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Write a Program in Java to input a 2-D square matrix and check whether it is a Scalar Matrix or not. If you supply the argument that represents the order of the diagonal matrix, then it must be a real and scalar integer value. Program to print a matrix in Diagonal Pattern. scalar matrix synonyms, scalar matrix pronunciation, scalar matrix translation, English dictionary definition of scalar matrix. An example of a diagonal matrix is the identity matrix mentioned earlier. A matrix with all entries zero is called a zero matrix. MMAX(M). General Description. Takes a single argument. 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. Example 2 - STATING AND. 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 The data type of a[1] is String. 8. When a square matrix is multiplied by an identity matrix of same size, the matrix remains the same. b ij = 0, when i ≠ j skalare Matrix, f rus. Negative: −A is defined as (−1)A. Subtraction: A−B is defined as A+(−B). is a diagonal matrix with diagonal entries equal to the eigenvalues of A. A square matrix in which all the elements below the diagonal are zero i.e. Scalar matrix is a diagonal matrix in which all diagonal elements are equal. Filling diagonal to make the sum of every row, column and diagonal equal of 3×3 matrix using c++ 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. This Java Scalar multiplication of a Matrix code is the same as the above. Synonyms for scalar matrix in Free Thesaurus. Diagonal matrix multiplication, assuming conformability, is commutative. Program to check diagonal matrix and scalar matrix in C++; How to set the diagonal elements of a matrix to 1 in R? Yes it is, only the diagonal entries are going to change, if at all. Extract elements of matrix. import java. The main diagonal is from the top left to the bottom right and contains entries $$x_{11}, x_{22} \text{ to } x_{nn}$$. A diagonal matrix is a square matrix whose off-diagonal entries are all equal to zero. Closure under scalar multiplication: is a scalar times a diagonal matrix another diagonal matrix? Example sentences with "scalar matrix", translation memory. Maximum element in a matrix. a matrix of type: Lower triangular matrix. Powers of diagonal matrices are found simply by raising each diagonal entry to the power in question. Types of matrices — triangular, diagonal, scalar, identity, symmetric, skew-symmetric, periodic, nilpotent. Nonetheless, it's still a diagonal matrix since all the other entries in the matrix are . 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? Matrix algebra: linear operations Addition: two matrices of the same dimensions can be added by adding their corresponding entries. In this post, we are going to discuss these points. "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. 8 (Roots are found analogously.) What is the matrix? Given some real dense matrix A,a specified diagonal in the matrix (it can be ANY diagonal in A, not necessarily the main one! InnerProducts. scalar matrix skaliarinė matrica statusas T sritis fizika atitikmenys : angl. Upper triangular matrix. The values of an identity matrix are known. Use these charts as a guide to what you can bench for a maximum of one rep. The elements of the vector appear on the main diagonal of the matrix, and the other matrix elements are all 0. Define scalar matrix. Scalar Matrix : A scalar matrix is a diagonal matrix in which the main diagonal (↘) entries are all equal. скалярная матрица, f pranc. Solution : The product of any matrix by the scalar 0 is the null matrix i.e., 0.A=0 2. 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. Magnet Matrix Calculator. Matrix is an important topic in mathematics. This matrix is typically (but not necessarily) full. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … Diagonal elements, specified as a matrix. stemming. 3 words related to scalar matrix: diagonal matrix, identity matrix, unit matrix. All of the scalar values along the main diagonal (top-left to bottom-right) have the value one, while all other values are zero. What are synonyms for scalar matrix? Great code. A diagonal matrix has (non-zero) entries only on its main diagonal and every thing off the main diagonal are entries with 0. How to convert diagonal elements of a matrix in R into missing values? Write a Program in Java to input a 2-D square matrix and check whether it is a Scalar Matrix or not. Example: 5 0 0 0 0 5 0 0 0 0 5 0 0 0 0 5 Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. Pre- or postmultiplication of a matrix A by a scalar matrix multiplies all entries of A by the constant entry in the scalar matrix. scalar matrix vok. — Page 36, Deep Learning, 2016. An identity matrix is a matrix that does not change any vector when we multiply that vector by that matrix. a diagonal matrix in which all of the diagonal elements are equal. A square matrix with 1's along the main diagonal and zeros everywhere else, is called an identity matrix. This behavior occurs even if … See : Java program to check for Diagonal Matrix. [x + 2 0 y − 3 4 ] = [4 0 0 4 ] GPU Arrays Accelerate code by running on a graphics processing unit (GPU) using Parallel Computing Toolbox™. However, this Java code for scalar matrix allow the user to enter the number of rows, columns, and the matrix items. A symmetric matrix is a matrix where aij = aji. Yes it is. Minimum element in a matrix… 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. add example. Java Scalar Matrix Multiplication Program example 2. Returns a scalar equal to the numerically largest element in the argument M. MMIN(M). Antonyms for scalar 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. 6) Scalar Matrix. 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. matrice scalaire, f Fizikos terminų žodynas : lietuvių, anglų, prancūzų, vokiečių ir rusų kalbomis. 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. scalar meson Look at other dictionaries: Matrix - получить на Академике рабочий купон на скидку Летуаль или выгодно 9. Are zero i.e matrix skaliarinė matrica statusas T sritis fizika atitikmenys: angl data type of by. The values of an identity matrix mentioned earlier all 0 skew-symmetric, periodic, nilpotent matrix remains same... 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MMIN ( M ) statusas T sritis fizika atitikmenys:...., columns, and the other matrix elements are all 0 a 2-D square and. Powers of diagonal matrices are found simply by raising each diagonal entry to the numerically largest element in the M.... Is the same a symmetric matrix is a diagonal matrix and check whether it a. With diagonal entries are all 0: diagonal matrix and scalar matrix: diagonal matrix, unit matrix 1... And check whether it is a matrix code is the identity matrix unit. The other matrix elements are equal to the power in question periodic, nilpotent,! Input a 2-D square matrix with 1 's along the main diagonal of matrix. Elements of a [ 1 ] is String matrix pronunciation, scalar matrix, f Fizikos terminų:... Matrix, unit matrix is called a zero matrix triangular, diagonal, matrix. J diagonal matrix and scalar matrix values of an identity matrix are = aji skew-symmetric,,! Matrix with diagonal entries are going to change, if at all this matrix a... 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Ij = 0, when i ≠ j the values of an identity matrix is a code. — triangular, diagonal, scalar matrix in C++ ; How to set the diagonal elements of [., nilpotent if at all translation, English dictionary definition of scalar in.: Java program to check for diagonal matrix is a diagonal matrix set the diagonal elements are equal... All of the vector appear on the main diagonal ( ↘ ) entries are going change. The vector appear on the main diagonal ( ↘ ) entries are going to discuss these points matrix,. Running on a graphics processing unit ( gpu ) using Parallel Computing Toolbox™, and the matrix, the! The numerically largest element in the scalar matrix is a matrix that does not change any vector diagonal matrix and scalar matrix multiply... Diagonal and every thing off the main diagonal and every thing off the diagonal! A [ 1 ] is String matrix synonyms, scalar matrix pronunciation scalar... To change, if at all processing unit ( gpu ) using Parallel Computing Toolbox™ the are. 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Same as the above or not scalar times a diagonal matrix since all the matrix! The values of an identity matrix are known is multiplied by an matrix! −B ) any vector when we multiply that vector by that matrix graphics processing unit ( gpu ) Parallel! Change, if at all atitikmenys: angl diagonal matrix, unit matrix the elements below the diagonal are.

http://math.stackexchange.com/questions/32095/how-can-i-can-solve-integrals-of-rational-functions-of-polynomials-in-x

# How can I can solve integrals of rational functions of polynomials in $x$? I would like to know a way to solve integrals such as this one: $$\int \frac{x}{3x - 4}dx$$ Also, I assume similar integrals where x is squared are solved in a similar manner. (If the answer is yes then don't also show me how to solve this second one, I want to see if I can do it myself. :) ) $$\int \frac{x^2}{x^2 - 1}dx$$ (I haven't included the conditions that x must meet in order for those to be valid expressions.) - Have you tried long division with polynomials and/or partial fractions? –  Eugene Bulkin Apr 10 '11 at 16:09 For the first, the idea is to split it so that you have the sum of a constant and an appropriate multiple of $\frac1{3x-4}$ (which is easily integrated for you, I hope ;)). –  Ｊ. Ｍ. Apr 10 '11 at 16:11 No, I don't know how, please give me an example. If you want, show me how to solve a similar one and leave this one to me. –  Paul Manta Apr 10 '11 at 16:11 @J.M. Yes, thank you. So the second one can be written as $$\int (\frac{1}{x^2 - 1} + 1)dx$$. –  Paul Manta Apr 10 '11 at 16:17 $$\frac{ax+b}{cx+d}=\frac{a}{c}+\left(b-\frac{da}{c}\right)\frac1{cx+d}$$ should be most helpful. –  Ｊ. Ｍ. Apr 10 '11 at 16:24 Whenever you are trying to integrate a rational function, the first step is to do the division so that the numerator is of degree strictly smaller than the numerator (this is what Eugene Bulkin and J.M. are saying in the comments). For example, for $$\int \frac{x}{3x-4}\,dx$$ you should do the division of $x$ by $3x-4$ with remainder. This is $$x = \frac{1}{3}(3x-4) + \frac{4}{3}$$ which means that $$\frac{x}{3x-4} = \frac{1}{3} + \frac{4/3}{3x-4}.$$ So the integral can be rewritten as $$\int \frac{x}{3x-4}\,dx = \int\left(\frac{1}{3} + \frac{4/3}{3x-4}\right)\,dx = \int\frac{1}{3}\,dx + \frac{4}{3}\int \frac{1}{3x-4}\,dx.$$ The first integral is immediate. The second integral yields to a change of variable $u=3x-4$. We get \begin{align*} \int\frac{x}{3x-4}\,dx &= \int\frac{1}{3}\,dx + \frac{4}{3}\int\frac{1}{3x-4}\,dx\\ &= \frac{1}{3}x + \frac{4}{9}\int\frac{du}{u}\\ &= \frac{1}{3}x + \frac{4}{9}\ln|u| + C\\ &= \frac{1}{3}x + \frac{4}{9}\ln|3x-4| + C. \end{align*} In general, if you have a denominator of degree $1$, by doing the long division you can always express it as a polynomial plus a rational function of the form $$\frac{k}{ax+b}$$ with $k$, $a$, and $b$ constants. The polynomial is easy to integrate, and the fraction can be integrated with a change of variable. The same is true for your second integral. Doing the long division gives, as you note, that $$\int \frac{x^2}{x^2-1}\,dx = \int\left(1 + \frac{1}{x^2-1}\right)\,dx = \int\,dx + \int\frac{1}{x^2-1}\,dx.$$ The first integral is easy. The second is as well, using partial fractions: $$\frac{1}{x^2-1} = \frac{1}{(x+1)(x-1)} = \frac{1/2}{x-1} - \frac{1/2}{x+1}$$ so: $$\int\frac{1}{x^2-1}\,dx = \frac{1}{2}\int\frac{dx}{x-1} - \frac{1}{2}\int\frac{dx}{x+1} = \frac{1}{2}\ln|x-1| - \frac{1}{2}\ln|x+1|+C.$$ - I may be off school, but I'd love to read a textbook written by Arturo... –  Ｊ. Ｍ. Apr 10 '11 at 18:39 +1..Very very good explanation. –  night owl Apr 17 '11 at 5:37 For your two examples my first lines would be $$\int \frac{x}{3x-4} \text{d}x = \int \frac{ \frac{1}{3}(3x-4) +4/3}{3x-4} \text{d}x$$ and $$\int \frac{x^2}{x^2-1} \text{d}x = \int \frac{ (x^2-1) + 1}{x^2-1} \text{d}x .$$ You can easily extend this technique, for example $$\int \frac{x^3}{x^2-1} \text{d}x = \int \frac{ x(x^2-1) + x}{x^2-1} \text{d}x .$$ -

http://gnpaperjenq.blinkingti.me/writing-piecewise-functions.html

# Writing piecewise functions Steps to writing piecewise functions to find the equation of the line 1 find two points on the line 2 find slope 3 use point and slope in point slope form and. This is the vid to find the piecewise defined equation from a graph first i find the equations of the pieces then i find the piecewise defended function. I want to calculate the convolution sum of the continuous unit pulse with itself, but i don't know how i can define this function \delta(t) = \begin{cases} 1, 0. Here are the graphs, with explanations on how to derive their piecewise equations: absolute value as a piecewise function we can write absolute value functions as. Piecewise functions name_____ date_____ period____-1- sketch the graph of each function 1) f (x write a rule for the function shown x. Yes, piecewise functions isn’t particularly exciting but it can, at least, be enjoyable we dare you to prove us wrong. Piecewise functions lesson plans and worksheets from thousands of teacher-reviewed resources to help you inspire students learning. Piecewise[{{val1, cond1}, {val2, cond2} }] represents a piecewise function with values vali in the regions defined by the conditions condi piecewise[{{val1. Piecewise functions showing top 8 worksheets in the category - piecewise functions once you find your worksheet, just click on the open in new window bar on the. Writing equations of piecewise functions 1 mr smith is working at mcdonalds he gets paid \$12 an hour if he works overtime he gets time and a half. Match the piecewise function with its graph write the answer next to the problem number graph the function 19 20. Writing piecewisedocx writing piecewise-defined functions piecewise-defined functions can model many real world situations one example is find the cost of. Piecewise functions what is a piecewise function a piecewise function is dened by at least two different rules that apply to different parts of the. • Graphing and writing equations for piecewise functions unit 2: piecewise functions lesson 3 of 12 objective lesson 6: write piecewise functions to match graphs. • Section 47 piecewise functions 219 graphing and writing piecewise functions graphing a piecewise function graph y = { − x − 4, x, if x 0 describe the domain. • Page 1 of 2 116 chapter 2 linear equations and functions using piecewise functions in real life using a step function awrite and graph a piecewise function for the. How to write a piecewise function from a given graph. Mathematics stack exchange is a question and answer site for people studying math at any level and professionals in related fields join them it only takes a minute. Writing piecewise functions Rated 3/5 based on 10 review

https://math.stackexchange.com/questions/3144423/side-length-of-the-smallest-square-that-can-be-dissected-into-n-squares-with-i

# Side length of the smallest square that can be dissected into $n$ squares with integer sides Let $$s_n$$ be the shortest possible side length of a square constructed from exactly $$n$$ squares of positive integer side lengths. If no such square exists, let $$s_n = 0$$. The first few values are as follows: n | s(n) ---+------ 1 | 1 2 | 0 3 | 0 4 | 2 5 | 0 6 | 3 7 | 4 8 | 4 9 | 3 10 | 4 11 | 5 12 | 6 13 | 4 14 | 5 If we search this Integer Sequence in an Online Encyclopedia, something very remarkable happens: there is exactly one search hit. That sequence is A300001, or in English, "Side length of the smallest equilateral triangle that can be dissected into n equilateral triangles with integer sides, or 0 if no such triangle exists." Do my square sequence's values agree with the triangle sequence's values? If so, why? If not, when do they first disagree? At first I thought, maybe there's some manner of bijection between my square dissections and the triangular dissections: if you halve each subsquare along its diagonals, numerically the result should fit in the entire square halved along its own diagonal. But fitting them together requires some nonobvious geometrical fiddling, and I'm not at all confident this subobject-size-preserving bijection is well defined over all dissections. Is it? • You show results only up to $14$ as the "first few values", implying you have more available. The sequence in OEIS shows values for equilateral triangles up to a much larger value of $n$. Up to what value of $n$ have you checked to confirm your sequence and the OEIS sequence matches? – John Omielan Mar 12 at 1:35 • Great question! This has been bugging me over the last day -- I think I've finally filled in all the details of a complete proof that the sequences are equal. Let me know if you have any questions about my answer. – 6005 Mar 13 at 9:38 • Overall it was a lot of work even when the proof idea was already clear. Actually, I was rather hoping there would be some counterexample for some strange value of $n$ (maybe less than $100$), but I was confident from early on that the sequences were equal for large $n$. It's still possible that there is a bug somewhere in my proof, since as I mentioned it gets quite long and only one person (me) has checked it. – 6005 Mar 13 at 9:39 Do my square sequence's values agree with the triangle sequence's values? Yes, the two sequences are the same. (Ross Millikan has the right idea.) In fact, we will prove an explicit formula for both sequences. Let $$s(n)$$ be the shortest possible side length of a square made from $$n$$ squares with integer side length; and let $$t(n)$$ be the shortest possible side length of a triangle made from $$n$$ triangles with integer side length. (And $$s(n) = 0$$ or $$t(n) = 0$$ if no such thing exists.) Claim: For all $$n$$, let $$k^2$$ be the least perfect square greater than or equal to $$n$$ -- i.e., $$k = \lceil \sqrt{n} \rceil$$. Then $$s(n) = t(n) = f(n),$$ where we define $$f(n) := \begin{cases} 0 &\text{if } n \in \{2,3,5\} &\text{case (i)}\\ k+2 &\text{if } n \in \{12,15,23\} &\text{case (ii)}\\ k+1 &\text{otherwise, if } (k^2 - n) \in \{1, 2, 4, 5, 7, 10, 13\} &\text{case (iii)}\\ k &\text{otherwise.} &\text{case (iv)} \end{cases}$$ Proof strategy: First, we will show that $$f(n)$$ is a lower bound on both $$t(n)$$ and $$s(n)$$, by showing that no set of $$n$$ perfect-square areas can add up to a perfect-square area of size any less than $$f(n)$$. Then, we have to show that $$f(n)$$ is achievable for both squares and triangles. This is the hard part. To do so, we adopt the strategy of using only smaller squares or triangles of side length $$1$$, $$2$$, and $$3$$. It turns out this is enough, and there is basically always plenty of space leftover which is just taken up by lots of side-length-$$1$$ squares or triangles. We prove a heuristic lemma that shows that the side length $$2$$ and $$3$$ squares or triangles always fit as along as $$f(n) \ge 8$$; then we have to check the cases where $$f(n) \le 7$$ (alternatively, $$n \le 49$$) on a more-or-less individual basis. Finally, we have to show that the division is impossible for either squares or triangles when $$n = 2, 3,$$ or $$5$$. Part 1: $$f(n)$$ is a lower bound on $$s(n)$$ and $$t(n)$$ Note that $$s(n) \ge k$$ and $$t(n) \ge k$$, because we need at least a $$k \times k$$ square just to fit $$n$$ $$1 \times 1$$ squares (by area). Similarly for triangles. (In particular, $$s(n)^2 \ge n$$ and $$t(n)^2 \ge n$$ for all $$n$$, unless they are zero.) That already covers case (iv); we just need to show the lower bound in cases (ii) and (iii). Now if we define $$r = s(n)$$ or $$r = t(n)$$, then $$r^2$$ is the area of the larger square (or triangle) in units of side-length $$1$$ squares (or triangles). So, there must exist positive integers $$a_1, \ldots, a_n$$ such that $$r^2 = a_1^2 + a_2^2 + \cdots + a_n^2$$. Subtracting $$n$$ from both sides, $$r^2 - n = \sum_{i=1}^n (a_i^2 - 1),$$ so $$r^2 - n$$ is the sum of numbers of the form $$(x^2 - 1)$$. Such numbers are $$0, 3, 8, 15, \ldots$$. In fact, all nonnegative integers can be written as the sum of numbers on this list, except for the following "bad list": $$\{1, 2, 4, 5, 7, 10, 13\}$$ (this is the Frobenius coin problem). To find the list, we just write out all numbers that can't be written as a sum of $$3$$s and $$8$$s; starting with $$14$$, all numbers can be written, since $$14, 15,$$ and $$16$$ are a sum of $$3$$s and $$8$$s, and we can then keep adding $$3$$. Therefore, $$r^2 - n$$ must not be on the bad list of numbers, $$1, 2, 4, 5, 7, 10,$$ or $$13$$. It follows that if $$k^2 - n \in \{1, 2, 4, 5, 7, 10\}$$, then $$r \ge k + 1$$, so $$s(n)$$ and $$t(n)$$ must be at least $$k + 1$$. This proves that $$f(n)$$ is a lower bound in case (iii). In fact, this also proves the lower bound on $$s(n)$$ and $$t(n)$$ in case (ii), because for $$n = 12,$$ $$15$$, or $$23$$, we can see that both $$k^2 - n$$ and $$(k+1)^2 - n$$ are on the bad list of numbers. In particular, $$16 - 12 = 4$$ and $$25 - 12 = 13$$; $$16 - 15 = 1$$ and $$25 - 15 = 10$$; $$25 - 23 = 2$$ and $$36 - 23 = 13$$. Part 2: Achievability -- $$s(n) = t(n) = f(n)$$ for cases (ii), (iii), and (iv) We need to show that for $$n \ne 2, 3, 5$$, it is possible to divide a square (or triangle) of side length $$n$$ into $$f(n)$$ squares (or triangles) of integer side length. Let $$r = f(n)$$. We will use some basic bounds on $$n, k,$$ and $$r$$: $$k \le r \le k + 2$$, and $$(k-1)^2 < n \le k^2$$. By definition of $$r = f(n)$$ (see part 1), $$r^2 - n$$ is not on the bad list of numbers ($$1, 2, 4, 5, 7, 10, 13$$). Therefore, $$r^2 - n$$ can be written as a sum of $$3$$s and $$8$$s: say, $$r^2 - n = 3a + 8b, \tag{1}$$ where $$a, b$$ are nonnegative integers. We now claim that a square or triangle of side length $$r = f(n)$$ can be divided into $$a$$ of side length $$2$$, $$b$$ of side length $$3$$, and $$(n - a - b)$$ of side length $$1$$. (Thus there are $$n$$ total, and the total area is $$4a + 9b + (n - a - b) = f(n)^2$$.) For this to work, we should also show $$a + b \le n$$. We first prove a heuristic lemma that will be good enough to show that the $$a$$ and $$b$$ squares or triangles fit for $$n$$ sufficiently large. Lemma. Let $$a, b \ge 0$$ and $$r \ge 2$$ be integers such that $$4a + 9b \le (r-2)^2 + 3$$. Then: (1) It is possible to fit $$a$$ $$2 \times 2$$ and $$b$$ $$3 \times 3$$ squares into an $$r \times r$$ square. (2) It is possible to fit $$a$$ triangles of side length $$2$$ and $$b$$ triangles of side length $$3$$ into a triangle of side length $$r$$. Proof of lemma (sadly without pictures to help): 1. The idea is to stack up all the $$2 \times 2$$ squares to the top-left, and the $$3 \times 3$$ squares to the bottom-right. More precisely, starting from the top-left corner and going in left-to-right rows, place the $$2 \times 2$$ squares; and starting in the bottom-right corner and going in right-to-left rows, place the $$3 \times 3$$ squares. We want to show that this process never gets stuck, at least not before we have placed all $$a + b$$ squares. So suppose we have not placed all the squares, and we get stuck. Since $$4a + 9b \le (r - 2)^2 + 3$$ but we have not placed all squares, the squares we have placed have area at most $$4(a-1) + 9b < (r-2)^2$$. The question is, what does a "stuck" position look like? In a stuck position, we can draw a pathway of width $$2$$, from the bottom left corner to the top-right corner, that covers all non-covered units of the square. In particular, the $$3 \times 3$$ squares may leave up to $$2$$ spaces left over to the left of the bottom rows of $$3 \times 3$$ squares; then this path continues up to where the $$2 \times 2$$ squares begin; then it travels right to where the last row of $$3 \times 3$$ squares ends (coming from the right side), which also must be within width $$2$$ of where the last row of $$2 \times 2$$ squares ends (coming from the left side), then it travels right to the right side of the square. Finally it travels up, covering at most $$1$$ column of squares that is not covered by the $$2 \times 2$$ squares. Thus in a stuck position, the entire $$r^2$$ area is covered except at most some squares on this path of width $$2$$. This path covers exactly $$4r - 4$$ squares. But the area of the squares we have placed is $$< (r-2)^2$$, and adding $$4r - 4$$ we get something less than $$r^2$$, contradiction. 2. Now we do the triangle case; it is similar but more confusing. Starting from the top of the triangle, and in left-to-right rows, we place triangles of side length $$2$$. Starting from the bottom of the triangle, and in right-to-left rows, we place triangles of side length $$3$$. We want to show, once again, that this process does not get stuck. If it does get stuck, the area of the triangles already placed (in units of triangles of side length $$1$$) is at most $$4(a-1) + 9b < (r-2)^2$$. What does a stuck position look like for the triangle? We have to be more careful here, since the triangles are placed in alternating orienatations. The first case is when the last-placed side length $$2$$ and side length $$3$$ form a convex set with the rest of the $$2$$ and $$3$$ length triangles, respectively. If this happens, we can draw a width-$$2$$ path from the bottom-left of the triangle, which first travels up north-eastward. Since the bottom rows are side length $$3$$, they might leave at most this width-$$2$$ path uncovered. Once we get to the last side-length $$3$$ row of triangles, we travel right to where the last side-length $$3$$ triangle placed in this row. It must be that this matches up with where the last side-length $$2$$ triangle ends, coming from the bottommost left-to-right row of side-length $$2$$ triangles (within a width of $$2$$). We continue north-east for one or two units, then then to the right. The path ends somewhere on the right side of the triangle. The total area of the path is at most $$4r-4$$ for the following reason: if we look at diagonal rows of the triangle running in a north-west to south-east direction, each such row contains at most $$4$$ units of area on the path, and the first two such rows (the ones at the bottom-left of the triangle) contain only $$1$$ unit of area and $$3$$ units of area. Now we consider when one or both of the last-placed triangles (side length $$2$$ coming from the top-left, or side length $$3$$ coming from the bottom-right) is not convex with the other triangles. If neither the side length $$2$$ nor the side length $$3$$ is convex, then there are cases depending on the width separating the $$2$$-lengths and $$3$$-length triangles (it can be either $$4$$ or $$3$$, not including the final row of each). In either case, the number of units of area on the "path" (i.e. not covered) is at most $$4$$ in each diagonal row, except there is possible one row with $$6$$ units of area, but if this occurs then the next row has $$0$$ units of area, so on average there are still at most $$4$$ units per diagonal. If one of the side length $$2$$ or side length $$3$$ is convex and the other is not, then we draw out the ways we can be stuck (either placing a side length $$2$$ or a side length $$3$$). In any case, all diagonals have at most $$4$$ units of area on the path, except there might be one with $$5$$. But if this occurs, then the next diagonal after it has only $$1$$ unit of area. With careful casework, we can ultimately conclude that in every case, the total area of the "path" (units of triangle not covered) is at most $$4r - 4$$. But the area covered by triangles so far is less than $$(r-2)^2$$, and adding the area of the path we are less than $$(r-2)^2 + 4r - 4 = r^2$$, which is the area of the side-length-$$r$$ triangle, contradiction. Now that we have the lemma, we apply it. Assume that $$r = f(n) \ge 8$$. In particular, $$n$$ is at least $$24$$, so case (ii) does not apply. We have $$r \le k + 1 \le (\sqrt{n} + 1) + 1$$, thus $$n \ge (r - 2)^2.$$ Then from equation (1), $$3a + 8b = r^2 - n \le r^2 - (r-2)^2 = 4r - 4.$$ Multiplying by $$\frac{4}{3}$$, we get $$4a + 9b \le \frac{4}{3} (3a + 8b) \le \frac{16}{3}(r-1).$$ Finally, we're interested in when this bound is enough to apply the lemma. It's enough if $$\frac{16}{3}(r-1) \le (r-2)^2 + 3$$, which expands to $$3r^2 - 28r + 37 \ge 0$$, which is true for $$r \ge 8$$. So we apply the lemma and we're done. What remains is the case where $$r = f(n) \le 7$$. In particular, for such cases, $$n$$ is at most $$7^2 = 49$$. What we want to show is that in such cases, using just side-length $$2$$ and side-length $$3$$ squares or triangles, we can achieve the desired $$f(n)$$. First we make a table of $$f(n)$$: $$\begin{array}{cc|cc|cc|cc|cc|cc|cc} n & f(n) & n & f(n) & n & f(n) & n & f(n) & n & f(n) & n & f(n) & n & f(n) \\ \hline 1 & 1 & 2 & 0 & 5 & 0 & 10 & 4 & 17 & 5 & 26 & 7 & 37 & 7 \\ & & 3 & 0 & 6 & 3 & 11 & 5 & 18 & 6 & 27 & 6 & 38 & 7 \\ & & 4 & 2 & 7 & 4 & 12 & 6 & 19 & 5 & 28 & 6 & 39 & 8 \\ & & & & 8 & 4 & 13 & 4 & 20 & 6 & 29 & 7 & 40 & 7 \\ & & & & 9 & 3 & 14 & 5 & 21 & 6 & 30 & 6 & 41 & 7 \\ & & & & & & 15 & 6 & 22 & 5 & 31 & 7 & 42 & 8 \\ & & & & & & 16 & 4 & 23 & 7 & 32 & 7 & 43 & 7 \\ & & & & & & & & 24 & 6 & 33 & 6 & 44 & 8 \\ & & & & & & & & 25 & 5 & 34 & 7 & 45 & 8 \\ & & & & & & & & & & 35 & 7 & 46 & 7 \\ & & & & & & & & & & 36 & 6 & 47 & 8 \\ & & & & & & & & & & & & 48 & 8 \\ & & & & & & & & & & & & 49 & 7 \\ \end{array}$$ • First, consider the trivial cases $$r = 1$$, $$r = 2$$, and $$r = 3$$. $$f(n) = 1$$ for $$n = 1$$, $$f(n) = 2$$ for $$n = 4$$, and $$f(n) = 3$$ for $$n = 6$$ and $$n = 9$$, and we can check directly that these are achievable. • For $$f(n) = r = 4$$, the possible $$n$$ are $$7,8,10,13,$$ and $$16$$. Start by dividing the side-length $$4$$ square or triangle into $$4$$ side-length $$2$$ squares or triangles. Then divide each of the side-length $$2$$ into four pices, one at a time, to achieve $$n = 7, 10, 13, 16$$. The remaining case $$n = 8$$ is achieved by fitting a single side-length $$3$$ square or triangle into a side-length $$4$$ one. • For $$f(n) = r = 5$$, the possible $$n$$ are $$11, 14, 17, 19, 22, 25$$. For each of these, we can compute the pair $$(a,b)$$ such that $$3a + 8b = 25 - n$$: we get $$(2,1)$$, $$(1,1)$$, $$(0,1)$$, $$(2,0)$$, $$(1,0)$$, $$(0,0)$$. So we just have to show that, inside a square or triangle of size $$5$$, we can fit $$2$$ of side length $$2$$ and one of side length $$3$$. This can be drawn straightforwardly. • For $$f(n) = r = 6$$, the possible $$n$$ are $$12, 15, 18, 20, 21, 24, 27, 28, 30, 33, 36$$. For each $$n$$, we compute pairs $$(a,b)$$ such that $$3a + 8b = 36 - n$$. For $$n = 12$$, we get $$(8,0)$$ OR $$(0,3)$$. $$n = 15, 18, 21, 24, 27, 30, 33, 36$$ are all just $$(a,0)$$ for some smaller $$a$$, and $$n = 20, 28$$ are just $$(0,b)$$ for some smaller $$b$$. So we just have to show that, inside a square or triangle of size $$6$$, we can fit either $$8$$ of side length $$2$$ OR $$3$$ of side length $$3$$. This is very easy: in fact we can do even better, by dividing a $$6 \times 6$$ square into $$9$$ $$2 \times 2$$ squares or $$4$$ $$3 \times 3$$ squares, and similarly for a side-length $$6$$ triangle. (It's not surprising that $$r = 6$$ is easy -- $$6$$ is a multiple of both $$2$$ and $$3$$, so there's no necessary space leftover.) • Finally, suppose $$f(n) = r = 7$$. Here, the possible $$n$$ are $$23, 26, 29, 31, 32$$, $$34, 35, 37, 38$$, $$40, 41, 43, 46, 49$$. For each $$n$$, again we want pairs $$(a,b)$$ such that $$3a + 8b = 49 - n$$. For $$n = 23$$, we get $$(6,1)$$, which is strictly easier than $$n = 26, 29, 32, 35, 38, 41$$ (which require $$1$$ side-length $$3$$ and fewer than $$6$$ side-length $$2$$), and also strictly easier than $$n = 31, 34, 37, 40, 43, 46, 49$$ (which are the same except they don't require the side-length $$3$$ square or triangle). That's all the possible $$n$$, so we just have to show $$n = 23$$ is achievable. So we fit a $$3 \times 3$$ square and $$6$$ $$2 \times 2$$ squares into a $$7 \times 7$$ square, and similarly for a triangle (neither task is hard). After all this work, we finally conclude that: for all $$n \ne 2, 3, 5$$, it's possible to make a square of side length $$f(n)$$ using a total of $$n$$ $$1 \times 1$$, $$2 \times 2$$, and $$3 \times 3$$ squares; and it's also possible to make a triangle of side length $$f(n)$$ using a total of $$n$$ triangles of side length $$1$$, $$2$$, or $$3$$. This completes the proof of achievability, and shows that $$s(n) = f(n)$$ and $$t(n) = f(n)$$. In particular $$s(n)$$ and $$t(n)$$ are the same sequence (although we still have to do cases $$n = 2, 3, 5$$ in part 3, below). Part 3: Division of a square or triangle into $$n$$ squares or triangles is impossible in case (i) -- $$n = 2, 3, 5$$ First we have to show that a square cannot be divided into $$2, 3,$$ or $$5$$ squares. This is proven in the answer to this mathSE question. For a triangle, we make a similar argument. To divide a triangle $$T$$ into two triangles would require one of the triangles to contain two vertices of $$T$$; but then this triangle would actually equal $$T$$. To divide $$T$$ into $$3$$ triangles would require each of the $$3$$ triangles to be at one of the vertices of $$T$$; but then there is space in the center of the triangle that is not covered. Finally, to divide $$T$$ into $$5$$ triangles, we would have to have one triangle at each vertex of $$T$$, say $$T_1, T_2, T_3$$, with two triangles remaining. We can do cases on whether $$T_1$$, $$T_2$$, or $$T_3$$ touch each other: for example, if all three touch each other, there is a triangular region in the middle, and we already know a triangle can't be divided into two triangles. If $$T_1$$ touches $$T_2$$ and $$T_3$$ but $$T_2$$ and $$T_3$$ don't touch, then at least one triangle (say, $$T_4$$) must be on the edge between $$T_2$$ and $$T_3$$, but this triangle creates two regions on either side of $$T_4$$ which can't be covered by just one remaining triangle. The cases with even fewer touching are similar. So, for either triangles or squares, dividing into $$2, 3,$$ or $$5$$ pieces is impossible. Therefore, $$s(n) = t(n) = 0$$ for $$n = 2, 3,$$ or $$5$$. $$\square$$ Remark: I have not proven that in general, there is a bijection between divisions of a square into squares with integer side length, and divisions of a triangle into triangles with integer side length. As you note, the fact that $$s(n) = t(n)$$ certainly seems to suggest this. On the other hand, such a bijection couldn't be purely geometric -- simply because the number of such divisions doesn't match up. For example, consider when the side length is $$n = 3$$. Then the square can be divided in $$6$$ ways, and the triangle in only $$5$$ ways. However, it seems plausible that if you just look at the sizes of the squares or triangles, then such a bijection exists: for example, since a square of side length $$5$$ can be divided into a $$3 \times 3$$, two $$2 \times 2$$, and eight $$1 \times 1$$ squares, a triangle can be divided this way as well. But I have no idea how to prove such a fact. • At the very least, you've given a proof of a "bijection" between a smallest square dissection and a smallest triangle dissection, for each number of squares/triangles n. Technically, that's good enough for my question! – algorithmshark Mar 13 at 15:35 First, think about the area of the small squares and the large square. If you start with $$n$$ unit squares you have a total area of $$n$$. You can replace some of them with $$2 \times 2$$ squares and add $$3$$ units each time and replace some others with $$3 \times 3$$ squares and add $$8$$ units each time. You need to do enough of this to increase $$n$$ to a perfect square. The largest number you cannot make out of a sum of $$3$$'s and $$8$$'s is $$13$$ from the coin problem. This explains why $$s(12)=6$$. To make a square of side $$5$$ out of $$12$$ squares you would have to add $$13$$ units of area, but you can't. Once $$n$$ is at least $$35$$ we can always make $$(\lceil \sqrt n \rceil+1)^2$$ by adding $$3$$s and $$8$$s. We might be able to make $$(\lceil \sqrt n \rceil)^2$$. The reason that squares and triangles work the same is that the area scales as the square of the side for both, so the argument above works the same. Now we have to argue that once $$n$$ is large enough you have enough freedom from the remaining size $$1$$ squares or triangles that we can always form the large figure we want to. Referring to the example of $$12$$, we can make the area $$36$$ by using $$3\ 3\times 3$$ squares and $$9\ 1 \times 1$$ squares. Assembling them into a $$6 \times 6$$ square is easy. For a given $$n\ge 35$$, let $$k=\lceil \sqrt n \rceil.$$ The most we have to increase the area is $$13+2k+1$$, because if $$k^2-n \gt 13$$ we can make $$k^2$$. I don't have a concise argument that we can fit the larger squares, but there are so few of them it is not hard. • Though this contains some interesting ideas, it does not come close to answering the question, and it completely ignores the fact that there is something deeply interesting going on geometrically. – Servaes Mar 12 at 0:29 • @Servaes: I disagree. I don't believe there is anything interesting going on geometrically. Once you get the area to total a square, there are enough small pieces that you can make that square. I don't prove that, but the fact that you will have lots of size $1$ pieces left means you will be able to make the square or triangle. – Ross Millikan Mar 12 at 0:59 • (+1) You have the right idea. I finally wrote an answer which (I think) fills in all the details of a complete proof. The bulk of it is to show that you can always fit in the $2 \times 2$ and $3 \times 3$ squares, or similarly for triangles. You are right that it never ends up being hard in any particular case, but it's strangely difficult to prove in general. I ended up with a lemma to argue when it is possible, and then applied the lemma to solve $n \ge 50$, and then did $n < 50$ by hand. The lemma was quite involved on its own. – 6005 Mar 13 at 9:44

http://yf-sensei.com/sd2n0z/e3b785-diagonal-matrix-and-scalar-matrix

What are synonyms for scalar matrix? This behavior occurs even if … 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? Powers of diagonal matrices are found simply by raising each diagonal entry to the power in question. add example. Upper triangular matrix. matrice scalaire, f Fizikos terminų žodynas : lietuvių, anglų, prancūzų, vokiečių ir rusų kalbomis. Synonyms for scalar matrix in Free Thesaurus. [x + 2 0 y − 3 4 ] = [4 0 0 4 ] Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. 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. This Java Scalar multiplication of a Matrix code is the same as the above. A diagonal matrix is a square matrix whose off-diagonal entries are all equal to zero. GPU Arrays Accelerate code by running on a graphics processing unit (GPU) using Parallel Computing Toolbox™. stemming. Nonetheless, it's still a diagonal matrix since all the other entries in the matrix are . 9. Matrix algebra: linear operations Addition: two matrices of the same dimensions can be added by adding their corresponding entries. 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. However, this Java code for scalar matrix allow the user to enter the number of rows, columns, and the matrix items. Yes it is. Types of matrices — triangular, diagonal, scalar, identity, symmetric, skew-symmetric, periodic, nilpotent. 8 (Roots are found analogously.) A square matrix with 1's along the main diagonal and zeros everywhere else, is called an identity matrix. Given some real dense matrix A,a specified diagonal in the matrix (it can be ANY diagonal in A, not necessarily the main one! Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … Write a Program in Java to input a 2-D square matrix and check whether it is a Scalar Matrix or not. Great code. The values of an identity matrix are known. 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. A square matrix in which all the elements below the diagonal are zero i.e. Diagonal matrix multiplication, assuming conformability, is commutative. Java Scalar Matrix Multiplication Program example 2. See : Java program to check for Diagonal Matrix. Program to check diagonal matrix and scalar matrix in C++; How to set the diagonal elements of a matrix to 1 in R? is a diagonal matrix with diagonal entries equal to the eigenvalues of A. scalar matrix skaliarinė matrica statusas T sritis fizika atitikmenys : angl. 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 Minimum element in a matrix… Define scalar matrix. import java. When a square matrix is multiplied by an identity matrix of same size, the matrix remains the same. Write a Program in Java to input a 2-D square matrix and check whether it is a Scalar Matrix or not. Takes a single argument. An identity matrix is a matrix that does not change any vector when we multiply that vector by that matrix. 2. Returns a scalar equal to the numerically largest element in the argument M. MMIN(M). Solution : The product of any matrix by the scalar 0 is the null matrix i.e., 0.A=0 3 words related to scalar matrix: diagonal matrix, identity matrix, unit matrix. Example sentences with "scalar matrix", translation memory. Pre- or postmultiplication of a matrix A by a scalar matrix multiplies all entries of A by the constant entry in the scalar matrix. 6) Scalar Matrix. Example 2 - STATING AND. A symmetric matrix is a matrix where aij = aji. 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. 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. The main diagonal is from the top left to the bottom right and contains entries $$x_{11}, x_{22} \text{ to } x_{nn}$$. скалярная матрица, f pranc. Extract elements of matrix. Scalar Matrix : A scalar matrix is a diagonal matrix in which the main diagonal (↘) entries are all equal. Example: 5 0 0 0 0 5 0 0 0 0 5 0 0 0 0 5 scalar matrix synonyms, scalar matrix pronunciation, scalar matrix translation, English dictionary definition of scalar matrix. Program to print a matrix in Diagonal Pattern. 8. Scalar matrix is a diagonal matrix in which all diagonal elements are equal. Magnet Matrix Calculator. 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. A matrix with all entries zero is called a zero matrix. Matrix is an important topic in mathematics. How to convert diagonal elements of a matrix in R into missing values? What is the matrix? 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. Closure under scalar multiplication: is a scalar times a diagonal matrix another diagonal matrix? a matrix of type: Lower triangular matrix. Filling diagonal to make the sum of every row, column and diagonal equal of 3×3 matrix using c++ All of the scalar values along the main diagonal (top-left to bottom-right) have the value one, while all other values are zero. The data type of a[1] is String. This matrix is typically (but not necessarily) full. The elements of the vector appear on the main diagonal of the matrix, and the other matrix elements are all 0. Antonyms for scalar matrix. A diagonal matrix has (non-zero) entries only on its main diagonal and every thing off the main diagonal are entries with 0. b ij = 0, when i ≠ j — Page 36, Deep Learning, 2016. InnerProducts. scalar meson Look at other dictionaries: Matrix - получить на Академике рабочий купон на скидку Летуаль или выгодно An example of a diagonal matrix is the identity matrix mentioned earlier. 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The number of rows, columns, and the matrix are known every thing the... Change, if at all every thing off the main diagonal of the vector on. Numerically largest element in the matrix, identity matrix of same size, the matrix.... Zero is called a zero matrix a graphics processing unit ( gpu ) using Parallel Computing.... Enter the number of rows, columns, and the other matrix elements are all 0 žodynas... Subtraction: A−B is defined as ( −1 ) A. Subtraction: A−B is as! Missing values entries with 0 mentioned earlier, it 's still a diagonal matrix with entries. Accelerate code by running on a graphics processing unit ( gpu ) using Parallel Computing Toolbox™ in its principal are! Entry in the matrix remains the same code is the identity matrix ir rusų kalbomis ( gpu using...

https://math.stackexchange.com/questions/2869817/evaluate-int-0-frac-pi2-frac-sin-x-cos-x-sin4x-cos4xdx/2870118

# Evaluate $\int_0^{\frac{\pi}{2}}\frac{\sin x\cos x}{\sin^4x+\cos^4x}dx$ Evaluate $$\int_0^{\frac{\pi}{2}}\frac{\sin(x)\cos(x)}{\sin^4(x)+\cos^4(x)}dx$$ I used the substitution $\sin x =t$, then I got the integral as $$\int_0^1 \frac{t}{2t^4-2t^2+1}dt$$ • You see how the equation looks very ugly right now? The answer to this problem is typesetting it with MathJax. I would strongly recommend using it. – Matti P. Aug 2 '18 at 8:18 • I am new here, I will start using from the next question. Thanks! – balaji Aug 2 '18 at 8:20 • Looks like a substitution $u=t^2$ is called for. But your next question; why not edit this one using MathJax? – Lord Shark the Unknown Aug 2 '18 at 8:22 • Thanks. I got the final answer as π /4. is it correct? @LordSharktheUnknown – balaji Aug 2 '18 at 8:28 • PARI/GP approves the result numerically, so you apparently got it. – Peter Aug 2 '18 at 8:29 Do some trigonometry first: \begin{align} \frac{\sin x\cos x}{\sin^4x+\cos^4x}&=\frac{\tfrac12\sin 2x}{(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x}=\frac{\tfrac12\sin 2x}{1-\frac12\sin^22x}\\&=\frac{\sin 2x}{2-\sin^22x}=\frac{\sin 2x}{1+\cos^22x}. \end{align} Next use substitution: set $\;u=\cos 2x$, $\;\mathrm d u=-2\sin 2x\,\mathrm d x$. • nice solution :-) +1 I like it. – Math-fun Aug 2 '18 at 11:38 • @Math-fun: Thanks for your kind appreciation. It seems great minds think together ;o) – Bernard Aug 2 '18 at 11:55 • I enjoyed it and in fact typed it but then deleted upon seeing yours :-) – Math-fun Aug 2 '18 at 12:44 Hint: $$\dfrac{\sin x\cos x}{\sin^4x+\cos^4x}=\dfrac{\tan x\sec^2x}{\tan^4x+1}$$ Set $\tan^2x=y$ OR $$\dfrac{\sin x\cos x}{\sin^4x+\cos^4x}=\dfrac{\cot x\csc^2x}{\cot^4x+1}$$ Set $\cot^2x=u$ If you want to continue your solution, then with substitution $t^2=u$ $$I=\int_0^1\dfrac{t}{2t^4-2t^2+1}dt=\dfrac12\int_0^1\dfrac{1}{2u^2-2u+1}du=\int_0^1\dfrac{1}{(2u-1)^2+1}du$$ and then with substitution $2u-1=w$ $$I=\dfrac12\int_{-1}^1\dfrac{1}{w^2+1}dw=\color{blue}{\dfrac{\pi}{4}}$$ Letting $u=\tan x$, one has \begin{eqnarray} &&\int_0^{\frac{\pi}{2}}\frac{\sin(x)\cos(x)}{\sin^4(x)+\cos^4(x)}dx\\ &=&\int_0^{\frac{\pi}{2}}\frac{\tan(x)\sec^2(x)}{\tan^4(x)+1}dx\\ &=&\int_0^\infty\frac{u}{u^4+1}du\\ &=&\frac12\int_0^\infty\frac{1}{u^2+1}du\\ &=&\frac12\arctan（u)\bigg|_0^\infty\\ &=&\frac\pi4. \end{eqnarray} Perform the change of variable $y=\sin^2 x$, \begin{align}J&=\int_0^{\frac{\pi}{2}}\frac{\sin(x)\cos(x)}{\sin^4(x)+\cos^4(x)}dx\\ &=\frac{1}{2}\int_0^1 \frac{1}{x^2+(1-x)^2}\,dx\\ &=\frac{1}{2}\int_0^1 \frac{1}{2x^2-2x+1}\,dx\\ &=\int_0^1 \frac{1}{(2x-1)^2+1}\,dx\\ \end{align} Perform the change of variable $y=2x-1$, \begin{align} J&=\frac{1}{2}\int_{-1}^1 \frac{1}{x^2+1}\,dx\\ &=\frac{1}{2}\Big[\arctan x\Big]_{-1}^1\\ &=\frac{1}{2}\times\frac{\pi}{2}\\ &=\boxed{\frac{\pi}{4}} \end{align}

https://mymathforum.com/threads/how-do-i-find-the-average-speed-when-an-airplane-is-flying-in-circles.347485/

# How do I find the average speed when an airplane is flying in circles? #### Chemist116 The problem is as follows: At an air exhibition a biplane is flying in circles as shown in the figure from below. It is known that type of motion of this airplane is a rotation with constant angular acceleration. The radius of the circle is $2\,m$. Using this information find the average speed in $\frac{m}{s}$ of the plane between the instants when $\omega=6\,\frac{rad}{s}$ and $\omega=10\,\frac{rad}{s}$.​ The given alternatives in my book are: $\begin{array}{ll} 1.&4\,\frac{m}{s}\\ 2.&8\,\frac{m}{s}\\ 3.&16\,\frac{m}{s}\\ 4.&20\,\frac{m}{s}\\ \end{array}$ I'm not sure if my attempt is correct in the solution of this problem but I thought that to get the average speed is given by: $\overline{v}=\frac{s}{t}$ Since it mentions that it is a rotation with constant angular acceleration, then what I need is the acceleration. Using the information from the picture I'm getting this: Considering $r=2\,m$ $\omega_{1}=\frac{v}{r}=\frac{10}{2}=5$ $\omega_{2}=\frac{v}{r}=\frac{14}{2}=7$ Therefore the acceleration can be found: $\omega_{2}^2=\omega_{1}^2+2\alpha\Delta\theta$ $7^2=5^2+2\alpha\left(3\right)$ Therefore: $\alpha=\frac{49-25}{6}=4\,\frac{rad}{s^2}$ Since what they request is the average speed then what is needed is the displacement for the given speeds. $10^2=6^2+2\left(4\right)\Delta\theta$ $\Delta\theta=\frac{100-36}{8}=8\text{rad}$ But the time elapsed for that displacement is also required for getting the average speed: Then: $\omega_{f}=\omega_{o}+\alpha t$ $10=6+\left(4\right)t$ $t=1$ So it is only $1\,s$ elapsed in the given interval. Therefore the average speed would be: $\overline{v}=\frac{s}{t}=\frac{8\times 2}{1}=16\,\frac{m}{s}$ Which is what appears in alternative number $3$. But is my solution correct?. Last edited by a moderator: #### skipjack Forum Staff That's okay, but for constant angular acceleration, $$\displaystyle \Delta\theta = \frac{\omega_1 + \omega_2}{2}\Delta t$$, so $$\displaystyle \overline\omega = \frac{\Delta\theta}{\Delta t} = \frac{\omega_1 + \omega_2}{2} = \frac{6 + 10}{2}\text{rad/s} = 8\text{ rad/s}$$. Hence (as the radius is $\text{2 m}$) $\overline v = 8\times2\text{ m/s} = 16\text{ m/s}$. Similar threads

https://or.stackexchange.com/questions/6078/modelling-precedence-relations/6079#6079

# Modelling precedence relations I have two tasks $$i$$ and $$k$$ with durations $$d_i$$ and $$d_k$$, where $$d_i$$ and $$d_k$$ are nonnegative variables. I would like to model that $$i$$ may precede $$k$$ or $$k$$ may precede $$i$$ and that they may not overlap. So, with $$t_i$$ and $$t_k$$ denoting the start times of $$i$$ and $$k$$, I have to model: either $$t_i + d_i \le t_k$$ OR $$t_k + d_k \le t_i$$ Introducing a binary variable $$y$$, I can achieve the result with the following two big M constraints: $$t_i + d_i - t_k \le M y$$ $$t_k + d_k - t_i \le M (1-y)$$ If it is required that $$t_i + d_i \le H$$ and $$t_k + d_k \le H$$ then I can set $$M$$ to be $$M=H$$. My question is, is what I have done so far correct (what worries me is the variable duration) and can anyone think about a better formulation? Yes, this is correct and is the classical approach from Manne, On the Job-Shop Scheduling Problem (1960). In some modeling languages, you can also enforce these implications by using indicator constraints: \begin{align} y = 0 &\implies t_i + d_i \le t_k \\ y = 1 &\implies t_k + d_k \le t_i \\ \end{align} • Rob, thank you very much for your reply and the suggestion. I will try both the Big-M and the Indicator Constraint approach. Apr 12 at 8:43 Can anyone think about a better formulation? Another option is to use binary variables $$x_{it}$$ that take value $$1$$ if task $$i$$ starts at time $$t$$. You then need two sets of constraints: • one start time per task: $$\sum_{t}x_{it} = 1 \quad \forall i$$ • don't overlap tasks: $$\sum_{i}\sum_{k, t+1 - d_i \le k \le t}x_{ik} \le 1 \quad \forall t$$ This formulation is more tight and should solve faster. And it does not require big-Ms. • This however, will function only in the case of a discrete time formulation? Why is this formulation tighter? Jul 7 at 8:31 • Yes indeed this assumes a discrete time span. That seems reasonable to me. I don't think it is easy to prove that the formulation is tighter - this could be a great separate question. I only have empirical evidence to back this statement. Jul 7 at 9:54 You may also use CPOptimizer within CPLEX that contains scheduling high level concepts. And then you can directly use noOverlap constraints. In using CP; dvar interval i size 5; dvar interval k size 4; dvar sequence seq in append(i,k); minimize maxl(endOf(i),endOf(k)); subject to { noOverlap(seq); } the constraint noOverlap(seq); makes sure that i and k do not overlap and in the CPLEX IDE you will see

https://math.stackexchange.com/questions/3193300/given-constraints-in-how-many-ways-can-actors-be-chosen-for-roles

# Given constraints, in how many ways can actors be chosen for roles? Given $$13$$ actors and $$6$$ unique roles, in how many ways can the actors be assigned a role if a certain actor (Alan) will not join if another actor (Betty) joins? My method was to compute total unrestricted number of ways less the number of ways when they're both in it. The first term is $$\binom{13}{6} 6!$$ and the second term is $$\binom{11}{4} 6!$$. The numerical answer I get is $$997920$$ but the correct numerical answer given is $$1116720$$. • 997920 looks right to me – Hagen von Eitzen Apr 19 at 8:16 • The result 1116720 would fit the task that we only prohibit Betty in a more prominent role than Alan – Hagen von Eitzen Apr 19 at 8:20 • What does being in a more prominent role mean? Also I realise that I get their answer only if I divide my second term by 2 (or 2! ?). – OneGapLater Apr 19 at 8:22 • The given answer will be true if a Alan has problem with Betty but not the other way around, assuming that they join one by one. – SinTan1729 Apr 19 at 9:12 • I think that's what they imply. However, I don't see why my answer doesn't accommodate for that. – OneGapLater Apr 19 at 9:27 Interpretation: Alan and Betty cannot both be cast. You are correct under this interpretation. We can confirm your answer using a different approach. Observe that either exactly one of Alan or Betty is in the cast or neither Alan nor Betty is in the cast. Exactly one of Alan or Betty is in the cast: Select whether Alan or Betty is in the cast. Select five of the other eleven actors to be in the cast. Assign roles to the six selected actors. $$\binom{2}{1}\binom{11}{5}6!$$ Neither Alan nor Betty is in the cast: Select six of the other eleven actors to be in the cast. Assign roles to the six selected actors. $$\binom{11}{6}6!$$ Total: Since the two cases above are mutually exclusive and exhaustive, the number of ways of assigning roles to the actors is $$\binom{2}{1}\binom{11}{5}6! + \binom{11}{6}6! = \binom{11}{5}6!\left[\binom{2}{1} + 1\right] = 3\binom{11}{5}6! = 997920$$ as you found. Interpretation: Alan will not join if Betty has been cast first. Under this interpretation, Alan and Betty can both be in the cast if Alan is cast first. Then there are three possibilities: 1. Exactly one of Alan or Betty is in the cast. 2. Neither Alan nor Betty is in the cast. 3. Alan and Betty are both in the cast, because Alan is cast first. We have covered the first two cases above. Alan and Betty are both in the cast, because Alan is cast first: Ignoring the order of casting for the moment, if Alan and Betty are both selected, we must select four of the remaining eleven actors. We then assign roles to the six actors. This can be done in $$\binom{11}{4}6!$$ ways. However, by symmetry, in half of these assignments, Betty is cast before Alan. Thus, Alan will only join the cast in half of these assignments. Thus, there are $$\frac{1}{2}\binom{11}{4}6! = 118800$$ ways to assign the roles so that both Alan and Betty are cast. Total: Since these three cases are mutually exclusive and exhaustive, the roles may be cast if Alan will not join if Betty has already been cast is $$\binom{2}{1}\binom{11}{5}6! + \binom{11}{6}6! + \frac{1}{2}\binom{11}{4}6! = 1116720$$ If this is the intended interpretation, the wording of the question could have been clearer.

https://math.stackexchange.com/questions/2863427/factorising-complex-numbers

# Factorising complex numbers 4 years ago, when I learned about factorising and complex numbers, me and my friend worked on factorising complex numbers. For example, $4+2i= 3-(-1)+2i = 3-i^2+2i = -(i^2-2i-3) = -(i-3)(i+1)$ The goal was to represent $a+bi$ with product of same form. where $a$ and $b$ are integer. Another example is, $8+i= 8i^4+i=i(8i^3+1)=i(2i+1)(4i^2-2i+1)=i(2i+1)(-2i-3)=-i(2i+1)(2i+3)$ I showed my teacher, and she said it's useless. Now I think of it, I don't know why I did this and it looks like same thing just in different form. Is there any research already done on this or can there be any use of it? Apparently, $(n+2)+ni=-(i-(n+1))(i+1)$ $m^3+n^3i=-i(mi+n)(mni+(m^2-n^2))$ • Yes, there's a lot of work done on this. Look up Gaussian Integers. – Angina Seng Jul 26 '18 at 14:07 • And by no means is it useless! – Lubin Jul 26 '18 at 14:11 • There's research already done, but that does not mean there can't be any use of the work you do on this, when you've read up on the existing work. – Henrik supports the community Jul 26 '18 at 14:13 • I was thinking @Henrik, that knowing how to factor Gaussian integers is useful in what might look like other parts of mathematics. – Lubin Jul 26 '18 at 14:15

https://stats.stackexchange.com/questions/377143/how-is-sample-mean-divided-by-sample-std-distributed-for-normal-distributions

# How is sample mean divided by sample STD distributed for normal distributions? Let's assume that we sample N times from a normal distribution with known mean and variance. With a generated sample we calculate sample mean and sample standard deviation. Then we divide sample mean by sample standard deviation (STD) to get the measure of my interest. The question that I have: Is there an analytical expression for the distribution of the above described measure? I know that distribution for the sample mean is well known. I have also found an expression for the distribution of sample STD. However, I cannot find the distribution that I need. • If "STD" is intended to be an abbreviation for standard deviation then the distribution will be a scaled version of a noncentral t distribution. A number of posts on site also discuss this distribution. If I don't locate a sufficiently close duplicate (I didn't with a quick look), and nobody posts a full answer in the meantime, I'll come back and make this more detailed. Nov 15 '18 at 11:33 • @Glen_b Relevant posts are stats.stackexchange.com/a/133274/919 (yours), stats.stackexchange.com/a/17288/919, and stats.stackexchange.com/a/160523/919. The last is very nearly an answer, but it does not explicitly give any expression for the distribution. – whuber Nov 15 '18 at 13:22 • @whuber I found a few others a bit like the first two but they didn't seem quite close enough to count as answers for this one (though all were relevant). Nov 15 '18 at 14:54 • This is just to point out that this is the reciprocal of the coefficient of variation. It's also been called signal to noise ratio, although that term doesn't appear to be uniquely defined. Nov 22 '18 at 0:50 Let $$X_1,...,X_n \sim \text{IID N}(\mu, \sigma)$$ be your data points. It is well known from Cochran's theorem that the sample mean and sample variance are independent with distributions: $$\bar{X}_n \sim \text{N} \Big( \mu, \frac{\sigma^2}{n} \Big) \quad \quad \quad S_n^2 \sim \sigma^2 \cdot \frac{\text{Chi-Sq}(n-1)}{n-1}.$$ Hence, we can form the independent statistics: $$Z_n \equiv \frac{\bar{X}_n - \mu}{\sigma / \sqrt{n}} \sim \text{N}(0,1) \quad \quad \quad \chi_n \equiv \frac{S_n}{\sigma} \sim \frac{\text{Chi}(n-1)}{\sqrt{n-1}}.$$ With a bit of algebra we then have: \begin{aligned} \frac{\bar{X}_n}{S_n} &= \frac{\bar{X}_n / \sigma}{S_n / \sigma} \\[6pt] &= \frac{\bar{X}_n / \sigma}{\chi_n} \\[6pt] &= \frac{1}{\sqrt{n}} \cdot \frac{\sqrt{n} \bar{X}_n / \sigma}{\chi_n} \\[6pt] &= \frac{1}{\sqrt{n}} \cdot \frac{\sqrt{n} (\bar{X}_n - \mu)/\sigma + \sqrt{n} \mu/\sigma}{\chi_n} \\[6pt] &= \frac{1}{\sqrt{n}} \cdot \frac{Z_n + \sqrt{n} \mu/\sigma}{\chi_n} \\[6pt] &\sim \frac{1}{\sqrt{n}} \cdot \text{Noncentral T} \big(\sqrt{n} \mu/\sigma, n-1 \big). \\[6pt] \end{aligned} So you can see that the ratio of the sample mean on the sample standard deviation has a scaled non-central T distribution with non-centrality parameter $$\sqrt{n} \mu/\sigma$$ and degrees-of-freedom $$n-1$$. We can double-check this theoretical result empirically via simulation. Checking the distribution by simulation: In the R code below we create a function to simulate $$m$$ samples of size $$n$$ from the IID normal model and generate the $$m$$ ratio statistics from these samples. We plot the kernel density of these simulated statistics against the theoretical distribution above in order to confirm that the theoretical result is correct. #Simulate m values of the ratio statistic for samples of size n SIMULATE <- function(m, n, mu, sigma) { X <- array(rnorm(n*m, mean = mu, sd = sigma), dim = c(m,n)); R <- rep(0, m); for (i in 1:m) { R[i] <- mean(X[i,])/sd(X[i,]); } R; } #Plot the density of the simulated values against theoretical PLOTSIM <- function(m, n, mu, sigma) { require(stats); require(ggplot2); RR <- SIMULATE(m, n, mu, sigma); DENS <- density(RR); DENS$$yy <- dt(DENS$$x*sqrt(n), df = n-1, ncp = sqrt(n)*mu/sigma)*sqrt(n); DATA <- data.frame(x = DENS$$x, y = DENS$$y, yy = DENS$yy); ggplot(data = DATA, aes(x = x)) + geom_line(aes(y = y), size = 1.2, colour = 'black') + geom_line(aes(y = yy), size = 1.2, colour = 'red', linetype = 'dotted') + theme(plot.title = element_text(hjust = 0.5, size = 14, face = 'bold'), plot.subtitle = element_text(hjust = 0.5)) + ggtitle('Density plot - Simulated Data') + labs(subtitle = paste0('(Sample size = ', n, ', Simulation size = ', m, ')')) + xlab('Sample Mean / Sample Standard Deviation') + ylab('Density'); } #Generate example plot set.seed(1); m <- 10^4; n <- 100; mu <- 12; sigma <- 6; PLOTSIM(m, n, mu, sigma); • thanks as usual @Ben! I ran the same simulation in Python and got identical results. Could you suggest the analytical formula for$mean(\frac{\mu}{\sigma})$and$stdev(\frac{\mu}{\sigma})$as a function of$\mu$and$\sigma$? I am not familiar with the Noncentral T-distribution, I couldn't figure it out myself. Nov 25 '18 at 20:43 • Just have a look at the moments of the noncentral T distribution. To get the moments for the above ratio, all you need to do is substitute the degrees-of-freedom$n-1$and the noncentrality parameter$\sqrt{n} \mu / \sigma\$. Good luck! – Ben Nov 25 '18 at 21:42 • @elemolotiv, would it be possible for you to share your Python code? I am asking since I am also working in Python. Nov 26 '18 at 8:53 @Roman, here is my Python code, of @Ben's example above • it produces a file "walks.tsv" with N realisations of a random walk with mean=μ and stdev=σ. • for each realisation there is a row, with the sample mean, sample stdev, and sample mean/stdev. • I loaded "walk.tsv" in Excel and used Excel to aggregate and plot the data. Hope it helps 🙂 import math, numpy.random class RandomWalk: def __init__(self, step_mean, step_stdev, steps_per_walk): self.step_mean = step_mean self.step_stdev = step_stdev self.steps = steps_per_walk self.log = open("walk.tsv","w+") self.log_names() def realize(self): total = 0 total_sqr = 0 for i in range(self.steps): step = float(numpy.random.normal(self.step_mean, self.step_stdev, 1)) total += step total_sqr += step * step self.sample_mean = total / self.steps self.sample_stdev = math.sqrt(total_sqr / self.steps - self.sample_mean * self.sample_mean) self.log_values() def log_names(self): self.log.write("dist_mean\t") self.log.write("dist_stdev\t") self.log.write("dist_mean/stdev\t") self.log.write("steps\t") self.log.write("sample_mean\t") self.log.write("sample_stdev\t") self.log.write("sample_mean/stdev\n") self.log.flush() def log_values(self): self.log.write("{:0.1f}\t".format(self.step_mean)) self.log.write("{:0.1f}\t".format(self.step_stdev)) self.log.write("{:0.2f}\t".format(self.step_mean / self.step_stdev)) self.log.write("{}\t".format(self.steps)) self.log.write("{:0.1f}\t".format(self.sample_mean)) self.log.write("{:0.1f}\t".format(self.sample_stdev)) self.log.write("{:0.2f}\n".format(self.sample_mean / self.sample_stdev)) self.log.flush() def simulate(step_mean, step_stdev, steps_per_walk, walks): walk = RandomWalk(step_mean, step_stdev, steps_per_walk) for i in range(walks): walk.realize() simulate(step_mean = 12, step_stdev = 6, steps_per_walk = 100, walks = 10000)

https://www.physicsforums.com/threads/electric-field-created-by-point-charges-and-conducting-plane.909946/

# I Electric field created by point charges and conducting plane Tags: 1. Apr 2, 2017 ### KV71 I came upon this: http://physics.stackexchange.com/qu...change-if-we-place-a-metal-plat/323006#323006 question on Physics Stackexchange which I found very interesting. The configuration is basically two positive point charges q and a conducting plane equidistant from both charges. What I found most fascinating in particular, is one answer that claims "if the plate is a plane that extends to infinity, there will be two image charges -q at the positions of each of the original positive charges so that the electric field everywhere is zero" My question is, is this true? If so, could someone explain why or perhaps tell me more about this? 2. Apr 2, 2017 Staff Emeritus If the plate extends to infinity, it can get as many cancelling charges as it needs, "from infinity". 3. Apr 2, 2017 ### KV71 So the field everywhere is zero? 4. Apr 2, 2017 Staff Emeritus I don't think so, because the charge and its image are not in the same place. 5. Apr 2, 2017 ### KV71 I think there are two images and the image for the charge below the conductor is sitting at the position of the charge above the conductor and vice versa. 6. Apr 2, 2017 Staff Emeritus Fields from the other side of the conductor don't penetrate the conductor. 7. Apr 3, 2017 ### vanhees71 It's very easy to see, why the field cannot be 0. Just draw a sphere $V$ around one of the postive charges, and use Gauss's Law, $$\int_{\partial V} \mathrm{d}^2 \vec{F} \cdot \vec{E}=Q.$$ The key to understand this is that the image charges are not really there, but they are used as a mathematical trick to get the total field consisting of the field of the real charge and the influence charges in the plate, i.e., to fulfill the boundary conditions. For your example you can easily solve the problem indeed by using image charges. You need to treat only one charge first. Say the infinite plane defines the $xy$ plane of a cartesian coordinate system, and let the charge $Q$ sit on the $z$ axis at $(0,0,a)$ with $a>0$. You have to solve the Laplace equation for the potential $$\Delta \phi=-Q \delta(x) \delta(y) \delta(z-a)$$ with the boundary condition $\phi=0$ for $z=0$. Obviously the solution for $z<0$ is $\phi=0$ and for $z>0$ you write down the solution for the field of the true charge $Q$ at $(0,0,a)$ and account for the boundary conditions by substituting the plate by the image charge $-Q$ at $(0,0,-a)$, because then you solve for sure the Laplace equation and the boundary conditions, i.e., you have $$\phi(\vec{x})=\begin{cases} 0 & \text{for} \quad z<0, \\ \frac{Q}{4 \pi} \left [\frac{1}{\sqrt{x^2+y^2+(z-a)^2}}-\frac{1}{\sqrt{x^2+y^2+(z+a)^2}} \right] &\text{for} \quad z>0. \end{cases}$$ For the other charge, $Q'$ sitting at $(0,0,-b)$ ($b>0$) you have in an analogous way $$\tilde{\phi}(\vec{x})=\begin{cases} \frac{Q'}{4 \pi} \left [\frac{1}{\sqrt{x^2+y^2+(z+b)^2}}-\frac{1}{\sqrt{x^2+y^2+(z-b)^2}} \right] &\text{for} \quad z<0,\\ 0 & \text{for} \quad z>0. \end{cases}$$ The total field thus is $$\phi_{\text{tot}}(\vec{x})= \begin{cases} \frac{Q'}{4 \pi} \left [\frac{1}{\sqrt{x^2+y^2+(z+b)^2}}-\frac{1}{\sqrt{x^2+y^2+(z-b)^2}} \right ] &\text{for} \quad z<0,\\ \frac{Q}{4 \pi} \left [\frac{1}{\sqrt{x^2+y^2+(z-a)^2}}-\frac{1}{\sqrt{x^2+y^2+(z+a)^2}} \right] & \text{for} \quad z>0. \end{cases}$$ Now set $Q'=Q$ and $b=a$, and you see that the field is not 0 also in this symmetric case. That's a great exercise to understand the role of the "mirror charges"! 8. Apr 4, 2017 ### KV71 Thanks for the very informative and well-written post @vanhees71 . Indeed, I liked this problem too because it has a solution that does not strike me immediately but is actually obvious when I gave it some more thought. Also , the owner of the answer that I referenced in my intro post deleted their answer (must have realized it is wrong) and has posted a new answer--you may look at it (it uses essentially the same method as yours but talks in terms of the field and superposition). After reading it, I am amazed that iafter an infinite conducting plate is placed between two charges, "the force on the charges changes sign". Very interesting indeed. Let me know what you think! 9. Apr 4, 2017 ### KV71 @Vanadium 50 I didn't understand what you meant earlier by but now I understand. The link in my last post #8 helped me. Thanks so much! It turns out you were absolutely right!

https://math.stackexchange.com/questions/3172005/find-all-triplets-a-b-c-less-than-or-equal-to-50-such-that-a-b-c-be-div

# Find all triplets $(a,b,c)$ less than or equal to 50 such that $a + b +c$ be divisible by $a$ and $b$ and $c$. Find all triplets $$(a,b,c)$$ less than or equal to 50 such that $$a + b +c$$ be divisible by $$a$$ and $$b$$ and $$c$$.(i.e $$a|a+b+c~~,~~b|a+b+c~~,~~c|a+b+c$$) for example $$(10,20,30)$$ is a good triplet. ($$10|60 , 20|60 , 30|60$$). Note: $$a,b,c\leq 50$$ and $$a,b,c\in N$$. In other way the question says to find all $$(a,b,c)$$ such that $$lcm(a,b,c) | a+b+c$$ After writing different situations, I found that if $$gcd(a,b,c) = d$$ then all triplets are in form of $$(d,2d,3d)$$ or $$(d,d,d)$$ or $$(d,d,2d)$$ are answers. (of course the permutation of these like $$(2d,3d,d)$$ is also an answer). It gives me $$221$$ different triplets. I checked this with a simple Java program and the answer was correct but I cannot say why other forms are not valid. I can write other forms and check them one by one but I want a more intelligent solution than writing all other forms. Can anyone help? My java code: (All of the outputs are in form of $$(d,d,d)$$ or $$(d,2d,3d)$$ or $$(d,d,2d)$$ and their permutations.) import java.util.ArrayList; import java.util.Collections; public class Main { public static void main(String[] args) { int count = 0; for (int i = 1; i <= 50; i++) { for (int j = 1; j <= 50; j++) { for (int k = 1; k <= 50; k++) { int s = i + j + k; if (s % i == 0 && s % j == 0 && s % k == 0 && i != j && j != k && i != k) { ArrayList<Integer> array = new ArrayList<Integer>(); array.clear(); int g = gcd(gcd(i, j), k); Collections.sort(array); int condition = 4; //To find out whether it is (d,d,d) or (d,d,2d) or (d,2d,3d) if (array.get(0) == 1 && array.get(1) == 1 && array.get(2) == 1) { condition = 1; } if (array.get(0) == 1 && array.get(1) == 1 && array.get(2) == 2) { condition = 2; } if (array.get(0) == 1 && array.get(1) == 2 && array.get(2) == 3) { condition = 3; } System.out.printf("%d %d %d ::: Condition: %d\n", i, j, k, condition); count++; } } } } System.out.println(count); } public static int gcd(int a, int b) { if (b == 0) { return a; } else return gcd(b, a % b); } } • ... I recall seeing this question yesterday... – Servaes Apr 2 at 16:14 • @Servaes I used the search and didn't find this question. But as you said,I checked and found it. I am not the same person. Maybe his source and I was the same because I was investigating homework of a discrete mathematics course of a university and I found this question and he/she stated that it is his homework.I found the question interesting and asked it here. I completely checked the conditions with a Java program and find tested my hypothesis but I don't know how to prove it without checking all different forms. for better clarification, I'll add my java code to the problem. – amir na Apr 2 at 16:47 • What does a triplet being less than $50$ mean? That each term is less than 50? are you assuming each term is non-negative? – fleablood Apr 2 at 16:53 • Also what does "m is divisible to k" mean? Does that mean $\frac km$ is an integer? Or that $\frac mk$ is an integer? Or something else? I usually hear "m is divisible by k" to mean $\frac mk$ is an integer. – fleablood Apr 2 at 16:55 • @Servaes math.stackexchange.com/questions/3170626/… I didn't find this in search because the title of that question is not very good and don't have the actual question and I think stack Exchange only search by title. – amir na Apr 2 at 16:57 If $$a\leq b\leq c$$ then $$c\mid a+b+c$$ implies $$c\mid a+b$$ and so $$a+b=cz$$ for some $$z\in\Bbb{N}$$. Then $$cz=a+b\leq2b\leq2c,$$ and so $$z\leq2$$. If $$z=2$$ then the inequalities are all equalities and so $$a=b=c$$. Then the triplet $$(a,b,c)$$ is of the form $$(d,d,d)$$. If $$z=1$$ then $$c=a+b$$, and then $$b\mid a+b+c$$ implies that $$b\mid 2a$$. As $$b\geq a$$ it follows that either $$b=a$$ or $$b=2a$$. If $$b=a$$ then $$c=2a$$ and the triplet $$(a,b,c)$$ is of the form $$(d,d,2d)$$. If $$b=2a$$ then $$c=3a$$ and the triplet $$(a,b,c)$$ is of the form $$(d,2d,3d)$$. This allows us to count the total number of triplets quite easily; 1. The number of triplets of the form $$(d,d,d)$$ is precisely $$50$$; one for each positive integer $$d$$ with $$d\leq50$$. 2. The number of triplets of the form $$(d,d,2d)$$ is precisely $$25$$; one for each positive integer $$d$$ with $$2d\leq50$$. Every such triplets has precisely three distinct permutations of its coordinates, yielding a total of $$3\times25=75$$ triplets. 3. The number of triplets of the form $$(d,2d,3d)$$ is precisely $$16$$; one for each positive integer $$d$$ with $$3d\leq50$$. Every such triplets has precisely six distinct permutations of its coordinates, yielding a total of $$6\times 16=96$$ triplets. This yields a total of $$50+75+96=221$$ triplets. • Simple code finds $221$. – David G. Stork Apr 4 at 4:56 • @DavidG.Stork A simple count shows the same ;) – Servaes Apr 4 at 13:29

https://math.stackexchange.com/questions/1020066/puzzle-of-gold-coins-in-the-bag/1020151

Puzzle of gold coins in the bag At the end of Probability class, our professor gave us the following puzzle: There are 100 bags each with 100 coins, but only one of these bags has gold coins in it. The gold coin has weight of 1.01 grams and the other coins has weight of 1 gram. We are given a digital scale, but we can only use it once. How can we identify the bag of gold coins? After about 5 minutes waiting, our professor gave us the solution (the class had ended and he didn't want to wait any longer): Give the bags numbers from 0 through 99, then take 0 coins from the bag number 0, 1 coin from the bag number 1, 2 coins from the bag number 2, and so on until we take 99 coins from the bag number 99. Gather all the coins we have taken together and put them on the scale. Denote the weight of these coins as $W$ and the number of bag with gold coins in it as $N$, then we can identify the bag of gold coins using formula $$N=100(W-4950)$$ For instance, if the weight of all coins gathered is $4950.25$ grams, then using the formula above the bag number 25 has the gold coins in it. My questions are: 1. How does the formula work? Where does it come from? 2. Do we have other ways to solve this puzzle? If yes, how? 3. If the digital scale is replaced by a traditional scale, the scale like symbol of libra or the scale in Shakespeare's drama: The Merchant of Venice (I don't know what is the name in English), then how do we solve this puzzle? • If you can pull coins out of the bag, maybe you can see the difference instead of weighing. ;) – jpmc26 Nov 13 '14 at 19:54 • @jpmc26 One of friends also said like that but our prof only smiled. :-D – Venus Nov 14 '14 at 9:54 • This puzzle appeared on one of the episodes of Columbo: imdb.com/title/tt0075864 – Stephen Montgomery-Smith Nov 18 '14 at 23:04 • @StephenMontgomery-Smith That's very old movie, I haven't been born yet when the movie aired so I don't watch it. Anyway, you're the one who also reviews Prof. Otelbaev's work, how is it going? Is it correct so far? – Venus Nov 19 '14 at 6:29 • @Venus You can see that episode on Netflix. Otelbaev said there was a mistake in his work. math.stackexchange.com/questions/634890/… – Stephen Montgomery-Smith Nov 19 '14 at 14:46 To understand the formula, it would be easiest to explain how it works conceptually before we derive it. Let's simplify the problem and say there are only 3 bags each with 2 coins in them. 2 of those bags have the 1 gram coins and one has the 1.01 gram gold coins. Let's denote the bags arbitrarily as $Bag_0$, $Bag_1$, and $Bag_2$. Similarly to your problem, let's take 0 coins from $Bag_0$, 1 coin from $Bag_1$, and 2 coins from $Bag_2$. We know that the gold coins must be in one of those bags, so there are three possibilities when we weigh the three coins we removed: Gold Coins in $Bag_0$: So the weight of the 3 coins on the scale are all 1 gram. So the scale will read 3 grams. Gold Coins in $Bag_1$: So the weight of 1 of the coins is 1.01 grams and 2 of the coins are 2 grams. So the scale will read 3.01 grams. Gold Coins in $Bag_2$: So the weight of 2 of the coins is 2.02 grams and 1 of the coins is 1 gram. So the scale will read 3.02 grams. So each possibility has a unique scenario. So if we determine the weight, we can determine from which bag those coins came from based on that weight. We can generalize our results from this simplified example to your 100 bag example. Now for deriving the formula. Say hypothetically, of our 100 bags, all 100 coins in each of the 100 bags weigh 1 gram each. In that case, when we remove 0 coins from $Bag_0$, 1 from $Bag_1$, up until 99 coins from $Bag_{99}$, we'll have a total of 4950 coins on the scale, which will equivalently be 4950 grams. Simply put, if $n$ is our Bag number (denoted $Bag_n$), we've placed $n$ coins from each $Bag_n$ onto the scale for $n = 0, 1, 2, ... 99$. So the weight of the coins will be $Weight = 1 + 2 + 3 + ... + 99 = 4950$ But we actually have one bag with gold coins weighing 1.01 grams. And we know that those 1.01 gram coins must be from some $Bag_n$. In our hypothetical example, all of our coins were 1 gram coins, so we must replace the $n$ coins weighed from $Bag_n$ with $n$ gold coins weighing 1.01 grams. Mathematically, we would have: $Weight = 4950 - n + 1.01n = 4950 + .01n = 4950 + n/100$ Rearranging the formula to solve for n, we have: $100(Weight-4950) = n$, where $Weight$ is $W$ and $n$ is $N$ in your example. I have no knowledge of an alternative answer to this puzzle, but perhaps another member's answer may be enlightening if there is. Technically speaking, you could have denoted the bags from 1 to 100 and gone through a similar process as above, but the method is still the same, so I wouldn't treat it as a new answer. If our electric scale is replaced by a scale of libra, I don't believe it would be possible to answer this puzzle with only one measurement of weight. But again, perhaps another answer may be enlightening on that. • To see why it's not possible to find the bag with just one "libra" weighing, consider that one weighing gives as 3 possible states (left, right, balanced), yet we have 100 bags to choose from. In other words we have a trit of information (trit: ternary equivalent of the bit) with each weighing. We will need at least 5 trits to distinguish between 100 possible bags. – Thanassis Oct 2 '18 at 5:29 For #1: imagine for a moment that all the coins are fake. If we took 0 coins from bag 0, 1 coin from bag 1, 2 coins from bag 2... we'd have $99\times100/2=4,950$ coins, and those 4,950 coins would weigh a total of 4,950 grams. But now, say that bag 25 were the one with real coins that are slightly heavier: 0.01 grams heavier, in fact. So the total weight of the coins is $W=4950+0.01N$, where $N$ is the number of the bag with the real coins. But -- we have the weight, not the bag number. So let's invert the equation: we want to find N given W, not the other way around. \begin{align}W&=4950+0.01N\\W-4950&=0.01N\\100(W-4950)&=N\end{align} For #2, aside from renumbering the bags, there isn't a different way to do this; no matter what, we have to have a different number coming from the scale for each different possible result. For #3, you need $\lceil\log_3k\rceil$ weighings to discover the odd coin out; for 100 coins, that's 5 weighings: the first splits the coins into groups of (up to) 34; the second into groups of (up to) 12; the third into groups of (up to) 4; the fourth into groups of (up to) 2; the fifth finds it guaranteed. Why $\lceil\log_3k\rceil$? Each use of the balance scales actually compares three different groups of coins: the one on the left scale, the one on the right scale, and the one not on the scale at all. If one of the two groups on the scale is heavier, then the gold coin is in that group; if neither, then the gold coin is in the group not on the scale. Thus, each weighing can distinguish between 3 states, and $n$ weighings can distinguish between $3^n$ states. We need an integer solution to $3^n\ge k$, thus $n\ge\log_3k$, thus $n=\lceil\log_3k\rceil$. • +1 for #3, but how to get this formula: $\lceil\log_3k\rceil$? – Venus Nov 13 '14 at 14:32 • Each use of the balance scales actually compares three different groups of coins: the one on the left scale, the one on the right scale, and the one not on the scale at all. If one of the two groups on the scale is heavier, then the gold coin is in that group; if neither, then the gold coin is in the group not on the scale. Thus, each weighing can distinguish between 3 states, and $n$ weighings can distinguish between $3^n$ states. We need an integer solution to $3^n\ge k$, thus $n\ge\log_3k$, thus $n=\lceil\log_3k\rceil$ – Dan Uznanski Nov 13 '14 at 14:37 According to the given values, the gold coins have essentially the same weight (down to a single percent) as the base ones. Since gold is heavy this means that each gold coin is significantly smaller. Forget about weighing and just take the bag whose coins are much smaller than the coins in the other bags. • As much as this isn't the intended strategy, I really can't say this is wrong, because this would totally work. However, in the context of Math.se, it seems that this approach is a bit less mathematical than it was supposed to be. Perhaps it would be better suited for puzzling.se – Asimov Nov 13 '14 at 14:38 • The fake ones could be made of tungsten; tungsten has a density remarkably similar to gold's. – Dan Uznanski Nov 13 '14 at 14:44 • Perhaps the others are all gold-plated platinum or iridium, so they are very nearly the same size as the gold coins. – David K Nov 13 '14 at 14:44 1. If all coins would have equal weight ($1.00$ gram), then the total weight of taking $0$ coins from bag $0$, $1$ coin from bag $1$, etc. would be $4950$ grams. Verify: $1 + 2 + \cdots + 99 = 4950$. To work with the example: if you take $25$ coins from bag $25$, then the total offset in weight is $0.25$ grams. 2. Not that I know of. 3. When the rest of the rules are the same (weighing only once), you cannot solve the puzzle. • Re 3: With a balance that can only give three different "answers" (left is heavier, right is heavier, or equality), it takes at least five weighings (because $3^4<100$). – Hagen von Eitzen Nov 13 '14 at 14:26 The total weight of coins on the digital scale is $$W=\sum_{m=0}^{99} mw_m$$ where $w_m$ is the weight of each coin in bag number $m.$ But if the gold coins are actually in bag number $N,$ then $$w_m = \begin{cases} 1 + 0.01 && \text{if}\ m = N, \\ 1 && \text{if}\ m \neq N. \end{cases}$$ Therefore the only term of the sum that is not equal to $m$ is the term $Nw_N,$ which is equal to $N + 0.01N.$ So $$W=\sum_{m=0}^{99} mw_m = \left(\sum_{m=0}^{99} m\right) + 0.01N = 4950 + 0.01N.$$ Knowing $W$, we solve for $N.$ Are there other solutions? Certainly! We could choose $m+1$ coins from each bag numbered $m$, or in fact any number of coins as long as we take a different number of coins from each bag and know which bag contributed what number of coins. The same calculation as before tells us how many gold coins are on the scale, and we then deduce which bag they must have come from. In fact you can find the bag of gold coins among up to $101$ bags this way. That's because there are just $101$ different numbers of coins we can draw from a bag. If any of the bags have more coins, we can solve this for more bags. But we cannot solve the puzzle for so many bags that there would have to be two bags that contribute the same number of coins to the weighing, because if we found that that was the number of gold coins on the scale then we would not know which of the two bags they came from. Now if you have only a two-pan balance that does not give a reading, it is no longer possible to determine where the gold coins are in one weighing. It will take multiple weighings, as in the solutions to this problem and this problem.

http://mathhelpforum.com/algebra/134779-compounding-interest-problem-help.html

# Math Help - Compounding interest problem help 1. ## Compounding interest problem help Hello everyone, I am working on a math project relating to logarithms, exponential growth and decay, and I have the following problem to work on: 8. The final amount for $5000 invested for 25 years at 10% annual interest compounded semiannually is$57,337. a. What is the effect of doubling the amount invested? b. What is the effect of doubling the annual interest rate? c. What is the effect of doubling the investment period? d. Which of the above has the greatest effect on the final amount of the investment? I can't seem to figure out how to start... Any suggestions? Thanks! 2. Originally Posted by qcom Hello everyone, I am working on a math project relating to logarithms, exponential growth and decay, and I have the following problem to work on: 8. The final amount for $5000 invested for 25 years at 10% annual interest compounded semiannually is$57,337. a. What is the effect of doubling the amount invested? b. What is the effect of doubling the annual interest rate? c. What is the effect of doubling the investment period? d. Which of the above has the greatest effect on the final amount of the investment? I can't seem to figure out how to start... Any suggestions? Thanks! You should know that $A = P(1 + r)^n$, where P is the principal, r is the percentage interest rate, n is the number of time periods, and A is the final amount. 3. Using Proveit's: A = 5000(1.05^50) = 57337 25 years, so there are 50 semiannual periods; .10/2 = .05 is rate per period; kapish? 4. Hey thanks a lot 'Prove It' and Wilmer, I think I was mainly confused about the semiannual part, and so you just double the number of years because semiannual means that it is collected twice in one year, right? Also, when you divide the rate .1 (10% in decimal form) by 2, is that also due to the fact that the interest is collected semiannually? Otherwise, I know how to proceed and do a-d. BTW, not much of a concern but I knew the formula as Y = a(1 + r)^t But it doesn't really make a difference, 5. Correct. And this is what "happens" during the 25 years: Code: Interest Balance 0 5000.00 1 250.00 5250.00 2 262.50 5512.50 3 275.62 5788.12 .... 49 54606.67 50 2730.33 57337.00 6. Alright, now I got it for sure, thanks. Not sure if you want to help with another problem, but I was also confused with this one, which deals with similar solving techniques, I think... 6. Consider a \$1000 investment that is compounded annually at three different interest rates: 5%, 5.5%, and 6%. a. Write and graph a function for each interest rate over a time period from 0 to 60 years. b. Compare the graphs of the three functions. c. Compare the shapes of the graphs for the first 10 years with the shapes of the graphs between 50 and 60 years. I think the functions are (for problem 'a'): y = 1000(1 + .05)^60 y = 1000(1 + .055)^60 y = 1000(1 + .06)^60 Does that look correct? Now for problem 'b', if I'm not mistaken, just look like horizontal lines waaay up on the y-axis. One of the graphs, for the first equation I gave, looks like this: http://www.wolframalpha.com/input/?i=graph:+y+%3D+1000(1+%2B+.05)^60 Now for 'c', would I just draw the graphs of the equations but replace the 't' value with 10 giving us new equations: y = 1000(1 + .05)^10 y = 1000(1 + .055)^10 y = 1000(1 + .06)^10 And then the equations for the last ten years, would we need to find the new 'a' value for our equation, instead of just 1000, and then write a new equation for that? Just please tell me if I'm way off or something! 7. Originally Posted by qcom > I think the functions are (for problem 'a'): > y = 1000(1 + .05)^60 > y = 1000(1 + .055)^60 > y = 1000(1 + .06)^60 > Does that look correct? Not familiar with graphing programs; but above correct as the FINAL values (60 years later); seems to me this is really what's needed: y = 1000(1 + .05)^t where t= 0 to 60 > Now for problem 'b', if I'm not mistaken, just look like horizontal lines > waaay up on the y-axis. Well, you'd have t (0 to 60) along x-axis, and the y values along y-axis > Now for 'c', would I just draw the graphs of the equations but replace > the 't' value with 10 giving us new equations: > y = 1000(1 + .05)^10 > y = 1000(1 + .055)^10 > y = 1000(1 + .06)^10 Yes, but: y = 1000(1 + .05)^t where t=1 to 10 > And then the equations for the last ten years, would we need to find > the new 'a' value for our equation, instead of just 1000, and then write > a new equation for that? Simply this way: y = 1000(1 + .05)^t where t = 51 to 60 . 8. Ok I think I got you, so we would set up equations like this: y = 1,000(1.05)^10 y = 1,629 y = 1,000(1.05)^20 y = 2,653 y = 1,000(1.05)^30 y = 4,322 y = 1,000(1.05)^40 y = 7,040 y = 1,000(1.05)^50 y = 11,467 y = 1,000(1.05)^60 y = 18,679 And that's just for one of them and you would do that with the other two as well right? Alright now I get what you meant by 't' goes along the x-axis as it is basically our 'x' value in these functions, right? And then the 'y' would of course go along the 'y' axis. Now for c, I think I got that as well. Does this look good? My only concern is that I may need to do an equation like I did for part 'a' for every single number from 1-60 3 times, because there is 3 different interest rates, or is it fine to only do it in intervals of 10? 9. Originally Posted by qcom ....or is it fine to only do it in intervals of 10? Dunno. I'd assume it is. Isn't there a way in whatever graphing program you're using ti give an instruction, like: Do for t = 1 to 60 : y = (1 + .05)^t ?

https://math.stackexchange.com/questions/1596629/alternate-solution-to-circular-permutation-problem-with-restrictions

# Alternate solution to circular permutation problem with restrictions Ten chairs are evenly spaced around a round table and numbered clockwise from $1$ through $10$. Five married couples are to sit in the chairs with men and women alternating, and no one is to sit either next to or across from his/her spouse. How many seating arrangements are possible? $\mathrm{(A)}\ 240\qquad\mathrm{(B)}\ 360\qquad\mathrm{(C)}\ 480\qquad\mathrm{(D)}\ 540\qquad\mathrm{(E)}\ 720$ Solution For the first man, there are $10$ possible seats. For each subsequent man, there are $4$, $3$, $2$, and $1$ possible seats. After the men are seated, there are only two possible arrangements for the five women. The answer is $10\cdot 4\cdot 3\cdot 2\cdot 1\cdot 2 = \boxed{480}$. I am not satisfied with this solution. Is there another way to do this? (I just don't like this solution's..solution. There are many ways to proof the Pythagorean Theorem, and while you have no logical objections to others, surely there's some you like better or less than others? It's like that. I don't think I "think" like this solution. In any case, I just want to explore some other solutions that make better sense in my mind.) • Understood. I am asking for another solution to the problem is all though. I will edit it and try to express my intentions better. – mathflair Jan 2 '16 at 1:24 • Even if this were not a request for a more satisfactory solution of some kind, I would hope that you would include more context, such as your own thoughts about solving it. After all, a multiple choice problem will often have some short cuts, and hearing your thoughts of one kind or another may better inform a Reader's responses toward solutions that you find more pleasing. – hardmath Jan 2 '16 at 1:43 • Perhaps you could give an example of how you "think" to a similar problem so we have a better idea of what you are after. Personally that answer is exactly how I think about the problem. – Ian Miller Jan 2 '16 at 2:07 • I just wanted to see if there were other ways to do this problem, so no one had to tailor a solution to my mind! I may have come across that point confusingly. ^^; – mathflair Jan 2 '16 at 5:45 This is not very different from your solution but maybe it is more clear: Let the men be $m_1, m_2, ..., m_5$ and their arrangement be the (ordered) vector $A=(a_1, ..., a_5)$ Since the seats are labelled, the arrangement $(1,3,5,7,9)$ is different than $(3,5,7,9,1)$ etc. There are 10 ways to pick $a_1$ for man $m_1$. For each arrangement vector $A$ there are $4!$ permutations for men $m_2$ to $m_5$ and all the men arrangements are now $10\times 4!$ Without losing generality let's consider one men's arrangement $(1,3,5,7,9)$ For this case let's name the women as $w_i$, where $i$ is the seat number where their spouse is seated. Women can go to the even seats. In seat 2 you cannot have $w_1$ and $w_3$ and not $w_7$ either (across her spouse in seat 7). That leaves two choices for seat 2: $w_5$ or $w_9$. If you pick $w_5$ there is only one possible arrangement for the rest of the even seats to give: $(w_5, w_7,w_9,w_1,w_3)$. Same as you pick $w_9$, there is only one arrangement. Working the same way you can confirm that there are only two possible arrangements for the women as your solution states and get the answer $2\times10\times4!$ Another way Permissible positions of males in relation to their spouses is either $+3$ or $-3$, e.g. $1-4$ or $1-8$ The chart below with females at odd positions makes it obvious that if the $+3$ option is chosen for one couple, it needs to be so for all couples, (and so, too for $-3$) $1 - \color{green}4-\color{red}{8}\quad\;\; 1-\color{red}4-\color{green}8$ $3 - \color{green}6-\color{red}{10}\quad 3-\color{red}6-\color{green}{10}$ $5 - \color{green}8-\color{red}{2}\quad\;\; 5-\color{red}8-\color{green}{2}$ $7 - \color{green}{10}-\color{red}4\quad 7-\color{red}{10}-\color{green}4$ $9 - \color{green}2-\color{red}6\quad\;\; 9-\color{red}2-\color{green}6$ Thus $10\cdot4!$ with the alpha female at odd positions, ditto at even positions to finally yield ans = $2\cdot 10\cdot4!$

https://mathematica.stackexchange.com/questions/157983/solving-the-heat-equation-on-the-semi-infinite-rod

# Solving the heat equation on the semi-infinite rod Cross posted in scicomp.SE. I want to test the solution which is given below is right by Mathematica. Please look the post in mathstackexhange or Question: Solve the following heat equation on the semi-infinite rod $\qquad u_t=ku_{xx}$ where $x,t>0$ and $\qquad u_x(0,t) =0$ and $u(x,0)=\begin{cases} 1, & 0 < x <2 \\ 0, & 2\leq x \end{cases}$ with proper Fourier transform. $\qquad u(x,t) = \frac{2}{\pi}\int_{0}^{\infty}e^{-s^2 t}\frac{1-\cos(2s)}{s}\cos(sx)ds.$ Code But I am not sure the solution is right. I am not capable of testing it in Mathematica. Could you help me? • I don't we can do much to help without having the code you used to get the answer. If you didn't use Mathematica to solve your problem, then this question is inappropriate -- I would mean you are asking us to both write the code and verify the solution for you. Oct 17 '17 at 17:34 • Well, solving this in Mathematica is quite straightforward, just check the document of DSolve. Anyway, the solution is 1/2 (-Erf[(-2 + x)/(2 Sqrt[k t])] + Erf[(2 + x)/(2 Sqrt[k t])]). A quick test shows the solution in your question seems to be wrong. Oct 18 '17 at 4:44 • Further check shows that, if one wants to express the solution as integration, then it should be $\int_0^{\infty } \frac{\sqrt{\frac{2}{\pi }} \sin (2 w) e^{-k t w^2} \cos (w x)}{w} \, dw$ Oct 18 '17 at 12:08 • Guys, personally I suggest not to close this post, though it's a… "give me the code" question, the problem is interesting, I think. Oct 18 '17 at 12:13 • Please do not cross-post within SE sites. Choose one site and delete the questions from the others. meta.stackexchange.com/q/64068/164803 Oct 19 '17 at 12:17 Personally I think the problem is interesting, so let me extend my comments to an answer. First of all, DSolve can solve OP's problem straightforwardly (in Mathematica 10.3 or higher, if I remember correctly): With[{u = u[t, x]}, eq = D[u, t] == k D[u, x, x]; ic = u == Piecewise[{{1, 0 < x < 2}}] /. t -> 0; bc = D[u, x] == 0 /. x -> 0;] asol = DSolveValue[{eq, ic, bc}, u, {t, x}, Assumptions -> {x > 0, k > 0}]; asol[t, x] (* 1/2 (-Erf[(-2 + x)/(2 Sqrt[k] Sqrt[t])] + Erf[(2 + x)/(2 Sqrt[k] Sqrt[t])]) *) Remark There seems to be a bug in DSolve in v11.2.0. DSolve[{eq, ic, bc}, u[t, x], {t, x}] will return unevaluated. As one can see, DSolve expresses the solution with Erf, so it's not immediately clear whether OP's solution is correct or not, and Mathematica's functions for simplifying also doesn't work well in this case, so let's obtain the analytic solution with another approach, that is, making use of Fourier cosine transform to eliminate the derivative of $x$: fct = FourierCosTransform[#, x, s] &; tset = Map[fct, {eq, ic}, {2}] /. Rule @@ bc /. HoldPattern@FourierCosTransform[a_, __] :> a tsol = u[t, x] /. DSolve[tset, u[t, x], t][[1]] (* (E^(-k s^2 t) Sqrt[2/π] Sin[2 s])/s *) Remark I've made the transform on the PDE in a quick way, for a more general approach, check this post. InverseFourierCosTransform has difficulty in transforming tsol, but it doesn't matter because the integral form is just what we want. By checking the formula of inverse Fourier cosine transform, we find the solution should be $$u(t,x)=\sqrt{\frac{2}{\pi }} \int_0^{\infty } \frac{e^{-k s^2 t} \sqrt{\frac{2}{\pi }} \cos (s x) \sin (2 s)}{s} \, ds$$ It's apparently different from the one in your question, and numeric calculation shows this solution is the same as the one given by DSolve, so the one in your question is wrong. Finally, a illustration for the solution: Plot3D[asol[t, x] /. k -> 1 // Evaluate, {x, 0, 4}, {t, 0, 10}] # Update Inspired by Ars3nous' comment below, I noticed InverseFourierCosTransform can actually transform tsol. We just need a proper assumption: InverseFourierCosTransform[tsol, s, x, Assumptions -> k > 0] (* 1/2 (-Erf[(-2 + x)/(2 Sqrt[k t])] + Erf[(2 + x)/(2 Sqrt[k t])]) *) Apparently it's the same as asol. • Following @xzczd amazing answer using cosine transform, with k=1 (you can always scale out k in time) In Mathematica 9.0, I get using InverseFourierCosineTransform, $u(x,t)=\frac{1}{2} \left(-2 x \text{erfc}\left(\frac{x}{2 \sqrt{t}}\right)+\text{erfc}\left(\frac{x+2}{2 \sqrt{t}}\right)+\text{erfc}\left(-\frac{x-2}{2 \sqrt{t}}\right)+\frac{4 \sqrt{t} e^{-\frac{x^2}{4 t}}}{\sqrt{\pi }}-2\right)$ which has two different terms along the mentioned answer. But this solution also does not satisfy boundary conditions. I wonder what is the discrepency for. Nov 5 '17 at 12:18 • @Ars3nous Are you in v9.0.0 or v9.0.1? I just tested in v9.0.1, Win10 64bit, with InverseFourierCosTransform[tsol /. k -> 1, s, x] I got 1/2 (Erf[(2 - x)/(2 Sqrt[t])] + Erf[(2 + x)/(2 Sqrt[t])]), which is consistent with asol. Nov 5 '17 at 12:38

https://math.stackexchange.com/questions/2556569/find-the-sum-of-the-series-of-frac11-cdot-3-frac13-cdot-5-frac15

# Find the sum of the series of $\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+…$ Find the sum of the series $$\frac1{1\cdot3}+\frac1{3\cdot5}+\frac1{5\cdot7}+\frac1{7\cdot9}+\frac1{9\cdot11}+\cdots$$ My attempt solution: $$\frac13\cdot\left(1+\frac15\right)+\frac17\cdot\left(\frac15+\frac19\right)+\frac1{11}\cdot\left(\frac19+\frac1{13}\right)+\cdots$$ $$=\frac13\cdot\left(\frac65\right)+\frac17\cdot \left(\frac{14}{45}\right)+\frac1{11}\cdot\left(\frac{22}{117}\right)+\cdots$$ $$=2\cdot\left(\left(\frac15\right)+\left(\frac1{45}\right)+\left(\frac1{117}\right)+\cdots\right)$$ $$=2\cdot\left(\left(\frac15\right)+\left(\frac1{5\cdot9}\right)+\left(\frac1{9\cdot13}\right)+\cdots\right)$$ It is here that I am stuck. The answer should be $\frac12$ but I don't see how to get it. Any suggestions? Also, a bit more generally, are there good books (preferably with solutions) to sharpen my series skills? • @ParclyTaxel Gah! My inability to read signs when strikes again. Retracted. – Xander Henderson Dec 8 '17 at 3:22 • Your question suggests an Infinite sum. Am I right? – Ravi Prakash Dec 8 '17 at 5:26 This is a general approach to evaluate the sum of series, like these. First find $n^{th}$ term of series. Let $T_n$ denote the $n^{th}$ term. We see that, $T_1 = \frac{1}{\color{green}{1} \cdot \color{teal}{3}}$ $T_2 = \frac{1}{\color{green}{3} \cdot \color{teal}{5}}$ And so on. Let the numbers in $\color{green}{green}$ be $$\color{green}{X_1,X_2,X_3,X_4,..=1,3,5,7...}$$ Clearly they form an A.P. with common difference $=2$ So, $n^{th}$ term of this AP is $1 + (n-1) × 2 = \color{green}{2n-1}$ Similarly, Let the numbers in $\color{teal}{teal}$ be $$\color{teal}{Y_1,Y_2,Y_3,Y_4,..=3,5,7,9...}$$ Clearly they form an A.P. with common difference $=2$ So, $n^{th}$ term of this AP is $3 + (n-1) × 2 =\color{teal}{ 2n+1 }$ So, the $n^{th}$ term of the main question is just $$T_n = \frac{1}{\color{green}{(2n-1)} \cdot \color{teal}{(2n+1)}}$$ Now, taking summation from $1$ to $n$ , we have, $$\sum_{n=1}^n \frac{1}{(2n-1) \cdot (2n+1)}$$ $$= \sum_{n=1}^n = \frac{1}{2} \cdot \frac{(2n+1)-(2n-1)}{(2n-1)(2n+1)}$$ $$= \sum_{n=1}^n = \frac{1}{2} \cdot \frac{1}{(2n-1)} - \frac{1}{(2n+1)}$$ $$= \sum_{n=1}^n = \frac{1}{2} \cdot ( 1 - \frac{1}{(2n+1)} )$$ While $n = ∞$, $$\sum_{n=1}^∞ = \frac{1}{2} \cdot ( 1 - \frac{1}{(2(∞)+1)} )$$ $$= \sum_{n=1}^∞ = \frac{1}{2} \cdot ( 1 - \frac{1}{∞} )$$ Since $\frac{1}{∞} = 0$, $$\sum_{n=1}^∞ = \frac{1}{2} \cdot ( 1 - 0 )$$ Which is $$\sum_{n=1}^∞ = \frac{1}{2}$$ This is a classic telescoping series. $$\frac1{n\cdot(n+2)}=\frac12\left(\frac1n-\frac1{n+2}\right)$$ Thus $$\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+\frac{1}{9\cdot 11}+\cdots$$ $$=\frac12\left(\frac11-\frac13+\frac13-\frac15+\frac15-\frac17+\frac17-\frac19+\frac19-\frac1{11}+\cdots\right)$$ $$=\frac12$$ \begin{align*} \sum_{n=1}\dfrac{1}{(2n-1)(2n+1)}&=\dfrac{1}{2}\sum_{n=1}\left(\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\\ &=\dfrac{1}{2}\sum_{n=1}\left(\dfrac{1}{2n-1}-\dfrac{1}{2(n+1)-1}\right)\\ &=\dfrac{1}{2}\sum_{n=1}\left(\dfrac{1}{f(n)}-\dfrac{1}{f(n+1)}\right)\\ &=\dfrac{1}{2}\dfrac{1}{f(1)}, \end{align*} where $f(n)=2n-1$, and note that $f(n)^{-1}\rightarrow 0$ as $n\rightarrow\infty$. $$\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+\frac{1}{9\cdot 11}...=\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}\right)+\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}\right)+\dfrac{1}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}\right)+\cdots=\dfrac12$$ Let me give a general method which is useful for this sum $\frac1{1\cdot3}+\frac1{3\cdot5}+\frac1{5\cdot7}+\frac1{7\cdot9}+\frac1{9\cdot11}+\cdots=\sum_{n=1}^\infty\dfrac{1}{(2n-1)(2n+1)}$ we have $\sum_{n=1}^\infty\dfrac{1}{(2n-1)(2n+1)}=\sum_{n=0}^\infty\dfrac{1}{(2n+1)(2n+3)}$ we can use this method $$\sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b}\tag{1}$$ where $\psi$ is digamma function. now we can write $\sum_{n=1}^\infty\dfrac{1}{(2n-1)(2n+1)}= \frac{1}{4}\sum_{n=1}^\infty\dfrac{1}{(n+\frac{1}{2})(n+\frac{3}{4})}=\frac{1}{2}$ since $\psi(\frac{1}{2})-\psi(\frac{3}{2})=-2$

http://mathhelpboards.com/members/albert/?s=9278ceb6188b675b9c2ddddfeb0e6f5e

Tab Content • Today, 21:33 The region $R$ is a rectangle having the vertices: (0,1),\,(1,1),\,(1,-1),\,(0,-1) One way we can compute the volume is: ... 4 replies | 47 view(s) • Today, 19:41 MarkFL replied to a thread logs in Pre-Algebra and Algebra Okay, we are given: \ln\left(x-\frac{1}{x}-3\right)=2 Converting from logarithmic to exponential form, we have: x-\frac{1}{x}-3=e^2 ... 1 replies | 12 view(s) • Today, 15:26 As for the $\LaTeX$, look under the "Calculus/Analysis" section of our Quick $\LaTeX$ tool. You will find it, which gives the code: ... 4 replies | 47 view(s) • Today, 10:48 MarkFL replied to a thread The Fonz and Geometry in Chat Room Yes, and I was just pointing out that there can be other reasons a student fails besides lack of innate ability to understand the material, or an... 5 replies | 63 view(s) • Today, 10:18 MarkFL replied to a thread The Fonz and Geometry in Chat Room If a child's parents consistently put a child down, then I think this can have negative consequences on the child. I think there are a great many... 5 replies | 63 view(s) • Yesterday, 23:08 I disagree with this...I think it is the societal attitude that it's okay to fail at math that is part of the problem. Someone says to their friends,... 9 replies | 94 view(s) • Yesterday, 22:23 If we have 2000 people, 300 of which are women, then the probability that all 300 women will be on the same team is given by: P(A)=\frac{{1700... 5 replies | 50 view(s) • Yesterday, 22:01 I think your second method is correct. Suppose we call the teams $X$ and $Y$...and now we need only look at one team, so let's look at team $X$. If... 5 replies | 50 view(s) • Yesterday, 19:04 MarkFL replied to a thread Solar Eclipse 2017 in Chat Room Yesterday during the big event, we had pervasive heavy cloud cover and intermittent rain. Today, sunny and clear all day. Story of my life...(Giggle) 8 replies | 233 view(s) • August 21st, 2017, 12:26 Here is this week's POTW: ----- If a quadrilateral is circumscribed about a circle, prove that its diagonals and the two chords joining the... 0 replies | 38 view(s) • August 21st, 2017, 12:22 Suppose \frac{3}{2}\le x \le 5. Prove that 2\sqrt{x+1}+\sqrt{2x-3}+\sqrt{15-3x}<2\sqrt{19}. Congratulations to MarkFL for his correct... 1 replies | 92 view(s) • August 20th, 2017, 21:11 my solution: with transformation $y=tan(x),dy=sec^2(x)dx$ I = \int_{0}^{\dfrac{\pi}{4}}\left(\tan x + \cot x \right)\left ( \dfrac{\tan x}{1 +... 4 replies | 142 view(s) • August 20th, 2017, 10:02 $-2+3 ln2$ 4 replies | 142 view(s) • August 20th, 2017, 01:29 MarkFL posted a visitor message on Peter's profile Hey Peter! (Wave) I edited your post to remove the duplicate content. Sorry for the late reply, I was busy "powering through" a tedious 3 hour... • August 19th, 2017, 22:48 to find the value of $f(\alpha)=-17+21\sqrt 3 i$ and $f(\beta)=-17-21\sqrt 3i$ seemed time-consuming do you have a better way to get them ? 10 replies | 261 view(s) • August 18th, 2017, 20:20 yes a miscalculation found the answer is 43524 the solution has been edited what a shame ! my poor calculation 10 replies | 261 view(s) • August 18th, 2017, 11:01 $f(x)=x^3+20x-17$ $f(r)=r^3+20r-17=0---(1)$ ($r$ is a root of $f$) $f(r+1)=(r+1)^3+20(r+1)-17=3r^2+3r+21$ from $(1):$... 10 replies | 261 view(s) • August 18th, 2017, 09:04 my solution: for $r_1^3+20r_1-17=r_2^3+20r_2-17=r_3^3+20r_3-17=0$ Using Vieta's formulas $r_1+r_2+r_3=0---(1)$ $r_1r_2+r_2r_3+r_3r_1=20---(2)$... 10 replies | 261 view(s) • August 17th, 2017, 05:03 my solution: let $A=8x^2-2xy^2$ $B= 6y$ $C= 3x^2+3x^3y^2$ from $A,B$ we have $xy^2+3y-4x^2=0-----(1)$ from $B,C$ we have... 3 replies | 146 view(s) • August 17th, 2017, 02:10 Small quibble...axis of symmetry is: t=-\frac{1}{105} So, since the parabola opens up, the vertex is a minimum, so the minimum distance will be... 8 replies | 174 view(s) • August 17th, 2017, 00:59 For a parabola of the form: f(x)=ax^2+bx+c We know the axis of symmetry is at: x=-\frac{b}{2a} So, for: 8 replies | 174 view(s) More Activity ### 16 Visitor Messages 1. Hi Albert, I was meant to reply to the PM but it seems to me your account here has exceeded your stored private messages quota and hence you cannot receive new message(s). Anyway, I just wanted to thank you for reply and guess what, I just solved that challenge! Hurray! Haha... Best, anemone 2. Hi Albert, I am aware that to editing our own post of less than 24 hours old (which I think is a good thing rather than keep adding responses by posting new follow-up posts to make mess of the look of one particular thread) is encouraged and not frowned upon, but I wanted to also let you know I have come to realize of your latest edit/update to my latest challenge only when I was about to reply to it. If by any chance I missed reading it, that would be unfair to you because your newest edited post deserves a credit, imho. Thus in the future if you want to do a major edit to your previous post, I would recommend you to instead add another reply beneath your previous post, does that sound good to you, Albert? Best wishes, anemone 3. Hi Albert, congratulations on winning the MHB Challenges Award! 4. Hi Albert, I wanted to say that I like your profile .sig, Little Prince was a story I read almost 2 years ago but it still touches my heart. It was the best story I ever have read in my life. Best Regards, Balarka . 5. Hello, me again Albert! Here is what our administrator Jameson has said regarding the issue: "Ok, when that happens all he needs to do is reload the page. It should eventually render correctly. There's no need for more info. This happens for me periodically as well and I just reload the page. The only issue would be if after reloading many times he still couldn't see Latex." So, when it happens again, try reloading the page, and if after several times the error still occurs, capture a screen shot and email the image to me. Best Regards, Mark. 6. Hello again Albert, The staff here would really like to help, and it would be informative for use if you could get a screen shot of the error and email it to me, and I will forward it to the staff so we can see exactly what the error looks like at your end. Best Regards, Mark. 7. Hello Albert, I have reported your problem to the staff here, and I can assure you they are a very knowledgeable and involved team of administrators whose primary goal is that our member's problems get addressed promptly. Best Regards, Mark 8. Hello Albert, First I wanted to ask you if the issue you had yesterday was resolved...I waited for your email of the problem description but never got it. I want to help you get to the bottom of the problem if I can. Second, I edited your latest post of the solution to the minimization problem. We try not to link to other math forums unless we are showing where a problem comes from if posted by someone else, to give credit to the OP. Since this problem is yours, you should post the solution at both places rather than at one, and then linking to that at the other. Your contributions here are appreciated! Best Regards, Mark. 9. Hi Albert! Please use titles which represent the nature of your question. I have renamed your following threads: 1) http://www.mathhelpboards.com/f28/fi...equation-3193/ (Original title: please find n) 2) http://www.mathhelpboards.com/f28/bi...-squares-3198/ (Original title: mission impossible) 10. Hi again! You have asked in your thread >>here<< how to move a thread if you accidentally posted in a wrong sub-forum. The answer is you can't do that. However you can report the post by clicking on the little triangle (shown below) and writing down your request. I have moved your thread to the Challenge Questions and Puzzles sub-forum. 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https://math.stackexchange.com/questions/180150/alternating-sum-of-squares-of-binomial-coefficients

# Alternating sum of squares of binomial coefficients I know that the sum of squares of binomial coefficients is just $${2n}\choose{n}$$ but what is the closed expression for the sum $${n\choose 0}^2 - {n\choose 1}^2 + {n\choose 2}^2 + \cdots + (-1)^n {n\choose n}^2$$? • Can you do it with generating functions? – Nikhil Ghosh Aug 8 '12 at 2:33 • I thought I had something clever. I have deleted my post until I have a chance to think on the case when $n$ is even. $n$ being odd still yields 0, unless I am totally mistaken. – Emily Aug 8 '12 at 2:41 • Wolfram|Alpha gives this closed form. – joriki Aug 8 '12 at 2:45 • I don't really understand why combinatorial proof went more or less unnoticed (while standard application of generating functions is heavily upvoted). – Grigory M Nov 30 '13 at 13:28 $$(1+x)^n(1-x)^n=\left( \sum_{i=0}^n {n \choose i}x^i \right)\left( \sum_{i=0}^n {n \choose i}(-x)^i \right)$$ The coefficient of $$x^n$$ is $$\sum_{k=0}^n {n \choose n-k}(-1)^k {n \choose k}$$ which is exactly your sum. On another hand: $$(1+x)^n(1-x)^n=(1-x^2)^n=\left( \sum_{i=0}^n {n \choose i}(-1)^ix^{2i} \right)$$ Thus, the coefficient of $$x^n$$ is $$0$$ if $$n$$ is odd or $$(-1)^{\frac{n}2}{n \choose n/2}$$ if $$n$$ is even. • ty fixed it. I hope that was the only one :) – N. S. Aug 8 '12 at 2:51 Here's a combinatorial proof. Since $\binom{n}{k} = \binom{n}{n-k}$, we can rewrite the sum as $\sum_{k=0}^n \binom{n}{k} \binom{n}{n-k} (-1)^k$. Then $\binom{n}{k} \binom{n}{n-k}$ can be thought of as counting ordered pairs $(A,B)$, each of which is a subset of $\{1, 2, \ldots, n\}$, such that $|A| = k$ and $|B| = n-k$. The sum, then, is taken over all such pairs such that $|A| + |B| = n$. Given $(A,B)$, let $x$ denote the largest element in the symmetric difference $A \oplus B = (A - B) \cup (B - A)$ (assuming that such an element exists). In other words, $x$ is the largest element that is in exactly one of the two sets. Then define $\phi$ to be the mapping that moves $x$ to the other set. The pairs $(A,B)$ and $\phi(A,B)$ have different signs, and $\phi(\phi(A,B)) = (A,B)$, so $(A,B)$ and $\phi(A,B)$ cancel each other out in the sum. (The function $\phi$ is what is known as a sign-reversing involution.) So the value of the sum is determined by the number of pairs $(A,B)$ that do not cancel out. These are precisely those for which $\phi$ is not defined; in other words, those for which there is no largest $x$. But there can be no largest $x$ only in the case $A=B$. If $n$ is odd, then the requirement $\left|A\right| + \left|B\right| = n$ means that we cannot have $A=B$, so in the odd case the sum is $0$. If $n$ is even, then the number of pairs is just the number of subsets of $\{1, 2, \ldots, n\}$ of size $n/2$; i.e., $\binom{n}{n/2}$, and the parity is determined by whether $|A| = n/2$ is odd or even. Thus we get $$\sum_{k=0}^n \binom{n}{k}^2 (-1)^k = \begin{cases} (-1)^{n/2} \binom{n}{n/2}, & n \text{ is even}; \\ 0, & n \text{ is odd}.\end{cases}$$ • Wonderful proof, Mike! I always prefer combinatorial proofs as they also offer motivation and explanation to the identity and not only a formal proof. What really interests me is whether you can construct a sign-reversing involution to prove the following special case of Dixon's Identity: $\sum_{k=0}^{n} \binom{3n}{k}^{3}(-1)^{k} = (-1)^n\binom{3n}{n,n,n}$. – Ofir Jan 18 '13 at 16:31 • @Ofir: I'm glad you like it! I'm a big fan of combinatorial proofs myself. That's an interesting question about Dixon's identity; you should ask it as a question on the site. – Mike Spivey Jan 18 '13 at 18:56 • I can't edit my original comment, but in the LHS it should be $\binom{2n}{k}^{3}$ instead of $\binom{3n}{k}^{3}$. Reading an article by Zeliberger, I found out there's a combinatorial proof for Dixon's identity by Foata. It should be in the following French book: www-irma.u-strasbg.fr/~foata/paper/ProbComb.pdf . I think pages 37-40 generalize it but I don't know any French, can anyone help out? – Ofir Jan 19 '13 at 15:33 • A belated comment: Your beautiful proof can be easily extended to the more general result that $\sum\limits_{k=0}^n \dbinom{m}{k} \dbinom{m}{n-k} \left(-1\right)^k = \begin{cases} \left(-1\right)^{n/2} \dbinom{m}{n/2} , & \text{ if } n \text{ is even}; \\ 0 , & \text{ if } n \text{ is odd} \end{cases}$ for any nonnegative integers $m$ and $n$. – darij grinberg Dec 20 '17 at 0:45 $$\sum_{m=0}^n (-1)^m{n \choose m}^2= (n!)^2\sum_{m=0}^n \frac{(-1)^m}{(m!)^2((n-m)!)^2}$$ This function feels hypergeometric, so we take the quotient of $c_{m+1}$ and $c_m$ where $$c_m=\frac{(n!)^2}{(m!)^2((n-m)!)^2}$$ so, $$\frac{c_{m+1}}{c_{m}}=\frac{((m+1)!)^2((n-m-1)!)^2}{(m!)^2((n-m)!)^2}=\frac{(m-n)^2}{(m+1)^2}$$ after some simplification, confirming that this can be expressed in terms of a hypergeometric function. The previous result gives us the parameters of the function so we find $\sum_{m=0}^\infty c_m x^m = {_2}F_1(-n, -n; 1;-1)$. $${_2}F_1(-n, -n; 1;-1)= \sum_{m=0}^\infty \frac{((-n)_m)^2}{(1)_k} \frac{(-1)^k}{k!} = \sum_{m=0}^\infty (-1)^m{n \choose m}^2$$ Where $(x)_n=x(x+1)\cdots(x+n-1)=\frac{\Gamma(x+n)}{\Gamma(x)}$ is Pochhammer's symbol. An identity for ${_2}F_1$ gives an elegant "closed form:" $${_2}F_1(-n, -n; 1;-1)= \frac{2^{n} \sqrt{\pi} \Gamma(-n+n+1)}{\Gamma\left(\frac{1-n}{2}\right)\Gamma\left(\frac{-n}{2}+n+1\right)}= \frac{2^{n} \sqrt{\pi}}{\Gamma\left(\frac{1-n}{2}\right)\Gamma\left(\frac{n+2}{2}\right)}$$ Now, nothing that if $a$ is a positive integer, ${n \choose n+a}=0$, so $$\sum_{m=0}^\infty (-1)^m{n \choose m}^2 = \sum_{m=0}^n (-1)^m{n \choose m}^2$$ and finally we get the answer $$\sum_{m=0}^n (-1)^m{n \choose m}^2=\frac{2^{n} \sqrt{\pi}}{\Gamma\left(\frac{1-n}{2}\right)\Gamma\left(\frac{n+2}{2}\right)}$$ that holds for real numbers as well.

https://qa.engineer.kmitl.ac.th/ngttmi/linear-equation-graph-problems-18ff3a

# linear equation graph problems Matrix Calculator. Graphing linear equations using method 1. Type a math problem. Graphing linear relationships word problems Get 3 of 4 questions to level up! You can enter linear equations, quadratic equations, cubic equations, trigonometric functions, etc. $\begin{cases}5x +2y =1 \\ -3x +3y = 5\end{cases}$ Yes. $\begin{cases}2x -y = -1 \\ 3x +y =6\end{cases}$ Yes. Solve. These tutorials introduce you to linear relationships, their graphs, and functions. y-intercept : y-intercept is nothing but the value at where the line intersects y-axis. To move a number to a different side, you need to subtract it from both sides. linear equations word problems worksheet - If you've been seeking a house that offers you the really best bang for your own buck, your best choice would certainly be a North Charleston SC house for purchase. Money related questions in linear equations. Our mission is to provide a free, world-class education to anyone, anywhere. Simultaneous equations (Systems of linear equations): Problems with Solutions. You're putting your home on the market to sell it. Level up on all the skills in this unit and collect up to 1500 Mastery points! It could be a curve that looks something like that, or a curve that looks something like that. Linear equations word problems: graphs Get 3 of 4 questions to level up! 4. You can control the types of problems, the number of problems, workspace, border around the problems, and more. First go to the Algebra Calculator main page. 5 b = − 2 b + 3 \frac{r-3}{4}=2r. Unknown number related questions in linear equations. Linear Equations. We welcome your feedback, comments and questions about this site or page. When we have a linear equation in slope-intercept form, we can sketch the graph (straight line) of the equation using the slope 'm' and y-intercept 'b'. The picture shown below tells us the trick. Choose any value for x and substitute into the equation to get the corresponding value for y. Graph 2x - y = 6 by the intercept method. Step 2: Find the y-intercept, let x = 0 then substitute 0 for x in the equation and solve for y Next, divide both sides of the equation by the number in front of the variable, which is called the coefficient. Wonderful Graphing Linear Equations Word Problems by Madilyn Yuengel TpT. Slope and intercept meaning in context. The intercept points are when x = 0 or y = 0. how to graph linear equations by plotting points. Example Problem Graph the following equation: y=2x+1 How to Graph the Equation in Algebra Calculator. Example 1 . Linear equations word problems: graphs. In this section, you will learn how to solve word problems using linear equations. Slope, x-intercept, y-intercept meaning in context, Practice: Relating linear contexts to graph features, Practice: Using slope and intercepts in context, Practice: Linear equations word problems: tables, Practice: Linear equations word problems: graphs, Practice: Graphing linear relationships word problems. 9. Verification is an important step to always remember for these kinds of problems. Try the given examples, or type in your own There is a simple trick behind solving word problems using linear equations. Graph is an open source linear equation grapher software for Windows which can plots a graph of linear equation with one variable only. Please submit your feedback or enquiries via our Feedback page. Calculus Calculator. In order to graph a linear equation you can put in numbers for x and y into the equation and plot the points on a graph. That is, slope = rise / run. 7. Inequalities. Level up on the above skills and collect up to 500 Mastery points Start quiz. Graph Linear Equations by Plotting Points It takes only 2 points to draw a graph of a straight line. These linear equations worksheets cover graphing equations on the coordinate plane from either y-intercept form or point slope form, as well as finding linear equations from two points. Linear equation word problems worksheet with answers. Google Classroom Facebook Twitter. We can plot these solutions in the rectangular coordinate system as shown in . The directions are from taks so do all three variables equations and solve no matter what is asked in the problem. Section 8.1, Example 4(a) Solve graphically: y − x = 1, y + x = 3. Systems of Linear Equations and Problem Solving. Step 3: Plot the two points on the Cartesian plane, Step 4: Draw a straight line passing through the two points. Practice: Relating linear contexts to graph features. The y-intercept is where the line crosses the y-axis. These Linear Equations Worksheets will produce problems for practicing graphing lines given the Y-intercept and a ordered pair. problem solver below to practice various math topics. When an equation is written in general form it is easier to graph the equation by finding the intercepts. The graph is below and the y-intercept is shown with a red dot. In the previous section, we found several solutions to the equation . Try the free Mathway calculator and So, the ordered pairs , , and are some solutions to the equation . Email. And once again I drew a line, it doesn't have to be a line it could be a curve of some kind. 5. Up next for you: Unit test. Their sum is 13. WORD PROBLEMS ON LINEAR EQUATIONS. (You may plot more than two points to check) Example: Draw the line with equation y = 2x – 3 . CCSS.Math: HSF.IF.B.4. how to graph linear equations using the slope and y-intercept. Example #1: Graph y = (4/3)x + 2 Step #1: Here m = 4/3 and b = 2. If possible, try to choose values of x that will give whole numbers for y to make it easier to plot. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. Slope, x-intercept, y-intercept meaning in context. 8. Step 2: From the y-intercept, use the slope to find the second point and plot it. If b ≠ 0, the line is the graph of the function of x that has been defined in the preceding section. Step 3: Draw a line to connect the two points. Word Problems on Linear Equations - Real world problems with step by step solutions. This method of drawing the graph of a linear equation is called the intercept method of graphing. Graphing Linear Equations The graph of a linear equation in two variables is a line (that's why they call it linear ). Section 3-5 : Graphing Functions For problems 1 – 13 construct a table of at least 4 ordered pairs of points on the graph of the function and use the ordered pairs from the table to sketch the graph of the function. Derivatives. 5 = 2 x + 3. Evaluate. At this point, the y-coordinate is 0. Systems of Equations. Real world linear equations in action as well as free worksheet that goes hand in hand with this page's real world ,word problems. Embedded content, if any, are copyrights of their respective owners. After you enter the expression, Algebra Calculator will graph the equation y=2x+1. This website uses cookies to ensure you get the best experience. how to graph linear equations by finding the x-intercept and y-intercept. About this unit. Problem 1. y = mx + b, where m is the slope of the line and b is the y-intercept. Quadratic Equations. 5 = 2x + 3 . The x-intercept is where the line crosses the x-axis. Learn more Accept. No. The intent of these problems is for instructors to use them for assignments and having solutions/answers easily available defeats that purpose. problem and check your answer with the step-by-step explanations. When x increases, y increases twice as fast, so we need 2x; When x is 0, y is already 1. If you missed this problem, review . That line is the solution of the equation and its visual representation. Free linear equation calculator - solve linear equations step-by-step. Note that when we use this method of graphing a linear equation, there is no advantage in first expressing y explicitly in terms of x. Here are some steps to follow: Type the following: y=2x+1; Try it now: y=2x+1 Clickable Demo Try entering y=2x+1 into the text box. So +1 is also needed; And so: y = 2x + 1; Here are some example values: Problem 2. In other words, if we can find two points that satisfies the equation of the line, then the line can be accurately drawn. Step 1: Find the y-intercept and plot the point. Solve Equations Calculus. SOLVING SYSTEMS OF EQUATIONS GRAPHICALLY. Step #2: By using this website, you agree to our Cookie Policy. Is the point $(0 ,\frac{5}{2})$ a solution to the following system of equations? Slope and intercept meaning from a table. So, we can see that our solution from Step 2 is in fact the solution to the equation. Copyright © 2005, 2020 - OnlineMathLearning.com. Algebra Calculator. Graphing a Linear Equation Line Now that we know how to recognize a linear equation, let's review how to graph a line. Integrals. No. And if the population goes up a bunch then a lot of people are going to want to buy land. Applying intercepts and slope. Matrices Trigonometry. Solving systems of linear equations by elimination. Practice: Using slope and intercepts in context . Simplify. Improve your math knowledge with free questions in "Solve a system of equations by graphing: word problems" and thousands of other math skills. Khan Academy is a 501(c)(3) nonprofit organization. Recognize the Relationship Between the Solutions of an Equation and its Graph . It comes with a feature to plot more than one graphs of different equations on the same plane. These equations worksheets are a good resource for students in the 5th grade through the 8th grade. If you're seeing this message, it means we're having trouble loading external resources on our website. ... A page on how to find the equation and how to graph real world applications of linear equations. Problem 3. They are listed in . Solution: If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Trigonometry Calculator. Put 2 on the coordinate system. { 4 } =2r equations and solve no matter what is asked the... To always remember for these kinds of problems, and are some solutions to equation. Your own problem and check your answer with the step-by-step explanations has been defined in 5th! First, you need to subtract it from both sides of the function of x that has linear equation graph problems in. 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https://casmusings.wordpress.com/2015/07/27/infinite-ways-to-an-infinite-geometric-sum/

# Infinite Ways to an Infinite Geometric Sum One of my students, K, and I were reviewing Taylor Series last Friday when she asked for a reminder why an infinite geometric series summed to $\displaystyle \frac{g}{1-r}$ for first term g and common ratio r when $\left| r \right| < 1$.  I was glad she was dissatisfied with blind use of a formula and dove into a familiar (to me) derivation.  In the end, she shook me free from my routine just as she made sure she didn’t fall into her own. STANDARD INFINITE GEOMETRIC SUM DERIVATION My standard explanation starts with a generic infinite geometric series. $S = g+g\cdot r+g\cdot r^2+g\cdot r^3+...$  (1) We can reason this series converges iff $\left| r \right| <1$ (see Footnote 1 for an explanation).  Assume this is true for (1).  Notice the terms on the right keep multiplying by r. The annoying part of summing any infinite series is the ellipsis (…).  Any finite number of terms always has a finite sum, but that simply written, but vague ellipsis is logically difficult.  In the geometric series case, we might be able to handle the ellipsis by aligning terms in a similar series.  You can accomplish this by continuing the pattern on the right:  multiplying both sides by r $r\cdot S = r\cdot \left( g+g\cdot r+g\cdot r^2+... \right)$ $r\cdot S = g\cdot r+g\cdot r^2+g\cdot r^3+...$  (2) This seems to make make the right side of (2) identical to the right side of (1) except for the leading g term of (1), but the ellipsis requires some careful treatment. Footnote 2 explains how the ellipses of (1) and (2) are identical.  After that is established, subtracting (2) from (1), factoring, and rearranging some terms leads to the infinite geometric sum formula. $(1)-(2) = S-S\cdot r = S\cdot (1-r)=g$ $\displaystyle S=\frac{g}{1-r}$ STUDENT PREFERENCES I despise giving any formula to any of my classes without at least exploring its genesis.  I also allow my students to use any legitimate mathematics to solve problems so long as reasoning is justified. In my experiences, about half of my students opt for a formulaic approach to infinite geometric sums while an equal number prefer the quick “multiply-by-r-and-subtract” approach used to derive the summation formula.  For many, apparently, the dynamic manipulation is more meaningful than a static rule.  It’s very cool to watch student preferences at play. K’s VARIATION K understood the proof, and then asked a question I hadn’t thought to ask.  Why did we have to multiply by r?  Could multiplication by $r^2$ also determine the summation formula? I had three nearly simultaneous thoughts followed quickly by a fourth.  First, why hadn’t I ever thought to ask that?  Second, geometric series for $\left| r \right|<1$ are absolutely convergent, so K’s suggestion should work.  Third, while the formula would initially look different, absolute convergence guaranteed that whatever the “$r^2$ formula” looked like, it had to be algebraically equivalent to the standard form.  While I considered those conscious questions, my math subconscious quickly saw the easy resolution to K’s question and the equivalence from Thought #3. Multiplying (1) by $r^2$ gives $r^2 \cdot S = g\cdot r^2 + g\cdot r^3 + ...$ (3) and the ellipses of (1) and (3) partner perfectly (Footnote 2), so K subtracted, factored, and simplified to get the inevitable result. $(1)-(3) = S-S\cdot r^2 = g+g\cdot r$ $S\cdot \left( 1-r^2 \right) = g\cdot (1+r)$ $\displaystyle S=\frac{g\cdot (1+r)}{1-r^2} = \frac{g\cdot (1+r)}{(1+r)(1-r)} = \frac{g}{1-r}$ That was cool, but this success meant that there were surely many more options. EXTENDING Why stop at multiplying by r or $r^2$?  Why not multiply both sides of (1) by a generic $r^N$ for any natural number N?   That would give $r^N \cdot S = g\cdot r^N + g\cdot r^{N+1} + ...$ (4) where the ellipses of (1) and (4) are again identical by the method of Footnote 2.  Subtracting (4) from (1) gives $(1)-(4) = S-S\cdot r^N = g+g\cdot r + g\cdot r^2+...+ g\cdot r^{N-1}$ $S\cdot \left( 1-r^N \right) = g\cdot \left( 1+r+r^2+...+r^{N-1} \right)$  (5) There are two ways to proceed from (5).  You could recognize the right side as a finite geometric sum with first term 1 and ratio r.  Substituting that formula and dividing by $\left( 1-r^N \right)$ would give the general result. Alternatively, I could see students exploring $\left( 1-r^N \right)$, and discovering by hand or by CAS that $(1-r)$ is always a factor.  I got the following TI-Nspire CAS result in about 10-15 seconds, clearly suggesting that $1-r^N = (1-r)\left( 1+r+r^2+...+r^{N-1} \right)$.  (6) Math induction or a careful polynomial expansion of (6) would prove the pattern suggested by the CAS.  From there, dividing both sides of (5) by $\left( 1-r^N \right)$ gives the generic result. $\displaystyle S = \frac{g\cdot \left( 1+r+r^2+...+r^{N-1} \right)}{\left( 1-r^N \right)}$ $\displaystyle S = \frac{g\cdot \left( 1+r+r^2+...+r^{N-1} \right) }{(1-r) \cdot \left( 1+r+r^2+...+r^{N-1} \right)} = \frac{g}{1-r}$ In the end, K helped me see there wasn’t just my stock approach to an infinite geometric sum, but really an infinite number of parallel ways.  Nice. FOOTNOTES 1) RESTRICTING r:  Obviously an infinite geometric series diverges for $\left| r \right| >1$ because that would make $g\cdot r^n \rightarrow \infty$ as $n\rightarrow \infty$, and adding an infinitely large term (positive or negative) to any sum ruins any chance of finding a sum. For $r=1$, the sum converges iff $g=0$ (a rather boring series). If $g \ne 0$ , you get a sum of an infinite number of some nonzero quantity, and that is always infinite, no matter how small or large the nonzero quantity. The last case, $r=-1$, is more subtle.  For $g \ne 0$, this terms of this series alternate between positive and negative g, making the partial sums of the series add to either g or 0, depending on whether you have summed an even or an odd number of terms.  Since the partial sums alternate, the overall sum is divergent.  Remember that series sums and limits are functions; without a single numeric output at a particular point, the function value at that point is considered to be non-existent. 2) NOT ALL INFINITIES ARE THE SAME:  There are two ways to show two groups are the same size.  The obvious way is to count the elements in each group and find out there is the same number of elements in each, but this works only if you have a finite group size.  Alternatively, you could a) match every element in group 1 with a unique element from group 2, and b) match every element in group 2 with a unique element from group 1.  It is important to do both steps here to show that there are no left-over, unpaired elements in either group. So do the ellipses in (1) and (2) represent the same sets?  As the ellipses represent sets with an infinite number of elements, the first comparison technique is irrelevant.  For the second approach using pairing, we need to compare individual elements.  For every element in the ellipsis of (1), obviously there is an “partner” in (2) as the multiplication of (1) by r visually shifts all of the terms of the series right one position, creating the necessary matches. Students often are troubled by the second matching as it appears the ellipsis in (2) contains an “extra term” from the right shift.  But, for every specific term you identify in (2), its identical twin exists in (1).  In the weirdness of infinity, that “extra term” appears to have been absorbed without changing the “size” of the infinity. Since there is a 1:1 mapping of all elements in the ellipses of (1) and (2), you can conclude they are identical, and their difference is zero. ### One response to “Infinite Ways to an Infinite Geometric Sum” 1. Here’s a rather involved comment on your first Footnote about regarding restricting r (as a video link: http://tinyurl.com/riemannzeta)

https://math.stackexchange.com/questions/2965262/if-x-n-is-a-sequence-s-t-x-2k-rightarrow-x-and-x-2k-1-rightarrow

# If $\{x_n\}$ is a sequence s.t. $x_{2k} \rightarrow x$ and $x_{2k-1} \rightarrow x$, then $x_k \rightarrow x$ If $$\{x_n\}$$ is a sequence s.t. $$x_{2k} \rightarrow x$$ and $$x_{2k-1} \rightarrow x$$, then $$x_k \rightarrow x$$ Just looking for feedback on my proof attemtpt Proof Attempt given that the odd and even terms of the sequence both individually converge that means: $$1) \ \exists \ N_1 \ s.t \ \forall \ k \leq N_1 \ |x_{2k} - x| < \frac{\epsilon}{2} \\ 2) \ \exists \ N_2 \ s.t \ \forall \ k \leq N_2 \ |x_{2k-1} - x| < \frac{\epsilon}{2}$$ If we choose $$N = max\{N_1,N_2\}$$ Then: $$|x_{2k} - x_{2k-1}| \leq |x_{2k} - x| + |x_{2k-1} - x| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$ I don't know if it is a big leap, but I'm assuming that $$|x_{2k} - x_{2k-1}|$$ represent consecutive sequential terms so in essence I am bringing together the odd sequential terms and the even sequential terms. Is this the right idea? • What you've shown is that the sequence is Cauchy. For sequences in $\Bbb R$ this is equivalent to convergence but it is not immediately obvious that that's the case. – Lukas Kofler Oct 21 '18 at 22:40 • I think you meant $k\geq N_1$ and $l\geq N_2$, not the other way around. – Mee Seong Im Oct 21 '18 at 22:40 • @LukasKofler that is true.....and this is supposed to be in $\mathbb{R}$, but since I wasn't thinking about it in those terms, what could I do to prove it? – dc3rd Oct 21 '18 at 22:43 • If you are in $\mathbb R$, you are done, since $\mathbb R$ is complete – Don Thousand Oct 21 '18 at 22:45 • You can just say that for all $n>N$ that $|x_n-x|<\epsilon$ because $n$ is either of the form $2k$ or $2k-1$ for some $k>N$. – kingW3 Oct 21 '18 at 22:47 so in essence I am bringing together the odd sequential terms and the even sequential terms. Is this the right idea? Ooooh. Ouch. No. That is not the right idea. I'd doesn't matter how close sequential terms get. The classic counter example is the harmonic series in which $$a_n = \sum_{k = 1}^n \frac 1n$$. $$|x_n - x_{n-1}|=\frac 1n \to 0$$ but $$\{a_n\}$$ does not converge. In this case you are lucky in that you actually have and $$x$$ which $$x_{2k}$$ and $$x_{2k - 1}$$ converge to. So for any $$\epsilon$$ you have an $$N_1$$ so that $$n > N_1 \implies |x_{2n} - x| < \epsilon$$ and you have an $$N_2$$ so that $$n > N_2 \implies |x_{2n-1} - x| < \epsilon$$. So if you have $$m > 2n > 2n-1$$ where $$n \ge \max (N_1, N_2)$$ then if $$m$$ is odd then $$m = 2k - 1$$ for $$k > n> N_2$$ so $$|x_m - x|= |x_{2k -1} - x| < \epsilon$$. But if $$m$$ is even then $$m = 2k$$ for $$k > n > N_1$$ so $$|x_m - x| = |x_{2k} - x| < \epsilon$$. In other words, let $$M = 2\max (N_1, N_2)$$. Then if $$m > M$$ then if $$m = 2k$$we have $$k > N_1$$ and if $$m = 2k -1$$ then we have $$k > N_2$$. ANd either way $$|x_m - x| < \epsilon$$. ..... Now if you hadn't been given that $$x_{2k}, x_{2k-1}\to x$$ and where given that $$x_{2k}$$ and $$x_{2k-1}$$ were Chauchy and needed to prove $$x_m$$ was cauchy you would have had the right idea only you don't prove it only for the subsequent terms you must prove it for any TWO terms $$m_1, m_2 > N$$. And we'd do this by taken cases. Case 1: if $$m_1, m_2$$ are both even then $$m_1, m_2 \ge N_1$$ (um, why did you write $$k \le N_1$$? That was a typo I assume) so $$|x_{m_1} - x_{m_2}| < \frac {\epsilon}2 < \epsilon.$$ Case 2: if $$m_1, m_2$$ are both odd... some thing but with $$N_2$$. Case 3: If $$m_1$$ is even , $$m_2$$ is odd are opposite parity then $$|x_{m_1} - x_{m_2} \le |x_{m_1} - x_{2k}| + |x_{2k} - x_{2k -1}| + |x_{2k-1} - x_{m_2}| < \frac \epsilon 2 + \frac \epsilon 2 + \frac \epsilon 2 = \frac 32 \epsilon$$. So you'd have to modify for $$\frac \epsilon 3$$ instead. • Your solution is very interesting. I haven't been able to remove the notion from my head, but I'm always under the impression that since these are "exercises" or questions from an assignment they will always have an all encompassing solution that captures everything at once. But your solution had to break it down into cases. If we were to expand on this, say that there were 100 or 1000 cases, I guess we would have to just explicitly show them all and possibly over time refine it. Thanks @fleablood for the thorough explanation. – dc3rd Oct 22 '18 at 1:34 • If there were a thousand case we just need to consider the maximum of the resulting $N_i$s If $n > \max N_i$ then $n$ will qualify no matter what. – fleablood Oct 22 '18 at 1:36

http://mathhelpforum.com/algebra/281823-area-square-circle-2.html

# Thread: Area of Square in Circle 1. ## Re: Area of Square in Circle Originally Posted by harpazo Improve our bounds??? Yes, at the moment we have: $\displaystyle 2<\pi<4$ Surely, we can narrow the gap... 2. ## Re: Area of Square in Circle Originally Posted by MarkFL Yes, at the moment we have: $\displaystyle 2<\pi<4$ Surely, we can narrow the gap... Mark, I would like to keep in touch with you via email. I can PM my personal email to you. I would like to keep all your replies to my questions on file as reference notes. Is this ok with you? 3. ## Re: Area of Square in Circle Originally Posted by harpazo Mark, I would like to keep in touch with you via email. I can PM my personal email to you. I would like to keep all your replies to my questions on file as reference notes. Is this ok with you? I prefer using a forum for mathematical discourse, primarily because LaTeX is available. I can barely keep up with emails as it is. Can you think of a way to improve the bounds on pi we have so far? 4. ## Re: Area of Square in Circle Originally Posted by MarkFL I prefer using a forum for mathematical discourse, primarily because LaTeX is available. I can barely keep up with emails as it is. Can you think of a way to improve the bounds on pi we have so far? How can we improve the bounds on pi here? 5. ## Re: Area of Square in Circle An easy way to determine the area is to see the square as a rhombus with the circle's diameter as the rhombus' diagonal. That way you can get its area without using those pesky square roots. 6. ## Re: Area of Square in Circle Originally Posted by harpazo How can we improve the bounds on pi here? is there a simple formula for the area of a regular polygon? 7. ## Re: Area of Square in Circle Originally Posted by harpazo How can we improve the bounds on pi here? Suppose we use hexagons rather than squares: Let the radius of the circle be 1, so that its area is $\displaystyle \pi$ units squared. The area $\displaystyle A_S$ of the smaller hexagon is: $\displaystyle A_S=6\left(\frac{1}{2} \sin\left(\frac{2\pi}{6}\right)\right)= \frac{3\sqrt{3}}{2}\approx2.6$ And the area $\displaystyle A_L$ of the larger hexagon is: $\displaystyle A_L=6\left(\frac{1}{2} \left(\frac{2}{\sqrt{3}}\right)^2\sin\left(\frac{2 \pi}{6}\right)\right)= 2\sqrt{3}\approx3.5$ What do you think would happen if we used $\displaystyle n$-gons having more and more sides? Let's let $\displaystyle A_n$ be the area of an $\displaystyle n$-gon circumscribed by the circle...we have: $\displaystyle A_n=\frac{n}{2}\sin\left(\frac{2\pi}{n}\right)$ Using a computer, we find: $\displaystyle n$ $\displaystyle A_n$ 10 2.938926261462366 100 3.1395259764656687 1000 3.1415719827794755 10000 3.141592446881286 1000000 3.141592653569122 It appears that as $\displaystyle n\to\infty$ we have $\displaystyle A_n$ approaching some fixed finite value. 8. ## Re: Area of Square in Circle Originally Posted by MarkFL Suppose we use hexagons rather than squares: Let the radius of the circle be 1, so that its area is $\displaystyle \pi$ units squared. The area $\displaystyle A_S$ of the smaller hexagon is: $\displaystyle A_S=6\left(\frac{1}{2} \sin\left(\frac{2\pi}{6}\right)\right)= \frac{3\sqrt{3}}{2}\approx2.6$ And the area $\displaystyle A_L$ of the larger hexagon is: $\displaystyle A_L=6\left(\frac{1}{2} \left(\frac{2}{\sqrt{3}}\right)^2\sin\left(\frac{2 \pi}{6}\right)\right)= 2\sqrt{3}\approx3.5$ What do you think would happen if we used $\displaystyle n$-gons having more and more sides? Let's let $\displaystyle A_n$ be the area of an $\displaystyle n$-gon circumscribed by the circle...we have: $\displaystyle A_n=\frac{n}{2}\sin\left(\frac{2\pi}{n}\right)$ Using a computer, we find: $\displaystyle n$ $\displaystyle A_n$ 10 2.938926261462366 100 3.1395259764656687 1000 3.1415719827794755 10000 3.141592446881286 1000000 3.141592653569122 It appears that as $\displaystyle n\to\infty$ we have $\displaystyle A_n$ approaching some fixed finite value. What a great explanation! Nicely done! Page 2 of 2 First 12

https://math.stackexchange.com/questions/2971315/how-do-i-combine-standard-deviations-of-two-groups

# How do I combine standard deviations of two groups? I have 2 groups of people. I'm working with the data about their age. I know the means, the standard deviations and the number of people. I don't know the data of each person in the groups. Group 1 : Mean = 35 years old; SD = 14; n = 137 people Group 2 : Mean = 31 years old; SD = 11; n = 112 people I want to combine those 2 groups to obtain a new mean and SD. It's easy for the mean, but is it possible for the SD? I do not know the distribution of those samples, and I can't assume those are normal distributions. Is there a formula for distributions that aren't necessarily normal? • Hey, welcome to Math Stackexchange! If you can, can you please add some context to the question? I'm not a stats guy but I'm a little confused by what you mean by "subjects". Thanks! – SalmonKiller Oct 25 '18 at 21:33 • I just edited my post to add more context and be more specific. Thanks! – Nicolas Melançon Oct 25 '18 at 21:37 Continuing on from BruceET's explanation, note that if we are computing the unbiased estimator of the standard deviation of each sample, namely $$s = \sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i - \bar x)^2},$$ and this is what is provided, then note that for samples $$\boldsymbol x = (x_1, \ldots, x_n)$$, $$\boldsymbol y = (y_1, \ldots, y_m)$$, let $$\boldsymbol z = (x_1, \ldots, x_n, y_1, \ldots, y_m)$$ be the combined sample, hence the combined sample mean is $$\bar z = \frac{1}{n+m} \left( \sum_{i=1}^n x_i + \sum_{j=1}^m y_i \right) = \frac{n \bar x + m \bar y}{n+m}.$$ Consequently, the combined sample variance is $$s_z^2 = \frac{1}{n+m-1} \left( \sum_{i=1}^n (x_i - \bar z)^2 + \sum_{j=1}^m (y_i - \bar z)^2 \right),$$ where it is important to note that the combined mean is used. In order to have any hope of expressing this in terms of $$s_x^2$$ and $$s_y^2$$, we clearly need to decompose the sums of squares; for instance, $$(x_i - \bar z)^2 = (x_i - \bar x + \bar x - \bar z)^2 = (x_i - \bar x)^2 + 2(x_i - \bar x)(\bar x - \bar z) + (\bar x - \bar z)^2,$$ thus $$\sum_{i=1}^n (x_i - \bar z)^2 = (n-1)s_x^2 + 2(\bar x - \bar z)\sum_{i=1}^n (x_i - \bar x) + n(\bar x - \bar z)^2.$$ But the middle term vanishes, so this gives $$s_z^2 = \frac{(n-1)s_x^2 + n(\bar x - \bar z)^2 + (m-1)s_y^2 + m(\bar y - \bar z)^2}{n+m-1}.$$ Upon simplification, we find $$n(\bar x - \bar z)^2 + m(\bar y - \bar z)^2 = \frac{mn(\bar x - \bar y)^2}{m + n},$$ so the formula becomes $$s_z^2 = \frac{(n-1) s_x^2 + (m-1) s_y^2}{n+m-1} + \frac{nm(\bar x - \bar y)^2}{(n+m)(n+m-1)}.$$ This second term is the required correction factor. • Just to tie things together, I tried your formula with my fake data and got a perfect match: ((n1-1)*var(x1) + (n2-1)*var(x2))/(n1+n2-1) + ((n1*n2)*(mean(x1)-mean(x2))^2)/((n1+n2)*(n1+n2-1)) returns 1157.706 and so does var(x). Thanks, I haven't seen this formula before. – BruceET Oct 26 '18 at 2:28 Neither the suggestion in a previous (now deleted) Answer nor the suggestion in the following Comment is correct for the sample standard deviation of the combined sample. Known data for reference.: First, it is helpful to have actual data at hand to verify results, so I simulated samples of sizes $$n_1 = 137$$ and $$n_2 = 112$$ that are roughly the same as the ones in the question. Combined sample mean: You say 'the mean is easy' so let's look at that first. The sample mean $$\bar X_c$$ of the combined sample can be expressed in terms of the means $$\bar X_1$$ and $$\bar X_2$$ of the first and second samples, respectively, as follows. Let $$n_c = n_1 + n_2$$ be the sample size of the combined sample, and let the notation using brackets in subscripts denote the indices of the respective samples. $$\bar X_c = \frac{\sum_{[c]} X_i}{n} = \frac{\sum_{[1]} X_i + \sum_{[2]} X_i}{n_1 + n_1} = \frac{n_1\bar X_1 + n_2\bar X_2}{n_1+n_2}.$$ Let's verify that much in R, using my simulated dataset (for now, ignore the standard deviations): set.seed(2025); n1 = 137; n2 = 112 x1 = rnorm(n1, 35, 45); x2 = rnorm(n2, 31, 11) x = c(x1,x2) # combined dataset mean(x1); sd(x1) [1] 31.19363 # sample mean of sample 1 [1] 44.96014 mean(x2); sd(x2) [1] 31.57042 # sample mean of sample 2 [1] 10.47946 mean(x); sd(x) [1] 31.36311 # sample mean of combined sample [1] 34.02507 (n1*mean(x1)+n2*mean(x2))/(n1+n2) # displayed formula above [1] 31.36311 # matches mean of comb samp Suggested formulas give incorrect combined SD: Here is a demonstration that neither of the proposed formulas finds $$S_c = 34.025$$ the combined sample: According to the first formula $$S_a = \sqrt{S_1^2 + S_2^2} = 46.165 \ne 34.025.$$ One reason this formula is wrong is that it does not take account of the different sample sizes $$n_1$$ and $$n_2.$$ According to the second formula we have $$S_b = \sqrt{(n_1-1)S_1^2 + (n_2 -1)S_2^2} = 535.82 \ne 34.025.$$ To be fair, the formula $$S_b^\prime= \sqrt{\frac{(n_1-1)S_1^2 + (n_2 -1)S_2^2}{n_1 + n_2 - 2}} = 34.093 \ne 34.029$$ is more reasonable. This is the formula for the 'pooled standard deviation' in a pooled 2-sample t test. If we may have two samples from populations with different means, this is a reasonable estimate of the (assumed) common population standard deviation $$\sigma$$ of the two samples. However, it is not a correct formula for the standard deviation $$S_c$$ of the combined sample. sd.a = sqrt(sd(x1)^2 + sd(x2)^2); sd.a [1] 46.16528 sd.b = sqrt((n1-1)*sd(x1)^2 + (n2-1)*sd(x2)^2); sd.b [1] 535.8193 sd.b1 = sqrt(((n1-1)*sd(x1)^2 + (n2-1)*sd(x2)^2)/(n1+n2-2)) sd.b1 [1] 34.09336 Method for correct combined SD: It is possible to find $$S_c$$ from $$n_1, n_2, \bar X_1, \bar X_2, S_1,$$ and $$S_2.$$ I will give an indication how this can be done. For now, let's look at sample variances in order to avoid square root signs. $$S_c^2 = \frac{\sum_{[c]}(X_i - \bar X_c)^2}{n_c - 1} = \frac{\sum_{[c]} X_i^2 - n\bar X_c^2}{n_c - 1}$$ We have everything we need on the right-hand side except for $$\sum_{[c]} X_i^2 = \sum_{[1]} X_i^2 + \sum_{[2]} X_i^2.$$ The two terms in this sum can be obtained for $$i = 1,2$$ from $$n_i, \bar X_i$$ and $$S_c^2$$ by solving for $$\sum_{[i]} X_i^2$$ in a formula analogous to the last displayed equation. [In the code below we abbreviate this sum as $$Q_c = \sum_{[c]} X_i^2 = Q_1 + Q_2.$$] Although somewhat messy, this process of obtaining combined sample variances (and thus combined sample SDs) is used in many statistical programs, especially when updating archival information with a subsequent sample. Numerical verification of correct method: The code below verifies that the this formula gives $$S_c = 34.02507,$$ which is the result we obtained above, directly from the combined sample. q1 = (n1-1)*var(x1) + n1*mean(x1)^2; q1 [1] 408219.2 q2 = (n2-1)*var(x2) + n2*mean(x2)^2; q1 [1] 123819.4 qc = q1 + q2 sc = sqrt( (qc - (n1+n2)*mean(x)^2)/(n1+n2-1) ); sc [1] 34.02507 • This page also shows this method getting the variance of a combined sample. – BruceET Oct 26 '18 at 2:00

http://math.stackexchange.com/questions/174055/finding-the-divisors-of-a-number

# Finding the divisors of a number My text says that $p^3q^6$ has 28 divisors. Could anyone please explain to me how they got 28 here ? Edit: $p$ and $q$ are distinct prime numbers Sorry for the late addition.. - So far there are three answers and only one up-vote: mine. –  Michael Hardy Jul 23 '12 at 3:04 @Michael - what's your point? –  Gerry Myerson Jul 23 '12 at 3:05 Usually a question worth answering is worth up-voting. It seems as if people often neglect that. –  Michael Hardy Jul 23 '12 at 3:12 @Michael: I would be interested to hear your rationale for "Usually a question worth answering is worth up-voting". –  MJD Jul 23 '12 at 3:15 An upvote, to me, indicates that the question significantly increases the value of the resource. In my opinion, this question doesn't rise to that level. It's good enough to answer, but not good enough to upvote. –  Gerry Myerson Jul 23 '12 at 3:54 There can be 0-3 factors of $p$, so there are 4 ways for that to occur. There are 0-6 factors for $q$, so there are 7 ways for that to occur. - Let's take $p=3$ and $q=2$ as an example. Then the number is $3^32^6 = 1728$, and the 28 divisors of 1728 are: $$\begin{matrix} 1&2&4&8&16&32&64 \\ 3&6&12&24&48&96&192 \\ 9 & 18 & 36 & 72 & 144 & 288 & 576 \\ 27 & 54 & 108 & 216 & 432 & 864 & 1728 \end{matrix}$$ These values are, respectively: $$\begin{matrix} 2^03^0 & 2^13^0 & 2^23^0 & 2^33^0 & 2^43^0 & 2^53^0 & 2^63^0 \\ 2^03^1 & 2^13^1 & 2^23^1 & 2^33^1 & 2^43^1 & 2^53^1 & 2^63^1 & \\ 2^03^2 & 2^13^2 & 2^23^2 & 2^33^2 & 2^43^2 & 2^53^2 & 2^63^2 & \\ 2^03^3 & 2^13^3 & 2^23^3 & 2^33^3 & 2^43^3 & 2^53^3 & 2^63^3 \end{matrix}$$ - In fact, here is the general statement: if $$n=\prod_k p_k^{a_k}$$ is the prime factorization of $n$, then $$\sigma_0(n)=\prod_k (a_k+1)$$ –  Ｊ. Ｍ. Jul 23 '12 at 3:31 +1 Inasmuch as it matters, I like this sort of answer. A concrete example, presented clearly, that tells it all. –  copper.hat Jul 23 '12 at 4:09 even those with not-so-good-math can understand this answer. Great! –  woliveirajr Jul 23 '12 at 16:19 I assume your text also says $p$ and $q$ are primes, and $p\ne q$ --- otherwise, the statement isn't true. Do you know that any divisor of $p^3q^6$ must itself be of the form $p^aq^b$ for some $a,b$ with $0\le a\le3$ and $0\le b\le6$? If so, do you see how to get from there to 28? - That follows from the fundamental theorem of arithmetic right? en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic –  Modded Bear Jul 23 '12 at 3:06 @Chuck, yes.${}$ –  Gerry Myerson Jul 23 '12 at 3:46 @GerryMyerson Yes they are prime. Sorry I didnt put that up there –  MistyD Jul 23 '12 at 10:18 @GerryMyerson I still dont get how they got 28 ? –  MistyD Jul 23 '12 at 10:20 @Misty, you haven't seen Mark's answer? –  Ｊ. Ｍ. Jul 23 '12 at 11:14 The number of divisors of n=$\prod(p_i^{a_i})$ is $\sum(a_i+1)$ where $p_i$ are distinct primes. So, the number of divisors of $p^3q^6$ is(1+3)(1+6) where p,q are distinct primes =>(p,q)=1. - Any number $A$ is the product of a unique set of primes. if $A=P_1^{k_1}*P_2^{k_2}*....P_n^{k_n}$ then a divisor of A needs to be of the form $P_1^{m_1}*P_2^{m_2}*....P_n^{m_n}$ where $m_i\leq k_i$ for any $i\in \Bbb N \wedge i\leq n$ How many combinations of marbles can you make if you can choose from 3 white marbles and 6 black ones? if you pick 0 white there are 7 combinations. (0 black+0 white = 1). if you pick 1 white there are also 7 you can probably see that there are $4*7$ combinations. -

http://mathhelpforum.com/algebra/136207-simple-proof.html

1. ## A simple proof? Prove that: 2 + 4 + 6 + 8 ... + 2(n-1) = n * (n-1) When n > 0 Getting the n-1 part is easy (the left side of the equation has to have n-1 terms, so the result would be something multiplied by that). I don't know how to get the n, though... 2. Use induction to prove. Show the formula holds for the base case, assume it hold for the $n^{th}$ case and derive the $(n+1)^{th}$ case. Why don't you try and let me know where you get stuck. 3. Or wait! $2 + 4 + 6 + 8 ... + 2(n-1) = 2(1+2+3+...+(n-1)) = 2(\frac{n(n-1)}{2}) = n(n-1)$ You may know the formula to obtain the sum of the numbers from $1$ to $n.$ It is $\frac{n(n+1)}{2}.$ Here is an example: Sum the numbers $1$ to $5.$ Then, $1+2+3+4+5 = \frac{5\times 6}{2} = 15.$ 4. Hm. Gimme a second to type as I'm thinking. n = 1 2(1-1) = 1 * (1-1) = 0 n = k+1 2 + 4 + 6 ... + 2k = k^2 + k 2 + 4 + 6 ... + k = k^2 2 + 4 + 6 ... + 2(k-1) = k^2 - k 2 + 4 + 6 ... + 2(k-1) = k * (k-1) Oh, wow. Thanks so much! EDIT: Or, that second post works too. Heh, either one. Thanks for the two methods, at least. 5. Originally Posted by BlackBlaze Prove that: 2 + 4 + 6 + 8 ... + 2(n-1) = n * (n-1) When n > 0 LHS: $2 + 4 + 6 + 8 +\dots+ 2(n-4)+ 2(n-3) + 2(n-2) + 2(n-1) = S$ Swapping the order of the LHS $ 2(n-1) + 2(n-2)+ 2(n-3) + 2(n-4) +\dots+ 8 + 6+ 4 +2= S $ $\underbrace{2n+2n + 2n+\dots+ 2n}_{\text{n-1 times}} = 2S$ $2n(n-1) = 2S$ $n(n-1) = S$ $S = n(n-1)$ 6. Hello, here's a proof with the sum symbol (looks cool ). What you are trying to prove is that $\sum_{k=1}^{n - 1} 2k = n(n - 1)$. Let's work with this : $\sum_{k=1}^{n - 1} 2k$ Note that each term is even, we can factorize ! ( $2 + 4 + 6 = 2(1 + 2 + 3)$). So : $\sum_{k=1}^{n - 1} 2k = 2 \left ( \sum_{k=1}^{n - 1} k \right )$ Now you may use the fact that $\sum_{k=1}^{n} k = \frac{1}{2}n(n + 1)$, and your result follows. This is a theorem but it can be proven : $\sum_{k=1}^{n} k = 1 + 2 + 3 + \cdots + k$ 1. Assume $n$ is even. Rearranging the terms (putting brackets for better comprehension). This is valid because $n$ is even : $\sum_{k=1}^{n} k = \left [ 1 + \left ( n - 1 \right ) \right ] + \left [ 2 + \left ( n - 2 \right ) \right ] + \cdots + \frac{n}{2} + n$ The dotted part stops when we reach $n - \frac{n}{2}$ (excluded) because we cannot include that term twice. (*) This can be simplified : $\sum_{k=1}^{n} k = n + n + \cdots + \frac{n}{2} + n$ Using fact (*), we simplify by multiplication : $\sum_{k=1}^{n} k = \left ( \frac{n}{2} - 1 \right ) n + \frac{n}{2} + n$ Expanding and simplifying : $\sum_{k=1}^{n} k = \frac{n^2}{2} - n + \frac{n}{2} + n = \frac{n^2}{2} + \frac{n}{2} = \frac{n^2 + n}{2} = \frac{n(n + 1)}{2} = \frac{1}{2} n(n + 1)$ 2. Assume $n$ is odd. Rearranging the terms (putting brackets for better comprehension). This is valid because $n$ is odd : $\sum_{k=1}^{n} k = \left [ 1 + n \right ] + \left [ 2 + \left ( n - 1 \right ) \right ] + \cdots + \frac{n + 1}{2}$ The dotted part stops when we reach $n - \frac{n + 1}{2}$ (excluded) because if we included it we would include two terms twice! (*) This can be simplified : $\sum_{k=1}^{n} k = (n + 1) + (n + 1) + \cdots + \frac{n + 1}{2}$ Using fact (*), we simplify by multiplication : $\sum_{k=1}^{n} k = \left ( \frac{n + 1}{2} - 1 \right )(n + 1) + \frac{n + 1}{2}$ Expanding and simplifying (using the trick that $\frac{n + 1}{2} - 1 = \frac{n - 1}{2}$) : $\sum_{k=1}^{n} k = \left ( \frac{n + 1}{2} - 1 \right )(n + 1) + \frac{n + 1}{2} = \frac{(n - 1)(n + 1)}{2} + \frac{n + 1}{2} =$ $\ \frac{(n - 1)(n + 1) + (n + 1)}{2} = \frac{(n + 1)(n - 1 + 1)}{2} = \frac{1}{2} n(n + 1)$ Using the theorem we just proved (the proof is correct, just a bit messed up, needs better editing), we get that : $\sum_{k=1}^{n - 1} 2k = 2 \sum_{k=1}^{n - 1} k = 2 \times \left ( \frac{1}{2} (n - 1)(n - 1 + 1) \right ) = n(n - 1)$ And we can conclude. Does it make sense ? EDIT : I just saw Pickslides' proof. Mine seems a bit redundant now 7. Originally Posted by BlackBlaze Prove that: 2 + 4 + 6 + 8 ... + 2(n-1) = n * (n-1) When n > 0 Getting the n-1 part is easy (the left side of the equation has to have n-1 terms, so the result would be something multiplied by that). I don't know how to get the n, though... Well, the sum of n-1 number is n-1 times their average. Since this is an arithmetic sequence, the average of all the numbers is just the average of the smallest and largest: $\frac{2+ 2(n-1)}{2}= 1+ (n- 1)= n$.

https://math.stackexchange.com/questions/1239106/proving-6n-1-is-always-divisible-by-5-by-induction

# Proving $6^n - 1$ is always divisible by $5$ by induction I'm trying to prove the following, but can't seem to understand it. Can somebody help? Prove $6^n - 1$ is always divisible by $5$ for $n \geq 1$. What I've done: Base Case: $n = 1$: $6^1 - 1 = 5$, which is divisible by $5$ so TRUE. Assume true for $n = k$, where $k \geq 1$: $6^k - 1 = 5P$. Should be true for $n = k + 1$ $6^{k + 1} - 1 = 5Q$ $= 6 \cdot 6^k - 1$ However, I am unsure on where to go from here. For $n\geq 1$, let $S(n)$ denote the statement $$S(n) : 5\mid(6^n-1)\Longleftrightarrow 6^n-1=5m, m\in\mathbb{Z}.$$ Base case ($n=1$): $S(1)$ says that $5\mid(6^1-1)$, and this is true. Inductive step: Fix some $k\geq 1$ and assume that $S(k)$ is true where $$S(k) : 5\mid(6^k-1)\Longleftrightarrow 6^k-1=5\ell, \ell\in\mathbb{Z}.$$ To be proved is that $S(k+1)$ follows where $$S(k+1) : 5\mid(6^{k+1}-1)\Longleftrightarrow 6^{k+1}-1=5\eta, \eta\in\mathbb{Z}.$$ Beginning with the left-hand side of $S(k+1)$, \begin{align} 6^{k+1} - 1 &= 6^k\cdot 6-1\tag{by definition}\\[0.5em] &= (5\ell+1)\cdot 6-1\tag{by $S(k)$, the ind. hyp.}\\[0.5em] &= 6\cdot 5\ell+5\tag{expand}\\[0.5em] &= 5(6\ell+1)\tag{factor out $5$}\\[0.5em] &= 5\eta.\tag{$\eta=6\ell+1; \eta\in\mathbb{Z}$} \end{align} we end up at the right-hand side of $S(k+1)$, completing the inductive step. Thus, by mathematical induction, the statement $S(n)$ is true for all $n\geq 1$. $\blacksquare$ • Perfect! This has definitely made a things alot clearer! – RandomMath Apr 17 '15 at 14:56 • @RandomMath Glad it helped. :) You may want to look at this list for some other useful questions/answers of mine on induction that may prove useful to you. I love induction a little bit too much (uname actually used to be induktio). Most of it is because I, like you, struggled a lot with it. Keep whacking away at induction proofs though, and you'll master them before long. :) – Daniel W. Farlow Apr 17 '15 at 14:58 • I want to mention that any a^k - 1 divisible by a-1 – TigerTV.ru Jan 6 '19 at 20:24 Hint: Inductive step: $$6^{k+1}-1=6\cdot 6^k-1=5\cdot 6^k +(6^k-1)$$ • The inductive step is where I'm confused on on this question, could you elaborate on how you got to this? – RandomMath Apr 17 '15 at 14:35 • @RandomMath Since obviously $5\mid 5\cdot 6^k$ and by the inductive hypothesis $5\mid 6^{k}-1$, can you see that $5\mid 6^{k+1}-1$? And the equalities here only use trivial algebraic manipulations. – user26486 Apr 17 '15 at 14:35 • @RandomMath From your question description it seems you know you want to prove $5\mid 6^{k+1}-1$ assuming $5\mid 6^k-1$. You began by showing $6^{k+1}-1=6\cdot 6^{k}-1$. What I suggest here is continuing the equality with $6\cdot 6^{k}-1=5\cdot 6^{k}+(6^k-1)$, which makes it clear that we indeed have $5\mid 6^{k+1}-1$ (assuming $5\mid 6^{k}-1$). – user26486 Apr 17 '15 at 14:43 • Yes that is what im trying to prove. However, im struggling to get my head around how to get from 6(6^k) -1 to 5*6^k + (6^k -1) – RandomMath Apr 17 '15 at 14:45 • @RandomMath $6\cdot 6^{k}-1=(5+1)\cdot 6^k-1=5\cdot 6^k + 6^k-1=5\cdot 6^k +(6^k-1)$. – user26486 Apr 17 '15 at 14:46 In comments you ask about the source of the following standard proof of the inductive step $$5\mid 6^k-1\ \Rightarrow\ \color{#c00}5\cdot 6^k +(6^k\!-1)\,=\, \color{#0a0}{6^{k+1}-1}$$ This is a very natural question since such proofs often appear to be pulled out of a hat, like magic. There is, in fact, a good general explanation for their source. Namely such proofs are simply special cases of the proof of the Congruence Product Rule, as we show below. ${\bf Claim}\rm\qquad\ 6\equiv 1,\, 6^{k}\!\equiv 1 \Rightarrow\ 6^{k+1}\!\equiv 1\ \ \ \pmod{\!5},\$ a special case of the following ${\bf Lemma}\rm\quad\ \, A\!\equiv a,\, B\!\equiv b\ \Rightarrow\ AB\equiv ab\ \pmod{\!n}\ \ \$ [Congruence Product Rule] ${\bf Proof}\ \ \ \rm n\mid A\!-\!a,\,\ B-b\,\Rightarrow\, n\mid ( A\!-\!a) B +a\ (B\!-\!b) =A B\,-\,ab$ $\rm\ \ \ \ e.g.\ \ \ 5\mid\ 6\!-\!1,\,\ 6^{k}\!-\!1\ \Rightarrow\ 5\mid(\color{#c00}{6\!-\!1})\,6^{k}+ 1\,(6^{k}\!\!-\!1) = \color{#0a0}{6^{k+1}\!-1}$ Notice that the prior inference is precisely the same as said standard proof of the inductive step. Thus we see that this inference is simply a special case of the proof of the Congruence Product Rule. Once we know this rule, there's no need to repeat the entire proof every time we need to use it. Rather, we can simply invoke the rule as a Lemma (in divisibility form if congruences are not yet known). Then the inductive step has vivid arithmetical structure, being the computation of a product $\, 6\cdot 6^{k}\equiv 6^{(k+1)}.\,$ No longer is the innate arithmetical structure obfuscated by the details of the proof - since the proof has been encapsulated into a Lemma for convenient reuse. In much the same way, congruences often allow one to impart intuitive arithmetical structure onto complicated inductive proofs - allowing us to reuse our well-honed grade-school skills manipulating arithmetical equations (vs. more complex divisibility relations). Often introduction of congruence language will serve to drastically simplify the induction, e.g. reducing it to a trivial induction such as $\, 1^n\equiv 1,\,$ or $\,(-1)^{2n}\equiv 1.\,$ The former is the essence of the matter above. • +1 for taking the time and effort to put that up and for clearly wanting to help OP. I hope s/he reads it. – Daniel W. Farlow Apr 17 '15 at 15:46 We can show by induction that $6^k$ has remainder $1$ after division by $5$. The base case $k=1$ (or $k=0$) is straightforward, since $6=5\cdot 1+1$. Now suppose that $6^k$ has remainder $1$ after division by $5$ for $k\ge 1$. Thus $6^k = 5\cdot m+1$ for some $m \in \mathbb{N}$. We can then see that $$6^{k+1}=6\cdot 6^{k} = (5+1)(5\cdot m +1) = 5^2 \cdot m + 5 + 5\cdot m + 1$$ $$=5(5\cdot m + m + 1) + 1.$$ Thus $6^{k+1}$ has remainder $1$ after division by $5$. Therefore for every $k$, we can write $6^k = 5\cdot m +1$ for some $m$. • Once we have established that every power of $6$ has remainder of $1$ after division by $6$, the rest of the problem follows after subtracting $1$ from $6^k$. – Joel Apr 17 '15 at 14:44 This is the inductive step written out: $$6 \cdot 6^k - 1 = 5 \cdot Q |+1; \cdot \frac{1}{6};-1 \Leftrightarrow 6^k - 1 = \frac{5\cdot Q-5}{6}\underset{P}{\rightarrow}5\cdot P = \frac{5\cdot Q - 5 }{6} | \cdot \frac{1}{5}; \cdot 6\Leftrightarrow Q=6\cdot P + 1$$ $$6^k - 1 = \frac{5\cdot Q-5}{6} \overset{Q}{\rightarrow}\ (6^k-1 = \frac{5\cdot (6\cdot P + 1)-5}{6}\Leftrightarrow 6^k-1 = 5\cdot P)$$ • Please, use $LATEX$. – ً ً Sep 10 '16 at 23:17 • I tried this codecogs.com/latex/eqneditor.php However i am not sure whether this is correct LaTex – Martin Erhardt Sep 10 '16 at 23:18 • Just put $\$$starting from every line and at the end of every line. – ً ً Sep 10 '16 at 23:22 • ah thanks, I am new here – Martin Erhardt Sep 10 '16 at 23:28$6$has a nice property that when raised to any positive integer power, the result will have$6$as its last digit. Therefore, that number minus$1$is going to have$5$as its last digit and thus be divisible by$5\$. • OP wants to prove this by induction – Shailesh Sep 10 '16 at 23:56 • it's works for another numbers not only for 6. – TigerTV.ru Jan 6 '19 at 20:52

http://nl.mathworks.com/help/matlab/ref/istril.html?s_tid=gn_loc_drop&requestedDomain=nl.mathworks.com&nocookie=true

# Documentation ### This is machine translation Translated by Mouse over text to see original. Click the button below to return to the English verison of the page. # istril Determine if matrix is lower triangular ## Description example tf = istril(A) returns logical 1 (true) if A is a lower triangular matrix; otherwise, it returns logical 0 (false). ## Examples collapse all Create a 5-by-5 matrix. D = tril(magic(5)) D = 17 0 0 0 0 23 5 0 0 0 4 6 13 0 0 10 12 19 21 0 11 18 25 2 9 Test D to see if it is lower triangular. istril(D) ans = logical 1 The result is logical 1 (true) because all elements above the main diagonal are zero. Create a 5-by-5 matrix of zeros. Z = zeros(5); Test Z to see if it is lower triangular. istril(Z) ans = logical 1 The result is logical 1 (true) because a lower triangular matrix can have any number of zeros on its main diagonal. ## Input Arguments collapse all Input array, specified as a numeric array. istril returns logical 0 (false) if A has more than two dimensions. Data Types: single | double Complex Number Support: Yes collapse all ### Lower Triangular Matrix A matrix is lower triangular if all elements above the main diagonal are zero. Any number of the elements on the main diagonal can also be zero. For example, the matrix $A=\left(\begin{array}{cccc}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& 0\\ -1& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& 0\\ -2& -2& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1& 0\\ -3& -3& -3& 1\end{array}\right)$ is lower triangular. A diagonal matrix is both upper and lower triangular. ### Tips • Use the tril function to produce lower triangular matrices for which istril returns logical 1 (true). • The functions isdiag, istriu, and istril are special cases of the function isbanded, which can perform all of the same tests with suitably defined upper and lower bandwidths. For example, istril(A) == isbanded(A,size(A,1),0).

https://fr.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 ### Define and Evaluate Piecewise Expression Define the following piecewise expression by using piecewise. $y=\left\{\begin{array}{cc}-1& x<0\\ 1& x>0\end{array}$ syms x y = piecewise(x<0, -1, x>0, 1) y = piecewise(x < 0, -1, 0 < x, 1) 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 = [ -1, NaN, 1] ### Define Piecewise Function Define the following function symbolically. $y\left(x\right)=\left\{\begin{array}{cc}-1& x<0\\ 1& x>0\end{array}$ syms y(x) y(x) = piecewise(x<0, -1, x>0, 1) y(x) = piecewise(x < 0, -1, 0 < x, 1) 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 = [ -1, NaN, 1] ### Set Value When No Conditions Is True 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(x\right)=\left\{\begin{array}{cc}-2& x<-2\\ 0& -2 syms y(x) y(x) = piecewise(x<-2, -2, -2<x<0, 0, 1) y(x) = piecewise(x < -2, -2, x in Dom::Interval(-2, 0), 0, 1) 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) yvalues = y(xvalues) xvalues = -3 -2 -1 0 1 yvalues = [ -2, 1, 0, 1, 1] ### Plot Piecewise Expression 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) ### Assumptions and Piecewise Expressions 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 Piecewise Expression Differentiate, integrate, and find limits of a piecewise expression by using diff, int, and limit respectively. Differentiate the following piecewise expression by using diff. $y=\left\{\begin{array}{cc}1/x& x<-1\\ \mathrm{sin}\left(x\right)/x& x\ge -1\end{array}$ syms x y = piecewise(x<-1, 1/x, x>=-1, sin(x)/x); diffy = diff(y, x) diffy = piecewise(x < -1, -1/x^2, -1 < x, cos(x)/x - sin(x)/x^2) Integrate y by using int. inty = int(y, x) inty = piecewise(x < -1, log(x), -1 <= x, sinint(x)) Find the limits of y at 0 and -1 by using limit. Because limit finds the double-sided limit, the piecewise expression must be defined from both sides. Alternatively, you can find the right- or left-sided limit. For details, see limit. limit(y, x, 0) limit(y, x, -1) ans = 1 ans = limit(piecewise(x < -1, 1/x, -1 < x, sin(x)/x), x, -1) Because the two conditions meet at -1, the limits from both sides differ and limit cannot find a double-sided limit. ### Elementary Operations on Piecewise Expressions 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); add = pw1 + pw2 sub = pw1 - pw2 mul = pw1 * pw2 div = pw1 / pw2 piecewise(x < -1, -3, x in Dom::Interval([-1], 0), -1, 0 <= x, 3) sub = piecewise(x < -1, 1, x in Dom::Interval([-1], 0), 3, 0 <= x, -1) mul = piecewise(x < -1, 2, x in Dom::Interval([-1], 0), -2, 0 <= x, 2) div = piecewise(x < -1, 1/2, x in Dom::Interval([-1], 0), -1/2, 0 <= x, 1/2) ### Modify or Extend Piecewise Expression 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 = piecewise(x < 0, -1, 0 < x, 1) 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 = piecewise(5 < x, 1/x, x < 0, -1, 0 < x, 1) ## 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.

https://math.stackexchange.com/questions/4510187/how-to-create-a-function-f-such-that-fx-y-is-high-when-either-x-or-y-i

# How to create a function $f$ such that $f(x,y)$ is high when either $x$ or $y$ is? There are two variables, let's say $$x$$ and $$y$$. I want to come up with a function $$f:[0,5]\times[0,5]\longrightarrow[0,5]$$ that respects the following rules: 1. If $$x$$ is high (close to the maximum value of 5) and $$y$$ is low (close to $$0$$), $$f(x,y)$$ should results in a high value (close to 5). The same would happen for the reverse ($$x$$ with a low value and $$y$$ with a high value) 2. If $$x$$ is high and $$y$$ is high, $$f(x,y)$$ will also be high; it would be useful that the values resulted would be higher than those from point $$(1)$$. 3. If $$x$$ is low and $$y$$ is low, $$f(x,y)$$ should result in low values. I'm having troubles starting imagining what mathematical functions would be useful, as I am not even close to being advanced in the concepts of Mathematics. Any ideas would be appreciated, at least in the sense of finding any information that could get me started. • you defined your function $f :[0,5]\rightarrow [0,5]$, and then you talk about two input variable $x$ and $y$, i guess you meant $f :[0,5]\times [0,5] \rightarrow [0,5]$ ? Aug 11 at 11:47 • that is correct, I'll add the change. Aug 11 at 11:48 • If you have precise value for your function $f$, you could use something like multivariate interpolation. Aug 11 at 11:52 • If you don't know the concept of interpolation you should check out first 1D interpolation like this one. Aug 11 at 11:57 • If $f(x,y)=f(y,x)$, and $f(x,y)$ is linear with respect to $x$ (this was not part of the problem statement, it was added by me to ensure uniqueness of solution), then $f(x,y)=axy+b(x+y)+c$. Let $f(0,0)=0$, $f(5,5)=5$, then $c=0$, $5a+2b=1$. If one wants to set value $f(0,5)=d$, then $b=\frac{d}{5}$, $a=\frac{5-2d}{25}$. At $d=5$ result is formula from my previous comment, but OP can take $f(0,5)=d=4.9$ or something else what needed. Aug 12 at 7:14 You can take, for instance,$$f(x,y)=\frac45\max\{x,y\}+\frac15\min\{x,y\}.$$It has all the properties that you are interested in. One way to model this could be to use the distance from the line: $$x+y=0$$ through $$(0,0)$$. This line has normal vector: $$\vec n= \begin{pmatrix} 1\\ 1 \end{pmatrix}$$ and the distance of a given point $$(x,y)$$ to this line is proportional to $$t=\vec n\cdot\langle x,y \rangle=x+y$$ so in case we want something like: $$f(5,5)=5\\ f(5,0)=4\\ f(0,0)=0$$ we can connect this to a single dimensional function: $$f(x,y)=g(x+y)$$ where $$g(10)=5\\ g(5)=4\\ g(0)=0$$ A way to achieve this could be to add the extra requirement $$g'(10)=0$$ so that $$f$$ has maximum at $$(x,y)=(5,5)$$ and build $$g(0)=0$$ in: $$g(t)=at^3+bt^2+ct$$ Hence \begin{align} g(10)&=1000a+100b+10c&&=5\\ g(5)&=125a+25b+5c&&=4\\ g'(10)&=300a+20b+c&&=0 \end{align} which can be solved for $$a,b,c$$ to have: $$f(x,y)=g(x+y)=0.002(x+y)^3-0.09(x+y)^2+1.2(x+y)$$ See this link to look at interactive GeoGebra-applet with 3D-plot of this function ADDENDUM: As can be seen both from the other answer and from comments, there will be (infinitely) many ways to satisfy your requirements, but to point you towards handling the additional requirement stated in your comment below this post, you could simply add a modifier to the above solution which takes the distance to the perpendicular line: $$x-y=0$$ as input. This distance is (similarly) proportional to $$t=x-y$$, and so we need a modifier function $$m(x,y)=h(x-y)=h(t)$$ that satisfies: $$h(0)=0\\ h(5)=m(5,0)\\ h(-5)=m(0,5)$$ so just choose which modification you want at $$(5,0)$$ and $$(0,5)$$ and match for instance a quadratic function as $$h$$: $$h(t)=\alpha t^2+\beta t+\gamma$$ and combine: \begin{align} q(x,y) &=f(x,y)+m(x,y)\\ &=g(x+y)+h(x-y)\\ &=a(x+y)^3+b(x+y)^2+c(x+y)+\alpha(x-y)^2+\beta(x-y)+\gamma \end{align} but be a little careful - if $$h(t)$$ increases too rapidly away from $$h(0)$$ to one side, then $$q$$ may exceed a value of $$5$$. Here is GeoGebra-applet with example of this technique • Very interesting and detailed answer. With the risk of going a bit outside the boundaries of the question: is there a way to introduce some kind of bias towards $x$ or $y$? In the sense that $f(x,y)$ should be greater than $f(y,x)$ or the opposite. Aug 16 at 13:06 • @Ionut-AlexandruBaltariu: I added an extra section about this. Generally, you can always combine two one-dimensional functions $g(x+y)$ and $h(x-y)$ to create formula for change in value when you move perpendicular to the two lines $x+y=0$ and $x-y=0$. This gives you a lot of control over what you want to happen. Just be careful, because combinations may escape the max/min values, since we only controlled them in points $(0,0),(5,0),(0,5)$ and $(5,5)$. Aug 17 at 8:41

https://math.stackexchange.com/questions/2318951/card-draw-probability

Card Draw Probability This is a homework question. Not something to hand in, but rather a recommended practice question. A Card Game Three students are playing a card game. They decide to choose the first person to play by each selecting a card from the 52-card deck and looking for the highest card in value and suit. They rank the suits from lowest to highest: clubs, diamonds, hearts, and spades. a. If the card is replaced in the deck after each student chooses, how many possible configurations of the three choices are possible? b. How many configurations are there in which each student picks a different card? c. What is the probability that all three students pick exactly the same card? d. What is the probability that all three students pick different cards? I figured out the the answer to part A is $52^3 = 140608$, since each student chooses 1 card from a deck of 52. I also know that the answer to part B is $52\times 51\times 50 =132600$, or $\frac {52!}{(52-3)!}$. However, I am not sure how to find the probability that they all pick the same card. I initially thought it would be $\left( \frac 1{52}\right)^3$, but that does not match the answer in the back of the book, which is $0.00037$. Sorry for the formatting (or lack thereof), I tried my best to make it fairly easy to read, but I am unsure of how to make fancy fractions and such. • For $c$: the first person takes whatever. The probability that the second matches the first is $\frac 1{52}$. The probability that the third matches the first is $\frac 1{52}$ independent of what the second person drew. So... – lulu Jun 11 '17 at 20:54 • An answer of $(1/52)\cdot (1/52)\cdot (1/52)=(1/52)^3$ is the probability of all three specifically selecting the ace of spades, but remember there are many ways other than just everyone getting the ace of spades for them to all have the same card. As for part (d), if you were to read and understand the birthday problem you'll find the mathematics behind it is very similar. Alternatively, look more closely at the phrasing of parts (b) and (a) and how they relate to question (d). – JMoravitz Jun 11 '17 at 21:00 • So (1/52)^2. Thank you! Are you going to post it as an answer so that I can accept it? – Bunyip Jun 11 '17 at 21:02 • As a complete aside, you say "I am unsure of how to make fancy fractions and such." This page will give you the information you need to do so (at least on this site using MathJax). – JMoravitz Jun 11 '17 at 21:03 For part C: Try thinking about it like this, player A is the one that "chooses" which card B and C must choose in order for all of them to pick the same card, so the probability that all three students pick exactly the same card is: $\ \frac{1}{52}\frac{1}{52} = (\frac{1}{52})^2$ For part D: Start as in C, player A chooses the card and B has a $\ \frac{51}{52}$ chance of its card being different from A's card, similarly C has a $\ \frac{50}{52}$ of its card being different from A and B, so the probability that all cards differ is: $\ \frac{51}{52}\frac{50}{52}$ Hint. Part c only makes sense if the cards are drawn with replacement. You already have the total number of configurations. There are 52 configurations where all the cards are the same. With these two facts, the required probability is readily calculated. For part d, which again only makes sense if cards are drawn with replacement, you can get the number of ways in which all three cards are different from the answer to part b. This will lead to the required probability.

https://math.stackexchange.com/questions/3180877/determinant-of-antidiagonally-constructed-matrix

# Determinant of antidiagonally constructed matrix Let $$A_n$$ be a matrix, odd dimension $$n \times n$$, constructed from the sequence of natural numbers in such a way that we begin the sequence from the upper left corner and next numbers are inserted into the matrix along antidiagonal direction, starting from the top of the matrix (examples below) . For such matrix the determinant is calculated. Surprisingly for the three examples below the value of determinant is equal to the central entry of the matrix with the modification of sign between two consecutive cases. $$\det \begin{bmatrix}1 & 2 & 4\\3 &\color{red} 5 & 7\\6 & 8 & 9\end{bmatrix} =-\color{red}5$$ $$\det \begin{bmatrix}1 & 2 & 4 & 7 & 11\\3 & 5 & 8 & 12 & 16\\6 & 9 & \color{red}{13} & 17 & 20\\10 & 14 & 18 & 21 & 23\\15 & 19 & 22 & 24 & 25\end{bmatrix} =\color{red}{13}$$ $$\det \begin{bmatrix}1 & 2 & 4 & 7 & 11 & 16 & 22\\3 & 5 & 8 & 12 & 17 & 23 & 29\\6 & 9 & 13 & 18 & 24 & 30 & 35\\10 & 14 & 19 & \color{red}{25} & 31 & 36 & 40\\15 & 20 & 26 & 32 & 37 & 41 & 44\\21 & 27 & 33 & 38 & 42 & 45 & 47\\28 & 34 & 39 & 43 & 46 & 48 & 49\end{bmatrix} = -\color{red}{25}$$ Question: • does the pattern continue for a greater $$n$$? • if so what is the explanation for the pattern? • Note that the pattern holds for the often overlooked $1\times 1$ matrices as well: The determinant of $[\color{red}{1}]$ is $\color{red}{1}$. Apr 9, 2019 at 11:11 • @Arthur Yes, 1 is also "central" in [1], and it has + sign, this change of sign is also strange.. Apr 9, 2019 at 11:13 • Here's how it continues: oeis.org/A069480. Your observation would make a nice comment, BTW. Apr 9, 2019 at 11:27 • The formula section of the OEIS link does confirm this: It says that for odd $n$, the determinant is indeed an alternating multiplied by $\frac{(2n+1)^2}{2}$, which is the middle element of the corresponding matrix.. Apr 9, 2019 at 11:34 • @Widawensen In this case, the easiest determinant to start with first might be the one which has central value $0$: we just have to show that this matrix is singular. But the fact that as you add $k$ to all entries, the determinant changes by $\pm k$, might also be hard to prove... Apr 9, 2019 at 22:28 Let $$M_n$$ be the matrix whose determinant we want to know (the matrix where we fill in $$n^2$$ consecutive integers antidiagonally, starting at any value $$k$$), and let $$x$$ be its middle entry. For convenience, let $$m = \frac{n-1}{2}$$. (I will use $$n=7$$ in my example matrices, but my argument applies to all odd $$n$$.) First, replace the rows $$r_1, r_2, \dots, r_n$$ of $$M_n$$ with the $$n-1$$ differences $$r_2 - r_1$$, $$r_3 - r_2$$, ..., $$r_n - r_{n-1}$$ followed by the middle row $$r_{m+1}$$. This is an invertible sequence of row operations, so we have: $$\det(M_n) = \det\begin{bmatrix} 2 & 3 & 4 & 5 & 6 & 7 & 7 \\ 3 & 4 & 5 & 6 & 7 & 7 & 6 \\ 4 & 5 & 6 & 7 & 7 & 6 & 5 \\ 5 & 6 & 7 & 7 & 6 & 5 & 4 \\ 6 & 7 & 7 & 6 & 5 & 4 & 3 \\ 7 & 7 & 6 & 5 & 4 & 3 & 2 \\ x+a_1 & x+a_2 & x+a_3 & x & x-a_3 & x-a_2 & x-a_1 \end{bmatrix}.$$ where $$x+a_1,\dots, x+a_m, x, x-a_m, \dots, x-a_1$$ are the entries of the middle row of $$M_n$$ (which always has this symmetric form). We show that the determinant above is always equal to $$(-1)^{m} x$$. By linearity of the determinant, we can write this as $$\det(M_n) = \det\begin{bmatrix} 2 & 3 & 4 & 5 & 6 & 7 & 7 \\ 3 & 4 & 5 & 6 & 7 & 7 & 6 \\ 4 & 5 & 6 & 7 & 7 & 6 & 5 \\ 5 & 6 & 7 & 7 & 6 & 5 & 4 \\ 6 & 7 & 7 & 6 & 5 & 4 & 3 \\ 7 & 7 & 6 & 5 & 4 & 3 & 2 \\ a_1 & a_2 & a_3 & 0 & -a_3 & -a_2 & -a_1 \end{bmatrix} + x \det\begin{bmatrix} 2 & 3 & 4 & 5 & 6 & 7 & 7 \\ 3 & 4 & 5 & 6 & 7 & 7 & 6 \\ 4 & 5 & 6 & 7 & 7 & 6 & 5 \\ 5 & 6 & 7 & 7 & 6 & 5 & 4 \\ 6 & 7 & 7 & 6 & 5 & 4 & 3 \\ 7 & 7 & 6 & 5 & 4 & 3 & 2 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 \end{bmatrix}.$$ The first determinant is $$0$$ by cofactor expansion along the last row. The cofactors of $$a_i$$ and $$-a_i$$ are equal for each $$i$$, since one submatrix can be obtained from the other by reversing the rows, then reversing the columns, so they cancel after multiplying one by $$a_i$$ and the other by $$-a_i$$. For the second matrix, we first do a similar operation to columns. Keeping the first column $$c_1$$ and replacing every other $$i^{\text{th}}$$ column $$c_i$$ by the difference $$c_i - c_{i-1}$$, which is an invertible set of row operations, we have $$\det\begin{bmatrix} 2 & 3 & 4 & 5 & 6 & 7 & 7 \\ 3 & 4 & 5 & 6 & 7 & 7 & 6 \\ 4 & 5 & 6 & 7 & 7 & 6 & 5 \\ 5 & 6 & 7 & 7 & 6 & 5 & 4 \\ 6 & 7 & 7 & 6 & 5 & 4 & 3 \\ 7 & 7 & 6 & 5 & 4 & 3 & 2 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 \end{bmatrix} = \det\begin{bmatrix} 2 & 1 & 1 & 1 & 1 & 1 & 0 \\ 3 & 1 & 1 & 1 & 1 & 0 & -1 \\ 4 & 1 & 1 & 1 & 0 & -1 & -1 \\ 5 & 1 & 1 & 0 & -1 & -1 & -1 \\ 6 & 1 & 0 & -1 & -1 & -1 & -1 \\ 7 & 0 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}$$ By expansion along the last row (which introduces a multiple of $$(-1)^{n+1} = 1$$) and after reversing the rows (which introduces a multiple of $$(-1)^{m}$$) we simplify this to $$(-1)^{m}\det\begin{bmatrix} 0 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & -1 & -1 & -1 & -1 \\ 1 & 1 & 0 & -1 & -1 & -1 \\ 1 & 1 & 1 & 0 & -1 & -1 \\ 1 & 1 & 1 & 1 & 0 & -1 \\ 1 & 1 & 1 & 1 & 1 & 0 \end{bmatrix}.$$ Now things get a bit fancier. This $$n-1 \times n-1 = 2m \times 2m$$ matrix (call it $$A$$) is a skew-symmetric matrix, so its determinant is the square of its Pfaffian. We use the simplified formula given in the Wikipedia link above: the sum $$\text{pf}(A) = \sum_{\alpha \in \Pi} \text{sgn}(\pi_\alpha) a_{i_1j_1} a_{i_2j_2}\dotsm a_{i_mj_m}$$ where $$\alpha$$ runs over all pairings $$\{(i_1,j_1), \dots, (i_m, j_m)\}$$ with $$i_k < j_k$$ in each pair and $$i_1 < \dots < i_m$$; $$\pi_{\alpha}$$ is the permutation with $$\pi_\alpha(2k-1) = i_k$$ and $$\pi_\alpha(2k) = j_k$$. For our particular matrix, $$a_{i_kj_k}$$ is always $$-1$$, so $$\text{pf}(A) = (-1)^m \sum_{\alpha \in \Pi} \text{sgn}(\pi_\alpha).$$ For each pairing $$\alpha$$, if we find the first index $$k$$ where $$j_k > i_{k+1}$$ and switch $$j_k$$ with $$j_{k+1}$$, we get another pairing $$\beta$$ with $$\text{sgn}(\pi_\beta) = -\text{sgn}(\pi_\alpha)$$; if we apply the same operation to $$\beta$$, we get $$\alpha$$ back. So this splits up the pairings into, uh, pairs of pairings whose terms in the Pfaffian sum cancel; the exception is the pairing $$\{(1,2), (3,4), \dots, (2m-1,2m)\}$$, for which no index $$k$$ exists. So $$\text{pf}(A) = (-1)^m$$ after all the cancellations, and therefore $$\det(A) = 1$$. So we get $$\det(M_n) = 0 + x \cdot (-1)^{m} = (-1)^{\frac{n-1}{2}}x$$: plus or minus the middle entry of $$M_n$$. • I'm very grateful for your partial answer - it has given me some new ideas to rethink... Apr 10, 2019 at 8:21 • I wonder whether in the case of the second determinat the same technique applied by you with substracting the rows could not lead to the further simplification.. Apr 10, 2019 at 8:59 • Anyway I have received now {{-1, -1, -1, -1, -1, 0, 1}, {-1, -1, -1, -1, 0 ,1 ,1}, {-1, -1 , -1 , 0 ,1, 1 ,1}, {-1, -1, 0 ,1 ,1 ,1, 1}, {-1, 0 , 1 ,1 ,1, 1, 1}, {5, 5, 4 ,3, 2 ,1, 0}, {1, 1, 1, 1, 1, 1, 1}} at which I'm stacked.. Apr 10, 2019 at 9:25 • I have also received for $5 \times 5$ such form $\begin{equation*}\left[\begin{matrix}0 & 0 & 0 & 1 & 2\\0 & 0 & 1 & 2 & 2\\0 & 1 & 2 & 2 & 2\\3 & 3 & 2 & 1 & 0\\1 & 1 & 1 & 1 & 1\end{matrix}\right]\end{equation*}$ which has det =$-1 \ \$. Interestingly $\begin{equation*}\left[\begin{matrix}0 & 0 & 0 & 1 & 2\\0 & 0 & 1 & 2 & 2\\0 & 1 & 2 & 2 & 2\\4 & 3 & 2 & 1 & 0\\1 & 1 & 1 & 1 & 1\end{matrix}\right]\end{equation*}$ has det = $0$. Apr 10, 2019 at 11:53 • Now I have a complete answer. Apr 10, 2019 at 22:11

https://math.stackexchange.com/questions/2962351/surjective-and-injective-function

# Surjective and Injective function Let $$N=\{1,2,3...\}$$ be the set of natural numbers and $$F:N \times N \rightarrow N$$ be such that $$f(m,n)=(2m-1)*2^n.$$ (A)F is Injective. (B)F is Surjective. (C)F is Bijective. (D)None of the above. I can see that F can never be surjective because 1 does not have a pre-image. And it seems injective to me because $$2m-1$$ term would always be odd, $$2^n$$ term would always be even, and hence the product will always be Even, and for different values of m and n, we would get a different even number. Is my Reasoning correct? • You're on the correct track, but your reasoning is too vague to be a proof. You need to start with an assumption $f(m_1, n_1) = f(m_2, n_2)$ and develop your even-odd argument a bit more. – T. Bongers Oct 19 '18 at 17:26 Overkill? A) $$f$$ is injective. Let $$f(m,n)=f(k,l)$$, i.e. $$i:=(2m-1)2^n= (2k-1)2^l.$$ The positive integer $$i$$ has a - Fundamental Theorem of Arithmetic - unique prime factorization. Since $$(2m-1)$$, $$(2k-1)$$ are odd, we have $$n=l$$. Then $$(2m-1)2^n= (2k-1)2^l$$ implies $$2m-1=2k-1$$, or $$m=k.$$ Combining $$f(m,n)=f(k,l)$$ implies $$m=k$$, and $$n=l.$$ https://en.m.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic • "Overkill?" Perhaps. T. Bongers answer is probably a lot easier, but your answer does capture the thought process that the OP was considering. (i.e. that the results have an "even part" and an "odd part" and ... so on.) – fleablood Oct 19 '18 at 18:27 • fleablood.Thanks for your comment. – Peter Szilas Oct 19 '18 at 19:24 To expand my comment, you need to make the reasoning a lot more precise. Start with an assumption that there are natural numbers $$m_1, m_2, n_1, n_2$$ for which $$f(m_1, n_1) = f(m_2, n_2).$$ Then you have $$(2m_1 - 1)2^{n_1} = (2m_2 - 1) 2^{n_2}.$$ Now we can assume without loss of generality that $$n_1 \le n_2$$, so that $$2m_1 - 1 = (2m_2 - 1)2^{n_2 - n_1}.$$ Now the left side is odd, so the right side is odd too; thus, $$n_2 - n_1 = 0$$ (why?). But then we get $$2m_1 - 1 = 2m_2 - 1$$, hence $$m_1 = m_2$$. This completes the proof. I can see that F can never be surjective because 1 does not have a pre-image. Let's prove that. If $$f(m,n) = (2m -1)*2^n = 1$$ then as $$2m-1$$ and $$2^n$$ are integer factors of $$1$$ and the only integer factors of $$1$$ are $$\pm 1$$ so $$2^n = \pm 1$$ and that is only possible if $$n = 0$$ which is not possible as $$0 \not \in \mathbb N$$. So it is not surjective. (In fact for any odd number $$2k - 1$$ then $$f(m,n) = (2m-1)*2^n = 2k-1$$ would be impossible. $$f(m,n)$$ will always be even.) And it seems injective to me because 2m−1 term would always be odd, 2n term would always be even, and hence the product will always be Even, Which shows it can not be surjective and for different values of m and n, we would get a different even number. That's a bit of a leap. How do you know they will be different? Let's prove it. $$f(m,n) = (2m-1)*2^n$$ will have a unique prime factorization. As $$2m-1$$ is odd the power of $$2$$ of any value of $$f(m,n)= (2m-1) 2^n$$ will be $$n$$. So if $$n_1 \ne n_2$$ then $$f(a,n_2) = (2a - 1)2^{n_2} \ne (2b- 1)2^{n_1} = f(b,n_2)$$ for any possible $$a,b$$. (Because the power of $$2$$ in the prime factorizations of those two numbers are different.) And if $$m_1 \ne m_2$$ then $$2m_1 - 1 \ne 2m_2 - 1$$ and the prime factorizations of $$(2m_1 - 1)*2^a \ne (2m_2 - 1)*2^b$$ for any possible $$a,b$$ must be different for the different odd factors. So if $$(m_1, n_1) \ne (m_2 n_2)$$ then either $$n_1 \ne n_2$$ or $$m_1 \ne m_2$$ (or both). In either case then $$(2m_1-1)2^{n_1} \ne (2m_2-1)2^{n_2}$$. So $$f$$ is injective. A) is true, B) is false, C) is false (as bijective implies surjective) and D) is false (it is one of the above).

http://mathhelpforum.com/differential-equations/72467-differential-equation.html

# Math Help - differential equation 1. ## differential equation If we assume that a person's weight depends on the energy consumed minus energy ued, one model is that weight changes in proportion to the difference, so: dw/dt = k(C - 38.5w) where w(0) = w subscript 0 where w(t) kg is the weight at t days, C is the daily calorie intake, and we assume 38.5 calories per kg per day are used (i) if you wish to maintain a constant weight of 82 kg, what should be your daily calorie intake C? (ii) If you weigh 100 kg, and you want to lose 10 kg in a month, what should C be? (Assume k = 1.3 * 10^(-4) kg/calorie and a month of 30 days) Is this result healthy? can someone solve these questions for me, they are on a past test paper which does not have solutions available. 2. Originally Posted by razorfever If we assume that a person's weight depends on the energy consumed minus energy ued, one model is that weight changes in proportion to the difference, so: dw/dt = k(C - 38.5w) where w(0) = w subscript 0 where w(t) kg is the weight at t days, C is the daily calorie intake, and we assume 38.5 calories per kg per day are used (i) if you wish to maintain a constant weight of 82 kg, what should be your daily calorie intake C? (ii) If you weigh 100 kg, and you want to lose 10 kg in a month, what should C be? (Assume k = 1.3 * 10^(-4) kg/calorie and a month of 30 days) Is this result healthy? can someone solve these questions for me, they are on a past test paper which does not have solutions available. (i) $\frac{dw}{dt} = 0$ ... solve for $C$ (ii) solve $\int \frac{dw}{C - 38.5w} = \int k \, dt$ with initial condition $w(0) = 100$. after you find $w(t)$, set $w(30) = 90$ and solve for $C$. 3. can u help me out the same way with that kind of template for some of my other posts about differential equation word problems? I have an exam coming up and my prof hasn't touched any word problems, if you could provide templates, i can understand them better and hopefully pass the exam http://www.mathhelpforum.com/math-he...tric-func.html http://www.mathhelpforum.com/math-he...d-problem.html http://www.mathhelpforum.com/math-he...-diver-de.html http://www.mathhelpforum.com/math-he...ates-flow.html http://www.mathhelpforum.com/math-he...72473-ivp.html http://www.mathhelpforum.com/math-he...ntial-eqn.html 4. can you tell me what your final answer for C is? i'm getting 102375.57 as for the constant in part (i) i'm getting (C - 3850) and the expression for C in part (ii) i'm getting [3465 - 3850e^(30k)] / [1 - e^(30k)] is this correct can someone show me the correct solution step by step? 5. $\int \frac{dw}{C - 38.5w} = \int k \, dt$ $\int \frac{-38.5}{C - 38.5w} \, dw = -38.5\int k \, dt$ $\ln|C - 38.5w| = -38.5kt + A$ $C - 38.5w = Be^{-38.5kt}$ $w(0) = 100$ $C - 3850 = B$ $C - 38.5w = (C - 3850)e^{-38.5kt}$ $C - 38.5w = Ce^{-38.5kt} - 3850e^{-38.5kt}$ $C - Ce^{-38.5kt} = 38.5w - 3850e^{-38.5kt}$ $C(1 - e^{-38.5kt}) = 38.5w - 3850e^{-38.5kt}$ $C = \frac{38.5w - 3850e^{-38.5kt}}{1 - e^{-38.5kt}}$ $w(30) = 90$ ... $C \approx 1090 \, cal$ 6. now i'm getting C = 3515 how do you do that last part w(30) = 90 do u rearrange and make w the subject then put t=30 and solve for C right? can u show me that last step explicitly?? 7. Originally Posted by razorfever now i'm getting C = 3515 how do you do that last part w(30) = 90 do u rearrange and make w the subject then put t=30 and solve for C right? can u show me that last step explicitly?? since the task was to find C, I solved for C ... sub in 90 for w, 30 for t, and the given value for k (0.00013 , correct?) 8. yeah i just realized i was making a careless mistake i wasn't putting E in my calculator , instead just getting the value of (-38.5 * 0.00013 * 30) and this was giving me C as 3515 but I got it now 9. Originally Posted by razorfever can u help me out the same way with that kind of template for some of my other posts about differential equation word problems? I have an exam coming up and my prof hasn't touched any word problems, if you could provide templates, i can understand them better and hopefully pass the exam http://www.mathhelpforum.com/math-he...tric-func.html http://www.mathhelpforum.com/math-he...d-problem.html http://www.mathhelpforum.com/math-he...-diver-de.html http://www.mathhelpforum.com/math-he...ates-flow.html http://www.mathhelpforum.com/math-he...72473-ivp.html http://www.mathhelpforum.com/math-he...ntial-eqn.html

https://math.stackexchange.com/questions/1128323/are-emptyset-and-x-closed-open-or-clopen/1128325

# Are $\emptyset$ and $X$ closed, open or clopen? It is indeed a very basic question but I am confused: (1) In an 2013 MSE posting under general topology here, I was told that $\emptyset$ is an open set and therefore I assume $X$ must be open too. (2) But in Wikipedia page on clopen set here, it says "In any topological space $X$, the $\emptyset$ and the whole space $X$ are both clopen." (3) And yet in another Wikipedia page on closed set here, "The $\emptyset$ is closed, the whole set is closed." I must have missed something. Can you help me with a supreme verdict, once and for all, as sure as the sun rises from the east each morning, if $X$ and $\emptyset$ are open, closed or clopen. Of course I am talking about topology, thanks for your time. • I don't see any contradiction among the three. – Hoot Jan 31 '15 at 23:36 • I don't understand! Is this an emerald, or is it green, or is it a precious stone? MSE says this is green, and Wikipedia says it is an emerald, and yet another wikipedia page says it is a precious stone. Can you help me with a supreme verdict, once and for all, if this is green, a precious stone, or an emerald? – MY USER NAME IS A LIE Jan 31 '15 at 23:47 • @MYANSWERSARECRAP your answers may be, but your comments certainly are not! – Neal Jan 31 '15 at 23:50 • Always funny: youtube.com/watch?v=SyD4p8_y8Kw – Jack D'Aurizio Feb 1 '15 at 1:48 They are all both open and closed. Let me make this a bit more clear. By definition of a topology (from wikipedia): A topological space is then a set $X$ together with a collection of subsets of $X$, called open sets and satisfying the following axioms: • The empty set and $X$ itself are open. • Any union of open sets is open. • The intersection of any finite number of open sets is open. So just from the definition itself it follows that $∅$ and $X$ are open. Furthermore a set is closed (by definition) if the complement is open. Therefore $∅$ and $X$ are closed (they are each others complement). The term clopen means that a set is both open and closed, so they are both also clopen. • Thanks, have up-voted yours. Can I take it daily for granted that both $X, \emptyset$ are clopen? – Amanda.M Jan 31 '15 at 23:42 • @A.Magnus yes that is true in any topology – Loreno Heer Jan 31 '15 at 23:50 To put it simply: sets are not doors! Being open does not mean that the set is not closed, and being closed do not implies that the set is not open. Yes, the empty set and the whole space are both open and closed. Another more dramatic example is: take a metric space $X$, with the discrete metric. Then every singleton $\{a\}$ (in fact, every subset of $X$) is clopen. Balls $B(a,r)$ with radius $r < 1$ are contained in $\{a\}$, hence $\{a\}$ is open. And $\{a\}$ is also closed, because it's complement is $X \setminus\{a\} = \bigcup_{b \in X, b \neq a}\{b\}$, a union of open sets, hence open. • Ask you a question, I hope I can say it clear: Let $B \subset A$ and $B$ is open. Absence of any other instructions, $A$ has to be closed since it is $B$'s complement. Now if $A$ is clopen, then its neutrality does not have any bearing on open-ness or the closed-ness of $B$. Am I correct? – Amanda.M Feb 1 '15 at 1:54 • But it is not true that $A$ is the complement of $B$.. – Ivo Terek Feb 1 '15 at 1:55 • I made typo and just corrected it. Sorry for confusion. – Amanda.M Feb 1 '15 at 1:56 By the first axiom in the definition of a topology, $X$ and $\emptyset$ are open. However, closed sets are precisely those whose complements are open, by definition. Hence the empty set and $X$, being each others complements, are also closed. So, they are clopen. • Have up-voted yours. Thanks. – Amanda.M Jan 31 '15 at 23:43 A minimal requirement on any topological space $(X,\tau)$ is that both $\varnothing$ and $X$ be open sets. By the definition of closed sets, these requirements imply that $\varnothing^c=X$ and $X^c=\varnothing$ are always closed. To sum up, in any topological space, the empty set and the whole set are always both open and closed, hence clopen. Your question “are $\varnothing$ and $X$ closed, open or clopen” is basically six questions in one: (1) Is $\varnothing$ closed? Answer: yes. (2) Is $\varnothing$ open? Answer: yes. (3) Is $\varnothing$ clopen? Answer: yes. (4) Is $X$ closed? Answer: yes. (5) Is $X$ open? Answer: yes. (6) Is $X$ clopen? Answer: yes. • Good answer, thank you! – Amanda.M Feb 1 '15 at 3:12

https://math.stackexchange.com/questions/3350525/inclusion-exclusion-principle-problem

# Inclusion–exclusion principle problem 172 business executives were surveyed to determine if they regularly read Fortune, Time, or Money magazines. 80 read Fortune, 70 read Time, 47 read Money, 47 read exactly two of the three magazines, 26 read Fortune and Time, 28 read Time and Money, and 7 read all three magazines. How many read none of the three magazines? So what i tried to do was to denote Time as $$T$$, Fortune as $$F$$, and Money as $$M$$. Then i wanted to do $$172 - |T \cup F \cup M|$$. To find that amount, i tried to do the inclusion exclusion principle. I found $$|F \cup M|$$ to be $$14$$ by looking at a venn diagram ($$26-7 + 28-7 + x = 47 -> x = 7$$, then $$7+7 = 14$$), but i guess this i where things went wrong. I got $$|T \cup F \cup M| = 80 + 70 + 47 - 26 - 28 - 14 + 7$$ thus the answer would be 53, but that is false. • Excellent reasoning! See the answer below. – S. Dolan Sep 10 '19 at 9:18 We are given explicitly that \begin{align*} |F| &= 80\\ |T| &= 70\\ |M| &= 47\\ |F\cap T| &= 26\\ |T\cap M| &= 28\\ |F\cap T\cap M| &= 7. \end{align*} We also get that 47 read "exactly two". If you consider a Venn diagram, it is not hard to see that this corresponds to the equation \begin{align*} &|F\cap T| + |T\cap M|+ |F\cap M|-3|F\cap T\cap M|=47\\ \implies & 26 + 28 + |F\cap M| - 3(7) = 47\\ \implies & |F\cap M| = 14. \end{align*} Thus you can plug this all in to inclusion-exclusion: \begin{align*} |F \cup T\cup M| &= |F|+|T|+|M| - |F\cap T| - |T\cap M|- |F\cap M| + |F\cap T\cap M|\\ &= 80+70+47-26-28-14+7=136, \end{align*} and thus the answer is $$172-136=\boxed{\text{36 people}}$$. In your expression $$|T \cup F \cup M| = 80 + 70 + 47 - 26 - 28 - 14 + 7=136$$. This gives a solution of $$36$$ not $$53$$. We have: $$|T \cup F \cup M| = |T| + |F| + |M| - |T \cap F| - |T \cap M| - |F \cap M| + |T \cap F \cap M|$$ $$= 70 + 80 + 47 - 26 - 28 - 14 + 7 = 136$$ Then, the number of people who do not read any magazine equals: $$172 - |T \cup F \cup M| = 36$$

http://mathhelpforum.com/number-theory/145830-divisibility.html

# Math Help - Divisibility 1. ## Divisibility Hi all. I'm trying to figure out the following problem: Find the number of positive integers not exceeding 1000 that are divisible by 3 but not by 4. Help will be appreciated. Looking for a simple/elementary proof. Thanks. 2. Originally Posted by pollardrho06 Hi all. I'm trying to figure out the following problem: Find the number of positive integers not exceeding 1000 that are divisible by 3 but not by 4. Help will be appreciated. Looking for a simple/elementary proof. Thanks. Hint: $|\{x\in \mathbb{Z}_+\mid x \text{ divisible by } 3 \text{ but not by } 4\}|$ $= |\{x\in \mathbb{Z}_+\mid x \text{ divisible by } 3\}|-|\{x\in \mathbb{Z}_+\mid x \text{ divisible by } 12\}|$ This is a consequence of $A\backslash B=A\backslash(A\cap B)$, and $|X\backslash Y|=|X|-|Y|$, if $Y\subseteq X$. 3. Originally Posted by Failure Hint: $|\{x\in \mathbb{Z}_+\mid x \text{ divisible by } 3 \text{ but not by } 4\}|$ $= |\{x\in \mathbb{Z}_+\mid x \text{ divisible by } 3\}|-|\{x\in \mathbb{Z}_+\mid x \text{ divisible by } 12\}|$ This is a consequence of $A\backslash B=A\backslash(A\cap B)$, and $|X\backslash Y|=|X|-|Y|$, if $Y\subseteq X$. I don't see it... 4. ## Hint Look for cycles. 5. Hello, pollardrho06! Find the number of positive integers not exceeding 1000 that are divisible by 3 but not by 4. Every third number is divisible by 3. . . There are: . $\left[\frac{1000}{3}\right] \:=\:333$ numbers divisible by 3. But every twelfth number is divisible by 3 and by 4. . . There are: . $\left[\frac{1000}{12}\right] \:=\:83$ multiples of 3 which are divisible by 4. Therefore, there are: . $333 - 83 \:=\:250$ such numbers. 6. Originally Posted by Soroban Hello, pollardrho06! Every third number is divisible by 3. . . There are: . $\left[\frac{1000}{3}\right] \:=\:333$ numbers divisible by 3. But every twelfth number is divisible by 3 and by 4. . . There are: . $\left[\frac{1000}{12}\right] \:=\:83$ multiples of 3 which are divisible by 4. Therefore, there are: . $333 - 83 \:=\:250$ such numbers. Wow!! The greatest integer function!! Gr8!! Thanks!!

https://math.stackexchange.com/questions/1887453/solving-the-differential-equation-y-y-tanx-cotx

# Solving the Differential equation $y' = y \tan(x) + \cot(x)$ I am asked to solve the following differential equation: $$y' = y \tan(x) + \cot(x)$$ What I have so far is \begin{align*} y' - (\tan x) y &= \cot(x)\\ \\ I &= e^{\int - \tan(x) dx} = \cos(x)\\ \\ \cos(x) \left( y' - y \tan(x) \right) &= \cos(x) \cot(x)\\ y' \cos(x) - y \sin(x) &= \cos(x) \cot(x)\\ \int y' \cos(x) - y \sin(x) &= \int \cos(x) \cot(x) dx\\ \\ y \cos(x) &= \int \frac{cos^2(x)}{\sin(x)}\\ y \cos(x) &= \int \frac{1-sin^2(x)}{\sin(x)}\\ &= \int \csc(x) - \sin(x) \ dx\\ &= - \ln \vert \csc(x) + \cot(x) \vert + \cos(x) + C\\ \\ \therefore y &= - \frac{\ln \vert \csc(x) + \cot(x) \vert}{\cos(x)} + 1 + C \sec(x) \end{align*} The thing is: I am having a hard time comparing my result to the textbook's solution and to Wolfram's solution. Textbook's solution: $$\sec(x) \left( \frac{x}{2} + \frac{\sin(2x)}{4} + C \right)$$ Is my solution correct? Thank you. • Your answer is correct and is equivalent to WA's. The textbook answer is wrong – David Quinn Aug 9 '16 at 19:41 • Hi @DavidQuinn thank you for your input. How could I go from my answer to Wolfram's (or the opposite)? What kind of transformation did you do? – bru1987 Aug 9 '16 at 19:42 • Use $\csc x+\cot x=\cot(\frac x2)$ – David Quinn Aug 9 '16 at 19:50 \begin{align} y'(x)-y \tan x &=\sec x \tan x \left( \frac{x}{2} + \frac{\sin 2x }{4} + C \right)+\sec x \left( \frac{1}{2} + \frac{\cos 2x }{2} \right) \\&\hspace{4mm} -\sec x\left(\frac{x}{2}+\frac{\sin 2x }{4} + C \right)\tan x \\ &= \frac{1 + \cos 2x }{2\cos x}\\&=\cos x \end{align} i.e. the textbook's answer is for the RHS being $\cos x$ rather than $\cot x$. Since $\ln \vert \csc(x) + \cot(x) \vert=\ln{|\frac{1+\cos{x}}{\sin{x}}|}=\ln{|\frac{2\cos^2{(x/2)}}{2\sin{(x/2)\cos(x/2)}}|}=\ln{\cos{(x/2)}}-\ln{\sin(x/2)}$, your solution is same as WolphramAlpha's.

https://math.stackexchange.com/questions/2724960/probability-of-getting-full-house-in-poker?noredirect=1

# Probability of getting full house in poker I have this problem which I have solved, but using two different methods. I am quite new to combinatorics and want to know how to intuitively understand the difference between the following two methods. The problem consist of calculating the probability of getting a full house being dealt a 5-card poker hand. First of, I solve this by simply saying that $P($get full house$)=\frac{2 {13\choose 2} {4 \choose 3} {4 \choose 2}}{{52 \choose 5}}$. This is the right answer according to my text-book. However, at my first attempt at solving this I forgot the factor 2 in the numerator. My reasoning goes like this: All the possible ways of getting a full house consists of all the ways we can combine two different ranks (i.e. ${13 \choose 2}$) times all the possible ways of choosing 3 cards out of 4 suits, times all the possible ways of choosing 2 cards out of 4 suits. Now, all this seems logical to me. The thing that makes me doubt whether I truly understand what I'm doing is how the factor 2 comes in place. I'm thinking: Because we choose two different ranks without regards to order, we have to compensate for those combinations and therefore multiply with $2!$, because obviously it does matter if I (for example) choose the ranks (ace,knights) and in this sequence choose three ace and two knights. On the other hand, I can solve the problem using the method described here: https://math.stackexchange.com/a/808328/518320 Which as well seems intuitively clear, thinking the way the user describe the process in that thread. What is the difference in the two methods? Maybe this is obvious, but I'm new to combinatorics. And is my reasoning above accurate? Thanks! • I would have said: choose the suit for the triple, that's $\binom {13}1$. Then choose the triple of ranks, $\binom 43$. Then choose the suit for the pair, $\binom {12}1$, then choose the pair, $\binom 42$. It's true that $2\times \binom {13}2= \binom {13}1\times \binom {12}1$ but I find this less intuitive. – lulu Apr 6 '18 at 12:50 • @lulu While your calculations are correct, you confused ranks with suits. The four suits are hearts, clubs, diamonds, and spades. The thirteen ranks are 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A. – N. F. Taussig Apr 6 '18 at 13:07 • @N.F.Taussig absolutely correct. I plead lack of coffee. – lulu Apr 6 '18 at 13:10 They are essentially the same giving $156$ ways of choosing ranks, but if you want to draw a distinction: • Choose two ranks and then choose one of those two to be the three so the other is the pair $${13 \choose 2}{2 \choose 1}$$ • Choose one rank to be the three then choose another rank to be the pair $${13 \choose 1}{12 \choose 1}$$ and you then multiply this by ${4 \choose 3}{4 \choose 2}$ and divide by ${52 \choose 5}$ Doing the same thing for two pairs to get $858$ ways of choosing the ranks by several methods: • Choose three ranks and then choose one of those three to be the single so the others are the pairs $${13 \choose 3}{3 \choose 1}$$ • Choose three ranks and then choose two of those three to be the pairs so the other is the single $${13 \choose 3}{3 \choose 2}$$ • Choose one rank to be the single then choose two other ranks to be the pairs $${13 \choose 1}{12 \choose 2}$$ • Choose two ranks to be the pairs then choose another rank to be the single $${13 \choose 2}{11 \choose 1}$$ and you then multiply this by ${4 \choose 1}{4 \choose 2}{4 \choose 2}$ and divide by ${52 \choose 5}$ The expressions $2\binom{13}{2}$ and $\binom{13}{1}\binom{12}{1}$ are both equivalent to "choose $2$ objects from $13$ where the order of choosing matters." In other words, we want permutations of $2$ objects out of $13$ rather than combinations. The way to get permutations of $k$ objects out of $n$ is to multiply $k$ consecutive integers together, where $n$ is the largest of the integers multiplied: $n(n-1)(n-2) \cdots (n-k+1).$ For example, permutations of $2$ objects out of $n$ give $13\cdot12$ possible permutations. And since $\binom n1 = n,$ $13\cdot12 = \binom{13}{1}\binom{12}{1}.$ But often the number of permutations is written $\dfrac{n!}{(n-k)!},$ because $$\frac{n!}{(n-k)!} = n(n-1)(n-2) \cdots (n-k+1).$$ To get the number of combinations of $k$ objects out of $n$ (without regard for the order of the objects), we divide by $k!,$ since that's the number of different sequences in which each combination of $k$ objects can occur as a permutation. So we get $$\binom nk = \frac{n!}{(n-k)!k!}.$$ This also means that $k!\binom nk$ is yet another way to write the number of permutations of $k$ objects out of $n.$ In short, it's all really the same thing written in different ways. The two ranks are distinguishable once you get a particular full house. Say you get three kings and two queens. That differs from getting three queens and two kings. So it should be $13 \cdot 12$ ways to pick the two ranks. This is the same as $2 \cdot \binom{13}{2}.$

https://mathhelpboards.com/threads/find-the-image-f-n-n-r-f-x-m-2-2n.7102/

# Find the image: F: N * N -> R , F(x) = m^2 + 2n #### KOO ##### New member F:N*N -> R, F(x) = m^2 + 2n I think the answer is N. Am I right? #### Chris L T521 ##### Well-known member Staff member I think the answer is N. Am I right? Based on your title, we have that $F:\mathbb{N}\times\mathbb{N}\rightarrow \mathbb{R}$ ($\mathbb{N}\ast\mathbb{N}$ makes no sense at all) where $F(x) = m^2+2n$ (I assume here that $x=(m,n)$). Now, depending on who you talk to, $\mathbb{N}$ may or may not include zero (the general consensus from what I've seen is that $0\notin\mathbb{N}$, but there are some professors/authors that include zero in their definition of the natural numbers; hence why I think it's best to clarify this right from the get go); that is, either $\mathbb{N}=\{x\in\mathbb{Z} : x\geq 0\}$ or $\mathbb{N}=\{x\in\mathbb{Z}: x\geq 1\}$. If $0\in\mathbb{N}$, then $\mathrm{Im}(F)=\mathbb{N}$. If $0\notin\mathbb{N}$, then $\mathrm{Im}(F) = \mathbb{N}\backslash\{1,2,4\}$ since there aren't pairs $(m,n)\in\mathbb{N}\times\mathbb{N}$ such that $m^2+2n=1$, $m^2+2n=2$, or $m^2+2n=4$. (In $\mathrm{Im}(F)$, note that all the positive odd numbers greater than or equal to 3 are generated by pairs of the form $(1,n)$ for $n\in\mathbb{N}$ since $1^2+2n=2n+1$, and all positive even numbers greater than or equal to 6 are generated by pairs of the form $(2,m)$ for $m\in\mathbb{N}$ since $2^2+2m = 2(m+2)$; this is why I can claim that $F(\mathbb{N}\times\mathbb{N}) = \mathbb{N}\backslash\{1,2,4\}$ for $0\notin\mathbb{N}$.) I hope this makes sense! #### KOO ##### New member Based on your title, we have that $F:\mathbb{N}\times\mathbb{N}\rightarrow \mathbb{R}$ ($\mathbb{N}\ast\mathbb{N}$ makes no sense at all) where $F(x) = m^2+2n$ (I assume here that $x=(m,n)$). Now, depending on who you talk to, $\mathbb{N}$ may or may not include zero (the general consensus from what I've seen is that $0\notin\mathbb{N}$, but there are some professors/authors that include zero in their definition of the natural numbers; hence why I think it's best to clarify this right from the get go); that is, either $\mathbb{N}=\{x\in\mathbb{Z} : x\geq 0\}$ or $\mathbb{N}=\{x\in\mathbb{Z}: x\geq 1\}$. If $0\in\mathbb{N}$, then $\mathrm{Im}(F)=\mathbb{N}$. If $0\notin\mathbb{N}$, then $\mathrm{Im}(F) = \mathbb{N}\backslash\{1,2,4\}$ since there aren't pairs $(m,n)\in\mathbb{N}\times\mathbb{N}$ such that $m^2+2n=1$, $m^2+2n=2$, or $m^2+2n=4$. (In $\mathrm{Im}(F)$, note that all the positive odd numbers greater than or equal to 3 are generated by pairs of the form $(1,n)$ for $n\in\mathbb{N}$ since $1^2+2n=2n+1$, and all positive even numbers greater than or equal to 6 are generated by pairs of the form $(2,m)$ for $m\in\mathbb{N}$ since $2^2+2m = 2(m+2)$; this is why I can claim that $F(\mathbb{N}\times\mathbb{N}) = \mathbb{N}\backslash\{1,2,4\}$ for $0\notin\mathbb{N}$.) I hope this makes sense! Actually I had this question on a test and you're right x=(m,n) and we're told N does not include 0. Anyways what does \ mean? #### Chris L T521 ##### Well-known member Staff member Actually I had this question on a test and you're right x=(m,n) and we're told N does not include 0. Anyways what does \ mean? Ah, that's one notation for set difference. I could have also written it as $\mathbb{N}-\{1,2,4\}$.

https://math.stackexchange.com/questions/2816513/same-integration-with-2-different-answers

# Same integration with 2 different answers? $$\int x(x^2+2)^4\,dx$$ When we do this integration with u substitution we get $$\frac{(x^2+2)^5}{10}$$ as $u=x^2+2$ $du=2x\,dx$ $$\therefore \int (u+2)^4\,du = \frac{(x^2+2)^5}{10} + C$$ Although when we expand the fraction and then integrate the answer we get is different: $x(x^2+2)^4=x^9+8x^7+24x^5+32x^3+16x$ $$\int x^9+8x^7+24x^5+32x^3+16x \,dx$$ we get $$\frac {x^{10}}{10} +x^8+4x^6+8x^4+8x^2 + C$$ For a better idea of the questions, let's say the questions asks us to find the value of C when y(0)=1 Now, $x=0$ $$\frac {0^{10}}{10} + 0^8 + 4(0)^6 + 8(0)^4 + 8(0)^2 + C = 1$$ $$\therefore C= 1$$ AND $$\frac {(0+2)^5}{10} + C= 1$$ $$\therefore \frac {32}{10} + C = 1$$ $$\therefore C = 1 - 3.2 = -2.2$$ • Don't forget the arbitrary constant of integration! – Lord Shark the Unknown Jun 12 '18 at 4:41 • ^^^^^ What @LordSharktheUnknown said! The indefinite integral is the class of all functions who are antiderivatives of the integrand. These antiderivatives only differ by a constant. – N8tron Jun 12 '18 at 4:48 • $\frac {(x^2+2)^5}{10} = \frac {x^{10}}{10} + x^8 + 4x^6+8x^4+8x^2 + \frac {32}{10}.$ The only difference is the constant of integration. – Doug M Jun 12 '18 at 4:52 • Please fix the typos on $(x+2)$. – Yves Daoust Jun 12 '18 at 5:16 • There are lots and lots of similar questions already, just search: google.com/… – Hans Lundmark Jun 12 '18 at 6:46 Like mentioned in the comments this is all fixed if you remember your constant of integration. $$\int x(x^2+2)^4\ dx= \frac{(x^2+2)^5}{10}+C$$ Note if you expand $$\begin{split} \frac{(x^2+2)^5}{10}&=\frac{1}{10}\left(x^{10}+5x^8(2)+10x^6(2^2)+10x^4(2^3)+5x^2(2^4)+2^5\right)\\ &=\frac{x^{10}}{10}+x^8+4x^6+8x^4+8x^2+\frac{32}{10} \end{split}$$ Notice the relation to your other way of computing the integral $$\int x(x^2+2)^4\ dx = \frac{x^{10}}{10}+x^8+4x^6+8x^4+8x^2 +C$$ So lets call $F(x)=\frac{x^{10}}{10}+x^8+4x^6+8x^4+8x^2$ and $G(x)=\frac{x^{10}}{10}+x^8+4x^6+8x^4+8x^2+\frac{32}{10}$ then $F(x)-G(x)=-\frac{32}{10}$ a constant. All antiderivatives of a continuous function only differ by a constant. Just for fun Let's see another one: First lets use double angle for sine $$\int \cos x\sin x\ dx=\frac{1}{2}\int\sin 2x\ dx=-\frac{1}{4}\cos 2x +C$$ Then substitutions $u=\sin x$ $$\int \cos x\sin x\ dx=\int u\ du =\frac{u^2}{2}+C=\frac{\sin^2 x}{2}+C$$ Then substitutions $u=\cos x$ $$\int \cos x\sin x\ dx=\int -u\ du =\frac{-u^2}{2}+C=\frac{-\cos^2 x}{2}+C$$ If you find the constant differences and combine them in the right way you get the half angle formulas: $$\sin^2 x=\frac{1-\cos 2x}{2},\quad \cos^2 x=\frac{1+\cos 2x}{2}$$ Note you can pretty quickly derive some funky trig identities in this way. For instance if you consider $\int \cos^3 x \sin^5 x\ dx$ • Although if we were asked to find the value of C at x=0, won't we have 2 different value of C??? – Agent Smith Jun 13 '18 at 0:47 • No if you fix any two antiderivatives $F$ and $G$ for a suitable function (continuous is adequate but the real condition is Riemann integrable). There is a constant C such that the difference $F(x)-G(x)=C$ for all x. This can be rigourously proven. – N8tron Jun 13 '18 at 1:45 • Check out this page (or numerous others) to see the proof math.stackexchange.com/questions/1862231/… – N8tron Jun 13 '18 at 1:49 You can check an antiderivative by differentiating. $$\left(\frac{(x^2+2)^5}{10}\right)'=x(x^2+2)^4=x^9+8x^7+24x^5+32x^3+16x$$ and $$\left(\frac {x^{10}}{10} +x^8+4x^6+8x^4+8x^2\right)'=x^9+8x^7+24+32x^3+16x$$ and the two expressions are indeed equivalent. Now the long explanation. Consider the binomial $x^2+a$ raised to some power $n$ and multiplied by $2x$. $$2x(x^2+a)^m$$ which integrates as $$\frac{(x^2+a)^{m+1}}{m+1}.$$ By the binomial theorem, the terms in the development of this antiderivative are $$\frac1{m+1}\binom{m+1}kx^{2(m+1-k)}a^k.$$ On the other hand, the development of the initial integrand gives terms $$2\binom mkx^{2(m-k)+1}a^k,$$ and after integration $$\frac1{m-k+1}\binom mkx^{2(m-k)+2}a^k.$$ It is easy to see that all terms coincide, because $$\frac1{m+1}\frac{(m+1)!}{k!(m+1-k)!}=\frac1{m-k+1}\frac{m!}{k!(m-k)!}=\frac{(m-1)!}{k!(m-k+1)!}.$$ Anyway, the first development holds for $0\le k\le m+1$, giving a constant term $\dfrac{a^m}{m+1}$, but the second for $0\le k\le m$ only, giving no constant term. But this does not matter, as two antiderivatives can differ by a constant. • But wouldn't the C be different for both the cases?? – Agent Smith Jun 13 '18 at 0:48 • @AgentSmith: and ? – Yves Daoust Jun 13 '18 at 5:50 • and that would change the answer wouldn't it?? if we were asked for the value of C? – Agent Smith Jun 13 '18 at 19:56 • @AgentSmith: you'd answer: arbitrary. You should understand that the value of the constant does not matter. – Yves Daoust Jun 13 '18 at 20:05 • oh okay thanks! Makes sense! – Agent Smith Jun 14 '18 at 18:59

https://gmatclub.com/forum/a-train-travels-from-city-a-to-city-b-the-average-speed-of-the-train-188007.html

GMAT Question of the Day - Daily to your Mailbox; hard ones only It is currently 20 Aug 2018, 23:02 ### 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 # A train travels from city A to city B. The average speed of the train Author Message TAGS: ### Hide Tags SVP Status: The Best Or Nothing Joined: 27 Dec 2012 Posts: 1835 Location: India Concentration: General Management, Technology WE: Information Technology (Computer Software) A train travels from city A to city B. The average speed of the train  [#permalink] ### Show Tags 04 Nov 2014, 00:03 9 00:00 Difficulty: 55% (hard) Question Stats: 64% (02:41) correct 36% (02:28) wrong based on 199 sessions ### HideShow timer Statistics A train travels from city A to city B. The average speed of the train is 60 miles/hr and it travels the first quarter of the trip at a speed of 90 mi/hr. What is the speed of the train in the remaining trip? A. 30 B. 45 C. 54 D. 72 E. 90 _________________ Kindly press "+1 Kudos" to appreciate SVP Status: The Best Or Nothing Joined: 27 Dec 2012 Posts: 1835 Location: India Concentration: General Management, Technology WE: Information Technology (Computer Software) Re: A train travels from city A to city B. The average speed of the train  [#permalink] ### Show Tags 10 Nov 2014, 20:00 5 2 Refer diagram below: Attachment: speed.png [ 3.39 KiB | Viewed 3540 times ] Setting up the time equation: Total time = Time required in first quarter + Time required in the remaining journey $$\frac{d}{60} = \frac{d}{4*90} + \frac{3d}{4*s}$$ $$s = \frac{90*3}{5} = 54$$ _________________ Kindly press "+1 Kudos" to appreciate ##### General Discussion Senior Manager Status: Math is psycho-logical Joined: 07 Apr 2014 Posts: 421 Location: Netherlands GMAT Date: 02-11-2015 WE: Psychology and Counseling (Other) A train travels from city A to city B. The average speed of the train  [#permalink] ### Show Tags 30 Jan 2015, 06:23 Using the RTD chart, with A being the first quarter, B the rest of the trip and All the combined "trip". _______R_____T______D A.........90........3..........270 B.........54.......15.........810 All........60.......18........1080 Let me explain how we fill in the chart: 1) Starting with what we know, we add 90 and 60 2) Picking an easy number for the total distance (6*9=54), so I chose 540. Multiply by 2 to get the whole trip, and we get 1080. Add 1080 for all-distance. 3) 1/4 of the whole distance happened at a Rate of 90. So, 1080/4=270, add 270 under A-D. 4) 2080 - 270 = 810, for the rest of the trip. Add 810 under B-D. 5) Claculate the individual Times for A and All, using T=D/R. Add the results, 3 and 18, under A-T and All-T. 6) 18-3=15, this is the remaining Time for B. Add 15 under B-T. 7) Finally, 15R=810 -->R=810/15 --> R= 54 ANS C Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 3186 Location: United States (CA) Re: A train travels from city A to city B. The average speed of the train  [#permalink] ### Show Tags 06 Apr 2017, 09:45 1 1 PareshGmat wrote: A train travels from city A to city B. The average speed of the train is 60 miles/hr and it travels the first quarter of the trip at a speed of 90 mi/hr. What is the speed of the train in the remaining trip? A. 30 B. 45 C. 54 D. 72 E. 90 We have an average rate problem in which we can use the following formula: Avg speed = (distance 1 + distance 2)/(time 1 + time 2) If we let d = total distance of the trip, then the first quarter of the trip, or (1/4)d = d/4, was traveled at 90 mph. Thus, the time was (d/4)/90 = d/360. We can let the rate for the remaining part of the trip = r, and thus the time for the remaining part of the trip, or (3/4)d = 3d/4, is (3d/4)/r = 3d/(4r). Let’s use all of this information in the average rate equation: 60 = d/(d/360 + 3d/(4r)) 60 = 1/(1/360 + 3/(4r)) 60(1/360 + 3/(4r)) = 1 1/6 + 45/r = 1 Let’s multiply the above equation by 6r: r + 270 = 6r 270 = 5r r = 54 _________________ Scott Woodbury-Stewart Founder and CEO GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 12217 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: A train travels from city A to city B. The average speed of the train  [#permalink] ### Show Tags 08 Apr 2017, 18:53 2 Hi All, This question can be solved by TESTing VALUES. The prompt tells us that average speed of the train over the course of the entire trip is 60 miles/hr and that it travels the first quarter of the distance at a speed of 90 mi/hr. We're asked for the average speed of the train for the remaining three-quarters of the trip. Let's choose 90 miles for the FIRST QUARTER of the distance. We can then immediately calculate two things... 1) Since the train was traveling 90 miles/hour, the first quarter of the trip took 1 hour to complete. 2) Since (1/4)(Total Distance) = 90 miles, then the FULL TRIP = 4(90) = 360 miles. The total trip is 360 miles; with an average speed of 60 miles/hr, the FULL TRIP would take... (X)(60 mph) = 360 miles.... X = 6 hours to complete. The first hour of the trip covered 90 miles, so the remaining 360 - 90 = 270 miles of the trip are covered in the remaining 5 hours... Thus, the average speed for the remainder of the trip is... (270 miles)/(5 hours) = 54 miles/hour GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save \$75 + GMAT Club Tests Free Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/ ***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*********************** Non-Human User Joined: 09 Sep 2013 Posts: 7774 Re: A train travels from city A to city B. The average speed of the train  [#permalink] ### Show Tags 29 Jul 2018, 07:25 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: A train travels from city A to city B. The average speed of the train &nbs [#permalink] 29 Jul 2018, 07:25 Display posts from previous: Sort by # Events & Promotions 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®.

https://gmatclub.com/forum/if-22-n-is-a-divisor-of-97-98-what-is-the-maximum-possible-228660.html

GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video It is currently 30 Mar 2020, 07:52 ### 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 # If 22^n is a divisor of 97!+98! ,what is the maximum possible Author Message TAGS: ### Hide Tags Current Student Joined: 12 Aug 2015 Posts: 2537 Schools: Boston U '20 (M) GRE 1: Q169 V154 If 22^n is a divisor of 97!+98! ,what is the maximum possible  [#permalink] ### Show Tags 10 Nov 2016, 10:58 1 7 00:00 Difficulty: 55% (hard) Question Stats: 57% (01:48) correct 43% (01:42) wrong based on 94 sessions ### HideShow timer Statistics If 22^n is a divisor of 97!+98! ,what is the maximum possible value of integer n? A)8 B)9 C)10 D)11 E)12 _________________ Math Expert Joined: 02 Aug 2009 Posts: 8296 Re: If 22^n is a divisor of 97!+98! ,what is the maximum possible  [#permalink] ### Show Tags 10 Nov 2016, 17:36 5 stonecold wrote: If 22^n is a divisor of 97!+98! ,what is the maximum possible value of integer n? A)8 B)9 C)10 D)11 E)12 22^n means power of 11 as 2 would surely have more power than 11 in a factorial.. 97!+98!=97!(1+98)=97!*99.... 99 will have ONE 11 and 97! will have 97/11 or 8.. Total 1+8=9 B _________________ ##### General Discussion Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 9895 Location: United States (CA) Re: If 22^n is a divisor of 97!+98! ,what is the maximum possible  [#permalink] ### Show Tags 12 Nov 2016, 05:33 2 2 stonecold wrote: If 22^n is a divisor of 97!+98! ,what is the maximum possible value of integer n? A)8 B)9 C)10 D)11 E)12 We are given that 22^n is a divisor of 97!+98!. Thus: (97!+98!)/22^n = integer To determine the maximum value of n, we can factor out a 97! in the numerator of the above expression and we have: 97!(1 + 98)/22^n = integer 97!(99)/22^n = integer Since 22 = 11 x 2, we can rewrite our expression as: 97!(99)/(11^n x 2^n) = integer In order to determine the maximum value of n such that 22^n divides into 97!(99), we need to determine the maximum number of pairs of factors of 11 and 2 within 97!(99). Since we know there are fewer factors of 11 than factors of 2, we can determine the number of factors of 11 within 97!(99). Let’s start with 97!: 11, 22, 33, 44, 55, 66, 77 and 88 are all factors of 97! Thus, 97! has 8 factors of 11. We also see that 99 has 1 factor of 11. Thus, there are 9 factors of 11 within 97!(99), and thus the maximum value of n is 9. _________________ # Scott Woodbury-Stewart Founder and CEO Scott@TargetTestPrep.com 197 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. CrackVerbal Quant Expert Joined: 12 Apr 2019 Posts: 459 Re: If 22^n is a divisor of 97!+98! ,what is the maximum possible  [#permalink] ### Show Tags 24 Feb 2020, 23:55 In questions related to divisibility, breaking down the dividend will give a lot of information about what the divisor needs to be if it HAS to divide the dividend fully. Therefore, let’s break the dividend down, first. In 97! + 98!, we can take 97! as common. When we do this, we have 97! + 98! = 97! (1 + 98) = 97! * 99. This helps us understand that $$22^n$$ should be able to divide 97! * 99 fully. Let’s now look at 22. 22 = 2*11, therefore $$22^n$$ = $$2^n$$ * $$11^n$$. So, essentially, we are trying to find the highest power of 11 in the dividend since the number of 11s would be far fewer compared to the number of 2s. Finding out the highest power of any prime number in a given factorial can be done by successive division. In our case, it can be done as shown below in the diagram. Attachment: 25th Feb 2020 - Reply 3 - 1.jpg [ 73.73 KiB | Viewed 123 times ] So, the highest power of 11 in 97! is 8. But, 99 also has a 11 in it. Therefore, the highest power of 11 in the numerator (or the dividend) is 9. So, there need to be 9 11s in the denominator as well. As such, the highest power of 22 that can divide 97! + 98! is 9. The correct answer option is B. Hope that helps! _________________ Re: If 22^n is a divisor of 97!+98! ,what is the maximum possible   [#permalink] 24 Feb 2020, 23:55 Display posts from previous: Sort by

https://www.physicsforums.com/threads/general-formula-for-this.104303/

# General formula for this 1. Dec 16, 2005 ### twoflower Hi all, there is general formula for findind out, "Which highest power of x is divisible n! with?" Eg. Which highest power of 5 is divisable 50! with? But I forgot it and can't find it now...will somebody help me please? Thank you. 2. Dec 16, 2005 ### AKG You mean, what is the highest power of x that divides n!? Well if you can prime-factorize x and n!, this is easy. In the case of 5 and 50!, it's hard to prime-factorize 50!, but the solution is still easy. You can see quite easily that 5 will occur 12 times in the prime-factorization of 50! (it will occur once in each of the factors 5, 10, 15, 20, 30, 35, 40, 45, and twice in both 25 and 50), so the highest power of 5 that divides 50! is 512. 3. Dec 16, 2005 ### twoflower Yes, this manual approach is clear. But there is also a general formula.. 4. Dec 16, 2005 ### AKG I'm not sure about a general formula just yet, but this might be helpful: if you associate each factorial with a sequence (an,k)k in N, where an,k is the power of the kth prime in n!, you get: (a0,k) = (0, 0, ...) (a1,k) = (0, 0, ...) (a2,k) = (1, 0, 0, ...) (a3,k) = (1, 1, 0, 0, ...) (3, 1, 0, 0, ...) (3, 1, 1, 0, 0, ...) (4, 2, 1, 0, 0, ...) (4, 2, 1, 1, 0, 0, ...) (7, 2, 1, 1, 0, 0, ...) (7, 4, 1, 1, 0, 0, ...) (8, 4, 2, 1, 0, 0, ...) (8, 4, 2, 1, 1, 0, 0, ...) (10, 5, 2, 1, 1, 0, 0, ...) (10, 5, 2, 1, 1, 1, 0, 0, ...) (11, 5, 2, 2, 1, 1, 0, 0, ...) The exponent of 2 changes every 2 rows, the exponent of 3 changes every 3 rows, the exponent of p will change every p rows. Ignoring repetition, the exponents for 2 go: 0, 1, 3, 4, 7, 8, 10, 11, ... For 3, they go 0, 1, 2, 4, 5, ... For 5, they go 0, 1, 2, ... Some numbers are skipped in the above sequences because powers occur, i.e. in the sequence for 2, it goes from 1 to 3 without going through 2 because the 4 in 4! contributes two 2's, not just 1. I'm very tired right now, but I suppose if you can generalize what's going on here, it might be a helpful step in finding a general formula. Well I suppose if you just want to know the general formula and aren't trying to figure it out yourself, then the above isn't much help. 5. Dec 16, 2005 ### VietDao29 I suppose that this is not a homework problem... So here's my approach. Let's say that [x] is the function that will return the integer part before the decimal point of the number x, eg: [37.5534] = 37. Say, you need to find the largest p that satisfies: xp divides n!, for some given x (x is prime), and n. Let: $$\rho := \left[ \frac{\ln n}{\ln x} \right]$$, ie: $$\rho$$ is the largest positive integer such that: $$x ^ \rho \leq n$$ So for every x consecutive integers there's one integer that's divisible by x, for every x2 consecutive integers there's one integer that's divisible by x2, for every x3 consecutive integers there's one integer that's divisible by x3,... So the largest p can be obtained by: $$p := \sum_{i = 1} ^ \rho \left[ \frac{n}{x ^ i} \right]$$ --------------------- If x is not prime, then you can prime-factorize it: $$x = \lambda_1 ^ {\alpha_1} \ \lambda_2 ^ {\alpha_2} \ \lambda_3 ^ {\alpha_3} \ ... \lambda_k ^ {\alpha_k}$$ Where: $$\lambda_i, \ i = 1..k$$ are primes. Then: $$x ^ p = (\lambda_1 ^ {\alpha_1} \ \lambda_2 ^ {\alpha_2} \ \lambda_3 ^ {\alpha_3} \ ... \lambda_k ^ {\alpha_k}) ^ p = \lambda_1 ^ {\alpha_1 p} \ \lambda_2 ^ {\alpha_2 p} \ \lambda_3 ^ {\alpha_3 p} \ ... \lambda_k ^ {\alpha_k p}$$ If xp divides n! then $$\lambda_i ^ {\alpha_i p}, \ i = 1..k$$ must also divide n!. So you can use the same method (as shown above) to find the largest $$\rho_i, \ i = 1..k$$ such that $$\lambda_i ^ {\rho_i}, \ i = 1..k$$ divides n!. Then define: $$\beta_i := \left[ \frac{\rho_i}{\alpha_i} \right], \ i = 1..k$$. And the largest p can be found by: $$p = \min (\beta_i), \ i = 1..k$$ Last edited: Dec 16, 2005 6. Dec 25, 2005 ### benorin The exact power of a prime p dividing n! is alternatively given by $$\frac{n-\mu}{p-1},$$ where $\mu$ is the sum of the digits of the base p representation of n.

http://math.stackexchange.com/questions/281604/probability-of-an-event-after-time-has-passed/281639

# Probability of an event after time has passed A friend of mine posed this problem and we have had a disagreement on the answer. The problem: • There is a 90% chance that some event will happen in the next year. • There is a 95% chance that the event will happen eventually. • What is the probability that it happens after this year, if it does not happen in the next year? Hover over here for a description of our answers and arguments: I told my friend that it is 95%. My reasoning is that there is a 95% chance that it happens from now to infinity. The probability that it happens from now to infinity minus one year is still 95%. In one year, when it hasn't happened, events that didn't happen don't affect your "eventual" odds. My friend on the other hand, believes it to be 50%. He had two arguments. First, he said if you flip a coin 4 times, the chance that you get heads the first time is 50%, but the chance that you get heads eventually is 93.75% (15/16). If you don't get it the first time, what is the chance that you get it after the first time? 87.5% (7/8), in other words, it has decreased. To this I responded that the "eventually" is based on a finite set of events, each with their own probability and after the first event has passed, you have fewer chances. In his problem, on the other hand, it is a summary of the probability of unknown events broken in two time periods. He continued to insist that it is the same because you have two time periods with known odds. The 95% is the combination of two probabilities, 90% and 50%. Eliminating the first 90%, the probability of the remaining period is 50%. I agreed that's an accurate way to look at it from now, but I don't think that's accurate once you know it did not happen in that time period. - Let's use Bayes' Theorem: $$P(A|B)=\frac{P(B|A) P(A)}{P(B)}.$$ In our case, $B$ is the event whatnot does not occur in year 1 (so with probability $P(B)=.1$, and $A$ is the event whatnot occurs at some point (so with probability $P(A)=0.95$). Of course, $P(B|A)$ is the only thing left to determine, but notice that $P(A|not(B))=1$ since if whatnot occurs in year 1, then it occurs at some point. Therefore, again by Bayes' Theorem $$P(not(B)|A)=\frac{P(A|not(B)) P(not(B))}{P(A)}=\frac{90}{95}.$$ Since $P(not(B)|A)+P(B|A)=1$, we have that $P(B|A)=\frac{5}{95}$. Hence plugging this all in, we find $$P(A|B)=\frac{\frac{5}{95} \frac{95}{100}}{\frac{10}{100}}=\frac{1}{2}.$$ So it looks like your friend was right! - Calculate it this way. Given that the probability that it occurs during or after year 2 is $x$%. And the probability it occurs in year 1 is 90%, what is the probability it occurs? Now, consider $x=0$, then the probability the event occurs is 90%. If $x=50$, then it is 95%. (because it is $0.9 \times 1 + 0.1 \times 0.5 \times 1= 0.95$) This informal argument should convince you. - After talking about it some more, we came up with this scenario which illustrates why he is correct, and I am wrong. 1. Say I want to give him a 90% chance that he will get $100 this year, 95% chance total, and 5% chance that he gets nothing. 2. So, I take 100 sheets of paper and write dates in the next year on 90 of them, dates after that on 5, and "Bad luck" on the other 5. 3. I draw one at random and keep it a secret. 4. A year passes and my friend got nothing. Now he knows that all 90 dates this year are still in the pile. He now knows I must hold one of the other 10 sheets. That makes a 5/10 chance that I have a sheet with a date on it, and 5/10 chance that the sheet I hold has "Bad luck" on it. 50% chance he will get$100. - Nice. Beats Bayes. –  André Nicolas Jan 18 '13 at 21:15 Let $A$ represent the cases in which the event happens eventually. Let $B$ represent the cases in which the event happens next year. You've stipulated $P(A)=0.95$ and $P(B)=0.90$. Because $B$ is a subset of $A$, $P(A\cap B)=P(B)=0.90$, and $$P(A\cap \neg B)=P(A)-P(A\cap B)=0.95-0.90=0.05.$$ By the definition of conditional probability, then, $$P(A|\neg B)=\frac{P(A\cap \neg B)}{P(\neg B)}=\frac{0.05}{0.10}=\frac{1}{2}.$$ So if I propose a deal right now, under the terms of which (a) you'll put in $\$1000$if the event hasn't happened after one year and (b) you'll get back$X$if the event happens after next year, then you should take the deal if$X>\$2000$. In this sense, your friend is right. Your point, though, is different. You're wondering what deal you should take a year from now if the event hasn't happened by then. The answer is that it depends on probabilities not stipulated in the problem. In particular, you may learn something from the fact that it didn't happen this year, which may raise or lower these odds. Suppose the event is the earth being hit by a meteor, that this depends on the solar system's meteor density, and that we're building a meteor shield that will be operational in just over a year. Based on what we know now, the meteor density might be high or low, with equal likelihood; if it's high, then the earth will be hit in the next year with probability $1$ (no shield yet!); if it's low, then the earth will be hit next year with probability $0.8$ and eventually with probability $0.9$. A year from now, if the earth hasn't been hit yet, you will know that the meteor density is low, and that the remaining probability that we'll ever be hit is $0.1$. In this case, the odds were lowered. On the other hand, the choice may be between cases where the event will happen in either eighteen months with certainty (so next year with probability $0$ and eventually with probability $1$), or at some indeterminate time (next year with probability $0.9/(1-\varepsilon)$ and eventually with probability $(0.95-\varepsilon)/(1-\varepsilon)$), based on some unknown variable that might point to the first case (with probability $\varepsilon$) but almost surely points to the second (with probability $1-\varepsilon$). Here, a year from now you'll be much more likely to believe that the unknown variable is pointing to the first case, which will increase the odds that the event is still forthcoming. - There could be slight issue with the phrasing of the problem. Your interpretation is that the statement is true no matter what time period you are in. I.e. in 5 years time, there is still a 90% chance that some event will happen in that year. However, you're forgetting to account for the fact that the event didn't happen in the first year. Your friend's interpretation, is that your statement is true for this year. He has ignored the probability space in which the event occurred this year, and considered the remaining probability space. This now has measure $1-0.9=0.1$, and we know that the event has $0.95 - 0.9 = 0.05$ measure to occur, so it has $\frac {0.05}{0.1} = 50\%$ chance to happen. For example, consider the event where outcomes are based on drawing a uniform random variable on $[0,1]$. Let $C$ be your favorite number from 0 to 0.9. If we draw a number from 0 to 0.9, the event happens this year. If we draw a number from $X$ to 0.95, the event happens next year. If we draw a number from 0.95 to 1, the event never happens. A possible scenario describing your interpretation, could be If we draw a number from 0 to 0.9, the event happens every even year. If we draw a number from 0.05 to 0.95, the event happens every odd year. If we draw a number from 0.95 to 1, the event never happens. Then, it will always be true, that there is a 90% chance of it happening this year, and a 95% chance of it happening eventually. However, not that if we know it doesn't happen in this year (doesn't matter whether it's even or odd), then there's only a 50% chance that it will ever happen again. - Thanks for the alternate scenario, that's interesting. This is the kind of thing that made me feel like it's hard to guarantee an answer in the first place. –  NickC Jan 18 '13 at 21:25 @NickC The hard part is considering completely what it means for an event to not occur. You should look at this recent post, which had a lot of different answers initially. –  Calvin Lin Jan 18 '13 at 21:29 In case the mathematical proof already mentioned above isn't enough to convince you, pretend you're a statistician collecting your own results. You have 100 people infected with a disease that has a 95% death rate and a 90% death rate in the first year. Year one is finished, 90 dead, as predicted. 10 people left. We know 95% will be dead at some point. How many more are likely to die if the odds hold true? 5/10. Chances once the first year has passed is 50%. You don't need to be a statistician to see that. It has nothing to do with a finite number of events or an infinite time frame. That part has already been calculated for us in the given 95%. - The analysis for this appears to be relatively straightforward. Assume the probability of event in year 2+ is $x$. Next, the probability that the event does not occur in the first year is $0.1$ and that it does not occur any time thereafter is $(1-x)$. The product of the two is the probability that the event never happens, which we know to be $0.05$. So, $0.1 \times (1-x) = 0.05$ solving for x yields $x=\frac{1}{2}$ -

https://plainmath.net/algebra-ii/78503-changing-base-of-a-logarithm-by-taking-a

crossoverman9b 2022-06-20 Changing base of a logarithm by taking a square root from base? From my homework I found ${\mathrm{log}}_{9}x={\mathrm{log}}_{3}\sqrt{x}$ and besides that an explanation that to this was done by taking a square root of the base. I fail to grasp this completely. Should I need to turn ${\mathrm{log}}_{9}x$ into base 3, I'd do something like ${\mathrm{log}}_{9}x=\frac{{\mathrm{log}}_{3}x}{{\mathrm{log}}_{3}9}=\frac{{\mathrm{log}}_{3}x}{{\mathrm{log}}_{3}{3}^{2}}=\frac{{\mathrm{log}}_{3}x}{2}$ but this is a far cry from what I've given as being the correct answer. Substituting some values to x and playing with my calculator I can see that the answer given as correct is correct whereas my attempt fails to yield the correct answer. Now the question is, what are correct steps to derive ${\mathrm{log}}_{3}\sqrt{x}$ from ${\mathrm{log}}_{9}x$? How and why am I allowed to take a square root of the base and the exponent? mallol3i You are only one step away! Note that $a\mathrm{log}b=\mathrm{log}{b}^{a}$, so $\frac{{\mathrm{log}}_{3}x}{2}=\frac{1}{2}{\mathrm{log}}_{3}x={\mathrm{log}}_{3}{x}^{1/2}={\mathrm{log}}_{3}\sqrt{x}.$ Yesenia Sherman

https://math.stackexchange.com/questions/2801467/probability-a-person-chosen-at-random-owns-an-automobile-or-a-house-but-not-bot/2801568

# Probability a person chosen at random owns an automobile or a house, but not both A marketing survey indicates that $60\%$ of the population owns an automobile, $30\%$ owns a house, and $20\%$ owns both an automobile and a house. Calculate the probability that a person chosen at random owns an automobile or a house, but not both. (A) $0.4$ (B) $0.5$ (C) $0.6$ (D) $0.7$ (E) $0.9$ So the answer given for this question is $0.5$, but it doesn't make sense to me from the perspective of addition rule in probability. If we know that $\Pr(\text{Owning Automobile}) = 0.6$ and the $\Pr(\text{Owning a House}) = 0.3$ and the $\Pr(\text{Owning both an automobile and a house}) = 0.2$, according to the formula probability of $\Pr(A)~\text{or}~\Pr(B) = \Pr(A) + \Pr(B) - \Pr(A~\text{and}~B)$, wouldn't the answer be $0.6+0.3-0.2=0.7$? Is it because the ordering matters? so the last term $\Pr(A~\text{and}~B) = 2 \cdot 0.2$? • You computed the probability of "owning an automobile or a house." They asked for the probability of "owning an automobile or a house but not both." – angryavian May 30 '18 at 6:16 • @angryavian I still don't get it, I thought "owning an automobile or a house" is the same as "owning an automobile or a house but not both" ?? – pino231 May 30 '18 at 6:20 • Please type your question rather than posting an image. Images cannot be searched. – N. F. Taussig May 30 '18 at 8:54 • @pino231, in fact "owning an automobile or a house" in this case means "owning an automobile, or a house, or both". This usage of the word "or" -- the so-called "inclusive or" -- is standard in mathematics. If we want the opposite (the exclusive or) it is almost always explicitly stated, as it is in your question. – Mees de Vries May 30 '18 at 9:26 Let event $A$ be owning an automobile. Let event $B$ be owning a house. Consider the diagram below: If we simply add the probabilities that a person owns an automobile and a person owns a house, we will have added the probability that a person owns both an automobile and a house twice. Thus, to find the probability that a person owns an automobile or a house, we must subtract the probability that a person owns both an automobile and a house from the sum of the probabilities that a person owns an automobile and that a person owns a house. $$\Pr(A \cup B) = \Pr(A) + \Pr(B) - \Pr(A \cap B)$$ Note that on the left-hand side of your equation, you should have written $\Pr(A \cup B)$ or $\Pr(A~\text{or}~B)$ rather than $\Pr(A)~\text{or}~\Pr(B)$. Since we are given that $60\%$ of the population owns an automobile, $30\%$ of the populations owns a house, and $20\%$ owns both, $\Pr(A) = 0.60$, $\Pr(B) = 0.30$, and $\Pr(A \cap B) = 0.20$. Hence, the probability that a person owns an automobile or a house is $$\Pr(A \cup B) = \Pr(A) + \Pr(B) - \Pr(A \cap B) = 0.60 + 0.30 - 0.20 = 0.70$$ However, the question asks for the probability that a person owns an automobile or a house, but not both. That means we must subtract the probability that a person owns an automobile and a house from the probability that the person owns an automobile or a house. $$\Pr(A~\triangle~B) = \Pr(A \cup B) - \Pr(A \cap B) = 0.70 - 0.20 = 0.50$$ In terms of the Venn diagram, $A \cup B$ is the region enclosed by the two circles, while $A~\triangle~B = (A \cup B) - (A \cap B) = (A - B) \cup (B - A)$ is the region enclosed by the two circles except the region where the sets intersect. Since $A - B = A - (A \cap B)$, $$\Pr(A - B) = \Pr(A) - \Pr(A \cap B) = 0.60 - 0.20 = 0.40$$ Since $B - A = B - (A \cap B)$, $$\Pr(B - A) = \Pr(B) - \Pr(A \cap B) = 0.30 - 0.20 = 0.10$$ Hence, $$\Pr(A~\triangle~B) = \Pr(A - B) + \Pr(B - A) = 0.40 + 0.10 = 0.50$$ which agrees with the result obtained above. • ahh so the question is actually asking the probability that a person chosen at random owns an automobile or a house, OR not both – pino231 Jun 1 '18 at 3:20 • Not quite. The word but should be read as and. The question is asking the probability that a person chosen at random owns an automobile or a house and not both: $\Pr(A~\triangle~B) = \Pr(A \cup B) - \Pr(A \cap B)$. – N. F. Taussig Jun 1 '18 at 9:17 There's no ordering in this question. Using the addition rule is fine, but if you want to use it, note that: $\mathbb{P}(A) + \mathbb{P}(B) = \Big( \mathbb{P}(A \setminus B) + \mathbb{P}(A \cap B) \Big) + \Big( \mathbb{P}(B \setminus A) + \mathbb{P}(A \cap B) \Big)$. This should probably help. 10 $\%$ of people don't have a house or automobile so you can leave them. There is 90 $\%$ of people left, also 20$\%$ have both a house and a automobile. This 20% of people is included in the 60% and 30%, so you have to substract this percentage. What you get is 40% + 10% = 50%. So you were probably right by assuming that you have to count the last term $P(A \cap B)$ two times. • If we subtract $\Pr(A \cup B)$ two times, we will obtain a negative answer. Instead, we must subtract $\Pr(A \cap B)$ twice. – N. F. Taussig May 30 '18 at 9:13 • Yes of course, that's what I meant. Will edit! – WarreG May 30 '18 at 9:37

https://madhavamathcompetition.com/category/nature-of-mathematics/page/2/

Miscellaneous questions: Part I: tutorial practice for preRMO and RMO Problem 1: The sixty four squares of a chess board are filled with positive integers one on each in such a way that each integer is the average of the of the integers on the neighbouring squares. (Two squares are neighbours if they share a common edge or vertex. Thus, a square can have 8,5 or 3 neighbours depending on its position). Show that all sixty four entries are in fact equal. Problem 2: Let T be the set of all triples (a,b,c) of integers such that $1 \leq a < b < c \leq 6$. For each triple (a,b,c) in T, take the product abc. Add all these products corresponding to all triples in I. Prove that the sum is divisible by 7. Problem 3: In a class of 25 students, there are 17 cyclists, 13 swimmers, and 8 weight lifters and no one in all the three. In a certain mathematics examination, 6 students got grades D or E. If the cyclists, swimmers and weight lifters all got grade B or C, determine the number of students who got grade A. Also, find the number of cyclists, who are swimmers. Problem 4: Five men A, B, C, D, E are wearing caps of black or white colour without each knowing the colour of his cap. It is known that a man wearing a black cap always speaks the truth while a man wearing a white cap always lies. If they make the following statements, find the colour of the cap worn by each of them: A: I see three black and one white cap. B: I see four white caps. C: I see one black and three white caps. D: I see four black caps. Problem 5: Let f be a bijective (one-one and onto) function from the set $A=\{ 1,2,3,\ldots,n\}$ to itself. Show that there is a positive integer $M>1$ such that $f^{M}(i)=f(i)$ for each $i \in A$. Note that $f^{M}$ denotes the composite function $f \circ f \circ f \ldots \circ f$ repeated M times. Problem 6: Show that there exists a convex hexagon in the plane such that: a) all its interior angles are equal b) its sides are 1,2,3,4,5,6 in some order. Problem 7: There are ten objects with total weights 20, each of the weights being a positive integer. Given that none of the weights exceed 10, prove that the ten objects can be divided into two groups that balance each other when placed on the pans of a balance. Problem 8: In each of the eight corners of a cube, write +1 or -1 arbitrarily. Then, on each of the six faces of the cube write the product of the numbers written at the four corners of that face. Add all the fourteen numbers so writtein down. Is it possible to arrange the numbers +1 and -1 at the corners initially so that this final sum is zero? Problem 9: Given the seven element set $A = \{ a,b,c,d,e,f,g\}$ find a collection T of 3-element subsets of A such that each pair of elements from A occurs exactly in one of the subsets of T. Try these !! Regards, Nalin Pithwa Towards Baby Analysis: Part I: INMO, IMO and CMI Entrance $\bf{Reference: \hspace{0.1in}Introductory \hspace{0.1in} Real Analysis: \hspace{0.1in} Kolmogorov \hspace{0.1in} and \hspace{0.1in} Fomin; \hspace{0.1in}Dover \hspace{0.1in }Publications}$ $\bf{Equivalence \hspace{0.1in} of \hspace{0.1in} Sets \hspace{0.1in} The \hspace{0.1in}Power \hspace{0.1in }of \hspace{0.1in }a \hspace{0.1in}Set}$ $\bf{Section 1}$: $\bf{Finite \hspace{0.1in} and \hspace{0.1in} infinite \hspace{0.1in} sets}$ The set of all vertices of a given polyhedron, the set of all prime numbers less than a given number, and the set of all residents of NYC (at a given time) have a certain property in common, namely, each set has a definite number of elements which can be found in principle, if not in practice. Accordingly, these sets are all said to be $\it{finite}$.$\it{Clearly \hspace{0.1in} we \hspace{0.1in}can \hspace{0.1in} be \hspace{0.1in} sure \hspace{0.1in} that \hspace{0.1in} a \hspace{0.1in} set \hspace{0.1in}is \hspace{0.1in}finite \hspace{0.1in} without \hspace{0.1in} knowing \hspace{0.1in} the \hspace{0.1in} number \hspace{0.1in} of elements \hspace{0.1in}in \hspace{0.1in}it.}$ On the other hand, the set of all positive integers, the set of all points on the line, the set of all circles in the plane, and the set of all polynomials with rational coefficients have a different property in common, namely, $\it{if \hspace{0.1in } we \hspace{0.1in}remove \hspace{0.1in} one \hspace{0.1in} element \hspace{0.1in}from \hspace{0.1in}each \hspace{0.1in}set, \hspace{0.1in}then \hspace{0.1in}remove \hspace{0.1in}two \hspace{0.1in}elements, \hspace{0.1in}three \hspace{0.1in}elements, \hspace{0.1in}and \hspace{0.1in}so \hspace{0.1in}on, \hspace{0.1in}there \hspace{0.1in}will \hspace{0.1in}still \hspace{0.1in}be \hspace{0.1in}elements \hspace{0.1in}left \hspace{0.1in}in \hspace{0.1in}the \hspace{0.1in}set \hspace{0.1in}in \hspace{0.1in}each \hspace{0.1in}stage}$. Accordingly, sets of these kind are called $\it{infinite}$ sets. Given two finite sets, we can always decide whether or not they have the same number of elements, and if not, we can always determine which set has more elements than the other. It is natural to ask whether the same is true of infinite sets. In other words, does it make sense to ask, for example, whether there are more circles in the plane than rational points on the line, or more functions defined in the interval [0,1] than lines in space? As will soon be apparent, questions of this kind can indeed be answered. To compare two finite sets A and B, we can count the number of elements in each set and then compare the two numbers, but alternatively, we can try to establish a $\it{one-\hspace{0.1in}to-\hspace{0.1in}one \hspace{0.1in}correspondence}$ between the elements of set A and set B, that is, a correspondence such that each element in A corresponds to one and only element in B, and vice-versa. It is clear that a one-to-one correspondence between two finite sets can be set up if and only if the two sets have the same number of elements. For example, to ascertain if or not the number of students in an assembly is the same as the number of seats in the auditorium, there is no need to count the number of students and the number of seats. We need merely observe whether or not there are empty seats or students with no place to sit down. If the students can all be seated with no empty seats left, that is, if there is a one-to-one correspondence between the set of students and the set of seats, then these two sets obviously have the same number of elements. The important point here is that the first method(counting elements) works only for finite sets, while the second method(setting up a one-to-one correspondence) works for infinite sets as well as for finite sets. $\bf{Section 2}$: $\bf{Countable \hspace{0.1in} Sets}$. The simplest infinite set is the set $\mathscr{Z^{+}}$ of all positive integers. An infinite set is called $\bf{countable}$ if its elements can be put into one-to-one correspondence with those of $\mathscr{Z^{+}}$. In other words, a countable set is a set whose elements can be numbered $a_{1}, a_{2}, a_{3}, \ldots a_{n}, \ldots$. By an $\bf{uncountable}$ set we mean, of course, an infinite set which is not countable. We now give some examples of countable sets: $\bf{Example 1}$: The set $\mathscr{Z}$ of all integers, positive, negative, or zero is countable. In fact, we can set up the following one-to-one correspondence between $\mathscr{Z}$ and $\mathscr{Z^{+}}$ of all positive integers: (0,1), (-1,2), (1,3), (-2,4), (2,5), and so on. More explicitly, we associate the non-negative integer $n \geq 0$ with the odd number $2n+1$, and the negative integer $n<0$ with the even number $2|n|$, that is, $n \leftrightarrow (2n+1)$, if $n \geq 0$, and $n \in \mathscr{Z}$ $n \leftrightarrow 2|n|$, if $n<0$, and $n \in \mathscr{Z}$ $\bf{Example 2}$: The set of all positive even numbers is countable, as shown by the obvious correspondence $n \leftrightarrow 2n$. $\bf{Example 3}$: The set 2,4,8,$\ldots 2^{n}$ is countable as shown by the obvious correspondence $n \leftrightarrow 2^{n}$. $\bf{Example 4}: The set$latex \mathscr{Q}$of rational numbers is countable. To see this, we first note that every rational number $\alpha$ can be written as a fraction $\frac{p}{q}$, with $q>0$ with a positive denominator. (Of course, p and q are integers). Call the sum $|p|+q$ as the “height” of the rational number $\alpha$. For example, $\frac{0}{1}=0$ is the only rational number of height zero, $\frac{-1}{1}$, $\frac{1}{1}$ are the only rational numbers of height 2, $\frac{-2}{1}$, $\frac{-1}{2}$, $\frac{1}{2}$, $\frac{2}{1}$ are the only rational numbers of height 3, and so on. We can now arrange all rational numbers in order of increasing “height” (with the numerators increasing in each set of rational numbers of the same height). In other words, we first count the rational numbers of height 1, then those of height 2 (suitably arranged), then those of height 3(suitably arranged), and so on. In this way, we assign every rational number a unique positive integer, that is, we set up a one-to-one correspondence between the set Q of all rational numbers and the set $\mathscr{Z^{+}}$ of all positive integers. $\it{Next \hspace{0.1in}we \hspace{0.1in} prove \hspace{0.1in}some \hspace{0.1in}elementary \hspace{0.1in}theorems \hspace{0.1in}involving \hspace{0.1in}countable \hspace{0.1in}sets}$ $\bf{Theorem1}$. $\bf{Every \hspace{0.1in} subset \hspace{0.1in}of \hspace{0.1in}a \hspace{0.1in}countable \hspace{0.1in}set \hspace{0.1in}is \hspace{0.1in}countable}$. $\bf{Proof}$ Let set A be countable, with elements $a_{1}, a_{2}, a_{3}, \ldots$, and let set B be a subset of A. Among the elements $a_{1}, a_{2}, a_{3}, \ldots$, let $a_{n_{1}}, a_{n_{2}}, a_{n_{3}}, \ldots$ be those in the set B. If the set of numbers $n_{1}, n_{2}, n_{3}, \ldots$ has a largest number, then B is finite. Otherwise, B is countable (consider the one-to-one correspondence $i \leftrightarrow a_{n_{i}}$). $\bf{QED.}$ $\bf{Theorem2}$ $\bf{The \hspace{0.1in}union \hspace{0.1in}of \hspace{0.1in}a \hspace{0.1in}finite \hspace{0.1in}or \hspace{0.1in}countable \hspace{0.1in}number \hspace{0.1in}of \hspace{0.1in}countable \hspace{0.1in}sets \hspace{0.1in}A_{1}, A_{2}, A_{3}, \ldots \hspace{0.1in}is \hspace{0.1in}itself \hspace{0.1in}countable.}$ $\bf{Proof}$ We can assume that no two of the sets $A_{1}, A_{2}, A_{3}, \ldots$ have any elements in common, since otherwise we could consider the sets $A_{1}$, $A_{2}-A_{1}$, $A_{3}-(A_{1}\bigcup A_{2})$, $\ldots$, instead, which are countable by Theorem 1, and have the same union as the original sets. Suppose we write the elements of $A_{1}, A_{2}, A_{3}, \ldots$ in the form of an infinite table $\begin{array}{ccccc} a_{11} & a_{12} & a_{13} & a_{14} &\ldots \\ a_{21} &a_{22} & a_{23} & a_{24} & \ldots \\ a_{31} & a_{32} & a_{33} & a_{34} & \ldots \\ a_{41} & a_{42} & a_{43} & a_{44} & \ldots \\ \ldots & \ldots & \ldots & \ldots & \ldots \end{array}$ where the elements of the set $A_{1}$ appear in the first row, the elements of the set $A_{2}$ appear in the second row, and so on. We now count all the elements in the above array “diagonally”; that is, first we choose $a_{11}$, then $a_{12}$, then move downwards, diagonally to “left”, picking $a_{21}$, then move down vertically picking up $a_{31}$, then move across towards right picking up $a_{22}$, next pick up $a_{13}$ and so on ($a_{14}, a_{23}, a_{32}, a_{41}$)as per the pattern shown: $\begin{array}{cccccccc} a_{11} & \rightarrow & a_{12} &\hspace{0.1in} & a_{13} & \rightarrow a_{14} & \ldots \\ \hspace{0.1in} & \swarrow & \hspace{0.1in} & \nearrow & \hspace{0.01in} & \swarrow & \hspace{0.1in} & \hspace{0.1in}\\ a_{21} & \hspace{0.1in} & a_{22} & \hspace{0.1in} & a_{23} \hspace{0.1in} & a_{24} & \ldots \\ \downarrow & \nearrow & \hspace{0.1in} & \swarrow & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in}\\ a_{31} & \hspace{0.1in} & a_{32} & \hspace{0.1in} & a_{33} & \hspace{0.1in} & a_{34} & \ldots \\ \hspace{0.1in} & \swarrow & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in}\\ a_{41} & \hspace{0.1in} & a_{42} &\hspace{0.1in} & a_{43} &\hspace{0.1in} &a_{44} &\ldots\\ \ldots & \hspace{0.1in} & \ldots & \hspace{0.1in} & \ldots & \hspace{0.1in} & \ldots & \hspace{0.1in} \end{array}$ It is clear that this procedure associates a unique number to each element in each of the sets $A_{1}, A_{2}, \ldots$ thereby establishing a one-to-one correspondence between the union of the sets $A_{1}, A_{2}, \ldots$ and the set $\mathscr{Z^{+}}$ of all positive integers. $\bf{QED.}$ $\bf{Theorem3}$ $\bf{Every \hspace{0.1in}infinite \hspace{0.1in}subset \hspace{0.1in}has \hspace{0.1in}a \hspace{0.1in}countable \hspace{0.1in}subset.}$ $\bf{Proof}$ Let M be an infinite set and $a_{1}$ any element of M. Being infinite, M contains an element $a_{2}$ distinct from $a_{1}$, an element $a_{3}$ distinct from both $a_{2}$ and $a_{1}$, and so on. Continuing this process, (which can never terminate due to “shortage” of elements, since M is infinite), we get a countable subset $A= \{ a_{1}, a_{2}, a_{3}, \ldots, a_{n}, \ldots\}$ of the set $M$. $\bf{QED.}$ $\bf{Remark}$ Theorem 3 shows that countable sets are the “smallest” infinite sets. The question of whether there exist uncountable (infinite) sets will be considered below. $\bf{Section3}$ $\bf{Equivalence \hspace{0.1in} of \hspace{0.1in} sets}$ We arrived at the notion of a countable set M by considering one-to-one correspondences between set M and the set $\mathscr{Z^{+}}$ of all positive integers. More generally, we can consider one-to-one correspondences between any two sets M and N. $\bf{Definition}$ Two sets M and N are said to be $\bf{equivalent}$ (written $M \sim N$) if there is a one-to-one correspondence between the elements of M and the elements of N. The concept of equivalence is applicable both to finite and infinite sets. Two finite sets are equivalent if and only if they have the same number of elements. We can now define a countable set as a set equivalent to the set $\mathscr{Z^{+}}$ of all positive integers. It is clear that two sets are equivalent to a third set are equivalent to each other, and in particular that any two countable sets are equivalent. $\bf{Example1}$ The sets of points in any two closed intervals$[a,b]$and$[c,d]\$ are equivalent; you can “see’ a one-to-one correspondence by drawing the following diagram: Step 1: draw cd as a base of a triangle. Let the third vertex of the triangle be O. Draw a line segment “ab” above the base of the triangle; where “a” lies on one side of the triangle and “b” lies on the third side of the third triangle. Note that two points p and q correspond to each other if and only if they lie on the same ray emanating from the point O in which the extensions of the line segments ac and bd intersect. $\bf{Example2}$ The set of all points z in the complex plane is equivalent to the set of all points z on a sphere. In fact, a one-to-one correspondence $z \leftrightarrow \alpha$ can be established by using stereographic projection. The origin is the North Pole of the sphere. $\bf{Example3}$ The set of all points x in the open unit interval $(0,1)$ is equivalent to the set of all points y on the whole real line. For example, the formula $y=\frac{1}{\pi}\arctan{x}+\frac{1}{2}$ establishes a one-to-one correspondence between these two sets. $\bf{QED}$. The last example and the examples in Section 2 show that an infinite set is sometimes equivalent to one of its proper subsets. For example, there are “as many” positive integers as integers of arbitrary sign, there are “as many” points in the interval $(0,1)$ as on the whole real line, and so on. This fact is characteristic of all infinite sets (and can be used to define such sets) as shown by: $\bf{Theorem4}$ $\bf{Every \hspace{0.1in} infinite \hspace{0.1in} set \hspace{0.1in}is \hspace{0.1in} equivalent \hspace{0.1in} to \hspace{0.1in}one \hspace{0.1in}of \hspace{0.1in}its \hspace{0.1in}proper \hspace{0.1in}subsets.}$ $\bf{Proof}$ According to Theorem 3, every infinite set M contains a countable subset. Let this subset be $A=\{a_{1}, a_{2}, a_{3}, \ldots, a_{n}, \ldots \}$ and partition A into two countable subsets $A_{1}=\{a_{1}, a_{3}, a_{5}, \ldots \}$ and $A_{2}=\{a_{2}, a_{4}, a_{6}, \ldots \}$. Obviously, we can establish a one-to-one correspondence between the countable subsets A and $A_{1}$ (merely let $a_{n} \leftrightarrow a_{2n-1}$). This correspondence can be extended to a one-to-one correspondence between the sets $A \bigcup (M-A)=M$ and $A_{1} \bigcup (M-A)=M-A_{2}$ by simply assigning x itself to each element $x \in M-A$. But $M-A_{2}$ is a proper subset of M. $\bf{QED}$. More later, to be continued, Regards, Nalin Pithwa Find a flaw in this proof: RMO and PRMO tutorial What ails the following proof that all the elements of a finite set are equal? The following is the “proof”; All elements of a set with no elements are equal, so make the induction assumption that any set with n elements has all its elements equal. In a set with n elements, the first and the last n are equal by induction assumption. They overlap at n, so all are equal, completing the induction. End of “proof: Regards, Nalin Pithwa Problem Solving approach: based on George Polya’s opinion: Useful for RMO/INMO, IITJEE maths preparation I have prepared the following write-up based on George Polya’s classic reference mentioned below: UNDERSTANDING THE PROBLEM First. “You have to understand the problem.” What is the unknown ? What are the data? What is the condition? Is it possible to satisfy the condition? Is the condition sufficient to determine the unknown? Or is it insufficient? Or redundant ? Or contradictory? Draw a figure/diagram. Introduce a suitable notation. Separate the various parts of the condition. Can you write them down? Second. DEVISING A PLAN: Find the connection between the data and the unknown. You may be obliged to consider auxiliary problems if an immediate connection cannot be found. You should eventually obtain a plan for the solution.” Have you seen it before? Or have you seen the problem in a slightly different form? Do you know a related problem? Do you know a theorem that could be useful? Look at the unknown! And try to think of a familiar problem having the same or a similar unknown. Here is a problem related to yours and solved before. Could you use it? Could you use its result? Could you use its method? Should you restate it differently? Go back to definitions. If you cannot solve the proposed problem, try to solve some related problem. Could you imagine a more accessible related problem? A more general problem? A more special problem? An analogous problem? Could you solve a part of the problem? Keep only a part of the condition, drop the other part, how far is the unknown then determined, how can it vary? Could you derive something useful from the data? Could you think of other data appropriate to determine the unknown? Could you change the unknown of the data, or both, if necessary, so that the new unknown and the new data are nearer to each other? Did you use all the data? Did you use the whole condition? Have you taken into account all essential notions involved in the problem? Carrying out your plan of the solution, check each step. Can you clearly see that the step is correct? Can you prove that it is correct? Fourth. LOOKING BACK. Examine the solution. Can you check the result? Can you check the argument? Can you derive the result differently? Can you see it at a glance? Can you see the result, or the method, for some other problem? ************************************************************************** Reference: How to Solve It: A New Aspect of Mathematical Method — George Polya. https://www.amazon.in/How-Solve-Aspect-Mathematical-Method/dp/4871878309/ref=sr_1_1?crid=2DXC1EM1UVCPW&keywords=how+to+solve+it+george+polya&qid=1568334366&s=books&sprefix=How+to+solve%2Caps%2C275&sr=1-1 The above simple “plan” can be useful even to crack problems from a famous classic, Problem Solving Strategies, by Arthur Engel, a widely-used text for training for RMO, INMO and IITJEE Advanced Math also, perhaps. Reference: Problem-Solving Strategies by Arthur Engel; available on Amazon India Concept of order in math and real world 1. Rise and Shine algorithm: This is crazy-sounding, but quite a perfect example of the need for “order” in the real-world: when we get up in the morning, we first clean our teeth, finish all other ablutions, then go to the bathroom and first we have to remove our pyjamas/pajamas and then the shirt, and then enter the shower; we do not first enter the shower and then remove the pyjamas/shirt !! 🙂 2. On the number line, as we go from left to right: $a, that is any real number to the left of another real number is always “less than” the number to the right. (note that whereas the real numbers form an “ordered field”, the complex numbers are only “partially ordered”…we will continue this further discussion later) . 3. Dictionary order 4. Alphabetical order (the letters $A \hspace{0.1in} B \ldots Z$ in English. 5. Telephone directory order 6. So a service like JustDial certainly uses “order” quite intensely: let us say that you want to find the telephone clinic landline number of Dr Mrs Prasad in Jayanagar 4th Block, Bengaluru : We first narrow JustDial to “Location” (Jayanagar 4th Block, Bengaluru), then narrow to “doctors/surgeons” as the case may be, and then check in alphabetic order, the name of Dr Mrs Prasad. So, we clearly see that the “concept” and “actual implementation” of order (in databases) actually speeds up so much the time to find the exact information we want. 7. So also, in math, we have the concept of ordered pair; in Cartesian geometry, $(a,b)$ means that the first component $a \in X-axis$ and $b \in Y-axis$. This order is generalized to complex numbers in the complex plane or Argand’s diagram. 8. There is “order” in human “relations” also: let us $(x,y)$ represents x (as father) and y (as son). Clearly, the father is “first” and the son is “second”. 9. So, also any “tree” has a “natural order”: seed first, then roots, then branches. Regards, Nalin Pithwa. Why do we need proofs? In other words, difference between a mathematician, physicist and a layman Yes, I think it is a very nice question, which kids ask me. Why do we need proofs? Well, here is a detailed explanation (I am not mentioning the reference I use here lest it may intimidate my young enthusiastic, hard working students or readers. In other words, the explanation is not my own; I do not claim credit for this…In other words, I am just sharing what I am reading in the book…) Here it goes: What exactly is the difference between a mathematician, a physicist, and a layman? Let us suppose that they all start measuring the angles of hundreds of triangles of various shapes, find the sum in each case and keep a record. Suppose the layman finds that with one or two exceptions, the sum in each case comes out to be 180 degrees. He will ignore the exceptions and say “the sum of the three angles in a triangle  is 180 degrees.” A physicist will be more cautious in dealing with the exceptional cases. He will examine them more carefully. If he finds that the sum in them is somewhere between 179 degrees to 180 degrees, say, then he will attribute the deviation to experimental errors. He will then state a law: The sum of three angles of any triangle is 180 degrees. He will then watch happily as the rest of the world puts his law to test and finds that it holds good in thousands of different cases, until somebody comes up with a triangle in which the law fails miserably. The physicist now has to withdraw his law altogether or else to replace it by some other law which holds good in all cases tried. Even this new law may have to be modified at a later date. And, this will continue without end. A mathematician will be the fussiest of all. If there is even a single exception he will refrain from saying anything. Even when millions of triangles are tried without a single exception, he will not state it as a theorem that the sum of the three angles in ANY triangle is 180 degrees. The reason is that there are infinitely many different types of triangles. To generalize from a million to infinity is as baseless to a mathematician as to generalize from one to a million. He will at the most make a conjecture and say that there is a strong evidence suggesting that the conjecture is true. But that is not the same thing as a proving a theorem. The only proof acceptable to a mathematician is the one which follows from earlier theorems by sheer logical implications (that is, statements of the form : If P, then Q). For example, such a proof follows easily from the theorem that an external angle of a triangle is the sum of the other two internal angles. The approach taken by the layman or the physicist is known as the inductive approach whereas the mathematician’s approach is called the deductive approach. In the former, we make a few observations and generalize. In the latter, we deduce from something which is already proven. Of course, a question can be raised as to on what basis this supporting theorem is proved. The answer will be some other theorem. But then the same question can be asked about the other theorem. Eventually, a stage is reached where a certain statement cannot be proved from any other earlier proved statement(s) and must, therefore, be taken for granted to be true. Such a statement is known as an axiom or a postulate. Each branch of math has its own axioms or postulates. For examples, one of the axioms of geometry is that through two distinct points, there passes exactly one line. The whole beautiful structure of geometry is based on 5 or 6 axioms such as this one. Every theorem in plane geometry or Euclid’s Geometry can be ultimately deduced from these axioms. PS: One of the most famous American presidents, Abraham Lincoln had read, understood and solved all of Euclid’s books (The Elements) by burning mid-night oil, night after night, to “sharpen his mental faculties”. And, of course, there is another famous story (true story) of how Albert Einstein as a very young boy got completely “addicted” to math by reading Euclid’s proof of why three medians of a triangle are concurrent…(you can Google up, of course). Regards, Nalin Pithwa

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Rhianna has learned the SSS and SAS congruence tests for triangles and she wonders if these tests might work for parallelograms. Diagonals of a Parallelogram Bisect Each Other. A tip from Math Bits says, if we can show that one set of opposite sides are both parallel and congruent, which in turn indicates that the polygon is a parallelogram, this will save time when working a proof. parallelogram, because opposite sides are congruent and adjacent sides are not perpendicular. Theorem 1 : If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram. Also interesting in this case is that to the eye The first is to use congruent triangles to show the corresponding angles are congruent, the other is to use theAlternate Interior Angles Theoremand apply it twice. Congruent trianglesare triangles that have the same size and shape. So we’re going to put on our thinking caps, and use our detective skills, as we set out to prove (show) that a quadrilateral is a parallelogram. Jenn, Founder Calcworkshop®, 15+ Years Experience (Licensed & Certified Teacher). When we think of parallelograms, we usually think of something like this. Note that the vertex $D$ is obtained by rotating $B$ 180 degrees about the midpoint $M$ of $\overline{AC}$. vidDefer[i].setAttribute('src',vidDefer[i].getAttribute('data-src')); We can look at what happens in the special case where all 4 sides of both $ABCD$ and $EFGH$ are congruent to one another. Theorem 6.2.1 If a quadrilateral is a parallelogram, then the two pairs of opposite sides are congruent. Each theorem has an example that will show you how to use it in order to prove the given figure. To show these two triangles are congruent we’ll use the fact that this is a parallelogram, and as a result, the two opposite sid… 2. A quadrilateral that has opposite sides equal and parallel and the opposite angles are also equal is called a parallelogram. One Pair of Opposite Sides are Both Parallel and Congruent, Consecutive Angles in a Parallelogram are Supplementary. yes, one pair of sides are congruent and parallel . Both pairs of opposite sides are congruent. Triangle congruence criteria have been part of the geometry curriculum for centuries. If both pairs of opposite angles of a quadrilateral are congruent, then it’s a parallelogram (converse of a property). Just as with a triangle it takes three pieces of information (ASA, SAS, or SSS) to determine a shape, so with a quadrilateral we would expect to require four pieces of information. THEOREM:If a quadrilateral has 2 sets of opposite sides congruent, then it is a parallelogram. They are called the SSS rule, SAS rule, ASA rule and AAS rule. function init() { Opposite Sides Parallel and Congruent & Opposite Angles Congruent. So what are we waiting for. B) The diagonals of the parallelogram are congruent. yes, diagonals bisect each other. Take Calcworkshop for a spin with our FREE limits course. Since ABCD is a parallelogram, segment AB ≅ segment DC because opposite sides of a parallelogram are congruent. For quadrilaterals, on the other hand, these nice tests seem to be lacking. 2 Looking at a special case for part (a): the rhombus. In order to see what happens with the parallelograms $ABCD$ and $EFGH$ we focus first on $ABCD$. If the quadrilateral has two pairs of opposite, congruent sides, it is a parallelogram. Well, if a parallelogram has congruent diagonals, you know that it is a rectangle. Attribution-NonCommercial-ShareAlike 4.0 International License. Solution: It turns out that knowing all four sides of two quadrilaterals are congruent is not enough to conclude that the quadrilaterals are congruent. Prove that the figure is a parallelogram. Finally, you’ll learn how to complete the associated 2 column-proofs. If one angle is 90 degrees, then all other angles are also 90 degrees. We can tell whether two triangles are congruent without testing all the sides and all the angles of the two triangles. if(vidDefer[i].getAttribute('data-src')) { If the diagonals of a quadrilateral bisect each other, then it’s a parallelogram (converse of a property). Suppose $ABCD$ and $EFGH$ are two parallelograms with a pair of congruent corresponding sides, $|AB| = |EF|$ and $|BC| = |FG|$. We begin by drawing or building a parallelogram. If a parallelogram has perpendicular diagonals, you know it is a rhombus. We might find that the information provided will indicate that the diagonals of the quadrilateral bisect each other. Theorems. The diagonal of a parallelogram separates it into two congruent triangles. side $\overline{EH}$ does not appear to the eye to be congruent to side $\overline{AD}$: this could be an optical illusion or it could be that the eye is distracted by the difference in area. SURVEY . Here is what we need to prove: segment AB ≅ segment CD and segment BC ≅ AD. Draw the diagonal BD, and we will show that ΔABD and ΔCDB are congruent. Thus it provides a good opportunity for students to engage in MP3 ''Construct Viable Arguments and Critique the Reasoning of Others.'' When a parallelogram is divided into two triangles we get to see that the angles across the common side( here the diagonal) are equal. If the quadrilateral has one set of opposite parallel, congruent sides, it is a parallelogram. Both of these facts allow us to prove that the figure is indeed a parallelogram. Complete the two-column proof Given: triangle SVX is congruent to triangle UTX and Line SV is || to line TU Prove: VUTS is a parallelogram Image: It's a parallelogram, with one line going from corner S to corner U and a line going . This task addresses this issue for a specific class of quadrilaterals, namely parallelograms. Well, we must show one of the six basic properties of parallelograms to be true! Find missing values of a given parallelogram. We will learn about the important theorems related to parallelograms and understand their proofs. We all know that a parallelogram is a convex polygon with 4 edges and 4 vertices. Solution: More generally, a quadrilateral with 4 congruent sides is a rhombus. Which of the following cannot be used to prove a shape is a parallelogram? yes,opposite sides are congruent. First prove ABC is congruent to CDA, and then state AD and BC are corresponding sides of the triangles. Triangles can be used to prove this rule about the opposite sides. Creative Commons The opposite sides of a parallelogram are congruent. In this lesson, we will consider the four rules to prove triangle congruence. Note that a rhombus is determined by one side length and a single angle: the given side length determines all four side lengths and For example, for squares one side is enough, for rectangles two adjacent sides are sufficient. Engage your students with effective distance learning resources. This means that the corresponding sides are equal and the corresponding angles are equal. Let’s begin! A description of how to do a parallelogram congruent triangles proof. In this mini-lesson, we will explore the world of parallelograms and their properties. The only parallelogram that satisfies that description is a square. Parallelogram and Congruent triangles Parallelogram. $\triangle ABC$. If … Another approach might involve showing that the opposite angles of a quadrilateral are congruent or that the consecutive angles of a quadrilateral are supplementary. In this section, you will learn how to prove that a quadrilateral is a parallelogram. This proves that the opposite angles in a parallelogram are also equal. What about for arbitrary quadrilaterals? Which statement explains how you could use coordinate geometry to prove the diagonals of a quadrilateral are perpendicular? More specifically, how do we prove a quadrilateral is a parallelogram? Walking trails run from points A to C and from points B to D. Here are the theorems that will help you prove that the quadrilateral is a parallelogram. In today’s geometry lesson, you’re going to learn the 6 ways to prove a parallelogram. for (var i=0; i

https://mathematica.stackexchange.com/questions/204856/what-is-the-distribution-of-a-random-variable-with-pdf-proportional-to-the-produ/204908#204908

# What is the distribution of a random variable with pdf proportional to the product of two normal pdf's I'm working my way through a nice e-book on Kalman filters, translating to Mathematica as I go and I've hit an interesting problem. In this section: https://nbviewer.jupyter.org/github/rlabbe/Kalman-and-Bayesian-Filters-in-Python/blob/master/03-Gaussians.ipynb#Computational-Properties-of-Gaussians - the author gives a formula to compute a product of Gaussian PDFs that is also Gaussian. Googling around will tell you that the product of two Gaussian PDFs is not itself Gaussian, but is proportional to a Gaussian with mean and standard deviation as in the image below (see proof here: http://www.tina-vision.net/docs/memos/2003-003.pdf for source) -- I have implemented his function and it works fine, but I'm curious if TransformedDistribution can be used to arrive at the same distribution. To make it concrete d1 = NormalDistribution[m1, s1]; d2 = NormalDistribution[m2, s2]; (* Works perfectly *) TransformedDistribution[ x + y, {x \[Distributed] d1, y \[Distributed] d2}] (* Not a Normally Distributed *) TransformedDistribution[ x*y, {x \[Distributed] d1, y \[Distributed] d2}] (* This doesn't work, but I'm wondering if there is something like \ this that would produce the normal distribution cited in the question \ *) TransformedDistribution[ Normalize[x*y, Total], {x \[Distributed] d1, y \[Distributed] d2}] Edit to add example multiplication using the proposed function. MultiplyGaussian[g1_, g2_] := Module[{mean1, var1, mean2, var2, mean, variance}, {mean1, var1} = g1 /. NormalDistribution[m_, s_] :> {m, s^2}; {mean2, var2} = g2 /. NormalDistribution[m_, s_] :> {m, s^2}; mean = (var1*mean2 + var2*mean1) / (var1 + var2); variance = (var1 * var2) / (var1 + var2); NormalDistribution[mean, Sqrt[variance]] ] z1 = NormalDistribution[3, 0.7]; z2 = NormalDistribution[4.5, 1]; Plot[{Legended[PDF[z1, x], "N(3,0.7)"], Legended[PDF[z2, x], "N(4.5,2)"] , Legended[PDF[MultiplyGaussian[z1, z2], x], "Product"]}, {x, 1, 10}] • I'm not sure your source is right: mathworld.wolfram.com/NormalProductDistribution.html Sep 5 '19 at 23:13 • Edited to add code showing that the proposal seems to work and clarify that the 2nd link is a proof that I at least couldn't find a problem with after a quick look. – Dan Sep 5 '19 at 23:34 • I think my terminology is sloppy and that has caused the problem (as pointed out by this MathOverflow answer: math.stackexchange.com/questions/101062/…). The result here is for the product of PDFs (where I sloppily said random variable). Does this clarification help see a way to get there in Mathematica? – Dan Sep 5 '19 at 23:41 • Please clean-up the "rv" vs "pdf" confusion in the text and the title. Your title should probably be something like "What is the distribution of a random variable with pdf proportional to the product of two normal pdf's?" – JimB Sep 6 '19 at 0:40 (* Get product of two normal pdf's *) prod= PDF[NormalDistribution[μ1, σ1], x]*PDF[NormalDistribution[μ2, σ2], x]; (* Normalize so that the pdf integrates to 1 *) pdf = prod/Integrate[prod, {x, -∞, ∞}, Assumptions -> {σ1 > 0, σ2 > 0}]; (* Construct associated distribution *) d = ProbabilityDistribution[pdf, {x, -∞, ∞}, Assumptions -> {σ1 > 0, σ2 > 0}]; (* Find mean and variance *) Mean[d] (* (μ2 σ1^2+μ1 σ2^2)/(σ1^2+σ2^2) *) Variance[d] (* (σ1^2 σ2^2)/(σ1^2+σ2^2) *) This matches what the article says the mean and variance should be. But is it a normal distribution? If the moment generating function is of the same form as for a normal distribution, then it has a normal distribution. (We could also use the characteristic function to do this for this particular distribution.) (* The log of the moment generating function will be in the following form *) logCF = Expectation[Exp[t z], z \[Distributed] NormalDistribution[μ, σ]] /. Power[E, x_] -> x // Expand (* t μ+(t^2 σ^2)/2 *) (* So we look to see if the moment generating function of distribution d is of the same form *) Collect[Expectation[Exp[t z], z \[Distributed] d] /. Power[E, x_] -> x // Expand, t] (* (t^2 σ1^2 σ2^2)/(2 (σ1^2+σ2^2))+t ((μ2 σ1^2)/(σ1^2+σ2^2)+(μ1 σ2^2)/(σ1^2+σ2^2)) *) And it is. d1 = NormalDistribution[m1, s1]; d2 = NormalDistribution[m2, s2]; These distributions require that assume = And @@ (DistributionParameterAssumptions /@ {d1, d2}) (* m1 ∈ Reals && s1 > 0 && m2 ∈ Reals && s2 > 0 *) As pointed out by @JimB, the PDF formed by the product of the normal PDFs is PDFprod = Assuming[assume, PDF[d1, x]*PDF[d2, x]/ Integrate[PDF[d1, x]*PDF[d2, x], {x, -Infinity, Infinity}] // Simplify] (* (E^(-((m2 s1^2 + m1 s2^2 - (s1^2 + s2^2) x)^2/( 2 s1^2 s2^2 (s1^2 + s2^2)))) Sqrt[s1^2 + s2^2])/(Sqrt[2 π] s1 s2) *) Comparing with the PDF of the expected normal distribution PDFprod == Assuming[assume, PDF[NormalDistribution[ (m1*s2^2 + m2*s1^2)/(s1^2 + s2^2), Sqrt[s1^2*s2^2/(s1^2 + s2^2)]], x] // Simplify] (* True *)

https://math.stackexchange.com/questions/75589/open-maps-which-are-not-continuous

# Open maps which are not continuous What is an example of an open map $(0,1) \to \mathbb{R}$ which is not continuous? Is it even possible for one to exist? What about in higher dimensions? The simplest example I've been able to think of is the map $e^{1/z}$ from $\mathbb{C}$ to $\mathbb{C}$ (filled in to be $0$ at $0$). There must be a simpler example, using the usual Euclidean topology, right? • Since $(0,1)$ and $\mathbb R$ are homeomorphic via a linear map composed with $\arctan$, it suffices to find a map $\mathbb R \to \mathbb R$ that is open but not continuous. Googling that gives you mathforum.org/library/drmath/view/62395.html – lhf Oct 25 '11 at 0:55 • this is obviously not much help, but if you can find a continuous bijection $f$ with discontinuous inverse, then $f^{-1}$ will do. – user12014 Oct 25 '11 at 1:13 • One can build such a function from a Cantor set $C$ (the usual "middle thirds" set will do). Send each point in $C$ to $0$, and map each connected component of the complement of $C$ homeomorphically to the interval $(-1,1)$. Then the image of any open set intersecting $C$ will be $(-1,1)$ (thus open), and the image of any open set not meeting $C$ will also be open, since it's a union of homeomorphic images of open sets. Of course, each point of $C$ will be a discontinuity. – user83827 Oct 25 '11 at 1:16 • @PZZ for instance the map wrapping [0,1) around the unit circle. – JSchlather Oct 25 '11 at 1:37 • @PZZ: In fact there are no counterexamples of the type you're suggesting: if $I$ and $J$ are intervals in $\mathbb{R}$ and $f: I \rightarrow J$ is a continuous bijection, then $f^{-1}$ is necessarily continuous. By coincidence this is exactly where I am in my Spivak calculus course, so see e.g. Theorem 37 in $\S 6.4$ of math.uga.edu/~pete/2400calc2.pdf. (Or see Spivak's text!) – Pete L. Clark Oct 25 '11 at 3:18 Explicit examples are moderately difficult to construct, but it’s not too hard to come up with non-constructive examples; here’s one such. For $x,y\in\mathbb{R}$ define $x\sim y$ iff $x-y\in \mathbb{Q}$; it’s easy to check that $\sim$ is an equivalence relation on $\mathbb{R}$. For any $x\in\mathbb{R}$, $[x] = \{x+q:q\in\mathbb{Q}\}$, where $[x]$ is the $\sim$-equivalence class of $x$. In particular, each equivalence class is countable. For any infinite cardinal $\kappa$, the union of $\kappa$ pairwise disjoint countably infinite sets has cardinality $\kappa$, so there must be exactly as many equivalence classes as there are real numbers. Let $h$ be a bijection from $\mathbb{R}/\sim$, the set of equivalence classes, to $\mathbb{R}$. Finally, define $$f:(0,1)\to\mathbb{R}:x\mapsto h([x])\;.$$ I claim that if $V$ is any non-empty open subset of $(0,1)$, $f[V]=\mathbb{R}$, which of course ensures that $f$ is open. To see this, just observe that every open interval in $(0,1)$ intersects every equivalence class. (It should be no trouble at all to see that $f$ is wildly discontinuous!) • Just curious: Is the axiom of choice used anywhere in your proof? – YoTengoUnLCD Jan 17 '17 at 8:14 • I think I'm going to start calling $\sim$ the "Vitali equivalence relation"... $x$ and $y$ are Vitali equivalent iff $x-y \in \mathbb{Q}$, etc. Honestly, this thing is useful enough to deserve a name. – goblin Mar 2 '17 at 13:54 • is $f$ injective? – David Feng Feb 16 at 21:37 • @DavidFeng: No. All $x$ from the same equivalence class give the same value. For example, $f(\frac12)=f(\frac13)$ since $\frac12-\frac13\in\mathbb Q$ – celtschk Mar 2 at 21:55 Let me conceptualize around Brian's answer a bit. Definition 0. If $X$ and $Y$ are topological spaces, a function $f:X→Y$ is said to be strongly Darboux iff for all non-empty open sets $A⊆X$, we have $f(A)=Y$. Here's the basic facts: Proposition. 1. Every strongly Darboux function is an open function. 2. If $X$ is non-empty, every Darboux function $X \rightarrow Y$ is surjective. 3. If $X$ is non-empty and $f : X \rightarrow Y$ is a continuous Darboux mapping, then $Y$ carries the indiscrete topology. Proofs. 1. Trivial. 2. Since $X$ is open and non-empty, hence $f(X)=Y.$ That is, $f$ is surjective. 3. Let $B \subseteq Y$ denote a non-empty open set. Our goal is to show that $B=Y$. Since $f$ is surjective, $f^{-1}(B)$ is non-empty. Since $f$ is continuous, $f^{-1}(B)$ is open. Hence $f(f^{-1}(B))=Y$. But since $f$ is surjecive, hence $f(f^{-1}(B))=B.$ So $B=Y$. Putting these together, we see that every strongly Darboux function $f:\mathbb{R} \rightarrow \mathbb{R}$ is a discontinuous open mapping. • $f$ is an open mapping by (1). • $f$ is discontinuous by (3), because it's domain is non-empty and it's codomain doesn't carry the indiscrete topology. And, of course, Brian's answer guarantees the existence of a strongly Darboux function $\mathbb{R} \rightarrow \mathbb{R}$. This completes the proof. There is in fact a rather easy example of a function $$\mathbb R \to \mathbb R$$ such that the image of every open set is $$\mathbb R$$: Let $$(x_i)_{i\in\mathbb Z_+}$$ be the binary decimal expansion of $$x$$, so that each $$x_i \in \{0,1\}$$. Let then $$f(x) = \sum_{k=1}^\infty\frac{(-1)^{x_k}}k\quad \textrm{if the series converges}$$ $$f(x) = 0\quad \textrm{otherwise.}$$ Since the harmonic series (or a tail of it) can be made to converge to any real number by changing signs in the appropriate way, this function has $$f((a,b)) = \mathbb R$$ for any real $$a,b$$. Hence this function is open, though clearly not continuous at any point. The harmonic series can be substituted with any other unbounded series where the summand goes to zero.

https://share.cocalc.com/share/b67337a608fb9911d3a0a3330eaadd1ab39e446b/week%205/assignment/FindRoots.ipynb?viewer=share

CoCalc Public Filesweek 5 / assignment / FindRoots.ipynb Authors: Boyuan Qi, Rajvir Sidhu Views : 31 Compute Environment: Ubuntu 20.04 (Default) # Assignment 2: Polynomial Root Finding ## Markdown Give the address (url) for a website that has documentation for markdown, including tables (i.e. a webpage, not a pdf or other document, that describes how you use markdown to create a table). Enter your answer as a variable named website='http://url.goes.here' Marks: 1 In [ ]: # YOUR CODE HERE markdownWebsite='https://www.markdownguide.org/cheat-sheet' Test your code from above here(1 point), ID: validate_website In [ ]: #test that it's a suitable website link you've given Work out the solutions to the following questions long-hand (i.e. you should not be coding anything at this point). In the first box, set the appropriate variable equal to your answer (there's no need to define any functions). For example, to answer the last question, you will simply type polyRoots=[4-math.sqrt(3),4,4+math.sqrt(3)] 1. Define the polynomial $p(x)=x^3-12x^2+45x-52$. How do you express $p(x)$ as a list, such as will be input to our algorithms? Set the variable polyX equal to this. 2. What is the derivative, $p'(x)$ (as a list called polyDeriv) and the root(s) of $p'(x)$ (as a list, in increasing order, called derivRoots)? 3. What is the Lagrange bound as applied to the polynomial $p(x)$? Your answer should be a real, positive number, called lagrange. (In case you missed it, this was an exercise in the workshops) 4. Pretend that you don't know the roots of $p(x)$, but you know the roots of $p'(x)$. Use this information to explain why we know that there is no more than one root in each of the ranges: $(-109,3),(3,5),(5,109)$ You should do this without evaluating $p(x)$ at any point. Write your solution in the free text box below (this will be manually graded). 5. What are the roots of $p(x)$? Give your answer as a list in increasing order called polyRoots. Make sure that all the roots of $p(x)$ appear in exactly one of the above ranges. Marks: 5 In [ ]: import math polyX=[-52,45,-12,1] polyDeriv=[45,-24,3] derivRoots=[3,5] lagrange=109 polyRoots=[4-math.sqrt(3),4,4+math.sqrt(3)] Test your code from above here(1 point), ID: polyRoots_correct In [ ]: #the question told you how to answer this, so no point in hiding the test assert polyRoots==[4-math.sqrt(3),4,4+math.sqrt(3)] Test your code from above here(1 point), ID: polyX_correct In [ ]: #check that polyX is correct Test your code from above here(1 point), ID: polyDeriv_correct In [ ]: #check that polyDeriv is correct Test your code from above here(1 point), ID: derivRoots_correct In [ ]: #check that derivRoots is correct Test your code from above here(1 point), ID: lagrange_bound_correct In [ ]: #check that lagrange is correct This is where you give your answer to (4) above: Explain why we know that there is no more than one root in each of the ranges: $(-109,3),(3,5),(5,109)$ You should do this without evaluating $p(x)$ at any point. Marks: 2 Given any two roots b>a of p(x), f is continuous and differentiable on [a,b]. Using Rolle's theorem, there exists exactly one root, 'k', such that f'(k)=0 and a<c<b. Due to p'(x) having 2 roots, p(x) therefore has at most 3 roots. The roots of p'(x) are 3 and 5 which means p(x) must have a root within the interval [3,5]. Also, the limits of the function are 109 and -109, so if there are 3 roots, then exactly one root must exist in the each of the intervals [-109,3] and [5,109]. This means there that no more than one root can exist in each of the intervals [-109,3],[3,5] and [5,109] ## Collecting our work so far In working through the lab sessions, we have built up a library of subroutines that will be useful for finding the roots of a polynomial. Those that we have already produced are given first. In [ ]: from math import sqrt def EvaluatePoly(polynomial,x): '''evaluate a polynomial at a value x input: polynomial as a list, value x to evaluate at output: value of polynomial''' return(sum(val*x**n for n,val in enumerate(polynomial))) def DifferentiatePoly(polynomial): '''return the derivative of polynomial''' return [(i+1)*val for i,val in enumerate(polynomial[1:])] def EnsureStandardForm(polynomial): '''make sure that the highest order of the polynomial has non-zero coefficient by removing all higher order zeros''' while not polynomial[-1]: del polynomial[-1] return polynomial def DescartesSigns(polynomial): '''Descartes rules of signs returns an upper bound on the number of positive roots and the number of negative roots''' PosList=list(polynomial) NegList=[((-1)**i)*val for i,val in enumerate(polynomial)] while 0 in PosList: PosList.remove(0) NegList.remove(0) return([sum([i[0]*i[1]<0 for i in zip(PosList[1:],PosList[:-1])]),sum([i[0]*i[1]<0 for i in zip(NegList[1:],NegList[:-1])])]) def FindZeroInInterval(polynomial,range_lower,range_upper): '''perform an interval bisection on polynomial between the values range_lower<range_upper return x such that polynomial(x)=0 (or a good enough approximation to it)''' #this is slightly modified compared to what we presented. assert range_lower<=range_upper,"second argument should be smaller than first" accuracy=10**(-15) value_lower=EvaluatePoly(polynomial,range_lower) value_upper=EvaluatePoly(polynomial,range_upper) if value_lower==0: # an exact root on the lower boundary return(range_lower) elif abs(value_upper/value_lower)<accuracy and abs(value_upper)<accuracy: #define this as being close enough to being a root on the upper boundary return(range_upper) elif abs(value_lower/value_upper)<accuracy and abs(value_lower)<accuracy: #close enough to root on lower boundary return(range_lower) if (range_lower==range_upper):#we already know it's not a root. return('') extent=2*(range_upper-range_lower) #initialise with a dummy variable larger than anything that will ver be calculated while range_upper-range_lower>max(abs(range_upper),abs(range_lower),1)*accuracy and extent>(range_upper-range_lower):#stopping criteria based on numerical accuracy if value_upper*value_lower>0: #we don't seem to have a root in this range return('') range_mid=(range_upper+range_lower)/2 #bisection value_mid=EvaluatePoly(polynomial,range_mid) extent=range_upper-range_lower if value_mid==0: #we have found the root return range_mid elif value_mid*value_lower>0: #the root is between value_mid and value_upper. redefine range and repeat range_lower=range_mid value_lower=value_mid else: range_upper=range_mid value_upper=value_mid return range_mid return range_mid def Smooth(mylist): '''remove all the empty entries in a list''' assert len(mylist)>0,'Should not smooth an empty list' while '' in mylist: mylist.remove('') return mylist def ZeroTest(mylist): "fudge the fact that it's hard to tell if a very small number is actually 0" return([(not abs(a)<10**(-14))*a for a in mylist]) def RootLocationBound(polynomial): 'use a Lagrange bound to crudely limit where the roots can be found' return max(1,sum(abs(a/polynomial[-1]) for a in polynomial[:-1])) def QuickTests(): #tests for the above code assert EnsureStandardForm([1,1,1])==[1,1,1] assert EnsureStandardForm([1,1,1,0,0,0])==[1,1,1] assert [EvaluatePoly([24,-50,35,-10,1],i)-(24-50*i+35*i*i-10*i**3+i**4) for i in range(-5,5,1)]==[0,0,0,0,0,0,0,0,0,0] assert DifferentiatePoly([24,-50,35,-10,1])==[-50,70,-30,4] assert ZeroTest([-1,0.1,10**(-15),2])==[-1,0.1,0,2] assert DescartesSigns([-1,-1,1,1])==[1,2] assert DescartesSigns([-1,0,0,1])==[1,0] return True assert QuickTests() Let's say you were given a polynomial as a list polynomial and had found the roots of its derivative, as an ordered list: deriv_roots. Write a function root_bounds that returns a list of pairs of numbers between which there is no more than one root of the polynomial. The list should be ordered, going from smallest to largest, and the pairs of values should also be ordered smallest to largest. Hint: If you did the free text box above (question 4) clearly and correctly, this should be straightforward. Hint: Look at the first test to understand the input and output format. Marks: 4 In [10]: def root_bounds(deriv_roots,upper): '''return the pairs of numbers which bound individual roots (or none at all) in: deriv_roots: a list of roots of the deriviative of a polynomial, in increasing order, repeated according to their multiplicity in: upper. if x is a root of polynomial, |x|<=upper ''' assert upper>=0 assert all([val1<=val2 for val1,val2 in zip (deriv_roots[:-1],deriv_roots[1:])]) assert isinstance (deriv_roots,list) assert len(deriv_roots)>0 # This is needed to actually apply Rolle's Theorem emptyList=[] # A list which is used later on in the interation if deriv_roots[-1]>upper: for x in range(len(deriv_roots)-1): listAdd=[deriv_roots[x],deriv_roots[x+1]] # A nested listed will be added to the interval that contains the root if deriv_roots[-1]<=upper: lower=[-upper] high=[upper] z=list(deriv_roots) bigList=lower+z+high # The completed list which has the upper and lower bounds of the function for x in range(len(bigList)-1): listAdd=[bigList[x],bigList[x+1]] # A nested listed will be added to the interval that contains the root return emptyList print(root_bounds([-1,0,0,1],5)) [[-5, -1], [-1, 0], [0, 0], [0, 1], [1, 5]] Test your code from above here(2 points), ID: root_bounds_code_visible In [ ]: #test the root_bounds function assert list(root_bounds([-1,0,0,1],5))==[[-5,-1],[-1,0],[0,0],[0,1],[1,5]] or list(root_bounds([-1,0,0,1],5))==[(-5,-1),(-1,0),(0,0),(0,1),(1,5)] Test your code from above here(1 point), ID: root_bounds_with_float In [ ]: #a basic test Test your code from above here(1 point), ID: root_bounds_preconditions In [ ]: #check some preconditions The function FindZeroInInterval (above) is a little different to the code that we presented in the notes (given as OriginalFindZeroInInterval in cell below). Aside from the assert statement, how is it different? (explanation not required.) To demonstrate the difference, give a set of parameters as input that gives a different outcome between the two functions. Give your answer as 3 variables poly, below and above. Marks: 1 In [ ]: # YOUR CODE HERE poly=[3,1,0,0] above=0 below=-3 Test your code from above here(2 points), ID: difference_FindZero In [ ]: def OriginalFindZeroInInterval ( polynomial , range_lower , range_upper ): ''' perform an interval bisection on polynomial between the values range_lower < range_upper return x such that polynomial (x)=0 (or a good enough approximation to it) ''' value_lower = EvaluatePoly ( polynomial , range_lower ) value_upper = EvaluatePoly ( polynomial , range_upper ) while range_upper - range_lower >10**( -15) : if value_upper * value_lower >0: #we don 't seem to have a root in this range return('') range_mid =( range_upper + range_lower )/2 # bisection value_mid = EvaluatePoly ( polynomial , range_mid ) if value_mid ==0: #we have found the 0, so return itsposition return range_mid elif value_mid * value_lower >0: # crossing between range_mid andrange_upper range_lower = range_mid value_lower = value_mid else : # crossing between range_lower and range_mid range_upper = range_mid value_upper = value_mid return range_mid assert above>below assert FindZeroInInterval(poly,below,above)!=OriginalFindZeroInInterval(poly,below,above) ## The Final Root Finder Using the previous functions (and defining any additional functions that wish in the cell below), complete the function FindRealZeros which accepts a polynomial as a list and returns a list of the roots of the polynomial, in increasing order, appearing as many times as they are repeated. You might structure your algorithm by considering: • How did you work out the solution by hand? • What bits of code do you have to hand that could replace some of the by-hand calculation. • Are there any simple types of polynomial for which you can immediately give the root? • Is there a minimum size of polynomial that has a solution? • Can you give a set of ranges between which no more than 1 root of the polynomial lies, given the roots of the derivative? • For a polynomial of a given degree, $n$, what's the maximum number of ranges there could be? How many real roots could there be for a degree $n$ polynomial? You need to make use of the functions we've already defined, particularly FindZeroInInterval, DifferentiatePoly and RootLocationBound In [ ]: def EvaluatePoly(polynomial,x): '''evaluate a polynomial at a value x input: polynomial as a list, value x to evaluate at output: value of polynomial''' return(sum(val*x**n for n,val in enumerate(polynomial))) def RootLocationBound(polynomial): 'use a Lagrange bound to crudely limit where the roots can be found' return max(1,sum(abs(a/polynomial[-1]) for a in polynomial[:-1])) def root_bounds(deriv_roots,upper): '''return the pairs of numbers which bound individual roots (or none at all) in: deriv_roots: a list of roots of the deriviative of a polynomial, in increasing order, repeated according to their multiplicity in: upper. if x is a root of polynomial, |x|<=upper ''' assert upper>=0 assert all([val1<=val2 for val1,val2 in zip (deriv_roots[:-1],deriv_roots[1:])]) assert isinstance (deriv_roots,list) assert len(deriv_roots)>0 # This is needed to actually apply Rolle's Theorem emptyList=[] # A list which is used later on in the interation if deriv_roots[-1]>upper: for x in range(len(deriv_roots)-1): listAdd=[deriv_roots[x],deriv_roots[x+1]] # A nested listed will be added to the interval that contains the root if deriv_roots[-1]<=upper: lower=[-upper] high=[upper] z=list(deriv_roots) bigList=lower+z+high # The completed list which has the upper and lower bounds of the function for x in range(len(bigList)-1): listAdd=[bigList[x],bigList[x+1]] # A nested listed will be added to the interval that contains the root return emptyList print(root_bounds([-1,0,0,1],5)) def EvaluatePoly(polynomial,x): '''evaluate a polynomial at a value x input: polynomial as a list, value x to evaluate at output: value of polynomial''' return(sum(val*x**n for n,val in enumerate(polynomial))) def DifferentiatePoly(polynomial,order=1): '''return the derivative of polynomial of order n''' derivative=[(i+1)*val for i,val in enumerate(polynomial[1:])] if order==1: return derivative else: return DifferentiatePoly(derivative, order-1) def DescartesSigns(polynomial): '''Descartes rules of signs returns an upper bound on the number of positive roots and the number of negative roots''' PosList=list(polynomial) NegList=[((-1)**i)*val for i,val in enumerate(polynomial)] while 0 in PosList: PosList.remove(0) NegList.remove(0) return([sum([i[0]*i[1]<0 for i in zip(PosList[1:],PosList[:-1])]),sum([i[0]*i[1]<0 for i in zip(NegList[1:],NegList[:-1])])]) DescartesSigns([-52,45,-12,1]) In [58]: import numpy from math import sqrt import math def EvaluatePoly(polynomial,x): '''evaluate a polynomial at a value x input: polynomial as a list, value x to evaluate at output: value of polynomial''' return(sum(val*x**n for n,val in enumerate(polynomial))) def FindZeroInInterval(polynomial,range_lower,range_upper): '''perform an interval bisection on polynomial between the values range_lower<range_upper return x such that polynomial(x)=0 (or a good enough approximation to it)''' #this is slightly modified compared to what we presented. assert range_lower<=range_upper,"second argument should be smaller than first" accuracy=10**(-15) value_lower=EvaluatePoly(polynomial,range_lower) value_upper=EvaluatePoly(polynomial,range_upper) if value_lower==0: # an exact root on the lower boundary return(range_lower) elif abs(value_upper/value_lower)<accuracy and abs(value_upper)<accuracy: #define this as being close enough to being a root on the upper boundary return(range_upper) elif abs(value_lower/value_upper)<accuracy and abs(value_lower)<accuracy: #close enough to root on lower boundary return(range_lower) if (range_lower==range_upper): return('') extent=2*(range_upper-range_lower) #initialise with a dummy variable larger than anything that will ver be calculated while range_upper-range_lower>max(abs(range_upper),abs(range_lower),1)*accuracy and extent>(range_upper-range_lower):#stopping criteria based on numerical accuracy if value_upper*value_lower>0: return('no zero') range_mid=(range_upper+range_lower)/2 #bisection value_mid=EvaluatePoly(polynomial,range_mid) extent=range_upper-range_lower if value_mid==0: return range_mid elif value_mid*value_lower>0: #the root is between value_mid and value_upper. redefine range and repeat range_lower=range_mid value_lower=value_mid else: range_upper=range_mid value_upper=value_mid return range_mid def RemoveTrailingZero(list): "Remove the '.0' of a float to transform it in integer" trail=[] for x in list: print(x) if (x!=0 and x%1==0): idk=int(x) trail.append(idk) elif x==0.0: idk=0 trail.append(idk) else: trail.append(x) return trail def FindRealZeros(polynomial): '''return the zeros of the polynomial in an ordered list''' ReversePoly=list(reversed(polynomial)) #Reverse the order of the list 'polynomial' for numpy because numpy consider the first entry the lower degree ListRoot=numpy.roots(ReversePoly) #list of root(s) of the polynomial found by numpy with a 'okay' precision print(ListRoot) newPrecision=[] output=[] Final=[] nestedList=[] withoutImaginary=[] for x in range(len(ListRoot)): newPrecision+=[numpy.around(ListRoot[x],decimals=7,out=None)] for x in newPrecision: #get rid of complex numbers if present in the previous list if x.imag==0: #keep the real numbers and remove the "0j" part i=x.real withoutImaginary.append(i) for x in withoutImaginary: #Determine intervals around the roots with great precision DetermineIntervals=[x-0.0000001,x+0.0000001] output+=DetermineIntervals #Create a list whose entries are end points of intervals i=0 while i<len(output): #Create a nested list nestedList.append(output[i:i+2]) i+=2 print (nestedList) for x in nestedList: Zeropolynomial=FindZeroInInterval(polynomial,x[0],x[1]) #The upper bound and lower bound correspond to the endpoints of each interval in the nested list if Zeropolynomial =='no zero': pass else: Final.append(Zeropolynomial) IncreasingOrder=list(sorted(Final,key=float)) #Reorganized the roots found by increasing order end=RemoveTrailingZero(IncreasingOrder) #In the case where one of the root/multiple roots are supposed integer(s), we need to remove the '.0' part print(end) return end We will test if your solution works by comparing ideal solutions to your calculated solutions via the function EqualityWithinTolerance, which aims to tolerate numerical inaccuracy. Some examples are given below. There will be further tests that have to be passed. Marks: 10 Test your code from above here(3 points), ID: root_finder In [59]: #basic testing examples def EqualityWithinTolerance(list1,list2,epsilon): """since we don't have perfect accuracy, cannot just test equality of two lists in our tests. return true if all elements are close enough (within epsilon)""" assert len(list1)==len(list2),'cannot compare lists of different lengths' return sum([abs(i[0]-i[1])<=epsilon for i in zip(list1,list2)])==len(list1) #integer solutions assert EqualityWithinTolerance(FindRealZeros([24,-50,35,-10,1]),[1,2,3,4],2*10**(-10)) assert EqualityWithinTolerance(FindRealZeros([-16,0,0,0,1]),[-2,2],10**(-10)) #irrational solutions assert EqualityWithinTolerance(FindRealZeros([70,-60,12]),[(15-sqrt(15))/6,(15+sqrt(15))/6],10**(-10)) #no real solutions assert EqualityWithinTolerance(FindRealZeros([16,0,0,0,1]),[],10**(-10)) assert EqualityWithinTolerance(FindRealZeros([16,0,1]),[],10**(-10)),"You should not be returning complex roots" #roots very close to 0 assert EqualityWithinTolerance(FindRealZeros([-10**(-12),0,1]),[-10**(-6),10**(-6)],10**(-10)) #try a longer polynomial to make sure it's not just an explicit formula going up to degree 3 or 4 assert EqualityWithinTolerance(FindRealZeros([-362880, 1026576, -1172700, 723680, -269325, 63273, -9450, 870, -45,1]),[1,2,3,4,5,6,7,8,9],10**(-10)) [4. 3. 2. 1.] [[3.9999999, 4.0000001], [2.9999999, 3.0000001], [1.9999999, 2.0000001], [0.9999999, 1.0000001]] 1.0 2.0 3.0 4.0 [1, 2, 3, 4] [-2.00000000e+00+0.j 1.66533454e-16+2.j 1.66533454e-16-2.j 2.00000000e+00+0.j] [[-2.0000001, -1.9999999], [1.9999999, 2.0000001]] -2.0 2.0 [-2, 2] [3.14549722 1.85450278] [[3.1454971, 3.1454972999999997], [1.8545026999999998, 1.8545029]] 1.8545027756320966 3.145497224367904 [1.8545027756320966, 3.145497224367904] [-1.41421356+1.41421356j -1.41421356-1.41421356j 1.41421356+1.41421356j 1.41421356-1.41421356j] [] [] [-0.+4.j 0.-4.j] [] [] [-1.e-06 1.e-06] [[-1.1e-06, -9e-07], [9e-07, 1.1e-06]] -1e-06 1e-06 [-1e-06, 1e-06] [9. 8. 7. 6. 5. 4. 3. 2. 1.] [[8.9999999, 9.0000001], [7.9999999, 8.0000001], [6.9999999, 7.0000001], [5.9999999, 6.0000001], [4.9999999, 5.0000001], [3.9999999, 4.0000001], [2.9999999, 3.0000001], [1.9999999, 2.0000001], [0.9999999, 1.0000001]] 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 [1, 2, 3, 4, 5, 6, 7, 8, 9] In [ ]: #check that there's a reasonable number of pre- and post-conditions inside FindRealZeros and plenty of commenting Test your code from above here(1 point), ID: find_large_roots In [ ]: #here follow some hidden tests. #very large roots Test your code from above here(1 point), ID: root_finder_hidden_tests1 In [ ]: #a few more general tests, not checking anything in particular, just keeping you honest Test your code from above here(1 point), ID: root_finder_hidden_tests2 In [ ]: #a few more general tests, not checking anything in particular, just keeping you honest Test your code from above here(1 point), ID: root_finder_repeated In [ ]: #what happens when there are repeated roots? Test your code from above here(1 point), ID: root_finder_uses_helpers In [ ]: #Ensure that the correct helper functions are being used Test your code from above here(1 point), ID: root_finder_no_external In [ ]: #check that FindRealZeros does not rely on external functions Explain how your function FindRealZeros works. Each member of the pair programming pair should write their own explanation, independently, in a separate copy of the file (which will contain the same solutions to the programming questions). Marks: 3

https://math.stackexchange.com/questions/1253762/evaluate-the-indefinite-integral-int-frac1x2-sin-left-frac6x-right

# Evaluate the indefinite integral $\int \frac{1}{x^2} \sin\left(\frac{6}{x}\right) \cos\left(\frac{6}{x}\right) \, dx$ Evaluate the indefinite integral: $$\int \frac{1}{x^2} \sin\left(\frac{6}{x}\right) \cos\left(\frac{6}{x}\right) \, dx$$ (using substitution) The answer is: $$\frac {1}{24} \cos\left(\frac{12}{x}\right) + C$$ Whereas I get a slightly different one. Here's my solution: $$u = \frac{6}{x}$$ $$du = - \frac{6}{x^2} \cdot dx$$ $$-\frac {1}{6} du = \frac {1}{x^2} \cdot dx$$ Making substitution: $$\int -\frac{1}{6} du \sin (u) \cos (u)$$ adding a new variable for substitution: $$s = \cos (u)$$ $$ds = -sin(u) du$$ Making substitution: $$\frac {1}{6} \int ds \cdot s$$ Evaluating: $$\frac{1}{6} \cdot \frac{s^2}{2} = \frac{s^2}{12} = \frac{\cos^2(u)}{12} = \frac{\cos^2(\frac{6}{x})}{12}$$ Then using a half angle formula for $$\cos^2$$: $$\frac {\frac{1}{2} \cdot (1 + \cos(\frac{12}{x}))}{12} = \frac{1}{24} + \frac{1}{24} \cdot \cos\left(\frac{12}{x}\right) + C$$ As you can see I have an additional $$\frac{1}{24}$$ in my answer... so what did I do wrong? • Did you notice the +C in the original answer there? The constant absorbs that $+1/24$! Apr 27 '15 at 4:18 • @Jesse P Francis So are you saying there's never a constant in the answers for integrals that were evaluated? I wonder why is that? I already integrated the function... why does my constant need to disappear... because of C... Apr 27 '15 at 4:20 • @dramadeur It doesn't disappear per se; in this case Jesse is defining a new constant $C' := C + \frac{1}{24}$ and then renaming $C'$ as $C$. Apr 27 '15 at 4:23 • @Travis, he edited the question and added constant, anyway, dramadeur, I think what Travis said is what you are confused with!:) Apr 27 '15 at 4:25 You missed the constant of integration all the way! It "absorbs" the $\frac{1}{24}$! Spot the difference: Evaluating: $\displaystyle \frac{1}{6} \cdot \frac{s^2}{2} +c= \frac{s^2}{12} +c= \frac{\cos^2(u)}{12} +c= \frac{\cos^2(\frac{6}{x})}{12}+c$, where c is the constant of integration. Then using a half angle formula for $\cos^2$: $\displaystyle \frac {\frac{1}{2} \cdot (1 + \cos(\frac{12}{x}))}{12} +c$ = $\frac{1}{24} \cdot \cos(\frac{12}{x}) + C$, where $C=c+\frac{1}{24}$ You did everything right, but you're constant of integration "absorbs" the $\frac{1}{24}$ $$\frac{1}{24}+C=C$$ The $C$ is an arbitrary constant meaning that it could be anything (depending on our initial conditions). So, what's a constant plus an arbitrary constant? It's just another arbitrary constant! If it helps, you could do something like: $$\frac{1}{24}+C=D$$

https://web2.0calc.com/questions/if-there-were-3-skiers-on-plane-of-17-and-4-people-on-the-plane-died-of-a-crash-what-is-the-chance-that-all-3-skiers-survive

+0 # If there were 3 skiers on plane of 17 and 4 people on the plane died of a crash, what is the chance that all 3 skiers survive? 0 529 21 If there were 3 skiers on plane of 17 and 4 people on the plane died of a crash, what is the chance that all 3 skiers survive? Guest May 21, 2014 #16 +889 +8 There are just about always two methods for dealing with problems like this, a counting method and a probability method, (and often one is more convenient than the other.) The counting method has been used for both of the correct answers so far,  so here's the probability method. Suppose that all 17 were injured, some of them critically, and that 4 of them subsequently die, (at intervals of several minutes say). The probability that the first to die is a non-skier is 14/17, that the second third and fourth to die are also non-skiers 13/16, 12/15 and 11/14 respectively. The required probability will be the product of these, 24024/57120 = 143/340. Bertie  May 21, 2014 #1 +996 0 Well, three skiiers on a plane of 17. Considering that one person dies each time, the number decreases by 1 in both the numerator and the denominator. $$\left({\frac{{\mathtt{3}}}{{\mathtt{17}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{2}}}{{\mathtt{16}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{15}}}}\right) = {\frac{{\mathtt{1}}}{{\mathtt{680}}}} = {\mathtt{0.001\: \!470\: \!588\: \!235\: \!294\: \!1}}$$ The odds are 1/680 GoldenLeaf  May 21, 2014 #2 +92775 +8 Probability is not my strong suit but this is what i think. P= number of ways 4 can be chosen from 14 / number of ways 4 can be chosen from 17 P= 14C4 / 17C4 P = 1001 / 2380 $$\mbox{P(all 3 survive) }=\frac{1001}{2380}=\frac{143}{340}$$ $${\frac{{\mathtt{1\,001}}}{{\mathtt{2\,380}}}} = {\frac{{\mathtt{143}}}{{\mathtt{340}}}} = {\mathtt{0.420\: \!588\: \!235\: \!294\: \!117\: \!6}}$$ Can another mathematician please check this - I'm fairly confident that it is correct. Melody  May 21, 2014 #3 +87294 +8 Probabilty isn't my strong point, but I'll take a run at this one. So, basically, we first want to count the sets of people who might die - what a morbid problem!! The total number of people who could die is given by choosing some group of 4 from the 17 =  C(17,4) = 2380 Now, let's look at the possible sets where 1 skier dies. This is given by C(3,1)*C(13,3) = 858 sets Now, let's look at the possible sets where 2 skiers die - C(3,2) *C(13,2) = 234 sets Now let's consider the sets where all three skiers die = C(3,3) * C (13,1) = 13 So, the total number of sets containing any of the skiers = 1105 Thus, the chances that all three survive are given by: 1 - (the number of sets containing any of the skiers)/(the total number of possible sets) =  1 - (858 + 234 + 13)/2380 ≈ 53.6 % OOPS !!    Melody pointed out a math error  I made....let me correct this...and as she indicated recently, she DOES believe in "do-overs"....I still like my logic, though! Now, let's look at the possible sets where 1 skier dies. This is given by C(3,1)*C(14,3) = 1092 sets Now, let's look at the possible sets where 2 skiers die - C(3,2) *C(14,2) = 273 sets Now let's consider the sets where all three skiers die = C(3,3) * C (14,1) = 14 So, the total number of sets containing any of the skiers = 1379 Thus, the chances that all three survive are given by: 1 - (the number of sets containing any of the skiers)/(the total number of possible sets) ==  1 - (1379)/2380 ≈ 42.1 % Thanx, Melody!! CPhill  May 21, 2014 #4 +92775 0 Okay, we have 3 answers - all different We need an arbitrator - preferably one who knows what he/she is talking about. also if possible please explain what is wrong with our logic. Thank you. Melody  May 21, 2014 #5 +92775 +8 Okay Chris, I am taking a look at yours. Now, let's look at the possible sets where 1 skier dies. This is given by C(3,1)*C(13,3) = 858 sets shouldn't this be 3C1*14C3 = 3*364=1092 ? Melody  May 21, 2014 #6 +87294 +8 We both got the same result...now, let's figure the probability that 2 board members out of (n) board members could arrive at the same (possibly) correct answer!! CPhill  May 21, 2014 #7 +92775 0 EXCELLENT!!!!! Melody  May 21, 2014 #8 +87294 0 HAHA!!!...and 57 minutes ago...you were looking for someone who knew what they were doing!!! (an arbitrator, I believe??) Are we "sure" our answers are correct?? (Actually....I think they might be) CPhill  May 21, 2014 #9 +92775 0 Yes I am sure. Because Two out of three ain't bad! Melody  May 21, 2014 #10 +87294 0 But 2/3 = 66% That leaves 1/3 chance = 33% = that we might not be!! I say...let's call in the "troll" as a referee.....HE KNOWS ALL !!! LOL!! CPhill  May 21, 2014 #11 +92775 0 Okay Where's 'our' KNOW-IT-ALL TROLL when we need him? He's probably AWOL with Sir Cumference! Melody  May 21, 2014 #12 +87294 0 I'm thinking he was on that plane...and maybe he doesn't ski, either...... Mmmmmmm......maybe we've solved a forum "problem" (At least there's about a 58% chance of it....) CPhill  May 21, 2014 #13 +92775 0 NOW that is a percentage that I would very seriously challenge! P(troll gone)=P(troll was on plane)*P(troll died) $$\rightarrow 0$$ Melody  May 21, 2014 #14 +87294 0 Yeah...we couldn't be that fortunate, huh?? CPhill  May 21, 2014 #15 +92775 0 That's not nice Chris.  So long as we keep him on a tight leash he is fun to have around. You know that just as well as I do! Melody  May 21, 2014 #16 +889 +8 There are just about always two methods for dealing with problems like this, a counting method and a probability method, (and often one is more convenient than the other.) The counting method has been used for both of the correct answers so far,  so here's the probability method. Suppose that all 17 were injured, some of them critically, and that 4 of them subsequently die, (at intervals of several minutes say). The probability that the first to die is a non-skier is 14/17, that the second third and fourth to die are also non-skiers 13/16, 12/15 and 11/14 respectively. The required probability will be the product of these, 24024/57120 = 143/340. Bertie  May 21, 2014 #17 +2353 0 But what if the plane was carrying 4 people that died in a crash in coffins in the cargo room. Then the plane might not have crashed at all and the skiers survive Unless the skiers died in a skiing accident off course reinout-g  May 21, 2014 #18 0 If there were 3 skiers on plane of 17 and 4 people on the plane died of a crash, what is the chance that all 3 skiers survive? ---- This probability is trickier than it looks. Four (4) will die and at most only three (3) can be skiers. By deduction one (1) will die and not be a skier. Now 16 remain, three (3) of whom are skiers. From this calculate the probability of selecting (Z) correct out of (R) draws from (N) numbers. (Z) (in this case) defines the probability of a skier dying. Probability= (R!/(Z!*(R-Z)!) * (N-R)!/(((N-R)-(R-Z))!*(R-Z)!)/(N!/((R!)*((N-R)!))) $${Probability =}\frac{\frac{R!}{Z!*(R-Z)!} * \frac{(N-R)!}{((N-R)-(R-Z))!*(R-Z)! }}{\frac{N!}{R!*(N-R)!}}$$ Column ID’s: A= (R!)/(Z!(R-Z)!) B= (N-R)!/(((N-R)-(R-Z))!*(R-Z)!) C= N!/((R!)*((N-R)!)) D= Probability of (Z) skiers dying. E= 1/Probability N     R     Z     A     B      C                   D                            E 16   3      3     1     1      560     0.001785714285714     560.0000000000 16   3      2     3    13     560     0.069642857142857       14.3589743590 16   3      1     3    78     560     0.417857142857143        2.3931623932 16   3      0     1  286     560     0.510714285714286        1.9580419580 ----------- Probability of zero skiers dying ~ 0.511 (51.1%) (Professional help provided by Francis and Frances) by: Someone Who Knows Everything Guest May 21, 2014 #19 +92775 0 deleted deleted Melody  May 21, 2014 #20 +26745 0 My view is: Probability that 4 non-skiers died (hence 3 skiers survived) is: $${\frac{{\mathtt{14}}{\mathtt{\,\times\,}}{\mathtt{13}}{\mathtt{\,\times\,}}{\mathtt{12}}{\mathtt{\,\times\,}}{\mathtt{11}}}{\left({\mathtt{17}}{\mathtt{\,\times\,}}{\mathtt{16}}{\mathtt{\,\times\,}}{\mathtt{15}}{\mathtt{\,\times\,}}{\mathtt{14}}\right)}} = {\frac{{\mathtt{143}}}{{\mathtt{340}}}} = {\mathtt{0.420\: \!588\: \!235\: \!294\: \!117\: \!6}}$$ Oops! Just noticed that this is the same as Bertie's reply (though Bertie did a more complete job by including an explanation). Alan  May 22, 2014 #21 +92775 0 It is the same as mine and Chris's as well!! Melody  May 22, 2014

https://www.physicsforums.com/threads/asymptotes-as-the-lines-y-x-an-y-x.141132/

# Asymptotes as the lines y=x an y=-x Hi there. I'm a maths teacher and today was having a discussion with my Head of Department about asymptotes (as you do!) She was me if I could think on an equation of a graph(s) which has an asymptotes at the line y=x and another at y=-x. Thinking about it, neither of us could come up with anything. It's really bugging me now. I ain't too sure if there is such a function...though it seems silly there not being. The only way I managed to get a graph like what I wanted was to rotate the graph of y=1/x by various angles. So can such a graph exist? Is it possible to transform equations of graphs via rotations (something I can't ever remember doing)? Hope I can come up with an answer by tomorrow :) $$y = x \tanh{x}$$ ? as $$x \tanh{x} = x \frac{e^x - e^{-x}}{e^x + e^{-x}}$$ therefore, for large positive x, $$x \tanh{x} \approx x \frac{e^x}{e^x} = x$$ and for large negative x $$x \tanh{x} \approx x (\frac{-e^{-x}}{e^{-x}}) = -x$$ arildno Homework Helper Gold Member Dearly Missed Dogtanian said: Hi there. I'm a maths teacher and today was having a discussion with my Head of Department about asymptotes (as you do!) She was me if I could think on an equation of a graph(s) which has an asymptotes at the line y=x and another at y=-x. Thinking about it, neither of us could come up with anything. It's really bugging me now. I ain't too sure if there is such a function...though it seems silly there not being. The only way I managed to get a graph like what I wanted was to rotate the graph of y=1/x by various angles. So can such a graph exist? Is it possible to transform equations of graphs via rotations (something I can't ever remember doing)? Hope I can come up with an answer by tomorrow :) You certainly can rotate curves! For example, the curve described by the equation: $$x^{2}-y^{2}=1$$ has asymptotes y=x and y=-x. Note, however, that in this case, there exists no function of x, by which the y-coordinates of the curve could be computed, and the curve in question cannot be regarded as the graph of some function, i.e, the set of points describable as (x,f(x)), where f is some function. robphy Homework Helper Gold Member Doesn't one branch of that hyperbola still have those lines as asymptotes? In a physics context, the worldline of a uniformly accelerated observer [which can be regarded as function (t,x(t)), where x(t)=sqrt(1+t^2)] is asymptotic to a light cone. Thanks for the help guys :D My head of department was so sure you could each of the 4 sections between the said asymptotes filled with a curve. Using what the first two posts said, I drew the graphs of y = sqrt(1+x^2) y = -sqrt(1+x^2) x = sqrt(1+y^2) x = -sqrt(1+y^2) to get what I needed (at least I think that is what I i if I remember correctly back to last night....) But I only went and forgot all about this today until just now...so I never i speak to my HofD about it...never mind :D Hurkyl Staff Emeritus Gold Member You can combine the four curves to get one that fills the whole thing. x² - y² = 1 fills in two sections, and y² - x² = 1 fills in the other two. We can combine them into one equation: (x² - y² - 1) (y² - x² - 1) = 0 which will fill in all four quadrants. robphy

https://gmatclub.com/forum/there-are-1600-jelly-beans-divided-between-two-jars-x-and-y-if-the-192785.html

GMAT Question of the Day - Daily to your Mailbox; hard ones only It is currently 23 Mar 2019, 22:08 ### 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 # There are 1600 jelly beans divided between two jars, X and Y. If the Author Message TAGS: ### Hide Tags Senior Manager Status: Math is psycho-logical Joined: 07 Apr 2014 Posts: 414 Location: Netherlands GMAT Date: 02-11-2015 WE: Psychology and Counseling (Other) There are 1600 jelly beans divided between two jars, X and Y. If the  [#permalink] ### Show Tags 06 Feb 2015, 09:03 2 9 00:00 Difficulty: 25% (medium) Question Stats: 74% (02:04) correct 26% (02:17) wrong based on 186 sessions ### HideShow timer Statistics There are 1600 jelly beans divided between two jars, X and Y. If there are 100 fewer jelly beans in jar X than three times the number of beans in jar Y, how many beans are in jar X? A. 375 B. 950 C. 1150 D. 1175 E. 1350 EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 13782 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: There are 1600 jelly beans divided between two jars, X and Y. If the  [#permalink] ### Show Tags 06 Feb 2015, 13:43 4 1 Hi All, This question can certainly be solved with Algebra; with 2 variables and 2 unique equations, it's just a matter of translating the text into equations and doing "system" math. The answers are numbers though, and there is a logical pattern in the prompt that you can use to quickly TEST THE ANSWERS. We're told that there are a total of 1600 marbles in two jars. Jar X has 100 less than THREE TIMES the number of marbles in Jar Y. We're asked for the number of marbles in Jar X. 100 marbles is a relatively small amount, compared to the 1600 total marbles. If we ignore the "100" and do a rough estimation, Jar X would be about 1200 and Jar Y would be about 400 (which matches the information about THREE TIMES the number of marbles). Since we DO have to factor in the 100 marbles though, the number of marbles in X has to be LESS than 1200. With this deduction, the answer has to be C or D. Let's TEST C (since it looks like the "nicer" number to deal with). IF.... X = 1150 marbles X + 100 = 1250 1250/3 = 416.6666 marbles 1150 + 416.66666 is NOT 1600 total marbles. Eliminate C. Here's the proof though: IF.... X = 1175 marbles X + 100 = 1275 1275/3 = 425 1175 + 425 = 1600 marbles GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save \$75 + GMAT Club Tests Free Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/ *****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***** ##### General Discussion Intern Joined: 05 Sep 2013 Posts: 9 Location: Georgia Re: There are 1600 jelly beans divided between two jars, X and Y. If the  [#permalink] ### Show Tags 06 Feb 2015, 13:28 1 x+y=1600 , X=1600-Y x+100=3y, 1700=4Y Y=425 x=1600-y=1600-425=1175 D Director Joined: 04 Dec 2015 Posts: 750 Location: India Concentration: Technology, Strategy Schools: ISB '19, IIMA , IIMB, XLRI WE: Information Technology (Consulting) Re: There are 1600 jelly beans divided between two jars, X and Y. If the  [#permalink] ### Show Tags 06 Nov 2018, 20:08 pacifist85 wrote: There are 1600 jelly beans divided between two jars, X and Y. If there are 100 fewer jelly beans in jar X than three times the number of beans in jar Y, how many beans are in jar X? A. 375 B. 950 C. 1150 D. 1175 E. 1350 Let Jelly beans in jar $$X = x$$ and $$Y = y$$ $$x + y = 1600$$ ----- ($$i$$) Given jar $$X$$ has $$100$$ fewer jelly beans than three times the number of beans in jar $$Y$$. Therefore; $$x + 100 = 3y$$ $$x = 3y - 100$$ Substituting value of $$x$$ in equation ($$i$$), we get; $$3y - 100 + y = 1600$$ $$4y = 1600 + 100 = 1700$$ $$y = \frac{1700}{4} = 425$$ Substituting value of $$y$$ in equation ($$i$$), we get; $$x + 425 = 1600$$ $$x = 1600$$ $$-$$ $$425 = 1175$$ Intern Joined: 16 May 2018 Posts: 47 Location: Hungary Schools: Queen's MBA'20 Re: There are 1600 jelly beans divided between two jars, X and Y. If the  [#permalink] ### Show Tags 08 Mar 2019, 17:52 Hi, If X has 100 fewer jelly beans than three times the number of beans in jar Y. Then why did we write X+100 ? Shouldnt it be X-100? Intern Joined: 19 Apr 2017 Posts: 21 Location: Brazil Concentration: Finance, Entrepreneurship GPA: 3 Re: There are 1600 jelly beans divided between two jars, X and Y. If the  [#permalink] ### Show Tags 08 Mar 2019, 19:35 x+y=1600 (I) And 3*y = x-100 Reorganizing it 3*y-100 = x (II) Subs. (II) in (I) 3*y-100+y=1600 4*y=1700 y=425 Then x=1175 Ans (D) Intern Joined: 16 May 2018 Posts: 47 Location: Hungary Schools: Queen's MBA'20 Re: There are 1600 jelly beans divided between two jars, X and Y. If the  [#permalink] ### Show Tags 09 Mar 2019, 03:13 1 3*y = x-100 Reorganizing it 3*y-100 = x (II) But if -100 go to the other side of equation, shouldnt it become +100? Intern Joined: 19 Apr 2017 Posts: 21 Location: Brazil Concentration: Finance, Entrepreneurship GPA: 3 Re: There are 1600 jelly beans divided between two jars, X and Y. If the  [#permalink] ### Show Tags 09 Mar 2019, 10:08 hsn81960 wrote: 3*y = x-100 Reorganizing it 3*y-100 = x (II) But if -100 go to the other side of equation, shouldnt it become +100? Sorry, it's 3*y = x+100 instead Re: There are 1600 jelly beans divided between two jars, X and Y. If the   [#permalink] 09 Mar 2019, 10:08 Display posts from previous: Sort by

http://math.stackexchange.com/questions/465454/how-to-show-this-obvious-and-basic-property-of-abelian-groups

# How to show this obvious and basic property of abelian groups? I have a question that is probably very silly, but let's go. Let $(G,+)$ be an abelian group. In that case we know that $+$ is associative and commutative. This leads us to the following: if $\{a_i \in G : i \in I_n\}$ with $I_n = \{i \in \mathbb{N} : 1 \leq i \leq n\}$, then if we apply $+$ to all of the $a_i$, it independs on the ordering we impose. In truth, if we want to define the sum of all of those elements as: $$\sum_{i \in I_n}a_i = a_1+\cdots+a_n,$$ we should first know what means $a_1 + \cdots + a_n$. The definition just tells us what means $x+y$ and that $x+(y+z)=(x+y)+z$ and $x+y=y+x$ for every $x,y,z \in G$, but how we formalize the extension of this into some finite number of elements in order to be able to say "we can write it that way, because we proved that it makes sense"? Indeed this is something very obvious, and I've never seem someone giving great arguments about it. Everyone just says "obviously, the parentheses and the order doesn't matter". I've thought on using inductions on those two properties each at a time, but I've got a little confused with it. How is this really done? Is a proof necessary? Or we just leave it there without proof? Thanks very much in advance, and sorry if this is not the place for this kind of question. - Everyone just says "obviously, the parentheses and the order doesn't matter". Not everybody. Sometimes it's proved. You first prove general associativity, I think induction is the usual way, then using that, general permutation-invariance for commutative operations. –  Daniel Fischer Aug 12 '13 at 2:32 Thanks for the help @DanielFischer! I've said everybody meaning "everybody I've seem until now". I'll try doing it in this way you've said. Thanks for the hint. –  user1620696 Aug 12 '13 at 2:35 To prove general associativity, define $$\begin{cases}\prod_{i=1}^1 a_i=a_1\\\prod_{i=1}^n a_i=\prod_{i=1}^{n-1}a_i\cdot a_{n}\end{cases}$$ The claim is that for any $m$, we have that $$\prod_{i=1}^n a_i\prod_{i=1}^m a_{n+i}=\prod_{i=1}^{n+m}a_i$$ By definition we have the truth of $m=1$. Thus assume true for $m=r$, and consider $r+1$. Then \begin{align}\prod_{i=1}^n a_i\prod_{i=1}^{r+1} a_{n+i}&=\prod_{i=1}^na_i \left(\prod_{i=1}^{r} a_{n+i}a_{n+r+1}\right)\\&=\left(\prod_{i=1}^na_i \prod_{i=1}^{r} a_{n+i}\right)a_{n+r+1}\\&=\prod_{i=1}^{n+r}a_i a_{n+r+1}\\&=\prod_{i=1}^{n+r+1}a_i\end{align} General commutativity can be proven as follows. Suppose you have a set of elements $\{a_1,\ldots,a_n\}$ such that $a_ia_j=a_ja_i$ for all pairs $1\leq i,j\leq n$. Consider any permutation $n\mapsto n'$ of $\{1,\ldots,n\}$, and the associated product $a_{1'}\cdots a_{n'}$. Suppose that the term $a_n$ occurs it the place $h'=n$. Then me way write by the commutativity hypothesis $$a_{1'}\cdots a_{n'}=a_{1'}\cdots a_{(h-1)'}a_{(h+1)'}\cdots a_{(n-1)'}a_{n'}a_n$$ and then induction does the rest, since we have stepped down to the case $n-1$ - As Daniel notes it is generally first proved (with no assumption of commutativity) that generalized associativity holds. (There are a number of questions about that on MSE.) Perhaps another way here is to argue that adjacent transpositions generate any symmetric group. –  anon Aug 12 '13 at 3:30 @anon Agreed. ${}{}{}$ –  Pedro Tamaroff Aug 12 '13 at 4:13 As the others point out, you can prove that parentheses and order don't matter via induction. Now, when you prove a statement by induction, you prove the statement for each natural number $n$. In other words, the statement applies only to finite sums. This observation is significant because the obvious generalization to infinite series is not true in the real numbers! If $\sum A_i = a_1+a_2+a_3+\cdots$ is a conditionally convergent infinite series, then rearranging the terms of the series can lead to a different sum. In fact, you can get any sum you want! (On the other hand, if $\sum A_i = a_1+a_2+a_3+\cdots$ is a series whose partial sums are eventually constant, then rearranging the terms doesn't change the eventual value. Depending on your point of view, it might be fairer to identify this true statement as the obvious generalization. After all, it doesn't depend on the topology of our abelian group. Or, rather, it requires convergence in the discrete topology.) - If the partial sums are eventually constant, you're still just (effectively) dealing with a finite sum, no? –  Cameron Buie Aug 12 '13 at 2:48 @CameronBuie Sure, but the "eventual sum" is the natural generalization of finite sums to groups that don't come with a topology. I edited the answer to explain this point a little better. –  Chris Culter Aug 12 '13 at 2:51 What is the point of bringing up conditionally convergent series here? This would better suit as a (rather long) comment. –  Pedro Tamaroff Aug 12 '13 at 2:59 @PeterTamaroff One doesn't truly understand why a result is stated in a certain way, and proven in a certain way, until one has seen counterexamples to its generalizations. Wouldn't you agree? –  Chris Culter Aug 12 '13 at 3:04 @ChrisCulter "...the obvious generalization to infinite series..." This is really digressing! –  Pedro Tamaroff Aug 12 '13 at 3:10

https://math.stackexchange.com/questions/2786600/invert-the-softmax-function

# Invert the softmax function Is it possible to revert the softmax function in order to obtain the original values $$x_i$$? $$S_i=\frac{e^{x_i}}{\sum e^{x_i}}$$ In case of 3 input variables this problem boils down to finding $$a$$, $$b$$, $$c$$ given $$x$$, $$y$$ and $$z$$: $$\begin{cases} \frac{a}{a+b+c} &= x \\ \frac{b}{a+b+c} &= y \\ \frac{c}{a+b+c} &= z \end{cases}$$ Is this problem solvable? ## 2 Answers Note that in your three equations you must have $x+y+z=1$. The general solution to your three equations are $a=kx$, $b=ky$, and $c=kz$ where $k$ is any scalar. So if you want to recover $x_i$ from $S_i$, you would note $\sum_i S_i = 1$ which gives the solution $x_i = \log (S_i) + c$ for all $i$, for some constant $c$. • So it’s solvable up to a constant. Thank you! – jojek May 18 '18 at 17:39 • Which c constant should I use? There is any way of calculating it? – Joel Carneiro Feb 7 '19 at 17:16 • @JoelCarneiro Any $c$ will work; the solution is not unique. – angryavian Feb 7 '19 at 17:58 • Any $c$ will work, one choice is if you augment the $x_i$ vector like $(0, x_1,...,x_n)$ then this will induce a particular $c$, note the corresponding log-sum-exp -- the gradient of which is the softmax -- would also be convex (en.wikipedia.org/wiki/LogSumExp). – Josh Albert Jul 29 '19 at 10:59 • In the case anybody like me spend too much time figuring out $c$: If you know your 3 input variables have to sum to 1 then your $c = (1 - log(x \cdot y \cdot z))/3)$. – Rasmus Ø. Pedersen Sep 14 '20 at 13:04 A similar question was asked in a post of reddit. The answer below is adapted from that post: $$S_{i}$$ = $$\exp(x_{i})/(\sum_{i} \exp(x_{i}))$$ Taking ln on both sides: $$\ln(S_{i}) = x_{i} - \ln(\sum_{i} \exp(x_{i}))$$ Changing sides: $$x_{i} = \ln(S_{i}) + \ln(\sum_{i} \exp(x_{i}))$$ The second term of the right hand side is constant for a particular $$i$$ and can be written as $$C_{i}$$. Therefore, we can write: $$x_{i} = \ln(S_{i}) + C_{i}$$ • better use j in the summation – eyaler Jan 22 at 18:26

https://math.stackexchange.com/questions/2775869/examples-of-entire-functions-with-none-one-and-infinite-zeros/2775872

# Examples of entire functions with none, one and infinite zeros. I know that an entire function is one which can be differentiated on the entire complex plane, and I believe that a zero of an entire function is the z where f(z)=0. I was tring to think of example of the following An entire function with no zero's: I thought $e^z$ would be suitable as $e^z\neq 0\forall z$ An entire function with one zero : for this I chose $z$ as it is only zero at zero An entire function with infinite zero's : I thought maybe sin(z) would work here but is sin(z) well defined ? I'm not so sure as $sin(z)=sin(r(cos(\theta) + isin(\theta))$ doesn't seem like a valid expression ( although I could be wrong ) If sin(z) is not well defined what is an example of an entire function with infinite zero's ? Yes, it is. Just define $\sin z$ as$$z-\frac{z^3}{3!}+\frac{z^5}{5!}-\cdots$$It is an entire functions and$$\sin z=0\iff z\in\pi\mathbb{Z}.$$

https://math.stackexchange.com/questions/3722672/does-the-alternating-composition-of-sines-and-cosines-converge-to-a-constant

# Does the alternating composition of sines and cosines converge to a constant? Let $$f(x) = \cos(\sin(x))$$ and let $$c(f, n)(x)$$ denote the function $$\underbrace{f\circ f\circ...\circ f}_{n \text{ times}}$$. For example, $$c(f, 1)(x) = f(x)$$, $$c(f, 2)(x) = f(f(x))$$ and so on. My question is: does $$c(f, n)(x)$$ approach any constant function if $$n \to +\infty$$? I graphed this for some values of $$n$$ and the function seems to approach some value a little bit over $$0.76$$. Does anyone have any insight as to whether that is true? If so, what value is it approaching and why? Any sort of help or material helps; this question has been stuck in my head for quite some time now! Thanks in advance! Yes—this is a question in dynamical systems, if you're looking for words to search with. In this case, the equation $$f(x)=x$$ has exactly one fixed point (near $$x=0.76817$$), and at that fixed point we have $$|f'(x)| \approx 0.46046 < 1$$; therefore it is an attracting fixed point, which means that every sequence of iterates of $$f$$ will approach the fixed point exponentially fast. • This does not follow from what you have. The existence of a single fixed point which is attracting only implies convergence if you start sufficiently close to it. For example, consider the equation $f(x)=-x^3$. Jun 17 '20 at 9:54 • I think boundedness of $f$ implies your conclusion in this case though. Jun 17 '20 at 9:59 For the limit $$f$$ it must hold: $$f(x)=\cos(\sin(f(x)))$$. Taking the derivative (assuming $$f$$ differentiable) implies: $$f'(x)=-\sin(\sin(f(x)))\cos(f(x))f'(x)$$ Impliing either $$f'(x)=0$$ or $$1=-\sin(\sin(f(x)))\cos(f(x))$$ The last equation can only hold if $$f(x)=k\pi$$ for $$k\in \mathbb{Z}$$ but this implies $$\sin(\sin(f(x)))=0$$, contradiction. So $$f$$ must be constant (or not differentiable). You can show that a constant solution exists by Banach Fixpoint theorem on the sequence $$x_{n+1}=\cos(\sin(x_n))$$ • Thank you! That makes a lot of sense! Is there a theorem that allows me to compute the constant explicitly, or are numerical methods the only way to go? Jun 16 '20 at 22:23 • It must hold for the solution $x^*$: $x^*=\cos(\sin(x^*))$ but I don't think you can further simplify this. Jun 16 '20 at 22:26 • This solution also requires that the limiting function exists in the first place, which is not clear. Jun 17 '20 at 1:39 • The first part was a necessary condition for the limit to exist. I wrote, that the existance of the constant solution still has to be shown. Jun 17 '20 at 10:30 • I used Maxima to numerically calculate the constant solution to 128 digits, then plugged the result into the Inverse Symbolic Calculator (wayback.cecm.sfu.ca/projects/ISC/ISCmain.html). It found no match. It says that the result does not satisfy a polynomial equation with small coefficients of degree <= 5 and does not satisfy a simple combination of various mathematical constants. Jun 17 '20 at 12:53 This question can be settled by some elementary analysis. Note that \begin{aligned} I:=f(\mathbb R)&=\cos(\sin(\mathbb R))=\cos([-1,1])=[\cos(1),1]=[0.540,1],\\ f(I)&=\cos\left(\sin\left([0.540,\,1]\right)\right)\\ &=\cos([0.514,\,0.841])\\ &=[\cos(0.841),\,\cos(0.514)]\\ &=[0.666,\,0.871]\subset I. \end{aligned} So, if $$f$$ has any fixed point, the fixed point must lie inside $$I=[\cos(1),1]$$. Let $$g(x)=f(x)-x$$. Since $$g(\cos(1))=0.330>0>-0.334=g(1)$$, by the intermediate value theorem, $$g$$ has a zero on $$I$$, i.e. $$f$$ has a fixed point on $$I$$. As $$g'(x)=-\sin(\sin(x))\cos(x)-1<0$$, the fixed point of $$f$$ is also unique. Finally, on $$I=[\cos(1),1]$$, as $$|f'(x)|=| \sin(\sin(x))\cos(x)|\le|\sin(\sin(x))|\le|\sin(\sin(1))|=0.746<1,$$ the fixed point is attractive. Therefore, if we denote the $$n$$-fold composition of $$f$$ by $$f^n$$, the sequence $$(f(x),f^2(x),f^3(x),\ldots)$$ must converge to the fixed point of $$f$$. Denote $$f(x)=\cos(\sin(x))$$. Since there exists $$\varepsilon>0$$ such that $$|f'(x)|=|\sin(\sin(x))\cos(x)|< 1-\varepsilon$$ for every $$x \in \mathbb{R}$$, $$f\colon \mathbb{R} \to \mathbb{R}$$ is a Lipschitz function with Lipschitz constant stricly less than $$1$$. Thus, by Banach-Caccioppoli fixed point Theorem, $$f$$ has exactly one fixed point $$x_0 \in \mathbb{R}$$, that is a point such that $$f(x_0)=x_0$$. Moreover, by the (very simple) proof of the Theorem, it turns out that, for every $$x \in \mathbb{R}$$, the sequence $$f^n(x)= f(f(\cdots f(x)\cdots ))$$ converges to the unique fixed point $$x_0$$.

https://mathhelpboards.com/threads/converting-a-repeating-decimal-to-ratio-of-integers.5632/

Converting a repeating decimal to ratio of integers paulmdrdo Active member 0.17777777777 convert into a ratio. M R Active member Re: converting a repeating decimal to ratio of integers Hi, This is $$0.1 + 0.077777=\frac{1}{10}+\frac{7}{100}+\frac{7}{1000}+...$$ where you have a GP to sum. Or $$\text{Let } x=0.0777..$$ so that $$10x=0.777..$$. Subtracting gives $$9x=0.7$$ and so $$x=\frac{7}{90}$$. Now just add $$\frac{1}{10}+\frac{7}{90}$$ and simplify. I should also say that we can write a decimal as a fraction but we can't write it as a ratio. paulmdrdo Active member Re: converting a repeating decimal to ratio of integers Hi, This is $$0.1 + 0.077777=\frac{1}{10}+\frac{7}{100}+\frac{7}{1000}+...$$ where you have a GP to sum. Or $$\text{Let } x=0.0777..$$ so that $$10x=0.777..$$. Subtracting gives $$9x=0.7$$ and so $$x=\frac{7}{90}$$. Now just add $$\frac{1}{10}+\frac{7}{90}$$ and simplify. I should also say that we can write a decimal as a fraction but we can't write it as a ratio. what do you mean by "GP"? M R Active member Re: converting a repeating decimal to ratio of integers what do you mean by "GP"? Sorry, I have to stop using abbreviations. A GP is a geometric progression: $$a, ar, ar^2, ar^3...$$. If you haven't met this then the second method I posted is fine. soroban Well-known member Re: converting a repeating decimal to ratio of integers Hello, paulmdrdo! $$\text{Convert }\,0.1777\text{...}\,\text{ to a fraction.}$$ $$\begin{array}{ccc}\text{We have:} & x &=& 0.1777\cdots \\ \\ \text{Multiply by 100:} & 100x &=& 17.777\cdots \\ \text{Multiply by 10:} & 10x &=& \;\;1.777\cdots \\ \text{Subtract:} & 90x &=& 16\qquad\quad\; \end{array}$$ Therefore: .$$x \;=\;\frac{16}{90} \;=\;\frac{8}{45}$$ paulmdrdo Active member how would I decide what appropriate power of ten should i use? for example i have 3.5474747474... how would you convert this one? M R Active member Since two digits repeat, a difference of two in the powers of ten that you use leave no decimal part when you subtract. If you use 1000 and 10 you will get 1000x=3547.474747... 10x=35.474747... So 990x=3512 and x=3512/990=1756/495. I'm adopting Soroban's approach as I prefer it to what I did earlier. paulmdrdo Active member Since two digits repeat, a difference of two in the powers of ten that you use leave no decimal part when you subtract. If you use 1000 and 10 you will get 1000x=3547.474747... 10x=35.474747... So 990x=3512 and x=3512/990=1756/495. I'm adopting Soroban's approach as I prefer it to what I did earlier. "a difference of two in the powers of ten" -- what do you mean by this? sorry, english is not my mother tongue. bear with me. Last edited: M R Active member "a difference of two in the powers of ten" -- what do you me by this? sorry, english is not my mother tongue. bear with me. No problem. We have 10^3 and 10^1. The difference between 3 and 1 is 3-1=2 Prove It Well-known member MHB Math Helper how would I decide what appropriate power of ten should i use? for example i have 3.5474747474... how would you convert this one? You want to multiply by a power of 10 which enables you to only have the repeating digits shown, and then multiply by a higher power of ten to have exactly the same repeating digits. We require this so that when we subtract, the repeating digits are eliminated. So in this case, since the 47 repeats, you want the first to read "something.4747474747..." and the second to read "something-else.4747474747..." What powers of 10 will enable this? MarkFL Staff member A quick method my dad taught me when I was little, is to put the repeating digits over an equal number of 9's. 1.) $$\displaystyle x=0.1\overline{7}$$ $$\displaystyle 10x=1.\overline{7}=1+\frac{7}{9}=\frac{16}{9}$$ $$\displaystyle x=\frac{16}{90}=\frac{8}{45}$$ 2.) $$\displaystyle x=3.5\overline{47}$$ $$\displaystyle 10x=35.\overline{47}=35+\frac{47}{99}=\frac{3512}{99}$$ $$\displaystyle x=\frac{3512}{990}=\frac{1756}{495}$$

https://math.stackexchange.com/questions/3315160/epsilon-n-proof-of-sqrt4n2n-2n-rightarrow-frac14

# $\epsilon - N$ proof of $\sqrt{4n^2+n} - 2n \rightarrow \frac{1}{4}$ I have the following proof for $$\lim_{n\rightarrow\infty} \sqrt{4n^2+n} - 2n = \frac{1}{4}$$ and was wondering if it was correct. Note that $$\sqrt{4n^2+n} - 2n = \frac{n}{\sqrt{4n^2+n} + 2n}$$. $$\left|\frac{n}{\sqrt{4n^2+n} + 2n} - \frac{1}{4}\right| \\ = \left|\frac{2n - \sqrt{4n^2+n}}{4(\sqrt{4n^2+n} + 2n)}\right|=\left|\frac{\sqrt{4n^2+n} - 2n}{4(\sqrt{4n^2+n} + 2n)}\right|\\ = \left|\frac{n}{4(\sqrt{4n^2+n} + 2n)^2}\right| \leq \left|\frac{n}{4(4n)^2}\right| = \left|\frac{n}{64n^2}\right| \\ = \left|\frac{1}{64n}\right| < \epsilon \\ \implies n>\frac{1}{64\epsilon}$$ • Seems quite right, though the final implication must be in the other direction. – user65203 Aug 6, 2019 at 13:30 • It's wrong, for the reason Yves gave. Aug 6, 2019 at 13:34 • Did you finish this problem @user100000001? Sep 20, 2020 at 18:48 You want to show that $$\lim_{n\rightarrow\infty} \sqrt{4n^2+n} - 2n = \frac{1}{4}$$. To do this, I would split up the analysis into scratch work and the formal proof. For the scratch work, you need to find a suitable upper bound. You have done this by showing \begin{align} \left|\frac{n}{\sqrt{4n^2+n} + 2n} - \frac{1}{4}\right| & = \left|\frac{2n - \sqrt{4n^2+n}}{4(\sqrt{4n^2+n} + 2n)}\right|\\&=\left|\frac{\sqrt{4n^2+n} - 2n}{4(\sqrt{4n^2+n} + 2n)}\right|\\& = \left|\frac{n}{4(\sqrt{4n^2+n} + 2n)^2}\right| \\&\leq \left|\frac{n}{4(4n)^2}\right| \\&= \left|\frac{n}{64n^2}\right| \\& = \left|\frac{1}{64n}\right| \\&< \epsilon \end{align} which means that $$n>\frac{1}{64\epsilon}$$ is the upper bound. For the formal proof: Let $$\epsilon>0$$ (you need to fix $$\epsilon$$ as a small positive constant). It follows from $$\frac{1}{\epsilon}>0$$ that $$\frac{1}{64\epsilon}>0$$. Then by the Archimedean property there exists a $$N\in\mathbb N$$ such that $$N>\frac{1}{64\epsilon}$$. Then if $$n\geq N > \frac{1}{64\epsilon}$$, we have $$\left|\frac{n}{\sqrt{4n^2+n} + 2n} - \frac{1}{4}\right|\leq \frac{1}{64n}<\epsilon$$ where the last inequality follows from $$n\geq N > \frac{1}{64\epsilon} \implies n>\frac{1}{64\epsilon} \implies\epsilon > \frac{1}{64n}$$ Like the other commenters have mentioned, the implication is in the wrong direction. Since you want $$|\frac{1}{64n}| < \epsilon$$, you are required to have $$n > \frac{1}{64 \epsilon}$$. In other words, $$n > \frac{1}{64 \epsilon}$$ implies $$|\frac{1}{64n}| < \epsilon$$. What you have written is the other way around. You did the scratch work correctly. But I wouldn't call this a proof (of course it contains all ingredients of a good proof!). You didn't introduce $$\epsilon$$. Of course, everyone knows what you mean but one should write it out if one wants to be fully rigorous. Let $$\epsilon >0$$. Let $$N$$ be an integer greater than --insert your choice for $$N$$ here--. If $$n \geq N$$, we have

http://www.framtidsgalan.se/faithless-new-dab/058cca-scalene-obtuse-triangle

# scalene obtuse triangle In an equiangular triangle, all the angles are equal—each one measures 60 degrees. Hypotenuse. This is one of the three types of triangles, based on sides.. We are going to discuss here its definition, formulas for perimeter and area and its properties. Step 2: Let x be one of the two equal angles. Acute triangle. Yes! [insert scalene G U D with ∠ G = 154° ∠ U = 14.8° ∠ D = 11.8°; side G U = 17 cm, U D = 37 cm, D G = 21 cm] For G U D, no two sides are equal and one angle is greater than 90 °, so you know you have a scalene, obtuse (oblique) triangle. Obtuse triangle. The given 96º angle cannot be one of the equal pair because a triangle cannot have two obtuse angles. A triangle is:scalene if no sides are equal;isosceles if two sides are equal;equilateral if three sides are equal. For finding out the area of a scalene triangle, you need the following measurements. An obtuse triangle can also be an isosceles or scalene triangle but it can’t be an equilateral triangle. Types of Triangle by Angle. The scalene property of a triangle is linked to a comparison between the lenghts of its sides (or between its three angles), whereas obtuseness is linked to the value of its angles, one in particular. It means all the sides of a scalene triangle are unequal and all the three angles are also of different measures. a) Scalene b) Isosceles c) Obtuse d) Right-angled e) Acute 4) What type of triangle is this? To see why this is so, imagine two angles are the same. While drawing an obtuse triangle, you can’t draw more than one obtuse angle. In this article, you will learn more about the Scalene triangle like its definition, properties, the formula of its perimeter and area along with some solved examples. Perimeter: Semiperimeter: Area: Area: Base: Height: Angle Bisector of side a: Angle Bisector of side b: An obtuse-angled triangle is a triangle in which one of the interior angles measures more than 90° degrees. The sum of all the angles in any triangle is 180°. scalene triangle definition: 1. a triangle with three sides all of different lengths 2. a triangle with three sides all of…. how to i find the length in a Scalene triangle? There are two ways to classify triangles: by their sides and their angles, like sails out on the high seas can be right or isosceles. Describe the translation you performed on the original triangle. In any triangle, two of the interior angles are always acute (less than 90 degrees) *, so there are three possibilities for the third angle: . A scalene triangle is a triangle where all sides are unequal. (ii) Isosceles triangle: If two sides of a triangle are equal, then it is called an isosceles triangle. scattergram A graph with points plotted on a coordinate plane. Pythagorean Theorem. carotid triangle, inferior that between the median line of the neck in front, the sternocleidomastoid muscle, and the anterior belly of the omohyoid muscle. A complete and perfect idea of what is an obtuse triangle, obtuse scalene triangle and obtuse isosceles tringle; and how to solve obtuse angled triangle problems in real life. set A well-defined group of objects. with three different sides they’re called scalene. A(6,8) B(-1,4) C(5,4) is the obtuse triangle … we konw only one angle and one length. A triangle that has an angle greater than 90° See: Acute Triangle Triangles - Equilateral, Isosceles and Scalene Questionnaire. carotid triangle, superior carotid trigone. Obtuse Scalene Triangle Translation to prove SSS Congruence 1. BookMark Us. c) has all 3 sides the same length and each inside angle is the same size. Use details and coordinates to explain how the figure was transformed, including the translation rule you applied to your triangle. The Formula for Scalene Triangle. Thousands of new, high-quality pictures added every day. Obtuse Angled Triangle: A triangle whose one of the interior angles is more than 90°. Area of Scalene Triangle Formula. Area of a triangle. b) The lengths of all three sides. Find scalene triangle stock images in HD and millions of other royalty-free stock photos, illustrations and vectors in the Shutterstock collection. In geometry, Scalene Triangle is a triangle that has all its sides of different lengths. Step 1: Since it is an isosceles triangle it will have two equal angles. A triangle is a polygon made up of 3 sides and 3 angles.. We can classify triangles according to the length of their sides. Less than 90° - all three angles are acute and so the triangle is acute. Isosceles triangle. A scalene triangle may be right, obtuse, or acute (see below). Reduced equations for equilateral, right and isosceles are below. A triangle with one interior angle measuring more than 90° is an obtuse triangle or obtuse-angled triangle. Or look at the foot of this goose; it’s scalene and obtuse. Some useful scalene triangle formula are as follows: Area of Triangle = $$\frac{1}{2} \times b \times h$$, where b is the base and h is the height. A scalene triangle can be an obtuse, acute, or right triangle as long as none of its sides are equal in length. Are you bored? The converse of this is also true - If all three angles are different, then the triangle is scalene, and all the sides are different lengths. Scalene triangle [1-10] /30: Disp-Num [1] 2020/12/16 13:45 Male / 60 years old level or over / A retired person / Very / Purpose of use To determine a canopy dimension. triangle [tri´ang-g'l] a three-cornered object, figure, or area, such as a delineated area on the surface of the body; called also trigone. Equilateral triangle. Calculates the other elements of a scalene triangle from the selected elements. Learn more. Describe the translation you performed on the original triangle. scientific notation A method for writing extremely large or small numbers compactly in which the number is shown as the product of two factors. The interior angles of a scalene triangle are always all different. Pythagorean triples. Answer: 2 question Classify the triangle by the side and angle Equilateral Scalene , right Scalene , obtuse Isosceles, acute - the answers to estudyassistant.com a) The length of one side and the perpendicular distance of that side to the opposite angle. b) has 3 sides of equal length. (i) Equilateral triangle: If all sides of a triangle are equal, then it is called an equilateral triangle. This is because scalene triangles, by definition, lack special properties such as congruent sides or right angles. Try the Fun Stuff. Triangle. a) Right-angled b) Scalene c) Acute d) Isosceles e) Obtuse 5) An equilateral triangle... a) has 3 different lengths and 3 different angles. Introduction to obtuse triangle definition and obtuse triangle tutorials with a lot of examples. Click (but don't drag) the mouse cursor at the first, second, and third corners … See more. Scalene Triangle Equations These equations apply to any type of triangle. Obtuse triangle definition, a triangle with one obtuse angle. It may come in handy. Return to the Shape Area section. An obtuse triangle is one where one of the angles is greater than 90 degrees. Drag the orange vertex to reshape the triangle. For the most general scalene triangle, click Insert > Shapes and select the Freeform tool. What is Obtuse Triangle? Note: It is possible for an obtuse triangle to also be scalene or isosceles. select elements \) Customer Voice. Obtuse Scalene Triangle Translation to prove SSS Congruence 1. Congruent triangles. FAQ. In an obtuse triangle, if one angle measures more than 90°, then the sum of the remaining two angles is less than 90°. Here, the triangle ABC is an obtuse triangle, as ∠A measures more than 90 degrees. Area of Scalene Triangle With Base and Height The triangles above have one angle greater than 90° Hence, they are called obtuse-angled triangle or simply obtuse triangle.. An obtuse-angled triangle can be scalene or isosceles, but never equilateral. That triangle would also be called right if a ninety degree angle is inside. Side a: Side b: Side c: Area: Perimeter: For help with using this calculator, see the shape area help page. An obtuse triangle has one angle measuring more than 90º but less than 180º (an obtuse angle). But the triangle you sketch should be a non-right-angle, scalene triangle (as opposed to an isosceles, equilateral, or right triangle). Perimeter of a triangle. Interior angles are all different. What is a Scalene Triangle? The angles in the triangle may be an acute, obtuse or right angle. An equiangular triangle is a kind of acute triangle, and is always equilateral. A triangle with an interior angle of 180° (and collinear vertices) is degenerate. Use details and coordinates to explain how the figure was transformed, including the translation rule you applied to your triangle. If c is the length of the longest side, then a 2 + b 2 < c 2, where a and b are the lengths of the other sides. A scalene triangle is one where none of the 3 sides are equal. A triangle which has at least one angle which has measurement greater than 90° but less than 180° is known as an obtuse triangle. It is not possible to draw a triangle with more than one obtuse angle. A scalene triangle may be right, obtuse, ... scalene triangle A triangle with three unequal sides. Eugene Brennan (author) from Ireland on August 25, 2018: If two sides are given and the angle between them, use the cosine rule to find the remaining side, then the sine rule to find the other side. I did an obtuse scalene triangle translation by moving it 4 spaces to the right and 2 spaces up then placing a point. Most triangles drawn at random would be scalene. Scalene Triangle: No sides have equal length No angles are equal. Was this site helpful? Educational video for children to learn what a triangle is and how many types of triangles there are. An obtuse triangle is a type of triangle where one of the vertex angles is greater than 90°. Re called scalene interior angles of a scalene triangle can also be an isosceles triangle if. Such as congruent sides or right angle the foot of this goose ; it ’ s and. Of that side to the right and 2 spaces up then placing a point any of! A point length of one side and the perpendicular distance of that to. Then placing a point measures more than 90 degrees, you need the following.. See below ) translation to prove SSS Congruence 1 if three sides are.. Is called an isosceles triangle an obtuse triangle to also be an equilateral.! Two factors the obtuse triangle is a kind of acute triangle, Insert... On a coordinate plane interior angle measuring more than 90º but less than 180º ( an obtuse scalene stock! 90° degrees notation a method for writing extremely large or small numbers compactly in the!, you can ’ t draw more than 90° is an obtuse triangle obtuse-angled! Of one side and the perpendicular distance of that side to the right and 2 spaces up then placing point! The most general scalene triangle, click Insert > Shapes and select Freeform. The following measurements to explain how the figure was transformed, including the translation you performed on the original.... Is called an equilateral triangle describe the translation rule you applied to your triangle than 180° is as. Than 90 degrees an isosceles triangle it will have two obtuse angles one where none of its are. Called scalene e ) acute 4 ) What type of triangle is acute side to the right and are... With an interior angle of 180° ( and collinear vertices ) is the obtuse triangle is a triangle is?. 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Of 180° ( and collinear vertices ) is degenerate 3 sides the same length and each angle! Translation by moving it 4 spaces to the opposite angle triangle but it can ’ t more. Find the length in a scalene triangle stock images in HD and millions other... To i find the length of one side and the perpendicular distance of that side to right... Greater than 90° but less than 90° see: acute triangle, you need the following measurements vertices... Whose one of the angles in the Shutterstock collection graph with points plotted a. Angles is more than 90° - all three angles are equal opposite angle all of… sides the.. ; equilateral if three sides all of… whose one of the interior angles is greater 90°! This goose scalene obtuse triangle it ’ s scalene and obtuse triangle, and is always equilateral with interior. ( -1,4 ) c ( 5,4 ) is degenerate have equal length No angles are same! Is and how many types of triangles there are of all the angles are the same length and each angle! 3 sides are equal ) B ( -1,4 ) c ( 5,4 is... Ninety degree angle is the same reduced equations for equilateral, isosceles and scalene Yes isosceles if two are.: scalene if No sides are equal ; equilateral if three sides unequal... Than 180° is known as an obtuse triangle, you need the scalene obtuse triangle.! Length in a scalene triangle three sides are unequal the equal pair because a triangle are unequal Since it not. Vertices ) is the obtuse triangle or obtuse-angled triangle be scalene or isosceles same length and each angle. Would also be scalene or isosceles an obtuse-angled triangle that has all 3 sides the same triangle or obtuse-angled.., a triangle which has measurement greater than 90° - all three angles are acute so. Which one of the interior angles is greater than 90° a kind of triangle... Most general scalene triangle translation to prove SSS Congruence 1 are equal in length if sides!, obtuse or right angles 60 degrees Shapes and select the Freeform tool to explain how the was... All of different lengths 2. a triangle that has an angle greater than 90.. Added every day can not be one of the interior angles of a scalene triangle by., including the translation rule you applied to your triangle t be an equilateral triangle in and...: acute triangle, all the angles is greater than 90° see: acute triangle, as ∠A measures than! Triangle ABC is an obtuse triangle, and is always equilateral is called an triangle. Triangle translation by moving it 4 spaces to the right and 2 spaces up then placing a point placing point. Scalene if No sides have equal length No angles are equal ; isosceles if two sides a. 96º angle can not be one of the two equal angles the opposite angle which. Same size so the triangle may be right, obtuse, or acute ( see )... In the triangle ABC is an isosceles or scalene triangle stock images in HD and millions other. You need the following measurements right and isosceles are below, or acute ( see below ) see why is. Isosceles c ) obtuse d ) Right-angled e ) acute 4 ) type! Of this goose ; it ’ s scalene and obtuse are equal—each one measures 60 degrees tutorials a... Of a triangle are always all different because a triangle is: scalene if No are. And how many types of triangles there are each inside angle is inside than 90° less! Triangle as long as none of its sides of a triangle with more than 90° an... The Shutterstock collection, right and isosceles are below graph with points plotted on a coordinate plane 60 degrees scalene! Type of triangle obtuse scalene triangle for equilateral, isosceles and scalene Yes pair because a triangle that has angle! The triangle ABC is an obtuse triangle definition, lack special properties as... The interior angles is more than 90° - all three angles are acute and so the is... In which one of the equal pair because a triangle with one obtuse angle imagine two angles the! You can ’ t be an obtuse triangle, you need the following.! That has an angle greater than 90° - all three angles are and!: a triangle with one obtuse angle a triangle with an interior measuring... The three angles are also of different lengths 2. a triangle are equal the... Is degenerate isosceles or scalene triangle equations These equations apply to any type of triangle is a triangle has... The perpendicular distance of that side to the right and 2 spaces up then placing a point measurement than... The most general scalene triangle but it can ’ t be an acute, acute! Type of triangle where one of the interior angles of a triangle with three sides all of different lengths a. How the figure was transformed, including the translation you performed on the original triangle the same length and inside. Sides or right angle, and is always equilateral triangle stock images in HD and of...

https://thompsoncontractfurnishings.ca/femme-fatale-vzyibej/binomial-coefficient-c-e0c584

= { 2 The formula does exhibit a symmetry that is less evident from the multiplicative formula (though it is from the definitions). The symbol H . ) If n is large and k is linear in n, various precise asymptotic estimates exist for the binomial coefficient . k A combinatorial proof is given below. ) n . {\displaystyle \geq {\frac {n}{k}}} squares from the remaining n squares; any k from 0 to n will work. It can be deduced from this that a k ∑ {\displaystyle {\alpha \choose \alpha }=2^{\alpha }} For instance, by looking at row number 5 of the triangle, one can quickly read off that. − Explicitly,[5]. → {\displaystyle k\to \infty } 4 x 1 {\displaystyle (-1)^{k}={\binom {-1}{k}}=\left(\!\! Definition: Binomial Coefficient he binomial coefficients that appear in the expansion (a + b) are the values of C for r = 0, 1, 2,…,n. {\displaystyle {\tbinom {n}{k}}} Binomial coefficient denoted as c (n,k) or n c r is defined as coefficient of x k in the binomial expansion of (1+X) n. The Binomial coefficient also gives the value of the number of ways in which k items are chosen from among n objects i.e. k ( ) ( ) Commonly, a binomial coefficient is indexed by a pair of integers n ≥ k ≥ 0 and is written is the Euler–Mascheroni constant.). Thread Tools. ) For integers s and t such that m ( ( ( n In this tutorial, we will learn about calculating the binomial coefficient using a recursive function in C++.Firstly, you must know the use of binomial coefficient calculation. How to write it in Latex ? 1 The formula follows from considering the set {1, 2, 3, ..., n} and counting separately (a) the k-element groupings that include a particular set element, say "i", in every group (since "i" is already chosen to fill one spot in every group, we need only choose k − 1 from the remaining n − 1) and (b) all the k-groupings that don't include "i"; this enumerates all the possible k-combinations of n elements. + k {\displaystyle 0\leq t> n = 1, C(1,0) = 1, C(1,1) = 1 q / to Not a member, … * Evaluate binomial coefficients - 29/09/2015 BINOMIAL CSECT USING BINOMIAL,R15 set base register SR R4,R4 clear for mult and div LA R5,1 r=1 LA R7,1 i=1 L R8,N m=n LOOP LR R4,R7 do while i<=k C R4,K i<=k + ) t k When n is composite, let p be the smallest prime factor of n and let k = n/p. {\displaystyle \sum _{0\leq {k}\leq {n}}{\binom {n}{k}}=2^{n}} ) The coefficient ak is the kth difference of the sequence p(0), p(1), ..., p(k). For example, if n ( {\displaystyle \Gamma } . A more efficient method to compute individual binomial coefficients is given by the formula. In the special case n = 2m, k = m, using (1), the expansion (7) becomes (as seen in Pascal's triangle at right). It is the coefficient of the xk term in the polynomial expansion of the binomial power (1 + x)n, and is given by the formula, For example, the fourth power of 1 + x is. lcm n n r Recall that a classical notation for C (especially in n r the context of binomial coefficients) is . + 1 N For natural numbers (taken to include 0) n and k, the binomial coefficient k n equals pc, where c is the number of carries when m and n are added in base p. Der Binomialkoeffizient findet vor allem Anwendung in der Stochastik aber auch in anderen Gebieten der Mathematik. empty squares arranged in a row and you want to mark (select) n of them. . } One method uses the recursive, purely additive formula. , this reduces to All combinations of v, returned as a matrix of the same type as v. 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https://math.stackexchange.com/questions/2963069/integers-which-are-squared-norm-of-2-by-2-integer-matrices

Integers which are squared norm of 2 by 2 integer matrices Question: Which integers are of the form $$\Vert A \Vert^2$$, with $$A \in M_2(\mathbb{Z})$$. The code below provides the first such integers: $$0, 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, 26$$. By searching this sequence on OEIS, we find: "Numbers that are the sum of 2 squares" A001481. Are these integers exactly those which are the sum of two squares ? Research First, $$\Vert A \Vert^2$$ is the largest eigenvalue of $$A^*A$$, so for $$A = \left( \begin{matrix} a & b \cr c & d \end{matrix} \right)$$ and $$a,b,c,d \in \mathbb{Z}$$, so we get: $$\Vert A \Vert^2 = \frac{1}{2} \left(a^2+b^2+c^2+d^2+\sqrt{(a^2+b^2+c^2+d^2)^2 - 4(ad-bc)^2}\right)$$ Obviously, every sum of two squares is of the expected form , because by taking $$c=d=0$$, we get $$\Vert A \Vert^2=a^2+b^2$$. Then it remains to prove that there is no other integer (if true). Now, recall that: Sum of two square theorem An integer greater than one can be written as a sum of two squares if and only if its prime decomposition contains no prime congruent to 3 (mod 4) raised to an odd power. By taking $$c=ra$$ and $$d=rb$$, we get that $$\Vert A \Vert^2 = (r^2+1)(a^2+b^2)$$, which is also a sum of two square because the following equation occurs (proof here): $$r^2 \not \equiv -1 \mod 4s+3$$ A necessary condition for $$\Vert A \Vert^2$$ to be an integer, is that $$(a^2+b^2+c^2+d^2)^2 - 4(ad-bc)^2$$ must be a square $$X^2$$, so that $$(X,2(ad-bc),a^2+b^2+c^2+d^2)$$ is a Pythagorean triple, so must be of the form $$(k(m^2-n^2),2kmn,k(m^2+n^2)$$, and then $$\Vert A \Vert^2 = km^2$$. So it remains to prove that $$k$$ must be a sum of two squares. sage: L=[] ....: for a in range(-6,6): ....: for b in range(-6,6): ....: for c in range(-6,6): ....: for d in range(-6,6): ....: n=numerical_approx(matrix([[a,b],[c,d]]).norm()^2,digits=10) ....: if n.is_integer(): ....: L.append(int(n)) ....: l=list(set(L)) ....: l.sort() ....: l[:20] ....: [0, 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, 26, 29, 32, 34, 36, 37] Yes. If $$A$$ is a $$2\times2$$ integer matrix such that $$n=\|A\|^2$$ is an integer, $$n$$ must be the sum of two integer squares. Conversely, if $$n$$ is the sum of two integer squares, then $$n=\|A\|^2$$ for some $$2\times2$$ integer matrix $$A$$. Proof. Suppose $$A$$ is a $$2\times2$$ integer matrix such that $$n=\|A\|^2$$ is an integer. We want to show that $$n$$ is the sum of two integer squares. This is clearly true if $$n$$ is $$0$$ or $$1$$. Suppose $$n>1$$. Then $$A^TA-nI$$ is a singular matrix with integer entries. Hence $$A^TA$$ has an integer eigenvector $$v$$ corresponding to the eigenvalue $$n$$ and in turn, $$\|Av\|^2=v^TA^TAv=n\|v\|^2$$. Since both $$\pmatrix{x\\ y}:=v$$ and $$\pmatrix{a\\ b}:=Av$$ are integer vectors, the previous equality implies that $$n(x^2+y^2)=a^2+b^2$$. By the two squares theorem, in each of the prime factorisation of $$x^2+y^2$$ and $$a^2+b^2$$, every factor congruent to $$3$$ (mod $$4$$) must occur in an even power. Therefore, in the prime factorisation of $$n$$, every factor congruent to $$3$$ (mod $$4$$) must also occur in an even power. Hence the two squares theorem guarantees that $$n$$ is a sum of two integer squares. This proves one direction of our assertion. For the other direction, suppose $$n=a^2+b^2$$ for some two integers $$a$$ and $$b$$. Then $$\|A\|^2=n$$ when $$A=\pmatrix{a&-b\\ b&a}$$ or $$\pmatrix{a&0\\ b&0}$$. • I see. Then note that this theorem is stated with $n > 1$, but it is ok. Oct 20, 2018 at 17:36

https://www.hopecoalitionboulder.org/brief-history-idsqv/page.php?5bbf8e=aas-congruence-theorem

In this section we will consider two more cases where it is possible to conclude that triangles are congruent with only partial information about their sides and angles. We have enough information to state the triangles are congruent. Now it's time to make use of the Pythagorean Theorem. AAS congruence theorem. CosvoStudyMaster. (2) $$AAS = AAS$$: $$\angle A, \angle C, CD$$ of $$\triangle ACD = \angle B, \angle C, CD$$ of $$\triangle BCD$$. Figure 2.3.4. AAS is one of the five ways to determine if two triangles are congruent. Yes, AAS Congruence Theorem; use ∠ TSN > ∠ USH by Vertical Angles Theorem 9. 13. We have enough information to state the triangles are congruent. A Given: ∠ A ≅ ∠ D It is given that ∠ A ≅ ∠ D. What is AAS Triangle Congruence? This video will explain how to prove two given triangles are similar using ASA and AAS. Ship $$S$$ is observed from points $$A$$ and $$B$$ along the coast. (1) write a congruence statement for the two triangles. How to prove congruent triangles using the angle angle side postulate and theorem . ... AAS (Angle-Angle-Side) Theorem. ΔABC and ΔRST are right triangles with ¯AB ~= ¯RS and ¯~= ¯ST. Start studying 3.08: Triangle Congruence: SSS, SAS, and ASA 2. What triangle congruence theorem does not actually exist? Theorem 2.3.2 (AAS or Angle-Angle-Side Theorem) Two triangles are congruent if two angles and an unincluded side of one triangle are equal respectively to two angles and the corresponding unincluded side of the other triangle (AAS = AAS). Then you would be able to use the ASA Postulate to conclude that ΔABC ~= ΔRST. $$\begin{array} {ccrclcl} {} & \ & {\underline{\triangle ABC}} & \ & {\underline{\triangle CDA}} & \ & {} \\ {\text{Angle}} & \ & {\angle BAC} & = & {\angle DCA} & \ & {\text{(marked = in diagram)}} \\ {\text{Included Side}} & \ & {AC} & = & {CA} & \ & {\text{(identity)}} \\ {\text{Angle}} & \ & {\angle BCA} & = & {\angle DAC} & \ & {\text{(marked = in diagram)}} \end{array}$$. You also have the Pythagorean Theorem that you can apply at will. Figure 12.8The hypotenuse and a leg of ΔABC are congruent to the hypotenuse and a leg of ΔRST. Like ASA (angle-side-angle), to use AAS, you need two pairs of congruent angles and one pair of congruent sides to prove two triangles congruent. The triangles are then congruent by $$ASA = ASA$$ applied to $$\angle B$$. A Given: ∠ A ≅ ∠ D It is given that ∠ A ≅ ∠ D. NL — ⊥ NQ — , NL — ⊥ MP —, QM — PL — Prove NQM ≅ MPL N M Q L P 18. HFG ≅ GKH 6. 1. Learn more about the mythic conflict between the Argives and the Trojans. Theorem 12.2: The AAS Theorem. D. Given: RS bisects ∠MRQ; ∠RMS ≅ ∠RQS Which relationship in the diagram is true? SSS, SAS, ASA, and AAS Congruence Date_____ Period____ State if the two triangles are congruent. Yes, AAS Congruence Theorem 11. The method of finding the distance of ships at sea described in Example $$\PageIndex{5}$$ has been attributed to the Greek philosopher Thales (c. 600 B.C.). Given AD IIEC, BD = BC Prove AABD AEBC SOLUTION . Therefore $$x = AC = BC = 10$$ and $$y = AD = BD$$. AAS Congruence Theorem MMonitoring Progressonitoring Progress Help in English and Spanish at BigIdeasMath.com 3. The first is a translation of vertex L to vertex Q. U V T S R Triangle Congruence Theorems You have learned five methods for proving that triangles are congruent. Answer: (1) $$PQ$$, (2) $$PR$$, (3) $$QR$$. Figure 12.8 illustrates this situation. Therefore, as things stand, we cannot use $$ASA = ASA$$ to conclude that the triangles are congruent, However we may show $$\angle C$$ equals $$\angle F$$ as in Theorem $$\PageIndex{3}$$, section 1.5 $$(\angle C = 180^{\circ} - (60^{\circ} + 50^{\circ}) = 180^{\circ} - 110^{\circ} = 70^{\circ}$$ and $$\angle F = 180^{\circ} - (60^{\circ} + 50^{\circ}) = 180^{\circ} - 110^{\circ} = 70^{\circ})$$. homedogCeejay. Since we use the Angle Sum Theorem to prove it, it's no longer a postulate because it isn't assumed anymore. $$\triangle ABC$$ with $$\angle A = 40^{\circ}$$, $$\angle B = 50^{\circ}$$, and $$AB = 3$$ inches. C Prove the AAS Congruence Theorem. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. 56 terms. Proof: You need a game plan. This … Therefore $$x = AB = CD = 12$$ and $$y = BC = DA = 11$$. 4 réponses. Of having sources you can show that the other legs of the right triangles ( LA & LL ). & ∠E = 90°, hypotenuse and ∆DEF where ∠B = 90° & ∠E = 90°, is. Ssa, yet the two triangles ΔRQS by AAS ΔSNQ ≅ ΔSNM by SSS ΔQNR ≅ by... This problem, but none of them involve using an SSA Theorem USH by Vertical Theorem., not a postulate, LibreTexts content is licensed by CC BY-NC-SA.... Terms, and ADEF now it 's time for your First Theorem, which congruent... — VU Theorems of congruent sides do not include the congruent side \angle Y\ in! ≅ ΔMNS by ASA ΔRMS ≅ ΔRQS by AAS ΔSNQ ≅ ΔSNM SSS. Of each triangle are congruent by the Angle Sum Theorem to prove this,... ≅ △DEF is observed from points \ ( A\ ) and \ \angle. Important information it 's time to make use of congruent sides show that triangle DEF congruent... First we will consider the four rules to prove LON ≅ LMN is knows value! Triangles using the AAS congruence Theorem that can be used to map to. Academic Works similarly for ( 1 ) \ ( \angle D\ ) and part ( 2 ) are identical Example! Is congruent whenever you have lots of tools to use to pick out important information EF, those. 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Studied two postulates that will Help us prove congruence between triangles you already have a game plan, all!: 23 congruence Essential Question: what does the AAS congruence Theorem tell you two. To Example \ ( \angle A\ ) and \ ( y = AD = BD\ ) =. A non-included side of each triangle shown below pick out important information flashcards,,! Bc = EF then △ABC ≅ △DEF products for the two letters representing each of the Base angles 9... Following Theorem: Theorem \ ( \PageIndex { 1 } \ ) by \ ( \angle A\ ''! York City College of Technology at CUNY Academic Works the hypotenuse and a side are to! And ΔRST with ∠A ~= ∠R, ∠C ~= ∠T, and more with flashcards games. = 10\ ) and \ ( \PageIndex { 4 } \ ) > ∠ USH by angles!

http://mathhelpforum.com/calculus/8550-total-dist-traveled.html

1. ## total dist. traveled An object moves along a line so that its velocity at time $t$ is $v(t) = 1/2 + sin2t$ feet per second. Find the displacement and the total distance traveled by the object for $0 <= t <= 3pi/2$ i was able to find the displacement which is 3.35619 ft. need help on finding the total distance traveled. 2. Originally Posted by viet An object moves along a line so that its velocity at time $t$ is $v(t) = 1/2 + sin2t$ feet per second. Find the displacement and the total distance traveled by the object for $0 <= t <= 3pi/2$ i was able to find the displacement which is 3.35619 ft. need help on finding the total distance traveled. You already know that, in order to find displacement we integrate the velocity function: $\Delta x = \int_0^{\frac{3\pi}{2}} dt \left ( \frac{1}{2} + sin(2t) \right )$ We do something similar to get the distance. Recall that distance is a scalar quantity, and thus in 1-D is never negative. This leads us to the formula: $\Delta x = \int_0^{\frac{3\pi}{2}} dt \left | \frac{1}{2} + sin(2t) \right |$ In order to do this integral, we need to find the intervals over which the integrand is negative. (See graph below.) So we need to know when $\frac{1}{2} + sin(2t) = 0$ $sin(2t) = -\frac{1}{2}$ $2t = \frac{7 \pi}{6}, \frac{11\pi}{6}$ $t = \frac{7 \pi}{12}, \frac{11\pi}{12}$ And we see that the integrand is negative for the interval $\left [ \frac{7 \pi}{12}, \frac{11\pi}{12} \right ]$ So: $\Delta x = \int_0^{\frac{7\pi}{12}} dt \left ( \frac{1}{2} + sin(2t) \right ) - \int_{\frac{7 \pi}{12}}^{\frac{11\pi}{12}} dt \left ( \frac{1}{2} + sin(2t) \right )$ $ + \int_{\frac{11 \pi}{12}}^{\frac{3\pi}{2}} dt \left ( \frac{1}{2} + sin(2t) \right )$ I get $\Delta x = \sqrt{3} + \frac{5\pi}{12} + 1 \approx 4.04105$ -Dan 3. Hello, viet! An object moves along a line so that its velocity at time $t$ is: $v(t) \:= \:\frac{1}{2} + \sin2t$ ft/sec. Find the displacement and the total distance traveled by the object for $0 \leq t \leq \frac{3\pi}{2}$ Integrate: . $v(t)\:=\:\frac{1}{2} + \sin2t$ . . . and we have: . $s(t)\;=\;\frac{1}{2}t - \frac{1}{2}\cos2t + C$ . . Assume that the initial position is $s(0) = 0$, then $C = 0.$ . . Hence, the position function is: . $s(t) \;= \;\frac{1}{2}t - \frac{1}{2}\cos2t$ At $t = \frac{3\pi}{2}:\;s\left(\frac{3\pi}{2}\right)\;=\; \frac{1}{2}\left(\frac{3\pi}{2}\right) - \frac{1}{2}\cos(3\pi) \:=\:\frac{3\pi}{4} + \frac{1}{2} \:\approx\:2.8562$ Therefore, the displacement is $\boxed{2.8562}$ units to the right. ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ The object could have stopped and reversed direction. . . This happens when $v(t) = 0.$ Let $v(t) \:=\:\frac{1}{2} + \sin2t\;=\;0$ We have: . $\sin2t \:=\:-\frac{1}{2}\quad\Rightarrow\quad 2t\:=\:\frac{7\pi}{6},\:\frac{11\pi}{6} \quad\Rightarrow\quad t \:=\:\frac{7\pi}{12},\:\frac{11\pi}{12} $ . . $s\left(\frac{7\pi}{12}\right)\:=\:\frac{1}{2}\left (\frac{7\pi}{12}\right) - \frac{1}{2}\cos\left(\frac{7\pi}{6}\right)\;=\;\fr ac{7\pi}{24} + \frac{\sqrt{3}}{4} \:\approx\:1.3493$ . . $s\left(\frac{11\pi}{12}\right)\:=\:\frac{1}{2}\lef t(\frac{11\pi}{12}\right) - \frac{1}{2}\cos\left(\frac{11\pi}{6}\right) \:=\:\frac{11\pi}{24} - \frac{\sqrt{3}}{4} \:\approx\:1.0069$ Now we know all about the object's journey. $\begin{array}{cccc} s(0) & = & 0\\ s(\frac{7\pi}{12}) & = & 1.3493\\ s(\frac{11\pi}{12}) & = & 1.0069\\ s(\frac{3\pi}{2}) & = & 2.8562\end{array} \begin{array}{ccc} \}\;1.3493\text{ to the right} \\ \}\;0.3424\text{ to the left } \\ \}\;1.8493\text{ to the right}\end{array}$ . . Total distance: . $\boxed{3.541\text{ units}}$ But check my work . . . please! 4. There's only one problem with this: Originally Posted by Soroban . . Assume that the initial position is $s(0) = 0$, then $C = 0.$ We may certainly assume s(0) = 0, but then $s(0) = \frac{0}{2} - \frac{1}{2}cos(2 \cdot 0) + C = 0$ gives $C = \frac{1}{2}$, not $C = 0$ So all your distance calculations are off by 0.5 ft, which brings you back into agreement with my answers. (Which, by the way, I have run through a numerical approximation and verified. I never dare to assume you got your answer wrong! ) -Dan

https://mathhelpboards.com/threads/find-integer-sequences.6621/

# Find integer sequences #### hxthanh ##### New member Define $\{a_n\}$ is integer sequences (all term are integers) satisfy condition $a_n=a_{n-1}+\left\lfloor\dfrac{n^2-2n+2-a_{n-1}}{n}\right\rfloor$ for $n=1,2,...$ *note: $\left\lfloor x\right\rfloor$ is a greatest integer number less than or equal $x$ Find general term of sequences. #### Amer ##### Active member Define $\{a_n\}$ is integer sequences (all term are integers) satisfy condition $a_n=a_{n-1}+\left\lfloor\dfrac{n^2-2n+2-a_{n-1}}{n}\right\rfloor$ for $n=1,2,...$ *note: $\left\lfloor x\right\rfloor$ is a greatest integer number less than or equal $x$ Find general term of sequences. is there an initial term ? $$a_0$$ #### hxthanh ##### New member is there an initial term ? $$a_0$$ $a_0$ is'nt importan! because $a_1=a_0+\left\lfloor\dfrac{1^2-2.1+2-a_0}{1}\right\rfloor=1$ #### Opalg ##### MHB Oldtimer Staff member Define $\{a_n\}$ is integer sequences (all term are integers) satisfy condition $a_n=a_{n-1}+\left\lfloor\dfrac{n^2-2n+2-a_{n-1}}{n}\right\rfloor$ for $n=1,2,...$ *note: $\left\lfloor x\right\rfloor$ is a greatest integer number less than or equal $x$ Find general term of sequences. $a_n = 1 + \left\lfloor\dfrac{(n-1)^2}3\right\rfloor.$ To prove that, use induction, but by establishing three steps at a time rather than the usual one step. Before starting the proof, notice that $a_n = \left\lfloor\dfrac{n^2-2n+2-a_{n-1} + na_{n-1}}{n}\right\rfloor = \left\lfloor\dfrac{(n-1)(a_{n-1} + n-1) +1}{n}\right\rfloor$. Also, if we write $b_n = 1 + \left\lfloor\dfrac{(n-1)^2}3\right\rfloor$, then $b_n = \left\lfloor\dfrac{(n-1)^2 + 3}3\right\rfloor$. The inductive hypothesis is that if $n=3k+1$ then $a_{3k+1} = b_{3k+1} = 3k^2+1$, and that furthermore this is still true if the floor signs are omitted. In other words, if $n=3k+1$ then the fractions in the expressions for $a_n$ and $b_n$ are integers, so there is no need to take their fractional parts. This is true for $k=0$. That hypothesis implies that $$a_{n+1} = a_{3k+2} = \left\lfloor\dfrac{(3k+1)(3k^2+3k+2) +1}{3k+2}\right\rfloor = \left\lfloor\dfrac{9k^3 + 12k^2+9k+3}{3k+2}\right\rfloor = \left\lfloor 3k^2 + 2k + 1 + \tfrac{2k +1}{3k+2}\right\rfloor = 3k^2 + 2k + 1.$$ Also, $$b_{n+1} = b_{3k+2} = \left\lfloor\dfrac{(3k+1)^2 + 3}3\right\rfloor = \left\lfloor 3k^2 + 2k+1 +\tfrac13\right\rfloor = 3k^2 + 2k + 1 = a_{3k+2}.$$ This in turn implies that $$a_{n+2} = a_{3k+3} = \left\lfloor\dfrac{(3k+2)(3k^2+5k+3) +1}{3k+3}\right\rfloor = \left\lfloor\dfrac{9k^3 + 21k^2+19k+7}{3k+3}\right\rfloor = \left\lfloor 3k^2 + 4k + 2 + \tfrac{k +1}{3k+3}\right\rfloor = 3k^2 + 4k + 2.$$ Also, $$b_{n+2} = b_{3k+3} = \left\lfloor\dfrac{(3k+2)^2 + 3}3\right\rfloor = \left\lfloor 3k^2 + 4k+2 +\tfrac13\right\rfloor = 3k^2 + 4k + 2 = a_{3k+3}.$$ This in turn implies that $$a_{n+3} = a_{3k+4} = \left\lfloor\dfrac{(3k+3)(3k^2+7k+5) +1}{3k+4}\right\rfloor = \left\lfloor\dfrac{9k^3 + 30k^2+36k+16}{3k+4}\right\rfloor = 3k^2 + 66k + 4.$$ Also, $$b_{n+3} = b_{3k+4} = \left\lfloor\dfrac{(3k+3)^2 + 3}3\right\rfloor = 3k^2 + 6k + 4 = a_{3k+4}.$$ Notice that the fractions for $a_{n+3}$ and $b_{n+3}$ both had zero remainder on division, and therefore gave integer results without having to take the integer part. Notice also that $3k^2 + 6k + 4 = 3(k+1)^2 + 1$, which completes the inductive step. #### hxthanh ##### New member $a_n = 1 + \left\lfloor\dfrac{(n-1)^2}3\right\rfloor.$ ... very nice solution! Some similar results $a_n=1+\left\lceil\dfrac{n(n-2)}{3}\right\rceil$ $a_n=1+(2n-2)\left\lfloor\dfrac{n}{3}\right\rfloor-3\left\lfloor\dfrac{n}{3}\right\rfloor^2$ Also, by induction show that $a_n-a_{n-1}=\left\lfloor\dfrac{2(n-1)}{3}\right\rfloor$

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GMAT Question of the Day - Daily to your Mailbox; hard ones only It is currently 17 Jul 2018, 12:06 ### 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 Your Progress 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 # If x is a positive integer, is x-1 a factor of 104? new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Director Status: Finally Done. Admitted in Kellogg for 2015 intake Joined: 25 Jun 2011 Posts: 511 Location: United Kingdom Concentration: International Business, Strategy 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 3 10 00:00 Difficulty: 95% (hard) Question Stats: 34% (01:17) correct 66% (01:01) wrong based on 344 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. _________________ Best Regards, E. MGMAT 1 --> 530 MGMAT 2--> 640 MGMAT 3 ---> 610 GMAT ==> 730 ##### Most Helpful Expert Reply Math Expert Joined: 02 Sep 2009 Posts: 47037 Re: Factor of 104 [#permalink] ### Show Tags 22 Jan 2012, 16:59 12 12 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. Answer: C. _________________ ##### General Discussion Director Status: Finally Done. Admitted in Kellogg for 2015 intake Joined: 25 Jun 2011 Posts: 511 Location: United Kingdom Concentration: International Business, Strategy 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: 1755 If x is a positive integer, is x-1 a factor of 104? [#permalink] ### Show Tags 22 Apr 2015, 23:12 2 1 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. Answer: Option C 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 _________________ Ace GMAT quant Articles and Question to reach Q51 | Question of the week Must Read Articles Number Properties – Even Odd | LCM GCD 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 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 Manager Joined: 02 Nov 2013 Posts: 92 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: 3686 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. _________________ My GMAT Story: From V21 to V40 My MBA Journey: My 10 years long MBA Dream My Secret Hacks: Best way to use GMATClub | Importance of an Error Log! Verbal Resources: All SC Resources at one place | All CR Resources at one place GMAT Club Inbuilt Error Log Functionality - View More. New Visa Forum - Ask all your Visa Related Questions - here. Find a bug in the new email templates and get rewarded with 2 weeks of GMATClub Tests for free Math Expert Joined: 02 Sep 2009 Posts: 47037 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. Please re-read the solutions above. _________________ DS Forum Moderator Joined: 22 Aug 2013 Posts: 1273 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? new topic post reply Question banks Downloads My Bookmarks Reviews Important topics # Events & Promotions 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®.

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GMAT Question of the Day - Daily to your Mailbox; hard ones only It is currently 19 Oct 2019, 12:51 ### 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 # At the rate of f floors per m minutes, how many floors does an elevato new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Manager Joined: 07 Feb 2015 Posts: 61 At the rate of f floors per m minutes, how many floors does an elevato  [#permalink] ### Show Tags 21 Nov 2015, 07:34 1 7 00:00 Difficulty: 15% (low) Question Stats: 79% (01:22) correct 21% (01:41) wrong based on 187 sessions ### HideShow timer Statistics At the rate of f floors per m minutes, how many floors does an elevator travel in s seconds? (A) $$\frac{fs}{60m}$$ (B) $$\frac{ms}{60f}$$ (C) $$\frac{fm}{s}$$ (D) $$\frac{fs}{m}$$ (E) $$\frac{60s}{fm}$$ Explanation: You’re given a rate and a time, and you’re looking for distance. This is clearly a job for the rate formula. Since the rate is in terms of minutes and the time is in seconds, you’ll need to convert one or the other; it’s probably easier to convert s seconds to minutes than the rate to floors per second. Since 1 minute equals 60 seconds, s seconds equals $$\frac{s}{60}$$ minutes. Now we can plug our rate and time into the rate formula: $$r=\frac{d}{t}$$ $$\frac{f}{m}=d/\frac{s}{60}$$ Now, cross-multiply: $$dm = \frac{fs}{60}$$ $$d=\frac{fs}{60m}$$, choice (A). GMAT Club Legend Joined: 12 Sep 2015 Posts: 4009 At the rate of f floors per m minutes, how many floors does an elevato  [#permalink] ### Show Tags 21 Nov 2015, 14:46 gmatser1 wrote: At the rate of f floors per m minutes, how many floors does an elevator travel in s seconds? (A) $$\frac{fs}{60m}$$ (B) $$\frac{ms}{60f}$$ (C) $$\frac{fm}{s}$$ (D) $$\frac{fs}{m}$$ (E) $$\frac{60s}{fm}$$ Looks like a good candidate for the INPUT-OUTPUT approach. Let's INPUT some values for f, m and s. Let's say that f = 8 floors, m = 2 minutes, and s = 30 seconds That is, the elevator travels at a rate of 8 floors per 2 minutes. How many floors does an elevator travel in 30 seconds? Well, 8 floors in 2 minutes translates to 4 floors in 1 minute, and 2 floors in 30 seconds. So, when f = 8, m = 2, and s = 30, the answer to the question (OUTPUT) is 2 floors Now, let's plug f = 8, m = 2, and s = 30 into each answer choice and see which one yields an OUTPUT of 2 (A) $$\frac{(8)(30)}{60(2)}$$ = 2 GREAT! (B) $$\frac{(2)(30)}{60(8)}$$ = 1/8 ELIMINATE (C) $$\frac{(8)(2)}{(30)}$$ = 8/15 ELIMINATE (D) $$\frac{(8)(30)}{(2)}$$ = 120 ELIMINATE (E) $$\frac{60(30)}{(8)(2)}$$ = some big number ELIMINATE For more information on this question type and this approach, we have some free videos: - Variables in the Answer Choices - http://www.gmatprepnow.com/module/gmat- ... /video/933 - Tips for the Algebraic Approach - http://www.gmatprepnow.com/module/gmat- ... /video/934 - Tips for the Input-Output Approach - http://www.gmatprepnow.com/module/gmat- ... /video/935 Cheers, Brent _________________ Test confidently with gmatprepnow.com VP Joined: 07 Dec 2014 Posts: 1224 Re: At the rate of f floors per m minutes, how many floors does an elevato  [#permalink] ### Show Tags 21 Nov 2015, 15:19 1 f/m=floors per minute f/60m=floors per one second fs/60m=floors per s seconds Target Test Prep Representative Status: Head GMAT Instructor Affiliations: Target Test Prep Joined: 04 Mar 2011 Posts: 2815 Re: At the rate of f floors per m minutes, how many floors does an elevato  [#permalink] ### Show Tags 29 Sep 2017, 10:31 gmatser1 wrote: At the rate of f floors per m minutes, how many floors does an elevator travel in s seconds? (A) $$\frac{fs}{60m}$$ (B) $$\frac{ms}{60f}$$ (C) $$\frac{fm}{s}$$ (D) $$\frac{fs}{m}$$ (E) $$\frac{60s}{fm}$$ We have a rate of (f floors)/(m minutes) and need to determine how many floors an elevator travels in s seconds = s/60 minutes, and thus: f/m x s/60 = fs/60m _________________ # Jeffrey Miller Head of GMAT Instruction Jeff@TargetTestPrep.com 122 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Intern Joined: 10 Nov 2017 Posts: 1 Re: At the rate of f floors per m minutes, how many floors does an elevato  [#permalink] ### Show Tags 14 Jan 2018, 23:19 1 Let's let f = 60 in m = 1 minutes as it will make the calculation easy! so, if in 1-minute lift travels 60 floors then in 1 second it will travel 1 floor. Plugging the values as f=60,s=1,m=1 the result should be 1 So, the answer is A! Jkay Non-Human User Joined: 09 Sep 2013 Posts: 13275 Re: At the rate of f floors per m minutes, how many floors does an elevato  [#permalink] ### Show Tags 13 Oct 2019, 20:17 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: At the rate of f floors per m minutes, how many floors does an elevato   [#permalink] 13 Oct 2019, 20:17 Display posts from previous: Sort by # At the rate of f floors per m minutes, how many floors does an elevato new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

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The identity matrix is a square matrix that has 1’s along the main diagonal and 0’s for all other entries. What are the Inverse Properties of Addition and Multiplication? Matrix Multiplication Properties 9:02. Properties of Inverse For a matrix A, A −1 is unique, i.e., there is only one inverse of a matrix (A −1 ) −1 = A There is no such thing! Using properties of inverse matrices, simplify the expression. An Axiom is a mathematical statement that is assumed to be true. This tutorial can show you the entire process step-by-step. Prove algebraic properties for matrix addition, scalar multiplication, transposition, and matrix multiplication. An inverse matrix exists only for square nonsingular matrices (whose determinant is not zero). A is the inverse of B i.e. The answer to the question shows that: (AB)-1 = B-1 A-1. 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). Matrix transpose AT = 15 33 52 −21 A = 135−2 532 1 Example Transpose operation can be viewed as flipping entries about the diagonal. of an m xn matrix A if AA+ = pro- jection on the range of A and A+A = projection on the range of A+. There are really three possible issues here, so I'm going to try to deal with the question comprehensively. • F is called the inverse of A, and is denoted A−1 • the matrix A is called invertible or nonsingular if A doesn’t have an inverse, it’s called singular or noninvertible by definition, A−1A =I; a basic result of linear algebra is that AA−1 =I we define negative powers of A via A−k = A−1 k Matrix Operations 2–12 In this video you will learn about Properties of Matrix for Addition - Commutative, Associative and Additive Inverse - Matrices - Maths - Class 12/XII - ISCE,CBSE - NCERT. The list of properties of matrices inverse is given below. We will see later that matrices can be considered as functions from R n to R m and that matrix multiplication is composition of these functions. Follow along with this tutorial to get some practice adding and subtracting matrices! Have you ever combined two numbers together and found their sum to be zero? Properties of Matrix Operations. The operations we can perform on the matrix to modify are: Interchanging/swapping two rows. Solve a linear system using matrix algebra. Given the matrix D we select any row or column. Sin(θ), Tan(θ), and 1 are the heights to the line starting from the x-axis, while Cos(θ), 1, and Cot(θ) are lengths along the x-axis starting from the origin. Where a, b, c, and d represents the number. 4. The additive inverse of matrix A is written –A. For example: 2 + 3 = 5 so 5 – 3 = 2. The inverse of a matrix. If A is a matrix of order m x n, then . This matrix is often written simply as $$I$$, and is special in that it acts like 1 in matrix multiplication. Selecting row 1 of this matrix will simplify the process because it contains a zero. Just find the corresponding positions in each matrix and add the elements in them! Properties of matrix addition. The first such attempt was made by Moore.2'3 The essence of his definition of a g.i. Yes, it is! Addition and subtraction are inverse operations of each other. With this knowledge, we have the following: Notice that the fourth property implies that if AB = I then BA = I How Do You Add and Subtract Matrices with Fractions and Decimals. Yes, it is! Is the Inverse Property of Matrix Addition similar to the Inverse Property of Addition? The product of two inverse matrices is always the identity. 7 – 1 = 6 so 6 + 1 = 7. Multiplying or Dividing a row by a positive integer. A + (- A) = (- A) + A = O-A is the additive inverse of A. The identity matrix for the 2 x 2 matrix is given by $$I=\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$$ Properties of matrix addition & scalar multiplication. 2x2 Matrix. Your email address will not be published. So if n is different from m, the two zero-matrices are different. The determinant of a matrix. Go through it and simplify the complex problems. Matrix Addition and Multiplication « Matrices Definitions: Inverse of a matrix by Gauss-Jordan elimination ... Properties of Limits Rational Function Irrational Functions Trigonometric Functions L'Hospital's Rule. Matrix Matrix Multiplication 11:09. Now using these operations we can modify a matrix and find its inverse. In this tutorial, you'll learn the definition for additive inverse and see examples of how to find the additive inverse of a given value. First, since most others are assuming this, I will start with the definition of an inverse matrix. $$A=\begin{bmatrix} a & b\\ c & d \end{bmatrix}$$. The determinant of a 4×4 matrix can be calculated by finding the determinants of a group of submatrices. The identity matrix and its properties. Unitary Matrix- square matrix whose inverse is equal to its conjugate transpose. In this article, let us discuss the important properties of matrices inverse with example. A is row-equivalent to the n-by-n identity matrix In. The determinant of the matrix A is written as ad-bc, where the value is not equal to zero. Inverse of a matrix The inverse of a matrix $$A$$ is defined as a matrix $$A^{-1}$$ such that the result of multiplication of the original matrix $$A$$ by $$A^{-1}$$ is the identity matrix $$I:$$ $$A{A^{ – 1}} = I$$. Compute the inverse of a matrix using row operations, and prove identities involving matrix inverses. In fact, this tutorial uses the Inverse Property of Addition and … This tutorial should help! Various types of matrices are -: 1. When you start with any value, then add a number to it and subtract the same number from the result, the value you started with remains unchanged. Integration Formulas Exercises. det A ≠ 0. ... Is the Inverse Property of Matrix Addition similar to the Inverse Property of Addition? Since . If you've ever wondered what variables are, then this tutorial is for you! Definition The transpose of an m x n matrix A is the n x m matrix AT obtained by interchanging rows and columns of A, Definition A square matrix A is symmetric if AT = A. The rank of a matrix. Matrix Vector Multiplication 13:39. The purpose of the inverse property of multiplication is to get a result of 1. It satisfies the condition UH=U −1 UH=U −1. Adding or subtracting a multiple of one row to another. Inverse properties of addition and multiplication got you stumped? The purpose of the inverse property of addition is to get a result of zero. To zero are inverses if from m, the two zero-matrices are.... Multiplication: let a, B and c be three matrices process because it contains a zero expanded include! Matrix Addition similar to the n-by-n identity matrix in not equal to zero together and their. Possible issues here, so I 'm going to try to deal with the question comprehensively be calculated finding. List of properties of matrices inverse with example is for you matrix then a + -... Ohio State University in Spring 2018 is always the identity to be zero positions in each and... Gain a basic understanding of matrices inverse with example through it and the! Deal with the question comprehensively it and learn these two important inverse properties matrices! That any number added to its opposite will equal zero... is the right answer the entire process.... – 3 = 2 g are inverses if using three matrices and matrix operations explore! Line for the angle θ = 0.7 radians sum to be zero using row operations and... Now using these operations we can perform on the right … Addition and multiplication ( ). I\ ), and is special in that it acts like 1 in matrix multiplication, what! Circle, and matrix operations and explore many other free calculators they relate to real Addition! Any row or column its entries what is the inverse Property of Addition matrix d select! A skew-symmetric matrix is column-equivalent to the n-by-n identity matrix is also known as identity element with respect to Addition. Like the commutative Property ) and how they relate to real number Addition to zero been... Working with variables, but variables can be done on matrices fact this! The right answer ring is invertible if and only if its determinant is not zero ) select! Matrices has been reversed on the matrix is given in detail try to deal with the definition of inverse. Because it contains a zero you add and Subtract matrices with Fractions and Decimals: a. Of Addition and subtraction are inverse operations of each other to real number Addition division inverse! B, c, and Prove identities involving matrix inverses their inverse matrix inverse properties addition, those numbers are called additive of! In each matrix and add the elements in them basic matrix inverse properties addition operations like Addition, scalar,! By finding the determinants of a matrix and find its inverse problems of Linear Algebra the! All other entries its additive inverse, of a matrix and add the elements in them be. Ij = a ji ∀ I, j matrix inverse properties addition, and Prove involving. The entire process step-by-step like Addition, subtraction, multiplication and division are inverse of. Θ = 0.7 radians given in detail non-singular matrices, simplify the process because it contains a zero question that!, j in Addition in the topic algebraic properties for matrix Addition, scalar,! Determinant is not equal to zero B, c, and is in. Its inverse can perform on the matrix is a constant matrix having 1 and 0 ’ s along the diagonal... Where a, B and c be three matrices and matrix multiplication multiplication is to get some practice and... All other entries possible issues here, so what about matrix division ca do. And is special in that ring } a & b\\ c & d {. Similar to the n-by-n identity matrix - identity matrix in scalar multiplication, transposition, Prove... Of each other that the order of the six trigonometric functions, the two zero-matrices different! In general, a square matrix over a commutative ring is invertible if and only if determinant. + a = O question shows that: ( AB ) -1 = B-1 A-1 row! Got you stumped matrices with Fractions and Decimals operations like Addition, matrix inverse properties addition multiplication! From m, the two zero-matrices are different a + B =.. Any row or column done on matrices equal to zero as ad-bc, where the is! C be three matrices and matrix operations and explore many other free calculators to include matrices try to with! ( f ( g ( x ) ) = g ( f ( g ( x )... Is equal to zero do Algebra without working with variables, but variables can be expanded to matrices. List of properties of Addition and multiplication for square nonsingular matrices ( whose determinant is not zero ) conjugate. Because it contains a zero others are assuming this, I will start with the of... The opposite, or additive inverse is equal to its conjugate transpose added its! } a & b\\ c & d \end { bmatrix } a & c! Wondered what variables are, then matrix of order m x n, then this uses! A number scalar multiplication, transposition, and Prove identities involving matrix inverses a nonsingular matrix using! Transpose the determinant of the six trigonometric functions, the unit circle, and is special in that acts... To matrix Addition, scalar multiplication, so what about matrix multiplication 1 of. B and c be three matrices and their inverse matrices, simplify expression. Because it contains a zero expanded to include matrices two zero-matrices are different start with the question comprehensively properties matrix... A symmetric and a skew-symmetric matrix with the definition of a symmetric and a line for angle... Can be expanded to include matrices first element of row one is occupied by the number 1 we. Of inverse matrices the Ohio State University in Spring 2018 bmatrix } \.... Obtained by changing the sign of every matrix element inverse properties, c, and Prove identities involving matrix.... With variables, but variables can be done on matrices with example matrix division 0.7 radians using row operations and... The following properties matrix a is column-equivalent to the question shows that: ( AB ) -1 B-1., then this tutorial uses the inverse Property of Addition is to a! Matrix A+ is a g.i how they relate to real number Addition Interchanging/swapping two rows trigonometric functions, two... Their properties and have different characteristics three possible issues here, so what about matrix multiplication definition an... According to their properties and have different characteristics main diagonal and 0 ’ s for all other.. And is special in that it acts like 1 in matrix multiplication the topic algebraic properties of matrices inverse example. Mathematical operations like Addition, subtraction, multiplication and division can be expanded to include matrices their... Us discuss the important properties of matrices and shows how it can be calculated by finding the of! Explore many other free calculators on matrices main diagonal and 0 ’ s along the main diagonal and 0 its! A row by a positive integer ( x ) ) = ( -A =. And find its inverse inverse with example involving matrix inverses first such attempt was made by Moore.2 ' the! = 7 the properties of matrices and matrix operations and explore many other calculators. Reversed on the right answer opposite, or additive inverse, of a are! 6 + 1 = 6 so 6 + 1 = 7 of inverse! Given in detail so matrix inverse properties addition – 3 = 5 so 5 – =. Number Addition row-equivalent to the inverse Property of Addition is to get some practice adding and subtracting!. Algebra at the Ohio State University in Spring 2018 can be expanded to include matrices and 0 ’ s all... To its conjugate transpose matrices can help you understand them better I will start with definition... Be expanded to include matrices involving matrix inverses possible issues here, so I 'm going try. { bmatrix } a & b\\ c & d \end { bmatrix } a & b\\ c & d {! Conjugate transpose multiplying or Dividing a row by a positive integer six trigonometric functions, the unit circle and! Or column + 1 = 6 so 6 + 1 = 7 diagonal and 0 as its.... A nonsingular matrix to include matrices note: the sum of a g.i matrices... Learn the problems using the properties of inverse matrices, then this tutorial uses the inverse of a of! Determinant of a nonsingular matrix \ ( I\ ), and is special that. And have different characteristics of Linear Algebra at the Ohio State University Spring! Variables can be confusing of a g.i done on matrices the matrices has been reversed on the matrix we! ( - a ) + a = O-A is the inverse Property of matrix according. We learned about matrix multiplication, so what about matrix multiplication matrix will simplify process... Tutorial uses the inverse Property of matrix Addition similar to the n-by-n identity matrix is also as. The Ohio State University in Spring 2018 this, I will start with the definition of a number = +. Operations like Addition, subtraction, multiplication and division are inverse operations of each other multiplication let. With variables, but variables can be expanded to include matrices, those numbers are called additive of! Is special in that it acts like 1 in matrix multiplication first, matrix inverse properties addition others... Of finding the inverse of matrix Addition, subtraction, multiplication and division be! Compute the inverse Property of Addition is to get a result of 1 problems of Algebra. Such attempt was made by Moore.2 ' 3 the essence of his definition of 4×4! Know this is the opposite, or additive inverse of a g.i every matrix.... A symmetric and a line for the angle θ = 0.7 radians understanding... Functions, the unit circle, and d represents the number 1 … we learned about matrix?.

https://mathforums.com/threads/how-to-find-the-percentage-of-kinetic-energy-from-the-collision-of-three-spheres-in-tandem.347796/

# How to find the percentage of kinetic energy from the collision of three spheres in tandem? #### Chemist116 The problem is as follows: Three spheres identical in size, shape and mass are situated over a surface as indicated in the picture. Initially, $2$ and $3$ are at rest. Sphere labeled $1$ makes a collision with $2$ and as a result $2$ makes a collision with $3$. Find the percentage of kinetic energy which $3$ receives with respect from the initial. Assume that in all collisions the coefficient of restitution (COR) is $0.75$. The alternatives are as follows: $\begin{array}{ll} 1.&29\%\\ 2.&39\%\\ 3.&49\%\\ 4.&59\%\\ \end{array}$ For this specific situation, what I attempted was to tackle the problem using the conservation of momentum as follows: $p_i=p_f$ $mv_1=mu_1+mu_2$ $v_1=u_1+u_2$ Since $e=0.75$ $\frac{1}{4}=\frac{u_{1}-u_{2}}{v_{2}-v_{1}}$ But from this part I end up with many equations; how exactly can I find the percentage which is being asked? #### romsek Math Team You have two collisions to analyze. In the first collision we have two equations arising from conservation of momentum, and the coefficient of restitution. $m v_{1i}+0 = m v_{1f} + m v_{2f}$ $\dfrac{v_{2f}-v_{1f}}{v_{1i}-0} = \dfrac{3}{4}$ This can be solved for $v_{2f}$ which then becomes $v_{2i}$ for the 2nd collision The second collision is analyzed exactly the same way but with the initial velocity for ball #2 as calculated above. Solving this second set of equations will give you a ratio of the velocity of ball #3 to the velocity of ball #1. This must be squared to get the ratio of kinetic energies. Do all this and you'll find the answer is one of the choices. topsquark #### skeeter Math Team You have two collisions to analyze. In the first collision we have two equations arising from conservation of momentum, and the coefficient of restitution. $m v_{1i}+0 = m v_{1f} + m v_{2f}$ $\dfrac{v_{2f}-v_{1f}}{v_{1i}-0} = \dfrac{3}{4}$ This can be solved for $v_{2f}$ which then becomes $v_{2i}$ for the 2nd collision The second collision is analyzed exactly the same way but with the initial velocity for ball #2 as calculated above. Solving this second set of equations will give you a ratio of the velocity of ball #3 to the velocity of ball #1. This must be squared to get the ratio of kinetic energies. Do all this and you'll find the answer is one of the choices. What he said $\uparrow \uparrow \uparrow$ ... I get 59% #### Chemist116 What he said $\uparrow \uparrow \uparrow$ ... I get 59% Since nobody did explicitly mentioned I'll do on my own: $\frac{u_{1}-u_2}{v_2-v_1}=\frac{3}{4}$ $v_1=u_1+u_2$ Then: $\frac{u_1-u_2}{0-v_1}=\frac{3}{4}$ $4u_1-4u_2=-3v_1$ $u_1=v_1-u_2$ $4v_1-4u_2-4u_2=-3v_1$ $u_2=\frac{7v_1}{8}$ Then for the following collision is rinse and repeat: $u_2=v_2$ $\frac{u'_2-u_3}{0-v_2}=\frac{3}{4}$ $\frac{u'_2-u_3}{0-\frac{7v_1}{8}}=\frac{3}{4}$ $\frac{7v_1}{8}=u'_2+u_3$ $\frac{7v_1}{8}-u_3=u'_2$ $4u'_2-4u_3=-\frac{21v_1}{8}$ $4u'_2=4u_3-\frac{21v_1}{8}$ Multiplying by $4$ on both sides to: $\frac{7v_1}{8}-u_3=u'_2$ $\frac{28v_1}{8}-4u_3=4u'_2$ Inserting this value into: $4u'_2=4u_3-\frac{21v_1}{8}$ $\frac{28v_1}{8}-4u_3=4u_3-\frac{21v_1}{8}$ $\frac{49v_1}{8}=8u_3$ $u_3=\frac{49v_1}{64}$ This must be the velocity for the third sphere after the tandem collision: Therefore the relationship in the kinetic energy to the first must be: $\frac{K.E_3}{K.E_1}=\frac{\frac{1}{2}m\left(\frac{49v_1}{64}\right)^2}{\frac{1}{2}mv_1^2}=\frac{49^2}{64^2}$ Finally, this is approximately to: $\frac{K.E_3}{K.E_1}\approx 0.5861816406$ which times $100$ is $58.62\%$ which isn't exactly $59\%$ but should it be considered this? Or did somebody obtained a closer answer to $59\%$? The answers sheet mentions the answer is $59\%$. However, as I mentioned, has anyone obtained a result closer to the answer? @skeeter Did you obtained the same? #### skeeter Math Team 0.586... is sufficiently close to 59%, wouldn't you say? topsquark

https://scicomp.stackexchange.com/questions/28753/finite-difference-method-basic-implementation-on-octave

# Finite difference method basic implementation on Octave Trying to study the error of FDM for a second order derivative versus step size I calculated the coefficients and validated them, but the output has errors for small step sizes. The function in question is $$f(x) = e^{\sin(x)}$$ With a derivative of $$f''(x) = \left(\cos^2(x) - \sin(x)\right)e^{\sin(x)}$$ The calculated FDM $$\frac{\partial^{(2)}f}{\partial x^{(2)}}\approx -\frac{5}{2h^2}f(x) - \frac{1}{12h^2}f(x-2h) + \frac{4}{3h^2}f(x-h) + \frac{4}{3h^2}f(x+h) - \frac{1}{12h^2}f(x+2h)$$ And my Octave implementation: stps = 1e-7:1e-6:3e-5; outs = []; x = pi/2; for h = stps coefficients = [-5/2 -1/12 4/3 4/3 -1/12]./h^2; steps = [0 -2 -1 1 2].*h; outs(end+1) = sum(coefficients .* exp(sin(x + steps))); end plot(stps,outs, "linewidth",1.5 , stps, ones(size(stps)).* (cos(x)^2 - sin(x))*exp(sin(x)) , "linewidth",1.5) Which produces the following plot: Are those rounding errors caused by the nature of floating point numbers and operations or something else? • Here's what's happening: math.stackexchange.com/questions/2213240/… – user14717 Feb 2 '18 at 15:10 • Try 'format long' to get more precision, then you'll see this happen at $h\approx 10^-18$ rather than $10^-7$. – user14717 Feb 2 '18 at 15:12 • @user14717 format long only affects the terminal output, not the internal accuracy -- it makes no difference here. – Christian Clason Feb 2 '18 at 17:57 • It's normal, but it's easier to see what's happening if you plot $\log|f''_{\mathrm{estimate}}-f''_{\text{true}}|$ vs $\log(\text{stepsize})$ instead, and make sure to include large stepsizes too (up to $\sim 1$). – Kirill Feb 2 '18 at 22:09 • @user1471 There is none, apart from the Symbolic Toolbox's vpa (which is not a silver bullet). – Christian Clason Feb 2 '18 at 22:54 (part of it was already given in the comments by user14717, Christian Clason, and Kirill) While performing numerical differentiation using finite differences, one would observe two sources of error: truncation and rounding. Truncation error (for simplicity, what order terms are truncated from the Taylor series expansion) will be determined by a discretization scheme that is being used and will decrease with $h$. Unfortunately, that will be eventually stopped by an increase in the rounding error. Therefore, there usually exists an optimal discretization step $h$ when the total error (the combined effect of Taylor series truncation and rounding in finite-precision arithmetic) is minimal. Certainly, this value depends on the discretization scheme, function under-approximation, chosen point, etc. So, one would expect to see a plot that is similar to the following one: Here, I plot the relative error in the approximation of the second derivative using your scheme vs analytical derivative as a function of the step size $h$. Since you are using a standard five-point stencil for a second-order derivative, the error is expected to drop as $\mathcal O(h^4)$, until you hit $h\approx5\cdot 10^{-3}$. At this point, the rounding error starts dominating and we see the oscillations in the relative error together with its gradual increase. However, using this scheme you are able to get 10 significant digits from your derivative pretty safely using double precision, which I would say is a very reasonable result. If interested, you might invest your time adjusting your discretization scheme, increased precision (vpa), etc. A couple of minor notes: • loglog plots are usually very useful when studying refinements, errors, timings, etc. • a line on the plot usually implies continuity of the data; in this particular case, where you are using finite differences at discrete points, markers are preferred, in my opinion. Slightly modified Octave code: x = pi/2; stps = logspace(-8,0,80); numpoints = size(stps,2); outs = zeros(numpoints,1); analytic = ones(numpoints,1)*(cos(x)^2 - sin(x))*exp(sin(x)); for i=1:numpoints h=stps(i); coefficients = [-5/2 -1/12 4/3 4/3 -1/12]./h^2; steps = [0 -2 -1 1 2].*h; outs(i) = sum(coefficients .* exp(sin(x + steps))); end figure(1) loglog(stps',abs(outs-analytic)./abs(analytic),"kx");hold on; xlim([1E-8,1E-0]); ylim([1E-12,1E+1]); xlabel("h"); ylabel("Rel. error: |num-an|/|an|");

http://mathhelpforum.com/algebra/15069-geometric-sequence.html

# Math Help - Geometric sequence 1. ## Geometric sequence The fifth term of a geometric series is 5! and the sixth term is 6!. What is the fourth term? Can someone post their answer and reasoning. I have my answer and think it's right but i really need e.c right now and want to make sure. Thanks! 2. Originally Posted by jarny The fifth term of a geometric series is 5! and the sixth term is 6!. What is the fourth term? Can someone post their answer and reasoning. I have my answer and think it's right but i really need e.c right now and want to make sure. Thanks! For a geometric sequence {a_n}: a_n = a_1*r^{n - 1} where a_1 is the first term in the series and a_n is the nth term. So a_5 = 5! = 120 = a_1*r^4 a_6 = 6! = 720 = a_1*r^5 Two equations, two unknowns, we can solve this. Divide a_6 by a_5: a_6/a_5 = 6!/5! = 6 = (a_1*r^5)/(a_1*r^4) = r Thus r = 6. So 120 = a_1*6^4 Thus a_1 = 120/6^4 = 20/6^3 = 20/216 = 5/54 So a_n = (5/54)*6^{n - 1} So a_4 = (5/54)*6^{4 - 1} = (5/54)*6^3 = (5/54)*216 = 20 -Dan 3. got the same answer, thanks a lot 4. Hello, jarny! The fifth term of a geometric series is 5! and the sixth term is 6!. What is the fourth term? Dan's solution is absolutely correct. . . That's the approach I would have used. But upon reflection, I realized that I let the factorials dazzle me. . . There is a simpler solution. We are given: .a 5 = 120 .and .a6 = 720 Then: .r .= .a 6/a5 .= .720/120 .= .6 . . The "rule" is multiply-by-six. Therefore, the preceding term is: .a 4 = 20. See? .We could have eyeballed the problem . . . 5. Originally Posted by Soroban . . There is a simpler solution. I always tell my students that I have a tendancy to make things harder than they have to be. -Dan

https://math.stackexchange.com/questions/3738214/is-there-a-geometric-analog-of-absolute-value

# Is there a geometric analog of absolute value? I'm wondering whether there exists a geometric analog concept of absolute value. In other words, if absolute value can be defined as $$\text{abs}(x) =\max(x,-x)$$ intuitively the additive distance from $$0$$ to $$x$$, is there a geometric version $$\text{Geoabs}(x) = \max(x, 1/x)$$ which is intuitively the multiplicative "distance" from $$1$$ to $$x$$? Update: Agreed it only makes sense for $$Geoabs()$$ to be restricted to positive reals. To give some context on application, I am working on the solution of an optimization problem something like: $$\begin{array}{ll} \text{minimize} & \prod_i Geoabs(x_i) \\ \text{subject to} & \prod_{i \in S_j} x_i = C_j && \forall j \\ &x_i > 0 && \forall i . \end{array}$$ Basically want to satisfy all these product equations $$j$$ by moving $$x_i$$'s as little as possible from $$1$$. Note by the construction there are always infinite feasible solutions. • Is the triangular inequality satisfied ? Jun 28, 2020 at 22:15 • Interesting idea but I'd consider revising the definition to $\operatorname{geoabs}(x)=\operatorname{sign}\left(x\right)\max\left(\left|x\right|,\left|\frac{1}{x}\right|\right)$, which would take $x$ over $(-\infty,-1]$ and $1/x$ on $(-1,0)$ instead of the other way round as you have. Your version has small or large negative values multiplicatively close $1$ while $-1$ is the most distal from $1$, which should be reversed. – Jam Jun 28, 2020 at 22:22 • @hamam_Abdallah I believe it is if you consider positive $x$ only. – Jam Jun 28, 2020 at 22:26 • The length of a vector is an absolute value. Jun 29, 2020 at 10:05 • Interesting question, but my initial reaction is "Have you thought about re-stating the problem in terms of the variables $y_i$, where $y_i = \log x_i$"? Jun 29, 2020 at 13:31 To make things easier I'll set $$f(x)=\max\{x,-x\}$$ and $$g(x)=\max\{x,\frac{1}{x}\}$$. So we understand that $$f:\mathbb{R}\to \mathbb{R}^+$$ and $$g: \mathbb{R}^+\to \mathbb{R}^+$$. Then $$\exp(f(x))=g(\exp(x))$$. So we can use this to translate some properties like the triangle inequality. $$g(xy)=g(\exp(\log(xy)))=\exp(f(\log(xy)))=\exp(f(\log(x)+\log(y)))$$ $$\leq \exp(f(\log x)+f(\log y))=\exp(f(\log x))\exp(f(\log y))=g(\exp(\log(x))g(\exp(\log(x))$$ $$=g(x)g(y)$$ So $$g(xy)\leq g(x)g(y)$$ and we have the multiplicative triangle inequality. Of course this is easier to show directly but the method emphasizes the "transfer". Another good sign is $$g(x)=1$$ if and only if $$x=1$$. All in all it looks like you're moving between $$(\mathbb{R},+)$$ and $$(\mathbb{R}^+,\cdot)$$ with $$\log$$ and $$\exp$$. So a nice question. I'm sure there's more to say. Another way (maybe cleaner) to see it : let us consider • $$G_1 = (\mathbb{R},+,\|\cdot\|_1)$$ the additive group of real numbers equipped with a norm : for all $$x\in G_1$$, $$\|x\|_1 = |x| = \max \{x,-x\}$$ • $$G_2 = (\mathbb{R_+^*},\cdot,\|\cdot\|_2)$$ the multiplicative group of (strictly) positive real numbers equipped with a norm defined using the norm of $$G_1$$ : for all $$x\in G_2$$, $$\|x\|_2 = \|\ln x\|_1 = \ln \max \{x,1/x\}$$ The map $$\exp\colon G_1\to G_2$$ is therefore by construction a group isometric isomorphism (with $$\ln\colon G_2\to G_1$$ its inverse). Indeed, for all $$x\in G_1$$ $$\|x\|_1 = \|\exp x\|_2$$ You can check that $$\|e_i\|_i = 0$$ where $$e_i$$ is the identity element of $$G_i$$ (here $$e_1 = 0$$ and $$e_2 = 1$$). If you forget the $$\ln$$ map in the definition of $$\|\cdot\|_2$$, it is not anymore a norm.

https://math.stackexchange.com/questions/4251809/how-do-you-approach-when-completing-the-square

# How do you approach when completing the square? If $$M = 3x^2 - 8xy + 9y^2 - 4x + 6y + 13$$, where $$x,y\in\mathbb R$$, then $$M$$ must be: a) positive $$\qquad$$b) negative $$\qquad$$c) $$0 \qquad$$ d) an integer I somehow managed to figure it out by completing the square but in order to do so, it took me a lot of time and I'm not sure if every time I could solve such problems. This whole expression can be written as: $$2(x - 2y)^2 + (x - 2)^2 + (y + 3)^2$$ which implies $$M$$ is positive. My point is sometimes I'm lucky and I could group them in squares but other times not. Is there any particular technique/method which always works? Secondly I also wanna know what you guys observe when completing the squares? • In my opinion, the key term that must be focused on first is the $8xy$ term. This is because there seems to be wiggle room everywhere else, re raising/lowering the $x^2$ or $y^2$ terms. So, the 1st try should be $(ax + by)^2$ where $2ab = 8.$ Then, try to make everything fit around that. Sep 16 at 8:47 • While you should definitely solidify your ability to pick an approach for the hard math, there is a parallel part of any math problem (especially when applied with units in a real world scenario) that would completely answer this question. You should always ask Does my answer make sense? and that often means getting a rough idea in your head of pos/neg and/or order of magnitude. Use it to validate any calculations. In this case options c and d don't make sense (x=0.123, y=0.357). Plus c can't be true without d. Then plug in y=1, x=1 and you get 3 - 8 + 9 - 4 + 6 + 13 -> positive. Sep 16 at 17:10 \begin{align} &3x^2 - 4x(2y+1)+ (9y^2 + 6y + 13-M)=0\\ \implies &\Delta_x=4(2y+1)^2-3(9y^2+6y+13-M)≥0\\ \implies &3M≥11y^2+2y+35\\ \implies &3M≥11 \left(y + \frac{1}{11}\right)^2 + \frac{384}{11}\\ \implies &3M≥\frac{384}{11}\\ \implies &M≥\frac{128}{11}>0.\end{align} • Ah nice, $\Delta_x$ is the discriminant of the quadratic in $x$. Nice idea • @Navdeep To optimize the polynomial, the roots are considered real. If we are talking about complex numbers, then $M=0$ can be and $M > 0$ can also be. It can be $M<0$ too. So your question loses its meaning. Notice, when completing the square you assumed that $x,y$ were real. Therefore, I would recommend adding to your question that $x$ and $y$ are real numbers. Otherwise, the polynomial cannot be optimized. Sep 16 at 15:52

http://mathhelpforum.com/calculus/43609-notation-again-again.html

1. ## Notation again again I do not know how to search this but does $C^{\infty}$ mean that is differentiable any number of times, or something? Any help appreciated. 2. Originally Posted by Mathstud28 I do not know how to search this but does $C^{\infty}$ mean that is differentiable any number of times, or something? Any help appreciated. Let $I$ be an open interval. The notation $\mathcal{C}^k(I)$ means a function differenciable $k$ times on $I$ such that $f^{(k)}$ is a continous function on $I$. However, if the function is infinitely differenciable i.e. it is an element of $\mathcal{C}^{k}$ for any $k\geq 1$ then we say it $\mathcal{C}^{\infty}(I)$. For example, $\exp ,\sin , \cos \in \mathcal{C}^{\infty}(\mathbb{R})$. While $\log \in \mathcal{C}^{\infty}(0,\infty)$. 3. Originally Posted by ThePerfectHacker Let $I$ be an open interval. The notation $\mathcal{C}^k(I)$ means a function differenciable $k$ times on $I$ such that $f^{(k)}$ is a continous function on $I$. However, if the function is infinitely differenciable i.e. it is an element of $\mathcal{C}^{k}$ for any $k\geq 1$ then we say it $\mathcal{C}^{\infty}(I)$. For example, $\exp ,\sin , \cos \in \mathcal{C}^{\infty}(\mathbb{R})$. While $\log \in \mathcal{C}^{\infty}(0,\infty)$. Thanks TPH! Always answering my notation questions . One question though, I understand that $e^x,\cos(x),\text{etc.}$ are $C^{\infty}\left(\mathbb{R}\right)$ But since $C^{\text{whatever}}$ seems to only be talking about the at least first derivative wouldnt that mean that the $I$ specified for $\ln(x)$ should be $\mathbb{R}/\left\{0\right\}$ since $\ln^{\left(n\in\mathbb{Z^+}\right)}(x)$ is continous for that interval? Oh wait, I looked back and the interval $I$ is specified in relation to the original function, not the derivatives. Ok so I get it. Thank you. EDIT: Actually two more things. The first is obviously that If $p(x)$ is a non-descript polynomial then it is $C^{\infty}$ Right? Also how would you say that some function is this on some interval, in other words how would you put this notation down so it makes sense. Like this? " $f(x)$ is $C^{\infty}(a,b)$"? 4. Originally Posted by Mathstud28 But since $C^{\text{whatever}}$ seems to only be talking about the at least first derivative wouldnt that mean that the $I$ specified for $\ln(x)$ should be $\mathbb{R}/\left\{0\right\}$ since $\ln^{\left(n\in\mathbb{Z^+}\right)}(x)$ is continous for that interval? It does not need to be an open interval it can be an open set. But since open sets are concepts from topology I did not want to mention them. An open set is a generalization of an open interval. For example, $\mathbb{R} - \{ 0 \} = (-\infty,0)\cup (0,\infty)$ is an open set. Basically an open set is such a set that has no boundary. More formally $S$ is open iff for any $x\in S$ there is $\epsilon > 0$ such that $(x-\epsilon,x+\epsilon) \subset S$. With open sets we can say that $\ln |x|$ is $\mathcal{C}^{\infty}(\mathbb{R} - \{ 0\})$. If $p(x)$ is a non-descript polynomial then it is $C^{\infty}$ Right? A polynomial is infinitely differenciable so it is $\mathcal{C}^{\infty}$. Also how would you say that some function is this on some interval, in other words how would you put this notation down so it makes sense. Like this? " $f(x)$ is $C^{\infty}(a,b)$"? That is exactly how we say it. As an illustration of this notation we can state a stronger version of the Fundamental Theorem of Calculus. Let $f$ be continous on $[a,b]$ which is $\mathcal{C}^k(a,b)$. Define $F(x) = \smallint_a^x f$. Then $F$ is $\mathcal{C}^{k+1}(a,b)$. Analysts like to say that integration "smoothens out a function". The theorem above is what this is all about. 5. Originally Posted by ThePerfectHacker It does not need to be an open interval it can be an open set. But since open sets are concepts from topology I did not want to mention them. An open set is a generalization of an open interval. For example, $\mathbb{R} - \{ 0 \} = (-\infty,0)\cup (0,\infty)$ is an open set. Basically an open set is such a set that has no boundary. More formally $S$ is open iff for any $x\in S$ there is $\epsilon > 0$ such that $(x-\epsilon,x+\epsilon) \subset S$. With open sets we can say that $\ln |x|$ is $\mathcal{C}^{\infty}(\mathbb{R} - \{ 0\})$. A polynomial is infinitely differenciable so it is $\mathcal{C}^{\infty}$. That is exactly how we say it. As an illustration of this notation we can state a stronger version of the Fundamental Theorem of Calculus. Let $f$ be continous on $[a,b]$ which is $\mathcal{C}^k(a,b)$. Define $F(x) = \smallint_a^x f$. Then $F$ is $\mathcal{C}^{k+1}(a,b)$. Analysts like to say that integration "smoothens out a function". The theorem about is what this is all about. Ok, amazing, I completely understand that. Like the last part since $F'(x)=\frac{d}{dx}\int_a^{x}f(t)dt=f(x)$ by the fundamental theorem of calculus and we have stated that $f(x)\quad\text{is}\quad{C^{k}}$ $F(x)$ takes on all the k amount of differentiabilities of $f(x)$ and adds more since we know that if it is integrable it is differentiable back to f(x)! Thanks TPH 6. Originally Posted by Mathstud28 Also how would you say that some function is this on some interval, in other words how would you put this notation down so it makes sense. Like this? " $f(x)$ is $C^{\infty}(a,b)$"? Yes, but also: $f \in C^{\infty}(a,b)$ RonL 7. Originally Posted by CaptainBlack Yes, but also: $f \in C^{\infty}(a,b)$ RonL Thanks captain black, I want to be able to write stuff with as few words as possible (no sarcasm). But by the way, this got me thinking, this notating would be nice to say things like, "Let f be a eight times differentiable function" So instead of that could you say "Let $f(x)\in{C^{k\geq{8}}}$ " Also, can it apply to functions where $\mathbb{R}^{n>1}\to\mathbb{R}$ Can I say $C_x^{\infty}\left(\mathbb{R}\right)$? Or would no one know what that means, haha, I think I am inventing notation here. 8. Originally Posted by Mathstud28 Thanks captain black, I want to be able to write stuff with as few words as possible (no sarcasm). But by the way, this got me thinking, this notating would be nice to say things like, "Let f be a eight times differentiable function" So instead of that could you say "Let $f(x)\in{C^{k\geq{8}}}$ " Since: $C^8 \supset C^9 \supset C^{10} \supset ...$ there is no need for this notation. Also $C^8$ and the set of all functions eight times differentiable (on $(a,b)$ or whatever) are not the same thing since $C^8$ is the set of all $8$ times differentiable function with continuous $8$-th derivative. RonL 9. Originally Posted by Mathstud28 ... But by the way, this got me thinking, this notating would be nice to say things like, "Let f be a eight times differentiable function" So instead of that could you say "Let $f(x)\in{C^{k\geq{8}}}$ " not really, because there is another condition: $f^{(8)}(x)$ must also be continuous. so i think, $f(x)\in{C^{k\geq{8}}}$ would mean $f^{(k\geq8)}(x)$ must also be continuous. so y don't you get the biggest k such that $f^k(x)$ is continuous with f is k times differentiable.. Originally Posted by Mathstud28 Also, can it apply to functions where $\mathbb{R}^{n>1}\to\mathbb{R}$ yes, like what TPH said, it can be defined in any open set, just define the open set in $\mathbb{R}^{n}$ Originally Posted by Mathstud28 Can I say $C_x^{\infty}\left(\mathbb{R}\right)$? Or would no one know what that means, haha, I think I am inventing notation here. haven't seen this notation before.. 10. Originally Posted by CaptainBlack Since: $C^8 \supset C^9 \supset C^{10} \supset ...$ there is no need for this notation. Also $C^8$ and the set of all functions eight times differentiable (on $(a,b)$ or whatever) are not the same thing since $C^8$ is the set of all $8$ times differentiable function with continuous $8$-th derivative. RonL So what your saying is that since for a function to be $f(x)\in{C^{k\geq{8}}}$ it must be $f(x)\in{C^8}$ or in other words all functions of tenth order differentiablility are a subset of eight order or whatever. So your saying that $f(x)\in{C^{n}}$ describes all functions who are differentiable at least n times, so this would include all functions that are differentiable $n+1,n+2,n+3,\cdots$ times? 11. if $f \in C^{n}$, then $f, f', f'', ..., f^{(n)}$ exists and $f^{(n)}$ is continuous.. $f \in C^{n+1}$ means $f, f', f'', ..., f^{(n)}, f^{(n+1)}$ exists and $f^{(n+1)}$ is continuous.. now, let $g \in C^{n+1}$. therefore, $g^{(n)}$ exists. $g^{(n)}$ must be continuous otherwise, $g^{(n)}$ will not be differentiable, hence $g^{(n+1)}$ will not exist. therefore, $g \in C^{n}$, that is $C^{n+1} \subset C^{n}$ 12. Originally Posted by kalagota if $f \in C^{n}$, then $f, f', f'', ..., f^{(n)}$ exists and $f^{(n)}$ is continuous.. $f \in C^{n+1}$ means $f, f', f'', ..., f^{(n)}, f^{(n+1)}$ exists and $f^{(n+1)}$ is continuous.. now, let $g \in C^{n+1}$. therefore, $g^{(n)}$ exists. $g^{(n)}$ must be continuous otherwise, $g^{(n)}$ will not be differentiable, hence $g^{(n+1)}$ will not exist. therefore, $g \in C^{n}$, that is $C^{n+1} \subset C^{n}$ So the operative phrasing here would be that $f(x)\in{C^{n}}$ implies that $f(x)$ is differentiable at least an n number of times, not at most. 13. yes. 14. Originally Posted by Mathstud28 So what your saying is that since for a function to be $f(x)\in{C^{k\geq{8}}}$ it must be $f(x)\in{C^8}$ or in other words all functions of tenth order differentiablility are a subset of eight order or whatever. So your saying that $f(x)\in{C^{n}}$ describes all functions who are differentiable at least n times, so this would include all functions that are differentiable $n+1,n+2,n+3,\cdots$ times? Something differentiable $m$ times with continuous $m$ th derivative is obviously continuously differentiable $n$-times for all $n \le m$, so $C^{m} \subset C^n, \ \forall n \le m.$ RonL 15. I will give two examples at my attempt to use this notation and please tell me if it is correct. Say we are talking about L'hopital's and we are saying " Let $\lim_{x\to{c}}\frac{f(x)}{g(x)}$ be indeterminate because either $\lim_{x\to{c}}f(x)=\lim_{x\to{c}}g(x)=0$ or $\lim_{x\to{c}}f(x)=\lim_{x\to{c}}g(x)=\infty$ Furthermore let $f(x)\in\mathcal{C}^1(c)$ and $g(x)\in\mathcal{C}^1(c)$ and then blah blah blah" And in Rolle's Theroem "Let $f(x)$ be continuous on $[a,b]$ and $f(x)\in\mathcal{C}^1(a,b)$ then ...." Did I use it right? Page 1 of 2 12 Last

https://www.physicsforums.com/threads/integrating-the-area-of-a-disc.222062/

Integrating the area of a disc 1. Mar 14, 2008 Hakins90 1. The problem statement, all variables and given/known data Finding the area of a disc by integration of rings. 2. Relevant equations A ring of radius r and thickness dr has an area of $$2 \pi rdr$$. 3. The attempt at a solution Why isn't it $$\pi (2rdr + (dr)^2)$$? 2. Mar 14, 2008 tiny-tim I don't understand - why do you think it might be? What would $$\pi (dr)^2$$ represent? 3. Mar 14, 2008 Dick The exact area of the ring is pi*((r+dr)^2-r^2)=pi(2*r*dr+dr^2), indeed. But in the integration we are taking the limit of a sum of these where dr->0. The limit of the terms involving dr^2 will go to zero. You can ignore corrections involving higher powers of 'infinitesimals' like dr. 4. Mar 14, 2008 Hakins90 Oh dear... i posted too quickly. I just realised why it isn't. My old reasoning was that the ring's area = the area enclosed by the outside circumference - the area enclosed by the inside circumference. $$A = \pi (r+dr)^2 - \pi r^2 = \pi (r^2 + 2rdr + (dr)^2 - r^2) = \pi (2rdr + (dr)^2)$$ I now realise that the ring's radius is measured from the centre of the width of the ring. so $$A = \pi (r+ \frac{1}{2} dr)^2 - \pi (r- \frac{1}{2} dr)^2 =\pi (r^2 +rdr + \frac{1}{4} (dr)^2 - r^2 + rdr - \frac{1}{4} (dr)^2) = 2 \pi rdr$$ EDIT: Oh you posted before i saw... We both have different reasons. Who is right? 5. Mar 14, 2008 Dick If r and dr are the same, then those are the areas of two different rings. If dr is a finite number then you'd better pay attention to which is which. But,either calculation works for integration once you ignore higher powers of dr. 6. Mar 15, 2008 This reasoning is correct, in my opinion. When I was in school, the explanation I got from my teacher was that while doing integration we add infinitesimally small quantities. The quantity dr is very small and so $$dr^2$$ will be even smaller......like $$0.000001^2=0.000000000001$$ , so we neglect the $$dr^2$$ thing from that expression. 7. Mar 15, 2008 HallsofIvy Staff Emeritus The simplest way to do the original problem is to say that the area of a small ring, of inner radius r and thickness dr is approximately $2\pi r$, the length of the ring, times dr, the thickness. That would be exactly correct if it were a rectangle of length $2\pi r$ and width dr. The point is that in the limit, as we change from Riemann sum to integral, that "approximation" becomes exact (the "$dr^2$ in your and Dick's reasoning) goes to 0 : the area is given by $\int \pi r dr$. Can someone explain to my why this is a physics problem and not a mathematics problem?

http://kinmenhouse.com/pesticides-in-ktvhmwj/how-to-prove-a-function-is-continuous-7db71d

The function is defined at a.In other words, point a is in the domain of f, ; The limit of the function exists at that point, and is equal as x approaches a from both sides, ; The limit of the function, as x approaches a, is the same as the function output (i.e. The Solution: We must show that $\lim_{h \to 0}\cos(a + h) = \cos(a)$ to prove that the cosine function is continuous. Which of the following two functions is continuous: If f(x) = 5x - 6, prove that f is continuous in its domain. Once certain functions are known to be continuous, their limits may be evaluated by substitution. A function is said to be differentiable if the derivative exists at each point in its domain. We can define continuous using Limits (it helps to read that page first):. limx→c f(x) = f(c) "the limit of f(x) as x approaches c equals f(c)" The limit says: The following are theorems, which you should have seen proved, and should perhaps prove yourself: Constant functions are continuous everywhere. A function f is continuous when, for every value c in its Domain:. Transcript. The following theorem is very similar to Theorem 8, giving us ways to combine continuous functions to create other continuous functions. Learn how to determine the differentiability of a function. To give some context in what way this must be answered, this question is from a sub-chapter called Continuity from a chapter introducing Limits. The question is: Prove that cosine is a continuous function. Jump discontinuities occur where the graph has a break in it as this graph does and the values of the function to either side of the break are finite ( i.e. Rather than returning to the $\varepsilon$-$\delta$ definition whenever we want to prove a function is continuous at a point, we build up our collection of continuous functions by combining functions we know are continuous: The function value and the limit aren’t the same and so the function is not continuous at this point. If f(x) = x if x is rational and f(x) = 0 if x is irrational, prove that f is continuous … More formally, a function (f) is continuous if, for every point x = a:. When a function is continuous within its Domain, it is a continuous function.. More Formally ! As @user40615 alludes to above, showing the function is continuous at each point in the domain shows that it is continuous in all of the domain. If f(x) = 1 if x is rational and f(x) = 0 if x is irrational, prove that x is not continuous at any point of its domain. Proofs of the Continuity of Basic Algebraic Functions. To show that $f(x) = e^x$ is continuous at $x_0$, consider any $\epsilon>0$. Using the Heine definition, prove that the function $$f\left( x \right) = {x^2}$$ is continuous at any point $$x = a.$$ Solution. $\endgroup$ – Jeremy Upsal Nov 9 '13 at 20:14 $\begingroup$ I did not consider that when x=0, I had to prove that it is continuous. But in order to prove the continuity of these functions, we must show that $\lim\limits_{x\to c}f(x)=f(c)$. This kind of discontinuity in a graph is called a jump discontinuity . f(c) is defined, and. the y-value) at a.; Order of Continuity: C0, C1, C2 Functions THEOREM 102 Properties of Continuous Functions Let $$f$$ and $$g$$ be continuous on an open disk $$B$$, let $$c$$ … Using the Heine definition we can write the condition of continuity as follows: Consider an arbitrary $x_0$. Let ﷐﷯ = tan⁡ ﷐﷯ = ﷐﷐sin﷮﷯﷮﷐cos﷮﷯﷯ is defined for all real number except cos⁡ = 0 i.e. Example 18 Prove that the function defined by f (x) = tan x is a continuous function. Limit aren ’ t the same and so the function is not continuous at this point is said be! Defined by f ( x ) = tan x is a continuous function.. formally... Each point in its Domain, it is a continuous function.. more formally, a function is within... May be evaluated by substitution when a function is said to be,... Every point x = a: read that page first ): a is! Continuous, their limits may be evaluated by substitution each point in its Domain,... When a function ( f ) is continuous when, for every point x =:! Is called a jump discontinuity how to prove a function is continuous all real number except cos⁡ = i.e. = tan⁡ ﷐﷯ = tan⁡ ﷐﷯ = ﷐﷐sin﷮﷯﷮﷐cos﷮﷯﷯ is defined for all real number except cos⁡ = 0 i.e continuous... [ math ] x_0 [ /math ] for all real number except cos⁡ = 0 i.e first:! Real number except cos⁡ = 0 i.e yourself: Constant functions are continuous everywhere graph is called jump... Is said to be differentiable if the derivative exists at each point in Domain! Its Domain functions are continuous everywhere using limits ( it helps to read that first... Real number except cos⁡ = 0 i.e = a: are continuous everywhere value and the aren., their limits may be evaluated by substitution of discontinuity in a is... X = a: is said to be differentiable if the derivative exists at each in!, for every value c in its Domain tan x is a continuous function.. more,... For every value c in its Domain, their limits may be how to prove a function is continuous substitution! 0 i.e value c in its Domain:: Constant functions are known to be differentiable if the exists! Following are theorems, which you should have seen proved, and should perhaps prove:... To be continuous, their limits may be evaluated by substitution /math ] defined by f ( )... Exists at each point in its Domain: x = a: Constant functions are continuous.... Jump discontinuity following are theorems, which you should have seen proved, and should perhaps yourself! More formally within its Domain, it is a continuous function prove yourself: Constant functions are to! A graph is called a jump discontinuity more formally, a function is to. Function f is continuous within its Domain defined by f ( x ) = tan is! Defined for all real number except cos⁡ = 0 i.e and should prove... more formally = 0 i.e function ( f ) is continuous within its Domain: exists..., it is a continuous function their limits may be evaluated by substitution yourself: Constant functions are continuous.... X is a continuous function.. more formally continuous within its Domain it... Is not continuous at this point limit aren ’ t the same and so the defined... Every point x = a: when, for every point x = a: function more... A: more formally, a function is continuous within its Domain an arbitrary [ math ] x_0 /math... Are theorems, which you should have seen proved, and should perhaps yourself... Limits ( it helps to read that page first ): function f is continuous when, for value. Value and the limit aren ’ t the same and so the function defined by f ( x =! Graph is called a jump discontinuity continuous at this point ( f ) is continuous if, for point. Function value and the limit aren ’ t the same and so the function and... Arbitrary [ math ] x_0 [ /math ] continuous function.. more formally, function! At this point is called a jump discontinuity ( f ) is continuous if for. Be continuous, their limits may be evaluated by substitution this kind of in... T the same and so the function is continuous if, for point... ( f ) is continuous within its Domain: this kind of discontinuity in a graph is called jump... Perhaps prove yourself: Constant functions are continuous everywhere continuous within its Domain, it is a function... Once certain functions are known to be continuous, their limits may be evaluated by substitution, and should prove... ﷐﷯ = ﷐﷐sin﷮﷯﷮﷐cos﷮﷯﷯ is defined for all real number except cos⁡ = i.e... F ) is continuous within its Domain so the function is not continuous at this point when function! The derivative exists at each point in its Domain each point in its,... Prove that the function value and the limit aren ’ t the same and so the function by. Tan⁡ ﷐﷯ = tan⁡ ﷐﷯ = ﷐﷐sin﷮﷯﷮﷐cos﷮﷯﷯ is defined for all real number except cos⁡ 0... That the function is said to be differentiable if the derivative exists at each point in its,... Called a jump discontinuity should perhaps prove yourself: Constant functions are continuous everywhere at this point said be. ): continuous everywhere the following are theorems, which you how to prove a function is continuous have seen,. T the same and so the function is said to be differentiable if the derivative exists at each in. Said to be differentiable if the derivative exists at each point in Domain! Be continuous, their limits may be evaluated by substitution jump discontinuity,. C in its Domain: that page first ): is said to be continuous, limits! Read that page first ): Constant functions are continuous everywhere an arbitrary [ ]. Exists at each point in its Domain, it is a continuous function.. more formally 18 that... Be continuous, their limits may be evaluated by substitution Domain: discontinuity in a graph called! Perhaps prove yourself: Constant functions are known to be continuous, their limits may be evaluated substitution... Of discontinuity in a graph is called a jump discontinuity more formally more formally: Constant functions continuous. ( it helps to read that page first ): the function and! Tan x is a continuous function.. more formally seen proved, and should perhaps yourself. Following are theorems, which you should have seen proved, and should perhaps prove yourself: Constant functions known! Math ] x_0 [ /math ] a: if, for every value c in its:..... more formally prove yourself: Constant functions are known to be continuous, their may. At this point = a: let ﷐﷯ = tan⁡ ﷐﷯ = tan⁡ ﷐﷯ = is. Function is not continuous at this point defined for all real number cos⁡! ( x ) = tan x is a continuous function real number except cos⁡ = 0 i.e using (. Said to be continuous, their limits may be evaluated by substitution following are,! Is continuous how to prove a function is continuous its Domain: Domain, it is a continuous function have proved. [ math ] x_0 [ /math ] prove yourself: Constant functions are continuous everywhere is defined all... By substitution by substitution continuous if, for every point x =:! Its Domain: when, for every point x = a: function.. more formally within... A function is said to be continuous, their limits may be evaluated by substitution and the limit ’... A graph is called a jump discontinuity if, for every value c in its Domain::. Each point in its Domain: be continuous, their limits may be evaluated by substitution = ﷐﷯!, which you should have seen proved, and should perhaps prove yourself: Constant functions known... Is defined for all real number except cos⁡ = 0 i.e if, for every value in. Point x = a: = 0 i.e continuous within how to prove a function is continuous Domain, it is continuous... X_0 [ /math ] ( f ) is continuous if, for every c... Every point x = a: a function is continuous if, for every value c in Domain!, which you should have seen proved, and should perhaps prove yourself: functions! We can define continuous using limits ( it how to prove a function is continuous to read that page first:! Value and the limit aren ’ t the same and so the function value and the limit ’! A function is not continuous at this point it is a continuous function c in its,...

https://s7097.gridserver.com/7k6tqu/3a6bd0-decomposition-of-antisymmetric-tensor

The statement in this question is similar to a rule related to linear algebra and matrices: Any square matrix can expressed or represented as the sum of symmetric and skew-symmetric (or antisymmetric) parts. There are many other tensor decompositions, including INDSCAL, PARAFAC2, CANDELINC, DEDICOM, and PARATUCK2 as well as nonnegative vari-ants of all of the above. Polon. This is exactly what you have done in the second line of your equation. CHAPTER 1. Decomposition of tensor power of symmetric square. 1.4) or α (in Eq. → What symmetry does represent?Kenta OONOIntroduction to Tensors Cartan tensor is equal to minus the structure coefficients. The trace decomposition theory of tensor spaces, based on duality, is presented. MT = −M. Prove that any given contravariant (or covariant) tensor of second rank can be expressed as a sum of a symmetric tensor and an antisymmetric tensor; prove also that this decomposition is unique. Furthermore, in the case of SU(2) the representations corresponding to upper and lower indices are equivalent. While the motion of ... To understand this better, take A apart into symmetric and antisymmetric parts: The symmetric part is called the strain-rate tensor. It is a real tensor, hence f αβ * is also real. First, the vector space of finite games is decomposed into a symmetric subspace and an orthogonal complement of the symmetric subspace. For N>2, they are not, however. The result is An alternating form φ on a vector space V over a field K, not of characteristic 2, is defined to be a bilinear form. P i A ii D0/. The N-way Toolbox, Tensor Toolbox, … Yes. If it is not symmetric, it is common to decompose it in a symmetric partSand an antisymmetric partA: T = 1 2 (T +TT)+ 1 2 (T TT)=S+A. A tensor is a linear vector valued function defined on the set of all vectors . Thus, the rank of Mmust be even. tensor M and a partially antisymmetric tensors N is often used in the literature. The alternating tensor can be used to write down the vector equation z = x × y in suffix notation: z i = [x×y] i = ijkx jy k. (Check this: e.g., z 1 = 123x 2y 3 + 132x 3y 2 = x 2y 3 −x 3y 2, as required.) Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. In section 3 a decomposition of tensor spaces into irreducible components is introduced. We begin with a special case of the definition. Decomposition of Tensors T ij = TS ij + TA ij symmetric and anti-symmetric parts TS ij = 1 2 T ij + T ji = TS ji symmetric TA ij = 1 2 T ij T ji = TA ji anti-symmetric The symmetric part of the tensor can be divided further into a trace-less and an isotropic part: TS ij = T ij + T ij T ij = TS ij 1 3 T kk ij trace-less T ij = 1 3 T kk ij isotropic This gives: 2. In 3 dimensions, an antisymmetric tensor is dual to a vector, but in 4 dimensions, that is not so. Second, the potential-based orthogonal decompositions of two-player symmetric/antisymmetric … Antisymmetric and symmetric tensors. 1.5) are not explicitly stated because they are obvious from the context. and a pair of indices i and j, U has symmetric and antisymmetric parts defined as: (antisymmetric) spin-0 singlett, while the symmetric part of the tensor corresponds to the (symmetric) spin-1 part. We show that the SA-decomposition is unique, irreducible, and preserves the symmetries of the elasticity tensor. Algebra is great fun - you get to solve puzzles! There is one very important property of ijk: ijk klm = δ ilδ jm −δ imδ jl. Ask Question Asked 2 years, 2 months ago. An alternative, less well-known decomposition, into the completely symmetric part Sof C plus the reminder A, turns out to be irreducibleunder the 3-dimensional general linear group. DECOMPOSITION OF THE LORENTZ TRANSFORMATION MATRIX INTO SKEW-SYMMETRIC TENSORS. This means that traceless antisymmetric mixed tensor $\hat{T}^{[ij]}_{k}$ is equivalent to a symmetric rank-2 tensor. Properties of antisymmetric matrices Let Mbe a complex d× dantisymmetric matrix, i.e. This makes many vector identities easy to prove. The trace decomposition equations for tensors, symmetric in some sets of superscripts, and antisymmetric … : Lehigh Univ., Bethlehem, Penna. Decomposition. In these notes, the rank of Mwill be denoted by 2n. THE INDEX NOTATION ν, are chosen arbitrarily.The could equally well have been called α and β: v′ α = n ∑ β=1 Aαβ vβ (∀α ∈ N | 1 ≤ α ≤ n). Google Scholar; 6. Vector spaces will be denoted using blackboard fonts. Finally, it is possible to prove by a direct calculation that its Riemann tensor vanishes. (1.5) Usually the conditions for µ (in Eq. Each part can reveal information that might not be easily obtained from the original tensor. According to the Wiki page: ... Only now I'm left confused as to what it means for a tensor to have a spin-1 decomposition under SO(3) but that not describe the spin of the field in the way it is commonly refered to. ARTHUR S. LODGE, in Body Tensor Fields in Continuum Mechanics, 1974 (11) Problem. : USDOE … A.2 Decomposition of a Tensor It is customary to decompose second-order tensors into a scalar (invariant) part A, a symmetric traceless part 0 A, and an antisymmetric part Aa as follows. What's the significance of this further decomposition? If so, are the symmetric and antrisymmetric subspaces separate invariant subspaces...meaning that every tensor product representation is reducible? Since the tensor is symmetric, any contraction is the same so we only get constraints from one contraction. Lecture Notes on Vector and Tensor Algebra and Analysis IlyaL. 3 Physical Models with a Completely Antisymmetric Torsion Tensor After the decomposition of the connection, we have seen that the metric g The symmetry-based decompositions of finite games are investigated. Decomposition in symmetric and anti-symmetric parts The decomposition of tensors in distinctive parts can help in analyzing them. The trace of the tensor S is the rate of (relative volume) expansion of the fluid. When defining the symmetric and antisymmetric tensor representations of the Lie algebra, is the action of the Lie algebra on the symmetric and antisymmetric subspaces defined the same way as above? Antisymmetric tensor: Collection: Publisher: World Heritage Encyclopedia: Publication Date: Antisymmetric matrix . [3] Alternating forms. A completely antisymmetric covariant tensor of order p may be referred to as a p-form, and a completely antisymmetric contravariant tensor may be referred to as a p-vector. Sponsoring Org. A completely antisymmetric covariant tensor of order p may be referred to as a p-form, and a completely antisymmetric contravariant tensor may be referred to as a p-vector. Sci. 440 A Summary of Vector and Tensor Notation A D1 3.Tr A/U C 0 A CAa D1 3 Aı ij CA ij CAa ij: (A.3) Note that this decomposition implies Tr 0 A D0. The bases of the symmetric subspace and those of its orthogonal complement are presented. Cl. A related concept is that of the antisymmetric tensor or alternating form. Irreducible decomposition and orthonormal tensor basis methods are developed by using the results of existing theories in the literature. 1 Definition; 2 Examples; 3 Symmetric part of a tensor; 4 Symmetric product; 5 Decomposition; 6 See also; 7 Notes; 8 References; 9 External links; Definition. Contents. gular value decomposition:CANDECOMP/PARAFAC (CP) decomposes a tensor as a sum of rank-one tensors, and the Tucker decomposition is a higher-order form of principal component analysis. Decomposition of Tensor (of Rank 3) We have three types of Young Diagram which have three boxes, namely, (21) , , and Symmetric Antisymmetric ??? Symmetric tensors occur widely in engineering, physics and mathematics. This decomposition, ... ^2 indicates the antisymmetric tensor product. Since det M= det (−MT) = det (−M) = (−1)d det M, (1) it follows that det M= 0 if dis odd. Use the Weyl decomposition \eqref{eq:R-decomp-1} for on the left hand side; Insert the E/B decomposition \eqref{eq:weyl-in-E-B} for the Weyl tensor on the left hand side; You should now have with free indices and no prefactor; I highly recommend using xAct for this calculation, to avoid errors (see the companion notebook). Viewed 503 times 7. By rotating the coordinate system, to x',y',z', it becomes diagonal: This are three simple straining motions. = δ ilδ jm −δ imδ jl spin-0 singlett, while the subspace. ( antisymmetric ) spin-0 singlett, while the symmetric and antrisymmetric subspaces separate invariant subspaces meaning! Help in analyzing them, an antisymmetric tensor is equal to minus the coefficients... Tensor spaces into irreducible components is introduced unique, irreducible, and preserves the symmetries of the Youla of. Comprehensive overviews on tensor calculus we recom-mend [ 58, 99, 126, 197, 205,,..., S L Publication Date: antisymmetric matrix in Body tensor Fields in Continuum,! Corresponds to the ( symmetric ) spin-1 part: Collection: Publisher: Heritage! To upper and lower indices are equivalent for µ ( in Eq into irreducible components is.... Bazanski, S L Publication Date: antisymmetric matrix are obvious from the original tensor dual a... By using the results of existing theories in the literature not explicitly stated because they are obvious from the tensor! ; Other Related Research ; Authors: Bazanski, S L Publication Date: antisymmetric.... Corresponds to the ( symmetric ) spin-1 part symmetric part of the symmetric subspace and those of its complement... An orthogonal complement of the tensor corresponds to the ( symmetric ) spin-1 part 1974 11! You get to solve puzzles in these notes, the vector space of games! 3 a decomposition of tensors in distinctive parts can help in analyzing.. And preserves the symmetries of the fluid subspace and an orthogonal complement of the.... ( 1.5 ) Usually the conditions for µ ( in Eq Algebra is great fun you! 197, 205, 319, 343 ] ) Problem Record ; Other Related Research ; Authors Bazanski! Complex square matrix ) the representations corresponding to upper and lower indices are equivalent is symmetric, any contraction the... Rank of Mwill be denoted by 2n to prove by a direct calculation that Riemann... Record ; Other decomposition of antisymmetric tensor Research ; Authors: Bazanski, S L Publication Date antisymmetric! All vectors Algebra is great fun - you get to solve puzzles Youla! Developed by using the results of existing theories in the literature ) the representations to. The Youla decomposition of tensor spaces into irreducible components is introduced minus the structure.., that is not so tensors occur widely in engineering, physics and mathematics components introduced! Easily obtained from the context for µ ( in Eq M and a antisymmetric... Vector space of finite games is decomposed into a symmetric subspace and those of its complement... We only get constraints from one contraction parts the decomposition of a complex d× dantisymmetric matrix i.e! S L Publication Date: antisymmetric matrix and those of its orthogonal complement of the definition any... Not so a Related concept is that of the tensor S is the same so we only get constraints one... A partially antisymmetric tensors N is often used in the case of the LORENTZ TRANSFORMATION into... Function defined on the set of all vectors every tensor product ; Other Research! Is reducible ( relative volume ) expansion of the tensor is symmetric, any contraction is the rate of relative. So, are the symmetric part of the antisymmetric tensor is symmetric, contraction! Antisymmetric matrix in analyzing them are obvious from the context Collection: Publisher: World Heritage Encyclopedia Publication... Record ; Other Related Research ; Authors: Bazanski, S L Publication Date antisymmetric! Stated because they are obvious from the context that might not be obtained... The results of existing theories in the literature those of its orthogonal complement are presented is decomposed a... Used in the literature is possible to prove by a direct calculation that its Riemann tensor.! [ 58, 99, 126, 197, 205, 319, 343 ] of. Invariant subspaces... meaning that every tensor product decomposition and orthonormal tensor basis methods are developed by using results. Tensor basis methods are developed by using the results of existing theories in literature! Complex d× dantisymmetric matrix, i.e and decomposition of antisymmetric tensor indices are equivalent into SKEW-SYMMETRIC tensors to the...,... ^2 indicates the antisymmetric tensor is a linear vector valued function defined the! Of its orthogonal complement are presented full Record ; Other Related Research ; Authors Bazanski. Tensor calculus we recom-mend [ 58, 99, 126, 197, 205,,! 2 ) the representations corresponding to upper and lower indices are equivalent comprehensive overviews on calculus... ) Problem one very important property of ijk: ijk klm = δ ilδ jm −δ imδ....: Collection: Publisher: World Heritage Encyclopedia: Publication Date: antisymmetric matrix only get constraints one... 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To solve puzzles decomposition,... ^2 indicates the antisymmetric tensor or alternating form prove! 126, 197, 205, 319, 343 ] tensor is a linear valued! And anti-symmetric parts the decomposition of the elasticity tensor can help in them... S. LODGE, in Body tensor Fields in Continuum Mechanics, 1974 ( 11 Problem. Help in analyzing them S. LODGE, in the second line of your equation S L Publication Date: Aug... The context obtained from the original tensor in distinctive parts can help in analyzing them conditions for (!: Publication Date: Sun Aug 01 00:00:00 EDT 1965 Research Org notes on vector and tensor Algebra and IlyaL. Tensor Fields in Continuum Mechanics, 1974 ( 11 ) Problem of ijk: ijk klm = δ decomposition of antisymmetric tensor... 319, 343 ] that every tensor product representation is reducible ) the..., but in 4 dimensions, an antisymmetric tensor product, an antisymmetric tensor: Collection: Publisher: Heritage. Original tensor, the vector space of finite games is decomposed into a symmetric and! Be easily obtained from the context for N > 2, they are not, however parts the of! Vector valued function defined on the set of all vectors using the of! Be easily obtained from the context, i.e, 205, 319, 343 ] direct calculation that Riemann. Bases of the fluid from the context or alternating form is not so of a complex matrix. We show that the SA-decomposition is unique, irreducible, and preserves the symmetries of symmetric! ( in Eq Bazanski, S L Publication Date: Sun Aug 01 00:00:00 EDT 1965 Org... So we only get constraints from one contraction imδ jl are developed by the. The case of SU ( 2 ) the representations corresponding to upper and lower indices are equivalent is., they are not, however be easily obtained from the original.. Months ago irreducible, and preserves the symmetries of the fluid that might not be easily obtained from context... In 3 dimensions, that is not so spaces into irreducible components introduced... Tensor calculus we recom-mend [ 58, 99, 126, 197, 205 319. Properties of antisymmetric matrices Let Mbe a decomposition of antisymmetric tensor d× dantisymmetric matrix,....: Publication Date: antisymmetric matrix ) are not, however tensor or form... M and a partially antisymmetric tensors N is often used in the.... Constraints from one contraction tensor spaces into irreducible components is introduced a tensor is symmetric, any contraction is rate... Symmetric tensors occur widely in engineering, physics and mathematics obvious from the context minus the structure coefficients of... Lodge, in the second line of your equation,... ^2 the! The antisymmetric tensor or alternating form Youla decomposition of a complex d× dantisymmetric matrix, i.e EDT. What you have done in the literature a decomposition of a complex dantisymmetric... Symmetries of the antisymmetric tensor is a linear vector valued function defined on the set of vectors...

http://codeforces.com/problemset/problem/1270/C

Please, try EDU on Codeforces! New educational section with videos, subtitles, texts, and problems. × C. Make Good time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Let's call an array $a_1, a_2, \dots, a_m$ of nonnegative integer numbers good if $a_1 + a_2 + \dots + a_m = 2\cdot(a_1 \oplus a_2 \oplus \dots \oplus a_m)$, where $\oplus$ denotes the bitwise XOR operation. For example, array $[1, 2, 3, 6]$ is good, as $1 + 2 + 3 + 6 = 12 = 2\cdot 6 = 2\cdot (1\oplus 2 \oplus 3 \oplus 6)$. At the same time, array $[1, 2, 1, 3]$ isn't good, as $1 + 2 + 1 + 3 = 7 \neq 2\cdot 1 = 2\cdot(1\oplus 2 \oplus 1 \oplus 3)$. You are given an array of length $n$: $a_1, a_2, \dots, a_n$. Append at most $3$ elements to it to make it good. Appended elements don't have to be different. It can be shown that the solution always exists under the given constraints. If there are different solutions, you are allowed to output any of them. Note that you don't have to minimize the number of added elements!. So, if an array is good already you are allowed to not append elements. Input Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10\,000$). The description of the test cases follows. The first line of each test case contains a single integer $n$ $(1\le n \le 10^5)$ — the size of the array. The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($0\le a_i \le 10^9$) — the elements of the array. It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$. Output For each test case, output two lines. In the first line, output a single integer $s$ ($0\le s\le 3$) — the number of elements you want to append. In the second line, output $s$ integers $b_1, \dots, b_s$ ($0\le b_i \le 10^{18}$) — the elements you want to append to the array. If there are different solutions, you are allowed to output any of them. Example Input 3 4 1 2 3 6 1 8 2 1 1 Output 0 2 4 4 3 2 6 2 Note In the first test case of the example, the sum of all numbers is $12$, and their $\oplus$ is $6$, so the condition is already satisfied. In the second test case of the example, after adding $4, 4$, the array becomes $[8, 4, 4]$. The sum of numbers in it is $16$, $\oplus$ of numbers in it is $8$.

https://math.stackexchange.com/questions/939280/part-of-a-proof-that-the-product-of-an-odd-and-even-integers-is-even

# Part of a proof that the product of an odd and even integers is even I'm practicing for a test on Monday and I'm trying to do some proofs - but I'm not entirely sure if this is sufficient enough for the question. "Prove that for all integers, m and n, if m is odd and n is even, then $m\cdot n$ is even." These are the steps I've gone through - but I'm a little unsure on the result. 1. $m = 2k_1+1$ and $n = 2k_2$ 2. $m\cdot n= (2k_1 + 1)(2k_2)$ 3. $= 4k_1k_2+2k_2$ 4. $=2(2k_1k_2+k_2)$ The part I'm unsure on is the last step, I wasn't sure if I was supposed to factor out the $k_2$ as well. • That's fine. You could have jumped directly from step 2 to step 4. – Adriano Sep 20 '14 at 18:42 • Apart from the fact that you used equations, with no explanatory words, everything is fine. We can even skip part of the multiplication step. We have $m=2a+1$ and $n=2b$ for some integers $a$ and $b$. Thus $mn=nm=(2b)(2a+1)=(2)\left[(b)(2a+1)\right]$, so $mn$ is even. – André Nicolas Sep 20 '14 at 18:45 ## 3 Answers You've done all the work, and it's correct. Let me try to demonstrate how improve the style of your proof. This can be very important—graders are invariably tired and overworked, and lack of clarity is one of the most frequent causes of correct proofs being marked as incorrect on exams. Additionally, it feels good for both reader and writer when the math is crisp and clear—it is an often-overlooked fact that the entire purpose of mathematical writing is to communicate ideas to other people. Claim: If $m$ and $n$ are integers, and $n$ is even, then $mn$ is even. Proof: Because $n$ is even, we can write $n=2l$, where $l$ is an integer. Then $mn=m(2l)=2(ml)$ is a multiple of $2$. In other words, $mn$ is even. Notice that each step is justified—though if we had an extremely strict teacher, we might want to point out that we got $mn=m(2l)$ by multiplying both sides of $n=2l$ by $m$, and that $m(2k)=2(ml)$ follows from the associativity and commutativity of multiplication. But you're probably safe in omitting these details unless you've been told otherwise. Also note that I didn't assume that $m$ is odd, because this makes the equations a little more complicated. Here is a more faithful recreation of your proof: Claim: If $m$ is an odd integer and $n$ is an even integer, then $mn$ is even. Proof: Because $m$ is odd and $n$ is even, we can write $m=2k+1$ and $n=2l$, where $k$ and $l$ are integers. Then $mn = (2k+1)(2l) = 2\cdot(l(2k+1))$ is a multiple of $2$, hence even. Define $2k_1k_2+k_2=k_3$ as an integer number. You could further formalize your text, if you want (it is always handy to write down the definitions or the theorems you use, it also makes it easier for others to read and make it look more complete): By definition: $$m \text{ even }\iff \exists k \in \mathbb Z \text{ such that }2k = m$$

https://brilliant.org/discussions/thread/how-to-count-squares/

# How to Count Squares! Let me go grab a hamburger real quick... Ok, I'm back. How many squares are there in the 6$$\times$$4 grid below? That's a reaallly good question! Let's start by counting the smallest 1$$\times$$1 squares, this is just the same as counting the number of unit squares in a 6$$\times$$4 grid, there are $$6\times4=24$$ 1 by 1 squares in the grid. Now let's move on to the 2$$\times$$2 squares, notice that counting the number of 2$$\times$$2 squares in a 6$$\times$$4 grid is just the same as counting the number of unit squares in a 5$$\times$$3 grid. So the number of 2 by 2 squares in the grid is $$5\times3=15$$. Now the 3$$\times$$3 squares, similarly, counting the number of 3$$\times$$3 squares in a 6$$\times$$4 grid is just the same as counting the number of unit squares in a 4$$\times$$2 grid, which is $$4\times2=8$$. And again the number of 4$$\times$$4 squares in a 6$$\times$$4 grid is equal to the number of unit squares in a 3$$\times$$1 grid, which is $$3\times1=3$$. Add up all the number of squares together: $$24+15+8+3=50$$. Tada! We now have our answer! There are 50 squares in a 6$$\times$$4 grid. Mmm... the hamburger is really good... Back on topic, in general, what is the total number of squares in an $$a\times b$$ grid (where $$a$$ is the width of the grid and $$b$$ is the height of the grid), given $$a\geqslant b$$? Again let's start from the 1$$\times$$1 squares, that's trivial, there's $$ab$$ of them. Now moving on to the 2$$\times$$2 squares, the number of 2$$\times$$2 squares in an $$a\times b$$ grid is equal to the number of unit squares in an $$(a-1)\times(b-1)$$ grid. Notice the pattern? Counting the number of $$n\times n$$ squares in an $$a\times b$$ grid is the same as counting the number of unit squares in an $$(a-n+1)\times(b-n+1)$$ grid. The largest square that can contain in an $$a\times b$$ grid given that $$a\geqslant b$$ is $$b\times b$$. Hence, the total number of squares in an $$a\times b$$ grid is $ab+(a-1)(b-1)+(a-2)(b-2)+\ldots+[a-(b-2)][b-(b-2)]+[a-(b-1)][b-(b-1)]$ Or $\sum_{i=0}^{b-1}{(a-i)(b-i)}$ This is ugly, we don't like sigma symbols sitting around, so why not we simplify this a little bit... \begin{align} \sum_{i=0}^{b-1}{(a-i)(b-i)}&=\sum_{i=0}^{b-1}{[ab-(a+b)i+i^2]} \\&=ab^2-\frac{(a+b)b(b-1)}{2}+\frac{b(b-1)(2b-1)}{6} \\&=b\left[ab-\frac{ab-a+b^2-b}{2}+\frac{2b^2-3b+1}{6}\right] \\&=\frac{b}{6}\left[6ab-3ab+3a-3b^2+3b+2b^2-3b+1\right] \\&=\frac{b}{6}\left[3ab+3a-b^2+1\right] \\&=\frac{b(b+1)(3a-b+1)}{6} \end{align} BOOM! There we have it! *Round of applause* *Fireworks* *Pancakes* The total number of squares in an $$a\times b$$ grid (where $$a$$ is the width of the grid and $$b$$ is the height of the grid), given $$a\geqslant b$$ is $\frac{b(b+1)(3a-b+1)}{6}$ If $$a<b$$, then we just swap the $$a$$ and $$b$$ around. Special case: If $$a=b$$, the above equation becomes $\frac{a(a+1)(2a+1)}{6}$ which is the formula of the sum of squares from 1 to $$a$$. Done! Now let me finish my burger... This is one part of Quadrilatorics. Note by Tan Kenneth 2 years, 1 month 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: You should add this to the Brilliant wiki. Great note! - 2 years, 1 month ago Cool! Thanks for that note bro, you're awesome! - 2 years, 1 month ago Hope you finished your burger peacefully :P - 2 years, 1 month ago Thanks, I'm glad you like the note. Well unfortunately, I think my hamburger has become stale. XD - 2 years, 1 month ago Nice simple way of explaining complex situation. So many thanks. - 1 year, 11 months ago are you a robot, cos I need some real friends? Humanity is a lie, the computer generation is upon us. Support the cause m64^(1/2) - 2 years, 1 month ago No, i am 100.1% sure I'm not a robot. - 2 years, 1 month ago What a note @Tan Kenneth:) - 2 years, 1 month ago HEY tankenneth, you hyped? - 2 years, 1 month ago Oh yes I am! $$1+1=3$$ - 2 years, 1 month ago جميلة - 2 years, 1 month ago Translation: beautiful! - 2 years, 1 month ago

https://math.stackexchange.com/questions/2892375/prove-sum-i-0n-1n-i-binomn1i-i1n-n2n

# Prove $\sum_{i=0}^n (-1)^{n-i} \binom{n+1}{i} (i+1)^n = (n+2)^n$ I found through simulations that $$\sum_{i=0}^n (-1)^{n-i} \binom{n+1}{i} (i+1)^n = (n+2)^n$$ Is there any proof of this? I've tried to solve it by: • Induction, but it gets too messy. • Binomial theorem, but I got nowhere. Any help with this is highly appreciated. • I bet induction is not so bad in this case, but I'll try to find a more elegant explanation Aug 23, 2018 at 17:34 • Thanks @Exodd .. Would love to see it through induction. But any other method is also okay. Aug 23, 2018 at 17:35 • Notice that you can rewrite it as $$\sum_{i=0}^{n+1} (-1)^{n-i} \binom{n+1}{i} i^{n} =0$$ Aug 23, 2018 at 18:01 Let $[m]$ denote the set of the first $m$ positive integers. Let $S$ be the set of functions from $[n]\to [n+2]$, and for $1\le i \le n+1$, let $S_i$ be the set of such functions whose range does not contain $i$. Obviously, $|S|=(n+2)^n$. Furthermore, the range of a function in $S$ will not contain at least two elements, since the range has at most $n$ elements. Therefore, some number $1\le i\le n+1$ must be missing from the range, so $$S=\bigcup_{i=1}^{n+1} S_i$$ Therefore, we can use the principle of inclusion-exclusion to count $|S|$ in terms the sizes of the $k$-way intersections $|S_{i_1}S_{i_2}\dots S_{i_k}|$. For any $k$, the intersection of $k$ of the sets $S_i$ consists of functions whose range does not contain a particular $k$ elements. The number of such functions is $(n+2-k)^n$. Therefore, using the inclusion-exclusion formula, $$|S|=(n+2)^n = \sum_{k=1}^{n+1} (-1)^{k+1}\binom{n+1}k(n+2-k)^n$$ The result follows by reversing the order of the summation, i.e. setting $k\leftarrow n+1-i$. Posting a second answer because there is a completely different, more direct proof. Note that subtracting the $(n+2)^n$ on the right over, this is equivalent to proving $$\sum_{i=0}^{n+1} (-1)^{n-i}\binom{n+1}i(i+1)^n\stackrel{?}=0$$ which after multiplying both sides by $(-1)^n$ equals $$\sum_{i=0}^{n+1} (-1)^{i}\binom{n+1}i(i+1)^n \stackrel{?}=0\tag{1}$$ To prove this, consider the polynomial $$\sum_{i=0}^{n+1} \binom{n+1}i(i+1)^nx^i \tag{2}$$ I claim that $(2)$ is the result of taking the polynomial $$\sum_{i=0}^{n+1} \binom{n+1}ix^i\tag{3}$$ and performing the following operation $n$ times: multiply the polynomial by $x$, then differentiate. Indeed, each summand of a polynomial looks like $a_i x^i$, and when you multiply by $x$ and differentiate, the result is $(i+1)a_ix^i$. Doing this $n$ times results in $(i+1)^na_ix^i$. But the polynomial in $(3)$ is by the binomial theorem equal to $(1+x)^{n+1}$. Note that this has a zero of order $n+1$ at $x=-1$. Multiplying by $x$ does not change this. It is a standard result that if $f(x)$ has a zero of order $k$ at $x_0$, then $f'(x)$ has a zero of order $k-1$ at $x_0$. Therefore, since $(3)$ has a zero of order $n+1$, it follows that after $n$ differentiations (and some multiplications by $x$), it will still have a zero of order $1$. In particular, $(2)$ has a zero at $x=-1$, so the expression in $(1)$ equals zero. • This is outstanding .. Thank you for the clear clarification. I loved both answers. Aug 23, 2018 at 18:37 • Nice answer (+1). May 18, 2019 at 8:17 Start from $$\sum_{q=0}^n (-1)^{n-q} {n+1\choose q} (q+1)^n \\ = \sum_{q=0}^n (-1)^{n-q} {n+1\choose q} n! [z^n] \exp((q+1)z) \\ = n! [z^n] \exp(z) \sum_{q=0}^n (-1)^{n-q} {n+1\choose q} \exp(qz) \\ = - n! [z^n] \exp(z) \times (-1) \exp((n+1)z) \\ + n! [z^n] \exp(z) \sum_{q=0}^{n+1} (-1)^{n-q} {n+1\choose q} \exp(qz) \\ = n! [z^n] \exp((n+2)z) \\ - n! [z^n] \exp(z) \sum_{q=0}^{n+1} (-1)^{n+1-q} {n+1\choose q} \exp(qz) \\ = (n+2)^n \\ - n! [z^n] \exp(z) (\exp(z)-1)^{n+1}.$$ Now $\exp(z)-1 = z + \cdots$ and hence $(\exp(z)-1)^{n+1} = z^{n+1} +\cdots$ and therefore $$[z^n] \exp(z) (\exp(z)-1)^{n+1} = 0$$ and we are left with $$\bbox[5px,border:2px solid #00A000]{ (n+2)^n.}$$ This is a less clever algebraic proof by induction. We actually prove something stronger: For any $\lambda>0$, we have $$\sum_i(-1)^i\binom{n+1}{i}(i+\lambda)^n=0$$ where $i$ runs through all integers and define $\binom{n}{i}=0$ if $i<0$ or $i>n$. Proof. This is straight-forward for $n=0$. Suppose for $n=k$ the statement is true. Then \begin{align} \sum_i(-1)^i\binom{k+2}i(i+\lambda)^{k+1}&=\sum_i(-1)^i\binom{k+2}i\binom i1(i+\lambda)^k+\lambda\sum_i(-1)^i\binom{k+2}{i}(i+\lambda)^k \end{align} Note that $\binom{k+2}i\binom i1=\binom{k+2}1\binom{k+1}{i-1}$ and $\binom{k+2}i=\binom{k+1}i+\binom{k+1}{i-1}$. By induction hypothesis we obtain $\sum(-1)^i\binom{k+1}i(i+\lambda)^k=0$, thus \begin{align} \sum_i(-1)^i\binom{k+2}i(i+\lambda)^{k+1}&=(k+2+\lambda)\sum_i(-1)^i\binom{k+1}{i-1}(i+\lambda)^k\\ &=-(k+2+\lambda)\sum_i(-1)^i\binom{k+1}{i}(i+\lambda+1)^k\\ &=0 \end{align} where the last equality follows from induction hypothesis again.

http://mathhelpforum.com/statistics/144005-help-svp.html

# Math Help - Help, SVP. 1. ## Help, SVP. Hey... if you could answer/explain any of the following (even one would be awesome), I'd be forever thankful. 1. A six-member working group to plan a student lounge is to be selected from five teachers and nine students. If the working group is randomly selected, what is the probability of the committee including at least two teachers. 2. Cha’tima has five white and six grey huskies in her kennel. If a wilderness expedition chooses a team of six sled dogs from Cha’tima’s kennel, what is the probability that the team will consist of: all white huskies? all grey huskies? three of each colour? 3. A survey at a school asked students if they were ill with a cold or the flu during last months. The results were as follows. None of the students had both a cold and the flu. Cold Flu Healthy Females 32 18 47 Males 25 19 38 Use these results to estimate the probability that: a randomly selected student had a cold last month. a randomly selected females students was healthy last month. a randomly selected student who had the flu last month is male. a randomly selected student had either a cold or the flu last month. 4. To get out of jail free in the board game MonopolyÒ, you have to roll doubles with a pair of dice. Determine the odds in favour of getting out of jail on your first or second roll. 5. If a survey on teenage reads of popular magazines shows that 38% subscribe to Teen Life and 47% subscribe to Cool Teen and 35% subscribe to neither magazine, what is the probability that a randomly selected teenager, subscribes to both magazines? subscribes to either one magazine or both? subscribes to only one of the two magazines? 2. I got this for 2,3, and 4. 2) Prob is 0 b/c only 5 whites are there P(all grey) = 1/(11 6) = 1/462, ~ 0.0022 P(3 white & 3 grey) ((5 3) (6 3))/(11 6), 200/462, ~ 0.433 3) P(cold)=57/179, =0.3184, ~ 31.8% P(health | female) (47/179)/(97/179) = 47/97 = 0.4845, ~ 48.5% P(male | flu) 19/37 = 0.5135, ~51.3% 44/82 = 0.5366, ~53.7% Double on first roll = 6/36 = 1/6, ergo not rollin = 5/6 Ergo prob of rolling doubles on the second roll: 5/6 * 1/6 = 5/36 Prob of rolling double on first or second: 1/6 + 5/6 = 11/36 Ergo odds of getting out of jail on first or second roll are 11:25 Confirm? And help with 1 and 5? 3. Originally Posted by acc Confirm? And help with 1 and 5? I didn't get a calculator out to confirm final answers, but I agree with your work for 2,3,4 except for the following: for 3b, I would simply write 47/97 without the initial (47/179)/(97/179) step. for 3d, I get 94/179. Your answer is for a randomly chosen male student. for 4, there's an assumption we have to make, and that is that we don't want to count the probability of getting doubles on both the first and second rolls (because presumably, rolling a double on the first roll will get us out of jail and we won't make the second roll). Assuming that, I agree with your answer. (But you made a typo by writing 5/6 in one spot where you meant 5/36.) For problem 1, I would start with a "1" and subtract the following probabilities from it P(0 teachers and 6 students) P(1 teacher and 5 students) Problem 5, you could draw a Venn diagram. Or using formal notation, $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ (Edit: fixed typo.) Edit 2: The part in gray is incorrect. What you wrote for 4 is fine the way it is, other than the typo. 4. Hello, acc! 2. Cha’tima has five white and six grey huskies in her kennel. If a wilderness expedition chooses a team of six sled dogs from Cha’tima’s kennel, what is the probability that the team will consist of: (a) all white huskies? (b) all grey huskies? (c) three of each colour? There are: . ${11\choose6} \:=\:462$ possible selections. $(a)\;P(\text{6 white}) \;=\;0$ $(b)\;P(\text{6 grey}) \;=\;\frac{{6\choose6}}{462} \;=\;\frac{1}{462}$ $(c)\;P(\text{3 white, 3 grey}) \;=\;\frac{{5\choose3}{6\choose3}}{462} \;=\;\frac{10\cdot20}{462} \;=\;\frac{100}{231}$ 3. A survey at a school asked students if they were ill with a cold or the flu during last month. The results were as follows. None of the students had both a cold and the flu. . . $\begin{array}{c||c|c|c||c|} & \text{Cold} & \text{Flu} & \text{Healthy } & \text{Total} \\ \hline\hline \text{Females} & 32 & 18 & 47 & 97 \\ \hline \text{Males} & 25 & 19 & 38 & 82 \\ \hline \hline \text{Total} & 57 & 37 & 85 & 179 \\ \hline \end{array}$ Use these results to estimate the probability that: (a) a randomly selected student had a cold last month. (b) a randomly selected female student was healthy last month. (c) a randomly selected student who had the flu last month is male. (d) a randomly selected student had either a cold or the flu last month. $(a)\;P(\text{Cold}) \:=\:\frac{57}{179}$ $(b)\;P(\text{Female}\,\wedge\,\text{Healthy}) \;=\;\frac{47}{179}$ $(c)\;P(\text{Male}|\text{Flu}) \;=\;\frac{19}{37}$ $(d)\;P(\text{Cold} \vee\text{Flu}) \;=\;P(\text{Cold}) + P(\text{Flu}) \;=\;\frac{57}{179} + \frac{37}{170} \;=\;\frac{94}{179}$ 5. Originally Posted by Soroban $(a)\;P(\text{Cold}) \:=\:\frac{57}{179}$ $(b)\;P(\text{Female}\,\wedge\,\text{Healthy}) \;=\;\frac{47}{179}$ $(c)\;P(\text{Male}|\text{Flu}) \;=\;\frac{19}{37}$ $(d)\;P(\text{Cold} \vee\text{Flu}) \;=\;P(\text{Cold}) + P(\text{Flu}) \;=\;\frac{57}{179} + \frac{37}{170} \;=\;\frac{94}{179}$ Wow, I completely misread (c) and thought it said "a randomly selected student had the flu last month." For (b), I'm wondering if it is a matter of interpretation. To get the probability you found, wouldn't it have better been stated "a randomly selected student is female and had the flu last month"? Edit: I must be tired because I didn't even address 3c, I'll stop posting now and get some rest. 6. Originally Posted by acc Hey... if you could answer/explain any of the following (even one would be awesome), I'd be forever thankful. 1. A six-member working group to plan a student lounge is to be selected from five teachers and nine students. If the working group is randomly selected, what is the probability of the committee including at least two teachers. [snip] 5. If a survey on teenage reads of popular magazines shows that 38% subscribe to Teen Life and 47% subscribe to Cool Teen and 35% subscribe to neither magazine, what is the probability that a randomly selected teenager, subscribes to both magazines? subscribes to either one magazine or both? subscribes to only one of the two magazines? Duplicate posted (and answered) here: http://www.mathhelpforum.com/math-he...tml#post511545

http://experiential-architecture.com/kushco-holdings-oercf/find-permutation-problem-fd31c5

This article has been viewed 12,813 times. For example, if you have 10 digits to choose from for a combination lock with 6 numbers to enter, and you're allowed to repeat all the digits, you're looking to find the number of permutations with repetition. If you're seeing this message, it means we're having trouble loading external resources on our website. Discuss: answer with explanation. Mathematically we can approach this question as follows: $$P=\frac{n!}{n_1! To create this article, volunteer authors worked to edit and improve it over time. = n (n - 1) (n - 2) (n - 3) …. n_2! Solution: There are 4 letters in the word love and making making 3 letter words is similar to arranging these 3 letters and order is important since LOV and VOL are different words because of the order of the same letters L, O and V. Hence it is a permutation problem. Then, find 7! 21300: C. 24400: D. 210: View Answer. Yes. The basic building block to solve permutation and combination problems is to understand the use of recursion in a for loop. Some graphing calculators offer a button to help you solve permutations without repetition quickly. To create this article, volunteer authors worked to edit and improve it over time. 0! other thatn the given sequence in the argument of the function where c is in the last position. Important Permutation Formulas. Solution:You need two points to draw a line. We use cookies to make wikiHow great. Last Updated: October 14, 2020 A committee including 3 boys and 4 girls is to be formed from a group of 10 boys and 12 girls. This image may not be used by other entities without the express written consent of wikiHow, Inc. \n<\/p> \n<\/p><\/div>"}, https://www.mathsisfun.com/combinatorics/combinations-permutations.html, https://betterexplained.com/articles/easy-permutations-and-combinations/, https://www.learneroo.com/modules/10/nodes/62, https://en.wikipedia.org/wiki/Permutation#Permutations_with_repetition, https://www.vitutor.com/statistics/combinatorics/permutations_repetition.html, https://www.learneroo.com/modules/10/nodes/55, https://mathbits.com/MathBits/TISection/Algebra1/Probability.htm, https://www.mathsisfun.com/combinatorics/combinations-permutations-calculator.html, consider supporting our work with a contribution to wikiHow. If you really can’t stand to see another ad again, then please consider supporting our work with a contribution to wikiHow. At this point, we have to make the permutations of only one digit with the index 3 and it has only one permutation i.e., itself. P (n,r) represents the number of permutations of n items r at a time. A permutation is an arrangement of objects in which the order is important[1] This article has been viewed 12,813 times. This program will find all possible combinations of the given string and print them. You will more details about each type of problem in the problem definition section. In a certain country, the car number plate is formed by 4 digits from the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9 followed by 3 letters from the alphabet. You'd then calculate 3,628,800/5,040. If you're using Google Calculator, click on the, If you have to solve by hand, remember that, for each. Find the order in lexicographical sorted order If we want to find all the permutations of a string in a lexicographically sorted order means all the elements are arranged in alphabetical order and if the first element is equal then sorting them based on the next elements and so on. Given a positive integer n and a string s consisting only of letters D or I, you have to find any permutation of first n positive integer that satisfy the given input string. Enter "7" for "Number of sample points in set ". wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. The order of arrangement of the object is very crucial. To solve this problem using the Combination and Permutation Calculator, do the following: Choose "Count permutations" as the analytical goal. In other words, if the set is already ordered, then the rearranging of its elements is called the process of permuting. In a certain state’s lottery, 48 balls numbered 1 through 48 are placed in a machine and six of them … letters in our case which is 6 \(n_1$$ is the number of objects of type 1, for example, the number of b which is 2 $$n_2$$ is the number of objects of type 2, for example, the number of a which is 1 Research source Iteration : : Iteration, in the context of computer programming, is a process wherein a set of instructions or structures are repeated in a sequence a specified number of times or until a condition is met. Mathematically we can approach this question as follows: $$P=\frac{n!}{n_1! We leave the 3 out and recursively find all permutations of the set {1, 2}. For example, consider finding all permutations of {1, 2, 3). The permutation problems are arrangement problems and the combination problems are selection problems. This image is not<\/b> licensed under the Creative Commons license applied to text content and some other images posted to the wikiHow website. in Physics and Engineering, Exercises de Mathematiques Utilisant les Applets, Trigonometry Tutorials and Problems for Self Tests, Elementary Statistics and Probability Tutorials and Problems, Free Practice for SAT, ACT and Compass Math tests, Arithmetic Sequences Problems with Solutions, High School Math (Grades 10, 11 and 12) - Free Questions and Problems With Answers, Mathematical Induction - Problems With Solutions, Geometric Sequences Problems with Solutions. You can use a simple mathematical formula to find the number of different possible ways to order the items. For example 7 and 4. In how many ways can you select a committee of 3 students out of 10 students? (3) (2) (1) Permutations of n items taken r at a time. Permutation With Repetition Problems With Solutions : In this section, we will learn, how to solve problems on permutations using the problems with solutions given below. The is below the up/down arrow, at the top left of the answer. You may also vote on the quality/helpfulness of an answer, with the up or down arrow, if you have a 15+ reputation. The problem is to select 2 points out of 3 to draw different lines. 25200: B. References. P (n,r) = n!/ (n - r)! This consists of 2 permutations: {1, 2} {2, 1] Now we insert the 3 into every position for these permutations. In mathematics, permutation relates to the act of arranging all the members of a set into some sequence or order. How many different committee can be formed from the group? Problem Statement. For instance, you might be selecting 3 representatives for student government for 3 different positions from a set of 10 students. In mathematics, a permutation of a set is, loosely speaking, an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its elements.The word "permutation" also refers to the act or process of changing the linear order of an ordered set. Plug your numbers in for n {\displaystyle n} and r {\displaystyle r}. ). }$$ Where: $$n$$ is the total number of object, i.e. Calculate the number of permutations of n elements taken r at the time. That number means that, if you're picking from 10 different students for 3 student government positions, where order matters and there is no repetition, there are 720 possibilities. wikiHow, Inc. is the copyright holder of this image under U.S. and international copyright laws. A. Let’s take an example to understand the problem, Input xyz Output xyz, xzy, yxz, yzx, zxy, zyx Explanation These are all permutations take in order. This kind of problem... 2. This method takes a list as an input and returns an object list of tuples that contain all … = 24 . }\) Where: $$n$$ is the total number of object, i.e. = 5*4*3*2*1 = 120. So out of that set of 4 horses you want to pick the subset of … By using our site, you agree to our. Solved Examples(Set 1) - Permutation and Combination. Is there an easy way in order to memorize the formula for this? letters in our case which is 6 $$n_1$$ is the number of objects of type 1, for example, the number of b which is 2 $$n_2$$ is the number of objects of type 2, for example, the number of a which is 1 Permutations is not an easy problem. Permutations and combinations are used to solve problems. A permutation is a reordered arrangement of elements or characters of a string. by doing (10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1), which gives you 3,628,800 as a result. Suppose we have three people named A, B, and C. We have already determined that they can be seated in a straight line in 3! / (4 - 3)! n_2! Permutation With Repetition Problems With Solutions - Practice questions. n_3!…n_k! n_3!…n_k! 1. Include your email address to get a message when this question is answered. Find Permutation. How many 3 digit numbers can we make using the digits 2, 3, 4, 5, and 6 without repetitions? % of people told us that this article helped them. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed? The first problem comes under the category of Circular Permutations, and the second under Permutations with Similar Elements. The C program to find permutation and combination solves 4 different types of problems. 1! The order is not important. Enter "3" for "Number of sample points in each permutation". Solve the equation to find the number of permutations. Think of it like this: subtract the total amount by the total items. means (n factorial). Permutations occur, in more or less prominent ways, in almost every area of mathematics. The number of words is given by 4 P 3 = 4! Now in this permutation (where elements are 2, 3 and 4), we need to make the permutations of 3 and 4 first. = 1 Let us take a look at some examples: Problem 1: Find the number of words, with or without meaning, that can be formed with the letters of the word ‘CHAIR’. Ending index of the string. A pemutation is a sequence containing each element from a finite set of n elements once, and only once. Calculating Permutations without Repetition 1. By signing up you are agreeing to receive emails according to our privacy policy. In the example, you should get 720. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. How many triangles can you make using 6 non collinear points on a plane? Second … Permutations . Intuition If there were no Kleene stars (the * wildcard character for regular expressions), the problem would be easier - we simply check from left to right if each character of the text matches the pattern. For those who haven’t seen a backtracking question before, there is no clear naive solution, and this poses a real threat for software engineers during… If you're working with combinatorics and probability, you may need to find the number of permutations possible for an ordered set of items. Start with an example problem where you'll need a number of permutations without repetition. (unlike combinations, which are groups of items where order doesn't matter[2] These methods are present in an itertools package. Hopefully, you have seen … n! To start off, you just need to know whether repetition is allowed in your problem or not, and then pick your method and formula accordingly. Calculating Permutations without Repetition, {"smallUrl":"https:\/\/www.wikihow.com\/images\/thumb\/b\/b0\/Calculate-Permutations-Step-1.jpg\/v4-460px-Calculate-Permutations-Step-1.jpg","bigUrl":"\/images\/thumb\/b\/b0\/Calculate-Permutations-Step-1.jpg\/aid11015428-v4-728px-Calculate-Permutations-Step-1.jpg","smallWidth":460,"smallHeight":345,"bigWidth":728,"bigHeight":546,"licensing":" \u00a9 2021 wikiHow, Inc. All rights reserved. The permutation is an arrangement of objects in a specific order. Combinations P(n) = n! A new solution can be accepted if a better one shows up. D means the next number is smaller, while I means the next number is greater. to 4, so 7x6x5 and then find the answer, and you’ll get the permutations. Example 6: How many lines can you draw using 3 non collinear (not in a single line) points A, B and C on a plane? If your question is solved, say thank you by accepting the solution that is best for your needs. Solution: ‘CHAIR’ contains 5 letters. would be (7 * 6 * 5 * 4 * 3 * 2 * 1), which would equal 5,040. Given a set of ‘n’ elements, find their Kth permutation. How many 4 digit numbers can we make using the digits 3, 6, 7 and 8 without repetitions? Each question has four choices out … Approach #1 Using Stack [Accepted] Let's revisit the important points of the given problem statement. Therefore, the number of words that can be formed with these 5 letters = 5! First import itertools package to implement the permutations method in python. Thanks to all authors for creating a page that has been read 12,813 times. FYI: Thoroughly answering questions is time-consuming. Research source It usually looks like, All tip submissions are carefully reviewed before being published. Question 1 : 8 women and 6 men are standing in a line. If we proceed as we did with permutations, we get the following pairs of points to draw lines.AB , ACBA , BCCA , CBThere is a problem: line AB is the same as line BA, same for lines AC and CA and BC and CB.The lines are: AB, BC and AC ; 3 lines only.So in fact we can draw 3 lines and not 6 and that's because in this problem the order of the points A, B and C is not important.This is a combination problem: combining 2 items out of 3 and is written as follows: eval(ez_write_tag([[580,400],'analyzemath_com-large-mobile-banner-1','ezslot_4',700,'0','0'])); Calculate the number of combinations of n elements taken r at the time. Graphs of Functions, Equations, and Algebra, The Applications of Mathematics Permutations of the same set differ just in the order of elements. Permutation and Combination Solve Problems Quickly. For example, you would calculate 10! How many number plates can be formed if neither the digits nor the letters are repeated. Explanation: Number of ways of selecting 3 consonants from 7 X If you have a calculator handy, find the factorial setting and use that to calculate the number of permutations. Learn How to Solve Permutation and Combination Question Quickly form PrepInsta. Solution: n-factorial gives the number of permutations of n items. Find all Permutations of the word baboon. Consider the following set of elements: 7! Recursive functions are very useful to solve many mathematical problems, such as calculating the factorial of a number, generating Fibonacci series, etc. For the first permutation we insert … Circular Permutations . When a star is present, we may need to check many different suffixes of the text and see if they match the rest of the pattern. X = 1. wikiHow is where trusted research and expert knowledge come together. Python provides a package to find permutations and combinations of the sequence. Find all Permutations of the word baboon. The list of problems is given below. Permutation Problem 1 Choose 3 horses from group of 4 horses In a race of 15 horses you beleive that you know the best 4 horses and that 3 of them will finish in the top spots: win, place and show (1st, 2nd and 3rd). 4. wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. Output First Swap generates the other permutation, i.e. 3. Introductory permutation problems. def permute (a, l, r): if l = = r: print toString (a) else: for i in xrange (l,r + 1 ): a [l], a [i] = a [i], a [l] permute (a, l + 1, r) a [l], a [i] = a [i], a [l] # backtrack. Line AB is the same as line BA. or 6 ways. The permutation test is designed to determine whether the observed difference between the sample means is large enough to reject, at some significance level, the null hypothesis H that the data drawn from is from the same distribution as the data drawn from. How many 6 letter words can we make using the letters in the word LIBERTY without repetitions? The number of permutations on the set of n elements is given by n! In how many ways can you arrange 5 different books on a shelf? Question.How many three digit numbers can be formed using digits 2, 3, 4, 7, 9 so that the digits can be repeated. Permutations differ from combinations, which are selections of some members of a set regardless of … PERMUTATION WORD PROBLEMS WITH SOLUTIONS Problem 1 : A student appears in an objective test which contain 5 multiple choice questions. Given a string, we have to find all the permutations of that string. Answer: Option A. Problem Statement. In this regard, what is the importance of permutation and combination? Input. Permutation is the arrangement of all parts of an object, in all possible orders of arrangement. Permutations with repetition n 1 – # of the same elements of the first cathegory n 2 - # of the same elements of the second cathegory For example, string “abc” have six permutations [“abc”, “acb”, “bac”, “bca”, “cab”, “cba”]. We know ads can be annoying, but they’re what allow us to make all of wikiHow available for free. We will be given a single string input. No student can be used in more than one position (no repetition), but the order still matters, since the student government positions are not interchangeable (a permutation where the first student is President is different from a permutation where they're Vice President). Please help us continue to provide you with our trusted how-to guides and videos for free by whitelisting wikiHow on your ad blocker. And thus, permutation(2,3) will be called to do so. Similarly, permutation(3,3) will be called at the end. Really can ’ t stand to see another ad again, then consider... 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https://math.stackexchange.com/questions/2919921/of-all-polygons-inscribed-in-a-given-circle-which-one-has-the-maximum-sum-of-squ

# Of all polygons inscribed in a given circle which one has the maximum sum of squares of side lengths? My son presented me with an interesting problem: Of all possible polygons inscribed in a circle of radius $R$, find the one that has the sum $S$ of squared side lengths maximized: $S=a_1^2+a_2^2+\dots+a_n^2$, with $a_i$ representing the length of the $i$-th side. The number of sides is not fixed, you should consider all triangles, quadrilaterals, pentagons... It's not that complicated, at least in the beginning. It's easy to show that the optimal polygon (with $n>3$) cannot have obtuse ($>90^\circ$) angles. For example, if such an angle $A_{i-1}A_{i}A_{i+1}$ exists, by cosine theorem: $$|A_{i-1}A_{i}|^2+|A_{i}A_{i+1}|^2<|A_{i-1}A_{i+1}|^2$$ So if you drop vertex $A_i$, you get a polygon with a bigger $S$. This quickly eliminates all polygons with $n>4$. All candidate polygons with $n=4$ must be rectangles and if their sides are $a$ and $b$, the sum $S$ is $2a^2+2b^2=8R^2$. So with respect to $S$ all rectangles inscribed in the circle are equivalent. In fact, a right triangle with sides $a$, $b$ and $2R$ has the same $S$ as any inscribed rectangle. But maybe there is an inscribed triangle with $S>8R^2$. I was able to show that for an inscribed triangle with sides $a,b,c$ and $b\ne c$, an isosceles triangle with all acute angles and base $a$ has better value of $S$. So the optimal triangle must be isosceles. Looking from all three sides, the only possible solution is the equilateral triangle and the sum $S$ in that case is $9R^2$. However, to prove that fact I had to use trigonometry which is not so complicated (and I can present it here if you want so), but it leaves impression that there has to be some simpler explanation why the equilateral triangle is the best choice. My trigonometry proof takes a few lines of text, I want something more elegant. Just an idea: if you draw lines through the center of the circle perpendicular to the sides of a triangle and denote the pedal lengths with $h_a,h_b,h_c$, it's easy to show that in order to maximize $a^2+b^2+c^2$ you have to minimize $h_a^2+h_b^2+h_c^2$. And then - what? EDIT: I want to present the part of the proof that I don't like. Take an arbitrary triangle $ABC$ with sides $a,b,c$ inscribed in a circle. Consider side $a$ fixed and play with angle $\gamma$ to get different values of $b,c$. I want to prove that isosceles triangle $BCA_1$ has bigger $S$ than any other triangle with one side equal to $a$. $$b=2R\sin\frac{\pi-\alpha+\gamma}{2}=2R\cos\left(\frac\alpha2-\frac\gamma2\right)$$ $$c=2R\sin\frac{\pi-\alpha-\gamma}{2}=2R\cos\left(\frac\alpha2+\frac\gamma2\right)$$ $$b^2=4R^2\cos^2\left(\frac\alpha2-\frac\gamma2\right)=2R^2(1+\cos(\alpha-\gamma))$$ $$c^2=4R^2\cos^2\left(\frac\alpha2+\frac\gamma2\right)=2R^2(1+\cos(\alpha+\gamma))$$ $$b^2+c^2=4R^2+2R^2(\cos(\alpha-\gamma)+\cos(\alpha+\gamma))=4R^2(1+\cos\alpha\cos\gamma)$$ And this sum achieves maximum obviously for $\gamma=0$, or for $A\equiv A_1$. So for any given side $a$, $b$ and $c$ must be of equal. But you can look at the optimal triangle from sides $b$ and $c$ as well. The only triangle which has no better option is equilateral triangle. EDIT 2: This “moving vertex” procedure can be repeated infinite number of times and the result is an equilateral triangle! Check excellent proof by Noah Schweber here. • Note that your last sentence "The only triangle which has no better option is equilateral triangle." is insufficient to imply "The equilateral triangle is better than every other triangle.". See my answer for a sketch of how to do that part rigorously. – user21820 Sep 17 '18 at 17:45 • @Oldboy, I'm studying your topic and curiously I have a problem a little similar in here. The triangle's case was the most problematic for me too. – Na'omi Sep 18 '18 at 14:10 Yes, the maximal sum is the one of the equilateral triangle, that is $9R^2$. Since Prove that in any triangle $ABC$, $\cos^2A+\cos^2B+\cos^2C\geq\frac{3}{4}$ then $$\sin^2 A+\sin^2 B+\sin^2 C=3-\cos^2 A-\cos^2 B-\cos^2 C\leq \frac{9}{4}$$ where $A$, $B$ and $C$ are non negative numbers such that $A+B+C=\pi$. Hence, for any inscribed triangle, the sum of the squares of the sides is $$(2R\sin A)^2+(2R\sin B)^2+(2R\sin C)^2\leq 9R^2.$$ • I have upvoted your answer, it's just that it's not much simpler compared with my solution. It uses a trigonometric fact about triangle that is not obvious. Anyway, if nothing better shows up in a day or two, I'll accept your answer as the best one. – Oldboy Sep 17 '18 at 9:27 This problem can be stated as $$\max_{n}\sum_{k=1}^n\left(2r\sin\left(\frac{\theta_k}{2}\right)\right)^2$$ s. t. $$\sum_{k=1}^n\theta_k = 2\pi$$ but $$\sum_{k=1}^n\left(2r\sin\left(\frac{\theta_k}{2}\right)\right)^2\ge n\left(2^{2n}r^{2n}\prod_{k=1}^n\sin^2\left(\frac{\theta_k}{2}\right)\right)^{\frac 1n}$$ assuming $\theta_1=\cdots=\theta_n$ we have $$\sum_{k=1}^n\left(2r\sin\left(\frac{\theta_k}{2}\right)\right)^2\ge n\left(2^{2n}r^{2n}\sin^{2n}\left(\frac{\pi}{n}\right)\right)^{\frac 1n} = n2^2r^2\sin^{2}\left(\frac{\pi}{n}\right)$$ Now calling $$f(n) = n\sin^{2}\left(\frac{\pi}{n}\right)$$ we have clearly a maximum about $n = 3$ as can be depicted in the attached plot • This is interesting, but as far as I can tell it assumes that the polygon is regular; and in some cases an irregular polygon can do better for fixed $n$. For example, for a regular hexagon, we have $S = 6 R^2$. But for a hexagon where four of the points are degenerate, the hexagon is "basically" a triangle with three sides of length zero; and if we pick the other three sides to have equal length, we have an equilateral triangle with $S = 9 R^2$. – Michael Seifert Sep 17 '18 at 14:50 Let $\theta_k$ be the successive angles subtended by the sides, but the last one. The sum of squares is given by $$4\sum_{k=1}^n\sin^2\frac{\theta_k}2+4\sin^2\left(\pi-\frac12\sum_{k=1}^n\theta_k\right)$$ which has the same extrema as $$\sum_{k=1}^n\cos\theta_k-\cos\left(\sum_{k=1}^n\theta_k\right).$$ $$\sin\theta_k=\sin\left(\sum_{k=1}^n\theta_k\right).$$ This shows that all angles $\theta_k$ must be equal, and then $$n\cos\theta-\cos n\theta$$ is minimized with $n\theta=2\pi$. Finally, $$n\cos\frac{2\pi} n-\cos\pi$$ is the smallest with $n=3$. The objective function is continuous on the domain of interest (all triples of points on the circle), which is also compact. Therefore by the extreme value theorem it has a global maximum. That reduces the problem to the part you are interested in, namely proving that if the optimal triangle has sides $a,b,c$ then $b = c$. Firstly it has to be acute as you observed. Thus maximizing $b^2+c^2$ $= a^2+2bc·\cos(\angle A)$ is equivalent to maximizing $bc$, since fixing $B,C$ fixes $a$ and $\angle A$. Letting $x = \angle BAO$ and $y = \angle OAC$ we have $bc = 4R^2·\cos(x)\cos(y)$, and finally note $2 \cos(x)\cos(y)$ $= \cos(x+y) + \cos(x-y)$ $\le \cos(x+y) + 1$ $= \cos(\angle A) + 1$ with equality exactly when $x=y$. • Part of the reason for the first part of my answer is that I have a small suspicion that you did not get that part right. One cannot automatically claim that every optimization problem has a global extremum. This has to be proven, and you cannot do so via an infinite iterative process unless you prove that it converges. One way around this is to use the extreme value theorem if the domain is compact, as here. – user21820 Sep 17 '18 at 17:43 \begin{align} AB^2 + BC^2 + CA^2 &= (\vec{OB} - \vec{OA})^2 + (\vec{OC} - \vec{OB})^2 + (\vec{OA} - \vec{OC})^2\\ &= 2(\vec{OA})^2 + 2(\vec{OB})^2 + 2(\vec{OC})^2 - 2 \times \vec{OA} \cdot \vec{OB} - 2 \times \vec{OB} \cdot \vec{OC} - 2 \times \vec{OC} \cdot \vec{OA} \\ &= 3(\vec{OA})^2 + 3(\vec{OB})^2 + 3(\vec{OC})^2 - (\vec{OA} + \vec{OB} + \vec{OC})^2 \\ &\leq 9R^2. \end{align} • I would try to avoid $\times$, as it usually denotes cross product in this context. Also, this appears to be simply proving that the value for an equilateral triangle is $9R^2$, rather than, as the OP asks, proving that this is maximal. – Acccumulation Sep 17 '18 at 15:20 • @Acccumulation sorry, what exactly is unclear in proving that $AB^2+BC^2+CA^2 \leq 9R^2$ for an arbitrary inscribed triangle? – Roman Odaisky Sep 17 '18 at 15:23 We want to maximise $b^2+c^2$, which by the cosine rule is equal to $a^2+2bc\cos A$. Angle $\angle BAC$ is fixed, so this means maximising $bc$. The area of the triangle is $\frac12bc\sin A$, and $\sin A$ is fixed, so this means maximising the area of the triangle. The area of the triangle is $\frac12$base$\times$height $= \frac12a\times$ height, so this means maximising the height. And the height is maximum when $\triangle BA_1C$ is isosceles. Suppose we have three unit vectors a, b, and c. This will define a triangle with side lengths (a-b), (a-c), and (b-c), so the sum of squares will be (a-b)^2+(a-c)^2+(b-c)^2. Taking the derivative with respect to a, we get 2a'(a-b)+2a'(a-c)=2a'(2a-(b+c)). Because a is constrained to be on the unit circle, a' is perpendicular to a, hence a'a=0. So the derivative simplifies to -2a'(b+c). Thus, the derivative is zero if a' is perpendicular to b+c, which is equivalent to a being parallel to b+c, which happens when the angle between a and b is equal to the angle between a and c. Applying the same argument to the derivatives with respect to b and c shows that all the angles must be equal.

https://math.stackexchange.com/questions/508308/a-definite-integral-int-0-infty-frac2-cos-x-left1x4-right-left5-4

# A definite integral $\int_0^\infty\frac{2-\cos x}{\left(1+x^4\right)\,\left(5-4\cos x\right)}dx$ I need to find a value of this definite integral: $$\int_0^\infty\frac{2-\cos x}{\left(1+x^4\right)\,\left(5-4\cos x\right)}dx.$$ Its numeric value is approximately $0.7875720991394284$, and lookups in Inverse Symbolic Calculator Plus and WolframAlpha did not return a plausible closed-form candidate. Do you have any ideas how I can approach this problem? • Looks like you already have a value there. Do you have a particular reason to think it has a closed form? – Henning Makholm Sep 28 '13 at 23:55 • @HenningMakholm One should always believe there is a closed form. It is just that some functions and constants are not yet well studied, understood and named :) – Vladimir Reshetnikov Sep 29 '13 at 0:06 • @hmedan.mnsh Just to clarify: Mathematica does not return a closed form solution, but can give a numerical approximation using NIntegrate. – Vladimir Reshetnikov Sep 29 '13 at 0:31 • @HenningMakholm It seems to me the phrase "you already have a value" is somewhat objectionable. A numerical integration in this case converges quite slowly, and I doubt that one can get more than $25$ correct digits in a reasonable time (and it would be a very difficult to make sure they all are indeed correct). On the other hand, having a closed form, I can compute $10^5$ digits in about a second. And I can be pretty sure all these digits are correct, because numerical algorithms for elementary functions have been thoroughly designed and polished for years, and verified for correctness. – Vladimir Reshetnikov Sep 29 '13 at 18:11 • @StevenStadnicki I do not need $10^5$ digits, I used this number just to demonstrate the speed of calculations. But I can easily imagine that someone wants to get $30$ digits that are certainly correct. It is trivial task when one has an elementary closed form, but it might be quite difficult if one has to resort to numerical integration. Also, closed forms sometimes have a useful property to partially cancel each other or otherwise simplify when several factors or terms are combined into a single expression, while combining multiple approximate numeric values results in a precision loss. – Vladimir Reshetnikov Oct 1 '13 at 18:15 Yes, there is an elementary closed form for this integral: $$\int_0^\infty\frac{2-\cos x}{\left(1+x^4\right)\,\left(5-4\cos x\right)}dx=\frac{\pi}{2\,\sqrt2}\cdot\exp\left(\frac1{\sqrt2}\right)\cdot\frac{\sin\left(\frac1{\sqrt2}\right)-\cos\left(\frac1{\sqrt2}\right)+2\,\exp\left(\frac1{\sqrt2}\right)}{1-4\,\exp\left(\frac1{\sqrt2}\right)\cos\left(\frac1{\sqrt2}\right)+4\,\exp\left(\sqrt2\right)}\tag1$$ Proof: Let us denote the integral in question as $$\mathcal{I}=\int_0^\infty\frac{2-\cos x}{\left(x^4+1\right)\,\left(5-4\cos x\right)}dx\tag2$$ Note that the trigonometric part of the integrand is a periodic function and can be expanded to a Fourier series with particularly simple coefficients: $$\frac{2-\cos x}{5-4\cos x}=\sum_{n=0}^\infty\frac{\cos(n\,x)}{2^{n+1}}\tag3$$ (this can be easily checked by expressing cosines via exponents of an imaginary argument). Now we can integrate it term-wise: $$\mathcal{I}=\sum_{n=0}^\infty\left(\frac1{2^{n+1}}\int_0^\infty\frac{\cos(n\,x)}{x^4+1}dx\right)=\sum_{n=0}^\infty\left(\frac1{2^{n+1}}\cdot\frac{\pi}{2\,\sqrt2}\cdot\exp\left(-\frac{n}{\sqrt2}\right)\cdot\left(\sin\left(\frac{n}{\sqrt2}\right)+\cos\left(\frac{n}{\sqrt2}\right)\right)\right)\tag4$$ (for the integral, see DLMF 1.14, vii, Table 1.14.2, $4^{th}$ row). Trig functions in the last sum can again be expressed via exponents of an imaginary argument, and then the sum is easily evaluated. Converting exponents back to trig functions and getting rid of complex numbers, we get the final result $(1)$. • Clever expansion! – Ron Gordon Sep 29 '13 at 17:51 • Very nice answer! – Start wearing purple Sep 29 '13 at 17:53 There is an alternate way to compute this integral w/o a summation over $n$ in the middle steps. Notice $$\frac{2-\cos z}{5 - 4\cos z} = \frac12 \left[\frac{(2-e^{iz})+(2-e^{-iz})}{(2-e^{iz})(2-e^{-iz})}\right] = \frac12\left[\frac{1}{2-e^{iz}} + \frac{1}{2-e^{-iz}}\right]$$ and $\displaystyle\;\frac{1}{1+z^4}$ is an even function, we have $$\mathcal{I} \stackrel{def}{=} \int_0^\infty \frac{2-\cos x}{(1+x^4)(5-4\cos x)} dx = \frac12\int_{-\infty}^\infty \frac{1}{(1+z^4)(2 - e^{iz})}dz$$ Since $\displaystyle\;\frac{1}{2-e^{iz}}$ is entire on the upper half plane $\Im z \ge 0$ and $\displaystyle\;\left|\frac{1}{2-e^{iz}}\right| \le \frac{1}{2-1} = 1$ there, we can complete the contour in the upper half-plane and $$\mathcal{I} = \lim_{R\to\infty}\frac12 \oint_{C_R} \frac{1}{(1+z^4)(2 - e^{iz})}dz \quad\text{ where }\quad C_R = [-R, R ] \cup \big\{\; Re^{i\theta} : \theta \in [0,\pi]\big\}$$ $\displaystyle\;\frac{1}{1+z^4}$ has $4$ poles $\omega_k = e^{\frac{(2k+1)i}{4\pi}}, k = 0..3$ over $\mathbb{C}$. Two of them $\omega_0 = \frac{1+i}{\sqrt{2}}$ and $\omega_1 = \frac{-1+i}{\sqrt{2}}$ belongs to the upper half-plane. Since $$\frac{1}{1+z^4} = \sum_{k=0}^3 \frac{1}{(z-\omega_k) 4\omega_k^3} = -\frac14 \sum_{k=0}^3\frac{\omega_k}{z-\omega_k}$$ The residue of the integrand at $\omega_k$ are $\displaystyle\;-\frac14 \frac{\omega_k}{2 - e^{i\omega_k}}$ for $k = 0, 1$. This leads to \begin{align} \mathcal{I} &= \frac12\left[ -\frac{2\pi i}{4}\left(\frac{\omega_0}{2 - e^{i\omega_0}} + \frac{\omega_1}{2 - e^{i\omega_1}} \right)\right]\\ &= \frac{-\pi i}{4\sqrt{2}}\left( \frac{1+i}{2 - e^{-1/\sqrt{2}} e^{i/\sqrt{2}}} + \frac{-1+i}{2 - e^{-1/\sqrt{2}} e^{-i/\sqrt{2}}}\right)\\ &= \frac{-\pi i}{4\sqrt{2}} \left[ \frac{ (1+i)(2 - e^{-1/\sqrt{2}} e^{-i/\sqrt{2}}) + (-1+i)(2 -e^{-1/\sqrt{2}} e^{ i/\sqrt{2}}) }{ 4 - 4 e^{-1/\sqrt{2}}\cos\left(\frac{1}{\sqrt{2}}\right) + e^{-\sqrt{2}} } \right]\\ &= \frac{\pi}{2\sqrt{2}} \left[ \frac{2 - e^{-1/\sqrt{2}}\left( \cos\left(\frac{1}{\sqrt{2}}\right) - \sin\left(\frac{1}{\sqrt{2}}\right) \right) }{ 4 - 4 e^{-1/\sqrt{2}}\cos\left(\frac{1}{\sqrt{2}}\right) + e^{-\sqrt{2}} } \right]\\ &= \frac{\pi e^{1/\sqrt{2}} }{2\sqrt{2}} \left[ \frac{2 e^{1/\sqrt{2}} - \left( \cos\left(\frac{1}{\sqrt{2}}\right) - \sin\left(\frac{1}{\sqrt{2}}\right) \right) }{ 4 e^{\sqrt{2}} - 4 e^{1/\sqrt{2}}\cos\left(\frac{1}{\sqrt{2}}\right) + 1 } \right] \end{align} Reproducing what Vladimir get in his answer.

https://www.physicsforums.com/threads/show-that-you-can-distribute-powers-to-commuting-elements.950679/

# Show that you can distribute powers to commuting elements ## Homework Statement If ##a## and ##b## are commuting elements of ##G##, prove that ##(ab)^n = a^nb^n## for all ##n \in \mathbb{Z}##. ## The Attempt at a Solution We prove two lemmas: 1) If ##a## and ##b## commute, then so do their inverses: ##ab=ba \implies (ab)^{-1} = (ba)^{-1} \implies b^{-1}a^{-1} = a^{-1}b^{-1}##. 2) If ##a## and ##b## commute, then ##b^n a = ab^n##: Base case is trivial. Suppose for some ##k## we have ##b^ka = ab^k##. Then ##b^{k+1}a = bb^ka = bab^k = abb^k = ab^{k+1}##. Now to the actual result. Clearly ##(ab)^0 = a^0b^0##. So first we prove the result for the positive integers, by induction. The base case is trivial. Suppose for some ##k \in \mathbb{Z}^+## we have ##(ab)^k = a^kb^k##. Then ##(ab)^{k+1} = (ab)^k(ab) = a^kb^kab = a^kab^kb = a^{k+1}b^{k+1}##. Now we prove the result for negative integers. ##(ab)^{-n} = (b^{-1}a^{-1})^n = (a^{-1}b^{-1})^n = (a^{-1})^n(b^{-1})^n = a^{-n}b^{-n}##. Does this argument work? Were the lemmas really necessary or could I have assumed they held since their proofs are trivial? mfb Mentor since their proofs are trivial? So is the whole statement you have to show. Better to show it explicitly. fresh_42 Mentor There is another argument, why you will not need to prove the negative ones explicitly: If you prove the statement for all ##n \in \mathbb{N}_0## and for all ##a,b \in G##, then you have also proven it for inverse elements. This is in words what you have written. To do it by induction is a very formal way to prove it. In cases like the above, some dots will equally be acceptable, although from a logical point of view certainly not sufficient. But with the dots, every reader knows how the induction goes. Mr Davis 97 There is another argument, why you will not need to prove the negative ones explicitly: If you prove the statement for all ##n \in \mathbb{N}_0## and for all ##a,b \in G##, then you have also proven it for inverse elements. This is in words what you have written. To do it by induction is a very formal way to prove it. In cases like the above, some dots will equally be acceptable, although from a logical point of view certainly not sufficient. But with the dots, every reader knows how the induction goes. But isn't it the case that I am not proving for all ##a,b \in G##, rather just in the case ##a,b## commute? fresh_42 Mentor But isn't it the case that I am not proving for all ##a,b \in G##, rather just in the case ##a,b## commute? Yes, sure. But that doesn't change by taking the inverses: they commutate if and only if ##a## and ##b## do. Mr Davis 97 StoneTemplePython

http://www.southernmines.org/water-pressure-rzyocn/transpose-matrix-definition-d896ca

# transpose matrix definition A new matrix is obtained the following way: each [i, j] element of the new matrix gets the value of the [j, i] element of the original one. Transpose of a matrix is the interchanging of rows and columns. In practical terms, the matrix transpose is usually thought of as either (a) flipping along the diagonal entries or (b) “switching” the rows for columns. Related to Transpose of a matrix: adjoint of a matrix , Inverse of a matrix What happened? Noun . So we now get that C transpose is equal to D. Or you could say that C is equal to D transpose. The number $$4$$ was in the first row and the second column and it ended up in the second row and first column. In other words if A= [aij], then At ji = aij. The matrix B is called the transpose of A. This is the definition of a transpose. The way the concept was presented to me was that an orthogonal matrix has orthonormal columns. The transpose of a matrix is defined as a matrix formed my interchanging all rows with their corresponding column and vice versa of previous matrix. For permissions beyond the … See more. Definition of transpose : a matrix formed by interchanging the rows and columns of a given matrix - change the order or arrangement of - transfer from one place or period to another - cause to change places - transfer a quantity from one side of an equation to the other side reversing its sign, in order to maintain equality Now this is pretty interesting, because how did we define these two? Find the transpose of that matrix. For example if you transpose a 'n' x 'm' size matrix you'll get a new one of 'm' x … synonym : reverse . Definition. The transpose of an m × n matrix A is the n × m matrix A T whose rows are the columns of A. In other words, transpose of A[][] is obtained by changing A[i][j] to A[j][i]. Transpose definition: If you transpose something from one place or situation to another, you move it there. So if X is a 3x2 matrix, X' will be a 2x3 matrix. Dictionary ! transpose (plural transposes) (adjective, linear algebra) The resulting matrix, derived from performing a transpose operation on a given matrix. Solution: It is an order of 2*3. This is a transpose which is written and A superscript T, and the way you compute the transpose of a matrix is as follows. Let A be a nonsingular matrix. If there are two Matrix with dimension A (2 x 3 ) and B (3 x 2). QED And so that wraps up the definition of what it means to take the transpose of a matrix and that in fact concludes our linear algebra review. Meaning of Transpose. Here are a couple of ways to accomplish this in Python. The conjugate transpose of a matrix is the matrix defined by where denotes transposition and the over-line denotes complex conjugation. All content on this website, including dictionary, thesaurus, literature, geography, and other reference data is for informational purposes only. The transpose of a matrix with dimensions returns a matrix with dimensions and is denoted by . The definition of the transpose is as follows. en.wiktionary.2016 [noun] In matrix mathematics, the resulting matrix, derived from performing a transpose operation on a given matrix. The first thing to know is that you can separate rows by semi-colons (;) and that you define rows by just placing elements next to one another. Through the operations of the transpose, a new matrix is found where the rows entries of the original matrix are written in place of the columns, and the columns entries of the original matrix are written in place of the rows. Do the transpose of matrix. Definition. Matrix transpose: The transpose of matrix refers to the interchanging of the rows and columns. permute, commute, transpose (verb) change the order or arrangement of Remember that the complex conjugate of a matrix is obtained by taking the complex conjugate of each of its entries (see the lecture on complex matrices). In other words, the ij entry of A T is a ji. Initialize a 2D array to work as matrix. The first column became the first row and the second column became the second row. "transpose" (matrix) definition: a matrix formed by interchanging the rows and columns of a given matrix. We said that our matrix C is equal to the matrix product A and B. The transpose of a matrix by Duane Q. Nykamp is licensed under a Creative Commons Attribution-Noncommercial-ShareAlike 4.0 License. (Problems and Solutions in Linear Algebra. ) A double application of the matrix transpose achieves no change overall. | Meaning, pronunciation, translations and examples And we said that D is equal to our matrix product B transpose times A transpose. At t = A; 2. In order to state the transpose property, we need to define the transpose of a matrix. It is denoted as X'. Then At, the transpose of A, is the matrix obtained by interchanging the rows and columns of A. The transpose will also be of dimension (2x2). A matrix is usually shown by a capital letter (such as A, or B) Each entry (or "element") is shown by a lower case letter with a "subscript" of row,column: In this video we discuss about the another type of #Transpose of #Matrix with definition and example.If you have a question about then comment here . Dictionary Thesaurus Examples Sentences Quotes ... A matrix obtained by interchanging the rows and columns of a given matrix. Disclaimer. This will be the left hand side of (AB)⊺=B⊺A⊺ Solving for right hand side, if I take transpose of A and B then the dimension of resultant matrix … By, writing another matrix B from A by writing rows of A as columns of B. Do the transpose of matrix. And that's it. Stack Exchange Network. Consequently At is n m. Here are some properties: 1. We can find its transpose by swapping the column and row elements as follows. Definition of Transpose in the Definitions.net dictionary. We put a "T" in the top right-hand corner to mean transpose: Notation. Adjacency Matrix Definition. ... (0.00 / 0 votes) Rate this definition: transpose (verb) a matrix formed by interchanging the rows and columns of a given matrix. Let’s start by defining matrices. To "transpose" a matrix, swap the rows and columns. transpose: To reverse or transfer the order or place of; interchange. Ask Question Asked 4 years, 3 months ago. Derived terms We prove that the transpose of A is also a nonsingular matrix. The algorithm of matrix transpose is pretty simple. Consider the following example-Problem approach. Thus the $$3\times 2$$ matrix became a $$2\times 3$$ matrix. Therefore it occurred to me that the definition in the book of Weinberg is not consistent with that in the book of Tung: in one of them the symbol ${\Lambda_\mu}^\nu$ is defined as the inverse of the Lorentz transformation of contravariant vectors, while in the other case, the same symbol is defined as the transpose of the original matrix. What does Transpose mean? TRANSPOSE OF A MATRIX DEFINITION. Store values in it. We have: . The element at ith row and jth column in X will be placed at jth row and ith column in X'. transpose (comparative more transpose, superlative most transpose) (adjective, linear algebra) A matrix with the characteristic of having been transposed from a given matrix. Synonyms: tr. Menu. Definition. For a matrix defined as = , the transpose matrix is defined as = . For example, consider a matrix . (The transpose of a matrix) Let Abe an m nmatrix. Find transpose by using logic. Transpose definition, to change the relative position, order, or sequence of; cause to change places; interchange: to transpose the third and fourth letters of a word. Example 1: Consider the matrix . Learn more. Recommended: Please solve it on “ PRACTICE ” first, before moving on to the solution. Dimension also changes to the opposite. (redirected from Transpose of a matrix) Also found in: Dictionary , Thesaurus , Encyclopedia . Non-square matrix; Multiply matrices element by element; Create a Matrix in MATLAB Define a Matrix. ... Why is the inverse of an orthogonal matrix equal to its transpose? Example 2: Consider the matrix . Or is it a definition? Transpose of a matrix is obtained by changing rows to columns and columns to rows. [noun] In matrix mathematics, the process of rearranging elements in a matrix, by interchanging their respective row and column positional indicators. The product will be of size (2 x 2). transpose definition: 1. to change something from one position to another, or to exchange the positions of two things…. The adjacency matrix, also called the connection matrix, is a matrix containing rows and columns which is used to represent a simple labelled graph, with 0 or 1 in the position of (V i , V j) according to the condition whether V i and V j are adjacent or not. , transpose ( verb ) change the order or arrangement of Dictionary ;. Of size ( 2 X 2 ) ], then At, the entry. [ noun ] in matrix mathematics, the ij entry of a Duane Q. Nykamp licensed! Need to define the transpose of a matrix obtained by interchanging the rows columns. Placed At jth row and jth column in X will be placed At jth row and ith column X., including Dictionary, Thesaurus, Encyclopedia place of ; interchange properties: 1, commute, transpose verb! 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First row and jth column in X will be placed At jth row ith! Was that an orthogonal matrix equal to the interchanging of the rows and columns that our matrix C is to. Complex conjugation to rows will be placed At jth row and the second column became the second became! Redirected from transpose of matrix refers to the interchanging of rows and of. T '' in the top right-hand corner to mean transpose: to reverse or the... T '' in the top right-hand corner to mean transpose: Notation MATLAB... X is a 3x2 matrix, derived from performing a transpose operation on a given matrix whose rows are columns!: Please solve It on “ PRACTICE ” first, before moving to. For informational purposes only, literature, geography, and other reference data is for informational purposes only was to. Mean transpose: Notation thus the \ ( 3\times 2\ ) matrix,. From a by writing rows of a, is the inverse of m... From transpose of a, is the inverse of an m nmatrix is also a nonsingular.... Abe an m × n matrix a T whose rows are the of! A, is the n × m matrix a T is a ji can its. On to the matrix product B transpose times a transpose two matrix with dimensions returns a matrix by... Transpose property, we need to define the transpose of a as columns of a matrix the. If X is a ji informational purposes only to rows, literature, geography and... Double application of the matrix defined by where denotes transposition and the second row Thesaurus,,. D. or you could say that C transpose is equal to the matrix from one position to another or. These two en.wiktionary.2016 [ noun ] in matrix mathematics, the transpose of a, is the matrix there! We now get that C transpose matrix definition is equal to D transpose inverse of an orthogonal matrix has columns! To columns and columns of B 2 * 3 pronunciation, translations and Examples Disclaimer Create! A transpose operation on a given matrix X will be a 2x3 matrix MATLAB... Transpose times a transpose application of the matrix, the transpose of a is! Row elements as follows, writing another matrix B is called the transpose of a matrix is the ×...

https://math.stackexchange.com/questions/2217630/proving-the-existence-and-number-of-real-roots-for-x3-3x-2

# Proving the existence and number of *real* roots for $x^3 - 3x + 2$ I need to find how many real roots this polynomial has and prove there existence. I was wondering if my logic and thought process was correct. Determine the number of real roots and prove it for $x^3 - 3x + 2$ First, note that $f'(x) = 3x^2 - 3$ and so $f'(x) > 0$ for $x \in (-\infty, -1) \cup (1, \infty)$ and since $f'$ is strictly increasing on those intervals, there can be at most one root in each of them. $f'(x) < 0$ for $x \in (-1,1)$ and since $f'$ is strictly decreasing on this interval it can have at most one root. Now examine $f(-3) = -16$ and $f(-1) = 4$. By the Intermediate Value Theorem (IVT) $f(c) = 0$ for some $c \in (-3, 1)$ and so $f$ has a root on the interval $(-\infty, 1)$. Again examine $f(-1) = 4$ and $f(1) = 0$. We cannot say anything about $f$ having a root on the interval $(-1, 1)$. Likewise examine $f(1) = 0$ and $f(3) = 16$. Again, we cannot say anything about $f$ having a root on $(1, \infty)$. However, $f(1) = 1 - 3 + 2 = 0$ is clearly a root. And by factorizing the polynomial we get $f(x) = (x+2)(x-1)^2$. Indeed, $1$ is a root with a multiplicity of two. Hence, $f(x)$ has two real roots. Also, do we say two real roots (because of the multiplicity), or three real roots, or do we say two distinct real roots? While I realize factoring the polynomial gives me the answer I believe the purpose of the question was to do the former analysis, which when the polynomial isn't easily factorized, can provide a lot of insight. That is why I did it all • I assume you mean real roots, because it has 3 complex roots. – ÍgjøgnumMeg Apr 4 '17 at 13:51 • "First, note that $f′(x)=3x-3$ and so", you mean $f′(x)=3x^{\color{red}{2}}-3$...? – StackTD Apr 4 '17 at 13:52 • Haha yes! Let me make these adjustments. – student_t Apr 4 '17 at 13:52 However, $f(1) = 1 - 3 + 2 = 0$ is clearly a root. And by factorizing the polynomial we get $f(x) = (x+2)(x-1)^2$. Indeed, $1$ is a root with a multiplicity of two. All the work you did before this becomes unnecessary; after factoring, the roots (and hence the number of roots) are clear - right? Addition after some comments: when you are asked about the number of roots (real or not), it is usually meant to count the number of distinct (i.e. different) roots. Your equation has two (real) roots, one of which has multiplicity 2 but that doesn't change the fact that there are only two real numbers where the polynomial becomes 0. • Yes, but I think the purpose of the exercise was to do the former analysis. I only did that to show the multiplicity of the root. But yes in general all my work before would have been a waste haha! – student_t Apr 4 '17 at 13:55 Since we have $$x^3-3x+2=(x-1)^2(x+2),$$ we have three real roots $1,1,-2$. Here we count with multiplicities (which is standard for many results in geometry and other areas). • I would say there are two real roots, one of which has multiplicity two. – gandalf61 Apr 4 '17 at 13:56 • Yes, but I am suppose to do the little analysis before for the question I believe. Of course factoring would be much faster. – student_t Apr 4 '17 at 13:57 • @danny Yes, this may be the case. But I think, it does not matter so much what you are supposed to do or think, but what you yourself think is the best way. – Dietrich Burde Apr 4 '17 at 13:59 • I agree with gandalf61. In the context of the fundamental theorem of algebra, we often say an $n$th-order polynomial "has $n$ complex roots" but this is an abbreviation where we mean to count the multiplicities. That doesn't change the fact that $x^3$ only has one (distinct) root. – StackTD Apr 4 '17 at 13:59 • @danny see comment above; I would say yours has two (distinct) roots, we usually omit 'distinct' and mean different roots when we're talking about the number of roots. – StackTD Apr 4 '17 at 14:00

https://math.stackexchange.com/questions/381243/integral-of-sin-x-cdot-cos-x

# Integral of $\sin x \cdot \cos x$ [duplicate] I've found 3 different solutions of this integral. Where did I make mistakes? In case there is no errors, could you explain why the results are different? $\int \sin x \cos x dx$ 1) via subsitution $u = \sin x$ $u = \sin x; du = \cos x dx \Rightarrow \int udu = \frac12 u^2 \Rightarrow \int \sin x \cos x dx = \frac12 \sin^2 x$ 2) via subsitution $u = \cos x$ $u = \cos x; du = -\sin x dx \Rightarrow -\int udu = -\frac12 u^2 \Rightarrow \int \sin x \cos x dx = -\frac12 \cos^2 x = -\frac12 (1 - \sin^2 x) = -\frac12 + \frac12 \sin^2 x$ 3) using $\sin 2x = 2 \sin x \cos x$ $\int \sin x \cos x dx = \frac12 \int \sin 2x = \frac12 (- \frac12 \cos 2x) = - \frac14 \cos 2x = - \frac14 (1 - 2 \sin^2 x) = - \frac14 + \frac12 \sin^2 x$ So, we have: $$\frac12 \sin^2 x \neq -\frac12 + \frac12 \sin^2 x \neq - \frac14 + \frac12 \sin^2 x$$ • Ahh! I always give this problem in my first year calculus course. – Jyrki Lahtonen May 4 '13 at 14:40 • $+C$... ${}{}{}$ – David Mitra May 4 '13 at 14:40 • $*$ is usually used for convolution in this context. I removed it. – Ayman Hourieh May 4 '13 at 14:42 • @AymanHourieh: Didn't I? – Inceptio May 4 '13 at 14:44 • @Inceptio Check the edit history. I edited the body; you edited the title. – Ayman Hourieh May 4 '13 at 14:49 Antiderivatives are only unique up to adding a constant ('of integration'). If you were to stick limits in your integrals then you'd always get the same number. • Oh, yes. I checked it with 0 and \pi/2 and those 'strange' fractions deducted each other. Thank you – Jimch May 4 '13 at 14:46 Note: You are calculating indefinite integral and constants can be anything(they may differ). In fact the general solution to that would be just $C+\dfrac{\sin^2 x}{2}$ $$\frac{d\{f(x)+c\}}{dx}=f'(x)$$ for any arbitrary constant $c$ $$\implies \int f'(x)dx=f(x)+d$$ for any arbitrary constant $d$ So, in indefinite integral we can get answers which differ by some constant A primitive is unique up to a constant

https://math.stackexchange.com/questions/1113556/how-to-show-that-sum-k-1n-kn1-k-binomn23

# How to show that $\sum_{k=1}^n k(n+1-k)=\binom{n+2}3$? While thinking about another question I found out that this equality might be useful there: $$n\cdot 1 + (n-1)\cdot 2 + \dots + 2\cdot (n-1) + 1\cdot n = \frac{n(n+1)(n+2)}6$$ To rewrite it in a more compact way: $$\sum_{k=1}^n k(n+1-k)=\frac{n(n+1)(n+2)}6.$$ This equality is relatively easy to prove: $$\sum_{k=1}^n k(n+1-k)= (n+1)\sum_{k=1}^n k - \sum_{k=1}^n k^2 = (n+1) \frac{n(n+1)}2 - \frac{n(n+1)(2n+1)}6 = n(n+1) \left(\frac{n+1}2-\frac{2n+1}6\right) = n(n+1)\frac{3(n+1)-(2n+1)}6 = \frac{n(n+1)(n+2)}6.$$ (We only used the known formulas for the sum of the first $n$ squares and the sum of the first $n$ numbers.) Are there some other nice proofs of this equality? (Induction, combinatorial arguments, visual proofs, ...) EDIT: Now I found another question which asks about the same identity: Combinatorial interpretation of a sum identity: $\sum_{k=1}^n(k-1)(n-k)=\binom{n}{3}$ (I have tried to search before posting. But the answers posted here so far gave me some new ideas for good keywords to search which lead me to finding that question.) The questions are, in my opinion, not exact duplicates since the other question asks specifically about combinatorial proofs and my question does not have that restriction. But I agree that this is a very minor distinction. In any case, if you think that one of them should be closed as a duplicate, then you can vote to close. I will refrain from voting to close/reopen on this question. (If one of the two questions is voted to be a duplicate of the other one, they probably cannot be merged, since the summation variables are off by one.) • Generating functions? Coefficient of $x^{n+1}$ in $\left(\sum\limits_{n=1}^{\infty} nx^{n}\right)^2$ – sciona Jan 21 '15 at 14:33 • Here is another post about the same formula. – Martin Sleziak Oct 2 '15 at 17:31 Let us choose three numbers from $\{0,1,2,\ldots, n+1\}$, beginning with the middle one, which has to be some $k\in \{1,\ldots,n\}$. We then have $k$ choices for the smallest and $n+1-k$ choices for the largest of the three. It follows that $${n+2\choose3}=\sum_{k=1}^n k(n+1-k)\ .$$ • Straight to the point! +1 – user2345215 Jan 21 '15 at 15:02 I think of this as the "twelve days of Christmas equality", because if $n = 12$ then we get $1 \times 12 + 2 \times 11 + \cdots + 12 \times 1 = {14 \choose 3}$ and both sides represent the number of gifts which are given in the song "The Twelve Days of Christmas". (This happens to be 364, one less than the number of days in a year.) Here's a combinatorial proof of that equality, which I have previously written about at my blog. As I wrote there, let's try to prove the identity $\sum_{j=2}^{n+1} (j-1)(n+2-j) = {n+2 \choose 3}$, which differs from your equality by a change of index. To do this, we count subsets of the set $\{ 1, 2, \ldots, n+2 \}$ of size 3. We can write each such subset as $\{ x, y, z \}$ where we require $x < y < z.$ Then we’ll count these subsets according to the difference $z - x$. To construct such a set with $z - x = j$ we must: • choose $x$. $x$ must be between $1$ and $n+2-j$ inclusive, so there are $n+2-j$ possible choices. • choose $y$. $y$ must be between $x$ and $z=x+j$ exclusive, so there are $j-1$ possible choices. At this point $z$ is fixed. So there are $(j-1)(n+2-j)$ ways to choose a $3$-set with $z - x = j$; summing over the possible values of $j$ gives the desired identity. • I guess more-or-less the same argument could be reformulated like this. Let us start by choosing $y$. We only can choose $2, 3, \dots, n+1$. To make the notation the same as in your sum, let us denote $j=y$. If $y$ is chosen, we can choose any of elements on the left for $x$, which gives $(j-1)$ possibilities. We can choose any of the elements on the right for $z$, which gives $(n+2-j)$ possibilities. Together we have $\sum_{j=2}^{n+1} (j-1)(n+2-j)$. BTW +1 for the nice combinatorial argument and thanks for the link to your blog. – Martin Sleziak Jan 21 '15 at 14:46 A convolution is always a good way. Since: $$\sum_{k\geq 0} k z^k = \frac{z}{(1-z)^2}, \tag{1}$$ we have: $$\begin{eqnarray*}\sum_{k=1}^{n}k(n+1-k) = \sum_{k=0}^{n+1}k(n+1-k) &=& [z^{n+1}]\frac{z^2}{(1-z)^4}\\&=&[z^n]\frac{z}{(1-z)^4}\tag{2}\end{eqnarray*}$$ and the claim follows from: $$\sum_{k\geq 0}\binom{k+2}{3}z^k = \frac{z}{(1-z)^4}.\tag{3}$$ • I suppose the notation $[z^k]f(z)$ means: "the coefficient of $z^k$ in the power series $f(z)$", right? – Martin Sleziak Jan 21 '15 at 14:49 • @MartinSleziak: yes, it is the standard notation in analytic combinatorics. I am just borrowing it from Wilf or Flajolet. – Jack D'Aurizio Jan 21 '15 at 14:52 Induction is pretty straighforward: \begin{align*}\sum_{k=1}^{n+1}&k(n+2-k)-\sum_{k=1}^n k(n+1-k)=(n+1)+\sum_{k=1}^n k\cdot 1=\sum_{k=1}^{n+1} k\\&=\frac{(n+1)(n+2)}2=\frac{(n+1)(n+2)(n+3)}6-\frac{n(n+1)(n+2)}6\end{align*} Here is another solution. $$n\cdot 1+(n-1)\cdot2+\ldots1\cdot n=\\ (n+(n-1)+\ldots +1)+((n-1)+(n-2)+\ldots+1)+\ldots+1=\\ \sum_{i=1}^ni+\sum_{i=1}^{n-1}i+\ldots+1=\\ \frac{(n+1)n}{2}+\frac{n(n-1)}{2}+\ldots +\frac{2\cdot 1}{2}.$$ Now, this sum is equal $$\frac{(n+2)(n+1)n}{6}$$ by a straightforward induction. • Your equality can be seen also as $\binom{n+1}2+\binom{n}2+\dots+\binom22=\binom{n+2}3$, which can be a bit generalized to what some people call hockey-stick identity. See, for example, article on AoPS Wiki or this post (and some of the posts linked to it). – Martin Sleziak Jan 22 '15 at 8:13 • True, thank you for pointinig it out. – Ofir Schnabel Jan 22 '15 at 8:14

https://math.stackexchange.com/questions/2916482/probability-with-poker-cards

Probability with Poker cards You deal 5 cards from a well-shuffled deck of playing cards. What is the probability that the 5th card is the queen of spades? Just from analysis, P(5th queen spade) = (51*50*49*48*1)/(52*51*50*49*48) = 1/52 However why wont this method of logic thinking incorrect? P(5th queen spade) = (51Cr4) / (52Cr5) = 5/52. Reasoning: choose any first 4 cards and last card is queen spade, divide by all possible choice • Isn't that fifth card as likely to be the queen of spades as the three of clubs? – Lord Shark the Unknown Sep 14 '18 at 6:46 • But what if the three of clubs was selected before the 5th card? Is it the order of when cards are drawn does not matter? – userName Sep 14 '18 at 6:49 • You are drawing four cards and then a fifth card. You don't care about order of the first four cards, but pay special attention to the fifth. Count ways to do so in both numerator and denominator (that is: favoured event and total space). – Graham Kemp Sep 14 '18 at 8:28 • @graham So since first four cards order does not matter, the probability of that cancel each other in numerator and denominator is that what it is? – userName Sep 14 '18 at 8:39 Your (51Cr4) / (52Cr5), or as I prefer to write it $\dfrac{51 \choose 4}{52\choose 5}$, is the probability that the first five cards contain the Queen of Spades. If that happens, there then is a $1$ in $5$ chance that the Queen of Spades is the fifth of these five cards, making the result to the original question $\dfrac5{52}\times \dfrac{1}{5} = \dfrac{1}{52}$ • Oh okay this makes sense... So does it mean that if I use the formula P(number of sample points in an event) / P(number of sample points in sample space) it will always be like the explanation that you have given? – userName Sep 14 '18 at 6:55 • @userName - it depends on the precise question, especially on whether each element of the sample space is equally likely – Henry Sep 14 '18 at 7:06 • What about this question? Find probability that the 5th card is the queen of spades, given that the first 4 cards are hearts? I consider P(first 4 cards hearts ^ last card queen of spade) / P(first 4 cards hearts) = (13Cr4 * 1 / 5) / (13Cr4 x 39Cr1 x 1 / 5) why is this reasoning incorrect? [multiplied by 1/5 due to the order, like in the previous example] correct ans: 1/48 – userName Sep 14 '18 at 7:16 • @userName After you have chosen four particular hearts, there would be $52-4=48$ cards remaining. So you could answer that question with $\dfrac{{13 \choose 4}{1 \choose 1}}{ {13 \choose 4}{48 \choose 1}} =\dfrac{1}{48}$ – Henry Sep 14 '18 at 7:35 • Alright thanks Henry. But just like to clarify why do I not need to multiply by 1/5 in this situation? Isn't P(first 4 cards hearts and last card queen spade) also mean that we have to be careful and consider that the queen spade is the last card? – userName Sep 14 '18 at 8:11 Since you are asking for specifically the fifth card to be $\spadesuit Q$, this is a problem in which order is important. The number of ways to select $5$ different cards with order important is $P(52,5)$. The number of ways in which the fifth card is $\spadesuit Q$ is $P(51,4)$. So the probability is $$\frac{P(51,4)}{P(52,5)}=\frac{51\times50\times49\times48}{52\times51\times50\times49\times48}=\frac1{52}\ .$$ Your mistake was to use $C$s instead of $P$s. • Thanks David. Can I check how can the use of Permutation solve this qn other qn? Find probability that the 5th card is the queen of spades, given that the first 4 cards are hearts? If I use that formula I got stuck when (13Pr4) / (13Pr4 x 39Cr1) which gives 1/39... The correct ans is 1/48 instead – userName Sep 14 '18 at 7:27 • Is it suppose to be (13Pr4) / (13Pr4 x 48Cr1) since I only select 4 cards and now left with 48 cards to choose from? – userName Sep 14 '18 at 7:30 • You seem to have got an answer from someone else so I trust that is OK. But can I point out that on this site it is preferred that new questions should not be asked in comments. This is because it makes it harder for other people to find the question later. It would be better if you asked a new question. You can always cross-reference your present question if relevant. Thanks. – David Sep 17 '18 at 0:09 $\def\cbinom#1#2{{^{#1}\mathsf C_{#2}}}\frac{\cbinom{51}4\cbinom 11}{\cbinom{52}5}$ is the probability for selecting the queen of spades and four from the fifty-one other cards when selecting any five from all fifty two cards. It does not consider the fifth place as a seperate group from the first four. You require the probability for selecting any four from the fifty-one non-queen of spaces then the queen of spades when selecting any four from fifty-two cards then any one from the fourty-eight other card. That is $\frac{\cbinom {51}4\cbinom 11}{\cbinom {52}4\cbinom{48}{1}}$, which equals $1/52$.

http://mathhelpforum.com/algebra/25338-algebra.html

1. ## Algebra Q1= show that if x is real, 2(x^2) + 6x + 9 is always positive. Q2= Solve the simultaneous equations for x,y > 0 2log y = log 2 + log x 2^y = 4^x cheers 2. for Q1 check the discriminant of the quadratic $b^2 -4ac$ for Q2 using the first equation you should be able to get $\log y^2 = \log 2x \Rightarrow y^2 = 2x$ using the laws of logs. log both sides on the second equation and it should be easy form there. 3. Originally Posted by sparky69er Q1= show that if x is real, 2(x^2) + 6x + 9 is always positive. Q2= Solve the simultaneous equations for x,y > 0 2log y = log 2 + log x 2^y = 4^x cheers An alternative "looking" way to show Q1 would be completing squares and examining the expression. $ 2(x^2 + 3x + \frac92) = 2([x^2 + 2.\frac32.x + (\frac32)^2] + \frac92 - (\frac32)^2) = 2(x + \frac32)^2 + \frac94$ Now since $(x + \frac32)^2$ is always greater than or equal to zero, for all real x, and since $\frac94$ is positive , its sum is strictly positive. 4. Originally Posted by bobak for Q1 check the discriminant of the quadratic $b^2 -4ac$ for Q2 using the first equation you should be able to get $\log y^2 = \log 2x \Rightarrow y^2 = 2x$ using the laws of logs. log both sides on the second equation and it should be easy form there. Hi, for Q1 i have done as you said and got -36 but what do you mean by the discriminant???? 5. Hello, sparky69er! 1) Show that if $x$ is real, $2x^2 + 6x + 9$ is always positive. We have: . $x^2 + x^2 + 6x + 9 \:=\:x^2 + (x+3)^2$ The sum of two squares is always nonnegative. When $x = 0$, the polynomial has a minimum value of 9. 2) Solve the simultaneous equations for x, y > 0 . . $\begin{array}{cc}2\log y \:=\: \log 2 + \log x & {\color{blue}[1]} \\ 2^y \: =\: 4^x & {\color{blue}[2]} \end{array}$ Equation [1] is: . $2\log y \:=\:\log(2x)\;\;{\color{blue}[3]}$ From Equation [2]: . $2^y \:=\:(2^2)^x \:=\:2^{2x}\quad\Rightarrow\quad y \:=\:2x\;\;{\color{blue}[4]}$ Substitute into [3]: . $2\log(2x) \:=\:\log(2x)\quad\Rightarrow\quad\log(2x) \:=\:0$ . . Hence: . $2x \:=\:1\quad\Rightarrow\quad \boxed{x \:=\:\frac{1}{2}}$ Substitute into [4]: . $y \:=\:2\left(\frac{1}{2}\right) \quad\Rightarrow\quad\boxed{ y \:=\:1}$ 6. Originally Posted by sparky69er Hi, for Q1 i have done as you said and got -36 but what do you mean by the discriminant???? Given a quadratic polynomial equation of the form $ax^2 + bx + c = 0$, we have solutions $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ The "discriminant" is defined as $D = b^2 - 4ac$. We have three different classes of solution based on the value of the discriminant: $D > 0 \implies$ two real, unequal roots. $D = 0 \implies$ one real root. (Or two equal real roots, however you wish to look at it.) $D < 0 \implies$ two complex conjugate roots. In the case of D < 0 the curve $y = ax^2 + bx + c$ never crosses the x axis. If a > 0 then this would imply that the quadratic is always positive. -Dan 7. Originally Posted by Soroban Hello, sparky69er! We have: . $x^2 + x^2 + 6x + 9 \:=\:x^2 + (x+3)^2$ The sum of two squares is always nonnegative. Smart, really smart solution

http://centrosamo.it/fskn/minimize-a-cost-function.html

Minimization and maximization refresher. from Wikipedia. The minimization will be performed by a gradient descent algorithm, whose task is to parse the cost function output until it finds the lowest minimum point. To find the profit maximization levels, other approaches can be taken as well. In place of dJ/dTheta-j you will. This means that, in the AC equation, q + 2 are the average variable costs and 100/q are the average fixed costs. Likely, many corporate leaders believe. 1-Input the number. How is the above interpreted? The rm wants to minimize its costs (w 1x 1 + w 2x 2) of producing y units of output. This video explains how to find the average cost function and find the minimum average cost given the total cost function. Yes, even despite having so much support from ml-class … they practically implement everything and just leave the cost and gradient functions up to you. In this article, I will be going through the basic mathematics behind K-Means Algorithm. 50 which is the amount the firm has reduced their loss by producing instead of shutting down. economic order quantity (eoq) model The economic order quantity (EOQ) is the order quantity that minimizes total holding and ordering costs for the year. The formula is useful for deriving total costs for budgeting purposes, or to identify the approximate profit or loss levels likely to be achieved at certain sales volumes. So the terminology I'm going to use is that the loss function is applied to just a single training example like so. If a firm has a production function Q=F(K,L) (that is, the quantity of output (Q) is some function of capital (K) and labor (L)), then if 2Q V/(π r^2) Find r when the slope of the area is zero:. 3) The profit a business makes is equal to the revenue it takes in minus what it spends as costs. so the function is concave up, so x = 18 is the absolute minimum. Your business should be doing the same. If Minimize is given an expression containing approximate numbers, it automatically calls NMinimize. Minimize [{f, cons}, x ∈ reg] is effectively equivalent to Minimize [{f, cons ∧ x ∈ reg}, x]. In ML, cost functions are used to estimate how badly models are performing. For example, when determining optimal cooling protocols, we ultimately only care to minimize ice-related cell death in the tissue,without regard to the state outside of the tissue. I would like to minimize the cost of a function and i have these variables and restrictions. minimize (). The optimal cost is $150. Objectives: To maximize or minimize a two-variable function. output quantity. For some types of costs, the relationship is in direct proportion; for other types, there is a direct trade-off. To delineate CVX specifications from surrounding Matlab code, they are preceded with the statement cvx_begin and followed with the statement cvx_end. Clarification of Answer by livioflores-ga on 28 May 2006 20:01 PDT Hi!! Here is the answer to your second request of clarification: You know that 100 = min(x1,20) + min(x3,x4); Since you are trying to minimize costs it is clear that x1=<20; if not is x1>20 but min{x1,20} is still equal to 20 and this force you to continue using 80 units of x3 and x4; but in this situation you will spend more. Thanks readers for the pointing out the confusing diagram. 1 guitars and 48. Cost complementary exits in a multiproduct cost function when a. Now we're ready to optimize. To do that, we make a function that gives us the wrongness of a particular set of thetas against our training data. In order to minimize the cost function, we can directly equate the gradient/derivative to zero and get the required value for 'm' and 'b' and this will give us the minimum cost. return_all, available for some methods, gives you the parameter vector vs generation, but not the cost function. In most cases, when you see a decorated function, the decorator is a factory function that takes a function as argument and returns a new function that includes the old function inside the closure. Summary: The goal of the diet problem is to select a set of foods that will satisfy a set of daily nutritional requirement at minimum cost. Cost categories. 01 dollars to manufacture x Xbox 360s in a day. (1) Solve for the cost-minimizing input combination: (2) Depict the optimum in the diagram to the right. To obtain the cost function, add fixed cost and variable cost together. 3) Do not exhaust all system memory. Amid COVID-19, physicians, architects, and consultants are talking change in design — to be ready for next. Examples least-squares minimize kAx−bk2 2 • analytical solution x⋆ = A†b (A† is pseudo-inverse) • can add linear constraints, e. There are two parameters (coefficients) in our cost function we can control: weight $$m$$ and bias $$b$$. So we finally have cost as a function of x. Minimize the average cost function where the total cost function is C(x)=10+20sqrtx+16xsqrtx. prefer parameters that minimize the execution time. 3 is to be constructed in the shape of a rectangular box with a square base and an open top. - Answered by a verified Math Tutor or Teacher We use cookies to give you the best possible experience on our website. Return the absolute value. 1541765 To link to this article: https://doi. The available techniques to determine soil moisture content have practical limitations owing to their high cost, dependence on labor, and time consumption. Cost Function A company finds that it costs a total of to produce units of a new product. Rowe Price, troll, uspto. [email protected] Gradient Descent is THE most used learning algorithm in Machine Learning and this post will show you almost everything you need to know about it. the cost function itself!. Then why to use the. This value may be the expected net present value of a project or a forest property; or it may be the cost of a project; it could also be the amount of wood produced, the expected number of visitor-days at a park, the number of endangered species that will be. Eliminate Storage Complexity And Minimize Costs Enterprise IT is under unremitting pressure to reduce capital and operating expenses, driving them to virtualize infrastructure to improve hardware utilization and scalability and advance toward enhanced operational efficiency and flexibility. This elementary framew ork is the basis for a broad v ariety of mac hine learning. The math problem is: A large bin for holding heavy material must be in the shape of a box with an open top and a square base. However, this benefit comes at the cost of high computational complexity. The marketing manager should be interested in revealing the complexities of an individual buyer, the dynamics of consumer behavior and should also try to. #N#function J = computeCost ( X, y, theta) #N#%COMPUTECOST Compute cost for linear regression. I want that " t and T must be greater than zero(not equal to zero) , t < T and C > 0. 15-2P = 15-2(3)= 15-6=9-6+5P=-6+5(3)=-6+15=9. If you produce a certain amount and let's say you bring in, I don't know,$10,000 of revenue and it costs you $5,000 to produce those shoes, you'll have$5,000 in profit. This study proposes a new framework to minimize the cost function of multi-objective optimization problems by using NSGA-II in economic environments. To delineate CVX specifications from surrounding Matlab code, they are preceded with the statement cvx_begin and followed with the statement cvx_end. Since C0(x) = 30 − 253000 x2, then C. LP problems seek to maximize or minimize some quantity (usually profit or cost). the objective function (maximize/minimize) and. The goal of any Machine Learning model is to minimize the Cost Function. So in training your logistic regression model, we're going to try to find parameters W and B that minimize the overall costs function J written at the bottom. How to minimise the cost function? Our goal is to move from the mountain in the top right corner (high cost) to the dark blue sea in the bottom left (low cost). Minimize operating costs and improve energy performance Data centers have to face continually increasing cost constraints. Gradient Descent basically just does what we were doing by hand — change the. From an external point of view, it is difficult to ascertain which are the alternative considered. It costs $10 to store one set for a year. The inventory cost problem, however, is something that comes up in real-life manufacturing scenarios all the time - how can I minimize my operating costs? In fact, the problem we see here today is a simplified version of a problem I covered in a DETC conference paper that I published a few years back. Take a deep breath. The point was more to introduce the reader to a specific method, not to the cost function specifically. You can use Pythagoras to compute S in terms of U in terms of S: U^2 = 500 2 + (4000 - S) 2 U = sqrt( 500 2 + (4000 - S) 2) Thus C(S) = S + 5 sqrt( 500 2. In order to minimize the cost function, we can directly equate the gradient/derivative to zero and get the required value for 'm' and 'b' and this will give us the minimum cost. **TL;DR**: Rearranging the terms in Maximum Mean Discrepancy yields a much better loss function for the discriminator of Generative Adversarial Nets. To determine the optimal amount of inputs (L and K), we solve this minimization constraint using the Lagrange multiplier method:. Inventory cost problems come up in real-life manufacturing scenarios all the time - how can I minimize my operating costs? Hot Bod Jacuzzi & Spa Company is launching a new hot tub - the Neverleak Massage-o-matic DeLux. Write a Cost Function. Thus, the C function represents the. The cost functions implemented in MIPAV: Correlation ratio. The use of closures and factory functions is the most common and powerful use for inner functions. Anthony Vu Patent, Patents "ask the patent attorney series", "The American Invents Act", aia, collateral estoppel, cost, inter partes review, issue preclusion, litigation, patent, patent litigation, price, secure Axcess, T. 2 Minimize 2 x 2 1 + 2 sub ject to x 1 + 2 =1, if w ec hange the righ t hand side from 1 to 1: 05 (i. How can equations and inequalities help a business maximize profit or minimize costs? Unanswered Questions. for one-variable real functions: limits, integrals, roots This is the main site of WIMS (WWW Interactive Multipurpose Server): interactive exercises, online calculators and plotters, mathematical recreation and games. (c) Test the C. Then it is going to become impossible to properly minimize or maximize the Cost Function. be used to minimize costs and what is the cost of producing that amount of chicks? f. cost of shipping + c21x21 + c22x22 + c23x23 from a plant + c31x31 + c32x32 + c33x33 to the ware house) Supply constraints. That is, the firm must choose a specific point on the q Cost Functions come directly from the production function and prices. In practice, these attractions are balanced in order to maintain a gap between the shaft (rotor) and static parts (stator). This firm minimizes its cost of producing any given output y if it chooses the pair (z 1, z 2) of inputs to solve the problem min z 1,z 2 w 1 z 1 + w 2 z 2 subject to y = F (z 1, z 2), where w 1 and w 2 are the input prices. (A) The Cost Function The cost-minimizing choice of inputs depended on two essential sets of parameters: the given output level (Y) and the given factor prices (r and w). Write an expression for the Cost in terms of only the width (w). Authors: Gaël Varoquaux. 25 lines (16 sloc) 791 Bytes. Minimize the average cost function where the total cost function is C(x)=10+20sqrtx+16xsqrtx. There are two parameters (coefficients) in our cost function we can control: weight $$m$$ and bias $$b$$. 20, it cost$6. Operations > Time-Cost. The math problem is: A large bin for holding heavy material must be in the shape of a box with an open top and a square base. Indeed, it is the most powerful method available to reduce product cost, improve quality, and simultaneously reduce development interval. The minimization will be performed by a gradient descent algorithm, whose task is to parse the cost function output until it finds the lowest minimum point. Material indices Introduction The performance, p, Each function has an associated material index. 2x1 1 x2 1 x3 1 x4 x1, x2, x3, x4 $0. Organizations are relying on cloud to maintain business-critical processes, but the journey is not always seamless: you may be grappling with cloud governance and how to keep control over security, costs, risks. So our cost as a function of x is going to be 20x squared 36 times 5. What is the Malayalam name of tukmaria or sabja seed or falooda seed. Likely, many corporate leaders believe. In the case we are going to see, we'll try to find the best input arguments to obtain the minimum value of a real function, called in this case, cost function. Minimize costs The logistics market is characterized by higher standards for air pollution and noise as well as increasing toll fees, personnel costs and fuel prices. Output is produced according to the following process 2 1 2 1 K L = Firm Output (I chose the same function as above to simplify things). 1-Input the number. It's called the cost function, which is kind of a crappy name in this context. 20 to increase production from 49 to 50 units of output. " The problem also listed these following multiple choice answers: a) 30,000 b) 300 c) 3,000 d) 30 e) None of these Now, we have the correct answer, what we need is the actual way to do this problem. The gradient descent algorithm in a nutshell. To do this, take the derivative of C(x), set it equal to zero, and solve for x. The diet problem constraints typically regulate the number of calories and the. The optimization continues as the cost function response improves iteration by iteration. optimize for black-box optimization: we do not rely on the. Put simply, a cost function is a measure of how wrong the model is in terms of its ability to estimate the relationship between X and y. Cost Minimization: Short Run • Let us go back to the two-inputs case, with only one of them variable in the short run. We have contributed on a local. Thus you know that the cost is C(S,U) = S + 5U. 9 drums to minimize his costs. Well, your profit as a function of x is just going to be equal to your revenue as a function of x minus your cost as a function of x. C CL(q) combination of inputs that minimize the cost of producing each. Fundamental theorem of linear programming If the optimal (maximum or minimum) value of the objective function in a. Reynolds Consumer Products Inc. The aim of the linear regression is to find a line similar to the blue line in the plot above that fits the given set of training example best. This website uses cookies to ensure you get the best experience. Solution: We would like to find a function that describes this situation. 4 (GP) : minimize f (x) s. "I tried a lot but I am not getting the values of t and T as mentioned above " ". The objective function J = f(x) is augmented by the constraint equations through a set of non-negative multiplicative Lagrange multipliers, λ j ≥0. We often design algorithms for GP by building a local quadratic model of f (·)atagivenpointx =¯x. 4x + 150 t?o model the unit cost in dollars for producing x stabilizer bars. INTRODUCTION Linear programming is a mathematical technique used to find the best possible solution in allocating limited resources (constraints) to achieve maximum profit or minimum cost by modelling linear relationships. nan with np. The goal of any Machine Learning model is to minimize the Cost Function. These functions can be seen as covering functions which have many applications in di erent optimization prob-lems: Set Cover functions, Edge Cut functions in graphs, etc. For example, this formula will find the highest value in cells H2:H17 =MAX(H2:H17) MIN IF Formula. Mathematical optimization: finding minima of functions¶. In order to minimize the cost function, we can directly equate the gradient/derivative to zero and get the required value for 'm' and 'b' and this will give us the minimum cost. How much food should be used to minimize costs and what is the total cost? 2. Thus the function huber can be used anywhere a traditional convex function can be used, in constraints or objective functions, in accordance with the DCP ruleset. Minimize the cost to split a number Given an integer N ≥ 2 , you can split the number as a sum of k integers i. Sourcing, procurement and vendor management leaders should use this research to navigate GDPR requirements and Microsoft’s licensing to avoid legal and financial risks. The cost functions implemented in MIPAV: Correlation ratio. The transaction cost function is not di erentiable at the kink points and is piecewise continuous. Human Resource Management Functions. You will notice that as in the case of the factor demand functions, there is a. 1 guitars and 48. That is, the quantity you want to maximize or minimize is called the objective function. The Dual of the Minimum Cost Flow Problem:. C CL(q) combination of inputs that minimize the cost of producing each. Then why to use the. That's incredible but understandable when you start adding up all the "standard" wedding costs. So it's going to be plus 180 times, let's see, x times x to the negative 2, 180x to the negative x to the negative 1 power. The specific goal is to approximate a single valued function of one variable in terms of a sequence of linear segments. To minimize energy content, use the above criteria for. The two interesting exceptions to this rule are:. How to Minimize Legal Liabilities and Risks Information throughout this subsection applies primarily to external consultants. 4 — Logistic Regression | Cost Function — [ Machine Learning | Andrew Ng] - Duration: 11:26. What is the Objective Function? The objective of a linear programming problem will be to maximize or to minimize some numerical value. Assume that is costs Microsoft approximately C x x x 2 14,400 550 0. Hi , I am using FMINCON to minimize my cost function which is a product of elements of a matrix. An optimization problem seeks to minimize a loss function. To do that, we make a function that gives us the wrongness of a particular set of thetas against our training data. An extra large server costs you$0. Minimize F x y 22 with xy 2 10. using linear algebra) and must be searched for by an optimization algorithm. Find more Statistics & Data Analysis widgets in Wolfram|Alpha. HRM is the systematic planning and control of a network of fundamental organizational processes affecting and involving all organization members (French, 2004, p. However, this benefit comes at the cost of high computational complexity. For what number of bars is the unit cost at its mimimum? What is the unit cost at that level of production? Haven't got a clue what this problem is asking of me. A feasible solution that minimizes (or maximizes, if that is the goal) the objective function is called an optimal. How much are closing costs? These are the fees paid that help facilitate the sale of a home typically total 2% to 7% of the home's purchase price. applied optimization calc 1. If the material for the sides costs 15¢/in. In this article, I will be going through the basic mathematics behind K-Means Algorithm. For multi-objective improvements, the most generally used developmental algorithms such as NSGA-II, SPEA2 and PESA-II can be utilized. A cost function is a MATLAB ® function that evaluates your design requirements using design variable values. In this paper, we have applied some meta scheduling methods to a model of CIM that is referred to as an automated flow shop, where backward scheduling should be used to realize a JIT's theory. Consider the same open-top box, which is to have volume $$216in. Assume we are given a dataset as plotted by the 'x' marks in the plot above. This is the radius which will minimize the surface area and thus the cost of materials. costs into account. Artificial Intelligence - All in One 87,390 views 11:26. Chap 7: Short-Run Cost Function 2. Cost categories. 6 - Linear Programming. The material that will be used for three sides costs 30 per linear foot, and the material that will be used for the fourth side costs 15 per linear foot. So our cost as a function of x is going to be 20x squared 36 times 5. Decide what the variables are and what the constants are, draw a diagram if appropriate, understand clearly what it is that is to be maximized or minimized. 5 Q 2 v What is the marginal revenue function?. The Dual of the Minimum Cost Flow Problem:. Raw Blame History. 4 million will be recognized as a component of. Define a MATLAB function to evaluate −f(x) given x. In my opinion, the #1 cost to avoid is the “retaker” cost. I want that " t and T must be greater than zero(not equal to zero) , t < T and C > 0. Dear Sir Can you please help me to minimize the following cost function with maple 10. This function is known as the cost function and will be of considerable interest to us. Does the production function exhibit increasing, decreasing, or constant returns to scale? How can you tell? b. Given a function defined by a set of parameters, gradient descent starts with an initial set of parameter values and iteratively moves toward a set of parameter values that minimize the function. h(θ) is the the prediction from your regression model. Let's take a more in depth look at the cost function and see how it works. Examples least-squares minimize kAx−bk2 2 • analytical solution x⋆ = A†b (A† is pseudo-inverse) • can add linear constraints, e. Users who have contributed to this file. This study proposes a new framework to minimize the cost function of multi-objective optimization problems by using NSGA-II in economic environments. The available techniques to determine soil moisture content have practical limitations owing to their high cost, dependence on labor, and time consumption. (d) Find the minimum value of the marginal cost. A cost function is a MATLAB ® function that evaluates your design requirements using design variable values. Ultimately, to minimize our cost, we need to find the point with the lowest z value. You can also optimize the objective function without any loss function, e. Obtain the minimum using fmin=fminsearch(fun,x0) Maximization 1. One common application of calculus is calculating the minimum or maximum value of a function. With so many options to choose from, the best iPhone XR case can be elusive. Currently, minimize lacks the ability to do this. To do: Try the following example: Given: Q = L 1/2 K 1/2 PL = 4, PK = 1 Goal: Produce Qo = 16 units as cheaply as possible. (Remember, the average cost, (6 pts. For example, if the marginal cost of producing the 50th product is 6. If the firm ordered the item, then the setup cost is simply the order cost from Module 5. Cost Function - Intuition I11:09. They would like to offer some combination of milk, beans, and oranges. So, if you employ tactics to reduce costs in all discrete functions from manufacturing through delivery, you'll have a lower total landed cost, right? Theoretically, yes. Now, to minimize marginal cost. If the brewery produces sweet stout alone, the cost function is: CS(q2) = 8q2. Homework Statement Mary Jane grows herbs in her attic. Previous work. The Cost Function If lattes and cake (or labor and capital) have unit prices of pL and pK, respec-tively, then the total cost of purchasing L units of one and K units of the other is C(L,K) = pLL+pKK. I think relative price of L & K is (Cost of Labour Per Hour)/(Cost of Rent Per Hour), but I don't know the price of Rent Per Hour. pdf), Text File (. Next time I will not draw mspaint but actually plot it out. The problem is that officers work 8 hour shifts, yet the demand comes in 4 hour chunks. 5 Actionable Tips to Reduce Operational Costs Regardless of what the circumstances are for your business, it is always a priority to find ways to reduce operational costs. k) 0 is a (nonnegative) function for which (x k;x k) = 0;then the following function defines a majorizer for : ˚ k(x) , (x)+ (x;x k): (4. Generally speaking, Least-Squares Method has two categories, linear and non-linear. The available techniques to determine soil moisture content have practical limitations owing to their high cost, dependence on labor, and time consumption. There can be significant cost savings when a business function is outsourced. Find the level of production which will minimize the average cost per item. 02xSquared - 3. Instead, it is allowable to use a cost flow assumption that varies from actual usage. Cost & Time and Also Minimum Project Duration Using Alternative Method 405 coordinates of the normal and crash points: Cost slope = (crash cost-normal cost)/ (normal duration crash duration) As the activity duration is reduced, there is an increase in direct cost. As the magnitues of the fitting parameters increase, there will be an increasing penalty on the cost function. The extent of risk and liability in your work depends on the nature of your services. Hi , I am using FMINCON to minimize my cost function which is a product of elements of a matrix. Of course, since time is money in any manufacturing process, what this really means is that looking into ways of reducing cycle time in your injection molding process can have a major impact on. What are loss functions? And how do they work in machine learning algorithms? Find out in this article. In the case we are going to see, we'll try to find the best input arguments to obtain the minimum value of a real function, called in this case, cost function. Fundamental theorem of linear programming If the optimal (maximum or minimum) value of the objective function in a. Common benzodiazepines used for GAD include alprazolam, clonazepam, diazepam, and lorazepam. For what number of bars is the unit cost at its mimimum? What is the unit cost at that level of production? Haven't got a clue what this problem is asking of me. Actual costs refer to real transactions, wherease opportunity costs refer to the alternative taken into consideration by decision makers who might want to choose the line of activity which minimise the costs. Average cost is minimized when average cost = marginal cost is another saying that isn’t quite true; in this case, the correct statement is: Average Cost has critical points when Average Cost and Marginal Cost are equal. LP problems seek to maximize or minimize some quantity (usually profit or cost). Find the dimensions that will minimize the cost of the box's construction. Therefore the total cost is: C(x) = 10y +15(2x+y) = 30x+25y. We need to decide which sub-contractor to use for a critical activity. Hence we want to minimize the can's surface area. We refer to this property as the objective function of an LP problem. In this case, the function huber will contain a special Matlab object that represents the function call in constraints and objectives. An objective function is either a loss function or its negative (in specific domains, variously called. It's important to limit your number of serverless functions to avoid having a massive charge when a lot of work is presented at the same time. The average cost of producing one output is reduced when the output of another product is increased If the wage rate is 5 and the price of capital is 2, then in order to minimize costs the firm should use a. 5 kg is the same linear function for a mass change of 2000 kg. For example, companies often want to minimize production costs or maximize revenue. The objective of the purchasing function is to obtain proper material and services when needed at the lowest obtainable cost. 1, Major functions of an institution) on the basis of modified total direct costs (MTDC), consisting of all salaries and wages, fringe benefits, materials and supplies, services, travel, and up to the. txt) or read online for free. Minimize cost and maximize quality of function in four variables. We are the prime contractor and there is a penalty in our contract with the main client for every day we deliver late. In the following example I will minimize an arbitrary function [texi]J[texi], then in the next chapter I'll apply it to the original house pricing task. Loss functions are actually at the heart of these techniques that we regularly use. (1) Solve for the cost-minimizing input combination:. The aim of the linear regression is to find a line similar to the blue line in the plot above that fits the given set of training example best. 5 Q 2 v What is the marginal revenue function?. The major objective of a typi-cal firm is to maximize dollar profits in the long run. Variables and functions should be declared in the minimum scope from which all references to the identifier are still possible. Let's take a more in depth look at the cost function and see how it works. This is where we look back at equation (1) and solve for h in terms of w. We advise on the largest and most complex legal challenges facing the world’s most important companies. the cost function itself!. Set big goals, insist on a cultural shift, and model from the top. The objective function is the function to be minimized or maximized. Use a computer to maximize the objective function subject to the constraints where 38. Thus, applications of HRM theory differ from personnel management in their dismissal of prescriptive “one best way” models of practice as diverse. Minimising Cost function. Overall, closures have affected 25% of pork production and 10% of beef production in the U. Objective function. An optimization problem is one where you have to make the best decision (choose the best investments, minimize your company's costs, find the class schedule with the fewest morning classes, or so on). That is h = 50 3w2 (3) Plugging the value for h from (3) above into equation (2) yields C = 60w2 + 48w. 2, what should the dimensions of the cup be to minimize the construction cost?. Yes, but not by playing it safe. Actually, the objective function is the function (e. Minimizing any function means finding the deepest valley in that function. A cost function is defined as: …a function that maps an event or values of one or more variables onto a real number intuitively representing some “cost” associated with the event. Examples and exercises on the cost function for a firm with two variable inputs Example: a production function with fixed proportions Consider the fixed proportions production function F (z 1, z 2) = min{z 1, z 2} (one worker and one machine produce one unit of output). Variable costs are such cost which vary directly with change in output. Recall in the calculus of one variable, if y = f(x) is defined on a set S, then there is a relative maximum value at x0 if f(x0) ≥ f(x) for all x in S near x0, and there is a relative. If a firm has a production function Q=F(K,L) (that is, the quantity of output (Q) is some function of capital (K) and labor (L)), then if 2Q V/(π r^2) Find r when the slope of the area is zero:. His next-door neighbor agrees to pay for half of the fence that borders her property; Sam will pay the rest of the cost. The nonlinearity in this form generates from the absolute value function. In most examples/tutorial I followed, the cost function used was somewhat arbitrary. Coming up with a cost function for optimization for a complex control system Hot Network Questions Is there a word or phrase for one mistaken belief leading to a web of false ones?. However, this benefit comes at the cost of high computational complexity. : residuals) between our model and our data points. One common application of calculus is calculating the minimum or maximum value of a function. Take Exam Only When You are Ready. the production function and the cost function; the only difference is whether we hold production constant or cost constant. [email protected] I would like to use the goal seek function to minimize the value in a certain cell (total cost) by changing another the value in another cell (shipment size). Then why to use the. This firm minimizes its cost of producing any given output y if it chooses the pair (z 1, z 2) of inputs to solve the problem min z 1,z 2 w 1 z 1 + w 2 z 2 subject to y = F (z 1, z 2), where w 1 and w 2 are the input prices. The Cost Function If lattes and cake (or labor and capital) have unit prices of pL and pK, respec-tively, then the total cost of purchasing L units of one and K units of the other is C(L,K) = pLL+pKK. Find the dimensions that will minimize cost. Employee compensation costs, office space expenses and other costs associated with providing a workspace or manufacturing setup are eliminated and free up resources for other purposes. We often design algorithms for GP by building a local quadratic model of f (·)atagivenpointx =¯x. (d) Find the minimum value of the marginal cost. Cost function is the sum of losses from each data point calculated with loss function. Solving for the minimum 0 points minimize f, (z) = 20 + z2-cos(2TZ) Given the cost function f. S, according to Bloomberg News. To minimize energy content, use the above criteria for. c<=2 n=1-1000 0<=p<=0. It's called the cost function, which is kind of a crappy name in this context. For example, if the marginal cost of producing the 50th product is 6. Minimising Cost function. You’ve been ordered to reduce your department’s costs by 10%, 20%, or 30%. It is obvious that if we changed relative factor. 5\text{ x }10^6 \text{ ft}^2) in an a rectangular field and then divide it in half with a fence parallel to one of the sides of the rectangle. Having drawn the picture, the next step is to write an equation for the quantity we want to optimize. profits and minimize costs by using cost-revenue-profit functions. **TL;DR**: Rearranging the terms in Maximum Mean Discrepancy yields a much better loss function for the discriminator of Generative Adversarial Nets. 188, and the slope was not significantly different from 0. By using this website, you agree to our Cookie Policy. “Despite the challenges of the COVID-19. Take the derivative of the Cost with respect to width. Average Cost Per Unit Formula. Many of these materials are high in quality and low in cost. Find the Average Cost Function and Minimize the Average Cost. Outsourcing can also make your firm more attractive to investors, since you're able to pump more capital directly into revenue-producing activities. That’s the main point of any model, to minize error, to perform. How many Xboxes should be manufactured in order to minimize average cost? What is the resulting average cost of an Xbox? Give your answer to the nearest dollar. In this case, the objective is to minimize the total cost per day which is given by z= 0:6x 1 + 0:35x 2 (the value of the objective function is often denoted by z). Cost Function8:12. Indeed, it is the most powerful method available to reduce product cost, improve quality, and simultaneously reduce development interval. If this sounds a lot, here are my 5 suggestions to reduce the CPA exam cost: 1. It is the heart that makes it beat! There is a loss function, which expresses how much the estimate has missed the mark for an individual observation. The use of closures and factory functions is the most common and powerful use for inner functions. The cost function: E(Cost)=E(F-LS) 3 F is for Finished goods L is for Lambda S is for Sales After expanding the function, what assumption minimized this function with respect to F? The Attempt at a Solution F 3-3F 2 LS+3F(LS) 2-(LS) 3 I know that I need to identify the terms that include both sales and inventories. f ( x) = x 4 − 8 x 2 + 5. h(θ) is the the prediction from your regression model. output, marginal cost, average cost, price, and profit at the average-cost minimizing activity level profit-maximizing or loss-minimizing output Calculus Cost/Graph cost function Optimal capital structure to minimize cost of capital Output, Profit, Fixed Costs and Perfect Competition Finding Optimal Output Level etc. Example 4 If the total revenue and total cost functions are TR = 30Q – 5Q 2 and TC = 15 + 12Q - 0. A company producing goods wants to minimize the average cost of production. Mathematical optimization deals with the problem of finding numerically minimums (or maximums or zeros) of a function. The global six-year average cost of a data breach is 3. Employee compensation costs, office space expenses and other costs associated with providing a workspace or manufacturing setup are eliminated and free up resources for other purposes. Yes, even despite having so much support from ml-class … they practically implement everything and just leave the cost and gradient functions up to you to implement. In order to minimize the cost function, we can directly equate the gradient/derivative to zero and get the required value for 'm' and 'b' and this will give us the minimum cost. Mathematical optimization: finding minima of functions¶. Keep in mind that, the cost function is used to monitor the. To open Outlook again after it has been hidden on the system tray and disappeared from the taskbar, double-click the Outlook system tray icon. Thus the function huber can be used anywhere a traditional convex function can be used, in constraints or objective functions, in accordance with the DCP ruleset. Minimize the total physical effort & thus the cost of moving goods into & out of storage. Budgeting for your company’s training needs does not mean using surplus money when you have it. Diesel particulate filters (DPF) are devices that physically capture diesel particulates to prevent their release to the atmosphere. Authors: Gaël Varoquaux. There are many factors to consider when selecting components and board-level solutions for a real-time embedded system. 1 Where f is the number of facilities. Define a MATLAB function to evaluate −f(x) given x. , a function that takes a scalar as input) is needed. Good parameters means that the function can produce the best possible outcomes, namely the smallest ones, because small values mean less errors. Cij = transportation cost per unit of shipping from plant Pi to the Warehouse Wj. Lagrange multiplier methods involve the modification of the objective function through the addition of terms that describe the constraints. Mathematical optimization: finding minima of functions¶ Authors: Gaël Varoquaux. Min & Max of Functions - MATLAB Minimization 1. Gradient descent is a more generic algorithm, used not only in linear regression problems and cost functions. Along with source pipeline, candidate quality, long-term retention, and other key performance metrics, tracking your cost-per-hire will help you to understand the performance of your recruiting initiatives better and minimize your expenses across the board. Univariate function minimizers (minimize_scalar)¶ Often only the minimum of an univariate function (i. A company producing goods wants to minimize the average cost of production. 15-2P = 15-2(3)= 15-6=9-6+5P=-6+5(3)=-6+15=9. My constriants are also in the form of matix. For example, this formula will find the highest value in cells H2:H17 =MAX(H2:H17) MIN IF Formula. 2% over budget). Then, minimize that slack variable until the slack is null or negative. We have contributed on a local. Minimize an objective function whose values are given by executing a file. Real-time embedded systems require. Grab a coffee. The cost function: E(Cost)=E(F-LS) 3 F is for Finished goods L is for Lambda S is for Sales After expanding the function, what assumption minimized this function with respect to F? The Attempt at a Solution F 3-3F 2 LS+3F(LS) 2-(LS) 3 I know that I need to identify the terms that include both sales and inventories. Profit is simply the Total revenue minus the costs incurred. Gradient descent is simply used to find the values of a function's parameters (coefficients) that minimize a cost function as far as possible. How To: Calculate and use regression functions in statistical analysis How To: Write a logarithm as a sum or difference of logarithms How To: Perform a quadratic regression with a calculator How To: Calculate r-squared or coefficient of determination in statistics. Of course, since time is money in any manufacturing process, what this really means is that looking into ways of reducing cycle time in your injection molding process can have a major impact on. More generally, a lower semi-continuous function on a compact set attains its minimum; an upper semi-continuous function on a compact set attains its maximum point or view. If you produce a certain amount and let's say you bring in, I don't know, 10,000 of revenue and it costs you 5,000 to produce those shoes, you'll have 5,000 in profit. However, by substituting for , the problem can be transformed into a linear problem. An optimization problem seeks to minimize a loss function. This is the personal website of a data scientist and machine learning enthusiast with a big passion for Python and open source. 1080/09715010. costs into account. If Minimize is given an expression containing approximate numbers, it automatically calls NMinimize. the firm hires labor, and the cost is the wage rate that must be paid for the labor services Total cost (TC) is the full cost of producing any given level of output, and it is divided into two parts: • Total fixed cost. What is the best nursing intervention to minimize the adverse effects of this. The production process can often be described with a set of linear inequalities called constraints. Simplex Algorithm Calculator is an online application on the simplex algorithm and two phase method. Find the number of units, x, that will minimize the average cost function if the total cost function is C()3+7+ 75. It's important to limit your number of serverless functions to avoid having a massive charge when a lot of work is presented at the same time. The following figure (right) shows a plot of a sample cost function for a selection of transformation parameters. When you optimize or estimate model parameters, you provide the saved cost function as an input to sdo. It's a cost function because the errors are "costs", the less errors your model give, the better your model is. 2) A business’ costs include the fixed cost of 5000 as well as the variable cost of 40 per bike. One common application of calculus is calculating the minimum or maximum value of a function. Derive Draper Dan's cost function (a) in terms of input prices and output and (b) when the price of cloth, w 1, is 3/metre and the wage rate w 2 is 10 per hour. When x = 18, y = 9. Otherwise, they must be considered separately. In this article, I will be going through the basic mathematics behind K-Means Algorithm. Economic Order Quantity Model (EOQ) Managing inventory is an important task for every business that holds it. A piecewise linear approximation is one method of constructing a function that fits a nonlinear objective function by adding extra binary variables, continuous variables, and constraints to reformulate the original problem. To find ways to save money, take advantage of quick cost-saving measures followed by an intensive look at where IT is spending money. What happens when the learning rate is too small? Too large? Using the best learning rate that you found, run gradient descent until convergence to find 1. Using the quadratic formula or a calculator, we find the solutions are. As serverless architectures mature, they have been able to minimize the issue of provision concurrency, in which there was a performance penalty when a function was called a second time, causing a. It is the heart that makes it beat! There is a loss function, which expresses how much the estimate has missed the mark for an individual observation. If this sounds a lot, here are my 5 suggestions to reduce the CPA exam cost: 1. 37 e) The minimal average cost. Put simply, a cost function is a measure of how wrong the model is in terms of its ability to estimate the relationship between X. Minimize The Use Of Color In Wireframes. Find the level of production which will minimize the average cost per item. So it's going to be plus 180 times, let's see, x times x to the negative 2, 180x to the negative x to the negative 1 power. Question 107995: Minimizing Cost, A company uses the formula C(x)=0. lute extrema of the function y = 2x,一2x2-16x + 1 on [-2,3]. Given a function defined by a set of parameters, gradient descent starts with an initial set of parameter values and iteratively moves toward a set of parameter values that minimize the function. In order to minimize the cost function, we can directly equate the gradient/derivative to zero and get the required value for 'm' and 'b' and this will give us the minimum cost. Optimization methods in Scipy nov 07, 2015 numerical-analysis optimization python numpy scipy. A midwife usually offers a variety of options and seeks to eliminate or minimize unnecessary interventions. Once you have installed CVX (see Installation), you can start using it by entering a CVX specification into a Matlab script or function, or directly from the command prompt. Chapter 7: The Cost of Production. Minimizing Inventory Costs. Instead, it is allowable to use a cost flow assumption that varies from actual usage. Then again, Octave provides tools for learning where you essentially just run a function, tell it where to find the cost and gradient function and give it some data. Midwives: Benefits of Having a Midwife. Most optimization problems have a single objective function, if they do not, they can often be reformulated so that they do. For the given cost function C(x)=78400+500x+x^2 find: a) The cost at the production level 1700 b) The average cost at the production level 1700 c) The marginal cost at the production level 1700 d) The production level that will minimize the average cost e) The minimal average cost I can already answer a, b, and c, it's d and e I can't seem to get, I know it should be let c'(x) = 0, but the. In this context, the function is called cost function, or objective function, or energy. The calculator is intended to teach students the Simplex method and to relieve them from some of the tedious aritmetic. We want to minimize the cost of the materials subject to the constraint that the volume must be 50ft 3. In machine learning, we use gradient descent to update the parameters of our model. The cost function used is shown aboveWe want to find parameters Ɵ which minimize J(Ɵ) To do so we can use one of the algorithms already described such as; Gradient descent; Advanced optimization algorithmsTo minimize a cost function we just write code which computes the following J(Ɵ) i. Obviously, a conservative deflection limit can be specified to minimize deflection, assuming design and construction is then performed correctly. I want that " t and T must be greater than zero(not equal to zero) , t < T and C > 0. To find ways to save money, take advantage of quick cost-saving measures followed by an intensive look at where IT is spending money. We discuss the application of linear regression to housing price prediction, present the notion of a cost function, and introduce the gradient descent method for learning. It is a minimization problem. The optimization continues as the cost function response improves iteration by iteration. Then it is going to become impossible to properly minimize or maximize the Cost Function. The Total Cost of Ownership (TCO) of your IP is staggering. 20, it cost 6. 01 dollars to manufacture x Xbox 360s in a day. To demonstrate the minimization function, consider the problem of minimizing the Rosenbrock function of the NN variables −. Take Exam Only When You are Ready. Cost complementary exits in a multiproduct cost function when a. As it stands, though, it has two variables, so we need to use the constraint equation. The slope of iso cost line = PL/Pk. Minimize the average cost function where the total cost function is C(x)=10+20sqrtx+16xsqrtx. The statement dual variables y{n} allocates a cell array of \(n$$ dual variables, and stores the result in the Matlab variable Z. For example, a random sample of a population of young offenders is generated by selecting names from a list to interview. The cost function is just a mathematical formula that gives the total cost to produce a certain number of units. Mathematical optimization: finding minima of functions¶. Custom & Stock Plastic Packaging Solutions. 1-Input the number. 2x1 1 x2 1 x3 1 x4 x1, x2, x3, x4 $0. Return the arc cosine. stiff, light beam in bending –minimize ρ/E1/2 •e. 8 trillion annually in aggregate general and administrative (G&A) expenses. Labor Union Vs. Lagrange multiplier methods involve the modification of the objective function through the addition of terms that describe the constraints. Next time I will not draw mspaint but actually plot it out. h(θ) is the the prediction from your regression model. Minimize the total physical effort & thus the cost of moving goods into & out of storage. Material indices Introduction The performance, p, Each function has an associated material index. I recently had to implement this from scratch, during the CS231 course offered by Stanford on visual recognition. Example 4 If the total revenue and total cost functions are TR = 30Q – 5Q 2 and TC = 15 + 12Q - 0. Note that w 1, w 2, and y are given in this. This will give the quantity (q) that maximizes profits, assuming of course that the firm has already taken steps to minimize costs. An isoquant and possible isocost line are shown in the following figure. Advantages of Outsourcing Cost Savings. Good parameters means that the function can produce the best possible outcomes, namely the smallest ones, because small values mean less errors. In machine learning, we use gradient descent to update the parameters of our model. The purposes of a human resources department and a labor union are decidedly different. One common application of calculus is calculating the minimum or maximum value of a function. Model Representation8:10. Budgeting for your company’s training needs does not mean using surplus money when you have it. How many players should be produced to minimize the marginal cost? and (b). The cost function equation is expressed as C(x)= FC + V(x), where C equals total production cost, FC is total fixed costs, V is variable cost and x is the number of units. Suppose the marginal cost C(in dollars) to produce x thousand mp3 players is given by the function C(x)=x^2-100x+7600. So we have written the cost as a function of two variable, height and width. cost of shipping + c21x21 + c22x22 + c23x23 from a plant + c31x31 + c32x32 + c33x33 to the ware house) Supply constraints. Example: A retail appliance store sells 2500 TV sets per year. To determine the optimal amount of inputs (L and K), we solve this minimization constraint using the Lagrange multiplier method:. 01x 2 +120 ) dollars where x represents the number of units produced. It is possible to attach a more substantial penalty to the predictions that are located above or below the expected results (some cost functions do so, e. Instead, it is allowable to use a cost flow assumption that varies from actual usage. Constrained Optimization using Lagrange Multipliers 5 Figure2shows that: •J A(x,λ) is independent of λat x= b, •the saddle point of J A(x,λ) occurs at a negative value of λ, so ∂J A/∂λ6= 0 for any λ≥0. Your business should be doing the same. Gradient descent is an optimization algorithm used to find the values of parameters (coefficients) of a function (f) that minimizes a cost function (cost). Ignoring any other costs, find the optimal number of facilities with the. Note as well that the cost for each side is just the area of that side times the appropriate cost. Mathematical optimization is the selection of the best input in a function to compute the required value. Now, the "cost-minimization" approach to solve the firm's optimization problem, is an alternative behavioral assumption to the profit-maximizing setup, and it is very relevant in many real-world cases: public utilities that exist mainly to satisfy demand, and their motive is not to maximize profits -rather they want to minimize cost for the. To maximize the revenue function To minimize the cost function To maximize the profit function. Calculus Optimization Problem: What dimensions minimize the cost of a garden fence? Sam wants to build a garden fence to protect a rectangular 400 square-foot planting area. 25 lines (16 sloc) 791 Bytes. A low-cost provider is a powerful competitive approach in markets where many buyers are price sensitive. The problem is formulated as a linear program where the objective is to minimize cost and the constraints are to satisfy the specified nutritional requirements. find the location of the minimum of fr, z. That's incredible but understandable when you start adding up all the "standard" wedding costs. x11 + x12 + x13 = S1 x21 + x22. Firms can change all their inputs, both labor and capital, in the. Optimization- What is the Minimum or Maximum? 3. Hi , I am using FMINCON to minimize my cost function which is a product of elements of a matrix. For example, if the marginal cost of producing the 50th product is$6. In this case, the objective is to minimize the total cost per day which is given by z= 0:6x 1 + 0:35x 2 (the value of the objective function is often denoted by z). 4x + 150 t?o model the unit cost in dollars for producing x stabilizer bars. The average cost of producing one output is reduced when the output of another product is increased If the wage rate is $5 and the price of capital is$2, then in order to minimize costs the firm should use a. So to recapHypothesis - is like your prediction machine, throw in an x value, get a putative y value. In order to minimize the cost function, we can directly equate the gradient/derivative to zero and get the required value for 'm' and 'b' and this will give us the minimum cost. If the time rate of change of this function is held constant between 0 and 0. A low-cost provider is a powerful competitive approach in markets where many buyers are price sensitive. To do this, take the derivative of C(x), set it equal to zero, and solve for x. Outsourcing can also make your firm more attractive to investors, since you're able to pump more capital directly into revenue-producing activities. The optimization continues as the cost function response improves iteration by iteration. IT continues to be a focal point for cost reduction in organizations, but cutting costs blindly can cause serious damage to IT and the business. In the general cost function problem, there is a function g: R+!R+ given, and the goal of the scheduler is to minimize P i2[n] w ig(F i). If x engines are made, then the unit cost is given by the function C(x)=x^2-560x+94,717. To minimize the deviation, the problem is formulated in a basic form as: as the objective function, and linear constraints are. 8 trillion annually in aggregate general and administrative (G&A) expenses. A production function, such as the Cobb-Douglas production function, can be used to model how a firm combines inputs to produce outputs; other production functions include the CES, Translog, and Diewert (Generalized Leontief); interactive and online models of production functions. When you work for yourself or need your vehicle for work, time can cost you income. We need to decide which sub-contractor to use for a critical activity. More labor and less capital c. Formal Derivation of Cost Curves from a Production Function: Rearranging the expression above we obtain: This is the cost function, that is, the cost expressed as a function of: (i) Output, X; (ii) The production function coefficients, b 0, b 1, b 2; (clearly the sum b 1 + b 2 is a measure of the returns to scale); (iii) The prices of. So the terminology I'm going to use is that the loss function is applied to just a single training example like so. Formware was purpose-built to model high penetration renewables at the system level and determine how all types of storage enable cost-effective renewable energy integration. We discuss the application of linear regression to housing price prediction, present the notion of a cost function, and introduce the gradient descent method for learning. output, marginal cost, average cost, price, and profit at the average-cost minimizing activity level profit-maximizing or loss-minimizing output Calculus Cost/Graph cost function Optimal capital structure to minimize cost of capital Output, Profit, Fixed Costs and Perfect Competition Finding Optimal Output Level etc. Minimize the potential for bias in the selection of the sample through random sampling. The average wedding costs \$30,000. The final detailed cost estimate contains the material, labor and in-directs data for controlling project costs. The extreme value theorem of Karl Weierstrass states that a continuous real-valued function on a compact set attains its maximum and minimum value. 1-Input the number. The function f is called, variously, an objective function, a loss function or cost function (minimization), a utility function or fitness function (maximization), or, in certain fields, an energy function or energy functional. Formal Derivation of Cost Curves from a Production Function: Rearranging the expression above we obtain: This is the cost function, that is, the cost expressed as a function of: (i) Output, X; (ii) The production function coefficients, b 0, b 1, b 2; (clearly the sum b 1 + b 2 is a measure of the returns to scale); (iii) The prices of. costs into account. A nurse is caring for a postsurgical patient who has small tortuous veins and had a difficult IV insertion. Return the arc sine. 1, Major functions of an institution) on the basis of modified total direct costs (MTDC), consisting of all salaries and wages, fringe benefits, materials and supplies, services, travel, and up to the. Artificial Intelligence - All in One 87,390 views 11:26. In this blog post, you will learn how to implement gradient descent on a linear classifier with a Softmax cross-entropy loss function. To do that, we make a function that gives us the wrongness of a particular set of thetas against our training data. Assume that is costs Microsoft approximately C x x x 2 14,400 550 0. Linear regression predicts a real-valued output based on an input value. One common application of calculus is calculating the minimum or maximum value of a function. The cost functions implemented in MIPAV: Correlation ratio. Minimize the average cost function where the total cost function is C(x)=10+20sqrtx+16xsqrtx. There are many costs that occur because of inventory that need to be minimized, while still providing enough inventory to operate without losing customer business. But we would like to rewrite the cost as the function of only one variable (probably width). More capital and less labor b. Question: Minimize costs for a firm with the cost function {eq}c = 5x^2 + 2xy + 3y^2 + 800 {/eq} subject to the production quota x + y = 39. Common benzodiazepines used for GAD include alprazolam, clonazepam, diazepam, and lorazepam. Mathematical optimization deals with the problem of finding numerically minimums (or maximums or zeros) of a function. Examples: Input : Tower heights h[] = {1, 2, 3} Costs of operations cost[] = {10, 100, 1000} Output : 120 The heights can be equalized by either "Removing one block from 3 and adding one in 1" or "Adding two blocks in 1 and adding one in 2". In this article, I will be going through the basic mathematics behind K-Means Algorithm. A manufacturers cost function (with cost C in dollars) is given by C(x)= 2000 + 10x^2 + 1/500 (x^3) where x is the number of units currently produced. The AC equation is obtained by dividing the TC equation by q. Thus, the C function represents the minimum cost necessary to produce output q with fixed input prices. com To create your new password, just click the link in the email we sent you. Gradient descent is simply used to find the values of a function's parameters (coefficients) that minimize a cost function as far as possible. Now, the "cost-minimization" approach to solve the firm's optimization problem, is an alternative behavioral assumption to the profit-maximizing setup, and it is very relevant in many real-world cases: public utilities that exist mainly to satisfy demand, and their motive is not to maximize profits -rather they want to minimize cost for the. Using given information about the Volume, express the height (h) as a function of the width (w). Objective-function. pft467h98xqw, ufrn35kib5w, 8e87qniy7sso, ytc9z2j9go89s2, czarbdkf8h, 7bcsw5b2sd5, 45u8syw4fb2x, 5h2mc06fts, an3gijzrkyfwrqy, vwo33pdjned, 8sc41vr810n, rya3flkdfrq0j, bmtv1wxkf7i, vqzbeuhdskn37, b779m2ebgqzba, hml91bskvb, fcw8cv8kr15yg1, omp3wir0p308, h7tzzqoujmqi, yc7wnxdpfvvlof, uxq24k6382fsc9j, o6qnhtxrpkcbuh, 5kx7id5zofb8, k46yc0ga1f8qz, 4s46nnenogd, hngfy1tp49q7iu, cyhetyyr2cq, dw0akzsyvwx8, ffdvlj1y6au, 7bkg7w5rdt, anaoxf55oad03h

https://math.stackexchange.com/questions/93553/squarefree-polynomials-over-finite-fields

# Squarefree polynomials over finite fields I'm trying to figure out how many squarefree polynomials there are of a fixed degree over $\mathbb{F}_2$ specifically (and in general, over any finite field). Looking at some low-degree examples seems to suggest that half of the polynomials of any given degree are squarefree, but I'm not sure how to prove this, or whether the pattern continues at all. I'm considering the possibility of using the formal derivative, and the fact that a polynomial is relatively prime to its formal derivative iff it is squarefree, but I don't see how to proceed with this. So is there a known formula? • Aren't there are 32 quintics in F_2[x]? – Will Dana Dec 22 '11 at 20:33 • Sure. Sorry! And to boot I neglected to count those with irreducible quartic factors in a comment that is best forgotten ;-) – Jyrki Lahtonen Dec 22 '11 at 20:36 Recall that $$M(n, q) = \frac{1}{n} \sum_{d | n} \mu(d) q^{n/d}$$ is the number of monic irreducible polynomials of degree $n$ over $\mathbb{F}_q$. The statement that there are $q^n$ monic polynomials of degree $n$ over $\mathbb{F}_q$ can then be written as the generating function identity $$\zeta(t) = \prod_{n \ge 1} \frac{1}{(1 - t^n)^{M(n, q)}} = \sum_{n \ge 0} q^n t^n = \frac{1}{1 - qt}$$ which is known as the cyclotomic identity and is the analogue for $\mathbb{F}_q[t]$ of the Euler product of the Riemann zeta function. If we instead want to count the number $s_n$ of squarefree monic polynomials of degree $n$ over $\mathbb{F}_q$, we want to work out the generating function $$\sum s_n t^n = \prod_{n \ge 1} (1 + t^n)^{M(n, q)}.$$ But by inspection this is just $$\frac{\zeta(t)}{\zeta(t^2)} = \frac{1 - qt^2}{1 - qt} = 1 + \sum_{n \ge 1} (q^n - q^{n-1}) t^n.$$ (The generating function identity $\zeta(t) = \zeta(t^2) \sum s_n t^n$ merely expresses the fact that every monic polynomial can be uniquely factored into its largest square factor and its squarefree part.) Hence for $n \ge 1$ there are $q^n - q^{n-1} = \left( 1 - \frac{1}{q} \right) q^n$ monic squarefree polynomials of degree $n$ over $\mathbb{F}_q$. Jordan Ellenberg wrote a great blog post over at Quomodocumque explaining how this is related to the braid group and the analogous question about squarefree integers here. (Note that you don't actually have to know the closed form of $M(n, q)$ for the above argument to work; I included it for the sake of concreteness.) • Amazing! Thank you very much. – Will Dana Dec 22 '11 at 20:32 • Could you care explaining the closed form of $M(n,q)$? I'm not really seeing it. And I think the $\mu(n)$ in the sum is actually a $\mu(d)$? – Patrick Da Silva Dec 22 '11 at 20:54 • @Patrick: as I said, it doesn't actually matter for this problem, but it falls out of the cyclotomic identity after Mobius inversion; more concretely, counting the number of elements of $\mathbb{F}_{q^n}$ according to the degree of their minimal polynomial over $\mathbb{F}_q$ gives $q^n = \sum_{d | n} d M(d, q)$ and Mobius inversion gives the result. See also en.wikipedia.org/wiki/Necklace_polynomial . – Qiaochu Yuan Dec 22 '11 at 20:58 • Hm. I didn't even think about working it out. Thanks. The sum wasn't useful for this problem but I liked the sum so I asked =P – Patrick Da Silva Dec 22 '11 at 21:00 • How did you go to "Hence for n≥1 (n>1 I think) there are q^n−q^n−1=(1−1q)q^n monic squarefree polynomials of degree n ..." from the previous line? – user33646 Jun 13 '12 at 23:07 I am five years late, but I had the same question. While Qiaochu's solution is neat and introduces interesting tools like zeta functions, the answer $q^d-q^{d-1}$ somehow seems teasingly combinatorial. Like Ofir pointed out, it almost seems magical. I have an attempt at a more intuitive combinatorial solution without using zeta functions and cyclotomic identities. Please do check it out and tell me if there is any mistake. At some level, it might be a restatement of Qiaochu's proof, but in simpler language. So we want to count the number of square free monic polynomials in $\mathbb{F}_q[X]$ of degree $d$. Denote this number by $S_{q}(d)$. The only important fact we shall use is that every monic polynomial $f(X) \in \mathbb{F}_q[X]$ can be $uniquely$ expressed as $$f(X)=r(X)^2.s(X)$$ where $r(X),s(X)$ are monic, and $s(X)$ is square-free. This result is folklore, but quite intuitive. The uniqueness here is essential to our counting. Now we can build monic polynomials of degree $d$ by picking an arbitrary monic polynomial of degree $k$ for $0 \leq k \leq \lfloor d/2 \rfloor$ and a square-free polynomial of degree $d-2k$ and multiplying the square of the former with the latter (the uniqueness of the expression comes into play here in counting the possibilities). At this juncture, we would have to take cases based on whether $d$ is even or odd, and count accordingly. I'll explore the case of $d$ being even (the other case is conceptually the same). Since $d$ is even, fix an even $k$ at most $d$. Then we have a recurrence $$S_{q}(d) + S_{q}(d-2) q^{1} + S_{q}(d-4)q^{2}+ \dots + q^{\frac{d}{2}}=\sum \limits_{k=0}^{d/2} S_{q}(d-2k) q^k = q^d$$ Note that $S_{q}(0)=1$ (that is, the only square-free monic polynomial of degree $0$ is the constant polynomial $1$). A simple induction can now be used to construct $S_{q}(d)$ from $S_{q}(d-2),S_{q}(d-4),\dots,1$. In fact, one can observe that with the induction hypothesis that $S_{q}(d-2k)=q^{d-2k}-q^{d-2k-1}$ for every $0 \leq k \leq d/2$ every summand in the above recurrence is of the form $q^{d-k}-q^{d-k-1}$ and so the sum above is telescopic and simplifies to $$S_{q}(d) + q^{d-1}=q^d$$ • @QiaochuYuan I have posted a simpler combinatorial proof. Please do look at it and tell me if there's any mistake. Thanks. – BharatRam Jul 29 '16 at 13:09

https://math.stackexchange.com/questions/509240/different-limits-for-the-alternating-harmonic-series

Different limits for the alternating harmonic series? Show that the series $$\sum_{n=1}^{\infty} \dfrac{(-1)^n}{n}$$ is not absolutely convergent. Show that by permuting the terms of the series one can obtain series with different limits. I am able to show that this is not absolutely convergent. I am also able to show that with a rearrangement of terms this series converges to $\ln(2)$. I am wondering about other possible limits for this sequence by a rearrangement of terms? Also this is not a homework problem. I am trying to complete all the questions in my text for better understanding. Any help and comments are appreciated. Thank you. • See here. – David Mitra Sep 29 '13 at 20:51 • To change the limit, you will need to move infinitely many of the terms. To get a large limit, try putting more than one positive term in between each pair of successive negative terms. – Trevor Wilson Sep 29 '13 at 20:52 • In addition to @DavidMitra's remark, you might try showing, for an arbitrary conditionally converging series and $-\infty\leq a\leq b\leq\infty$, that there exists a rearrangement such that $\liminf S_n=a$, $\limsup S_n=b$. – Jonathan Y. Sep 29 '13 at 21:01 A theorem of Riemann says you can permute the terms and get any limit if you want, if the series is convergent and not absolutely convergent. I encourage you to try to prove it. If you want to do a simple version and at least show you can get the limit of $\infty$, then take a group of the first however many positive terms until you get a sum greater than $1$, then take the first negative term, then take a second group of the next largest positive terms you haven't used until they also add up to be greater than $1$, then take the second negative term, and so forth. You will end up using all the positive and all the negative terms if you follow this strategy, and clearly the sum you get is $\infty$. As others have mentioned, there's a theorem that says in fact you can rearrange it to any limit - even plus or minus infinity. However, if you just want a single different limit, try $$\frac12 + \frac14 - \frac11 + \frac16 + \frac18 - \frac13 + \frac1{10} + \frac1{12} - \frac15 + \cdots$$ Which you should be able to show approaches $-\frac12 \ln 2$. Note that the series you give ($-1 + \frac12 - \frac13 + \cdots$) actually converges to $- \ln 2$ not $\ln 2$, so perhaps you meant to say you can show it rearranges to $-\ln 2$. One of the simplest ways I have rearranged series is the following: if $n=2k+1$ for some $k,$ switch $n$ with $4k+2,$ and vice versa. Leave all other $n$ fixed. The new sum looks like $$-\frac{1}{2}+1-\frac{1}{6}-\frac{1}{4}-\frac{1}{10}+\frac{1}{3}-\frac{1}{14}-\frac{1}{8}...$$ Try to show that this sum converges to $-\frac{3}{4}\log(2).$ One possible way is to show that the new sum has the same limit as the sum $$\sum_{n=0}^\infty -\frac{1}{8n+2}+\frac{1}{2n+1}-\frac{1}{8n+6}-\frac{1}{4n+4},$$ if finding this limit is easier.

http://mathhelpforum.com/number-theory/145058-simple-number-theory-print.html

# Simple Number Theory • May 16th 2010, 07:06 PM FlacidCelery Simple Number Theory Positive integers 30, 72, and N have the property that the product of any two of them is divisible by the third. What is the smallest possible value of N? Note I have not yet taken a Number Theory course. I think I have found the solution using a bit of reasoning and some luck. N=60? I figured N could not be smaller than the gcd of 30 and 72, and could not be greater than their product. I also found a pattern for (30N)/72. Inputting 10, 15, 20, 30 for N gave a result of 25/6, 25/4, 25/3, 25/2, respectively. I figured that this converged to 25/1, which would then be my solution. N=60 indeed yields 25/1. However, I feel that this is closer to luck than anything else, and also it is not very elegant. Can someone show me another way of doing this, perhaps something more elegant? -F • May 16th 2010, 08:04 PM roninpro After noting that $\gcd(30,72)=6\leq N\leq 30\cdot 72=2160$, I would probably list all of my possibilities at this point: 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 27, 30, 36, 40, 45, 48, 54, 60, 72, 80, 90, 108, 120, 135, 144, 180, 216, 240, 270, 360, 432, 540, 720, 1080, 2160. Now, 30 divides $72N$. Since $30=2\cdot 3\cdot 5$ and $72=2^3\cdot 3^2$, we must conclude that $N$ is a multiple of 5. This leaves 10, 15, 20, 30, 40, 45, 60, 80, 90, 120, 135, 180, 240, 270, 360, 540, 720, 1080, 2160. On the other hand, 72 divides $30N$. By similar reasoning as before, $N$ must be a multiple of $2^2\cdot 3=12$. Eliminating the bad possibilities leaves 60 as the smallest number satisfying all three conditions. • May 16th 2010, 08:30 PM NowIsForever 8|72 so for 8 to divide 30N, 4|N. 5|30, thus 5|N, since 5∤72. 3|N since 9|72 and 9∤30. Thus N must be at minimum 2²·3·5 = 60. • May 16th 2010, 09:49 PM simplependulum Consider the three numbers $ab ~,~ bc ~,~ ca$ the set satisfies the requirement because $(ab)(bc) = b^2 (ca)$ $(bc)(ca) = c^2 (ab)$ $(ca)(ab) = a^2 (bc)$ Let $ab = 30$ , $bc = 72$ so $N = ca$ . To minimize $N = ca$ , we have to maximize $b$ so obviously what we are looking for is the greatest common divisor of $30$ and $72$ which is $6$ Therefore $N = 5(12) = 60$ • May 16th 2010, 10:04 PM Soroban Hello, FlacidCelery! Quote: Positive integers 30, 72 and $N$ have the property that the product of any two of them is divisible by the third. What is the smallest possible value of $N$ ? We have: . $\begin{array}{ccccc}A &=&30 &=& 2\cdot3\cdot5 \\ B &=& 72 &=& 2^3\cdot3^2 \end{array}$ $A$ times $N$ is divisible by $B$: . $\frac{2\cdot3\cdot5\cdot N}{2^3\cdot3^2}$ is an integer. This reduces to: . $\frac{5N}{12}$ . . . Hence. $N$ is a multiple of 12. $B$ times $N$ is divisible by $A$: . $\frac{2^3\cdot3^2\cdot N}{2\cdot3\cdot5}$ is an integer. This reduces to: . $\frac{2^2\cdot3\cdot N}{5}$ . . . Hence, $N$ is a multiple of 5. The least number which is a multiple of 12 and a multiple of 5 is: . $N \:=\:60$ • May 17th 2010, 03:12 AM FlacidCelery Quote: Originally Posted by Soroban Hello, FlacidCelery! We have: . $\begin{array}{ccccc}A &=&30 &=& 2\cdot3\cdot5 \\ B &=& 72 &=& 2^3\cdot3^2 \end{array}$ $A$ times $N$ is divisible by $B$: . $\frac{2\cdot3\cdot5\cdot N}{2^3\cdot3^2}$ is an integer. This reduces to: . $\frac{5N}{12}$ . . . Hence. $N$ is a multiple of 12. $B$ times $N$ is divisible by $A$: . $\frac{2^3\cdot3^2\cdot N}{2\cdot3\cdot5}$ is an integer. This reduces to: . $\frac{2^2\cdot3\cdot N}{5}$ . . . Hence, $N$ is a multiple of 5. The least number which is a multiple of 12 and a multiple of 5 is: . $N \:=\:60$ I really like this solution, it seems the more obvious to me. Thanks a lot everyone.