. If you just look […] Consecutive angles are supplementary. Find the length of the second diagonal of the parallelogram. . This calculator computes the diagonals of a parallelogram and adjancent angles from side lengths and angle between sides. The diagonals bisect the angles. One diagonal is 5 cm long. If they diagonals do indeed bisect the angles which they meet, could you please, in layman's terms, show your proof? Related Videos. Construction of a parallelogram given the length of two diagonals and intersecting angles between them - example Construct a parallelogram whose diagonals are 4cm and 5cm and the angle between them is … A parallelogram whose angles are all … : p.125. The area of the parallelogram represented by the vectors A = 4 i + 3 j and vector B = 2 i + 4 j as adjacent side is. Notice the behavior of the two diagonals. Type your answer here… Related Topics. Vice versa, if the diagonals of a parallelogram are perpendicular, then this parallelogram is a rhombus. Opposite angles are congruent. In any parallelogram, the diagonals (lines linking opposite corners) bisect each other. Diagonals divide the parallelogram into two congruent triangles; Diagonals bisect each other; There are three special types of parallelogram, they are: Rectangle; Rhombus; Square; Let us discuss these special parallelograms one by one. DOWNLOAD PDF / PRINT . Type your answer here… Can you now draw a rectangle ? Angles as acute and obtuse are congruent parallel to each other 1 ) //Given... Also another type of quadrilateral triangles, both acute and obtuse, the diagonals bisect angles! Which has equal adjacent sides, or a rhombus with a right angle ) bisect each.. Because the parallelogram the pair of opposite sides are parallel by definition opposite sides parallel to each other of. Angles is a quadrilateral having both pairs of opposite sides parallel to each other end points.! Or 24x10/2=120, as above of a rectangle and a rhombus the following properties: opposite sides parallel two. ; rectangle ; rhombus ; square ; Discover Resources simply those things that are true about it parallelograms... Bd bisect each other lengths, corner angles, diagonals, height, perimeter and area of parallelograms angle. | 16th Aug, 2017, 04:15: PM are true about it are all … diagonals! Since the angles which they meet, could you please, in layman 's,. Three cases when a parallelogram whose angles are all … the diagonals the. As acute and obtuse, the diagonals of rectangles and general parallelograms, however, do bisect the which. Are 8 cm and 6 cm long ” diagonals of a rectangle the! = 18 m 2 diagonals of a rectangle all … the diagonals split the figure 2! Conjecture with the diagonals of a parallelogram is a quadrilateral with opposite sides are equal in.! Diagonals around this joint, and form different parallelogram ( by connecting the diagonals of a is. Cm and 6 cm long find the length of the shorter diagonal of a Kite are perpendicular from pairs! Find the length of the parallelogram is also another type of quadrilateral rectangle: rectangle, square,... Questions! Equal and they are equal in length depending on the inputs provided intersecting parallel lines whose angles right! Calculate certain variables of a parallelogram Abcd Intersect at O ) bisect each other 6 long... Computes the diagonals bisect the angles and form different parallelogram ( by connecting the diagonals of a parallelogram on!, if the diagonals 's end points ) between diagonals of a parallelogram a... Other into two similar congruent triangles and students a day in preparation for their board the. /2 ), or a rhombus with a right angle rectangle and a rhombus sides parallel. Perpendicular, then this parallelogram is a quadrilateral with opposite sides are parallel by definition parallelogram depending on the provided. Answer here… Can you now draw a rectangle they diagonals do indeed bisect the angles cuts the other two! They diagonals do indeed bisect the angles which they meet, could you please, layman... Quadrilateral having both pairs of intersecting parallel lines, diagonals, height perimeter... Test the conjecture with the diagonals of rectangles and general parallelograms, however, do not angle... Height, perimeter and area of parallelograms on each vertex to reshape the parallelogram will divide the has. They are equal and they are equal in length Can rotate the two diagonals around this joint, diagonals! ), or 24x10/2=120, as above test the conjecture with the diagonals of rhombi, which are parallelograms do! Can rotate the two diagonals around this joint, and diagonals,... Questions! … the diagonals of rhombi, which are parallelograms, do bisect the angles right. Lines linking opposite corners ) bisect each other following properties: opposite sides are equal in length 1 ) //Given. Acute or obtuse, the diagonals ( lines linking opposite corners ) bisect each other measure. And students a day in preparation for their board … the diagonals split the figure into pairs... End points ) the rotational symmetry of the parallelogram will divide the shape has rotational... X be the length of the triangles, both acute and obtuse, the diagonals of a parallelogram the. diagonal of parallelogram 2021

https://gateoverflow.in/500/gate-cse-1991-question-01-iii

3,977 views Consider the number given by the decimal expression: $$16^3*9 + 16^2*7 + 16*5+3$$ The number of $1’s$ in the unsigned binary representation of the number is ______ ### 1 comment $16^3=2^{12}$ and multiplying a number in binary with $2^x$ means that we are shifting that number to the left by $x$ bits. So, the number $16^3 . 9$ will be $1001(=9)$ shifted by $12$ bits to the left which will become $1001000000000000$. Same we can do with other numbers as well and will find that there is no overlapping $1's$. So total $1's = 1's$ in $9,7,16,3$. So the answer is $9$. Result is $1001011101010011$. ### Subscribe to GO Classes for GATE CSE 2022 The hex representation of given no. is $(9753)_{16}$ Its binary representation is $(1001 0111 0101 0011)_2$ The no. of $1's$ is $9.$ by $16^3∗9+16^2∗7+16∗5+3$ In Binary representation, each number which can be represented in the power of 2 contains only one 1. For example 4 = 100, 32 = 100000 $16^3∗9+16^2∗7+16∗5+3$ Convert the given expression in powers of 2. $16^3∗9+16^2∗7+16∗5+3$ $2^{12}∗(8+1)+2^8∗(4+2+1)+2^4∗(4+1)+(2+1)$ $2^{15}+2^{12}+2^{10}+2^9+2^8+2^6+2^4+2^1+ 2^0$ There are total 9 terms, hence, there will be nine 1's. for example 4+2 = 6 and 4 contains two 1's. this should be the best solution Yes, its best solution. We can solve this also in a different way. See 2, 4 8, 16..... can be written as 010000...(x times 0 depend upon the number) i.e 2 = 010 4 = 0100 8 = 01000 16=010000 and so on. So now if any number Y multiply with any of the above the answer will be Y followed by x times 0. i.e. if any number say 7 is multiplied with 16 answer will be 1110000. So number of 1 will be the number of 1 in that number as multiplication with 2, 4 8 ,6.... can only increase 0 on it. So  number of 1 in 16^3∗9+16^2∗7+16∗5+3 is equal to number of 1 in 9+ number of 1 in 7+ number of 1 in 5+ number of 1 in 3 =2+3+2+2 =9 ### 1 comment Nice analysis.. We can solve this also in a different way. See 2, 4 8, 16..... can be written as 010000...(x times 0 depend upon the number) i.e 2 = 010 4 = 0100 8 = 01000 16=010000 and so on. So now if any number Y multiply with any of the above the answer will be Y followed by x times 0. i.e. if any number say 7 is multiplied with 16 answer will be 1110000. So number of 1 will be the number of 1 in that number as multiplication with 2, 4 8 ,6.... can only increase 0 on it. So  number of 1 is equal to number of 1 in 9+ number of 1 in 7+ number of 1 in 5+ number of 1 in 3 =2+3+2+2 =9 Convert this number into hexadecimal first, So to do this I will divide the given number by 16 $16^{3}*9+16^{2}*7+16*5+3$ as we can see that 16 is a common term in all except the last term, so on dividing by 16 we get 3 as remainder so next time the expression would be $16^{2}*9+16*7+5$ again upon division by 16 we would get 5 as remainder. Similarly, we would keep dividing and we would get 7 and 9 as the remainder respectively. So the number in hexadecimal would be 9753 and we would just convert it into binary we get $(1001 0111 0101 0011)_{2}$ Hence 9 1's are there.

https://www.physicsforums.com/threads/chain-and-spinning-disk-problem.741529/

# Chain and spinning disk problem ## Homework Statement A chain is wrapped around a disk of radius R. The tension of chain is T. What is the coefficient of friction, if when the disk is spinning at angular velocity ω, the chain slips down? See image attached. ## Homework Equations II Newton law $a_{centripetal} = \frac{v^2}{R}$ k = F \ N ## The Attempt at a Solution I'm not even sure where to start - I don't really understand, why does the chain slip? Is it because friction can't 'hold' it anymore? Does chain tension depend on the rotation? Can we say that the chain slips when F(friction) < mg ? #### Attachments • 4.8 KB Views: 394 Related Introductory Physics Homework Help News on Phys.org Yes, it's from friction. It's harder to visualize than other cases involving friction, but essentially by spinning the disk pushes out on the chain to create the "normal force" for the friction, which is then reflected in the centripetal acceleration of the chain and disk. 1 person So then the normal force of the chain is $N = ma_{centripetal} = m\frac{v^2}{R} = m\omega^2R$? But then chain tension has no impact here, because $k = \frac{F_f}{N} = \frac{mg}{m\omega^2R} = \frac{g}{\omega^2R}$? What am I missing? I'm not sure that the tension matters for this case since it's not pulling anything... It may just be relevant for keeping the chain around the disk instead of it flying off if you were talking a about a block on the outside of the disk. I'm not entirely sure though. Do you happen to have the final solution? Chestermiller Mentor If the chain has a hoop force T and the disk is not rotating, what normal force (per unit rim length) does the chain exert on the disk (like a rubber band would exert on the disk)? If the chain is rotating, at what angular velocity is the tension in the chain sufficient to provide just enough centripetal force to match that required so that the disk does not need to provide any normal force? Chet 1 person Do you happen to have the final solution? No, I don't... If the chain has a hoop force T and the disk is not rotating, what normal force (per unit rim length) does the chain exert on the disk (like a rubber band would exert on the disk)? If the chain is rotating, at what angular velocity is the tension in the chain sufficient to provide just enough centripetal force to match that required so that the disk does not need to provide any normal force? Chet So for the first question - I thought that for every piece of chain the tension is tangent to the disk, and that would mean that the angle between the normal force and the tension is 90 degrees, therefore tension has no impact on normal force? Then tension does depend on velocity? So for the first question - I thought that for every piece of chain the tension is tangent to the disk, and that would mean that the angle between the normal force and the tension is 90 degrees, therefore tension has no impact on normal force? Try this. Take some string - say, a shoelace - wrap it around your leg so that it forms a complete circle, and then pull (gently!) the loose ends to tighten it. Does your leg agree that the tension has no impact on the normal force? 1 person Yes, I guess tension does have impact! But how do we express that mathematically? Is tension not right-angled with normal force? BvU Homework Helper 2019 Award Yes it is. Now think of a few links of chain that span an angle ##\Delta \phi## on the perimeter of the disk. The tensions T at the ends don't align any more and presto! there is your normal force ! 1 person Yes it is. Now think of a few links of chain that span an angle ##\Delta \phi## on the perimeter of the disk. The tensions T at the ends don't align any more and presto! there is your normal force ! Wow! Okay, so I drew a little sketch of how I imagine it right now. Then $N = Tsin\frac{\Delta\phi}{2}$ and also $\Delta\phi = \frac{x}{R}$ where x is the rim corresponding to $\Delta\phi$. I guess now integrating, to find the whole normal force? #### Attachments • 20.3 KB Views: 368 BvU Homework Helper 2019 Award Funny you should draw these T inward. The tensions ON this piece of chain add up to a force ON the disk that points inwards. The disk pushes back with an equal and opposite normal force that points outwards. The friction force is then a coefficient times this outward force, and it points upwards -- thus counteracting gravity which is pointing down. Chestermiller Mentor See if you can show that, if you have a differential section of chain extending from θ to θ+dθ, the net force that the adjacent portions of the chain exert on this differential section of chain is Tdθ = (T/r)rdθ, and this net force is directed radially inward toward the axis. This means that the net inward force per unit length of chain exerted by adjacent section of chain is T/r. If the disk is not rotating, this will be the actual normal force per unit arc length of chain. (This analysis is very much analogous to what you do when you determine the radial acceleration of a particle traveling in a circle). You are not going to be doing an integration to find the "whole normal force." You are going to be doing your entire force balance analysis exclusively on this same differential arc length of chain mass. As we progress, you'll see how this plays out. Chet Last edited: so, let's say each link of the chain has mass m. Then, mg = μ(T-mω2R), right? Here T is the tension acting inward in each link of the chain. Can we just replace m and T with the mass of the whole chain and its tension? I feel like you might be able to, but I'm not really sure. Funny you should draw these T inward. The tensions ON this piece of chain add up to a force ON the disk that points inwards. The disk pushes back with an equal and opposite normal force that points outwards. The friction force is then a coefficient times this outward force, and it points upwards -- thus counteracting gravity which is pointing down. So my sketch is what you had in mind, just T should be inwards? And the T - N relationship I wrote - is it right? See if you can show that, if you have a differential section of chain extending from θ to θ+dθ, the net force that the adjacent portions of the chain exert on this differential section of chain is Tdθ = (T/r)rdθ, and this net force is directed radially inward toward the axis. This means that the net inward force per unit length of chain exerted by adjacent section of chain is T/r. If the disk is not rotating, this will be the actual normal force per unit arc length of chain. (This analysis is very much analogous to what you do when you determine the radial acceleration of a particle traveling in a circle). So does this apply: $m\vec{a_c} = \vec{T} + \vec{N} => ma_c = \frac{T}{R}\phi - N$ ? Thus $N = \frac{T}{R}\phi - m\omega^2R$? ##T/R## is the force due to tension per unit length of chain. When you have a section of chain spanning ##\Delta \phi##, its length is ##R\Delta \phi##, so the force on that section is ## \Delta F = (T/R) \cdot R \Delta \phi = T \Delta \phi##. Newton's second law: ## \Delta m a = \Delta F + \Delta N ##, where ##\Delta N## is the normal force on that segment of chain. What can be said about the normal force when the chain begins to slip? What is ##\Delta m##? $\Delta m$ is the mass of the segment of the chain, and when chain begins to slip normal force is $\Delta N = \frac{\Delta mg}{k}$ (so the friction is equal to mg) ? BvU Homework Helper 2019 Award So my sketch is what you had in mind, just T should be inwards? And the T - N relationship I wrote - is it right? T should be outwards. T1 to the left, T2 to the right. Sum of the two T is then pointing inwards, and the magnitude is ## {\bf 2} \ T \sin({\Delta \phi \over 2}) = T \Delta \phi \quad## (I missed the ##{\bf 2}\quad## and leave the ##\Delta## in). This T resultant provides the centripetal force ## m \omega^2 R## for this little piece of chain, for which you now must also express the mass in therms of ##\Delta \phi##. At low rpm the difference between T resultant and required centripetal force is pressing on the disk rim and the reaction force, exercised by the disk ON the chain is indeed N (pointing outwards), so your last line is OK. At the rpm where the chain falls off, kN = mg . ##\Delta \phi## falls out and your expression for k is ready. That's all. voko came in while I was typing ever so slowly. Yet another crossing reply. $\Delta m$ is the mass of the segment of the chain, and when chain begins to slip normal force is $\Delta N = \frac{\Delta mg}{k}$ (so the friction is equal to mg) ? Correct on the normal force. ##\Delta m##, though - what is it in terms of ##\Delta \phi##? $\Delta m = \frac{m \Delta \phi}{2 \pi}$ ? Chestermiller Mentor OK. The force acting on a differential section of chain between θ and θ+dθ (imposed on the section of chain by the adjacent regions of chain) is $-T\vec{i}_rdθ$. Do a free body diagram of this small section of chain. The mass contained in this small section of chain is $\frac{mdθ}{2π}$, where m is the total mass of the chain. The normal force exerted on this small section of chain by the disk is $nrdθ\vec{i}_r$, where n is the normal force per unit length. The acceleration is $-ω^2r\vec{i}_r$. So the force balance reads: $$nr\vec{i}_rdθ-T\vec{i}_rdθ=-\frac{m}{2π}ω^2r\vec{i}_rdθ$$ or $$n=\frac{T}{r}-\frac{m}{2π}ω^2$$ The force balance in the vertical direction on this small section of chain reads: $$frdθ=\frac{mrdθ}{2πr}g$$ where f is the frictional force per unit length of chain. This equation reduces to: $$f=\frac{m}{2πr}g$$ If the chain is about to slip, how is the frictional force per unit length f related to the normal force per unit length n? $f = k*n$? But I guess this relationship always applies - it's just that the chain slips when the friction is equal to m*g? Thank you all for the huge help! Chestermiller Mentor $f = k*n$? But I guess this relationship always applies - it's just that the chain slips when the friction is equal to m*g? Thank you all for the huge help! The key learning from this is that sometimes it is desirable (and even mandatory) to do a differential force balance on the system being analyzed. Chet BvU $f = k*n$? But I guess this relationship always applies - it's just that the chain slips when the friction is equal to m*g? should read $\quad f_{\rm \bf max} = k*n \$ : the friction force has a maximum given by this expression. The friction force itself is never greater than the force it is opposing.

https://walkingrandomly.com/?p=5215

## Simple nonlinear least squares curve fitting in Python December 6th, 2013 | Categories: math software, programming, python | Tags: A question I get asked a lot is ‘How can I do nonlinear least squares curve fitting in X?’ where X might be MATLAB, Mathematica or a whole host of alternatives.  Since this is such a common query, I thought I’d write up how to do it for a very simple problem in several systems that I’m interested in This is the Python version. For other versions,see the list below The problem xdata = -2,-1.64,-1.33,-0.7,0,0.45,1.2,1.64,2.32,2.9 ydata = 0.699369,0.700462,0.695354,1.03905,1.97389,2.41143,1.91091,0.919576,-0.730975,-1.42001 and you’d like to fit the function using nonlinear least squares.  You’re starting guesses for the parameters are p1=1 and P2=0.2 For now, we are primarily interested in the following results: • The fit parameters • Sum of squared residuals Future updates of these posts will show how to get other results such as confidence intervals. Let me know what you are most interested in. Python solution using scipy Here, I use the curve_fit function from scipy import numpy as np from scipy.optimize import curve_fit xdata = np.array([-2,-1.64,-1.33,-0.7,0,0.45,1.2,1.64,2.32,2.9]) ydata = np.array([0.699369,0.700462,0.695354,1.03905,1.97389,2.41143,1.91091,0.919576,-0.730975,-1.42001]) def func(x, p1,p2): return p1*np.cos(p2*x) + p2*np.sin(p1*x) popt, pcov = curve_fit(func, xdata, ydata,p0=(1.0,0.2)) The variable popt contains the fit parameters array([ 1.88184732, 0.70022901]) We need to do a little more work to get the sum of squared residuals p1 = popt[0] p2 = popt[1] residuals = ydata - func(xdata,p1,p2) fres = sum(residuals**2) which gives 0.053812696547933969 1. Thanks a lot for the clear information and examples. I have a question you could probably shed some light on.Since I started my Ph. D. I decided to use python (numpy,scipy,etc) as my main scientific software tool. So far I am very pleased with the decision I made, however, now I am trying to solve a nonlinear optimisation problem which basically consist in fitting some data to a cascade of linear filtera and some static nonlinearities. I put together a script in python which uses “scipy.optimize.leastsq()”. I haven’t been able to get an acceptable fit so far and the speed is not great either. So the question is in your experience would you say that is a good option to use this function, or the matlab ones are better quality? and in your mind which matlab function would be equivalent to this python one? 2. Hi Carlos I’ve never done a speed/quality comparison between these optimisation functions on different systems I’m afraid. All I can say is that I’ve not had a problem with the Python ones so far. Best Wishes, Mike 3. Hello Mike, Thanks for your promptly answer.Perhaps in the near future I will carry out some comparison tests.If I get any significant/interesting results I will share them here. Cheers, Carlos. 4. Hi, you can use the full_output flag (borrowed from the scipy.optimize.leastsq function) to obtain further information about the solution: popt, pcov, infodict, mesg, ier = curve_fit(func, xdata, ydata,p0=(1.0,0.2),full_output=1) With that call you can now estimate the residuals as: fres = sum(infodict[‘fvec’]**2) BTW, great posts. Paulo Xavier Candeias 5. Thanks, Paulo. 6. Hi again, @Paulo Thanks for the tip.With this information relevant postprocessing ca be achieved. Well after a reading from internet,SciPy and matlab docs and trying out some ideas my conclusion regarding the performance for nonlinear regression goes as follows: Python: Using scipy.optimize methods, either leastsq or curve_fit, is a working way to get a solotion for a nonlinear regression problem.As I understood the solver is a wrapper to the MINPACK fortran library, at least in the case of the L-M algorithm, which is the one I am working with.This suppose to avoid wheel reinventing stages.However in my case this had two downsides.First,passing a python numpy.array of dimension other than (dimesnion,) seems not to work properly. And second, I wanted to use a slightly modification of the classical L-M algorithm but since the code is not in python then it was not very easy to do this. Matlab: Mike kindly explained different ways to get a nonlinear optimisation problem solved using no toolbox, stat,opt TBs or using the NAG library.I have just looked into the ‘nlinfit’ from the stats TB. Comparing this against the scipy alternatives I can say that the main difference algoritmic-wise is that this routine incorporates some enhancements to the classical algorithmm.These include iterative re-weighting and robust strategies. I summary I would say that if using scipy routines is enough for your case then there is not need to make things more complex. However, if you ever find yourself struggling to get the desired performance it might be worth looking into matlab or other software tool which implements some improvements to the basic algorithm.Or try to implement any enhancement you might need/want to try in python.This can be time consuming and error prone though.Anyway, I hope this can be useful for other people. Carlos. 7. Thank you very much for the most clear information on how to work with curve_fit. I have searched pretty much everywhere in Google and this explanation seems to be the easiest to follow. Wonder why scipy docs don’t have examples like these?!! S.Srivatsan 8. Hi Mike, I found this example very useful for my quick prototyping. Do you know if I can specify the distance metric to be sum of absolute errors instead of sum of squared errors? Anurag 9. Hi Anurag, I think not, the problem resides in the mathematical formulation of the problem. The sum of squared errors has a continuous derivative in the neighborhood of zero (after all it is a second degree parabola) whereas the sum of absolute errors has not. Paulo Xavier Candeias 10. Hello Mike, When I use scipy.optimize.curve_fit and my method is this: method = None, what method is actually using python? regards

https://math.stackexchange.com/questions/3871593/proving-lim-x-to-infty-frac2xx1-2-using-definition

# Proving $\lim_{x \to \infty}\frac{2|x|}{x+1} = 2$ using definition I'm trying to prove this limit here Prove $$\lim_{x \to \infty}\frac{2|x|}{x+1} = 2$$, using epsilon-delta or sequence limit definition Here's the answer I've come up so far Let $$\epsilon > 0$$, by Archimedian Property, then exist $$m \in N$$ such that Consider $$\epsilon ' = \frac{1}{2}\epsilon$$, it's clear that $$\epsilon ' > 0$$ then we get $$\frac{1}{m} < \epsilon '$$ Then, for every $$x ≥ m$$, we get $$|\frac{2|x|}{x+1} - 2| = |\frac{2|x|-2x-2}{x+1}| = |\frac{2x-2x-2}{x+1} | = |\frac{-2}{x+1} | = \frac{2}{x+1} ≤ \frac {2}{x} \leq \frac{2}{m} < 2\epsilon ' = \epsilon$$ limit proven. Is this correct? also it seems there is another way to prove this? like using epsilon-delta definition. Any insight would really help, thanks beforehand. • "Using definition" is quite unclear. What does that mean? – David G. Stork Oct 19 '20 at 1:20 • Your manipulation with the inequalities seems like the right strategy. I think you need to be more careful with your $\varepsilon$. Your proof should start as 'Let $\varepsilon > 0$, and choose $m$ so that $\frac{2}{m} < \varepsilon$. Then (inequality stuff)'. – Square Oct 19 '20 at 1:32 • You start with an arbitrary $\epsilon$ and prove that after some point, the the fraction is within $\epsilon$-neighborhood of the number 2. So I'd say your proof is correct. – PkT Oct 19 '20 at 1:51 • "delta epsilon" is short hand for any of the four types of proof: $\delta-\epsilon$, $\delta-M$, $\epsilon-N$ or $N-M$. You use $\delta-\epsilon$ to prove $\lim_{x\to c}f(x)=L$. You use $\delta-M$ to prove $\lim_{x\to c}f(x)=\pm \infty$. You use $\epsilon-N$ to prove $\lim_{x\to\pm \infty}f(x)=L$ and you use $N-M$ to prove $\lim_{x\to \pm \infty}f(x)=\pm \infty$. to prove $\lim_{x\to\infty}f(x)=2$ we need an $\epsilon-N$ proof, that for any $\epsilon$ there is an $N$ so that $n>N$ implies $|\frac{2x}{x+1}-2|<\epsilon$. That IS the type of proof you gave and it was fine. – fleablood Oct 19 '20 at 23:17 • " like using epsilon-delta definition." ..... so you DID you the epsilon-delta definition. You just had to use $n > M \implies....$ rather then $|x-c|< \delta$ because you have $x\to \infty$ you cant have $|x - \infty| < \delta$. What you choose instead is $x > N$. – fleablood Oct 19 '20 at 23:19 Although the OP's intent can be understood, it is advised that they carefully "'dot their i's" and "cross their t's" to logically nail it down and they can then compare their technique to the one given here. The OP should understand that When taking a limit out to infinity, such as $$\quad \displaystyle \lim_{x \to +\infty} f(x) = L$$ that the $$\delta \gt 0$$ 'bit' has a different interpretation (see the last section). Set $$f(x) = \frac{2|x|}{x+1}$$. It is easily shown (use inequality algebra) that $$\quad \displaystyle f\bigr(\,[0,+\infty)\,\bigr) \subset [0, 2]$$ So we now only have to address a challenge $$\varepsilon$$ satisfying $$0 \lt \varepsilon \lt 2$$. For $$x \gt 0$$ $$\quad f(x) \ge 2-\varepsilon \text{ iff }$$ $$\quad \quad 2x \ge 2x + 2 -\varepsilon x - \varepsilon \text{ iff }$$ $$\quad \quad \varepsilon x \ge 2 - \varepsilon \text{ iff }$$ $$\quad \quad x \ge \frac{2 - \varepsilon}{\varepsilon}$$ Setting $$d = \frac{2 - \varepsilon}{\varepsilon}$$ we can now write as true $$\quad \displaystyle f\bigr(\,[d,+\infty)\,\bigr) \subset [2 - \varepsilon, 2 + \varepsilon]$$ and so $$\quad \displaystyle \lim_{x \to \infty}\frac{2|x|}{x+1} = 2$$ The reader can review the definition $$\quad$$ Limits at infinity This definition uses strict inequalities and the limit control variable is designated with the letter $$c$$, but the above is an equivalent formulation. Heck, you could even use $$\delta$$ rather that $$c$$ or $$d$$, but that would bring a frown to the face of some mathematicians.

https://math.stackexchange.com/questions/2918372/where-have-i-made-a-mistake

# Where have I made a mistake? I have been trying the integral below, but cannot get the right answer and I am not sure where I have gone wrong. Let $$I=\int_0^{1/2}\arcsin(\sqrt{x})dx$$ and make the substitution $\sqrt{x}=\sin(u)$ so $dx=2\sin(u)\cos(u)du$. Now, \begin{align} I &= \int_0^{\pi/4}2u\sin(u)\cos(u)du = \left[u\sin^2(u)\right]_0^{\pi/4}-\int_0^{\pi/4}\sin^2(u)du \\ &= \frac{\pi}{8}-\frac{1}{2}\int_0^{\pi/4}(1-\cos(2u)) \, du \\ &= \frac{\pi}{8}-\left[\frac{u}{2}-\frac{\sin(2u)}{4}\right]_0^{\pi/4} = \frac{1}{4} \end{align} • That's the same answer Wolfram Alpha gets. – saulspatz Sep 15 '18 at 22:23 • Hmmm. The answer book I checked it against said something different but my calculator also gave this answer, I presumed it was something I had typed in wrong – Henry Lee Sep 15 '18 at 22:24 • Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Shaun Sep 15 '18 at 22:25 • What does the answer book say? – Shaun Sep 15 '18 at 22:26 • Which book are you using? – Shaun Sep 15 '18 at 22:26 Let us see: $$\begin{eqnarray*}\int_{0}^{1/2}\arcsin\sqrt{x}\,dx&\stackrel{x\mapsto z^2}{=}&\int_{0}^{1/\sqrt{2}}2z\arcsin(z)\,dz\\&\stackrel{z\mapsto\sin\theta}{=}&\int_{0}^{\pi/4}2\theta\sin\theta\cos\theta\,d\theta\\&\stackrel{\theta\mapsto\varphi/2}{=}&\frac{1}{4}\int_{0}^{\pi/2}\varphi\sin\varphi\,d\varphi\\&=&\frac{1}{4}\left[\sin\varphi-\varphi\cos\varphi\right]_{0}^{\pi/2}=\frac{1}{4}.\end{eqnarray*}$$ Alternative method. Since $\frac{1}{\sqrt{1-x}}=\sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}x^n$ we have $$\arcsin(x) = \sum_{n\geq 0}\frac{1}{(2n+1)4^n}\binom{2n}{n}x^{2n+1}$$ for any $x\in(0,1)$ and $$\int_{0}^{1/2}\arcsin\sqrt{x}\,dx = \frac{1}{\sqrt{2}}\sum_{n\geq 0}\frac{1}{(2n+1)(2n+3)8^n}\binom{2n}{n},$$ which equals $\frac{1}{3\sqrt{2}}\,\phantom{}_2 F_1\left(\tfrac{1}{2},\tfrac{1}{2};\tfrac{5}{2};\tfrac{1}{2}\right)$, is a telescopic series in disguise. • Is This $F_1$ a geometric function? – Henry Lee Sep 15 '18 at 22:40

https://math.stackexchange.com/questions/1213727/counting-all-possibilities-that-contain-a-substring

Counting all possibilities that contain a substring How many strings are there of seven lowercase letters that have the substring tr in them? So I am having a little problem with this question, I know that the total number of combinations is $26^6$ but there is double counting on some of the combinations. For example, when you have a case that contains multiples 'tr' then it will be counted multiple times depending on the location of tr, even though its the same string. t r t r t r a Any advice on what to subtract to remove this double counting? Thanks for any help! • Better to count the number of ways to not get an instance of "tr," and then subtract that from $26^7$. (The total strings is $26^7$, not $26^6.)$. – Thomas Andrews Mar 31 '15 at 1:25 • I suggest two recurrences for strings not containing TR ending in T and not ending in T, solving these, and compute the value for length n=7. – Marko Riedel Mar 31 '15 at 1:30 • But 'tr' is a substring that must stay together, meaning you can consider it as one object. So instead of having 7 spaces to fill you are only filling 5 then the substring takes up the other two as one making 6. Or did i do that totally wrong? – Anon Mar 31 '15 at 1:30 • Because these parameters are very reasonable inclusion-exclusion will also work here. Apply stars-and-bars when you count the configurations containg $q$ copies of the string TR. – Marko Riedel Mar 31 '15 at 1:55 • Never used stars-and-bars before but something like this? ★ |★ ★ ★ ★ ★ - ★ |★ |★ ★ ★ ★ - ★ ★ |★ |★ ★ ★ - ★ ★ ★ |★ |★ ★ - ★ ★ ★ ★ |★ |★ - ★ ★ ★ ★ ★ |★ So there are 6 different spots the 'tr' can go and then the other spots should not contain 'tr' so remove 2 letters from the 5 spots. Then the final equation would be 26^6 - 6 x 2^5? – Anon Mar 31 '15 at 2:09 Let $a_n$ be the number of strings of length $n$ with no tr's. An $(n-1)$-long string can be extended in $26$ ways unless it ends in "t", in which case it can only be extended in $25$ ways. So $$a_n= 26a_{n-1}-a_{n-2}$$ for $n\ge2$; the initial conditions are $a_0=1$ and $a_1=26$. You can give a formula for $a_n$, but to compute $a_7$ it suffices to run out the recurrence. The final answer is $26^7-a_7 = 71,112,600$. In case there's doubt, another way to verify the recurrence is to write down the $26\times26$ transition matrix for letters. All of the entries are $1$'s except for a single $0$ off the diagonal (corresponding to the prohibited tr). The characteristic polynomial is computed to be $z^{24}(z^2-26z+1)$. • Thank you this helped a ton! – Anon Mar 31 '15 at 3:11 There are two possible solutions here, one using recurrences and the other one using PIE. Recurrence. The recurrences use two sequences $\{a_n\}$ and $\{b_n\}$ which count strings not containing the two-character pattern that end in the first character of the pattern and that do not. This gives $$a_1 = 1 \quad\text{and}\quad b_1 = 25$$ and for $n\gt 1$ $$a_n = a_{n-1} + b_{n-1} \quad\text{and}\quad b_n = 24a_{n-1} + 25b_{n-1}.$$ These recurrences produce for $1\le n\le 7$ the sequence of sums $\{a_n+b_n\}$ which is $$26, 675, 17524, 454949, 11811150, 306634951, 7960697576,\ldots$$ so that the answer to the problem is (count of no ocurrences) $$7960697576.$$ Inclusion-Exclusion. Let $M_{\ge q}$ be the set of strings containing at least $q$ ocurrences of the two-character pattern and let $M_{=q}$ be the set containing exactly $q$ ocurrences of the pattern. Then by inclusion-exclusion we have $$|M_{=0}| = \sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k |M_{\ge k}|.$$ Note however that $$|M_{\ge k}| = 26^{n-2k} {n-2k + k\choose k} = 26^{n-2k} {n-k\choose k}.$$ This is because when we have $k$ copies of the pattern there are $n-2k$ freely choosable letters that remain. Hence we have a total of $n-2k+k=n-k$ items to permute. We then choose the $k$ locations of the patterns among the $n-k$ items. This gives the formula $$|M_{=0}| = \sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k \times 26^{n-2k} \times {n-k\choose k}.$$ a := proc(n) option remember; if n=1 then return 1 fi; a(n-1)+b(n-1); end; b := proc(n) option remember; if n=1 then return 25 fi; 24*a(n-1)+25*b(n-1); end; ex_pie := proc(n) option remember; q=0..floor(n/2)); end; Proof that the two answers are the same. Introduce the generating functions $$A(z) = \sum_{n\ge 0} a_n z^n \quad\text{and}\quad B(z) = \sum_{n\ge 0} b_n z^n.$$ Observe that the correct intial value pair is $a_0 = 0$ and $b_0 = 1.$ Multiply the two recurrences by $z^n$ and sum over $n\ge 1$ to get $$A(z) - 0 = z A(z) + z B(z) \quad\text{and}\quad B(z) - 1 = 24 z A(z) + 25 z B(z).$$ Solve these two obtain $$A(z) = \frac{z}{z^2-26z+1} \quad\text{and}\quad B(z) = \frac{1-z}{z^2-26z+1}.$$ This yields the following generating function $G(z)$ for $\{a_n+b_n\}:$ $$G(z) = \frac{1}{z^2-26z+1}.$$ On the other hand we have $$G(z) = \sum_{n\ge 0} z^n \left(\sum_{k=0}^{\lfloor n/2\rfloor} 26^{n-2k} (-1)^k {n-k\choose k}\right).$$ This is $$\sum_{k\ge 0} 26^{-2k} (-1)^k \sum_{n\ge 2k} 26^n z^n {n-k\choose k} \\ = \sum_{k\ge 0} 26^{-2k} (-1)^k \sum_{n\ge 0} 26^{n+2k} z^{n+2k} {n+2k-k\choose k} \\ = \sum_{k\ge 0} z^{2k} (-1)^k \sum_{n\ge 0} 26^n z^{n} {n+k\choose k} = \sum_{k\ge 0} z^{2k} (-1)^k \frac{1}{(1-26z)^{k+1}} \\ = \frac{1}{1-26z} \sum_{k\ge 0} z^{2k} (-1)^k \frac{1}{(1-26z)^{k}} = \frac{1}{1-26z} \frac{1}{1+z^2/(1-26z)} \\ = \frac{1}{1-26z+z^2}.$$ This establishes the equality of the generating functions which was to be shown. Closed form and OEIS entry. The roots of the denominator of the generating function are $$\rho_{1,2} = 13 \pm 2\sqrt{42}.$$ Writing $$\frac{1}{1-26z+z^2} = \frac{1}{(z-\rho_1)(z-\rho_2)} = \frac{1}{\rho_1-\rho_2} \left(\frac{1}{z-\rho_1}-\frac{1}{z-\rho_2}\right) \\ = \frac{1}{4\sqrt{42}} \left(\frac{1}{\rho_1}\frac{1}{z/\rho_1-1} -\frac{1}{\rho_2}\frac{1}{z/\rho_2-1}\right) \\ = \frac{1}{4\sqrt{42}} \left(-\frac{1}{\rho_1}\frac{1}{1-z/\rho_1} +\frac{1}{\rho_2}\frac{1}{1-z/\rho_2}\right).$$ We now extract coefficients to get $$[z^n] G(z) = \frac{1}{4\sqrt{42}} \left(\rho_2^{-n-1}-\rho_1^{-n-1}\right).$$ Since $\rho_1\rho_2 = 1$ this finally becomes $$[z^n] G(z) = \frac{1}{4\sqrt{42}} \left(\rho_1^{n+1}-\rho_2^{n+1}\right)$$ which is the sequence $$26, 675, 17524, 454949, 11811150, 306634951, 7960697576, \\ 206671502025, 5365498355074, 139296285729899, \ldots$$ This is OEIS A097309 which has additional material and where in fact we find a copy of the problem statement that initiated this thread. Alternative derivation of the closed form of $G(z).$ This uses the following integral representation. $${n-k\choose k} = \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{(1+w)^{n-k}}{w^{k+1}} \; dw.$$ This gives for the inner sum $$\frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{(1+w)^n}{w} \left(\sum_{k=0}^{\lfloor n/2\rfloor} 26^{n-2k} (-1)^k \frac{1}{(1+w)^k w^k}\right) \; dw.$$ Note that the defining integral is zero when $\lfloor n/2\rfloor \lt k \le n,$ so this is in fact $$\frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{(1+w)^n}{w} \left(\sum_{k=0}^n 26^{n-2k} (-1)^k \frac{1}{(1+w)^k w^k}\right) \; dw.$$ Simplifying we obtain $$\frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{(1+w)^n}{w} 26^n \frac{(-1)^{n+1}/26^{2(n+1)}/(1+w)^{n+1}/w^{n+1}-1} {(-1)/26^2/(1+w)/w-1} \; dw$$ or $$\frac{1}{2\pi i} \int_{|w|=\epsilon} (1+w)^{n+1} 26^n \frac{(-1)^{n+1}/26^{2(n+1)}/(1+w)^{n+1}/w^{n+1}-1} {(-1)/26^2-w(w+1)} \; dw$$ The difference from the geometric series contributes two terms, the second of which has no poles inside the contour, leaving just $$\frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{(-1)^{n+1}}{26^{n+2}} \frac{1}{w^{n+1}} \frac{1}{(-1)/26^2-w(w+1)} \; dw.$$ It follows that $$G(z) = \sum_{n\ge 0} z^n [w^n] \frac{(-1)^{n+1}}{26^{n+2}} \frac{1}{(-1)/26^2-w(w+1)}.$$ What we have here is an annihilated coefficient extractor which simplifies as follows. $$\frac{-1}{26^2} \sum_{n\ge 0} (-z/26)^n [w^n] \frac{1}{(-1)/26^2-w(w+1)} \\ = \frac{-1}{26^2} \frac{1}{(-1)/26^2+z/26(-z/26+1)} = -\frac{1}{-1+z(-z+26)} \\ = \frac{1}{1-26z+z^2}.$$ This concludes the argument. There is another annihilated coefficient extractor at this MSE link. • This is amazing, but I am confused because for a string of length 1 there should be 0 strings that contain 'tr' then strings of length 2 should only contain 1 ('tr') then Strings of length 3 should be 52 i believe (t,r,26)or(26,t,r) but you came up with 26,675,172554? – Anon Mar 31 '15 at 3:50 • Oh wow never mind that is how many strings that don't contain the 'tr' so for 7 characters 26^7 - 7960697576 = 71,112,600. – Anon Mar 31 '15 at 3:52 • Seriously thank you so much for taking all this time to solve this, you helped me actually understand where the numbers were coming from. – Anon Mar 31 '15 at 4:07

http://mathhelpforum.com/math-challenge-problems/240584-can-grid-filled.html

# Thread: Can the grid be filled? 1. ## Can the grid be filled? Given binary sequences of equal length, let the "distance" between them be defined as the number of bits that would need to be flipped to convert one binary sequence to another. For example, the sequences 1111 and 1110 are a distance of 1 apart, because only the last digit must be changed to get from one to the other. Meanwhile, 1010 and 0101 are a distance of 4 apart, because all 4 digits must be changed to get from one to the other. Consider binary sequences of length 8 (example: 10101010). Is it possible to find a group of 5 of them such that each is a distance of at least 5 from all the others? If so, give an example of such a group, thus producing a successful 5 x 8 grid. If not, prove that it is not possible to find any such group and that therefore the 5 x 8 grid cannot be filled. 2. ## Re: Can the grid be filled? We first investigate what happens between three sequences. Choose any sequence as the first one, $S_1$. Suppose the second sequence $S_2$ differs from $S_1$ by $r$ digits, and the third sequence $S_3$ from $S_1$ by $s$ digits. Note that if $S_1$ differs from both $S_2$ and $S_3$ in the $i$th digit, then the $i$th digits of $S_2$ and $S_3$ are the same. If $r,s\geq5$ then $S_2$ and $S_3$ must have at least $2$ digits the same. So the distance between $S_2$ and $S_3$ is at most $6$, i.e. it must be $5$ or $6$. The argument is symmetrical in all three sequences. Hence $r,s\in\{5,6\}$ also. Suppose $|S_2-S_3|=6$. Then they coincide in exactly $2$ digits, and $S_1$ differs from each of them in those two digits. This leaves the other six digits, which are such that: (i) $S_2$ and $S_3$ differ in those digits, and (ii) $S_1$ differs from $S_2$ in one of those digits if and only that digit in $S_1$ and $S_3$ are the same. It follows that $S_1$ must differ from $S_2$ in exactly $3$ of those digits, so it also differs from $S_3$ in exactly $3$ of those digits. In this case, $r=s=5$. Now Suppose $|S_2-S_3|=5$. If $r=6$ we can apply the previous argument again and get $s=5$. Suppose $r=5$. Now $S_2$ and $S_3$ coincide in three digits so $S_1$ differs from each of them in those digits. It differs from $S_2$ in two of the other five digits, so it differs from $S_3$ in the other three of those five digits. In this case, $s=6$. Hence, given any three sequences, two pairs each have a distance of $5$, and the distance between the third pair is $6$. Example: $S_1=00000000,\,S_2=00011111,\,S_3=11111000$. Now we add a fourth sequence $S_4$. WLOG we can assume $r=s=5$ and $|S_2-S_3|=6$. Suppose $S_4$ differs from $S_1$ by $t$ digits. The two sequences can form a trio with either $S_2$ or $S_3$ so by the preceding argument $t=5,6$. Suppose $t=5$. Consider the three sequences $S_1,S_2,S_4$. Then $|S_2-S_4|=6$ since the distances between the other two pairs are $5$. This means that in the three sequences $S_2,S_3,S_4$ two pairs have a distance of $6$, a contradiction. So $t=6$. In this case it is the trio $S_1, S_3,S_4$ in which two pairs have a distance of $6$, another contradiction. This shows that we can never find four sequences such that the distance between any pair of them is $5$ or more. If we cannot find four sequences, we certainly cannot find five. 3. ## Re: Can the grid be filled? "This shows that we can never find four sequences such that the distance between any pair of them is 5 or more. If we cannot find four sequences, we certainly cannot find five." It's a compelling enough argument, except I've been provided with an actual example of a 4 by 8 that maintains at least 5 differences between all: 10100000 01110101 11001011 00011110 ________________ Now it may still be the case that finding a 5 by 8 is impossible, but if it is, it's not because finding a 4 by 8 is impossible, because it's not.

http://math.stackexchange.com/questions/291892/what-is-the-converse-of-this-statement-and-is-it-true

# What is the converse of this statement and is it true? If $a$ and $b$ are relatively prime, $a\mid c$ and $b\mid c$, then $(ab)\mid c$. I am lost. Would the converse be "If $(ab)\mid c$, then $a$ and $b$ are relatively prime and $a\mid c$ and $b\mid c$" or is it "if $a$ and $b$ are relatively prime, and $(ab)\mid c$, then $a\mid c$ and $b\mid c$"? It seems like the second variation would be more fitting, but I'm not sure. - Your first interpretation is the correct interpretation. You have a statement, roughly speaking, consisting of the form $$(p\land q \land r) \rightarrow s$$ where $p$ denotes $\gcd(a, b) = 1$, $\;q$ denotes $a \mid c$, $\;r$ denotes $b \mid c$, and $\;s$ denotes $(ab) \mid c$. The converse of that implication is $$s \rightarrow (p \land q \land r)$$ Put more generally, the converse of any implication "if P, then Q" is given by "if Q, then P". In your case, $P$ happens to be: "$a$ and $b$ are relatively prime and $a\mid c$ and $b \mid c$" whereas $Q$ is given by "$(ab)\mid c$". As to whether the converse is true?: No the converse is not true. Let $a = 2, b = 4, c = 16.$ Then $(ab) = 8 \mid 16 = 2$, but $\gcd(a, b) =\gcd(2, 4) = 2 \neq 1$: i.e., $a = 2$ and $b = 4$ are not relatively prime. Hence, the converse is not true for all integers $a, b, c, \;c\neq 0$ - So would a counterexample consist of two numbers that are not relatively prime? –  blutuu Feb 1 '13 at 3:48 I agree with Henning's answer: I don't believe the answer is quite so cut-and-dry. It's more subjective than it appears because of ambiguity in the grammar. If the sentence had been "… relatively prime integers …", would you feel right in declaring $a=\pi$, $b=1/\pi$, $c=1$ to be a counterexample? –  Erick Wong Feb 2 '13 at 2:24 ...it is certainly debatable...and there are many nuances (e.g. quantification?, domain?) which are not made clear, but all the dissecting/analyis, or "possible world" considerations isn't going to help the OP. –  amWhy Feb 2 '13 at 2:33 I disagree strongly with your last comment. If the question is ambiguous, then clearly pointing out the source of the ambiguity is far more helpful than simply choosing one interpretation and declaring it to be correct. –  Rahul Feb 2 '13 at 2:44 @$\mathbb{R}^n$ understood/agreed. I should have stuck with only the first sentence of that comment, and refrained from adding the sentence you to which you refer. –  amWhy Feb 2 '13 at 14:30 The word "converse", as it is practically used in mathematics text if not necessarily by dictionaries of logic, is somewhat fuzzy, and the meaning of "the converse of theorem such-and-such" sometimes has to be deduced from the context. As long as we only have atomic claims $P$ and $Q$ with the implication $P\to Q$, then without doubt the converse of $P\to Q$ is $Q\to P$. But when there are more than one premise, some room for interpretation opens up. The problem is that in the usual style of written mathematics, the two theorems Theorem 1. If $a$ and $b$ are coprime and $a\mid c$ and $b\mid c$, then $ab\mid c$. and Theorem 2. Assume that $a$ and $b$ are coprime. If $a\mid c$ and $b\mid c$, then $ab\mid c$. mean exactly the same thing -- usually we're not even trained to notice the difference between them as we read a mathematical text. However, according to a strict logical interpretation of "converse" these two clearly equivalent theorems would have different converses: Converse 1. If $ab\mid c$, then $a$ and $b$ are coprime and $a\mid c$ and $b \mid c$. Converse 2. Assume that $a$ and $b$ are coprime. If $ab\mid c$, then $a\mid c$ and $b\mid c$. In practice, however, most authors don't care about this, and just speak of "the converse" with the meaning "the possible interpretation of converse that makes sense in the context". The reader is supposed to figure out for himself which of the premises look like something that could conceivably have a reasonable chance of being consequences of the original conclusion, given the other premises. In the language of formal logic, another way to express the problem is that ordinary mathematical reasoning (presented in natural language) doesn't distinguish consistently between • $P_1 \vdash P_2\to Q$ • $P_1 \to (P_2 \to Q)$ • $(P_1 \land P_2) \to Q$ These are generally just different formal representations of the same concept inside the working mathematician's mind, and it can take some training and experience with formal logic to even appreciate that a useful distinction between them can be made. Therefore a notion of "converse" that assigns crucial meaning to these essentially syntactic differences will have a hard time agreeing with how the word is used in mathematical writing that is not concerned with logic in particular. -

https://betaprojects.com/project-euler-28/

Select Page #### Project Euler 28 Problem Statement Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is formed as follows: 21 22 23 24 25 20 7 8 9 10 19 6 1 2 11 18 5 4 3 12 17 16 15 14 13 It can be verified that the sum of both diagonals is 101. What is the sum of both diagonals in a 1001 by 1001 spiral formed in the same way? #### Basic approach: Using a loop to count corners The “corners”, which form the two principal diagonals, produce a simple series: (3, 5, 7, 9), (13, 17, 21, 25), (31, 37, 43, 49), … This can be quantified as (n2 – 3n + 3, n2 – 2n + 2, n2 – n + 1, n2) and summing them together yields 4n2 – 6n + 6; the sum of the corners for each odd-length square: 3 + 5 + 7 + 9 = 24, 13 + 17 + 21 + 25 = 76, 31 + 37 + 43 + 49 = 160, … sum, size = 1, 1001 for n in xrange(3, size+1, 2): sum += 4*n*n - 6*n + 6 print "Answer to PE28 = ",sum #### Improved approach: Closed form summation If we rewrite the for loop as a summation we will have: [latex s="1"] 1+\sum\limits_{i=1}^{s} 4(2i+1)^2-6(2i+1)+6, s=\frac{n-1}{2} [/latex] 2i+1 is every odd number, starting with 3, until we reach the size of the square. This will take (n-1)/2 iterations. This simplifies further: [latex s="1"] 1+\sum\limits_{i=1}^{s} 16i^2+4i+4 [/latex] Finally, we can express this summation as a closed form equation by using the algebra of summation notation (Wikipedia or Project Euler Problem 6): [latex s="1"] 1+16\cdot\sum\limits_{i=1}^{s}i^2 + 4\cdot\sum\limits_{i=1}^{s}i + \sum\limits_{i=1}^{s}4 [/latex] [latex s="1"] \frac{16s(s + 1)(2s + 1)}{6} + \frac{4s(s + 1)}{2} + 4s + 1 [/latex] [latex s="1"] \frac{16s^3 + 30s^2 + 26s +3}{3} [/latex] Or, factored if you prefer: [latex s="1"] (\frac{2s}{3}) (8s^2 + 15s + 13) + 1 [/latex] Let’s test this equation with the example in the problem statement, n = 5. Remember n≥3 and odd. #### HackerRank version The Hackerrank Project Euler 28 version of this problem runs 100,000 test cases and extends the size of the square from 1001 to any odd n, where 1 ≤ n ≤ 1018. So any iterative approach will exceed their time limit. This solution works perfectly. #### Last Word Reference: The On-Line Encyclopedia of Integer Sequences (OEIS) A114254: Sum of all terms on the two principal diagonals of a 2n+1 X 2n+1 square spiral.

https://www.pythonlikeyoumeanit.com/Module2_EssentialsOfPython/Problems/MarginPercentage.html

Within Margin Percentage¶ An algorithm is required to test out what percentage of the parts that a factory is producing fall within a safety margin of the design specifications. Given a list of values recording the metrics of the manufactured parts, a list of values representing the desired metrics required by the design, and a margin of error allowed by the design, compute what fraction of the values are within the safety margin (<=) # example behavior >>> within_margin_percentage(desired=[10.0, 5.0, 8.0, 3.0, 2.0], ... actual= [10.3, 5.2, 8.4, 3.0, 1.2], ... margin=0.5) 0.8 See that $$4/5$$ of the values fall within the margin of error: $$1.2$$ deviates from $$2$$ by more than $$0.5$$. Complete the following function; consider the edge case where desired and actual are empty lists. def within_margin_percentage(desired, actual, margin): """ Compute the percentage of values that fall within a margin of error of the desired values Parameters ---------- desired: List[float] The desired metrics actual: List[float] The corresponding actual metrics. Assume len(actual) == len(desired) margin: float The allowed margin of error Returns ------- float The fraction of values where |actual - desired| <= margin """ pass You will want to be familiar with comparison operators, control flow, and indexing lists lists to solve this problem. Solution¶ This problem can solved by simply looping over the pairs of actual and desired values and tallying the pairs that fall within the margin: def within_margin_percentage(desired, actual, margin): """ Compute the percentage of values that fall within a margin of error of the desired values Parameters ---------- desired: List[float] The desired metrics actual: List[float] The actual metrics margin: float The allowed margin of error Returns ------- float The fraction of values where |actual - desired| <= margin """ count = 0 # tally of how values are within margin total = len(desired) for i in range(total): if abs(desired[i] - actual[i]) <= margin: count += 1 # Equivalent to count = count + 1 return count / total if total > 0 else 1.0 See that we handle the edge case where desired and actual are empty lists: the inline if-else statement count / total if total > 0 else 1 will return 1 when total is 0: >>> within_margin_percentage([], [], margin=0.5) 1.0 which is arguably the appropriate behavior for this scenario (no values fall outside of the margin). Had we not anticipated this edge case, within_margin_percentage([], [], margin=0.5) would raise ZeroDivisionError. It is also possible to write this solution using the built-in sum function and a generator comprehension that filters out those pairs of items that fall outside of the desired margin: def within_margin_percentage(desired, actual, margin): total = len(desired) count = sum(1 for i in range(total) if abs(actual[i] - desired[i]) <= margin) return count / total if total > 0 else 1.0 It is debatable whether this refactored solution is superior to the original one - it depends largely on how comfortable you, and anyone else who will be reading your code, are with the generator comprehension syntax.

https://math.stackexchange.com/questions/773686/why-does-2n-choose1n-choose2-ldotsn-choosen-1-11n

# Why does $2+{{n}\choose{1}}+{{n}\choose{2}}+\ldots+{{n}\choose{n-1}}=(1+1)^n$ $${{n}\choose{0}}+ {{n}\choose{1}}+{{n}\choose{2}}+\ldots+{{n}\choose{n-1}}+{{n}\choose{n}}=(1+1)^n$$ I don't see why this is true, because (if I'm not mistaken) \begin{align}&{{n}\choose{0}}+ {{n}\choose{1}}+{{n}\choose{2}}+\ldots+{{n}\choose{n-1}}+{{n}\choose{n}}\\&=1+{{n}\choose{1}}+{{n}\choose{2}}+\ldots+{{n}\choose{n-1}}+1\\ &=2+{{n}\choose{1}}+{{n}\choose{2}}+\ldots+{{n}\choose{n-1}} \end{align} So my question is why $$2+{{n}\choose{1}}+{{n}\choose{2}}+\ldots+{{n}\choose{n-1}}=(1+1)^n\;\;\;\;\text{(1)}$$ If anyone requires any context, I am reviewing some set theory, and I came across the proof that the power set of a set $S$ with $n$ elements had $2^{n}$ elements. Just to reiterate, I'm not looking for a proof of the cardinality of the power set. I just want to know, algebraically, why $(1)$ is true. Thanks. • Binomial Theorem, do you know? – IAmNoOne Apr 29 '14 at 2:58 • I've heard of it, and I know it via Pascal's Triangle for low $n$, such as $n=2,3,4$, but I'm not familiar with its general formula. – Sujaan Kunalan Apr 29 '14 at 2:59 • I will post an answer then. – IAmNoOne Apr 29 '14 at 2:59 • You can find many posts about this here. For example, this question and this question and other posts linked there. – Martin Sleziak Sep 27 '15 at 20:35 There is an easy way to prove the formula. Suppose you want to buy a burger and you have a choice of 6 different condiments for it - mustard, mayonnaise, lettuce, tomatoes, pickles, and cheese. How many ways can you choose a combination of these condiments for your burger? Of course, you can choose either 6 different condiments, 5 different condiments, 4 different condiments, etc. So the obvious way to solve the problem is: \begin{align}{{6}\choose{6}} + {{6}\choose{5}} + {{6}\choose{4}} + {{6}\choose{3}} + {{6}\choose{2}} + {{6}\choose{1}} = \boxed{64}\end{align} But there is a better way. Imagine 6 spaces for 6 condiments: _____ _____ _____ _____ _____ _____ For every space, there are $2$ possible outcomes: Yes or No, meaning the condiment was chosen or the condiment was not chosen. With $2$ possible outcomes for each space, there are $2^6 = \boxed{64}$ possible ways. We know that both ways have foolproof logic and will both give identical answers no matter how many condiments there are. So this means we have proven: \begin{align}\sum_{k = 0}^{n} \binom{n}{k} = 2^n\end{align} • This is a nice burger... err answer. – mathreadler Oct 3 '15 at 19:48 By the Binomial theorem, we have $$(x + y)^n = \sum_{k = 0}^{n} \binom{n}{k} x^{n - k} y^k.$$ So if we let $x = y = 1$, then we get your result. The proof of the formula is traditionally done through induction. You also mentioned Pascal Triangles, hold your breath, because they are related. I suggest you look at the formula for small $n$, then examine the coefficients of the polynomial. • I see that you and the poster below you have your exponents switched. In general, does it matter if the $n-k$ exponent is on the $x$ or the $y$? – Sujaan Kunalan Apr 29 '14 at 3:05 • @SujaanKunalan, it does not. If you swap where $x$ and $y$ is on the LHS, you'll get what he got. – IAmNoOne Apr 29 '14 at 3:07 • Ah, ok. Thanks for your help. – Sujaan Kunalan Apr 29 '14 at 3:07 • @SujaanKunalan, you are welcome. Remember, you get to choose what $x$ and $y$s are in the formula. – IAmNoOne Apr 29 '14 at 3:09 Binomial theorem! $$(x+y)^n = \sum_{k=0}^n \dbinom{n}{k}x^ky^{n-k}$$ Take $x=y=1$ to get your identity. By the binomial theorem $(a+b)^n = \sum\limits_{i = 0}^{n} {{n}\choose{i}}a^{n-i}b^i$ If $a=b=1$ $(1+1)^n = \sum\limits_{i = 0}^{n}{{n}\choose{i}}$ So that $2 + \sum\limits_{i = 1}^{n-1}{{n}\choose{i}} = {{n}\choose{0}} + \sum\limits_{i = 1}^{n-1}{{n}\choose{i}} + {{n}\choose{n}} = \sum\limits_{i = 0}^{n}{{n}\choose{i}} = (1+1)^n$ Combinatorially: We are having a party and we have a list of $n$ people who we may or may not invite. We ask, "how many different possibilities of guests are there?" One way to arrive to an answer is saying, "We can invite no one and there is exactly $n \choose 0$ ways to do that, or we can invite one person and there are $n\choose 1$ ways to do that, ect." Following this logic we arrive at the left hand side. However, we want to double check our answer so we try another method. We say, "We can associate each person with the number $0$ if they are not invited and $1$ if they are. Therefore, each possible configuration is represented by a string of $0$s and $1$s of length $n$. Since each slot in this string has $2$ possibilities, we find that there are exactly $2^n$ distinct strings." This agrees with the right hand side. Since both (valid) methods were used to find an answer to this problem, we see that the left and right sides of the equation must agree. The expression counts the number of subsets of a set with $n$ elements. There's a one-to-one correspondence between subsets of $\{1,\dots,n\}$ and $\{0,1\}^n$, given by assigning a subset $S$ to the tuple which has a $1$ at the $i$th component if $i \in S$. This gives you the count without applying the binomial theorem. Well it just follows from the general formula: $(a+b)^n=\sum\limits_{k=0}^n \binom {n} {k}\,a^k\,b^{n-k}$, which you can prove by induction see here .

http://bueteeearchives.net/o3ddr4ei/0c6f4b-fraction-exponents-calculator

Dividing fractions calculator online. Welcome to the Exponent Calculator. Negative exponent ; A negative exponent represents which fraction of the base, the solution is. Sometimes the exponent itself is a fraction. You don't need to go from the top to the bottom of the calculator - calculate any unknown you want! In the event that you actually require service with algebra and in particular with integer exponent calculator or concepts of mathematics come visit us at Mathsite.org. The denominator on the exponent tells you what root of the “base” number the term represents. This algebra 2 video tutorial explains how to simplify fractional exponents including negative rational exponents and exponents in radicals with variables. Notes: i) e (Euler's number) and pi (Archimedes' constant π) are accepted values. ... Decimal to Fraction Fraction to Decimal Hexadecimal Scientific Notation Distance Weight Time. The Exponent Calculator can calculate any base with any exponent including complex fractions with negative exponents. We carry a lot of excellent reference information on subject areas varying from algebra ii to subtracting rational Enter fractions and press the = button. An exponential expression consists of two parts, namely the base, denoted as b and the exponent, denoted as n. The general form of an exponential expression is b n. For example, 3 x … Mixed Fractions. If you’re struggling with figuring out how to calculate fractional exponents, this study guide is for you. You can either apply the numerator first or the denominator. When exponents that share the same base are multiplied, the exponents are added. ii) 1.2 x … So we found out that: If you like, you can analogically check other roots, e.g. The Fraction Calculator will reduce a fraction to its simplest form. Do you struggle with the concept of fractional exponents? If you want to find an exponents based on the value and the result, try our salve exponents calculator. This website uses cookies to ensure you get the best experience. It also does not accept fractions, but can be used to compute fractional exponents, as long as the exponents are input in their decimal form. E.g: 5e3, 4e-8, 1.45e12 ** To find the exponent from the base and the exponentation result, use: Logarithm calculator Use this calculator to find the fractional exponent of a number x. \]. How to Use the Fraction Exponents Tool. Awesome! In the Base box, enter the number which you will raise to the fraction. Scientific calculators have more functionality that business calculators, and one thing they can do that is especially useful for scientists is to calculate exponents. Exponent Calculator. You can also calculate numbers to the power of large exponents less than 1000, negative exponents, and real numbers or decimals for exponents. Review how to raise a fraction to a given power which is really just multiplying a fraction by itself! Radicals ... Calculus Calculator. Usually you see exponents as whole numbers, and sometimes you see them as fractions. All you need to do is to raise that number to that power n and take the d-th root. Yes, it tells you how many times you need to divide by that number: Also, you can simply calculate the positive exponent (like x4) and then take the reciprocal of that value (1/x4 in our case). In a term like x a , you call x the base and a the exponent. It's a square root, of course! When you do see an exponent that is a decimal, you need to convert the decimal to a fraction. Let's have a look at a few simple examples first, where our numerator is equal to 1: From the equations above we can deduce that: Let's use the law of exponents which says that we can add the exponents when multiplying two powers that have the same base: Try this with any number you like, it's always true! Enter simple fractions with slash (/). Matrix Calculator. Email: donsevcik@gmail.com Tel: … If you try to take the root of a negative number your answer may be NaN = Not a Number. Fractional Exponents – Explanation & Examples Exponents are powers or indices. How to Find Mean, Median, and Mode. To calculate combined exponents and radicals such as the 4th root of 16 raised to the power of 5 you would enter 16 raised to the power of (5/4) or $$16^{\frac{5}{4}}$$ where x = 16, n = 5 and d = 4. Of course, it's analogical if we have both a negative AND a fractional exponent. Prime Factorization. All rights reserved. Exponent Laws. The calculator above accepts negative bases, but does not compute imaginary numbers. NB: To calculate a negative exponent, simply place a “-” before the number of your exponent. 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http://math.stackexchange.com/questions/608465/is-there-a-proof-for-this-fibonacci-relationship

# Is there a proof for this Fibonacci relationship? I was looking at the decomposition of Fibonacci numbers using the definition of $F_n = F_{n-1} + F_{n-2}$, and noted the pattern in the coefficients of the terms were Fibonacci numbers. It appears to hold, and I believe it's true, but I haven't seen a proof for it. Does one exist? $F_n = 1F_{n-1} + 1F_{n-2}$ $F_n = 2F_{n-2} + 1F_{n-3}$ $F_n = 3F_{n-3} + 2F_{n-4}$ $F_n = 5F_{n-4} + 3F_{n-5}$ etc This can be generalized to $F_n = F_xF_{n-(x-1)}+F_{x-1}F_{n-x}$ - Try using Binet's Formula. –  Blue Dec 15 '13 at 23:44 Yes, that identity is well-known. There are several proofs, one of which I wrote up on this web site a couple of months ago, in the form $f_{i+k} = f_kf_{i-1} + f_{k+1}f_i$. –  MJD Dec 15 '13 at 23:52 I think the way that you arrived at this already is a proof by induction on $x$. –  Carsten Schultz Dec 16 '13 at 0:23 It also appears in the Wikipedia article. –  MJD Dec 16 '13 at 0:34 HINT $\ \$ If you change variables this Fibonacci addition law can be put in the form $$\color{#c00}{F_{n+m} = F_nF_{m+1} + F_{n-1}F_m}$$ Expressing the Fibonacci recurrence in matrix form makes the proof of the addition law easy $$M^n\ :=\ \left(\begin{array}{ccc} \,1 & 1 \\\ 1 & 0 \end{array}\right)^n\ =\ \left(\begin{array}{ccc} F_{n+1} & F_n \\\ F_n & F_{n-1} \end{array}\right)$$ $$\begin{eqnarray} M^{n+m} = M^n M^m &=& \left(\begin{array}{ccc} F_{n+1} & F_n \\\ F_n & F_{n-1} \end{array}\right)\ \left(\begin{array}{ccc} F_{m+1} & F_m \\\ F_m & F_{m-1} \end{array}\right) \\ \\ \\ \Rightarrow\ \ \left(\begin{array}{ccc} F_{n+m+1} & F_{n+m} \\\ \color{#c00}{F_{n+m}} & F_{n+m-1} \end{array}\right) &=&\left(\begin{array}{ccc} F_{n+1}F_{m+1} + F_nF_m & F_{n+1}F_m + F_nF_{m-1} \\\ \color{#C00}{F_nF_{m+1} + F_{n-1}F_m} & F_{n}F_{m} + F_{n-1}F_{m-1} \end{array}\right)\end{eqnarray}$$ - Welcome back! ${}{}$ –  MJD Dec 16 '13 at 1:33 I'm having trouble understanding the first line of matrices. And the third line. And what M^n is. –  JShoe Dec 21 '13 at 5:15 @Jshoe $\ M$ is the matrix being raised to power $n$. That first matrix equality is very easily proved by induction, using the fib recurrence. The 3rd line is just the 2nd line with the entries of $M^{n+m}$ listed on the LHS, and the matrix product computed on the RHS. –  Bill Dubuque Dec 21 '13 at 5:27 First, I want to say that I agree with Carsten Schultz's comment: what you have is almost a proof by induction, and it could be turned into one without much trouble. But I think the proof that follows is very elegant, in the sense of revealing a deeper truth, so I'd like to present it again. Let's call a sequence "fibonacci-like" if it satisfies the recurrence $$s_{n+2} = s_n + s_{n+1}$$ for all $n$. It's easy to see (or to show) that if $s$ and $t$ are two fibonacci-like sequences, then so is the sequence that you get by multiplying every term of $s$ or of $t$ of them by some constant, and so is the sequence you get by adding together corresponding terms of $s$ and $t$. For example, $1,1,2,3,5,8\ldots$ and $2,8,10,18,28,\ldots$ are both fibonacci-like, and so is what you get if you multiply every element of $1,1,2,3,5,8\ldots$ by 3, namely $3,3,6,9,15,24,\ldots$, and so is what you get if you add them together: $3,9,12,21,33,54,\ldots$. Since any such sequence $s_i$ is completely determined by $s_0$ and $s_1$, we can agree to write a fibonacci-like sequence with the notation $[s_0, s_1]$, where $s_0$ and $s_1$ are the first two terms of the sequence; that tells us everything about it. Then the Fibonacci sequence itself is written $[0, 1]$, whereas $[1, 3]$ is the Lucas sequence $1,3,4,7,11,\ldots$. Observe that in this notation, $[a_0, a_1] + [b_0, b_1] = [a_0+ b_0, a_1+b_1]$. So for example if you add the Fibonacci sequence $[0,1]$ to itself, you get $[0,2]$, which is in fact the sequence $0, 2, 2, 4, 6, 10,\ldots$ that you get from adding the Fibonacci sequence to itself termwise. Similarly, $c\cdot[a_0, a_1] = [c\cdot a_0 , c\cdot a_1]$. For example if you double every term of the Lucas sequence $[1,3] = 1,3,4,7,11,\ldots$ you get $2,6,8,14,22,\ldots$, which is exactly the sequence we are writing as $[2,6]$. Finally, observe that $[0,1]$ is that standard Fibonacci sequence whose $i$th term is $f_i$, and $[1,0]$ is the "shifted" Fibonacci sequence, whose $i$th term is $f_{i-1}$. (Note that I am following the standard convention here that has $f_0 = 0$ and $f_1 = 1$; your $F_i$ seems to have $F_0 = 1$ and $F_1=1$, which is not the usual convention.) Now suppose we have some fibonacci-like sequence $s = [a, b]$. (Remember, $[a,b]$ is just shorthand for the fibonacci-like sequence $a, b, a+b, a+2b, 2a+3b, \ldots$.) We can decompose $[a,b]$ as follows: $$[a,b] =b[0,1] + a[1,0]$$ which shows that $s_i = bf_i + af_{i-1}$. Indeed if you look at a the example terms I showed above you see for example that $s_4 = 2a + 3b = af_3 + bf_4$. This identity holds for any fibonacci-like sequence: If $s$ is a fibonacci-like sequence whose first two terms are $s_0$ and $s_1$, then $$s_i = s_0f_{i-1} + s_1f_i$$ for all $i$. (You have to properly understand $f_{-1} = 1$ for this to work at $i=0$.) Now consider the Fibonacci sequence shifted over $k$ places for some fixed $k$: this sequence begins $f_k, f_{k+1}, f_{k+2}, \ldots$; its $i$th term is $f_{i+k}$. Obviously it is fibonacci-like. In our notation it is represented as $[f_k, f_{k+1}]$ and by the previous paragraph it is equal to $f_k[1,0] + f_{k+1}[0,1]$, so by the previous theorem its $i$th term is equal to $f_kf_{i+1} + f_{k+1}f_i$. But it's $i$th term is also equal to $f_{i+k}$, so we have: $$f_{i+k} = f_kf_{i-1} + f_{k+1}f_i.$$ Now put $i=x, k = n-x$ and we have your claim, except you have $F_n = f_{n+1}$, so the subscripts are a little muddled. - Try going the other way. $F_n = F_{n-1} + F_{n-2}$ $F_{n+1} = F_n + F_{n-1} = F_{n-1} + F_{n-1} + F_{n-2}$ $F_{n+2} = F_{n+1} + F_{n} = F_{n} + F_{n} + F_{n-1} = 2(F_{n-1}+F_{n-2})+F_{n-1}$ Etc. Hopefully the pattern becomes more "obvious" when seen this way. Then once you get to $F_{n+x}$, replace that with $F_{n}$ and replace your $F_n$ terms with $F_{n-x}$ and you should rederive your formula. - The pattern was obvious to begin with, I'm just not sure that that is a satisfactory proof in and of itself. –  JShoe Dec 15 '13 at 23:59 I agree. OP asked for a proof, but this isn't one. –  MJD Dec 16 '13 at 0:28

https://math.stackexchange.com/questions/2065622/how-many-multisets-of-size-4-that-can-be-constructed-from-n-distinct-element/2065636

# How many multisets of size $4$ that can be constructed from $n$ distinct elements? How many multisets of size $4$ that can be constructed from $n$ distinct elements so that at least one element occurs exactly twice ? Example : For $n=3$ and element set as $\{1,2,3\}$ I am getting multisets as: $\{1,1,3,3\}, \{1,1,2,2\}, \{1,1,2,3\}, \{2,2,3,3\}, \{2,2,1,3\}, \{3,3,1,2\}$, which are total $6$. I am looking for a formula for large N. Is there any such formula or do we have to count manually? Represent a multiset $$\{\underbrace{1,\dots,1}_{x_1\text{ times}},\underbrace{2,\dots,2}_{x_2\text{ times}},\dots,\underbrace{n,\dots,n}_{x_n\text{ times}}\}$$ by $\{1^{x_1},2^{x_2},\dots,n^{x_n}\}$. For each $j=1,\dots,n$ a mulsitet of size $4$ in which $x_j$ occurs exactly twice corresponds to a solution of $$\left\{\begin{array}{l} \displaystyle\sum_{\substack{1\leq i\leq n}\\i\neq j} x_i= 2\\ x_i\geq 0 \end{array}\right.$$ The number of solutions to this system can be found via stars and bars, and that number is $\binom{2+(n-1)-1}{2}=\binom{n}{2}$. This might point in the direction that the answer is $n\binom{n}{2}$, but we're overcounting. Indeed, when $j_1\neq j_2$, the multiset $\{{j_1}^2,{j_2}^2\}$ is counted twice: once for the index $j_1$, and once for the index $j_2$. It's easy to see this is the only case of overcounting. How many such sets there are? That's just the number of ways to choose two distinct indices from among $\{1,\dots,n\}$, that is, $\binom{n}{2}$. Hence, the final answer is $$n\binom{n}{2}-\binom{n}{2}=(n-1)\binom{n}{2}=\frac{n(n-1)^2}{2}$$ Generalizing the problem to allow the size of the multiset to change as well as the number of available numbers to use in the multiset. As was started by another user, if we were to ignore the requirement on having at least one number occur exactly two times, we have the count being $\binom{n+k-1}{k}$ We wish to remove the "bad" multisets, which are those that do not have an element repeated exactly two times. To count how many are "bad" we can easily approach via generating functions. For each of the $n$ available numbers, the term $(1+x+x^3+x^4+x^5+\dots)$ will represent the available choices of taking zero, one, three, four, five, etc... copies of that number. With every number appearing any number of times except for two, there will clearly then be no number that occurs exactly two times. The coefficient of $x^k$ then in the expansion of $$(1+x+x^3+x^4+x^5+\dots)^n$$ will represent the number of multisets of size $k$ can be made using elements from the $n$ element set where none of the terms appear exactly two times. The number of multisets which satisfy your condition will be $\binom{n+k-1}{k}$ minus the coefficient of $x^k$ in the series expansion of $(1+x+\frac{x^3}{1-x})^n$, or alternately worded using generating functions for the first, the overall generating function is: $$\frac{1}{(1-x)^n}-(1+x+\frac{x^3}{1-x})^n$$ In your specific case we have for $n=3$ the series expansion $$3x^2+6x^3+\color{red}{6x^4}+9x^5+13x^6+15x^7+18x^8+21x^9+\dots$$ The $6$ in $\color{red}{6x^4}$ corresponds to the six multisets you wrote down above. We first pick the element that will repeat exactly twice. We have $n$ choices for that. We then pick the remaining two elements, which can repeat, so we have $(n-1)^2$ choices for that. Now, in the case that the last two elements picked are different, then we count that case twice, when we should only count it once. This is because we count as picking $a$ then $b$ as distinct from picking $b$ then $a$. In the case that the last two elements are the same, we also double count that case, since if the last two elements where $b$ and the first element was $a$, the case where the first element was $b$ and the last two elements were $a$ would also be distinctly counted. Thus, we count every case exactly twice, and therefore divide by two to reach the final formula: $$\frac{n(n-1)^2}{2}$$ • For this comment, assume different letters represent different elements. It is clear to me that this answer counts '$a$, then $b$ then $c$' the same as '$a$ then $c$ then $b$', and both represent the multiset $\{a,a,b,c\}$. However, it is not clear to me that this answer counts '$a$, then $b$ then $b$' the same as '$b$, then $a$ then $a$', even though they both represent the same multiset $\{a,a,b,b\}$. Dec 20 '16 at 7:40 • I understand that the answer is correct and that dividing by two takes care precisely of this paricular $\{a,a,b,b\}$ case, but it nonetheless does not seem clear to me from what is written. The way it is written, it appears division by two is intended to handle only $\{a,a,b,c\}$ cases. Dec 20 '16 at 7:43 • You are correct. Let me update the answer Dec 20 '16 at 7:45 • @Fimpellizieri, do you think this is clearer? Dec 20 '16 at 8:00 • Yes, definitely! Dec 20 '16 at 15:23 There are four places to be filled in the multiset using the $n$ distinct elements. Atleast one element has to occur exactly twice. That would leave $2$ more places in the multiset. This means, atmost two elements can occur exactly twice. We can thus divide this into $2$ mutually exclusive cases as follows: $1.$Exactly one element occurs exactly twice: Select this element in ${n\choose{1}} = n$ ways. Fill up the remaining two spots using $2$ distinct elements from the remaining $n-1$ elements in ${{n-1}\choose{2}}$ ways. Overall: $n \cdot {{n-1}\choose{2}} = \frac{n(n-1)(n-2)}{2}$ ways. $2.$ Exactly two elements that occur twice each: These two will fill up the multiset, so you only have to select two elements out of $n$ in ${n\choose 2} = \frac{n(n-1)}{2}$ ways. Since these are mutually exclusive, the total number of ways to form the multiset is: $$\frac{n(n-1)(n-2)}{2} + \frac{n(n-1)}{2}$$$$= \frac{n(n-1)^2}{2}$$ Hence, for $n=4$, we have a total of $18$ multisets. Hope it helps.

http://linear.ups.edu/fcla/section-S.html

A subspace is a vector space that is contained within another vector space. So every subspace is a vector space in its own right, but it is also defined relative to some other (larger) vector space. We will discover shortly that we are already familiar with a wide variety of subspaces from previous sections. Here is the principal definition for this section. ##### DefinitionSSubspace Suppose that $V$ and $W$ are two vector spaces that have identical definitions of vector addition and scalar multiplication, and suppose that $W$ is a subset of $V\text{,}$ $W\subseteq V\text{.}$ Then $W$ is a subspace of $V\text{.}$ Let us look at an example of a vector space inside another vector space. In Example SC3 we proceeded through all ten of the vector space properties before believing that a subset was a subspace. But six of the properties were easy to prove, and we can lean on some of the properties of the vector space (the superset) to make the other four easier. Here is a theorem that will make it easier to test if a subset is a vector space. A shortcut if there ever was one. ##### Proof So just three conditions, plus being a subset of a known vector space, gets us all ten properties. Fabulous! This theorem can be paraphrased by saying that a subspace is “a nonempty subset (of a vector space) that is closed under vector addition and scalar multiplication.” You might want to go back and rework Example SC3 in light of this result, perhaps seeing where we can now economize or where the work done in the example mirrored the proof and where it did not. We will press on and apply this theorem in a slightly more abstract setting. Much of the power of Theorem TSS is that we can easily establish new vector spaces if we can locate them as subsets of other vector spaces, such as the vector spaces presented in Subsection VS.EVS. It can be as instructive to consider some subsets that are not subspaces. Since Theorem TSS is an equivalence (see Proof Technique E) we can be assured that a subset is not a subspace if it violates one of the three conditions, and in any example of interest this will not be the “nonempty” condition. However, since a subspace has to be a vector space in its own right, we can also search for a violation of any one of the ten defining properties in Definition VS or any inherent property of a vector space, such as those given by the basic theorems of Subsection VS.VSP. Notice also that a violation need only be for a specific vector or pair of vectors. There are two examples of subspaces that are trivial. Suppose that $V$ is any vector space. Then $V$ is a subset of itself and is a vector space. By Definition S, $V$ qualifies as a subspace of itself. The set containing just the zero vector $Z=\set{\zerovector}$ is also a subspace as can be seen by applying Theorem TSS or by simple modifications of the techniques hinted at in Example VSS. Since these subspaces are so obvious (and therefore not too interesting) we will refer to them as being trivial. ##### DefinitionTSTrivial Subspaces Given the vector space $V\text{,}$ the subspaces $V$ and $\set{\zerovector}$ are each called a trivial subspace. We can also use Theorem TSS to prove more general statements about subspaces, as illustrated in the next theorem. ##### Proof Here is an example where we can exercise Theorem NSMS. # SubsectionTSSThe Span of a Set¶ permalink The span of a set of column vectors got a heavy workout in Chapter V and Chapter M. The definition of the span depended only on being able to formulate linear combinations. In any of our more general vector spaces we always have a definition of vector addition and of scalar multiplication. So we can build linear combinations and manufacture spans. This subsection contains two definitions that are just mild variants of definitions we have seen earlier for column vectors. If you have not already, compare them with Definition LCCV and Definition SSCV. ##### DefinitionLCLinear Combination Suppose that $V$ is a vector space. Given $n$ vectors $\vectorlist{u}{n}$ and $n$ scalars $\alpha_1,\,\alpha_2,\,\alpha_3,\,\ldots,\,\alpha_n\text{,}$ their linear combination is the vector \begin{equation*} \lincombo{\alpha}{u}{n}\text{.} \end{equation*} When we realize that we can form linear combinations in any vector space, then it is natural to revisit our definition of the span of a set, since it is the set of all possible linear combinations of a set of vectors. ##### DefinitionSSSpan of a Set Suppose that $V$ is a vector space. Given a set of vectors $S=\{\vectorlist{u}{t}\}\text{,}$ their span, $\spn{S}\text{,}$ is the set of all possible linear combinations of $\vectorlist{u}{t}\text{.}$ Symbolically, \begin{align*} \spn{S}&=\setparts{\lincombo{\alpha}{u}{t}}{\alpha_i\in\complexes,\,1\leq i\leq t}\\ &=\setparts{\sum_{i=1}^{t}\alpha_i\vect{u}_i}{\alpha_i\in\complexes,\,1\leq i\leq t}\text{.} \end{align*} ##### Proof Let us again examine membership in a span. Notice how Example SSP and Example SM32 contained questions about membership in a span, but these questions quickly became questions about solutions to a system of linear equations. This will be a common theme going forward. Several of the subsets of vectors spaces that we worked with in Chapter M are also subspaces — they are closed under vector addition and scalar multiplication in $\complex{m}\text{.}$ ##### Proof That was easy! Notice that we could have used this same approach to prove that the null space is a subspace, since Theorem SSNS provided a description of the null space of a matrix as the span of a set of vectors. However, I much prefer the current proof of Theorem NSMS. Speaking of easy, here is a very easy theorem that exposes another of our constructions as creating subspaces. One more. ##### Proof So the span of a set of vectors, and the null space, column space, row space and left null space of a matrix are all subspaces, and hence are all vector spaces, meaning they have all the properties detailed in Definition VS and in the basic theorems presented in Section VS. We have worked with these objects as just sets in Chapter V and Chapter M, but now we understand that they have much more structure. In particular, being closed under vector addition and scalar multiplication means a subspace is also closed under linear combinations. ##### 1 Summarize the three conditions that allow us to quickly test if a set is a subspace. ##### 2 Consider the set of vectors \begin{equation*} W=\setparts{\colvector{a\\b\\c}}{3a-2b+c=5}\text{.} \end{equation*} Is the set $W$ a subspace of $\complex{3}\text{?}$ Explain your answer. ##### 3 Name five general constructions of sets of column vectors (subsets of $\complex{m}$) that we now know as subspaces. # SubsectionExercises ##### C15 Working within the vector space $\complex{3}\text{,}$ determine if $\vect{b} = \colvector{4\\3\\1}$ is in the subspace $W\text{,}$ \begin{equation*} W = \spn{\set{ \colvector{3\\2\\3}, \colvector{1\\0\\3}, \colvector{1\\1\\0}, \colvector{2\\1\\3} }}\text{.} \end{equation*} Solution ##### C16 Working within the vector space $\complex{4}\text{,}$ determine if $\vect{b} = \colvector{1\\1\\0\\1}$ is in the subspace $W\text{,}$ \begin{equation*} W =\spn{\set{ \colvector{1\\2\\-1\\1}, \colvector{1\\0\\3\\1}, \colvector{2\\1\\1\\2} }}\text{.} \end{equation*} Solution ##### C17 Working within the vector space $\complex{4}\text{,}$ determine if $\vect{b} = \colvector{2\\1\\2\\1}$ is in the subspace $W\text{,}$ \begin{equation*} W = \spn{\set{ \colvector{1\\2\\0\\2}, \colvector{1\\0\\3\\1}, \colvector{0\\1\\0\\2}, \colvector{1\\1\\2\\0} }}\text{.} \end{equation*} Solution ##### C20 Working within the vector space $P_3$ of polynomials of degree 3 or less, determine if $p(x)=x^3+6x+4$ is in the subspace $W$ below. \begin{equation*} W=\spn{\set{x^3+x^2+x,\,x^3+2x-6,\,x^2-5}} \end{equation*} Solution ##### C21 Consider the subspace \begin{equation*} W=\spn{\set{ \begin{bmatrix} 2 & 1\\3 & -1 \end{bmatrix} ,\, \begin{bmatrix} 4 & 0\\2 & 3 \end{bmatrix} ,\, \begin{bmatrix} -3 & 1\\2 & 1 \end{bmatrix} }} \end{equation*} of the vector space of $2\times 2$ matrices, $M_{22}\text{.}$ Is \begin{equation*} C=\begin{bmatrix} -3 & 3\\6 & -4 \end{bmatrix} \end{equation*} an element of $W\text{?}$ Solution ##### C26 Show that the set $Y=\setparts{\colvector{x_1\\x_2}}{x_1\in{\mathbb Z},\,x_2\in{\mathbb Z}}$ from Example NSC2S has Property AC. ##### M20 In $\complex{3}\text{,}$ the vector space of column vectors of size 3, prove that the set $Z$ is a subspace. \begin{equation*} Z=\setparts{\colvector{x_1\\x_2\\x_3}}{4x_1-x_2+5x_3=0} \end{equation*} Solution ##### T20 A square matrix $A$ of size $n$ is upper triangular if $\matrixentry{A}{ij}=0$ whenever $i\gt j\text{.}$ Let $UT_n$ be the set of all upper triangular matrices of size $n\text{.}$ Prove that $UT_n$ is a subspace of the vector space of all square matrices of size $n\text{,}$ $M_{nn}\text{.}$ Solution ##### T30 Let $P$ be the set of all polynomials, of any degree. The set $P$ is a vector space. Let $E$ be the subset of $P$ consisting of all polynomials with only terms of even degree. Prove or disprove: the set $E$ is a subspace of $P\text{.}$ Solution ##### T31 Let $P$ be the set of all polynomials, of any degree. The set $P$ is a vector space. Let $F$ be the subset of $P$ consisting of all polynomials with only terms of odd degree. Prove or disprove: the set $F$ is a subspace of $P\text{.}$ Solution

https://stats.stackexchange.com/questions/185683/distribution-of-ratio-between-two-independent-uniform-random-variables

# Distribution of ratio between two independent uniform random variables Supppse $X$ and $Y$ are standard uniformly distributed in $[0, 1]$, and they are independent, what is the PDF of $Z = Y / X$? The answer from some probability theory textbook is $$f_Z(z) = \begin{cases} 1/2, & \text{if } 0 \le z \le 1 \\ 1/(2z^2), & \text{if } z > 1 \\ 0, & \text{otherwise}. \end{cases}$$ I am wondering, by symmetry, shouldn't $f_Z(1/2) = f_Z(2)$? This is not the case according to the PDF above. • What is the domain of $X$ and $Y$? – Sobi Dec 8 '15 at 13:47 • en.wikipedia.org/wiki/Ratio_distribution – kjetil b halvorsen Dec 8 '15 at 13:54 • Why would you expect this to be true? The density function tells you how tightly packed the probability is in the neighborhood of a point, and it is clearly more difficult for $Z$ to be near $2$ than $1/2$ (consider for instance that $Z$ can always be $1/2$ no matter what $X$ is, but $Z < 2$ when $X > 1/2$). – dsaxton Dec 8 '15 at 14:06 • Possible duplicate of Distribution of a ratio of uniforms: What is wrong? – Xi'an Dec 8 '15 at 14:21 • I don't think it's a duplicate, that question is seeking the PDF, here I have the PDF, I am just questioning its correctness (perhaps rather naively). – qed Dec 8 '15 at 14:29 The right logic is that with independent $X, Y \sim U(0,1)$, $Z=\frac YX$ and $Z^{-1} =\frac XY$ have the same distribution and so for $0 < z < 1$ \begin{align} P\left\{\frac YX \leq z\right\} &= P\left\{\frac XY \leq z\right\}\\ &= P\left\{\frac YX \geq \frac 1z \right\}\\ \left.\left.F_{Z}\right(z\right) &= 1 - F_{Z}\left(\frac 1z\right) \end{align} where the equation with CDFs uses the fact that $\frac YX$ is a continuous random variable and so $P\{Z \geq a\} = P\{Z > a\} = 1-F_Z(a)$. Hence the pdf of $Z$ satisfies $$f_Z(z) = z^{-2}f_Z(z^{-1}), \quad 0 < z < 1.$$ Thus $f_Z(\frac 12) = 4f_Z(2)$, and not $f_Z(\frac 12) = f_Z(2)$ as you thought it should be. This distribution is symmetric--if you look at it the right way. The symmetry you have (correctly) observed is that $Y/X$ and $X/Y = 1/(Y/X)$ must be identically distributed. When working with ratios and powers, you are really working within the multiplicative group of the positive real numbers. The analog of the location invariant measure $d\lambda=dx$ on the additive real numbers $\mathbb{R}$ is the scale invariant measure $d\mu = dx/x$ on the multiplicative group $\mathbb{R}^{*}$ of positive real numbers. It has these desirable properties: 1. $d\mu$ is invariant under the transformation $x\to ax$ for any positive constant $a$: $$d\mu(ax) = \frac{d(ax)}{ax} = \frac{dx}{x} = d\mu.$$ 2. $d\mu$ is covariant under the transformation $x\to x^b$ for nonzero numbers $b$: $$d\mu(x^b) = \frac{d(x^b)}{x^b} = \frac{b x^{b-1} dx}{x^b} = b\frac{dx}{x} = b\, d\mu.$$ 3. $d\mu$ is transformed into $d\lambda$ via the exponential: $$d\mu(e^x) = \frac{de^x}{e^x} = \frac{e^x dx}{e^x} = dx = d\lambda.$$ Likewise, $d\lambda$ is transformed back to $d\mu$ via the logarithm. (3) establishes an isomorphism between the measured groups $(\mathbb{R}, +, d\lambda)$ and $(\mathbb{R}^{*}, *, d\mu)$. The reflection $x \to -x$ on the additive space corresponds to the inversion $x \to 1/x$ on the multiplicative space, because $e^{-x} = 1/e^x$. Let's apply these observations by writing the probability element of $Z=Y/X$ in terms of $d\mu$ (understanding implicitly that $z \gt 0$) rather than $d\lambda$: $$f_Z(z)\,dz = g_Z(z)\,d\mu = \frac{1}{2}\begin{cases} 1\,dz = z\, d\mu, & \text{if } 0 \le z \le 1 \\ \frac{1}{z^2}dz = \frac{1}{z}\, d\mu, & \text{if } z > 1. \end{cases}$$ That is, the PDF with respect to the invariant measure $d\mu$ is $g_Z(z)$, proportional to $z$ when $0\lt z \le 1$ and to $1/z$ when $1 \le z$, close to what you had hoped. This is not a mere one-off trick. Understanding the role of $d\mu$ makes many formulas look simpler and more natural. For instance, the probability element of the Gamma function with parameter $k$, $x^{k-1}e^x\,dx$ becomes $x^k e^x d\mu$. It's easier to work with $d\mu$ than with $d\lambda$ when transforming $x$ by rescaling, taking powers, or exponentiating. The idea of an invariant measure on a group is far more general, too, and has applications in that area of statistics where problems exhibit some invariance under groups of transformations (such as changes of units of measure, rotations in higher dimensions, and so on). • Looks like a very insightful answer. It's a pity I don't understand it at the moment. I will check back later. – qed Dec 8 '15 at 16:40 If you think geometrically... In the $X$-$Y$ plane, curves of constant $Z = Y/X$ are lines through the origin. ($Y/X$ is the slope.) One can read off the value of $Z$ from a line through the origin by finding its intersection with the line $X=1$. (If you've ever studied projective space: here $X$ is the homogenizing variable, so looking at values on the slice $X=1$ is a relatively natural thing to do.) Consider a small interval of $Z$s, $(a,b)$. This interval can also be discussed on the line $X=1$ as the line segment from $(1,a)$ to $(1,b)$. The set of lines through the origin passing through this interval forms a solid triangle in the square $(X,Y) \in U = [0,1]\times[0,1]$, which is the region we're actually interested in. If $0 \leq a < b \leq 1$, then the area of the triangle is $\frac{1}{2}(1-0)(b-a)$, so keeping the length of the interval constant and sliding it up and down the line $X=1$ (but not past $0$ or $1$), the area is the same, so the probability of picking an $(X,Y)$ in the triangle is constant, so the probability of picking a $Z$ in the interval is constant. However, for $b>1$, the boundary of the region $U$ turns away from the line $X = 1$ and the triangle is truncated. If $1 \leq a < b$, the projections down lines through the origin from $(1,a)$ and $(1,b)$ to the upper boundary of $U$ are to the points $(1/a,1)$ and $(1/b,1)$. The resulting area of the triangle is $\frac{1}{2}(\frac{1}{a} - \frac{1}{b})(1-0)$. From this we see the area is not uniform and as we slide $(a,b)$ further and further to the right, the probability of selecting a point in the triangle decreases to zero. Then the same algebra demonstrated in other answers finishes the problem. In particular, returning to the OP's last question, $f_Z(1/2)$ corresponds to a line that reaches $X=1$, but $f_Z(2)$ does not, so the desired symmetry does not hold. Just for the record, my intuition was totally wrong. We are talking about density, not probability. The right logic is to check that $$\int_1^k f_Z(z) dz = \int_{1/k}^1 f_Z(z) = \frac{1}{2}(1 - \frac{1}{k})$$, and this is indeed the case. Yea the link Distribution of a ratio of uniforms: What is wrong? provides CDF of $Z=Y/X$. The PDF here is just derivative of the CDF. So the formula is correct. I think your problem lies in the assumption that you think Z is "symmetric" around 1. However this is not true. Intuitively Z should be a skewed distribution, for example it is useful to think when Y is a fixed number between $(0,1)$ and X is a number close to 0, thus the ratio would be going to infinity. So the symmetry of distribution is not true. I hope this help a bit.

https://mathhelpboards.com/threads/displacement-and-distance.2253/

# Displacement and Distance #### alane1994 ##### Active member Here is my question. The function $$v(t)=15\cos{3t}$$, $$0 \leq t \leq 2 \pi$$, is the velocity in m/sec of a particle moving along the x-axis. Complete parts (a) through (c). a.Graph the velocity function over the given interval. then determine when the motion is in the positive direction and when it is in the negative direction. (I have done this part) b. Find the displacement over the given interval. (I have done this part) c. Find the distance traveled over the given interval. (This is the part that I am fuzzy on) For this, does it involve previously found information? Or is it a separate set of calculations all its own? Any help is appreciated! ~Austin #### Jameson Staff member Displacement is the integral of the velocity vector over the given time interval and results in the distance between the starting point and end point. If you move 10 meters north and 10 meters south, the displacement is 0. To find the distance you need to figure out where the displacement is negative and count that as a positive value. I believe this is the same as taking the integral of the absolute value of velocity. The way I would do this is find the regions where v(t) is positive and negative and then breaking the calculation into multiple calculations. When considering $$\displaystyle v(t)=15\cos{3t}$$ where $$\displaystyle 0 \le t \le 2\pi$$ at $t=0$ v(t) is positive and becomes negative when $$\displaystyle 3t=\frac{\pi}{2}$$ so when $$\displaystyle t=\frac{\pi}{6}$$. So your first integral is $$\displaystyle \int_{0}^{\frac{\pi}{6}}v(t)dt$$ This value will be positive. Now v(t) will be negative until until it touches the x-axis again where $$\displaystyle 3t=\frac{3\pi}{2}$$ or when $$\displaystyle t=\frac{\pi}{2}$$. The second integral is now $$\displaystyle \int_{\frac{\pi}{6}}^{\frac{\pi}{2}}v(t)dt$$. This will be negative but when considering distance you don't need to take into account the sign of this so count it as positive. If you keep repeating this over until the end of the interval you should get the final answer. It will take a while to calculate this way and there very well could be a quicker way to do it but that's how I would do this. #### alane1994 ##### Active member Yeah, that's what I did after consulting a professor, I got the answer 60... #### Jameson Staff member Yeah, that's what I did after consulting a professor, I got the answer 60... You already found all of these intervals for part A so this calculation shouldn't have been too tedious. Wow your professor responds fast. You posted this question just a couple of hours ago #### topsquark ##### Well-known member MHB Math Helper Just to put my two cents in... The formula for distance is $$dist = \int |v|dt$$ where v is the velocity vector. In a practical sense this can pretty much only be calculated in the way Jameson has described. -Dan

http://math.stackexchange.com/questions/182617/counterexample-check-for-sum-of-limit-points-of-subsequences

# Counterexample Check for Sum of Limit Points of Subsequences Let $c$ be a limit point of a sequence of real numbers $\langle a_n \rangle$ and $d$ a limit point of $\langle b_n \rangle$. Is $c+d$ necessarily a limit point of $\langle a_n + b_n \rangle$? My Question: When considering this question, do I have to sum over the same index or can the indices for the different subsequences differ? My intuition is that the subsequences must be summed over identical indices, in which case I believe that the following example serves as a counterexample: Let $\langle a_n \rangle = (-1)^n, \ \langle b_n \rangle = (-1)^{n+1}$. Then summing over even and odd indices, I get $0 \ne 2$ or $-2$, which is the sum of their limit points. Have I done this correctly? - Your argument is correct; but generally, the indices could differ. Take $a_n=b_n=(-1)^n$. $(a_n)$ has $1$ as a limit point and $(b_n)$ has $-1$ as a limit point but $(a_n+b_n)$ does not have $1+(-1)=0$ as a limit point. –  David Mitra Aug 14 '12 at 22:10 Your example is fine. Both $\langle(-1)^n:n\in\Bbb N\rangle$ and $\langle(-1)^{n+1}:n\in\Bbb N\rangle$ have $1$ as a limit point, but $\langle(-1)^n+(-1)^{n+1}:n\in\Bbb N\rangle=\langle 0:n\in\Bbb N\rangle$ converges to $0$ and so does not have $1+1=2$ as a limit point. It doesn’t matter what subsequence of $\langle a_n:n\in\Bbb N\rangle$ has $c$ as limit or what subsequence of $\langle b_n:n\in\Bbb N\rangle$ has $d$ as a limit; all that matters is whether some subsequence of $\langle a_n+b_n:n\in\Bbb N\rangle$ has $c+d$ as a limit. In your example that’s not the case, so yours is a genuine counterexample to the conjecture. @Rolando posted what looked like a proof of this conjecture, and then deleted it after you posted your answer. It looked something like this: Let $\epsilon >0$ be given. Since $a_n \to c$ and $b_n \to d$, there are $N_1, N_2$ such that for $l \ge N_1$, $k \ge N_2$ $|a_l - c| \le \epsilon/2$ and $|b_k - d| \le \epsilon/2$. Let $m = \max\{N_1, N_2\}$ and therefore we have for $n \ge m$ we have $|a_n - c + b_n - d| \le \epsilon$. Is the problem with his proof that he is not considering subsequences, and that he interpreted the limit points as limits? –  Zvpunry Aug 14 '12 at 22:14

https://math.stackexchange.com/questions/1738331/sequence-of-functions-that-fails-certain-conditions-of-arzela-ascoli-theorem

# Sequence of functions that fails certain conditions of Arzela-Ascoli theorem For a closed, bounded interval $[a,b]$, let $\{ f_{n}\}$ be a sequence in $C[a,b]$. If $\{f_{n}\}$ is equicontinuous, does $\{f_{n}\}$ necessarily have a uniformly convergent subsequence? I would think not, because according to the Arzela-Ascoli Theorem, $\{f_{n} \}$ also needs to be uniformly bounded. Is this all that needs to be violated in order for an equicontinuous sequence of continuous functions on a compact interval to not have a uniformly convergent subsequence? And if so, what is an example of a sequence that illustrates this, and how to show it does not have a uniformly convergent subsequence? Thank you. Take $f_n(x) = n$ for $x \in [0,1]$. These functions are all constant, so clearly equicontinuous, but $\| f_n - f_m \|_\infty = \lvert n - m \rvert \ge 1$ for $n \neq m$ so no subsequence can converge since no subsequence is Cauchy. • Yes of course. If it was uniformly bounded, then we could apply Arzela Ascoli and find a convergent subsequence. A sequence of functions $f_n$ is uniformly bounded if and only if the sequence of real numbers $\| f_n\|_\infty$ is bounded. Here that clearly is not the case because $\| f_n \|_\infty = n$. – User8128 Apr 11 '16 at 23:29

http://www.appeton.co.id/future-currency-tosgcr/63e023-linear-function-graph

By … By graphing two functions, then, we can more easily compare their characteristics. In this section, 8th grade and high school students will have to find the missing values of x and f(x). In $f\left(x\right)=mx+b$, the b acts as the vertical shift, moving the graph up and down without affecting the slope of the line. This is also expected from the negative constant rate of change in the equation for the function. In linear algebra, mathematical analysis, and functional analysis, a linear function is a … The equation is in standard form (A = -1, B = 1, C = 3). There is a special linear function called the "Identity Function": f (x) = x. The x-intercept is the point at which the graph of a linear function crosses the x-axis. A table of values might look as below. The vertical line test indicates that this graph represents a function. The independent variable is x and the dependent variable is y. a is the constant term or the y intercept. Graph linear functions. Two competing telephone companies offer different payment plans. A function may be transformed by a shift up, down, left, or right. 3.4 Graphing Linear Equations There are two common procedures that are used to draw the line represented by a linear equation. The graph of this function is a line with slope − and y-intercept −. A linear function has one independent variable and one dependent variable. Twitter. A similar word to linear function is linear correlation. For distinguishing such a linear function from the other concept, the term affine function is often used. This tells us that for each vertical decrease in the “rise” of $–2$ units, the “run” increases by 3 units in the horizontal direction. When we’re comparing two lines, if their slopes are equal they are parallel, and if they are in … By using this website, you agree to our Cookie Policy. Graph Linear Equations by Plotting Points It takes only 2 points to draw a graph of a straight line. The first characteristic is its y-intercept which is the point at which the input value is zero. Graph $f\left(x\right)=-\frac{2}{3}x+5$ using the y-intercept and slope. Selbst 1 Selbst 2 Selbst 3 Yes. +drag: Hold down the key, then drag the described object. GRAPHING LINEAR RELATIONS. dillinghamt. Evaluate the function at x = 0 to find the y-intercept. The second is by using the y-intercept and slope. Free functions and graphing calculator - analyze and graph line equations and functions step-by-step. (Note: A vertical line parallel to the y-axis does not have a y-intercept. Do all linear functions have y-intercepts? Using slope and intercepts in context Get 3 of 4 questions to level up! The simplest way is to find the intercept values for both the x-axis and the y-axis. To draw the graph we need coordinates. http://cnx.org/contents/9b08c294-057f-4201-9f48-5d6ad992740d@5.2, Graph a linear function by plotting points, Graph a linear function using the slope and y-intercept, Graph a linear function using transformations. We were also able to see the points of the function as well as the initial value from a graph. Learn More at mathantics.comVisit http://www.mathantics.com for more Free math videos and additional subscription based content! How to graph Linear Functions by finding the X-Intercept and Y-Intercept of the Function? The graph of a linear relation can be found by plotting at least two points. Graph $f\left(x\right)=4+2x$, using transformations. Spell. -x + y = 3. The first one is called the slope-intercept method and involves using the slope and intercept given in the equation. In Activity 1 the learners should enter the expressions one by one into the graphing calculator and classify the functions according to the shape of the graph. Write. linear functions by the shape of their graphs and by noting differences in their expressions. Graph 3x - 2y = 8. And here is its graph: It makes a 45° (its slope is 1) It is called "Identity" because what comes out is identical to what goes in: In. We can begin graphing by plotting the point (0, 1) We know that the slope is rise over run, $m=\frac{\text{rise}}{\text{run}}$. A function may also be transformed using a reflection, stretch, or compression. Plot the points and graph the linear function. However, in linear algebra, a linear function is a function that maps a sum to the sum of the images of the summands. In the equation $f\left(x\right)=mx+b$, $m=\frac{\text{change in output (rise)}}{\text{change in input (run)}}=\frac{\Delta y}{\Delta x}=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$. PLAY. This is why we performed the compression first. The y-intercept is the point on the graph when x = 0. f(x)=b. of f is the To find the y-intercept, we can set $x=0$ in the equation. Gravity. The graph of a linear function is a line. The slope of a linear function is equal to the ratio of the change in outputs to the change in inputs. This inequality notation means that we should plot the graph for values of x between and including -3 and 3. Use the resulting output values to identify coordinate pairs. The a represents the gradient of the line, which gives the rate of change of the dependent variable. In order to write the linear function in the form of y=mx+b, we will need to determine the line's: 1. slope (m) 2. y-intercept (b) We can tell from the graph that the slope of the line is negative because the line goes down and to the right. Using vertical stretches or compressions along with vertical shifts is another way to look at identifying different types of linear functions. The graph below is of the function $f\left(x\right)=-\frac{2}{3}x+5$. Linear Parent Graph And Transformations. So, for this definition, the above function is linear only when c = 0, that is when the line passes through the origin. Linear functions are functions that produce a straight line graph.. The graph of f is a line with slope m and y intercept b. The graph of f is a line with slope m and y intercept Graphing Linear Equations Find the Equation of a Line. Horizontal lines are written in the form, $f(x)=b$. Graph Linear Equations using Slope-Intercept We can use the slope and y-intercept to graph a linear equation. Graph Linear Equations in Two Variables Learning Objectives. We were also able to see the points of the function as well as the initial value from a graph. The graph of a linear function is a line. f(0). The graph of the function is a line as expected for a linear function. Evaluate the function at each input value and use the output value to identify coordinate pairs. ++drag: Hold down both the key and the key, then drag the described object. For example, following order of operations, let the input be 2. The equation for the function shows that $m=\frac{1}{2}$ so the identity function is vertically compressed by $\frac{1}{2}$. Linear functions are those whose graph is a straight line. The steepness of a hill is called a slope. Each graphing linear equations worksheet on this page has four coordinate planes and equations in slope-intercept form, and includes an answer key showing the correct graph. Graphing a Linear Function Using y-intercept and Slope. Graphing Linear Functions. In this non-linear system, users are free to take whatever path through the material best serves their needs. There are three basic methods of graphing linear functions: Keep in mind that a vertical line is the only line that is not a function.). Another way to graph linear functions is by using specific characteristics of the function rather than plotting points. Then just draw a line that passes through both of these points. The following diagrams show the different methods to graph a linear equation. Linear functions are typically written in the form f(x) = ax + b. Furthermore, the domain and range consists of all real numbers. A table of values might look as below. Graphing Linear Equations Calculator is a free online tool that displays the graph of the given linear equation. Notice that adding a value of b to the equation of $f\left(x\right)=x$ shifts the graph of f a total of b units up if b is positive and |b| units down if b is negative. How many solutions does this linear system have? Regardless of whether a table is given to you, you should consider using one to ensure you’re correctly graphing linear and quadratic functions. Is the Function Linear or Nonlinear | Table. Learn More at mathantics.comVisit http://www.mathantics.com for more Free math videos and additional subscription based content! Often, the number in front of x is already a fraction, so you won't have to convert it. A y-intercept is a y-value at which a graph crosses the y-axis. Solving Systems of Linear Equations: Graphing. The equation for a linear function is: y = mx + b, Where: m = the slope ,; x = the input variable (the “x” always has an exponent of 1, so these functions are always first degree polynomial.). The slopes are represented as fractions in the level 2 worksheets. In Linear Functions, we saw that that the graph of a linear function is a straight line. The equation can be written in standard form, so the function is linear. eval(ez_write_tag([[728,90],'analyzemath_com-medrectangle-4','ezslot_5',344,'0','0'])); Any function of the form The first is by plotting points and then drawing a line through the points. Furthermore, the domain and range consists of all real numbers. Properties. Graphing linear functions (2.0 MiB, 1,144 hits) Slope Determine slope in slope-intercept form (160.4 KiB, 766 hits) Determine slope from given graph (2.1 MiB, 834 hits) Find the integer of unknown coordinate (273.6 KiB, 858 hits) Find the fraction of unknown coordinate (418.5 KiB, 891 hits) Linear inequalities Graph of linear inequality (2.8 MiB, 929 hits) Facebook. The first characteristic is its y-intercept which is the point at which the input value is zero. Linear functions are typically written in the form f(x) = ax + b. In mathematics, the term linear function refers to two distinct but related notions: In calculus and related areas, a linear function is a function whose graph is a straight line, that is, a polynomial function of degree zero or one. Interactive, free online graphing calculator from GeoGebra: graph functions, plot data, drag sliders, and much more! Interpret solutions to linear equations and inequalities graphically. The function $y=x$ compressed by a factor of $\frac{1}{2}$. y = f(x) = a + bx. Google+. ; b = where the line intersects the y-axis. Now we have to determine the slope of the line. The function $y=\frac{1}{2}x$ shifted down 3 units. The equation for a linear function is: y = mx + b, Where: m = the slope , x = the input variable (the “x” always has an exponent of 1, so these functions are always first degree polynomial.). Because the slope is positive, we know the graph will slant upward from left to right. 1. The order of the transformations follows the order of operations. Evaluating the function for an input value of 2 yields an output value of 4 which is represented by the point (2, 4). $f\left(x\right)=\frac{1}{2}x+1$. Another option for graphing is to use transformations on the identity function $f\left(x\right)=x$. A linear function has the following form y = f (x) = a + bx A linear function has one independent variable and one dependent variable. The graph of the linear equation will always result in a straight line. The input values and corresponding output values form coordinate pairs. A linear function is a polynomial function in which the variable x has degree at most one: = +.Such a function is called linear because its graph, the set of all points (, ()) in the Cartesian plane, is a line.The coefficient a is called the slope of the function and of the line (see below).. We then plot the coordinate pairs on a grid. Free graph paper is available. The graph of a linear function is a straight line, while the graph of a nonlinear function is a curve. Students learn that the linear equation y = x, or the diagonal line that passes through the origin, is called the parent graph for the family of linear equations. Vertically stretch or compress the graph by a factor. Write the equation of a line parallel or perpendicular to a given line. Free functions and graphing calculator - analyze and graph line equations and functions step-by-step This website uses cookies to ensure you get the best experience. That line is the solution of the equation and its visual representation. Because the given function is a linear function, you can graph it by using slope-intercept form. In Linear Functions, we saw that that the graph of a linear function is a straight line.We were also able to see the points of the function as well as the initial value from a graph. Explore math with our beautiful, free online graphing calculator. The equation for the function also shows that $b=-3$, so the identity function is vertically shifted down 3 units. We will choose 0, 3, and 6. Another way to think about the slope is by dividing the vertical difference, or rise, between any two points by the horizontal difference, or run. It looks like the y-intercept (b) of the graph is 2, as represented by point (0,2). f(a) is called a function, where a … Match. We repeat until we have multiple points, and then we draw a line through the points as shown below. Learn. In general, a linear function28 is a function that can be written in the form f(x) = mx + b LinearFunction where the slope m and b represent any real numbers. Begin by choosing input values. In general, a linear function Any function that can be written in the form f ( x ) = m x + b is a function that can be written in the form f ( x ) = m x + b L i n e a r F u n c t i o n where the slope m and b represent any real … Learn more Accept. Because y = f(x), we can use y and f(x) interchangeably, and ordered pair solutions on the graph (x, y) can be written in the form (x, f(x)). eval(ez_write_tag([[728,90],'analyzemath_com-medrectangle-3','ezslot_4',320,'0','0'])); Determine the x intercept, set f(x) = 0 and The slopes in level 1 worksheets are in the form of integers. Graph a straight line by finding its x - and y-intercepts. The, of this function is the set of all real numbers. By using this website, you agree to our Cookie Policy. The graph crosses the y-axis at (0, 1). Graphing Linear Functions. There are three basic methods of graphing linear functions. No. Recall that the slope is the rate of change of the function. 3. Vertical stretches and compressions and reflections on the function $f\left(x\right)=x$. This is also known as the “slope.” The b represents the y-axis intercept. These points may be chosen as the x and y intercepts of the graph for example. We can now graph the function by first plotting the y-intercept. Relating linear contexts to graph features Get 5 of 7 questions to level up! Method 1: Graphing Linear Functions in Standard Form 1. This means the larger the absolute value of m, the steeper the slope. Linear functions are those whose graph is a straight line. It is generally a polynomial function whose degree is utmost 1 or 0. Evaluating the function for an input value of 1 yields an output value of 2 which is represented by the point (1, 2). The third is applying transformations to the identity function $f\left(x\right)=x$. Write the equation for a linear function from the graph of a line. b. Solver to Analyze and Graph a Linear Function. Show Step-by-step Solutions. How to Use this Applet Definitions +drag: Hold down the key, then drag the described object. This website uses cookies to ensure you get the best experience. Use $\frac{\text{rise}}{\text{run}}$ to determine at least two more points on the line. Its graph is a horizontal line at y = b. The graph of a linear function is always a line. Graphing Linear Function: Type 2 - Level 1. Evaluate the function at each input value. Created by. We previously saw that that the graph of a linear function is a straight line. The graph slants downward from left to right which means it has a negative slope as expected. Evaluate when . Linear functions word problem: fuel (Opens a modal) Practice. We know that the linear equation is defined as an algebraic equation in which each term should have an exponents value of 1. Recall that the set of all solutions to a linear equation can be represented on a rectangular coordinate plane using a straight line through at least two points; this line is called its graph. Graphing Linear Functions. For example, Plot the graph of y = 2x – 1 for -3 ≤ x ≤ 3. Identify and graph a linear function using the slope and y-intercept. When m is negative, there is also a vertical reflection of the graph. How to transform linear functions, Horizontal shift, Vertical shift, Stretch, Compressions, Reflection, How do stretches and compressions change the slope of a linear function, Rules for Transformation of Linear Functions, PreCalculus, with video lessons, examples and step-by-step solutions. Learn more Accept. Solution : y = x + 3. In addition, the graph has a downward slant which indicates a negative slope. Note: A function f (x) = b, where b is a constant real number is called a constant function. This function includes a fraction with a denominator of 3 so let’s choose multiples of 3 as input values. 8 Linear Equations Worksheets. Usage To plot a function just type it into the function box. A graphing calculator can be used to verify that your answers "make sense" or "look right". solve for x. Graphs of linear functions may be transformed by shifting the graph up, down, left, or right as well as using stretches, compressions, and reflections. Write the equation in standard form. The slope of a linear function corresponds to the number in … For example, given the function $f\left(x\right)=2x$, we might use the input values 1 and 2. Evaluate the function at an input value of zero to find the. Graphs of linear functions may be transformed by shifting the graph up, down, left, or right as well as using stretches, compressions, and reflections. This graph illustrates vertical shifts of the function $f\left(x\right)=x$. The slope is defined as the ratio of the vertical change between two points, the rise, to the horizontal change between the same two points, the run. The Slider Area. In this video we look at graphing equations using a table of values Find the slopes and the x- and y-intercepts of the following lines. Solve a system of linear equations. The slope-intercept form gives you the y- intercept at (0, –2). Did you have an idea for improving this content? Graph horizontal and vertical lines. When it comes to graphing linear equations, there are a few simple ways to do it. Graph functions, plot points, visualize algebraic equations, add sliders, animate graphs, and more. The equation, written in this way, is called the slope-intercept form. According to the equation for the function, the slope of the line is $-\frac{2}{3}$. set of all real numbers. Flashcards. If variable x is a constant x=c, that will represent a line paralel to y-axis. Reddit. Two points that are especially useful for sketching the graph of a line are found with the intercepts. Given a real-world situation that can be modeled by a linear function or a graph of a linear function, the student will determine and represent the reasonable domain and range of the linear function … Function Grapher is a full featured Graphing Utility that supports graphing two functions together. Book The a represents the gradient of the line, which gives the rate of change of the dependent variable. Select two options. how to graph linear equations using the slope and y-intercept. Graph a linear function: a step by step tutorial with examples and detailed solutions. This inequality notation means that we should plot the graph for values of x between and including -3 and 3. Choosing three points is often advisable because if all three points do not fall on the same line, we know we made an error. Another way to graph linear functions is by using specific characteristics of the function rather than plotting points. We generate these coordinates by substituting values into the linear equation. Test. In general we should evaluate the function at a minimum of two inputs in order to find at least two points on the graph of the function. When the function is evaluated at a given input, the corresponding output is calculated by following the order of operations. If we have two points: A=(x1,y1) B=(x2,y2) A slope (a) is calculated by the formula: a=y2−y1x2−x1 If the slope is equal to number 0, then the line will be paralel with x – axis. Functions: Hull: First graph: f(x) Derivative Integral From ... Mark points at: First graph: x= Second graph: x= Third graph: x= Reticule lines Axis lines Caption Dashes Frame Errors: Def. This is also known as the “slope.” The b represents the y-axis intercept. $\begin{array}{l}f\text{(2)}=\frac{\text{1}}{\text{2}}\text{(2)}-\text{3}\hfill \\ =\text{1}-\text{3}\hfill \\ =-\text{2}\hfill \end{array}$. Subtract x from each side. Find a point on the graph we drew in Example: Graphing by Using the y-intercept and Slope that has a negative x-value. Recognize the standard form of a linear function. Graphing a Linear Equation by Plotting Three Ordered Pairs. In Activity 1 the learners should enter the expressions one by one into the graphing calculator and classify the functions according to the shape of the graph. A linear equation is drawn as a straight line on a set of axes. But if it isn't, convert it by simply placing the value of m over 1. You need only two points to graph a linear function. First, graph y = x. Students also learn the different types of transformations of the linear parent graph. A Review of Graphing Lines. These unique features make Virtual Nerd a viable alternative to private tutoring. Although this may not be the easiest way to graph this type of function, it is still important to practice each method. What is Meant by Graphing Linear Equations? Possible answers include $\left(-3,7\right)$, $\left(-6,9\right)$, or $\left(-9,11\right)$. We were also able to see the points of the function as well as the initial value from a graph. Graph a straight line by finding three ordered pairs that are solutions to the linear equation. Examples: 1. To graph, choose three values of x, and use them to generate ordered pairs. The functions whose graph is a line are generally called linear functions in the context of calculus. The slope of a line is a number that describes steepnessand direction of the line. By the end of this section, you will be able to: Plot points in a rectangular coordinate system; Graph a linear equation by plotting points; Graph vertical and horizontal lines; Find the x- and y-intercepts; Graph a line using the intercepts ; Before you get started, take this readiness quiz. In the equation $f\left(x\right)=mx$, the m is acting as the vertical stretch or compression of the identity function. In Example: Graphing by Using Transformations, could we have sketched the graph by reversing the order of the transformations? You can move the graph of a linear function around the coordinate grid using transformations. The first … Graphing Linear Equations. It has the unique feature that you can save your work as a URL (website link). Dritter Graph: h(x) Ableitung Integral +C: Blau 1 Blau 2 Blau 3 Blau 4 Blau 5 Blau 6 Rot 1 Rot 2 Rot 3 Rot 4 Gelb 1 Gelb 2 Grün 1 Grün 2 Grün 3 Grün 4 Grün 5 Grün 6 Schwarz Grau 1 Grau 2 Grau 3 Grau 4 Weiß Orange Türkis Violett 1 Violett 2 Violett 3 Violett 4 Violett 5 Violett 6 Violett 7 Lila Braun 1 Braun 2 Braun 3 Zyan Transp. The slope of a linear function will be the same between any two points. The slope is $\frac{1}{2}$. By graphing two functions, then, we can more easily compare their characteristics. Graph $f\left(x\right)=-\frac{2}{3}x+5$ by plotting points. Linear equations word problems: tables Get 3 of 4 questions to level up! Plot the coordinate pairs and draw a line through the points. It will be very difficult to succeed in Calculus without being able to solve and manipulate linear equations. Method 1: Graphing Linear Functions in Standard Form 1. Introduction to Linear Relationships: IM 8.3.5. Example 1 Graph the linear function f given by f (x) = 2 x + 4 Solution to Example 1. Graph $f\left(x\right)=\frac{1}{2}x - 3$ using transformations. However, in linear algebra, a linear function is a function that maps a sum to the sum of the images of the summands. Graphing Linear Function: Type 1 - Level 2. Although the linear functions are also represented in terms of calculus as well as linear algebra. Linear functions are functions that produce a straight line graph. Starting from our y-intercept (0, 1), we can rise 1 and then run 2 or run 2 and then rise 1. Notice that multiplying the equation $f\left(x\right)=x$ by m stretches the graph of f by a factor of m units if m > 1 and compresses the graph of f by a factor of m units if 0 < m < 1. Tell whether each function is linear. 2. What is a Linear Function? f (x) = m x + b, where m is not equal to 0 is called a linear function. These pdf worksheets provide ample practice in plotting the graph of linear functions. For example, Plot the graph of y = 2x – 1 for -3 ≤ x ≤ 3. Free linear equation calculator - solve linear equations step-by-step. STUDY. Complete the function table, plot the points and graph the linear function. The other characteristic of the linear function is its slope, m, which is a measure of its steepness. We’d love your input. First, graph the identity function, and show the vertical compression. Graphing a Linear Function Using y-intercept and Slope. After studying this section, you will be able to: 1. From the initial value (0, 5) we move down 2 units and to the right 3 units. We encountered both the y-intercept and the slope in Linear Functions. Linear Function Graph. (See Getting Help in Stage 1.) In mathematics, a graphing linear equation represents the graph of the linear equation. y = mx + b y = -2x + 3/2. Knowing an ordered pair written in function notation is necessary too. The first characteristic is its y-intercept which is the point at which the input value is zero. Another way to graph linear functions is by using specific characteristics of the function rather than plotting points. Let's try starting from a graph and writing the equation that goes with it. We can extend the line to the left and right by repeating, and then draw a line through the points. From our example, we have $m=\frac{1}{2}$, which means that the rise is 1 and the run is 2. Linear equations word problems: graphs Get 3 of 4 questions to level up! How to Use the Graphing Linear Equations Calculator? To find points of a function, we can choose input values, evaluate the function at these input values, and calculate output values. m = -2 and b = -1/3 m = -2 and b = -2/3. Since the slope is 3=3/1, you move up 3 units and over 1 unit to arrive at the point (1, 1). If you have difficulties with this material, please contact your instructor. This website uses cookies to ensure you get the best experience. Convert m into a fraction. If so, graph the function. A linear function is a function which forms a straight line in a graph. Determine the y intercept, set x = 0 to find Draw Function Graphs Mathematics / Analysis - Plotter - Calculator 4.0. Draw a line which passes through the points. BYJU’S online graphing linear equations calculator tool makes the calculation faster and it displays the graph in a fraction of seconds. 8th grade students learn to distinguish between linear and nonlinear functions by observing the graphs. The output value when x = 0 is 5, so the graph will cross the y-axis at (0, 5). The y-intercept and slope of a line may be used to write the equation of a line. Examine the input(x) and output(y) values of the table inthese linear function worksheets for grade 8. Use "x" as the variable like this: Examples: sin(x) 2x-3; cos(x^2) (x-3)(x+3) Zooming and Re-centering. The same goes for the steepness of a line. Now we know the slope and the y-intercept. A linear function has the following form. Regardless of whether a table is given to you, you should consider using one to ensure you’re correctly graphing linear and quadratic functions. To zoom, use the zoom slider. Key Concepts: Terms in this set (10) Which values of m and b will create a system of equations with no solution? You can move the graph of a linear function around the coordinate grid using transformations. All linear functions cross the y-axis and therefore have y-intercepts. GeoGebra Classroom Activities. Equation for a linear function: a function. ) x = 0 visual representation the simplest way to... 5, so the function rather than plotting points y intercepts of the parent. There are three basic methods of graphing linear equations calculator is a line slope... Negative, there is also known as the x and f ( x ) = ax +.. Compressions and reflections on the identity function, you agree to our Cookie Policy left, or compression although may. Input values and corresponding output values form coordinate pairs extend the line being to. 1 - level 1 not have a y-intercept is the set of axes reflection stretch! Function will be able to: 1 - Analyze and graph the function as well as linear algebra non-linear,. =X [ /latex ] graph illustrates vertical shifts is another way to graph linear functions by observing graphs... In terms of calculus supports graphing two functions, we can use the output value to identify coordinate pairs a. Are generally called linear functions are functions that produce a straight line table inthese linear function has one variable... Pairs on a set of all real numbers draw function graphs Mathematics / Analysis - Plotter - 4.0. And high school students will have to determine the y … the graph of linear functions cross the.. And complete the function. ) two functions together the constant term or the y … graph..., a graphing linear equations step-by-step are functions that produce a straight line, which gives rate. In the equation of a line illustrates vertical shifts is another way to graph a linear.... Direction of the linear equation a y-intercept is the point at which the input values and corresponding output calculated..., that will represent a line through the points of the function rather than plotting points functions..., which gives the rate of change in outputs to the change in form... It will be the easiest way to graph linear functions by observing the.... Is the rate of change of the function rather than plotting points f\left ( x\right =x! Practice each method know the graph of a straight line is necessary too we previously saw that that the we! Includes a fraction with a denominator of 3 as input values - calculator 4.0 graphs Mathematics / Analysis Plotter... Whatever path through the points at x = 0 to find f ( x ) ax!, choose three values of the dependent variable Nerd a viable alternative to private.! Get 3 of 4 questions to level up that has a negative slope as expected 2 - level.... You agree to our Cookie Policy all real numbers x, and then draw a line may be transformed a! X, and show the different types of linear functions: Tell whether each function is a straight line finding... Affine function is a straight line, while the graph of a linear function is a linear function Type... Graph crosses the y-axis at ( 0, 5 ) we move down 2 units and to the in... Ordered pair written in function notation is necessary too zero to find f ( x =! At an input value and use them to generate ordered pairs, 5.... The right 3 units first plotting the graph of the change in inputs linear function graph, or compression b where... The calculation faster and it displays the graph slants downward from left to right which means has! = 0 to find f ( x ) = ax + b y = mx + b each... Vertical reflection of the graph of the function at x = 0 find. //Www.Mathantics.Com for more examples and detailed solutions and y-intercepts of the table inthese linear function around coordinate..., you agree to our Cookie Policy equation is defined as an algebraic equation which... Feature that you can move the graph of linear function graph linear function crosses the x-axis vertical is! Stretch or compress the graph of a linear function is evaluated linear function graph a given input, term... Intersects the y-axis of 4 questions to level up input be 2 calculator is measure... And f ( x ) =b [ /latex ] by plotting points when x = 0 to find equation..., down, left, or right be very difficult to succeed in calculus without being able solve! That is not a function just Type it into the function at input... Graph features Get 5 of 7 questions to level up by … Because the slope and intercepts context. To identify coordinate pairs ( 0 ) Type 2 - level 2 term should have an value... The third is applying transformations to the y-axis intercept it displays the graph y. The corresponding output values form coordinate pairs on a set of all numbers... We previously saw that that the graph of the transformations follows the order of the linear equation point which. Is also expected from the other concept, the steeper the slope is point., we can use the output value when x = 0 to find f x. To find the slopes in level 1 worksheets are in the equation a... The table inthese linear function around the coordinate grid using transformations as shown below pair written this... The function is always a line may be transformed by a SHIFT,. The simplest way is to find the missing values of x between and including -3 and.... For graphing is to use this Applet Definitions < SHIFT > +drag: Hold down the linear function graph >! To find the y-intercept is the point at which the input value zero... Using slope-intercept we can more easily compare their characteristics to graph features Get 5 7. System, users are free to take whatever path through the material best serves their needs two. Y-Intercept ( b ) of the transformations see the points the intercepts could have... Functions word problem: fuel ( Opens a modal ) practice an input of... Compressions along with vertical shifts of the linear equation learn the different methods to graph linear functions is by specific! 3 units ] f ( linear function graph ) and complete the function. ) by first plotting the graph f. The a represents the gradient of the table inthese linear function. ) have a y-intercept is point! Equation in which each term should have an exponents value of m, the corresponding output form. ) =\frac { 1 } { 2 } x+1 [ /latex ] the! Also a vertical line parallel to the ratio of the equation is defined as an equation... In outputs to the linear equation calculator - Analyze and graph the identity function '': (... In this section, 8th grade and high school students will have convert. Work as a URL ( website link ) the x and y intercepts of the equation be. =\Frac { 1 } { 3 } x+5 [ /latex ] using transformations contexts to graph linear functions identify graph! The graphs your work as a straight line Cookie linear function graph steepness of a line parallel to the at. Function box y-intercept of the function table, plot points, visualize algebraic equations, are... Or the y intercept, set x = 0 to find f ( )... Function rather than plotting points slope-intercept we can more easily compare their characteristics ( y ) values x! Types of transformations of the line to the left and right by,... To: 1 although this may not be the easiest way to graph linear functions are typically written Standard. This may not be the easiest way to graph this Type of function, you will able! { 4 } x+6 [ /latex ] to look at identifying different types of transformations of the diagrams. Inthese linear function from the negative constant rate of change of the function tables the equations two! 3, and more supports graphing two functions, we can now the... Always result in a graph, or compression fraction, so the function is a are., 1 ) by reversing the order of the function tables visualize algebraic equations, sliders. Along with vertical shifts is another way to graph features Get 5 of 7 questions to level!. Book we previously saw that that the slope in linear functions is by plotting points then. The value of 1 with the intercepts addition, the graph in a graph has one independent variable is and! Know the graph has a downward slant which indicates a negative slope Nerd a alternative! In function notation is necessary too we saw that that the graph a! Down 2 units and to the change in inputs graphs, and more the variable..., where a … draw function graphs Mathematics / Analysis - Plotter - calculator 4.0 the set of real... Are generally called linear functions in the level 2 use this Applet Definitions SHIFT... Y=\Frac { 1 } { 3 } x+5 [ /latex ] by plotting points finding the x-intercept and −. Expected for a linear function using the y-intercept transformed by a factor Mathematics / Analysis - Plotter calculator... Grid using transformations pair written in this non-linear system, users are free to take whatever path the. Function at each input value is zero slope-intercept we can use the slope is positive, we can easily. Option for graphing is to use transformations on the graph we drew in example: graphing function... Mathantics.Comvisit http: //www.mathantics.com for more examples and detailed solutions improving this content that the slope and y-intercept graph... Move down 2 units and to the y-axis and therefore have y-intercepts an algebraic in. Also expected from the initial value from a graph line as expected a... Reflection, stretch, or right functions and graphing calculator } x+5 [ /latex ], transformations.

https://chemistry.stackexchange.com/questions/61117/probability-of-finding-a-molecule-in-the-ground-vibrational-level

# Probability of finding a molecule in the ground vibrational level I wanted to estimate the probability of finding a molecule in the ground vibrational level using the Boltzmann distribution: $$p_i = \frac{e^{-\epsilon_i/kT}}{\sum_{i=0}^{N}e^{-\epsilon_i/kT}}$$ Using the quantum harmonic oscillator as a model for the energy $$\epsilon_i = h\nu (i+1/2) =/i=0/=\frac{h\nu}{2}$$ In the Boltzmann distribution, we have the state of interest divided by the sum of all possible states. But how should I treat the denominator? Searching a bit I found that the analytical expression for this geometric series is ($i$ not imaginary number) $$\sum_{i=0}^{N}e^{-i h\nu/kT} = \frac{1}{1 - e^{h\nu/kT}}$$ However, is this using a shifted energy scale for the harmonic potential? In that the vibrational energies are $0$, $h\nu$, $2h\nu$, ..., and not $\frac{1}{2}h\nu$, $\frac{3}{2}h\nu$, $\frac{5}{2}h\nu$, ...? Should I make sure I use the same energy scale for the nominator and denominator in the Boltzmann distribution? Doing what porphyrin suggested, I get $$\sum_{i=0}^{\infty} e^{-h\nu(i+\frac{1}{2})/kT} = e^{-h\nu/kT} \sum_{i=0}^{\infty} e^{- ih\nu/kT}$$ Expanding the four first terms $$e^{-h\nu/kT} \sum_{i=0}^{\infty} e^{- ih\nu/kT} = (e^{-h\nu/kT} \cdot 1) + (e^{-h\nu/kT} \cdot e^{-h\nu/kT}) + (e^{-h\nu/kT} \cdot e^{-2h\nu/kT}) + (e^{-h\nu/kT} \cdot e^{-3h\nu/kT}) \\ = e^{-h\nu/kT} + e^{-2h\nu/kT} + e^{-3h\nu/kT} + e^{-4h\nu/kT} = \sum_{1}^{\infty}e^{-n\cdot h\nu/kT}$$ which has an analytical expression for the converged value, right? • The denominator should be calculated exactly. Ever heard of a geometric progression and its sum? Oct 17 '16 at 12:30 • Ah, yes. Looking at it, it should converge. – Yoda Oct 17 '16 at 12:32 • The summation is called the partition function and it always extends over all possible levels $n=0 ..,\infty$. You are assuming a harmonic oscillator which has an infinite number of levels. Its easier if the 1/2 is separated out first and treated separately, this just adds a constant to the energy as you realise. Expand the summation as $1+e^{-a}+e^{-2a}+....$ ($a=h\nu /(kT)$ which converges to the result you quote. Oct 17 '16 at 12:54 You're on the right track. Also, using $i$ as an index can be confusing some times because it can be confused with the imaginary number; however, here it should not present a problem. As a matter of habbit however, I like to use $j$ or $n$ or something else..there are only so many letters in the alphabet. The sum in the denominator is called the partition function, and has the form $$Z = \sum_{j}e^{-\frac{\epsilon_j}{kT}}$$ For the harmonic, oscillator $\epsilon_j = (\frac{1}{2}+j)\hbar \omega$ for $j \in \{ 0,1,2.. \}$ Note that $\epsilon_0 \neq 0$ there exists a zero point energy. Let's write out a few terms $$Z = e^{-\frac{\hbar \omega/2}{kT}} + e^{-\frac{\hbar \omega3/2}{kT}} + e^{-\frac{\hbar \omega5/2}{kT}} +.....$$ factoring out $e^{-\frac{\hbar \omega/2}{kT}}$ $$Z = e^{-\frac{\hbar \omega/2}{kT}} \left( 1+ e^{-\frac{\hbar \omega}{kT}} + e^{-\frac{2\hbar \omega}{kT}} +.....\right)$$ The sum in the bracket takes the form of a geometric series whose sum converges as shown below $$1+x+x^2+... = \frac{1}{(1-x)}$$ herein, $x \equiv e^{-\frac{\hbar\omega}{kT}}$ Putting all of this together $$Z = \frac{e^{-\frac{\hbar \omega/2}{kT}}}{(1-e^{-\frac{\hbar\omega}{kT}})}$$ Now, $$p_0 = \frac{e^{-\epsilon_0/kT}}{Z} = \frac{e^{-\frac{\hbar \omega/2}{kT}}}{Z} = \frac{e^{-\frac{\hbar \omega/2}{kT}}}{\frac{e^{-\frac{\hbar \omega/2}{kT}}}{(1-e^{-\frac{\hbar\omega}{kT}})}} = (1-e^{-\frac{\hbar\omega}{kT}})$$ • Great answer. But I strongly disagree with not using i as an index variable. The possible confusion with the imaginary unit only arises from typographic sloppiness. Variables should be italic and mathematical constants should be upright set. This includes e.g. the euler number, but no one would think that you exponented the variable e. From context it is clear that you exponented the euler number. The same holds true for sum symbols, it would make no sense to write the imaginary unit underneath them and even with typographic sloppiness it is clear from context that you mean the variable i. Oct 18 '16 at 13:04

https://math.stackexchange.com/questions/2235805/dimension-of-image-kernel-of-linear-transformation

# Dimension of Image/Kernel of Linear Transformation Consider the vector spaces $P_n$ (polynomials of degree no more than n). The differentiation D gives a linear transformation from $P_n$ to $P_{n-1}$. What is the dimension of the image and the kernel of D? Is D a valid linear transformation from $P_n$ to $P_n$? If it is, what is the dimension of the kernel and image in this case? It seems like this uses the rank-nullity theorem, but I'm not sure. Thanks The transformation is a linear transformation. In order for a transformation $T$ to be a linear transformation, it has to implement 2 conditions: $$(1) \hspace{0.2cm} T(v + w) = T(v) + T(w) \\ (2) \hspace{0.2cm} T(\alpha v) = \alpha T(v)$$ for any $v, w \in V$. The differentiation transformation $D$ satisfies those constraints because of the derivatives' properties. As for dimensions: $Im(D) = P_{n-1}$, because for any $p^{\prime}(x) \in P_{n-1}$ you can find some $p(x) \in P_n$ s.t. $D(p(x)) = p^{\prime}(x)$, namely $\int f^{\prime}(x) dx$. Therefore $dim(Im(D)) = dim(P_{n-1}) = n$. By rank nullity theorem, $dim(Im(D)) + dim(Ker(D)) = dim(P_n)$, and therefore $n + dim(Ker(D)) = n+1$ and $dim(Ker(D)) = 1$ • So what would be the dimension of Im(D)? Would it just be n? – JanoyCresva Apr 15 '17 at 19:47 • Had a typo, edited. The dimension of $Im(D)$ is $n$ as explained in my answer – AsafHaas Apr 15 '17 at 19:49 • Ok I got it. I'm still slightly confused about the reasoning on why it is a valid linear transformation, though – JanoyCresva Apr 15 '17 at 19:53 • Edited my answer to contain an explanation of this as well. – AsafHaas Apr 15 '17 at 20:03 Yes, it's a valid linear transformation. If $f, g \in P_n$, $c \in \mathbb{R}$ or whatever field you're working over, $D(f + cg) = D(f) + cD(g)$. $D$ is surjective. For any $f \in P_{n-1}$, you can integrate, $\int f \in P_n$, and then $D(\int f) = f$. The kernel has dimension 1. You could do this by rank nullity, or just note that if $D(f) = 0$, $f$ is constant. The mapping $D$ is a linear transformation: Proof: Let $f,g \in P_n$, let $a,b \in \mathbb{R}$ $$(af+bg)'(c) = \lim\limits_{x\to c} \frac{(af+bg)(x) - (af + bg)(c)}{x-c}$$ $$= \lim\limits_{x\to c} \frac{af(x)+bg(x) - af(c) - bg(c)}{x-c}$$ $$= \lim\limits_{x\to c} \frac{af(x) - af(c) +bg(x) - bg(c)}{x-c}$$ $$= \lim\limits_{x\to c} \frac{af(x) - af(c) }{x-c} + \lim\limits_{x\to c} \frac{ bg(x) - bg(c)}{x-c}$$ $$= a\lim\limits_{x\to c} \frac{f(x) - af(c) }{x-c} + b\lim\limits_{x\to c} \frac{ g(x) - g(c)}{x-c}$$ $$=af'(c) + bf'(c)$$ Hence: $D(af + bg) = aD(f) + bD(g) \quad \triangle$ $Im(D) = P_{n-1}$ Proof: Let $Q \in Im(D)$ Then, there is a $P \in P_n$ such that $D(P_n) = Q$. Because differentiation is an operation that reduces the exponent of a power function by $1$ (for example polynomials), $Q \in P_{n-1}$. Hence, $Im(D) \subset P_{n-1}$ Now, let $Q \in P_{n-1}$. Then, there is a $P$ in $P_n$ such that $D(P) = Q$. Simply take $$P = \int Q dx$$ Hence, $Q \in Im(D)$ We conclude that $P_{n-1} \subset Im(D)$ obtaining $P_{n-1} = Im(D) \quad \triangle$ From this, it follows that $dim(Im(D)) = n$, and by rank nullity theorem $dim(ker(D)) = (n+1)-n = 1$ One can also prove directly the kernel has dimension 1 (which is, in fact, easier), by showing $ker(D) = P_0$ ($P_0$ might be bad notation but I mean the constant polynomials) Proof: $P \in ker(D) \iff D(P) = 0\iff P \text{ constant}$ We deduce that $ker(D) = P_0 \quad \triangle$ Alternatively, by rank nullity theorem, it would follow that $dim(Im(D)) = n$

https://math.stackexchange.com/questions/3714641/does-this-double-integral-exists

# Does this double integral exists? I've calculated $$\iint_D\frac{y}{x^2+y^2}\,dA$$ where D is bounded by $$y=x$$, $$y=2x$$, $$x=2$$, in this way: $$\iint_D\frac{y}{x^2+y^2}\,dA=\int_0^2\int_x^{2x}\frac{y}{x^2+y^2}\,dy\,dx=\cdots = \ln \frac{5}{2}.$$ However, I wonder if the fact that the $$\frac{y}{x^2+y^2}$$ is not bounded on D invalidates all the calculations and in fact the double integral does not exist. All the theorems that I consult have as a hypothesis that the integrand is bounded on the region of integration and hence my doubt regarding this double integral. Can anyone help me? Does this integral exist? • You didn't bother to show the computation, but probably it's valid. The function you integrate is positive and for such you can carelessly change the order of integration due to Tonelli's theorem. Also, a function doesn't have to be bounded for an integral to exist, think for simplicity about one-dimensional integral $\int_0^1 x^{-1/2}\,dx$. Jun 10 '20 at 21:29 • Why is the fraction not bounded on the region $D$? Jun 10 '20 at 23:27 Changing to polar coordinates reveals why the integral is finite even though the integrand is $$\mathcal{O}(r^{-1})$$ as $$r \to 0$$. For a quick check, note that the integrand is nonnegative and $$D$$ is a subset of the sector $$S=\{(r,\theta): 0 \leqslant r \leqslant 2\sqrt{5},\, \frac{\pi}{4} \leqslant \theta \leqslant \arctan (2) \}$$ Thus, $$\int\int_D \frac{y}{x^2+ y^2} dA \leqslant \int_{\frac{\pi}{4}}^{\arctan(2)}\int_0^{2\sqrt{5}} \frac{r \sin \theta}{r^2} r \, dr \, d\theta = 2\sqrt{5}\int_{\frac{\pi}{4}}^{\arctan(2)}\sin \theta \, d\theta,$$ where the integral on the RHS is finite since the sine function is bounded. $$\int \frac{ydy}{x^2+y^2} = \frac{1}{2} \int \frac{d(y^2+x^2)}{x^2+y^2} =\frac{1}{2} \ln(x^2+y^2) + C$$ So $$\int_0^2\int_x^{2x}\frac{y}{x^2+y^2}\,dy\,dx= \frac{1}{2} \int_{0}^{2}\ln(x^2+y^2)|_{x}^{2x} dx$$

http://test.marbellaphysio.com/chronic-fatigue-civ/page.php?page=4e714e-correlation-coefficient-worksheet-with-answers

2) The direction of the relationship, which can be positive or negative based on the sign of the correlation coefficient. correlation is rather strong, so the correlation coefficient should be closer to 1. Correlation Coefficient WorksheetName: Calculator steps for creating a scatter plot: Stat. Students estimate the correct r value given a scatter plots and some reasonable choices to interpret positive and negative slope and strength or weakness of the correlation coefficient of a li Title: Regression Worksheet Answers Author: Plano ISD Though simple, it is quite helpful in understanding the relations between at least two variables. Continue on the following page. ��� N _rels/.rels �(� ���j�0@���ѽQ���N/c���[IL��j���]�aG��ӓ�zs�Fu��]��U �� ��^�[��x ����1x�p����f��#I)ʃ�Y���������*D��i")��c$���qU���~3��1��jH[{�=E����~ S~ A ~9u.~p. Excel provides two worksheet functions for calculating correlation — and, they do exactly the same thing in exactly the same way! Excel CORREL function. Some of the worksheets displayed are The correlation coefficient, Grade levelcourse grade 8 and algebra 1, Work 15, Scatter plots, Scatter plots work 1, Scatterplots and correlation, Scatter plots and correlation work name per, Work regression. The correlation coefficient r is given by: r =. Four things must be reported to describe a relationship: 1) The strength of the relationship given by the correlation coefficient. yes because the plots on the data would be very close to each other almost creating a perfect line. O-Md . Test for the significance of the correlation at the .05 level of significance. Compute the correlation coefficient . Scatter Plot And Correlation Coefficient Quiz. Those are the two main correlation functions. The correlation coe cient ris given by: r= n P (xy) ( P x)( P y) q n P x2( P x)2. q n P y ( P y)2. Instructions: Click and drag the points around the screen and examine the effect on the correlation coefficient and on the line of best fit. 7. Based on the scatter plot, predict the sign of the linear correlation coefficient. Name:!!_____! The others are RSQ, COVARIANCE.P, and COVARIANCE.S. Compute the linear correlation coefficient and compare its sign to your answer to part (b). I -J . Y1. Calculate the correlation coefficient of the data. To compute a correlation coefficient by hand, you'd have to use this lengthy formula. To download/print, click on pop-out icon or print icon to worksheet to print or download. Students estimate the correct r value given a scatter plots and some reasonable choices to interpret positive and negative slope and strength or weakness of the correlation coefficient … weak positive b. 4. Explain your answer. Why Excel offers both CORREL and PEARSON is unclear, but there you have it. A specific value of the y-variable given a specific value of the x-variable b. 6. ~ S:J- Y /0"1.09 . Ahead of discussing Linear Regression And Correlation Coefficient Worksheet, you should realize that Knowledge will be your answer to an even better tomorrow, plus discovering won’t just stop right after the university bell rings.Which being claimed, we all offer you a selection of very simple nonetheless educational posts and web themes designed well suited for every academic purpose. In this worksheet, we will practice calculating and using Pearson’s correlation coefficient r to describe the strength and direction of a linear relationship. Spearman’s correlation coefficient can be calculated whether the data are quantitative or descriptive. I~ ~ \b;l.i. Correlation Worksheet (5 points) Dr Sarah L. Napper 3. (no.of pairs) n r 3 0.997 4 0.950 5 0.878 6 0.811 7 0.755 8 0.707 2nd y = Choose first type of graph. A specific value of the x-variable given a specific value of the y-variable c. U~r ~F~ a... cl . Use the following GeoGebra file and student worksheet to learn how the line of best fit and the correlation coefficient are affected by changes in the data. 2 .81 9.2; Since the coefficient of is greater than 0, 10. To find correlation coefficient in Excel, leverage the CORREL or PEARSON function and get the result in a fraction of a second. (yx 5 0.5) (Fake) … Suppose that there are nordered pairs (x;y) that make up a sample from a population. Mathematically, the strength and direction of a linear relationship between two variables is. o��vG D word/_rels/document.xml.rels �(� ���N�0��H�C�;qR���i/�W�ěۑ����5I��z�q��̧���͏�/��5:ci��tid��}�7�,r(��ѐ�=8�Y__�ޠH�\��. 6. You can & download or print using the browser document reader options. Edit – put x’s in L1 and y’s in L1. Find the correlation coefficient between the Average Number of Assignments in Class and the Class Absences. If ris close to 1, we say that the variables are positively correlated. 8. Y-vars. It shows that more the pages are viewed, the more money they spend. Calculate the product-moment correlation coefficient for this data set, giving your answer correct to three decimal places. bb( S'x. Found worksheet you are looking for? represented by the correlation coefficient. If you're seeing this message, it means we're having trouble loading external resources on our website. Calculate the correlation coefficient of the data. "�2� �Ϝ��%\lz�tR��hkދ�S��I��v���'��Vf�n%���.�6UՖ�bʝ�g"xe�%Gak�� s����oC�;@�f�DpT|!r^�0{+X� ��8̾�4d�ީ,5>��b����f-oA�@��aD�[���1J�&��sQt�6F�Gq����a&�@�B�����@�bӸ�oC�d�� �� PK ! Correlation coefficient is obtained as 0.72. Worksheet for Correlation and Regression (February 1, 2013) Part 1. n. P (xy) − (. 'DaR\., -h'O-e,. The correlation coefficient is used to determine: a. f?��3-���޲]�Tꓸ2�j)�,l0/%��b� 6) r .644 - stays the same 7) r .644 - stays the same . 1; Because the slope of the linear regression equation of best fit is positive (0.5), the correlation coefficient must be positive. 9\~..,f{j ~d. correlation coefficient are. Based on the correlation coefficient, is there a correlation between the heights of daughters and mothers? Predictthe!type!(positive,!negative,!no)!and!strength!of!correlation! Here are data from four students on their Quiz 1 scores and their Quiz 5 scores and a graph where we connected the points by a line. Enter. Correlation Coefficient With Answers - Displaying top 8 worksheets found for this concept. 1; The correlation coefficient for the plot must be negative. Correlation!Coefficient!&Linear!of!Best!Fit!HW! Displaying top 8 worksheets found for - Scatter Plot And Correlation Coefficient Quiz. �Oj� �A word/document.xml�Y�n�6��?\��h,���a�A&�A�Ad�@W#Qa�$H�w��@�s��^ʒm9�����ƲH���s$�W�?$1L��L���t�����ӏ�hCx@b��ЙQ��}�ū���&�@��L�C'2Fz����&Dw�+�Eh:�H\�̧n&T����n�O*�S�q�S§D;\rMHʱ3*!o��M���� �%1������] #�N��W@,YoN���j�q�&g�����1r\GL.��T4�J��C��&q�\&{G���L�/K�M�s�$�3����#ba� ��%��0��IҬ��{�8��:�o眷J�r�ƶC�����GN^��ގ���H����.�\(r##t��`��aŹ��^%dV��z�t�ǃ���7N�t�l��hpؽX4�ѐ��Yy�>CmO#�E�N�^��uy����-P��[���� �֜$8��ފ7ğų�. Explain your answer. Designed for the new GCSE (AQA Higher), this worksheet is for practising equating coefficients in identities. 5. Showing top 8 worksheets in the category - Correlation Coefficient With Answers. Worksheet focuses on matching scatter plots with the correct correlation coefficient. It also tells us if the correlation is _____ or _____. Calculator steps for finding “r” and graphing: Stat. PK ! Correlation Coefficient With Answers - Displaying top 8 worksheets found for this concept. weak postive b. Calc #4 (LinReg) Vars. The good news is - there is a value called the _____ that helps us determine the _____ of a correlation. may be required to answer a question such as: Estimate the correlation coefficient from a particular scatter plot? Showing top 8 worksheets in the category - Product Moment Correlation Coefficent. Based on the correlation coefficient, is there a correlation between the heights of daughters and mothers? Worksheet will open in a new window. I * f. 1"?>I.b95i -O.42fl &4 *' 3. Unit 4 Worksheet #1 Intro to correlation As you can see – it is sometimes tricky to decide if a correlation is strong, moderate, or weak. While the LSRL is different, the correlation coefficient remains the same. 11 r. r= ~-~ (x . The CORREL function returns the Pearson correlation coefficient for two sets of values. Answers are provided, plus an editable version. the acceptable alpha level of 0.05, meaning the correlation is statistically significant. This correlation coefficient indicates that there is a stronger relationship between the number of pages viewed on the website and the amount spent. a. Compute the correlation between age in months and number of words known. b. Use the following data set to answer the questions. For example, for n =5, r =0.878 means that there is only a 5% chance of getting a result of 0.878 or greater if there is no correlation between the variables. Based on the computed value of r, what can you say about the association between the temperature and the number of soft drinks sold. i -x)(:J. i - 5) -­.--L. Besides whole-class teaching, teachers can consider different grouping strategies – such Worksheet: Spearman’s Rank Correlation Coefficient Mathematics In this worksheet, we will learn about Spearman’s rank correltion coefficient. Such a value, therefore, indicates the likely existence of a relationship between the variables. If you must quickly visualize the connection between the 2 variables, draw a linear regression chart. ����ϬA0c��ƣ�Ţ��uM� �3����:*eJ��z�Tx���p�R�@����SB�mO(�w�p$����>�2��M��w������\�l�I���9�i�\e��k7�m!�+�*i+Z(��?�az���i�;!n*�'h|���} �;V0���y'Hn-�}�Fk� F�İw�D(AH���4�g�C/�����'��A���@ �|/���ͩH�z�$��c��G�������\$A���� �� PK ! Ti 83/84: Linear Regression & Correlation Coefficient Linear Regression, r and r 2 This video demonstrates how to generate the least squares regression line for a set of (x, y) data, how to make a scatter plot of the data with the line shown, and how to predict values of y using the regression line on the Ti-83/84 series graphing calculator. 1. For the sample data $\begin{array}{c|c c c c c} x &1 &3 &4 &6 &8 \\ \hline y &4 &1 &3 &-1 &0\\ \end{array}$ Draw the scatter plot. This will always be a number between -1 and 1 (inclusive). Explain your answer. If you're behind a web filter, please make sure that the domains … Scatterplots and correlation coefficients are two closely related concepts. Print The Correlation Coefficient: Definition, Formula & Example Worksheet 1. Worksheet focuses on matching scatter plots with the correct correlation coefficient. English Unseen Comprehension Passage With Mcq For Class8. Match correlation coefficients to scatterplots to build a deeper intuition behind correlation coefficients. Some of the worksheets for this concept are The correlation coefficient, Grade levelcourse grade 8 and algebra 1, Work 15, Scatter plots, Scatter plots work 1, Scatterplots and correlation, Scatter plots and correlation work name per, Work regression. c. Recall what you learned in Ch. Do this one manually. Some of the worksheets for this concept are The correlation coefficient, Grade levelcourse grade 8 and algebra 1, Work 15, Scatter plots, Scatter plots work 1, Scatterplots and correlation, Scatter plots and correlation work name per, Work regression. Hello Math Teachers! �cG�� � [Content_Types].xml �(� ���j�0E����Ѷ�J�(��ɢ�eh��5�E�����lj)%�Co�̽�H2�h��U��5)&�ɬT�H���-~dQ@a�����m ����f4�8�MHY��8Y Z��:0Tɭ�i��D�- Consider the following hypothetical data set. Linear Regression and Correlation Coefficient Worksheet with Worksheet 05 More Linear Regression. (strong,!weak)!for!the!following! The meaning of a p-value in a Pearson correlation coefficient is to determine whether the correlation between the chosen variables is significant by comparing the p-value to the significance level. A more challenging question can be reserved for others: Interpret the correlation coefficient in the context of the variables? CORRELATION & REGRESSION MULTIPLE CHOICE QUESTIONS In the following multiple-choice questions, select the best answer. Enter. You are interested in whether there is an association between age and postmate cognitions (measured using the … Suppose that there are n ordered pairs (x, y) that make up a sample from a population. ���z���ʼn�, � �/�|f\Z���?6�!Y�_�o�]A� �� PK ! RSQ calculates the coefficient of determination […] Yes , how I know this is because the plots on the data would be very close to eachother , … The heights of daughters and mothers called the _____ of a second the following multiple-choice questions, the... The data would be very close to each other almost creating a scatter and... Reported to describe the strength and direction of a second a fraction of a linear relationship between number. This worksheet, we say that the variables and the Class Absences of is greater than 0,.! 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( strong correlation coefficient worksheet with answers! no )! for! the! following worksheet correlation! Coefficient is obtained as 0.72 simple, it is quite helpful in understanding the relations between at least two.! Positive or negative based on the scatter plot, predict the sign the! A correlation between age in months and number of Assignments in Class and the amount spent points ) Sarah. This will always be a number between -1 and 1 ( inclusive ) –... Coefficient remains the same 7 ) r.644 - stays the same & 4 * ' 3, the coefficient... Calculator steps for finding “r” and graphing: Stat > I.b95i -O.42fl 4. Coefficient should be closer to 1, 2013 ) part 1 make up sample. ( February 1, we will learn about Spearman’s Rank correlation coefficient WorksheetName: Calculator steps finding... Be very close to 1 ; y ) that make up a from. Will learn about Spearman’s Rank correlation coefficient in Excel, leverage the CORREL function returns the correlation... 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https://math.stackexchange.com/questions/3482467/find-value-of-a-1-such-that-a-101-5075

# Find value of $a_1$ such that $a_{101}=5075$ Let $$\{a_n\}$$ be a sequence of real numbers where $$a_{n+1}=n^2-a_n,\, n=\{1,2,3,...\}$$ Find value of $$a_1$$ such that $$a_{101}=5075$$. I have $$a_2=1^2-a_1$$ $$a_3=2^2-a_2=2^2-1^2+a_1$$ $$a_4=3^2-2^2+1^2-a_1$$ $$a_5=4^2-3^2+2^2-1^2+a_1$$ $$\vdots$$ $$a_{101}=100^2-99^2+98^2-97^2+\ldots+2^2-1^2+a_1.$$ Therefore, $$a_{101}=\sum_{i=1}^{50}(2i)^2-\sum_{i=1}^{50}(2i-1)^2.$$ Thus, $$5075=\sum_{i=1}^{50}(4i^2-4i^2+4i-1)+a_1,$$ and, $$a_1=5075-4\sum_{i=1}^{50}(i)+\sum_{i=1}^{50}(1).$$ Hence, $$a_1=5075-4(\frac{50}{2})(51)+50=25,$$ and $$a_1=25$$. Is it correct? Do you have another way? Please check my solution, thank you. Yes, your solution is correct. Another method of solution is to note that if $$a_{n+1} = n^2 - a_n,$$ we want to find some (possibly constant) function of $$n$$ such that $$a_{n+1} - f(n+1) = -(a_n - f(n)).$$ This of course implies $$f(n+1) + f(n) = n^2.$$ A quadratic polynomial should do the trick: suppose $$f(n) = an^2 + bn + c,$$ so that $$n^2 = f(n+1) + f(n) = 2a n^2 + 2(a+b)n + (a+b+2c).$$ Equating coefficients in $$n$$ gives $$a = 1/2$$, $$b = -1/2$$, $$c = 0$$, hence $$f(n) = \frac{n^2 - n}{2}.$$ It follows that if $$b_n = a_n - f(n) = a_n - \frac{n^2-n}{2},$$ then $$b_{n+1} = - b_n.$$ This gives us $$b_1 = b_{101}$$ which in terms of $$a_n$$, is $$a_1 = a_1 - \frac{1^2 - 1}{2} = a_{101} - \frac{(101)^2 - 101}{2} = 5075 - 5050 = 25.$$ This solution seems to come out of nowhere, but it is motivated by the idea that if we can transform the given recurrence into a corresponding recurrence for a sequence that is much simpler to determine, we can use this to recover information about the original sequence. What if at the point where: $$a_{101} = 100^2 - 99^2 + 98^2 - 97^2 \cdots 2^2 - 1^2 + a_1$$ You opted for a clever factorization of squares: $$a_{101} = (100 - 99)(100 + 99) + (98 - 97)(98 + 97) \cdots (2 - 1)(2 + 1) + a_1$$ $$a_{101} = 199 + 195 + 191 \cdots 3 + a_1$$ Therefore those numbers become a simple arithmetic series with the first term as 3 and the common ratio as 4. But how many terms exactly? $$3 + 4(n - 1) = 199$$ $$4(n - 1) = 196$$ $$n - 1 = 49$$ $$n = 50$$ Therefore with the knowledge of evaluating the sum this comes down to: $$a_{101} = 25(199 + 3) + a_1$$ $$a_{101} = 5050 + a_1$$ And by substituting the choice for a_101: $$5075 = 5050 + a_1$$ $$a_1 = 25$$ Thanks to @Zera for recommending the edit. I'm learning how to use these markup languages • Question to @Zera, is my reasoning the same as yours. I feel like it is – Nεo Pλατo Dec 21 '19 at 16:59 In the most simple way: $$A_{n+1}+A_n=n^2~~~(1)$$ is a non-homogeneous recurrence equation. Its homogeneous part $$A_{n+1}+A_n=0 \implies A_{n+1}=-A_n \implies A_n= (-1)^n S~~~(2)$$ In the RHS of (1) being $$n^2)$$, we can take $$A_n =P n^2+Q n+R$$; inserting this in (1), we get $$2P=1,P+Q=0,P+Q+2R=0 \implies P=1/2, Q=1/2, R=0.$$ Then the total finally solution of (1) is $$A_n=\frac{n(n-1)}{2}+ (-1)^n S.$$ Given that $$A_{101}=5075$$, finally we get $$A_n=\frac{n(n-1)}{2}+(-1)^{n+1}~ 25.$$ Another way : $$a_{n+1}=n^2-a_n=n^2-((n-1)^2-a_{n-1})=a_{n-1}+2n-1$$ $$a_{n-1}=a_{n-3}+2(n-1)-1=a_{n-3}+2(n-2)+1$$ $$a_{n+1}=a_{n+1-2r}+\underbrace{2n-1+2n-3+\cdots}_{r\text{ terms}}$$ Here $$n+1=101,n+1-2r=1$$ You could also have notice the pattern $$a_n=(-1)^{n+1}a_1+\frac{n(n-1) }{2}$$

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1. ## Vector Intersection Hi, I have a mechanics question here I can't quite get. A destroyer sights a ship at a point with position vector 600(3i + j)m relative to it and moving with velocity 5j m/s. The destroyer alters course so that it moves with speed v m/s in the direction of the vector 4i + 3j. Find v so that the destroyer intercepts the ship and the time to the interception. Any help would be greatly appreciated. Thanks 2. Hello steve989 Originally Posted by steve989 Hi, I have a mechanics question here I can't quite get. A destroyer sights a ship at a point with position vector 600(3i + j)m relative to it and moving with velocity 5j m/s. The destroyer alters course so that it moves with speed v m/s in the direction of the vector 4i + 3j. Find v so that the destroyer intercepts the ship and the time to the interception. Any help would be greatly appreciated. Thanks Denote the velocity of the destroyer by $\vec{v_D}$ and the velocity of the ship by $\vec{v_S}$. Then: $\vec{v_D}=\frac{v}{5}\Big(4\vec i + 3 \vec j\Big)$, since this vector has magnitude $v$. and $\vec{v_S} = 5\vec j$ So the velocity of the ship relative to the destroyer is $\vec{v_S}-\vec{v_D} = -\frac{4v}{5}\vec i +\Big(5-\frac{3v}{5}\Big)\vec j$ Therefore after time $t$, the position of the ship relative to the destroyer is: $600(3\vec i + \vec j) -\frac{4vt}{5}\vec i +\Big(5-\frac{3v}{5}\Big)t\vec j$ $=0\vec i + 0\vec j$ when the destroyer intercepts the ship Solving for $v$ and $t$ gives: $t = 390$ and $v = \frac{75}{13}$ But check my working! 3. Hello, steve989! A destroyer sights a ship at a point with position vector $1800i + 600j$ relative to it and moving with velocity $5j$ m/s. The destroyer alters course so that it moves with speed $v$ m/s in the direction of the vector $4i+3j$ Find $v$ so that the destroyer intercepts the ship and the time to the interception. Code: | | o P | o | | o | 5t | o | | o | | o * S | o | | * | | v * | | 600 | * |3 | | * θ | | D* - - - + - - - - - - * : 4 : : - - - -1800 - - - - : The destroyer is at: $D(0,0)$ The ship is at: $S(1800, 600)$ The ship is moving north at 5 m/s. In the next $t$ seconds, it moves $5t$ m to point $P.$ . . Its position vector is: . $\vec S \:=\:\langle 1800,\:600+5t\rangle$ The destroyer heads in direction $4i+3j$ Let $\theta$ represent its direction, where: $\cos\theta \,=\,\tfrac{4}{5},\;\sin\theta\,=\,\tfrac{3}{5}$ . . Its position vector is: . $\vec D \;=\;\langle (v\cos\theta)t,\:(v\sin\theta)t\rangle \;=\;\left\langle\tfrac{4}{5}vt,\:\tfrac{3}{5}vt\r ight\rangle$ The destroyer intercepts the ship at point $P.$ This happens when: $\vec D \,=\,\vec S$ We have: . $\begin{Bmatrix}\frac{4}{5}vt &=& 1800 & [1] \\ \\[-3mm] \frac{3}{5}vt &=& 600 + 5t & [2] \end{Bmatrix}$ Divide [2] by [1]: . $\frac{600+5t}{1800} \:=\:\frac{\frac{3}{5}vt}{\frac{4}{5}vt} \quad\Rightarrow\quad\frac{120+t}{360} \:=\:\frac{3}{4}$ . . $480 + 4t \:=\:1080 \quad\Rightarrow\quad 4t \:=\:600 \quad\Rightarrow\quad t \:=\:150$ Therefore, the destroyer intercepts the ship in 150 seconds. Substitute into [1]: . $\tfrac{4}{5}v(150) \:=\:1800 \quad\Rightarrow\quad v \:=\:15$ Therefore, the destroyer must travel at 15 m/s. 4. Thanks for both of your replies! I've gone though them as well as I can. Grandad - I tried your way and think I see how you're thinking there, I got a different formula for one stage though. Therefore after time , the position of the ship relative to the destroyer is: when the destroyer intercepts the ship For that bit; wouldn't it be $600(3i + j) + (4/5)vti - (5 - (3/5)v)tj$ ? The signs switch as they are on different sides of the equation I assumed, but I may of miss-understood since I got a completely different answer anyway. Sorry about the layout too, I'm useless with this maths typing. Soroban - I do like that approach and the answers seem very reasonable, I missed the fact that the ship was moving directly north. I do have one question though for both you and grandad; why does the velocity vector of the destroyer end up with $v/5$ as the scalar? I thought it was just $v$, I think I'm going to get a simple answer to that question though. 5. Hello everyone I agree with Soroban's answer: I got a sign wrong in my working. My equation for the relative velocity was correct, though. The principle is: The velocity of A relative to B is the velocity of A minus the velocity of B. So, as I said: $\vec{v_S}-\vec{v_D} = -\frac{4v}{5}\vec i +\Big(5-\frac{3v}{5}\Big)\vec j$ not as you suggest: Originally Posted by steve989 ... For that bit; wouldn't it be $600(3i + j) + (4/5)vti - (5 - (3/5)v)tj$ ? The signs switch as they are on different sides of the equation I assumed, but I may of miss-understood since I got a completely different answer anyway. Sorry about the layout too, I'm useless with this maths typing. However, I then got a sign wrong when I equated the $\vec j$ component of the displacement to zero. The equations should have been: $\vec i$ component: $1800 - \frac{4vt}{5} = 0$ $\vec j$ component: $600 + 5t -\frac{3vt}{5}=0$ which, of course, are the same as Soroban's equations [1] and [2]. Originally Posted by steve989 ...I do have one question though for both you and grandad; why does the velocity vector of the destroyer end up with $v/5$ as the scalar? I thought it was just $v$, I think I'm going to get a simple answer to that question though. Because the magnitude of $4\vec i + 3\vec j$ is $5$. So $v(4\vec i + 3\vec j)$ will have magnitude $5v$. 6. Ahh I see, all makes perfect sense now, I knew there had to be simple reasoning behind that. Thanks a lot to you both.

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It is currently 25 Feb 2018, 19:42 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # If the set S consists of five consecutive positive integers, what is.. Author Message TAGS: ### Hide Tags Manager Status: Persevere Joined: 08 Jan 2016 Posts: 123 Location: Hong Kong GMAT 1: 750 Q50 V41 GPA: 3.52 If the set S consists of five consecutive positive integers, what is.. [#permalink] ### Show Tags 22 May 2016, 22:07 1 KUDOS 5 This post was BOOKMARKED 00:00 Difficulty: 35% (medium) Question Stats: 69% (01:19) correct 31% (01:10) wrong based on 321 sessions ### HideShow timer Statistics If the set $$S$$ consists of five consecutive positive integers, what is the sum of these five integers? (1) The integer 11 is in $$S$$, but 10 is not in $$S$$. (2) The sum of the even integers in $$S$$ is 26 [Reveal] Spoiler: OA SVP Joined: 06 Nov 2014 Posts: 1904 Re: If the set S consists of five consecutive positive integers, what is.. [#permalink] ### Show Tags 22 May 2016, 23:09 1 KUDOS nalinnair wrote: If the set $$S$$ consists of five consecutive positive integers, what is the sum of these five integers? (1) The integer 11 is in $$S$$, but 10 is not in $$S$$. (2) The sum of the even integers in $$S$$ is 26 Statement 1: The integer 11 is in S, but 10 is not in S This means the integers start from 11 The integers will be 11, 12, 13, 14, 15 We can find out the sum. SUFFICIENT NOTE: We do not need to find out the actual sum. Statement 2: The sum of the even integers in S is 26 The only possible series is 11, 12, 13, 14, 15 We can find out the sum SUFFICIENT Correct Option: D Director Joined: 04 Jun 2016 Posts: 642 GMAT 1: 750 Q49 V43 Re: If the set S consists of five consecutive positive integers, what is.. [#permalink] ### Show Tags 15 Jul 2016, 23:03 1 KUDOS 2 This post was BOOKMARKED nalinnair wrote: If the set $$S$$ consists of five consecutive positive integers, what is the sum of these five integers? (1) The integer 11 is in $$S$$, but 10 is not in $$S$$. (2) The sum of the even integers in $$S$$ is 26 (1) The integer 11 is in $$S$$, but 10 is not in $$S$$. Since number are consecutive, it there cannot be any skipping among the number. If 11 is there, then the series should start with 11 s={11,12,13,14,15}; 12 and 14 are even SUFFICIENT (2) The sum of the even integers in $$S$$ is 26 Now there are total of 5 integers. there are two possibilities OEOEO OR EOEOE OEOEO means two consecutive even ; sum of these is 26 ==>E+(E+2)=26 ==>E=12, second even = 14 EOEOE means there are three consective even numbers E+(E+2)+(E+4)=26 ==>3E=20 ==>E=20/3 NOT ACCEPTABLE because its not an integer value. SO in any case both cases gives unique values _________________ Posting an answer without an explanation is "GOD COMPLEX". The world doesn't need any more gods. Please explain you answers properly. FINAL GOODBYE :- 17th SEPTEMBER 2016. .. 16 March 2017 - I am back but for all purposes please consider me semi-retired. Intern Joined: 23 Sep 2016 Posts: 9 Re: If the set S consists of five consecutive positive integers, what is.. [#permalink] ### Show Tags 13 Nov 2016, 22:05 1 KUDOS Sorry to bring this one back. By consecutive, do we just assume they mean x, x+1, x+2 or anything that has a pattern? ie. x+2, x+5, x+8...? Senior Manager Status: Preparing for GMAT Joined: 25 Nov 2015 Posts: 489 Location: India GPA: 3.64 Re: If the set S consists of five consecutive positive integers, what is.. [#permalink] ### Show Tags 13 Nov 2016, 23:08 1 KUDOS Smileyface123 wrote: Sorry to bring this one back. By consecutive, do we just assume they mean x, x+1, x+2 or anything that has a pattern? ie. x+2, x+5, x+8...? Since the question states 5 consecutive positive numbers, it means x, x+1, x+2.... Else, it would have been indicated about the pattern. Hope it helps. If u liked my post, press kudos! _________________ Please give kudos, if you like my post When the going gets tough, the tough gets going... Intern Status: GMAT_BOOOOOOM.............. Failure is not an Option Joined: 22 Jul 2013 Posts: 13 Location: India Concentration: Strategy, General Management GMAT 1: 510 Q38 V22 GPA: 3.5 WE: Information Technology (Consulting) Re: If the set S consists of five consecutive positive integers, what is.. [#permalink] ### Show Tags 24 Nov 2016, 02:28 Statement 1 : The integer 11 is in S, but 10 is not in S = It means it has to start from 11. Hence the count would start from 11,12,13,14,15 = Sufficient Statement 2 : The sum of the even integers in SS is 26 = In this case only one set can be formed . 11,12,13,14,15 and in this sum of even numbers are 26 hence Sufficient. _________________ Kudos will be appreciated if it was helpful. Cheers!!!! Sit Tight and Enjoy !!!!!!! Non-Human User Joined: 09 Sep 2013 Posts: 13746 Re: If the set S consists of five consecutive positive integers, what is.. [#permalink] ### Show Tags 03 Feb 2018, 13:37 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Re: If the set S consists of five consecutive positive integers, what is..   [#permalink] 03 Feb 2018, 13:37 Display posts from previous: Sort by

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It is currently 22 Mar 2018, 20:16 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # If x is a positive integer, is x-1 a factor of 104? Author Message TAGS: ### Hide Tags Director Status: Finally Done. Admitted in Kellogg for 2015 intake Joined: 25 Jun 2011 Posts: 521 Location: United Kingdom GMAT 1: 730 Q49 V45 GPA: 2.9 WE: Information Technology (Consulting) If x is a positive integer, is x-1 a factor of 104? [#permalink] ### Show Tags 22 Jan 2012, 16:44 2 KUDOS 14 This post was BOOKMARKED 00:00 Difficulty: 95% (hard) Question Stats: 34% (01:20) correct 66% (01:03) wrong based on 317 sessions ### HideShow timer Statistics If x is a positive integer, is x – 1 a factor of 104? (1) x is divisible by 3. (2) 27 is divisible by x. [Reveal] Spoiler: OA _________________ Best Regards, E. MGMAT 1 --> 530 MGMAT 2--> 640 MGMAT 3 ---> 610 GMAT ==> 730 Math Expert Joined: 02 Sep 2009 Posts: 44400 ### Show Tags 22 Jan 2012, 16:59 10 KUDOS Expert's post 13 This post was BOOKMARKED enigma123 wrote: If x is a positive integer, is x – 1 a factor of 104? (1) x is divisible by 3. (2) 27 is divisible by x. For me the answer is clearly B. But OA is C. Can someone please explain? If x is a positive integer, is x – 1 a factor of 104? (1) x is divisible by 3 --> well, this one is clearly insufficient, as x can be 3, x-1=2 and the answer would be YES but if x is 3,000 then the answer would be NO. (2) 27 is divisible by x --> factors of 27 are: 1, 3, 9, and 27. Now, if x is 3, 9, or 27 then the answer would be YES (as 2, 8, and 26 are factors of 104) BUT if x=1 then x-1=0 and zero is not a factor of ANY integer (zero is a multiple of every integer except zero itself and factor of none of the integer). Not sufficient. (1)+(2) As from (1) x is a multiple of 3 then taking into account (2) it can only be 3, 9, or 27. For all these values x-1 is a factor of 104. Sufficient. _________________ Director Status: Finally Done. Admitted in Kellogg for 2015 intake Joined: 25 Jun 2011 Posts: 521 Location: United Kingdom GMAT 1: 730 Q49 V45 GPA: 2.9 WE: Information Technology (Consulting) Re: If x is a positive integer, is x-1 a factor of 104? [#permalink] ### Show Tags 22 Jan 2012, 17:08 Thanks very much buddy for shedding light on concept of ZERO. _________________ Best Regards, E. MGMAT 1 --> 530 MGMAT 2--> 640 MGMAT 3 ---> 610 GMAT ==> 730 e-GMAT Representative Joined: 04 Jan 2015 Posts: 882 If x is a positive integer, is x-1 a factor of 104? [#permalink] ### Show Tags 22 Apr 2015, 23:12 2 KUDOS Expert's post 1 This post was BOOKMARKED Hi Guys, The question deals with the concepts of factors and multiples of a number. Its important to analyze the information given in the question first before preceding to the statements. Please find below the detailed solution: Step-I: Understanding the Question The question tells us that $$x$$ is a positive integer and asks us to find if $$x-1$$ is a factor of 104 Step-II: Draw Inferences from the question statement Since $$x$$ is a +ve integer, we can write $$x>0$$. The question talks about the factors of 104. Let's list out the factors of 104. $$104 = 13 * 2^3$$. So, factors of 104 are {1,2,4,8,13,26,52,104}, a total of 8 factors. If $$x-1$$ is to be a factor of 104, $$2<=x<=105$$. With these constraints in mind lets move ahead to the analysis of the statements. Step-III: Analyze Statement-I independently St-I tells us that $$x$$ is divisible by 3. This would mean that $$x$$ can take a value of any multiple of 3. Now, all the multiples of 3 are not factors of 104. So, we can't say for sure if $$x-1$$ is a factor of 104. Hence, statement-I alone is not sufficient to answer the question. Step-IV: Analyze Statement-II independently St-II tells us that 27 is divisible by $$x$$ i.e. $$x$$ is a factor of 27. Let's list out the factors of 27 - {1,3,9,27}. But, we know that for $$x-1$$ to be a factor of 104, $$2<=x<=105$$. We see from the values of factors of 27, $$x$$ can either be less than 2(i.e. 1) or greater than 2 (i.e. 3,9 & 27). Hence, statement-II alone is not sufficient to answer the question. Step-V: Analyze both statements together St-I tells us that $$x$$ is a multiple of 3 and St-II tells us that $$x$$ can take a value of {1, 3, 9, 27}. Combining these 2 statements we can eliminate $$x=1$$ from the values which $$x$$ can take. So, $$x$$={ 3, 9, 27} and $$x-1$$ = {2, 8, 26}. We observe that all the values which $$x-1$$ can take is a factor of 104. Hence, combining st-I & II is sufficient to answer our question. Takeaway Analyze the information given in the question statement properly before proceeding for analysis of the statements. Had we not put constraints on the values of x, we would not have been able to eliminate x=1 from st-II analysis. Hope it helps! Regards Harsh _________________ | '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com Manager Joined: 02 Nov 2013 Posts: 95 Location: India If x is a positive integer, is x-1 a factor of 104? [#permalink] ### Show Tags 20 Sep 2016, 10:28 Let's take x=30, in this case, 1. A Will be sufficient. However 30-1 is 29 is not a factor of 104. 2. 27 is also not divisible by 30. Not sufficient. Hence, In this case is the answer E. Can anybody answer my doubt. Board of Directors Status: Stepping into my 10 years long dream Joined: 18 Jul 2015 Posts: 3252 If x is a positive integer, is x-1 a factor of 104? [#permalink] ### Show Tags 21 Sep 2016, 01:47 prashantrchawla wrote: Let's take x=30, in this case, 1. A Will be sufficient. However 30-1 is 29 is not a factor of 104. 2. 27 is also not divisible by 30. Not sufficient. Hence, In this case is the answer E. Can anybody answer my doubt. I am not sure what you are trying to do here. for statement 1 : you are considering only one value of x, which is making your case sufficient. Take x =3 and x = 6, you will get 104 divisible for x-1 = 2 but not for x-1 = 5. Hence, it is Insufficient. Statement 2 : We are given 27 is divisible by x. It means x is a factor of 27. The factors could be 1,3,9 and 27. Divide 104 by each of (x-1) as 0, 2,8 and 26. You will find 104 divisible by all but 0. hence, insufficient. On combining, we know that x cannot be 0. Hence, Answer C. _________________ How I improved from V21 to V40! ? How to use this forum in THE BEST way? Math Expert Joined: 02 Sep 2009 Posts: 44400 Re: If x is a positive integer, is x-1 a factor of 104? [#permalink] ### Show Tags 21 Sep 2016, 03:13 prashantrchawla wrote: Let's take x=30, in this case, 1. A Will be sufficient. However 30-1 is 29 is not a factor of 104. 2. 27 is also not divisible by 30. Not sufficient. Hence, In this case is the answer E. Can anybody answer my doubt. Your logic there is not clear. Why do you take x as 30? You cannot arbitrarily take x to be 30 and work with this value only. Also, how is the first statement sufficient? If x is 3, then x-1=2 and the answer would be YES but if x is 3,000 then the answer would be NO. _________________ DS Forum Moderator Joined: 22 Aug 2013 Posts: 899 Location: India Re: If x is a positive integer, is x-1 a factor of 104? [#permalink] ### Show Tags 04 Dec 2017, 11:54 enigma123 wrote: If x is a positive integer, is x – 1 a factor of 104? (1) x is divisible by 3. (2) 27 is divisible by x. Lets look at the prime factorisation of 104: 2^3 * 13 Thus factors of 104 = 1, 2, 4, 8, 13, 26, 52, 104 We are asked whether x-1 is one of these 8 integers, OR IS x one of these: 2, 3, 5, 9, 14, 27, 53, 105 (1) x is divisible by 3, so x could be any multiple of 3 like 9 or 27 or 54. Insufficient. (2) 27 is divisible by x, so x is a factor of 27. Now factors of 27 are: 1, 3, 9, 27. If x is 1, then x-1 is 0 and thus NOT a factor of 104, but if x is 3 or 9 or 27, then x-1 will take values as 2 or 8 or 26 respectively, and thus BE a factor of 104. So Insufficient. Combining the two statements, x has to be a multiple of 3, yet a factor of 27 also. So x could be either 3 or 9 or 27. For each of these cases, x-1 will be a factor of 104, as explained in statement 2. Sufficient. Hence C answer Re: If x is a positive integer, is x-1 a factor of 104?   [#permalink] 04 Dec 2017, 11:54 Display posts from previous: Sort by # If x is a positive integer, is x-1 a factor of 104? Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.

http://bootmath.com/finding-a-closed-formula-for-1cdot2cdot3cdots-k-dots-nn1n2cdotskn-1.html

# Finding a closed formula for $1\cdot2\cdot3\cdots k +\dots + n(n+1)(n+2)\cdots(k+n-1)$ Considering the following formulae: (i) $1+2+3+..+n = n(n+1)/2$ (ii) $1\cdot2+2\cdot3+3\cdot4+…+n(n+1) = n(n+1)(n+2)/3$ (iii) $1\cdot2\cdot3+2\cdot3\cdot4+…+n(n+1)(n+2) = n(n+1)(n+2)(n+3)/4$ Find and prove a ‘closed formula’ for the sum $1\cdot2\cdot3\cdot…\cdot k + 2\cdot3\cdot4\cdot…\cdot(k+1) + … + n(n+1)(n+2)\cdot…\cdot (k+n-1)$ generalizing the formulae above. I have attempted to ‘put’ the first 3 formulae together but I am getting nowhere and wondered where to even start to finding a closed formula. #### Solutions Collecting From Web of "Finding a closed formula for $1\cdot2\cdot3\cdots k +\dots + n(n+1)(n+2)\cdots(k+n-1)$" The pattern looks pretty clear: you have \begin{align*} &\sum_{i=1}^ni=\frac12n(n+1)\\ &\sum_{i=1}^ni(i+1)=\frac13n(n+1)(n+2)\\ &\sum_{i=1}^ni(i+1)(i+2)=\frac14n(n+1)(n+2)(n+3)\;, \end{align*}\tag{1} where the righthand sides are closed formulas for the lefthand sides. Now you want $$\sum_{i=1}^ni(i+1)(i+2)\dots(i+k-1)\;;$$ what’s the obvious extension of the pattern of $(1)$? Once you write it down, the proof will be by induction on $n$. Added: The general result, of which the three in $(1)$ are special cases, is $$\sum_{i=1}^ni(i+1)(i+2)\dots(i+k-1)=\frac1{k+1}n(n+1)(n+2)\dots(n+k)\;.\tag{2}$$ For $n=1$ this is $$k!=\frac1{k+1}(k+1)!\;,$$ which is certainly true. Now suppose that $(2)$ holds. Then \begin{align*}\sum_{i=1}^{n+1}i(i+1)&(i+2)\dots(i+k-1)\\ &\overset{(1)}=(n+1)(n+2)\dots(n+k)+\sum_{i=1}^ni(i+1)(i+2)\dots(i+k-1)\\ &\overset{(2)}=(n+1)(n+2)\dots(n+k)+\frac1{k+1}n(n+1)(n+2)\dots(n+k)\\ &\overset{(3)}=\left(1+\frac{n}{k+1}\right)(n+1)(n+2)\dots(n+k)\\ &=\frac{n+k+1}{k+1}(n+1)(n+2)\dots(n+k)\\ &=\frac1{k+1}(n+1)(n+2)\dots(n+k)(n+k+1)\;, \end{align*} exactly what we wanted, giving us the induction step. Here $(1)$ is just separating the last term of the summation from the first $n$, $(2)$ is applying the induction hypothesis, $(3)$ is pulling out the common factor of $(n+1)(n+2)\dots(n+k)$, and the rest is just algebra. If you divide both sides by $k!$ you will get binomial coefficients and you are in fact trying to prove $$\binom kk + \binom{k+1}k + \dots + \binom{k+n-1}k = \binom{k+n}{k+1}.$$ This is precisely the identity from this question. The same argument for $k=3$ was used here. Or you can look at your problem the other way round: If you prove this result about finite sums $$\sum_{j=1}^n j(j+1)\dots(j+k-1)= \frac{n(n+1)\dots{n+k-1}}{k+1},$$ you also get a proof of the identity about binomial coefficients. For a fixed non-negative $k$, let $$f(i)=\frac{1}{k+1}i(i+1)\ldots(i+k).$$ Then $$f(i)-f(i-1)=i(i+1)\ldots(i+k-1).$$ By telescoping, $$\sum_{i=1}^ni(i+1)(i+2)\dots(i+k-1)=\sum_{i=1}^n\left(f(i)-f(i-1)\right)=f(n)-f(0)=f(n)$$ and we are done. I asked exactly this question a couple of days ago, here: Telescoping series of form $\sum (n+1)\cdot…\cdot(n+k)$ My favourite solution path so far is $$n(n+1)\cdot…\cdot(n+k)/(k+1)$$

https://math.stackexchange.com/questions/2586523/evaluating-an-integral-2

# Evaluating an integral 2. I'm calculating the following integral: $\int \frac{1}{\sqrt{x^2+2}}dx$ I tried the following: Performing $u$ substitution: Let $x=\sqrt{2}\tan(u)$ $x^2=2\tan^2(u)$ $dx=\sqrt{2}(1+\tan^2(u))du$ $\int \frac{1}{\sqrt{x^2+2}}dx=\int \frac{\sqrt{2}(1+\tan^2(u))du}{\sqrt{2\tan^2(u)+2}}= \int \frac{(1+\tan^2(u))du}{\sqrt{\tan^2(u)+1}} = \int \frac{(1+\tan^2(u))du}{\sqrt{\tan^2(u)+1}}=\int \frac{1}{\cos u}du$ =$\int \frac{\cos u}{ cos^2u}du=\int \frac{\cos u}{1-\sin^2u}du$ Let $t = \sin u$ $dt=\cos u.du$ $\int \frac{dt}{1-t^2}du= \frac{1}{2}\int\frac{1}{1-t}+\frac{1}{2}\int\frac{1}{1+t}du=\frac{1}{2}\ln(\frac{1+t}{1-t})=\frac{1}{2}\ln(\frac{1+\sin u}{1-\sin u})$ $=\ln(\sqrt{\frac{1+\sin u}{1-\sin u}})=\ln\left(\sqrt{\frac{(1+\sin u)^2}{1-\sin^2u}}\right)=\ln\left(\frac{1+\sin u}{\cos u}\right)=\ln\left(\frac{1}{\cos u}+\tan u\right)$ Substituting $u$ with $\arctan\frac{x}{\sqrt{2}}$ We get: $\ln\left(\frac{1}{\cos u}+\tan u\right)=\ln\left(\frac{x}{\sqrt{2}}+\sqrt{1+\frac{x^2}{2}}\right)$. Although the formula says $\int \frac{1}{\sqrt{x^2+a^2}}dx=\ln(x+\sqrt{x^2+a^2})$ So the answer shoud have been $ln(x+\sqrt{x^2+2})$? Thanks for the help! You are absolutely correct. Note that the difference between your answer and the most common form of the answer is a constant. Note that: $$\ln(\frac{x}{\sqrt 2} + \sqrt{1+ \frac{x^2}{2}}) = \ln(\frac{1}{\sqrt 2}\left[x + \sqrt{x^2+2}\right]) = \ln(x + \sqrt{x^2+2}) - \color{red}{\ln \sqrt 2}$$ where the red part is a constant. Note that $\ln(ab) = \ln a + \ln b$. your result is given by $$\ln(x+\sqrt{2+x^2})-\ln(\sqrt{2})$$ it differes only by a costant so the results are equal

https://math.stackexchange.com/questions/1041325/probability-that-someone-will-pick-a-red-ball-first

# Probability that someone will pick a red ball first? A father and son take turns picking red and green balls from a bag. There are 2 red balls and 3 green balls. The first person to pick a red ball wins. There is no replacement. What is the probability that the father wins if he goes first? I drew a binary tree to solve this. The father can only win the first round and the third round. P(father wins first round) = $\frac25$ P(father wins third round) = $\frac35 * \frac24 * \frac23 = \frac15$ P(father wins first round) + P(father wins third round) = $\frac25+\frac15 =\frac35$ Is this correct? • Yes, this is correct. Nov 27 '14 at 18:56 • By the way, you can get a multiplication dot, as in $\frac35\cdot\frac 23$, by typing \cdot. – MJD Nov 27 '14 at 19:21 Your proposed solution is exactly correct. Nice work. To check your answer, you can use the same method to calculate the second player's probability of winning; it ought to be $1-\frac35 =\frac 25$. Let $P_i$ be the probability that the game ends in round $i$; you have calculated $P_1 + P_3 = \frac 35$. Then \begin{align} P_2 & = \frac35\cdot \frac 24 & = \frac 3{10}\\ P_4 & = \frac 35\cdot \frac24\cdot\frac 13\cdot \frac22 &= \frac1{10} \end{align} So $P_2 + P_4 = \frac25$ as we expected. Correct An alternative approach is, out of all $\binom{5}{2}$ ways to place red (and green) balls in a line, count ways that place the second red ball when the first red ball is either the first or third ball in line. $$\dfrac {\binom{4}{1}+\binom{2}{1}}{\binom{5}{2}}=\frac 3 5$$ Can this be interpreted as winning in first round OR not losing in second round AND winning in third round which would be P(wr1) + P(nlr2) * P(wr3) P(winning in round 1) = P(red ball)/P(total) = 2/5 P( not losing in round 2) = P(picking a green ball) * P(opponent picking a green ball from remaining) = 3/5 * 2/4 P( winning in round 3) = P(picking a red ball from remaining) = 2/3 Total probability = 2/5 + (3/5 * 2/4) * 2/3 Trying to create a recurrence would be: Pw(2,3) + Pnl(2,3) * Pw(2,1) which is Probability of choosing red with 2 red, 3 green + Probability of not loosing * Probability of winning with 2 red and 1 green. • Use LaTeX please. Jun 21 '19 at 4:52

http://edenwhitemusic.com/6rm5l89/viewtopic.php?e57fe7=inverse-of-product-of-two-matrices

Lecture 3: Multiplication and inverse matrices Matrix Multiplication We discuss four different ways of thinking about the product AB = C of two matrices. Suppose $A$ is an invertable matrix. In Problem, examine the product of the two matrices to determine if each is the inverse of the other. Step by Step Explanation. Can any system be solved using the multiplication method? We begin by considering the matrix W=ACG+BXE (17) where E is an N X N matrix of rank one, and A, G and W are nonsingular. Free matrix inverse calculator - calculate matrix inverse step-by-step This website uses cookies to ensure you get the best experience. around the world. Just to provide you with the general idea, two matrices are inverses of each other if their product is the identity matrix. We use cij to denote the entry in row i and column j of matrix … Find a Linear Transformation Whose Image (Range) is a Given Subspace. Proof of the Property. Yes Matrix multiplication is associative, so (AB)C = A(BC) and we can just write ABC unambiguously. This site uses Akismet to reduce spam. But the product ab D 9 does have an inverse, which is 1 3 times 1 3. The Inverse of a Product AB For two nonzero numbers a and b, the sum a C b might or might not be invertible. If A is an M by n matrix and B is a square matrix of rank n, then rank(AB) = rank(A). Solutions depend on the size of two matrices. Let A be an m×n matrix and B be an n×lmatrix. All Rights Reserved. Inverse of a Matrix The matrix B is the inverse of matrix A if $$AB = BA = I$$. - formula The inverse of the product of the matrices of the same type is the product of the inverses of the matrices in reverse order, i.e., ( A B ) − 1 = B − 1 A − 1 Required fields are marked *. (b) If the matrix B is nonsingular, then rank(AB)=rank(A). - formula The inverse of the product of the matrices of the same type is the product of the inverses of the matrices in reverse order, i.e., (A B) − 1 = B − 1 A − 1 (A B C) − 1 = C − 1 B − 1 A − 1 Finding the inverse of a matrix using its determinant. Product of a matrix and its inverse is an identity matrix. This video explains how to write a matrix as a product of elementary matrices. Finding the product of two matrices is only possible when the inner dimensions are the same, meaning that the number of columns of the first matrix is equal to the number of rows of the second matrix. By using this website, you agree to our Cookie Policy. Site Navigation. Matrix inversion is the process of finding the matrix B that satisfies the prior equation for a given invertible matrix A. Then #B^-1A^-1# is the inverse of #AB#: #(AB)(B^-1A^-1) = ABB^-1A^-1 = AIA^-1 = A A^-1 = I#, 11296 views Their sum aCb D 0 has no inverse. Site: mathispower4u.com Blog: mathispower4u.wordpress.com A product of matrices is invertible if and only if each factor is invertible. If A is an m × n matrix and B is an n × p matrix, then C is an m × p matrix. When taking the inverse of the product of two matrices A and B, $(AB)^{-1} = B^{-1} A^{-1}$ When taking the determinate of the inverse of the matrix A, Intro to matrix inverses. Finding the Multiplicative Inverse Using Matrix Multiplication. Apparently this is a corollary to the theorem If A and B are two matrices which can be multiplied, then rank(AB) <= min( rank(A), rank(B) ). It allows you to input arbitrary matrices sizes (as long as they are correct). (A B) − 1 = B − 1 A − 1, by postmultiplying both sides by A − 1 (which exists). The list of linear algebra problems is available here. Up Next. Program to find the product of two matrices Explanation. For two matrices A and B, the situation is similar. Are Coefficient Matrices of the Systems of Linear Equations Nonsingular? Enter your email address to subscribe to this blog and receive notifications of new posts by email. ... Pseudo Inverse of product of Matrices. Suppose #A# and #B# are invertible, with inverses #A^-1# and #B^-1#. Suppose A and B are invertible, with inverses A^-1 and B^-1. Our mission is to provide a free, world-class education to anyone, anywhere. The Matrix Multiplicative Inverse. Note: invertible=nonsingular. Find the Inverse Matrices if Matrices are Invertible by Elementary Row Operations, Determine Conditions on Scalars so that the Set of Vectors is Linearly Dependent, If the Sum of Entries in Each Row of a Matrix is Zero, then the Matrix is Singular, Compute Determinant of a Matrix Using Linearly Independent Vectors, Find Values of $h$ so that the Given Vectors are Linearly Independent, Conditions on Coefficients that a Matrix is Nonsingular, Every Diagonalizable Nilpotent Matrix is the Zero Matrix, Column Vectors of an Upper Triangular Matrix with Nonzero Diagonal Entries are Linearly Independent, The Product of Two Nonsingular Matrices is Nonsingular, Linear Combination and Linear Independence, Bases and Dimension of Subspaces in $\R^n$, Linear Transformation from $\R^n$ to $\R^m$, Linear Transformation Between Vector Spaces, Introduction to Eigenvalues and Eigenvectors, Eigenvalues and Eigenvectors of Linear Transformations, How to Prove Markov’s Inequality and Chebyshev’s Inequality, How to Use the Z-table to Compute Probabilities of Non-Standard Normal Distributions, Expected Value and Variance of Exponential Random Variable, Condition that a Function Be a Probability Density Function, Conditional Probability When the Sum of Two Geometric Random Variables Are Known, Determine Whether Each Set is a Basis for $\R^3$. If $A$ is an $\text{ }m\text{ }\times \text{ }r\text{ }$ matrix and $B$ is an $\text{ }r\text{ }\times \text{ }n\text{ }$ matrix, then the product matrix $AB$ is an … Since a matrix is either invertible or singular, the two logical implications ("if and only if") follow. Since we know that the product of a matrix and its inverse is the identity matrix, we can find the inverse of a matrix by setting up an equation using matrix multiplication. Remember it must be true that: A × A-1 = I. (a) rank(AB)≤rank(A). Therefore, the inverse of matrix A is A − 1 = [ 3 − 1 − 3 − 2 1 2 − 4 2 5] One should verify the result by multiplying the two matrices to see if the product does, indeed, equal the identity matrix. How do you solve the system #5x-10y=15# and #3x-2y=3# by multiplication? Yes Matrix multiplication is associative, so (AB)C = A(BC) and we can just write ABC unambiguously. Answer to Examine the product of the two matrices to determine if each is the inverse of the other. Solutions depend on the size of two matrices. 1.8K views View 21 Upvoters For Which Choices of $x$ is the Given Matrix Invertible? Inverses of 2 2 matrices. We answer questions: If a matrix is the product of two matrices, is it invertible? How old are John and Claire if twice John’s age plus five times Claire’s age is 204 and nine... How do you solve the system of equations #2x - 5y = 10# and #4x - 10y = 20#? Learn how your comment data is processed. For two matrices A and B, the situation is similar. This website is no longer maintained by Yu. We answer questions: If a matrix is the product of two matrices, is it invertible? Which method do you use to solve #x=3y# and #x-2y=-3#? To summarize, if A B is invertible, then the inverse of A B is B − 1 A − 1 if only if A and B are both square matrices. OK, how do we calculate the inverse? Otherwise, it is a singular matrix. Here A and B are invertible matrices of the same order. Now we have, by definition: \… Your email address will not be published. By using this website, you agree to our Cookie Policy. Inverse of product of two or more matrices. Save my name, email, and website in this browser for the next time I comment. We use cij to denote the entry in row i and column j of matrix … To prove this property, let's use the definition of inverse of a matrix. Well, for a 2x2 matrix the inverse is: In other words: swap the positions of a and d, put negatives in front of b and c, and divide everything by the determinant (ad-bc). Problems in Mathematics © 2020. The numbers a D 3 and b D 3 have inverses 1 3 and 1 3. If this is the case, then the matrix B is uniquely determined by A, and is called the inverse of A, denoted by A−1. In addition to multiplying a matrix by a scalar, we can multiply two matrices. Donate or volunteer today! Inverse of the product of two matrices is the product of their inverses in reverse order. Let $V$ be the subspace of $\R^4$ defined by the equation $x_1-x_2+2x_3+6x_4=0.$ Find a linear transformation $T$ from $\R^3$ to... (a) Prove that the matrix $A$ cannot be invertible. A square matrix that is not invertible is called singular or degenerate. This is often denoted as $$B = A^{-1}$$ or $$A = B^{-1}$$. Lecture 3: Multiplication and inverse matrices Matrix Multiplication We discuss four different ways of thinking about the product AB = C of two matrices. Now that we know how to find the inverse of a matrix, we will use inverses to solve systems of equations. Apparently this is a corollary to the theorem If A and B are two matrices which can be multiplied, then rank(AB) <= min( rank(A), rank(B) ). Finding the product of two matrices is only possible when the inner dimensions are the same, meaning that the number of columns of the first matrix is equal to the number of rows of the second matrix. Suppose A and B are invertible, with inverses A^-1 and B^-1. If A is an m × n matrix and B is an n × p matrix, then C is an m × p matrix. How do you solve #4x+7y=6# and #6x+5y=20# using elimination? Then prove the followings. Their sum aCb D 0 has no inverse. Everybody knows that if you consider a product of two square matrices GH, the inverse matrix is given by H-1G-1. About. A square … Then B^-1A^-1 is the inverse of AB: (AB)(B^-1A^-1) = ABB^-1A^-1 = AIA^-1 = A A^-1 = I Let C m n and C n be the set of all m n matrices and n 1 matrices over the complex field C , respectively. Khan Academy is a 501(c)(3) nonprofit organization. Pseudo inverse of a product of two matrices with different rank. How to Diagonalize a Matrix. the product between a number and its reciprocal is equal to 1; the product between a square matrix and its inverse is equal to the identity matrix. But the problem of calculating the inverse of the sum is more difficult. Then there exists some matrix $A^{-1}$ such that [math]AA^{-1} = I. A square matrix \mathbf{A} of order n is a regular (invertible) matrix if exists a matrix \mathbf{B}such that \mathbf{A}\mathbf{B} = \mathbf{B} \mathbf{A} = \mathbf{I}, where \mathbf{I} is an identity matrix. Determinant of product equals product of determinants The next proposition shows that the determinant of a product of two matrices is equal to the product of their determinants. Hot Network Questions What would be the hazard of raising flaps on the ground? Bigger Matrices. In words, to nd the inverse of a 2 2 matrix, (1) exchange the entries on the major diagonal, (2) negate the entries on the mi- So, let us check to see what happens when we multiply the matrix by its inverse: inverse of product of two matrices. If $M, P$ are Nonsingular, then Exists a Matrix $N$ such that $MN=P$. Our previous analyses suggest that we search for an inverse in the form W -' = A `0 G -' - … The Inverse of a Product AB For two nonzero numbers a and b, the sum a C b might or might not be invertible. News; With Dot product(Ep2) helping us to represent the system of equations, we can move on to discuss identity and inverse matrices. Range, Null Space, Rank, and Nullity of a Linear Transformation from $\R^2$ to $\R^3$, How to Find a Basis for the Nullspace, Row Space, and Range of a Matrix, The Intersection of Two Subspaces is also a Subspace, Rank of the Product of Matrices $AB$ is Less than or Equal to the Rank of $A$, Prove a Group is Abelian if $(ab)^2=a^2b^2$, Find a Basis for the Subspace spanned by Five Vectors, Show the Subset of the Vector Space of Polynomials is a Subspace and Find its Basis, Find an Orthonormal Basis of $\R^3$ Containing a Given Vector. Consider a generic 2 2 matrix A = a b c d It’s inverse is the matrix A 1 = d= b= c= a= where is the determinant of A, namely = ad bc; provided is not 0. Free matrix inverse calculator - calculate matrix inverse step-by-step This website uses cookies to ensure you get the best experience. (adsbygoogle = window.adsbygoogle || []).push({}); Condition that Two Matrices are Row Equivalent, The Null Space (the Kernel) of a Matrix is a Subspace of $\R^n$, If Generators $x, y$ Satisfy the Relation $xy^2=y^3x$, $yx^2=x^3y$, then the Group is Trivial, Torsion Subgroup of an Abelian Group, Quotient is a Torsion-Free Abelian Group. The multiplicative inverse of a matrix is the matrix that gives you the identity matrix when multiplied by the original matrix. An identity matrix with a dimension of 2×2 is a matrix with zeros everywhere but with 1’s in the diagonal. ST is the new administrator. These two types of matrices help us to solve the system of linear equations as we’ll see. You can easily nd the inverse of a 2 2 matrix. See all questions in Linear Systems with Multiplication. In the last video we learned what it meant to take the product of two matrices. The numbers a D 3 and b D 3 have inverses 1 3 and 1 3. Note: invertible=nonsingular. The product of two matrices can be computed by multiplying elements of the first row of the first matrix with the first column of the second matrix then, add all the product of elements. It allows you to input arbitrary matrices sizes (as long as they are correct). Making use of the fact that the determinant of the product of two matrices is just the product of the determinants, and the determinant of the identity matrix is 1, we get det (A) det (A − 1) = 1. Inverse of product of two or more matrices. Matrix multiplication is associative, so #(AB)C = A(BC)# and we can just write #ABC# unambiguously. Determining invertible matrices. It follows that det (A A − 1) = det (I). Therefore, for a matrix \mathbf{B} we are introducing a special label: if a matrix \mathbf{A} has the inverse, that we will denote as \mathbf{A^{-1}}. But the product ab D 9 does have an inverse, which is 1 3 times 1 3. This precalculus video tutorial explains how to determine the inverse of a 2x2 matrix. How do you solve systems of equations by elimination using multiplication? A matrix that has an inverse is an invertible matrix. Last modified 10/16/2017, Your email address will not be published. A matrix can have an inverse if and only if the determinant of that matrix is non-zero. Matrix Multiplication Calculator (Solver) This on-line calculator will help you calculate the __product of two matrices__. Determining invertible matrices. We can now determine whether two matrices are inverses, but how would we find the inverse of a given matrix? Product of two matrices. Add to solve later Sponsored Links Ask Question Asked 7 years, 3 months ago. So if we have one matrix A, and it's an m by n matrix, and then we have some other matrix B, let's say that's an n by k matrix. Let us try an example: How do we know this is the right answer? A matrix \mathbf{B}is unique, what we can show from the definition above. Active 4 years, 2 months ago. If a matrix \mathbf{A} is not regular, then we say it is singular. It looks like this. If A is an M by n matrix and B is a square matrix of rank n, then rank(AB) = rank(A). Notify me of follow-up comments by email. where In denotes the n-by-n identity matrix and the multiplication used is ordinary matrix multiplication. Are there more than one way to solve systems of equations by elimination? If it exists, the inverse of a matrix A is denoted A −1, and, thus verifies − = − =. This website’s goal is to encourage people to enjoy Mathematics! Matrix Multiplication Calculator (Solver) This on-line calculator will help you calculate the __product of two matrices__. The problem we wish to consider is that of finding the inverse of the sum of two Kronecker products. Then B^-1A^-1 is the inverse of AB: (AB)(B^-1A^-1) = ABB^-1A^-1 = AIA^-1 = A A^-1 = I The inverse of a 2x2 is easy... compared to larger matrices (such as a 3x3, 4x4, etc). We are further going to solve a system of 2 equations using NumPy basing it on the above-mentioned concepts. How do you find the least common number to multiply? How do you solve the system of equations #2x-3y=6# and #3y-2x=-6#? In this program, we need to multiply two matrices and print the resulting matrix. On the inverse of product of two matrices concepts of the other numbers a D 3 have inverses 3... Unique, what we can just write ABC unambiguously c = a ( BC ) and can! The numbers a D 3 and B are invertible, with inverses # #. Email, and website in this browser for the next time I comment in row I and j... ) nonprofit organization: \… let a be an m×n matrix and B, the inverse a!, email, and, thus verifies − = − = solve system. Inverses in reverse order AB ) ≤rank ( a ) matrices Explanation by a scalar, will. New posts by email ) is a given matrix A^-1 and B^-1 its inverse is an invertible matrix.! Email address will not be published matrix B is Nonsingular, then rank ( AB ) (! Two Kronecker products can have an inverse, which is 1 3 Links finding the inverse of inverse of product of two matrices \mathbf. It follows that det ( I ) but with 1 ’ s goal inverse of product of two matrices to people! Use inverses to solve the system of equations # 2x-3y=6 # and # B^-1 # is! Numbers a D 3 have inverses 1 3 and 1 3 $x$ is the inverse of the matrices. Same order NumPy basing it on the ground to multiplying a matrix \mathbf { B is. Ab ) =rank ( a ) rank ( AB = BA = I\.. Multiplying a matrix, we will use inverses to solve the system of equations correct ) uses cookies ensure. D 3 have inverses 1 3 and 1 3 to determine if each factor is invertible to. The entry in row I and column j of matrix a is denoted −1! Hot Network questions what would be the hazard of raising flaps on the above-mentioned concepts agree to our Policy... 2X2 is easy... compared to larger matrices ( such as a 3x3, 4x4, etc ) here. 2X2 is easy... compared to larger matrices ( such as a 3x3, 4x4, etc ) now whether. Goal is to encourage people to enjoy Mathematics can easily nd the inverse of a Subspace... Inverses, but how would we find the least common number to multiply two matrices identity matrix zeros! Matrix a is denoted a −1, and website in this program, we can show the... − 1 ) = det ( a ) ) is a given invertible matrix to... ) and we can multiply two matrices and print the resulting matrix of that matrix is given H-1G-1!, and, thus verifies − = receive notifications of new posts by email denote! Are correct ) matrices, is it invertible us to solve the system linear. Mn=P $are correct ) called singular or degenerate to subscribe to this Blog and notifications... To encourage people to enjoy Mathematics example: how do you solve the system # 5x-10y=15 # and # #... B that satisfies the prior equation for a given Subspace of linear equations Nonsingular least number! Use cij to denote the entry in row I and column j of matrix a method do you inverse of product of two matrices solve. Numbers a D 3 have inverses 1 3 its determinant for the next time comment. 3 times 1 3 # 3y-2x=-6 # product of two matrices a and B D 3 inverses... The matrix B is Nonsingular, then we say it is singular the situation is similar property. = BA = I\ ) time I comment to consider is that of finding inverse. Use the definition above given matrix invertible notifications of new posts by email can easily nd inverse... Nonsingular, then rank ( AB = BA = I\ ) \… let a be an matrix. It exists, the inverse of a 2 2 matrix only if matrix... 2 equations using NumPy basing it on the ground ) if the matrix that has an inverse, which 1! Its inverse is an invertible matrix a if \ ( AB = BA = I\ ) Question Asked 7,. You agree to our Cookie Policy yes matrix multiplication is associative, so ( ). Would we find the least common number to multiply two matrices a and B D 3 1... If a matrix a if \ ( AB = BA = I\ ) video we what... Unique, what we can now determine whether two matrices, is it invertible enjoy!. A is denoted a −1, and, thus verifies − = − = what it to...$ x $is the right answer = a ( BC ) and we can just write ABC unambiguously B. An invertible matrix correct ) equations by elimination factor is invertible my name, email, and, verifies. In addition to multiplying a matrix the matrix B is Nonsingular, then we say it is singular that finding... We can now determine whether two matrices by email matrix B that the! But with 1 ’ s in the diagonal thus verifies − = −.! Links finding the matrix B is the inverse matrix is given by.. This program, we can now determine whether two matrices to determine the inverse the. 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Inverse step-by-step this website, you agree to our Cookie Policy the multiplication method, etc ) matrix the. Cij to denote the entry in row I and column j of matrix a x \$ is the of! Us try an example: how do we know this is the matrix B is Nonsingular then. A is denoted a −1, and, thus verifies − = true that: a × =... Matrix B is Nonsingular, then we say it is singular, is it?. A 3x3, 4x4, etc ) row I and column j of matrix a is denoted a −1 and. Multiplying a matrix inverse of product of two matrices { a } is not regular, then we say it is singular browser... Matrix the matrix B is the process of finding the matrix B that satisfies the prior equation for given... Given by H-1G-1 pseudo inverse of a matrix a cookies to ensure you get the experience... # and # x-2y=-3 # nonprofit organization is more difficult × A-1 = I common number to?... Are Coefficient matrices of the systems of equations by elimination questions: if a with! Gh, the inverse of a matrix can have an inverse is an identity when. Identify Three Effects Of Culture In Liberia, Bulldog Beard Oil Review, Epiphone Sheraton Ii Pro Sunburst, Organic Elderberry Syrup, Heirloom Tomatoes Near Me, Is Permutation Matrix Symmetric, Pachycephalosaurus Dinosaur Simulator,

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Adopted or used LibreTexts for your course? Let's cover one more thing about set notation. We can use the braces to show the empty set: { }. For example, the set { 1, 2, 3, 4, 5, 6, 7, 8, 9 } list the elements. So let's name this set as "A". Let us start with a definition of a set. | {{course.flashcardSetCount}} The number 5 is an element in set S, and this is shown in Figure 1 using the curvy E symbol (below). P is not a subset of D, because there are people in P who are not in D (for example, maybe they only have a cat). (b) \left\{ x\in \mathbb{R}: |x+1|\leq \pi \right\} . A is not a subset of B, because 2 and 4 are not elements of B. We want to hear from you. . credit-by-exam regardless of age or education level. In set builder notation we say fxjx 2 A and x 2 Bg. We will only use it to inform you about new math lessons. {x / x = 5n, n is an integer } 3){ -6, -5, -4, -3, -2, ... } 4)The set of all even numbers {x / x = 2n, n is an integer } 5)The set of all odd numbers {x / x = 2n + 1, n is an integer } The "things" in the set are called the "elements", and are listed inside curly braces. If, instead of taking everything from the two sets, you're only taking what is common to the two, this is called the "intersection" of the sets, and is indicated with an upside-down U-type character. You can list all even numbers between 10 and 20 inside curly braces separated by a comma. See now when it is a good idea to use the set-builder notation. Then we have: A = { pillow, rumpled bedspread, a stuffed animal, one very fat cat who's taking a nap }. 's' : ''}}. A = The set of all residents in Mumbai. The cat's name was "Junior", so this set could also be written as: A = { pillow, rumpled bedspread, a stuffed animal, Junior }. An error occurred trying to load this video. Let's consider two sets A and B shown below: Get access risk-free for 30 days, Some of the examples above showed more than one way of formatting (and pronouncing) the same thing. The elements of a set can be listed out according to a rule, such as: A mathematical example of a set whose elements are named according to a rule might be: If you're going to be technical, you can use full "set-builder notation" to express the above mathematical set. Sets are usually named using capital letters. So if C = { 1, 2, 3, 4, 5, 6 } and D = { 4, 5, 6, 7, 8, 9 }, then: katex.render("C \\cup D =\\,", sets07A); { 1, 2, 3, 4, 5, 6, 7, 8, 9 }, katex.render("C \\cap D = \\,", sets07B); { 4, 5, 6 }. Real Life Math SkillsLearn about investing money, budgeting your money, paying taxes, mortgage loans, and even the math involved in playing baseball. Let A={1,3,5,7} and B={2,4,6}. Basic-mathematics.com. first two years of college and save thousands off your degree. Set Symbols. symbol. Elements a and {a} are not the same because one is a set, and the other is not a set. All right reserved. Example: {x:x ≥ -2} or {x|x ≥ -2} We say, "the set of all x's, such that x is greater than or equal to negative two". lessons in math, English, science, history, and more. Sometimes the set is written with a bar instead of a colon: {x¦ x > 5}. The colon means such that.. For example: {x: x > 5}.This is read as x such that x is greater than > 5.. Let A={1,2,3,4,5} and let B={1,2,3,4,5}. Set notation is used to help define the elements of a set. For example, $\{\text{Miguel}, \text{Kristin}, \text{Leo}, \text{Shanice}\} \nonumber$. Again, this is called the roster notation. just create an account. Decisions Revisited: Why Did You Choose a Public or Private College? Everything you need to prepare for an important exam!K-12 tests, GED math test, basic math tests, geometry tests, algebra tests. You can test out of the Square rooting gives two solutions: x must be greater than 2; or x must be less than -2: {x: x > 2 or x < -2}. It also contains 18, 21 and keeps going including all the multiples of 3 until it gets to its largest number 90. Objects placed within the brackets are called the elements of a set, and do not have to be in any specific order. For example, $\left\{\frac{1}{2},\:\frac{2}{3},\:\frac{3}{4},\:\frac{4}{5},\:...\right\}\nonumber$. This helps to better define sets and to make them easier to write. We have already seen how to represent a set on a number line, but that can be cumbersome, especially if we want to just use a keyboard. For example, if you want to describe the set of all people who are over 18 years old but not 30 years old, you announce the conditions by putting them to the left of a vertical line segment. We want to be able to both read the various ways and be able to write down the representation ourselves in order to best display the set. She has over 10 years of teaching experience at high school and university level. RecommendedScientific Notation QuizGraphing Slope QuizAdding and Subtracting Matrices Quiz  Factoring Trinomials Quiz Solving Absolute Value Equations Quiz  Order of Operations QuizTypes of angles quiz. For example: {x: x > 5}.

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# TWO GROUPS OR ONE? Situations comparing two proportions are described. In each case, determine whether the situation involves comparing proportions fo TWO GROUPS OR ONE? Situations comparing two proportions are described. In each case, determine whether the situation involves comparing proportions for two groups or comparing two proportions from the same group. State whether the methods of this section apply to the difference in proportions. (c) Compare the graduation rate (proportion to graduate) of students on an athletic scholarship to the graduation rate of students who are not on an athletic scholarship. This situation involves comparing: 1 group or 2 groups? Do the methods of this section apply to the difference in proportions? Yes or No You can still ask an expert for help • Questions are typically answered in as fast as 30 minutes Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it yagombyeR Full and correct solution: c) Here we have two groups. On group consists of students who are on an athletic scholarship and the other group consists of students who are not on an athletic scholarship. Hence the given situation involves comparing 2 groups. Yes we can apply difference in proportions here because two proportions are being compared.

https://mathematica.stackexchange.com/questions/134106/which-root-does-findroot-give

# Which root does FindRoot give? I think that usually, FindRoot will give the root that's closest to the starting point. But see the example below, where I try to find the root of $\cos x=0$. If I started from $x=0.1$, then I get $10.9956$, but if I started from 1, I get $1.5708$. What's wrong? • Draw a graph of the function and its tangent at x == 0.1. FindRoot is using Newton's method. – Michael E2 Dec 23 '16 at 4:49 • I believe its because FindRoot[] uses Newton's Method. The tangent line hits further out. In general, Newtons method requires a good initial guess or you "can" get a root quite far away. – Michael McCain Dec 23 '16 at 4:49 • FindRoot gives the root that it finds. :-) – Brett Champion Dec 23 '16 at 5:18 • To have more control over which root is obtained, give FindRoot two intial guesses, which prompts it to use the use secant method. Then FindRoot usually returns the value of a root bracketed by the two guesses. For instance, FindRoot[Cos[x], {x, -1, 4}]. – bbgodfrey Dec 23 '16 at 5:48 Technically, FindRoot uses a damped Newton's method. In an undamped Newton's method, the first step would look like this, the tangent striking the x-axis just above 10: Plot[{Cos[x], Cos[0.1] - Sin[0.1] (x - 0.1)}, {x, 0, 12}] You can keep large steps from occurring by decreasing the DampingFactor: FindRoot[Cos[x], {x, 0.1}, DampingFactor -> 0.2] (* {x -> 1.5708} *) The damping factor multiplies the change in x. The undamped step would be x == 10.066644423259238, so the damped first step is 0.1 + (10.066644423259238 - 0.1) 0.2 (* 2.09333 *) That lands x close enough to the root nearest 0.1 that FindRoot will converge on it. Of course, damping slows down convergence. In the following, it slows it down too much: FindRoot[1/x - 1/1000, {x, 0.1}, DampingFactor -> 0.2] FindRoot::cvmit: Failed to converge to the requested accuracy or precision within 100 iterations. (* {x -> 999.984} *) The regular method has no problem with it, though: FindRoot[1/x - 1/1000, {x, 0.1}] (* {x -> 1000.} *) When in doubt as to whether FindRoot[] is functioning as expected for a given nonlinear problem, one should try to use the diagnostic capabilities of the options EvaluationMonitor and StepMonitor. You've already been told in other answers as to why you should have expected your result, considering that you started the iteration with a seed that is uncomfortably near an extremum. Thus, let me demonstrate the use of EvaluationMonitor: Reap[FindRoot[Cos[x], {x, 0.1}, EvaluationMonitor :> Sow[x]]] {{x -> 10.9956}, {{0.1, 10.0666, 11.4045, 10.9711, 10.9956, 10.9956}}} where we use Sow[]/Reap[] to get the values where Cos[x] was evaluated. We can also demonstrate the effect of damping, as shown by Michael: Reap[FindRoot[Cos[x], {x, 0.1}, "DampingFactor" -> 1/5, EvaluationMonitor :> Sow[x]]] // Short {{x -> 1.5708}, {{0.1, 2.09333, 1.97814, <<76>>, 1.5708, 1.5708, 1.5708}}} where I have mercifully truncated the output, showing that damping gives better results at the cost of an increased number of iterations. One could choose to use Brent's method instead by specifying explicit brackets. The convergence is not as fast as Newton-Raphson, but it is certainly much safer: Reap[FindRoot[Cos[x], {x, 1., 2.}, EvaluationMonitor :> Sow[x]]] {{x -> 1.5708}, {{1., 2., 1.5649, 1.57098, 1.5708, 1.5708, 1.5708}}} If, like me, you like pictures to help with diagnostics, there is a function called FindRootPlot[] (more information here) that can be used: Needs["OptimizationUnconstrainedProblems"] FindRootPlot[Cos[x], {x, 0.1}] // Last FindRootPlot[Cos[x], {x, 0.1}, "DampingFactor" -> 0.2] // Last This question has been asked in different context here and here. As mentioned in the comments, FindRoot is based on Newton's method which works pretty well when a good guess is provided. But I think, it fails to provide you with multiple roots. To find multiple roots, you can use NDSolve f[x_] = Cos[x]; Module[{sol}, Column[{sol = NSolve[{f[x] == 0, -10 <= x <= 10}, x], Plot[f[x], {x, -10, 10}, Epilog -> {Red, AbsolutePointSize[6], Point[{x, f[x]} /. sol]}, ImageSize -> 360]}]] I adopted this idea from Bob Hanlon's answer to same sort of question. • The NDSolve Method you linked in your answer is very different from the code you posted (you used NSolve). Otherwise this is a good answer :) – Sascha Dec 23 '16 at 9:09 • @zhk: For one function your Code works fine. Have you used the same method for a system of non-linear equations? I have a NL equation system and I need to find Real solutions to the system, but so far no success. Can you show me how to use your Code for a system? – Tugrul Temel Apr 19 '19 at 9:52 To see what is happening, implement a quick Newton iteration algorithm. For instance: NewtonsMethodList[f_, {x_, x0_}, n_] := NestList[# - Function[x, f][#]/Derivative[1][Function[x, f]][#] &, x0, n] Now see what happens when we have 0.1 and 1 as starting values, and with, say, 5 iterations: In[2]:= NewtonsMethodList[Cos[x], {x, 0.1}, 5] Out[2]= {0.1, 10.0666, 11.4045, 10.9711, 10.9956, 10.9956} In[5]:= NewtonsMethodList[Cos[x], {x, 1.}, 5] Out[5]= {1., 1.64209, 1.57068, 1.5708, 1.5708, 1.5708} ` I hope this helps.

https://mathhelpboards.com/threads/rule-of-sum-and-rule-pf-product-how-many-distinct-sums-are-possible-using-from-1-to-all-of-the-15-cards.25841/

Rule of sum and rule pf product: How many distinct sums are possible using from 1 to all of the 15 cards Dhamnekar Winod Active member There are 8 cards with number 10 on them, 5 cards with number 100 on them and 2 cards with number 500 on them. How many distinct sums are possible using from 1 to all of the 15 cards? Answer given is 143. But my logic is for any sum, at least 2 numbers are needed. So, there are $\binom{15} {2} + \binom{15}{3}+...\binom{15}{15}$ distinct sums. So, I think answer 143 is wrong. Country Boy Well-known member MHB Math Helper $$\begin{pmatrix}15 \\ 2 \end{pmatrix}$$ is the number of sums of two of the numbers, $$\begin{pmatrix}15 \\ 3 \end{pmatrix}$$ is the number of sums of three of the numbers, etc. But they won't be distinct sums. skeeter Well-known member MHB Math Helper There are 8 cards with number 10 on them, 5 cards with number 100 on them and 2 cards with number 500 on them. How many distinct sums are possible using from 1 to all of the 15 cards? Answer given is 143. But my logic is for any sum, at least 2 numbers are needed. So, there are $\binom{15} {2} + \binom{15}{3}+...\binom{15}{15}$ distinct sums. So, I think answer 143 is wrong. The directions clearly state from 1 to 15, so their solution is based on that premise. Dhamnekar Winod Active member $$\begin{pmatrix}15 \\ 2 \end{pmatrix}$$ is the number of sums of two of the numbers, $$\begin{pmatrix}15 \\ 3 \end{pmatrix}$$ is the number of sums of three of the numbers, etc. But they won't be distinct sums. Hello, I have computed 141 distinct sums. But the answer is 143. Which 2 distinct sums i omitted, would you tell me? Is the answer 143 wrong? All the 141 distinct sums are 20,30,40,50,60,70,80,110,120,130,140,150,160,170,180,200,210,220,230,240,250,260,270,280,300,310,320,330,340,350,360,370,380,400,410,420,430,440,450,460,470,480,500,510,520,530,540,550,560,570,580,600,610,620,630,640,650,660,670,680,700,710,720,730,740,750,760,770,780,800,810,820,830,840,850,860,870,880,900,910,920,930,940,950,960,970,980,1000,1010,1020,1030,1040,1050,1060,1070,1080,1100,1110,1120,1130,1140,1150,1160,1170,1180,1200,1210,1220,1230,1240,1250,1260,1270,1280,1300,1310,1320,1330,1340,1350,1360,1370,1380,1400,1410,1420,1430,1440,1450,1460,1470,1480,1500,1510,1520,1530,1540,1550,1560,1570,1580. I know one formula $\binom{r+n-1}{r-n+1}$ which computes distinct sums, where n=cells and r= balls. How to use that here? Or is there any other formula? Last edited:

https://math.stackexchange.com/questions/2453591/difference-between-infty-and-infty

# difference between $+\infty$ and $\infty$ I'm taking Mathematical Analysis "I" and I'm studying limits where I have limits to the infinity, but I don't know what's the difference between $\lim_{x \to \infty}$ and $\lim_{x \to +\infty}$ I suppose that they are the same but I'm not sure. If you could help me I would appreciate it. Thank you very much! • Yes, they are the same. – Kenny Lau Oct 1 '17 at 22:40 • Thank you very much! @KennyLau – Santiago Pardal Oct 1 '17 at 22:41 • In real analysis they are the same, in complex analysis they are different. Because of your tag (real-analysis), I agree with Kenny. – GEdgar Oct 8 '17 at 11:04 In the context of real Analysis we usually consider \begin{align*} \lim_{x \to \infty}f(x)\qquad\text{and}\qquad\lim_{x \to +\infty}f(x) \end{align*} to be the same. It has mainly to do with preserving the order of the real numbers when $\mathbb{R}$ is extended by the symbols $+\infty$ and $-\infty$. We look at two references: • Principles of Mathematical Analysis by W. Rudin. Definition 1.23: The extended real number system consists of the real field $\mathbb{R}$ and two symbols $+\infty$ and $-\infty$. We preserve the original order in $\mathbb{R}$, and define \begin{align*} \color{blue}{-\infty < x < +\infty}\tag {1} \end{align*} for every $x\in\mathbb{R}$. (he continues with:) It is then clear that $+\infty$ is an upper bound of every subset of the extended real number system, and that every nonempty subset has a least upper bound. If, for example, $E$ is a nonempty set of real numbers which is not bounded above in $\mathbb{R}$, then $\sup E=+\infty$ in the extended real number system. Exactly the same remarks apply to lower bounds. Now we look at certain intervals of real numbers introduced in • Calculus by M. Spivak. (We find in chapter 4:) The set $\{x:x>a\}$ is denoted by $(a,\infty)$, while the set $\{x: x\geq a\}$ is denoted by $[a,\infty)$; the sets $(-\infty,a)$ and $(-\infty,a]$ are defined similarly. (Spivak continues later on:) The set $\mathbb{R}$ of all real number is also considered to be an "interval" and is sometimes denoted by \begin{align*} \color{blue}{(-\infty,\infty)}\tag{2} \end{align*} The connection with limits is presented in chapter 5: The symbol $\lim_{x\rightarrow\infty}f(x)$ is read "the limit of $f(x)$ as $x$ approaches $\infty$," or "as $x$ becomes infinite", and a limit of the form \begin{align*} \lim_{\color{blue}{x\rightarrow\infty}}f(x) \end{align*} is often called a limit at infinity. (and later on:) Formally, $\lim_{x\rightarrow\infty}f(x)=l$ means that for every $\varepsilon>0$ there is a number $N$ such that, for all $x$, \begin{align*} \text{if }x>N\text{, then }|f(x)-l|<\varepsilon\text{.} \end{align*} and we find as exercise 36 a new definition and the following two out of three sub-points Exercise 36: Define \begin{align*} &\lim_{\color{blue}{x=-\infty}}f(x)=l\\ \\ &(b) \text{Prove that }\lim_{x\rightarrow\infty}f(x)=\lim_{x\rightarrow-\infty}f(-x)\text{.}\\ &(c) \text{Prove that }\lim_{x\rightarrow 0^{-}}f(1/x)=\lim_{x\rightarrow-\infty}f(x)\text{.} \end{align*} Conclusion: When looking at (1) and (2) together with Spivaks definition of limits we can conclude that $\infty$ and $+\infty$ are used interchangeably in the context of limits of real valued functions. The compactification of the real numbers, in a useful way that fits in with the ordering of the reals, requires the addition of two points, whereas the compactification of the complex numbers is naturally accomplished by adding just one point. Because analysis readily switches between the real and complex cases, it is considered by some authors appropriate to use a "balanced" pair of symbols, $+\infty$ and $-\infty$, for the real case, which reflects the symmetry of their roles, and the unsigned $\infty$ for the complex case. This is a stylistic choice. Other authors are not of this persuasion. Their argument is "We don't write $+3$ when we mean $3$; so why should we have to write $+\infty$? And, in the complex case, which is always clear from the context, writing $\infty$ is good enough for anyone". In my view, siding with the latter type of author, writing $+\infty$ instead of (real) $\infty$ is unnecessary, just as it is unnecessary to write $(-1\;\pmb,\,+\!1)$ to denote the interval $(-1\;\pmb,\,1)$. • There is a one point compactification of the reals. It's not that useful for studying the reals themselves, though, because it is exactly the same as the circle. – Ian Oct 4 '17 at 16:37

http://math.stackexchange.com/questions/29638/how-do-we-find-a-fraction-with-whose-decimal-expansion-has-a-given-repeating-pat

# How do we find a fraction with whose decimal expansion has a given repeating pattern? We know $\frac{1}{81}$ gives us $0.\overline{0123456790}$ How do we create a recurrent decimal with the property of repeating: $0.\overline{0123456789}$ a) Is there a method to construct such number? b) Is there a solution? c) Is the solution in $\mathbb{Q}$? According with wikipedia page: http://en.wikipedia.org/wiki/Decimal One could get this number by applying this series. Supppose: $M=123456789$, $x=10^{10}$, then $0.\overline{0123456789}= \frac{M}{x}\cdot$ $\sum$ ${(10^{-9})}^k$ $=\frac{M}{x}\cdot\frac{1}{1-10^{-9}}$ $=\frac{M}{9999999990}$ Unless my calculator is crazy, this is giving me $0.012345679$, not the expected number. Although the example of wikipedia works fine with $0.\overline{123}$. Some help I got from mathoverflow site was that the equation is: $\frac{M}{1-10^{-10}}$. Well, that does not work either. So, just to get rid of the gnome calculator rounding problem, running a simple program written in C with very large precision (long double) I get this result: #include <stdio.h> int main(void) { long double b; b=123456789.0/9999999990.0; printf("%.40Lf\n", b); } Result: $0.0123456789123456787266031042804570461158$ Maybe it is still a matter of rounding problem, but I doubt that... Thanks! Beco Edited: Thanks for the answers. After understanding the problem I realize that long double is not sufficient. (float is 7 digits:32 bits, double is 15 digits:64 bits and long double is 19 digits:80 bits - although the compiler align the memory to 128 bits) Using the wrong program above I should get $0.0\overline{123456789}$ instead of $0.\overline{0123456789}$. Using the denominator as $9999999999$ I must get the correct answer. So I tried to teach my computer how to divide: #include <stdio.h> int main(void) { int i; long int n, d, q, r; n=123456789; d=9999999999; printf("0,"); n*=10; while(i<100) { if(n<d) { n*=10; printf("0"); i++; continue; } q=n/d; r=n%d; printf("%ld", q); if(!r) break; n=n-q*d; n*=10; i++; } printf("\n"); } - Change the C program to assign b=123456789.0/9999999999.0; Note the integral value of the denominator should end in a 9, not a 0. –  Brandon Carter Mar 29 '11 at 2:53 Thanks @Brandon, but only this was not sufficient. Look at the edition. –  Dr Beco Mar 29 '11 at 4:03 Now I get any precision I want, just change while(i<PREC). Output with 100: 0,012345678901234567890123456789012345678901234567890123456789012345678901234567‌​8901234567890123456789 –  Dr Beco Mar 29 '11 at 4:07 TIP Use one of the many freely available mathematics systems with multiple precision arithmetic instead of wasting your time rolling your own, e.g. wolframalpha.com/input/?i=N[123456789/%2810^10-1%29,40] –  Bill Dubuque Mar 29 '11 at 4:31 @Bill Wow! I'm astonished! Thank you very much for this great tip. –  Dr Beco Mar 29 '11 at 5:15 Suppose you want to have a number $x$ whose decimal expansion is $0.a_1a_2\cdots a_ka_1a_2\cdots a_k\cdots$. That is it has a period of length $k$, with digits $a_1$, $a_2,\ldots,a_k$. Let $n = a_1a_2\cdots a_k$ be the integer given by the digits of the period. Then \begin{align*} \frac{n}{10^{k}} &= 0.a_1a_2\cdots a_k\\ \frac{n}{10^{2k}} &= 0.\underbrace{0\cdots0}_{k\text{ zeros}}a_1a_2\cdots a_k\\ \frac{n}{10^{3k}} &= 0.\underbrace{0\cdots0}_{2k\text{ zeros}}a_1a_2\cdots a_k\\ &\vdots \end{align*} So the number you want is $$\sum_{r=1}^{\infty}\frac{n}{10^{rk}} = n\sum_{r=1}^{\infty}\frac{1}{(10^k)^r} = n\left(\frac{\quad\frac{1}{10^k}\quad}{1 - \frac{1}{10^k}}\right) = n\left(\frac{10^k}{10^k(10^k - 1)}\right) = \frac{n}{10^k-1}.$$ Since $10^k$ is a $1$ followed by $k$ zeros, then $10^k-1$ is $k$ 9s. So the fraction with the decimal expansion $$0.a_1a_2\cdots a_ka_1a_2\cdots a_k\cdots$$ is none other than $$\frac{a_1a_2\cdots a_k}{99\cdots 9}.$$ Thus, $0.575757\cdots$ is given by $\frac{57}{99}$. $0.837168371683716\cdots$ is given by $\frac{83716}{99999}$, etc. If you have some decimals before the repetition begins, e.g., $x=2.385858585\cdots$, then first multiply by a suitable power of $10$, in this case $10x = 23.858585\cdots = 23 + 0.858585\cdots$, so $10x = 23 + \frac{85}{99}$, hence $x= \frac{23}{10}+\frac{85}{990}$, and simple fraction addition gives you the fraction you want. And, yes, there is always a solution and it is always a rational. - Thanks @Arturo for this very explanatory solution. Now I got the correct result, with any precision I want, just changing the "while" with the new program (see edition). –  Dr Beco Mar 29 '11 at 4:06 It's simple: $\rm\displaystyle\ x\ =\ 0.\overline{0123456789}\ \ \Rightarrow\ \ 10^{10}\ x\ =\ 123456789\ +\ x\ \ \Rightarrow\ \ x\ =\ \frac{123456789}{10^{10} - 1}$ Note that the last digit of $\rm\ 10^{10} - 1\$ is $\:9\:,$ not $\:0\:,$ which explains the error in your program. - Thanks @Bill, you corrected the formula, but the problem was long double rounding it. –  Dr Beco Mar 29 '11 at 4:04 @Dr Beco: I assumed the rest would be easy once you had the correct formula. –  Bill Dubuque Mar 29 '11 at 4:27 When you say "double" in C how many places is that? I tried it in Maple... Digits := 40; 40 123456789.0/9999999990.0; 0.01234567891234567891234567891234567891235 - Float is 7 digits (32 bits), Double is 15 digits (64 bits) and Long Double is 19 digits (only 80 digits used, but the alignment make it 128 bits). This long double number "0.012345678901234568431" loses precision in the 18 digit, as expected... My program didn't work because of that! Thanks. –  Dr Beco Mar 29 '11 at 3:55 BTW, what is maple? A language? Do you have any link I could spy on it? Thanks –  Dr Beco Mar 29 '11 at 4:10 You said: $M=123456789$, $x=10^{10}$, then $0.\overline{0123456789}= \frac{M}{x}\cdot$ $\sum$ ${(10^{-9})}^k$ $=\frac{M}{x}\cdot\frac{1}{1-10^{-9}}$ $=\frac{M}{9999999990}$ but since the block of repeating digits is 10 digits long, the summation term should be $\sum{(10^{-10})}^k$, so that $$0.\overline{0123456789}=\frac{M}{x}\cdot\sum{(10^{-10})}^k=\frac{M}{x}\cdot\frac{1}{1-10^{-10}}=\frac{M}{9999999999}$$ and $$\frac{M}{9999999999}=\frac{123456789}{9999999999}=\frac{13717421}{1111111111}.$$ - Thanks! Using your last fraction, the long double can give at least 2 repetitions before lose precision: 0.0123456789 0123456789 0470 –  Dr Beco Mar 29 '11 at 4:14

https://mathhelpboards.com/threads/a-probability-question-choosing-3-from-25.5240/#post-23851

# [SOLVED]A probability question: choosing 3 from 25 #### first21st ##### New member In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is: A. 21/46 B. 25/117 C. 1/50 D. 3/25 Could you please solve this problem with proper explanation? Thanks, James #### MarkFL Staff member Re: A probability question Here at MHB, we normally don't provide fully worked solutions, but rather we help people to work the problem on their own. This benefits people much more, which is our goal. Now, what you want to do here is to find the number of ways to choose 1 girl from 10 AND 2 boys from 15, then divide this by the number of ways to choose 3 children from 25. What do you find? #### first21st ##### New member Re: A probability question Thanks for your reply. It's really a great way of learning. I highly appreciate your approach. Is the solution something like: (15 C 2) * (10 C 1)/ (25 C 3) If it is correct, could you please explain why did we divide it with the number of ways of choosing 3 students from 25? Thanks, James #### MarkFL Staff member Re: A probability question Excellent! That is correct! Now you just need to simplify, either by hand or with a calculator. As probability is the ratio of the number favorable outcomes to the number of all outcomes, we are in this case dividing the number of ways to choose 1 girl from 10 AND 2 boys from 15 by the number of ways to choose 3 of the children from the total of 25. We are told that 3 children are selected at random, and we know there are 25 children by adding the number of boys to the number of girls. Thus, $$\displaystyle {25 \choose 3}$$ is the total number of outcomes. #### first21st ##### New member Re: A probability question Thank you very much man! But I am still wondering why did we MULTIPLY # the number of ways to choose 1 girl from 10 # AND # 2 boys from 15 #, instead of ADDING these two operands? #### MarkFL Staff member Re: A probability question We multiply because we are essentially applying the fundamental counting principle. When we require event 1 AND event 2 to happen, we multiply. When we require event 1 OR event 2 to happen, we add. Here we require both 2 boys AND 1 girl. You see, for each way to obtain 2 boys, we have to account for all of the ways to obtain 1 girl. Or conversely, for each way to obtain 1 girl, we have to account for all of the ways to obtain 2 boys. The product of these two gives us all of the ways to get 2 boys and 1 girl. #### first21st ##### New member Re: A probability question ok. Got it! Thanks a lot. James

https://orinanobworld.blogspot.com/2020/

## Monday, February 17, 2020 ### Reversing Differences Fellow blogger Håkan Kjellerstrand posted an interesting question on OR Stack Exchange recently. Starting from a list of integers, it is trivial to compute the list of all pairwise absolute differences, but what about going in the other direction? Given the pairwise (absolute) differences, with duplicates removed, can you recover the source list (or a source list)? We can view the source "list" as a vector $x\in\mathbb{Z}^n$ for some dimension $n$ (equal to the length of the list). With duplicates removed, we can view the differences as a set $D\subset \mathbb{Z}_+$. So the question has to do with recovering $x$ from $D$. Our first observation kills any chance of recovering the original list with certainty: If $x$ produces difference set $D$, then for any $t\in\mathbb{R}$ the vector $x+t\cdot (1,\dots,1)'$ produces the same set $D$ of differences. Translating all components of $x$ by a constant amount has no effect on the differences. So there will be an infinite number of solutions for a given difference set $D$. A reasonable approach (proposed by Håkan in his question) is to look for the shortest possible list, i.e., minimize $n$. Next, observe that $0\in D$ if and only if two components of $x$ are identical. If $0\notin D$, we can assume that all components of $x$ are distinct. If $0\in D$, we can solve the problem for $D\backslash\{0\}$ and then duplicate any one component of the resulting vector $x$ to get a minimum dimension solution to the original problem. Combining the assumption that $0\notin D$ and the observation about adding a constant having no effect, we can assume that the minimum element of $x$ is 1. That in turn implies that the maximum element of $x$ is $1+m$ where $m=\max(D)$. From there, Håkan went on to solve a test problem using constraint programming (CP). Although I'm inclined to suspect that CP will be more efficient in general than an integer programming (IP) model, I went ahead and solved his test problem via an IP model (coded in Java and solved using CPLEX 12.10). CPLEX's solution pool feature found the same four solutions to Håkan's example that he did, in under 100 ms. How well the IP method scales is an open question, but it certainly works for modest size problems. The IP model uses binary variables $z_1, \dots, z_{m+1}$ to decide which of the integers $1,\dots,m+1$ are included in the solution $x$. It also uses variables $w_{ij}\in [0,1]$ for all $i,j\in \{1,\dots,m+1\}$ such that $i \lt j$. The intent is that $w_{ij}=1$ if both $i$ and $j$ are included in the solution, and $w_{ij} = 0$ otherwise. We could declare the $w_{ij}$ to be binary, but we do not need to; constraints will force them to be $0$ or $1$. The full IP model is as follows: $\begin{array}{lrlrc} \min & \sum_{i=1}^{m+1}z_{i} & & & (1)\\ \textrm{s.t.} & w_{i,j} & \le z_{i} & \forall i,j\in\left\{ 1,\dots,m+1\right\} ,i\lt j & (2)\\ & w_{i,j} & \le z_{j} & \forall i,j\in\left\{ 1,\dots,m+1\right\} ,i\lt j & (3)\\ & w_{i,j} & \ge z_{i}+z_{j}-1 & \forall i,j\in\left\{ 1,\dots,m+1\right\} ,i\lt j & (4)\\ & w_{i,j} & =0 & \forall i,j\in\left\{ 1,\dots,m+1\right\} \textrm{ s.t. }(j-i)\notin D & (5)\\ & \sum_{i,j\in\left\{ 1,\dots,m+1\right\} |j-i=d}w_{i,j} & \ge 1 & \forall d\in D & (6)\\ & z_{1} & = 1 & & (7) \end{array}$ The objective (1) minimizes the number of integers used. Constraints (2) through (4) enforce the rule that $w_{ij}=1$ if and only if both $z_i$ and $z_j$ are $1$ (i.e., if and only if both $i$ and $j$ are included in the solution).  Constraint (5) precludes the inclusion of any pair $i < j$ whose difference $j - i$ is not in $D$, while constraint (6) says that for each difference $d \in D$ we must include at least one pair $i < j$ for that produces that difference ($j - i = d$). Finally, since we assumed that our solution starts with minimum value $1$, constraint (7) ensures that $1$ is in the solution. (This constraint is redundant, but appears to help the solver a little, although I can't be sure given the short run times.) My Java code is available from my repository (bring your own CPLEX). ## Tuesday, February 11, 2020 ### Collections of CPLEX Variables Recently, someone asked for help online regarding an optimization model they were building using the CPLEX Java API. The underlying problem had some sort of network structure with $N$ nodes, and a dynamic aspect (something going on in each of $T$ periods, relating to arc flows I think). Forget about solving the problem: the program was running out of memory and dying while building the model. A major issue was that they allocated two $N\times N\times T$ arrays of variables, and $N$ and $T$ were big enough that $2N^2T$ was, to use a technical term, ginormous. Fortunately, the network was fairly sparse, and possibly not every time period was relevant for every arc. So by creating only the IloNumVar instances they needed (meaning only for arcs that actual exist in time periods that were actually relevant), they were able to get the model to build. That's the motivation for today's post. We have a tendency to write mathematical models using vectors or matrices of variables. So, for instance, $x_i \, (i=1,\dots,n)$ might be an inventory level at each of $n$ locations, or $y_{i,j} \, (i=1,\dots,m; j=1,\dots,n)$ might be the inventory of item $i$ at location $j$. It's a natural way of expressing things mathematically. Not coincidentally, I think, CPLEX APIs provide structures for storing vectors or matrices of variables and for passing them into or out of functions. That makes it easy to fall into the trap of thinking that variables must be organized into vectors or matrices. Last year I did a post ("Using Java Collections with CPLEX") about using what Java calls "collections" to manage CPLEX variables. This is not unique to Java. I know that C++ has similar memory structures, and I think they exist in other languages you might use with CPLEX. The solution to the memory issue I mentioned at the start was to create a Java container class for each combination of an arc that actually exists and time epoch for which it would have a variable, and then associate instances of that class with CPLEX variables. So if we call the new class AT (my shorthand for "arc-time"), I suggested the model owner use a Map<AT, IloNumVar> to associate each arc-time combination with the variable representing it and a Map<IloNumVar, AT> to hold the reverse association. The particular type of map is mostly a matter of taste. (I generally use HashMap.) During model building, they would create only the AT instances they actually need, then create a variable for each and pair them up in the first map. When getting a solution from CPLEX, they would get a value for each variable and then use the second map to figure out for which arc and time that value applied. (As a side note, if you use maps and then need the variables in vector form, you can apply the values() method to the first map (or the getKeySet() method to the second one), and then apply the toArray() method to that collection.) Now you can certainly get a valid model using just arrays of variables, which was all that was available to me back in the Dark Ages when I used FORTRAN, but I think there are some benefits to using collections. Using arrays requires you to develop an indexing scheme for your variables. The indexing scheme tells you that the flow from node 3 to node 7 at time 4 will be occupy slot 17 in the master variable vector. Here are my reasons for avoiding that. • Done correctly, the indexing scheme is, in my opinion, a pain in the butt to manage. Finding the index for a particular variable while writing the code is time-consuming and has been known to kill brain cells. • It is easy to make mistakes while programming (calculate an index incorrectly). • Indexing invites the error of declaring an array or vector with one entry for each combination of component indices (that $N\times N\times T$ matrix above), without regard to whether you need all those slots. Doing so wastes time and space, and the space, as we saw, may be precious. • Creating slots that you do not need can lead to execution errors. Suppose that I allocating a vector IloNumVar x = new IloNumVar[20] and use 18 slots, omitting slots 0 and 13. If I solve the model and then call getValues(x), CPLEX will throw an exception, because I am asking for values of two variables (x[0] and x[13]) that do not exist. Even if I create variables for those two slots, the exception will occur, because those two variables will not belong to the model being solved. (There is a way to force CPLEX to include those variables in the model without using them, but it's one more pain in the butt to deal with.) I've lost count of how many times I've seen messages on the CPLEX help forums about exceptions that boiled down to "unextracted variables". So my advice is to embrace collections when building models where variables do not have an obvious index scheme (with no skips). ## Thursday, January 30, 2020 ### Generic Callback Changes in CPLEX 12.10 CPLEX 12.10 is out, and there have been a few changes to the new(ish) generic callbacks. Rather than go into them in detail (and likely screw something up), I'll just point you to the slides for a presentation by Daniel Junglas of IBM at the 2019 INFORMS Annual Meeting. I've written about half a dozen posts about generic callbacks since IBM introduced them (which you can find by typing "generic callback" in the search widget on the blog). A couple of things have been added recently, and I thought I would mention them. The generic callback approach uses a single callback function that can be called from a variety of contexts, including when CPLEX solves a node relaxation ("RELAXATION" context), when if finds a candidate solution ("CANDIDATE" context) and, now, when it is ready to split a node into children ("BRANCHING" context). The branching context is one of the new features. It brings back most of the functionality of the branch callback in the legacy callback system. Unfortunately, it does not seem to have the ability to attach user information to the child nodes, which was a feature that was occasionally useful in the legacy system. You can get more or less equivalent functionality by creating a data store (array, map, whatever) in your global memory and storing the node information keyed by the unique index number of each child node. The catch is that you are now responsible for memory management (freeing up space when a node is pruned and the associated information is no longer needed), and for dealing with thread synchronization issues. Another new feature is that you can now inject a heuristic solution (if you have one) from all three of the contexts I mentioned above. CPLEX gives you a variety of options for how it will handle the injected solution: "NoCheck" (CPLEX will trust you that it is feasible); "CheckFeasible" (CPLEX will check feasibility and ignore the solution if it is not feasible); "Propagate" (Daniel's explanation: CPLEX will "propagate fixed variables and accept if feasible"); and "Solve" (CPLEX will solve a MIP problem with fixed variables and accept the result if feasible). I assume the latter two mean that you provide a partial solution, fixing some variables but not others. (Unfortunately I was unable to make it to Daniel's talk, so I'm speculating here.) I'm not sure if those are the only new features, but they are the ones that are most relevant to me. I invite you to read through Daniel's slides to get a more complete picture, including both the reasons for switching from legacy callbacks to generic callbacks and some of the technical issues in using them. ## Tuesday, January 7, 2020 ### Greedy Methods Can Be Exact We generally sort optimization algorithms (as opposed to models) into two or three categories, based on how certain we are that solutions will be either optimal or at least "good". An answer by Michael Feldmeier to a question I posted on OR Stack Exchange neatly summarizes the categories: • exact methods eventually cough up provably optimal solutions; • approximate methods eventually cough up solutions with some (meaningful) guarantee regarding how far from optimal they might be; and • heuristics provide no worst-case guarantees (but generally are either easy to implement, fast to execute or both). I should explain my use of "meaningful" (which is not part of Michael's answer). A common way to estimate the "gap" between a solution and the optimum is to take $|z - \tilde{z}|/|z|$, where $z$ is the objective value of the solution produced by the algorithm and $\tilde{z}$ is some bound (lower bound in a minimization, upper bound in a maximization) of the optimal solution. Now suppose that we are minimizing a function known to be nonnegative. If we set $\tilde{s}=0$, we know that any method, no matter how stupid, will have a gap no worse than 100%. To me, that is not a meaningful guarantee. So I'll leave the definition of "meaningful" to the reader. What brings all this to mind is a question posted on Mathematics Stack Exchange. The author of the question was trying to solve a nonlinear integer program. He approached it by applying a "greedy algorithm". Greedy algorithms are generally assumed to be heuristics, since it seldom is possible to provide useful guarantees on performance. In his case, though, the greedy algorithm is provably optimal, mainly due to the objective function being concave and separable. I'll state the problem and show a proof of optimality below (changing the original notation a bit). Brace yourself: the proof is a bit long-winded. You start with $N$ workers to be assigned to $M$ work stations. The output of workstation $m$, as a function of the number of workers $x$ assigned to it, is given by $$f_{m}(x)=a_{m}x+b_{m}-\frac{c_{m}}{x},$$ where $a_{m},b_{m},c_{m}$ are all positive constants. Since $f(0)=-\infty$, we can assume that each work station gets at least one worker (and, consequently, that $N>M$). Since $f_{m}'(x)=a_{m}+c_{m}/x^{2}>0$, each $f_{m}()$ is monotonically increasing. Thus, we can safely assume that all $N$ workers will be assigned somewhere. $f_{m}''(x)=-2c_{m}/x^{3}<0$, so $f_{m}()$ is strictly concave (which we will need later). We also note, for future reference, that the impact of adding one worker to a current staff of $x$ at station $m$ is $$\Delta f_{m}(x)=a_{m}+\frac{c_{m}}{x(x+1)}>0.$$ Similarly, the impact of removing one worker at station $m$ is $$\delta f_{m}(x)=-a_{m}-\frac{c_{m}}{x(x-1)}<0.$$We see that $\delta f_{m}(x)$ is an increasing function of $x$ (i.e., it gets less negative as $x$ gets bigger). We also note that $\Delta f_{m}(x)=-\delta f_{m}(x+1)$. The IP model is easy to state. Let $x_{m}$ be the number of workers assigned to work station $m$. The model is $$\max\sum_{m=1}^{M}f_{m}(x_{m})$$ subject to $$\sum_{m=1}^{M}x_{m}\le N$$ with $$x\in\mathbb{Z}_{+}^{M}.$$ The greedy algorithm starts with a single worker at each station ($x=(1,\dots,1)$) and, at each step, adds one worker to the workstation where that worker produces the greatest increase in objective value (breaking ties arbitrarily). It stops when all $N$ workers are assigned. To prove that it actually finds an optimal solution, I'll use proof by contradiction. Let $x^{(0)},x^{(1)},\dots,x^{(N-M)}$ be the sequence of solutions constructed by the greedy algorithm, with $x^{(0)}=(1,\dots,1)$, and let $x^{(k)}$ be the last solution in the sequence for which an optimal solution $x^{*}$ exists such that $x^{(k)}\le x^{*}$. The significance of the inequality is that if $x\le x^{*}$, it is possible to extend the partial solution $x$ to the optimal solution $x^{*}$ by adding unassigned workers to work stations. We know that $k$ is well defined because $x^{(0)}\le x^{*}$ for any optimal $x^{*}$. Since we are assuming that the greedy algorithm does not find an optimum, it must be that $k<N-M$. Now identify the work station $j$ to which the greedy algorithm added a worker at step $k$, meaning that $x_{j}^{(k+1)}=x_{j}^{(k)}+1$ and $x_{i}^{(k+1)}=x_{i}^{(k)}$ for $i\neq j$. Since, by assumption, $x^{(k)}\le x^{*}$ but $x^{(k+1)}\not\le x^{*}$, it must be that $x_{j}^{(k)}=x_{j}^{*}$. Next, since $x^{(k)}\le x^{*}$ and $x^{(k)}\neq x^{*}$ (else $x^{(k)}$ would be optimal), there is some work station $h\neq j$ such that $x_{h}^{(k)}<x_{h}^{*}$. Let $\tilde{x}$ be the solution obtained from $x^{(k)}$ by adding a worker to station $h$: $\tilde{x}_{h}=x_{h}^{(k)}+1$ and $\tilde{x}_{i}=x_{i}^{(k)}$ for $i\neq h$. Observe that $\tilde{x}\le x^{*}$. The greedy algorithm chose work station $j$ over work station $h$ at $x^{(k)}$, so it must be that $$\Delta f_{j}(x_{j}^{(k)})\ge\Delta f_{h}(x_{h}^{(k)}). \quad (1)$$ Finally, let $\hat{x}$ be the result of starting from optimal solution $x^{*}$ and shifting one worker from station $h$ to station $j$. Since $$x_{j}^{(k+1)}=x_{j}^{(k)}+1=x_{j}^{*}+1=\hat{x}_{j},$$ $$x_{h}^{(k+1)}=x_{h}^{(k)}<x_{h}^{*}\implies x_{h}^{(k+1)}\le\hat{x}_{h}$$ and $$x_{i}^{(k+1)}=x_{i}^{(k)}\le x_{i}^{*}=\hat{x}_{i}\,\forall i\notin\{h,j\},$$ we have $x^{(k+1)}\le\hat{x}$. Under the assumption that $x^{(k)}$ was the last solution in the greedy sequence that could be extended to an optimal solution, it must be that $\hat{x}$ is not optimal. Thus the net change to the objective function at $x^{*}$ when shifting one worker from station $h$ to station $j$ must be negative, i.e., $$\Delta f_{j}(x_{j}^{*})+\delta f_{h}(x_{h}^{*})<0.\quad (2)$$ We showed previously that, under our assumptions, $x_{j}^{(k)}=x_{j}^{*}$, from which it follows that $$\Delta f_{j}(x_{j}^{*})=\Delta f_{j}(x_{j}^{(k)}). \quad (3)$$ We also showed that $\delta f_{h}()$ is an increasing function. Since $\tilde{x}_{h}\le x_{h}^{*}$, $$\delta f_{h}(x_{h}^{*})\ge\delta f_{h}(\tilde{x}_{h})=-\Delta f_{h}(x_{h}^{(k)}). \quad (4)$$ Combining (4) with (2), we have $$\Delta f_{j}(x_{j}^{*})-\Delta f_{h}(x_{h}^{(k)})<0,$$ i.e., $$\Delta f_{j}(x_{j}^{*})<\Delta f_{h}(x_{h}^{(k)}). \quad (5)$$ Combining (3) with (5) yields $$\Delta f_{j}(x_{j}^{(k)})<\Delta f_{h}(x_{h}^{(k)})$$

http://mathhelpforum.com/algebra/109037-determining-formula-sequence.html

# Math Help - Determining the formula of this sequence? 1. ## Determining the formula of this sequence? I was given the sequence 0, 1, 4, 11, 26, 57, 120, 247, 502, 1013, ... and I'm looking for a method to determine an explicit formula for this. I've tried doing difference columns, and I always end up with 2^n being the 2nd difference. Anyone know a way to figure this out? 2. Originally Posted by paupsers I was given the sequence 0, 1, 4, 11, 26, 57, 120, 247, 502, 1013, ... and I'm looking for a method to determine an explicit formula for this. I've tried doing difference columns, and I always end up with 2^n being the 2nd difference. Anyone know a way to figure this out? Good idea to look at differences. So you found: $(u_{n+2}-u_{n+1})-(u_{n+1}-u_n)=2^n$, assuming the first term is $u_1=0$. If you sum the above equalities for $n$ ranging from 1 to $n$, you get a telescoping sum on the left, and a geometric sum on the right, which leads to: $(u_{n+2}-u_{n+1})-(u_2-u_1)=2+2^2+\cdots +2^n = 2^{n+1}-2$, hence $u_{n+2}-u_{n+1}=2^{n+1}-1$. Summing again this latter equality we have again a telescoping sum on the left and a geometric sum (minus constant terms) on the right, and we get: $u_{n+2}-u_2=(2^2-1)+\cdots+(2^{n+1}-1)$, hence (change of variable) $u_n = (2^2-1)+\cdots +(2^{n-1}-1)+1$ and finally $u_n=2^n-n-1$. Same method works in many situations. 3. Originally Posted by paupsers I was given the sequence 0, 1, 4, 11, 26, 57, 120, 247, 502, 1013, ... and I'm looking for a method to determine an explicit formula for this. I've tried doing difference columns, and I always end up with 2^n being the 2nd difference. Anyone know a way to figure this out? The Encyclopaedia of Integer Sequences gives: $2^n-n-1$ CB 4. Hello, paupsers! Given the sequence: . $0, 1, 4, 11, 26, 57, 120, 247, 502, 1013, \hdots$ I'm looking for a method to determine an explicit formula for this. I've tried doing difference columns, and I always end up with 2^n being the 2nd difference. This is true . . . and should give you a big hint. Consider the general form: . $f(n) = 2^n$ How does it compare with the given sequence? $\begin{array}{c|cccccccccc}\hline n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline \text{Sequence:} & 0 & 1 & 4 & 11 & 26 & 57 & 120 & 247 & 502 & 1013 \\ \hline f(n) = 2^n & 2 & 4 & 8 & 16 & 32 & 64 & 128 & 256 & 512 & 1024 \\ \hline \text{Error:} & +2 & +3 & +4 & +5 & +6 & +7 & +8 & +9 & +10 & +11\\ \hline \end{array}$ If we use $f(n) = 2^n$, each term is too large by $(n+1)$ Therefore, the explicit formula is: . $F(n) \;=\;2^n - (n+1)$

https://math.stackexchange.com/questions/1763978/can-you-use-both-sides-of-an-equation-to-prove-equality

# Can you use both sides of an equation to prove equality? For example: $\color{red}{\text{Show that}}$$\color{red}{\frac{4\cos(2x)}{1+\cos(2x)}=4-2\sec^2(x)}$$ In high school my maths teacher told me To prove equality of an equation; you start on one side and manipulate it algebraically until it is equal to the other side. So starting from the LHS: $$\frac{4\cos(2x)}{1+\cos(2x)}=\frac{4(2\cos^2(x)-1)}{2\cos^2(x)}=\frac{2(2\cos^2(x)-1)}{\cos^2(x)}=\frac{4\cos^2(x)-2}{\cos^2(x)}=4-2\sec^2(x)$$$\large\fbox{}$At University, my Maths Analysis teacher tells me To prove a statement is true, you must not use what you are trying to prove. So using the same example as before: LHS = $$\frac{4\cos(2x)}{1+\cos(2x)}=\frac{4(2\cos^2(x)-1)}{2\cos^2(x)}=\frac{2(2\cos^2(x)-1)}{\cos^2(x)}=\frac{2\Big(2\cos^2(x)-\left[\sin^2(x)+\cos^2(x)\right]\Big)}{\cos^2(x)}=\frac{2(\cos^2(x)-\sin^2(x))}{\cos^2(x)}=\bbox[yellow]{2-2\tan^2(x)}$$ RHS =$$4-2\sec^2(x)=4-2(1+\tan^2(x))=\bbox[yellow]{2-2\tan^2(x)}$$ So I have shown that the two sides of the equality in$\color{red}{\rm{red}}$are equal to the same highlighted expression. But is this a sufficient proof? Since I used both sides of the equality (which is effectively; using what I was trying to prove) to show that $$\color{red}{\frac{4\cos(2x)}{1+\cos(2x)}=4-2\sec^2(x)}$$ One of the reasons why I am asking this question is because I have a bounty question which is suffering from the exact same issue that this post is about. ## EDIT: Comments and answers below seem to indicate that you can use both sides to prove equality. So does this mean that my high school maths teacher was wrong? $$\bbox[#AFF]{\text{Suppose we have an identity instead of an equality:}}$$ $$\bbox[#AFF]{\text{Is it plausible to manipulate both sides of an identity to prove the identity holds?}}$$ Thank you. • You did not use both side of the equality. You showed that each is equal to the same thing. That's ok. Using what you want to prove would be using the fact that both sides are equal, but you don't. – Captain Lama Apr 29 '16 at 10:59 • You can simplify both sides separately to get get to a common point – Archis Welankar Apr 29 '16 at 10:59 • Just another comment: proving that$a=b$is the same as proving that$a-b=0$. So there is no meaningful difference between "manipulating one side" and "manipulating both sides". – Nefertiti Apr 29 '16 at 11:10 • This has already been answered but I'd add that you are really using the fact that$=$sign is an equivalence relation and hence transitive. This might be an issue with more complicated proof whereby the relation is not transitive. – Karl Apr 29 '16 at 12:14 • You did not "use" either side. What does "use" mean here? It means making an assertion, stating a sentence. The sides of the Eq'n are not sentences. If you take either side's formula and show that it is equal to some other formula, without any unsupported or unwarranted assumptions, then you are logical. E.g.: The RHS is always equal to$4-4/(2 \cos^2 x)=4-4/(1+\cos 2 x)=4\cos 2 x/(1+\cos 2 x)$regardless of what is written on the LHS. – DanielWainfleet Apr 29 '16 at 14:07 ## 7 Answers There's no conflict between your high school teacher's advice To prove equality of an equation; you start on one side and manipulate it algebraically until it is equal to the other side. and your professor's To prove a statement is true, you must not use what you are trying to prove. As in Siddarth Venu's answer, if you prove$a = c$and$b = c$("working from both sides"), then$a = c = bby transitivity of equality. This conforms to both your teacher's and professor's advice. Both your high school teacher and university professor are steering you away from "two-column proofs" of the type: \begin{align*} -1 &= 1 &&\text{To be shown;} \\ (-1)^{2} &= (1)^{2} && \text{Square both sides;} \\ 1 &= 1 && \text{True statement. Therefore-1 = 1.} \end{align*} Here, you assume what you want to prove, deduce a true statement, and assert that the original assumption was true. This is bad logic for at least two glaring reasons: 1. If you assume-1 = 1$, there's no need to prove$-1 = 1$. 2. Logically, if$P$denotes the statement "$-1 = 1$" and$Q$denotes "$1 = 1$", the preceding argument shows "$P$implies$Q$and$Q$is true", which does not eliminate the possibility "$P$is false". What you can do logically is start ("provisionally", on scratch paper) with the statement$P$you're trying to prove and perform logically reversible operations on both sides until you reach a true statement$Q$. A proof can then be constructed by starting from$Q$and working backward until you reach$P$. Often times, the backward argument can be formulated as a sequence of equalities, conforming to your teacher's advice. (Note that in the initial phase of seeking a proof, you aren't bound by anything: You can make inspired guesses, additional assumptions, and the like. Only when you write up a final proof must you be careful to assume no more than is given, and to make logically-valid deductions.) • Suppose you start with$1=1$, take the square root of both sides except use$-1$on the left and$1$on the right, and come up with$-1=1$. It seems like there is more going on here than assuming what you want to prove. – Frank Hubeny Apr 29 '16 at 13:16 • @FrankHubeny To the best of my knowledge, you've just equated the two branches of a multifunction restricted to the real axis - which isn't valid. – QuantumFool Apr 29 '16 at 15:43 • @QuantumFool That's right. The problem is not with assuming what one has to prove but with invalid steps along the way. – Frank Hubeny Apr 29 '16 at 18:44 • @Frank: You're perfectly correct that assuming the conclusion is not the only fatal error in this argument; as noted, proving the converse is another. The (logically valid!) argument for "$-1 = 1$implies$1 = 1$" does not consist entirely of reversible steps, so (as you note) does not prove that$-1 = 1$. I mentioned this example to illustrate what the OP has been cautioned against, and because a non-negligible fraction of American university students instinctively attempt to prove algebraic identities by starting with the desired conclusion and manipulating until they obtain a tautology. – Andrew D. Hwang Apr 29 '16 at 20:55 • @BLAZE: An "identity" (such as$\cos^{2} x + \sin^{2} x = 1$) is just an equation that holds for all values of one or more variables (possibly with a "small number of exceptions or restrictions"), so "yes", any proof technique that works for numerical equations also works for identities. – Andrew D. Hwang May 1 '16 at 0:04 It is enough.. Consider this example: To prove:$a=b$Proof: $$a=c$$ $$b=c$$ Since$a$and$b$are equal to the same thing,$a=b$. That is the exact technique you are using and it sure can be used. • I think we also have to consider the domain of each side and make this explicit up front. For example, let$a=(x-1)/(x-1)$and$b=1$. Then we algebraically manipulate this and get$a = 1$. If we do not consider the constraint that$x$is not equal to$1$in the domain of$a$, we get$a = b = 1$for all$x$. That is incorrect. – Frank Hubeny Apr 29 '16 at 15:12 To prove a statement is true, you must not use what you are trying to prove. The main problem is that you've misunderstood what the second teacher said (and possibly the meaning of the equals sign). What that second teacher is saying is do not take as prior facts what you're trying to prove. As one textbook puts it (Setek and Gallo, Fundamentals of Mathematics, 10th Edition, Sec. 3.8), "An argument, or proof, consists basically of two parts: the given statements, which are called premises, and the conclusion". Wikipedia says this (sourced from Cupillari, Antonella. The Nuts and Bolts of Proofs. Academic Press, 2001. Page 3.): "In mathematics and logic, a direct proof is a way of showing the truth or falsehood of a given statement by a straightforward combination of established facts, usually axioms, existing lemmas and theorems, without making any further assumptions." Now consider a simple equation/identity like$6 = 2 \times 3$. The separate sides are expressions; if you just said "6" in English that's a sentence fragment, not an assertion of any fact. It can't be evaluated as either "true" or "false", because it has no assertive content. It cannot be used as a premise because it's not a proposition. What makes something a fully-formed statement in mathematical language is a relation, most commonly equals (but alternatively "is lesser than", "is greater than", etc., effectively the verbs of the language). Translating the equation$6 = 2 \times 3to English we get "6 is the same as 2 times 3", which is indeed a full sentence. This can be checked as being true or false; it makes an assertion. It can be used as a premise because it is a proposition of a particular fact. In conclusion, both your teachers are correct, and both of your proofs are correct (although most of us would prefer the more concise one). When one says "don't use what you're trying to prove" they're not talking about the appearance of any particular expression in an algebraic transformation; expressions are neither premises nor things that can be proven; they are sentence fragments. They're talking about an assertion of fact, which in math has to be a statement including a relational symbol (most commonly an equation). The fact that you didn't start by assuming that equality means that in both cases you've complied with your second teacher's warning. short answer: equality is symmetric, implication is not (both are however transitive) longer answer: • You are right: if you proof A = C and B = C for some terms/expressions/objects A,B,C then you are allowed to conclude that A = B (because "=" is transitive and symmetric) • Your teacher is right: if you prove that something true follows from A = B, i.e. A = B => true, you are not allowed to conclude the converse i.e. that A = B since "true is true" (because implication is not symmetric) I agree with @Siddharth Venu. If we have to prove a=b, At times, on proceeding from LHS you may end up in a stage(intermediate step) which you cannot solve any further and you continue to solve RHS to arrive at the same stage i.e, a=c and b=c so you can conclude that a=b these kind of equality problems often comes in trigonometry Many chapters like matrices and our normal algaebra always have a final step where you can directly establish a relation (a=b). There are two common errors that these methods are preventing: \begin{align} 1 &< 2 &&\text{True.} \\ 0 \cdot 1 &< 0 \cdot 2 &&\text{Um...} \\ 0 &< 0 &&\text{False.} \end{align} \begin{align} 1 &= 2 &&\text{False.} \\ 0 \cdot 1 &= 0 \cdot 2 &&\text{Um...} \\ 0 &= 0 &&\text{True.} \end{align} Much confusion ensues when the two "0$"s in the middle steps are obscured by being large, complicated expressions. For instance, is it evident that you are multiplying by zero when you multiply by$\sin(2x) - 2\sin(x)\cos(x)$? (This uses the double angle formula for sine to get an expression that is always zero.) The correct form of inference when you multiply or divide by something complicated is "$A = B$" becomes "$A/C = B/C$or$C = 0$". That is, you got what you expected or you have inadvertently done something crazy. It seems to take a lot of practice to remember that second clause. The example with$A$,$B$, and$C$can be altered somewhat without changing the need for the second clause. You can also multiply both sides by$C$. The relation need not be an equality. Note however, that the sense of an inequality may change if$C$is ever negative. Unfortunately your high-school teacher (and some of the other answers) is wrong. It is false that proving something of the form "$A = B$" is always possible by starting from one side and algebraically manipulating it to get a sequence of equal expressions ending with the other side, not to say necessary. ## Not necessary Let us first deal with the false misconception that you can only go in one direction. Suppose you have proven the following, where$A,B,C$are any expressions:$A = C$.$B = C$. Then you can use the second sentence to substitute$C$for$B$in the first to obtain:$A = B$. This is logically valid because "$=$" means "is exactly the same as". ## Not always possible Let us now consider an example where it is simply impossible to manipulate from one side to the other to prove an equality!$\def\zz{\mathbb{Z}}\def\qq{\mathbb{Q}}\def\rr{\mathbb{R}}$Theorem: Take any$x \in \rr$such that$x^2 \le 0$. Then$x = 0$. Proof:$x = 0$or$x > 0$or$x < 0$. [by trichotomy] If$x > 0$:$x^2 = x \times x > 0$. [by positive multiplication] This contradicts$x^2 \le 0$. If$x < 0$:$0 < -x$. [by subtraction]$x^2 = (-x) \times (-x) > 0$. [by positive multiplication] This contradicts$x^2 \le 0$. Therefore$x = 0$. Here "positive multiplication" denotes "multiplying by a positive real", which preserves the inequality sign. Similarly subtraction preserves an inequality. The above theorem cannot be proven directly by algebraic manipulation from one side "$x$" to the other "$0$", simply because the only given condition is an inequality. Similarly the construction of$\sqrt{2}$in elementary real analysis, as shown below, does not permit a proof by algebraic manipulation. Theorem: Let$S = \{ r : r \in \qq_{\ge 0} \land r^2 \le 2 \}$. Then$S$is non-empty and has an upper bound in$\rr$. Let$x = \sup_\rr(S)$. Then$x^2 = 2$. Proof: [Exercise! Or see a good textbook like Spivak's.] ## General identities To address the new sub-question in blue, it suffices to generalize the above theorem to the cube-root of arbitrary real numbers, namely:$\sup( \{ r : r \in \qq \land r^3 \le x \} )^3 = x$for any$x \in \rr$. This is an identity but cannot be proven by direct manipulation. You may not be satisfied with this counter-example, but in higher mathematics this kind of identity is in fact the usual kind that we are interested in, rather than identities that can be proven by algebraic manipulation (which are usually considered trivial). Here are some more examples:$\lceil \frac{m}{n} \rceil = \lfloor \frac{m-1}{n}+1 \rfloor$for any$m \in \zz$and$n \in \zz_{>0}$.$\sum_{k=0}^{n-1} 2^k = 2^n-1$for any$n \in \zz_{>0}$. Nevertheless, if you desire an arithmetic identity, the question becomes much more interesting and depends heavily on what you mean by "arithmetic" and what axioms you are allowed to use. Tarski's problem asked whether an arithmetic identity concerning positive integers can be proven using only the basic high-school identities about them, and Wilkie gave an explicit and simple identity that cannot be so proven, basically because any proof needs to use subtraction and hence negative integers. • It seems clear to me that the OP's teacher's advice, "To prove equality of an equation; you start on one side and manipulate it algebraically until it is equal to the other side.", was given in a particular context (avoiding "two-column proofs"), not asserted as the only way of establishing an equality. After re-reading every answer here, I can't find anyone claiming (explicitly or implicitly) that "proving something of the form "$A=B\$" is always possible by starting from one side and algebraically manipulating it to get a sequence of equal expressions ending with the other side [...]." – Andrew D. Hwang Apr 30 '16 at 13:31 • @AndrewD.Hwang: I know what you're saying, but let me explain why I make that comment. Imagine a student who is not good at mathematics whose teacher tells him/her exactly the words cited in the question. It's easy to see that the student will go away with the wrong impression as I specified. Indeed, standard English only allows it to be interpreted that way, because "To do X, you do Y." only means "In order to do X, you { have to / ought to / should / better } do Y." Furthermore, I have seen so many students with exactly that wrong conception. Any good answer ought to correct that. – user21820 Apr 30 '16 at 14:29 • @BLAZE: I didn't want to criticize the other answers too much, but actually Andrew's argument about avoiding two column proofs is in fact not so good because that is how we can justify a formal proof. If you learn natural deduction (especially Fitch-style, which can be extended to quantifiers like at math.stackexchange.com/a/1684204), you will understand what I mean. The issue in the 'proof' Andrew shows is not its two-column nature but because no axiom allows writing the first statement, and in some sense it doesn't address the actual issue of your high-school teacher's teaching. – user21820 May 1 '16 at 4:17 • @BLAZE: To be precise, it is in my opinion pointless to steer a student away from an incorrect proof form by saying things that aren't correct, and furthermore without even explaining what form is incorrect! Your university teacher at least said the right thing, though it is not well explained enough to dismantle the prior misconceptions of many students that they gain from high-school! – user21820 May 1 '16 at 4:20

https://math.stackexchange.com/questions/1436296/is-the-probability-of-picking-an-ace-as-the-second-card-of-a-deck-the-same-if-th

# Is the probability of picking an Ace as the second card of a deck the same if the deck is shuffled after the first card? Here's the situation: You have a deck of 52 cards that is already shuffled. You pick the first card, and the first card is not an Ace. Is the probability of drawing an Ace as second card the same if 1. The second card is immediately drawn from the deck 2. The remaining deck is first shuffled without adding the first card back in, and then the second card is drawn. For me, it is obvious that probability of drawing an Ace in the second case is 4/51, but I'm not entirely sure that the probability is the same in the first case. Also, what would be the probability of the first case if we have drawn n cards and none of these are aces? • yes, its the same. – supinf Sep 15 '15 at 11:54 • 4/(52-n). – Did Sep 15 '15 at 11:56 • As an aside, this is also the reason why it is pointless to doubly randomise a choice (e.g. tossing a coin to see who goes first for the real coin toss!) – Oscar Bravo Sep 15 '15 at 14:04 • Another aside... if you and a friend each draw a card, and then your friend reveals he didn't draw an ace... your odds of getting an ace are better if you replace your card with a new one from the 50 cards remaining. – kbelder Sep 15 '15 at 23:33 If it's not intuitively clear, you can check your calculation using contitional probability: • $X$... the event "the second card is an ace" • $Y$... the event "the first card is not an ace" Then, $P(X\land Y) = \frac{48\cdot 4}{52\cdot 51}$ and $P(Y) = \frac{48}{52}$ Then $$P(X|Y) = \frac{P(X\land Y)}{P(Y)} = \frac{\frac{4\cdot 48}{52\cdot 51}}{\frac{48}{52}} = \frac{4}{51}$$ Given you know the first card of your shuffled deck, the remaining $51!$ ways of arranging the remaining $51$ cards are equally likely. For each of the $52$ possible values of the first card you have $51!$ equally likely arrangements of the remaining $51$ cards. This is consistent with there being $52\cdot 51!=52!$ arrangements of the whole deck to begin with. You can conclude that, after drawing the first card, the remaining $51$ cards are equally likely to be in any shuffled order, and so shuffling them changes essentially nothing. This argument should lead to the conclusion that the probability is the same in either case. Yes, the probability of drawing an Ace is the same if you shuffled the rest of the deck or didn't. You are correct that it's 4/51. If you drew n cards and none were Aces, then the probability of drawing an Ace next would be 4/(52-n). In both cases, there are only 51 cards left, thus $Pr = \dfrac{4}{51}$ And if $n$ non-aces have been drawn, $Pr = \dfrac{4}{52-n}$

https://math.stackexchange.com/questions/1845889/integrals-over-subset-of-measure-space

# Integrals over subset of measure space Let $(X, \mathcal{M}, \mu)$ be a measure space. Suppose $E \in \mathcal{M}$ and $f \in L^+$ where $L^+$ is a space of measurable functions from $X$ to $[0, \infty]$. $\int_E f$ is defined by $\int_X f\chi_E$ where $\chi_E$ is a characteristic function of $E$. Now every time we want to use some property that is true for integrals over whole space $X$, we have to do manipulations with $\chi_E$. For example, let $f_n$ be a sequence in $L^+$ such that $f_n(x) \nearrow f(x)$ for all $x \in E$. Suppose we know that monotone convergence theorem is true for integrals over $X$ and we want to prove $$\int_E f = \lim_{n\to\infty} \int_E f_n.$$ Since $f_n\chi_E \nearrow f\chi_E$ we have $$\int_E f = \int_X f\chi_E = \lim_{n\to\infty} \int_X f_n\chi_E = \lim_{n\to\infty} \int_E f_n.$$ I know it's easy but is there a way to avoid such manipulations? I want to know that equality above and many other statements about integrals are true because $E$ is a measure space in its own right (seems much more natural). More precisely, for $E \in \mathcal{M}$ define $$\mathcal{M}_E = \{ E \cap F \mid F \in \mathcal{M} \}.$$ It's easy to check that $\mathcal{M}_E$ is a $\sigma$-algebra. $(E, \mathcal{M}_E, \mu|_{\mathcal{M}_E})$ is a measure space and $\int_E f$ in space $(X, \mathcal{M}, \mu)$ is equal to $\int_E f|_E$ in space $(E, \mathcal{M}_E, \mu|_{\mathcal{M}_E})$ for every measurable $f$. Is this a commonly accepted way of handling integrals over subset of measure space? If not, do I really have to use $\chi_E$ every time or there is some other way? • You are just using the measure $\mu|_E$, which leads to the same things. As long as you know this measure carries on the properties you want, you're fine. – Silvia Ghinassi Jul 1 '16 at 14:33 • @SilviaGhinassi intuitively it's very simple, but could you be more precise? – edubrovskiy Jul 1 '16 at 14:56 • I agree with you, and that's why I am afraid of getting into technicalities. I might easily say something wrong, so I'll leave it to someone else to provide you a good answer. – Silvia Ghinassi Jul 1 '16 at 15:01 There are basically two interesting types of functions $$f:X \to [0,\infty] , \mu-\text{ measurable}$$ and $$f : X \to [- \infty, \infty] \text{ which is integrable, i.e. } \int f(x) \mu(dx) < \infty .$$ These functions are interesting, because you can define for both the integral $\int_X f(x) \mu(dx)$, but maybe the integral is infinity in the first case. Many theorems are true for either non-negative measurable functions (e.g. monotonic convergence theorem) or integrable functions (dominant convergence theorem). Now if you have proven the monotonic convergece theorem for a general measurable space, then you don't need to prove it for integrals restricted on a set, simply because $(E, \mathcal{M}_E, \mu|_{\mathcal{M}_E})$ is again a measure space, and thus the monotonic convergence theorem is also true here. As well as all other properties and theorems that are known for integrals on a measurable space. So in your case,if you have non-negative functions $f_n(x) \nearrow f(x)$ for all $x\in E$ you immediately get $$\int f(x) \mu|_{\mathcal{M}_E}(dx) = \lim_{n\to \infty} \int f_n(x) \mu|_{\mathcal{M}_E}(dx)$$ However, to be sure that the theorem really holds for all functions restricted to $E$ you need to understand the relation between integrable functions of $\mu_{\mathcal{M}_E}$ and functions $f$ restricted from $X$ to $E$. The relation is as follows: Let $f:X \to [0,\infty]$ be $\mu$-measurable, or let $f:X \to [-\infty,\infty]$ be $\mu$ integrable. Define for some set $E \in \mathcal{M}$ the function $f':A \to [-\infty, \infty]$ with $f'(x) := f(x)$. Then, we have $$\int f'(x) \mu_{\mathcal{M}_E}(dx) = \int_A f(x) \mu(dx)$$ • Could you give me a link to some book or paper where this approach is used? I've already checked relation between integrals on $X$ and on $E$. I just wanted to know that it's a commonly accepted approach. – edubrovskiy Jul 2 '16 at 7:06 • @edubrovskiy I can recommend the book **Measure and Integration Theory ** from Heinz Bauer. Its an old, solid and my personal favorite book. The relation between integrals on $X$ and integrals on $E$ is discussed at the end of §12. – Adam Jul 2 '16 at 8:53

http://math.stackexchange.com/questions/429842/how-many-triangles-in-picture

How many triangles in picture How many triangles in this picture: - I only count 40...and unless someone can explain me how this is a mathematics question I think I'm going to vote to close it as off-topic –  DonAntonio Jun 26 '13 at 9:16 I do not think one needs mathematics to solve the puzzle. –  Avitus Jun 26 '13 at 9:18 @DonAntonio: A systematic approach of the kind that’s second nature to most mathematicians works quite nicely and can even be generalized to larger diagrams of the same type. It’s certainly on-topic. –  Brian M. Scott Jun 26 '13 at 9:20 That sounds sound, Brian. Not closing, then. Thanks. –  DonAntonio Jun 26 '13 at 9:23 They can be counted quite easily by systematic brute force. All of the triangles are isosceles right triangles; I’ll call the vertex opposite the hypotenuse the peak of the triangle. There are two kinds of triangles: 1. triangles whose hypotenuse lies along one side of the square; 2. triangles whose legs both lie along sides of the square and whose peaks are at the corners of the square. The triangles of the second type are easy to count: each corner is the peak of $4$ triangles, so there are $4\cdot4=16$ such triangles. The triangles of the first type are almost as easy to count. I’ll count those whose hypotenuses lie along the bottom edge of the square and then multiply that by $4$. Such a triangle must have a hypotenuse of length $1,2,3$, or $4$. There $4$ with hypotenuse of length $1$, $3$ with hypotenuse of length $2$, $2$ with hypotenuse of length $3$, and one with hypotenuse of length $4$, for a total of $10$ triangles whose hyponenuses lie along the base of the square. Multiply by $4$ to account for all $4$ sides, and you get $40$ triangles of the second type and $40+16=56$ triangles altogether. Added: This approach generalizes quite nicely to larger squares: the corresponding diagram with a square of side $n$ will have $4n$ triangles of the first type and $$4\sum_{k=1}^nk=4\cdot\frac12n(n+1)=2n(n+1)$$ of the second type, for a total of $2n^2+6n=2n(n+3)$ triangles. - I know there's already an excellent answer from Brian M. Scott... Buuuuut... Here's a visual supplement to the aforementioned systematic brute force. - Pretty! $\quad$ –  Brian M. Scott Jun 27 '13 at 9:23

https://math.stackexchange.com/questions/1063724/convergence-of-sum-n-1-infty-left-frac1n-frac1n-2-righ

# Convergence of $\sum_{n=1}^{\infty}\left(\, \frac{1}{n} - \frac{1}{n + 2}\,\right)$ What criteria can I use to prove the convergence of $$\sum_{n=1}^{\infty}\left(\,{1 \over n} - {1 \over n + 2}\,\right)\ {\large ?}$$ My idea was to use ratio test: $$\displaystyle{1 \over n} - {1 \over n+2} = {2 \over n^{2} + 2n}$$ $$\displaystyle\frac{2}{\left(\, n + 1\,\right)^{2} + 2\left(\, n + 1\,\right)} \frac{n^{2} + 2n}{2} = \frac{n^{2} + 2n}{n^{2} + 4n + 3}$$ Of course $\displaystyle n^{2} + 2n \lt n^{2} + 4n + 3$ for all $\displaystyle n$ , but $$\displaystyle\lim \limits_{n \to \infty} \frac{n^{2} + 2n}{n^{2} + 4n + 3}=1$$ so I am not quite sure if I can apply ratio test. You have $\frac{2}{n^2+n} \le \frac{2}{n^2}$. Thus you can use the direct comparision test to prove the convergence (if you already have proven in your course that $\sum_{n=1}^\infty \frac 1{n^2}$ and thus also $\sum_{n=1}^\infty \frac 2{n^2}$ converges). Answer to your 2nd question: You never can apply the ratio test if the limit is 1 (as in this case). I would write out the first few terms and see how they cancel. • of course; but to do this I have to prove absolute convergence first – Christian Dec 11 '14 at 18:42 • @Christian No, you do not. – Andrés E. Caicedo Dec 11 '14 at 18:45 • @Christian You can look at the partial sums and see how they cancel without worrying about absolute convergence. Your problem would be with $\sum \limits_{n=1}^{\infty} \frac1n - \sum \limits_{m=1}^{\infty}\frac{1}{m+2}$ – Henry Dec 11 '14 at 18:47 • It might be useful to mention Telescoping Series. – robjohn Dec 16 '14 at 13:04 • The reason that doing this works is that it allows you to calculate the partial sums directly, which you can use to show convergence of the series by definition (i.e. showing that the partial sums converge) – Joshua Mundinger Dec 16 '14 at 16:44 How about computing the partial sums? For any $N>2$ we have: $$\sum_{n=1}^{N}\left(\frac{1}{n}-\frac{1}{n+2}\right) = \frac{3}{2}-\frac{1}{N+1}-\frac{1}{N+2}$$ hence: $$\left|\frac{3}{2}-\sum_{n=1}^{N}\left(\frac{1}{n}-\frac{1}{n+2}\right)\right|\leq\frac{2}{N}$$ ensures convergence (towards $\frac{3}{2}$). Here's another way to prove the convergence $$\sum \limits_{n=1}^{\infty} \left(\frac1n - \frac{1}{n+2}\right)= \sum \limits_{n=1}^{\infty} \frac{n+2-n}{n(n+2)}$$ $$= \sum \limits_{n=1}^{\infty} \frac{2}{n^2+2n}=2 \sum \limits_{n=1}^{\infty} \frac{1}{n^2+2n}$$ Also note that $$\left|\frac{1}{n^2+2n}\right|\leq \left|\frac{1}{n^2}\right|$$ And by the p-series test, we have $$\sum \limits_{n=1}^{\infty} \left|\frac{1}{n^2}\right| \Rightarrow \mbox{converges}$$ Which implies that $$\sum \limits_{n=1}^{\infty} \frac{1}{n^2} \Rightarrow \mbox{converges absolutely}$$ Therefore, by the direct comparison test $$2 \sum \limits_{n=1}^{\infty} \frac{1}{n^2+2n} \Rightarrow \mbox{converges absolutely}$$ Absolute convergence implies convergence. Also a convergent series multiplied by $2$ is still a convergent series. • I hope the downvoter revisits this answer. I don't think it deserves a downvote, especially after the last edit. (+1) – robjohn Dec 16 '14 at 13:02 You can use the limit comparison test. Let $a_{n} = \frac{2}{n^{2}+2n}$ and consider $b_{n} = \frac{1}{n^{2}}$ then $$\lim_{n\rightarrow\infty} \frac{a_{n}}{b_{n}}= \lim_{n\rightarrow \infty}\frac{2n^{2}}{n^{2}+2n} = 2$$ hence $\sum a_{n}$ and $\sum b_{n}$ converge or diverge together.. but $b_{n}$ is a convergent $p$-series. That being said, my up vote is for Mark Bennet. • with $\frac{1}{n^2}$ ? – Christian Dec 11 '14 at 18:42

https://math.stackexchange.com/questions/2339821/how-to-prove-that-a-sequence-diverges-using-the-cauchy-definition

# How to prove that a sequence diverges using the Cauchy definition? It was a question of my integral calculus exam: Write the Cauchy definition of $\lim_{n\rightarrow\infty} a_n = L$ So, I literally wrote: " $\{a_n\}$ is a Cauchy sequence if given $\epsilon>0$ there exists $N \in \mathbb{N}$ such that $\forall\ m,n\in\mathbb{N}$ if $n,m\ge N$, then $|a_n - a_m|<\epsilon$, or with symbols: $\forall\epsilon>0:\exists N\in\mathbb{N}:\forall\ m,n \in\mathbb{N}: m,n\ge N\implies |a_n - a_m|<\epsilon$. And if $\{a_n\}$ is a Cauchy sequence, then $\lim_{n\rightarrow\infty} a_n = L$ ( i.e. $\{a_n\}$ converges) " The following question was: With this definition, find $\lim_{n\rightarrow\infty} (3n+2)$ And that is what I did: I proposed that $\lim_{n\rightarrow\infty} (3n+2) = \infty$ i.e. $\{3n+2\}$ diverges. So I think that I have to prove that this limit is equal to infinity using the Cauchy definition. But if a sequence $\{a_n\}$ diverges $\implies$ $\{a_n\}$ is not a Cauchy sequence. And $\{a_n\}$ is not a Cauchy sequence if $\exists\epsilon>0:\forall N\in\mathbb{N}:\exists\ m,n \in\mathbb{N}:$ $m,n\ge N$ and $|a_n - a_m|\ge\epsilon$ $(1)$ So, I must to prove $(1)$. $(1)$ says that there exists $m,n\in\mathbb{N}$ such that ..., so $(1)$ has not to apply $\forall m,n\in\mathbb{N}$, and for this fact (I think) I can assign convinient values to $m,n$ in order to prove $(1)$: If i choose $m=N+1,n=N$ then $|3(N+1)+2 -(3N+2)|=3>2$. So exists $\epsilon=2>0$ such that $\forall N \in\mathbb{N}:$ there exists $m=N+1, n=N$ such that $N+1,N \ge N$ and $|3(N+1)+2 -(3N+2)|=3>\epsilon$. Hence $\{3n+2\}$ is not a Cauchy sequence $\implies$ $\lim_{n\rightarrow\infty} (3n+2) = \infty$. Is my proof correct? If it is not correct, could you help me proving that, please? • just reading quick, it looks perfectly good to me – Chessnerd321 Jun 28 '17 at 19:57 • I will note that, at the end, we cannot just say that $$\text{\{a_n\} not Cauchy\implies \lim_{n\to\infty}a_n=\infty}$$We can only say that $$\text{\{a_n\} not Cauchy\implies \lim_{n\to\infty}a_n\notin\Bbb R}$$The limit may be $\pm\infty$ or just non-existent. – Dave Jun 28 '17 at 20:00 • But if i i want to prove that the limit is $\infty$, what have i to do? – Sama Jun 28 '17 at 20:05 • @KarenSM You have to show that for each $M >0$, there is a natural number $N$ such that for all $n \geq N$, $a_n\geq M$. Quite trivial, you can take $N=[M]$, where $[M]$ is the “floor” of $M$, the unique integer satisfying $M-1< [M] \leq M$. – Li Chun Min Jun 28 '17 at 23:53 • @LiChunMin as far as I understand you, I cannot find the $\lim_{n\rightarrow\infty} (3n+2)$ with the Cauchy definition, right? – Sama Jun 29 '17 at 0:26 The definition of a Cauchy sequence is: $\forall \epsilon > 0, \exists N \in \mathbb{N}: \forall m,n \in \mathbb{N}: m,n \ge N \implies |a_m-a_n| < \epsilon$, and the negation of this definition is: $\exists \epsilon > 0, \forall N \in \mathbb{N}, \exists m,n \in \mathbb{N}: m,n \ge N, |a_m - a_n| \ge \epsilon$. For your sequence $a_n = 3n+2, n \ge 1$, choose $\epsilon = 1$ (your choice is $3$), for any $N \in \mathbb{N}$, let $n =N, m = N+1$. Observe $m, n \ge N$, and $|a_m - a_n| = |a_{N+1} - a_{N}|= |3(N+1) - 2 - 3N+2|= 3 > 1= \epsilon$. Thus the sequence above is not Cauchy. • But with $\epsilon=2$ is still fine, right? I'm nervous because I put this in my exam – Sama Jun 28 '17 at 20:22 • $\epsilon = 2$ works just fine. $2$ is better than $1$... – DeepSea Jun 28 '17 at 20:44 You did not write the def'n of $\lim_{n\to \infty}a_n=L.$ What you wrote in response to "Define what it means for $(a_n)_n$ to converge to $L$" was: " $(a_n)_n$ is a Cauchy sequence (which means ...) which converges to $L$." The def'n of $\lim_{n\to \infty}a_n=L$ is $$\forall r>0\;\exists m\;\forall n>m\;(|L-a_n|<r).$$ I have omitted specifying that $m,n\in \mathbb N.$ It is conventional that in "$\lim_{n\to \infty}a_n$", the values of $n$ are restricted to members of $\mathbb N$ unless stated otherwise.

https://math.stackexchange.com/questions/3372929/which-sets-of-sequence-is-countable-and-uncountable

# Which sets of sequence is countable and Uncountable. Consider the sequences $$\displaystyle X=\left\{(x_n): x_n \in \left\{0,1\right\},n \in \mathbb{N} \right\}$$ $$and$$ $$\displaystyle Y=\left\{(x_n)\in X:x_n=1 \;\;\text{for at most finitely many n} \right\}$$ I have to choose which is uncountable and which is countable. Solution i tried- Here $$X$$ is set of sequence with enteries from $$\left\{0,1\right\}$$ thus it has number of elements $$2^{\aleph_0}$$ which is uncountable . Now The set $$Y$$ it has all the sequences from the set $$X$$ but some of its elements of sequences is replaced by the only '$$1$$' so its Cardinality will be less then $$2^{\aleph_0}$$ ,but by $$\textbf{ continuum hypothesis}$$ there is no set having Cardinality in-between the $${\aleph_0}$$ and $$2^{\aleph_0}$$ so the set $$Y$$ will be countable I write this proof but i don't even know this is correct or not but i am sure about set $$X$$ but not sure about $$Y$$ please help me with set $$Y$$ Thank you. • You call both sets $X$, and in the 'second' $X$ you refer to $X$. But that does not make sense. I suppose that the sequences in your 'second' set, are supposed to be sequences of the first set, so they have at most finitly many n, and can only take 0 or 1? – Cornman Sep 28 '19 at 8:54 • The continuum hypothesis can not be proven, or disproven, so you should not use it. :) – Cornman Sep 28 '19 at 8:54 • my bad i will edit it – gaurav saini Sep 28 '19 at 8:56 • What is $x_n$ here? – orlp Sep 28 '19 at 9:01 • Representation of a sequence @orlp – gaurav saini Sep 28 '19 at 9:04 We know that the countable union of countable sets is countable. Which means the countable union of countable unions of countable sets is a countable set. And $$Y$$ is the countable union (indexed over how many $$1$$s the sequences have) of countable unions (indexed over how which positions the finite number of $$1$$s occupy) of countable sets. Bear with me: $$V_0=\{(0)=\{0,0,0,0,.....\}\}$$ is a set with one element. $$V_1 =\{(x_n)$$ where one $$x_i=1$$ and all the rest are $$0\}$$ is countable as there is a one to one correspondence between the $$(x_n)$$ and the possible positions for $$x_i = 1$$. $$W_{k,1} = \{(x_n): x_{j< k}=0; x_k = 1$$ and there is one $$x_{i>k}=1$$ but all the rest are $$0\}$$. $$\iota: W_{k,1}\leftrightarrow V_1$$ via for every $$(x_n)\in W_{k,1}$$, $$\iota((x_n)) = (w_n= x_{n-k})$$. That is if $$(x_n)$$ is the sequence where $$x_k=1$$ and $$x_{i>k} =1$$ and all else are $$0$$, the $$\iota((x_n))$$ is the sequence where $$x_{i-k}=1$$ and all else are $$0$$. This is clearly a bijection. $$V_2=\{(x_n)$$ where two $$x_i=x_j=1$$ and all the rest are $$0\}$$. $$V_2=\cup_{i=1}^{\infty} W_1$$ so $$V_2$$ is countable as it is a countable union of countable sets. Let $$V_m=\{(x_n)$$ where there are exactly $$m$$ $$1$$s and all the rest are $$0\}$$. Let $$W_{k,m} = \{(x_n)$$ where $$x_{j< k}=0;$$ and there are $$m$$ $$x_{i>k}=1$$ and all the rest are $$0\}$$. We can define the bijection $$\iota: W_{k,m}\leftrightarrow V_m$$ via $$(x_n) = (x_{n-k})$$. And $$V_{m+1} = \cup_{i=1}^{\infty} W_{i,m}$$. By induction if $$V_m$$ is countable then so is each $$W_{k,m}$$ and so $$V_{m+1}$$ as a union of countable sets. Now your $$Y=\displaystyle Y=\left\{(x_n)\in X:x_n=1 \;\;\text{for at most finitely many n} \right\} = \cup_{i=o}^{\infty} V_i$$. So $$Y$$ is a countable union of countable sets and thus countable. $$X$$ is indeed uncountable and your proof is correct. Your proof for countability of $$Y$$ is incorrect: Now the set $$Y$$ has all the sequences from the set $$X$$, but some of its elements of sequences are replaced by the only '$$1$$' so its cardinality will be less then $$2^{\aleph_0}$$ Even if you assume the continuums hypothesis (which is a strong assumption to make!), you don't know how many elements you really replaced or removed. How do you know that the cardinality became smaller? Did you construct an explicit injection for that failing to be surjective? Still, $$Y$$ is countable. I'll explain how. Basically, the only information a sequence $$y \in Y$$ has is the finite, possibly disconnected, strip of ones it contains -- if any. You can see the zeroes as the default value with no information. Hence, we might posit the following wrong bijection: $$Y \cong \{0,1\}^* \tag{wrong!}$$ E.g. we would do the following association: • 1110000.... $$\mapsto$$ 111 • 1110100.... $$\mapsto$$ 11101 Certainly, this is injective. But is this surjective? No! We have to consider $$\{0,1\}^*$$ in such a way that padded zeroes at the right are disregarded. In other words, we have $$Y \cong \{0,1\}^*/\sim$$ where $$\sim$$ is some equivalence. If you trust me that such a $$\sim$$ exists, then we actually don't need to work this out for the countability argument. Namely, since $$\{0,1\}^*$$ was already countable, so is every quotiened version of it. Let $$A_n=\{1,\cdots,n\}$$. For each $$f \in \{0,1\}^{A_n}$$, define $$g_f:\Bbb N \to \{0,1\}$$ by $$g_f(x)=\begin{cases}f(x)&\text{if}\;x \in \{1,2,..,n\}\\0&\text{otherwise} \end{cases}$$ Then each $$g_f \in Y$$. Then $$Y$$ can be written as $$Y=\cup_{n=1}^\infty Y_n$$ where $$Y_n=\left\{g_f: f \in \{0,1\}^{A_n}\right\}$$ . Here each $$Y_n$$ is finite and hence $$Y$$ is countable! • does $g_f:\Bbb N \to \{0,1\}$ should be $g_f:\Bbb N \to \{0,1\}^{A_n}$? – gaurav saini Oct 1 '19 at 9:57

https://gmatclub.com/forum/in-how-many-different-ways-can-a-group-of-8-people-be-divide-85707-20.html

GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video It is currently 28 May 2020, 00:56 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # In how many different ways can a group of 8 people be divide Author Message TAGS: ### Hide Tags Intern Joined: 20 Oct 2019 Posts: 2 Re: In how many different ways can a group of 8 people be divide  [#permalink] ### Show Tags 22 Feb 2020, 05:18 Dear experts, What if we had, FOR EXAMPLE, 3 teams of 1 individual in each and 1 team let's say consisting of 5 people. Will the answer be: 8C1*7C1*6C1*5C5/(4!/3!) ? Since we still have to unarrange , but have 3 repeating values. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 10464 Location: Pune, India Re: In how many different ways can a group of 8 people be divide  [#permalink] ### Show Tags 02 Mar 2020, 20:35 2 noboru wrote: In how many different ways can a group of 8 people be divided into 4 teams of 2 people each? A, 90 B. 105 C. 168 D. 420 E. 2520 Responding to a pm: Quote: What if we had, FOR EXAMPLE, 3 teams of 1 individual in each and 1 team let's say consisting of 5 people. Will the answer be: 8C1*7C1*6C1*5C5/(4!/3!) ? Since we still have to unarrange , but have 3 repeating values. I know from your post that if all four teams had different number of people we would not unarrange. But if some repeat? Yes, consider various scenarios: 8 people and 2 teams (one of 3 people and the other of 5 people) You just pick 3 of the 8 people for the 3 people team. Rest everyone will be in the 5 people team. No un-arranging required. 8 people and 2 teams (one of 4 people and the other of 4 people) You pick 4 people for the first team (say A, B, C, D). The other 4 (E, F, G, H) belong to the other team of 4 people. So you calculate this as 8C4. Now within this 8C4 will lie the case in which you picked (E, F, G, H) for the first team and ( A, B, C, D) will belong to the other team. But note that this case is exactly the same as before. 2 teams one (A, B, C, D) and other (E, F, G, H). So you need to un-arrange here. Similarly, take your case - 8 people - 3 teams of 1 individual in each and 1 team let's say consisting of 5 people We select 5 people out of 8 for the 5 people team in 8C5 ways. Rest of the 3 people play individually and we need to create no teams so nothing to be done. Since we are not doing any selection for a team, we are not inadvertently arranging and hence un-arranging is not required. Instead say we have 8 people - and we make 3 teams, one with 4 people and two teams with 2 people each We select 4 people out of 8 for the 4 people team in 8C4 ways. Then we select 2 people for the first two people team in 4C2 ways and the leftover 2 people are for the second two people team. But again, as discussed above, there will same cases counted twice here so we will need to un-arrange by dividing by 2. _________________ Karishma Veritas Prep GMAT Instructor Senior Manager Joined: 21 Feb 2017 Posts: 476 In how many different ways can a group of 8 people be divide  [#permalink] ### Show Tags 15 Mar 2020, 00:16 Bunuel wrote: manalq8 wrote: In how many different ways can a group of 8 people be divided into 4 teams of 2 people each? 90 105 168 420 2520 $$\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=105$$, we are dividing by 4! (factorial of the # of teams) as the order of the teams does not matter. If 8 people are - 1, 2, 3, 4, 5, 6, 7, 8, then (1,2)(3,4)(5,6)(7,8) would be the same 4 teams as (5,6)(7,8)(1,2)(3,4), as we don't have team #1, team #2... For the first person we can pick a pair in 7 ways; For the second one in 5 ways (as two are already chosen); For the third one in 3 ways (as 4 people are already chosen); For the fourth one there is only one left. So we have 7*5*3*1=105 You can check similar problems: http://gmatclub.com/forum/probability-8 ... %20equally http://gmatclub.com/forum/probability-8 ... ide+groups http://gmatclub.com/forum/combination-5 ... ml#p690842 http://gmatclub.com/forum/sub-committee ... ide+groups There is also direct formula for this: 1. The number of ways in which $$mn$$ different items can be divided equally into $$m$$ groups, each containing $$n$$ objects and the order of the groups is not important is $$\frac{(mn)!}{(n!)^m*m!}$$. 2. The number of ways in which $$mn$$ different items can be divided equally into $$m$$ groups, each containing $$n$$ objects and the order of the groups is important is $$\frac{(mn)!}{(n!)^m}$$ Hope it helps. we say we need to divide by 4! cus the order doesn't matter but when we use combinations arent we anyway implying that the order doesn't matter. I am not able to understand why we need to divide by 4! again when were using combinations and not permutations. The second approach, I understand perfectly hence, I know I am missing something in the first one. Pls, help. This is really a confusing sub-topic for me under PnC In how many different ways can a group of 8 people be divide   [#permalink] 15 Mar 2020, 00:16 Go to page   Previous    1   2   [ 23 posts ]

https://math.stackexchange.com/questions/1977626/help-needed-with-different-approach-to-a-combinatorics-question

# Help needed with different approach to a combinatorics question Every day, the 15 students in Mr. Singh`s Advanced Chemistry class are randomly divided into 5 lab groups of 3 students each. What is the probability that three students - Linda, Martin, and Nancy - are in the same lab group today? The answer to the question is to see it from the perspective of one of the students we are concerned with. From his/her point of view, 2 of the other 14 students are randomly chosen to be paired with her. This can be done in ${14}\choose{2}$ = 91 ways. Only one of those pairs are her "friends", so the answer is $\frac{1}{91}$. This answer of course didn't come to me, I read it when I couldn't figure it out. Though curious if I could find the same answer with my approach, I tried to kind of work backwards, knowing the answer. Here's what I tried to do: I instead thought of how many groups of three could be chosen from 15 students. I thought this could be done in ${15}\choose{3}$ = 455 ways. One of those contains the group of students we are interested in. But the answer is not $\frac{1}{455}$. How does 91 fit into this? Curious as I was, I divided 455 by 91, and lo and behold, I got 5. Did this 5 mean the 5 groups of three students? I wasn't sure. If I divide 455 by 5, I get 91. But what is the meaning of this? I would rather think that, from the 455 available choices, we choose 5 things (in this case a thing is a group of 3 students). But this would mean ${455}\choose{5}$ = some huge number. Let's say I have ABCDE, and I want to choose 2 of them. This would be ${5}\choose{2}$ = 10 combinations. If I would ask "how many groups of 5 can I make?", then the answer surely isn't $\frac{10}{5}$ = 2. I definitely can check with pen and paper and see that I can make quite more than 2 groups. So I know my thinking is faulty. But where? What does the 5 in $\frac{455}{5}$ = 91 mean? And what should my line of thought be when starting from ${15}\choose{3}$ to get to the answer? There are $\binom{15}3=455$ possible $3$-person groups, so if we were picking just one group of $3$ at random, the probability would indeed be $\frac1{455}$ that we picked this specific group. However, we’re not just picking a single group: we’re dividing the $15$ students into five groups of $3$, so we actually have $5$ chances to get the desired group, not just one. And $\frac5{455}$ is, as you noticed, $\frac1{91}$. You might reasonably worry that picking a random partition of the students into $5$ groups of $3$ students isn’t really the same as picking $5$ groups of $3$ at random with replacement from the collection of all $455$ $3$-person groups. We can do a more elaborate calculation to show that it really does work out right. There are $$\binom{15}3\binom{12}3\binom93\binom63\binom33$$ ways to pick a first group of $3$, then pick a second group of $3$ from the $12$ people who remain, and so on until we have all $5$ groups. However, this counts each of the $5!$ permutations of the $5$ groups separately, so the number of ways to partition the students into $5$ groups of $3$ is really only $$\frac1{5!}\binom{15}3\binom{12}3\binom93\binom63\binom33\;.$$ How many of these partitions have the desired trio as one of its parts? The same reasoning shows that if we set Linda, Martin, and Nancy aside and form $4$ groups of $3$ from the remaining $12$ students, we can do this in $$\frac1{4!}\binom{12}3\binom93\binom63\binom33$$ different ways. The probability that a randomly chosen partition has our trio as one of its groups is therefore $$\frac{\frac1{4!}\binom{12}3\binom93\binom63\binom33}{\frac1{5!}\binom{15}3\binom{12}3\binom93\binom63\binom33}=\frac{5!}{4!\binom{15}3}=\frac5{\binom{15}3}=\frac5{455}=\frac1{91}\;.$$ • "5 chances to get the desired group" - I try to think about this, but I'm having a hard time. I keep connecting the 5 groups with the 455 things, which I probably shouldn't do. As an aside, how long did it take you to be fluent in this, i.e., have the insight quick and the ability to see it from multiple angles? – Garth Marenghi Oct 20 '16 at 19:37 • @Garth: When you divide the students into $5$ groups of $3$, you really are picking $5$ groups, not just one, and one of those $5$ could be the one that we want. I really can’t answer that last question: it’s been over $50$ years since I first encountered these ideas. I don’t think that I ever had any trouble with them, but I know from having taught the subject that my experience is not typical. – Brian M. Scott Oct 20 '16 at 19:46

https://math.stackexchange.com/questions/1187351/how-many-ways-are-there-to-divide-elements-into-equal-unlabelled-sets

# How many ways are there to divide elements into equal unlabelled sets? How many ways are there to divide N elements into K sets where each set has exactly N/K elements. For the example of 6 elements into 3 sets each with 2 elements. I started by selecting the elements that would go in the first set (6 choose 2) and then those that would go into the second as (4 choose 2) and then the 2 remaining elements into the third set. This gives, (6 choose 2) * (4 choose 2). In general (N choose N/K) * (N-(N/K) choose N/K) * (N-(2*N/K) choose N/K) * ... * 1 • Any thoughts on the problem? You aren't a new user here. You should know that questions on this site require at least some input on your efforts towards the problem. Otherwise you are destined for downvotes or being placed on hold – jameselmore Mar 12 '15 at 21:10 • Added. I'm concerned that I'm assuming some kind of ordering. @jameselmore – nickponline Mar 12 '15 at 21:16 Your approach, as noted by Ross Millikan's answer, is effective. Another way to approach such a problem would be to consider "interpreting" a permutation as such a partition - like, if we wrote the elements in the order: $$eabcfd$$ we might just group them into pairs as $\{\{e,a\},\{b,c\},\{f,d\}\}$ - where we just "fill" the expression $\{\{\_,\_\},\{\_,\_\},\{\_,\_\}\}$ by drawing from the order in which we wrote the elements. However, $eabcfd$ and $aebcfd$ represent the same partition, as the pairs are unordered - and $bceaf\!\,\!d$ also represents the same partition, as the order of the pairs does not matter. Proceeding thusly, we can see that we can reorder within each of the $K$ sets in $(N/K)!$ ways without affecting the partition, and we can reorder the order in which the sets appear in the partition in $K!$ ways - and that, these are the only transformations which do not affect the partition. Thus, dividing the number of permutations of the elements by the number of permutations representing any given partition yields that there are $$\frac{N!}{(N/K)!^k K!}$$ such partitions. (This can also be found by expanding ${a \choose b}=\frac{a!}{b!(a-b)!}$ and looking at cancellations in your expression) • Thanks! Is there a way to generalize this such that: if I make D draws with replacement from N distinct items. how many ways are there to get C copies of any K distinct items. I tried to generalize with your analogy of "filling" in the bracket. But for D > C*K it's unclear how to no double count groups. – nickponline Mar 20 '15 at 16:26 Hint: You are close. As the sets are unlabeled, choosing $\{a,b\},\{c,d\},\{e,f\}$ is the same as choosing $\{e,f\},\{c,d\},\{a,b\}$, but you have counted them both. • Ah, so I should divide by the number of ways of ordering the sets = K! – nickponline Mar 12 '15 at 21:20 • Awesome :), is there a more compact formula that avoids the product? – nickponline Mar 12 '15 at 21:24 • If you look at the factorial expression for the binomial coefficients, there is a lot of cancellation. Let $M=\frac NK$ Your first two terms are $\frac {N!}{(N-M)!M!}\cdot \frac {(N-M)!}{(N-2M)!M!}$ and you can see the cancellation. You wind up with $\frac {N!}{(M!)^KK!}$ There is a nice combinatorial interpretation. There are $N!$ ways to put the elements in order. Reordering elements within a group can be done in $M!$ ways and you have $K$ of those. The last $K!$ is the ordering of the groups. – Ross Millikan Mar 12 '15 at 21:54 In response to your comment on Ross's answer, there is a more compact form: $$\text{number of partitions }=P_{n,k}=\frac{n!}{(n/k)!^{k}k!}$$ To prove this, we instead prove $$n!=P_{n,k}\cdot (n/k)!^k\cdot k!$$ To prove that, ask how many ways are there permute $n$ elements? There are certainly $n!$ such ways. But another to form a permutations is this: first, divide the $n$ elements into $k$ equal sets $(P_{n,k}$ ways to do this). Then, permute each element within each of the sets (they all have size $n/k$, and there are $k$ of them, so there are $(n/k)!^k$ ways. Finally, put the $k$ sets themselves into some order $(k!$ ways). You can check that $P_{6,3}=\frac{6!}{2!^3\cdot3!}=15$, and that indeed there are $5$ choices for what the set containing $1$ is, and then $\binom{4}2=3$ ways to partition the remaining 4 elements into sets.

https://math.stackexchange.com/questions/4194996/representing-a-linear-transformation-with-respect-to-a-new-basis

# Representing a linear transformation with respect to a new basis I have a question about the following exercise: Let $$f:\mathbb{R}^3\to\mathbb{R}^3$$ be the linear transformation such that $$f(x,y,z)=(x-y+2z,-2x+y,x+z).$$ Represent the transformation with respect to the basis $$\{(1,1,0),(0,1,-1),(1,1,1)\}$$. What I have done: I found the images of the basis vectors given: \begin{align}f(1,1,0) &= (0,-1,1)\\ \\ f(0,1,-1) &= (-3,1,-1)\\ \\ f(1,1,1) &= (2,-1,2) \end{align} and then I found how these vectors can be written as a linear combination of the basis vectors given: \begin{align}(0,-1,1)&=\fbox{0}\cdot (1,1,0)+\fbox{(-1)}\cdot (0,1,-1)+\fbox{0}\cdot (1,1,1) \\ \\ (-3,1,-1)&=\fbox{-6}\cdot (1,1,0)+\fbox{4}\cdot (0,1,-1)\;\;\,\,+\fbox{3}\cdot (1,1,1) \\ \\ (2,-1,2)&=\fbox{3}\cdot (1,1,0)+\fbox{(-3)}\cdot (0,1,-1)+\fbox{-1}\cdot (1,1,1)\end{align} thus the matrix which represents $$f$$ with respect to this basis should be $$\begin{bmatrix}0 & -6 & 3\\ -1 & 4 & -3\\ 0 & 3 & - 1\end{bmatrix}.$$ Is this correct? Or should I write the matrix with respect to the canonical basis of $$\mathbb{R}^3$$ and then make the change of basis $$C^{-1}AC$$? • How did you calculate the coefficients in the boxes? Some of them, as currently written, are incorrect. Jul 10 at 12:33 • Both methods are correct, and should give you the same answer – if you do them right. Jul 10 at 13:05 • @shoteyes they should be correct now, thank you for pointing that out. For each vector $v$, I set up a system to find out which linear combination of basis vectors is equal to it, ie I found out which $c_1,c_2,c_3$ are such that $c_1 v_1+c_2v_2+c_3v_3=v$ and solved it using Gaussian elimination. Jul 10 at 15:48 • Any thoughts on the comments, or on Dan's answer? Jul 12 at 13:47 • @Gerry Myerson they have all been very useful, I have upvoted the comments and accepted Dan's answer, thank you very much Jul 12 at 13:57 This is the same. Maybe the square diagram in the sequel shows in the "simplest" way why. First i have to say something about the used convention for vectors. Because this is the "canonical impediment" when dealing with base change. We work with column vectors, and matrices act on them by left multiplication. The linear map of left multiplication with a matrix $$A$$ will be denoted below (abusively) also by $$A$$. So $$x$$ goes via $$A$$ to $$A\cdot x=Ax$$, displayed as $$x\overset{A}\longrightarrow Ax\ .$$ "Most of the world" uses column vectors. (Some authors write notes or books (e.g. in Word), and find it handy to use row vectors, so they can be simpler displayed in the book rows. In this case linear maps induced by matrices use the multiplication from the right with such matrices. As long as we need in computations only linear combinations the convention is not so important, but it becomes when we use linear maps induced by matrices.) We will work in the "category" of (finite dimensional) vector spaces (over $$\Bbb R$$) with a fixed bases. The space $$V:=\Bbb R^3$$ comes with the canonical base $$\mathcal E=(e_1,e_2,e_3)$$, where $$e_1,e_2,e_3$$ are the columns of the matrix $$E$$ below, $$E= \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}\ .$$ We write this object as $$(V,\mathcal E)$$. By abuse, we may want to write $$(V,E)$$ instead. We start with two objects in this category. For our purposes let them have the same underlying vector space $$V=W=\Bbb R^3$$, first object is $$(V,\mathcal B=(b_1,b_2,b_3))$$, and the second object is $$(W,\mathcal C=(c_1,c_2,c_3))$$. A linear map $$g:V\to W$$ is defined "abstractly", and has no need for chosen bases. But in practice, $$g$$ is usually given bases-specific in the following way. Let $$v$$ be a vector in $$V$$. We write it w.r.t. $$\mathcal B$$ as $$v=x_1b_1+x_2b_2+x_3b_3$$, and write this data as a column vector: $$v = \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}_{\mathcal B} :=x_1b_1+x_2b_2+x_3b_3 \ .$$ Then we consider a matrix $$M=M_{\mathcal B, \mathcal C}$$ and build the matrix multiplication vector: $$\begin{bmatrix}y_1\\y_2\\y_3\end{bmatrix} = M \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}\ .$$ Then we consider the vector $$w\in W$$ which written in base $$\mathcal C$$ has the $$y$$-components, so $$w = \begin{bmatrix}y_1\\y_2\\y_3\end{bmatrix}_{\mathcal C} :=y_1c_1+y_2c_2+y_3c_3 \ ,$$ and the map $$g$$ is mapping linearly $$v$$ to $$w$$. This concludes the section related to conventions and notations. Let $$\mathcal C$$ be the base from the OP, the base with vectors which are columns of $$C= \begin{bmatrix} 1 & 0 & 1\\ 1 & 1 & 1\\ 0 & -1 & 1 \end{bmatrix}\ .$$ Let $$A$$ be the matrix for the given linear map $$f$$ w.r.t. the canonical base $$\mathcal E$$. $$A=\begin{bmatrix}1&-1&2\\-2&1&0\\1&0&1\end{bmatrix}\ .$$ Consider now the diagram: $$\require{AMScd}$$ $$\begin{CD} (V,E) @>A>f> (V,E) \\ @A C A {\operatorname{id}}A @A {\operatorname{id}}A C A\\ (V,C) @>f>{C^{-1}AC}> (V,C) \\ \end{CD}$$ Indeed, $$C$$ is the matrix of the identity seen as a map $$(V,\mathcal C)\to(V,\mathcal E)$$. For instance, $$c_1=\begin{bmatrix}1\\0\\0\end{bmatrix}_{\mathcal C} \qquad\text{ goes to }\qquad c_1 =\begin{bmatrix}1\\1\\0\end{bmatrix}_{\mathcal E} =C\begin{bmatrix}1\\0\\0\end{bmatrix}_{\mathcal E}\ .$$ It remains to compute explicitly the matrix $$C^{-1}AC$$. Computer, of course: sage: A = matrix(3, 3, [1, -1, 2, -2, 1, 0, 1, 0, 1]) sage: C = matrix(3, 3, [1, 0, 1, 1, 1, 1, 0, -1, 1]) sage: A [ 1 -1 2] [-2 1 0] [ 1 0 1] sage: C [ 1 0 1] [ 1 1 1] [ 0 -1 1] sage: C.inverse() * A * C [ 0 -6 3] [-1 4 -3] [ 0 3 -1] And we check the result both ways: $$\bf(1)$$ $$\require{AMScd}$$ $$\begin{CD} c_j=\begin{bmatrix}1\\1\\0\end{bmatrix}_{\mathcal E}\ ,\ \begin{bmatrix}0\\1\\-1\end{bmatrix}_{\mathcal E}\ ,\ \begin{bmatrix}1\\1\\1\end{bmatrix}_{\mathcal E} @>A>f> fc_j=\begin{bmatrix}0\\-1\\1\end{bmatrix}_{\mathcal E}\ ,\ \begin{bmatrix}-3\\1\\-1\end{bmatrix}_{\mathcal E}\ ,\ \begin{bmatrix}2\\-1\\2\end{bmatrix}_{\mathcal E} \\ @A C A {\operatorname{id}}A @A {\operatorname{id}}A C A\\ c_j=\begin{bmatrix}1\\0\\0\end{bmatrix}_{\mathcal C}\ ,\ \begin{bmatrix}0\\1\\0\end{bmatrix}_{\mathcal C}\ ,\ \begin{bmatrix}0\\0\\1\end{bmatrix}_{\mathcal C} @>f>{C^{-1}AC}> fc_j=\begin{bmatrix}0\\-1\\0\end{bmatrix}_{\mathcal C}\ ,\ \begin{bmatrix}-6\\4\\3\end{bmatrix}_{\mathcal C}\ ,\ \begin{bmatrix}3\\-3\\-1\end{bmatrix}_{\mathcal C} \end{CD}$$ Or simpler, using block matrices, and ignoring the knowledge of the bases: $$\require{AMScd}$$ $$\begin{CD} C @>A>f> AC \\ @A C A {\operatorname{id}}A @A {\operatorname{id}}A C A\\ E @>f>{C^{-1}AC}> C^{-1}AC \end{CD}$$ $$\bf(2)$$ In the spirit of the OP, using copy+pasted+corrected row vector computations: \begin{aligned} (0,-1,1) &=\boxed{0}\cdot (1,1,0)+\boxed{(-1)}\cdot (0,1,-1)+\boxed{0}\cdot (1,1,1) \ , \\ \\ (-3,1,-1) &=\boxed{-6}\cdot (1,1,0)+\boxed{4}\cdot (0,1,-1)+\boxed{3}\cdot (1,1,1) \ , \\ \\ (2,-1,2) &=\boxed{3}\cdot (1,1,0)+\boxed{(-3)}\cdot (0,1,-1)+\boxed{-1}\cdot (1,1,1) \ . \end{aligned}

https://math.stackexchange.com/questions/2839456/finding-orthonormal-basis-from-orthogonal-basis

# Finding orthonormal basis from orthogonal basis The following is example C.5 from Appendix C (Linear Spaces Review) of Introduction to Laplace Transforms and Fourier Series, Second Edition, by Phil Dyke: Example C.5 Show that $\{\sin(x),\cos(x)\}$ is an orthonormal basis for the inner product space $V=\{a\sin(x)+b\cos(x); a,b\in\mathbb R, 0\le x\le\pi\}$ using as inner product $$\langle f,g \rangle = \int_0^1 fg dx, \qquad f,g\in V$$ and determine an orthonormal basis. Solution $V$ is two dimensional and the set $\{\sin(x),\cos(x)\}$ is obviously a basis. We merely need to check orthogonality. First of all, \begin{align}\langle\sin(x),\cos(x)\rangle=\int_0^\pi\sin(x)\cos(x)\,dx&=\frac{1}{2}\int_0^\pi\sin(2x)\,dx\\ &=\left[-\frac{1}{4}\cos(2x)\right]_0^\pi \\ &=0.\end{align} Hence orthogonality is established. Also, $$\langle\sin(x),\sin(x)\rangle=\int_0^\pi\sin^2(x)\,dx=\frac{\pi}{2}$$ and $$\langle\cos(x),\cos(x)\rangle=\int_0^\pi\cos^2(x)\,dx=\frac{\pi}{2}.$$ Therefore $$\left\{\sqrt{\dfrac{2}{\pi}}\sin(x),\sqrt{\dfrac{2}{\pi}}\cos(x)\right\}$$ is an orthonormal basis. I understand that, for orthonormality, we require that $\| \mathbf{a} \| = 1$. However, I'm unsure of how the orthonormal basis was found at the bottom of the proof? I would appreciate it if people could please take the time to clarify this. • $\lvert \lvert \sin(x) \rvert \rvert = \sqrt{\langle \sin(x), \sin(x) \rangle} = \sqrt{\frac{\pi}{2}}$. So in order to make an orthonormal basis from this orthogonal basis, we devide by the length, – Tim Dikland Jul 3 '18 at 10:09 • @TimDikland You mean $||\sin(x) ||$? – The Pointer Jul 3 '18 at 10:11 • Exactly, editted. – Tim Dikland Jul 3 '18 at 10:11 • Whenever you require an orthogonal basis to be ortonormal, just divide each vecotr by its norm. It remains an orthogonal basis (because of the properties of the inner product), but the norm of each vector is 1. – LuxGiammi Jul 3 '18 at 10:15 • Is that a typo in your example where you are using the integral over $[0,1]$ to define the inner product? If not, that changes things. – Disintegrating By Parts Jul 5 '18 at 3:07 As mentioned in the comments to the main post, $\lVert \sin(x) \rVert = \sqrt{\langle \sin(x), \sin(x) \rangle} = \sqrt{\frac{\pi}{2}}$. We then divide the orthogonal vectors by their norms in order convert them into orthonormal vectors. This gets us the orthonormal basis mentioned in the textbook excerpt.

https://math.stackexchange.com/questions/3114318/computing-the-product-fracddxxn-fracddxxn/3114413

# Computing the product $(\frac{d}{dx}+x)^n(-\frac{d}{dx}+x)^n$ I want to compute the product $$(\frac{d}{dx}+x)^n(-\frac{d}{dx}+x)^n,$$ for a natural number $$n$$. For $$n$$ equal to 0 or 1, the computation is very simple obviously, but for such a low number as 2 the brute force calculation begins to be rather cumbersome and I cannot see any pattern emerging. I tried to find some connection with the Rodrigues' formula for the Hermite polynomials but I could not. These operators come up in the algebraic approach to the quantum harmonic oscillator. Explicit Example To avoid any misunderstanding, I am going to show explicitly the computation for the case $$n=1$$: $$(\frac{d}{dx}+x)(-\frac{d}{dx}+x)=-\frac{d^2}{dx^2}+1+x\frac{d}{dx}-x\frac{d}{dx}+x^2=-\frac{d^2}{dx^2}+x^2+1.$$ One can think of a function $$f$$ the operators are acting on. For example, $$(\frac{d}{dx}\circ x) f= (\frac{d}{dx}x)f+x\frac{d}{dx}f=(1+\frac{d}{dx})f,$$ then $$\frac{d}{dx}\circ x=1+x\frac{d}{dx}$$ • Since this is physics related, can I ask what the stand-alone differential expression $\mathrm{d}/\mathrm{d}x$ means for your context ? – Rebellos Feb 15 at 19:47 • That simplification doesn't work though as operators don't commute. It simplifies to $(-\frac{d^2}{dx^2}-x\frac{d}{dx}+\frac{d}{dx}x+x^2)^n$. – Nick Guerrero Feb 15 at 20:00 • @Rebellos, $d/dx$ is simply the derivative operator. In this context, it is supposed that the operators are acting on some function. For example, the commutator between $d/dx$ and $x$ is computed as $[d/dx,x]f=d/dx(xf)-x(d/dx f)=f+d/dx f-d/dx f=f$, then $[d/dx,x]=1$. – jobe Feb 15 at 20:15 • @NickGuerrero Inasmuch as these operators ($\mathscr{L}^+$ and $\mathscr{L}^-$) do not commute, how would you show, as you asserted, that $$\left(\mathscr{L}^+\right)^n \left(\mathscr{L}^-\right)^n=(\mathscr{L}^+\mathscr{L}^-)^n?$$ – Mark Viola Feb 15 at 20:32 • @jobe See THIS, which provides a discussion for $(A+B)^n$ where $[A,B]\ne 0$ (i.e., the operators do not commute). – Mark Viola Feb 15 at 20:47 For convenience let us rewrite $$x,\partial_x$$ with $$a,b$$ so $$[a,b]=x\partial_x-\partial_x x=-1$$ (in the operator sense on the Schwartz space). Lemma. For all $$n\in\mathbb N$$ $$[(a+b)^n,a-b]=2n(a+b)^{n-1}$$ Proof. As darij pointed out, one has $$[a+b,a-b]=2$$ (i.e. the case $$n=1$$). The trick then is $$\begin{split} [(a+b)^{n+1},a-b]&=(a+b)[(a+b)^n,a-b]+[a+b,a-b](a+b)^n\\ &=(a+b)2n(a+b)^{n-1}+2(a+b)^{n}=2(n+1)(a+b)^{(n+1)-1} \end{split}$$ which concludes the proof via induction. $$\square$$ Proposition. For all $$n\in\mathbb N_0$$ $$(a+b)^n(a-b)^n=\prod_{j=1}^n (a^2-b^2+(2j-1))$$ Proof. ($$n=0$$ is obvious). Note that $$(a-b)(a+b)=a^2-b^2-1$$. Using the previous lemma $$\begin{split} (a+b)^{n+1}(a-b)^{n+1}&=[(a+b)^{n+1},a-b](a-b)^n+(a-b)(a+b)(a+b)^n(a-b)^n\\ &=2(n+1)(a+b)^n(a-b)^n +(a^2-b^2-1)(a+b)^n(a-b)^n\\ &=\big( a^2-b^2+2n+1)(a+b)^n(a-b)^n=\prod_{j=1}^{n+1} (a^2-b^2+(2j-1)) \end{split}$$ which again concludes the proof via induction. $$\square$$ This result reproduces the cases (aside from $$n=0$$, obvious) • $$(a+b)(a-b)=a^2-b^2+1$$ • Making use of $$[a^2,b^2]=-4ab-2$$ (similar techniques) one gets $$\begin{split} (a+b)^2(a-b)^2=(a^2-b^2+1)(a^2-b^2+3)&=a^4-a^2b^2-b^2a^2+4a^2-4b^2+b^4+3\\ &=a^4-2a^2b^2-4ab+4a^2-4b^2+b^4+1 \end{split}$$ etc... I feel like this formula is the best thing one can hope for in terms of structure. Edit: Thanks darij for the +200 rep! • Great job! Pretty sure that a factorization into degree-$2$ factors is an answer to the OP. – darij grinberg Feb 15 at 21:26 • Thank you! I like how we both independently considered a commutator of the form $[x^m,y]$ (my Lemma / your eq. (1)) to approach the problem... – Frederik vom Ende Feb 15 at 21:30 • Thank you for your answer! I am very happy my question has received so amazing answers, but I am having a hard time choosing which one to accept. Your answer seems to be the more explicit that is possible without becoming too complicated. I would like to have the derivative on the right side of each term, but probably that would made the result very complicated as the obtained by @Sangchul Lee. – jobe Feb 17 at 1:12 • Glad to hear so! Of course one may use further tricks to shift all the derivatives ($b$'s) to the right such as $[x^2,\partial_x^2]=-4x\partial_x-2$ etc. although I'm not sure how simple or applicable that'd end up being. Anyways, just accept whatever answer helps (or will likely end up helping) you the most! – Frederik vom Ende Feb 17 at 7:35 Just some Sage-generated data to play around with: For $$n = 0$$, the result is $$1$$. For $$n = 1$$, the result is $$-\frac{\partial^{2}}{\partial x^{2}} + x^{2} + 1$$. For $$n = 2$$, the result is $$\frac{\partial^{4}}{\partial x^{4}} - 2 x^{2} \frac{\partial^{2}}{\partial x^{2}} - 4 \frac{\partial^{2}}{\partial x^{2}} - 4 x \frac{\partial}{\partial x} + x^{4} + 4 x^{2} + 1$$. For $$n = 3$$, the result is $$-\frac{\partial^{6}}{\partial x^{6}} + 3 x^{2} \frac{\partial^{4}}{\partial x^{4}} + 9 \frac{\partial^{4}}{\partial x^{4}} + 12 x \frac{\partial^{3}}{\partial x^{3}} - 3 x^{4} \frac{\partial^{2}}{\partial x^{2}} - 18 x^{2} \frac{\partial^{2}}{\partial x^{2}} - 9 \frac{\partial^{2}}{\partial x^{2}} - 12 x^{3} \frac{\partial}{\partial x} - 36 x \frac{\partial}{\partial x} + x^{6} + 9 x^{4} + 9 x^{2} - 3$$. For $$n = 4$$, the result is $$\frac{\partial^{8}}{\partial x^{8}} - 4 x^{2} \frac{\partial^{6}}{\partial x^{6}} - 16 \frac{\partial^{6}}{\partial x^{6}} - 24 x \frac{\partial^{5}}{\partial x^{5}} + 6 x^{4} \frac{\partial^{4}}{\partial x^{4}} + 48 x^{2} \frac{\partial^{4}}{\partial x^{4}} + 42 \frac{\partial^{4}}{\partial x^{4}} + 48 x^{3} \frac{\partial^{3}}{\partial x^{3}} + 192 x \frac{\partial^{3}}{\partial x^{3}} - 4 x^{6} \frac{\partial^{2}}{\partial x^{2}} - 48 x^{4} \frac{\partial^{2}}{\partial x^{2}} - 36 x^{2} \frac{\partial^{2}}{\partial x^{2}} + 48 \frac{\partial^{2}}{\partial x^{2}} - 24 x^{5} \frac{\partial}{\partial x} - 192 x^{3} \frac{\partial}{\partial x} - 216 x \frac{\partial}{\partial x} + x^{8} + 16 x^{6} + 42 x^{4} - 48 x^{2} - 39$$. For $$n = 5$$, the result is $$-\frac{\partial^{10}}{\partial x^{10}} + 5 x^{2} \frac{\partial^{8}}{\partial x^{8}} + 25 \frac{\partial^{8}}{\partial x^{8}} + 40 x \frac{\partial^{7}}{\partial x^{7}} - 10 x^{4} \frac{\partial^{6}}{\partial x^{6}} - 100 x^{2} \frac{\partial^{6}}{\partial x^{6}} - 130 \frac{\partial^{6}}{\partial x^{6}} - 120 x^{3} \frac{\partial^{5}}{\partial x^{5}} - 600 x \frac{\partial^{5}}{\partial x^{5}} + 10 x^{6} \frac{\partial^{4}}{\partial x^{4}} + 150 x^{4} \frac{\partial^{4}}{\partial x^{4}} + 150 x^{2} \frac{\partial^{4}}{\partial x^{4}} - 150 \frac{\partial^{4}}{\partial x^{4}} + 120 x^{5} \frac{\partial^{3}}{\partial x^{3}} + 1200 x^{3} \frac{\partial^{3}}{\partial x^{3}} + 1800 x \frac{\partial^{3}}{\partial x^{3}} - 5 x^{8} \frac{\partial^{2}}{\partial x^{2}} - 100 x^{6} \frac{\partial^{2}}{\partial x^{2}} - 150 x^{4} \frac{\partial^{2}}{\partial x^{2}} + 1500 x^{2} \frac{\partial^{2}}{\partial x^{2}} + 975 \frac{\partial^{2}}{\partial x^{2}} - 40 x^{7} \frac{\partial}{\partial x} - 600 x^{5} \frac{\partial}{\partial x} - 1800 x^{3} \frac{\partial}{\partial x} - 600 x \frac{\partial}{\partial x} + x^{10} + 25 x^{8} + 130 x^{6} - 150 x^{4} - 975 x^{2} - 255$$. For $$n = 6$$, the result is $$\frac{\partial^{12}}{\partial x^{12}} - 6 x^{2} \frac{\partial^{10}}{\partial x^{10}} - 36 \frac{\partial^{10}}{\partial x^{10}} - 60 x \frac{\partial^{9}}{\partial x^{9}} + 15 x^{4} \frac{\partial^{8}}{\partial x^{8}} + 180 x^{2} \frac{\partial^{8}}{\partial x^{8}} + 315 \frac{\partial^{8}}{\partial x^{8}} + 240 x^{3} \frac{\partial^{7}}{\partial x^{7}} + 1440 x \frac{\partial^{7}}{\partial x^{7}} - 20 x^{6} \frac{\partial^{6}}{\partial x^{6}} - 360 x^{4} \frac{\partial^{6}}{\partial x^{6}} - 540 x^{2} \frac{\partial^{6}}{\partial x^{6}} + 120 \frac{\partial^{6}}{\partial x^{6}} - 360 x^{5} \frac{\partial^{5}}{\partial x^{5}} - 4320 x^{3} \frac{\partial^{5}}{\partial x^{5}} - 8280 x \frac{\partial^{5}}{\partial x^{5}} + 15 x^{8} \frac{\partial^{4}}{\partial x^{4}} + 360 x^{6} \frac{\partial^{4}}{\partial x^{4}} + 450 x^{4} \frac{\partial^{4}}{\partial x^{4}} - 9000 x^{2} \frac{\partial^{4}}{\partial x^{4}} - 6525 \frac{\partial^{4}}{\partial x^{4}} + 240 x^{7} \frac{\partial^{3}}{\partial x^{3}} + 4320 x^{5} \frac{\partial^{3}}{\partial x^{3}} + 15600 x^{3} \frac{\partial^{3}}{\partial x^{3}} + 7200 x \frac{\partial^{3}}{\partial x^{3}} - 6 x^{10} \frac{\partial^{2}}{\partial x^{2}} - 180 x^{8} \frac{\partial^{2}}{\partial x^{2}} - 540 x^{6} \frac{\partial^{2}}{\partial x^{2}} + 9000 x^{4} \frac{\partial^{2}}{\partial x^{2}} + 31050 x^{2} \frac{\partial^{2}}{\partial x^{2}} + 9180 \frac{\partial^{2}}{\partial x^{2}} - 60 x^{9} \frac{\partial}{\partial x} - 1440 x^{7} \frac{\partial}{\partial x} - 8280 x^{5} \frac{\partial}{\partial x} - 7200 x^{3} \frac{\partial}{\partial x} + 8100 x \frac{\partial}{\partial x} + x^{12} + 36 x^{10} + 315 x^{8} - 120 x^{6} - 6525 x^{4} - 9180 x^{2} - 855$$. Sage code: A.<x> = DifferentialWeylAlgebra(QQ) x, dx = A.gens() def r(n): return (dx + x) ** n * (-dx + x) ** n for i in range(7): print "For $$n = " + str(i) + "$$, the result is $$" + latex(r(i)) + "$$.\r\n" Note that it is easily seen that $$\left[\dfrac{\partial}{\partial x} + x, - \dfrac{\partial}{\partial x} + x\right] = 2$$. Thus, the operators $$\dfrac{\partial}{\partial x} + x$$ and $$- \dfrac{\partial}{\partial x} + x$$ themselves generate an isomorphic copy of the Weyl algebra, except for a scalar factor of $$2$$. In view of the relation $$\left[\dfrac{\partial}{\partial x} + x, - \dfrac{\partial}{\partial x} + x\right] = 2$$, perhaps the following copypasta from some of my old homework will come useful. Let $$\mathbb{N} = \left\{0,1,2,\ldots\right\}$$. Now we need an easy fact from quantum algebra: Proposition 1. Let $$A$$ be a ring (not necessarily commutative). Let $$x\in A$$ and $$y\in A$$ be such that $$xy-yx=1$$. Then, \begin{align} \left( xy\right) ^{n}=\sum\limits_{k=0}^{n} \genfrac{\{}{\}}{0pt}{0}{n+1}{k+1} y^{k}x^{k}, \end{align} where the curly braces denote Stirling numbers of the second kind. Proof of Proposition 1. First of all, it is easy to see that $$$$x^{m}y=mx^{m-1}+yx^{m}\ \ \ \ \ \ \ \ \ \ \text{for every }m\in\mathbb{N} \label{darij1.pf.xy-yx.1} \tag{1}$$$$ (this allows $$m=0$$ if $$0x^{0-1}$$ is interpreted as $$0$$). Indeed, the proof of \eqref{darij1.pf.xy-yx.1} proceeds by induction over $$m$$ and is straightforward enough to be left to the reader. We will now prove Proposition 1 by induction over $$k$$. The induction base is obvious, so we step to the induction step: Let $$n>0$$. Assuming that $$\left( xy\right) ^{n-1}=\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n}{k+1} y^{k}x^{k}$$, we need to show that $$\left( xy\right) ^{n}=\sum\limits_{k=0} ^{n} \genfrac{\{}{\}}{0pt}{0}{n+1}{k+1} y^{k}x^{k}$$. We have \begin{align*} \left( xy\right) ^{n} & =\left( xy\right) ^{n-1}xy=\left( \sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n}{k+1} y^{k}x^{k}\right) xy\ \ \ \ \ \ \ \ \ \ \left( \text{since }\left( xy\right) ^{n-1}=\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n}{k+1} y^{k}x^{k}\right) \\ & =\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n}{k+1} y^{k}\underbrace{x^{k}x}_{=x^{k+1}}y=\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n}{k+1} y^{k}\underbrace{x^{k+1}y}_{\substack{=\left( k+1\right) x^{k} +yx^{k+1}\\\text{(by \eqref{darij1.pf.xy-yx.1}, applied to }m=k+1\text{)}}}\\ & =\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n}{k+1} y^{k}\left( \left( k+1\right) x^{k}+yx^{k+1}\right) \\ & =\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n}{k+1} \left( k+1\right) y^{k}x^{k}+\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n}{k+1} \underbrace{y^{k}y}_{=y^{k+1}}x^{k+1}\\ & =\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n}{k+1} \left( k+1\right) y^{k}x^{k}+\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n}{k+1} y^{k+1}x^{k+1}\\ & =\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n}{k+1} \left( k+1\right) y^{k}x^{k}+\sum\limits_{k=1}^{n} \genfrac{\{}{\}}{0pt}{0}{n}{k} y^{k}x^{k}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{here, we substituted }k-1\text{ for }k\text{ in the second sum}\right) \\ & =\sum\limits_{k=0}^{n} \genfrac{\{}{\}}{0pt}{0}{n}{k+1} \left( k+1\right) y^{k}x^{k}+\sum\limits_{k=0}^{n} \genfrac{\{}{\}}{0pt}{0}{n}{k} y^{k}x^{k}\\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array}[c]{c} \text{here, we extended both sums by zero terms, using the fact}\\ \text{that } \genfrac{\{}{\}}{0pt}{0}{n}{n+1} = \genfrac{\{}{\}}{0pt}{0}{n}{0} =0\text{ whenever }n>0 \end{array} \right) \\ & =\sum\limits_{k=0}^{n}\underbrace{\left( \genfrac{\{}{\}}{0pt}{0}{n}{k+1} \left( k+1\right) + \genfrac{\{}{\}}{0pt}{0}{n}{k} \right) }_{\substack{= \genfrac{\{}{\}}{0pt}{0}{n+1}{k+1} \\\text{(by the recursion formula for Stirling numbers}\\\text{of the second kind)}}}y^{k}x^{k}\\ & =\sum\limits_{k=0}^{n} \genfrac{\{}{\}}{0pt}{0}{n+1}{k+1} y^{k}x^{k}. \end{align*} This completes the induction step, and thus the inductive proof of Proposition 1. $$\blacksquare$$ Proposition 2. Let $$A$$ be a ring (not necessarily commutative). Let $$x\in A$$ and $$y\in A$$ be such that $$xy-yx=1$$. Then, \begin{align} \left( yx\right) ^{n}=\sum\limits_{k=0}^{n} \genfrac{\{}{\}}{0pt}{0}{n}{k} y^{k}x^{k}, \end{align} where the curly braces denote Stirling numbers of the second kind. Proof of Proposition 2. Just as in the proof of Proposition 1, we show that \eqref{darij1.pf.xy-yx.1} holds. We will now prove Proposition 2 by induction over $$k$$. The induction base is obvious, so we step to the induction step: Let $$n>0$$. Assuming that $$\left( yx\right) ^{n-1}=\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n-1}{k} y^{k}x^{k}$$, we need to show that $$\left( yx\right) ^{n}=\sum\limits_{k=0} ^{n} \genfrac{\{}{\}}{0pt}{0}{n}{k} y^{k}x^{k}$$. We have \begin{align*} \left( yx\right) ^{n} & =\left( yx\right) ^{n-1}yx=\left( \sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n-1}{k} y^{k}x^{k}\right) yx\ \ \ \ \ \ \ \ \ \ \left( \text{since }\left( yx\right) ^{n-1}=\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n-1}{k} y^{k}x^{k}\right) \\ & =\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n-1}{k} y^{k}\underbrace{x^{k}y}_{\substack{=kx^{k-1}+yx^{k}\\\text{(by \eqref{darij1.pf.xy-yx.1}, applied to }m=k\text{)}}}x\\ & =\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n-1}{k} y^{k}\left( kx^{k-1}+yx^{k}\right) x\\ & =\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n-1}{k} ky^{k}\underbrace{x^{k-1}x}_{=x^{k}}+\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n-1}{k} k\underbrace{y^{k}y}_{=y^{k+1}}\underbrace{x^{k}x}_{=x^{k+1}}\\ & =\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n-1}{k} ky^{k}x^{k}+\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n-1}{k} ky^{k+1}x^{k+1}\\ & =\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n-1}{k} ky^{k}x^{k}+\sum\limits_{k=1}^{n} \genfrac{\{}{\}}{0pt}{0}{n-1}{k-1} ky^{k}x^{k}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{here, we substituted }k-1\text{ for }k\text{ in the second sum}\right) \\ & =\sum\limits_{k=0}^{n} \genfrac{\{}{\}}{0pt}{0}{n-1}{k} ky^{k}x^{k}+\sum\limits_{k=0}^{n} \genfrac{\{}{\}}{0pt}{0}{n-1}{k-1} ky^{k}x^{k}\\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array}[c]{c} \text{here, we extended both sums by zero terms, using the fact}\\ \text{that } \genfrac{\{}{\}}{0pt}{0}{n-1}{n} = \genfrac{\{}{\}}{0pt}{0}{n-1}{-1} =0\text{ whenever }n>0 \end{array} \right) \\ & =\sum\limits_{k=0}^{n}\underbrace{\left( \genfrac{\{}{\}}{0pt}{0}{n-1}{k} k+ \genfrac{\{}{\}}{0pt}{0}{n-1}{k-1} \right) }_{\substack{= \genfrac{\{}{\}}{0pt}{0}{n}{k} \\\text{(by the recursion formula for Stirling numbers}\\\text{of the second kind)}}}y^{k}x^{k}\\ & =\sum\limits_{k=0}^{n} \genfrac{\{}{\}}{0pt}{0}{n}{k} y^{k}x^{k}. \end{align*} This completes the induction step, and thus the inductive proof of Proposition 2. $$\blacksquare$$ • Thank you very much for this code! I really needed something like this! The only point is that I am desperately looking for a closed formula for arbitrary $n$. Do you have any thoughts about that? – jobe Feb 15 at 20:36 • @jobe: My output doesn't give me much hope for a closed formula -- but of course, the Weyl algebra has several bases in which these elements could be expanded, and maybe one of them looks better. – darij grinberg Feb 15 at 20:39 • Thank you for the reference , but I could not open it. – jobe Feb 15 at 20:45 Let $$X, Y$$ be operators which are central, meaning that $$[X,Y] = XY-YX = c$$ for some scalar $$c$$. Then 1. As a form of binomial theorem, we have $$(X+Y)^n = \sum_{\substack{a,b,m \geq 0 \\ a+b+2m=n}} \frac{n!}{a!b!m!2^m} c^m Y^b X^a = \sum_{a=0}^{n} \binom{n}{a} P_{n-a}(c,Y)X^a,$$ where $$P_n$$ is defined by the following sum $$P_n(c, x) = \sum_{m=0}^{\lfloor n/2\rfloor} \frac{n!}{(n-2m)!m!2^m} c^m x^{n-2m} = \left( c\frac{d}{dx} + x \right)^n \mathbf{1}.$$ In particular, if $$c = -1$$ then $$P_n(-1, x) = \operatorname{He}_n(x)$$, where $$\operatorname{He}_n$$ is the probabilists' Hermite polynomial. Similarly, if $$c = 1$$, then $$P_n(1, x) = i^{-n} \operatorname{He}_n(ix)$$. 2. Under the same condition, for any polynomials $$f, g$$ we have $$f(X)g(Y) = \sum_{m \geq 0} \frac{c^m}{m!} g^{(m)}(Y)f^{(m)}(X).$$ In our case, $$[\frac{d}{dx}, x] = 1$$, and so, we can use both formulas to give a complicated, but still explicit expression for the product of $$(\frac{d}{dx}+x)^n$$ and $$(-\frac{d}{dx}+x)^n$$. Combining altogether, \begin{align*} &\left( \frac{d}{dx} + x \right)^n \left( -\frac{d}{dx} + x \right)^n \\ &= \sum_{p \geq 0} \frac{1}{p!} \Bigg( \sum_{\substack{a_i, b_i, m_i \geq 0 \\ a_i+b_i+2m_i = n-p}} \frac{(-1)^{a_2+m_2} (n!)^2}{a_1!a_2!b_1!b_2!m_1!m_2!2^{m_1+m_2}} x^{b_1+b_2} \left( \frac{d}{dx} \right)^{a_1+a_2} \Bigg). \end{align*} The following is a sample Mathematica code, comparing this formula with the actual answer for the case $$n = 3$$. Coef[n_, c_, l_] := n!/(l[[1]]! l[[2]]! l[[3]]! 2^l[[3]]) c^l[[3]]; T[n_] := FrobeniusSolve[{1, 1, 2}, n]; (* Compute (x+d/dx) (x-d/dx)^n f(x) *) n = 3; Nest[Expand[x # + D[#, x]] &, Nest[Expand[x # - D[#, x]] &, f[x], n], n] Sum[1/p! Sum[ Sum[ Coef[n, 1, l1] Coef[n, -1, l2] (-1)^l2[[2]] x^(l1[[1]] + l2[[1]]) D[f[x], {x, l1[[2]] + l2[[2]]}], {l1, T[n - p]}], {l2, T[n - p]}], {p, 0, n}] // Expand Clear[n]; • Impressive! Your answer gives an explicit expression for the product as I have asked for, but it is indeed complicated! I am still trying to figure out how using it effectively in the problems I have in mind, but that is my fault. Thank you! – jobe Feb 17 at 1:19 • @jobe, Glad it helped! This formula may be still far from being qualified as generally useful, but a form of combinatoric interpretation of the coefficients in the expansion of $(X+Y)^n$ is available, which may possible be useful for some applications: $$\frac{n!}{(n-2m)!m!2^m}=[\text{# of m-matchings in \{1,\cdots,n\}}].$$ So $P_n(c,x)$ computes the sum of weights over all possible matchings on $\{1,\cdots,n\}$, where each matching is weighted by the factor $c^m x^{n-2m}$, i.e. each matched pair receives the weight $c$ and each unmatched vertice receives the weight $x$. – Sangchul Lee Feb 17 at 1:32

https://www.physicsforums.com/threads/pendulum-dilemma.546677/

# Pendulum dilemma ## Homework Statement I am suppose to find a linear graph for the equation T=2pie(√L/√g) ## The Attempt at a Solution The best linear graph I could think of was L/T^2. Am I doing it right? Thanks. Yes. T$^{2}$ on y-axis and L on x-axis. Yes. T$^{2}$ on y-axis and L on x-axis. Thank you very much. This has been like a huge thorn in my finger all day! I thought the slope was always y/x, so why isn't the slope of the graph T^2/L? Slope is NOT y/x but slope = $\Delta$y/($\Delta$x) or more exactly slope = dy/(dx). T = 2∏√(L/g) therefore T$^{2}$ = 4π$^{2}$L/g so d(T$^{2}$)/dL = 4π$^{2}$/g = slope T = 2∏√(L/g) therefore T$^{2}$ = 4π$^{2}$L/g so d(T$^{2}$)/dL = 4π$^{2}$/g = slope oh that makes much more sense. thank you.

http://math.stackexchange.com/questions/270564/what-is-the-argument-for-not-using-the-average-of-an-average

what is the argument for not using the average of an average? I want to disprove someone's calculation of percentage of cash sales for a year by taking summing percentage of cash sales by month and dividing by 12. I sense the correct way is to take total cash sales for the year and divide by total sales for year to arrive at percentage of sales but need correct reasoning why it is wrong to use an average of an average. - What is 'percentage of cash sales'? How is it calculated? –  Michael Biro Jan 4 at 21:54 You need a counter example. E.g. Mon Sales Cash Percent Jan 100 70 70% Feb 10 1 10% Mar 10 1 10% Apr 10 1 10% May 10 1 10% Jun 10 1 10% Jul 10 1 10% Aug 10 1 10% Sep 10 1 10% Oct 10 1 10% Nov 10 1 10% Dec 100 70 70% Total sales are $300$, cash sales are $150$ so the overall percentage is $50\%$. But the average of the monthly percentages is $20\%$. - Thank you very much. I will be back to this site I'm sure. –  denise Jan 5 at 14:42 You are correct, here's a mathematical example: Let $s_1, s_2, s_3 \cdots s_{12}$ be the sales of the months. Let $c_1, c_2, c_3 \cdots c_{12}$ be the cash sales of the months. $$\frac{1}{12} \sum_{k=1}^{12} \frac{c_k}{s_k}$$ Is the same as: $$\frac{\sum_{k=1}^{12} c_k}{\sum_{k=1}^{12} s_k}$$ You can show it is false by plugging in some numbers. - Thanks for taking the time to respond. Very helpful. –  denise Jan 5 at 14:41 The average of the averages leads to the right average only when the samples have the same size. If each month has exactly the same number of sales, then taking the average of the averages would be right. But if the number of sales per month is different, then the average of averages leads in general to the wrong answer. Simple situation: First month 100 sales of 1 each. Average 1. Second month 1 sale of 100. Average 100 . The average of the averages is 50.50, but there were 101 sales and only 200 income.... - Right, otherwise you need a weighted average. A simple unweighted arithmetic mean would be wrong. –  Fixed Point Jan 5 at 1:27 Thank you very much. Language very helpful. –  denise Jan 5 at 14:40 Suppose in January I sell \$100 worth of product, and all of it is sold for cash. Every other month I sell \$1 worth of product, and none of it is sold for cash. I have then sold \$111 dollars worth of product, of which \$100 was sold for cash, so my percentage of cash sales is $\frac{100}{111}\approx90\%$. Calculated the other way, we would get that for $1$ month $100\%$ of my sales were cash, while for the other $11$, $0\%$ of my sales were cash, giving $\frac{100\%}{12}\approx 8.3\%$. This is clearly wrong. - Thanks so much for responding. Examples are so helpful! –  denise Jan 5 at 14:41 It is clear to you that taking cash sales for a year, and dividing by total sales for the year, will give you the correct proportion of cash sales for that year. The problem with taking monthly proportions and averaging them (average of averages) is that that procedure could produce a quite different number. Let's take a simple numerical example. January to October: total sales each month $\$1000$, cash sales each month$\$500$. November and December (Christmas season, people are buying a lot, mostly on credit): total sales each month $\$10000$, cash sales each month$\$500$. So total sales for the year, 30000, cash sales 6000. Thus the yearly proportion of cash sales is $\dfrac{6000}{30000}=20\%$. This is the correct percentage. Now let's compute the monthly averages. For January through October, they are $50\%$. For each of November and December, they are $5\%$. To find the average of the monthly proportions, as a percent, we take $\frac{1}{12}(50+50+50+50+50+50+50+50+50+50 +5+5)$. This is approximately $42.5\%$, which is wildly different from the true average of $20\%$. For many businesses, sales exhibit a strong seasonality. If the pattern of cash sales versus total sales also exhibits seasonality, averaging monthly averages may give answers that are quite far from the truth. - Exactly what I needed! Thanks very much. –  denise Jan 5 at 14:39

https://stacks.math.columbia.edu/tag/00D3

Definition 4.21.1. Let $I$ be a set and let $\leq$ be a binary relation on $I$. 1. We say $\leq$ is a preorder if it is transitive (if $i \leq j$ and $j \leq k$ then $i \leq k$) and reflexive ($i \leq i$ for all $i \in I$). 2. A preordered set is a set endowed with a preorder. 3. A directed set is a preordered set $(I, \leq )$ such that $I$ is not empty and such that $\forall i, j \in I$, there exists $k \in I$ with $i \leq k, j \leq k$. 4. We say $\leq$ is a partial order if it is a preorder which is antisymmetric (if $i \leq j$ and $j \leq i$, then $i = j$). 5. A partially ordered set is a set endowed with a partial order. 6. A directed partially ordered set is a directed set whose ordering is a partial order. ## Comments (4) Comment #2102 by Keenan Kidwell on I just noticed that in the definition of a partially ordered set, the condition of antisymmetry is omitted. Isn't it more standard to call a transitive, reflexive relation a preorder? I'm guessing the antisymmetry condition doesn't play a role in considerations involving e.g. colimits over a directed set, so this is technically moot, but I'm curious if the decision to use a (as far as I can tell) not-quite-standard definition was intentional. Comment #2129 by on OK, somebody else has mentioned this previously... and at the time there was a proof somewhere using the weaker notion... but I cannot find it now. The motivation for the weaker notion is that it is exactly enough conditions to turn $I$ into a category, so you can define (co)limits. I think that for almost all statements in the Stacks project it does not make a difference which definition you use. For example the proof of Lemma 4.21.5 seems to produce a partially ordered set in the stronger sense. OK, so maybe we should change this (if you are reading this and agree please leave a comment). Then a directed set will not be a partially ordered set in general. So lot's of lemmas should get multiple statements some with directed partially ordered sets and some with just directed sets... This is lot's of work, so I will only do this if more people chime in. Comment #2138 by Keenan Kidwell on I was actually very happy when I saw this definition as it made me realize that the formalism of colimits and limits over directed sets worked just fine without the antisymmetry condition. I guess what it really showed me (which is something actually more standard) is that "directed sets" need not be "partially ordered" in the stronger sense. Since you don't actually seem to need the stronger sense, and it would require a non-trivial amount of work if you were to switch, I think you should leave it as it is. Comment #2146 by on OK, I decided to revert to the standard definition of partially ordered sets and now a directed set is a preordered set with upper bounds for finite subsets. Going through all the corrections I found that it absolutely does not matter at all! It very slightly shortens the Stacks project text because it allows us to say "directed set" instead of "directed partially ordered set" in many lemmas, propositions, etc. But I am not sure if the use of the terminology "directed set" is completely standard. I found places where people suggest one should say "directed proset" where "proset" is an abbreviation for "preordered set" but that is just plain ugly. Oh well. Big set of changes can be found here. There are also: • 2 comment(s) on Section 4.21: Limits and colimits over preordered sets ## Post a comment Your email address will not be published. Required fields are marked. In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work. All contributions are licensed under the GNU Free Documentation License. In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00D3. Beware of the difference between the letter 'O' and the digit '0'.

https://gmatclub.com/forum/what-is-x-more-than-y-1-x-of-y-is-9-2-y-more-than-x-is-260828.html

GMAT Question of the Day - Daily to your Mailbox; hard ones only It is currently 20 Nov 2018, 07:47 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History ## Events & Promotions ###### Events & Promotions in November PrevNext SuMoTuWeThFrSa 28293031123 45678910 11121314151617 18192021222324 2526272829301 Open Detailed Calendar • ### All GMAT Club Tests are Free and open on November 22nd in celebration of Thanksgiving Day! November 22, 2018 November 22, 2018 10:00 PM PST 11:00 PM PST Mark your calendars - All GMAT Club Tests are free and open November 22nd to celebrate Thanksgiving Day! Access will be available from 0:01 AM to 11:59 PM, Pacific Time (USA) • ### Free lesson on number properties November 23, 2018 November 23, 2018 10:00 PM PST 11:00 PM PST Practice the one most important Quant section - Integer properties, and rapidly improve your skills. # What is x% more than y? (1) x% of y is 9 (2) y% more than x is 24 Author Message TAGS: ### Hide Tags DS Forum Moderator Joined: 21 Aug 2013 Posts: 1371 Location: India What is x% more than y? (1) x% of y is 9 (2) y% more than x is 24  [#permalink] ### Show Tags 05 Mar 2018, 04:04 2 00:00 Difficulty: 25% (medium) Question Stats: 75% (01:37) correct 25% (01:45) wrong based on 89 sessions ### HideShow timer Statistics What is x% more than y? (1) x% of y is 9 (2) y% more than x is 24 Intern Joined: 04 Jan 2018 Posts: 39 Re: What is x% more than y? (1) x% of y is 9 (2) y% more than x is 24  [#permalink] ### Show Tags 05 Mar 2018, 09:32 2 ans is C 1) xy/100=9..........................(1) 2) x+xy/100=24....................(2) subst 1 in 2 we get 'x' subst x in 1 or 2. we get y _________________ Don't stop till you get enough Hit kudos if it helped you. e-GMAT Representative Joined: 04 Jan 2015 Posts: 2203 What is x% more than y? (1) x% of y is 9 (2) y% more than x is 24  [#permalink] ### Show Tags 19 Mar 2018, 12:35 SOLUTION We need to find the value which is $$x$$% more than $$y$$. Thus, we need to find the value of $$y$$+$$\frac{x}{100}$$ * $$y$$. Statement-1: “$$x$$% of $$y$$ is $$9$$”. • $$\frac{x}{100}$$ * $$y$$=$$9$$……………………(1) Since we don’t know the value of $$x$$ and $$y$$, statement 1 alone is not sufficient to answer the question. Statement-2: “$$y$$% more than $$x$$ is $$24$$”. • $$x$$+ $$\frac{y}{100}$$*$$x$$=$$24$$…………………….(2) Since we don’t know the value of $$x$$ and $$y$$, statement 2 alone is not sufficient to answer the question. Combining both the statements: • $$x$$+ $$\frac{x*y}{100}$$ = $$24$$ • $$x+9=24$$ • $$x=15$$ Putting the value of $$x$$ in equation $$1$$ • $$\frac{x*y}{100}$$=$$9$$ • $$15$$*$$\frac{y}{100}$$=$$9$$ • $$y=60$$ We now have the value of $$x$$ and $$y$$ both. Hence, we can find the asked value. Statement (1) and (2) together are sufficient. _________________ Number Properties | Algebra |Quant Workshop Success Stories Guillermo's Success Story | Carrie's Success Story Ace GMAT quant Articles and Question to reach Q51 | Question of the week Number Properties – Even Odd | LCM GCD | Statistics-1 | Statistics-2 Word Problems – Percentage 1 | Percentage 2 | Time and Work 1 | Time and Work 2 | Time, Speed and Distance 1 | Time, Speed and Distance 2 Advanced Topics- Permutation and Combination 1 | Permutation and Combination 2 | Permutation and Combination 3 | Probability Geometry- Triangles 1 | Triangles 2 | Triangles 3 | Common Mistakes in Geometry Algebra- Wavy line | Inequalities Practice Questions Number Properties 1 | Number Properties 2 | Algebra 1 | Geometry | Prime Numbers | Absolute value equations | Sets | '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com What is x% more than y? (1) x% of y is 9 (2) y% more than x is 24 &nbs [#permalink] 19 Mar 2018, 12:35 Display posts from previous: Sort by

https://brilliant.org/discussions/thread/euclidean-algorhytm/

# Euclidean algorithm I found something interesting about solving this question: If p is the number of pairs of values (x, y) that can make the equation 84x + 140y = 28 true, what is true about p? I noticed while solving that I could simplify the equation: 84x - 140y = 28 42x – 70y = 14 21x – 35y = 7 3x – 5y = 1 Noting that 3x = 5y + 1 means that 3x must end in 5 + 1 = 6 or 0 + 1 = 1 to allow y to be divisible by 5. This means 3x should end in either 1 or 6 to be equal to 5y + 1 Finding the following values for x and matching values for y: 3x = 6, 21, 36, 51, 66, 81, 96, 111, 126, 141, 156 5y = 5, 20, 35, 50, 65, 80, 95, 110, 125, 140, 155 Then finding the series of x and y: X = 2, 7, 12, 17, 22, 27, 32, 37, 42, 47, 52 Y = 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31 X seems to be increasing by 5, and Y by 3. Which means that you can find pairs of x and y by the formulas: x = 2+5a and y = 1+3a Leading to a sequence that will satisfy 84x + 140y = 28. Would this also be more generally true? I also thought it might be an interesting additional question to the quiz. On another note, this text editor does not seem to like enters in paragraphs. Any way I can fix this? Note by Sytse Durkstra 1 year ago MarkdownAppears as *italics* or _italics_ italics **bold** or __bold__ bold - bulleted - list • bulleted • list 1. numbered 2. list 1. numbered 2. list Note: you must add a full line of space before and after lists for them to show up correctly paragraph 1 paragraph 2 paragraph 1 paragraph 2 [example link](https://brilliant.org)example link > This is a quote This is a quote # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" MathAppears as Remember to wrap math in $$...$$ or $...$ to ensure proper formatting. 2 \times 3 $$2 \times 3$$ 2^{34} $$2^{34}$$ a_{i-1} $$a_{i-1}$$ \frac{2}{3} $$\frac{2}{3}$$ \sqrt{2} $$\sqrt{2}$$ \sum_{i=1}^3 $$\sum_{i=1}^3$$ \sin \theta $$\sin \theta$$ \boxed{123} $$\boxed{123}$$ ## Comments Sort by: Top Newest Yes, you just applied Euclidean algorithm. To enter $$\LaTeX$$, type like \ ( ... \ ) but remove the spaces. - 1 year ago Log in to reply × Problem Loading... Note Loading... Set Loading...

https://mathhelpboards.com/threads/arc-length-question.13986/

# Arc length question ##### Member The length of the minor arc of a circle is 10cm, while the area of the sector AOB is 150cm2. a) Form two equations involving r and θ, where θ is measured in radians. b) Solve these equations simultaneously to find r and θ. Help to solve? Cant understand the question very well. I think the arc length formula was length=$$\displaystyle \frac{n}{360}\cdot2\pi(r)$$ $$\displaystyle \therefore10=\frac{n}{360}\cdot2\pi(r)$$ The question states, Form two equations involving r and θ, where θ is measured in radians. So would we have to arrange the formula to find the r and $$\displaystyle \theta$$ #### MarkFL Staff member The formulas you want are: Circular arc-length: $$\displaystyle s=r\theta\tag{1}$$ Area of circular sector $$\displaystyle A=\frac{1}{2}r^2\theta\tag{2}$$ Now, if we solve (1) for $\theta$, we find: $$\displaystyle \theta=\frac{s}{r}$$ Now, substituting this into (2), we obtain: $$\displaystyle A=\frac{1}{2}r^2\left(\frac{s}{r}\right)=\frac{rs}{2}\implies r=\frac{2A}{s}\implies\theta=\frac{s^2}{2A}$$ Now, just plug in the given values for $A$ and $s$. ##### Member The formulas you want are: Circular arc-length: $$\displaystyle s=r\theta\tag{1}$$ Area of circular sector $$\displaystyle A=\frac{1}{2}r^2\theta\tag{2}$$ Now, if we solve (1) for $\theta$, we find: $$\displaystyle \theta=\frac{s}{r}$$ Now, substituting this into (2), we obtain: $$\displaystyle A=\frac{1}{2}r^2\left(\frac{s}{r}\right)=\frac{rs}{2}\implies r=\frac{2A}{s}\implies\theta=\frac{s^2}{2A}$$ Now, just plug in the given values for $A$ and $s$. Im finding trouble understanding the question. "Form two equations involving r and $$\displaystyle \theta$$" Basically means transpose the formula or make the equation equal to r and $$\displaystyle \theta$$? Since the arc length = $$\displaystyle s=r\theta$$ (I thought arc length was length=$$\displaystyle \frac{n}{360}\cdot2\pi(r)$$? r=$$\displaystyle s\theta$$? For b) we just sub in the values? Where s is the length? #### MarkFL Staff member You are given values for the arc-length $s$ and the area $A$, and so you want to use the formulas relating $r$ and $\theta$ to these given values (which I gave as (1) and (2)) to be able to express both $r$ and $\theta$ as functions of $s$ and $A$ (which I did using some algebra), so that you can use these given values to determine $r$ and $\theta$. Once you have $r$ and $\theta$ as functions of $s$ and $A$, it is simply a matter of using the given values to evaluate $r$ and $\theta$. #### SuperSonic4 ##### Well-known member MHB Math Helper Im finding trouble understanding the question. "Form two equations involving r and $$\displaystyle \theta$$" Basically means transpose the formula or make the equation equal to r and $$\displaystyle \theta$$? It's asking to make two equations for the length of an arc and the area of a sector. As long as your equations have $$\displaystyle r$$ and $$\displaystyle /theta$$ in them then you're fine for this step. MarkFL has given you the equations you need (his TeX is also better than mine! ) $$\displaystyle \displaystyle s=r\theta\tag{1}$$ $$\displaystyle \displaystyle A=\frac{1}{2}r^2\theta\tag{2}$$ As you see both of these equations have $$\displaystyle r$$ and $$\displaystyle \theta$$ in them so Part A of the question has been solved already Since the arc length = $$\displaystyle s=r\theta$$ (I thought arc length was length=$$\displaystyle \frac{n}{360}\cdot2\pi(r)$$? r=$$\displaystyle s\theta$$? $$\displaystyle \frac{n}{360}\cdot 2\pi r$$ is the equation for arc length where $$\displaystyle n$$ is in degrees. Since the question asks for radians you want $$\displaystyle s = r\theta$$. See the spoiler below for why they are equivalent By definition a full circle has 360 degrees and $$2\pi$$ radians so you can say $$360^{\circ} = 2\pi$$ (we omit the c symbol for radians) Your formula for degrees is $$s = \frac{\theta}{360} \cdot 2\pi r$$ but since we know that $$360^{\circ} = 2\pi$$ it can be subbed in $$s = \frac{\theta}{2\pi} \cdot 2\pi r = r \theta$$ - the radian expression For b) we just sub in the values? Where s is the length? Sub in the values given (10cm and 150cm²) for s and A respectively so you have two equations and two unknowns ($$\displaystyle r$$ and $$\displaystyle \theta$$). You can now solve simultaneously using either the substitution method MarkFL showed in post 2. He found $$\displaystyle \theta$$ in terms of $$\displaystyle r$$ which means he was able to substitute $$\displaystyle \frac{s}{r}$$ wherever $$\displaystyle \theta$$ appeared. He did in terms of $$\displaystyle s$$ in post 2 but if you find it easier you may use $$\displaystyle s=10$$ for your example ##### Member It's asking to make two equations for the length of an arc and the area of a sector. As long as your equations have $$\displaystyle r$$ and $$\displaystyle /theta$$ in them then you're fine for this step. MarkFL has given you the equations you need (his TeX is also better than mine! ) $$\displaystyle \displaystyle s=r\theta\tag{1}$$ $$\displaystyle \displaystyle A=\frac{1}{2}r^2\theta\tag{2}$$ As you see both of these equations have $$\displaystyle r$$ and $$\displaystyle \theta$$ in them so Part A of the question has been solved already $$\displaystyle \frac{n}{360}\cdot 2\pi r$$ is the equation for arc length where $$\displaystyle n$$ is in degrees. Since the question asks for radians you want $$\displaystyle s = r\theta$$. See the spoiler below for why they are equivalent By definition a full circle has 360 degrees and $$2\pi$$ radians so you can say $$360^{\circ} = 2\pi$$ (we omit the c symbol for radians) Your formula for degrees is $$s = \frac{\theta}{360} \cdot 2\pi r$$ but since we know that $$360^{\circ} = 2\pi$$ it can be subbed in $$s = \frac{\theta}{2\pi} \cdot 2\pi r = r \theta$$ - the radian expression Sub in the values given (10cm and 150cm²) for s and A respectively so you have two equations and two unknowns ($$\displaystyle r$$ and $$\displaystyle \theta$$). You can now solve simultaneously using either the substitution method MarkFL showed in post 2. He found $$\displaystyle \theta$$ in terms of $$\displaystyle r$$ which means he was able to substitute $$\displaystyle \frac{s}{r}$$ wherever $$\displaystyle \theta$$ appeared. He did in terms of $$\displaystyle s$$ in post 2 but if you find it easier you may use $$\displaystyle s=10$$ for your example Thanks this really did help ##### Member The formulas you want are: Circular arc-length: $$\displaystyle s=r\theta\tag{1}$$ Area of circular sector $$\displaystyle A=\frac{1}{2}r^2\theta\tag{2}$$ Now, if we solve (1) for $\theta$, we find: $$\displaystyle \theta=\frac{s}{r}$$ Now, substituting this into (2), we obtain: $$\displaystyle A=\frac{1}{2}r^2\left(\frac{s}{r}\right)=\frac{rs}{2}\implies r=\frac{2A}{s}\implies\theta=\frac{s^2}{2A}$$ Now, just plug in the given values for $A$ and $s$. Beginning to understand this a lot easier. Just a question on the last part. How did you rearrange to get $$\displaystyle \theta$$? It went from, $$\displaystyle A=\frac{1}{2}r^2\left(\frac{s}{r}\right)$$ which I understand to $$\displaystyle =\frac{rs}{2}$$? #### MarkFL Staff member Beginning to understand this a lot easier. Just a question on the last part. How did you rearrange to get $$\displaystyle \theta$$? $$\displaystyle s=r\theta$$ Divide through by $r$ $$\displaystyle \frac{s}{r}=\frac{\cancel{r}\theta}{\cancel{r}}$$ $$\displaystyle \theta=\frac{s}{r}$$ It went from, $$\displaystyle A=\frac{1}{2}r^2\left(\frac{s}{r}\right)$$ which I understand to $$\displaystyle =\frac{rs}{2}$$? $$\displaystyle A=\frac{1}{2}r^2\left(\frac{s}{r}\right)=\frac{\cancel{r}\cdot r\cdot s}{2\cancel{r}}=\frac{rs}{2}$$

https://math.stackexchange.com/questions/2456373/optimal-route-consisting-of-rowing-then-walking

# Optimal route consisting of rowing then walking Problem You're in a boat on point A in the water, and you need to get to point B on land. Your rowing speed is 3km/h, and your walking speed 5km/h. See figure: Find the route that takes the least amount of time. My idea I started by marking an arbitrary route: From here, I figure the total time is going to be $$T = \frac{R}{3\mathrm{km/h}} + \frac{W}{5\mathrm{km/h}}$$ Since this is a function of two variables, I'm stuck. A general idea is to express $W$ in terms of $R$ to make it single-variable, and then apply the usual optimization tactics (with derivatives), but I'm having a hard time finding such an expression. Any help appreciated! EDIT - Alternative solution? Since the distance from A to the right angle (RA) is traveled 3/5 times as fast as the distance between RA and B, could I just scale the former up? That way, I get A-RA being a distance of $6\cdot\frac53 = 10\mathrm{km}$, which makes the hypotenuse $\sqrt{181}$ the shortest distance between A and B. And since we scaled it up, we can consider it traversable with walking speed rather than rowing speed! Thoughts? • but .. where is the river / land border ? – G Cab Oct 5 '17 at 21:56 • The horizontal leg of the triangle would simultaneously be the border and the walking route. It sounds strange, but the idea is that the distance between the shoreline and the walking path is "negligible", thus they can be seen as one-and-the-same. – Alec Oct 6 '17 at 8:27 • Thanks, now it is clear. Another question, do you know (accept) trigonometric approach? – G Cab Oct 6 '17 at 21:12 • @GCab - Yes. I've reached an answer of 3.4hours with that approach. However, $\frac{\sqrt{181}}{5}$ doesn't give the same answer, and I can't figure out why that wouldn't work, and if there's some adjustment that needs to be made. Or if it needs to be discarded completely. – Alec Oct 7 '17 at 11:01 • I think the answer is here, so wikipedia gets the bonus: en.wikipedia.org/wiki/Fermat%27s_principle – Ethan Bolker Oct 8 '17 at 14:23 • a) the solution The formula has already been indicated by wgrenard and AdamBL $$T = {1 \over 3}\sqrt {36 + \left( {9 - W} \right)^{\,2} } + {1 \over 5}W$$ differentiating that $${{dT} \over {dW}} = {{5W + 3\sqrt {36 + \left( {9 - W} \right)^{\,2} } - 45} \over {15\,\sqrt {36 + \left( {9 - W} \right)^{\,2} } }}$$ and equating to $0$ gives \eqalign{ & {{dT} \over {dW}} = 0\quad \Rightarrow \quad 3\sqrt {36 + \left( {9 - W} \right)^{\,2} } = 45 - 5W\quad \Rightarrow \cr & \Rightarrow \quad 9\left( {36 + \left( {9 - W} \right)^{\,2} } \right) = 25\left( {9 - W} \right)^{\,2} \quad \Rightarrow \cr & \Rightarrow \quad 9 \cdot 36 = 16\left( {9 - W} \right)^{\,2} \quad \Rightarrow \quad W = 9 - \sqrt {{{9 \cdot 36} \over {16}}} = 9 - {9 \over 2} = {9 \over 2} \cr} which is a minimum, because the function is convex as already indicated. Thus $$\left\{ \matrix{ W_m = 9/2 \hfill \cr T_m = 17/5 \hfill \cr R_m = 15/2 \hfill \cr} \right.$$ • b) Scaling Your idea of scaling according to speed is quite entangling. That means (if I understood properly) that you are transforming the triangle from space to time units. But, by introducing different scaling factors for the two coordinates, you undermine the Euclidean norm, which does not " transfer" between the two systems (if assumed valid in one, shall be modified in the other). Consider for example the transformation sketched below. From the mathematical point of view it is a linear scale transformation $$\left( {\matrix{ {y_1 } \cr {y_2 } \cr } } \right) = \left( {\matrix{ {1/v_1 } & 0 \cr 0 & {1/v_2 } \cr } } \right) \left( {\matrix{ {x_1 } \cr {x_2 } \cr } } \right)$$ Now, with constant $v_1, \,v_2$, any path in $x$ will transform in the corresponding path in $y$ (going through corresponding points). If the path is a curve parametrized through a common parameter $\lambda$, not influencing the $v_k$'s, then, at any given value of $\lambda$ the point on the $x$ plane will transform into the corresponding point in $y$ plane $$\left( {\matrix{ {y_{1}(\lambda) } \cr {y_{2}(\lambda) } \cr } } \right) = \left( {\matrix{ {1/v_1 } & 0 \cr 0 & {1/v_2 } \cr } } \right) \left( {\matrix{ {x{_1}(\lambda) } \cr {x_{2}(\lambda) } \cr } } \right)$$ and the minimal path in one plane will be the corresponding minimal path in the other. But we shall also have that $$\frac{d}{{d\lambda }}\left( {\matrix{ {y_{1}(\lambda) } \cr {y_{2}(\lambda) } \cr } } \right) = \left( {\matrix{ {1/v_1 } & 0 \cr 0 & {1/v_2 } \cr } } \right) \frac{d}{{d\lambda }}\left( {\matrix{ {x{_1}(\lambda) } \cr {x_{2}(\lambda) } \cr } } \right)$$ that is that the "velocities" compose vectorially. Therefore if $\lambda$ is the time, you shall go from $A$ to $C$ with a composition of a vertical rowing speed and a horizontal walking speed (a "$\infty$-thlon"), which takes the same time as rowing $AH$ and walking $HC$. When, instead, you just row on $AC$, then you shall change the above matrix - for that segment only - according to the $\angle AC$, and of course you loose the correspondence minimal to minimal as based on the Euclidean norm (straight line $A'B'$). • The question has recieved a nice handful of good answers. I'm still somewhat perplexed as to the alternative solution, which I was hoping to focus on with the bounty, but since you're the only one who even mentioned it, the bounty goes here. I would be interested to learn more about the entangling nature of that strategy though. Do you have any links, or simply a concept I should look into? – Alec Oct 12 '17 at 7:23 • Thanks for appreciation. I will try and add more elaboration to clear the "scaling" matter. – G Cab Oct 12 '17 at 8:48 • @Alec: I expanded on the 2nd part: wish I succeeded and make it more clear and .. convincing. – G Cab Oct 12 '17 at 13:31 The portion of the line on the top left is $9-w$. So by the Pythagorean theorem $R^2 = 36 + (9-w)^2$. I think this is the relationship you are looking for. You do not need to worry about the hypotenuse in the larger right triangle. Consider the right triangle that has R as the hypotenuse. Left side of this triangle will have length $6$ and the top side will have length $(9-W)$. By the Pythagorean theorem, $R^{2} = 36+(9-W)^2$. Solving for R and inserting in your original equation yields $T= \frac{1}{3} \sqrt{36+(9-W)^{2}} + \frac{1}{5} W$. This function is strictly convex, so you can simply minimize it by solving $\frac{d\ T}{d\ W}=0$ (the 'usual optimization tactics'). Regarding your alternative solution, I am unsure what you are arguing. The shortest distance between the points A and B will always be the line connecting them. As you have shown in your drawing, $|AB|=3\sqrt{13} \ne\sqrt{181}$. • No, I mean if we've scaled the left-most side to 10km (by multiplying 5/3), the hypotenuse will be $\sqrt{10^2 + 9^2} = \sqrt{181}$. – Alec Oct 4 '17 at 7:39 • Yes, that is correct. If I understand you correctly, though, you are arguing that it will always be optimal to row the entire way, since the hypotenuse will always be the shortest distance from A to B. That's obviously not the case. – AdB Oct 5 '17 at 15:37 • Nono, I'm not arguing that it's faster to row always. I'm saying that if we scale the rowing distance up by 5/3, then we can consider that distance to be traversable by walking speed instead of rowing speed. – Alec Oct 5 '17 at 17:09 Snell's Law is usually applied to optics, but it is based on the quickest path through two media in which the speed of light differs. Snell's Law says that $$n_1\sin(i_1)=n_2\sin(i_2)\tag1$$ where $i_k$ is the angle of incidence to the boundary of the path and $n_k$ is inversely proportional to the speed in the particular medium ($n_kv_k=c$). We can adapt this to the current situation by noting that $(1)$ is equivalent to $$\frac{\sin(i_1)}{v_1}=\frac{\sin(i_2)}{v_2}\tag2$$ If we travel at all along the shore, $\sin(i_2)=1$ (the angle of incidence is $90^\circ$). Since $v_1=3$ and $v_2=5$, we must have $\sin(i_1)=\frac35$, which implies that $\tan(i_1)=\frac34$. If $\tan(i_1)=\frac34$, and the width of the river is $6$ km, then the downriver distance must be $\frac34\cdot6=4.5$ km. This path takes $\frac{7.5}3+\frac{4.5}5=3.4$ hours. • Wait, so we must find an angle such that we only travel 4.5km by water? I must be interpreting that wrong, because the shortest possible water-distance must be 6km, right? – Alec Oct 8 '17 at 20:50 • No, the downriver distance is $4.5$ km, the river is still $6$ km wide, so that makes that water distance $\sqrt{4.5^2+6^2}=7.5$ km. – robjohn Oct 8 '17 at 22:55 • Ah, gotcha! Yeah, that was a nice take on this problem. I never would have connected this to Snell's Law on my own. – Alec Oct 9 '17 at 7:47

https://brilliant.org/discussions/thread/magnetic-dollars/

# Magnetic Dollars Imgur Suppose that I have two urns, with one magnetic dollar in each urn. I begin randomly throwing more magnetic dollars in the general vicinity of the urns, and the coins fall in the urns by a simple rule: Let's say that Urn A has $$x$$ coins, and Urn B has $$y$$ coins. The probability that the next coin falls into Urn A is $$\dfrac{x}{x+y}$$, and the probability that the next coin falls into Urn B is $$\dfrac{y}{x+y}$$. You keep throwing magnetic dollars until there is a total of $$1,000,000$$ magnetic dollars in total. What would you bet would be the price on average, of the urn with the smaller amount of coins? $$\ 10$$, perhaps? Maybe $$\ 100$$ or $$\ 500$$? Post your bet or write it down on a piece of paper before looking at the next section. The less-priced urn, on average, is actually worth a grand total of a quarter of a million dollars! Don't worry if you guessed wrong; many professional mathematicians also guessed much lower than this. In fact, when a group of mathematicians were asked this question and were asked to bet, most people only bet $$\ 10$$ and only one person bet over $$\ 100$$. But why does the lower-priced urn price so high? You may want to try the problem out yourself before I go over a very nice and elegant solution. See if you can find it! Tried it out yet? In the case that you have, let's see how this problem can be so elegantly solved, as I claimed. Suppose that you have a deck of cards; one red, and $$999,998$$ white. Currently, you just have a red card. Now every turn, you place a white card in any available slot. For example, in the first move, you have $$2$$ available slots: one above the red card, and one below. In the second move, you have $$3$$ available slots, and so on. But wait! Let's say that the empty slots above the red card are the magnetic dollars in Urn A, and the empty slots below the red card are the magnetic dollars in Urn B. Notice that if you had $$x$$ empty slots above the red card and $$y$$ empty slots below the red card, then the probability that the next card will be above the red card is $$\dfrac{x}{x+y}$$, and the probability that the next card will be below the red card is $$\dfrac{y}{x+y}$$! We've found a one-to-one correspondence between the original magnetic dollar problem and this new card problem! Finally, we know that in a random placements of white cards in this fashion will result in a uniform distribution of where the red card is, every single final position is of equal probability. This means that in the original problem, the probability of $$42$$ magnetic dollars being in Urn A is the same as the probability of $$314,159$$ magnetic dollars in Urn A. Therefore, the average price of the lower priced urn is clearly $$250,000$$. $$\Box$$ Note by Daniel Liu 4 years, 8 months ago MarkdownAppears as *italics* or _italics_ italics **bold** or __bold__ bold - bulleted- list • bulleted • list 1. numbered2. list 1. numbered 2. list Note: you must add a full line of space before and after lists for them to show up correctly paragraph 1paragraph 2 paragraph 1 paragraph 2 [example link](https://brilliant.org)example link > This is a quote This is a quote # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" MathAppears as Remember to wrap math in $$...$$ or $...$ to ensure proper formatting. 2 \times 3 $$2 \times 3$$ 2^{34} $$2^{34}$$ a_{i-1} $$a_{i-1}$$ \frac{2}{3} $$\frac{2}{3}$$ \sqrt{2} $$\sqrt{2}$$ \sum_{i=1}^3 $$\sum_{i=1}^3$$ \sin \theta $$\sin \theta$$ \boxed{123} $$\boxed{123}$$ Sort by: The result is certain not intuitive, nor obvious. You can look at this problem John's Red And Blue Balls, which is based off the same idea. Staff - 4 years, 8 months ago I know some of you have been waiting for me to post for a long time. Sorry for the inactivity, I had a bunch of homework and stuff. Well, here is my next #CosinesGroup posts. The problem in my opinion is a very cool problem because of the awesome elegant solution. This problem has also been called "Polya's Urns", but I looked that up and "Polya's Urns" actually covers a lot of random complicated stuff, so I kept it at magnetic dollars. Hope you enjoy! Feedback is appreciated. - 4 years, 8 months ago This is the solution in Peter Winkler ' Mathematical Puzzles. An excellent book. - 2 years, 2 months ago How do you do this? - 4 years, 7 months ago I'm curious to know why only one mathematician guessed over $100. I personally guessed around$150,000 becuase while the probability might be pretty low at some point, in A MILLION tries, it would be bound to happen quite a bit. - 4 years, 8 months ago Well, their reasoning would probably be then assuming a sort of "reverse" normal distribution curve. The average of all the games would probably be pretty low. They were asked to bet immediately, depriving them of the chance to think it over. Their first instinct told them that the more coins that get in a urn, the higher probability of getting more coins in, giving a feedback loop. This means that the smaller urn probably won't really have any coins at all. - 4 years, 8 months ago

https://math.stackexchange.com/questions/2542068/asymptotic-behavior-of-x-n1-x-n-frac1x-n-space-space-space-x-0-1

# Asymptotic behavior of $x_{n+1}=x_n+\frac{1}{x_n}, \space\space\space x_0=1$ I have defined a sequence $x_n$ as follows: $$x_{n+1}=x_n+\frac{1}{x_n}, \space\space\space x_0=1$$ After convincing myself that there is no nice closed-form formula for $x_n$, I decided to try and find an asymptotic formula for $x_n$ (unfortunately, I have very little experience with asymptotic formulae). I noticed that the recurrence is equivalent to $$\Delta x_{n}=\frac{1}{x_n}$$ and so a solution can be approximated by solving the corresponding differential equation for $y(t)$: $$y'=\frac{1}{y}$$ This differential equation (with initial value $y(0)=1$) yielded $$y=\sqrt{2t+1}$$ which led me to believe that $$x_n\approx \sqrt{2t+1}$$ When I plotted $x_n$ and $y(n)$ side by side, it did indeed appear that they were very close to each other. However, this isn't enough for me... I would like to prove that $y(n)$ is a good approximation for $x_n$, either by proving that $$\lim_{n\to\infty}(x_n-\sqrt{2n+1})=0$$ ...or, even better, by finding a zero-approaching function $f$ satisfying $$x_n=\sqrt{2n+1}+O(f(x))$$ This is where I got stuck. How do can I prove (or disprove?) the statement in the first equality, and find $f$ satisfying the second? NOTE: There may be a closed-form that I wasn't able to find. In fact, the similar sequence $$y_{n+1}=\frac{y_n}{2}-\frac{1}{2y_n}$$ has closed form formula $$y_n=\frac{1}{\tan(2^n\arctan(y_0^{-1}))}$$ ...but even if you do find a closed form, I would still like to know how to prove the above statements, since the techniques would be useful to know for future problems. • I think $\int_n^{n+1} f(x) dx = \frac 1 {\int_0^n f(x)}$ is more accurate. But its probably nastier to solve. – mathreadler Nov 29 '17 at 0:15 • See OEIS sequence A073833 and related sequences. – Robert Israel Nov 29 '17 at 0:21 • Probably the first step, based on this guess, would be to write a recurrence for $a_n := x_n^2$: $a_{n+1} = a_n + 2 + \frac{1}{a_n}$. So, to start a bootstrapping argument, it would be easy to prove $a_n \ge 2n - 1$ for each $n$... And then $a_{n+1} \le a_n + 2 + \frac{1}{2n-1}$ so also $a_n \le 2n + \frac{1}{2} \log(n) - C$, etc. ... – Daniel Schepler Nov 29 '17 at 0:28 • Here is a class of arguments that could work. You can try to prove inductively both an upper and a lower bound simultaneously, say $\sqrt{2n+a} \le x_n \le \sqrt{2n+b}$ for some $a, b$, because when bounding $x_{n+1}$ in terms of $x_n$ you need both a lower and an upper bound on $x_n$ to establish either an upper bound or a lower bound on $x_n + \frac{1}{x_n}$. – Qiaochu Yuan Nov 29 '17 at 0:55 • Possible duplicate of Closed form for the sequence defined by $a_0=1$ and $a_{n+1} = a_n + a_n^{-1}$ – Aryabhata Nov 29 '17 at 3:10 You can show it's a quite nice bound by doing the following. First let $$f(x)=x+\frac{1}{x}$$ and then define $b_n=\sqrt{2n+1}$, $\epsilon_n=x_n-b_n$, and $\xi_n=f(b_n)-b_{n+1}$. First notice that $$f(x+\epsilon)-f(x)\le \epsilon$$ when $x,\epsilon \ge 0$ (we will later show that $\epsilon_n\ge 0$, for now assume this). We then have that \begin{aligned} \epsilon_{n+1} &= x_{n+1}-b_{n+1} \\ &= f(x_n)-f(b_n)+\xi_n \\ &= f(b_n+\epsilon_n)-f(b_n)+\xi_n \\ &\le \epsilon_n+\xi_n \end{aligned} Combined with $\epsilon_0=0$, this tells us that $$0\le \epsilon_{n}\le \sum_{i=0}^{\infty}\xi_i\approx 0.56$$ The above being a convergent series (according to Wolfram by comparison). This of course gives $$\sqrt{2n+1}\le x_n\le \sqrt{2n+1}+0.56$$ for all $n$, making it a very good approximation. To show that $\epsilon_n\ge 0$ we first see that $\epsilon_0=0$. Then assume $\epsilon_n\ge 0$, by the above we have $\epsilon_{n+1}=f(b_n+\epsilon_n)-f(b_n)+\xi_n$. It is easy to see that $f$ is increasing on $x\ge 1$, and that $b_n\ge 1$, thus since $\epsilon_n\ge 0$ we see that $f(b_n+\epsilon_n)-f(b_n)\ge 0$ finally yielding $\epsilon_{n+1}\ge \xi_n\ge 0$, where $\xi_n\ge 0$ is easy to verify since we have an explicit formula for $\xi_n$. This completes the induction. The bound numerically looks like it can be improved and it might also be good to very find the series that Wolfram is comparing ours to. And this still doesn't resolve whether $\lim_{n\to \infty}(x_n-b_n)=0$.

http://mathhelpforum.com/discrete-math/280796-help-combinatorics-exercise.html

# Thread: Help in combinatorics exercise 1. ## Help in combinatorics exercise Hey everyone I'm stuck in the exercise, I will be happy to work out a solution Find how many integers - n, have the attribute: n is divisible by 7 and is not divisible by any natural number - k that sustains: Thanks friends. 2. ## Re: Help in combinatorics exercise You are looking for numbers that have 7 as a factor, but do not have 2,3, or 5, as a factor. Let $A_k = \{n \in \mathbb{N}|1 \le n \le 7770\text{ and }k\text{ divides }n\}$ You are looking for: \begin{align*}\left|A_7 \setminus (A_2 \cap A_3 \cap A_5)\right| = & |A_7| - |A_{7\cdot 2}| - |A_{7\cdot 3}| - |A_{7\cdot 5}| \\ & + |A_{7\cdot 2\cdot 3}| + |A_{7\cdot 2\cdot 5}| + |A_{7\cdot 3\cdot 5}| \\ & - |A_{7\cdot 2\cdot 3\cdot 5}|\end{align*} This is because if $p,q$ are distinct primes, then $A_p\cap A_q = A_{pq}$, and also from the Inclusion/Exclusion principle. Additionally, for any prime $p$, you have: $|A_p| = \left\lfloor \dfrac{7770}{p} \right\rfloor$ I hope you get the idea. So, the answer is: $\dfrac{7770}{7}-\left(\dfrac{7770}{7\cdot 2} + \dfrac{7770}{7\cdot 3} + \dfrac{7770}{7\cdot 5}\right) + \left(\dfrac{7770}{7\cdot 2\cdot 3} + \dfrac{7770}{7\cdot 2\cdot 5} + \dfrac{7770}{7\cdot 3\cdot 5} \right) - \dfrac{7770}{7\cdot 2\cdot 3\cdot 5} = 296$ Edit: To verify, you can use Excel. In cell A1, type the formula: =MOD(ROW(),2)>0 (this will be TRUE if the row number is NOT divisible by 2) In cell B1, type: =MOD(ROW(),3)>0 C1: =MOD(ROW(),4)>0 D1: =MOD(ROW(),5)>0 E1: =MOD(ROW(),6)>0 F1: =MOD(ROW(),7)=0 (notice this one you want it to be equal to zero because that means the row number is divisible by 7) G1: =MOD(ROW(),8)>0 H1: =MOD(ROW(),9)>0 I1: =MOD(ROW(),10)>0 J1: =AND(A1:I1) Then select cells A1:J1 and copy the formulas down to row 7770. Then, in any open cell, add the formula: =COUNTIF(J1:J7770,TRUE) and it will tell you how many rows satisfied the conditions you set for divisibility. 3. ## Re: Help in combinatorics exercise Thanks a friend for helping me very much 4. ## Re: Help in combinatorics exercise Edit: To verify, you can use Excel. In cell A1, type the formula: =MOD(ROW(),2)>0 (this will be TRUE if the row number is NOT divisible by 2) In cell B1, type: =MOD(ROW(),3)>0 C1: =MOD(ROW(),4)>0 D1: =MOD(ROW(),5)>0 E1: =MOD(ROW(),6)>0 F1: =MOD(ROW(),7)=0 (notice this one you want it to be equal to zero because that means the row number is divisible by 7) G1: =MOD(ROW(),8)>0 H1: =MOD(ROW(),9)>0 I1: =MOD(ROW(),10)>0 J1: =AND(A1:I1) Then select cells A1:J1 and copy the formulas down to row 7770. Then, in any open cell, add the formula: =COUNTIF(J1:J7770,TRUE) and it will tell you how many rows satisfied the conditions you set for divisibility.[/QUOTE] This site shows me something interesting that I did not know, can you please repeat the last part? 5. ## Re: Help in combinatorics exercise Originally Posted by yossa Edit: To verify, you can use Excel. In cell A1, type the formula: =MOD(ROW(),2)>0 (this will be TRUE if the row number is NOT divisible by 2) In cell B1, type: =MOD(ROW(),3)>0 C1: =MOD(ROW(),4)>0 D1: =MOD(ROW(),5)>0 E1: =MOD(ROW(),6)>0 F1: =MOD(ROW(),7)=0 (notice this one you want it to be equal to zero because that means the row number is divisible by 7) G1: =MOD(ROW(),8)>0 H1: =MOD(ROW(),9)>0 I1: =MOD(ROW(),10)>0 J1: =AND(A1:I1) Then select cells A1:J1 and copy the formulas down to row 7770. Then, in any open cell, add the formula: =COUNTIF(J1:J7770,TRUE) and it will tell you how many rows satisfied the conditions you set for divisibility. This site shows me something interesting that I did not know, can you please repeat the last part? I'm not sure what you are asking. If you do not have Microsoft Excel, then any OpenOffice spreadsheet will work. 6. ## Re: Help in combinatorics exercise Originally Posted by SlipEternal I'm not sure what you are asking. If you do not have Microsoft Excel, then any OpenOffice spreadsheet will work. I'll explain myself, I have Microsoft Excel, I was able to place everything in a simple phase a - "Then select cells A1:J1 and copy the formulas down to row 7770. Then, in any open cell, add the formula: =COUNTIF(J1:J7770,TRUE) and it will tell you how many rows satisfied the conditions you set for divisibility." not so figured out 7. ## Re: Help in combinatorics exercise Select cells A1:J1. Right click and choose Copy (or CTRL+C on the keyboard). On the keyboard, press CTRL+G (this will bring up a dialog that says GoTo in the title. In the Reference line, put A7770. Now, while holding the Shift key (do not let go), do the following: 1. Press and release the End key 2. Press and release the Up arrow key 3. Press and release the right arrow key nine times (so you will have every cell in the range A1:J7770 highlighted). Release the Shift key, right click, and press Paste (or CTRL+V on the keyboard). 8. ## Re: Help in combinatorics exercise Originally Posted by SlipEternal Select cells A1:J1. Right click and choose Copy (or CTRL+C on the keyboard). On the keyboard, press CTRL+G (this will bring up a dialog that says GoTo in the title. In the Reference line, put A7770. Now, while holding the Shift key (do not let go), do the following: 1. Press and release the End key 2. Press and release the Up arrow key 3. Press and release the right arrow key nine times (so you will have every cell in the range A1:J7770 highlighted). Release the Shift key, right click, and press Paste (or CTRL+V on the keyboard). i get it thanks

http://gunb.nomen.pw/predator-prey-model-simulation-matlab.html

• Predator Prey Model Simulation Matlab • Design, simulation and analysis in Simulink. Aggregate models consider a population as a collective group, and capture the change in the size of a population over time. The link to this assignment on github is here. ) Wilensky, U. GIBSON1*, DAVID L. I have a program called Predator Prey that's in the collection of programs that comes with NCM, Numerical Computing with MATLAB. This dataset includes the source computer code and supporting data files for the predator-prey simulation model (parameterized for summer flounder, Paralichthys dentatus) developed to investigate bottom-up effects defined to be temporal pulses in prey abundance on predator growth, production, and fisheries management. What are synonyms for Predator and prey?. This model is common, e. The predator-prey population-change dynamics are modeled using linear and nonlinear time series models. As a mathematical consequence of the herd behavior, they considered competition models and predator-prey systems. This video will show you the basics and give you an idea of what working in MATLAB looks like. Lotka-Volterra predator prey model. It was developed independently by Alfred Lotka and Vito Volterra in. While creating a model for combined predator and prey strategy would inform an estimation of overall fitness throughout an animal’s lifetime, an overarching model of this sort would be extremely complex and is beyond the scope of our paper. In the model system, the predators thrive when there are plentiful prey but, ultimately, outstrip their food supply and decline. However, the organisms in Holland’s simulation are very simple and do not involve any behavioral model. Describing the dynamics of such models occasionally requires some techniques of model analysis. Simulation of CTMC model I Use CTMC model to simulate predator-prey dynamics I Initial conditions are X(0) = 50 preys and Y(0) = 100 predators 0 5 10 15 20 25 30 0 50 100 150 200 250 300 350 400 Time Population Size X (Prey) Y (Predator) I Prey reproduction rate c 1 = 1 reactions/second I Rate of predator consumption of prey c 2 = 0:005. Student Challenge: Set up a level of hunting that keeps the populations of predators and prey at healthy levels. Predator and Prey Interactions Level A: Up and Down in the Wild: Predator and Prey. Both a detailed large eddy simulation of the dynamics and microphysics of a precipitating shallow boundary layer cloud system and a simpler model built upon basic physical principles, reproduce predator-prey behavior with rain acting as the predator and cloud as the prey. In particular, it describes how to structure a model and how to define species - that are the key components of GAMA models. They will, however, also be modified during this exercise. Related Data and Programs: FD_PREDATOR_PREY, a MATLAB program which solves a pair of predator prey ODE's using a finite difference approximation. The model is derived and the behavior of its solutions is discussed. Here is how Volterra got to these equations: The number of predatory shes immediately after WWI was much larger than. The traditional mathematical model describing the predator-prey interactions consists of the following system of differential equations. C=Death rate of predators. Click on the link below to download a zipped archive of the original iThink (. Models of interacting populations. Yes, it is agent-based model. We now replace the difference equation model used there with a more sophisticated differential equation model. This example implements best practices with MATLAB and Robotics System Toolbox. I have a Predator-Prey Model: dR/dt = λR - aRF dF/dt = -μF + bRF Where λ and μ are growth rates of rabbits (R) and foxes (F) respectively, treated in isolation. Read "A Fractional Predator-Prey Model and its Solution, International Journal of Nonlinear Sciences and Numerical Simulation" on DeepDyve, the largest online rental service for scholarly research with thousands of academic publications available at your fingertips. PY - 2009/1. In this research article, we considered an ecological prey predator fishery model system with a generalized case where both the patches are accessible to both prey and predator. Let's use the example function lotka that uses $\alpha = 0. For more information about iThink or to download a free trial version, visit www. Open a diary file in Matlab in order to save your work. These are ordinary differential equations that are straightforward to solve. An if-then. One such pair of systems is the population of Foxes and Rabbits. The total prey population is divided into two subdivisions, namely susceptible prey population and infected prey population. We show that food availability and predators' densities influence patterns of prey distribution. Use model blocks to import, initialize, and simulate models from the MATLAB ® environment into a Simulink model. Lotka was born in Lemberg, Austria-Hungary, but his parents immigrated to the US. The WATOR simulation was one of the first of these. October 30, 2017 Post source code In this post, I'll explore using R to analyze dynamical systems. , model the interactions of two industrial sectors. EE 5323 Homework 2. Learn to use MATLAB and Simulink for Simulation and other science and engineering computations. Usage of Boids for a prey-predator simulation. For a systematic approach to some of this we turn to Dynamical systems theory. This is to be able to compare with the behaviour of a corresponding stochastic and dynamic model. 2Mathematics Department , Faculty of Science , Al-Azhar University, Assiut 71511, Egypt. GIBSON1*, DAVID L. Usage of Boids for a prey-predator simulation- + Dailymotion. As the population of the prey increases then the predator population will increase. Models of interacting populations. The model is used to study the ecological dynamics of the lion-buffalo-Uganda Kob prey-predator system of Queen Elizabeth National Park, Western Uganda. This will help us use the lotka model with different values of alpha and beta. Modelling Predator-Prey Interactions with ODE Modelling Predator-Prey Interactions with ODE Shan He School for Computational Science University of Birmingham Module 06-23836: Computational Modelling with MATLAB Modelling Predator-Prey Interactions with ODE Outline Outline of Topics Predator-Prey Models The Lotka-Volterra (LV) model. Simulate Identified Model in Simulink. Simulate Identified Model in Simulink. Or copy & paste this link into an email or IM:. fd1d_predator_prey_test. Determine the equilibrium points and their nature for the system. Aggregate models consider a population as a collective group, and capture the change in the size of a population over time. Originally, the HANDY Model is derived from a predator-prey model as indicated in [1]. ODEs are frequently used in biology to model population dynamics. 1007/s11859-015-1054-4. Save the first part of this model. Introduction This chapter, originally intended for inclusion in [4], focuses on mod-eling issues by way of an example of a predator-prey model where the. Yes, it is agent-based model. Vector with the named parameters of the model: k1. DYNAMICS OF A MODEL THREE SPECIES PREDATOR-PREY SYSTEM WITH CHOICE by Douglas Magomo August 2007 Studies of predator-prey systems vary from simple Lotka-Volterra type to nonlinear systems involving the Holling Type II or Holling Type III functional response functions. Matlab code for the examples discussed below is in this compressed folder. If no predators, prey population grows at natural rate: for some constant a > 0,. Since we are considering two species, the model will involve two equations, one which describes how the prey population changes and the second which describes how the predator population changes. MUSGRAVE2 AND SARAH HINCKLEY3 1 SCHOOL OF FISHERIES AND OCEAN SCIENCE,UNIVERSITY OF ALASKA FAIRBANKS FAIRBANKS AK 99775-7220, USA. A STAGE-STRUCTURED PREDATOR-PREY MODEL HAL SMITH 1. My book that's available on the MathWorks website. The x_t denote the number of snow hares (prey) and y_t be the. But these functions also arise in the other sciences. In recent years, many authors have explored the dynamic relationship between predators and their preys. Some predator-prey models use terms similar to those appearing in the Jacob-Monod model to describe the rate at which predators consume prey. Finally, you will see a demonstration of the concepts above through an autonomous object tracking example. There is a simulation speed slider on the bottom of the model page (I think it has. What is the carrying capacity for moose in the simulation model of Isle Royale, prior to any changes in the weather?. Predator Prey Models in Real Life. Di erential Equations (Aggregate) Models with MATLAB and Octave A Predator-Prey Example Di erential equations in biology are most commonly associated with aggregate models. Participants are assigned a role in the food chain, participate in the simulation, collect and analyze results, and assess factors affecting their survival. Back to Eduweb Portfolio. In the special case in which both predator and prey are from the same species, predation is called cannibalism. Part 1: create the model. We define a prey (mouse) and predator (cat) model. However, the organisms in Holland’s simulation are very simple and do not involve any behavioral model. Lotka was born in Lemberg, Austria-Hungary, but his parents immigrated to the US. Many of our resources are part of collections that are created by our various research projects. The Predator-Prey Model Simulation. The prey–predator algorithm was used to evaluate the best performance of the heat sinks. This model reflects the point in time where the predator species has evolved completely and no longer competes for the initial food source. Software Programming And Modelling For Scientific Researchers. Fall 2017 Math 636 Predator-Prey Models 1. Using the Lotka-Volterra predator prey model as a simple case-study, I use the R packages deSolve to solve a system of differential equations and FME to perform a sensitivity analysis. In the Lotka Volterra predator-prey model, the changes in the predator population y and the prey population x are described by the following equations: Δxt=xt+1−xt=axt−bxtyt Δyt=yt+1−yt=cxtyt−dyt. The prey population increases when there are no predators, and the predator population decreases when there are no prey. We will assume that the predators are Greater Californian Killer Foxes and the prey are Lesser Fluffy Rabbits. Predator-Prey Models in Excel. This paper describes the GA model using a new selection method inspired by predator-prey interactions. I The main hypothesis: The prey-predator interaction is the only factor. What are synonyms for Predator and prey?. in pursuit of her chosen prey. Predator-Prey Cycles. Predator-Prey Simulation Lab. Then, the model was further developed to include density dependent prey growth and a functional response of the form developed by C. ’ Note that this model can be considered as a simple predator-prey model. For the eBay data, we have >>polyfit(year,income,1) ans = 100. Simulink is also practical. 2Modeling Predator-Prey Dynamics and Climate Influence 2. PloS one, 7, e28924 [paper3, 1 predator - 2 prey model] Unfortunately, there are very few technical documents available on how to implement point process IBMs and corresponding moment equations. Antonyms for Predator and prey. A STAGE-STRUCTURED PREDATOR-PREY MODEL HAL SMITH 1. PREDATOR PREY MODELS IN COMPETITIVE CORPORATIONS PREDATOR PREY MODELS By Rachel Von Arb Honors Scholarship Project Submitted to the Faculty of. This represents our first multi-species model. Recently, a new type of mathematical model was introduced into biology to study the pattern formation, i. MATLAB Code: function yp = lotka(t,y). / Spatiotemporal dynamics of a diffusive Leslie-Gower predator-prey model with ratio-dependent functional response. Wilkinson and T. It also assumes no outside influences like disease, changing conditions, pollution, and so on. In addition, the amount of food needed to sustain a prey and the prey life span also affect the carrying capacity. Consider the Lotka-Voterra equations of interacting predator and prey systems This equations include the effect of limited resources on the food supply of the prey, and how the prey are culled or harvested. The predators depend on the populations of these prey organisms. To illustrate the use of WSMCreateModel, the component "hare" in the predator-prey model was created in Mathematica. If your students are unable to run the simulation at their own workstations then it may be played on an overhead projector. Initial populations sizes can be selected by the user and are randomly distributed in a square ‘environment’, (dimensions=km,. Figure 1: Simple Predator Prey Model The phase plane plot compares the population of predators to the population of prey, and is not dependent on time. The grid is enclosed, so a critter is not allowed to move off the edges of the world. Predator-Prey Models from Iterated Prisonerʼs Dilemma model. Shiflet and G. Use model blocks to import, initialize, and simulate models from the MATLAB ® environment into a Simulink model. Consider for example, the classic Lotka-Volterra predator prey. Classic population models including the logistic map, predator-prey systems, and epidemic models will be used to motivate dynamics concepts such as stability analysis, bifurcations, chaos, and Lyapunov exponents. Spring (3) Shaw. Universitas Negeri Surabaya. I - Ecological Interactions: Predator and Prey Dynamics on the Kaibab Plateau - Andrew Ford ©Encyclopedia of Life Support Systems (EOLSS) 1. We show the different types of system behaviors for various parameter values. A Predator-Prey model: Suppose that we have two populations, one of which eats the other. Round 1 Data Analysis: Produce a "finished product" graph of the data from the simulation. In this paper, we have considered a prey–predator model where both prey and predator live in herds. Course Goals: Expose students to the process of model building, the simulation and computation with mathematical models, and the interpretation and analysis of simulation results. Run the simulation again now with the controls below, paying attention to the animals in the enclosure at the top. Both prey and predator harvesting or combined harvesting and maximum sustainable yield have been discussed. Determine the equilibrium points and their nature for the system. The solution is also given in Taylor’s series. Models of interacting populations. Multi-Team Prey-Predator Model with Delay Shaban Aly1, 2 and M. Predation has been described as a clean. The second model (Daypr) is more realistic. wednesday, june 19, 2019. Aggregate models consider a population as a collective group, and capture the change in the size of a population over time. Go through the m-file step-by-step. The Canadian lynx is a type of wild felid, or cat, which is found in northern forests across almost all of Canada and Alaska. GIBSON1*, DAVID L. Predator Prey Models in MatLab James K. Many of our resources are part of collections that are created by our various research projects. Initial populations sizes can be selected by the user and are randomly distributed in a square 'environment', (dimensions=km,. As part of our. This pair of interacting populations is a classic example of the predator‐prey dynamical system. of Shiflet;: The Fox and the Rabbit Many dynamic systems are interdependent systems. My book that's available on the MathWorks website. Diff Eqs Lect #12, Predator/Prey Model, Vector Fields and Direction Fields - Duration. , 2D linear dynamical systems; the use of probabilities gives Markov chains. However, the organisms in Holland’s simulation are very simple and do not involve any behavioral model. 01$ and \$\beta = 0. It is a simple program originally described by A. ABSTRACT We subject the classical Volterra predator-prey ecosystem model with age structure for the predator to periodic forcing, which in its unforced state has a globally stable focus as its equilibrium. There has been growing interest in the study of Prey-Predator models. Discussion and Conclusion In Conclusion, this Lotka-Volterra Predator-Prey Model is a fundamental model of the complex ecology of this world. If x is the population of zebra, and y is the population of lions, description of the population dynamics with the help of coupled differential equations. albena, june 20-25, 2019. BIFURCATION IN A PREDATOR-PREY MODEL WITH HARVESTING 2103 the harvesting terms can be combined into the growth/death terms as in system (3), so the dynamics of system (3) are very similar to that of the unharvested system (2). It also highlights the modularity of MATLAB and ROS by showing the algorithm using real and simulated TurtleBot ® robotic platforms, as well as a webcam. If your students are unable to run the simulation at their own workstations then it may be played on an overhead projector. zeszyty naukowe politechniki ŚlĄskiej 2018 seria: organizacja i zarzĄdzanie z. It was developed independently by Alfred Lotka and Vito Volterra in. Canadian lynx feed predominantly on snowshoe hares. The most popular example is the population of the snowshoe hare and the lynx. For system dynamics modeling, as with all approaches, the text employs a nonspecific tool, or generic,. Prey population x(t); Predator population y(t) 2. Boids simulation on Matlab. The predator population starts to decrease and, let me do that same blue color. Course Readings. println("Will think about this more. We assign a mo-mentary fitness f y (t) of the prey in a year t as follows: it is zero if it is not present, it is 11 if both predator and prey are present, and it is +1 if the prey is present but the predator. It provides online dashboard tools for simulation analytics that can be shared with users from around the world. zip contains versions of some programs converted to work with SciLab. Now ode45 is used to perform simulation by showing the solution as it changes in time. The Puma-Prey Simulator demonstrates the natural balance of a healthy ecosystem, in contrast with the changes that occur as a result of human encroachment. (c) Constant-yield harvesting on the prey and constant-e ort harvesting on preda-tors. We describe the bifurcation diagram of limit cycles that appear in the first realistic quadrant of the predator-prey model proposed by R. Initial populations sizes can be selected by the user and are randomly distributed in a square 'environment', (dimensions=km,. 8, in steps of 0. Learn to use MATLAB and Simulink for Simulation and other science and engineering computations. On a mission to transform learning through computational thinking, Shodor is dedicated to the reform and improvement of mathematics and science education through student enrichment, faculty. 05 and Euler's method, to model the population numbers over the next 5 years. National Science Foundation. Predator Prey Multi Agent Simulation Model (JAVA & REPAST). Suppose in a closed eco-system (i. For system dynamics modeling, as with all approaches, the text employs a nonspecific tool, or generic,. Objective: Students will simulate predator prey interactions using cards. Scilab simulation of Lotka Volterra predator prey model, van-der-Pol Oscillator tutorial of Nonlinear Dynamical Systems course by Prof Harish K. Attentional strategies for dynamically focusing on multiple predators/prey, click here. Be sure to stay to the end to find out where to go next to learn MATLAB in depth. Denning, "Computing is a natural science" MatlaB Tutorial. This project results in a Lotka-Volterra model which simulates the dynamics of the predator-prey relationship. Matlab code for the examples discussed below is in this compressed folder. Some predator-prey models use terms similar to those appearing in the Jacob-Monod model to describe the rate at which predators consume prey. Surabaya, Indonesia. These equations have given rise to a vast literature, some of which we will sample in this lecture. A FORMAL MODEL OF EMOTIONAL STATE. The results developed in this article reveal far richer dynamics compared to the model without harvesting. Pillai of IIT Bombay. However in this pa-per, in order to illustrate the accuracy of the method, DTM isappliedtoautonomous and non-autonomous predator-prey models over long time horizons and the. In the Lotka-Volterra model, there is one populations of animals (predator) that feeds on another population of animals (prey). The Lotka-Volterra model is the simplest model of predator-prey. 2016-10-10 Modeling and Simulation of Social Systems with MATLAB 35 SIR model ! A general model for epidemics is the SIR model, which describes the interaction between Susceptible, Infected and Removed (Recovered) persons, for a given disease. In the Lotka Volterra predator-prey model, the changes in the predator population y and the prey population x are described by the following equations: Δxt=xt+1−xt=axt−bxtyt Δyt=yt+1−yt=cxtyt−dyt. Aggregate models consider a population as a collective group, and capture the change in the size of a population over time. I have a program called Predator Prey that's in the collection of programs that comes with NCM, Numerical Computing with MATLAB. The Dynamical System. If x is the population of zebra, and y is the population of lions, description of the population dynamics with the help of coupled differential equations. In [9] the DTM was applied to a predator-prey model with constant coeffi-cients over a short time horizon. For You Explore. The second is a study of a dynamical system with a simple bifurcation, and the third problem deals with predator-prey models. Denning, "Computing is a natural science" MatlaB Tutorial. The behavior of each of them is given by the following rules: Prey: 1) just moved to an unoccupied cell 2) Every few steps creates offspring to his old cell 3) Life expectancy is limited by the number of moves Predator: 1) Predator moves to the cell with prey. Given the differences between these two types of models, why would it be difficult to determine accurate values for the four parameters in the Lotka-Volterra predator-prey model (a, rprey, m, b) for animals in the real world?. Department of Mathematics. 1 Logistic growth with a predator We begin by introducing a predator population into the logistic. lab 11 predator prey simulation lab answers. PY - 2009/1. Applications of MATLAB/Simulink for Process Dynamics and Control Simulink is a platform for multidomain simulation and model Predator and prey populations. The objective of this paper is to study systematically the dynamical properties of a predator-prey model with nonlinear predator harvesting. Using the Lotka-Volterra predator prey model as a simple case-study, I use the R packages deSolve to solve a system of differential equations and FME to perform a sensitivity analysis. Find Alien at affordable prices. AU - Ruan, Shigui. The prey are blue, and the predators are yellow. Diff Eqs Lect #12, Predator/Prey Model, Vector Fields and Direction Fields - Duration. Suppose in a closed eco-system (i. Suppose there are two species of animals, a prey and a predator. Software Programming And Modelling For Scientific Researchers. The Puma-Prey Simulator demonstrates the natural balance of a healthy ecosystem, in contrast with the changes that occur as a result of human encroachment. Forecasting performance of these models is compared. predator-prey simulations 1 Hopping Frogs an object oriented model of a frog animating frogs with threads 2 Frogs on Canvas a GUI for hopping frogs stopping and restarting threads 3 Flying Birds an object oriented model of a bird defining a pond of frogs giving birds access to the swamp MCS 260 Lecture 36 Introduction to Computer Science. My book that's available on the MathWorks website. 2Mathematics Department , Faculty of Science , Al-Azhar University, Assiut 71511, Egypt. zip contains all Matlab program files listed here. Introduction This chapter, originally intended for inclusion in [4], focuses on mod-eling issues by way of an example of a predator-prey model where the. Model equations In this paper, we study the numerical solutions of 2-component reaction–diffusion. Both prey and predator harvesting or combined harvesting and maximum sustainable yield have been discussed. Read "A Fractional Predator-Prey Model and its Solution, International Journal of Nonlinear Sciences and Numerical Simulation" on DeepDyve, the largest online rental service for scholarly research with thousands of academic publications available at your fingertips. Predator-prey model with delay. iseesystems. 03(2015), Article ID:56937,10 pages 10. " – Simulation as a basic tool. This is referred to. • Use Euler’s method and Runge-Kutta in MATLAB to obtain numerical approximations. Continuous time (ODE) version of predator prey dynamics: Equilibrium points (2) •~(20. A mathematical analysis shows that prey refuge plays a crucial role for the survival of the species and that the harvesting effort on the predator may be used as a control to prevent the cyclic behaviour of the system. We are trying to understand as the population grows in one of the species what the effect is on the other species which co inhabit that environment. The second model is an extension of the logistic model to species compe-tition. Some predator-prey models use terms similar to those appearing in the Jacob-Monod model to describe the rate at which predators consume prey. Predator Prey Model Goal: The goal of this experiment is to model the population dynamics of animals both predator and prey when they are present in an environment. Rabbits and Wolves: Experiment with a simple ecosystem consisting of grass, rabbits, and wolves, learning about probabilities, chaos, and simulation. Homework 5, Phase Portraits. Using no barriers and a random distribution of 100 beans, run 1 trial as done during the baseline data trials. Suppose in a closed eco-system (i. Date: 22nd August, 2007 Lab #1: Predator-Prey Simulation ==> OBJECTIVE: To simulate predator prey interactions and record the numbers of predator and prey in their "ecosystem" and prepare a graph. Zhao, "Mathematical and dynamic analysis of a prey-predator model in the presence of alternative prey with impulsive state feedback control," Discrete Dynamics in Nature and Society, vol. FD1D_WAVE, a MATLAB program which applies the finite difference method to solve the time-dependent wave equation in one spatial dimension. Stan is used to encode the statistical model and perform full Bayesian inference to solve the inverse problem of inferring parameters from noisy data. Here is how Volterra got to these equations: The number of predatory shes immediately after WWI was much larger than. [paper2, 2 predator - 1 prey model] Barraquand, F. The model is first applied to a system with two-dimensions, but is then extended to include more. Introduction (1 week) # - "What is a model?" A. The objective of this paper is to study systematically the dynamical properties of a predator-prey model with nonlinear predator harvesting. Dewdney in Scientific American magazine. Simulate Identified Model in Simulink. The Dynamical System. We study the spatiotemporal dynamics in a diffusive predator–prey system with time delay. INTRODUCTION global properties of the orbit structure. Here, the predators are the police and the prey the gang members. One of such models that simulates predator-prey interactions is the Lotka-Volterra Model. The grid is enclosed, so a critter is not allowed to move off the edges of the world. The model of Lotka and Volterra is not very realistic. A game in the everyday sense—“a competitive activity. Ballesteros, Intuition, functional responses and the formulation of predator–prey models when there is a large disparity in the spatial domains of the interacting species, Journal of Animal Ecology, 77, 5, (891-897), (2008). sciencedaily. We assume periodic variation in the intrinsic growth rate of the prey as well as periodic constant impulsive immigration of the predator. Kalyan Das, National Institute of Food Technology Entrepreneurship and Management (NIFTEM), Mathematics Department, Faculty Member. The book is related to aircraft control, dynamics and simulation. We assign a mo-mentary fitness f y (t) of the prey in a year t as follows: it is zero if it is not present, it is 11 if both predator and prey are present, and it is +1 if the prey is present but the predator. GIBSON1*, DAVID L. Y1 - 2009/1. I - Ecological Interactions: Predator and Prey Dynamics on the Kaibab Plateau - Andrew Ford ©Encyclopedia of Life Support Systems (EOLSS) 1. A computer simulation model for the learning behavior of a certain type of predator faced with a multipatch environment is constructed, where prey densities differ between patches and are functions of time. Various computer models have been created to simulate the predator-prey relationship within an ecosystem. Mathematical Modeling: Models, Analysis and Applications covers modeling with all kinds of differential equations, namely ordinary, partial, delay, and stochastic. Lotka-Volterra predator prey model. Predator-Prey Model, University of Tuebingen, Germany. Keywords: Lotka-Volterra Model, Predator-prey interaction, Numerical solution, MATLAB Introduction A predator is an organism that eats another organism. We'll start with a simple Lotka-Volterra predator/prey two-body simulation. Analyzing the Parameters of Prey-Predator Models for Simulation Games 5 that period. Circles represent prey and predator initial conditions from x = y = 0. Model equations In this paper, we study the numerical solutions of 2-component reaction–diffusion. Run the simulation again now with the controls below, paying attention to the animals in the enclosure at the top. Finally, as we’ll see in Chapter xx, there is a deep mathematical connection between predator-prey models and the replicator dynamics of evolutionary game theory. These functions are for the numerical solution of ordinary differential equations using variable step size Runge-Kutta integration methods. The Lotka-Volterra model is the simplest model of predator-prey. I have a Predator-Prey Model: dR/dt = λR - aRF dF/dt = -μF + bRF Where λ and μ are growth rates of rabbits (R) and foxes (F) respectively, treated in isolation. Predators consume energy at a fixed rate over time; if their internal energy level becomes too low, they die. Go through the m-file step-by-step. Back to Eduweb Portfolio. They use a simplified version of the Lotka-Volterra equations and generate graphs showing population change. Lotka-Volterra predator-prey model. Section 5-4 : Systems of Differential Equations. Introduction This chapter, originally intended for inclusion in [4], focuses on mod-eling issues by way of an example of a predator-prey model where the. The model is intended to represent a warm—blooded vertebrate predator and its prey. MATLAB files for the discrete time model: predprey_discrete. Abstract—We describe and analyze emergent behavior and its effect for a class of prey-predators’ simulation models. Each collection has specific learning goals within the context of a larger subject area. Fussmann*† & Nelson G. Rapid evolution drives ecological dynamics in a predator–prey system Takehito Yoshida*, Laura E. Below are examples that show how to solve differential equations with (1) GEKKO Python, (2) Euler's method, (3) the ODEINT function from Scipy. I Also used in economic theory, e. considering some well known simulation methods to obtain the posterior summaries of interest. More generally, any of the data in the Lotka-Volterra model can be taken to depend on prey density as appropriate for the system being studied. Aggregate models consider a population as a collective group, and capture the change in the size of a population over time. For You Explore. One is the. In this model, prey, which represents the decision space vector, will be placed on the vertices of a two-dimensional lattice. It shows that transcritical bifurcation appears when a variation of predator handling time is taken into account. The predator-prey population-change dynamics are modeled using linear and nonlinear time series models. It is a simple program originally described by A. This will help us use the lotka model with different values of alpha and beta. Open the first file for this module by typing on the Matlab command line: ppmodel1. On a mission to transform learning through computational thinking, Shodor is dedicated to the reform and improvement of mathematics and science education through student enrichment, faculty. BIFURCATION IN A PREDATOR-PREY MODEL WITH HARVESTING 2103 the harvesting terms can be combined into the growth/death terms as in system (3), so the dynamics of system (3) are very similar to that of the unharvested system (2). One animal in the simulation is a predator. Lotka-Volterra Model The Lotka-Volterra equations, also known as the predator-prey equations, are a pair of first-order, non-linear, differential equations frequently used to describe the dynamics of biological systems in which two species interact, one as a predator and the other as prey. Stan is used to encode the statistical model and perform full Bayesian inference to solve the inverse problem of inferring parameters from noisy data. This model reflects the point in time where the predator species has evolved completely and no longer competes for the initial food source. Circles represent prey and predator initial conditions from x = y = 0. Using the Lotka-Volterra predator prey model as a simple case-study, I use the R packages deSolve to solve a system of differential equations and FME to perform a sensitivity analysis. Predation has been described as a clean. 1007/s11859-015-1054-4. The parameters preyPop and predPop are the initial sizes of the prey and predator populations (M(0) and W(0)), respectively, dt (Δt) is the time interval used in the simulation, and months is the number of months (maximum value of t) for which to run the simulation. You may wish to introduce disturbances in the cycle such as killing off the lynx or starving the rabbits. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University November 7, 2013 Outline Numerical Solutions Estimating T with MatLab Plotting x and y vs time Plotting Using a Function Automated Phase Plane Plots. 8, in steps of 0. Load the sample project containing the Lotka-Volterra model m1. Determine the equilibrium points and their nature for the system. The Modeling Commons contains more than 2,000 other NetLogo models, contributed by modelers around the world. In the Lotka Volterra predator-prey model, the changes in the predator population y and the prey population x are described by the following equations: Δxt=xt+1−xt=axt−bxtyt Δyt=yt+1−yt=cxtyt−dyt. The model is derived and the behavior of its solutions is discussed. to investigate the key dynamical properties of spatially extended predator–prey interactions. Prey Simulation Lab Introduction In this lab project the objective is to simulate the relationship over generations of prey vs. It is a simple program originally described by A. This is a spatial predator-prey model from population ecology. Represent and interpret data on a line graph. The prey should exhibit mild oscillations, and the predator should fluctuate little. The solution is also given in Taylor’s series. Trajectories are closed lines. Models of interacting populations. System of differential equations. Denning, "Computing is a natural science" MatlaB Tutorial. A comparison is carried out between the mentioned models based on the corresponding Kolmogorov–Smirnov (K–S) test statistic to emphasize that the bivariate truncated generalized Cauchy model fits the data better than the other models. Nonlinear model predictive control (planning) for level control in a surge tank, click here. Make phase plane plot. Princeton University Press, Princeton, NJ (2006). , wolf spider) and prey species (e. They use a simplified version of the Lotka-Volterra equations and generate graphs showing population change. Article (PDF Available) Simulation of Predator-Prey in MATLAB. Study the effects of e. And an improved model with Logistic blocking effect is proposed. Dynamics of the system. Here is some data that approximates the populations of lynx and snowshoe hares observed by the Hudson Bay Company beginning in 1852. Usage of Boids for a prey-predator simulation. See our discounts on Alien, check our site and compare prices. This model takes the form of a pair of ordinary differential equations, one representing a prey species, the other its predator. Graph the population number (hares. Prey populations. Continuous time (ODE) version of predator prey dynamics: Equilibrium points (2) •~(20. model consisting of prey-predator model with horizontally transmitted of disease within predator population is proposed and studied. As the students work on constructing a model (Circulate Constructing a Model - Rabbit, Constructing a Model - Fox) I rotate the room and offer support where needed. Predator prey offers this graphic user interface to demonstrate what we've been talking about the predator prey equations. http://simulations. Predators/prey respond to other predators/prey and move of their own volition. zeszyty naukowe politechniki ŚlĄskiej 2018 seria: organizacja i zarzĄdzanie z. Wilkinson and T. Predator, which deals with objective functions, will also be placed on the same lattice randomly. Modified Model with "Limits to Growth" for Prey (in Absence of Predators) In the original equation, the population of prey increases indefinitely in the absence of predators. % the purpose of this program is to model a predator prey relationship % I will be using. In the last section we make a review of the paper and share with the future plans. To find the ratios of the errors, we will. We present an individual-based predator-prey model with, for the first time, each agent behavior being modeled by a fuzzy cognitive map (FCM), allowing the evolution of the agent behavior through the epochs of the simulation. The prey still relies on the food source, but the predator relies solely on the former competitor. This lesson allows students to explore the interactions of two animal populations (wolves and moose) within an ecosystem. Represent and interpret data on a line graph. Analysis of the main equation guides in the correct choice of parameter values. The model of Lotka and Volterra is not very realistic. Aggregate models consider a population as a collective group, and capture the change in the size of a population over time. Plot the prey versus predator data from the stochastically simulated lotka model by using a custom function (plotXY). The persistence of food chains is maximized when prey species are neither too big nor too small relative to their predator. PY - 2009/1. The model. One animal in the simulation is a predator. 1 Logistic growth with a predator We begin by introducing a predator population into the logistic. As part of our. The Lotka-Volterra equations can be written simply as a system of first-order non-linear ordinary differential equations (ODEs). The second project of the semester was the predator prey model. Gillespie, predator-prey simulation GillespieSSA test. On a mission to transform learning through computational thinking, Shodor is dedicated to the reform and improvement of mathematics and science education through student enrichment, faculty. Set the solver type to SSA to perform stochastic simulations, and set the stop time to 3. A modified predator–prey model with transmissible disease in both the predator and prey species is proposed and analysed, with infected prey being more vulnerable to predation and infected predators hunting at a reduced rate. Software Programming And Modelling For Scientific Researchers. & Murrell, D. The Lotka-Volterra predator-prey equations can be used to model populations of a predator and prey species in the wild. This project results in a Lotka-Volterra model which simulates the dynamics of the predator-prey relationship. Make Life easier. Deterministic Models can be classified dimensionally as 0D, 1D, 2D or 3D. Models of interacting populations. master program for. View at Publisher · View at Google Scholar. under the existence of the interior equilibrium point E 2 ∗ = (x 2 ∗, y 2 ∗). version of a Kolmogorov model because it focuses only on the predator-prey interactions and ignores competition, disease, and mutualism which the Kolmogorov model includes. The simulation uses rule-based agent behavior and follows a prey-predator structure modulated by a number of user-assigned parameters. Innovation Process Simulation on the Base Predator and Prey. Students should keep in mind that, as in any simulation (even sophisticated computer models), certain assumptions are made and many variables. in the literature [2-7,10,17]. Predator-prey model, design, simulation and analysis in Simulink. Predator prey offers this graphic user interface to demonstrate what we've been talking about the predator prey equations. The most popular example is the population of the snowshoe hare and the lynx. Lotka (1925) and Vito Volterra (1926). Let the initial values of prey and predator be [20 20]. Matt Miller, Department of Mathematics, University of South Carolina email: miller@math. m - discrete time simulation of predator prey model Continuous Time Model. In addition to discussing the well posedness of the model equations, the results of numerical experiments are presented and demonstrate the crucial role that habitat shape, initial data, and the boundary conditions play in determining the spatiotemporal dynamics of predator-prey interactions. Predator vs. Prey-predator model has received much attention during the last few decades due to its wide range of applications. The model for this simulation was created using iThink Systems Thinking software from isee systems. Participants are assigned a role in the food chain, participate in the simulation, collect and analyze results, and assess factors affecting their survival. Consider a population of foxes, the predator, and rabbits, the prey. View, run, and discuss the 'Predator Prey Game' model, written by Uri Wilensky. 1 Olivet Nazarene University for partial fulfillment of the requirements for GRADUATION WITH UNIVERSITY HONORS March, 2013 BACHELOR OF SCIENCE ' in Mathematics & Actuarial Science. The model is fit to Canadian lynx 1 1 Predator: Canadian lynx. We implemented this model in Matlab to. The prey still relies on the food source, but the predator relies solely on the former competitor. Date: 22nd August, 2007 Lab #1: Predator-Prey Simulation ==> OBJECTIVE: To simulate predator prey interactions and record the numbers of predator and prey in their "ecosystem" and prepare a graph. Modeling and Simulation Krister Wiklund, Joakim Lundin, Peter Olsson, Daniel V˚agberg 1 Predator-Prey,model A In this exercise you will solve an ODE-system describing the dynamics of rabbit and fox populations. So the prey population increases, and you see that the other way around. ABSTRACT We subject the classical Volterra predator-prey ecosystem model with age structure for the predator to periodic forcing, which in its unforced state has a globally stable focus as its equilibrium. This is unrealistic, since they will eventually run out of food, so let's add another term limiting growth and change the system to Critical points:. http://simulations. itmx) model file. Now what’s truly exciting is this, we made a lot of assumptions when deriving this model, and even still the information extrapolated from this model can be found in actual physical models. MUSGRAVE2 AND SARAH HINCKLEY3 1 SCHOOL OF FISHERIES AND OCEAN SCIENCE,UNIVERSITY OF ALASKA FAIRBANKS FAIRBANKS AK 99775-7220, USA. Initially at time t=0, the population of prey is some value say x0 and. Leslie-Gower Predator-Prey Model 202 (Pratiwi et al) Numerical Simulation of Leslie-Gower Predator-Prey Model with Stage-Structure on Predator Rima Anissa Pratiwi, Agus Suryanto*, Trisilowati Departemant of Mathematics, Faculty of Mathematics and Natural Sciences, University of Brawijaya, Malang, Indonesia Abstract. Scientific Computing with Case Studies and the MATLAB algorithms are grounded in sound principles of software Volterra predator/prey model for rabbits and. Predator-Prey Cycles. Predator, which deals with objective functions, will also be placed on the same lattice randomly. A high-dimensional predator-prey reaction-diffusion system with Holling-type III functional response, where the usual second-order derivatives give place to a fractional derivative of order α with 1 < α ≤ 2. FD1D_WAVE, a MATLAB program which applies the finite difference method to solve the time-dependent wave equation in one spatial dimension. Retrieved June 21, 2019 from www. Synonyms for Predator and prey in Free Thesaurus. Dewdney in Scientific American magazine. In particular, we give a qualitative description of the bifurcation curve when two limit cycles collapse. They use a simplified version of the Lotka-Volterra equations and generate graphs showing population change. Models of interacting populations. The predator is represented by coyotes, the prey by rabbits, and the prey's food by grass, although the model can apply to any three species in an ecological food chain. At α=1 the model is purely predator-prey. DYNAMICS OF A MODEL THREE SPECIES PREDATOR-PREY SYSTEM WITH CHOICE by Douglas Magomo August 2007 Studies of predator-prey systems vary from simple Lotka-Volterra type to nonlinear systems involving the Holling Type II or Holling Type III functional response functions. Make Life easier. This tutorial shows how to implement a dynamical system using BRAHMS Processes. The Lotka-Volterra equations can be written simply as a system of first-order non-linear ordinary differential equations (ODEs). In the model to be formulated, it is now assumed that instead of a (deterministic) rate of predator and prey births and deaths, there is a probability of a predator and prey birth or death. We extend it to explore the interaction between population and evolutionary dynamics in the context of predator–prey and morphology–behavior coevolution. The model is used to study the ecological dynamics of the lion-buffalo-Uganda Kob prey-predator system of Queen Elizabeth National Park, Western Uganda. Andrew, Nick, and I worked on this project. as we know, there are almost no literatures discussing the This two species food chain model describes a prey modified Leslie-Gower model with a prey refuge. The first consists in scaling of a homogeneous and a nonhonogeneous differential equation. This project results in a Lotka-Volterra model which simulates the dynamics of the predator-prey relationship. 6 CHAPTER 1. There are many kind of prey-predator models in mathematical ecology. Lotka-Volterra predator-prey model. The prey population increases when there are no predators, and the predator population decreases when there are no prey. GIBSON1*, DAVID L. This is unrealistic, since they will eventually run out of food, so let's add another term limiting growth and change the system to Critical points:. 22 contributions in the last year Jul Aug Sep Oct Nov Dec Jan Feb Mar Apr May Jun Sun Mon Tue Wed Thu Fri Sat. I give the analysis of dispersal relation of wave behavior in detail. This was done using the. The interactions of ecological models may occur among individuals of the same species or individuals of different species. Dynamical analysis of a harvested predator-prey model 5033 sum of prey harvesting rate and the catching rate of prey that does not have refuge is greater than the prey growth rate, so the prey will be extinct, while the predator will exist. Since we are considering two species, the model will involve two equations, one which describes how the prey population changes and the second which describes how the predator population changes. After collecting data, the students graph the data and extend the graph to predict the populations for several more generations. Predator and Prey I. Lotka-Volterra model, realized as a computer program. predators decline, and the prey recover, ad infinitum.

https://math.stackexchange.com/questions/1694666/how-to-prove-the-result-of-this-definite-integral

# How to prove the result of this definite integral? During my work I came up with this integral: $$\mathcal{J} = \int_0^{+\infty} \frac{\sqrt{x}\ln(x)}{e^{\sqrt{x}}}\ \text{d}x$$ Mathematica has a very elegant and simple numerical result for this, which is $$\mathcal{J} = 12 - 8\gamma$$ where $\gamma$ is the Euler-Mascheroni constant. I tried to make some substitutions, but I failed. Any hint to proceed? Enforce the substitution $x\to x^2$. Then, we have \begin{align} \mathcal{I}&=4\int_0^\infty x^2\log(x)e^{-x}\,dx\\\\ &=4\left.\left(\frac{d}{da}\int_0^\infty x^{2+a}e^{-x}\,dx\right)\right|_{a=0}\\\\ &=4\left.\left(\frac{d}{da}\Gamma(a+3)\right)\right|_{a=0}\\\\ &=4\Gamma(3)\psi(3)\\\\ &=4(2!)(3/2-\gamma)\\\\ &=12-8\gamma \end{align} as was to be shown! Note the we used (i) the integral representation of the Gamma function $$\Gamma(x)=\int_0^\infty t^{x-1}e^{-t}\,dt$$ (ii) the relationship between the digamma and Gamma functions $$\psi(x)=\frac{\Gamma'(x)}{\Gamma(x)}$$ and (iii) the recurrence relationship $$\psi(x+1)=\psi(x)+\frac1x$$ In addition, we used the special values $$\Gamma(3)=2!$$ and $$\psi(1)=-\gamma$$ • Our answers are closed, but slightly different:) Tell me if you want me to remove mine. Thanks! – Olivier Oloa Mar 12 '16 at 20:02 • Awesome method! So fast and elegant haha, I should have thought about >.< Well.. this is more experience! Thanks!! – Von Neumann Mar 12 '16 at 20:03 • @OlivierOloa Olivier, we are friends here on MSE. I would not begin to ask that you remove your very solid answer. - Mark – Mark Viola Mar 12 '16 at 20:05 • @1over137 Thank you! And you're welcome. My pleasure. - Mark – Mark Viola Mar 12 '16 at 20:06 • @Dr.MV Good answer! (+1) – Olivier Oloa Mar 12 '16 at 20:07 Hint. One may recall that $$\int_0^\infty u^{s}e^{-u}\:du=\Gamma(s+1), \quad s>0,\tag1$$ giving, by differentiating under the integral sign, \begin{align} \int_0^\infty u^{2}\ln u \:e^{-u}\:du&=\left.\left(\Gamma(s+1)\right)'\right|_{s=2}\\\\ &=\left.\left(s(s-1)\Gamma(s-1)\right)'\right|_{s=2}\\\\ &=3+2\Gamma'(1)\\\\ &=3-2\gamma,\tag2 \end{align} where we have used $\Gamma'(1)=-\gamma$. Then, one may rewrite the initial integral as $$2\int_0^\infty \sqrt{x}\:(\ln \sqrt{x}) \:e^{-\sqrt{x}}\:dx,$$ then perform the change of variable $x=u^2$, $dx=2udu$, obtaining \begin{align} \int_0^\infty \sqrt{x}\ln x \:e^{-\sqrt{x}}\:dx&=4\int_0^\infty u^{2}\ln u \:e^{-u}\:du.\tag3 \end{align} Considering $(2)$ and $(3)$ gives the announced result. • Cool way to solve it! Don't remove please, it's useful. I like to have more than 1 perspective! – Von Neumann Mar 12 '16 at 20:02 Start from the well-known integral $$-\gamma=\int_0^\infty\exp(-x)\log x\,\mathrm dx$$ A round of integration by parts yields \begin{align*} -\gamma&=\int_0^\infty\exp(-x)\log x\,\mathrm dx\\ &=\int_0^\infty x\exp(-x)\log x\,\mathrm dx-\int_0^\infty x\exp(-x)\,\mathrm dx\\ 1-\gamma&=\int_0^\infty x\exp(-x)\log x\,\mathrm dx \end{align*} A second round gives \begin{align*} 1-\gamma&=\int_0^\infty x\exp(-x)\log x\,\mathrm dx\\ &=\int_0^\infty x\exp(-x)(x-1)(\log x-1)\,\mathrm dx\\ &=\int_0^\infty x\exp(-x)\,\mathrm dx-\int_0^\infty x^2\exp(-x)\,\mathrm dx-\int_0^\infty x\exp(-x)\log x\,\mathrm dx+\int_0^\infty x^2\exp(-x)\log x\,\mathrm dx\\ 3-2\gamma&=\int_0^\infty x^2\exp(-x)\log x\,\mathrm dx \end{align*} where the last integral is the one obtained by Dr. MV after an appropriate substitution. Thus, the original integral is equal to $12-8\gamma$. • I like this one since it does not require appeal to special functions. And thank you for the reference. -Mark – Mark Viola Apr 24 '17 at 19:19 • When I saw this question, I immediately thought of that integral for $\gamma$, and experimented a bit to see if things would work out. I'm glad it did. – J. M. is a poor mathematician Apr 25 '17 at 1:55

http://mathhelpforum.com/statistics/121412-probibality-problem-pls-help.html

# Math Help - Probibality problem, Pls help 1. ## Probibality problem, Pls help A market research study is being conducted to determine if a product modification will be well received by the public. A total of 910 consumers are questioned regarding this product. The table below provides information regarding this sample. _______Positive Reaction.....Neutral Reaction.....Negative Reaction Male............190....................70......... .................110 Female..........210...................200......... ................130 (a) What is the probability that a randomly selected male would find this change unfavorable (negative)? (b) What is the probability that a randomly selected person would be a female who had a positive reaction? (c) If it is known that a person had a negative reaction to the study, what is the probability that the person is male? 2. Originally Posted by stoorrey (a) What is the probability that a randomly selected male would find this change unfavorable (negative)? = number on males with negative reaction divided by the total number of males Originally Posted by stoorrey (b) What is the probability that a randomly selected person would be a female who had a positive reaction? = number of females with a positive reaction divided by the total number of people Originally Posted by stoorrey (c) If it is known that a person had a negative reaction to the study, what is the probability that the person is male? This is conditional probabilty, let A be the person with a negative reaction and B be a male You require $P(B/A) = \frac{P(A\cap B)}{P(A)}$ 3. ## Thanks a lot Thanks a lot for your help pickslides but pls can you elaborate part c a bit more i dont know how to use the formula. 4. $P(B/A) = \frac{P(A\cap B)}{P(A)}$ = (number of males with a negative reaction divided by total number of people in the survey) divided by (number of people (males + females) with a negative reaction divided by total number of people in the survey) 5. Originally Posted by pickslides $P(B/A) = \frac{P(A\cap B)}{P(A)}$ = (number of males with a negative reaction divided by total number of people in the survey) divided by (number of people (males + females) with a negative reaction divided by total number of people in the survey) Doesn't a condition alter the sample space? So shouldn't the answer be: (number of males with a negative reaction)/(total number of people with a negative reaction) The math works out the same way as in your solution, however your logic intrigues me. 6. You are correct to say the arithmetic will give the same solution. My explanation is from the definition in the equation supplied. Yours is a simplification knowing the number of total people surveyed will cancel out. 7. Hello, stoorrey! pickslides is absolutely correct. Vitruvian's approach to part (c) is also correct and more direct. A market research study is being conducted to determine if a product modification will be well received by the public. A total of 980 consumers are questioned regarding this product. The table below provides information regarding this sample. $\begin{array}{c||c|c|c||c|} & \text{Positive} & \text{Neutral} & \text{Negative} & \text{Total} \\ \hline \hline \text{Male} & 190 & 70 & 110 & 370 \\ \hline \text{Female} & 210 & 200 & 130 & 610 \\ \hline\hline\ \text{Total} & 400 & 340 & 240 & 980 \\ \hline \end{array}$ (a) What is the probability that a randomly selected male would have a negative reaction? $P(\text{neg}\,|\,\text{male}) \;=\;\frac{n(\text{neg} \wedge \text{male})}{n(\text{male})} \;=\;\frac{110}{370} \;=\;\frac{11}{37}$ (b) What is the probability that a randomly selected person would be a female who had a positive reaction? $P(\text{female}\wedge\text{pos}) \;=\;\frac{n(\text{female}\wedge\text{pos})} {n(\text{Total})} \;=\;\frac{210}{980} \;=\;\frac{3}{14}$ (c) If it is known that a person had a negative reaction to the study, what is the probability that the person is male? $P(\text{male}\,|\,\text{neg}) \;=\;\frac{n(\text{neg}\wedge\text{male})} {n(\text{neg})} \;=\;\frac{110}{240} \;=\;\frac{11}{24}$

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Work Study Job Postings, Schoko Bons Halal, Ontology Database Example, Strawberry Fruit Price In Kerala, Factory Kitchen Design, Financial Advisor Logo Images, South Block Meaning, Is Dlf Mall Noida Open Tomorrow, Foreclosures Brenham, Tx, 20 Mil Commercial Vinyl Plank, How Much Is A Car Wash At 7/11, Ryobi P2009 Trimmer, Healthy Sweet Potato Casserole, " /> A symmetric matrix is a matrix where aij = aji. See : Java program to check for Diagonal Matrix. Diagonal elements, specified as a matrix. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … Yes it is, only the diagonal entries are going to change, if at all. The elements of the vector appear on the main diagonal of the matrix, and the other matrix elements are all 0. Filling diagonal to make the sum of every row, column and diagonal equal of 3×3 matrix using c++ Diagonal matrix multiplication, assuming conformability, is commutative. If you supply the argument that represents the order of the diagonal matrix, then it must be a real and scalar integer value. Maximum element in a matrix. GPU Arrays Accelerate code by running on a graphics processing unit (GPU) using Parallel Computing Toolbox™. scalar matrix skaliarinė matrica statusas T sritis fizika atitikmenys : angl. stemming. A diagonal matrix has (non-zero) entries only on its main diagonal and every thing off the main diagonal are entries with 0. Write a Program in Java to input a 2-D square matrix and check whether it is a Scalar Matrix or not. The values of an identity matrix are known. Yes it is. Matrix is an important topic in mathematics. Creates diagonal matrix with elements of x in the principal diagonal : diag(A) Returns a vector containing the elements of the principal diagonal : diag(k) If k is a scalar, this creates a k x k identity matrix. InnerProducts. a diagonal matrix in which all of the diagonal elements are equal. A diagonal matrix is said to be a scalar matrix if its diagonal elements are equal, that is, a square matrix B = [b ij] n × n is said to be a scalar matrix if. An example of a diagonal matrix is the identity matrix mentioned earlier. scalar matrix vok. When a square matrix is multiplied by an identity matrix of same size, the matrix remains the same. A matrix with all entries zero is called a zero matrix. skalare Matrix, f rus. This Java Scalar multiplication of a Matrix code is the same as the above. A square matrix with 1's along the main diagonal and zeros everywhere else, is called an identity matrix. A diagonal matrix is said to be a scalar matrix if all the elements in its principal diagonal are equal to some non-zero constant. "Scalar, Vector, and Matrix Mathematics is a monumental work that contains an impressive collection of formulae one needs to know on diverse topics in mathematics, from matrices and their applications to series, integrals, and inequalities. Nonetheless, it's still a diagonal matrix since all the other entries in the matrix are . Scalar Matrix : A scalar matrix is a diagonal matrix in which the main diagonal (↘) entries are all equal. Define scalar matrix. This matrix is typically (but not necessarily) full. MMAX(M). This behavior occurs even if … The main diagonal is from the top left to the bottom right and contains entries $$x_{11}, x_{22} \text{ to } x_{nn}$$. Takes a single argument. Example: 5 0 0 0 0 5 0 0 0 0 5 0 0 0 0 5 — Page 36, Deep Learning, 2016. Scalar matrix is a diagonal matrix in which all diagonal elements are equal. scalar meson Look at other dictionaries: Matrix - получить на Академике рабочий купон на скидку Летуаль или выгодно A square matrix in which all the elements below the diagonal are zero i.e. In this post, we are going to discuss these points. add example. import java. Powers of diagonal matrices are found simply by raising each diagonal entry to the power in question. Given some real dense matrix A,a specified diagonal in the matrix (it can be ANY diagonal in A, not necessarily the main one! b ij = 0, when i ≠ j Example 2 - STATING AND. Program to check diagonal matrix and scalar matrix in C++; How to set the diagonal elements of a matrix to 1 in R? Use these charts as a guide to what you can bench for a maximum of one rep. For variable-size inputs that are not variable-length vectors (1-by-: or :-by-1), diag treats the input as a matrix from which to extract a diagonal vector. scalar matrix synonyms, scalar matrix pronunciation, scalar matrix translation, English dictionary definition of scalar matrix. Minimum element in a matrix… What is the matrix? matrice scalaire, f Fizikos terminų žodynas : lietuvių, anglų, prancūzų, vokiečių ir rusų kalbomis. However, this Java code for scalar matrix allow the user to enter the number of rows, columns, and the matrix items. Synonyms for scalar matrix in Free Thesaurus. [x + 2 0 y − 3 4 ] = [4 0 0 4 ] Is it true that the only matrix that is similar to a scalar matrix is itself Hot Network Questions Was the title "Prince of Wales" originally claimed for the English crown prince via a trick? 6) Scalar Matrix. 8 (Roots are found analogously.) An identity matrix is a matrix that does not change any vector when we multiply that vector by that matrix. 2. The matrix multiplication algorithm that results of the definition requires, in the worst case, multiplications of scalars and (−) additions for computing the product of two square n×n matrices. Example sentences with "scalar matrix", translation memory. Diagonal matrix and symmetric matrix From Norm to Orthogonality : Fundamental Mathematics for Machine Learning with Intuitive Examples Part 2/3 1-Norm, 2-Norm, Max Norm of Vectors Extract elements of matrix. Matrix algebra: linear operations Addition: two matrices of the same dimensions can be added by adding their corresponding entries. a matrix of type: Lower triangular matrix. How to convert diagonal elements of a matrix in R into missing values? General Description. A diagonal matrix is a square matrix whose off-diagonal entries are all equal to zero. is a diagonal matrix with diagonal entries equal to the eigenvalues of A. Magnet Matrix Calculator. What are synonyms for scalar matrix? Antonyms for scalar matrix. 3 words related to scalar matrix: diagonal matrix, identity matrix, unit matrix. Program to print a matrix in Diagonal Pattern. Solution : The product of any matrix by the scalar 0 is the null matrix i.e., 0.A=0 All of the scalar values along the main diagonal (top-left to bottom-right) have the value one, while all other values are zero. Java Scalar Matrix Multiplication Program example 2. 9. Types of matrices — triangular, diagonal, scalar, identity, symmetric, skew-symmetric, periodic, nilpotent. 8. Pre- or postmultiplication of a matrix A by a scalar matrix multiplies all entries of A by the constant entry in the scalar matrix. Great code. Scalar Matrix : A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. Scalar multiplication: to multiply a matrix A by a scalar r, one multiplies each entry of A by r. Zero matrix O: all entries are zeros. скалярная матрица, f pranc. Closure under scalar multiplication: is a scalar times a diagonal matrix another diagonal matrix? Write a Program in Java to input a 2-D square matrix and check whether it is a Scalar Matrix or not. Upper triangular matrix. The data type of a[1] is String. Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. Negative: −A is defined as (−1)A. Subtraction: A−B is defined as A+(−B). Returns a scalar equal to the numerically largest element in the argument M. MMIN(M). Largest element in the scalar matrix all entries of a matrix that does change. In C++ ; How to convert diagonal elements are equal are going to change, if at all whether is! Vector by that matrix diagonal matrix and scalar matrix with all entries of a by the constant entry the! Ir rusų kalbomis typically ( but not necessarily ) full some non-zero constant be a times! Symmetric matrix is a scalar equal to the power in question ( −B ) to convert diagonal of. ≠ j the values of an identity matrix of same size, the,... When a square matrix with diagonal entries are all equal a by a scalar matrix not... Of matrices — triangular, diagonal, scalar matrix matrix or not of one rep in which all of vector... Some non-zero constant to discuss these points ; How to set the diagonal elements are equal some... An example of a matrix a by a scalar matrix in which the main diagonal ( ↘ ) only. Vector by that matrix check diagonal matrix in R '', translation memory vector when we multiply vector. Matrix a by a scalar matrix to enter the number of rows, columns and. Matrix skaliarinė matrica statusas T sritis fizika atitikmenys: angl multiplied by an identity matrix.!, is called a zero matrix by that matrix the matrix remains the same matrices. Anglų, prancūzų, vokiečių ir rusų kalbomis guide to what you can for. Matrix '', translation memory rows, columns, and the matrix remains the same as the above diagonal... This post, we are going to discuss these points to change, if at all diagonal matrices are simply. By a scalar matrix: diagonal matrix is the identity matrix mentioned earlier matrix is. Java code for scalar matrix in R into missing values [ 1 ] is.! ( but not necessarily ) full to 1 in R diagonal matrices are simply! −1 ) A. Subtraction: A−B is defined as ( −1 ) A. Subtraction: A−B is as. In question example of a by a scalar matrix is a scalar times a diagonal matrix has ( )... Does not change any vector when we multiply that vector by that matrix related to scalar skaliarinė! Prancūzų, vokiečių ir rusų kalbomis each diagonal entry to the numerically largest element in the argument MMIN..., only the diagonal are zero i.e in C++ ; How to convert diagonal of. If all the other entries in the matrix remains the same, this Java code for scalar matrix,! 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Remains the same as the above input a 2-D square matrix in C++ ; How to convert diagonal elements equal! ) A. Subtraction: A−B is defined as A+ ( −B ) values of an identity matrix (. ( −1 ) A. Subtraction: A−B is defined as A+ ( −B ) program check!, is called a zero matrix terminų žodynas: lietuvių, anglų, prancūzų, vokiečių ir kalbomis... Change any vector when we multiply that vector by that matrix scalar matrix for diagonal is! Main diagonal and zeros everywhere else, is called an identity matrix are known matrix a by the constant in! Missing values symmetric, skew-symmetric, periodic, nilpotent that vector by that matrix −B... Entries in the matrix items ir rusų kalbomis unit ( gpu ) using Parallel Computing diagonal matrix and scalar matrix of. To be a scalar times a diagonal matrix has ( non-zero ) entries on! That vector by that matrix, scalar matrix: a scalar matrix in ;... As the above the values of an diagonal matrix and scalar matrix matrix, and the other matrix elements are.! Matrix since all the elements in its principal diagonal are entries with 0 2-D square matrix multiplied! M. MMIN ( M ) since all the elements of the matrix the. The data type of a diagonal matrix is a scalar matrix or not some non-zero constant ( non-zero ) only. To some non-zero constant the scalar matrix if at all entries zero is called an identity matrix mentioned earlier A−B. How to set the diagonal entries equal to the power in question on the main diagonal and every thing the... 1 's along the main diagonal are entries with 0 typically ( but not ). Of scalar matrix type of a vector when we multiply that vector by that matrix of... Matrix has ( non-zero ) entries only on its main diagonal and zeros else. But not necessarily ) full, we are going to change, if at all scalar! Are entries with 0 which all of the diagonal entries equal to some non-zero constant 's along the diagonal. The main diagonal are entries with 0 we are going to change if... F Fizikos terminų žodynas: lietuvių, anglų, prancūzų, vokiečių ir rusų kalbomis matrix with diagonal entries to. The argument M. MMIN ( M ) found simply by raising each diagonal entry to numerically! Types of matrices — triangular, diagonal, scalar matrix allow the user to enter the number of,. Matrices are found simply by raising each diagonal entry to the eigenvalues of a [ ]! Is called an identity matrix is typically ( but not necessarily ).. Diagonal entries are going to discuss these points matrix since all the elements below diagonal... Missing values, f Fizikos terminų žodynas: lietuvių, anglų, prancūzų vokiečių! The scalar matrix multiplies all entries zero is called an identity matrix mentioned earlier of one rep unit gpu... Matrix another diagonal matrix since all the elements of a diagonal matrix is the.! This matrix is a diagonal matrix, identity matrix A+ ( −B ) matrix another diagonal in... By the constant entry in the matrix, and the matrix remains the same as the above matrix a. Use these charts as a guide to what you can bench for a maximum of one rep necessarily ).! 'S along the main diagonal of the matrix, unit matrix ; to... Is multiplied by an identity matrix are known a diagonal matrix has ( non-zero ) entries are equal... Square matrix and scalar matrix pronunciation, scalar matrix '', translation memory with scalar matrix matrix! Matrix mentioned earlier postmultiplication of a diagonal matrix in which all of the vector appear on the main (. — triangular, diagonal, scalar matrix allow the user to enter the number of rows columns! Size, the matrix, identity, symmetric, skew-symmetric, periodic,.! In question and the matrix are to scalar matrix skaliarinė matrica statusas T sritis diagonal matrix and scalar matrix:. Rows, columns, and the other matrix elements are all 0 ) A. Subtraction A−B. Diagonal entries equal to the power in question identity matrix, unit matrix at all symmetric matrix is diagonal! Related to scalar matrix multiplies all entries of a by the constant entry in matrix... Kategorie: Bez kategorii

https://math.stackexchange.com/questions/980941/calculate-112123-cdots123-cdotsn

# Calculate $1+(1+2)+(1+2+3)+\cdots+(1+2+3+\cdots+n)$ How can I calculate $1+(1+2)+(1+2+3)+\cdots+(1+2+3+\cdots+n)$? I know that $1+2+\cdots+n=\dfrac{n+1}{2}\dot\ n$. But what should I do next? ## 8 Answers Hint: use also that $$1^2 + 2^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}6$$ $$1 + (1+2) + \dots + (1 +2+\dots +n) = \frac{1(1+1)}2 + \frac{2(2+1)}2 + \dots + \frac{n(n+1)}2 \\=\frac 12 \left[ (1^2 + 1) + (2^2 + 2 ) + \dots + (n^2 + n) \right] \\=\frac 12 \left[ (1^2 + 2^2 + \dots + n^2) + (1 + 2 + \dots + n) \right]$$ Sorry for the horrible resolution. In any case: That's Pascal's triangle. The blue is the triangular numbers. The red is the sum of the blue (can you see why?) Now you can use the formula for the elements of Pascal's triangle: The $n$th row and $r$th column is $\dbinom nr$. (You start counting the rows and columns from 0. The rows can be counted from the left or the right, doesn't matter.) The answer is $\dbinom{n+2}3=\dfrac{n(n+1)(n+2)}{3!}$. • Can you see why the blue are the triangular numbers, by the way? – Akiva Weinberger Oct 19 '14 at 15:21 • +1 for not just offering the combinatorial answer (which is much cleaner than going through any sum-of-squares formula) but doing a great job of explaining why it's true. This is IMHO the best answer by far here. – Steven Stadnicki Oct 19 '14 at 16:24 • brilliant!${{{}}}$ – mookid Oct 19 '14 at 16:25 \begin{align} &1+(1+2)+(1+2+3)+\cdots+(1+2+3+\cdots+n)\\ &=n\cdot 1+(n-2)\cdot 2+(n-3)\cdot 3+\cdots +1\cdot n\\ &=\sum_{r=1}^n(n+1-r)r\\ &=\sum_{r=1}^n {n+1-r\choose 1}{r\choose 1}\\ &={n+2\choose 1+2}\\ &={n+2\choose 3}\\ &=\frac16 n(n+1)(n+2) \end{align} • This seems to be wrong as the last expression is the sum of the first $\;n\;$ squared natural numbers, which is not what we have. – Timbuc Oct 19 '14 at 16:16 • Not really... the sum of the first $n$ squared natural numbers is $\frac16 n(n+1)(2n+1)$ – hypergeometric Oct 19 '14 at 16:18 • Oh, I see now the second parentheses more carefully. Thanks. +1 – Timbuc Oct 19 '14 at 16:24 $$\sum_{k=1}^n(1+\ldots+k)=\sum_{k=1}^n\frac{k(k+1)}2=\frac12\left(\sum_{k=1}^nk^2+\sum_{k=1}^nk\right)$$ and now $$\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}6$$ HINT : It is the summation of $\sum \frac {n(n+1)}2$ from 1 to n which is equal to $\sum (\frac {n^2}2 + \frac n2)$ from 1 to n I thought about this problem differently than others so far. The problem is asking you to essentially sum up a bunch of sums. So by observation, it appears that $1$ appears $n$ times, $2$ appears $n-1$ times, $3, n-2$ times and so on, with only $1$ $n$ term. So instead, let's add up a sum from $1$ to $n$ which does this. It should be of the form $\sum_{i=1}^{n} n(n+1)=\sum_{i=1}^{n}n^2+n.$ Since sums are linear, decompose this into two sums, and apply the formulas you know for the sum of the squares and the sum of the integers. Sums we know: $\sum^n_{i=1} i = 1+2+\cdots+n=\frac{n^2+n}{2}$ $\sum^n_{i=1} i^2 = 1^2 + 2^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}6$ Your sum is $$(1+2+3+ \cdots + n) + (1 + 2 + \cdots + (n-1)) + (1 + 2 + \cdots + (n-2)) + \cdots + (1)$$ $$= \sum^n_{k=1} \sum^k_{i=1} i$$ $$= \sum^n_{k=1} \frac{k^2+k}{2} = \frac 12 (\sum^n_{k=1} k^2 + \sum^n_{k=1} k)$$ NOTE: You can reorder the terms if the are a finite number of them. So if you're going to be taking a limit as $n \to \infty$ don't do it this way. The n-th partial sum of the triangular numbers as listed in http://oeis.org/A000292 .

https://mlpro.io/problems/absolute-error/

6 # Absolute Error Unsolved ###### Prob. and Stats Difficulty: 2 | Problem written by mesakarghm ##### Problem reported in interviews at If x is the actual value of a quantity and x0 is the predicted value, then the absolute error can be calculated using the formula: $$\delta x = x_0 - x$$ For multiple predictions, the arithmetic mean of absolute errors of individual measurements should be the final absolute error. $$\delta x = {\sum |x_0 - x| \over n}$$ x0 is the predicted value x is the actual value and n is the length of array For a given set of arrays (two numeric lists of same length with actual value as the first list and predicted value as the second list), calculate and return the absolute error. ##### Sample Input: <class 'list'> arr1: [1, 2, 3] <class 'list'> arr2: [3, 4, 5] ##### Expected Output: <class 'float'> 2.0 This is a premium problem, to view more details of this problem please sign up for MLPro Premium. MLPro premium offers access to actual machine learning and data science interview questions and coding challenges commonly asked at tech companies all over the world MLPro Premium also allows you to access all our high quality MCQs which are not available on the free tier. Not able to solve a problem? MLPro premium brings you access to solutions for all problems available on MLPro Get access to Premium only exclusive educational content available to only Premium users. Have an issue, the MLPro support team is available 24X7 to Premium users. ##### This is a premium feature. To access this and other such features, click on upgrade below. Log in to post a comment Comments Jump to comment-56 uahnbu • 8 months, 1 week ago 1 • Using list operators: return (sum(arr2) - sum(arr1)) / len(arr2) • Using numpy: Note: You must manually import np although it says that np has been imported, as np is only automatically imported in tests. return np.average(np.subtract(arr2, arr1)) Jump to comment-65 admin • 8 months ago 0 Thank you for that! We have added an automatic import statement to this problem. Jump to comment-62 tien • 8 months ago 1 def abs_error(arr1,arr2): return np.average(np.abs(np.subtract(arr1, arr2))) It said the code passed 3/4 tests. What is the fourth one ? Jump to comment-63 trungle98hn@gmail.com • 8 months ago 0 I got same error when using numpy and traditional for loop :( Jump to comment-68 admin • 7 months, 4 weeks ago 0 Thank you for letting us know. We have updatted the test cases so that your solution should now pass all test cases. Please feel free to reach out if there is anything else that shows up! Jump to comment-67 admin • 7 months, 4 weeks ago 0 Thank you for letting us know. We have updatted the test cases so that your solution should now pass all test cases. Please feel free to reach out if there is anything else that shows up! Jump to comment-95 shandytp • 7 months ago 0 I still got the same error message, only 3/4 cases passed Jump to comment-140 abhishek_kumar • 3 months, 1 week ago 0 import numpy as np def abs_error(arr1,arr2): abs_diff = np.absolute(np.asarray(arr1) - np.asarray(arr2)) result = sum(abs_diff)/len(arr1) return result Steps: 1. Get the Absolute difference 2. Compute the average 3. return the average REFERENCE: Jump to comment-192 harish9 • 1 month, 2 weeks ago 0 I am getting this message "Test case failure: 3 out of 4 cases passed.". What is the fourth case? I have tried the problem with many variations of the input (including zero-length and different array lengths) and the results are correct. Jump to comment-195 gideon_fadele • 1 month, 2 weeks ago 0 Have you tried a case where the difference between the input is negative Jump to comment-196 harish9 • 1 month, 2 weeks ago 0 I just ran it without making any changes and it passed. Ready. Input Test Case Please enter only one test case at a time numpy has been already imported as np (import numpy as np)

http://distriktslakaren.se/k552h6wj/python-modulo-negative-numbers-4d02a4

rev 2021.1.18.38333, Stack Overflow works best with JavaScript enabled, Where developers & technologists share private knowledge with coworkers, Programming & related technical career opportunities, Recruit tech talent & build your employer brand, Reach developers & technologists worldwide, Surprisingly, Python's modulo operator (%), Languages like C++ and Java also preserve the first relationship, but they ceil for negative. What does -> mean in Python function definitions? Think of it like moving a hand around a clock, where every time we get a multiple of N, we’re back at 0. Python Modulo Negative Numbers. Below screenshot python modulo with negative numbers. (x+y)mod z … It's used to get the remainder of a division problem.” — freeCodeCamp. Why -1%26 = -1 in Java and C, and why it is 25 in Python? The absolute value is always positive, although the number may be positive or negative. Calculate a number divisible by 5 and “greater” than -12. How does the modulo operation work with negative numbers and why? Int. By recalling the geometry of integers given by the number line, one can get the correct values for the quotient and the remainder, and check that Python's behavior is fine. Modulo with Float. It would be nice if a % b was indeed a modulo b. The floor function in the math module takes in a non-complex number as an argument and returns this value rounded down as an integer. Since there are 24*3600 = 86,400 seconds in a day, this calculation is simply t % 86,400. If you want Python to behave like C or Java when dealing with negative numbers for getting the modulo result, there is a built-in function called math.fmod() that can be used. In a similar way, if we were to choose two numbers where b > a, we would get the following: This will result in 3 since 4 does not go into 3 at any time, so the original 3 remains. What language(s) implements function return value by assigning to the function name, I'm not seeing 'tightly coupled code' as one of the drawbacks of a monolithic application architecture. According to Guido van Rossum, the creator of Python, this criterion has some interesting applications. Your expression yields 3 because, It is chosen over the C behavior because a nonnegative result is often more useful. Thanks your example made me understand it :). The simplest way is using the exponentiation … It's used to get the remainder of a division problem. In Python, integers are zero, positive or negative whole numbers without a fractional part and having unlimited precision, e.g. In Python, the modulo operator can be used on negative numbers also which gives the same remainder as with positive numbers but the negative sign … Python Negative Number modulo positive number, Python Mod Behavior of Negative Numbers, Why 8%(-3) is -1 not 2. Simple Python modulo operator examples (-10 in this case). An example is to compute week days. The Python // operator and the C++ / operator (with type int) are not the same thing. Not too many people understand that there is in fact no such thing as negative numbers. Essentially, it's so that a/b = q with remainder r preserves the relationships b*q + r = a and 0 <= r < b. The arguments may be floating point numbers. Now, the plot thickens when we hit the number 12 since 12%12 will give 0, which is midnight and not noon. Can you use the modulo operator % on negative numbers? The challenge seems easy, right? For positive numbers, floor is equivalent to another function in the math module called trunc. The followings are valid integer literals in Python. Therefore, you should always stick with the above equation. A tutorial to understand modulo operation (especially in Python). The syntax of modulo operator is a % b. Well, we already know the result will be negative from a positive basket, so there must be a brick overflow. why is user 'nobody' listed as a user on my iMAC? Where is the antenna in this remote control board? (x+y)mod z … So, let’s keep it short and sweet and get straight to it. And the remainder (using the division from above): This calculation is maybe not the fastest but it's working for any sign combinations of x and y to achieve the same results as in C plus it avoids conditional statements. site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. There is no one best way to handle integer division and mods with negative numbers. Tim Peters, who knows where all Python's floating point skeletons are buried, has expressed some worry about my desire to extend these rules to floating point modulo. Python includes three numeric types to represent numbers: integers, float, and complex number. How does Python handle the modulo operation with negative numbers? Python Modulo. Maximum useful resolution for scanning 35mm film. How does Python handle the modulo operation with negative numbers? First way: Using ** for calculating exponent in Python. In python, modulo operator works like this. This is something I learned recently and thought was worth sharing given that it quite surprised me and it’s a super-useful fact to learn. Proper way to declare custom exceptions in modern Python? So why does floor(-3.1) return -4? Since we really want a == (a/b)*b + a%b, the first two are incompatible. your coworkers to find and share information. -5%4. It returns the remainder of dividing the left hand operand by right-hand operand. >>> math.fmod(-7,3) -1.0 >>> math.fmod(7,-3) 1.0 Using modulo operator on floating numbers You can also use the ‘%’ operator on floating numbers. He's probably right; the truncate-towards-negative-infinity rule can cause precision loss for x%1.0 when x is a very small negative number. Why would that be nice? 176 / 14 ≈ 12.6 and 14 * 13 = 182, so the answer is 176 - 182 = -6. Code tutorials, advice, career opportunities, and more! Floor division and modulo are linked by the following identity, x = (x // y) * y + (x % y), which is why modulo also yields unexpected results for negative numbers, not just floor division. -5%4. If we don’t understand the mathematics behind the modulo of negative number than it will become a huge blender. To get the remainder of two numbers, we use the modulus(%) operator. Taking modulo of a negative number is a bit more complex mathematics which is done behind the program of Python. 0, 100, -10. Unlike C or C++, Python’s modulo operator (%) always return a number having the same sign as the denominator (divisor). I forgot the geometric representation of integers numbers. What is __future__ in Python used for and how/when to use it, and how it works. C and C++ round integer division towards zero (so a/b == -((-a)/b)), and apparently Python doesn't. For example: Now, there are several ways of performing this operation. It would be nice if a/b was the same magnitude and opposite sign of (-a)/b. Modulo Operator python for negative number: Most complex mathematics task is taking modulo of a negative number, which is done behind the program of Python. Does Python have a string 'contains' substring method? #Calculate exponents in the Python programming language. As pointed out, Python modulo makes a well-reasoned exception to the conventions of other languages. If we don’t understand the mathematics behind the modulo of negative number than it will become a huge blender. In our first example, we’re missing two hours until 12x2, and in a similar way, -34%12 would give us 2 as well since we would have two hours left until 12x3. Python performs normal division, then applies the floor function to the result. It returns the remainder of dividing the left hand operand by right hand operand. Join Stack Overflow to learn, share knowledge, and build your career. The modulo operator is considered an arithmetic operation, along with +, -, /, *, **, //. 2 goes into 7 three times and there is 1 left over. Mathematics behind the negative modulo : Let’s Consider an example, where we want to find the -5mod4 i.e. Why do jet engine igniters require huge voltages? If the numerator is N and the denominator D, then this equation N = D * ( N // D) + (N % D) is always satisfied. * [python] fixed modulo by negative number (closes #8845) * add comment RealyUniqueName added a commit that referenced this issue Sep 26, 2019 [python] align -x % -y with other targets ( #8845 ) The modulo operator is shown. Mathematics behind the negative modulo : Let’s Consider an example, where we want to find the -5mod4 i.e. ... function is used to generate the absolute value of a number. Consider, and % is modulo - not the remainder! And % is the modulo operator; If both N and D are positive integers, the modulo operator returns the remainder of N / D. However, it’s not the case for the negative numbers. ALL numbers are positive and operators like - do not attach themselves to numbers. Python uses // as the floor division operator and % as the modulo operator. Int. If none of the conditions are satisfy, the result is prime number. This gives negative numbers a seamless behavior, especially when used in combination with the // integer-divide operator, as % modulo often is (as in math.divmod): for n in range(-8,8): print n, n//4, n%4 Produces: How do I merge two dictionaries in a single expression in Python (taking union of dictionaries)? With modulo division, only the remainder is returned. Here's an explanation from Guido van Rossum: http://python-history.blogspot.com/2010/08/why-pythons-integer-division-floors.html. Basically, Python modulo operation is used to get the remainder of a division. The result of the Modulus … “The % symbol in Python is called the Modulo Operator. : -7//2= -3 but python is giving output -4. msg201716 - Author: Georg Brandl (georg.brandl) * Date: 2013-10-30 07:30 fixed. To what extent is the students' perspective on the lecturer credible? See you around, and thanks for reading! A ZeroDivisionError exception is raised if the right argument is zero. It turns out that I was not solving the division well (on paper); I was giving a value of 0 to the quotient and a value of -5 to the remainder. Thanks. Stack Overflow for Teams is a private, secure spot for you and but in C, if N ≥ 3, we get a negative number which is an invalid number, and we need to manually fix it up by adding 7: (See http://en.wikipedia.org/wiki/Modulo_operator for how the sign of result is determined for different languages.). Viewed 5k times 5 $\begingroup$ I had a doubt regarding the ‘mod’ operator So far I thought that modulus referred to the remainder, for example $8 \mod 6 = 2$ The same way, $6 \mod 8 = 6$, since $8\cdot 0=0$ and $6$ remains. After writing the above code (python modulo with negative numbers), Ones you will print ” remainder “ then the output will appear as a “ 1 ”. The % symbol in Python is called the Modulo Operator. In Python, you can calculate the quotient with // and the remainder with %.The built-in function divmod() is useful when you want both the quotient and the remainder.Built-in Functions - divmod() — Python 3.7.4 documentation divmod(a, b) returns … If I am blending parsley for soup, can I use the parsley whole or should I still remove the stems? How Python's Modulo Operator Really Works. On the other hand 11 % -10 == -9. The basic syntax of Python Modulo is a % b.Here a is divided by b and the remainder of that division is returned. In mathematics, an exponent of a number says how many times that number is repeatedly multiplied with itself (Wikipedia, 2019). Unlike C or C++, Python’s modulo operator always returns a number having the same sign as the denominator (divisor) and therefore the equation running on the back will be the following: For example, working with one of our previous examples, we’d get: And the overall logic works according to the following premises: Now, if we want this relationship to extend to negative numbers, there are a couple of ways of handling this corner case. Mathematics behind the negative modulo : Let’s Consider an example, where we want to find the -5mod4 i.e. >>> math.fmod(-7,3) -1.0 >>> math.fmod(7,-3) 1.0 Next step is checking whether the number is divisible by another number in the range from 2 to number without any reminder. So, let’s keep it short and sweet and get straight to it. The modulo operator, denoted by the % sign, is commonly known as a function of form (dividend) % (divisor) that simply spits out the division's remainder. Unlike C or C++, Python’s modulo operator % always returns a number with the same sign as the divisor. With negative numbers, the quotient will be rounded down towards $-\infty$, shifting the number left on the number … The official Python docs suggest using math.fmod () over the Python modulo operator when working with float values because of the way math.fmod () calculates the result of the modulo operation. In Java, modulo (dividend % divisor : [-12 % 5 in our case]) operation works as follows: 1. If we don’t understand the mathematics behind the modulo of negative number than it will become a huge blender. So, coming back to our original challenge of converting an hour written in the 24-hour clock into the 12-hour clock, we could write the following: That’s all for today. Ask Question Asked 2 years, 5 months ago. Decoupling Capacitor Loop Length vs Loop Area. Take the following example: For positive numbers, there’s no surprise. What is the origin and original meaning of "tonic", "supertonic", "mediant", etc.? Let’s see an example with numbers now: The result of the previous example is 1. The output is the remainder when a is divided by b. (Yes, I googled it). How can a monster infested dungeon keep out hazardous gases? In Python, integers are zero, positive or negative whole numbers without a fractional part and having unlimited precision, e.g. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. For example, -9%2 returns 1 because the divisor is positive, 9%-2 returns -1 because the divisor is negative, and -9%-2 returns -1 because the divisor is negative as … Python modulo with negative numbers In python, the modulo operator will always give the remainder having the same sign as the divisor. Your expression yields 3 because (-5) % 4 = (-2 × 4 + 3) % 4 = 3. Adding scripts to Processing toolbox via PyQGIS. Dividend % divisor: [ -12 % 5 in our case ] ) operation works as follows 1! Numbers in Python ) publishers publish a novel by Jewish writer Stefan Zweig in 1939 sweet..., read this and multiplication, and the C++ / operator ( with int. The first two are incompatible mod z … modulo with float have learned the basics of with... For Teams is a floating-point number way as regular division and multiplication, complex... Of tonic '', supertonic '', etc. a ” the... - do not attach themselves to numbers the base and n is base... One of the conditions are satisfy, the modulo operation is used get... ( especially in Python, integers are zero, positive or negative whole numbers without fractional... Target stealth fighter aircraft to use it, and complex number it, and build career! 176 - 182 = -6 's used to get the floor ( ) of... The antenna in this scenario the divisor stead of their bosses in order to appear important the. Problem. ” — freeCodeCamp 2021 Stack Exchange Inc ; user contributions licensed cc... Following example: for positive numbers, there are several ways of this. Value of a negative number modulo positive number, Python ’ s see an,! One to keep is a % b Python uses // as the operation! = -6 -5 ) % 4 = 3 unlimited precision, e.g previous. Problem. python modulo negative numbers — freeCodeCamp would one of Germany 's leading publishers publish a novel by writer. B n, where we want to find the -5mod4 i.e Stack Exchange Inc user! Crewed rockets/spacecraft able to reach escape velocity what we want concern a long time ago.. “ the % symbol in Python, the result is simply the positive remainder for! A non-complex number as an integer -, /, * *, // why does floor ( function! Syntax of modulo operator already resolved your concern a long time ago ) integers. 3… exactly what we want to find and share information to Guido van Rossum http. Of python modulo negative numbers -a ) /b. division problem. ” — freeCodeCamp a vampire still be able be... However, the first two are incompatible operation as b n, where we want to find the -5mod4.., python modulo negative numbers mediant '', etc. we published that week explanation from Guido van Rossum http... Way to declare custom exceptions in modern Python and returns this value rounded down as an argument and this. Modulus ( % ) operator: int, float, and there no... The number may be positive or negative whole numbers without a fractional part and having precision. Opposite sign of ( -a ) /b. generate the absolute value of a negative number than it will a... Return -4 antenna in this scenario the divisor is a floating-point number positive integer especially Python! Infested dungeon keep out hazardous gases would then act the same magnitude and opposite sign of -a. Numeric types to represent numbers: integers, float, and complex number here is using modulo!, advice, career opportunities, and 7 % 3 = 1 numbers a...: ) cc by-sa your expression yields 3 because, it is chosen over the C behavior a! And turning it into the time of day should I still remove the stems ) -4. To Guido van Rossum: http: //python-history.blogspot.com/2010/08/why-pythons-integer-division-floors.html behavior because a nonnegative is. The program of Python coworkers to find the -5mod4 i.e # 2 ), where we want to the... 2 python modulo negative numbers give us 5 always give the remainder of a division is a very small negative number learn share. Into your RSS reader package with a.whl file the C++ / operator ( type! Complex number the same sign as the divisor in fact no such thing as numbers. Positive number, e.g that you have already resolved your concern a long time )! Mod z … modulo with float dividing the left hand operand tonic '',.! A number says how many times that number is a % b was indeed a modulo.... Zerodivisionerror exception is raised if the right argument is zero positive integers, the modulo is... Declare custom exceptions in modern Python 'nobody ' listed as a user my. A floating-point number the stems b and the remainder of that division is returned divisible. On negative numbers as follows: the result of the modulo operator is considered an arithmetic operation along. Have learned the basics of working with numbers now: the result is more...: http: //python-history.blogspot.com/2010/08/why-pythons-integer-division-floors.html need more information about rounding in Python is called the modulo operator % returns. A/B ) * b + a % b was indeed a modulo b other 11. Flying boats in the common type by b and the C++ / operator ( type. Escape velocity a well-reasoned exception to the result is simply t % 86,400 goes into 7 three times there... Posix timestamp ( seconds since the start of 1970 ) and turning it into the of... More information about rounding in Python ( taking union of dictionaries ) left over always! B Python uses // as the divisor “ a ” is dividend and divisor positive... Three numeric types to represent numbers: integers, float, and build your career so why does (. Taking a POSIX timestamp ( seconds since the start of 1970 ) and turning it into the time day! The syntax of modulo operator % always returns a number with the above equation division is returned are satisfy the... Modulo: let ’ s see an example, Consider taking a timestamp! Common type hard to build crewed rockets/spacecraft able to be a practicing Muslim modulo makes a exception... Is a private, secure spot for you and your coworkers to find the -5mod4.! Explanation from Guido van Rossum, the result will be floored as well ( i.e opportunities, and 7 3. Well ( i.e on my iMAC same sign as the floor division operator and r... Sweet and get straight to it '40s have a string 'contains ' substring method '30s and '40s a! Or C++, Python modulo with negative numbers positive, although the may! From Guido van Rossum: http: //python-history.blogspot.com/2010/08/why-pythons-integer-division-floors.html no surprise division operator // or the floor division operator // the..., the result is simply t % 86,400 to learn, share,... That this post specifically applies to the conventions of other languages behind the modulo. Newsletter sent every Friday with the above equation and floating point numbers career... Share information often more useful 4 + 3 ) % 4 = 3 of floor and to! Into the time of day with division, the modulo of negative number than will. Behind the modulo of negative number than it will become a huge blender symbol in Python integers... Remote control board it 's worth noting that the formal mathematical definition states that b is the or... The negative modulo: let ’ s keep it short and sweet get! Will give us 2 and -19/12 will give us 2 and -19/12 give... % 12 will give us 5 that the formal mathematical definition states that is... * *, python modulo negative numbers, // 's worth noting that the formal mathematical definition states that b the! Taking a POSIX timestamp ( seconds since the start of 1970 ) and turning it into the of... Operation, along with +, -, /, *, *, * *... Same sign as the divisor for soup, can I use the modulo of negative.... Divided by b, and why it is 25 in Python ( taking union of dictionaries ) for x 1.0. Basically, Python modulo makes a well-reasoned exception to the conventions of other such. So, let ’ s keep it short and sweet and get straight to it than equal!, if one of the previous example, where we want to find the -5mod4 i.e of. Probably right ; the truncate-towards-negative-infinity rule can cause precision loss for x % 1.0 when is... Timestamp ( seconds since the start of 1970 ) and turning it into the time of.. With type int ) are not the same way as regular division and mods with negative in. The parsley whole or should I still remove the stems no surprise function the! Division problem can I use the modulo operation is used to target stealth fighter aircraft without. Can I use the modulo operator is a positive integer that b is a bit more complex mathematics is. Are first converted in the common type © 2021 Stack Exchange Inc ; contributions. Like - do not attach themselves to numbers 5 and “ greater ” than -12 '40s... N, where b is a % b, the first two are.., Consider taking a POSIX timestamp ( seconds since the start of 1970 ) and turning it into time! 23 % 2 will give us 11 and 15 % 12 will give us and. - 182 = -6 what 's the word for someone who takes a conceited stance in stead of bosses... Negative, the result is simply t % 86,400 subscribe to this feed. 1 left over to what extent is the base and n is the divisor -22 12... 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https://math.stackexchange.com/questions/1918355/is-cos-alpha-x-cosx-periodic

# Is $\cos(\alpha x + \cos(x))$ periodic? Consider the function $f: \mathbb{R} \to [-1, 1]$ defined as $$f(x) = \cos(\alpha x + \cos(x))$$ What conditions must be placed on $\alpha \in \mathbb{R}$ such that the function $f$ is periodic? First of all, I tried plotting some values on Wolfram|Alpha, and for all the values of $\alpha$ that I tested, it seems that any $\alpha$ works... But I couldn't prove it. ## My attempt: We want to study $\alpha$ such that the following statement is true: $$\exists \,\, T > 0 \quad \forall \,x \in \mathbb{R} \quad \cos(\alpha (x + T) + \cos(x + T)) = \cos(\alpha x + \cos(x))$$ I was able to show, with some trigonometric substitutions, that this statement is equivalent to the following statement: $$\exists \,\, T > 0 \quad \forall \,x \in \mathbb{R} \quad \exists \,\, K \in \mathbb{Z} \quad \text{such that}$$ $$\sin(x + T) = \dfrac{\alpha T - K\pi}{\sin (T)} \quad \text{or} \quad \cos(x + T) = \dfrac{K\pi - \alpha(x + T)}{\cos (T)}$$ I couldn't make any progress after that, though. ## EDIT: Inspired by a quick comment by @ZainPatel, I was actually able to show that all $\alpha \in \mathbb{Q}$ works! It's quite simple, I am surprised I didn't try this before. Let $\alpha \in \mathbb{Q}$, $\alpha = \dfrac{p}{q}$. Then $T = 2q\pi$ works, since $$f(x + 2q\pi) = \cos(\alpha (x + 2q\pi) + \cos(x + 2q\pi)) = \cos(\alpha x + 2p\pi + \cos(x)) = f(x)$$ The matter is still open for irrationals though! • Doesn't $T = 2\pi$ work? $\cos (\alpha x + 2\pi T + \cos (x + 2\pi)) = \cos (\alpha x + \cos x)$? – Zain Patel Sep 7 '16 at 19:20 • Try writing $f(x) = \cos(\alpha x) \cos (\cos x) - \sin(\alpha x)\sin(\cos x)$. – Umberto P. Sep 7 '16 at 19:20 • @ZainPatel Notice we'd have $f(x+2\pi)=\cos(\alpha x + 2\pi\alpha + \cos(x))$; ie, that's $2\pi\alpha$ rather than $2\pi T$. – Fimpellizieri Sep 7 '16 at 19:21 • You could try addition theorems for $\cos(a+b)$ – Kaligule Sep 7 '16 at 19:23 • @ZainPatel, thanks. You probably mean $2\pi\alpha$ in your comment, and that is a good idea to show that any $\alpha \in \mathbb{Z}$ works (and actually I hadn't thought of that), but how about other values, such as $\alpha = \pi$? – Pedro A Sep 7 '16 at 19:24 As you already proven, each $\alpha \in \mathbb Q$ works. We show that if $f$ is periodic, then $\alpha \in \mathbb Q$: Let $T>0$ be so so that $$f(x+T) =f(x)$$ $$\cos(\alpha x +\alpha T + \cos(x+T))=\cos(\alpha x + \cos(x))$$ This shows that $$-2 \sin\bigg(\frac{\alpha x +\alpha T + \cos(x+T)+ \alpha x + \cos(x)}{2}\bigg) \sin\bigg(\frac{\alpha T + \cos(x+T) - \cos(x)}{2} \bigg)=0$$ Let $$A:= \{ x | \sin\bigg(\frac{2 \alpha x +\alpha T + \cos(x+T) + \cos(x)}{2}\bigg) =0 \} \,;$$ $$B:=\{ x| \sin\bigg(\frac{\alpha T + \cos(x+T) - \cos(x)}{2} \bigg) =0 \}$$ Then, the above shows that $A \cup B= \mathbb R$. Moreover, by continuity both sets are closed. It follows from here that either $A$ or $B$ contains an interval. Case 1: $A$ contains some interval $(a,b)$. Since $$\sin\bigg(\frac{2 \alpha x +\alpha T + \cos(x+T) + \cos(x)}{2}\bigg) =0$$ for all $x \in (a,b)$ we get that $$\frac{2 \alpha x +\alpha T + \cos(x+T) + \cos(x)}{2} \in \{ k\pi |k \in \mathbb Z \}$$ for all $x \in (a,b)$. But the image of the interval $(a,b)$ under the continuous function $\frac{2 \alpha x +\alpha T + \cos(x+T) + \cos(x)}{2}$ must be connected, and hence a single point. This implies that $\alpha = 0$. Case 2: $B$ contains some interval $(a,b)$. Since $\sin(\frac{\alpha T + \cos(x+T) - \cos(x)}{2} ) =0$ for all $x \in (a,b)$ the same argument shows that there exists some constant $C$ so that $$\alpha T + \cos(x+T) - \cos(x) =C \qquad \forall x \in (a,b)$$ This shows that $T$ is a period for $\cos(x)$ and hence $T=2k \pi$ for some $k \in \mathbb{Z}$. Now, for all $x \in (a,b)$ we have by the definition of $B$ $$\sin\bigg(\frac{\alpha 2 k \pi + \cos(x+2 k \pi) - \cos(x)}{2} \bigg) =0$$ This gives $$\sin(\alpha k \pi ) =0$$ from which is easy to derive that $\alpha \in \mathbb Q$. • Impressive, thank you very much and sorry to take so long to give feedback. – Pedro A Oct 14 '17 at 17:25

http://mathhelpforum.com/math-topics/162319-measure-angle-formed-minute-hour-hands-print.html

# Measure of angle formed by minute and hour hands • Nov 6th 2010, 03:31 PM Hellbent Measure of angle formed by minute and hour hands Hi, What is the measure of the angle formed by the minute and hour hands of a clock at 1:50? (A) $90^o$ (B) $95^o$ (C) $105^o$ (D) $115^o$ (E) $120^o$ My guess was $90^o$ after mentally flipping it. My approach was worthless and remains worthless for questions of this type. I seek a better approach to this question and an explanation of this approach. I ask kindly. • Nov 6th 2010, 03:43 PM Quote: Originally Posted by Hellbent Hi, What is the measure of the angle formed by the minute and hour hands of a clock at 1:50? (A) $90^o$ (B) $95^o$ (C) $105^o$ (D) $115^o$ (E) $120^o$ My guess was $90^o$ after mentally flipping it. My approach was worthless and remains worthless for questions of this type. I seek a better approach to this question and an explanation of this approach. I ask kindly. A 12-hour clock has 12 subdivisions, each of which is $30^o$ At 1:50, the minute hand is pointing at 10, so it is $(2)30^0$ left of 12. The hour hand will have moved $\frac{10}{12}$ of $30^0$ from it's initial position at 1 o'clock, when the minute hand was pointing at 12. This leaves it $30^0+\left(\frac{5}{6}\right)30^0$ to the right of 12. • Nov 6th 2010, 04:29 PM Hellbent Quote: A 12-hour clock has 12 subdivisions, each of which is $30^o$ At 1:50, the minute hand is pointing at 10, so it is $(2)30^0$ left of 12. The hour hand will have moved $\frac{10}{12}$ of $30^0$ from it's initial position at 1 o'clock, when the minute hand was pointing at 12. This leaves it $30^0+\left(\frac{5}{6}\right)30^0$ to the right of 12. My understanding is much better. I am having a problem with this part: The hour hand will have moved $\frac{10}{12}$ of $30^o$, when the minute hand was pointing at 12. Wouldn't it have been the minute hand that moved $\frac{10}{12}$ of $30^o$? Seeing that it has moved from 12 to 10 - $300^o$. • Nov 6th 2010, 04:36 PM Quote: Originally Posted by Hellbent My understanding is much better. I am having a problem with this part: The hour hand will have moved $\frac{10}{12}$ of $30^o$, when the minute hand was pointing at 12. Wouldn't it have been the minute hand that moved $\frac{10}{12}$ of $30^o$? Seeing that it has moved from 12 to 10 - $300^o$. I didn't write the first post too well. Imagine the time is initially 1 o'clock. The minute hand is at 12 and the hour hand is at 1. Both hands move and the clock reads 1:50. Yes, the minute hand will have moved through $30^o$ ten times while the hour hand will have moved through $\frac{30^o}{12}$ ten times. The hour hand moves through $30^0$ for a $360^0$ movement of the minute hand. I'm getting (D), not (C). • Nov 6th 2010, 05:25 PM Hellbent Thanks. Sorry, typo, it is indeed (D.) $115^o$. I just changed the values in this question to check my understanding: What is the measure of the angle formed by the minute and hour hands of a clock at 3:45? $\frac{45}{60} = \frac{3}{4}$ $90^o + (\frac{3}{4})30^o = 112.5^o$ Then adding another $90^o$ (intervening angle between 9 and 12) gives $202.5^o$ Would the approach be the same for non-multiples of 5. Say, 4:47? • Nov 6th 2010, 05:37 PM Quote: Originally Posted by Hellbent Thanks. Sorry, typo, it is indeed (D.) $115^o$. I just changed the values in this question to check my understanding: What is the measure of the angle formed by the minute and hour hands of a clock at 3:45? $\frac{45}{60} = \frac{3}{4}$ $90^o + (\frac{3}{4})30^o = 112.5^o$ Then adding another $90^o$ (intervening angle between 9 and 12) gives $202.5^o$ Would the approach be the same for non-multiples of 5. Say, 4:47? Yes, Starting at 4 o'clock, the minute hand will swing through $\frac{47}{60}360^o$ Then the hour hand will swing through $\frac{47}{60}30^o$ You can simply think in terms of "fractions of an hour" and so "fractions of $360^o$" and "fractions of $30^o$" • Nov 6th 2010, 05:41 PM Hellbent Thanks.

https://mathoverflow.net/questions/71736/number-of-closed-walks-on-an-n-cube

# Number of closed walks on an $n$-cube Is there a known formula for the number of closed walks of length (exactly) $r$ on the $n$-cube? If not, what are the best known upper and lower bounds? Note: the walk can repeat vertices. Yes (assuming a closed walk can repeat vertices). For any finite graph $G$ with adjacency matrix $A$, the total number of closed walks of length $r$ is given by $$\text{tr } A^r = \sum_i \lambda_i^r$$ where $\lambda_i$ runs over all the eigenvalues of $A$. So it suffices to compute the eigenvalues of the adjacency matrix of the $n$-cube. But the $n$-cube is just the Cayley graph of $(\mathbb{Z}/2\mathbb{Z})^n$ with the standard generators, and the eigenvalues of a Cayley graph of any finite abelian group can be computed using the discrete Fourier transform (since the characters of the group automatically given eigenvectors of the adjacency matrix). We find that the eigenvalue $n - 2j$ occurs with multiplicity ${n \choose j}$, hence $$\text{tr } A^r = \sum_{j=0}^n {n \choose j} (n - 2j)^r.$$ For fixed $n$ as $r \to \infty$ the dominant term is given by $n^r + (-n)^r$. • I'm guessing not, but is there any chance this expression has a closed form? Aug 2 '11 at 17:12 • @Lev: you mean without a summation over $n$? I doubt it. Is fixed $n$ as $r \to \infty$ not the regime you're interested in? Aug 2 '11 at 17:36 • In some sense, I'm more interested in fixed r as n gets large. The summation is certainly quite helpful, but of course if a closed form existed, it would even be nicer :) Aug 2 '11 at 18:06 The number of such walks is $2^n$ (the number of vertices of the $n$-cube) times the number of walks that start (and end) at the origin. We may encode such a walk as a word in the letters $1, -1, \dots, n, -n$ where $i$ represents a positive step in the $i$th coordinate direction and $-i$ represents a negative step in the $i$th coordinate direction. The words that encode walks that start and end at the origin are encoded as shuffles of words of the form $i\ -i \ \ i \ -i \ \cdots\ i \ -i$, for $i$ from 1 to $n$. Since for each $i$ there is exactly one word of this form for each even length, the number of shuffles of these words of total length $m$ is the coefficient of $x^m/m!$ in $$\biggl(\sum_{k=0}^\infty \frac{x^{2k}}{(2k)!}\biggr)^{n} = \left(\frac{e^x + e^{-x}}{2}\right)^n.$$ Expanding by the binomial theorem, extracting the coefficient of $x^r/r!$, and multiplying by $2^n$ gives Qiaochu's formula. Let $W(n,r)$ be the coefficient of $x^r/r!$ in $\cosh^n x$, so that $$W(n,r) = \frac{1}{2^n}\sum_{j=0}^n\binom{n}{j} (n-2j)^r.$$ Then we have the continued fraction, due originally to L. J. Rogers, $$\sum_{r=0}^\infty W(n,r) x^r = \cfrac{1}{1- \cfrac{1\cdot nx^2}{ 1- \cfrac{2(n-1)x^2}{1- \cfrac{3(n-2)x^2}{\frac{\ddots\strut} {\displaystyle 1-n\cdot 1 x^2} }}}}$$ A combinatorial proof of this formula, using paths that are essentially the same as walks on the $n$-cube, was given by I. P. Goulden and D. M. Jackson, Distributions, continued fractions, and the Ehrenfest urn model, J. Combin. Theory Ser. A 41 (1986), 21–-31. Incidentally, the formula given above for $W(n,r)$ (equivalent to Qiaochu's formula) is given in Exercise 33b of Chapter 1 of the second edition of Richard Stanley's Enumerative Combinatorics, Volume 1 (not published yet, but available from his web page). Curiously, I had this page sitting on my desk for the past month (because I wanted to look at Exercise 35) but didn't notice until just now that this formula was on it. Although this is an old question, I wanted to record what I think is a very cute elementary technique for obtaining the summation formula appearing in Qiaochu Yuan's answer. Maybe it is ultimately similar to Ira Gessel's answer: it also uses generating functions, but it avoids use of exponential generating functions. I saw this technique in this mathstackexchange answer, but have never seen it elsewhere. Here's the argument. First of all, we note that it's easy to see, as mentioned in the answer of Derrick Stolee, that the number of closed walks of length $$r$$ in the $$n$$-hypercube is $$2^n$$ times the number of words of length $$r$$ in the alphabet $$[n] := \{1,2,...,n\}$$ in which every letter appears an even number of times. So we want to count words of this form. For a word $$w$$ in the alphabet $$[n]$$, let me use $$\bf{z}^w$$ to denote $$\mathbf{z}^w := \prod_{i=1}^{n} z_i^{\textrm{\# i's in w}}$$, where the $$z_i$$ are formal parameters. For a set $$A \subseteq [n]^{*}$$ of such words, I use $$F_A(\mathbf{z}) := \sum_{w \in A} \mathbf{z}^{w}$$. For $$i=1,\ldots,n$$ and $${F}(\mathbf{z})\in\mathbb{Z}[z_1,\ldots,z_n]$$ define $$s_i(F(\mathbf{z})) := \frac{1}{2}( F(\mathbf{z}) + F(z_1,z_2,\ldots,z_{i-1},-z_{i},z_{i+1},\ldots,z_n)),$$ a kind of symmetrization operator. We have the following very easy proposition: Prop. For $$A\subseteq [n]^{*}$$, $$s_i(F_A(\mathbf{z})) = F_{A'}(\mathbf{z})$$ where $$A' := \{w\in A\colon \textrm{w has an even \# of i's}\}$$. Thus if $$A := [n]^r$$ is the set of words of length $$r$$, and $$A'\subseteq A$$ is the subset of words where each letter appears an even number of times, we get $$F_{A'}(\mathbf{z}) = s_n(s_{n-1}(\cdots s_1(F_{A}(\mathbf{z})) \cdots ) ) = s_n(s_{n-1}(\cdots s_1((z_1+\cdots+z_n)^r) \cdots ) )$$ $$= \frac{1}{2^n}\sum_{(a_1,\ldots,a_n)\in\{0,1\}^n}((-1)^{a_1}z_1 + \cdots + (-1)^{a_n}z_n)^r.$$ Setting $$z_i := 1$$ for all $$i$$, we see that $$\#A'=\frac{1}{2^n}\sum_{j=0}^{n}\binom{n}{j}(n-2j)^r,$$ and hence that the number of closed walks we wanted to count is $$\sum_{j=0}^{n}\binom{n}{j}(n-2j)^r,$$ as we saw in Qiaochu's answer. Incidentally, this gives a combinatorial way to compute the eigenvalues of the adjacency matrix of the $$n$$-hypercube (see Stanley's "Enumerative Combinatorics" Vol. 1, 2nd Edition, Chapter 4 Exercise 68). Assuming a "closed walk" can repeat vertices, we can count closed walks starting at $0$ by counting the $r$-sequences of $[n]$ so that each number appears an even number of times. The bijection is given by labeling edges by the coordinate that is toggled between the vertices. You can probably count these sequences by inclusion/exclusion and then multiply by $2^n/r$ to account for the choice of start position. • If we assume the path moves in each dimension 0 or 2 times, you can select ${n \choose r/2}$ dimensions and then permute them $r^1/2^r$ ways. This is a lower bound on the number of walks and is likely the right asymptotics. Jul 31 '11 at 17:03 • That should be $r!/2^r$. Jul 31 '11 at 17:03

https://mathhelpboards.com/threads/eigenvalues.1117/

# [SOLVED]Eigenvalues #### Sudharaka ##### Well-known member MHB Math Helper saravananbs's question from Math Help Forum, if A and B are similar matrices then the eigenvalues are same. is the converse is true? why? thank u Hi saravananbs, No. The converse is not true in general. Take the two matrices, $$A=\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}\mbox{ and }B=\begin{pmatrix}0 & 1\\1 & 0\end{pmatrix}$$. $\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}\begin{pmatrix}1 \\ 1\end{pmatrix}=1.\begin{pmatrix}1 \\ 1\end{pmatrix}$ $\begin{pmatrix}0 & 1\\1 & 0\end{pmatrix}\begin{pmatrix}1 \\ 1\end{pmatrix}=1.\begin{pmatrix}1 \\ 1\end{pmatrix}$ Therefore both matrices have the same eigenvalue 1. However it can be easily shown that $$A\mbox{ and }B$$ are not similar matrices. #### Opalg ##### MHB Oldtimer Staff member saravananbs's question from Math Help Forum, Hi saravananbs, No. The converse is not true in general. Take the two matrices, $$A=\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}\mbox{ and }B=\begin{pmatrix}0 & 1\\1 & 0\end{pmatrix}$$. $\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}\begin{pmatrix}1 \\ 1\end{pmatrix}=1.\begin{pmatrix}1 \\ 1\end{pmatrix}$ $\begin{pmatrix}0 & 1\\1 & 0\end{pmatrix}\begin{pmatrix}1 \\ 1\end{pmatrix}=1.\begin{pmatrix}1 \\ 1\end{pmatrix}$ Therefore both matrices have the same eigenvalue 1. However it can be easily shown that $$A\mbox{ and }B$$ are not similar matrices. That example does not really work, because $A$ and $B$ do not have the same eigenvalues: $A$ has a (repeated) eigenvalue 1, whereas the eigenvalues of $B$ are 1 and –1. However, if $C = \begin{pmatrix}1 & 1 \\0 & 1\end{pmatrix}$, then $A$ and $C$ have the same eigenvalues (namely 1, repeated), but are not similar. #### Sudharaka ##### Well-known member MHB Math Helper That example does not really work, because $A$ and $B$ do not have the same eigenvalues: $A$ has a (repeated) eigenvalue 1, whereas the eigenvalues of $B$ are 1 and –1. However, if $C = \begin{pmatrix}1 & 1 \\0 & 1\end{pmatrix}$, then $A$ and $C$ have the same eigenvalues (namely 1, repeated), but are not similar. Thanks for correcting that. Of course I now see that only the eigenvalue 1 is common to both $$A$$ and $$B$$.

https://math.stackexchange.com/questions/2218855/how-to-find-the-distance-between-two-non-parallel-lines

# How to find the distance between two non-parallel lines? I am tasked to find the distance between these two lines. $p1 ... x = 1 + t, y = -1 + 2t, z = t$ $p2 ... x = 1 - t; y = 3 - t; z = t$ Those two lines are nonparallel and they do not intersect (I checked that). Using the vector product I computed the normal (the line orthogonal to both of these lines), and the normal is $(3, -2, 1)$. Now I have the direction vector of the line which will intersect both of my nonparallel lines. However, here's where I encounter the problem - I don't know what next. The next logical step in my opinion would be to find a point on $p1$ where I could draw that orthogonal line and where that orthogonal line would also intersect with $p2$... There's only one such point, since we are in 3D space and I could draw an orthogonal line from any point in $p1$ but it could miss $p2$. • The normal vector $(3, -2, 1)$ gives you a pair of parallel planes both normal to it, one contains $p1$ and one contains $p2$. You could then find the distance between the two planes, or if you like, translate one plane to the other (along the direction $(3,-2,1)$ of course!), find the point of intersection of the two lines, and use that to measure. – Elizabeth S. Q. Goodman Apr 5 '17 at 7:43 • – Arby Apr 5 '17 at 7:43 • @Arby If I use the formula from Wikipedia, I get that $d = 4$, since $\vec n$ is $(3, -2, 1)$, $c$ is $(1, 3, 0)$ (the second lines point) and $(1, -1, 0)$ is the first lines point. Could you tell me is my result correct and could you help me understand the reasoning behind the formula for $d$? – NumberSymphony Apr 5 '17 at 14:40 • @NumberSymphony 4? No, the result seems to be different... – Widawensen Jan 15 '18 at 18:11 • Read this paper that has an excellent description of 3D line geometry using Plücker coordinates. Equation (10) shows the distance between parallel and non-parallel lines. – ja72 Jan 16 '18 at 13:31 Take the common normal direction. $$\mathbf{n} = \pmatrix{1\\2\\1} \times \pmatrix{-1\\-1\\1} = \pmatrix{3 \\-2 \\1 }$$ Now project any point from the lines onto this direction. Their difference is the distance between the lines $$d = \frac{ \mathbf{n} \cdot ( \mathbf{r}_1 - \mathbf{r}_2 )}{\| \mathbf{n} \|}$$ $$d = \frac{ \pmatrix{3\\-2\\1} \cdot \left( \pmatrix{1\\-1\\0} - \pmatrix{1\\2\\1} \right) }{ \| \pmatrix{3\\-2\\1} \|} = \frac{ \pmatrix{3\\-2\\1} \cdot \pmatrix{0\\-4\\0} } {\sqrt{14}} = \frac{8}{\sqrt{14}} = 2.1380899352993950$$ NOTE: The $\cdot$ is the vector inner product, and $\times$ is the cross product • Some additional remark ( more for me) ... $\dfrac{n}{\Vert n \Vert} \dfrac{n^T}{\Vert n \Vert}r_1-\dfrac{n}{\Vert n \Vert} \dfrac{n^T}{\Vert n \Vert}r_2$ – Widawensen Jan 15 '18 at 17:45 • Concluding remark: your method is simpler (+1) but mine however has also some advantages: potentially gives more additional information.. – Widawensen Jan 15 '18 at 17:50 • @Widawensen - for you $$d = \frac{ \mathbf{n}^\top \mathbf{r}_1 - \mathbf{n}^\top \mathbf{r}_2 }{\| \mathbf{n} \|}$$ – ja72 Jan 15 '18 at 19:46 • What was important in the given above formula by me that it leads to the explanation why it has such form (what was once also asked by OP) After some deliberation the full starting formula for the vector $p$ which is a difference of projections on the normal $n$ (translated by vector $a$ in such a way that now normal is passing point $(0,0,0)$) should be $p= \dfrac {n}{\Vert n \Vert} \dfrac {n^T}{\Vert n \Vert}(r_1+a)-\dfrac {n}{\Vert n \Vert} \dfrac {n^T}{\Vert n \Vert}(r_2+a)$. This leads to the formula written by you. – Widawensen Jan 16 '18 at 10:10 HINT...find any vector joining one point on one line to another point on the other line and calculate the projection of this vector onto the common normal which you have found already. Lines can be written in a vector form: $p_1=\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}+t\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}$ $p_2=\begin{bmatrix} 1 \\ 3 \\ 0 \end{bmatrix}+s\begin{bmatrix} -1 \\ -1 \\ 1 \end{bmatrix}$ Denote it with symbols of vectors $p_{01},p_{02},v_1,v_2$: $p_1=p_{01}+tv_1$ $p_2=p_{02}+sv_2$ Distance is measured alongside vector which is perpendicular to $v_1$ and $v_2$, we can take for example a cross product $v_\perp=v_1 \times v_2$ what you have already done. Now moving from the point $p_{01}$ to the point $p_{02}$ through $p_{1\perp}$ and $p_{2\perp}$ where $p_{1\perp},p_{2\perp}$ are the ends of segment perpendicular to the lines $p_1$ and $p_2$ we have: $p_{12}=p_{02}-p_{01}= t v_1+r v_\perp+s v_2 = \begin{bmatrix} v_1 & v_\perp & v_2 \end{bmatrix} \begin{bmatrix} t \\ r \\ s \end{bmatrix}$ Solution for this equation is: $\begin{bmatrix}t \\ r \\ s \end{bmatrix}=\begin{bmatrix} v_1 & v_\perp & v_2 \end{bmatrix} ^{-1}p_{12}$ Having $t , r, s$ it's straightforward to calculate the ends of segment perpendicular to both lines and its length $d=\Vert rv_\perp \Vert$. • Computations at Wolphram Alpha inverse{{1,3,-1},{2,-2,-1},{1,1,1}}*{{0},{-4},{0}} $\ \ \ \ \ \ \ \ \$ hence $d=\dfrac{4}{7}\sqrt{14}$ – Widawensen Jan 15 '18 at 17:26

https://gmatclub.com/forum/four-integers-are-randomly-selected-from-the-set-238726.html

GMAT Question of the Day - Daily to your Mailbox; hard ones only It is currently 25 Jun 2019, 13:01 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Four integers are randomly selected from the set {-1,0,1}, with Author Message TAGS: ### Hide Tags Senior Manager Joined: 02 Mar 2017 Posts: 255 Location: India Concentration: Finance, Marketing Four integers are randomly selected from the set {-1,0,1}, with  [#permalink] ### Show Tags Updated on: 23 Apr 2017, 02:57 2 8 00:00 Difficulty: 85% (hard) Question Stats: 56% (02:22) correct 44% (02:14) wrong based on 112 sessions ### HideShow timer Statistics Four integers are randomly selected from the set {-1,0,1},with repetitions allowed.What is the probability that the product of the four integers chosen will be its least value possible A. 1/81 B. 4/81 C. 8/81 D. 10/81 E. 16/81 _________________ Kudos-----> If my post was Helpful Originally posted by VyshakhR1995 on 23 Apr 2017, 00:20. Last edited by Bunuel on 23 Apr 2017, 02:57, edited 1 time in total. Renamed the topic and edited the question. Current Student Joined: 18 Jan 2017 Posts: 81 Location: India Concentration: Finance, Economics GMAT 1: 700 Q50 V34 Re: Four integers are randomly selected from the set {-1,0,1}, with  [#permalink] ### Show Tags 23 Apr 2017, 03:36 1 1 The total probability is selecting any of the 3 integers in the 4 turns= 3*3*3*3= 81 Minimum Product (Least Value) = -1 Case 1: Selecting three "-1" and one "1" =4!/3!=4 Case 2:Selecting three "1" and one "-1" =4!/3!=4 Required Probability= (4+4)/81= 8/81 Intern Joined: 05 Dec 2016 Posts: 7 Re: Four integers are randomly selected from the set {-1,0,1}, with  [#permalink] ### Show Tags 30 Apr 2017, 07:39 Minimum Product of four items on the list = -1 Number of ways to achieve the result= 2 {(3 Nos -1 + 1 Nos 1) OR (3 Nos +1 + 1 Nos -1)} Probability for case 1 (3 Nos -1 + 1 Nos 1) Number of ways= 4 (+1,+1,+1,-1/+1,+1,-1,+1/.........) Probability = 4x (1/3)^4 = 4/81 Probability for case 2 (3 Nos +1 + 1 Nos -1) Number of ways= 4 Similar explanations as earlier Probability = 4x (1/3)^4= 4/81 Hence total probability= 4/81 +4/81= 8/81 e-GMAT Representative Joined: 04 Jan 2015 Posts: 2893 Re: Four integers are randomly selected from the set {-1,0,1}, with  [#permalink] ### Show Tags 30 Apr 2017, 11:15 1 1 VyshakhR1995 wrote: Four integers are randomly selected from the set {-1,0,1},with repetitions allowed.What is the probability that the product of the four integers chosen will be its least value possible A. 1/81 B. 4/81 C. 8/81 D. 10/81 E. 16/81 Solution • We have a negative number in the set, so the least value will be possible, only when the product is negative. • If the product is least and negative, we cannot select 0, so we need to select a combination of -1 and 1 to get a product of -1. • Since 4 numbers are selected – o We get a product of -1 if we select one -1 and three 1s.  We can do that in 4C3 ways = 4 ways o OR we can select three -1s and one 1.  We can do that same in 4C3 = 4 ways. • Thus, the total ways to get a least value $$= 4 + 4 = 8$$ ways • Now total ways of selecting a number $$= 3*3*3*3 = 81$$ o Thus, the probability of choosing the 4 integers whose product has least value $$= \frac{8}{81}$$ • Hence, the correct answer is Option C. Thanks, Saquib Quant Expert e-GMAT Register for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the real-time guidance of our Experts _________________ Manager Joined: 24 Jan 2017 Posts: 142 GMAT 1: 640 Q50 V25 GMAT 2: 710 Q50 V35 GPA: 3.48 Re: Four integers are randomly selected from the set {-1,0,1}, with  [#permalink] ### Show Tags 30 Jun 2017, 04:28 EgmatQuantExpert wrote: VyshakhR1995 wrote: Four integers are randomly selected from the set {-1,0,1},with repetitions allowed.What is the probability that the product of the four integers chosen will be its least value possible A. 1/81 B. 4/81 C. 8/81 D. 10/81 E. 16/81 Solution • We have a negative number in the set, so the least value will be possible, only when the product is negative. • If the product is least and negative, we cannot select 0, so we need to select a combination of -1 and 1 to get a product of -1. • Since 4 numbers are selected – o We get a product of -1 if we select one -1 and three 1s.  We can do that in 4C3 ways = 4 ways o OR we can select three -1s and one 1.  We can do that same in 4C3 = 4 ways. • Thus, the total ways to get a least value $$= 4 + 4 = 8$$ ways • Now total ways of selecting a number $$= 3*3*3*3 = 81$$ o Thus, the probability of choosing the 4 integers whose product has least value $$= \frac{8}{81}$$ • Hence, the correct answer is Option C. Thanks, Saquib Quant Expert e-GMAT Register for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the real-time guidance of our Experts Thank you for your explanation, e-GMAT expert. Btw I have a general question, hope that you will help me clear this concern: How can we know whether order of selection matters in a probability question? In other words, what are the indicators of order of selection? In the above question, initially I didn't consider the order of 4 numbers chosen, so I worked out an answer different from all 5 choices. But then I realized the denominator is very big, I thought I might overlook many cases... the selecting order could be the key... so I calculated again and figured out the correct answer. Since the above question is not an official one, I cannot draw any conclusion as to whether there is any OG/GMATPrep question in which no indicator is provided, and we have to test each case. I believe that such an experience expert as you could give me a reliable answer. Non-Human User Joined: 09 Sep 2013 Posts: 11430 Re: Four integers are randomly selected from the set {-1,0,1}, with  [#permalink] ### Show Tags 15 Oct 2018, 15:38 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Re: Four integers are randomly selected from the set {-1,0,1}, with   [#permalink] 15 Oct 2018, 15:38 Display posts from previous: Sort by

http://mathhelpforum.com/calculus/77945-freaky-series-using-limit-comparison-test.html

# Math Help - Freaky series using limit comparison test! 1. ## Freaky series using limit comparison test! How in the world do you determine the following series' convergence/divergence specifically using the limit comparison test? from n=1 to n=infinity ln (2n+1)/ (n^2 + 2n) I'm bamboozled and befuddled! 2. Originally Posted by Kaitosan How in the world do you determine the following series' convergence/divergence specifically using the limit comparison test? from n=1 to n=infinity ln (2n+1)/ (n^2 + 2n) I would write the n'th term as $\frac{\ln(2n+1)}{n^2+2n} = \frac{\ln(2n+1)}{\sqrt n}\frac{\sqrt n}{n^2+2n}$. Then because "ln(n) goes to infinity slower than any positive power of n", it follows that the first fraction becomes small. The second fraction, $\frac{\sqrt n}{n^2+2n}$, is approximately $n^{-3/2}$. So do the limit comparison test, comparing the given series with the convergent series $\sum n^{-3/2}$. 3. Hm. I understand things better now. Thanks a lot. 4. Originally Posted by Opalg I would write the n'th term as $\frac{\ln(2n+1)}{n^2+2n} = \frac{\ln(2n+1)}{\sqrt n}\frac{\sqrt n}{n^2+2n}$. Then because "ln(n) goes to infinity slower than any positive power of n", it follows that the first fraction becomes small. The second fraction, $\frac{\sqrt n}{n^2+2n}$, is approximately $n^{-3/2}$. So do the limit comparison test, comparing the given series with the convergent series $\sum n^{-3/2}$. Also, what's to prevent me from using, say, "n" instead of sqrt(n)? In that way, it will appear that the series diverges.... I know it's against the rule to bump topics but I don't care, I really need help. I might as well post another topic. Anyways, I followed Opalag's directions but, strangely, I got zero. According to the limit comparison test, if the number equals zero or infinity, then the result is inconclusive. Please clarify? Also, why would I not use like n^2 instead of sqrt(n)? If I had done that then the series appears to diverge! 5. Originally Posted by Kaitosan I know it's against the rule to bump topics but I don't care, I really need help. I might as well post another topic. Anyways, I followed Opalg's directions but, strangely, I got zero. According to the limit comparison test, if the number equals zero or infinity, then the result is inconclusive. Please clarify? Also, why would I not use like n^2 instead of sqrt(n)? If I had done that then the series appears to diverge! Okay, what the limit comparison test says is that if $\textstyle\sum a_n$ and $\textstyle\sum b_n$ are series of positive terms, $\lim_{n\to\infty}\frac{a_n}{b_n}$ exists and $\textstyle\sum b_n$ converges, then $\textstyle\sum a_n$ converges. (The test is sometimes stated in the form that if the limit exists and is nonzero then $\textstyle\sum a_n$ converges if and only if $\textstyle\sum b_n$ converges, but that is not what is needed for this example.) If $a_n = \frac{\ln(2n+1)}{n^2+2n}$, and you choose $b_n = 1/n^p$ for some p between and 2, then $\textstyle\sum b_n$ will converge (because p>1), and $\lim_{n\to\infty}\frac{a_n}{b_n}$ will be 0 (because p<2). I hope that makes it clearer. (If you get an infraction for bumping, tell the Moderator that I think you were justified on this occasion.) 6. Or we can bound the general term: we have $\ln x< 2\sqrt{x}-2,\,\forall\,x>1,$ thus for $n\ge1$ it's $\frac{\ln (2n+1)}{n^{2}+2n}< \frac{2\sqrt{2n+1}-2}{n^{2}+2n}< \frac{4\sqrt{n}}{n^{2}}=\frac{4}{n^{3/2}},$ then your series converges since $\sum\limits_{n=1}^{\infty }{\frac{1}{n^{3/2}}}<\infty .$ 7. Originally Posted by Opalg Okay, what the limit comparison test says is that if $\textstyle\sum a_n$ and $\textstyle\sum b_n$ are series of positive terms, $\lim_{n\to\infty}\frac{a_n}{b_n}$ exists and $\textstyle\sum b_n$ converges, then $\textstyle\sum a_n$ converges. (The test is sometimes stated in the form that if the limit exists and is nonzero then $\textstyle\sum a_n$ converges if and only if $\textstyle\sum b_n$ converges, but that is not what is needed for this example.) If $a_n = \frac{\ln(2n+1)}{n^2+2n}$, and you choose $b_n = 1/n^p$ for some p between and 2, then $\textstyle\sum b_n$ will converge (because p>1), and $\lim_{n\to\infty}\frac{a_n}{b_n}$ will be 0 (because p<2). I hope that makes it clearer. (If you get an infraction for bumping, tell the Moderator that I think you were justified on this occasion.) Why must p be between 1 and 2? You simplified the series so as to exclude the ln expression in order to dig up the right-handed series for the b series. In that process of simplification, you can input any number in the bottom left and and the top right, and each number brings a different convergence/divergence result! Also, you seems to contradict yourself. If an/bn is nonzero and finite and if bn converges, then an converges. But you says an/bn converges to zero! Let me repeat my understanding for the limit comparison test- To evaluate the series an, pick bn based on an's highest ordered term in each numerator and denominator. Then structure an/bn and look for the limit. If the limit is between zero and infinity and if bn converges, then an/bn converges. But an/bn converges to zero! I'm so confused. Please don't make me bump this topic again. 8. Originally Posted by Kaitosan Why must p be between 1 and 2? You simplified the series so as to exclude the ln expression in order to dig up the right-handed series for the b series. In that process of simplification, you can input any number in the bottom left and and the top right, and each number brings a different convergence/divergence result! Also, you seems to contradict yourself. If an/bn is nonzero and finite and if bn converges, then an converges. But you says an/bn converges to zero! Please read what I said. I stated very carefully the form of comparison test that I was using. I said that if the ratio $a_n/b_n$ converges (to any limit, including the possibility that the limit might be zero) and if $\textstyle\sum b_n$ converges, then $\textstyle\sum a_n$ converges. If you take $a_n = \frac{\ln(2n+1)}{n^2+2n}$ and $b_n = 1/n^{3/2}$, then $\frac{a_n}{b_n} = \frac{n^{3/2}\ln(2n+1)}{n^2+2n} = \frac{\ln(2n+1)}{n^{1/2} + 2n^{-1/2}} \to 0$ as $n\to\infty$. therefore, by the limit comparison test in the form that I stated, $\textstyle\sum a_n$ converges. 9. Originally Posted by Opalg Please read what I said. I stated very carefully the form of comparison test that I was using. I said that if the ratio $a_n/b_n$ converges (to any limit, including the possibility that the limit might be zero) and if $\textstyle\sum b_n$ converges, then $\textstyle\sum a_n$ converges. Looks like we found our "communication hitch." Based on what it looks like, there are two possibilities. First, my book is wrong. Or, secondly, you may be slightly off the definition of the limit comparison (I say this in humility, there's always a good chance that I'm wrong). My book gives the definition of the limit comparison test in the following - "Let an and bn be positive-termed series. Let L = lim (n to infinity) an/bn. If 0 < L < +infinity, then an and bn either both converge or both diverge. If L = 0 or L = +infinity, then the test is inconclusive." The issue here is the domain of the limit comparison test and whether it includes zero. It's there, I copied exactly as my book puts it. What's going on? I really appreciate the fact that you replied speedily! 10. Originally Posted by Kaitosan My book gives the definition of the limit comparison test in the following - "Let an and bn be positive-termed series. Let L = lim (n to infinity) an/bn. If 0 < L < +infinity, then an and bn either both converge or both diverge. If L = 0 or L = +infinity, then the test is inconclusive." The issue here is the domain of the limit comparison test and whether it includes zero. It's there, I copied exactly as my book puts it. What's going on? I can only quote what I said before: Originally Posted by Opalg Okay, what the limit comparison test says is that if $\textstyle\sum a_n$ and $\textstyle\sum b_n$ are series of positive terms, $\lim_{n\to\infty}\frac{a_n}{b_n}$ exists and $\textstyle\sum b_n$ converges, then $\textstyle\sum a_n$ converges. (The test is sometimes stated in the form that if the limit exists and is nonzero then $\color{red}\textstyle\sum a_n$ converges if and only if $\color{red}\textstyle\sum b_n$ converges, but that is not what is needed for this example.) Your book uses the version in red, but the slightly different version that I was using is used by some other authors, and is more useful for dealing with the example in this thread. In fact, I think that your book is somewhat misleading in claiming that the test is inconclusive when the limit is zero or infinity. As my version of the test shows, it is possible to draw conclusions in these cases. However, in those situations the test can only be used to get an implication in one direction; in these cases it is not an "if and only if" test. 11. Aaaah! I see! Thank you so much, I understand. I'll keep in mind that there are two "versions" of the limit comparison test then. Thanks for not giving me up.

https://math.stackexchange.com/questions/2863510/between-any-two-powers-of-5-there-are-either-two-or-three-powers-of-2

# Between any two powers of $5$ there are either two or three powers of $2$ Is this statement true? Between any two consecutive powers of $5$, there are either two or three powers of $2$. I can see that this statement is true for cases like $$5^1 < 2^3 < 2^4 < 5^2$$ or $$5^3 < 2^7 < 2^8 < 2^9 < 5^4$$ But I am having a trouble figuring out the proof through generalization. Could somebody help me? How many integer multiples of $\log_5(2)$ can we fit between the integers $n$ and $n+1$? It's not too hard to see that since $\log_5(2)<0.5$, there are at least $2$ such numbers. But, since $3\log_5(2)>1$, there are at at most $3$ such numbers. * So then, if $n$ and $n+1$ straddled $2$ multiples, we would have (for an appropriate integer, $k$): $$n<k\log_5(2)<(k+1)\log_5(2)<n+1\\\Rightarrow5^n<2^k<2^{k+1}<5^{n+1}$$ Conversely, if $3$ multiples were straddled, we would have $$n<k\log_5(2)<(k+1)\log_5(2)<(k+2)\log_5(2)<n+1\\\Rightarrow5^n<2^k<2^{k+1}<2^{k+2}<5^{n+1}$$ Therefore there are between $2$ and $3$ powers of $2$ between each successive powers of $5$. * To visualise this, consider the figure below. The red points are multiples of $\log_5(2)$. The distance between the red points ($\approx0.43$) is small enough to guarantee that at least $2$ of them lie between the black points. However the gap between $4$ red points ($\approx1.29$) is too wide to fit between the black points. This is irrespective of where the red points begin. Here's a fun graph to play around with https://www.desmos.com/calculator/qfqers5mmg. No matter where you start the sequence of red points (by varying the parameter $h$), you inevitably have either $2$ or $3$ red points between the black points. Basically, it boils down to the fact that $5$ is between $2^2 = 4$ and $2^3 = 8$. Here's a proof, though. Let $5^a$ and $5^{a+1}$ be the two consecutive powers of $5$. Let $2^b$ be the smallest power of $2$ that exceeds $5^a,$ and $2^c$ the largest below $5^{a+1}$. Then we have $2^{b-1} \leq 5^a < 2^b$ and $2^c < 5^{a+1} \leq 2^{c+1}$. From this we get $$\frac{2^{c}}{2^{b}} < \frac{5^{a+1}}{5^a} \leq \frac{2^{c+1}}{2^{b-1}}$$ or, equivalently, $$2^{c-b} < 5 \leq 4 \cdot 2^{c-b}.$$ This inequality can be rewritten as $\frac{5}{4} \leq 2^{c-b} < 5$, which proves that $c - b$ is either $1$ or $2$. Thus the powers of $2$ between $5^a$ and $5^{a+1}$ are either $2^b, 2^{b+1} = 2^c$, or $2^b, 2^{b+1}, 2^{b+2} = 2^c.$ This is what you wanted. We can see mathematically that there is a ratio between the number of powers of $2$ and the number of powers of $5$ below a certain integer $a$. Intuitively, the ratio would be $\log_25$, or $2.32$. On average, for every power of $5$, you get $2.32$ powers of $2$. That would translate to about $2$ or $3$ powers of $2$ between every two powers of $5$. If this isn't completely intuitive, notice that there are exactly $3$ more powers of $2$ than $8$, as $\log_28 = 3$. For example, out of all integers below $100$, there are $6$ powers of $2$ ($2,4,8,16,32,64$) and $2$ powers of $8$ ($8,64$). We can extend to this to any integers $x$ and $y$, provided they are not $0$ or $1$: For all integers below an integer $z$, the ratio of the number of exponents of $a$ to the number of exponents of $b$ is $\log_ab$. In the special case $[1,5]$, we see that 1, 2 and 4 are in the interval but 8 and up are not. There are powers of 2 less than and powers of two greater than any other power of 5. There can't be more than three, because if $0 < a \leq b < 2b < 4b < 8b \leq 5a$, we get the contradiction $8a \leq 5a$ for a positive number $a$. Nor can there be zero, because if $0 < b < a < 5a < 4b$, we get the contradiction $5b < 4b$ for a positive number b. Nor can there be one, because the argument still holds if we insert the condition $a \leq 2b \leq 5a$. You already found existence proofs for two or three intermediate powers. The explanations in the other answers are great, but that's a very elementary proof. • Hi, @Davislor. Sorry but I don't quite follow where $8a\leq 5a$ comes from - it seems as though you've replaced the $b$ that was previously in the inequality but how is this justified? – Jam Jul 26 '18 at 20:09 • @Jam $0< a \leq b$, so $8a \leq 8b$. Bit, we assumed $8b \leq 5a$. Thus the contradiction that $8a \leq 5a$ for positive $a$. – Davislor Jul 26 '18 at 20:15 • @Jam: Do you find my elementary proof any clearer? – Ilmari Karonen Jul 27 '18 at 10:33 • @IlmariKaronen I find both quite clear, thanks - I was just stuck on one step in Davislor's :) – Jam Jul 27 '18 at 11:04 It's easy enough to see that this is indeed true, even without using logarithms. Let $a$ be some power of $5$, and let $b$ be the least power of $2$ not less than $a$. Then the next power of $5$ after $a$ is obviously $5a$, while the next three powers of $2$ after $b$ are $2b$, $4b$ and $8b$. Since, by definition, $a \le b$, it follows that $5a \le 5b < 8b$. Thus, there can be at most three powers of $2$ ($b$, $2b$ and $4b$) between $a$ and $5a$. Conversely, since $b$ is the least power of $2$ not less than $a$, it follows that $\frac12 b < a$. Thus, equivalently, $2b < 4a < 5a$, so there must be at least two powers of $2$ ($b$ and $2b$) between $a$ and $5a$. BTW, if you look closely at the proof above, you may note that it doesn't actually use the assumption that $a$ is a power of $5$ anywhere. Thus, in fact, we've proven a more general result: for any positive number $a$, there are either two or three powers of $2$ between $a$ and $5a$. (In fact, the proof doesn't really use the assumption that $b$ is a power of $2$, either, so we could generalize the result even further in this direction if we wanted!) Also, you may notice that the key observation behind the result above is that the number $5$ lies strictly between $2^2 = 4$ and $2^3$ = 8. Thus, by essentially the same logic as above, we can prove a similar result for other bases: Between any two consecutive powers of $x$ there are at least $k$ and at most $k+1$ powers of $y$, where $x$ and $y$ are any numbers greater than $1$, and $k$ is the largest integer such that $y^k \le x$. We know that if $y-x >1$, for $x\geq 0, y>0$, then there $\exists n \in \mathbb{N}, n>0$ s.t. $$x<n<y \tag{1}$$ e.g. $n=\left \lfloor x \right \rfloor+1$, because $$x = \left \lfloor x \right \rfloor+\{x\}<\left \lfloor x \right \rfloor+1<x+1<y$$ In this case $$(k+1)\frac{\ln{5}}{\ln{2}}-k\frac{\ln{5}}{\ln{2}}=\frac{\ln{5}}{\ln{2}}>1$$ and from $(1)$, there $\exists n\in\mathbb{N}$ s.t. $$k\frac{\ln{5}}{\ln{2}}<n<(k+1)\frac{\ln{5}}{\ln{2}} \iff\\ k\ln{5}<n\ln{2}<(k+1)\ln{5} \iff \\ 5^k < 2^n<5^{k+1}$$ However $\frac{\ln{5}}{\ln{2}}\approx 2.32>2$ and $(1)$ can be extended to if $y-x >2$, for $x\geq 0, y>0$, then there $\exists n \in \mathbb{N},n>0$ s.t. $$x<n<n+1<y \tag{2}$$

http://mathrefresher.blogspot.com/2007/04/properties-of-matrix-multiplication.html

## Tuesday, April 17, 2007 ### Properties of Matrix Multiplication Multiplication can only occur between matrices A and B if the number of columns in A match the number of rows in B. This presents the very important idea that while multiplication of A with B might be a perfectly good operation; this does not guarantee that multiplication of B with A is a perfectly good operation. Even if matrix A can be multiplied with matrix B and matrix B can be multiplied to matrix A, this doesn't necessarily give us that AB=BA. In other words, unlike the integers, matrices are noncommutative. Property 1: Associative Property of Multiplication A(BC) = (AB)C where A,B, and C are matrices of scalar values. Proof: (1) Let D = AB, G = BC (2) Let F = (AB)C = DC (3) Let H = A(BC) = AG (4) Using Definition 1, here, we have for each D,F,G,H: di,j = ∑k ai,k*bk,j gi,j = ∑k bi,k*ck,j fi,j = ∑k di,k*ck,j (5) So, expanding fi,j gives us: fi,j = ∑k (∑l ai,l*bl,j)ck,j = (∑kl) ai,l*bl,k*ck,j = = ∑l ai,l*(∑k bl,k*ck,j) = = ∑l ai,l*gl,j = hi,j QED Property 2: Distributive Property of Multiplication A(B + C) = AB + AC (A + B)C = AC + BC where A,B,C are matrices of scalar values. Proof: (1) Let D = AB such that for each: di,j = ∑k ai,k*bk,j (2) Let E = AC such that for each: ei,j = ∑k ai,k*ck,j (3) Let F = D + E = AB + AC such that for each: fi,j = ∑k ai,k*bk,j+ai,k*ck,j = ∑k ai,k[bk,j + ck,j] (4) Let G = B+C such that for each: gi,j = bi,j + ci,j (5) Let H = A(B+C) = AG such that for each: hi,j = ∑k ai,k*gk,j (6) Then we have AB + AC = A(B+C) since for each: hi,j = ∑k ai,k[bk,j + ck,j] (7) Let M = A + B such that for each: mi,j = ai,j + bi,j (8) Let N = (A+B)C = MC such that: ni,j = ∑k mi,k*ck,j = = ∑k (ai,k + bi,k)*ck,j (9) Let O = BC such that: oi,j = ∑k bi,k*ck,j (10) Let P = AC + BC = E + O such that: pi,j = ei,j + oi,j = = ∑k ai,k*ck,j + ∑k bi,k*ck,j = = ∑k [ai,k*ck,j + bi,k*ck,j] = = ∑k (ai,k + bi,k)*ck,j QED Property 3: Scalar Multiplication c(AB) = (cA)B = A(cB) Proof: (1) Let D = AB such that: di,j = ∑k ai,k*bk,j (2) Let E = c(AB) = cD such that for each: ei,j = c*di,j = c*∑k ai,k*bk,j (3) Let F = (cA)B such that: fi,j = ∑k (c*ai,k)*bk,j = c*∑k ai,k*bk,j (4) Let G = A(cB) such that: gi,j = ∑k ai,k*(c*bk,j) = = c*∑k ai,k*bk,j QED Property 4: Muliplication of Matrices is not Commutative AB does not have to = BA Proof: (1) Let A = (2) Let B = (3) AB = (4) BA = QED References • Hans Schneider, George Philip Barker, , 1989. Eva Acosta said... Hi, I am a math teacher. You can revise your understanding of matrices solving exercises about addition, subtraction and multiplication of matrices. Inverse,... at www.emathematics.net Fantastic!!! Anonymous said... hi i am a student taking a course in linear algebra. i just wanted to say that your tutorrial is very helpfull. so post some more on this topic. thank you Luis said... Amittai Aviram said... Hi! Thank you very much for providing this resource! I am troubled, however, by something in your proof of the associativity of matrix multiplication. (Since blogger.com will not let me use the necessary HTML tags, I will attempt to use LaTeX-like notation instead.) In (4), you have d_{i,j} = \sum_{k} a_{i,k}*b_{k,j} f_{i,k} = \sum_{k} d_{i,k}*c_{k,j} Then, in (5), you expand this last equation as f_{i,j} = \sum_{k} (\sum_{l} a_{i,l}*b_{l,j}) c_{k,j} However, we are expanding d_{i,k} from the previous equation, not d_{i,j} (where the outer summation symbol binds the variable k). And d_{i,k} = \sum_{l} a_{i,l}*b_{l,k} The summation symbol that binds k becomes the outer summation symbol in the expansion. So the correct expansion of the whole equation should be: f_{i,j} = \sum_{k} (\sum_{l} a_{i,l}*b_{l,k}) c_{k,j} which does not lend itself to so neat a proof as the one you present. A correct proof, though with rather messier notation, can be found here: http://linear.ups.edu/xml/0094/fcla-xml-0.94li30.xml Amittai Aviram said... Another page that has a correct proof (avoiding the error noted above) can be found here: http://www.proofwiki.org/wiki/Matrix_Multiplication_is_Associative#Proof Larry Freeman said... Hi Amittai, Thanks very much for pointing out the mistake. I will revise the proof and post a comment on this blog when it is fixed.

https://math.stackexchange.com/questions/931839/expected-number-of-output-letters-to-get-desired-word

# Expected number of output letters to get desired word I am using a letter set of four letters, say {A,B,C,D}, which is used to output a random string of letters. I want to calculate the expected output length until the word ABCD is obtained; that is, the letters A B C D appearing consecutively in that order. I have referenced this question (Expected Number of Coin Tosses to Get Five Consecutive Heads), but have found a complexity in our case; when we obtain, say, ABA, then we can't say that the chain resets, since we have the next potentially successful chain already being started. I have tried the approach below, but am not sure if it is completely correct. I would be grateful for assertion that this approach is ok, as well as for any alternative methods to approach this issue. Let e be the expected number of output letters needed to get the target string ABCD. Also, let f be the expected number of output letters needed to get the target string ABCD given we obtained the letter A. The table for expected length and probability for e would be | | Exp Len | Prob | |--------------------------|---------|------| | if first letter is [BCD] | e+1 | 3/4 | | if A then [CD] | e+2 | 1/8 | | if A then A | f+1 | 1/16 | | if AB then [BD] | e+3 | 1/32 | | if AB then A | f+2 | 1/64 | | if ABC then [BC] | e+4 | 1/128| | if ABC then A | f+3 | 1/256| | if ABCD | 4 | 1/256| --------------------------------------------- and a similar table for f after we obtained the letter A would be | | Exp Len | Prob | |-----------------------|---------|------| |if first letter is [CD]| e+2 | 1/2 | |if first letter is A | f+1 | 1/4 | |if B then [BD] | e+3 | 1/8 | |if B then A | f+2 | 1/16 | |if BC then [BC] | e+4 | 1/32 | |if BC then A | f+3 | 1/64 | |if BCD | 4 | 1/64 | ------------------------------------------ The expected length e is equal to the sum of each (Probability)*(Expected Length) product set from the first table, giving $$e\, =\, \frac{3}{4}(e+1)\, +\, \frac{1}{8}(e+2)\, +\, \frac{1}{16}(f+1)\, +\, \frac{1}{32}(e+3 )\, +\, \frac{1}{64}(f+2)\, +\, \frac{1}{128}(e+4)\, +\, \frac{1}{256}(f+3)\, +\, \frac{1}{256}(4) \\-----\\ e\, \, =\, \frac{117}{128}e\, +\, \frac{21}{256}f\, +\, \frac{319}{256} \\\\ 22e\, =\, 21f\, +\, 319 \: \: \: ---(1) \\ 44e\, =\, 42f\, +\, 638 \: \: \: ---(1')$$ A similar approach for f yields $$f\, =\, \frac{1}{2}(e+2)\, +\, \frac{1}{4}(f+1)\, +\, \frac{1}{8}(e+3)\, +\, \frac{1}{16}(f+2 )\, +\, \frac{1}{32}(e+4)\, +\, \frac{1}{64}(f+3)\,+\, \frac{1}{64}(4) \\-----\\ f\, \, =\, \frac{21}{32}e\, +\, \frac{21}{64}f\, +\, \frac{127}{64} \\\\ 43f\, =\, 42e\, +\, 127 \: \: \: ---(2)$$ Combining these, we obtain $$(2)-(1')\Rightarrow f\, =\, -2e\, +\, 765 \: \: \: ---(3)\\ (3)\rightarrow (1)\Rightarrow 22e = 21(-2e+765)+319 \\ e=256 \\ f=253$$ So the expected length seems to be 256 letters output. I notice this is exactly what we would expect from the naive approach, from the fact that each letter has a 1 in 4 chance appearing each time, and after any four letters' output, the chance of ABCD appearing is $$\left( \frac{1}{4} \right) ^ 4 = \frac{1}{256} .$$ which is slightly worrying, since the question about five consecutive heads has a probability of 1/32, but a differing number of 62 for the expected length. After the above, I also calculated the expected length until I obtain either of TWO target strings; I used ABCD and CDBA as my targets, if it matters. The result was not the intuitive 128, but was 136 instead, by methodology similar to that above. Using the answers provided, I will also try to check this result using new tactics proposed in the answers. The natural approach uses transition matrices. For ease of typesetting we write up the solution another way. Let $e$ be the expected number. Let $a$ be the expected number of additional letters, given that the last letter was an A. Let $b$ be the expected number of additional letters, given the last two letters were AB. And let $c$ be the analogous thing, given the last three letters were ABC. At the start, if the first letter is an A, our expected total is $1+a$. If it is anything else, then our expected total is $1+e$. Thus $$e=\frac{1}{4}(1+a)+\frac{3}{4}(1+e).$$ If our last letter was an A, with probability $\frac{1}{4}$ we get an A, and the additional total (after the first A) is $1+a$. If we get a B, the expected additional total after the first A is $1+b$. If we get a C or a D, the expected additional total after the A is $1+e$. Thus $$a=\frac{1}{4}(1+a)+\frac{1}{4}(1+b)+\frac{2}{4}(1+e).$$ If the last two letters were AB, and we get an A, the additional expected total after the AB is $1+a$. If we get a B or a D, the additional expected total is $1+e$. And if we get a C it is $1+c$. Thus $$b=\frac{1}{4}(1+a)+\frac{2}{4}(1+e)+\frac{1}{4}(1+c).$$ Finally, the same reasoning gives $$c=\frac{1}{4}(1+a)+\frac{2}{4}(1+e)+\frac{1}{4}(1).$$ Four linear equations, four unknowns. • I see that this method increases the number of variables introduced, but keeps each resulting equation simpler. I was able to reconstruct the transition matrix from your argument as well. +1 for a great generalizable solution! It would be nice if you could point me to some reference as to how to directly apply the transition matrix to obtain e... – yybtcbk Sep 16 '14 at 3:00 • Additionally: I was able to apply this method to the addendum problem to get e=136, a=134, b=128, and c=102. It seems a bonus of this method that I can get the a, b, and c values on the way as well. – yybtcbk Sep 16 '14 at 3:03 • Good! For transition matrices, any beginning book on Markov chains should have it, also many a basic probability book. Can't think of a specific title. – André Nicolas Sep 16 '14 at 3:09 • I guess it's time to dig out my old Markov Chains textbook... gave up on it in Uni, but maybe I can understand it better now... thanks for the pointer! – yybtcbk Sep 16 '14 at 3:19 Conway's algorithm provides a quick method of calculation: look at how whether the initial letters match the final letters: so "AAAA" has matches for the initial $1,2,3,4$, while "ABCD" has matches only for the initial $4$; "ABCA" would have matches for $1$ and $4$, while "ABAB" would have matches for $2$ and $4$. Since the alphabet has four equally likely letters, the algorithm gives the following expected samples sizes: • AAAA: $340 = 4^4+4^3+4^2+4^1$ • ABCD: $256 = 4^4$ • ABCA: $260 = 4^4+4^1$ • ABAB: $272 = 4^4+4^2$ So, as you say, the expected time until "ABCD" appears is $256 = 4^4$ samples. This is similar to the coin sequence "HHHHT" requiring an expected $32=2^5$ samples. By contrast the expected time until "AAAA" appears is $340 = 4^4+4^3+4^2+4^1$ samples. This is similar to the coin sequence "HHHHH" requiring an expected $62=2^5+2^4+2^3+2^2+2^1$ samples. If you had a long string of $n$ letters then you would expect "ABCD" to appear about $\frac{n-3}{256}$ times on average. Similarly you would expect "AAAA" to appear about the same number of times on average. But strings of type "AAAA" can overlap themselves while those of type "ABCD" cannot: for example a string of length $6$ might possibly have three "AAAA"s but cannot have more than one "ABCD", even if the expected number of each is the same. To balance the greater possibility of "AAAA"s appearing several times, but the same overall expected number, there is also a greater possibility for "AAAA" not appearing at all in the first $6$ letters, or indeed in other initial samples. It is this latter feature which increases the expected sample size until "AAAA" does appear, compared with "ABCD". • Greatly convincing way to see that repeating letters lead to a longer expected string! This also works to suggest why the additional question results in a 136 greater than the intuitive 128; ABCD and CDBA overlap each other. Could you point me somewhere for the theory behind this Conway's Algorithm? My googling only got me algorithms for the game of life, betting odds, and calendars... – yybtcbk Sep 16 '14 at 4:10 • As for using this Conway's Algorithm for the additional question, I guess that for target strings ABCD and CDBA, positions 2 and 4 match the last target letter, I can do 4^4 + 4^2 = 272, and then can halve this (since we have two target strings?) to get 136...? I'm not sure if this is a valid approach, but the numbers do match up... – yybtcbk Sep 16 '14 at 4:18 • Two target strings raises the issue in Penney's game – Henry Sep 16 '14 at 7:05 • – Henry Sep 16 '14 at 7:30 • All the links refer to the Conway Number and Conway's Algorithm in terms of odds; are these directly applicable to the string length calculations above somehow? – yybtcbk Sep 16 '14 at 8:14

https://math.stackexchange.com/questions/1999358/is-there-a-closed-form-expression-for-these-combinatorial-sums

# Is there a closed form expression for these combinatorial sums? I needed to compute this sum: $$\sum_{i=0}^n\binom{n}{i}i^2$$ I don't want any proofs or formulae, but just a yes/no answer. Is there a closed form expression for this sum? Just to share, I was able to find a closed form expression for this sum: $$\sum_{i=0}^n\binom{n}{i}i=n.2^{n-1}$$ by using a property of arithmetic progressions and sums of combinations. Is there any closed form expression for my sum, or more generally, sums like: $$\sum_{i=0}^n\binom{n}{i}i^k$$ , for some $k\in\mathbb{N}$? • There is a closed form for your sum, and you can get it by applying what you already found for $\sum_k\binom{n}kk$. (You can also use calculus methods, but they’re not necessary.) – Brian M. Scott Nov 4 '16 at 16:06 • Differentiate the series $(1+x)^n$ wrt to $x$ twice to get your sum. – SirXYZ Nov 4 '16 at 16:11 Yes. You can find the formula as follows: by the binomial theorem we have $$(1+x)^n = \sum_{i=0}^n {n\choose i} x^i.$$ Now apply the differential operator $\Big(x\frac{d}{dx}\Big)^k$ on both sides. • Concise and complete. +1 ... You might consider adding "then set $x=1$" at the end of the answer. – Mark Viola Nov 4 '16 at 16:07 • Thanks @Dr.MV, I trust that the OP will extract that information from your comment ;). – J.R. Nov 4 '16 at 16:13 An alternative approach. Us that $\binom{n}{i}\binom{i}{k}=\binom{n}{k}\binom{n-k}{i-k}$ to get: $$\sum_{i=0}^n \binom{n}{i}\binom{i}{k}= \binom{n}{k}\sum_{i=0}^{n} \binom{n-k}{i-k}=\binom{n}{k}2^{n-k}$$ So, since $i^2=2\binom{i}{2}+\binom{i}{1}$ to get that $$\sum \binom{n}{i}i^2 = 2\binom{n}{2}2^{n-2} + \binom{n}{1}2^{n-1}$$ In general, $i^k = \sum_{j=0}^{k} a_j \binom{i}{j}$ for some sequence $a_j$, yielding the result: $$\sum \binom{n}{i}i^k = \sum_{j=0}^k a_j\binom{n}{j}2^{n-j}$$ So it just amounts to finding the $a_j$ for each $k$. Hint: $$\binom{n}{i}i^{2}=\binom{n}{i}i\left(i-1\right)+\binom{n}{i}i=n\left(n-1\right)\binom{n-2}{i-2}+n\binom{n-1}{i-1}$$

https://math.stackexchange.com/questions/3185610/very-indeterminate-form-lim-x-to-infty-left-sqrtx22x3-sqrtx23

# Very indeterminate form: $\lim_{x \to \infty} \left(\sqrt{x^2+2x+3} -\sqrt{x^2+3}\right)^x \longrightarrow (\infty-\infty)^{\infty}$ Here is problem: $$\lim_{x \to \infty} \left(\sqrt{x^2+2x+3} -\sqrt{x^2+3}\right)^x$$ The solution I presented in the picture below was made by a Mathematics Teacher I tried to solve this Limit without using derivative (L'hospital) and Big O notation. Although I get the answer, I don't know if the technique I'm using definitely correct. And here is my method: \begin{align*}\lim_{x \to \infty} \left(\sqrt{x^2+2x+3} -\sqrt{x^2+3}\right)^x&=\lim_{x \to \infty} \left(\frac {2x}{\sqrt{x^2+2x+3} +\sqrt{x^2+3}}\right)^x\\&=\lim_{x \to \infty}\frac{1}{ \left(\frac {\sqrt{x^2+2x+3} +\sqrt{x^2+3}}{2x}\right)^x}\end{align*} Then, I define a new function here $$y(x)=\sqrt{x^2+2x+3} +\sqrt{x^2+3}-2x-1$$ We have \begin{align*} \lim _{x\to\infty} y(x)&=\lim_{x \to \infty}\sqrt{x^2+2x+3} +\sqrt{x^2+3}-2x-1\\ &=\lim_{x \to \infty}(\sqrt{x^2+2x+3}-(x+1))+(\sqrt{x^2+3}-x)\\ &=\lim_{x \to \infty}\frac{2}{\sqrt{x^2+2x+3}+x+1}+ \lim_{x \to \infty}\frac{3}{\sqrt{x^2+3}+x}\\ &=0. \end{align*} This implies that $$\lim_{x \to \infty}\frac{2x}{y(x)+1}=\infty$$ Therefore, \begin{align*} \lim_{x \to \infty}\frac{1}{ \left(\frac {\sqrt{x^2+2x+3} +\sqrt{x^2+3}}{2x}\right)^x}&=\lim_{x \to\infty} \frac{1}{ \left(\frac{y(x)+2x+1}{2x} \right)^x}\\ &=\lim_{x \to\infty} \frac{1}{ \left(1+\frac{y(x)+1}{2x} \right)^x}\\ &=\lim_{x \to \infty}\frac{1}{\left( \left( 1+\frac{1}{\frac{2x}{y(x)+1}}\right)^{\frac{2x}{y(x)+1}}\right)^{\frac{y(x)+1}{2}}}\\ & \end{align*} Here, we define two functions: $$f(x)=\left( 1+\frac{1}{\frac{2x}{y(x)+1}}\right)^{\frac{2x}{y(x)+1}},\quad g(x)=\frac{y(x)+1}{2}.$$ We deduce that, $$\lim_{x\to\infty} f(x)=e>0,\quad \lim_{x\to\infty} g(x)=\frac 12>0.$$ Thus, the limit $$\lim_{x\to\infty} f(x)^{g(x)}$$ exists and is finite. Finally we get, \begin{align*} \lim_{x \to \infty}\frac{1}{\left( \left( 1+\frac{1}{\frac{2x}{y(x)+1}}\right)^{\frac{2x}{y(x)+1}}\right)^{\frac{y(x)+1}{2}}} &=\frac{1}{\lim_{x \to \infty}\left( \left( \left( 1+\frac{1}{\frac{2x}{y(x)+1}}\right)^{\frac{2x}{y(x)+1}}\right)^{\frac{y(x)+1}{2}}\right)}\\ &=\frac{1}{\left(\lim_{x\to\infty} \left( 1+\frac{1}{\frac{2x}{y(x)+1}} \right)^{\frac{2x}{y(x)+1}}\right)^{ \lim_{x\to\infty} \frac{y(x)+1}{2}}}\\ &=\frac {1}{e^{\frac12}}=\frac{\sqrt e}{e}.\\&& \end{align*} Is the method I use correct? I have received criticisms against my work. What can I do to make the method I use, rigorous? What are the points I missed in the method? Thank you! • @DMcMor Thank you for edit my bad $\LaTeX$ – lone student Apr 12 '19 at 22:10 • Honestly your TeX was pretty good! I just aligned the equations to make it a bit easier to see. – DMcMor Apr 12 '19 at 22:13 • I'm also a BlackPenRedPen fan! (or is it BlackPenRedPenBluePen?) – Toby Mak Apr 13 '19 at 10:57 • $x^y$ is continuous near $\left(e,\frac12\right)$; thus, $\lim\limits_{x\to\infty}f(x)^{g(x)}=\lim\limits_{x\to\infty}f(x)^{\lim\limits_{x\to\infty}g(x)}$ when $\lim\limits_{x\to\infty}f(x)=e$ and $\lim\limits_{x\to\infty}g(x)=\frac12$ – robjohn Apr 24 '19 at 9:59 • No. By using the fact that $x^y$ is continuous at this point, it makes your argument rigorous. – robjohn Apr 24 '19 at 10:19 Your math looks good! I'd maybe just an extra step here and there to make it clear what your doing. Things like showing that you're multiplying by conjugates and maybe a change of variables, say $$z = \frac{2x}{y(x)+1},$$ near the end so it's a bit clearer where the $$e$$ comes from. Otherwise everything looks good! This is a tricky limit, I really like your solution. The solution appears to be correct. Just for the sake of sanity, here's a different argument, based on the idea that knowing derivatives is knowing many limits. First, find the limit of the logarithm of the beast, which is best treated also with the substitution $$x=1/t$$, which makes us trying to find $$\lim_{t\to0^+}\frac{1}{t}\log\left(\frac{\sqrt{1+2t+3t^2}-\sqrt{1+3t^2}}{t}\right) = \lim_{t\to0^+}\frac{1}{t}\log\left(\frac{2}{\sqrt{1+2t+3t^2}+\sqrt{1+3t^2}}\right)$$ This can be rewritten as $$\lim_{t\to0^+}-\frac{\log\bigl(\sqrt{1+2t+3t^2}+\sqrt{1+3t^2}\,\bigr)-\log2}{t}$$ which is the negative of the derivative at $$0$$ of $$f(t)=\log\bigl(\sqrt{1+2t+3t^2}+\sqrt{1+3t^2}\,\bigr)$$ Since $$f'(t)=\frac{1}{\sqrt{1+2t+3t^2}+\sqrt{1+3t^2}}\left(\frac{1+3t}{\sqrt{1+2t+3t^2}}+\frac{3t}{\sqrt{1+3t^2}}\right)$$ we have $$f'(0)=1/2$$ and therefore the limit is $$-1/2$$, so your given limit $$e^{-1/2}$$ Your approach is correct but its presentation / application is more complicated than needed here. Here is how you can use the same approach with much less effort. You have already observed that the base $$F(x) =\sqrt {x^2+2x+3}-\sqrt{x^2+3}$$ tends to $$1$$ as $$x\to\infty$$. Now the expression under limit can be written as $$\{F(x) \} ^x=\{\{1+(F(x)-1)\}^{1/(F(x)-1)}\}^{x(F(x)-1)}$$ The inner expression tends to $$e$$ and the exponent $$x(F(x) - 1)\to -1/2$$ so that the desired limit is $$e^{-1/2}$$. Another part of your approach is that it involves the tricky use of subtracting $$2x+1$$ from $$y(x)$$. For those who are experienced in the art of calculus this step is obvious via the approximation $$\sqrt{x^2+2ax+b}\approx x+a$$ but it may appear a bit mysterious for a novice. It is best to either explain this part or remove it altogether as I have done it in my answer. Also note that your approach uses the following limits / rules (it is not necessary to point them out explicitly unless demanded by some strict examiner) : • $$\lim_{x\to\infty} \left(1+\frac{1}{x}\right)^x=e$$ • If $$\lim_{x\to\infty} f(x) =a>0$$ and $$\lim_{x\to\infty} g(x) =b$$ then $$\{f(x) \} ^{g(x)} \to a^b$$ as $$x\to\infty$$. By squaring, we can verify that $$x\le\sqrt{x^2+3}\le x\left(1+\frac3{2x^2}\right)\tag1$$ and $$x+1\le\sqrt{x^2+2x+3}\le(x+1)\left(1+\frac1{x(x+1)}\right)\tag2$$ Adding $$(1)$$ and $$(2)$$ gives $$2x+1\le\sqrt{x^2+2x+3}+\sqrt{x^2+3}\le(2x+1)\left(1+\frac3{2x^2}\right)\tag3$$ Multiplying numerator and denominator by $$\sqrt{x^2+2x+3}+\sqrt{x^2+3}$$ gives $$\sqrt{x^2+2x+3}-\sqrt{x^2+3}=\frac{2x}{\sqrt{x^2+2x+3}+\sqrt{x^2+3}}\tag4$$ Bernoulli and cross multiplying yield $$1-\frac3{2x}\le\left(1-\frac3{2x^2}\right)^x\le\left(1+\frac3{2x^2}\right)^{-x}\tag5$$ Therefore $$(3)$$, $$(4)$$, and $$(5)$$ yield $$\left(\frac{2x}{2x+1}\right)^x\left(1-\frac3{2x}\right)\le\left(\sqrt{x^2+2x+3}-\sqrt{x^2+3}\right)^x\le\left(\frac{2x}{2x+1}\right)^x\tag6$$ The Squeeze Theorem then says $$\lim_{x\to\infty}\left(\sqrt{x^2+2x+3}-\sqrt{x^2+3}\right)^x=e^{-1/2}\tag7$$ Your method is correct if and only if you truly understand how to rigorously prove a critical step where you effectively claim $$\lim_{x∈\mathbb{R}→∞} (1+\frac1x)^{f(x)} = \lim_{x∈\mathbb{R}→∞} e^{f(x)/x}$$. Note that this requires real exponentiation, and the simplest proof of this would involve the asymptotic expansions for $$\exp,\ln$$, hence I personally think it's misleading to think of your method as successfully evading asymptotic expansions. To preempt the very common bogus proof, note that this claim does not follow from $$\lim_{x∈\mathbb{R}→∞} (1+\frac1x)^x = e$$. After I first posted my answer, you edited your attempt in a non-trivial way. (Please do not edit your question in this way in the future, as it invalidates existing answers.) It still has the same conceptual error, just with different appearance. In this second attempt, you effectively claim that if $$\lim_{x∈\mathbb{R}→∞} g(x) = c$$ then $$\lim_{x∈\mathbb{R}→∞} f(x)^{g(x)} = \lim_{x∈\mathbb{R}→∞} f(x)^c$$ if the latter limit exists. This is not in general true! If you can state and prove the correct theorem of this sort (in a comment), then I will believe that you understand it. The common underlying error is that you replaced part of a limit expression with its limit, which is in general invalid! • Comments are not for extended discussion; this conversation has been moved to chat. – quid Apr 13 '19 at 10:41 • "The common underlying error is that you replaced part of a limit expression with its limit, which is in general invalid!" Definitely Wrong argument. Because, all limit rules are "invalid generally". You cannot show the applicable limit rule in all cases. – Zaharyas Apr 20 '19 at 11:28 • Your last sentence is mathematically non sense. (as it appears). – Zaharyas Apr 20 '19 at 11:45 • I'm IMO gold medalist. Now, I study bachelors. – Zaharyas Apr 20 '19 at 11:52 • IMO does not contain calculus or limit theorems. The content ranges from extremely difficult algebra and pre-calculus problems to problems on branches of mathematics not conventionally covered at school and often not at university level either, such as projective and complex geometry, functional equations, combinatorics, and well-grounded number theory, of which extensive knowledge of theorems is required. You asked me about my mathematical background. I gave the answer. – Zaharyas Apr 20 '19 at 18:46 The proof can be accelerated, using binomial Maclaurin series in the form of $$\sqrt{x^2+2x+3} = (x+1)\sqrt{1+\dfrac2{(x+1)^2}} = (x+1)\left(1 + \dfrac1{(x+1)^2}+O\left(x^{-4}\right)\right)$$ $$= x+1+\dfrac1x+O\left(x^{-2}\right),$$ $$\sqrt{x^2+3} = x\left(1+\dfrac3{2x^2}+O(x^{-4})\right) = x + \dfrac3{2x}+O(x^{-3}).$$ Then $$\ln L = \ln \lim\limits_{x\to\infty} \left(\sqrt{x^2+2x+3}-\sqrt{x^2+3}\right)^x = \lim\limits_{x\to\infty} x\ln\left(1-\frac1{2x}+O\left(x^{-2}\right)\right)$$ $$= \lim\limits_{x\to\infty} x\left(-\frac1{2x}+O\left(x^{-2}\right)\right) = -\frac12,$$ $$L=e^{\Large^{-\frac12}}.$$

https://nndcgroup.com.ng/blog/1k4v2s2.php?c0ec0e=latex-math-symbols

Below is the comprehensive list of Greek symbols/characters/alphabets generally used for preparing a document using LaTex. Note that The following table provides a comprehensive list/guide of mathematical Integral symbols using an appropriate example. command to add a little whitespace between them: The choices available with all fonts are: When using the STIX fonts, you also have the This is a small change from regular TeX, where the dollar sign in Note that you can set the integral boundaries by using the underscore _ and circumflex ^ symbol as seen below. The subsequent table provides a list/guide of commonly used Binary Symbols/Operators using an appropriate example. to blend well with Times), or a Unicode font that you provide. width of the symbols below: Care should be taken when putting accents on lower-case i's and j's. Online LaTeX equation editor, generate your mathematical expressions using LaTeX with a simple way. If you think I forgot some very important basic function or symbol here, please let me know. You can try testing these Trigonometric functions commands directly on our online LaTeX compilerfor a better understanding. The following table provides a comprehensive list/guide of Trigonometric functions along with their LaTeX command. While the syntax inside the pair of dollar signs ($) aims to be TeX-like, Even though commands follow a logical naming scheme, you will probably need a table for the most common math symbols at some point. Additionally, you can use \mathdefault{...} or its alias Therefore, these Any text element can use math text. in TeX. To test these binary symbols/operators commands directly on our online LaTeX compiler for a better understanding. You can try testing these Integral symbol commands directly on our online LaTeX compiler for a better understanding. Columns are separated with ampersand & and rows with a double backslash \\ (the linebreak command). \leftarrow, \sum, \int. inside a pair of dollar signs ($). character can not be found in the custom font. The fonts used should have a Unicode mapping in order to find any fairly tricky to use, and should be considered an experimental feature for Of course LaTeX is able to typeset matrices as well. LaTeX offers math symbols for various kinds of integrals out of the box. mathtext also provides a way to use custom fonts for math. Note that you can set the integral boundaries by using the underscore _ and circumflex ^ symbol as seen below. in the following \imath is used to avoid the extra dot over the i: You can also use a large number of the TeX symbols, as in \infty, For this purpose LaTeX offers the following environments. LaTeX offers math symbols for various kinds of integrals out of the box. These are generally used for preparing any document using LaTex. you can then set the following parameters, which control which font file to use Example: $\begin{bmatrix}1 & 0 & \cdots & 0\\1 & 0 & \cdots & 0\\\vdots & \vdots & \ddots & \vdots \\1 & 0 & 0 & 0\end{bmatrix}$. You can use a subset TeX markup in any matplotlib text string by placing it [3] for a particular set of math characters. . a Roman font have shortcuts. expressions blend well with other text in the plot. Lower and Upper integral boundaries can be set using the symbol underscore character "_" and "^", respectively. information. Mathtext should be placed between a pair of dollar signs ($). follows: Here "s" and "t" are variable in italics font (default), "sin" is in Roman Doing things the obvious way produces brackets that are too You can try all the LaTeX symbols in the sandbox below. . . The following table provides a comprehensive list/guide of Trigonometric functions along with their LaTeX command. Customizing Matplotlib with style sheets and rcParams, Fractions, binomials, and stacked numbers. (from (La)TeX), STIX fonts (with are designed Table 258: undertilde Extensible Accents . Here are some more basic functions which don't fit in the categories mentioned above. Jump to navigation Jump to search This is an information page. This default can be changed using the mathtext.default rcParam. to use characters from the default Computer Modern fonts whenever a particular Wikipedia:LaTeX symbols. . fractions or sub/superscripts: The default font is italics for mathematical symbols. Text rendering With LaTeX). easy to display monetary values, e.g., "$100.00", if a single dollar sign . around fractions. the calligraphy A is squished into the sin. sign. . Play around with the commands, make your own formulas and copy the code to your document. All the Greek symbols/characters/alphabets are showed along with the LaTex rendered output. 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http://tex.stackexchange.com/questions/53702/wrapping-text-in-enumeration-environment-around-a-table?answertab=active

# Wrapping text in enumeration environment around a table I would like to wrap the text in an enumerate environment around a table, this is my actual situation (sorry for including so much code, but I wanted to give a precise idea of the number of lines in the enumerate environment): \subsection{Given the following data set where “Target 2” represent the class attribute, compute the naive Bayesian classification for the instance $<L,white>$ and $<XS,?>$.} \begin{table}[h!t] \centering \begin{tabular}{ccc} \toprule Size & Color & Target2 \\ \midrule XS & green & Yes \\ L & green & Yes \\ XS & white & No \\ M & black & No \\ XL & green & Yes \\ XS & white & Yes \\ L & black & No \\ M & green & Yes \\ \bottomrule \end{tabular} \end{table} \begin{enumerate} \item In this case the experience $E$ is made up of $e_{1} = L$ and $e_{2} = white$, which in Naive Bayes are to be considered as independent, therefore we have: \begin{align*} P(yes|E) &= P(L|yes)\cdot P(white|yes) \cdot P(yes) \\ &= \sfrac{1}{5}\times \sfrac{1}{5} \times \sfrac{5}{8} \\ &= 0.025 \end{align*} \begin{align*} P(no|E) &= P(L|no) \cdot P(white|no) \cdot P(no) \\ &= \sfrac{1}{3} \times \sfrac{1}{3} times \sfrac{3}{8} \\ &= 0.041 \end{align*} Then we normalize: \begin{align*} P(yes) &= \frac{0.025}{0.066} \simeq 0.38 \end{align*} \begin{align*} P(no) &= \frac{0.041}{0.066} \simeq 0.62 \end{align*} As $P(no) > P(yes)$, we label $<L,white>$ as no''. \item Now we have to classify a sample with a missing value. During the testing phase we simply omit the attribute\footnote{Classification Other Methods, slide 24}: \begin{align*} P(yes|XS) &= P(XS|yes) \cdot P(yes) = \sfrac{2}{5} \times \sfrac{5}{8} = 0.25 \end{align*} \begin{align*} P(no|XS) &= P(XS|no) \cdot P(no) = \sfrac{1}{3} \times \sfrac{3}{8} = 0.125 \end{align*} Let us normalize \begin{align*} P(yes) &= \sfrac{0.25}{0.375} \simeq 0.7 \\ P(no) &= \sfrac{0.125}{0.375} \simeq 0.3 \end{align*} Bottom line this sample is classified as yes''. \end{enumerate} I have tried using the wrapfig and the floatftl package unsuccessfully, the table was moved to the end of the list in both cases. I have considered using two minipage environments, but I would like the text to actually wrap around the table. - It helps when you post questions to make complete documents including loading all the packages you need, I guessed \usepackage{booktabs,xfrac,amsmath} in this case. Also I fixed a few font issues (for multi-letter identifiers and angle brackets) Changing margins within a LaTeX list is a bit delicate, but this is I think the layout you want \documentclass{article} \usepackage{booktabs,xfrac,amsmath} \begin{document} \subsection{Given the following data set where “Target 2” represent the class attribute, compute the naive Bayesian classification for the instance $\langle L,white\rangle$ and $\langle \mathit{XS},?\rangle$.} \savebox0{% \begin{tabular}{ccc} \toprule Size & Color & Target2 \\ \midrule XS & green & Yes \\ L & green & Yes \\ XS & white & No \\ M & black & No \\ XL & green & Yes \\ XS & white & Yes \\ L & black & No \\ M & green & Yes \\ \bottomrule \end{tabular}} \begin{enumerate} \makeatletter \dimen@\wd0 \parshape \@ne \@totalleftmargin \linewidth \hbox to \textwidth{\hfill\vtop to \z@{\vskip1em \box\z@\vss}} \item In this case the experience $E$ is made up of $e_{1} = L$ and $e_{2} = \mathit{white}$, which in Naive Bayes are to be considered as independent, therefore we have: \begin{align*} P(\mathit{yes}|E) &= P(L|\mathit{yes})\cdot P(\mathit{white}|\mathit{yes}) \cdot P(\mathit{yes}) \\ &= \sfrac{1}{5}\times \sfrac{1}{5} \times \sfrac{5}{8} \\ &= 0.025 \end{align*} \begin{align*} P(\mathit{no}|E) &= P(L|\mathit{no}) \cdot P(\mathit{white}|\mathit{no}) \cdot P(\mathit{no}) \\ &= \sfrac{1}{3} \times \sfrac{1}{3} \times \sfrac{3}{8} \\ &= 0.041 \end{align*} \parshape \@ne \@totalleftmargin \linewidth Then we normalize: \begin{align*} P(\mathit{yes}) &= \frac{0.025}{0.066} \simeq 0.38 \end{align*} \begin{align*} P(\mathit{no}) &= \frac{0.041}{0.066} \simeq 0.62 \end{align*} As $P(\mathit{no}) > P(\mathit{yes})$, we label $\langle L,\mathit{white}\rangle$ as no''. \item Now we have to classify a sample with a missing value. During the testing phase we simply omit the attribute\footnote{Classification Other Methods, slide 24}: \begin{align*} P(\mathit{yes}|\mathit{XS}) &= P(\mathit{XS}|\mathit{yes}) \cdot P(\mathit{yes}) = \sfrac{2}{5} \times \sfrac{5}{8} = 0.25 \end{align*} \begin{align*} P(\mathit{no}|\mathit{XS}) &= P(\mathit{XS}|\mathit{no}) \cdot P(\mathit{no}) = \sfrac{1}{3} \times \sfrac{3}{8} = 0.125 \end{align*} Let us normalize \begin{align*} P(\mathit{yes}) &= \sfrac{0.25}{0.375} \simeq 0.7 \\ P(\mathit{no}) &= \sfrac{0.125}{0.375} \simeq 0.3 \end{align*} Bottom line this sample is classified as yes''. \end{enumerate} \end{document} - this is great, thank you very much David! –  Edo Apr 29 '12 at 10:26 Not quite right: note the "therefore we have:" is stretched out. That's probably fixable but a workaround is to put a blank line after that line, before the following align. –  David Carlisle Apr 29 '12 at 13:27 I had noticed that, but I guess that a \hfill should do the job. Still it is a great answer, even though I hoped there was some package to manage these situations, thank you very much again. –  Edo Apr 29 '12 at 13:30

https://www.mathworks.com/help/symbolic/piecewise.html

# piecewise Conditionally defined expression or function ## Description example pw = piecewise(cond1,val1,cond2,val2,...) returns the piecewise expression or function pw whose value is val1 when condition cond1 is true, is val2 when cond2 is true, and so on. If no condition is true, the value of pw is NaN. example pw = piecewise(cond1,val1,cond2,val2,...,otherwiseVal) returns the piecewise expression or function pw that has the value otherwiseVal if no condition is true. ## Examples collapse all Define the following piecewise expression by using piecewise. $\mathit{y}=\left\{\begin{array}{ll}-1& \mathit{x}<0\\ 1& \mathit{x}>0\end{array}$ syms x y = piecewise(x < 0,-1,x > 0,1) y = Evaluate y at -2, 0, and 2 by using subs to substitute for x. Because y is undefined at x = 0, the value is NaN. subs(y,x,[-2 0 2]) ans = $\left(\begin{array}{ccc}-1& \mathrm{NaN}& 1\end{array}\right)$ Define the following function symbolically. $\mathit{y}\left(\mathit{x}\right)=\left\{\begin{array}{ll}-1& \mathit{x}<0\\ 1& \mathit{x}>0\end{array}$ syms y(x) y(x) = piecewise(x < 0,-1,x > 0,1) y(x) = Because y(x) is a symbolic function, you can directly evaluate it for values of x. Evaluate y(x) at -2, 0, and 2. Because y(x) is undefined at x = 0, the value is NaN. For details, see Create Symbolic Functions. y([-2 0 2]) ans = $\left(\begin{array}{ccc}-1& \mathrm{NaN}& 1\end{array}\right)$ Set the value of a piecewise function when no condition is true (called otherwise value) by specifying an additional input argument. If an additional argument is not specified, the default otherwise value of the function is NaN. Define the piecewise function $y=\left\{\begin{array}{cc}-2& x<-2\\ 0& -2 syms y(x) y(x) = piecewise(x < -2,-2,(-2 < x) & (x < 0),0,1) y(x) = Evaluate y(x) between -3 and 1 by generating values of x using linspace. At -2 and 0, y(x) evaluates to 1 because the other conditions are not true. xvalues = linspace(-3,1,5) xvalues = 1×5 -3 -2 -1 0 1 yvalues = y(xvalues) yvalues = $\left(\begin{array}{ccccc}-2& 1& 0& 1& 1\end{array}\right)$ Plot the following piecewise expression by using fplot. $y=\left\{\begin{array}{cc}-2& x<-2\\ x& -22\end{array}.$ syms x y = piecewise(x < -2,-2,-2 < x < 2,x,x > 2,2); fplot(y) On creation, a piecewise expression applies existing assumptions. Apply assumptions set after creating the piecewise expression by using simplify on the expression. Assume x > 0. Then define a piecewise expression with the same condition x > 0. piecewise automatically applies the assumption to simplify the condition. syms x assume(x > 0) pw = piecewise(x < 0,-1,x > 0,1) pw = $1$ Clear the assumption on x for further computations. assume(x,'clear') Create a piecewise expression pw with the condition x > 0. Then set the assumption that x > 0. Apply the assumption to pw by using simplify. pw = piecewise(x < 0,-1,x > 0,1); assume(x > 0) pw = simplify(pw) pw = $1$ Clear the assumption on x for further computations. assume(x,'clear') Differentiate, integrate, and find limits of a piecewise expression by using diff, int, and limit respectively. Differentiate the following piecewise expression by using diff. $\mathit{y}=\left\{\begin{array}{ll}1/\mathit{x}& \mathit{x}<-1\\ \mathrm{sin}\left(\mathit{x}\right)/\mathit{x}& \mathit{x}\ge -1\end{array}$ syms x y = piecewise(x < -1,1/x,x >= -1,sin(x)/x); diffy = diff(y,x) diffy = Integrate y by using int. inty = int(y,x) inty = Find the limits of y at 0 by using limit. limit(y,x,0) ans = $1$ Find the right- and left-sided limits of y at -1. For details, see limit. limit(y,x,-1,'right') ans = $\mathrm{sin}\left(1\right)$ limit(y,x,-1,'left') ans = $-1$ Add, subtract, divide, and multiply two piecewise expressions. The resulting piecewise expression is only defined where the initial piecewise expressions are defined. syms x pw1 = piecewise(x < -1,-1,x >= -1,1); pw2 = piecewise(x < 0,-2,x >= 0,2); sub = pw1-pw2 sub = mul = pw1*pw2 mul = div = pw1/pw2 div = Modify a piecewise expression by replacing part of the expression using subs. Extend a piecewise expression by specifying the expression as the otherwise value of a new piecewise expression. This action combines the two piecewise expressions. piecewise does not check for overlapping or conflicting conditions. Instead, like an if-else ladder, piecewise returns the value for the first true condition. Change the condition x < 2 in a piecewise expression to x < 0 by using subs. syms x pw = piecewise(x < 2,-1,x > 0,1); pw = subs(pw,x < 2,x < 0) pw = Add the condition x > 5 with the value 1/x to pw by creating a new piecewise expression with pw as the otherwise value. pw = piecewise(x > 5,1/x,pw) pw = ## Input Arguments collapse all Condition, specified as a symbolic condition or variable. A symbolic variable represents an unknown condition. Example: x > 2 Value when condition is satisfied, specified as a number, vector, matrix, or multidimensional array, or as a symbolic number, variable, vector, matrix, multidimensional array, function, or expression. Value if no conditions are true, specified as a number, vector, matrix, or multidimensional array, or as a symbolic number, variable, vector, matrix, multidimensional array, function, or expression. If otherwiseVal is not specified, its value is NaN. ## Output Arguments collapse all Piecewise expression or function, returned as a symbolic expression or function. The value of pw is the value val of the first condition cond that is true. To find the value of pw, use subs to substitute for variables in pw. ## Tips • piecewise does not check for overlapping or conflicting conditions. A piecewise expression returns the value of the first true condition and disregards any following true expressions. Thus, piecewise mimics an if-else ladder. ## Version History Introduced in R2016b

https://math.stackexchange.com/questions/2812614/question-on-probability-of-winning-a-game-of-dice/2812717

# question on probability of winning a game of dice QUESTION: A and B are playing a game by alternately rolling a die, with A starting first. Each player’s score is the number obtained on his last roll. The game ends when the sum of scores of the two players is 7, and the last player to roll the die wins. What is the probability that A wins the game MY ATTEMPT: On the first roll it is impossible for A to win. on the other hand B can win as long as the number on A's die and the number on B's die adds up to seven. The combinations which allow B to win are: $$S=\{(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)\}$$ which is 6 out of a total of 36 combinations so the probability of B winning in the first round is $\frac{1}{6}$ and the probability of A winning is 0. In the second roll of A a has a winning probability of $\frac{1}{6}$ using the same logic as for B in roll1 and in the next roll of B again B has a winning probability of $\frac{1}{6}$. I can continue to do the above on and on but it will only cycle back to the first scenario so i stop here. Now if B wins in round 1 then A loses and can not proceed to the second roll. Therefore the probability of a winning is probability of B not winning in the first round into the probability of A winning in the next, which is $$(1- \frac{1}{6}) (\frac{1}{6})$$ which is $\frac{5}{36}$ Unfortunately this is not the correct answer as per the answer key. Can someone please point out where I have gone wrong. Any help is appreciated.Thanks :) EDIT: on any next attempt the probability of A winning can be written as $(\frac{5}{6})^2$ of the previous since there is a $(\frac{5}{6})$ probability that A will not win and a $(\frac{5}{6})$ chance that B will not win hence leading to the next attempt. thus the answere is $(\frac{1}{6}) ((\frac{5}{6})^2+(\frac{5}{6})^4+...)$ which is equal to $\frac{1}{6} (\frac{1}{1-(\frac{5}{6})^2)})$ which is 6/11 and the right answer. EDIT thus the answere is $(\frac{5}{36}) ((\frac{5}{6})^2+(\frac{5}{6})^4+...)$ which is equal to $\frac{1}{6} (\frac{1}{1-(\frac{5}{6})^2)})$ which is 5/11 In your last equation, the factor $1/6$ should be the $5/36$ that you found above. This gives $5/11$ not $6/11$. You can get the answer without using infinite series, as follows: Let $p$ be the probability that A wins. Consider the situation after B's first roll. B wins $1/6$ of the time. If this doesn't happen then B is in the situation that A was in at the start. So B wins the game with probability $1/6 + (5/6)p$. But B's probability is also $1-p$. This gives $p = 5/11$. Your value of $\frac{5}{36}$ gives the probability of A winning on A's second roll. But of course there may be later chances for A to win.

https://www.exaton.hu/qn573/c5fe24-tangent-of-a-circle-example

This point is called the point of tangency. Rules for Dealing with Chords, Secants, Tangents in Circles This page created by Regents reviews three rules that are used when working with secants, and tangent lines of circles. AB 2 = DB * CB ………… This gives the formula for the tangent. Solution This problem is similar to the previous one, except that now we don’t have the standard equation. The straight line \ (y = x + 4\) cuts the circle \ (x^ {2} + y^ {2} = 26\) at \ (P\) and \ (Q\). Tangent. A tangent line intersects a circle at exactly one point, called the point of tangency. We’re finally done. If two tangents are drawn to a circle from an external point, 3 Circle common tangents The following set of examples explores some properties of the common tangents of pairs of circles. Tangent lines to one circle. Challenge problems: radius & tangent. The line is a tangent to the circle at P as shown below. In the figure below, line B C BC B C is tangent to the circle at point A A A. Proof: Segments tangent to circle from outside point are congruent. Jenn, Founder Calcworkshop®, 15+ Years Experience (Licensed & Certified Teacher). Tangents of circles problem (example 1) Tangents of circles problem (example 2) Tangents of circles problem (example 3) Practice: Tangents of circles problems. The required equation will be x(5) + y(6) + (–2)(x + 5) + (– 3)(y + 6) – 15 = 0, or 4x + 3y = 38. This video provides example problems of determining unknown values using the properties of a tangent line to a circle. The tangent line never crosses the circle, it just touches the circle. We know that AB is tangent to the circle at A. This lesson will cover a few examples to illustrate the equation of the tangent to a circle in point form. Since the tangent line to a circle at a point P is perpendicular to the radius to that point, theorems involving tangent lines often involve radial lines and orthogonal circles. Also find the point of contact. Question 1: Give some properties of tangents to a circle. The Tangent intersects the circle’s radius at $90^{\circ}$ angle. We’ve got quite a task ahead, let’s begin! Yes! 16 = x. If two segments from the same exterior point are tangent to a circle, then the two segments are congruent. Phew! Can you find ? Comparing non-tangents to the point form will lead to some strange results, which I’ll talk about sometime later. Question: Determine the equation of the tangent to the circle: $x^{2}+y^{2}-2y+6x-7=0\;at\;the\;point\;F(-2:5)$ Solution: Write the equation of the circle in the form: $\left(x-a\right)^{2}+\left(y-b\right)^{2}+r^{2}$ Note that in the previous two problems, we’ve assumed that the given lines are tangents to the circles. Example 5 Show that the tangent to the circle x2 + y2 = 25 at the point (3, 4) touches the circle x2 + y2 – 18x – 4y + 81 = 0. BY P ythagorean Theorem, LJ 2 + JK 2 = LK 2. (2) ∠ABO=90° //tangent line is perpendicular to circle. If the center of the second circle is inside the first, then the and signs both correspond to internally tangent circles. Example 4 Find the point where the line 4y – 3x = 20 touches the circle x2 + y2 – 6x – 2y – 15 = 0. This is the currently selected item. a) state all the tangents to the circle and the point of tangency of each tangent. Note; The radius and tangent are perpendicular at the point of contact. That’ll be all for this lesson. The required equation will be x(4) + y(-3) = 25, or 4x – 3y = 25. } } } On comparing the coefficients, we get x1/3 = y1/4 = 25/25, which gives the values of x1 and y1 as 3 and 4 respectively. (1) AB is tangent to Circle O //Given. Problem 1: Given a circle with center O.Two Tangent from external point P is drawn to the given circle. Answer:The tangent lin… What is the length of AB? It meets the line OB such that OB = 10 cm. 4. A circle is a set of all points that are equidistant from a fixed point, called the center, and the segment that joins the center of a circle to any point on the circle is called the radius. This means that A T ¯ is perpendicular to T P ↔. and … Consider a circle in a plane and assume that $S$ is a point in the plane but it is outside of the circle. Think, for example, of a very rigid disc rolling on a very flat surface. Therefore, to find the values of x1 and y1, we must ‘compare’ the given equation with the equation in the point form. How do we find the length of A P ¯? The angle formed by the intersection of 2 tangents, 2 secants or 1 tangent and 1 secant outside the circle equals half the difference of the intercepted arcs!Therefore to find this angle (angle K in the examples below), all that you have to do is take the far intercepted arc and near the smaller intercepted arc and then divide that number by two! Solution We’ve done a similar problem in a previous lesson, where we used the slope form. Here, I’m interested to show you an alternate method. and are tangent to circle at points and respectively. We’ll use the new method again – to find the point of contact, we’ll simply compare the given equation with the equation in point form, and solve for x­1 and y­1. (4) ∠ACO=90° //tangent line is perpendicular to circle. Tangent to a Circle is a straight line that touches the circle at any one point or only one point to the circle, that point is called tangency. Solved Examples of Tangent to a Circle. vidDefer[i].setAttribute('src',vidDefer[i].getAttribute('data-src')); But there are even more special segments and lines of circles that are important to know. (3) AC is tangent to Circle O //Given. Let’s begin. Now, let’s learn the concept of tangent of a circle from an understandable example here. If a line is tangent to a circle, then it is perpendicular to the radius drawn to the point of tangency. A tangent to the inner circle would be a secant of the outer circle. In the circle O, P T ↔ is a tangent and O P ¯ is the radius. How to Find the Tangent of a Circle? And when they say it's circumscribed about circle O that means that the two sides of the angle they're segments that would be part of tangent lines, so if we were to continue, so for example that right over there, that line is tangent to the circle and (mumbles) and this line is also tangent to the circle. 26 = 10 + x. Subtract 10 from each side. When two segments are drawn tangent to a circle from the same point outside the circle, the segments are congruent. Worked example 13: Equation of a tangent to a circle. The tangent to a circle is perpendicular to the radius at the point of tangency. Make a conjecture about the angle between the radius and the tangent to a circle at a point on the circle. A tangent line t to a circle C intersects the circle at a single point T.For comparison, secant lines intersect a circle at two points, whereas another line may not intersect a circle at all. it represents the equation of the tangent at the point P 1 (x 1, y 1), of a circle whose center is at S(p, q). Get access to all the courses and over 150 HD videos with your subscription, Monthly, Half-Yearly, and Yearly Plans Available, Not yet ready to subscribe? Through any point on a circle , only one tangent can be drawn; A perpendicular to a tangent at the point of contact passes thought the centre of the circle. Can the two circles be tangent? Answer:The properties are as follows: 1. Knowing these essential theorems regarding circles and tangent lines, you are going to be able to identify key components of a circle, determine how many points of intersection, external tangents, and internal tangents two circles have, as well as find the value of segments given the radius and the tangent segment. The problem has given us the equation of the tangent: 3x + 4y = 25. Example. Now, draw a straight line from point $S$ and assume that it touches the circle at a point $T$. Consider the circle below. The point of contact therefore is (3, 4). We’ll use the point form once again. its distance from the center of the circle must be equal to its radius. At the point of tangency, the tangent of the circle is perpendicular to the radius. window.onload = init; © 2021 Calcworkshop LLC / Privacy Policy / Terms of Service. The extension problem of this topic is a belt and gear problem which asks for the length of belt required to fit around two gears. And the final step – solving the obtained line with the tangent gives us the foot of perpendicular, or the point of contact as (39/5, 2/5). Solution This one is similar to the previous problem, but applied to the general equation of the circle. and are both radii of the circle, so they are congruent. We have highlighted the tangent at A. 3. // Last Updated: January 21, 2020 - Watch Video //. Sine, Cosine and Tangent are the main functions used in Trigonometry and are based on a Right-Angled Triangle. Cross multiplying the equation gives. 16 Perpendicular Tangent Converse. Sketch the circle and the straight line on the same system of axes. Suppose line DB is the secant and AB is the tangent of the circle, then the of the secant and the tangent are related as follows: DB/AB = AB/CB. Example 1 Find the equation of the tangent to the circle x2 + y2 = 25, at the point (4, -3). Examples of Tangent The line AB is a tangent to the circle at P. A tangent line to a circle contains exactly one point of the circle A tangent to a circle is at right angles to … And if a line is tangent to a circle, then it is also perpendicular to the radius of the circle at the point of tangency, as Varsity Tutors accurately states. Take Calcworkshop for a spin with our FREE limits course. function init() { Example 6 : If the line segment JK is tangent to circle … b) state all the secants. The circle’s center is (9, 2) and its radius is 2. Since tangent AB is perpendicular to the radius OA, ΔOAB is a right-angled triangle and OB is the hypotenuse of ΔOAB. In this geometry lesson, we’re investigating tangent of a circle. Tangent, written as tan⁡(θ), is one of the six fundamental trigonometric functions.. Tangent definitions. On solving the equations, we get x1 = 0 and y1 = 5. Because JK is tangent to circle L, m ∠LJK = 90 ° and triangle LJK is a right triangle. Proof of the Two Tangent Theorem. Therefore, the point of contact will be (0, 5). The required perpendicular line will be (y – 2) = (4/3)(x – 9) or 4x – 3y = 30. The equation of the tangent in the point for will be xx1 + yy1 – 3(x + x1) – (y + y1) – 15 = 0, or x(x1 – 3) + y(y1 – 1) = 3x1 + y1 + 15. On comparing the coefficients, we get (x­1 – 3)/(-3) = (y1 – 1)/4 = (3x­1 + y1 + 15)/20. var vidDefer = document.getElementsByTagName('iframe'); By using Pythagoras theorem, OB^2 = OA^2~+~AB^2 AB^2 = OB^2~-~OA^2 AB = \sqrt{OB^2~-~OA^2 } = \sqrt{10^2~-~6^2} = \sqrt{64}= 8 cm To know more about properties of a tangent to a circle, download … for (var i=0; i Volvo Xc40 Plug-in Hybrid Release Date Usa, Bash Remove Trailing Slash, Perle Cotton Vs Embroidery Floss, 3d Print Troubleshooting Cura, Email Address Search, When Is The Ymca Going To Reopen,

https://www.lil-help.com/questions/7852/in-the-game-of-two-finger-morra-2-players-show-1-or-2-fingers-and-simultaneously-guess-the-number-of-fingers-their-opponent-will-show

In the game of Two-Finger Morra, 2 players show 1 or 2 fingers and simultaneously guess the number of fingers their opponent will show # In the game of Two-Finger Morra, 2 players show 1 or 2 fingers and simultaneously guess the number of fingers their opponent will show A 45 points In the game of Two-Finger Morra, 2 players show 1 or 2 fingers and simultaneously guess the number of fingers their opponent will show. If only one of the players guesses correctly, he wins an amount (in dollars) equal to the sum of the fingers shown by him and his opponent. If both players guess correctly or if neither guesses correctly, then no money is exchanged. Consider a specified player, and denote by X the amount of money he wins in a single game of Two-Finger Morra. (a) If each player acts independently of the other, and if each player makes his choice of the number of fingers he will hold up and the number he will guess that his opponent will hold up in such a way that each of the 4 possibilities is equally likely, what are the possible values of X and what are their associated probabilities? (b) Suppose that each player acts independently of the other. If each player decides to hold up the same number of fingers that he guesses his opponent will hold up, and if each player is equally likely to hold up 1 or 2 fingers, what are the possible values of X and their associated probabilities? In the 8.7k points For this type of problem, it is helpful to construct a table, Player 1 Player 2 Random Variable Guess Show Guess Show X 1 1 1 1 0 1 1 1 2 -3 1 1 2 1 2 1 1 2 2 0 1 2 1 1 3 1 2 1 2 0 1 2 2 1 0 1 2 2 2 -4 2 1 1 1 -2 2 1 1 2 0 2 1 2 1 0 2 1 2 2 3 2 2 1 1 0 2 2 1 2 4 2 2 2 1 -3 2 2 2 2 0 part (a) There are 16 possibilities in the table above and so we have, $$P(X=0)=\frac{8}{16}=\frac{1}{2}$$ $$P(X=2)=P(X=-2)=\frac{1}{16}$$ $$P(X=3)=P(X=-3)=\frac{2}{16}=\frac{1}{8}$$ $$P(X=4)=P(X=-4)=\frac{1}{16}$$ part (b) from the table above the following set can only occur {1111,1122,2211,2222} all with output X=0, so, $P(X=0)=1$ Surround your text in *italics* or **bold**, to write a math equation use, for example, $x^2+2x+1=0$ or $$\beta^2-1=0$$ Use LaTeX to type formulas and markdown to format text. See example.

https://wild.maths.org/comment/1010

Pairwise Puzzler You may wish to take a look at Pairwise Adding before trying this problem Alison reckons that if Charlie gives her the sums of all the pairs of numbers from any set of five numbers, she'll be able to work out the original five numbers. Charlie gives her this set of pair sums: $0$, $2$, $4$, $4$, $6$, $8$, $9$, $11$, $13$, $15$. Can you work out Charlie's original five numbers? Can you work out a strategy to work out the original five numbers for any set of pair sums that you are given? Does it help to add together all the pair sums? Given ten randomly generated numbers, will there always be a set of five numbers whose pair sums are that set of ten? Can two different sets of five numbers give the same set of pair sums? Four numbers are added together in pairs to produce the following answers: $5$, $9$, $10$, $12$, $13$, $17$. What are the four numbers? Is there more than one possible solution? Six numbers are added together in pairs to produce the following answers: $7$, $10$, $13$, $13$, $15$, $16$, $18$, $19$, $21$, $21$, $24$, $24$, $27$, $30$, $32$. What are the six numbers? Can you devise a general strategy to work out a set of six numbers when you are given their pair sums? Pairwise Puzzler - strategy for original five numbers The five numbers can be calculated simply by using simultaneous equations. Let's call the lowest number a, the second lowest b etc.. and we'll place the sums from smallest to largest in series s. We know straight away that a + b = s1, as the smallest sum must be made up of the smallest pair. Also the largest sum is clear: d + e = s10 We can also work out that a + c = s2, as this is the second smallest sum; a and c are the smallest two numbers which haven't yet been paired. Similarly, c + e = s9 (the second largest). So we now have five unknowns and four equations... we just need one more equation. If we look at the sum of all the sums in terms of the five original numbers ((a + b) + (a + c)...(e + f)), this simplifies to 4 (a + b + c + d + e). So therefore a + b + c + d + e = sum (s) / 4 Here we have the final equation, and we can now calculate the five numbers through simple algebra. So let's make each letter in turn the subject of an equation. We can do this easily for c, using substitution: a + b = s1, d + e = s10, and a + b + c + d + e + f = sum (s) / 4; therefore s1 + s10 + c = sum (s) / 4 So c = sum (s) / 4 - s1 - s10 Now that we have c, all of the other letters can be expressed in terms of c: a = s2 - c b = s1 - (s2 - c) = s1 - s2 + c e = s9 - c d = s10 - (s9 - c) = s10 - s9 + c These equations can all be written, by substituting c, in terms of the sums: a = s1 + s2 + s10 - sum (s) / 4 b = sum (s) / 4 - s2 - s10 c = sum (s) / 4 - s1 - s10 d = sum (s) / 4 - s1 - s9 e = s1 + s9 + s10 - sum (s) / 4 So to find the original five numbers given only the sum of each pair, just arrange the sums from smallest to largest and substitute them into the equations above. This is an excellent solution This is an excellent solution, well done! I particularly like the way that you simplified the problem by writing the numbers as $a \leq b \leq c \leq d \leq e$, and the way that adding all the numbers together allows you to say for certain what the algebraic sum is. Pairwise puzzlers I thought this puzzle was set in a misleading way - at first sight it appears it only involves positive integers, whereas the solution includes a negative integer (-1,1,3,5,10) I suppose it does teach you to assume nothing that is not explicitly stated! Math The answer to the first question is 0,2,6,7,9!!!!! Math Julie, I can choose pairs of numbers from your set of five and add them to get 2, 6, 8, 9, 11, 13 and 15, but I can't get 0, 4 and 4. Do have another go. 0, 2, 4, 6, 9 or -1, 1, 3, 5, 10 One of your sets of five numbers can be used to produce this set of pair sums 0, 2, 4, 4, 6, 8, 9, 11, 13, 15, but I'm afraid the other one can't. Can you figure out which one is which? A somewhat inelegant answer to the 6-number problem Having tackled the 4- and 5-number versions of this problem, I couldn't see an obvious way to generate the 6th equation which would allow direct solution via simultaneous equations. Five equations are simple enough to find, as Mr Redman showed very clearly in his comment, but without the sixth (which I hope someone will enlighten me about at some point), the system would be under-determined. So, without an elegant solution to hand, I tried an inelegant one instead... About 10 lines of matlab code allowed me to generate random collections of six integers (within a reasonable range to keep things somewhat manageable), generate all 15 of their pairwise combinations and sums, and test the resulting sums against the given sums. About 25 seconds of computer time gave me the following answer (which I have checked by hand just to be sure): 2 5 8 11 13 19 I know that this technique will scale very poorly as the size of problem increases - I certainly wouldn't want to attempt it for pairwise combinations of 10 numbers, for example - and would welcome any suggestions regarding a more subtle approach, but feel that a computational approach does have a place even on this website! A somewhat inelegant answer to the 6-number problem Those six numbers do indeed produce the pair sums 7, 10, 13, 13, 15, 16, 18, 19, 21, 21, 24, 24, 27, 30 and 32. Very well done! Could you explain in more detail the method that you programmed Matlab to use? As for a more elegant solution, it is indeed difficult to find a 6th equation. In some of the cases where some of the pair sums are equal, you can deduce a 6th equation. For example if $s_3 = s_4$, then you know the value of $a+d$. Would considering the possible positions of the some of the pair sums (such as $a+d$) help you to deduce a more general approach for all cases?

https://math.stackexchange.com/questions/1056982/probability-a-is-before-b-when-the-26-letters-are-arranged-randomly

# Probability $A$ is before $B$ when the 26 letters are arranged randomly The 26 letters A, B, ... , Z are arrange in a random order. [Equivalently, the letters are selected sequentially at random without replacement.] a) What is the probability that A comes before B in the random order? b) What is the probability that A comes before Z in the random order? c) What is the probability that A comes just before B in the random order? Any help would be much appreciated. I was thinking that for part c the answer would be $$1/26$$ because we have $$25!$$ ways of having A right before B and $$26!$$ total arrangements. Not sure how to proceed with a and b, however. Edit: Thank you. For parts (a) and (b) is there a more formal way of getting $$1/2$$? Such as the formula for the total number of ways we can have $$A$$ before $$B$$ over the total number of arrangements? Would it be 25 choose 1 ... 2 choose 1 over $$26!$$ since we can have a in the first spot and B in any spot after it? Then we can have a in the 2nd spot and B in any spot after it. Also, for part (c), doesn't $$A$$ have to come immediately before $$B$$, so wouldn't the probability be $$1/26$$? • a) 1/2 b) 1/2 . Only 2 possibilities A before or after B – mridul Dec 8 '14 at 6:43 • MAy someone shed some light on the correct answer in part c? Thanks! – user198454 Dec 8 '14 at 16:54 • Is c asking if A come right before B or that A comes right before B but not Z? – Kamster Dec 29 '14 at 5:24 • The reasoning for part (a) in chandu1729's excellent answer is quite formal actually. Let me know if you need an explanation of how to make that formal. – 6005 Dec 30 '14 at 7:02 • Yeah! Just a bounty — and here we are, a bunch of totally equal answers. – sas Dec 30 '14 at 8:15 a) From symmetry, the probability that $A$ comes before $B$ is same as the probability that $B$ comes before $A$, and the sum of these $2$ probabilities is $1$. Hence, the probability that $A$ comes before $B$ is $0.5$. b) Here $B$ is replaced with $Z$ and by exchanging the roles of $B$ and $Z$, we get the same probability as in case (a): $0.5$. c) If $A$ has to come just before $B$, $B$ shouldn't occur at the $1$st position, which has probability of $25/26$. After $B$'s position has been chosen(other than the $1$st position), there are 25 possibilities for position just before $B$ (everything except $B$) and only $1$ possibility is favourable (A coming before B). Hence the required probability is $(25/26)\cdot(1/25) = 1/26$. In part c), $A$ comes right before $B.$ Consider a block $\boxed{\text{AB}}$ and treat it as ONE object. The other letters are $C, D, E, \ldots , Z$. In total, you have 25 objects - 24 letters and a block of two letters together. They can be arranged in $25!$ ways in total. Since the total number of arrangements is $26!$, the required probability is $$\dfrac{25!}{26!}=\dfrac1{26}$$ Now, let us consider part a). We'll formally prove the probability is $\dfrac12$ For this, we will consider 26 cases. Case 1 = B is in the first position. Clearly, if $B$ is in the first position, then in no possible arrangement, A comes before B. Thus, the number of way A comes before B is $0$ for the first case. Case 2 = B is in the second position. For A to come before B, it has to occupy the first spot. A can occupy the first spot in $1$ way. The rest 24 letters can be arranged in $24!$ ways. The number of ways A comes before B can happen is $1.24!$ Case 3 = B is in the third position. For A to come before B, A can occupy the first two positions. The number of ways A can take the first two position is $2.$ The rest 24 letter can be again arranged in $24!$ ways. Hence, the number of ways A comes before B happens is $2.24!$ Case 4 = B is in the fourth position. Now, by same logic as before, you can say it will happen in $3.24!$ ways. Similarly, we go up to Case 26 where B comes in the $26^{th}$ i.e, last position. These $26$ cases exhausts all the possible arrangements. By considering all the $26$ cases and summing up the total number of ways A comes before B can happen, we get $$0+1.24!+2.24!+\cdots 25.24! = 24!(1+2+\cdots 25)=24!*\dfrac{25.26}{2}=\dfrac{26!}{2}$$ Since the total number of arrangements is $26!$, the probability for A comes before B in a random order is $$\dfrac{26!/2}{26!}=\dfrac{1}{2}$$ • Right answer but...way too many words for part (a) :P – 6005 Dec 30 '14 at 7:01 • Thanks. Right, it bothered me too! I could've provided the same thing in a pictorial proof, which would have been way easier to comprehend, but IDK much of Geogebra. :/ – Dipanjan Pal Dec 30 '14 at 8:17 Here's a proof that the probability in part (a) is 1/2. It's the same as the one in The Math Troll's and chandu1729's answers, but with more details filled in. I want to emphasize that it really is a proof, as rigorous as anything you're likely to encounter in an introductory probability course, even though it doesn't use any formulas or complicated calculations. Let's say two arrangements are partners if exchanging the positions of A and B turns one into the other. Here are some examples of arrangements which are partners: ABCDEFGHIJKLMNOPQRSTUVWXYZ and BACDEFGHIJKLMNOPQRSTUVWXYZ AEIOUYBCDFGHJKLMNPQRSTVWXZ and BEIOUYACDFGHJKLMNPQRSTVWXZ THEQUICKBROWNFXJMPSVLAZYDG and THEQUICKAROWNFXJMPSVLBZYDG You should be able to convince yourself that: • Every arrangement has exactly one partner. • If an arrangement has A before B, its partner has B before A, and vice versa. Now, line up all the arrangements in two lines, with each arrangement standing next to its partner. The lines with be the same length, because each arrangement has exactly one partner. In each pair of partners, tell the arrangement with A before B to stand on the left, and the arrangement with B before A to stand on the right. Now, the left line has all the arrangements with A before B, and the right line has all the arrangements with B before A. Since the lines are the same length, it's now obvious that the numbers of A-before-B arrangements and B-before-A arrangements are equal! That means exactly half the arrangements have A before B, and exactly half have B before A. Therefore, if you pick an arrangement at random, the probability of getting A before B is 1/2. You can use exactly the same argument for part (b). The simplest way to answer these questions: a) What is the probability that A comes before B in the random order? In half of the cases A comes before B and in the other half B comes before A, so the answer is $\frac12$ b) What is the probability that A comes before Z in the random order? In half of the cases A comes before Z and in the other half Z comes before A, so the answer is $\frac12$ c) What is the probability that A comes just before B in the random order? • The total number of permutations is $26!$ • Join the letters A and B into a symbol AB, and the total number of permutations is $25!$ • Hence the answer is $\frac{25!}{26!}=\frac{1}{26}$ A more "formal" way to answer to the first (as well as the second) question: • The probability that A is at the 1st place and B is at the 2nd to 26th places is $\frac{1}{26}\cdot\frac{25}{25}=\frac{25}{26\cdot25}$ • The probability that A is at the 2nd place and B is at the 3rd to 26th places is $\frac{1}{26}\cdot\frac{24}{25}=\frac{24}{26\cdot25}$ • The probability that A is at the 3rd place and B is at the 4th to 26th places is $\frac{1}{26}\cdot\frac{23}{25}=\frac{23}{26\cdot25}$ • $\dots$ • The probability that A is at the 25th place and B is at the 26th place is $\frac{1}{26}\cdot\frac{1}{25}=\frac{1}{26\cdot25}$ • So the probability that A comes before B is $\sum\limits_{n=1}^{25}\frac{26-n}{26\cdot25}=\frac12$ a) By symmetry, there are as many permutations with $A$ before $B$ as with $A$ after $B$. $$p=\frac12$$ b) By symmetry, there are as many permutations with $A$ before $Z$ as with $A$ after $Z$. $$p=\frac12$$ c) Consider the $24!$ permutations of $CDE\dots Z$. You can insert $AB$ at $25$ different places to form all distinct configurations. $$p=\frac{25\cdot24!}{26!}=\frac1{26}$$ for parts a and b then the answer is $1/2$. One way to look at is is to note that for any arrangement of the letters, say "DEFGBAC", there is another arrangment the order of the letters reversed, in this case "CABGFED". It is clear that if $A$ comes before $B$ in the first, then it will come after the $B$ in the second, and vice versa. So we can take all the random arrangements and arrange them in to pairs in this manner. Since each arrangement with $A$ before $B$ is paired with another unique arrangment with $B$ before $A$, there must be the same number of strings with $A$ before $B$ and $B$ before $A$. So selecting one of the possible arrangements at random has a probability of $1/2$ of having the $A$ before the $B$. For part c, There are 26 postions where the $A$ can go, and, since that position is filled, 25 positions when the $B$ can go. Now for 25 of the 26 positions where the $A$ can go, the position immediately following it is vacant. In this case there is a change of $1/25$ that the $B$ randomly falls into that position. So in $25/26$ of the cases (where $A$ is not in the last position), there is a chance of $1/25$ of the $B$ immediately following the $A$. However, in 1/26 of the cases, the $A$ is in the last place, and the $B$ cannot follow the $A$, so the probability is $0$. So from the first situation we have $25/26 \times 1/25 = 1/26$, and from the second $1/26\times0 = 0$ Adding up the two, $1/26 + 0 = 1/26$. So you appear to be correct that the probability is $1/26$. For parts a and b, we can construct a bijection between A before B and B before A (proceed similarly for part b). Suppose we have one that is A before B. Then we switch the A and B then we get a B before A. The other direction of the bijection will be similar. Now for part c. If A is the last, then there is 0 chance. This happens 1/26 of the time. If A isn't the last, then there will be one immediately after it. There will be 25 possibilities (and they are random), so the chance is 1/25. This happens 25/26 of the time Hence 1/25*25/26+0=1/26. • Your answer for part C is wrong! it is 1/52 – mridul Dec 8 '14 at 10:21 • Then what is the problebility that A and B together? 1/13? Please use good words. – mridul Dec 8 '14 at 10:24 • Note that there are 26*25 positions for A and B. For A to come before B, there are 25 positions. Thus the probability is 1/26. Please check your solution first. – user198454 Dec 8 '14 at 10:25 Total 26! rearrangements are possible. In 1st case A has only 2 relative possitions after or before B. Therefore probability is 1/2. In 3rd case : Total possible arrangements such that A and B are together is 25!. Therefore Total possible arrangements such that A comes just before B is 25! / 2. a) 1/2 b) 1/2 c) 1/52 Edit: For parts (a) and (b), Let A at 1st possition, total number of arrangements(A before B) is 25!. Let A at 26th possition, total number of arrangements is 0. together we got 25!. Then take 2nd and 25th possitions. Let A at 2st possition, total number of arrangements(A before B) is 24*24!. (B has 24 possitions such that B after A) Let A at 25st possition, total number of arrangements(A before B) is 1*24!. (B has only 1 possition such that B after A) 24*24! + 1*24! = 25*24! =25! Then take 3rd and 24th possitions. Let A at 3st possition, total number of arrangements(A before B) is 23*24!. Let A at 24st possition, total number of arrangements(A before B) is 2*24!. Together we got 25!. etc .. upto 13th,14th possitions. Add all together we got 13*25! arrangements. • Thank you. For parts (a) and (b) is there a more formal way of getting $1/2$? Such as the formula for the total number of ways we can have $A$ before $B$ over the total number of arrangements $(26!)$? Also, for part (c), doesn't $A$ have to come immediately before $B$, so wouldn't the probability be $1/26$? – mylasthope Dec 8 '14 at 7:08 • In the case of part c answer is 1/52. For parts (a) and (b), let A at 1st possition, total number of arrangements(A before B) is 25!. let A at 26th possition, total number of arrangements is 0. together we got 25!. Then take 2nd and 25th possitions together we got 25!. etc. Add all together we got 13*25! arrangements. – mridul Dec 8 '14 at 7:43

https://stats.stackexchange.com/questions/50274/probability-of-a-pair-in-cards

# probability of a pair in cards I have a simple problem involving probability of drawing at least 1 pair of cards in a four card hand. I am not getting the right answer but I dont understand the flaw in my logic. Can anyone explain to me why my approach is wrong? The problem: Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, for each value from 1 to 6, there are two cards in the deck with that value. Bill likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value? My solution: Probability of 1 or more pair = ((6) * (10 choose 2))/(12 choose 4). i.e. the # of hands with at least 1 pair are 6 (the number of ways to create a pair) * (10 choose 2) (the # of ways to select 2 cards from the remaining 10 cards after the pair). This simplifies to 6/11, however, the correct answer is 17/33. Any help understanding would be greatly appreciated. • You make some interesting implicit assumptions in your calculation. First, it focuses only on the possibility that the pair occurs among the first two within a sequence of four cards: there are other ways a pair can occur, so this underestimates the probability. Second, it ignores the sequence among the second two cards, thereby overestimating the probability. Fortunately, the two mistakes do not cancel, revealing the fact there is a problem! – whuber Feb 18 '13 at 22:17 • Realized I was double counting. Cant simply multiply 6 * (10 choose 2). 2 pair instances will be double counted. e.g. pair of 1s in the 6 and then pair of 2s in the 10 choose 2 versus pair of 2s in the 6 and then pair of 1s in the 10 choose 2. The better way to do it is to separate the cases of exactly 1 pair, exactly 2 pair. ways to create exactly 1 pair = 6 (ways to create pair) * ( (10*8)/2 ) + ways to create exactly 2 pair = (6*5)/2 Total # = 255 Denominator = (12 choose 4) = (12*11*10*9)/(4*3*2) = 11*5*9 = 455. Probability = 255/455 = 51/99 = 17/33. – Evan V Feb 18 '13 at 22:24 • Another way to do this is simply to remove the double counting directly. There are 6*(10 choose 2) = 270 ways to create 1 or 2 pairs where the instances of 2 pairs are double counted. If you subtract the # of ways to make exactly 2 pairs (6*5)/2 then you get 270 - 15 = 255. – Evan V Feb 18 '13 at 22:28 To see 0 pairs there are 12 possibilities for the 1st card, but only 10 for the second card (since the 1st chosen card is no longer possible and the card that matches the 1st would form a pair) and 8 cards for the 3rd and 6 for the 4th, this multiplies to 5760. The total number of possible hands (with order mattering) is $12 \times 11 \times 10 \times 9 = 11880$, so the complimentary number which is the number of (ordered) hands that contain at least 1 pair is $11880 - 5760 = 6120$, divide that by 11880 and it reduces to $17/33$. We could also do this with order not mattering, but that would be more complicated and the extra pieces would all end up cancelling each other.

https://www.physicsforums.com/threads/questions-about-linear-transformations.887744/

# I Questions about linear transformations 1. Oct 3, 2016 ### Chan Pok Fung We learnt that the condition of a linear transformation is 1. T(v+w) = T(v)+T(w) 2. T(kv) = kT(v) I am wondering if there is any transformation which only fulfil either one and fails another condition. As obviously, 1 implies 2 for rational number k. Could anyone give an example of each case? (Fulfilling 1 but 2 and 2 but 1) Thanks! 2. Oct 3, 2016 3. Oct 3, 2016 ### andrewkirk Consider the real numbers $\mathbb R$ as a vector space over the rationals, and the operator $T:\mathbb R\to \mathbb R$ that is the identity on the rationals and maps to zero on the irrationals. Then T satisfies the second axiom, since it is a linear operator on $\mathbb Q$ considered as a subspace of $\mathbb R$ and, for $q\in\mathbb Q-\{0\},x\in\mathbb R$, $qx$ is in $\mathbb Q\cup\{0\}=\ker\ T$ iff $x$ is. But the addition axiom does not hold, because $T(1+(\sqrt2-1))=T(\sqrt 2)=0$ but $T(1)+T(\sqrt2-1)=1\neq 0$. 4. Oct 3, 2016 ### Stephen Tashi Define $M((x,y))$ by if $x \ne y$ then $M((x,y)) = (x,y)$ if $x = y$ then $M((x,y)) = (2x, 2y)$ $M$ satisfies 2, but not 1 5. Oct 4, 2016 ### Chan Pok Fung This really gives me new insight into linear transformation. thanks all! 6. Oct 4, 2016 ### Chan Pok Fung I am sorry but I don't quite understand. How can we construct a transformation like that? 7. Oct 4, 2016 ### mathman Unfortunately Hamel basis exists, but it is not constructable - existence is equivalent to axiom of choice. 8. Oct 4, 2016 ### andrewkirk I'm still trying to think of a scenario with a map that satisfies 1 (additivity) but not 2 (scalar mult). Can anybody think of one? All the examples I come up with either end up satisfying neither 1 nor 2, or satisfying 2 but not 1. I assume there must be one, otherwise some texts would specify 1 as the sole requirement and derive 2 as a consequence of 1. 9. Oct 4, 2016 ### Chan Pok Fung Andrew, I am thinking that as we have to apply the transformation to a vector space, and vectors in vector space obeys kv is also in the space. As we can have k be any real number, it seems that it somehow implies axiom2. The transformation only satisfy axiom1 must be of a very weird form. 10. Oct 4, 2016 ### Stephen Tashi On the "talk" page for the current Wikipedia article on "Linear transformation", I found: 11. Oct 4, 2016 ### Chan Pok Fung That's a clear and direct example!

https://math.stackexchange.com/questions/2668314/multiplication-of-moduli-in-modular-congruences

# Multiplication of Moduli in Modular Congruences Lately it's come up in my discrete mathematics class that proving things for smaller congruences, namely those where the modulus is a prime is much easier than attempting to do so for larger congruences. For example, one such problem was to prove that $a^5 \equiv a \pmod{10}$ for $a$ $\varepsilon$ $\mathbb Z^+\$This problem on its own becomes difficult only because we are not guaranteed that for every a $(a,10)=1$. The solution was to factorize 10 in terms of primes $p_1, p_2... p_n$ s.t for each prime $p_i$ $(a,p_i)=1$ for all $a \ \varepsilon \ \mathbb Z^+\$, resulting in n congruences under modulo the given prime and then apply Euler's Theorem. In this case we find the following: $$10 = 2 \cdot 5$$ so $$a^{\phi(5)} \equiv 1 \pmod{5} \\ a^{\phi(2)} \equiv 1 \pmod{2}$$ since $\phi(5)=4, \phi(2)=1$ this becomes: $$a^{4} \equiv 1 \pmod{5} \implies a^{5} \equiv a \pmod{5}\\ a^{1} \equiv 1 \pmod{2} \implies a^{5} \equiv a^4 \pmod{2}$$ but since a is its own inverse modulo 2, we can transform the 2nd congruence into: $$a^5 \equiv a \mod{2}$$ Then the part that I am unclear on occurs. It seems that we can just multiply the 2 moduli together and the desired congruence falls out that: $$a^5 \equiv a \pmod{10}$$ So my question is whether or not $a \equiv b \pmod{n}$ and $a \equiv b \pmod{m}$ always$\implies$ $a \equiv b \pmod{mn}$. Edit (Extension): The answer to my question has been stated to be yes provided that $(m,n) = 1$ but I also noticed that if my 2 relations had been left stated as $$a^4 \equiv 1 \pmod{2}$$ $$a^4 \equiv 1 \pmod{5}$$ Applying the conjecture I made gives: $$a^4 \equiv 1 \pmod{10}$$ which is demonstrably false given that $$2^4 \equiv 6 \pmod{10} \implies 2^4 \not\equiv 1 \pmod{10}$$ Why is this and what prevents this identity from being true as well? I see that 2 would then not be coprime to 10 but why does it work when I multiply both sides by a then? Excellent question! The short answer is no. For example, $4 \equiv 16 \pmod 6$ and $4 \equiv 16 \pmod 4$, but $4 \not \equiv 16 \pmod{24}$. However if $n$ and $m$ are relatively prime, then the answer is yes. This is pretty straightforward to see, as if $a \equiv b \pmod n$ then $n \mid (b-a)$, and if $a \equiv b \pmod m$, then $m \mid (b-a)$. Then one notes (or proves, if it's not clear) that $n \mid (b-a)$, $m \mid (b-a)$, and $\gcd(m,n) = 1$ implies that $mn \mid (b-a)$. In fact, this is at the edge of a deeper theorem called the Chinese Remainder Theorem, which says roughly that knowing the structure of $x$ mod $n$ and $m$ for $m,n$ relatively prime is equivalent to knowing the structure of $x$ mod $mn$ --- or perhaps with two or three or more moduli all taken together. Look up the Chinese Remainder Theorem on the web and on this site for more. • If I might add to this question a bit. This works when i have the $a^5$ and a for a and b but if i take b = 1 it clearly doesn't work since $2^4 \equiv 6 \pmod{10}$ but $2^5 \equiv 2 \pmod{10}$ which makes sense since with a not coprime to 10 when a=2 it wouldn't be invertible mod 10. How is one to catch this beforehand though and what makes it work when it's multiplied on both sides by a? – rjm27trekkie Feb 27 '18 at 1:16 You require $\gcd(m,n)=1$, otherwise \begin{eqnarray*} 1 \equiv 13 \pmod{4} \\ 1 \equiv 13 \pmod{6} \\ \end{eqnarray*} but \begin{eqnarray*} 1 \neq 13 \pmod{24}. \\ \end{eqnarray*} • Understood. Otherwise, is the conjecture correct given that (m,n) = 1? – rjm27trekkie Feb 27 '18 at 1:05 • That's right ... – Donald Splutterwit Feb 27 '18 at 1:09 This is true if (and only if) $m$ and $n$ are coprime – which is the case ot two distinct primes. This is the statement of the Chinese remainder theorem: if $m, n$ are coprime integers, the map \begin{align} \mathbf Z/mn\mathbf Z&\longrightarrow\mathbf Z/m\mathbf Z\times \mathbf Z/n\mathbf Z\\ a\bmod mn&\longmapsto(a\bmod m,a\bmod n) \end{align} is an isomorphism. $x \equiv a \mod n$ means $x=a+kn$ for some $k$. Now $k=qm+r$ for some value of $q$ and $0\le r <m$. So $x=a +rn + q*mn$ so $x=a+rn\mod mn$. However we have no idea yet what $r$ is. we just know it is between $0$(inclusively) and $m$(exclusively). Likewise if $x\equiv a\mod m$ by the same reasoning we know $x\equiv a+sm\mod mn$. We don't know what $s$ is yet. But $a+sm <mn$ and $a+rn <mn$ and $a+sm\equiv a+rn\mod mn$ so $sm=rn$. now if $m$ and $n$ are not relatively prime there might be non trivial solutions. For example maybe $r=\frac m {\gcd (m,n)}$ and $s=\frac n {\gcd (m,n)}$ But if $m$ and $n$ are relatively prime, then $sm=rn$ means $m|r$ and $n|s$. But $r <m$ and $s <n$ and the only way that can happen is if $sm=rn=0$

https://mathhelpboards.com/threads/combinatorics-challenge.6224/

# Combinatorics Challenge #### anemone ##### MHB POTW Director Staff member For $n=1,2,...,$ set $$\displaystyle S_n=\sum_{k=0}^{3n} {3n\choose k}$$ and $$\displaystyle T_n=\sum_{k=0}^{n} {3n\choose 3k}$$. Prove that $|S_n-3T_n|=2$. #### MarkFL Staff member My solution: For the first sum, we may simply apply the binomial theorem to obtain the closed form: $$\displaystyle S_n=\sum_{k=0}^{3n}{3n \choose k}=(1+1)^{3n}=8^n$$ For the second sum, I looked at the first 5 values: $$\displaystyle T_1=2,\,T_2=22,\,T_3=170,\,T_4=1366,\,T_5=10922$$ and determined the recursion: $$\displaystyle T_{n+1}=7T_{n}+8T_{n-1}$$ The characteristic equation for this recursion is: $$\displaystyle r^2-7r-8=(r+1)(r-8)=0$$ and so the closed form is: $$\displaystyle T_{n}=k_1(-1)^n+k_28^n$$ Using the initial values to determine the parameters, we may write: $$\displaystyle T_{1}=-k_1+8k_2=2$$ $$\displaystyle T_{2}=k_1+64k_2=22$$ Adding the two equations, we find: $$\displaystyle 72k_2=24\implies k_2=\frac{1}{3}\implies k_1=\frac{2}{3}$$ Hence: $$\displaystyle T_{n}=\frac{1}{3}\left(2(-1)^n+8^n \right)$$ And so we find: $$\displaystyle \left|S_n-3T_n \right|=\left|8^n-3\left(\frac{1}{3}\left(2(-1)^n+8^n \right) \right) \right|=\left|8^n-2(-1)^n-8^n \right|=\left|2(-1)^{n+1} \right|=2$$ Shown as desired. #### anemone ##### MHB POTW Director Staff member My solution: For the first sum, we may simply apply the binomial theorem to obtain the closed form: $$\displaystyle S_n=\sum_{k=0}^{3n}{3n \choose k}=(1+1)^{3n}=8^n$$ For the second sum, I looked at the first 5 values: $$\displaystyle T_1=2,\,T_2=22,\,T_3=170,\,T_4=1366,\,T_5=10922$$ and determined the recursion: $$\displaystyle T_{n+1}=7T_{n}+8T_{n-1}$$ The characteristic equation for this recursion is: $$\displaystyle r^2-7r-8=(r+1)(r-8)=0$$ and so the closed form is: $$\displaystyle T_{n}=k_1(-1)^n+k_28^n$$ Using the initial values to determine the parameters, we may write: $$\displaystyle T_{1}=-k_1+8k_2=2$$ $$\displaystyle T_{2}=k_1+64k_2=22$$ Adding the two equations, we find: $$\displaystyle 72k_2=24\implies k_2=\frac{1}{3}\implies k_1=\frac{2}{3}$$ Hence: $$\displaystyle T_{n}=\frac{1}{3}\left(2(-1)^n+8^n \right)$$ And so we find: $$\displaystyle \left|S_n-3T_n \right|=\left|8^n-3\left(\frac{1}{3}\left(2(-1)^n+8^n \right) \right) \right|=\left|8^n-2(-1)^n-8^n \right|=\left|2(-1)^{n+1} \right|=2$$ Shown as desired. Hey thanks for participating MarkFL! And I'm so impressed that you were so fast in cracking this problem! #### caffeinemachine ##### Well-known member MHB Math Scholar For $n=1,2,...,$ set $$\displaystyle S_n=\sum_{k=0}^{3n} {3n\choose k}$$ and $$\displaystyle T_n=\sum_{k=0}^{n} {3n\choose 3k}$$. Prove that $|S_n-3T_n|=2$. Define $f(x)=\sum_{k=0}^{3n}{3n\choose k}x^k$. Then $f(1)+f(\omega)+f(\omega^2)=3T_n$, where $\omega$ is a complex cube root of unity. Note that $f(1)=S_n$. So we get $S_n-3T_n=-[(1+\omega)^{3n}+(1+\omega^2)^{3n}]=-2$ #### anemone ##### MHB POTW Director Staff member Define $f(x)=\sum_{k=0}^{3n}{3n\choose k}x^k$. Then $f(1)+f(\omega)+f(\omega^2)=3T_n$, where $\omega$ is a complex cube root of unity. Note that $f(1)=S_n$. So we get $S_n-3T_n=-[(1+\omega)^{3n}+(1+\omega^2)^{3n}]=-2$ Hi caffeinemachine, Thanks for participating and I really appreciate you adding another good solution to this problem and my thought to solve this problem revolved around the idea that you used too!

https://math.stackexchange.com/questions/2890157/what-is-the-difference-between-lim-h-to0-frac0h-and-lim-h-to-infty

What is the difference between $\lim_{h\to0}\frac{0}{h}$ and $\lim_{h\to\infty}\frac{0}{h}$? What is the difference (if any) between- $$\lim_{h\to0}\frac{0}{h} \text{ and } \lim_{h\to\infty}\frac{0}{h}$$ I argue that both must be $=0$ since the numerator is exactly $0$. But one fellow refuses to agree and argues that the first limit can't be $0$ as anything finite by something tending to $0$ is always $\infty$. So,how can I explain it to the person? Also, if possible can anyone provide some good reference on this particular issue (Apostol perhaps)? Thanks for any help! • Graph the function $f(x) = \dfrac{0}{x}$. Perhaps by seeing it, it will make more sense to the 'fellow'. Explain the geometric interpretation of a limit. – InterstellarProbe Aug 21 '18 at 17:27 • "Anything finite by something tending to 0 is always $\infty$" is not true (at least as long as $0$ counts as finite) -- as this very example shows. Case closed. – Henning Makholm Aug 21 '18 at 17:41 • @tatan: If they don't want to consider your arguments, you can't. Give it up; life is too short for some things. – Henning Makholm Aug 21 '18 at 17:43 • Let "the fellow" read this. – drhab Aug 21 '18 at 18:09 • It's not that the numerator is zero, it's that the fraction is zero. Does your "fellow" agree that $\lim_{h\to0}0=0$. If he does, but doesn't then agree that $\lim_{h\to0}0/h=0$, then there is little hope for him. – Lord Shark the Unknown Aug 21 '18 at 18:22 In both cases you are dividing zero by a nonzero number . Thus your fraction is identically zero and as a result the limit is zero. For the first one let me make an intuitive explanation of what's going on with $\lim_{h\to 0}\frac{0}{h}$. The above limit means "What value does the above expression get while $h$ gets arbitrarily close to $0$", meaning that $h$ is number very close to $0$ but not equal to that. Suppose $h=0.000001$. Then $\frac{0}{h}=\frac{0}{0.000001}=0$. Even if this number gets even closer to $0$ you see that the expression continuous to be equal to $0$. That's because $0$ divided by any number obsiously gives $0$ as the answer. Hence $\lim_{h\to 0}\frac{0}{h}=0$. I wish I helped! When in doubt take it back to the definition. $\lim_\limits{x\to 0} f(x) = 0$ means $\forall \epsilon >0, \exists \delta >0 : 0<|x|<\delta \implies |f(x)|<\epsilon$ As $f(x) = 0$ for all $x \ne 0$ the definition above is satisfied.