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https://math.stackexchange.com/questions/414865/more-than-continuum-many-functions | # More than continuum many functions
Suppose we have a function $f_\alpha:\mathbb{N} \to o$ for each $\alpha<c^+$ (successor cardinal of $c=2^{\aleph_{0}}$), where $o$ is some ordinal. Show that there exists a set $S \subseteq c^+$ such that $|S|=c^+$ and for every $\alpha,\beta \in S$ where $\alpha<\beta$ we have that $f_\alpha(n)\leq f_\beta(n)$ for all $n \in \mathbb{N}$.
I was thinking maybe to color pairs $\alpha,\beta \in o$ in colors depending on the smallest index $n$ for which $f_\alpha(n)\leq f_\beta(n)$ doesn't hold (i.e countable number of colors), and maybe use some coloring theorem? The problem is I can't see an appropriate coloring theorem for the result I want, and also I don't see a contradiction in there being a homogeneous set not of color $-1$ (i.e no such index).
My second approach would be assuming the negation and somehow building an infinite decreasing sequence in $o$, $f_{\alpha_1}(n)>f_{\alpha_2}(n)>f_{\alpha_3}(n)...$ which would contradict well-orderedness, but I can't really see how to do that either.
• Do you have any conditions on $o$? – Sungjin Kim Jun 8 '13 at 18:12
• No, $o$ is just there to be able to compare $f_\alpha(n)$ and $f_\beta(n)$. – ctlaltdefeat Jun 8 '13 at 18:31
• Just to remind myself: what was the successor cardinal? – Sungjin Kim Jun 8 '13 at 18:53
• Well, it's the minimal cardinal (initial ordinal) that is greater than the specified cardinal. – ctlaltdefeat Jun 8 '13 at 19:00
I need the following generalization of (a special case of) the Dushnik-Miller theorem: If the set $[c^+]^2$ of $2$-element subsets of $c^+$ is partitioned into countably many pieces $P_n$ ($n\in\omega$), then either there is a set $H\subseteq c^+$ of order-type $c^+$ with $[H]^2$ (the set of $2$-element subsetes of $H$) included in $P_0$, or there is an infinite $H\subseteq c^+$ with $[H]^2$ included in $P_n$ for some $n>0$. Looking for a reference for this, the first I found was Theorem 3.10 of Chapter 2 ("Partition Relations" by András Hajnal and Jean Larson) of the Handbook of Set Theory. This is a theorem of Erdős and Rado, which (by specializing to $\kappa=\aleph_1$ and $\gamma=\omega$) gives more than I need; I expect there are easier proofs of what I actually need.
Given that result, and given functions $f_\alpha$ as in the question, define $P_0$ to consist of those $2$-element sets $\{\alpha<\beta\}$ for which $(\forall n)\,f_\alpha(n)\leq f_\beta(n)$, and define $P_{k+1}$ to consist of those $\{\alpha<\beta\}$ for which $k$ is the smallest integer with $f_\alpha(k)>f_\beta(k)$. These $P_n$'s constitute a partition of $[c^+]^2$, and a homogeneous set of order-type $c^+$ for piece $P_0$ is exactly what the question asks for. So all I need to do is to exclude the possibility of infinite homogeneous sets for any of the other pieces $P_{k+1}$. But such a homogeneous set would begin with an $\omega$-sequence of ordinals $\alpha(0)<\alpha(1)<\dots$ such that, for each $i$, $f_{\alpha(i)}(k)>f_{\alpha(i+1)}(k)$. That is, we'd have an infinite decreasing sequence of ordinals, a contradiction.
• I had the same idea (when I wrote coloring I was meaning partitioning), but must have missed that specific partition theorem because it looks like we did indeed learn it. Thanks! – ctlaltdefeat Jun 8 '13 at 21:25
• @user14111 An alternative to winning the lottery: Find the table of contents of the Handbook, go to the web sites of the authors of individual chapters, and look for downloadable pre-publication versions of the chapters. By the way, the handbook is in three volumes, any one of which already presupposes a very large hand, so I would not advise printing out what you download. – Andreas Blass Jun 9 '13 at 22:51
Here’s a proof that doesn’t use a big combinatorial hammer, though I got the idea from a proof of the Erdős-Rado theorem. Let $S=\{\alpha<\mathfrak{c}^+:\operatorname{cf}\alpha=\omega_1\}$. For each $\alpha\in S$ and $n\in\omega$ construct a finite sequence $\langle\beta(\alpha,n,k):k<\ell(\alpha,n)\rangle$ as follows. If $\{\gamma<\alpha:f_\gamma(n)>f_\alpha(n)\}\ne\varnothing$, let $$\beta(\alpha,n,0)=\min\{\gamma<\alpha:f_\gamma(n)>f_\alpha(n)\}\;;\tag{1}$$ otherwise let $\ell(\alpha,n)=0$. Given $\beta(\alpha,n,k)$, let
$$\beta(\alpha,n,k+1)=\min\{\gamma<\alpha:f_{\beta(\alpha,n,k)}(n)>f_\gamma(n)>f_\alpha(n)\}\tag{2}$$
if such an ordinal exists, and otherwise let $\ell(\alpha,n)=k+1$. Since $\beta(\alpha,n,k)>\beta(\alpha,n,k+1)$ whenever both ordinals are defined, $\ell(\alpha,n)\in\omega$ for each $\alpha\in S$ and $n\in\omega$. For each $\alpha\in S$ let $A_\alpha=\bigcup_{n\in\omega}\{\beta(\alpha,n,k):k<\ell(\alpha,n)\}$, and if $A_\alpha\ne\varnothing$, let $\eta_\alpha=\sup A_\alpha$; $A_\alpha$ is countable, so $\eta_\alpha<\alpha$. The map $\alpha\mapsto\eta_\alpha$ is a pressing-down function on $S_0=\{\alpha\in S:A_\alpha\ne\varnothing\}$.
If $S_0$ is stationary, then there are a stationary $S_1\subseteq S_0$ and an $\eta<\mathfrak{c}^+$ such that $\eta_\alpha=\eta$ for all $\alpha\in S_1$. There are only $\mathfrak{c}$ distinct possibilities for
$$\left\{\{\beta(\alpha,n,k):k<\ell(n)\}:n\in\omega\right\}\;,$$
so there is $S_2\subseteq S_1$ such that $|S_2|=\mathfrak{c}$, and
$$\left\{\{\beta(\alpha_0,n,k):k<\ell(n)\}:n\in\omega\right\}=\left\{\{\beta(\alpha_1,n,k):k<\ell(n)\}:n\in\omega\right\}$$
for all $\alpha_0,\alpha_1\in S_2$. Suppose that there are $\alpha_0,\alpha_1\in S_2$ and $n\in\omega$ such that $\alpha_0<\alpha_1$ and $f_{\alpha_0}(n)>f_{\alpha_1}(n)$. Then for all $k<\ell(\alpha_1,n)$ we have $f_{\beta(\alpha_1,n,k)}(n)>f_{\alpha_0}(n)>f_{\alpha_1}(n)$, and $\beta\big(\alpha_1,n,\ell(\alpha_1,n)\big)$ should have been defined: there was at least one ordinal available, $\alpha_0$, that met the requirements of whichever of $(1)$ and $(2)$ was appropriate. This contradiction shows that if $\alpha_0,\alpha_1\in S_2$ with $\alpha_0<\alpha_1$, then $f_{\alpha_0}(n)\le f_{\alpha_1}(n)$ for all $n\in\omega$, which is the desired result.
Suppose now that $S_0$ is not stationary, and let $S_1=S\setminus S_0$; $S_1$ is stationary, so $|S_1|=\mathfrak{c}^+$. Suppose that $\alpha_0,\alpha_1\in S_1$ with $\alpha_0<\alpha_1$. $A_{\alpha_1}=\varnothing$, so for each $n\in\omega$ we must have $f_{\alpha_0}(n)\le f_{\alpha_1}(n)$, which is again the desired result.
Here is a partial answer by transfinite induction. I'm not very used to ordinals so I hope it's correct.
We proceed by transfinite induction on $o$. We strenghten the problem by considering partial functions: $f_\alpha\leq f_\beta$ if for all $n$ where both are defined, we have $f_\alpha(n)\leq f_\beta(n)$, The base case with $o=\{1\}$ is clear.
If $o=o'+1$ is a successor ordinal, we will isolate the maximal possible output $M$ (i.e. $o'$). We consider the partial functions $f'_\alpha$ where $M$ is replaced by $\bot$ (undefined). By hypothesis, there is $S'\subseteq c^+$ verifying all the requirements on the $f'_\alpha$'s.
Now for each $g:\mathbb N\to \{o',\{M\},\{\bot\}\}$ we define the set $S_g=\{\alpha \in S' : \forall n\in\mathbb N,f_\alpha(n)\in g(n)\}$. Since the $S_g$'s form a $c$-partition of $S'$, and $|S'|=c^+$, there exists $g$ such that $|S_g|=c^+$. This $S_g$ verifies all the conditions required for the $S$ we are looking for.
We now have to treat the limit case. For each $o'<o$, we consider $f_\alpha^{o'}$ to be $f_\alpha$ where the value is undefined outside of $o'$. Let $S_{o'}$ be the solution to the problem on the $f_\alpha^{o'}$ (by induction hypothesis). Notice that if $\alpha\leq \beta$ then $S_\alpha\supseteq S_\beta$ Let $S=\bigcap_{o'<o} S_{o'}$, we just need to prove that $|S|=c^+$.
• Why does $f_\alpha$ have to have a maximal output? It could be that it doesn't achieve a maximal output. – ctlaltdefeat Jun 8 '13 at 19:31
• Here $M$ is just the maximal element of $o$, i.e. $o'$ in the standard definition of ordinals. – Denis Jun 8 '13 at 19:32
• Why can't $f_\alpha$ have the output $o$ at some $n$? The functions are built for $o$. – ctlaltdefeat Jun 8 '13 at 19:35
• $o$ is defined as the set of ordinals $\alpha<o$. So if it is the codomain of $f$, it is not a possible output value. – Denis Jun 8 '13 at 19:36
• Oh right I got it confused with $o'$ So $M$ is just $o'$ all the time? – ctlaltdefeat Jun 8 '13 at 19:36 | 2020-07-05T04:34:57 | {
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http://claudiaerografie.it/nxqd/matrix-multiplication-divide-and-conquer.html | # Matrix Multiplication Divide And Conquer
Divide-and-Conquer Matrix Factorization Lester Mackeya Ameet Talwalkara Michael I. The Karatsuba algorithm provides a striking example of how the \Divide and Conquer" technique can achieve an asymptotic speedup over an ancient algorithm. Divide and conquer is a way to break complex problems into smaller problems that are easier to solve, and then combine the answers to solve the original problem. We also study the impact of the o -diagonal compression on the accuracy of the eigenvalues when a matrix is approximated by an HSS form. Introduction. If the problem is small enough in size then we solve the problem in a. We implemented Ztune to autotune serial divide-and-conquer matrix-vector multiplication on machines with different hardware configurations, and found that Ztuneoptimized codes ran 1%-5% faster than the hand. Other problems such as the Tower of Hanoi are also simplified by this approach. In International Linear Algebra Society(ILAS) Symposium on Algorithms for Control, Signals and Image Processing, 1997. NASA Astrophysics Data System (ADS) James, S. The most obvious cause of the poor performance of matrix multiplication was the absence of spatial locality. Zigzag (or diagonal) traversal of Matrix; Divide and Conquer | Set 5 (Strassen's Matrix Multiplication) Print all possible paths from top left to bottom right of a mXn matrix; Count all possible paths from top left to bottom right of a mXn matrix; Printing brackets in Matrix Chain Multiplication Problem. In this tutorial, you'll learn how to implement Strassen's Matrix Multiplication in Swift. It turns out that Matrix multiplication is easy to break into subproblems because it can be performed blockwise. For a by $4 \times 4$ matrix you will have to intersect, multiply and accumulate 11 times. Do not be confused about these indexes. Rivest and Clifford Stein. Divide and Conquer Mergesort Quicksort Binary Search Selection Matrix Multiplication Convex Hull * * * * * * * * * Selection Find the kth smallest (largest) item in a list. •These are huge matrices, say n ≈50,000. Zima (SCS, UW) Module 4: Divide and Conquer Winter 20207/14. 7: Matrix multiplication using Strassen’s algorithm. Greedy Algorithms Idea: Find solution by always making the choice that looks. Merge sort, quicksort, and binary search use divide and conquer while matrix chain multiplication and optimal binary search tree use dynamic programming. 00003 https://dblp. To see what this means, carve into four,. Computer Programming - C++ Programming Language - A C++ Program to Multiply two Matrices. This calculator is designed to multiply and divide values of any Binary numbers. divide-and-conquer algorithms is new, to our knowledge: the only explicit exam-ples in [9,3] describe Karatsuba multiplication. Tree data structures. For a by $4 \times 4$ matrix you will have to intersect, multiply and accumulate 11 times. Divide-and-Conquer PowerPoint Presentation- CS 46101 Section 600. Get this from a library! An I/O-Complexity Lower Bound for All Recursive Matrix Multiplication Algorithms by Path-Routing. And, the element in first row, first column can be selected as X[0][0]. The main condition of matrix multiplication is that the number of columns of the 1st matrix must equal to the number of rows of the 2nd one. Y1 - 2016/10/1. , algorithms that compute less than O(N^3) operations--- are becoming attractive for two simple reasons: Todays software libraries are reaching the core peak performance (i. An overview of the algorithm. Conquer by sorting the two subarrays recursively using mergesort. Winner of the Standing Ovation Award for "Best PowerPoint Templates" from Presentations Magazine. • The basic Strassen’s algorithm, and a general understanding of how Strassen’s method managed to eliminate one multiplication from the “ordinary” computation (although you. Matrix multiplication. Matrix multiplication is particularly easy to break into subproblems, because it can be performed blockwise. Given a n x n matrix where each of the rows and columns are sorted in ascending order, find the kth smallest element in the matrix. I want to make a series in which I will discuss about some algorithms which follow divide and conquer strategy. BruteForce 2. In this post I will explore how the divide and conquer algorithm approach is applied to matrix multiplication. There is an extensive literature on such divide-and-conquer algorithms: the PRISM project algorithm [2, 6], the reduction of the symmetric eigenproblem to matrix multiplication by Yau and Lu [35], the matrix sign. 7 Determining thresholds 2. html db/journals/cacm/cacm41. Recall that when multiplying two matrices, A =aij and B =bjk, the resulting matrix C =cik is given by. ・n / bi = size of subproblem at level i. • His method uses. ) This is typical in the analysis of. An overview of the algorithm. Combine these results together. Matrix chain multiplication (or Matrix Chain Ordering Problem, MCOP) is an optimization problem that to find the most efficient way to multiply given sequence of matrices. I have 4 Years of hands on experience on helping student in completing their homework. 2 Strassen's algorithm for matrix multiplication Table of contents. Divide the problem (instance) into subproblems. CONTENTS List of topics that we cover in the article: What is Divide and conquer? How problems are solved using Divide and conquer? Running times for sorting algorithms. This calculator is designed to multiply and divide values of any Binary numbers. Tree data structures. 27) [20 points]. O(n) multiplication) to find out only one entry of the result Z •Total time will be O(n3). Divide And Conquer Algorithms Lecture 5 Recall in last lecture, we looked at one way of parallelizing matrix multiplication. Perkalian Matrix dengan Divide and Conquer dan Algoritma Strassen Otniel and 13508108 Program Studi Teknik Informatika Sekolah Teknik Elektro dan Informatika Institut Teknologi Bandung, Jl. 8 When not to use D & C Multiplying matrices example • Given: two matrices A and B the product C will have the same number of rows and columns. The time-complexity is O(n^(2. Pros of Divide and Conquer Strategy. Divide and Conquer 0 12 Young CS 331 D&A of Algo. complexity of matrix multiplication algorithm is crucial in many. , 90% of peak performance) and thus reaching the limitats of current systems. I want to know when to switch to another algoritm, which in this case is the regular matrix multiplication. Keywords—Matrix, Matrix Multiplication, Divide and Conquer, Freivald’s Algorithm I. org/rec/conf/ppopp. On the left is the normal 4-by-4 matrix multiplication. Basic idea. Applying a divide and conquer strategy recursively (view A i;j, B i;j and C i;j as matrices instead of scalars) allows matrix multiplication over n = 2N size matrices to be performed using only 7N = 7log 2 n= nlog 2 7 = O(n2:81) multiplications. This paper deals with parallels of the fast matrix multiplication strassen's algorithm, winograd's algorithm and analyzes empirical study of the matrix multiplication under the distributed environment in. Divide the problem into smaller problems. Problem 1: Divide and Conquer Matrix Multiplication (Taken from DPV 2. Divide And Conquer Algorithm sample code - Build a C++ Program with C++ Code Examples - Learn C++ Programming. Combine: add appropriate products using 4 matrix additions. It uses divide and conquer strategy, and thus, divides the square matrix of size n to n/2. In mathematics, matrix multiplication or matrix product is a binary operation that produces a matrix from two matrices with entries in a field, or, more generally, in a ring or even a semiring. Integer Multiplication 3. More on Recurrence Relations. Divide and Conquer is a recursive problem-solving approach which break a problem into smaller subproblems, recursively solve the subproblems, and finally combines the solutions to the subproblems to solve the original problem. Solve each part recursively. 4 The recursion-tree method for solving recurrences 4. Divide the problem (instance) into subproblems. A Divide-and-Conquer Parallel Pattern Implementation for Multicores Marco Danelutto, Tiziano De Matteis, Gabriele Mencagli, and Massimo Torquati Department of Computer Science, University of Pisa, Italy fmarcod, dematteis, mencagli, [email protected] 101-102 1998 41 Commun. BruteForce 2. Combine solutions to sub-problems into overall solution. It turns out that Matrix multiplication is easy to break into subproblems because it can be performed blockwise. Divide and Conquer example: Matrix Multiplication The normal procedure to multiply two n × n matrices requires n3 time. Algorithms designed with Divide and Conquer strategies are efficient when compared to its counterpart Brute-Force approach for e. The divide and conquer approach is an algorithm design paradigm which can be used to perform matrix multiplication. Big list of c program examples. The divide-and-conquer paradigm often helps in the discovery of efficient algorithms. Divide & Conquer 1. O(n) multiplication) to find out only one entry of the result Z •Total time will be O(n3). Strassens’s Matrix Multiplication • Strassen showed that 2x2 matrix multiplication can be accomplished in 7 multiplication and 18 additions or subtractions. From this, a simple algorithm can be constructed which loops over the indices i from 1 through n and j from 1 through p, computing the above using a nested loop:. Find the determinant of a larger matrix. Apr 28, 2020 - Lecture 13 : Recurrences and Divide and Conquer - PPT, Algorithms Notes | EduRev is made by best teachers of. It consists of rows and columns. Divide-and-Conquer Matrix Factorization Lester Mackeya Ameet Talwalkara Michael I. Shivakumar: Exploiting Geographical Location Information of Web Pages. analysis of a divide and conquer algorithm. [37:35] Divide and conquer algorithm for multiplying matrices. The Karatsuba algorithm for multiplication uses a 3-way divide and conquer to achieve a running time of O(3 n^log_2 3) which beats the O(n^2) for the ordinary multiplication algorithm (n is the number of digits in the numbers). Strassen’s matrix multiplication algorithm. Divide and Conquer. Matrix Multiplication • Simple Divide-and-conquer algorithm 𝑇 : ;=8𝑇 2 +Θ 2 # submatrices submatrix size work adding submatrices =2and =3 Case 1 (Master Method) Time Complexity: Θ 3 1) Divide matrices A and B in 4 sub-matrices of size 𝑁 2 ×𝑁 2. Contribute to saulmm/Divide-and-conquer development by creating an account on GitHub. Assuming that n is an exact power of 2, we divide each of A, B, and C into four n/2 × n/2 matrices, rewriting the equation C = AB as follows:. It enables us to reduce O(n^3) time complexity to O(n^2. Lecture 8 8-3 Strassen’s algorithm Divide and conquer algorithms can similarly improve the speed of matrix multiplication. But by using divide and conquer technique the overall complexity for multiplication two matrices is reduced. Merge sort, quicksort, and binary search use divide and conquer while matrix chain multiplication and optimal binary search tree use dynamic programming. Divide and conquer, and application to defective chessboard and min-max problem. This approach leads to an algorithm with running time , which is an improvement on the running time of the naive algorithm. Iterative algorithm. Question: Compute the n by n matrix product Z = XY. Recipe for solving common divide-and-conquer recurrences: Terms. The problem is not actually to perform the multiplications, but merely to decide the sequence of the matrix multiplications involved. T1 - Error-free transformation of matrix multiplication with a posteriori validation. Divide-and-conquer. Reading: Chapter 18 Divide-and-conquer is a frequently-useful algorithmic technique tied up in recursion. en stanford. Suppose, matrix A has p rows and q columns i. Classical matrix multiplication yields ω=3, and Strassen’s algorithm [Str69] achieves ω=log7/log2≈2. 6 Proof of the master theorem 4 Divide-and-Conquer In Section 2. BibTeX @INPROCEEDINGS{Pauca97architecture-efficientstrassen's, author = {Paul Pauca and Xiaobai Sun and Siddhartha Chatterjee and Alvin Lebeck}, title = {Architecture-efficient Strassen's Matrix Multiplication: A Case Study of Divide-and-Conquer Algorithms}, booktitle = {In International Linear Algebra Society(ILAS) Symposium on Algorithms for Control, Signals and Image Processing}, year = {1997}}. In this note, log will always mean log 2 (base-2 logarithm). The first row can be selected as X[0]. Towers of Hanoi 🗼 The Towers of Hanoi is a mathematical problem which compromises 3 pegs and 3 discs. Soumyottam Chatterjee. , Insertion Sort) •Transform & Conquer: Where we problems/its representations are transformed into a simplified problem or. T(n) arithmetic. Question: Compute the n by n matrix product Z = XY. Assume n is a power of 2. Computer Programming - C++ Programming Language - A C++ Program to Multiply two Matrices. A divide-and-conquer algorithm for this problem would proceed as follows: Let P = (n,a [i],…. Target array : All the sub arrays: 1-4. −multiplication by divide and conquer. Spring 2014 Divide-and-Conquer 22 Matrix Multiplication ! Given n x n matrices X and Y, wish to compute the product Z=XY. Summer 2012, at GSU. 3 The substitution method for solving recurrences 4. With divide-and-conquer multiplication, we split each of the numbers into two halves, each with n/2 digits. Greedy Algorithms Idea: Find solution by always making the choice that looks. Brief review of the tridiagonal DC method. Where the idea came from is unclear, however the goal was to reduce the number of multiplications needed to complete the algorithm. C program to find determinant of a matrix 12. CSE 6331 Algorithms. The Karatsuba algorithm provides a striking example of how the \Divide and Conquer" technique can achieve an asymptotic speedup over an ancient algorithm. Threaded Matrix Multiplication Tag: c++ , multithreading , c++11 , matrix-multiplication I'm working on a threaded implementation of matrix multiplication to work with my custom Matrix class, and I'm running into some issues with speed-up. To see what this means, carve A into four n/2 × n/2 blocks, and also square matrices B and C :. Who Should Enroll Learners with at least a little bit of programming experience who want to learn the essentials of. Probably most well-known technique in Computer Science. Such systems are able to support a large volume of parallel communication of various patterns in constant time. Simple Matrix Multiplication Method Divide and Conquer Method Strassen's Matrix Multiplication Method PATREON : https://www. Write a c program for scalar multiplication of matrix. Divide & Conquer: First Approach Assumption: n is always an exact power of 2. Graphs (Directed Graphs and. Consequence. Founded by Adam Hendricks, John Lang and Greg Gilreath. Divide and Conquer. Divide and Conquer Introduction. Matrix Chain Multiplication Dynamic Programming solves problems by combining the solutions to subproblems just like the divide and conquer method. Divide-and-Conquer Reading: CLRS Sections 2. divide-and-conquer algorithms is new, to our knowledge: the only explicit exam-ples in [9,3] describe Karatsuba multiplication. Introduction. Divide the problem into a number of sub-problems that are smaller instances of the same problem. – The above naturally leads to divide-and-conquer solution: ∗ Divide X and Y into 8 sub-matrices A, B, C, and D. The equation 4. 3 The D & C approach 2. ECE750-TXB Lecture 5: Veni, Divisi, Vici Todd L. Divide the problem into smaller problems. Find the determinant of a larger matrix. [50:00] Analysis of Strassen's algorithm. Easy Tutor says. [43:09] Strassen's matrix multiplication algorithm. De nition 4 (bilinear map). Divide and Conquer † A general paradigm for algorithm design; inspired by emperors and colonizers. ; The median of a finite list of numbers can be found by arranging all the numbers from lowest value to highest value and picking the middle one. Strassen's algorithm is a pretty smart algorithm which performs the same operation. On some problems, improving the running time makes interesting exercises, as will be duly mentioned. Not divide and conquer For a nice paper on this problem see J. • Dynamic programming is needed when subproblems are dependent; we don’t know where to partition the problem. Nipun Vats. GENERAL METHOD: Given a function to compute on n inputs the divide-and-conquer strategy suggests splitting the inputs into k distinct subsets, 1Running time satisfies T(1) = 1 and T(n) = 7 T(n / 2) + O(n2) Matrix Multiplication (out of scope). DIVIDE AND CONQUER II ‣ master theorem ‣ integer multiplication ‣ matrix multiplication ‣ convolution and FFTSECTION 5. Graphs (Depth-first Search) Chapter 3: pp. Iterative algorithm. In divide and conquer approach, a problem is divided into smaller problems, then the smaller problems are solved independently, and finally the solutions of smaller problems are combined into a solution for the large problem. It turns out that Matrix multiplication is easy to break into subproblems because it can be performed blockwise. Here the dimensions of matrices must be a power of 2. 1 of Introduction to Algorithms introduces Merge sort algorithm, Chapter 4 "Divide and conquer" introduces The maximum-subarray problem and Strassen's algorithm for matrix multiplication. To solve a given problem, it is subdivided into one or more subproblems each of which is similar to the given problem. Goal Implementing a large matrix-matrix multiplication on FPGA Approach Using divide and conquer techniques to describe the matrix multiplication algorithm and then using SDSoC for high-level synthesis Benefits High-performance implementation, short time-to-market design Credit This work has been done under the ENPOWER project (funded by EPSRC) at the University of Bristol. Divide: partition A and B into ½n-by-½n blocks. In the above divide and conquer method, the main component for high time complexity is 8 recursive calls. I will start with a brief introduction about how matrix multiplication is generally observed and implemented, apply different algorithms (such as Naive and Strassen) that are used in practice with both pseduocode and Python code, and then end with an analysis of their runtime. Simple Matrix Multiplication Method Divide and Conquer Method Strassen's Matrix Multiplication Method PATREON : https://www. Sorting Summary. Divide and conquer is a powerful algorithm design technique used to solve many important problems such as mergesort, quicksort, calculating Fibonacci numbers, and performing matrix multiplication. Divide and Conquer Idea: Divide problem instance into smaller sub-instances of the same problem, solve these recursively, and then put solutions together to a solution of the given instance. analysis •Decrease & Conquer: Very similar to divide & conquer, but reduce to 1 smaller sub-problem (e. , the shapes are 2 n × 2 n for some n. Strassens’s Matrix Multiplication • Strassen (1969) showed that 2x2 matrix multiplication can be accomplished in 7 multiplications and 18 additions or subtractions 𝑇𝑛= 7𝑇. On the left is the normal 4-by-4 matrix multiplication. Topic: Divide and Conquer 24 The Divide-and-Conquer way: Suppose x and y are large integers, divide x. Directed by Lana Wachowski, Lilly Wachowski. N2 - In this study, we examine the accurate matrix multiplication in floating-point arithmetic. 807) • This reduce can be done by Divide and Conquer Approach. Multiplication 21 −Strassen’s fast matrix multiplication (1969) (cont. 2 Strassen's algorithm for matrix multiplication 4. • Dynamic programming is needed when subproblems are dependent; we don’t know where to partition the problem. We'll see how it is useful in SORTING MULTIPLICATION A divide-and-conquer algorithm has three basic steps Divide problem into smaller versions of the same problem. Leetcode/F家，Linkedin -- 311. Each of these equations multiplies two. Matrix chain multiplication (or Matrix Chain Ordering Problem, MCOP) is an optimization problem that to find the most efficient way to multiply given sequence of matrices. 6: Normal matrix multiplication. This was the first matrix multiplication algorithm to beat the naive O(n³) implementation, and is a fantastic example of the Divide and Conquer coding paradigm — a favorite topic in coding interviews. Material in this lecture: What is the study of Algorithms all about? Why do we care about speci cations and proving guarantees? The Karatsuba multiplication algorithm. 3728639})[/math] time [1]. 2) Calculate following values recursively: • ae + bg • af + bh. g(x) results in a square matrix with elements f'(x). Consider again two n×n matrices A = X Y Z W. If the sub-problem sizes are small enough, however, just solve the sub-problems. Instead of changing the size at the beginning, we. When working over the integers and taking into account the growth of coefficients, the general bound for matrix multiplication specialises to. This is because there is an overhead of dividing each time, copying, adding, etc. With divide-and-conquer multiplication, we split each of the numbers into two halves, each with n/2 digits. Spring 2014 Divide-and-Conquer 22 Matrix Multiplication ! Given n x n matrices X and Y, wish to compute the product Z=XY. Use a divide-and-conquer approach as in Strassen’s algorithm, except that instead of getting 7 subproblems of size n/2, get 5 subproblems of size n/2 based on part (a). Classical matrix multiplication b. ・n / bi = size of subproblem at level i. Conquer the sub-problems by solving them recursively. −multiplication by divide and conquer. † Examples: Binary Search, Merge sort, Quicksort etc. 4 Quick sort (skip) 2. ie Abstract. Tiling is a key technique for data locality optimization and is widely used in high-performance implementations of dense matrix-matrix multiplication for multicore/manycore CPUs and GPUs. At the end of the lecture, we saw the reduce SUM operation, which divides the input into two halves, recursively calls itself to obtain the sum of these smaller inputs, and returns the sum of the results from those. 3 The D & C approach 2. In the previous post, we discussed some algorithms of multiplying two matrices. The result matrix has the number of rows of the first and the number of columns of the second matrix. I'm trying to implement vectorized matrix multiplication in Rust but there are a couple barriers I can't quite overcome. Combine solutions to sub-problems into overall solution. Case 1 of Master Method solution = Θ𝑛. Conclusion. 3 - Updated 12 days ago - 25 stars ndarray-complex. DIVIDE-AND-CONQUER: Mergesort, Quicksort, Binary Search, Binary Tree Traversals and Related Properties, Multiplication of large integers, Strassen‟s Matrix Multiplication. It uses divide and conquer strategy, and thus, divides the square matrix of size n to n/2. I've implemented the O(log_2 7) Strassen algorithm once (which should be really simple after implementing normal divide and conquer) and after benchmarking I've determined that for matrices smaller than 128x128 it's not worth to. On some problems, improving the running time makes interesting exercises, as will be duly mentioned. Examples: Mergesort, Quicksort, Strassen's algorithm, FFT. You can multiply a matrix A of p × q dimensions times a matrix B of dimensions q × r, and the result will be a matrix C with dimensions p × r. Because this algorithm is recursive, there are many method calls, and method returns. Since we divide A, B and C into 4 submatrices each, we can compute the resulting matrix C by • 8 matrix multiplications on the submatrices of A and B, • plus Θ(n2) scalar operations. 2), and finally. Else: Divide: Divide the problem into two or more disjoint subproblems Conquer: Use divide-and-conquer recursively to solve the subproblems Combine: Take the solutions to the subproblems and combine these solutions into a solution for the original problem Tiling: Divide-and-Conquer Tiling is a divide-and-conquer algorithm: Just do it trivially. The Karatsuba algorithm provides a striking example of how the \Divide and Conquer" technique can achieve an asymptotic speedup over an ancient algorithm. Jordana, b a Department of Electrical Engineeringand ComputerScience,UCBerkeley b Department of Statistics, UC Berkeley Abstract This work introduces Divide-Factor-Combine (DFC), a parallel divide-and-conquer framework for noisy matrix factorization. At the end of the lecture, we saw the reduce SUM operation, which divides the input into two halves, recursively calls itself to obtain the sum of these smaller inputs, and returns the sum of the results from those. The total time spent is then divided by the. We can treat each element as a row of the matrix. First example: matrix multiplication Matrix multiplication is one of the basic operations that you can do with matrices and a classic problem used in concurrent and parallel programming courses. CSC 210-12: Divide and Conquer: Multiplication of Large Integers and Strassen's Matrix Multiplication Based on slides prepared for the book: Anany Levitin, Introduction to The Design and Analysis Algorithms, 2nd edition, Addison Wesley, 2007 Strassen's Matrix Multiplication Let A. method for matrix multiplication. 3 The substitution method for solving recurrences 4. Divide-and-Conquer Algorithms for Computing Matrix Inverses By SHADY Sayed EL-OKUR Supervised by Dr. † Examples: Binary Search, Merge sort, Quicksort etc. divide and conquer (merge sort, exponentiation, matrix multiplication, Strassens algorithm, median finding, master method) Week 3. Divide and conquer is applied to many problems sorting matrix multiplication from CIS 502 at National Tsing Hua University, China. We improve the basic block. I will start with a brief introduction about how matrix multiplication is generally observed and implemented, apply different algorithms (such as Naive and Strassen) that are used in practice with both pseduocode and Python code, and then end with an analysis of their runtime. Conquer: multiply 8 ½n-by-½n recursively. – The above naturally leads to divide-and-conquer solution: ∗ Divide X and Y into 8 sub-matrices A, B, C, and D. For example. divide and conquer (median finding, closest pair problem) Week 4. Merge sort is a stable sort. 1 Divide: Partition A and B into submatrices; add and subtract to form terms. Upper triangular matrix in c 10. matrix it computes the Schur form. 2015/2016. Material in this lecture: What is the study of Algorithms all about? Why do we care about speci cations and proving guarantees? The Karatsuba multiplication algorithm. Divide-and-Conquer algorithsm for matrix multiplication A = A11 A12 A21 A22 B = B11 B12 B21 B22 C = A×B = C11 C12 C21 C22 Formulas for C11,C12,C21,C22: C11 = A11B11 +A12B21 C12 = A11B12 +A12B22 C21 = A21B11 +A22B21 C22 = A21B12 +A22B22 The First Attempt Straightforward from the formulas above (assuming that n is a power of 2):. 27) [20 points]. Divide-And-Conquer Approach. No longer only TWO subproblems Conquer: Solve each subproblem (directly or recursively), and Combine: Combine the solutions of the subproblems into a global solution. There are very many pushes and pops on the run-time stack. The divide and conquer strategy •A first example : sorting a set S of values sort (S) = if |S| ≤ 1 then return S else divide (S, S1, S2) fusion (sort (S1), sort (S2)) end if fusion is linear is the size of its parameter; divide is either in O(1) or O(n) The result is in O(nlogn). Hello Friends, I am Free Lance Tutor, who helped student in completing their homework. Zigzag (or diagonal) traversal of Matrix; Divide and Conquer | Set 5 (Strassen's Matrix Multiplication) Print all possible paths from top left to bottom right of a mXn matrix; Count all possible paths from top left to bottom right of a mXn matrix; Printing brackets in Matrix Chain Multiplication Problem. From here, I want to differentiate each column of the matrix with respect to g(x). AU - Ozaki, Katsuhisa. The total time spent is then divided by the. Divide X, Y and Z into four (n/2)×(n/2) matrices as represented below −. Divide and conquer approach has several advantages as follows: Solving conceptually difficult problems, it just require to divide them into sub problems. Binary Multiplication Rules. Algorithm Analysis techniques ----- Theory: Divide-and-Conquer Strategy: Given a function that has to compute on ‘n’ input the divide and conquer strategy suggest. delete() Creating a new Directory using File. Matrix multiplication. World's Best PowerPoint Templates - CrystalGraphics offers more PowerPoint templates than anyone else in the world, with over 4 million to choose from. Also, observe that divide and conquer ran twice as fast when ran using Threads. Conquer/Solve: This phase overcomes the subproblems by solving them recursively. Divide-and-Conquer Examples Sorting: mergesort and quicksort Binary tree traversals Multiplication of large integers Matrix multiplication: Strassen’s algorithm Closest-pair and convex-hull algorithms Binary search: decrease-by-half (or degenerate divide&conq. 2 Strassen's algorithm for matrix multiplication Table of contents. Graphs (Directed Graphs and. 1 The maximum-subarray problem 4. Consequence. O(n) multiplication) to find out only one entry of the result Z •Total time will be O(n3). For matrix multiplication, the number of columns in the first matrix must be equal to the number of rows in the second matrix. Divide and Conquer Idea: Divide problem instance into smaller sub-instances of the same problem, solve these recursively, and then put solutions together to a solution of the given instance. Multiplication: Karatsuba algorithm; Closest-Pair; Goddard: Part A - Divide and Conquer / Sorting and Searching. S into three categories: elements smaller than v, those equal to v (there might be duplicates), and those greater than v. Feb 1, 2020 - Explore anokair's board "MATRIX MULTIPLICATION " on Pinterest. Divide and conquer (D&C) is an algorithm design paradigm based on multi-branched recursion. Lower triangular matrix in c 9. Winner of the Standing Ovation Award for "Best PowerPoint Templates" from Presentations Magazine. // Matrix multiplication Divide & conquer order: int operations[16] =. The complexity for the multiplication of two matrices using the naive method is O(n 3), whereas using the divide and conquer approach (ie. Contribute to saulmm/Divide-and-conquer development by creating an account on GitHub. Divide and Conquer: 992 24 632 408 1600 272 720 1232 512 0 512 384 460 17 405 497 Could someone tell me what I am doing wrong for divide and conquer? All my matrices are int[][] and classical method is the traditional 3 for loop matrix multiplication. This method is introduced to reduce the complexity. Let's look at one more algorithm to understand how divide and conquer works. Strassen’s. Case 1 of Master Method solution = Θ𝑛. C program 2D matrix multiplication using malloc I then created a driver program to create Matrix C and fill in with Matrix A*B. † Examples: Binary Search, Merge sort, Quicksort etc. to the assembly processes can be captured through a matrix-matrix multiplication. The divide-and-conquer paradigm often helps in the discovery of efficient algorithms. length, but these values. Strassen’s Algorithm for Matrix Multiplication. For example, if the first bit string is "1100" and second bit string is "1010", output should be 120. The best currently known exponent ω<2. Input: array A[i, …, j] Ouput: sum of maximum-subarray, start. 5 Strassen' matrix multiplication 2. Divide and Conquer | Set 4 (Karatsuba algorithm for fast multiplication) Given two binary strings that represent value of two integers, find the product of two strings. Brief review of the tridiagonal DC method. It uses a divide-and-conquer approach along with a nice math trick to reduce the number of computations needed to calculate the product of two. The necessary condition: R2(Number of Rows of the Second Matrix) = C1(Number of Columns of the First Matrix). Strassen's matrix multiplication In order to obtain more accurate results, the algorithms should be tested with the same matrices of different sizes many times. Architecture-efficient Strassen's Matrix Multiplication: A Case Study of Divide-and-Conquer Algorithms By Paul Pauca, Xiaobai Sun, Siddhartha Chatterjee and Alvin Lebeck Abstract. But the algorithm is not very practical, so I recommend either naive multiplication, which runs in $\mathcal{O}(n^3)$, or S. Matrix multiplication, Selection, Convex Hulls. • Dynamic programming is needed when subproblems are dependent; we don’t know where to partition the problem. Divide & Conquer Many algorithms are recursive in nature to solve a given problem recursively dealing with sub-problems. Divide and Conquer Ming-Hwa Wang, Ph. Matrix Multiplication through Divide and Conquer Approach 6. Basis for many common algorithms. University. Perkalian Matrix dengan Divide and Conquer dan Algoritma Strassen Otniel and 13508108 Program Studi Teknik Informatika Sekolah Teknik Elektro dan Informatika Institut Teknologi Bandung, Jl. However, the irregular and matrix-dependent data access pattern of sparse matrix multiplication makes it challenging to use tiling to enhance data reuse. The matrix-chain multiplication problem can be stated as follows: given a chain A 1, A 2,. Logic: Divide the matrix, then use the Strassen's formulae:. Other problems such as the Tower of Hanoi are also simplified by this approach. In Recursive Matrix Multiplication, we implement three loops of Iteration through recursive calls. 1145/3293883. Also, observe that divide and conquer ran twice as fast when ran using Threads. Week 7: Divide and Conquer Example 3: Matrix multiplication: •Assume we are given two n ×n matrix X and Y to multiply. Add and shift to obtain result. 6 Multiplying large integers 2. Topic: Divide and Conquer 23 3. Introduction to Algorithms 6. No longer only TWO subproblems Conquer: Solve each subproblem (directly or recursively), and Combine: Combine the solutions of the subproblems into a global solution. Convex Hull algorithms (plus more on Mergesort, Quicksort, etc. The divide and conquer strategy •A first example : sorting a set S of values sort (S) = if |S| ≤ 1 then return S else divide (S, S1, S2) fusion (sort (S1), sort (S2)) end if fusion is linear is the size of its parameter; divide is either in O(1) or O(n) The result is in O(nlogn). Divide and Conquer. The efficient design of multiplierless implementations of con-stant matrix multipliers is. A divide and conquer algorithm works by recursively breaking down a problem into two or more sub-problems of the same or related type, until these become simple enough to be solved directly. 1145/3293883. A divide and conquer algorithm works by recursively breaking down a problem into two or more sub-problems of the same or related type, until these become simple enough to be solved directly. Again we are dealing with subproblems of sorting subarrays A[p. † Examples: Binary Search, Merge sort, Quicksort etc. This generalizes the products in size (2 2 2) used in the half-gcd algo-rithm or the Pad e approximant algorithm of [8]; often, nis small (say, a few dozens). Now, suppose we want to multiply three or more matrices: $$A_{1} \times A_{2} \times A_{3} \times A_{4}$$ Let A be a p by q matrix, let B be a q by r matrix. Matrix Multiplication through Divide and Conquer Approach 6. GENERAL METHOD: Given a function to compute on n inputs the divide-and-conquer strategy suggests splitting the inputs into k distinct subsets, 1Integer multiplication >Matrix multiplication >Fast Fourier Transform >Integer multiplication again Outline for Today >Processor provides ability to multiply small (<= 64 bit) numbers >Multiplying arbitrary-size integers is a classic problem. Divide and Conquer example: Matrix Multiplication The normal procedure to multiply two n × n matrices requires n3 time. 4-46 faster than the original versions and within 2-60% of a high-performance hand crafted implementation. Our goal is to reduce this total time for multiplying two polynomials to using Divide and Conquer. 5 Strassen' matrix multiplication 2. Optimisation of Constant Matrix Multiplication Operation Hardware Using a Genetic Algorithm Andrew Kinane, Valentin Muresan, and Noel O’Connor Centre for Digital Video Processing, Dublin City University, Dublin 9, Ireland kinanea@eeng. divide-and-conquer algorithms is new, to our knowledge: the only explicit exam-ples in [9,3] describe Karatsuba multiplication. The rather intricate details of this approach are strongly reminiscent of the Bunch-Hopcroft algorithm [4] for fast matrix inversion. Founded by Adam Hendricks, John Lang and Greg Gilreath. Try out: Matrix Multiplication Calculator. [50:00] Analysis of Strassen's algorithm. Divide-and-Conquer Reading: CLRS Sections 2. Divide, Combine and Conquer c. If A is a matrix, then AA is the square of A. This happens by decreasing the total number if multiplication performed at the expenses of a. 5 The master method for solving recurrences 4. Selection Procedure and Matrix Multiplication (in Hindi). This is a program to compute product of two matrices using Strassen Multiplication algorithm. In 1969, Strassen proposed a method of divide and conquer to try to break the q (n 3) barrier. This technique yields elegant, simple and quite often very efficient algorithms. This, as we shall see in a moment, is because of the way matrices are multiplied. The Divide and Conquer paradigm. ‣ integer multiplication ‣ matrix multiplication ‣ convolution and FFT. Matrix multiplication is particularly easy to break into subproblems, because it can be performed blockwise. World's Best PowerPoint Templates - CrystalGraphics offers more PowerPoint templates than anyone else in the world, with over 4 million to choose from. [35:45] Naive (standard) algorithm for multiplying matrices. Strassens's Matrix Multiplication • Strassen (1969) showed that 2x2 matrix multiplication can be accomplished in 7 multiplications and 18 additions or subtractions 𝑇𝑛= 7𝑇. This document is highly rated by students and has been viewed 263 times. This will culminate in the study of Strassen matrix multiplication algorithm. The BM displays a radiation pattern of 16 beams at different declinations (from -48, to +88 degrees). At the end of the lecture, we saw the reduce SUM operation, which divides the input into two halves, recursively calls itself to obtain the sum of these smaller inputs, and returns the sum of the results from those. Break up problem into several parts. The divide-and-conquer technique involves taking a large-scale problem and dividing it into similar sub-problems of a smaller scale, and recursively solving each of these sub-problems. ×) by min (resp. The approach taken can be analysed in the code written by me. C program to find determinant of a matrix 12. Finally, the detailed distributed experiment along with connectivity interface and implementation will be discussed. Mathematically we have or, Consider Divide and Conquer Remember Pascal’s Triangle Consider triangle shape Looks like a two dimensional array. Strassen in 1969 which gives an overview that how we can find the multiplication of two 2*2 dimension matrix by the brute-force algorithm. Divide and Conquer: The Karatsuba algorithm (multiplication of large integers) Instructor: L aszl o Babai Updated 01-13-2020 NOTATION. Conclusion. Quick sort algorithm. We could improve the required running time by the following Strassen's matrix multiplication algorithm. Combine these results together. Without communications the addition and subtraction of matrices can be computed in. S : 2 36 5 21 8 13 11 20 5 4 1. performs its own recursive divide and conquer approach as defined by strassen’s methodology[9][10] to obtain partitioned matrix multiplication. Such systems are able to support a large volume of parallel communication of various patterns in constant time. its just an example of parallel programming. id Penghitungan matrix sangat umum di bidang matematika. [50:00] Analysis of Strassen's algorithm. Each of these equations multiplies two. Disk/RAM di erences are a bottleneck for recursive algorithms, and PRAM assumes perfect scheduling. This paper discusses and compares several parallelization strategies for tree-structured computations. Which method yields the best asymptotic running time when used in a divide-and-conquer matrix-multiplication algorithm? Compare it with the running time for Strassen ' s algorithm. Strassen's Matrix multiplication can be performed only on square matrices where n is a power of 2. com/bePatron?u=20475192 U. renameTo() and File. De nition 4 (bilinear map). AU - Oishi, Shin'ichi. Solve the smaller (simpler) problem. This paper deals with parallels of the fast matrix multiplication strassen's algorithm, winograd's algorithm and analyzes empirical study of the matrix multiplication under the distributed environment in. Strassen's Matrix Multiplication Sibel KIRMIZIGÜL Basic Matrix Multiplication Suppose we want to multiply two matrices of size N x N: for example A x B = C. This problem is mostly used to teach recursion, but it has some real-world uses. • How matrix multiplication can be stated as a divide-and-conquer algorithm (i. Recur: solve the sub problems recursively Conquer: combine the solutions for S1, S2, …, into a solution for S The base case for the recursion are sub problems of constant size Analysis can be done using recurrence equations 5. •These are huge matrices, say n ≈50,000. Divide, Conquer and Combine The correct answer is: Divide, Conquer and Combine. January 2, 2013 January 3, 2013 saeediqbalkhattak How to multiply any two integer using divide & Conquer approach. And this is a super cool algorithm for two reasons. Split each matrix into 4 of size (n / 2) x (n / 2) 2. The multiply() method takes 3 matrices and their indexes and using the divide and conquer matrix multiplication algorithm where each matrice is divided into four parts and is multiplied to get the output. 3302576 https://doi. For example, if the first bit string is "1100" and second bit string is "1010", output should be 120. q] Initially, p= 1 and q= A. Both merge sort and quicksort employ a common algorithmic paradigm based on recursion. its deals with how parallel programming can be achieved. •The native algorithm will have to multiply one row of X by one column of Z (i. Divide and Conquer. Matrix Multiplication operation is associative in nature rather commutative. 1 Compute C = AB using the traditional matrix multiplication algorithm. Feb 1, 2020 - Explore anokair's board "MATRIX MULTIPLICATION " on Pinterest. Given two square matrices A and B of size n x n each, find their multiplication matrix. Top-Down Algorithms: Divide-and-Conquer. Solve the subproblems recursively and concurrently 3. Towers of Hanoi 🗼 The Towers of Hanoi is a mathematical problem which compromises 3 pegs and 3 discs. There are four types of algorithms: Iterative Algorithm; Divide and conquer algorithm; Sub-cubic algorithms. Following is simple Divide and Conquer method to multiply two square matrices. The name divide and conquer is because the problem is conquered by dividing it into several smaller problems. Suppose you wish to develop a matrix-multiplication algorithm that is asymptotically faster than Strassen's algorithm. It consists of rows and columns. Matrix multiplication is particularly easy to break into subproblems, because it can be performed blockwise. For better under understanding lets see one more sorting technique called quick sort. Architecture-efficient Strassen's Matrix Multiplication: A Case Study of Divide-and-Conquer Algorithms By Paul Pauca, Xiaobai Sun, Siddhartha Chatterjee and Alvin Lebeck Abstract. Consider two matrices: Matrix A have n rows and k columns; Matrix B have k rows and m columns (notice that number of rows in B is the same as number of columns in A). Divide and Conquer. \Divide and conquer" Giorgio Ottaviani Complexity of Matrix Multiplication and Tensor Rank. Graphs (Directed Graphs and. 3 Combine: Add and subtract terms to form C. Zima (SCS, UW) Module 4: Divide and Conquer Winter 20207/14. Explanation: The time complexity of recursive multiplication of two square matrices by the Divide and Conquer method is found to be O(n 3) since there are total of 8 recursive calls. CSC 210-12: Divide and Conquer: Multiplication of Large Integers and Strassen's Matrix Multiplication Based on slides prepared for the book: Anany Levitin, Introduction to The Design and Analysis Algorithms, 2nd edition, Addison Wesley, 2007 Strassen's Matrix Multiplication Let A. ,a [j]) denote an arbitrary instance of the problem. General Terms based on divide & conquer paradigm. Strassen’s Matrix Multiplication algorithm is the first algorithm to prove that matrix multiplication can be done at a time faster than O(N^3). Recur: solve the sub problems recursively Conquer: combine the solutions for S1, S2, …, into a solution for S The base case for the recursion are sub problems of constant size Analysis can be done using recurrence equations 5. Rivest and Clifford Stein. net/archives/V5/i3/IRJET-V5I3362. 7: Matrix multiplication using Strassen’s algorithm. Steve Lai. AU - Ozaki, Katsuhisa. Strassens's Matrix Multiplication • Strassen (1969) showed that 2x2 matrix multiplication can be accomplished in 7 multiplications and 18 additions or subtractions 𝑇𝑛= 7𝑇. Posts: 31,956; Joined: 06-March 08; Re: Divide and Conquer Matrix Multiplication. I assume from the question that the code has to cope with matrices of arbitrary size up to some reasonably sane limit. Description Presents the mathematical techniques used for the design and analysis of computer algorithms. To solve a given problem, it is subdivided into one or more subproblems each of which is similar to the given problem. Material in this lecture: What is the study of Algorithms all about? Why do we care about speci cations and proving guarantees? The Karatsuba multiplication algorithm. Veldhuizen tveldhui@acm. Write a c program to find out transport of a matrix. Matrix multiplication has attracted considerable attention for more than four decades and the challenge is whether or not matrix multiplication can be done in quadratic time. Divide and Conquer: Matrix Multiplication, Polynomial Multiplication, Closest Pair of Points August 2nd, 2019 Graph Data Structure: Definitions, Adjacency Matrix and List. Examples: Mergesort, Quicksort, Strassen's algorithm, FFT. Read on for Python implementations of both algorithms and a comparison of their running time. 2015/2016. Solve smaller instances recursively 3. The Strassen’s method of matrix multiplication is a typical divide and conquer algorithm. Academic year. This will culminate in the study of Strassen matrix multiplication algorithm. 5 The master method for solving recurrences 4. Multiplication: Karatsuba algorithm; Closest-Pair; Goddard: Part A - Divide and Conquer / Sorting and Searching. Divide and Conquer Mergesort Quicksort Binary Search Selection Matrix Multiplication Convex Hull * * * * * * * * * Selection Find the kth smallest (largest) item in a list. The necessary condition: R2(Number of Rows of the Second Matrix) = C1(Number of Columns of the First Matrix). mkdir() method Illustration of listFiles() and list() method of F Find SubArray having Maximum Sum using Divide and CREATING A NEW FILE USING java. ; The median of a finite list of numbers can be found by arranging all the numbers from lowest value to highest value and picking the middle one. In the previous post, we discussed some algorithms of multiplying two matrices. Find the determinant of a larger matrix. T1 - Error-free transformation of matrix multiplication with a posteriori validation. Attempt at solutions I have been able to get this divide and conquer code working for a 2x2 matrix and a 4x4 matrix. 7: Matrix multiplication using Strassen’s algorithm. partition and then direct block multiplication C= C 11 C 12 C 21 C 22 = A 11 A 12 A 21 A 22 B 11 B 12 B 21 B 22 = A 11B 11 + A 12B 21 A 11B 12 + A 12B 22 A 21B 11 + A 22B 21 A 21B 12 + A 22B 22 2. 2 Merge sort 2. Fast algorithms for matrix multiplication --- i. Lecture 3: The Polynomial Multiplication Problem A More General Divide-and-Conquer Approach Divide: Dividea givenproblemintosubproblems(ide-ally of approximately equal size). Divide and conquer (D&C) is an algorithm design paradigm based on multi-branched recursion. Efficiency also makes a difference between divide and conquer and dynamic programming. My problem is that I'm getting a Segmentation fault. Strassen's matrix multiplication method is based on a divide & conquer rule. Given a sequence of n matrices A 1, A 2, As we have remarked in the introduction that the dynamic programming is nothing but the fancy name for divide-and-conquer with a table. S into three categories: elements smaller than v, those equal to v (there might be duplicates), and those greater than v. Solve the smaller (simpler) problem. Divide and conquer algorithm. This is a very basic and very powerful algorithm design technique. Previous divide-and-conquer algorithms all. These subproblems must be solved and then a method must be found to combine subsolutions into a solution of a whole. In this note, log will always mean log 2 (base-2 logarithm). The divide and conquer strategy •A first example : sorting a set S of values sort (S) = if |S| ≤ 1 then return S else divide (S, S1, S2) fusion (sort (S1), sort (S2)) end if fusion is linear is the size of its parameter; divide is either in O(1) or O(n) The result is in O(nlogn). ) This is typical in the analysis of. Combine these results together. Differences between Dynamic programming 3. Let's look at one more algorithm to understand how divide and conquer works. Matrix multiplication arises in its own right in computing the results of such coordinate transformations as scaling, rotation, and translation for robotics and computer graphics. [50:00] Analysis of Strassen's algorithm. matrix multiplication case study divide-and-conquer algorithm architecture-efficient strassen efficient implementation storage space data locality high performance computer recent year memory hierarchy ubiquitous operation arithmetic complexity implementation issue automatic optimization strategy optimization scheme recursive algorithm. Divide-And-Conquer Approach.
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## MAT244--2019F => MAT244--Lectures & Home Assignments => Chapter 7 => Topic started by: nadia.chigmaroff on November 03, 2019, 12:04:50 PM
Title: Transforming a system of linear equations to a single higher-order equation
Post by: nadia.chigmaroff on November 03, 2019, 12:04:50 PM
Hi,
In the textbook, it says that a system of first-order equations can sometimes be transformed into a single higher-order equation (by the process given in problem 7 on the 7.1 problems).
Am I correct in saying that this is only possible to do in general when the determinant of the coefficient matrix associated with the system is nonzero?
I.e. if \left\{\begin{aligned} &x'_1= ax_1 + bx_2\\ &x'_2= cx_1 + dx_2 \end{aligned}\right., this can be converted into a singular equation iff $ad - bc$ $\neq 0$?
Thank you! :D
Title: Re: Transforming a system of linear equations to a single higher-order equation
Post by: aremorov on November 03, 2019, 06:12:48 PM
No, this is still possible to do if the determinant is 0.
For example:
$x_1' = x_1 + x_2$ (*)
$x_2' = x_1 + x_2$
has determinant 0 for the coefficients, however if we isolate for $x_1$ we get:
$x_1'' - 2x' = 0$ which has solution:
$x_1 = C_1 + C_2 e^{2t}$ and putting this into equation (*) gives us $x_2 = C_2 e^{2t} -C_1$ for arbitrary constants $C_1, C_2$.
Title: Re: Transforming a system of linear equations to a single higher-order equation
Post by: Victor Ivrii on November 05, 2019, 09:52:23 PM
Nothing to do with the determinant. In dimension 2 you can reduce if the matrix is not diagonal--obvious. What about the diagonal case? It is also possible unless a matrix is scalar, i.e. proportional to identity. To do this reduction, we make first a linear transform, so that after it the matrix is not diagonal anymore, and then reduce.
General criteria: System with constant coefficients could be reduced to a single equation iff each eigenspace is $1$-dimensional. To understand why we need to consider solutions to a homogeneous equation and to a homogeneous system.
For an equation one of the solutions is $t^{m-1} e^{kt}$ where $k$ is characteristic root, and $m$ is a multiplicity of $k$.
For system all solutions are in the form $P_{s-1}(t)e^{kt}$ where $P_{s-1}(t)$ are polynomials of degree $\le s-1$ with vector-coefficients and $s$ is the maximal dimension of the corresponding Jordan cells.
Therefore reduction can be done iff $s=m$ which means that for each eigenvalue $k$ there is just one cell, which in turn means, that there is only one linearly independent eigenvector.
Remark: If $s_1,...,s_j$ are dimensions of all cells, corresponding to $k$, then their sum $=m$ where $m$ is a multiplicity of $k$ as a root of characteristic equation, and also the dimension of the root subspace, and $j$ is a dimension of the corresponding eigenspace.
However, it is not important: we solve systems without reducing them to single equations.
Title: Re: Transforming a system of linear equations to a single higher-order equation
Post by: nadia.chigmaroff on November 06, 2019, 09:41:17 AM
Ok, thanks! | 2022-05-18T04:14:04 | {
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https://math.stackexchange.com/questions/1301978/questions-about-torsion-of-a-curve-in-mathbbr3-and-analogues-of-torsion-in | # Questions about torsion of a curve in $\mathbb{R}^3$ and analogues of torsion in higher dimensions
Suppose we have a curve $\alpha(s) : I \to \mathbb{R}^3$ parametrized by arc-length that has nowhere-vanishing second derivative, so that we are able to define the torsion $\tau(s)$ for every $s \in I$. It is clear to me that one can deform a curve in a way that can change the curvature by large amounts while keeping $\tau = 0$ everywhere. I think that it is possible to deform a curve in a way that we can vary torsion while keeping curvature fixed. I was wondering if anybody could give me a nice example of this.
My second question involves extending the idea of torsion. As far as I understand, torsion measures the planarity of a curve. For a curve that is embedded in $\mathbb{R}^n$ is there a quantity indicating how far the curve is from being embedded in $\mathbb{R}^{n-1}$, and is this even useful?
• I have no answer to the first question, but the answer to the first part of your second question is yes. It's called generalized curvature. See here: en.wikipedia.org/wiki/… – Randy E May 28 '15 at 2:57
• You would enjoy Kuhnel's differential geometry text. His Theorem 2.15 explains how you can take a set of curvatures and a given torsion plus some initial data and get back a curve. – James S. Cook May 28 '15 at 5:43
• I am aware of the fact that curves can be constructed given this data up to rigid motions which is why I was certain this could be done but I lacked a picture. – Memeozuki May 28 '15 at 5:59
• @AbrahamRabinowitz Since you're interested in a picture, I've added a small animation depicting the family of curves described in my answer. – Travis Willse May 28 '15 at 8:28
I think that it is possible to deform a curve in a way that we can vary torsion while keeping curvature fixed. I was wondering if anybody could give me a nice example of this.
Arguably the simplest class of examples are those for which the curvature $\kappa$ and torsion $\tau$ are constants for each curve in the family. Any such curve is a helix (including the degenerate case of zero torsion, which gives a circle) and can be parameterized by $$\alpha(t) := (r \cos t, r \sin t, bt)$$ for some parameters $r$ (the radius of the unique cylinder containing the image of the helix) and $b$ (a quantity which controls the component velocity in the direction of the axis of that cylinder). (Computing gives that $||\alpha'(t)||^2 = \sqrt{r^2 + b^2}$, and in particular $\alpha$ is a constant-speed parameterization, so it's no trouble to write down an arc length parameterization.) Now, direct computation gives that the curvature $\kappa$ and torsion $\tau$ of $\alpha$ are $$\kappa(t) = \frac{r}{r^2 + b^2} \qquad \text{and} \qquad \tau(t) = \frac{b}{r^2 + b^2}.$$
So, to produce a family of helices $\color{#bf0000}{\alpha_m(t)}$ with constant prescribed curvature $\kappa$ and varying torsion $\tau$, we need only pick (nonconstant) functions $r(m), b(m)$ that satisfy $$\kappa = \frac{r(m)}{r(m)^2 + b(m)^2}$$ for any prescribed constant $\kappa > 0$. Rearranging shows that this equation defines a circle $$\left(r - \frac{1}{2 \kappa}\right)^2 + b^2 = \left(\frac{1}{2 \kappa} \right)^2$$ in $rb$-space, and we can produce explicit functions $r(m), b(m)$ by parameterizing this circle (or more precisely, this circle less the point $(r, b) = (0, 0)$). The usual rational parameterization of the unit circle, for example, leads to the solution \begin{align} r(m) := \frac{1}{(1 + m^2) \kappa} \\ b(m) := \frac{m}{(1 + m^2) \kappa} \end{align} and hence to the parameterized family of helices defined by $$\color{#bf0000}{\alpha_m(t) = \left(\frac{\cos t}{(1 + m^2) \kappa}, \frac{\sin t}{(1 + m^2) \kappa}, \frac{m t}{(1 + m^2) \kappa}\right)}.$$ Substituting this parameterization in the above formulas for $\kappa$ and $\tau$ reveals that $$m = \frac{\tau}{\kappa};$$ in particular, when $\kappa = 1$ the parameter $m$ is nothing more than the torsion $\tau$ itself.
This animation shows how $\color{#bf0000}{\alpha_m}$ (with $\kappa = 1$) varies with prescribed torsion $\tau$.
This was generated by the following Maple code:
with(plots):
stau := [cos(t) / (tau^2 + 1), sin(t) / (tau^2 + 1), tau * t / (1 + tau^2)];
opts := color=black, numpoints = 400:
animate(spacecurve, [stau, t = -64..64, opts], tau = -4..4, view = [-2..2, -2..2, -8..8], scaling = constrained, frames = 192, axes = none);
One can generalize this example wildly, by the way, as given any functions $\kappa(s)$, $\tau(s)$ (say, with $\kappa > 0$) parameterized by arc length there is a curve with curvature $\kappa(s)$ and torsion $\tau(s)$, and this curve is unique up to Euclidean motions, though actually solving for such a curve requires integrating a differential equation.
For a curve that is embedded in $\Bbb R^n$ is there a quantity indicating how far the curve is from being embedded in $\Bbb R^n$, and is this even useful?
Yes, there are higher-order analogues of torsion for Euclidean spaces $\Bbb R^n$, $n > 3$. Recall that in in $\Bbb R^3$ (1) one can always choose (at least for curves with nonvanishing curvature) a unique adapted orthonormal frame $({\bf T}, {\bf N}, {\bf B})$ along a given smooth curve, and (2) derivatives of the curvature and torsion satisfy $$\begin{pmatrix} {\bf T} \\ {\bf N} \\ {\bf B} \end{pmatrix}' = \begin{pmatrix} 0 & \kappa & 0 \\ -\kappa & 0 & \tau \\ 0 & -\tau & 0\end{pmatrix}\begin{pmatrix} {\bf T} \\ {\bf N} \\ {\bf B} \end{pmatrix}.$$ (Here, $'$ denotes differentiation w.r.t. an arc length parameter.) Similarly, in $\Bbb R^4$, for sufficiently generic curves one can choose a unique adapted orthonormal frame $({\bf T}, {\bf N}, {\bf B}, \color{#0000ff}{{\bf U}})$ (here $\color{#0000ff}{{\bf U}}$ is sometimes called, predictably, the trinormal), and such a curve has three curvature quantities, $\kappa, \tau, \color{#00bf00}{\upsilon}$, and these satisfy $$\begin{pmatrix} {\bf T} \\ {\bf N} \\ {\bf B} \\ \color{#0000ff}{{\bf U}} \end{pmatrix}' = \begin{pmatrix} 0 & \kappa & 0 & 0\\ -\kappa & 0 & \tau & 0 \\ 0 & -\tau & 0 & \color{#00bf00}{\upsilon} \\ 0 & 0 & -\color{#00bf00}{\upsilon} & 0\end{pmatrix}\begin{pmatrix} {\bf T} \\ {\bf N} \\ {\bf B} \\ \color{#0000ff}{{\bf U}} \end{pmatrix}.$$ As you guessed, $\color{#00bf00}{\upsilon}$ measures the failure of the curve to be contained in the hyperplane $\langle {\bf T}, {\bf N}, {\bf B} \rangle$ to the appropriate order. (Preserving the analogy with the $3$-dimensional case, the triple $(\kappa, \tau, \color{#00bf00}{\upsilon})$ is a complete set of invariants for a generic curve in $\Bbb R^4$, in that their values generically determine a curve uniquely up to Euclidean motions.) The analogous statements for higher dimensions are the generalizations from the $\Bbb R^3$ and $\Bbb R^4$ cases that you'd guess; in particular generic curves in $\Bbb R^n$ have, and are generically determined by, $n - 1$ curvature functions. (Note that this general pattern captures the $2$-dimensional case, too, in which there is only a single invariant, just the usual (signed) curvature.)
• Maybe I've been working too long on this question, but, I think you want $r(u)/\kappa$ so you get back $\kappa = \kappa$ for the functions $r(u)$ and $c(u)$ you describe. Great answer. It's set me free to work on something else now. – James S. Cook May 28 '15 at 5:41
• Thanks, and I'm glad you found it useful. I've added an explicit family of helices with constant curvature but varying torsion (and a brief description of how to produce the family). – Travis Willse May 28 '15 at 6:08
• Thank you very much, your answer has been very illuminating. – Memeozuki May 28 '15 at 14:28
• @AbrahamRabinowitz You're welcome, I'm glad you found it helpful! – Travis Willse May 28 '15 at 14:38 | 2021-03-02T18:13:03 | {
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http://mathoverflow.net/questions/152750/periodic-orbit-property/153298 | # Periodic Orbit property
A topological space $X$ satisfies "Periodic orbit property", briefly POP, if for every continuous map
$f:X \to X$, there exist a natural number $n$ and a point $x_{0}\in X$ such that $f^{n}(x_{0})=x_{0}$
Obviously fixed point property(FPP) implies POP.
For a natural number $n$,a topological space $X$ is called $n-POP$ if for every continuous map $f$ on $X$, $f^{n}$ has a fixed point.(Ex: $\mathbb{S}^{2n}$ is a 2-POP manifold, using degree of maps)
The Question:
Is there an example of a manifold $M$ which satisfies POP but for every $n\in \mathbb{N}$, there is a continuous map $f$ on $M$ such that $f^{n}$ has no fixed point?
Namely: we search for a manifold for which every self map has a periodic orbit, but there is no any control on periods.
Equivalently:
Is there a manifold $M$ which is POP but not $n-POP$ for all $n\in \mathbb{N}$?
In particular, can we say:
"every compact POP manifold is necessarily a $n$-POP manifold, for some $n$"?
Motivating by Lefschetz fixed-point theorem, we ask that:
What algebraic topological criterion, can be introduced for consideration of this property(POP)?
Edit: According to the very interesting answer of Qiaochu Yuan, in the orientable case, the question is equivalent to the following:
Let M be a closed orientable manifold. Is it true that $M$ is not POP if and only if $\chi(M)=0$?
Note: I think the continuation of the argument of Qiaochu Yuan for his first statement is not easy, for arbitrary manifold. Because for the simplest case $S^{3}$ we had the famous conjecture of "existence of a vector field on $S^{3}$ without periodic orbit. In fact consideration of non vanishing vector fields is necessary but not sufficient. Periodic orbits of vector fields are important, too. Moreover, perhaps an approach which is not based on "vector fields" could be useful, for example consideration of oriention reversing diffeomorphisms.
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Let $R_{\alpha}:S^1\to S^1$ be an irrational rotation, then $R^n_{\alpha}$ has no fixed point for every $n\geqslant 1$ – Juan Valdez Dec 24 '13 at 20:08
So $S^{1}$ is not a POP manifold. But I search for a manifold which is POP but is not n-POP for all $n\in \mathbb{N}$. – Ali Taghavi Dec 25 '13 at 8:00
What are examples of manifolds that satisfy POP but not FPP? – Benjamin Dickman Jan 1 at 12:38
$S^{2n}$ is an example. The antipodal map has no fixed point, So $S^{2n}$ does not satisfy FPP. On the other hand, if $f$ is a self map on $S^{2n}$ without fixed point,Then deg(f)=-1. See Algebraic topology(Allen Hatcher). So deg($f^{2}$)=1. Then $f^{2}$ has fixed point. This shows that for every self map f on $S^{2n}$, $f^{2}$ has a fixed point. – Ali Taghavi Jan 1 at 13:23
Nice question! Here's what I can show.
Let $X$ be a smooth closed manifold. Then:
(1) If $\chi(X) = 0$, then $X$ is not $n$-POP for any $n$.
(2) If $\chi(X) \neq 0$ and $X$ is orientable, then $X$ is $\text{lcm}(1, 2, ... n)$-POP with respect to maps $f : X \to X$ of nonzero degree, where $n = \text{max}(b_0 + b_2 + ..., b_1 + b_3 + ...)$ (where $b_i$ is the $i^{th}$ Betti number of $X$).
Proof of 1. We will use the converse of the Poincaré-Hopf theorem: if $\chi(X) = 0$, then $X$ admits a nonvanishing vector field. Let $\varphi(t)$ denote the flow of this vector field. Let $t_{0}>0$ be small enough so that $\varphi(t_0)$ has no fixed points. Such $t_{0}$ exists, because there is a positive uniform lower bound for the period of all periodic orbits.(As a consequence of the flow box theorem, around regular points of a vector field). For a given $n \in \mathbb{N}$, let $f = \varphi \left( \frac{t_0}{n} \right)$. Then $f^n$ has no fixed points, hence $X$ is not $n$-POP. $\Box$
(I strongly suspect that in this case $X$ is not POP either; it seems like we should be able to consider a small flow of a sufficiently generic nonvanishing vector field. But I don't know how to finish this argument.)
Proof of 2. We will need the following two observations.
Lemma 1: Let $f_0, f_1$ be linear operators acting on two finite-dimensional vector spaces $V_0, V_1$. If $\text{tr}(f_0^k) = \text{tr}(f_1^k)$ for $k$ between $1$ and $\text{max}(\dim V_0, \dim V_1)$, then $f_0$ and $f_1$ have the same nonzero eigenvalues with the same multiplicities.
Proof. The above condition implies, using the Newton-Girard identities, that $f_0$ and $f_1$ have the same characteristic polynomial up to factors of $t$. $\Box$
Lemma 2: Let $X$ be an $n$-dimensional smooth closed oriented manifold and let $f : X \to X$ be a map of nonzero degree. Then every eigenvalue of $f$ acting on cohomology (with complex coefficients) is nonzero.
Proof. Let $e_1, ..., e_d$ be a basis of generalized eigenvectors for the action of $f$ on $H^k(X, \mathbb{C})$. By Poincaré duality the cup product $H^k \otimes H^{n-k} \to H^n$ is nondegenerate, so we can find a dual basis $e_1^{\ast}, ..., e_d^{\ast}$ of $H^{n-k}(X, \mathbb{C})$. Since $f$ acts by a nonzero scalar, namely $\deg f$, on $e_i \smile e_i^{\ast}$ for all $i$, the generalized eigenvalue of $e_i$ must also be nonzero. $\Box$
Now back to the proof of 2. With hypotheses as above, let $f_0$ denote the map induced by $f$ on the direct sum $V_0$ of the even-dimensional complex cohomology of $X$ and let $f_1$ denote the map induced by $f$ on the direct sum $V_1$ of the odd-dimensional complex cohomology of $X$, so that the Lefschetz trace of $f^k$ can be written
$$L(f^k) = \text{tr}(f_0^k) - \text{tr}(f_1^k).$$
By Lemma 2, the eigenvalues of $f_0$ and $f_1$ are all nonzero, so if $f_0$ and $f_1$ have the same nonzero eigenvalues then in particular $\dim V_0 = \dim V_1$. By the contrapositive of Lemma 1, if $\chi(X) = \dim V_0 - \dim V_1 \neq 0$, then there exists some $k$ between $1$ and $n = \text{max}(\dim V_0, \dim V_1)$ such that $L(f^k) \neq 0$, hence, by the Lefschetz fixed point theorem, such that $f^k$ has a fixed point. In particular, $f^{\text{lcm}(1, 2, ... n)}$ has a fixed point. $\Box$
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For $1$, it is sufficient to check that a sufficiently generic vector field has countably many loops. Then there is a real number that is not a rational multiple of the period of any loop. – Will Sawin Jan 2 at 3:18
@Will: cool. How would I check that? My differential geometry is quite poor. – Qiaochu Yuan Jan 2 at 4:11
For $\dim X=3$ you can always equip a contact 1-form $\lambda$ and speak about the Reeb vector field $R$, satisfying $\lambda(R)=1$ and $d\lambda(R,\cdot)=0$. "Loops" here are called Reeb orbits (and these exist by Taubes' proof of the Weinstein conjecture), and for generic $\lambda$ all Reeb orbits are "cut out transversely", and in particular are isolated. For example, the Hopf fibration depicts the Reeb flow on $(S^3,\lambda=\sum x_idy_i-y_idx_i)$ whose orbits are the $S^1$-fibers, and perturbing this contact form will break that foliation. – Chris Gerig Jan 3 at 21:08
For anyone who's still thinking about this, this seems silly but I can't think of an example of a endomap of degree $0$ without fixed points. All of the examples of maps of degree $0$ I can think of are projection maps $M \times N \to M \times \{ n \}$, which fix $n \in N$. Any ideas? – Qiaochu Yuan Jan 4 at 7:47
More generally, assume that $M$ is a manifold and $f$ is a map on $M$ without fixed point. Fix a point $y_{0}\in M$, Then $F:M\times M \to M\times M$ with $F=(f,y_{0})$ is another example – Ali Taghavi Jan 6 at 10:28 | 2014-11-23T22:29:59 | {
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http://math.stackexchange.com/questions/127415/real-vs-complex-for-integrals-int-0-infty-fracdx1-x3 | # Real VS Complex for integrals: $\int_0^\infty \frac{dx}{1 + x^3}$
The integral $$\int_0^\infty \frac{dx}{1 + x^4} = \frac{\pi}{2\sqrt2}$$ can be evaluated both by a complex method (residues) and by a real method (partial fraction decomposition). The complex method works also for the integral $$\int_0^\infty \frac{dx}{1 + x^3} = \frac{2\pi}{3\sqrt3}$$ but partial fraction decomposition does not give convergent integrals. I would like to know if there is some real method for evaluating this last integral.
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Make the substitution $x = \frac{1}{t}$ and you get
$$\int_{0}^{\infty} \frac{t}{1+t^3} \text{d}t$$
Write the one you want as
$$\int_{0}^{\infty} \frac{1}{1+t^3} \text{d}t$$
Now you can add both and cancel that pesky $1+t$ factor.
btw, a straightforward approach using partial fractions also works.
You consider
$$F(x) = \int_{0}^{x} \frac{1}{1+t^3} \text{d}t$$
Using partial fractions you can find that (I used Wolfram Alpha, I admit)
$$F(x) = \frac{1}{6}\left(2\log(x+1) - \log(x^2 - x -1) + 2\sqrt{3} \arctan\left(\frac{2x-1}{\sqrt{3}}\right)\right) + \frac{\pi}{6\sqrt{3}}$$
Now as $x \to \infty$, we have that $2\log(x+1) - \log(x^2 - x + 1) \to 0$ .
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See this answer: math.stackexchange.com/questions/34351/… to see that you can also evaluate the integral in that question to find your answer here, and then apply the other answers to that question (in particular, Eric's answer(s)). – Aryabhata Apr 2 '12 at 22:27
Thank you Aryabhata. This is a very elegant way of evaluation by reducing it to the $\int_0^\infty \frac{dx}{x^2 - x + 1}$. – Martin Apr 2 '12 at 22:45
@Martin: You are welcome. I have added another method, which does use partial fractions. – Aryabhata Apr 2 '12 at 22:47
Note that for $a > 0$, $$\int_0^N \frac{1}{x+a}\ dx = \ln(N+a) - \ln(a) = \ln(N) - \ln(a) + o(1)\ \text{as} \ N \to \infty$$ while \eqalign{\int_0^N \frac{x+a}{(x+a)^2 + b^2}\ dx &= \frac{1}{2} \left(\ln((N+a)^2+b^2) - \ln(a^2+b^2)\right)\cr &= \ln(N) - \ln(a^2+b^2) + o(1) \ \text{as} \ N \to \infty\cr} and (if $b > 0$) \eqalign{\int_0^N \frac{1}{(x+a)^2+b^2}\ dx = \frac{\arctan\left(\frac{N+a}{b}\right) - \arctan\left(\frac{a}{b}\right)}{b} = \frac{\pi}{2b} - \frac{\arctan\left(\frac{a}{b}\right)}{b} + o(1) \ \text{as} \ N \to \infty\cr} In particular, from the partial fraction decomposition $$\frac{1}{1+x^3} = \frac{1/3}{x+1} + \frac{(2-x)/3}{x^2 - x + 1} = \frac{1/3}{x+1} + \frac{1/2}{(x-1/2)^2+3/4} - \frac{(x-1/2)/3}{(x-1/2)^2 + 3/4}$$ you get $$\int_0^N \frac{1}{1+x^3} \ dx = \frac{\ln(N) - \ln(1))}{3} + \frac{\pi/2 + \arctan(1/\sqrt{3})}{\sqrt{3}} - \frac{\ln(N) - \ln((1/2)^2 + 3/4)}{3} + o(1)$$ i.e. $$\int_0^\infty \frac{1}{1+x^3} \ dx = \frac{\pi}{\sqrt{3}} + \frac{\arctan(1/\sqrt{3})}{\sqrt{3}} = \frac{2 \pi}{3 \sqrt{3}}$$
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Thank you Robert. Nice solution by taking asymptotics. Little typo: in the last integral the upper bound is $\infty$ – Martin Apr 3 '12 at 18:39
Thanks for spotting that, fixed it. – Robert Israel Apr 3 '12 at 21:01
For what is worth:
Your integral evaluates in terms of the sine function:
$$\int\limits_0^\infty \frac{1}{1+x^a}=\frac{\pi}{a}\sec\frac{\pi}{a}$$
refer to this question and the link in it.
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I would like to know if there is some real method for evaluating this last integral.
Actually, all integrals of the form $\displaystyle\int_0^\infty\frac{x^n}{1+x^m}dx$ can be solved by substituting $t=\dfrac1{1+x^m}$ , and then recognizing the expression of the beta function in the new integral, which can be written as a product of gamma functions. Then we use the reflection formula in order to finally arrive at the desired result, $I=\dfrac\pi m\cdot\csc\left[(n+1)\dfrac\pi m\right]$ — See my answer here for more information.
- | 2016-06-30T18:05:20 | {
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http://merganser.math.gvsu.edu/david/linear.algebra/ula/ula/knowl/activity-46.html | ##### Activity4.2.5
We will use Sage to find the eigenvalues and eigenvectors of a matrix. Let's begin with the matrix $$A = \left[\begin{array}{rr} 1 \amp 2 \\ 2 \amp 1 \\ \end{array}\right] \text{.}$$
1. We can find the characteristic polynomial of a matrix $$A$$ by writing A.charpoly('lam'). Notice that we have to give Sage a variable in which to write the polynomial; here, we use lam though you could just as well use x.
The factored form of the characteristic polynomial may be more useful since it will tell us the eigenvalues and their multiplicities. The factor chacteristic polynomial is found with A.fcp('lam').
2. If we only want the eigenvalues, we can use A.eigenvalues().
Notice that the multiplicity of an eigenvalue is the number of times it is repeated in the list of eigenvalues.
3. Finally, we can find eigenvectors by A.eigenvectors_right(). (We are looking for right eigenvalues since the vector $$\vvec$$ appears to the right of $$A$$ in the definition $$A\vvec=\lambda \vvec\text{.}$$)
At first glance, the result of this command can be a little confusing to interpret. What we see is a list with one entry for each eigenvalue. For each eigenvalue, there is a triple consisting of (i) the eigenvalue $$\lambda\text{,}$$ (ii) a basis for $$E_\lambda\text{,}$$ and (iii) the multiplicity of $$\lambda\text{.}$$
4. When working with decimal entries, which are called floating point numbers in computer science, we must remember that computers perform only approximate arithmetic. This is a problem when we wish to find the eigenvectors of such a matrix. To illustrate, consider the matrix $$A=\left[\begin{array}{rr} 0.4 \amp 0.3 \\ 0.6 \amp 0.7 \\ \end{array}\right] \text{.}$$
1. Without using Sage, find the eigenvalues of this matrix.
2. What do you find for the reduced row echelon form of $$A-I\text{?}$$
3. Let's now use Sage to determine the reduced row echelon form of $$A-I\text{:}$$
What result does Sage report for the reduced row echelon form? Why is this result not correct?
4. Because the arithmetic Sage performs with floating point entries is only approximate, we are not able to find the eigenspace $$E_1\text{.}$$ In this next chapter, we will learn how to address this issue. In the meantime, we can get around this problem by writing the entries in the matrix as rational numbers:
in-context | 2018-11-19T05:47:39 | {
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https://math.stackexchange.com/questions/500823/a-closed-form-for-int-0-infty-frac-sinx-operatornameerfi-left-sqrtx | # A closed form for $\int_0^\infty\frac{\sin(x)\ \operatorname{erfi}\left(\sqrt{x}\right)\ e^{-x\sqrt{2}}}{x}dx$
Let $\operatorname{erfi}(x)$ be the imaginary error function $$\operatorname{erfi}(x)=\frac{2}{\sqrt{\pi}}\int_0^xe^{z^2}dz.$$ Consider the integral $$I=\int_0^\infty\frac{\sin(x)\ \operatorname{erfi}\left(\sqrt{x}\right)\ e^{-x\sqrt{2}}}{x}dx.$$ Its numeric value is approximately $0.625773669454426\dots$
Is it possible to express $I$ in a closed form using only elementary functions, integers and constants $\pi$, $e$?
• Yes it is, and I am working on the derivation, but I get $$\tanh ^{-1}\left(\frac{\sqrt{2 \left(1-\sqrt{2}+\sqrt{4-2 \sqrt{2}}\right)}}{1+\sqrt{4-2 \sqrt{2}}}\right)$$ – Ron Gordon Sep 22 '13 at 1:45
• @Marty $$\ln\sqrt{\frac1{\frac12-\sqrt{\frac1{6\sqrt{4-2\sqrt2}+6\sqrt2-2}}}-1}$$ – Vladimir Reshetnikov Sep 22 '13 at 2:13
• @VladimirReshetnikov did you use some Computer Algebra System to get that value or did you get it by hand? If you get it by CAB, then please tell me how?? – Santosh Linkha Sep 22 '13 at 4:48
• @experimentX I derived the result in a semi-manual way with some help from Mathematica. Unfortunately, neither Maple nor Mathematica is able to evaluate this integral directly. – Vladimir Reshetnikov Sep 24 '13 at 1:05
My strategy here is to use Parseval's equality to express the integral in a simpler form. This requires a strategic splitting of the integrand into Fourier transforms.
Begin by writing
$$\text{erfi}({\sqrt{x}})=\frac{2}{\sqrt{\pi}} \sqrt{x} \int_0^1 dt \, e^{x t^2}$$
and consider the following Fourier Transform:
$$\int_{-\infty}^{\infty} dx \, \theta(x) \, \text{erfi}(\sqrt{x}) \, e^{-\sqrt{2} x} \, e^{i k x}$$
where $\theta(x)$ is the Heaviside step function, which is $1$ when $x \gt 0$ and $0$ when $x \lt 0$. Using a change in the order of integration, we may evaluate this Fourier transform in exact form:
\begin{align}\int_{0}^{\infty} dx \, \text{erfi}(\sqrt{x}) \, e^{-\sqrt{2} x} \, e^{i k x} &=\frac{2}{\sqrt{\pi}} \int_0^1 dt \, \int_{0}^{\infty} dx \, \sqrt{x} e^{x t^2} \, e^{-\sqrt{2} x} \, e^{i k x}\\ &= \frac{2}{\sqrt{\pi}} \int_0^1 dt \, \int_{0}^{\infty} dx \, \sqrt{x} e^{-(\sqrt{2}-t^2-i k) x}\\ &= \int_0^1 \frac{dt}{(\sqrt{2}-i k - t^2)^{3/2}} \\ &=\frac{1}{\sqrt{2}-i k} \frac{1}{\sqrt{\sqrt{2}-1-i k}}\end{align}
Note that the third line comes from the integral
$$\int_0^{\infty} dx \, \sqrt{x} e^{-a x} = \frac{\sqrt{\pi}}{2 a^{3/2}}$$
The result in the fourth line may be obtained using a trig substitution in the integral in the third line; the only trick is pretending that $\sqrt{2}-i k$ may be set to some $b^2$ parameter, and then proceeding with the usual trig substitution.
Now, the rest of the original integrand is $\sin{x}/x$, which Fourier transform is simply $\pi$ when $|k| \lt 1$ and $0$ otherwise. We may then invoke Parseval's equality, which states that, for functions $f$ and $g$ and their respective Fourier transforms $F$ and $G$, we have
$$\int_{-\infty}^{\infty} dx \, f(x) g(x) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} dk \, F(k) G(k)$$
Here,
$$f(x) = \text{erfi}(\sqrt{x}) \, e^{-\sqrt{2} x}\, \theta(x)$$ $$g(x) = \frac{\sin{x}}{x}$$ $$F(k) = \frac{1}{\sqrt{2}-i k} \frac{1}{\sqrt{\sqrt{2}-1-i k}}$$ $$G(k) = \begin{cases}\pi & |k| \lt 1 \\ 0 & k \gt 1\end{cases}$$
Thus, we have reduced the integral to the evaluation of the following:
$$\frac12 \int_{-1}^1 \frac{dk}{\sqrt{2}-i k} \frac{1}{\sqrt{\sqrt{2}-1-i k}}$$
Now sub $v^2=\sqrt{2}-1-i k$ and get the following integral
$$-i \int_{\sqrt{\sqrt{2}-1-i}}^{\sqrt{\sqrt{2}-1+i}} \frac{dv}{1+v^2}$$
which evaluates to
$$-i \left [\arctan{\sqrt{\sqrt{2}-1+i}} - \arctan{\sqrt{\sqrt{2}-1-i}} \right ]$$
which is equal to
$$\tanh ^{-1}\left(\frac{\sqrt{2 \left(1-\sqrt{2}+\sqrt{4-2 \sqrt{2}}\right)}}{1+\sqrt{4-2 \sqrt{2}}}\right)$$
or
$$\frac12 \log{\left [\frac{1+\sqrt{4-2 \sqrt{2}}+\sqrt{2 \left(1-\sqrt{2}+\sqrt{4-2 \sqrt{2}}\right)}}{1+\sqrt{4-2 \sqrt{2}}-\sqrt{2 \left(1-\sqrt{2}+\sqrt{4-2 \sqrt{2}}\right)}}\right ]} \approx 0.625774$$
• @achillehui: you are too kind. You're not bad yourself, by the way. – Ron Gordon Sep 22 '13 at 4:35
• @RonGordon This is great, thank you very much! – Marty Colos Sep 22 '13 at 18:41
• Just a question: How much time did it take you to find out this solution? +1 btw. – user93957 Feb 7 '14 at 13:01
• @Aðøbe: It's been long enough that I don't remember specifically, but it did take about a couple of hours to find the right path to the solution. – Ron Gordon Feb 7 '14 at 13:31
For $a>1$, $$\int_0^\infty\frac{\sin(x)\ \operatorname{erfi}\left(\sqrt x\right)\ e^{-a x}}x dx=\ln\sqrt{\frac{\sqrt{a^2-2a+2}+\sqrt2\sqrt{\sqrt{a^2-2a+2}-a+1}+1}{\sqrt{a^2-2a+2}-\sqrt2\sqrt{\sqrt{a^2-2a+2}-a+1}+1}}.$$
• +1. Is there an interpretation of the RHS when $a\leqslant1$? – Did Sep 22 '13 at 11:05
• @did: see my derivation above. There are convergence problems for $a \le 1$. – Ron Gordon Sep 22 '13 at 11:10
• This is precisely my point: the LHS does not exist when $a\lt1$ but the RHS does, for every $a$. – Did Sep 22 '13 at 11:15
• @Did Yes, there is a possible interpretation. Let $s_n(a)=\displaystyle\int_0^{2\pi n}\frac{\sin(x)\ \operatorname{erfi}\left(\sqrt x\right)\ e^{-a x}}x dx$ where $n\in\mathbb{N}$. The RHS gives the limit of the sequence $\lim\limits_{n\to\infty}s_n(a)$. Because $n$ increases discretely, in each wave of the integrand both half-waves can partially compensate each other, enabling the sequence to converge. I did not try to establish for which exactly values of $a$ the integral can be regularized this way. – Vladimir Reshetnikov Sep 22 '13 at 20:46
• Nice. Thanks. – Did Sep 22 '13 at 20:53 | 2019-08-22T18:00:57 | {
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http://mathhelpforum.com/statistics/124739-probability-52-card-deck.html | # Thread: probability of 52 card deck
1. ## probability of 52 card deck
I have a few simple probability questions regarding draws from a deck of cards.
1) If two cards are drawn face down, what is the probability that the second card is an ace?
2) If it is known the first card draw is an ace, how would that change the answer to (1)?
3) What is the probability that 2 randomly drawn cards are both aces?
4) If two cards are drawn from a deck, how many different combinations of the two cards are possible if the order is not considered and if the order is considered?
For (1), I think the answer is 4/51, but do I need to include the probability that the first card is not an ace?
For (2), I think it's 3/51.
For (3), I think it's: 4/52 * 3/51 = 1/221
For (4) - no idea
2. 1. You didn't learn anything about the 1st card since it's face down. So this is like choosing 1 card from 52, your probability is 4/52.
2. Correct
3. Correct
4. Order doesn't count - 52 choose 2.
Order does count - 52 x 51
3. Hello, chemekydie!
1) If two cards are drawn from a standard deck,
what is the probability that the second card is an ace?
I'll do this the Long Way.
There are 4 Aces and 48 Others.
We must consider both possibilities:
[1] The first card is an Ace: . $p(\text{1st Ace}) \:=\:\frac{4}{52}$
. . The second card is an Ace: . $P(\text{2nd Ace}) \:=\:\frac{3}{51}$
. . Hence: . $P(\text{Ace, then Ace}) \:=\:\frac{4}{52}\cdot\frac{3}{51} \:=\:\frac{12}{2652}$
[2] The first card is not an Ace: . $P(\text{1st not-Ace}) \:=\:\frac{48}{52}$
. . The second card is an Ace: . $P(\text{2nd Ace}) \:=\:\frac{4}{51}$
. . Hence: . $P(\text{not-Ace, then Ace}) \:=\:\frac{48}{52}\cdot\frac{4}{51} \:=\:\frac{192}{2652}$
Therefore: . $P(\text{2nd Ace}) \;=\;\frac{12}{2652} + \frac{192}{2652} \:=\:\frac{204}{2652} \;=\;\frac{1}{13}\;\;{\color{red}**}$
2) If it is known the first card draw is an ace,
how would that change the answer to (1)?
I think it's: . $\frac{3}{51}$ . . . . Right!
Just reduce it to $\frac{1}{17}$
3) What is the probability that 2 randomly drawn cards are both aces?
I think it's: . $\frac{4}{52}\cdot\frac{3}{51} \:=\:\frac{1}{221}$ . . . . Yes!
4) If two cards are drawn from a deck,
how many different combinations of the two cards are possible if:
(a) the order is not considered?
(b) the order is considered?
(a) If the order is not important, it is a Combination problem.
. . $_{52}C_2 \:=\:\frac{52!}{2!\,50!} \;=\;1326$
(b) If the order is important, it is a Permutation problem.
. . $_{52}P_2 \:=\:\frac{52!}{2!} \:=\:2652$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
**
The answer to #1 comes as a surprise ... until we think about it.
Suppose they asked about the $37^{th}$ card.
Do we have to consider what happened in the first 36 draws?
. . No, those events do not matter.
Spead the deck face down on the table.
Point to any card and ask "What is the probability that this is an Ace?"
Since there are 52 outcomes and 4 of them are Aces,
. . the probability is: . $P(\text{Ace}) \:=\:\frac{4}{52} \:=\:\frac{1}{13}$
So, you see, we don't care about the other cards.
Bottom line: qmech is absolutely correct! | 2017-12-18T13:00:50 | {
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http://mathhelpforum.com/calculus/169845-velocity-graph.html | 1. ## velocity graph
sorry the image is rather small. It shows the velocity of an object between time 0 and 9.
-When does the object obtain its greatest speed? I think it obtains its greatest speed at t=8, since time vs. velocity shows acceleration. Is this right
-The object was at its origin at t=3. When does it return to its origin? I thought it returned at t=6, but I got this problem wrong. Can someone guide me in the right direction because I have no clue!
2. I'm not sure what you mean by "time vs velocity shows acceleration". A "time versus velocity" graph shows velocity on its vertical axis. It is true that the slope of the graph is the accleration but that is not relevant to this problem. What is relevant is that "speed" is the absolute value of the (linear) velocity. The maximum value of the velocity shown on graph occurs at t= 5 and has value 2. The minimum value of the velocity occurs at t= 8 and is -4. But the maximum speed is |-4|= 4 so you are right about that.
The second question asks when the object is again at the origin. That means "when its position is 0", not it velocity. The information you need here is that the distance traveled is the area under the velocity graph. From t= 3 to t= 6, the velocity is positive so the object is moving in the positive direction (which I am going to call "to the right"). The graph, between t= 3 and t= 6, together with the "position= 0" axis, forms a triangle with base 6- 3= 3 and height 2. Its area is (1/2)(3)(2)= 3 so the object will move from the origin 3 places to the right. In order to get back to the origin, it must move 3 places to the left (-3).
The graph is below the axis to the right of t= 6 so the object is moving back to the right. The line passing through the points (5, 2) and (8, -4) is given by the equation
$y= \frac{-4- 2}{8- 5}= \frac{-6}{3}(t- 5)+ 2= -2(t- 5)+ 2= -2t+ 12$. If distance the object will have moved between t= 6 and t= x (which we would like to find) is the area of the right triangle having base of length x- 6, height y, and the line as hypotenuse. its area is (1/2)(x- 6)(y). Replace y with the formula above and set that area equal to -3 (it will be negative since for x> 6, y< 0- technically that's the "signed" area). Solve for x.
(Because the formula for y involves x, multiplying it by x- 5 gives a quadratic equation which may have two solutions. Of course, the one you want here is the one that is larger than 6.
3. thank you so much for that detailed response! That makes a lot of sense- thanks for covering up the gaps in my knowledge of physics | 2018-02-19T02:55:38 | {
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http://math.stackexchange.com/questions/275276/inspecting-the-function-fx-x-sqrt1-x2 | # Inspecting the function $f(x)=-x\sqrt{1-x^2}$
We are just wrapping up the first semester calculus with drawing graphs of functions. I sometimes feel like my reasoning is a bit shady when I am doing that, so I decided to ask you people from Math.SE.
I am supposed to draw a graph (and show my working) of the function $f(x)=-x\sqrt{1-x^2}$: Below is my work, I'd be very grateful for any comments on possible loopholes in my reasoning (the results should be correct), thanks!
$$\text{1. } f(x)=-x\sqrt{1-x^2}$$
Domain: We have a square-root function, therefore we need $1-x^2\geq0$. From that we get$x\in[-1,1]$. As this is the only necessary condition, $D(f)=[-1,1]$. The function is also continuous on this interval as there is nothing that would produce a discontinuity.
Symmetry: $f(x)$ is an odd function, as $-f(x)=f(-x)$, as demonstrated here: $$-f(x)=f(-x)$$ $$-(-x\sqrt{1-x^2})=-(-x)\sqrt{1-(-x)^2}$$ $$x\sqrt{1-x^2}=x\sqrt{1-x^2}$$
Therefore we are only interested in the interval $[0,1]$, as the function will behave symmetrically on $[-1,0]$.
x and y intercepts: $f(x)=0$ holds when $x=0,1$, those are then the x-intercepts. Thus also the y-intercept is at $(0,0)$
First derivative \begin{align} \ f'(x) &=(-x\sqrt{1-x^2})' \\ & = (-x)'\sqrt{1-x^2}+(-x)(\sqrt{1-x^2})' \\ & = -\sqrt{1-x^2}+(-x)\frac{1}{2\sqrt{1-x^2}}(-2x) \\ & = -\frac{1-x^2}{\sqrt{1-x^2}}+\frac{x^2}{\sqrt{1-x^2}} \\ & = \frac{2x^2-1}{\sqrt{1-x^2}} \\ \end{align}
Local minima and maxima: The equality $\frac{2x^2-1}{\sqrt{1-x^2}}=0$ yields $x=\frac{1}{\sqrt2}$. As $f(\frac{1}{\sqrt2})=-\frac{1}{2}$ and because we know the x-intercepts, we can conclude that this is the local minimum and the function is decreasing at $x\in [0,\frac{1}{\sqrt2})$ and increasing at $x\in(\frac{1}{\sqrt2},1]$. That can also be concluded from the fact that $f′≤0$ on the interval $[0,\frac{1}{\sqrt2}]$ and $f′≥0$ on the interval $[\frac{1}{\sqrt2},1]$.
Second derivative \begin{align} \ f''(x) &=(\frac{2x^2-1}{\sqrt{1-x^2}})' \\ & = \frac{(2x^2-1)'\sqrt{1-x^2}-(2x^2-1)(\sqrt{1-x^2})'}{1-x^2} \\ & = \frac{4x\sqrt{1-x^2}-(2x^2-1)\frac{-x}{\sqrt{1-x^2}}}{1-x^2} \\ & = \frac{4x\frac{1-x^2}{\sqrt{1-x^2}}-(2x^2-1)\frac{-x}{\sqrt{1-x^2}}}{1-x^2} \\ & = \frac{\frac{-2x^3+3x}{\sqrt{1-x^2}}}{1-x^2} \\ & = \frac{-2x^3+3x}{\sqrt{(1-x^2)^3}} \\ \end{align}
Inflection point and concavity: $f''(x)=0$ has only one result and that is $x=0$. If we look at the inequality for $x\in(0,1]$. \begin{align} \ 0&<f''(x) \\ 0&<\frac{-2x^3+3x}{\sqrt{(1-x^2)^3}} \\ 0&<-2x^3+3x \\ 2x^3&<3x \\ 2x^2&<3 \\ x^2&<3/2 \\ \end{align}
We can see that $x^2<3/2$ is satisfied on $x\in(0,1]$, thus the function is convex on this interval.
Conclusion: Using the symmetricity of the function, we can conclude that the function is increasing on $[-1,-\frac{1}{\sqrt2}]$ and $[\frac{1}{\sqrt2},1]$ and decreasing on $[-\frac{1}{\sqrt2},\frac{1}{\sqrt2}]$. It has local minimum at $x=\frac{1}{\sqrt2}$ and local maximum at $x=-\frac{1}{\sqrt2}$. It is concave on $[-1,0]$ and convex on $[0,1]$, with point of inflection at $x=0$.
-
This looks very nice. I have only two suggestions: 1) ephasize throughout that you are only considering the interval $[0,1]$. e.g, "in the interval $[0,1]$, $f''(x)=0$ has only the solution $x=0$. 2) When examining the local mins and maxes, it would be easier and more rigorous to just note $f'\le0$ on the interval $[0,1/\sqrt 2]$ and $f'\ge 0$ on the interval $[1/\sqrt2,1]$. – David Mitra Jan 10 '13 at 17:38
Sounds good to me, good job ! – Alan Simonin Jan 10 '13 at 17:39
(+1) for an archetypal question – The Chaz 2.0 Jan 10 '13 at 17:40
@DavidMitra Thanks for the suggestion. I've made a minor edit following your second suggestion. – Dahn Jahn Jan 10 '13 at 17:45
@TheChaz2.0 Actually, regarding that, what is the best way to name this question. I am asking because a) for purposes of easy findability for other users and b) in my language, Czech, we have a fixed word for this kind of a thing, whereas English doesn't seem to have one (or am I wrong?). – Dahn Jahn Jan 11 '13 at 1:01
Your answer and/or analysis of the function $$f(x)=-x\sqrt{1-x^2}$$ is accurate, thorough, well-justified, and consistent with its graph:
$\quad \quad f(x)=-x\sqrt{1-x^2}$.
$\quad$Source: Wolfram Alpha.
Kudos for the effort you've shown and your accurate and detailed analysis!
(Just don't forget to graph the function too!)
-
Haha, thanks, I've edited the question, as I am supposed to do all the working out. I am more interested in not giving claims I haven't explicitly proven (or at least demonstrated enough), rather than correct answers (which I can, as you pointed out, check with software/wolfram alpha) – Dahn Jahn Jan 10 '13 at 17:37
Absolutely - I like to back up claims as well, and your analysis is more accurate than simply trying to "eye-up"/estimating the intervals on which it is increasing, decreasing, etc. etc. Nice job with your assessment/analysis. – amWhy Jan 10 '13 at 17:40
Since it seems like Math.SE approves, I chose this as the accepted question, no point waiting for anything else. – Dahn Jahn Jan 10 '13 at 18:34 | 2014-08-23T11:55:58 | {
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http://mathhelpforum.com/calculus/91858-centroid-hemisphere.html | # Math Help - centroid of a hemisphere
1. ## centroid of a hemisphere
I must find the z centroid of a hemisphere with radius a. It's base is on the x-y plane and its dome extends up the z axis. I am using the following equations to determine the centroid.
$\overline{z}=\frac{\int_V\tilde{z} dV}{\int_V dV}$
I am using $dV=\pi a^2 dz$ and $\tilde{z}=z$ and integrating from 0 to a
$\overline{z}=\frac{\int_{0}^{a} z \pi a^2 dz}{\int_{0}^{a} \pi a^2 dz}$
$\overline{z}=\frac{\pi a^2 \int_{0}^{a} z dz}{\pi a^2\int_{0}^{a} dz}$
$\overline{z}=\frac{ \int_{0}^{a} z dz}{\int_{0}^{a} dz}$
$\overline{z}=\frac{\frac{z^2}{2}|_{0}^{a}}{z|_{0}^ {a}}$
$\overline{z}=\frac{a}{2}$
I know the answer is supposed to be $\overline{z}=\frac{3a}{8}$ Where did I mess up?
Thanks
2. These are supposed to be triple integrals.
You should switch to spherical co-ordinates with the region being
$0\le \rho\le a$, $0\le \theta \le 2\pi$ and $0\le \phi \le \pi/2$.
3. $dV = \pi r^{2}dz$, where r is the radius of each circular disc. Then you to write r in terms of z.
4. The hemisphere is center on the origin so I know that the x and y centroid are 0. Is this what you are refering to by triple integrals?
I also would like to be able to do this in cartesian if that is possible as that was that coordinates the problem specified.
If I write a in terms of z I get
$a^2=y^2 + z^2$
Then when ever I try to get the integrand in terms of one variable I wind back up at $a^2$
Thanks again
5. I found my mistake dV should be
$dV=\pi y^2 dz$
then
$dV= \pi (a^2 -z^2) dz$
thanks again
6. I read dV as dxdydz. Changing to spherical with $J=\rho^2\sin\phi$ and $z=\rho\cos\phi$ we have
${ \int_0^a\int_0^{2\pi}\int_0^{\pi /2} \rho^3\sin\phi \cos\phi d\phi d\theta d\rho\over \int_0^a\int_0^{2\pi}\int_0^{\pi /2} \rho^2 \sin\phi d\phi d\theta d\rho}$
$= { \int_0^a \rho^3d\rho \int_0^{2\pi}d\theta \int_0^{\pi /2} \sin\phi \cos\phi d\phi \over \int_0^a \rho^2 d\rho \int_0^{2\pi}d\theta \int_0^{\pi /2} \sin\phi d\phi }$
$= { (a^4/4)(2\pi)(1/2) \over (a^3/3)(2\pi)(1)}={3a\over 8}$.
7. I also did it this way
$hemisphere = a^2=y^2+x^2$
$\tilde{z}=\frac{\int_{V}\tilde{z} dV}{\int_{V} dV}$
$dV= \pi r^2$
$r=y$
$\tilde{z}=z$
$y^2=a^2-z^2$
$\tilde{z}=\frac{\int_{0}^{a}z \pi y^2 dz}{\int_{0}^{a}\pi y^2 dz}$
$\tilde{z}=\frac{\int_{0}^{a}z \pi (a^2-z^2) dz}{\int_{0}^{a}\pi (a^2-z^2) dz}$
$\tilde{z}=\frac{\pi \int_{0}^{a}z (a^2-z^2) dz}{\pi \int_{0}^{a}(a^2-z^2) dz}$
$\tilde{z}=\frac{\int_{0}^{a}z (a^2-z^2) dz}{ \int_{0}^{a}(a^2-z^2) dz}$
Your way looks much easier, so thanks
8. Originally Posted by manyarrows
The hemisphere is center on the origin so I know that the x and y centroid are 0. Is this what you are refering to by triple integrals?
I also would like to be able to do this in cartesian if that is possible as that was that coordinates the problem specified.
If I write a in terms of z I get
$a^2=y^2 + z^2$
Then when ever I try to get the integrand in terms of one variable I wind back up at $a^2$
Thanks again
You can do this via (x,y,z) but to solve the integrals you will need to various trig substitutions. It's smarter to switch to spherical immediately.
For example the bound for z would be $0\le z\le \sqrt{a^2-x^2-y^2}$. Then the (x,y) base is a circle of radius a, also screaming out for trig substitution.
The other bounds of integration would be $-\sqrt{a^2-x^2}\le y \le \sqrt{a^2-x^2}$ and $-a\le x \le a$. Which is begging for polar, i.e., trig substitution.
9. He did it correctly the second time. You don't have to set up a triple integral.
10. ## Re: centroid of a hemisphere
Here's the problem worked in a bit more detail, hope it reads:
Calculus: Centroid of a Hemisphere, Math 251 | 2015-05-25T04:56:37 | {
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https://www.mathworks.com/help/signal/ref/diric.html?requestedDomain=www.mathworks.com&nocookie=true | # Documentation
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# diric
Dirichlet or periodic sinc function
## Syntax
y = diric(x,n)
## Description
y = diric(x,n) returns a vector or array y the same size as x. The elements of y are the Dirichlet function of the elements of x. n must be a positive integer.
## Examples
collapse all
Compute and plot the Dirichlet function between and for N = 7 and N = 8.
x = linspace(-2*pi,2*pi,301); d7 = diric(x,7); d8 = diric(x,8); subplot(2,1,1) plot(x/pi,d7) ylabel('N = 7') title('Dirichlet Function') subplot(2,1,2) plot(x/pi,d8) ylabel('N = 8') xlabel('x / \pi')
The function has a period of for odd N and for even N.
The Dirichlet and sinc functions are related by . Illustrate this fact for .
xmax = 2; x = linspace(-xmax,xmax,1001)'; N = 6; yd = diric(x*pi,N); ys = sinc(N*x/2)./sinc(x/2); subplot(2,1,1) plot(x,yd) title('D_6(x*pi)') subplot(2,1,2) plot(x,ys) title('sinc(6*x/2) / sinc(x/2)')
Repeat the calculation for .
N = 9; yd = diric(x*pi,N); ys = sinc(N*x/2)./sinc(x/2); subplot(2,1,1) plot(x,yd) title('D_9(x*pi)') subplot(2,1,2) plot(x,ys) title('sinc(9*x/2) / sinc(x/2)')
## Diagnostics
If n is not a positive integer, diric gives the following error message:
Requires n to be a positive integer.
collapse all
### Dirichlet Function
The Dirichlet function, or periodic sinc function, is
${D}_{N}\left(x\right)=\left\{\begin{array}{ll}\frac{\mathrm{sin}\left(Nx/2\right)}{N\mathrm{sin}\left(x/2\right)}\hfill & x\ne 2\pi k,\text{ }k=0,±1,±2,±3,...\hfill \\ {\left(-1\right)}^{k\left(N-1\right)}\hfill & x=2\pi k,\text{ }k=0,±1,±2,±3,...\hfill \end{array}$
for any nonzero integer N. This function has period 2π for odd N and period 4π for even N. Its peak value is 1, and its minimum value is –1 for even N. The magnitude of the function is 1/N times the magnitude of the discrete-time Fourier transform of the N-point rectangular window. | 2017-03-01T20:45:18 | {
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http://math.stackexchange.com/questions/202335/did-i-prove-this-limit-correctly?answertab=active | Did I prove this limit correctly?
Given $\lim_{n \to \infty}a_n = L$ and $\lim_{n \to \infty}b_n = M$ implies that $\lim_{n \to \infty}2a_n + 3b_n = 2L + 3M$
Proof
Assume $\lim_{n \to \infty}a_n = L$ and $\lim_{n \to \infty}b_n = M$ and $\forall \epsilon > 0$, we have $\frac{\epsilon}{4}>0$ and $\frac{\epsilon}{6}>0$
$|a_n - L| < \frac{\epsilon}{4}$ when $n > N_1$
$|b_n - M| < \frac{\epsilon}{6}$ when $n > N_2$
Set $n>N= \max \left \{ N_1, N_2 \right \}$,we have
$|2a_n + 3b_n - (2L + 3M)| \leq |2a_n - 2L| + |3b_n - 3M| = |2||a_n - L| + |3||b_n - M| < 2\frac{\epsilon}{4} + 3\frac{\epsilon}{6} = \epsilon$
Question: I had to work backwards to choose my $\epsilon$, is that a "general" strategy? I omitted my work for finding my $\epsilon$
-
Perfect, almost too much so. You worked too hard in working backwards. I would have picked $\frac{\epsilon}{100}$ for both, then we get $\le \frac{2\epsilon}{100}+\frac{3\epsilon}{100}=\frac{5\epsilon}{100}\lt \epsilon$. – André Nicolas Sep 25 '12 at 18:15
Yes, this looks correct. Working backwards to choose $\epsilon$ is also the strategy I'd recommend in general.
-
But I've noticed that in my proof class that if we "include" the work of working backwards (with step-by-step explanation) then it is considered informal – Hawk Sep 25 '12 at 19:39
@jak It usually isn't included, but I wouldn't say it makes the proof "informal" in the sense I use the word (i.e. the proof is no less technically complete and correct). – Alex Becker Sep 25 '12 at 19:42
I asked because most books omit that important work. It took me a while to understand this concept. Thank you very much – Hawk Sep 25 '12 at 19:44 | 2015-05-28T06:26:22 | {
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https://math.stackexchange.com/questions/1475261/find-the-polynomial-of-the-fifth-degree-with-real-coefficients-such-that | Find the polynomial of the fifth degree with real coefficients such that…
Find the polynomial of the fifth degree with real coefficients such that the number 1 is a zero of the polynomial but to the second degree, the number $1+i$ is a zero but to the first degree and if divided by $(x+1)$ gives the remainder $10$, and if divided by $x-2$ gives the remainder $13.$
$$p(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5.$$
What I don't know is how to use these properties that are given in the question to find these coeficients, I'm pretty sure I'm going to have to solve a system of equations, but I just need some guidance as to how to get to that.
• A fifth degree polynomial has, counting multiplicities, five zeros in $\mathbb{C}$. How many zeros are determined by the given conditions? – Daniel Fischer Oct 11 '15 at 19:50
• Adding to the above comment: one needs six pieces of information to specify a fifth-degree polynomial. It looks like you're given only five pieces of information (counting the double zero as two pieces). How can you use the fact that the polynomial has real coefficients to get a sixth piece of information? (And once you have a partial list of zeros, you can write the possible $p(x)$ in semi-factored form, which will greatly cut down on the number of coefficient variables remaining.) – Greg Martin Oct 11 '15 at 20:07
If $1$ is a zero with multiplicity $2$, the polynomial is divisible by $(x-1)^2$.
If $1+\mathrm i\;$ is a complex root, as the polynomial has real coefficients, its conjugate is another root. Hence the polynomial is divisible by $$(x-1-\mathrm i)(x-1+\mathrm i)=x^2-2x+2.$$ Hence we have $$p(x)=(x-1)^2(x^2-2x+2)(ax+b)$$ and there remains to find the last (linear) factor $ax+b$.
The last two conditions will give a system of linear equations that will let you find the values of $a$ and $b$.
• Can you help me with which system of linear equations I am to look at using the fact that $p(x)$ when divided by $x+1$ gives remainder $10$ and when divided by $x-2$ gives remainder $13$? – Jerry West Oct 26 '15 at 13:00
• Dividing $p(x)$ by $x-\alpha$, with quotient $q(x)$ and remainder $r$ means that $$p(x)=(x-\alpha)q(x)+r.$$ Hence $p(\alpha)=?$. – Bernard Oct 26 '15 at 13:05 | 2019-06-19T03:29:47 | {
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http://math.stackexchange.com/questions/674621/how-can-you-find-the-cubed-roots-of-i?answertab=active | # How can you find the cubed roots of $i$?
I am trying to figure out what the three possibilities of $z$ are such that
$$z^3=i$$
but I am stuck on how to proceed. I tried algebraically but ran into rather tedious polynomials. Could you solve this geometrically? Any help would be greatly appreciated.
-
## 6 Answers
Using Euler's formula, which states $$e^{i \theta} = \cos \theta + i \sin \theta$$ we will see that $$i = 0 + i \cdot 1 = \cos \left( \frac{\pi}{2} + 2n \pi \right) + i \sin \left( \frac{\pi}{2} + 2n \pi \right) = e^{i \left(\frac{ \pi}{2} + 2n \pi \right)}$$ for all integers $n$. Thus, if $z^3 = i$, then $$z = \exp\left[ i \left(\frac{\pi}{6}+\frac{2n\pi}{3}\right)\right]$$ for all integers $n$.
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Thanks! This makes a lot more sense now. – RXY15 Feb 14 '14 at 2:01
As $\displaystyle i^2=-1,i=-i^3\implies z^3=i=-i^3=(-i)^3\iff z^3-(-i)^3=0$
Now, $\displaystyle a^3-b^3=(a-b)(a^2+ab+b^2),$
$\displaystyle\implies\{z-(-i)\}\{z^2+z(-i)+(-i)^2\}=\{z-(-i)\}(z^2-iz-1)$
If $z-(-i)=0, z=-i$
Else $z^2-iz-1=0\implies z=\dfrac{i\pm\sqrt{i^2-4(-1)}}2=\dfrac{i\pm\sqrt3}2$
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You can solve this geometrically if you know polar coordinates.
In polar coordinates, multiplication goes $(r_1, \theta_1) \cdot (r_2, \theta_2) = (r_1 \cdot r_2, \theta_1 + \theta_2)$, so cubing goes $(r, \theta)^3 = (r^3, 3\theta)$. The cube roots of $(r, \theta)$ are $\left(\sqrt[3]{r}, \frac{\theta}{3}\right)$, $\left(\sqrt[3]{r}, \frac{\theta+2\pi}{3}\right)$ and $\left(\sqrt[3]{r}, \frac{\theta+4\pi}{3}\right)$ (recall that adding $2\pi$ to the argument doesn't change the number). In other words, to find the cubic roots of a complex number, take the cubic root of the absolute value (the radius) and divide the argument (the angle) by 3.
$i$ is at a right angle from $1$: $i = \left(1, \frac{\pi}{2}\right)$. Graphically:
A cubic root of $i$ is $A = \left(1, \frac{\pi}{6}\right)$. The other two are $B = \left(1, \frac{5\pi}{6}\right)$ and $\left(1, \frac{9\pi}{6}\right) = -i$.
Recalling basic trigonometry, the rectangular coordinates of $A$ are $\left(\cos\frac{\pi}{6}, \sin\frac{\pi}{6}\right)$ (the triangle OMA is rectangle at M). Thus, $A = \cos\frac{\pi}{6} + i \sin\frac{\pi}{6} = \frac{\sqrt{3}}{2} + i\frac{1}{2}$.
If you don't remember the values of $\cos\frac{\pi}{6}$ and $\sin\frac{\pi}{6}$, you can find them using geometry. The triangle $OAi$ has two equal sides $OA$ and $Oi$, so it is isoceles: the angles $OiA$ and $OAi$ are equal. The sum of the angles of the triangle is $\pi$, and we know that the third angle $iOA$ is $\frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$; therefore $OiA = OAi = \dfrac{\pi - \frac{pi}{3}}{2} = \dfrac{\pi}{3}$. So $OAi$ is an equilateral triangle, and the altitude AN is also a median, so N is the midpoint of $[Oi]$: $\sin\frac{\pi}{6} = AM = ON = \frac{1}{2}$. By the Pythagorean theorem, $OM^2 + AM^2 = OA^2 = 1$ so $\cos\frac{\pi}{6} = \sqrt{1 - \left(\frac{1}{2}\right)^2} = \dfrac{\sqrt{3}}{2}$.
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I believe your "polynomial" approach would also have worked, if this is what you meant :
[In this, we are supposing that we knew nothing of the "Euler Identity", DeMoivre's Theorem, or roots of unity, all of which provide quite efficient devices]
If we (probably safely) assume that the solution(s) are complex numbers, and call $\ z \ = \ a + bi \ ,$ with $\ a \$ and $\ b \$ real, we can write the equation as
$$(a + bi)^3 \ = \ a^3 \ + \ 3a^2 b \cdot i \ + \ 3a b^2 \cdot i^2 \ + \ b^3 i^3 \ = \ (a^3 \ - \ 3ab^2) \ + \ (3a^2b \ - \ b^3) \cdot i \ \ = \ \ i \ ,$$
by applying the binomial theorem and "powers of $\ i \$ ". Since the right-hand side of the equation is a pure-imaginary number, this requires that
$$a^3 \ - \ 3ab^2 \ = \ a \ ( a^2 \ - \ 3b^2 ) \ = \ 0 \ \ \text{and} \ \ 3a^2b \ - \ b^3 \ = \ b \ (3a^2 \ - \ b^2) \ = \ 1 \ \ .$$
The first equation presents us with two cases:
I -- $\ a \ = \ 0 \$ :
$$a \ = \ 0 \ \ \Rightarrow \ \ b \ ( \ 0 \ - \ b^2 ) \ = \ -b^3 \ = \ 1 \ \ \Rightarrow \ \ b \ = \ -1 \ \ \Rightarrow \ \ z \ = \ 0 - i \ \ ;$$
II -- $\ a^2 \ - \ 3b^2 \ = \ 0$ :
$$a^2 \ = \ 3b^2 \ \ \Rightarrow \ \ b \ ( \ 3 \cdot [3b^2] \ - \ b^2 \ ) \ = \ 8b^3 \ = \ 1 \ \ \Rightarrow \ \ b \ = \ \frac{1}{2}$$
$$\Rightarrow \ \ a^2 \ = \ 3 \ \left( \frac{1}{2} \right)^2 \ = \ \frac{3}{4} \ \ \Rightarrow \ \ a \ = \ \pm \frac{\sqrt{3}}{2} \ \ \Rightarrow \ \ z \ = \ \frac{\sqrt{3}}{2} + \frac{1}{2}i \ , \ -\frac{\sqrt{3}}{2} + \frac{1}{2}i \ \ .$$
We have found three complex-number solutions to the equation. As Dan says, (one form of) the Fundamental Theorem of Algebra states that this third-degree polynomial with complex coefficients has, in all, three roots (counting multiplicities, which are each 1 here).
We probably wouldn't want to use this method for degrees higher than this, as the algebra would become more difficult to resolve. The techniques described by the other posters are far more generally used.
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Hmm I guess I made a mistake somewhere when using this method. Thank you for clearing things up! – RXY15 Feb 14 '14 at 2:02
Taking the absolute value of both sides: $|z^3| = |i|$, gives $|z| = 1$. So, $z = \cos (\theta) + i \sin (\theta)$ for some real $\theta$.
Using De Moivre's formula gives $z^3 = \cos(3\theta) + i \sin(3\theta)$. Given that $z^3 = i = 0 + 1i$, this means that $\cos(3\theta) = 0$ and $\sin(3\theta) = 1$. Solving this system gives $3\theta = \frac{\pi}{2} + 2\pi n$, or $\theta = \frac{\pi}{6} + \frac{2 \pi n}{3}$, for any $n \in \mathbb{Z}$.
Plugging in a few values for $n$ gives:
• $n = 0$ → $\theta = \frac{\pi}{6}$ → $z = \frac{\sqrt{3}}{2} + \frac{1}{2} i$
• $n = 1$ → $\theta = \frac{5\pi}{6}$ → $z = \frac{-\sqrt{3}}{2} + \frac{1}{2} i$
• $n = 2$ → $\theta = \frac{3\pi}{2}$ → $z = -i$
And we can stop there because this is a polynomial equation of degree 3, and the Fundamental Theorem of Algebra guarantees that it has at most 3 distinct roots. The solution set is thus $z \in \{ \frac{\sqrt{3}}{2} + \frac{1}{2} i, \frac{-\sqrt{3}}{2} + \frac{1}{2} i, -i \}$.
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The answer of @Petaro is best, because it suggests how to deal with such questions generally, but here’s another approach to the specific question of what the cube roots of $i$ are.
You know that $(-i)^3=i$, and maybe you know that $\omega=(-1+i\sqrt3)/2$ is a cube root of $1$. So the cube roots of $i$ are the numbers $-i\omega^n$, $n=0,1,2$.
- | 2016-05-01T15:46:54 | {
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https://math.stackexchange.com/questions/2029464/please-verify-my-induction-proof | # Please verify my induction proof.
I would like to show that the following statement is true by the principle of mathematical induction (I must only use induction, not other theorems to justify my answer)
If $n$ is odd natural number, then $n^3-n$ is divisible by 24.
My proof:
Base Case: For $n=1, n^3-n = 1-1 = 0$ which is divisible by $24$.
Induction hypothesis: Assume that that statement is true for $n=2k-1, k∈N$. This means that $(2k-1)^3 - (2k-1)$ is divisible by 24 and hence $(2k-1)^3 - (2k-1) = 24p, p∈N$.
$(2k-1) [(2k-1)^2-1] = 24p$
$(2k-1)[(2k-1-1)(2k-1+1)] = 24p$
$(2k-1)[(2k-2)(2k)] = 24p$
$8k^3-12k^2+4k=24p$
Inductive step: Show that the statement is true for $n=2k+1, k∈N$.
$n^3-n = (2k+1)^3-(2k+1)$
$= (2k+1)[(2k+1)^2-1]$
$=(2k+1)[(2k+1-1)(2k+1+1)]$
$=(2k+1)[(2k)(2k+2)]$
$=8k^3+12k^2+4k$
$=(8k^3-12k^2+4k)+24k^2$
$=24p+24k^2$ (Induction hypothesis)
$=24(p+k^2)$
Both expressions are divisible by $24$, hence the expression is divisible by 24.
We have shown that the statement is true for $n=2k+1$. Therefore, by induction, statement is true for all odd natural numbers n.
Please give me your suggestions. Thanks!
• Seems fine to me – RGS Nov 24 '16 at 23:18
• Your proof is right but there is a little mistake. It should be $(2k-1)(2k-2)(2k)=24p$. – Xam Nov 24 '16 at 23:24
• There is a tiny thing in your proof: $(2k-1)^2-1=4k^2-4k=4k(k-1)\neq (2k-1)2k$ – Ruzayqat Nov 24 '16 at 23:27
• @Ruzayqat where? – mathewconway Nov 24 '16 at 23:30
• okay, corrected @Charter – mathewconway Nov 24 '16 at 23:33
## 2 Answers
That's the right idea. Essentially you have verified the following equation
$$\color{}{(2k+1)^3 -(2k+1)}\, =\ \color{#0a0}{(2k-1)^3 - (2k-1)}\, +\, 24k^2$$
Therefore $\ 24\mid \color{#0a0}{\rm green}\,\Rightarrow\,24\mid\rm RHS\,\Rightarrow\,24\mid LHS,\,$ which yields the inductive step.
Remark $\$ A slicker way to prove that equality is to note that $\,f(k) = \rm RHS-LHS$ is at most quadratic (cubic terms cancel), so to verify that it is zero it suffices to show that it has $3$ roots, e.g. verify $\,0 = f(1) = f(2) = f(3).\,$ But this looks like the base of an inductive proof! Indeed, if you study the calculus of finite differences and/or telescopy you will learn even nicer ways to handle such inductive proofs. You can find some examples in my posts on telescopy.
• Nice job, Bill (really)! I like how you address the OP's question and also enriched it! – Namaste Nov 24 '16 at 23:49
Induction is not necessary.
$n^3-n=(n-1)n(n+1)$, so at least one of them is multipe of 3.
In addition, if $n$ is odd, both $n-1$ and $n+1$ are even. In fact one of then must be multiple of 4, so $(n-1)(n+1)$ is multiple of 8.
Hence, $n^3-n$ is multiple of 24.
• @amWhy, relax, Tito isn't hurting anyone, jeez. Also, your quote from the question is actually an addition made by Bill after Tito already answered. – tilper Nov 25 '16 at 0:18
• @tilper I'm relaxed. I didn't know that the sentence I quoted was edited in, after it's post. Tito's observations are fine (note, I did not nor will I downvote this answer). I just really hope that when a questioner asks whether their procedure/proof/solution is correct, that any answerer address that question first. The best then go on to also include alternative approaches, or how to simplify the OP's approach, etc. – Namaste Nov 25 '16 at 0:24 | 2019-05-23T03:35:27 | {
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https://math.stackexchange.com/questions/452075/not-both-2n-1-2n1-can-be-prime/452079 | # not both $2^n-1,2^n+1$ can be prime.
I am trying to prove that not both integers $2^n-1,2^n+1$ can be prime for $n \not=2$. But I am not sure if my proof is correct or not:
Suppose both $2^n-1,2^n+1$ are prime, then $(2^n-1)(2^n+1)=4^n-1$ have 2 precisely two prim factors. Now $4^n-1=(4-1)(4^{n-1}+4^{n-2}+ \cdots +1)=3A$. So one of $2^n-1, 2^n+1$ must be 3. which implies $n=1$ or $n=2$ (rejected by assumption). Putting $n=1$, we have $2^n-1=1$ which is not a prime. Hence the result follows.
I also wanna know if there is alternative proof, thank you so much.
• There is an alternative proof, $2^n - 1$ can only be prime if $n$ is prime, $2^n + 1$ can only be prime if $n$ is a power of $2$. And yes, your proof is correct. – Daniel Fischer Jul 25 '13 at 17:20
• Just found : math.stackexchange.com/questions/402603/… – lab bhattacharjee Jul 25 '13 at 18:22
• This is not a duplicate. This is a "check my proof" question, which cannot be a duplicate! – user1729 Jul 26 '13 at 9:49
• (As a side point, your proof is fine.) – user1729 Jul 26 '13 at 9:50
• @user1729: that's right. So, how can I recall my "close" vote? – mau Jul 26 '13 at 10:04
Of the three consecutive integers $2^n-1,2^n,2^n+1$, one must be divisble by 3, and it can't be $2^n$.
If $n$ is even $=2m,2^n-1=2^{2m}-1=4^m-1$ is divisible by $4-1=3$ and $4^m-1>3$ if $m\ge1\iff n\ge2$
If $n$ is odd $=2m+1,2^n+1=2^{2m+1}+1$ is divisible by $2+1=3$ and $2^{2m+1}+1>3$ if $m\ge1\iff n\ge3$
alternatively, $$(2^n-1)(2^n+1)=4^n-1$$ is divisible by $4-1=3$
So, at least one of $2^n-1,2^n+1$ is divisible by $3$
Now, $2^n+1>2^n-1>3$ for $n>2$
$\implies$ for $n>2,$ one of $2^n-1,2^n+1$ must be composite | 2019-04-23T12:27:57 | {
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https://math.stackexchange.com/questions/1263887/invertible-skew-symmetric-matrix | # Invertible skew-symmetric matrix
I'm working on a proof right now, and the question asks about an invertible skew-symmetric matrix. How is that possible? Isn't the diagonal of a skew-symmetric matrix always $0$, making the determinant $0$ and therefore the matrix is not invertible?
• Having vanishing diagonal entries means the trace is always zero, but the determinant need not necessarily be zero. Consider $\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}$. – Dustan Levenstein May 3 '15 at 0:37
• To jump a bit forward: odd-order skew-symmetric matrices are necessarily singular, but even-order ones don't have to be. – J. M. is a poor mathematician May 3 '15 at 0:46
The diagonal of a skew-symmetric matrix is always $0$ does not mean that its determinant be $0$. Look at following example:
$det\left[ \begin{array}{} 0 & 1 \\ -1 & 0 \\ \end{array} \right]=1$
Its inverse is: $\left[ \begin{array}{} 0 & -1 \\ 1 & 0 \\ \end{array} \right]$
No, the diagonal being zero does not mean the matrix must be non-invertible. Consider $\begin{pmatrix} 0 & 1 \\ -1 & 0 \\ \end{pmatrix}$. This matrix is skew-symmetric with determinant $1$.
Edit: as a brilliant comment pointed out, it is the case that if the matrix is of odd order, then skew-symmetric will imply singular. This is because if $A$ is an $n \times n$ skew-symmetric we have $\det(A)=\det(A^T)=det(-A)=(-1)^n\det(A)$. Hence in the instance when $n$ is odd, $\det(A)=-\det(A)$; over $\mathbb{R}$ this implies $\det(A)=0$. | 2019-06-25T07:23:41 | {
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https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_15&diff=prev&oldid=113094 | # Difference between revisions of "2000 AIME I Problems/Problem 15"
## Problem
A stack of $2000$ cards is labelled with the integers from $1$ to $2000,$ with different integers on different cards. The cards in the stack are not in numerical order. The top card is removed from the stack and placed on the table, and the next card is moved to the bottom of the stack. The new top card is removed from the stack and placed on the table, to the right of the card already there, and the next card in the stack is moved to the bottom of the stack. The process - placing the top card to the right of the cards already on the table and moving the next card in the stack to the bottom of the stack - is repeated until all cards are on the table. It is found that, reading from left to right, the labels on the cards are now in ascending order: $1,2,3,\ldots,1999,2000.$ In the original stack of cards, how many cards were above the card labeled $1999$?
## Solution
We try to work backwards from when there are 2 cards left, since this is when the 1999 card is laid onto the table. When there are 2 cards left, the 1999 card is on the top of the deck. In order for this to occur, it must be 2nd on the deck when there are 4 cards remaining, and this means it must be the 4th card when there are 8 cards remaining. This pattern continues until it is the 512th card on the deck when there are 1024 cards remaining. Since there are over 1000 cards remaining, some cards have not even made one trip through yet, 2(1024 - 1000) = 48, to be exact. Once these cards go through, 1999 will be the $512 - 48 = 464^\text{th}$ card on the deck. Since every other card was removed during the first round, it goes to show that 1999 was in position $464 \times 2 = 928$, meaning that there were $\boxed{927}$ cards are above the one labeled $1999$.
## Solution 2
To simplify matters, we want a power of $2$. Hence, we will add $48$ 'fake' cards which we must discard in our actual count. Using similar logic as Solution 1, we find that 1999 has position $1024$ in a $2048$ card stack, where the fake cards towards the front.
Let the fake cards have positions $1, 3, 5, \cdots, 95$. Then, we know that the real cards filling the gaps between the fake cards must be the cards such that they go to their correct starting positions in the $2000$ card case, where all of them are below $1999$. From this, we know that the cards from positions $1$ to $96$ alternate in fake-real-fake-real, where we have the correct order of cards once the first $96$ have moved and we can start putting real cards on the table. Hence, $1999$ is in position $1024 - 96 = 928$, so $\boxed{927}$ cards are above it. - Spacesam
## Solution 3 (Recursion)
Consider the general problem: with a stack of $n$ cards such that they will be laid out $1, 2, 3, ..., n$ from left to right, how many cards are above the card labeled $n-1$?
Let $a_n$ be the answer to the above problem.
As a base case, consider $n=2$. Clearly, the stack must be (top to bottom) $(1, 2)$, so $a_n=0$.
Next, let's think about how we can construct a stack of $n+1$ cards from a stack of $n$ cards. First, let us renumber the current stack of $n$ by adding $1$ to the label of each of the cards. Then we must add a card labeled "$1$".
Working backwards, we find that we must move the bottom card to the top, then add "$1$" to the top of the deck.
Therefore, if $a_n \ne n-1$ (meaning the card $n-1$ is not at the bottom of the deck and so it won't be moved to the top), then $a_{n+1} = a_n + 2$, since a card from the bottom is moved to be above the $n-1$ card, and the new card "$1$" is added to the top. If $a_n = n-1$ (meaning the card $n-1$ is the bottom card), then $a_{n+1}=1$ because it will move to the top and the card "$1$" will be added on top of it.
With these recursions and the base case we found earlier, we calculate $a_{2000} = \boxed{927}$. To calculate this by hand, a helpful trick is finding that if $a_n=1$, then $a_{2n-1}=1$ as well. Once we find $a_{1537}=1$, the answer is just $1+(2000-1537)\cdot2$. - Frestho | 2023-02-09T02:47:47 | {
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https://fr.mathworks.com/help/symbolic/sin.html | # sin
Symbolic sine function
## Description
example
sin(X) returns the sine function of X.
## Examples
### Sine Function for Numeric and Symbolic Arguments
Depending on its arguments, sin returns floating-point or exact symbolic results.
Compute the sine function for these numbers. Because these numbers are not symbolic objects, sin returns floating-point results.
A = sin([-2, -pi, pi/6, 5*pi/7, 11])
A =
-0.9093 -0.0000 0.5000 0.7818 -1.0000
Compute the sine function for the numbers converted to symbolic objects. For many symbolic (exact) numbers, sin returns unresolved symbolic calls.
symA = sin(sym([-2, -pi, pi/6, 5*pi/7, 11]))
symA =
[ -sin(2), 0, 1/2, sin((2*pi)/7), sin(11)]
Use vpa to approximate symbolic results with floating-point numbers:
vpa(symA)
ans =
[ -0.90929742682568169539601986591174,...
0,...
0.5,...
0.78183148246802980870844452667406,...
-0.99999020655070345705156489902552]
### Plot Sine Function
Plot the sine function on the interval from $-4\pi$ to $4\pi$.
syms x
fplot(sin(x),[-4*pi 4*pi])
grid on
### Handle Expressions Containing Sine Function
Many functions, such as diff, int, taylor, and rewrite, can handle expressions containing sin.
Find the first and second derivatives of the sine function:
syms x
diff(sin(x), x)
diff(sin(x), x, x)
ans =
cos(x)
ans =
-sin(x)
Find the indefinite integral of the sine function:
int(sin(x), x)
ans =
-cos(x)
Find the Taylor series expansion of sin(x):
taylor(sin(x), x)
ans =
x^5/120 - x^3/6 + x
Rewrite the sine function in terms of the exponential function:
rewrite(sin(x), 'exp')
ans =
(exp(-x*1i)*1i)/2 - (exp(x*1i)*1i)/2
### Evaluate Units with sin Function
sin numerically evaluates these units automatically: radian, degree, arcmin, arcsec, and revolution.
Show this behavior by finding the sine of x degrees and 2 radians.
u = symunit;
syms x
sinf = sin(f)
sinf =
[ sin((pi*x)/180), sin(2)]
You can calculate sinf by substituting for x using subs and then using double or vpa.
## Input Arguments
collapse all
Input, specified as a symbolic number, variable, expression, or function, or as a vector or matrix of symbolic numbers, variables, expressions, or functions.
collapse all
### Sine Function
The sine of an angle, α, defined with reference to a right angled triangle is
The sine of a complex argument, α, is
$\mathrm{sin}\left(\alpha \right)=\frac{{e}^{i\alpha }-{e}^{-i\alpha }}{2i}\text{\hspace{0.17em}}.$ | 2020-11-25T17:54:23 | {
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https://math.stackexchange.com/questions/3077803/is-there-anything-special-with-a-3x3-matrix-where-the-3rd-row-is-0-0-1/3077807 | # Is there anything special with a 3x3 matrix where the 3rd row is 0 0 1?
I'm coding using p5.js and I'm looking at this method https://p5js.org/reference/#/p5/applyMatrix
Using that method, I can multiply my current matrix with any matrix of the form:
$$\begin{pmatrix} a & c & e \\ b & d & f \\ 0 & 0 & 1 \\ \end{pmatrix}$$
by calling applyMatrix(a, b, c, d, e, f)
There is no method for multiplying any arbitrary matrix like: $$\begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \\ \end{pmatrix}$$
Is there anything special with a matrix of that form? Is it possible to convert any arbitrary matrix (like the bottom matrix) into a matrix of that form?
• Do you have a question about math? – John Douma Jan 18 '19 at 3:34
• My question is about matrices not the coding itself, I just put the link there for context. – DarkPotatoKing Jan 18 '19 at 3:39
• Your question appears to be about some programming language. – John Douma Jan 18 '19 at 3:39
• You could fit your $3 \times 3$ matrix into the larger matrix $$\pmatrix{1&2&3&0\\4&5&6&0\\7&8&9&0\\0&0&0&1}$$ which I would say is a "matrix of that form" – Ben Grossmann Jan 18 '19 at 3:42
• @DarkPotatoKing It is used to represent affine transformations. (This is also hinted at in the page you linked.) – Alex Provost Jan 18 '19 at 3:45
It is a standard way to represent an affine transformation of the plane; this is how it is used on the page you linked. The submatrix $$A = \begin{pmatrix} a & c \\ b & d \end{pmatrix}$$ in your question represents the linear part of the affine transformation, and the extra column $$t = \begin{pmatrix} e \\ f \end{pmatrix}$$ to the right corresponds to the translation part of the transformation. In full, the corresponding transformation maps a vector $$v$$ to the vector $$Av + t$$. | 2020-10-23T22:02:49 | {
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http://math.stackexchange.com/questions/54467/the-integral-int-08-sqrtx44x2-dx?answertab=votes | The integral $\int_0^8 \sqrt{x^4+4x^2}\,dx$
$\displaystyle \int_0^8 \sqrt{x^4+4x^2}\,dx$.
Alright, so I thought I had this figured out. Here's what I did:
1. I factor out an $x^2$ to get $\sqrt{x^2(x^2+4)}$.
2. I let $x = 2\tan(\theta)$, therefore the integrand is $\sqrt{4\tan^2(\theta) (4\tan^2(\theta) + 4)}$.
3. Factor out a 4 and it becomes $\sqrt{(16\tan^2(\theta) (\tan^2(\theta) + 1))}$
4. Which equals $\sqrt{16\tan^2(\theta) \sec^2(\theta)}$
5. This is easy to take the sqrt of. The integrand becomes $4\tan(\theta)\sec(\theta)$.
6. Now, the integral of this is $4\sec(\theta)$
7. And it's evaluated from $0$ to $\arctan(4)$ right? Because as $x$ goes to $0$, so does $\theta$, and as $x$ goes to $8$, $\theta$ goes to $\arctan(4)$...
8. But the end result $(4 (\sec(\arctan(4)) - 1) )$ isn't the correct answer
I put it into WolframAlpha and I get $(8/3) (17\sqrt{17} - 1)$, which is the right answer. How did they get that? (there's no "show steps" option)
Any help is greatly appreciated!
PS, what's the syntax for doing sqrts and exponentials?
-
By the way, you can right click on LaTex code and select "Show Source" to get the correct sytanx. – JavaMan Jul 29 '11 at 16:23
You started with $x$ in the integral (in the title), then switched to $\theta$; I rewrote it so your original integral agrees with the title, and $\theta$ is the new variable. – Arturo Magidin Jul 29 '11 at 16:54
@Silver: You changed the integrand, but you didn't change the $dx$... – Arturo Magidin Jul 29 '11 at 17:04
@Silver: I think I may have told you this before, but I really appreciate it when a poster shows their work up front, uses proper spelling, and responds. – mixedmath Jul 29 '11 at 17:47
If you put braces around the argument of sqrt the whole thing goes under the root sign. I did it on step 3. – Ross Millikan Jul 29 '11 at 19:12
There are other ways of doing this integral, but let me try to fix your attempt, which is certainly a fine idea as far as it goes.
The main problem I spot with your development is that you forgot to change the $dx$ when you did the change of variable. (And you should be able to evaluate $\sec(\arctan a)$ as well; we'll get to that shortly).
So: you start with $$\int_0^8 \sqrt{x^4+4x^2}\,dx = \int_0^8 \sqrt{x^2(x^2+4)}\,dx.$$
Then you do the change of variable $x=2\tan(\theta)$. If you do this, then $$dx = 2\sec^2\theta\,d\theta;$$ when $x=0$, you want $\theta=0$, and when $x=8$ you want $\theta=\arctan(4)$ (you are correct there). So the integral actually becomes, after changing integrand, limits, and the $dx$: \begin{align*} \int_0^8\sqrt{x^2(x^2+4)}\,dx &= \int_0^{\arctan(4)}\sqrt{4\tan^2\theta(4\tan^2\theta+4)}2\sec^2\theta\,d\theta\\ &= \int_0^{\arctan(4)} \sqrt{16\tan^2\theta(\tan^2\theta+1)}2\sec^2\theta\,d\theta\\ &= \int_0^{\arctan(4)}8\sec^2\theta\sqrt{\tan^2\theta\sec^2\theta}\,d\theta\\ &= 8\int_0^{\arctan(4)}\sec^2\theta|\tan\theta\sec\theta|\,d\theta. \end{align*} Now, on $[0,\arctan(4)]$, both tangent and secant are positive, so we can drop the absolute value signs (something else you were not careful with), and the integral becomes $$8\int_0^{\arctan(4)}\sec^3\theta\tan\theta\,d\theta.$$ Set $u=\sec\theta$. Then $du=\sec\theta\tan\theta$, so we have \begin{align*} 8\int_0^{\arctan(4)}\sec^3\theta\tan\theta\,d\theta &= 8\int_{\sec(0)}^{\sec(\arctan(4))}u^2\,du\\ &= \frac{8}{3}u^3\Biggm|_{\sec(0)}^{\sec(\arctan(4))}\\ &=\frac{8}{3}\left(\sec^3(\arctan(4)) - \sec^3(0)\right). \end{align*}
Now, $\sec(0) = 1$. What about $\sec(\arctan(4))$?
Say $\psi$ is an angle with $\tan(\psi)=4$. Take a right triangle with this angle; by scaling, we may assume the opposite side has length $4$ and the adjacent side has length $1$. Then the hypotenuse has length $\sqrt{17}$, so the cosine of $\psi$ is $\frac{1}{\sqrt{17}}$, hence the secant has value $\sqrt{17}$. So $\sec(\arctan(4)) = \sec(\psi) = \sqrt{17}$. Thus, the integral is: \begin{align*} \int_0^8\sqrt{x^4+4x^2}\,dx &= \frac{8}{3}\left(\sec^3(\arctan(4)) - \sec^3(0)\right)\\ &=\frac{8}{3}\left( \sqrt{17}^3 - 1^3\right)\\ &= \frac{8}{3}\left(17\sqrt{17} - 1\right). \end{align*}
In summary: your mistake was that when you did the change of variable, you forgot to change the differential as well; and at the end you could have simplified $\sec(\arctan(4))$.
Of course, the better way of doing this is to factor out $x$ from the square root, and then recognize that you can do $$\int_0^8\sqrt{x^4+4x^2}\,dx = \int_0^8x\sqrt{x^2+4}\,dx$$ with the change of variable $u=x^2+4$, like DJC suggested. But I thought you might like to know where exactly your approach went wrong (the substitution), and whether it could be brought to a correct conclusion (it could).
-
That was a very thorough explanation! Wow... thank you :) – user13327 Jul 29 '11 at 17:28
@Silver, there's more where that came from (literally)! – The Chaz 2.0 Jul 29 '11 at 17:59
Hint:
\begin{align} \int_0^8 \sqrt{x^4 + 4x^2}dx &= \int_0^8\sqrt{x^2(x^2 + 4)}dx \\ &= \int_0^8 |x| \sqrt{x^2 + 4}dx \\ &= \int_0^8 x \sqrt{x^2 + 4} dx \end{align}
Now try a $u$-substitution.
-
Or just spot an antiderivative directly. – Geoff Robinson Jul 29 '11 at 16:24
Alright, so that works and I got the right answer that way, but why didn't my trig substitution work? @Geoff: Well, it looks a lot like an arcsec.. if it were in the form of 1 / sqrt{x^4 + 4x^2}dx, then I'd know how to turn it into an arcsec, but... I don't know how from it's current state – user13327 Jul 29 '11 at 16:33
^^^ I did, to 0 to arctan(4) from 0 to 8 – user13327 Jul 29 '11 at 16:53
@Silver: Maybe a $u$ substitution was the way to go if you can't spot an antiderivative right off, but you still seem to be trying to make it more complicated than it is. You're looking for an antiderivative for $x(x^{2}+4)^{\frac{1}{2}}$, and what's before the bracket looks an awful lot like the derivative of what's inside the bracket. – Geoff Robinson Jul 29 '11 at 17:04
@ Silver: PS: I think your mistake in your working was that you didn't remember the factor $\frac{dx}{d \theta}$. – Geoff Robinson Jul 29 '11 at 17:09
If $x=2\tan\theta$, what is $\mbox{d}x$ in terms of $\theta$?
-
oooooooooooooooooooooooh! dang – user13327 Jul 29 '11 at 17:07 | 2014-08-29T08:13:27 | {
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https://math.stackexchange.com/questions/1614176/probability-of-urns | # Probability of urns
Four identical urns each contain 3 balls. In urn one, all three balls are black; urn two, 2 black 1 white; urn three, 1 black and 2 are white; urn four, all balls are whites.
One of the urn is picked at random, and a ball is chosen from the urn, which turns out to be white. What is the probability that all 3 balls in the selected urn are white?
This might be a silly question, but I am very confused. I know that there are 4 urns in total and only one urn contains three white balls. The part which I don't understand is if the probability of getting the urn with all white balls is 1/4 and it asks that all 3 balls in urn are white? So $1/4 * 1$?
Let $U_i$ be the $i$th urn just like you described, and let $A =\{\text{Picked a white ball}\},B_i = \{\text{Pick$U_i$}\}.$ Then the event "all 3 balls in the selected urn are white" is the same as event $B_4$. Thus, if we use conditioning, the probability we are interested in is \begin{align*} P(B_4|A)&=\frac{P(B_4A)}{P(A)}\\ &=\frac{P(A|B_4)P(B_4)}{P(A)}\\ &=\frac{P(A|B_4)P(B_4)}{P(A|B_1)P(B_1)+P(A|B_2)P(B_2)+P(A|B_3)P(B_3)+P(A|B_4)P(B_4)}\\ &=\frac{1\cdot \frac{1}{4}}{0(1/4)+(1/3)(1/4)+(2/3)(1/4)+1(1/4)}\\ &=\frac{1}{2}. \end{align*}
For this question you simply have to use bayes' rule: $$P(A \mid B) = \frac{P(B \mid A) \, P(A)}{P(B)}$$
First let us see what is the probability of selecting any of the urns. Since the urns are indistinguishable and there are 4 of them, the probabilities of selecting any of the urns is: $$P(U_1) = P(U_2) = P(U_3) = P(U_4) = \frac{1}{4}$$ where $U_i$ represents the $i^{th}$ urn.
We know one urn has all black balls, a second has 2 black and 1 white, a third has 2 white and 1 black, and the final one has 3 white ones. Counting how many white balls and black balls there are and looking at their proportion relative to the total number of balls, we find that the probability of selecting a white or black ball is: $$P(ball_{white}) = P(ball_{black}) = \frac{3 +2 + 1}{4 \cdot 3} = \frac{6}{12} = \frac{1}{2}$$
Your problem asks this: "Given that the ball selected was white, what is the probability that it came from the urn containing only white balls". Without loss of generality, let us assume that $U_4$ is the urn that contains white balls only. You want to find: $$P(U_4 \mid ball_{white})$$
Now we can use bayes' rule to rewrite the problem into one that is more suitable as follows: $$P(U_4 \mid ball_{white}) = \frac{P(ball_{white} \mid U_4) \, P(U_4)}{P(ball_{white})}$$
$P(ball_{white} \mid U_4) = 1$, because if you select a ball from an urn that only contains white balls, you are guaranteed to get a white ball hence the probability is 1.
And we have already computed the probabilities $P(U_4)$ and $P(ball_{white})$ above. Substituting these values into the bayes' rule form of our problem, we get: $$P(U_4 \mid ball_{white}) = \frac{1 \cdot \frac{1}{4}}{\frac{1}{2}} = \frac{1}{4}\cdot \frac{2}{1} = \frac{1}{2}$$
Therefore, the answer to your problem is $\frac{1}{2}$.
There are six white balls, which are equally likely to be the one picked.
Three of them were in the fourth urn, so the probability it was the fourth urn is $3/6$.
(Certainly, it wasn't the first urn, so the probabilities have changed from $1/4$ each.)
We want calculate $P(U_4|W)$. But $P(U_4|W)=P(W|U_4)\frac{P(U_4)}{P(W)}...(*)$.
Now, $P(W|U_4)=1$, cause $U_4=(W,W,W)$. Also, $P(U_4)=\frac{1}{4}$ cause the urn was selected randomly.
For $P(W)$ we have:
$P(W)=P(W|U_1)P(U_1)+P(W|U_2)P(U_2)+P(W|U_3)P(U_3)+P(W|U_4)P(U_4)$.
The first summand is 0, cause $P(W|U_1)=0$ since $U_1=(N,N,N)$. On the other hand, $P(W|U_2)=\frac{1}{3}$, $P(W|U_3)=\frac{2}{3}$ and $P(W|U_4)=1$, while $P(U_2)=P(U_3)=P(U_4)=\frac{1}{4}$.
Therefore have $P(W)=\frac{1}{3}\frac{1}{4}+\frac{2}{3}\frac{1}{4}+\frac{3}{3}\frac{1}{4}=\frac{6}{12}=\frac{1}{2}$. So, in $(*)$ we have $P(U_4|W)=\frac{1}{4}\times\frac{2}{1}=\frac{1}{2}$
• That reasoning is incorrect. The fact that we have 1 white ball in from the chosen urn changes the probability. Suppose we had 20 urns, 19 with 3 black balls and 1 with 3 white balls. Having chosen an urn, and taken 1 white ball from the urn, there is 100% chance that the urn contains 3 white balls, not 1/20 (original probability of having chosen the urn with 3 white balls). – Frentos Jan 16 '16 at 8:08
• Your new answer is correct :-). – Frentos Jan 16 '16 at 8:31
• Thanks @Frentos. Indeed I forgot the Monty Hall Problem – sinbadh Jan 16 '16 at 8:32
That problem can be generalized as follows, without removing any relevant information:
There is some number of urns containing balls of various colours. Of those balls, $n$ are white. While $m$ of the white balls are in urns that contain only white balls, all other white balls are in urns that contains also balls of other colours.
If you now draw one ball at random, and that ball turns out to be white, then what is the probability that the urn contains only white balls?
Formulated that way, it is immediately obvious that the answer is $m/n$: There are $n$ ways to draw a white ball, and $m$ of those are from a white-only urn.
Note that for this question, all other details (like the number of urns, or the exact number of non-white balls, or the fact that all urns have the same number of balls, or that no two urns have identical content) are irrelevant; all that matters is that you draw each ball with the same probability.
In your case, $m=3$ (there's one urn with only white balls, and that contains 3 balls), and $n=6$ ($0+1+2+3$), thus the probability is $3/6=1/2$.
One could also make a "tree diagram" for this. There are four "branches", presumably equally probable, with each urn appearing with probability $\ \frac{1}{4} \$ from the random selection. The probability of drawing a white ball from the urn selected is $\ 0 \ \cdot \ \frac{1}{4} \ \ , \ \ \frac{1}{3} \ \cdot \ \frac{1}{4} \ \ , \ \frac{2}{3} \ \cdot \ \frac{1}{4} \ \ , \ \ 1 \ \cdot \ \frac{1}{4} \$ at the end of each "branch". So the total probability of drawing a white ball is $\ 0 \ + \ \frac{1}{12} \ + \ \frac{2}{12} \ + \ \frac{3}{12} \$ . The "all-white-ball" urn accounts for the $\ \frac{3}{12} \$ term , which is then one-half of the total. So the conditional probability that this is the urn that has been drawn from upon obtaining a white ball is $\ \frac{1}{2} \$ . (This approach is just a way of viewing what the formal calculation from the definition of conditional probability tells us.) | 2019-05-20T18:32:35 | {
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https://nbviewer.jupyter.org/github/RobotLocomotion/drake/blob/nightly-release/tutorials/linear_program.ipynb | # Linear Program (LP) Tutorial¶
For instructions on how to run these tutorial notebooks, please see the README.
## Important Note¶
Please refer to mathematical program tutorial for constructing and solving a general optimization program in Drake.
## Linear Program¶
A linear program (LP) is a special type of optimization problem. The cost and constraints in an LP is a linear (affine) function of decision variables. The mathematical formulation of a general LP is \begin{align} \min_x \;c^Tx + d\\ \text{subject to } Ax\leq b \end{align}
A linear program can be solved by many open source or commercial solvers. Drake supports some solvers including SCS, Gurobi, Mosek, etc. Please see our Doxygen page for a complete list of supported solvers. Note that some commercial solvers (such as Gurobi and Mosek) are not included in the pre-compiled Drake binaries, and therefore not on Binder/Colab.
Drake's API supports multiple functions to add linear cost and constraints. We briefly go through some of the functions in this tutorial. For a complete list of functions, please check our Doxygen.
The easiest way to add linear cost is to call AddLinearCost function. We first demonstrate how to construct an optimization program with 2 decision variables, then we will call AddLinearCost to add the cost.
In [ ]:
from pydrake.solvers.mathematicalprogram import MathematicalProgram, Solve
import numpy as np
# Create an empty MathematicalProgram named prog (with no decision variables,
# constraints or costs)
prog = MathematicalProgram()
# Add two decision variables x[0], x[1].
x = prog.NewContinuousVariables(2, "x")
We can call AddLinearCost(expression) to add a new linear cost. expression is a symbolic linear expression of the decision variables.
In [ ]:
# Add a symbolic linear expression as the cost.
cost1 = prog.AddLinearCost(x[0] + 3 * x[1] + 2)
# Print the newly added cost
print(cost1)
# The newly added cost is stored in prog.linear_costs().
print(prog.linear_costs()[0])
If we call AddLinearCost again, the total cost stored in prog is the summation of all the costs. You can see that prog.linear_costs() will have two entries.
In [ ]:
cost2 = prog.AddLinearCost(2 * x[1] + 3)
print(f"number of linear cost objects: {len(prog.linear_costs())}")
If you know the coefficient of the linear cost as a vector, you could also add the cost by calling AddLinearCost(e, f, x) which will add a linear cost $e^Tx + f$ to the optimization program
In [ ]:
# We add a linear cost 3 * x[0] + 4 * x[1] + 5 to prog by specifying the coefficients
# [3., 4] and the constant 5 in AddLinearCost
cost3 = prog.AddLinearCost([3., 4.], 5., x)
print(cost3)
Lastly, the user can call AddCost to add a linear expression to the linear cost. Drake will analyze the structure of the expression, if Drake determines the expression is linear, then the added cost is linear.
In [ ]:
print(f"number of linear cost objects before calling AddCost: {len(prog.linear_costs())}")
# Call AddCost to add a linear expression as linear cost. After calling this function,
# len(prog.linear_costs()) will increase by 1.
cost4 = prog.AddCost(x[0] + 3 * x[1] + 5)
print(f"number of linear cost objects after calling AddCost: {len(prog.linear_costs())}")
We have three types of linear constraints
• Bounding box constraint. A lower/upper bound on the decision variable: $lower \le x \le upper$.
• Linear equality constraint: $Ax = b$.
• Linear inequality constraint: $lower <= Ax <= upper$.
The easiest way to add linear constraints is to call AddConstraint or AddLinearConstraint function, which can handle all three types of linear constraint. Compared to the generic AddConstraint function, AddLinearConstraint does more sanity will refuse to add the constraint if it is not linear.
In [ ]:
prog = MathematicalProgram()
x = prog.NewContinuousVariables(2, "x")
y = prog.NewContinuousVariables(3, "y")
# Call AddConstraint to add a bounding box constraint x[0] >= 1
print(f"number of bounding box constraint objects: {len(prog.bounding_box_constraints())}")
# Call AddLinearConstraint to add a bounding box constraint x[1] <= 2
print(f"number of bounding box constraint objects: {len(prog.bounding_box_constraints())}")
# Call AddConstraint to add a linear equality constraint x[0] + y[1] == 3
linear_eq1 = prog.AddConstraint(x[0] + y[1] == 3.)
print(f"number of linear equality constraint objects: {len(prog.linear_equality_constraints())}")
# Call AddLinearConstraint to add a linear equality constraint x[1] + 2 * y[2] == 1
linear_eq2 = prog.AddLinearConstraint(x[1] + 2 * y[2] == 1)
print(f"number of linear equality constraint objects: {len(prog.linear_equality_constraints())}")
# Call AddConstraint to add a linear inequality constraint x[0] + 3*x[1] + 2*y[2] <= 4
linear_ineq1 = prog.AddConstraint(x[0] + 3*x[1] + 2*y[2] <= 4)
print(f"number of linear inequality constraint objects: {len(prog.linear_constraints())}")
# Call AddLinearConstraint to add a linear inequality constraint x[1] + 4 * y[1] >= 2
linear_ineq2 = prog.AddLinearConstraint(x[1] + 4 * y[1] >= 2)
print(f"number of linear inequality constraint objects: {len(prog.linear_constraints())}")
AddLinearConstraint will check if the constraint is actually linear, and throw an exception if the constraint is not linear.
In [ ]:
# Add a nonlinear constraint square(x[0]) == 2 by calling AddLinearConstraint. This should
# throw an exception
try:
except RuntimeError as err:
print(err.args)
If the users know the coefficients of the constraint as a matrix, they could also call AddLinearConstraint(A, lower, upper, x) to add a constraint $lower \le Ax \le upper$. This version of the method does not construct any symbolic representations, and will be more efficient especially when A is very large.
In [ ]:
# Add a linear constraint 2x[0] + 3x[1] <= 2, 1 <= 4x[1] + 5y[2] <= 3.
# This is equivalent to lower <= A * [x;y[2]] <= upper with
# lower = [-inf, 1], upper = [2, 3], A = [[2, 3, 0], [0, 4, 5]].
A=[[2., 3., 0], [0., 4., 5.]],
lb=[-np.inf, 1],
ub=[2., 3.],
vars=np.hstack((x, y[2])))
print(linear_constraint)
If your constraint is a bounding box constraint (i.e. $lower \le x \le upper$), apart from calling AddConstraint or AddLinearConstraint, you could also call AddBoundingBoxConstraint(lower, upper, x), which will be slightly faster than AddConstraint and AddLinearConstraint.
In [ ]:
# Add a bounding box constraint -1 <= x[0] <= 2, 3 <= x[1] <= 5
bounding_box3 = prog.AddBoundingBoxConstraint([-1, 3], [2, 5], x)
print(bounding_box3)
If the variables share the same lower or upper bound, you could use a scalar lower or upper value in AddBoundingBoxConstraint. For example
In [ ]:
# Add a bounding box constraint 3 <= y[i] <= 5 for all i.
print(bounding_box4)
If your constraint is a linear equality constraint (i.e. $Ax = b$), apart from calling AddConstraint or AddLinearConstraint, you could also call AddLinearEqualityConstraint to be more specific (and slightly faster than AddConstraint and AddLinearConstraint).
In [ ]:
# Add a linear equality constraint 4 * x[0] + 5 * x[1] == 1
linear_eq3 = prog.AddLinearEqualityConstraint(np.array([[4, 5]]), np.array([1]), x)
print(linear_eq3)
### Solving Linear Program.¶
Once all the constraints and costs are added to the program, we can call Solve function to solve the program and call GetSolution to obtain the results.
In [ ]:
# Solve an optimization program
# min -3x[0] - x[1] - 5x[2] -x[3] + 2
# s.t 3x[0] + x[1] + 2x[2] = 30
# 2x[0] + x[1] + 3x[2] + x[3] >= 15
# 2x[1] + 3x[3] <= 25
# -100 <= x[0] + 2x[2] <= 40
# x[0], x[1], x[2], x[3] >= 0, x[1] <= 10
prog = MathematicalProgram()
# Declare x as decision variables.
x = prog.NewContinuousVariables(4)
# Add linear costs. To show that calling AddLinearCosts results in the sum of each individual
# cost, we add two costs -3x[0] - x[1] and -5x[2]-x[3]+2
# Add linear equality constraint 3x[0] + x[1] + 2x[2] == 30
prog.AddLinearConstraint(3*x[0] + x[1] + 2*x[2] == 30) | 2021-01-26T06:30:47 | {
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https://cs.stackexchange.com/questions/143563/prove-that-the-expected-length-e-n-hk-of-the-list-containing-key-k-is | # Prove that the expected length $E [n_{h(k)}]$ of the list containing key $k$ is at most $1 + \alpha$
Theorem: Suppose that a hash function $$h$$ is chosen from a universal collection of hash functions and is used to hash n keys into a table $$T$$ of size $$m$$, using chaining to resolve collisions. If key $$k$$ is not in the table, then the expected length $$E [n_{h(k)}]$$ of the list that key $$k$$ hashes to is at most $$\alpha$$. If key $$k$$ is in the table, then the expected length $$E [n_{h(k)}]$$ of the list containing key $$k$$ is at most $$1 + \alpha$$.
Solution: (Courtesy to Introduction to Algorithms book):
We note that the expectations here are over the choice of the hash function, and do not depend on any assumptions about the distribution of the keys. For each pair $$k$$ and $$l$$ of distinct keys, define the indicator random variable $$X_{kl}=I\{h(k) = h(l)\}$$. Since by definition, a single pair of keys collides with probability at most 1/m, we have $$Pr\{h(k) = h(l)\}\le \frac{1}{m}$$, so $$E[X_{kl}] \le \frac{1}{m}$$.
Next we define, for each key $$k$$, the random variable $$Y_k$$ that equals the number of keys other than $$k$$ that hash to the same slot as $$k$$, so that
$$Y_k=\sum_{\begin{array}{c} i\in T\\ l\ne k\\ \end{array}}{X_{kl}}$$
Thus we have $$E\left[ Y_k \right] =E\left[ \sum_{\begin{array}{c} i\in T\\ l\ne k\\ \end{array}}{X_{kl}} \right] \le \sum_{\begin{array}{c} i\in T\\ l\ne k\\ \end{array}}{\frac{1}{m}}$$
So, have 2 cases here,
1. When $$k$$ is NOT table $$T$$: If $$k \notin T$$ , then $$n_{h(k)} = Y_k$$ and $$|{l : l \in T ~and~ l \ne k}| = n$$. Thus $$E [n_{h(k)}] = E[Y_k] \le n/m = \alpha$$, where $$h(k)$$ is the hash function that hashes $$k$$ to a slot. $$h(k)$$ is a simple division method.
2. When $$k$$ in table $$T$$: If $$k \in T$$ , then because key $$k$$ appears in list $$T[h_{(k)}]$$ and the count $$Y_k$$ does not include key $$k$$, we have $$n_{h(k)} = Y_k +1$$ and $$|{l : l ∈ T ~and~ l \ne k}| = n −1$$. Thus $$E[n_{h(k)}] = E[Y_k]+1 \le (n−1)/m+1 = 1+ \alpha −1/m < 1 + \alpha$$.
Problem: I would like to discuss case 1 above please. Why this is the case that we have $$E [n_{h(k)}] = E[Y_k] \le n/m = \alpha$$ please?
• I think you wrote it wrong. When $k \in T$, $E[n_{h(k)}] = E[Y_k]+1 \leq 1+\alpha$. Please edit your question. Aug 31 at 16:39
• @InuyashaYagami. I edited it. Thanks for the reply.
– Avra
Aug 31 at 17:37
Let $$I$$ be the set of input keys. We have $$|I| = n$$.
If $$k$$ is not in the table, it means that $$k \notin I$$. If $$k$$ is in the table it means that $$k \in I$$.
Let $$S_h$$ be the slot in the table that key $$k$$ maps to, if the hash function $$h$$ is selected. Our aim is to find the expected length of $$S_h$$. Note that we are taking expectation over the choice of the hash function $$h$$ picked from the family $$\mathcal{H}$$, and not the expectation over $$I$$. The set of input keys $$I$$ is fixed for this problem.
Consider the case when $$k \in I$$. In this case, there are $$n-1$$ other keys in $$I$$. Whatever be the choice of hash function $$h$$, key $$k$$ always maps to $$S_{h}$$. Therefore, it accounts to a length of $$1$$ at $$S_h$$. To find the expected number of keys in $$I \setminus \{k\}$$ that maps to the same slot as $$k$$, you simply take the sum of expectation for every key in $$I \setminus \{k\}$$. It is simply at most $$\frac{|I \setminus \{k\} | }{m}$$ using linearity of expectation as stated in your question. Therefore, the expected length of $$S_h$$ is at most $$1 + \frac{|I \setminus \{k\} | }{m} = 1 + (n-1)/m \leq 1 + n/m = 1+\alpha$$.
Similarly, you can solve for the case when $$k \notin I$$. There you simply take the sum of expectation for every key in $$I \setminus \{k\} = I$$. Therefore, the expected length of $$S_h$$ is at most $$\frac{|I| }{m} = n/m = \alpha$$.
• @Avra I did not understand your comment. You said $n_{h(k)} = Y_k + 1$ when $k \in T$, so what is the issue here? :) Aug 31 at 18:12
• @Avra $1$ should be there because $k$ is in the list $T[h(k)]$. For the remaining keys, we have $Y_k$ term. Aug 31 at 18:21
• @Avra I do not think you are making sense. The argument is not that complicated. When $k \in I$ is equivalent to saying $k \in T$. All keys are mapped at the same time. Initially the table is empty. It is just the case that key $k$ always belongs to the universe $U$ but may or may not belong to the input set $I$. Aug 31 at 18:39
• @Avra Please do not upvote or accept any answer, if you are not convinced with it. We appreciate you clarify the doubts in comments, and even if your doubts are not cleared you should not upvote the answer. Aug 31 at 18:42
• I got it now! Thank you very much. The $1$ comes from $I_{i,i}$, which is the indicator function that the value $k$ hashes to the same slot in case $k \in T$ plus the list already there, $\frac{n}{m}$, so we have $1 + \alpha$.
– Avra
Aug 31 at 20:24 | 2021-12-07T03:21:31 | {
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https://math.stackexchange.com/questions/1812387/average-amplitude-of-the-sum-of-n-sines-with-random-phase-differences?noredirect=1 | # Average amplitude of the sum of N sines with random phase differences
• We have N functions of the form $asin(kx+c)$
• Their $a$ and $k$ values are the same
• their $c$ is a random number between $0$ and $2\pi$
• $f(x)$ is the sum all N functions
I know $f(x)=Asin(kx+c')$ will also be a sine function with the same period but most likely a different amplitude $A$ and phase $c'$. $A$ can be anything between $0$ and $Na$, so $0<A^2<N^2a^2$.
But what will be the average the value of $A^2$?
I got to this problem when I was studying sound inteference. I suspect from the physics behind it the answer is $Na^2$. I can prove this result for $N=2$. I suppose the harmonic addition theorem could be helpful but I don't quite know how to get to the average.
Your answer of $Na^2$ is correct. To get there, one can notice that this is equivalent to the following:
What is the expected value of $|A|^2$ if $A$ is the sum of $n$ points uniformly distributed on the unit circle?
This follows from "complexifying" the problem. In particular, the harmonic addition formula may be derived by noting that $$\sin(x)=\text{Im}(e^{ix})$$ where $\text{Im}$ is the imaginary part of a number. Note that, since $e^{ix}$ traces out a circle in the complex plane and $\text{Im}(x)$ is a projection onto the imaginary axis, the geometric interpretation of this is that if you move an angle of $x$ away from one axis on a unit circle, then project onto the other axis, you get $\sin(x)$.
Let us ignore $k$ and $a$ for now, as they just act to scale the problem. If we take a sum of $\sin(x+c_k)$ for various $c_k$ uniformly distributed in $[0,2\pi]$, we get $$\sum_{k=1}^{N}\sin(x+c_k)=\sum_{k=1}^N\text{Im}(e^{i(x+c_k)})=\sum_{k=1}^n\text{Im}(e^{ic_k}e^{ix})=\text{Im}\left(e^{ix}\cdot \sum_{k=1}^N e^{ic_k}\right)$$ Then, if we write things in terms of sines again, we get that if $z= \sum_{k=1}^N e^{ic_k}$, then the sum evaluates to $|z|\sin(x+\text{arg}(z))$, where $\text{arg}(z)$ is the argument of $z$.
To calculate the expected value of $|z|$, we first rewrite it as $z\overline{z}$. Then, noting that $\overline{e^{ic_k}}=e^{-ic_k}$, we expand $\mathbb E(|z|^2)$ to the following: $$\mathbb E(z\overline{z})=\mathbb E\left(\left(\sum_{k=1}^Ne^{ic_k}\right)\left(\sum_{j=1}^Ne^{-ic_j})\right)\right)=\sum_{k=1}^N\sum_{j=1}^N\mathbb E(e^{i(c_k-c_j)})$$ Then, we know that $\mathbb E(e^{i(c_k-c_j)})=0$ for $k\neq j$, since $c_k-c_j$ will be uniformly distributed mod $2\pi$ so $e^{i(c_k-c_j)}$ is uniformly distributed on the unit circle, so its expectation is the center of the unit circle, which is $0$. Then $\mathbb E(e^{i(c_k-c_j)})=1$ when $k=j$, since then $e^{i(c_k-c_k)}=1$. Thus, the above have $N$ non-zero terms, all of which are $1$, so $$\mathbb E(|z|^2)=N.$$ All of this algebra just expresses the more general fact that, if you take several independently and randomly distributed vectors, each distribution having the mean at the origin, then the expected value of $|z|^2$ for their sum equals the sum of the expected values of $|v|^2$ in each distribution.
Then, to get to your problem, one just multiplies the sum of sines by $a$, giving a new amplitude of $Na^2$. The value of $k$ does not affect the amplitude, so is irrelevant in the answer.
If one wants to do it directly via the harmonic addition theorem given as (for this case) $$A^2=\sum_{i=1}^N\sum_{j=1}^Na^2\cos(c_i-c_j)$$ we can just use linearity of expectation: $$\mathbb E(A^2)=\sum_{i=1}^N\sum_{j=1}^N\mathbb E(a^2\cos(c_i-c_j)).$$ Then, again, this sum vanishes for $i\neq j$, since $c_i-c_j$ is distributed uniformly mod $2\pi$ and $\cos(\theta)$ has a mean of $0$. For $i=j$, we of course have $\cos(c_i-c_j)=1$, so the above sum has $n$ non-zero summands, each of which is $a^2$. So, the sum is $$\mathbb E(A^2)=Na^2.$$ One should note that this calculation of expectation is precisely what happens when we take the real part of the equation we derived for $|z|^2$ before - indeed, the harmonic sum theorem follows from the work in this post.
• Excellent answer! Don't you think there is a simpler way to show it by quoting the harmonic addition theorem $A^2$ = link where $A_i$and $A_j$ are always $a$ and showing that the sum is on average $Na^2$ ? – mrk1357 Jun 4 '16 at 16:14
• @mrk1357 I added a note to that effect; the proof is about the same starting from there. – Milo Brandt Jun 4 '16 at 16:36 | 2020-03-29T04:04:10 | {
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https://proofwiki.org/wiki/Image_of_Set_Difference_under_Mapping | Image of Set Difference under Mapping
Theorem
Let $f: S \to T$ be a mapping.
The image of the set difference of two subsets of $S$ is a subset of the set difference of the images.
That is:
Let $S_1$ and $S_2$ be subsets of $S$.
Then:
$f \sqbrk {S_1} \setminus f \sqbrk {S_2} \subseteq f \sqbrk {S_1 \setminus S_2}$
where $\setminus$ denotes set difference.
Corollary 1
Let $f: S \to T$ be a mapping.
Let $S_1 \subseteq S_2 \subseteq S$.
Then:
$\relcomp {f \sqbrk {S_2} } {f \sqbrk {S_1} } \subseteq f \sqbrk {\relcomp {S_2} {S_1} }$
where $\complement$ (in this context) denotes relative complement.
Corollary 2
Let $f: S \to T$ be a mapping.
Let $X$ be a subset of $S$.
Then:
$\relcomp {\Img f} {f \sqbrk X} \subseteq f \sqbrk {\relcomp S X}$
where:
$\Img f$ denotes the image of $f$
$\complement_{\Img f}$ denotes the complement relative to $\Img f$.
This can be expressed in the language and notation of direct image mappings as:
$\forall X \in \powerset S: \relcomp {\Img f} {\map {f^\to} X} \subseteq \map {f^\to} {\relcomp S X}$
That is:
$\forall X \in \powerset S: \map {\paren {\complement_{\Img f} \circ f^\to} } X \subseteq \map {\paren {f^\to \circ \complement_S} } X$
where $\circ$ denotes composition of mappings.
Corollary 3
Let $f: S \to T$ be a surjection.
Let $A \subseteq S$ be a subset of $S$.
Then:
$T \setminus f \sqbrk A \subseteq f \sqbrk {S \setminus A}$
where $\setminus$ denotes set difference.
Proof 1
As $f$, being a mapping, is also a relation, we can apply Image of Set Difference under Relation:
$\mathcal R \sqbrk {S_1} \setminus \mathcal R \sqbrk {S_2} \subseteq \mathcal R \sqbrk {S_1 \setminus S_2}$
$\blacksquare$
Proof 2
$\displaystyle y$ $\in$ $\displaystyle f \sqbrk {S_1} \setminus f \sqbrk {S_2}$ $\displaystyle \leadsto \ \$ $\displaystyle \exists x \in {S_1}: x \notin {S_2}: \tuple {x, y}$ $\in$ $\displaystyle f$ Definition of Image of Subset under Mapping $\displaystyle \leadsto \ \$ $\displaystyle \exists x \in {S_1} \setminus {S_2}: \tuple {x, y}$ $\in$ $\displaystyle f$ Definition of Set Difference $\displaystyle \leadsto \ \$ $\displaystyle y$ $\in$ $\displaystyle f \sqbrk {S_1 \setminus S_2}$ Definition of Image of Subset under Mapping
$\blacksquare$ | 2019-08-19T12:24:38 | {
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http://math.stackexchange.com/questions/454594/calculate-probability-of-obtaining-at-least-a-sequence-of-numbers-for-a-given-nu | # Calculate probability of obtaining at least a sequence of numbers for a given number of dice rolls
I want to calculate the probability of rolling at least a sequence of values for a given set of dice. This sequence may have duplicate values.
I am aware of the Binomial Probability Equation used to compute the probability of rolling a single value k times in n attempts. However, since I am looking for a sequence of values, I have not found a way to apply this theorem correctly.
Here's an example:
Let $n = 3$ fair dice, each $3$-sided. What is the probability of obtaining at least one "$1$" and one "$2$"?
There are 27 possible outcomes, of which 12 satisfy the requirement:
111 121 131 211 221 231 311 321 331 112 122 132 212 222 232 312 322 332 113 123 133 213 223 233 313 323 333
Similarly, if $n = 4$ fair dice, the probability is $\frac{50}{81}$.
Is there a more elegant method of obtaining the solution? I have a hunch it involves combining permutations with Binomial Probability but I haven't found the pattern.
-
Imagine an $s$-sided die, say with all sides equally likely. One of these sides has $1$ written on it, and another side has $2$ written on it. We toss the die $n$ times, and want the probability of at least one $1$ and at least one $2$.
Let $A$ be the event "no $1$" and let $B$ be the event "no $2$." We will compute $\Pr(A\cup B)$, This is $\Pr(A)+\Pr(B)-\Pr(A\cap B)$.
The probability of $A$ is $\left(\frac{s-1}{s}\right)^m$. The probability of $B$ is the same.
The probability of $A\cap B$ is $\left(\frac{s-2}{s}\right)^m$.
So now we know $\Pr(A\cup B)$. You are looking for $1-\Pr(A\cup B)$.
Remark: The technique we have used is a special case of the method of Inclusion/Exclusion. When we add together $\Pr(A)$ and $\Pr(B)$, we have "double counted" the situations in which both $A$ and $B$ occur. The subtraction takes care of that. Inclusion/Exclusion can get much more complicated. It is a quite useful tool.
-
So if I were to use the Inclusion/Exclusion principle with a sequence of 3 values, say {1,2,3}, it would be Pr(A)+Pr(B)+Pr(C)−Pr(A∩B) −Pr(A∩C) −Pr(B∩C) + Pr(A∩B∩C) ... and then the answer for n=3 and s=3 turns out to be 6/27 as expected. Is it then possible to use this method if duplicate values are included in the sequence? i.e. probability of getting at least 2 "1"'s and 1 "2" ? – dbn Jul 29 '13 at 8:44
Yws, it can be handled. And yes, the expression you mentioned at the beginning is the right three-term Inclusion/Rxclusion. – André Nicolas Jul 29 '13 at 9:59 | 2015-04-28T02:07:17 | {
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https://www.hpmuseum.org/forum/thread-7151.html | Short & Sweet Math Challenge #21: Powers that be
11-02-2016, 01:13 AM (This post was last modified: 11-02-2016 09:48 PM by Valentin Albillo.)
Post: #1
Valentin Albillo Senior Member Posts: 486 Joined: Feb 2015
Short & Sweet Math Challenge #21: Powers that be
Short & Sweet Math Challenge #21: Powers that be
Hi all,
At the end of April 2008 and after 20 releases I concluded my "Short & Sweet Math Challenges" series which allowed you to dust off and show off both your favorite programmable HP calcs and your math/HP-programming skills while introducing and discussing interesting and unusual mathematical topics.
The series was well received and now, some 8 years later, I'm releasing "season 2", so to say, beginning with this brand-new S&SMC #21, in the tradition and style of the preceding ones. Iif this second installment gets a similarly good reception, I intend to eventually post another full 20-strong batch.
Now as for this present new challenge, first a short prelude. Many of you will surely have heard about Phi, the ubiquitous so-called Golden Ratio constant which can be promptly evaluated as (1+Sqrt(5))/2, namely:
Phi = 1.61803+
Phi is further characterized by being the largest root in absolute value of the polynomial x^2-x-1 (i.e.: Phi^2-Phi-1 = 0), which happens to be the lowest-degree polynomial for which Phi is a root and thus it gets called the "minimal polynomial" of Phi.
As it happens, Phi has a plethora of very interesting, sometimes unique properties, one of them being that its increasing powers are ever nearer to integer values, for instance:
Phi^20 = 15126.99993+
Phi^50 = 28143753122.99999999996+
Phi^100 = 792[...]126.999999999999999999998+
Phi^200 = 627[...]126.999999999999999999999999999999999999999998+
Alas, while certainly interesting, this quasi-integer property of the powers of Phi is far from unique, there are some other constants that boast this very property, as for instance (let's call it "Hpi" for the time being):
Hpi = 1.32471+
which happens to be the largest root in absolute value of the polynomial x^3-x-1 (i.e.: Hpi^3-Hpi-1 = 0), which is its minimal polynomial. Indeed, we do have:
Hpi^20 = 276.992+
Hpi^50 = 1276942.001+
Hpi^100 = 163[...]001.9999995+
Hpi^200 = 265[...]250.000000000001+
Hpi^500 = 115[...]876.9999999999999999999999999999994+
demonstrating that Hpi's powers also get nearer and nearer to integer values, like Phi's do.
That said, now go and get a programmable HP calc of your choice (say an HP-10C or better, hardware/software based HP-calc emulators also welcomed) and use that calc's language of your choice (say RPN, FOCAL, RPL, BASIC, FORTH, etc.) to try and meet:
The Challenge:
Write a program to find and output the constants that have the aforementioned quasi-integer-powers property, subject to the requirements that your program must:
- find ALL such constants in the range 1 < constant < 2, for minimal polynomials up to degree 8 having all their coefficients equal to either +1, 0, or -1, the leading coefficient in particular being always +1.
- output both each constant AND the minimal polynomial for which it is a root, sorted by increasing numerical value, with a final tally of how many constants were found (hint: more than 30) and, optionally, timing.
Of course, the faster your program runs the better, smaller program sizes being important but secondary. If your chosen HP calc runs too slowly you might consider, in order:
. using a better algorithm and/or optimizing your solution for speed,
. coding in a faster language (say FORTH or assembler, if available),
. using a faster HP calc,
. using an HP-calc emulator running on a faster platform, or
. just be patient and have a meal (or two) while it merrily runs.
That's all. Within a few days I'll post and comment my 12-line (60-statement, 585-byte) solution for the HP-71B, which runs like this (no spoilers):
>RUN
... some lines of output ...
1.57367896839 x^8-x^7-x^6+x^2-1
1.59000537390 x^7-x^5-x^4-x^3-x^2-x-1
1.60134733379 x^7-x^6-x^4-x^2-1
1.61803398875 x^2-x-1
... more lines of output ...
(number of constants found)
I'll also comment on the underlying theory and algorithm used as well as trivia and problems I found. Now for the small print:
- you can use any ROMs, libraries or binary files for your calc as long as they are readily available, preferably for free, as well as extra RAM if needed. For instance, for the HP41 and emulators you can use the Math ROM, Advantage ROM, card reader ROM, printer ROM and many others. My HP-71B solutions typically may use the Math ROM, HP-IL ROM, JPCROM, STRNGLEX binary and additional RAM modules.
- you can use any hard/soft emulator for a given HP calc model but solutions given for non-HP calcs (say written in Excel, Visual Basic, C#, for vintage SHARP or TI machines, etc) won't be considered to have met the Challenge and in fact may spoil the fun for all.
- ditto for using the web to search for solutions. You will ruin the fun for yourself and other people attempting the challenge so you'd better try and solve it by your own efforts first, not mercilessly scavenging the web.
Enough said, now let's see your solutions ! (and please do not use CODE sections in your posts, they don't print well).
Best regards.
V.
.
Find All My HP-related Materials here: Valentin Albillo's HP Collection
11-02-2016, 05:47 AM
Post: #2
Gerson W. Barbosa Senior Member Posts: 1,258 Joined: Dec 2013
RE: Short & Sweet Math Challenge #21: Powers that be
HP 50g, 181.5 bytes, 14 seconds. Only 6 constants, though.
Best regards,
Gerson.
%%HP: T(3)A(R)F(.);
\<< 1. 6.
FOR i i X SWAP ^ [ 1. -1. -1. ] X PEVAL * [ 1. 0. -1. ] X PEVAL + EVAL DUP 'X' ZEROS DUP TYPE
IF
THEN OBJ\-> DROP NIP
END
NEXT
\>>
TEVAL
'X^3.+0.*X^2.+-1.*X-1.'
1.32471795724
'X^4.+-1.*X^3.+0.*X^2.-1.'
1.3802775691
'X^5.+-1.*X^4.+-1.*X^3.+X^2.-1.'
1.44326879127
'X^6.+-1.*X^5.+-1.*X^4.+X^2.-1.'
1.50159480354
'X^7.+-1.*X^6.+-1.*X^5.+X^2.-1.'
1.54521564973
'X^8.+-1.*X^7.+-1.*X^6.+X^2.-1.'
1.57367896839
s:14.0446
11-02-2016, 07:16 AM
Post: #3
Massimo Gnerucci Senior Member Posts: 1,919 Joined: Dec 2013
RE: Short & Sweet Math Challenge #21: Powers that be
Welcome back Valentin and S&SMC!
Greetings,
Massimo
-+×÷ ↔ left is right and right is wrong
11-02-2016, 01:48 PM (This post was last modified: 11-02-2016 01:55 PM by J-F Garnier.)
Post: #4
J-F Garnier Senior Member Posts: 377 Joined: Dec 2013
RE: Short & Sweet Math Challenge #21: Powers that be
Hi Valentin,
I'm very happy to read you again in this forum !
One question:
From the output examples here:
(11-02-2016 01:13 AM)Valentin Albillo Wrote:
1.57367896839 x^8-x^7-x^6+x^2-1
1.59000537390 x^7-x^5-x^4-x^3-x^2-x-1
1.60134733379 x^7-x^6-x^4-x^2-1
it seems that null polynomial coefficients are allowed, not only +1 or -1,
Is it correct?
J-F
11-02-2016, 02:00 PM
Post: #5
Bill (Smithville NJ) Senior Member Posts: 408 Joined: Dec 2013
RE: Short & Sweet Math Challenge #21: Powers that be
Hello Valentin,
Welcome back and thanks for another very interesting challenge.
Some of the newer forum members may not be familiar with your past challenges. Back in 2005, I assembled the first eleven of these into a single PDF file. I never did update it. You are up to 21 now. I guess I need to spend a little time getting it updated to 21.
Please see the following forum post for a link to the PDF file of the first 11:
SSMC PDF
Bill
Smithville, NJ
11-02-2016, 10:08 PM
Post: #6
Valentin Albillo Senior Member Posts: 486 Joined: Feb 2015
RE: Short & Sweet Math Challenge #21: Powers that be
.
Hi, Gerson !
I'm truly glad to get your always valuable contributions to one of my S&SMC's, as in the good old times. A few comments:
(11-02-2016 05:47 AM)Gerson W. Barbosa Wrote: HP 50g, 181.5 bytes, 14 seconds. Only 6 constants, though.
Why only 6 ? Not being versed in RPL I don't fully understand your code but I also don't see any reference to the maximum degree for the polynomials, which should be 8.
Quote:'X^3.+0.*X^2.+-1.*X-1.' 1.32471795724
'X^4.+-1.*X^3.+0.*X^2.-1.' 1.3802775691
'X^5.+-1.*X^4.+-1.*X^3.+X^2.-1.' 1.44326879127
'X^6.+-1.*X^5.+-1.*X^4.+X^2.-1.' 1.50159480354
'X^7.+-1.*X^6.+-1.*X^5.+X^2.-1.' 1.54521564973
'X^8.+-1.*X^7.+-1.*X^6.+X^2.-1.' 1.57367896839
Also, even within this limited range, your program is missing two constants, one between 1.50.. and 1.54.. and another between 1.54.. and 1.57.., perhaps due to the typo that J-F pointed out (which I duly corrected) about the coefficients being -1, 0, or +1 (the '0' was missing in my OP).
Thnaks for your continued interest and best regards.
V.
.
Find All My HP-related Materials here: Valentin Albillo's HP Collection
11-02-2016, 10:14 PM
Post: #7
Valentin Albillo Senior Member Posts: 486 Joined: Feb 2015
RE: Short & Sweet Math Challenge #21: Powers that be
Hi, Massimo !:
(11-02-2016 07:16 AM)Massimo Gnerucci Wrote: Welcome back Valentin and S&SMC!
Thank you very much for your kind welcome, much appreciated.
Perhaps you'd consider contributing your very own solution to the present challenge ? It's just a warmer, there are many interesting ones ahead !
Best regards.
V.
Find All My HP-related Materials here: Valentin Albillo's HP Collection
11-02-2016, 10:22 PM
Post: #8
Valentin Albillo Senior Member Posts: 486 Joined: Feb 2015
RE: Short & Sweet Math Challenge #21: Powers that be
Hi, Jean-François !:
Thanks a lot for your kind comment and continued interest in my posts, much appreciated.
(11-02-2016 01:48 PM)J-F Garnier Wrote: One question:
From the output examples here:
(11-02-2016 01:13 AM)Valentin Albillo Wrote:
1.57367896839 x^8-x^7-x^6+x^2-1
1.59000537390 x^7-x^5-x^4-x^3-x^2-x-1
1.60134733379 x^7-x^6-x^4-x^2-1
it seems that null polynomial coefficients are allowed, not only +1 or -1,
Is it correct?
Yes, of course, my bad, thanks for pointing it out to me, I've already corrected it in my original post.
I was going to write that the absolute value of the integer coefficients had to be up to and including 1 but decided instead to just enumerate them and the '0' was simply left out.
Thanks and here's hoping for your own solution to this challenge, it would be an amazing way to start the "second season" ! ... 8-D
Best regards.
V.
Find All My HP-related Materials here: Valentin Albillo's HP Collection
11-02-2016, 10:39 PM
Post: #9
Valentin Albillo Senior Member Posts: 486 Joined: Feb 2015
RE: Short & Sweet Math Challenge #21: Powers that be
Hi, Bill !
Thank for your kind welcome and appreciation, I'm glad you like my S&SMC's
(11-02-2016 02:00 PM)Bill (Smithville NJ) Wrote: Back in 2005, I assembled the first eleven of these into a single PDF file. I never did update it. You are up to 21 now. I guess I need to spend a little time getting it updated to 21.
It would be great if you'd eventually find the time to update your PDF file with the whole collection but you might consider waiting till the "second season" is completed instead of doing incremental updates. Whatever suits you.
Quote:Please see the following forum post for a link to the PDF file of the first 11: SSMC PDF
I already downloaded your PDF file back then and found it extremely useful to me because I was missing some of the earliest challenges and your PDF put them back in my hands in a most convenient format so I was very obliged to you for it.
If you'd like to try your hand at providing your very own solution (or any comments) to the present challenge, it would be my pleasure to have a look at it.
Best regards.
V.
Find All My HP-related Materials here: Valentin Albillo's HP Collection
11-02-2016, 11:19 PM (This post was last modified: 11-02-2016 11:34 PM by Gerson W. Barbosa.)
Post: #10
Gerson W. Barbosa Senior Member Posts: 1,258 Joined: Dec 2013
RE: Short & Sweet Math Challenge #21: Powers that be
(11-02-2016 10:08 PM)Valentin Albillo Wrote: .
Hi, Gerson !
I'm truly glad to get your always valuable contributions to one of my S&SMC's, as in the good old times. A few comments:
(11-02-2016 05:47 AM)Gerson W. Barbosa Wrote: HP 50g, 181.5 bytes, 14 seconds. Only 6 constants, though.
Why only 6 ? Not being versed in RPL I don't fully understand your code but I also don't see any reference to the maximum degree for the polynomials, which should be 8.
Quote:'X^3.+0.*X^2.+-1.*X-1.' 1.32471795724
'X^4.+-1.*X^3.+0.*X^2.-1.' 1.3802775691
'X^5.+-1.*X^4.+-1.*X^3.+X^2.-1.' 1.44326879127
'X^6.+-1.*X^5.+-1.*X^4.+X^2.-1.' 1.50159480354
'X^7.+-1.*X^6.+-1.*X^5.+X^2.-1.' 1.54521564973
'X^8.+-1.*X^7.+-1.*X^6.+X^2.-1.' 1.57367896839
Hello, Valentin!
Thanks for starting Season 2. I'm looking forward for the next episodes. I've always appreciated your insightful and well-thought S&SMC series, even though most of the time I was able to solve only the easier items.
Regarding this particular problem, I remember Phi belongs to a special set of numbers named after a French mathematician which produce near-integers when raised to high powers. I misspelled his name, but Google pointed me to the right reference wherein I found three polynomials that generate such numbers. The RPL program uses only the first generating polynomial.
PEVAL creates a symbolic polynomial expression from a variable name and a coefficents vector. For instance,
[ 1 1 1 ] 'X' PEVAL --> '1+(1+X)*x' FACTOR --> 'X²+X+1'
PROOT might be a better alternative to ZEROS, but I couldn't find a built-in inverse to PEVAL so PROOT could be used. Anyway, I guess this is not the kind of solution your looking for, so I won't proceed with this approach any longer.
(11-02-2016 10:08 PM)Valentin Albillo Wrote: Also, even within this limited range, your program is missing two constants, one between 1.50.. and 1.54.. and another between 1.54.. and 1.57.., perhaps due to the typo that J-F pointed out (which I duly corrected) about the coefficients being -1, 0, or +1 (the '0' was missing in my OP).
Yes, I was aware of the missing constants. By using the third generating polynomial in the aforementioned reference, a few more can be found:
%%HP: T(3)A(R)F(.);
\<< 3. 8.
FOR n X n ^ X n 1. + ^ 1. - X 2. ^ 1. - / - DUP 'X' ZEROS DUP SIZE GET
NEXT
\>>
TEVAL
'X^3.-(X^4.-1.)/(X^2.-1.)'
1.46557123188
'X^4.-(X^5.-1.)/(X^2.-1.)'
1.53415774491
'X^5.-(X^6.-1.)/(X^2.-1.)'
1.5701473122
'X^6.-(X^7.-1.)/(X^2.-1.)'
1.5900053739
'X^7.-(X^8.-1.)/(X^2.-1.)'
1.60134733379
'X^8.-(X^9.-1.)/(X^2.-1.)'
1.60798272793
:s: 18.1289
Not in the required format, though. Also, the polynomial have yet to be simplified and checked whether the degrees are no greater than 8.
Best regards,
Gerson.
Edited to fix a typo
11-06-2016, 05:21 PM (This post was last modified: 11-06-2016 05:22 PM by J-F Garnier.)
Post: #11
J-F Garnier Senior Member Posts: 377 Joined: Dec 2013
RE: Short & Sweet Math Challenge #21: Powers that be
(11-02-2016 10:22 PM)Valentin Albillo Wrote: ... hoping for your own solution to this challenge, it would be an amazing way to start the "second season" ! ... 8-D
Well, I tried first on Emu71, then switched to Free42 to have higher computing accuracy, but I'm afraid I wasn't able to build a proper solution with either tool.
With Emu71, I was able to find the roots, but wasn't able to identify all the constants with the desired property due to the limited accuracy.
And with Free42, I had the right accuracy, but had difficulties to find the roots of the polynomials.
J-F
11-07-2016, 10:23 PM (This post was last modified: 11-07-2016 10:23 PM by Valentin Albillo.)
Post: #12
Valentin Albillo Senior Member Posts: 486 Joined: Feb 2015
RE: Short & Sweet Math Challenge #21: Powers that be
Hi, J-F:
(11-06-2016 05:21 PM)J-F Garnier Wrote: Well, I tried first on Emu71, then switched to Free42 to have higher computing accuracy, but I'm afraid I wasn't able to build a proper solution with either tool.
With Emu71, I was able to find the roots, but wasn't able to identify all the constants with the desired property due to the limited accuracy.
How many did you identify ? How do you know they aren't all ?
For the particular limits of this challenge, i.e.: minimal polynomials up to degree 8 (or less) and with coefficients -1,0,+1, I found no accuracy problems at all (though perhaps there are and I just didn't find them ... 8-D )
Quote:And with Free42, I had the right accuracy, but had difficulties to find the roots of the polynomials.
How so ? Details ?
I'll post it within two days, give or take a day. It does take a significant amount of time to write down the solution post and regrettably it seems there wasn't much interest at all, no one but you and Gerson made any attempt at a solution or posted comments, let alone post actual code.
Best regards.
V.
Find All My HP-related Materials here: Valentin Albillo's HP Collection
11-08-2016, 08:28 AM (This post was last modified: 11-08-2016 10:11 AM by J-F Garnier.)
Post: #13
J-F Garnier Senior Member Posts: 377 Joined: Dec 2013
RE: Short & Sweet Math Challenge #21: Powers that be
(11-07-2016 10:23 PM)Valentin Albillo Wrote: How many did you identify ? How do you know they aren't all ?
I didn't attempt to count them, but I know that I didn't identify all constants with Emu71, because I missed at least one: the 1.3802775691 value that Gerson reported.
When trying to check if the powers of this value get closer and closer to an integer, I got:
I 1.3802775691^I
51 13749532.9553
52 18978171.9238
53 26195145.0089
54 36156571.0751
55 49906104.0305
56 68884275.9545
57 95079420.9637
58 131235992.039
59 181142096.07
60 250026372.026
61 345105792.99
62 476341785.031
63 657483881.103
64 907510253.132
65 1252616046.13
66 1728957831.16
67 2386441712.27
68 3293951965.42
69 4546568011.56
70 6275525842.74
...
The closest values are around power 60, then the lack of accuracy makes the fractional part no more significant.
My criteria was that the value must be close to an integer with 0.01 tolerance for 3 successive power values, to have a good confidence.
I could have relax my criteria, but what if other values had an even worst behaviour?
With Free42, I had the right accuracy, with 35 digits.
I first solved the equation x^4-x^3-1=0 with the solver to have the root X with full accuracy, then the powers (frac part) are:
FP(X^75) = 0.98147851
FP(X^76) = 0.98907003
FP(X^77) = 0.01158426
FP(X^78) = 0.01475045
FP(X^79) = 0.99622896
FP(X^80) = 0.98529899
FP(X^81) = 0.99688326
FP(X^82) = 0.01163371
FP(X^83) = 0.00786267
FP(X^84) = 0.99316166
FP(X^85) = 0.99004492
FP(X^86) = 0.00167862
FP(X^87) = 0.00954129
FP(X^88) = 0.00270295
FP(X^89) = 0.99274787
FP(X^90) = 0.99442649
FP(X^91) = 0.00396778
FP(X^92) = 0.00667073
FP(X^93) = 0.99941859
FP(X^94) = 0.99384508
FP(X^95) = 0.99781286
Here my criteria is met around power 85.
But since there is no polynomial root solver on the HP-42S (and I didn't want to write one), I had to use the equation solver, and I wasn't sure that the reported root was the correct, largest one.
Regarding the participation to the challenge, maybe others met the same difficulties.
Unless there is another, better way to solve the challenge :-)
J-F
11-08-2016, 01:38 PM (This post was last modified: 11-08-2016 01:40 PM by Paul Dale.)
Post: #14
Paul Dale Senior Member Posts: 1,576 Joined: Dec 2013
RE: Short & Sweet Math Challenge #21: Powers that be
I'd figured out an approach to this problem. I've not implemented it on any hardware.
It is essentially a brute force approach to the problem:
For each length of polynomial (1 .. 8):
• Iterate over all polynomials for the given length that satisfy the specified constraints. That the leading coefficient is 1, the constant term is ±1 and the remaining term coefficients are from {-1, 0, 1}.
• Over all lengths there are 6560 such polynomials. If the constant term can also be 0, the count is 9840.
• Find the roots of each polynomial.
• If there is one real root > 1 and all of the remaining (possibly complex) roots have |root| < 1, then the polynomial is of the required form.
11-08-2016, 02:01 PM
Post: #15
J-F Garnier Senior Member Posts: 377 Joined: Dec 2013
RE: Short & Sweet Math Challenge #21: Powers that be
(11-08-2016 12:31 PM)Mike (Stgt) Wrote: ... and - regarding the polynom as Sum from i=0 to <=8 of a(i)*x^i - the coefficient a(0) always -1.
I'm suspecting that a(0) is always -1 (for solutions of the given challenge), but I don't understand your argument.
J-F
11-08-2016, 02:03 PM
Post: #16
J-F Garnier Senior Member Posts: 377 Joined: Dec 2013
RE: Short & Sweet Math Challenge #21: Powers that be
(11-08-2016 01:38 PM)Paul Dale Wrote: If there is one real root > 1 and all of the remaining (possibly complex) roots have |root| < 1, then the polynomial is of the required form.
It may be the element I was missing, but can you explain or justify this statement?
J-F
11-08-2016, 03:15 PM
Post: #17
J-F Garnier Senior Member Posts: 377 Joined: Dec 2013
RE: Short & Sweet Math Challenge #21: Powers that be
Here are the results of my search on HP71/Emu71:
Order 2
1.61803398875 ok x^2-x-1
Order 3
1.83928675521 ok x^3-x^2-x-1
1.46557123188 ok x^3-x^2-1
1.32471795724 ok x^3-x-1
Order 4
1.92756197548 ok x^4-x^3-x^2-x-1
1.75487766625 ok x^4-x^3-x^2-1
1.61803398875 ok x^4-x^3-x-1
1.46557123188 ok x^4-x^2-x-1
Order 5
1.61803398875 ok x^5-x^4-x^3+x^2-x-1
1.32471795724 ok x^5-x^4-1
1.32471795724 ok x^5-x^2-x-1
Order 6
1.83928675521 ok x^6-x^5-x^4-x^2-x-1
1.61803398875 ok x^6-x^5-x^4+x^2-x-1
1.46557123188 ok x^6-x^5-x^4+x^3-x^2+x-1
1.61803398875 ok x^6-x^5-x^3-x-1
1.46557123188 ok x^6-x^5-x^2-1
1.46557123188 ok x^6-x^5+x^4-x^3-x^2-x-1
1.32471795724 ok x^6-x^4-x-1
Order 7
1.83928675521 ok x^7-x^6-x^5-x^4+x^3-x^2-x-1
1.61803398875 ok x^7-x^6-x^5+x^2-x-1
...
1.32471795724 ok x^8-x^6-x^2-x-1
1.32471795724 ok x^8-x^5-x^4-x-1
884 candidates (1<root<2)
59 candidates (quasi-integer powers)
Quite disappointing since I got only ... 7 unique constants.
Here is my HP71 working program:
10 ! --- SSMC21 ---
20 OPTION BASE 0 @ DIM A(10)
30 C=0 @ C2=0
40 FOR D=2 TO 8
50 DISP "Order";D
60 DIM A(D) @ COMPLEX R(D-1)
70 A(0)=1
80 A(D)=-1 ! assumed...
90 K=3^(D-1) ! numbers of coefficient combinaisons
100 FOR J=0 TO K-1
110 L=J
120 ! build the coefficients
130 FOR I=D-1 TO 1 STEP -1
140 A(I)=MOD(L,3)-1 @ L=L DIV 3
150 NEXT I
160 ! find roots of polynomia A
170 MAT R=PROOT(A)
180 ! DISP "Polynomia";J
190 ! MAT DISP A
200 ! DISP "Roots"
210 ! MAT DISP R
220 X=REPT(R(D-1))
230 IF IMPT(R(D-1))=0 AND X>1.000001 AND X<2 THEN GOSUB 300
240 ! PAUSE
250 NEXT J ! next polynomia of order D
260 NEXT D ! next order polynomiae
270 DISP C;"candidates (1<root<2)"
280 DISP C2;"candidates (quasi-integer powers)"
290 END
300 ! evaluate candidate
310 'T':
320 C=C+1
330 ! DISP "Candidate x=";X
340 ! MAT DISP A
350 F=0 ! flag candidate found
360 N=20
370 Y=X^N
380 IF ABS(FP(Y)-.5)>.49 THEN F=F+1 ELSE F=0
390 N=N+1
400 IF Y<1E10 AND N<80 AND F<3 THEN 370 ! no need to go beyong 1E10 or power 80
410 ! IF X=1.38027756910 THEN PAUSE
420 IF F<3 THEN 530
430 C2=C2+1
440 FIX 11 @ DISP X;"ok"; @ STD
450 DISP " x^";STR$(D); 460 FOR I=1 TO D-1 470 IF A(I)=1 THEN DISP "+"; 480 IF A(I)=-1 THEN DISP "-"; 490 IF A(I)<>0 THEN DISP "x"; 500 IF A(I)<>0 AND D-I<>1 THEN DISP "^";STR$(D-I);
510 NEXT I
520 DISP STR\$(A(D))
530 ! PAUSE
540 RETURN
11-08-2016, 05:24 PM
Post: #18
J-F Garnier Senior Member Posts: 377 Joined: Dec 2013
RE: Short & Sweet Math Challenge #21: Powers that be
(11-08-2016 03:29 PM)Mike (Stgt) Wrote: And for a(0) = +1, well... then the restriction of the leading coefficient to +1 is thwarted as we look for the maximum _value_ of the root in the range ]1..2[. No?
Well, not necessarily. For instance, 1.32471795724 (one of the constants) is a solution of x^4-x^3-x^2+1.
Anyway, in my code above I assumed a(0)=-1:
80 A(D)=-1 ! assumed...
because all examples posted by Valentin and Gerson are so :-)
J-F
11-08-2016, 06:52 PM
Post: #19
Gerson W. Barbosa Senior Member Posts: 1,258 Joined: Dec 2013
RE: Short & Sweet Math Challenge #21: Powers that be
(11-08-2016 02:03 PM)J-F Garnier Wrote:
(11-08-2016 01:38 PM)Paul Dale Wrote: If there is one real root > 1 and all of the remaining (possibly complex) roots have |root| < 1, then the polynomial is of the required form.
It may be the element I was missing, but can you explain or justify this statement?
J-F
The following is a test for 4-th order polynomials, using Pauli's and Mike's conditions. I have assumed the solutions given by PROOT are ordered by magnitude, but I am not sure about that. But this implementation is probably wrong since it gives only two solutions (three when the program is modified for 3-rd order polynomials).
Hopefully no typing errors since the EMU71 version I have doesn't work on Windows 10 64-bit and I don't know how to copy and paste the listing in DosBox.
------------------
3 DESTROY ALL
5 INTEGER A,B,C,D,E,N,T
7 A=1 @ E=-1
10 OPTION BASE 1
15 INTEGER P(5)
20 COMPLEX R(4)
25 FOR B=-1 TO 1
30 FOR C=-1 TO 1
35 FOR D=-1 TO 1
40 P(1)=A @ P(2)=B @ P(3)=C @ P(4)=D @ P(5)=E
45 MAT R=PROOT(P)
50 IF IMPT(R(4))=0 THEN GOSUB 1000
55 NEXT D
60 NEXT C
70 NEXT B
75 END
1000 IF REPT(R(4))>1 THEN GOSUB 2000
1005 RETURN
2000 T=0
2005 FOR N=1 TO 3
2010 IF ABS(R(N))<1 THEN T=T+1
2015 NEXT N
2020 IF T=3 THEN PRINT REPT(R(4));A;B;C;D;E
2025 RETURN
>RUN
1.92756197548 1 -1 -1 -1 -1
1.3802775691 1 -1 0 0 -1
--------------
3-rd order polynomial solutions:
1.83928675521 1 -1 -1 -1
1.46557123188 1 -1 0 -1
1.32471795721 1 0 -1 -1
-------------
11-08-2016, 10:24 PM
Post: #20
Dwight Sturrock Member Posts: 125 Joined: Dec 2013
RE: Short & Sweet Math Challenge #21: Powers that be
(11-08-2016 01:38 PM)Paul Dale Wrote: I'd figured out an approach to this problem. I've not implemented it on any hardware.
It is essentially a brute force approach to the problem:
For each length of polynomial (1 .. 8):
• Iterate over all polynomials for the given length that satisfy the specified constraints. That the leading coefficient is 1, the constant term is ±1 and the remaining term coefficients are from {-1, 0, 1}.
• Over all lengths there are 6560 such polynomials. If the constant term can also be 0, the count is 9840.
• Find the roots of each polynomial.
• If there is one real root > 1 and all of the remaining (possibly complex) roots have |root| < 1, then the polynomial is of the required form.
My approach is similar to Paul's. One subroutine sequentially cycles through all coefficients - 1, 0, -1, starting with 1 1 1 1 1 1 1 1 1. The other routine uses the solver to check for roots in the proscribed range. Coded for the 15C, the routines appear to work independently, but not sure how they will behave together across the whole range.
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https://math.stackexchange.com/questions/1588344/complex-roots-of-a-polynomial | # Complex roots of a polynomial
Consider the equation: $(x-2)^6 + (x-4)^6 = 64$ This equation has two real roots $a, b$ and two pairs of complex conjugate roots $(p, q)$ and $(r, s)$. Is $p + q = r + s$ or $pq=rs$?
The trivial real roots of the equation are $(2,4)$. My question is, when there are two complex conjugate pairs of a polynomial, do the magnitude of the roots have any sort of relation? Is trivially trying to find them by substituting $x=a+ib$ (with $a=1$ in this case so that there remain no imaginary part on the LHS) the only method? This approach leads to an equation of the sixth degree in $b$.
Let's substitute $y = x-3$. Then, the equation becomes:
$$(y+1)^6+(y-1)^6 = 64$$
Expanding the left side using the Binomial Theorem and doing a bit of simplification gives us:
$$(y^6+6y^5+15y^4+20y^3+15y^2+6y+1)+(y^6-6y^5+15y^4-20y^3+15y^2-6y+1) = 64$$
$$2(y^6+15y^4+15y^2+1) = 64$$
$$y^6+15y^4+15y^2-31 = 0$$
We know that $x = 2$ and $x = 4$ are roots of the original equation, so $y = \pm 1$ are roots of the new equation. Hence, $y^2-1$ is a factor of $y^6+15y^4+15y^2-31$. Factoring this out gives us:
$$(y^2-1)(y^4+16y^2+31) = 0$$
So the complex roots are the roots of $y^4+16y^2+31$. By completing the square, we have:
$$y^4+16y^2+31 = 0$$
$$y^4+16y^2+64 = 33$$
$$(y^2+8)^2 = 33$$
$$y^2+8 = \pm\sqrt{33}$$
$$y^2 = -8\pm\sqrt{33}$$
$$y = \pm i \sqrt{8\pm \sqrt{33}}$$
Therefore, the complex roots to the original equation are $x = 3 \pm i \sqrt{8\pm \sqrt{33}}$.
By setting $p = 3+i\sqrt{8+\sqrt{33}}$, $q = 3-i\sqrt{8+\sqrt{33}}$, $r = 3+i\sqrt{8-\sqrt{33}}$, $s = 3-i\sqrt{8-\sqrt{33}}$, we see that $p+q = 6 = r+s$ but $pq = 17+\sqrt{33} \neq 17-\sqrt{33} = rs$. Therefore, $p+q=r+s$ is true while $pq=rs$ is false.
• If I only have paper and pen in the exam, how do I find the roots? – Sat D Dec 25 '15 at 4:35
• ^Oh, are you just trying to find the roots of the polynomial? Your question sounded like you wanted to know if there were any special relationships between the roots. – JimmyK4542 Dec 25 '15 at 4:36
• I believed that there existed a special relationship WHICH would give me the roots. The main question is finding which is correct, P+Q = R + S or PQ = RS – Sat D Dec 25 '15 at 4:38
• Ahh, so the problem asked you which of the two relationships were true for that specific polynomial, and you wanted to know how to figure that out. The second part of your question made it seem like you wanted to know if either relationship held for an all polynomials with 2 pairs of complex conjugate roots. I'll edit my answer so that it answers your actual question. – JimmyK4542 Dec 25 '15 at 4:55
Remark: if you want $p+q=r+s$ and $pq=rs$ to hold simultaneously, then you cannot have 2 distinct complex conjugate pairs. This is because if
$$p = a+bi, \ q=a-bi$$ and $$r = c+di, \ s=c-di,$$ then $$p+q=r+s \implies a=c,$$ and $$pq=rs \implies a^2+b^2=c^2+d^2,$$ and so $b=\pm d$.
• So which one of the conditions is applicable for the equation, and how? – Sat D Dec 25 '15 at 4:32
• @SatD Is the original question that you posted the whole question, or is it part of something else? When we use "and" in math we usually want both conditions to hold, and in this case they can't, unless the complex roots have multiplicity 2. – GaussTheBauss Dec 25 '15 at 4:39
• I apologize for the 'and'. It's an 'or'. – Sat D Dec 25 '15 at 4:41
• @SatD I think you changed the wrong "and" to an "or". You should change the one between the equations, not the pair of points. – GaussTheBauss Dec 25 '15 at 4:42 | 2019-07-19T14:55:06 | {
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http://math.stackexchange.com/questions/123684/how-does-one-prove-that-the-number-111-ldots-1-formed-by-3n-digits-all-e | # How does one prove that the number $111\ldots 1$ (formed by $3^{n}$ digits all equal to $1$) is divisible by $3^{n}?$
I am self-studying Discrete Mathematics (in Portuguese), and there is one exercise I was not able to solve.
Show that the number $111\ldots 1$ (formed by $3^{n}$ digits are equal to $1$) is divisible by $3^{n}.$
All I was able to do was to show the base case, but I do not know how to use the inductive step.
-
What is the value of $n$? – Martin Argerami Mar 23 '12 at 17:52
$n$ is natural! Meu colega! – checkmath Mar 23 '12 at 17:55
@RossMillikan: You are right! Sorry, I am ADHD. I mean it! – spohreis Mar 23 '12 at 18:15
@spohreis: No worries. – Ross Millikan Mar 23 '12 at 18:18
I’m sure that you meant to say that the number was formed by $3^n$ digits equal to $1$, not $3^3$ digits.
Let $m_n$ be the number whose decimal representation is a string of $3^n$ $1$’s. To be explicit, $m_n=\frac19(10^{3^n}-1)$. Suppose that $3^n$ divides $m_n$ for some $n$, say $m_n=3^nk$. Then \begin{align*} m_{n+1}&=10^{3^n}\left(10^{3^n}m_n+m_n\right)+m_n\tag{1}\\ &=\left(10^{2\cdot3^n}+10^{3^n}+1\right)m_n\\ &=\left(10^{2\cdot3^n}+10^{3^n}+1\right)3^nk\;, \end{align*}
and we need only show that $3$ divides $10^{2\cdot3^n}+10^{3^n}+1$. But $10\equiv 1\pmod 3$, so every power of $10$ is congruent to $1$ mod $3$, and therefore $10^{2\cdot3^n}+10^{3^n}+1$ is indeed a multiple of $3$.
If the step $(1)$ is a little puzzling at first, look at an example:
$$m_2=111111111=111000000+111000+111=1000(1000m_1+m_1)+m_1\;.$$
-
I think you meant
For $n\geq 1$ Show that the number $111 \cdots 1$ (formed by $3^n$ 1's) is divisible by $3^n$.
Base case is clear.Denote let $A_m:= \underbrace{111 \cdots 1}_{3^m}$. Suppose the result is true for integers $n=k$, then $$A_{k+1}= \frac{10^{3^{k+1}}-1}{9}=\frac{(10^{3^k})^3-1}{9}=\frac{10^{3^k}-1}{9}(10^{2\cdot3^n}+10^{3^n}+1)=A_k\times S$$
where $S=10^{2\cdot3^n}+10^{3^n}+1$, since $S$ is a multiple of $3$ the conclusion follows.
-
Hint $\$ By below, $\rm\:\ 3^k\: |\: f_n\ \Rightarrow\ 3^{k+1}\: |\: f_{n+1}.\:$ Yours is the special case $\rm\:c = 3.$ $$\rm\displaystyle\ f_n = \frac{(3c\!+\!1)^{3^n}\!-1}{3c}\ \ \Rightarrow\ \ f_{n+1} = \frac{(3c\:\!f_n\! +\! 1)^3-1}{3c}\: =\: \frac{9c\:f_n\:\!m}{3c}\: =\: 3\:f_n m$$ | 2014-03-13T22:15:09 | {
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https://math.stackexchange.com/questions/3992322/monty-hall-problem-with-multiple-players | # Monty Hall Problem with Multiple Players?
I understand the common Monty Hall Problem and why switching provides a 2/3 chance of winning, but I'm having trouble wrapping my head around how the probabilities work when multiple players are involved, as their probabilities seem to be contradictory.
Let's say we have two players and four doors. Each door has a 1/4 probability of hiding a prize. Let's say contestant 1 chooses door A and contestant 2 chooses door B. Door D is then revealed to be empty, leaving A, B, and C. From contestant 1's perspective the odds are 1/4 for door A, 3/8 for door B, and 3/8 for door C. But from contestant 2's perspective the odds are 3/8 for door A, 1/4 for door B, and 3/8 for door C.
What are the actual odds for each door hiding the prize, does it change depending on if they know each other's choices, and does it differ from the normal example of the problem? If so why, and if not, why not?
• A and B should both stay 1/4 while C should be 1/2, at least under the usual assumptions about the hosts behavior. 1/2 the time the car will not be selected by either contestant, and in that situation you will win if you adopt an always-switch strategy.
– guy
Jan 20, 2021 at 3:12
• Is that assuming both players are aware of each other's choices? Jan 20, 2021 at 3:14
• You might want to clarify whether the contestants are both aware of everything or not, especially if not. Jan 20, 2021 at 3:21
• I added that to the question. Jan 20, 2021 at 3:30
• For Monty, the probability is $1$ for one of the doors are $0$ for all others, which also is different from any contestant's probability calculations, but that usually doesn't bother us. Jan 20, 2021 at 4:14
In the case where both contestants can see everything, it's more or less the same as a single contestant getting to select two doors. The probabilities under both doors effectively freeze, and the the $$1/4$$ for door D flows to door C, giving C a $$1/2$$ chance of having the prize, leaving A and B at their original $$1/4$$.
In the case where contestants can't see what door the other contestant picks, then you have the situation where their door has $$1/4$$ chance, and the other two remaining doors both have $$3/8$$. It's fine that both contestants have different probabilities for the doors; it results from both the contestants having different information about the probability distribution, since only they know which door they selected.
• Ok, let's see if I understand it now. So if there was then a 3rd person observer who knew both of their choices, then they would know that the probability in my original example would be 1/2 for C, and then from each of the player's perspectives the 3/4 chance of it not being the one they originally chose would be divided unevenly (without their knowledge) as 1/2 for C and 1/4 for the other one they didn't choose. But the 1/4 and 1/2 still adds up to 3/4, leaving the odds properly distributed. Is that correct? Jan 20, 2021 at 14:36
If the players know what the other chose then the options are (and the actual probabilities that we know):
Prize Behind Door A: C is shown. $$1$$ in $$8$$. (Didn't happen)
Prize Behind Door A: D is shown. $$1$$ in $$8$$.
Prize Behind Door B: C is shown. $$1$$ in $$8$$. (Didn't happen)
Prize Behind Door B: D is shown. $$1$$ in $$8$$.
Prize Behind Door C: D is shown. $$1$$ in $$4$$.
Prize Behind Door D: C is shown. $$1$$ in $$4$$. (Didn't happen).
Of the ones that happened. Prize behind door A, or B are each $$1/4$$ and behind $$C$$ is $$1/2$$.
But if the players don't know what the other picked then from Player 1s perspective the probability as he knows is:
Prize behind Door A; other player choose A: D (one of 3 options) is shown: 1 in 48
Prize behind Door A; other player choose B: D (one of 2 options) is shown: 1 in 32
Prize behind Door A; other player choose C: D (one of 2) is shown: 1 in 32
Prize behind Door B; other player choose A; D (one of 2 options) is shown: 1 in 32
Prize behind Door B; other player choose B; D (one of 2 options) is shown:1 in 32
Prize behind Door B; other player choose D; D (only option) is shown: 1 in 16
Prize behind Door C; other player choose A; D (one of 2 options) is shown: 1 inn 32 Prize behind Door C; other player choose B; D (only option) is show: 1 in 16
Proze behind Door C; other chose C; D (one of two) is shown: 1 in 32
So the weighted options are A: 1 in 4
B: 3 in 8
C: 3 in 8
Why that differs from the actual probability is simply the player is lacking information we have.
Knowing information always changes probability.
• Ok, let's see if I understand it now. So if there was then a 3rd person observer who knew both of their choices, then they would know that the probability in my original example would be 1/2 for C, and then from each of the player's perspectives the 3/4 chance of it not being the one they originally chose would be divided unevenly (without their knowledge) as 1/2 for C and 1/4 for the other one they didn't choose. But the 1/4 and 1/2 still adds up to 3/4, leaving the odds properly distributed. Is that correct? Jan 20, 2021 at 14:31 | 2022-07-05T11:59:18 | {
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https://glassbox.tv/south-river-isfwre/3ae2aa-align-equations-latex | Use the ampersand character &, to set the points where the equations are vertically aligned. Using the multiline, aligned packages. Inside the equation environment, use the split environment to split the equations into smaller pieces, these smaller pieces will be aligned accordingly. In large equations or derivations which span multiple lines, we can use the \begin{align} and \end{align} commands to correctly display the aligned mathematics. Open an example of the amsmath package in Overleaf You can use flalign environment to get the equations flush with the margin, and the space precedding the flalign can be adusted by changing \abovedisplayskip:. As you see above, you can leave some columns blank. \begin{aligned} Mathematical Equations in LaTeX. Some of these equations include cases. LaTeX provides a feature of special editing tool for scientific tool for math equations in LaTeX. The align and align* environments, available through the amsmath package, are used for arranging equations of multiple lines. In this article, you will learn how to write basic equations and constructs in LaTeX, about aligning equations, stretchable horizontal lines, operators and ⦠As with matrices and tables, \\ specifies a line break, and & is used to indicate the point at which the lines should be aligned. The environment cases inside align results in that domains are not aligned at the same position. Example \begin{align} a_i &= \begin{dcases} b_i & i \leq 0 \\ c_i & i < 0 \end{dcases} \\ Related articles. Can I write a LaTeX equation over multiple lines? This is needed on at least one of the lines. Note: Note the trailing & in the flalign environment. As a style issue, notice that we start a new line in our source file after each \\. 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https://brilliant.org/discussions/thread/should-you/ | # Should you ?
(a) I give you an envelope containing a certain amount of money, and you open it. I then put into a second envelope either twice this amount or half this amount, with a fifty-fifty chance of each. You are given the opportunity to trade envelopes. Should you?
(b) I put two sealed envelopes on a table. One contains twice as much money as the other. You pick an envelope and open it. You are then given the opportunity to trade envelopes. Should you?
Note by Harnakshvir Singh Dhillon
3 years, 4 months ago
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## Comments
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These are variations of the Two envelopes problem, which has received a surprising amount of analysis over the years. I posted a note on the subject several months ago, which can serve as an introduction to the supposed paradox.
- 3 years, 3 months ago
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This reminds me of the Monty Hall problem which simply blew my mind the first time I learnt about it. For me, on both occasions, I would just choose randomly or based on the person's poker face because no new information came up to alter the probabilities. As far as I can see, there's equal probability for loss and gain at all times. So there's no advantage or disadvantage in switching or staying
- 3 years, 3 months ago
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Yes, that's the most practical approach. If no useful information is provided, (and the information provided in version (a) is of no practical use), then there is no reason to switch. In the Monty Hall problem we are provided with useful information and thus have a logical reason to switch so as to improve our odds of winning. It is edifying though to look further into the two envelopes problem and see why certain seemingly convincing perspectives are flawed, (which I deal with (somewhat simplistically) in the note I've linked to in my previous comment). It's also interesting to consider the problem with more, (and even infinite), envelopes, with different multiples of money in each envelope.
- 3 years, 3 months ago
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The subject of your note is whit in the scope of Game Theory. The choice depend of the criteria of the gambler then if he choose to keep the envelope he will be playing under a maximin policy since he is guaranteed the best of the worst scenario. I would recommend a textbook on Game Theory you may find resources in the web
- 3 years, 3 months ago
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https://mathematicsgre.com/viewtopic.php?f=1&t=646 | ## 0568 #48
Forum for the GRE subject test in mathematics.
juliaf
Posts: 2
Joined: Sun Oct 09, 2011 5:43 pm
### 0568 #48
Dear everyone,
Problem 48 says:
Consider the theorem: If f and f' are both strictly increasing real-valued functions on (0,infty), then $$\lim_{x\to\infty} f(x) = \infty$$. The following argument is suggested as a proof of this theorem.
(1) By the Mean Value Theorem, there is a $$c_1$$ in the interval (1,2) such that
$$f'(c_1) = \frac{f(2) - f(1)}{2-1} = f(2) - f(1) > 0$$
and then a bunch more steps.
According to the answer key, this is a valid argument. I must be missing something huge here though, because I was under the impression that the mean value theorem required f(x) to be continuous on [a,b] and f'(x) to be continuous on (a,b). Here all we have is that the functions are strictly increasing. What am I doing wrong?
talkloud
Posts: 17
Joined: Thu Apr 28, 2011 9:44 pm
### Re: 0568 #48
The fact that f' exists on (0, infty) implies that f is continuous there.
I missed this on the first pass too.
juliaf
Posts: 2
Joined: Sun Oct 09, 2011 5:43 pm
### Re: 0568 #48
Okay, thanks for that. I think what you said is half of it -- the other half is that if f' exists everywhere on (a,b), it must be continuous on (a,b). Does that seem right?
owlpride
Posts: 204
Joined: Fri Jan 29, 2010 2:01 am
### Re: 0568 #48
if f' exists everywhere on (a,b), it must be continuous on (a,b). Does that seem right?
It is if f' is monotone increasing. Monotonicity implies that the left and right limits of f'(x) as x -> x_0 exist - the only thing that could go wrong is that the left limit does not equal the right limit; but if that was the case, then f' would not be defined at x_0 at all.
In general, the existence of f' does not imply its continuity though. Here is a counterexample: http://planetmath.org/encyclopedia/Exam ... nuous.html
One more question: does the Mean Value Theorem actually require f' to be continuous? Wikipedia only uses differentiability as hypothesis. | 2021-08-03T13:03:48 | {
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https://www.physicsforums.com/threads/integration-over-a-discontinuity.672079/ | Integration over a discontinuity
1. Feb 16, 2013
nrivera1
1. The problem statement, all variables and given/known data
Evaluate ∫(1/x)sin^2(x)dx from -a to a
3. The attempt at a solution
Mathematica doesn't want to evaluate this because of the lack of convergence.
I think it is zero. When we consider non-zero values of x in the associated riemann sum, the integrand is odd and so contributions from x and -x cancel out. I can make the Δx in the riemann sum arbitrarily small, so when we consider the step near zero, sin^2 x is almost exactly equal to x^2 and that term in the sum gets replaced to xΔx where x is very nearly zero as is delta x. In other words, the integral is zero.
Is this kind of reasoning valid?
2. Feb 16, 2013
antibrane
Yea, it is zero by symmetry. The integrand is odd (it obeys $f(-x)=-f(x)$) so the "negative area" accumulated from -a to 0 cancels out the "positive area" accumulated from 0 to a.
3. Feb 16, 2013
jbunniii
Yes, your reasoning is OK. More formally, we may write the integrand as
$$\frac{\sin^2(x)}{x} = \left(\frac{\sin(x)}{x}\right)\sin(x)$$
Both factors on the right hand side are continuous on [-a,a] (we continuously extend $\sin(x)/x$ to equal 1 at $x = 0$), so their product is also continuous, hence integrable. Therefore, since the product is an odd function and we are integrating over [-a,a], the result is zero.
4. Feb 16, 2013
lurflurf
since f(0-)=f(0+)=0
this is no more of a problem than x^2/x would be
in fact x~sin(x) near zero
5. Feb 16, 2013
haruspex
That's not enough by itself. You also have to show that the integral is bounded on all subintervals, otherwise the symmetry argument might equate to saying ∞-∞ = 0. Other posts in this thread cover that.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook | 2017-08-23T14:39:08 | {
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https://math.stackexchange.com/questions/915345/a-question-on-the-order-of-an-element-involving-relatively-primes | # A question on the order of an element involving relatively primes
This question is based on an exercise that comes from the second chapter of Malik's Fundamentals of abstract algebra which states as follows (I paraphrase):
Let $(G, *)$ be a group and $x\in G$. Suppose $\circ(x) = n = n_1n_2\cdots n_k$, where for all $i\neq j$, $\gcd{(n_i, n_j)} = 1$ (that is, $n_i$ and $n_j$ are relatively primes). Show that there exists $x_i$ such that $\circ(x_i) = n_i$ for all $i = 1, 2, \dots, k$, $x=x_1*x_2*\cdots*x_k$ and $x_i*x_j = x_j*x_i$ for all $i$ and $j$.
In my attempt, I manage to get only so far:
Proof: Let's try and define $x_i$ as a power of $x$ (that is, let $x_i = x^{p_i}$). If we want $\circ(x_i) = \circ(x^{p_i}) = n_i$, we need $$\circ(x^{p_i})=\frac{n}{\gcd{(n,p_i)}}=n_i$$ Which can be achieved by defining $$p_i := \frac{n}{n_i}, p_i \in \mathbb{Z}$$
Because $*$ is closed in $G$, we know $x_i \in G$. And, also, $$x_i*x_j = x^{p_i}*x^{p_j} = x^{p_i + p_j} = x^{p_j + p_i} = x^{p_j}*x^{p_i} = x_j*x_i$$
But I fail to demonstrate that $x=x_1*x_2*\cdots*x_k=x^{\frac{n}{n_1}}*x^{\frac{n}{n_2}}*\cdots*x^{\frac{n}{n_k}}$. I notice that it is as easy as demonstrating that $$\frac{n}{n_1} + \frac{n}{n_2} + \cdots + \frac{n}{n_k} = \alpha n + 1, \alpha \in \mathbb{Z}$$ Because then $x^{\alpha n + 1} = x^{\alpha n}*x = e*x = x$.
So my questions are:
• Am I on the right road?
• Any tips on proving that $x= x_1*x_2*\cdots*x_k$?
EDIT: Let $n=6$, for example. We can write $6$ as $2\cdot 3$, and $\frac{6}{2}+\frac{6}{3} = 3 + 2 = 6 \neq \alpha6+1$. So it looks like I'm not on the right track.
• If it is true for any group, it must also be true for the cyclic group generated by x, which implies that powers of x are the only good candidates for the x_i. Which means you are certainly on the right track and moreover if you have insight into the subgroup structure of cyclic groups or use the fundamental theorem of f.g. abelian group, the result rolls out easily. (Although I suspect the purpose of this question is not to use that.) – Myself Aug 31 '14 at 22:27
• @Myself Exactly, I'm just starting the course. In the next lecture we'll only start talking about cyclic groups and Lagrange's theorem. – Miguelgondu Aug 31 '14 at 22:28
• Oh, btw, chinese remainder theorem might help here. To verify that $x = y \pmod n$, it it sufficient that $x = y \pmod{n_i}$ where $n = \prod_i n_i$ with the n_i coprime. – Myself Aug 31 '14 at 22:29
• @Myself That sounds useful indeed. Thanks. – Miguelgondu Aug 31 '14 at 22:29
(This got too long for the comment I was writing)
This is quite good. You just need to be a bit more careful with how you use notation, you don't even define these $p_i$ until later.
By the definition of order, each $x_i$ has order $n_i$, and since the $x_i$ commute, the order of their product is divisible by $n$, and is therefore equal to $n$. If you cannot see why $x_i*x_j=x_j*x_i$ note that both products are equal to $x^{p_i+p_j}=x^{p_j+p_i}$.
Now this doesn't quite give you $x=x_1*x_2*\ldots *x_k$, in fact, all you can really conclude with that is that if $\pi=x_1*x_2*\ldots *x_k$, then since $\langle\pi\rangle\subseteq\langle x\rangle$ and they have the same order, $\pi$ generates $\langle x\rangle$, not necessarily that $x=\pi$, but for that you can use the extended Euclidean algorithm to get tweaks of the $x_i$ with the same order properties such that the product is $x$.
In a specific example, let
$$G=\Bbb Z/3\Bbb Z\times\Bbb Z/2\Bbb Z\;(\cong \Bbb Z/6\Bbb Z)$$
then use the generator $x=(1,1)$ and $n_1=2, n_2=3$ so that $x_1=(0,1), x_2=(-1,0)$ then
$$x_1*x_2=(-1,0)+(0,1)=(-1,1)\ne x$$
since the group operation is component-wise addition.
On the other hand $x_1^1*x_2^2=x$, so that you can modify $x_1, x_2$ to produce $x_1', x_2'$ so that $x_1*x_2=x$.
• Thanks, but $\gcd{(p_i, n)}$ would equal $n_i$, and the order would be $\frac{n}{n_i}$, not $n_i$. Or am I just confused? – Miguelgondu Aug 31 '14 at 22:24
• @Miguelgondu oh duh, sorry. I did it backwards, and after thinking you were right from the start on that bit. I'll fix that. – Adam Hughes Aug 31 '14 at 22:26
The answer to your first question is : Yes, your reasonment is perfectly correct. The answer to your second question is: the statement $x=x_1*x_2*\ldots *x_k$ is erronous if we assume that the number of $x_i$ equals the number of $n_i$. Consider the case where $\circ(x)=6$ then $n=n_1n_2$ where $n_1=2$ and $n_2=3$. As you stated correctly $x_1=x^3$ has order $2$ and $x_2=x^2$ has order 3. But $x \neq x_1x_2$ since that would mean that $x=x^3x^2=x^5$ which would imply that $x$ has order $2$ and not order $6$. But what we do have, resulting from Bézouts lemma is that $-1.3+2.2=1$ which implies $x=x_1^{-1}x_2x_2$, where each factor has order $2$ or $3$. | 2020-05-27T06:12:59 | {
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https://math.stackexchange.com/questions/2828154/are-higher-order-b%C3%A9zier-curves-envelopes | Are higher-order Bézier curves envelopes?
I only realized from this question and the answers to it that quadratic Bézier curves are the envelopes of the lines used to compute them iteratively. That is, if a quadratic Bézier curve for points $P_0$, $P_1$, $P_2$ is constructed by linearly interpolating between $P_0$ and $P_1$ and between $P_1$ and $P_2$ and then again linearly interpolating along the line between the two resulting points, then the Bézier curve is the envelope of those lines.
I'm wondering whether this generalizes to higher Bézier curves. An $n$-th order Bézier curve can be analogously constructed by $n$ iterations of linear interpolation. Is it the envelope of any of the lines used in that construction? Or of the higher-order curves that are generated in the intermediate stages of the construction? Or both? If so, how can we see this most easily?
An answer for cubic Bézier curves would already be interesting, even if it doesn't generalize to higher orders.
(I tried searching for “Bézier” and “envelope” but didn't find anything useful, partly because the top hits are all about an Inkscape extension called Bezier Envelope.)
• Yes is the answer en.wikipedia.org/wiki/Bézier_curve#Higher-order_curves Jun 22 '18 at 15:43
• @Coolwater: I read that article, and I don't see where it says that. It shows that higher-order Bézier curves can be constructed by iterated interpolation, as I mentioned in the question. It doesn't mention envelopes at all, as far as I can tell. The animated diagrams certainly look as if the curves might be envelopes, but that might be a wrong impression. Jun 22 '18 at 15:49
• You'll find more useful stuff if you search for "de Casteljau" "curve tangent" Jul 21 '18 at 7:53
As you said, a quadratic Bezier curve is the envelope of lines whose end-points are moving along two straight lines.
In just the same way, a cubic Bezier curve is the envelope of lines whose end-points are moving along two quadratic curves. You can see this by looking at the last linear interpolation in the de Casteljau algorithm.
And so on. In general, a Bezier curve of degree $m$ is the envelope of lines whose end-points are moving along two curves of degree $m-1$.
Here are the details for the cubic case. Suppose the cubic curve has control points $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$, $\mathbf{D}$.
Then its equation is $$\mathbf{P}(t) = (1-t)^3\mathbf{A} + 3t(1-t)^2\mathbf{B} + 3t^2(1-t)\mathbf{C} + t^3\mathbf{D}$$ In the final step of the de Casteljau algorithm, this is expressed as $$\mathbf{P}(t) = (1-t)\mathbf{H}(t) + t\mathbf{K}(t)$$ where \begin{align} \mathbf{H}(t) &= (1-t)^2\mathbf{A} + 2t(1-t)\mathbf{B} + t^2\mathbf{C} \\ \mathbf{K}(t) &= (1-t)^2\mathbf{B} + 2t(1-t)\mathbf{C} + t^2\mathbf{D} \end{align} Clearly $\mathbf{H}$ and $\mathbf{K}$ are quadratic Bezier curves, which are shown in green and red respectively in the picture above.
Differentiating the original curve equation, and re-arranging, we get $$\mathbf{P}'(t) = 2\bigl[(1-t)^2(\mathbf{B} - \mathbf{A}) + 2t(1-t)(\mathbf{C} - \mathbf{B}) + t^2(\mathbf{D} - \mathbf{C}) \bigr]$$ Then, a little algebra confirms that, in fact $$\mathbf{P}'(t) = 2\bigl[\mathbf{K}(t) - \mathbf{H}(t) \bigr]$$ This shows that the line joining $\mathbf{H}(t)$ and $\mathbf{K}(t)$, which is constructed in the final step of the de Casteljau algorithm, is tangent to the cubic curve $\mathbf{P}(t)$.
• Thanks for the answer! You write: "You can see this by looking at the last linear interpolation in the de Casteljau algorithm." How can I see this? I can see that the higher-order Bézier curve is traced out by points on these lines -- but how can I see that it's the envelope of these lines? Jul 20 '18 at 15:54
• Because the lines are all tangent to the final curve Jul 20 '18 at 16:34
• Thanks for the nice graphic! I was apparently very unclear in formulating this question, as it keeps getting misunderstood :-) (This also happened with the deleted answer.) I'm aware that the property of being an envelope is characterized by the lines being tangent to the curve. I can also see that the lines appear to be tangent to the curve here. What I was after was a way to "see" this in the sense of proving it, not in the sense of looking at appearances on diagrams (as nice as they are). Jul 21 '18 at 5:10
• The tangency is a basic property of the de Casteljau algorithm. In fact, one of the nice things about the de Casteljau algorithm is that it gives you a point and first derivative vector from the same calculations. A little algebra shows that the first derivative of the curve at parameter value $t$ is parallel to the line constructed in the final step of the de Casteljau algorithm. I can write out the details, if you want, but I'm sure you could do the calculations yourself. Jul 21 '18 at 7:05
• There's an explanation here: pages.mtu.edu/~shene/COURSES/cs3621/NOTES/spline/Bezier/… Jul 21 '18 at 7:08
This answer merely elaborates on the computation shown by @bubba for cubic Bezier curves. It is an alternative derivation for the hodograph as defined in the link provided in the comments.
For a sequence of control points $a_0, a_1, \ldots$ and indices $0 \leq k \leq m$ let's write $$p(k,m; t)$$ for the Bezier curve of degree $m-k$ with control points $a_k, a_{k+1}, \ldots, a_m$. These can be defined recursively for $0\leq k < m$ by $$\begin{eqnarray} p(k, k; t) &=& a_k\\ \tag1 p(k, m; t) &=& (1-t)\, p(k, m-1;t) + t\, p(k+1, m; t) \end{eqnarray}$$
for $t\in [0,1]$. By induction on $m-k$ it follows that $$\tag2 p'(k, m; t) = (m-k) \left(p(k+1, m; t) - p(k, m-1; t)\right)$$ (where this is interpreted as simply $0$ if $k=m$). Indeed $p'(k, k; t) = 0$ and for $m > k$ by the recursive definition $(1)$ above: $$\begin{eqnarray} p'(k, m; t) &=& p(k+1, m; t) - p(k,m-1;t) + \\ && (1-t)\, p'(k,m-1; t) + t \,p'(k+1, m;t)\\ &=& p(k+1, m; t) - p(k,m-1;t) + \\ && (1-t)(m-k-1)\,(p(k+1,m-1;t)-p(k, m-2;t)) + \\ && t (m-k-1)\, (p(k+2, m;t) - p(k+1, m-1; t))\\ &=& p(k+1, m; t) - p(k,m-1;t) + \\ && (m-k-1)\,((1-t)\,p(k+1,m-1;t) + t \, p(k+2, m;t))-\\ && (m-k-1)\,((1-t)\, p(k, m-2; t) + t\, p(k+1, m-1;t))\\ &=& p(k+1, m; t) - p(k,m-1;t) + \\ && (m-k-1)\,(p(k+1, m; t)- p(k, m-1; t))\\ &=& (m-k)\,(p(k+1, m; t) - p(k,m-1;t)) \end{eqnarray}$$
Now $(1)$ and $(2)$ show that the segment $\left[p(k, m-1;t), p(k+1,m;t) \right]$ is indeed tangent to $p(k, m; s)$ at $s=t$. | 2021-12-05T17:26:11 | {
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https://stats.stackexchange.com/questions/257317/how-can-i-model-flips-until-n-successes | # How can I model flips until N successes?
You and I decide to play a game where we take turns flipping a coin. The first player to flip 10 heads in total wins the game. Naturally, there is an argument about who should go first.
Simulations of this game show that the player to flips first wins 6% more than the player who flips second (the first player wins approx 53% of the time). I'm interested in modelling this analytically.
This isn't a binomial random variable, as there are no fixed number of trials (flip until someone gets 10 heads). How can I model this? Is it the negative binomial distribution?
So as to be able to recreate my results, here is my python code:
import numpy as np
from numba import jit
@jit
def sim(N):
P1_wins = 0
P2_wins = 0
for i in range(N):
while True:
P1_wins+=1
break
P2_wins+=1
break
return P1_wins/N, P2_wins/N
a,b = sim(1000000)
• When you toss a coin until $r$ failures and then look at the distribution of the number of successes that happen before finishing such experiment, then this is by definition negative binomial distribution. – Tim Jan 20 '17 at 17:26
• I cannot reproduce the 2% value. I find that the first player wins $53.290977425133892\ldots\%$ of the time. – whuber Jan 20 '17 at 17:34
• @whuber yes, I believe you are right. I ran my simulation fewer times than I should. My results are commensurate with yours. – Demetri Pananos Jan 20 '17 at 17:35
• If one wins 53% of the time, the other should be 47%, so shouldn't the description read "the first player wins 6% more than the second player," or "3% more than half the time" ? Not (as it currently says) "3% more than the player who flips second" – JesseM Jan 20 '17 at 21:49
• Did you get this question from the FiveThirtyEight Riddler Express? – foutandabout Jan 20 '17 at 23:11
The distribution of the number of tails before achieving $10$ heads is Negative Binomial with parameters $10$ and $1/2$. Let $f$ be the probability function and $G$ the survival function: for each $n\ge 0$, $f(n)$ is the player's chance of $n$ tails before $10$ heads and $G(n)$ is the player's chance of $n$ or more tails before $10$ heads.
Because the players roll independently, the chance the first player wins with rolling exactly $n$ tails is obtained by multiplying that chance by the chance the second player rolls $n$ or more tails, equal to $f(n)G(n)$.
Summing over all possible $n$ gives the first player's winning chances as
$$\sum_{n=0}^\infty f(n)G(n) \approx 53.290977425133892\ldots\%.$$
That is about $3\%$ more than half the time.
In general, replacing $10$ by any positive integer $m$, the answer can be given in terms of a Hypergeometric function: it is equal to
$$1/2 + 2^{-2m-1} {_2F_1}(m,m,1,1/4).$$
When using a biased coin with a chance $p$ of heads, this generalizes to
$$\frac{1}{2} + \frac{1}{2}(p^{2m}) {_2F_1}(m, m, 1, (1 - p)^2).$$
Here is an R simulation of a million such games. It reports an estimate of $0.5325$. A binomial hypothesis test to compare it to the theoretical result has a Z-score of $-0.843$, which is an insignificant difference.
n.sim <- 1e6
set.seed(17)
xy <- matrix(rnbinom(2*n.sim, 10, 1/2), nrow=2)
p <- mean(xy[1,] <= xy[2,])
cat("Estimate:", signif(p, 4),
"Z-score:", signif((p - 0.532909774) / sqrt(p*(1-p)) * sqrt(n.sim), 3))
• Just as a note that may not be obvious at a glance, our answers agree numerically: (.53290977425133892 - .5) * 2 is essentially exactly the probability I gave. – Dougal Jan 20 '17 at 17:48
• @Dougal Thank you for pointing that out. I looked at your answer, saw the $6.6\%$, and knowing that it did not agree with the form of the answer requested in the question, I did not recognize that you had computed correctly. In general it's a good idea to frame an answer to any question in the form that is requested, if possible: that makes it easy to recognize when it's correct and easy to compare answers. – whuber Jan 20 '17 at 17:56
• @whuber I was responding to the phrase "Simulations of this game show that the player to flips first wins 2% (EDIT: 3% more after simulating more games) more than the player who flips second". I'd interpret "wins 2% more" as $\Pr(A\text{ wins}) - \Pr(B\text{ wins}) = 2\%$; the correct value is indeed 6.6%. I'm not sure of a way to interpret "wins 2% more" means "wins 52% of the time", though apparently that is what was intended. – Dougal Jan 20 '17 at 17:59
• @Dougal I agree that the OP's description is confusing and even wrong. However, the code and his result made it clear he meant "3% more than half the time" rather than "3% more than the other player." – whuber Jan 20 '17 at 18:06
• @whuber Agreed. Unfortunately, I answered the question before the code was posted, and didn't run a simulation myself. :) – Dougal Jan 20 '17 at 18:06
We can model the game like this:
• Player A flips a coin repeatedly, getting results $$A_1, A_2, \dots$$ until they get a total of 10 heads. Let the time index of the 10th heads be the random variable $$X$$.
• Player B does the same. Let the time index of the 10th heads be the random variable $$Y$$, which is an iid copy of $$X$$.
• If $$X \le Y$$, Player A wins; otherwise Player B wins. That is, \begin{align} \Pr(A\text{ wins})&= \Pr(X \ge Y) = \Pr(X > Y) + \Pr(X = Y)\\ \Pr(B\text{ wins})&= \Pr(Y > X) = \Pr(X > Y). \end{align}
The gap in the win rates is thus $$\Pr(X = Y) = \sum_k \Pr(X = k, Y = k) = \sum_k \Pr(X = k)^2 .$$
As you suspected, $$X$$ (and $$Y$$) are distributed essentially according to a negative binomial distribution. Notations for this vary, but in Wikipedia's parameterization, we have heads as a "failure" and tails as a "success"; we need $$r = 10$$ "failures" (heads) before the experiment is stopped, and success probability $$p = \tfrac12$$. Then the number of "successes," which is $$X - 10$$, has $$\Pr(X - 10 = k) = \binom{k + 9}{k} 2^{-10 - k},$$ and the collision probability is $$\Pr(X = Y) = \sum_{k=0}^\infty \binom{k + 9}{k}^2 2^{-2k - 20} ,$$ which Mathematica helpfully tells us is $$\frac{76\,499\,525}{1\,162\,261\,467} \approx 6.6\%$$.
Thus Player B's win rate is $$\Pr(Y > X) \approx 46.7\%$$, and Player A's is $$\frac{619\,380\,496}{1\,162\,261\,467} \approx 53.3\%$$.
• the heads need not be in a row, just 10 total. I assume that is what you are fixing. – Demetri Pananos Jan 20 '17 at 17:08
• (+1) I like this approach better than the one I posted because it is computationally simpler: it requires only the probability function, which has a simple expression in terms of binomial coefficients. – whuber Jan 20 '17 at 18:04
• I've submitted an edit replacing the last paragraph questioning the difference from the other answer with an explanation of how their results are actually the same. – Monty Harder Jan 20 '17 at 21:49
Let $$E_{ij}$$ be the event that the player on roll flips i heads before the other player flips j heads, and let $$X$$ be the first two flips having sample space $$\{ hh,ht,th,tt\}$$ where h means heads and t tails, and let $$p_{ij} \equiv Pr(E_{ij})$$.
Then $$p_{ij}=Pr(E_{i-1j-1}|X=hh)*Pr(X=hh)+Pr(E_{i-1j}|X=ht)*Pr(X=ht)+Pr(E_{ij-1}|X=th)*Pr(X=th)+Pr(E_{ij}|X=tt)*Pr(X=tt)$$
Assuming a standard coin $$Pr(X=*)=1/4$$ means that $$p_{ij}=1/4*[p_{i-1j-1}+p_{i-1j}+p_{ij-1}+p_{ij}]$$
solving for $$p_{ij}$$, $$= 1/3*[p_{i-1j-1}+p_{i-1j}+p_{ij-1}]$$
But $$p_{0j}=p_{00}=1$$ and $$p_{i0}=0$$, implying that the recursion fully terminates. However, a direct naive recursive implementation will yield poor performance because the branches intersect.
An efficient implementation will have complexity $$O(i*j)$$ and memory complexity $$O(min(i,j))$$. Here's a simple fold implemented in Haskell:
Prelude> let p i j = last. head. drop j \$ iterate ((1:).(f 1)) start where
start = 1 : replicate i 0;
f c v = case v of (a:[]) -> [];
(a:b:rest) -> sum : f sum (b:rest) where
sum = (a+b+c)/3
Prelude> p 0 0
1.0
Prelude> p 1 0
0.0
Prelude> p 10 10
0.5329097742513388
Prelude>
UPDATE: Someone in the comments above asked whether one was suppose to roll 10 heads in a row or not. So let $$E_{kl}$$ be the event that the player on roll flips i heads in a row before the other player flips i heads in a row, given that they already flipped k and l consecutive heads respectively.
Proceeding as before above, but this time conditioning on the first flip only, $$p_{k,l} = 1-1/2*[p_{l,k+1}+p_{l,0}]$$ where $$p_{il}=p_{ii}=1, p_{ki}=0$$
This is a linear system with $$i^2$$ unknowns and one unique solution.
To convert it into an iterative scheme, simply add an iterate number $$n$$ and a sensitivity factor $$\epsilon$$:
$$p_{k,l,n+1} = 1/(1+\epsilon)*[\epsilon*p_{k,l,n} +1-1/2*(p_{l,k+1,n}+p_{l,0,n})]$$
Choose $$\epsilon$$ and $$p_{k,l,0}$$ wisely and run the iteration for a few steps and monitor the correction term. | 2019-08-25T12:14:01 | {
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http://math.stackexchange.com/questions/286650/function-that-is-discontinuous-only-for-integer-fractions | # Function that is discontinuous only for integer fractions
I have this question:
Find a function $f :\mathbb R \to\mathbb R$ which is discontinuous at the points of the set $\{\frac1n : n \text{ a positive integer}\} \cup \{0\}$ but is continuous everywhere else.
I really don't know what to do. I was thinking maybe: $$f(x) = \begin{cases} 1 \quad&\text{if }x=0 \\ 0 &\text{if } x \text{ is in } \{\tfrac1n : n \text{ a positive integer}\}\\ x &\text{otherwise} \end{cases}$$ But that kind of seems like 'cheating'. Is there a better example?
EDIT: Would it be better to have:
$$f(x) = \begin{cases} 1 &\text{if } x \text{ is in } \{\tfrac1n : n \text{ a positive integer}\}\cup \{0\}\\ 0 &\text{otherwise} \end{cases}$$
-
Looks like Thomae's function: en.wikipedia.org/wiki/Thomae's_function – Austin Mohr Jan 25 '13 at 13:35
Actually a later part of this question seems to involve that function. – Joe Jan 25 '13 at 13:38
That's not cheating at all, as long as the function is well defined (it certainly is) and it verifies the requisites (does it?) – leonbloy Jan 25 '13 at 14:11
It does verify the requisites right? f(x) = 0 as x tends to 0 but f(0)=1 which is not the same, so it's discontinuous at 0, Same for all the 1/n as well, unless I've got this very wrong. – Joe Jan 25 '13 at 16:38
I fully agree with @leonbloy. Since the function satisfies the assumptions, it is a correct answer. It is good in that it is simple, so you immediately see what happens. That possibly makes it better than any "natural" example. – Feanor Apr 8 '13 at 8:11
I'm going to assume your true question is finding an answer that you do not consider "cheating."
Question/Problem
Find a function $f:\mathbb{R}\to\mathbb{R}$ such that $f$ is discontinuous at each point in $K\overset{\text{def}}{=}\{\frac{1}{n}:n\in\mathbb{N}\text{ and }n\ne 0\}\cup\{0\}$ and $f$ is continuous at each point in the complement of $K$ which is denoted $(\mathbb{R}\setminus K)$
Let $g:\mathbb{R}\to\mathbb{R}$ be an arbitrary continuous function. Let $\epsilon>0$ be an arbitrary positive real number.
Your edited answer has $g$ be the zero function and $\epsilon$ be $1$
Define $f:\mathbb{R}\to\mathbb{R}$ by $$f(x)=\begin{cases} g(x)+\epsilon&\text{if }x\in K\\ g(x)&\text{if }x\in(\mathbb{R}\setminus K) \end{cases}$$ for every $x\in\mathbb{R}$ where $K\overset{\text{def}}{=}\{\frac{1}{n}:n\in\mathbb{N}\text{ and }n\ne 0\}\cup\{0\}$.
The reason why I introduce "$K$" is because this method can be used for any given set where you want discontinuities. The set you were given is not special in any way for this problem. However it is a classic example of a compact set, but that's not relevant to your problem. | 2014-08-31T00:55:07 | {
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Mathematical modeling is the same - it simply refers to the creation of mathematical formulas to represent a real. Maths HL Type 2 IA. Practice exam 2014, Questions and answers - Maths IA, Eigenvalues, Example exam rn. the text Intro- duction to Demography by M. Math Journal 1, p. Neat homework can aid your comprehension and might make your teacher like you better. 4) Bad maths in court - how a misuse of statistics in the courtroom can lead to devastating miscarriages of justice. Math worksheets make learning engaging for your blossoming mathematician. Also available are Sadlier Math and Progress in Mathematics for use with Renaissance ® Star Math ®. Music uses similar strategies. Using the model, write down a second equation in a and b. Visual Math Learning offers free math lessons featuring an interactive on-line tutorial for teaching elementary mathematics and basic arithmetic for grades K-12 at the pre-algebra level. This example shows how to use Embedded Coder® Support Package for ARM® Cortex-A Processors to verify code using PIL simulation. The following sections describe an example model, Cruise Control Test Generation. Credit point value: 3CP. This example models a shunt compensation device that is increasingly used in modern grids: the Modular-Multi-level(MMC) STATCOM. A Linear Model Example and Technology Tips are provided in separate documents. The example problems on supplementary and complementary angles are given below: Example 1: Find the complement of 40 degrees. After the changed in academic structure in UP Board, it's very essential that students understand the examination pattern and style of questions to be asked under the latest exam structure. Mathematical studies SL teacher support material 17 Example 1. Two fixed points inside the ellipse, F1 and F2 are called the foci. Thinking With Mathematical Models: Homework Examples from ACE An innovative course that offers students an exciting new perspective on mathematics, Mathematical Models with Applications explores the same types of problems that math professionals encounter daily. [Optional] Math Review: Converting Units Teacher Resource (TG pp. Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more. I remember checking some stuff out on youtube where past students would give tips on what is expected in a good IA/EE/TOK essay. Mathematical modeling provides answers to real world questions like "Which recycling program is best for my city?" "How will a flu outbreak affect the US," o. 5x - 3y - 15 = 0. The examples are written primarily by scientists and engineers, and are available to teachers, students, and other interested parties via the PUMAS Web Site. Textbook solution for Precalculus: Mathematics for Calculus (Standalone… 7th Edition James Stewart Chapter 4. We live in the age of the algorithm. Contributed by. Develop analytical understanding: This is the ultimate goal, which is to apply the concept in real world situations. Solution: As the given angle is 40 degrees, then, Complement is 50 degrees. In other words, y is a function of x. For Santa Monica City College, the student's transfer course articulates to different math courses at PSUNV, depending on the number of units that a student takes for Math 20. UCLA Department of Mathematics. Joseph Crawley (000865-004). The second set of dependent variables represents the fraction of the total population in each of the three categories. The Common Core State Standards in mathematics were built on progressions: narrative documents describing the progression of a topic across a number of grade levels, informed both by research on children's cognitive development and by the logical structure of mathematics. Vacationers use math to find time of arrivals and departures to plan their trips. Games, Auto-Scoring Quizzes, Flash Cards, Worksheets, and tons of resources to teach kids the multiplication facts. 2 Interpret whole number quotients of whole numbers. All you need to do is learn all the formulas and you will understand math. com September 2018: 42,144 October 2018: 53,712 November 2018: 56,779. After the changed in academic structure in UP Board, it's very essential that students understand the examination pattern and style of questions to be asked under the latest exam structure. Wednesday, September 16, 2020 - 3:30pm. HL Criterion E: Use of mathematics. So far we’ve recreated 8 + 5 = 13 and 5 + 8 = 13 Since we know the inverse (opposite) of addition is subtraction, start with the sum (or the larger number), 13, and subtract one of the addends. See full list on mathsisfun. Teacher Resource Center. The resultant vector, which connects the first two vectors, can be found through mathematics, more specifically, graphic vector addition. g: 5e3, 4e-8, 1. The following sections describe an example model, Cruise Control Test Generation. The Guidelines link to examples of common errors, and demonstrate techniques that your instructors will love! In. Please read our Privacy Policy. Modeling no-shows, cancellations, overbooking, and walk-ins in restaurant revenue management Tony S. Lecturer(s) in 2020: Behrouz Taji; Email contact address: [email protected] 81% (1,613) of the reviews mentioned. net is a comprehensive math resource site for homeschooling parents, parents, and teachers that includes free math worksheets, lessons, online math games lists, ebooks, a curriculum guide, reviews, and more. Welcome to Big Ideas Math! Let's get you registered. LSU College of Science Geaux Teach Math & Science Baton Rouge, LA 70803 Telephone: (225) 578-6241 Fax: (225) 578-4522 ©. NET Numerics aims to provide methods and algorithms for numerical computations in science, engineering and every day use. Wednesday, September 16, 2020 - 3:30pm. In this unit, we will study quadratic functions and the relationships for which they provide suitable models. A simple sentence is one independent clause that has a subject and a verb and expresses a complete thought. Solved Examples. Rubric with Suggestions. Mathematics SL and HL teacher support material 5 Example 7: Annotated student work Substituting in the values for g and k we get: v = 9. Discretization techniques for variational problems, geometric integrators, advanced techniques in numerical discretization. example03_e. Mathematical Studies Internal Assessment: The correlation between shoe size and the length of forearm of an. The next example is from Chapter 2 of the book Caste and Ecology in Social Insects, by G. The example here scored 17/20. Ongoing Assessment: Mental Math and Reflexes Recognizing Student Achievement Use Mental Math and Reflexes to assess students’ ability to estimate. This becomes possible after a student achieves a strong definition-level understanding. Jan 4, 2019 - Plus One Maths Model Question Papers Paper 4 are part of Plus One Maths Previous Year Question Papers and Answers. For example, if you know anything about playing a piano, the note A above middle C produces a wave shaped like. Find the common factors of 10 and 30. To model many types of real-life quantities, such as the temperature inside and outside an igloo in Ex. Explores unfamiliar maths and devises own approach to area under curve. Model theory has strong relations with algebra. Motivation and Applications Numerical integration — the process of evaluating a definite integral numerically — is also known asquadrature. Oster and E. LOGIN New to Big Ideas Math? LOG IN. The diagram of information architecture of the web resource which is constructed correctly with necessary details presents to developers the resource in comprehensible and visual way. Computer Modeling. m, which calculates the full invariants in the noncommutative setting. Monthly, Vol 108 (2001), 512-521. For Santa Monica City College, the student's transfer course articulates to different math courses at PSUNV, depending on the number of units that a student takes for Math 20. Common Core for Mathematics. Notice how we get the same answer no matter what side we use to find an area. 227 (2011), no. In this paper we model the population growth of Rwanda using Verhulst model (logistic growth model). Several prints by M. 8% compounded monthly really pay 10. MATH 1001 (Quantitative Skills and Reasoning) Gordon College, Barnesville, GA. The combination of these traits makes it. The accompanying activities will help students to understand what a mathematical model is, and how we can build up increasingly complex models from the simplest of starting points. Krantz has turned his focus to modeling COVID-19 in hopes that better predictions can help lessen transmission and save lives. Roughly equivalent to:. " Breaking this down, nouns in math. Statistics and modelling 1) Traffic flow : How maths can model traffic on the roads. com September 2018: 42,144 October 2018: 53,712 November 2018: 56,779. UP Board Class 10 Mathematics Model Paper 2019 In this Article you will get UP Board class 10 th Mathematics Model paper 2019. An example of an SL type II modeling task in maths. Criterion D: Reflection D2—The reflection is meaningful, but not critical. Young students, for example, might notice that three and seven more is the same amount as seven and three more, or they may sort a collection of shapes according to how many sides the shapes have. Provide high-impact teaching and personalize instruction with Full Access for Mathematics. About the Example Model. When we attempt to engage students by using real-world examples, we often find that the colloquial or “street” language does not always map directly or correctly onto the mathematical syntax. The resultant vector, which connects the first two vectors, can be found through mathematics, more specifically, graphic vector addition. Topics include Algebra and Number (proof), Geometry, Calculus, Statistics and Probability, Physics, and links with other subjects. Keep in mind that this problem has many. 804 "verify" would As we continue working on the model, thus integrating it, we can prove the above result by graphing the equation found for velocity. Benjamin Chew's Mathematics SL Exploration. " Breaking this down, nouns in math. 1 are: Section 2. He could earn a 4 for creativity, a 5 for difficulty, a 3 for math content, a 3 for neatness and a 5 for presentation. Please select each answer from the buttons to the left of the answers. Criterion D: Reflection D2—The reflection is meaningful, but not critical. 2 Hamer: 1906 Hamer formulated and analyzed a discrete time. Murrieta Valley Unified School District / Overview. Climate models tend to agree on the twenty-year projections, both in regard to their sensitivity to variations in model physics as well as different emissions scenarios1. Gain fluency and confidence in math! IXL helps students master essential skills at their own pace through fun and interactive questions, built in support, and motivating awards. Penn Mathematics Colloquium. Please use all of our printables to make your day easier. The examples refer most frequently to the 6/49 lottery, which is the most widespread matrix. Considers modelling. Standard: CC. All you need to do is learn all the formulas and you will understand math. The y intercept is (0,-13) and the slope is 7. State the next point in the pattern. Modeling no-shows, cancellations, overbooking, and walk-ins in restaurant revenue management Tony S. Peter King's Football Morning In America column begins with an interview of Roger Goodell, and also includes a Super Bowl prediction, MVP picks + more. * Use e for scientific notation. Tsea and Yiu-Tung Poonb aSchool of Hotel and Tourism Management, The Hong Kong Polytechnic University, Kowloon, Hong Kong; bDepartment of Mathematics, Iowa State University, Ames, IA, USA ABSTRACT Few studies have examined overbooking in the. One of the simplest models of this type, used for projections over limited periods of time by demographers (cf. Academic Reading Duration: 60 minutes. A student can feel mathematically ready to attend College if he or she can get at least 33 out of the 36 problems correct. Home Helpful Links 2019-2020 Classes > All About Me Survey example01_e. • appreciate the international dimension in mathematics through an awareness of the universality of mathematics and its multicultural and historical perspectives • appreciate the contribution of mathematics to other disciplines, and as a particular “area of knowledge” in the TOK course. Math 2280 - Lecture 7: Population Models Dylan Zwick Fall 2013 Today we’re going to explore one of the major applications of differen-tialequations-population models. In electrical grids, shunt compensation is often used for reactive power and voltage control. To generate model ideas, take a look at: My 2019 TechSEO Boost presentation. Busch was recognized by the undergraduate Student Association for his commitment and dedication to students. Learn what the domain and range mean, and how to determine the domain and range of a given function. The math worksheets are randomly and dynamically generated by our math worksheet generators. A probability sample is a sample in which every unit in the population has a chance (greater than zero) of being selected in the sample, and this probability can be accurately determined. An example of an SL type II modeling task in maths. The Previous Year Model Papers are now available for Subject wise of English, Telugu, Hindi, Sanskrit, Maths, Physics, Chemistry, Botany, Zoology, Commerce, Economics. Topic 1 sample assessment instrument: Problem-solving and modelling task (PDF, 296. However, several different programs for instructing a machine to play chess well have been written. (It's a part of the whole 100%. For IB SL Maths. Search for wildcards or unknown words Put a * in your word or phrase where you want to leave a placeholder. Developed through research and field testing over 25 years, Everyday Mathematics is built for success, is built for the common core state standards, and is built for the 21st century learner. You got it: ANYONE can ace the Math IA if they put in adequate effort and use this eBook as a. Primary Science Electricity Worksheets. Shoppers use math to calculate change, tax, and sales prices. IA/GE/A2 (2007-17) Math A/B (1998-2010) REGENTS RESOURCES. Tuesday, September 22, 2020 - 4:30pm. The model is created using the SimPowerSystems™ toolbox, C28x peripherals, and DMC library blocks. Telangana (BIETS) Previous year Model Question Papers for Junior Inter Sample Papers Issued by the Board of Intermediate Education of Andra Pradesh State on its official website. For Santa Monica City College, the student's transfer course articulates to different math courses at PSUNV, depending on the number of units that a student takes for Math 20. Compares results and reflects on this. Good call! I remember spending my life on these websites while working on my IA/EE/TOK. Our Starter Kit is an example rich and quick introduction to teaching modeling based differential equations. Example #1: Minesweeper. Weekly computing practical classes teach simple programming via biological examples. Motivation and Applications Numerical integration — the process of evaluating a definite integral numerically — is also known asquadrature. Correlation range is -1 to +1. Stability of an equilibrium: Suppose that x∗ is. Certain integrals of “simple”. Wolfram offers advanced technical computing solutions for all fields of industry, education and technology. Introduction to the HL Math IA - Duration:. Washington University in St. Mathematical modeling provides answers to real world questions like “Which recycling program is best for my city?” “How will a flu outbreak affect the US,” o. School name: International School Bangkok. A representation of the model object 'Multiplicity Element'. Scalar models System models Equilibrium Model: xn+1 = f(xn), here f is a nonlinear function. One example is sound: whenever you play an instrument, or listen to your stereo, you're listening to sound waves. Industry Attractiveness-Business Strength Matrix. Serway, Chris Vuille. pdf: File Size: 468 kb: File Type: pdf: Download File. Also available are Sadlier Math and Progress in Mathematics for use with Renaissance ® Star Math ®. Free Sample papers for 7th Class Mathematics Maths Olympiad Model Test Paper-16. inTelangana Intermediate 1st Yr Model Question Papers 2020 @ […]. Apply geometric concepts in modeling situations. 400 E 14th Street | Des Moines, IA 50319-0146 Phone: 515-281-5294 | Fax: 515-242-5988. Fall 1999. We offer a comprehensive set of curricula in our disciplines, from introductory-level general education courses to doctoral dissertation direction and postdoctoral mentoring. net is a comprehensive math resource site for homeschooling parents, parents, and teachers that includes free math worksheets, lessons, online math games lists, ebooks, a curriculum guide, reviews, and more. Sample problems are under the links in the "Sample Problems" column and the corresponding review material is under the "Concepts" column. arantola,T Inverse Problem Theory and Methods for Model. Introduction Configure a Simulink model to run as a processor-in-the-loop (PIL) simulation. MATH 151 SI Exam 3 Review 4/17/14 Note: Exam 3 will cover more material than what is listed here and on the other sample test! Study your notes, homework, and textbook to cover all material from Chapter 4 1. Even fields such as number theory that are part of pure mathematics are now important in applications (such as cryptography ), though they are not generally. For Santa Monica City College, the student's transfer course articulates to different math courses at PSUNV, depending on the number of units that a student takes for Math 20. Modelling Zombies: Investigating how zombies survive. Simple Sentence Examples. Tasks for Linear Regression Model (LR) (LR-1) Describe your topic, provide your data, and cite your source. As the first step in the modeling process, we identify the independent and dependent variables. Climate models tend to agree on the twenty-year projections, both in regard to their sensitivity to variations in model physics as well as different emissions scenarios1. Providing a complete overview—beginning with the early history of. This math worksheet was created on 2008-01-16 and has been viewed 89 times this week and 52 times this month. dist (p, q) ¶ Return the Euclidean distance between two points p and q, each given as a sequence (or iterable) of coordinates. Here we have given Plus One Maths Model Question Papers Paper 3. Real-world connections that get students engaged. There are no ifs, ands, or buts about them. Mathematics Department Sample Placement Exam Note: This is a SAMPLE exam Instructions. © 2000-2005 Math. Multivariate, Reference) documentation. For example, there are lots of math worksheets, several multiplication charts and tables, plenty of number lines, and even a fraction calculator. Please note that difference between HL and SL IAs is the level of math expected of students which is reflected in slightly different rubrics and consequently the HL and SL grades differ slightly. Example #1: Minesweeper Minesweeper Rubric (5/20) Example #2: Florence Nightingale. The ACT mathematics test is a 60-question, 60-minute test designed to assess the mathematical skills students have typically acquired in courses taken up to the beginning of grade 12. 2013/2014. Questions that were explored were: What is the necessary width of the. Rubric with Suggestions. Mizoram Board 10th Model Paper 2021 Download with MBSE HSLC Sample Question Paper 2019, MBSE 10th Blueprint 2021 for New syllabus study material, practice papers, Mizoram Board Madhyamik Previous Paper Pdf with Bit Bank important question Questions Paper for Hindi Medium and English Medium suggestions with new question pattern with exam pattern along the paper style for all general and. Sample Estimation Question. In electrical grids, shunt compensation is often used for reactive power and voltage control. Yes it would work! A good example is the SIR model (and its alternatives) - it uses calculus within the SL syllabus, and offers lots of opportunities to show personal engagement. So, if N is the total population (7,900,000 in our example), we have. She then tosses a fair die the same number of times as she tossed the coin. The course develops geometric intuition and the language of di erential equations and is a fundamental part of the Calculus sequence we o er to all science, Engineering and math students. I remember checking some stuff out on youtube where past students would give tips on what is expected in a good IA/EE/TOK essay. Faculty collaborate by using experiments to motivate and validate mathematical modeling, as well as by spurring experiments to observe new phenomena predicted by the models. If you would like to contribute, please donate online using credit card or bank transfer or mail your tax-deductible contribution to: Mathematics Genealogy Project Department of Mathematics North Dakota State University P. Through participation, students experience what it’s like to work as a team to tackle a real-world problem under time and resource constraints, akin to those faced by professional mathematicians working in industry. Real-world connections that get students engaged. This information, along with its mechanism and the. Oxford Mathematicians are descendants of a long lineage from the Merton School of the 14th Century to Christopher Wren in the 17th Century and Hardy and Penrose in the 20th century. I’m obliged to answering this. Using the model, write down a second equation in a and b. If Shirley paid$18. For example, in a rectangle we find the area by multiplying the length times the width. Mathematical studies SL teacher support material 17 Example 1. Statistics and modelling 1) Traffic flow : How maths can model traffic on the roads. This math worksheet was created on 2008-01-16 and has been viewed 89 times this week and 52 times this month. Solve your math problems using our free math solver with step-by-step solutions. com has been designed by Tal. Here is a set of notes used by Paul Dawkins to teach his Differential Equations course at Lamar University. Practice exam 2014, Questions and answers - Maths IA, Eigenvalues, Example exam rn. Webpages in this section feature mathematics information supporting each part of Ohio’s educational system: Ohio’s Learning Standards, the model curriculum, assessments and additional resources to help teachers in the classroom. Find the increasing and decreasing intervals and any relative extrema for the function ( ) (5. DPMMS example sheets; DAMTP example sheets; Past exam papers; Schedules; Lecture lists; Course descriptions: Part IA, Part IB, Part II Study Skills in Mathematics booklet; DAMTP Part III example sheets. Find here details of companies selling Model Question Papers, for your purchase requirements. A person tosses a fair coin until she obtains 2 heads in a row. For example, from this viewpoint, an ecologist or geographer using population models and applying known mathematics would not be doing applied, but rather applicable, mathematics. Performing extrapolation relies strongly on the regression assumptions. MyScript Math Sample is a basic integration of MyScript Math Web Component and MyScript Cloud Development Kit to help developers understand how they can leverage MyScript Math technology into their own applications. Here are two more examples of what functions look like: $$y = 3x - 2$$ $$h = 5x + 4y$$ Let's examine the first example. Everybody uses math whether they realize it or not. Why you should learn it GOAL 2 GOAL 1 What you should learn 14. Washington University in St. Subjects covered include differential equations, compartmental analysis, coupled non-linear systems, probability, statistics, matrix algebra and ecological modelling. Joseph Crawley (000865-004). Real-world connections that get students engaged. From modern-day challenges such as balancing a checkbook, following the stock market, buying a home, and figuring out credit card finance charges to appreciating historical developments like the use of algebra by Mesopotamian mathematicians, this engaging resource addresses more than 1,000 questions relating to mathematics. When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. HomeschoolMath. 3D Drawing Example Net Example. Math in the Real Worldincludes lessons that range from “Break-Even Analysis” and “Profit Maximization” to payday loan expenses and building good credit. Phone: 319-335-0714 Fax: 319-335-0627 [email protected] Examples of Mathematical Models. Suggestions: Sample answers are given. Have students write number models to show their estimates. Exemplar definition is - one that serves as a model or example: such as. It can be used to model or describe an amazing variety of phenomena, in mathematics and science, art and nature. The vocabulary of math draws from many different alphabets and includes symbols unique to math. Peter King's Football Morning In America column begins with an interview of Roger Goodell, and also includes a Super Bowl prediction, MVP picks + more. Take your Free ASVAB and AFQT exam now! 4Tests. Types of bistability For an example of bistability, consider the lac operon. Assessed Maths IA examples. As an IBDP student, you hear this all the time: EXTENSION OF CURRENT KNOWLEDGE Now what’s peculiar about modeling in IA is that it’s done quite a lot, because you need that Exploration criterion, or to show enga. When the maximum height of the building is 36m, by thinking. Fourth Revision, July 2009. The following are examples of HL/SL IAs based on the current mark scheme with grader comments. LOGIN New to Big Ideas Math? LOG IN. com? Seemath. Gain fluency and confidence in math! IXL helps students master essential skills at their own pace through fun and interactive questions, built in support, and motivating awards. Webmath is a math-help web site that generates answers to specific math questions and problems, as entered by a user, at any particular moment. Featured Interactive. Notice that there are some important requirements for a simple sentence:. Free Sample papers for 6th Class Mathematics Maths Olympiad Model Test Paper-16. Example #1: Minesweeper. Real-World Example Bonanza! Scope of this Course Outline 1 Introducing Differential Equations! A First Look Real-World Example Bonanza! Scope of this Course 2 First Order Differential Equations What’s a First Order Equation? Slope Fields 3 Separable and Homogeneous Equations Separable Equations Homogeneous Equations W. Model theory has strong relations with algebra. Please note that difference between HL and SL IAs is the level of math expected of students which is reflected in slightly different rubrics and consequently the HL and SL grades differ slightly. The course develops geometric intuition and the language of di erential equations and is a fundamental part of the Calculus sequence we o er to all science, Engineering and math students. AP 10th Class Model Question Papers 2021, BIEAP SSC Question Papers. Examples & Applications. Peter King's Football Morning In America column begins with an interview of Roger Goodell, and also includes a Super Bowl prediction, MVP picks + more. The Common Core State Standards in mathematics were built on progressions: narrative documents describing the progression of a topic across a number of grade levels, informed both by research on children's cognitive development and by the logical structure of mathematics. Topic 1 sample assessment instrument: Problem-solving and modelling task (PDF, 296. inTelangana Intermediate 1st Yr Model Question Papers 2020 @ […]. Martin, Jr. By using regression analysis on the example data, you should be able to make conclusions about several things: Is the expression “if you shoot on goal you will score” true? If you can create model with a correlation coefficient (r-squared) close to 1 or -1 it is likely that the model. Computer Modelling of Waves can be used as part of designing breakwaters, marinas, light houses, oil rigs, ships, tourist resorts, water fun parks, and artificial surfing reefs. Please note that difference between HL and SL IAs is the level of math expected of students which is reflected in slightly different rubrics and consequently the HL and SL grades differ slightly. Here's another example of an equation as a mathematical model. Louis mathematician Steven G. The next example is from Chapter 2 of the book Caste and Ecology in Social Insects, by G. Combine searches Put "OR" between each search query. That is, if an item is x dollars, then the. 6 Understand how algebraic relationships can be represented in concrete models, pictorial models, and diagrams. On integral Stokes matrices Math-Physics Joint Seminar. 3 Use multiplication and division within 100 to solve word problems in situations involving equal groups, arrays, and measurement quantities. Math is a crucial subject to learning success, which students will continue through each grade of their school journey. Examples of such scales are the Mathematics Anxiety Rating Scale (Richardson & Suinn, 1972), the Mathematics Anxiety Rating Scale–Revised (Plake & Parker, 1982) and the Mathematics Anxiety Questionanaire (Wigfield & Meece, 1988). If the result is near -1, then correlation is negative. 5 EXPLORING DATA AND STATISTICS R E A L L I. Think Math! is a comprehensive kindergarten through fifth grade curriculum that provides high-quality mathematics for all children. Notice how we get the same answer no matter what side we use to find an area. Please close and relaunch it. This equation is in slope intercept form. A person tosses a fair coin until she obtains 2 heads in a row. For example, we can restrict x between 0 and 1 and t between 1 and 5, then set A = -1, B = 21, a = 2 to give: w(x,t) = (-x 2 + -4t) + 21 This begins to fit the behavior we want - at any fixed point x the density will decrease with time, and as we move further away from the initial point (x = 0) we have lower density. An important application of such functions is to describe the. Why you should learn it GOAL 2 GOAL 1 What you should learn 14. com--a website dedicated to Math lessons, demonstrations, interactive activities and online quizzes on all areas of geometry, algebra and trigonometry. For example, if you round 4,557 to 4,560 (the nearest ten), your estimation will be more accurate than if you round to 4,600 (the nearest hundred). If you figure out the period of this function (using the. Even homeowners use math to determine the cost of materials when doing projects. Examples of Mathematical Models. How to get the last two math facts of the family. North Carolina School of Science and Mathematics – Home. Please select each answer from the buttons to the left of the answers. Home Helpful Links 2019-2020 Classes > All About Me Survey example01_e. Number of monthly visits to simulation-math. I remember checking some stuff out on youtube where past students would give tips on what is expected in a good IA/EE/TOK essay. Maplesoft Welcome Center. 1 below), there is a clear disparity in the magnitude of the leading, first and second order terms. Criterion D: Reflection D2—The reflection is meaningful, but not critical. He could earn a 4 for creativity, a 5 for difficulty, a 3 for math content, a 3 for neatness and a 5 for presentation. Dice Number S uares. The mathematical and probability models of lottery provide information that players should know before they launch into a long-run play. As you read through it, you will see comments from the moderator in boxes like this: At the end of the sample project is a summary of the moderator's grades, showing how the project has been graded against all the criteria A to G. Arthur wasn't the best at Math; he received a 28% on his mocks. By learning multiplication and memorizing the times tables you provide yourself with essential building blocks to do higher learning math, like division, fractions and even algebra. HS Mathematics Item Specification C1 TD 1 Version 2. Maths and Time: Exploring ideas regarding time dilation; Plotting Planets: Using log functions to track planets! So there we have it, 50 IB Maths IA topic ideas to give you a head-start for attacking this piece of IB coursework! Still feeling confused? Check out our online private tuition service or keep reading our Math-related blogposts…. This example uses the model described in Model of the Yeast Heterotrimeric G Protein Cycle. Here we have given Plus One Maths Model Question Papers Paper 4. 122 Student Page Mental Math and Reflexes Write multiplication problems on the board. 330-275 BC) was the great expositor of Greek mathematics who brought together the work of generations in a book for the ages. Find here details of companies selling Model Question Papers, for your purchase requirements. Global Warming. Functions involving more than two variables also are common in mathematics, as can be seen in the formula for the area of a triangle, A = b h /2, which defines A as a function of both. Assessed Maths IA examples. ICME Master of Science Program Mathematical and Computational Finance Track This new track in the ICME M. the text Intro- duction to Demography by M. Math 6382/6383: Probability Models and Mathematical Statistics Sample Preliminary Exam Questions 1. As a nal project, students nd journal articles from partner disciplines and replicate the models with the numerical solution software. Explore different representations for fractions including improper fractions, mixed numbers, decimals, and percentages. Synonym Discussion of exemplar. Model theory has strong relations with algebra. )The PERCENT always goes over 100. Learn what the domain and range mean, and how to determine the domain and range of a given function. Everybody uses math whether they realize it or not. Find the common factors of 10 and 30. It continues to produce world-class mathematics research and is devoted to excellence in teaching. 1B Mathematical Studies 2008— Internal Assessment What is the relationship between the. 1 - 1, 8, 11, 16, 29 Population Models. Academic Reading Duration: 60 minutes. It may be printed, downloaded or saved and used in your classroom, home school, or other educational environment to help someone. Defining the term, Describing its essential characteristics, Providing examples of the idea, and. August 3rd-5th Learn More. Mathematics IA (MATHS 1011) Academic year. Derivative Estimates Introduction In this chapter we are mostly concerned with the numerical solution of ODEs. Adjust numerators and denominators to see how they alter the representations and models. Maes' World 'O Math. 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MANUEL MORALES, Department of Mathematics and Statistics, York University On the discounted penalty function for the generalized inverse Gaussian process We will review, from a historical point of view, the use of L´evy processes in ruin theory. The following sections describe an example model, Cruise Control Test Generation. 4x + 3y = 12. Does anyone have sample maths SL IAs of modelling infectious diseases with the SIR model? preferably with a note of what grade it got too. | 2020-12-02T13:21:41 | {
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https://math.stackexchange.com/questions/4512585/how-to-show-that-linear-map-is-surjective | # How to show that linear map is surjective?
I have the following linear map:
$$T:\mathcal{P}^2 \to \Bbb{R}^2$$
where $$\mathcal{P}^2$$ denotes the vector space of polynomials with real coefficients having degree at most $$2$$
$$T$$ is defined by $$T(ax^2+bx+c)=[a+b , b−c]^t$$
I do not know how to prove that $$T$$ is surjective.
I know its not injective.
Yet,i do not now how to formally show that is a surjection.
I tried following the answer at How to show that a linear map is surjective?
Using the formula described:
• $$\dim(V) = 2$$,
• $$\dim(\operatorname{range}T)=1$$
According to the formula "will be surjective if $$\dim V= \dim \operatorname{range} T$$" this map would not be surjective.
Yet, in the website i took this exercise from, it says this map is a surjective.
So, i am wondering what i did wrong, and if there are better ways to show that this map is a surjective.
I came across on this exercise on this website.
$$T:\mathcal{P}^2\to \Bbb{R}^2$$ defined by $$T(ax^2+bx+c) =\begin{pmatrix}a+b\\b-c\end{pmatrix}$$
$$\ker{N}(T)=\{p\in\mathcal{P}^2 :T(p) =0\}$$
To find the $$\ker T$$ we have to solve a system of two equations with three unknowns.
$$a+b=0\tag 1$$
$$b-c=0\tag 2$$
We get $$c=b=-a$$ and $$p(x)=ax^2+bx+c\in\ker T$$
$$\implies b=-a, c=b=-a$$
Hence null space basis is $$\{\begin{pmatrix}1\\-1\\-1\end{pmatrix}\}$$
Hence $$\dim\ker T=1$$ and by Rank-Nullity theorem :
$$\dim \ker T +\dim \operatorname{range} T =\dim \mathcal{P}^2$$
Hence $$\operatorname{range} T=3-1=2=\dim(\Bbb{R}^2)$$.
Hence $$\operatorname{range} T=\Bbb{R}^2$$ implies $$T$$ is surjective.
Note: $$\dim(\mathcal{P}^2) =3$$ as basis $$\{1, x, x^2\}$$.
For any $$(p,q)\in\mathbb R^2$$, $$T(px^2-q)=(p+0,0--q)=(p,q)$$. Since this is true for any pair, $$T$$ is surjective.
$$T(x^2+5x-3)=T(2x^2+4x+2)=(6,2)$$ so $$T$$ is not injective.
I recently make this figure to explain the concepts of injectivity and surjectivity and some related properties related to the rank and left/right inverses:
Although it is stated for matrices, it is valid for linear maps between spaces of finite dimension.
Your case corresponds to the green area in the above figure. Using the standard basis of $$\mathsf{P}_2({\rm I\!R})$$, that is, $$\{1, x, x^2\}$$, we have that your linear map is described by the matrix $$M_T\in{\rm I\!R}^{2\times 3}$$ $$M_T = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & -1 \end{bmatrix},$$ and it easy to see that $$\mathrm{rank} M_T = 2$$, which proves that $$M_T$$ (and $$T$$) is surjective.
@Ayla To fit more precisely to your 2 questions:
• "So, i am wondering what i did wrong": it is your computation of $$\dim(\operatorname{range}(T))$$ which is wrong. You claim (without proof) it is equal to $$1$$, whereas the correct result is $$2$$. You don't tell us how your got this wrong $$1$$ but I suppose it was by an incorrect computation of $$\dim(\ker(T))$$ or incorrect application of the rank-nullity theorem (for a correct version, see Sourav Ghosh's answer).
• "and if there are better ways to show that this map is a surjective": a direct computation is good enough. $$T(ax^2+bx+c) =\begin{pmatrix}u\\v\end{pmatrix}\Leftrightarrow\begin{cases}a+b&=u\\b-v&=c\end{cases}\Leftrightarrow\begin{cases}a&=u-b\\c&=-v+b\end{cases}$$ so any $$\begin{pmatrix}u\\v\end{pmatrix}\in\mathbb R^2$$ is the image (by $$T$$) of many polynomials, e.g. (choosing $$b=0$$) $$\begin{pmatrix}u\\v\end{pmatrix}=T(ux^2-v).$$ | 2022-10-07T12:17:52 | {
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http://www.physicsforums.com/printthread.php?t=355746 | Physics Forums (http://www.physicsforums.com/index.php)
- General Math (http://www.physicsforums.com/forumdisplay.php?f=73)
- - # of combinations (http://www.physicsforums.com/showthread.php?t=355746)
rsala004 Nov17-09 06:38 PM
# of combinations
Lets say you have a bunch of projects to do, says project 1,2,3,4,5.
You dont have to do them all, in fact you dont have to do any of them...and the order you did them in has no effect on how they come out.
how many ways can this be done? examples, 12, 1234, 234 or no projects at all
edit:
My quick assumption to solving this problem is that you either do the project or you dont...so i guess you have 2 choices 5 times.
2*2*2*2*2 = 32, is this the solution?
Mute Nov17-09 07:26 PM
Re: # of combinations
I would ask: how many ways can you do 0 of the projects? 1 project? 2, 3, 4, all of them?
The total number of ways should then be (# ways 0)(# ways 1)(# ways 2)(# ways 3)(# ways 4)(# ways 5), which is more than 32.
pbandjay Nov17-09 07:51 PM
Re: # of combinations
This one is easy enough to just list by hand, considering the number of combinations of picking zero to five of the projects. It's 32.
The long way: Number of ways to pick 0 + Number of ways to pick 1 + ... + Number of ways to pick 5 = 1 + 5 + 10 + 10 + 5 + 1 = 32.
Or: 25 = 32.
The connection is the beautiful theorem:
$$\displaystyle\sum_{k=0}^n\binom{n}{k}=2^n$$
Bingk Nov17-09 10:22 PM
Re: # of combinations
Just to be sure ... if the person decided to do project 3 first, then 5, and nothing else (i.e. 35), you're saying that would be the same as doing 5 first then 3 (i.e 53 = 35)?
If that's the case, then yes, I believe you have it correct, 32 outcomes.
But, if you mean that you can do the projects in any order, but they still count as a distinct way of doing it (i.e. 12345 and 54321 are two ways of doing it), then it would be more than 32 ...
arildno Nov18-09 02:08 AM
Re: # of combinations
Quote:
Quote by Bingk (Post 2447368) Just to be sure ... if the person decided to do project 3 first, then 5, and nothing else (i.e. 35), you're saying that would be the same as doing 5 first then 3 (i.e 53 = 35)? If that's the case, then yes, I believe you have it correct, 32 outcomes. But, if you mean that you can do the projects in any order, but they still count as a distinct way of doing it (i.e. 12345 and 54321 are two ways of doing it), then it would be more than 32 ...
True enough!
In that case, the total number will be:
$$\sum_{k=0}^{n}k!\binom{n}{k}$$
For n=5, this amounts to:
$$1*1+1*5+2*10+6*10+24*5+120*1=326$$
distinct ways.
CRGreathouse Nov18-09 08:04 AM
Re: # of combinations
I agree, 2^5 is the answer the OP wants.
Quote:
Quote by arildno (Post 2447554) In that case, the total number will be: $$\sum_{k=0}^{n}k!\binom{n}{k}$$ For n=5, this amounts to: $$1*1+1*5+2*10+6*10+24*5+120*1=326$$ distinct ways.
http://www.research.att.com/~njas/sequences/A000522
rsala004 Nov18-09 11:20 AM
Re: # of combinations
thank you
All times are GMT -5. The time now is 03:08 PM. | 2014-08-21T20:08:38 | {
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https://math.stackexchange.com/questions/2596894/does-fracx2-y2-yx-have-a-limit-at-0-0/2596901 | # Does $\frac{(x^2 + y^2) y}{x}$ have a limit at $(0,0)$?
Does $\frac{(x^2 + y^2) y}{x}$ have a limit at $(0,0)$?
Recently, someone asked whether a function from $\mathbb{R}^2$ to $\mathbb{R}$ had a limit at $(0,0)$. The question was easy and answered in the negative by showing that approaching $(0,0)$ on different lines led to different limits.
This prompted a question: is there such a function which has a limit when restricted to any straight line through $(0,0)$ and the limit is the same in all cases yet the function does not have a limit at $(0,0)$?
This led me to consider this function:
$$f(x, y) = \begin{cases} \frac{(x^2 + y^2) y}{x}, & \text{if x \neq 0} \\ 0, & \text{if x = 0} \end{cases}$$
This looks a bit nicer in polar coordinates with $x = r \sin \theta$ and $y = r \cos \theta$
$$f(x, y) = \begin{cases} r^2 \tan \theta, & \text{if \theta \neq \pm \frac{\pi}{2} } \\ 0, & \text{if \theta = \pm \frac{\pi}{2} } \end{cases}$$
So, if the function is restricted to a straight line through $(0,0)$ then the function clearly has the limit $0$ since $\tan \theta$ will be a constant.
However, it is not continuous at $(0,0)$ as within any radius of $(0,0)$, it takes arbitrarily large values.
So, here is my question: is the above right or have I made a mistake? (I am rather rusty in this area.)
I know that I don't need to restrict myself to straight lines when testing limits. In fact, that was the point of the exercise: to show that straight lines may disprove a limit but testing only straight lines will not prove a limit. I wanted an example that had a limit along all straight lines yet still failed to have a limit.
Simpler examples that demonstrate this would be welcome.
• You could also consider how the function behaves on a path like $y=ax^r$ and see if there’s any pair of parameters which don’t give a limit of zero. (This isn’t enough to confirm continuity, but it can be used to disprove it). – Semiclassical Jan 8 '18 at 14:35
• @Rick I am not sure what you are suggesting. Are you suggesting, as others have, that using non-straight paths will disprove the limit? – badjohn Jan 8 '18 at 16:01
Hint: try the limit along $y=x^{1/3}$. Is it also zero?
The easiest way is to rewrite $$\frac{(x^2+y^2)y}{x}=xy+\frac{y^3}{x}.$$ The first term goes to zero, so you need to study the second term only.
• Thanks. The motivation for my definition was to get the $\tan \theta$ term in the polar for as this would mean that the function attained arbitrarily large values on any circle around $(0,0)$. – badjohn Jan 8 '18 at 15:58
• @badjohn $\frac{y^2}{x}$ would probably be an easier example. Though the fact that you have to additionally define the function along the whole $y$-axis is a bit tense. Maybe $\frac{x^2y}{x^4+y^2}$ instead (and zero at the origin)? – A.Γ. Jan 8 '18 at 16:40
• Thanks. I mentioned my motivation for my function but I failed to consider simplifying it after finding it. Actually, I had $\frac{y}{x} \sqrt{x^2 + y^2}$ first as I was considering $r \tan \theta$ in polar form. My function above was a slight simplification of that as the square root was not required. – badjohn Jan 8 '18 at 17:42
What you wrote is correct.
Another way to see this function doesn't have a limit is to approach along the $y = x^\frac{1}{4}$ curve. (You don't have to limit yourself to approaching along a straight line!)
• I know that I don't have to restrict myself to a straight line. That was actually the point of the exercise: to show that although the straight line approach will often disprove a limit, it cannot be used to prove a limit. – badjohn Jan 8 '18 at 15:51
No, it does not have a limit at $(0,0)$ : you just have to use the sequence $(x_n,y_n):=(\tfrac{1}{n^2},\tfrac{1}{\sqrt{n}})$ to see it : $$f(x_n,y_n)=\Big(\frac{1}{n^4}+\frac{1}{n} \Big)\tfrac{1}{\sqrt{n}}\times n^2 = \frac{1}{n^2\sqrt{n}}+\sqrt{n}$$
• Thanks. I did not calculate the coordinates of large valued points near the origin because they clearly existed due to the $\tan \theta$ term. I could pick a circle around $(0,0)$ as small as I like but by picking a suitable $\theta$, I could make the value of $f$ as large as I liked. – badjohn Jan 8 '18 at 15:50 | 2019-12-14T13:43:51 | {
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https://www.physicsforums.com/threads/definition-of-liminf-of-sequence-of-functions.689470/ | # Definition of liminf of sequence of functions?
1. May 3, 2013
### mathmonkey
1. The problem statement, all variables and given/known data
Hi I've come across the term lim inf $f_n$ in my text but am not sure what it means.
$\lim \inf f_n = \sup _n \inf _{k \geq n} f_k$
In fact, I am not sure what is supposed to be the output of lim inf f? That is, is it supposed to return a real-valued number, or a function itself?
Generally for a real-valued function $\inf f$ refers to $\inf_x f(x)$. That is, it returns the largest real-valued number smaller than f(x) for all x. If that's the case, then it should follow that $\lim \inf f_n$ also returns a real-valued number? But I've always thought the implication of lim inf and lim sup is that if $f_n$ converges uniformly to $f$ then
$\lim \inf f_n = \lim \sup f_n = \lim f_n = f$
but that doesnt seem to hold if i use this definition? Any help or clarification would be greatly appreciated. Thanks!
2. May 3, 2013
### Mandelbroth
I really like this image, from Wikipedia. It helped get me started in understanding intuitively what limsup and liminf mean. Hopefully it does the same for you.
3. May 3, 2013
### mathmonkey
Hi Mandelbroth,
Thanks for your reply! However, that picture only describes the limsup/liminf of a sequence of points, which is itself a point, which is easier to intuit for me. But what I'm wondering is what is the limsup/liminf of a sequence of functions supposed to be? Is it to be a function itself? Or just a point? I am unclear as to the definition itself of limsup for sequences of functions....
So far I'm thinking it shouldnt be intuitive for limsup/liminf f_n to be a point, since as I mentioned earlier if $f_n$ converges to $f$ uniformly it ought to be the case that $\lim \sup f_n = \lim \inf f_n = \lim f_n$? Thanks again for any help
4. May 3, 2013
### Dick
Yes, it's a function. The value of lim inf $f_n$ at a point x is the lim inf of sequence of numbers $f_n(x)$ as n->infinity.
5. May 4, 2013
### micromass
Staff Emeritus
If you understand limsup/liminf of sequences of points, then this isn't too hard. Basically, you take a sequence of functions $(f_n)_n$. Now, if I take a fixed $x$, then $x_n = f_n(x)$ is a sequence of points. So the liminf makes sense. Now, we define
$$f(x) = \liminf x_n$$
And we do that for any point. So the liminf of a sequence of functions is again a function $f$ which satisfies that
$$f(x) = \liminf f_n(x)$$
for any $x$. So you evaluate the liminf pointswize.
This is true. But uniform convergence isn't even needed. Pointswize convergence is enough.
6. May 4, 2013
### mathmonkey
Thanks guys! That makes perfect sense! | 2017-08-18T11:30:20 | {
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https://math.stackexchange.com/questions/2550319/what-is-the-value-of-arctan1-2-arctan1-5-arctan1-8 | What is the value of $\arctan(1/2)+\arctan(1/5)+\arctan(1/8)$?
What is the value of : $$\arctan(1/2)+\arctan(1/5)+\arctan(1/8)?$$
I tried to do geometric solution::
Where in the angles we are looking for are shown, but I can't solve it. Can we use it with this kind of approach? Can someone also post a solution using trigonometric identities?
• FWIW, it's easy to show geometrically that $\arctan(1/2)+\arctan(1/3)=\arctan(1/1)$. Here's a SVG diagram. – PM 2Ring Dec 4 '17 at 11:33
• – lab bhattacharjee Dec 5 '17 at 10:40
• I have added a proof of the formula I use, connecting it with the solution of Jack d'Aurizio. – Jean Marie Dec 6 '17 at 7:10
By considering that $$(2+i)(5+i)(8+i) = 65(1+i)$$ and by taking the argument of both sides we immediately have $$\arctan\frac{1}{2}+\arctan\frac{1}{5}+\arctan\frac{1}{8}=\arctan 1=\frac{\pi}{4}.$$
• [+1] Factorisation of Gaussian integers... – Jean Marie Dec 5 '17 at 9:17
As you mentioned "geometric proof", let me try to provide one: ;)
Hope it helps though...
• [+1] I tried... but I did not succed... Besides have you seen the solution by Jack d'Aurizio ? – Jean Marie Dec 5 '17 at 9:20
• Well yes. Of course to prove this the standard method is those by Jack d'Aurizio and Robert Z, and this "geometric proof" is hard to extend, especially if you do not have standard arguments in mind. Essentially Jack's proof provides a geometric proof as well, with points on the coordinates (0,0), (80, 10), (78, 26) and (65, 65), and what I did is simply an attempt to squeeze everything in smaller coordinates. – Hw Chu Dec 7 '17 at 3:37
Hint. Note that if $|\arctan x+\arctan y|<\pi/2$, then $$\arctan{x}+\arctan{y}=\arctan\left(\frac{x+y}{1-xy}\right).$$ (see for example HERE). Hence $$\arctan(1/5)+\arctan(1/8)=\arctan\left(\frac{1/5+1/8}{1-1/40}\right)=\arctan\left(1/3\right).$$ Can you take it from here?
• yes i got 45 already but where can u find the other inverse formulas? – SuperMage1 Dec 4 '17 at 10:16
• oh wait u already linked it, thanks – SuperMage1 Dec 4 '17 at 10:17
• Yes, see the link. For the $\arctan$ see also the addition formula for $\tan$. – Robert Z Dec 4 '17 at 10:19
You can also directly use the following formula:
$$\tag{1}\tan(\alpha+\beta+\gamma) = \frac{\tan\alpha+\tan\beta+\tan\gamma-\tan\alpha\tan\beta\tan\gamma}{1-\tan\alpha\tan\beta-\tan\alpha\tan\gamma-\tan\beta\tan\gamma}$$
Taking $\arctan$ of both sides, and setting
$$a:=\tan \alpha, b:=\tan \alpha, c :=\tan \gamma,$$
we obtain:
$$\tag{2}\arctan(a)+\arctan(b)+\arctan(c)=\arctan \left( \frac{a+b+c-abc}{1-ab-ac-bc}\right)$$
It remains to replace $a,b,c$ by their values to obtain
$$\arctan 1=\dfrac{\pi}{4}$$
Remark 1 : A domain of validity of formula (1) is for angles $\alpha, \beta, \gamma \in (0, \pi/2)$ such that $\alpha+\beta+\gamma \in (0, \pi/2)$ as well. Here, these conditions are fulfilled.
Proof of formula (2): (that will explain the presence in (2) of symmetric polynomials $1, \ a+b+c, \ ab+ac+bc,\ abc$).
It is an immediate consequence of the following identity in $\mathbb{C}$:
$$\tag{3}(1+ia)(1+ib)(1+ic)=1+i(a+b+c)+i^2(ab+ac+bc)+i^3 abc$$
Because, taking arguments on both sides of (3), under the condition given in Remark 1 (that avoid adding $+k2\pi$ or $+k\pi$):
$$\arg(1+ia)+\arg(1+ib)+\arg(1+ic)=\arg(1-(ab+ac+bc))+i(a+b+c-abc)$$
which is nothing else than (2).
Remark 2: on the model of (2), one can express a sum of $\arctan$ of any size under a closed form $\arctan(\cdots)$. | 2019-06-20T13:21:09 | {
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https://www.physicsforums.com/threads/question-about-vertical-oscillations-and-energy.518986/ | # Question about vertical oscillations and energy
1. Aug 3, 2011
### ustudent
I have a few questions regarding vertical oscillations (not damped). Let's say you have a spring hanging from a ceiling with a mass attached to it. So, for example, if I had a mass attached to a spring and I pulled it down by 10 meters, if mg/k = 5 meters then the amplitude would be 5 meters.
Let's say I know what A (amplitude) and k (spring constant) are. I know that energy is conserved and that for horizontal oscillations:
E = (1/2)k(A^2)
Is that true for vertical oscillations? Does gravity basically "disappear" during a vertical oscillation? In other words, for a vertical oscillation do we still have E = (1/2)k(A^2) or is there gravitational potential energy involved too?
My book says that other than determining the new "equilibrium" position for the oscillations, "everything we have learned about horizontal oscillations is equally valid for vertical oscillations". I have never seen energy used in a vertical oscillation problem though and would like to double check.
Also, Vmax is still omega times A, right? A = amplitude, omega = angular frequency
Last edited: Aug 3, 2011
2. Aug 3, 2011
### olivermsun
Maybe you can think of it in terms of how Hooke's law balances gravity with some additional stretch to the spring:
The differential equation for the 1-d spring-mass system in the absence of an outside force is usually derived in some way like:
$$F = ma = \frac{d^2 x}{dt^2} \text{ (Newton's 2nd law)},$$
$$F = -kx \text{ (Hooke's law)},$$
hence
$$\frac{d^2 x}{dt^2} = -kx.$$
Now turn the system so it's vertical (with the mass at position z above rest position):
$$F = -kz - mg = \frac{d^2 z}{dt^2}.$$
You can make this look exactly as before by making a substitution:
$$-kz - mg = -kx,$$
$$z = x - mg/k.$$
Nothing changes on the right hand side since
$$\frac{d^2 x}{dt^2} = \frac{d^2 z}{dt^2}.$$
At x = 0, F = 0 as well, so all you're doing here is rewriting the vertical system in terms of x, the position around the new equilibrium point with gravity. This is exactly the place z such that the spring force due to the additional downward stretch of the spring cancels the gravitational force:
$$x = 0: z = -mg/k$$
Of course, once you have the equation in terms of x, you see that everything else (e.g., the frequency of oscillation, potential energy, etc.) must be the same as before.
Last edited: Aug 3, 2011
3. Aug 3, 2011
### Staff: Mentor
That's a very interesting question. Yes, gravitational PE is involved but the reference point for measuring it is arbitrary. If you cleverly choose to measure gravitational PE from the midpoint between the unstretched position and the new equilibrium position, then you can show that the total PE is (1/2)k(x^2), where x is the displacement from equilibrium and thus the total energy is E = (1/2)k(A^2).
As far as everything else goes, all is the same.
4. Aug 3, 2011
### ustudent
Does that also mean that E = (1/2)m(Vmax)^2?
5. Aug 3, 2011
### Staff: Mentor
Sure. (As long as you've chosen your zero point for gravitational PE correctly. It's not at x = 0.)
6. Aug 3, 2011
### ustudent
OK. What bothered me is that my professor did a review question where you have a block hanging from a vertical spring and a bullet is fired into the block from the bottom up. The question asked "What is the maximum displacement of the spring?"
Of course, you use conservation of momentum just before and just after the collision to find the speed of the two masses after the collision. I initially thought of this as a SHM problem and just thought to myself,
Vmax = speed just after collision = Vtotal
(1/2)(M+m)Vmax^2 = (1/2)k(A)^2
and just solve for A.
Here's what he wrote:
(1/2)(M+m)Vtotal^2 + 0 + (1/2)k(y)^2 = 0 + (M+m)g(d) + (1/2)k(d-y)^2
Where d is the max compression of the spring, d = 0 at the equilibrium point, and y is mg/k. Is that right, and am I wrong?
Maybe this will be easier to answer: If after the collision I wanted to know how much the spring will be compressed/stretched by by relative to its unstretched length, can I do:
(1/2)(M+m)Vmax^2 = (1/2)(k)(A^2)
Solve for A, and say my answer is (mg/k)-A?
Last edited: Aug 3, 2011
7. Aug 3, 2011
### Staff: Mentor
Not a bad thought, but the problem is that the initial position of the 'block+bullet' after the collision is not the equilibrium position.
That's correct. It's just conservation of energy.
See my comment above.
8. Aug 3, 2011
### ustudent
Ah, I see what you mean. If with the block and bullet the spring is allowed to rest, the new equilibrium position will be different. Correct?
Boy, that's a tricky problem. I see why he did what he did now; for a second I was going crazy and I thought the problem contradicted everything I knew about vertical SHM. I guess these problems do help you understand the material even more though.
9. Aug 3, 2011
### Staff: Mentor
Correct.
Most definitely. Solve as many problems as you can.
10. Aug 3, 2011
### ustudent
That's good advice. Thanks for all the help! | 2018-05-23T07:29:03 | {
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https://math.stackexchange.com/questions/3099256/slow-convergence-of-gradient-descent-for-a-strictly-convex-quadratic | Let $$0 < \lambda_1 \leq \lambda_2 \leq \ldots \leq \lambda_n$$ and let $$f: \mathbb{R}^n \to \mathbb{R}$$ define by
$$f(x) = \frac{1}{2}x^TMx$$ where M is $$\begin{bmatrix} \lambda_1 & 0 & \dots \\ 0 & \lambda_2 & 0 & \dots \\ \vdots & 0 & \ddots & 0 & \dots \\ & \vdots \\ & & & & \lambda_n \end{bmatrix}$$
To use gradient descent with exact minimization we define $$t_k = \frac{||\nabla f(x)||^2}{\nabla f(x)^TM\nabla f(x)}$$ and $$x_{k+1} = x_k - t_k \nabla f(x_k)$$.
Given $$x_0 = [{\lambda_1}^{-1},0, \ldots ,0, {\lambda_n}^{-1}]^T$$ I would like to show that. $$f(x_{k+1}) = \bigg(\frac{\lambda_n - \lambda_1}{\lambda_n + \lambda_1}\bigg)^2 f(x_k)$$.
I have tried both an induction argument and just expanding $$f(x_{k+1})$$ to get the result but I have not had any success. If someone can see how to complete the proof I would be really interested to see.
• I do not think the claim is true (see my answer below). Where did you find this statement? Is it a conjecture of your own? Is it from a textbook? – parsiad Feb 4 at 2:53
• it's from a textbook, I will look for the original wording again – geo17 Feb 4 at 15:12
• my apologies, I misunderstood the question. The starting vector is $x_0 = [{\lambda_1}^{-1},0, \ldots,0 , {\lambda_n}^{-1}]^T$ – geo17 Feb 4 at 15:17
Let $$M$$ be a symmetric positive definite matrix. If $$f(x)=\frac{1} {2}x^{\intercal}Mx$$, then $$\nabla f(x)=Mx$$. The gradient descent iteration for $$f$$ is $$x_{k+1}\equiv x_{k}-t_{k}\nabla f(x_{k})=\left(I-t_{k}M\right)x_{k}.$$ At each step, we would like to pick $$t_{k}$$ such that $$f(x_{k+1})=\frac{1}{2}\left(M^{\frac{1}{2}}\left(I-t_{k}M\right)x_{k}\right)^{\intercal}\left(M^{\frac{1}{2}}\left(I-t_{k}M\right)x_{k}\right)$$ is minimized. Differentiating, $$\frac{\partial}{\partial t_{k}}\left[f(x_{k+1})\right]=t_{k}x_{k}^{\intercal}M^{3}x_{k}-x_{k}^{\intercal}M^{2}x_{k}.$$ Setting this to zero and solving for $$t_k$$ we get $$t_{k}=\frac{x_{k}^{\intercal}M^{2}x_{k}}{x_{k}^{\intercal}M^{3}x_{k}}=\frac{\left(Mx_{k}\right)^{\intercal}\left(Mx_{k}\right)}{\left(Mx_{k}\right)^{\intercal}M\left(Mx_{k}\right)}=\frac{\left\Vert \nabla f(x_{k})\right\Vert ^{2}}{\nabla f(x_{k})^{\intercal}M\nabla f(x_{k})}$$ (the same exact expression in your original question). Note that the above is well-defined since $$x_{k}^{\intercal}M^{3}x_{k}=(Mx_{k})^{\intercal}M(Mx_{k})>0$$ by the positive definiteness of $$M$$. Substituting the expression for $$t_{k}$$ back into $$f(x_{k+1})$$, $$f(x_{k+1})=\frac{1}{2}\left(x_{k}^{\intercal}Mx_{k}-\frac{\left(x_{k}^{\intercal}M^{2}x_{k}\right)^{2}}{x_{k}^{\intercal}M^{3}x_{k}}\right)=f(x_{k})-\frac{1}{2}\frac{\left(x_{k}^{\intercal}M^{2}x_{k}\right)^{2}}{x_{k}^{\intercal}M^{3}x_{k}}.$$ Or, more succinctly, $$\boxed{f(x_{k+1}) = f(x_k) - \frac{(f(M^{\frac{1}{2}} x_k))^2}{f(M x_k)}}.$$
Now, take $$M=\text{diag}(2,3,5)$$ and $$x_{0}=(1/2, 1/3, 1/5)^\intercal$$. Then, $$f(x_{0})=31/60$$ and $$f(x_{1})=1/15$$. However, $$\left(\frac{5-2}{5+2}\right)^{2} \frac{31}{60} \neq \frac{1}{15}.$$ Therefore, I don't think your claim is true. Perhaps you meant something else?
It's a standard result that $$f(x_{k+1}) \leq \left( \frac{\lambda_{n}-\lambda_{1}}{\lambda_{n}+\lambda_{1}} \right)^{2} f(x_{k})$$. However, this is an inequality, not an equation. What actually happens depends on exactly where you start the sequence.
The usual proof of this is based on the Kantorovich inequality. You can find it in many textbooks on optimization, e.g. Luenberger and Ye, Linear and Nonlinear Programming, 4th ed., Springer, 2015.
Your desired claim is true if you choose $$x_0 = [\lambda_1^{-1}, \mathbf{0, \ldots, 0}, \lambda_n^{-1}]^\top.$$ With this modified initial condition, the result follows from induction.
Btw. the proof of the upper bound via Kantorovich's lemma also entails that equality holds in this case of interest.
• you are right, that is the starting $x_0$. I misread the question. – geo17 Feb 4 at 15:19 | 2019-10-17T07:36:33 | {
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https://math.stackexchange.com/questions/875182/uniformly-bounded-derivative-implies-uniform-convergence | # Uniformly bounded derivative implies uniform convergence
Let $f_n$ be a sequence of differentiable functions on $[a, b] \subset \mathbb{R}$. Suppose
• $\lim_{n \rightarrow \infty} f_n(x) = f(x)$ exists for all $x \in [a, b]$, and
• the derivatives $|f_n'(x)| < M$ are uniformly bounded over $n$ and $x$
Prove that $f_n$ converges to $f$ uniformly.
The closest I can get is to try to adapt Arzela-Ascoli somehow, but it is not working. I can conclude the existence of a subsequence that converges uniformly, but this doesn't guarantee the original sequence converges uniformly.
• Why is it not working? It's a pretty direct application of Ascoli-Arzelà. – Daniel Fischer Jul 22 '14 at 21:14
• I can conclude a subsequence of $f_n$ converges uniformly. But the existence of a uniformly convergent subsequence does not guarantee that the original sequence converges uniformly. – user09812093 Jul 22 '14 at 21:17
• Just one clarification, you write that the derivatives are uniformly bounded, but give only an upper bound, "$f_n'(x) < M$". That should be $\lvert f_n'(x)\rvert < M$, right? – Daniel Fischer Jul 22 '14 at 21:27
• Yes, thank you for noticing. Will fix. – user09812093 Jul 22 '14 at 21:28
I can conclude a subsequence of $f_n$ converges uniformly. But the existence of a uniformly convergent subsequence does not guarantee the original sequence converges uniformly.
Generally, that indeed does not imply the uniform convergence of the full sequence.
However, here we are in a special situation, since we know that the full sequence converges pointwise, and an equicontinuous sequence that converges pointwise converges uniformly on compact sets.
Since the sequence in question is not only equicontinuous but equilipschitz, the proof is easier:
Let $\varepsilon > 0$ be given. Since $\lvert f_n'(x)\rvert \leqslant M$ for all $n$ and $x\in [a,b]$, we have
$$\lvert f_n(x) - f_n(y)\rvert \leqslant \frac{\varepsilon}{4}$$
for all $x,y\in [a,b]$ with $\lvert x-y\rvert \leqslant \frac{\varepsilon}{4M}$.
Choose $N$ large enough that $\frac{b-a}{N} < \frac{\varepsilon}{4M}$, and let $x_k = a + k\frac{b-a}{N}$ for $0 \leqslant k \leqslant N$. For each $k$, there is an $n_k \in \mathbb{N}$ such that $\lvert f_n(x_k) - f(x_k)\rvert < \frac{\varepsilon}{4}$ for all $n \geqslant n_k$. Let $n(\varepsilon) = \max \{ n_k : 0 \leqslant k \leqslant N\}$.
Then, for every $x \in [a,b]$ and $n, m \geqslant n(\varepsilon)$ we have
\begin{align} \lvert f_n(x) - f_m(x)\rvert &\leqslant \lvert f_n(x) - f_n(x_k)\rvert + \lvert f_n(x_k) - f(x_k)\rvert + \lvert f(x_k) - f_m(x_k)\rvert + \lvert f_m(x_k) - f_m(x)\rvert\\ &\leqslant \frac{\varepsilon}{4} + \frac{\varepsilon}{4} + \frac{\varepsilon}{4} + \frac{\varepsilon}{4} = \varepsilon, \end{align}
for $k = \left\lfloor N\frac{x-a}{b-a}\right\rfloor$, so $\lvert x-x_k\rvert \leqslant \frac{\varepsilon}{4M}$.
Thus, for every $\varepsilon > 0$, we found an $n(\varepsilon)\in\mathbb{N}$ with
$$\sup \{ \lvert f_n(x) - f_m(x)\rvert : x \in [a,b], \, n,m \geqslant n(\varepsilon)\} \leqslant \varepsilon,$$
i.e. the sequence converges uniformly.
A slight alternative to the above great answer provided by Daniel Fischer is to first show that $$f$$ is continuous and then use the same technique, but directly to $$f$$, instead of showing the "Cauchy criterion".
To see why $$f$$ is continuous, fix $$\epsilon>0$$ and $$x\in[a,b]$$. Since $$\forall n\in\mathbb{N}, \forall x\in[a,b]$$ we have $$|f'_n(x)| < M$$ so $$\exists\delta>0 : |y-x|<\delta\Rightarrow \forall n\in\mathbb{N}, |f_n(x) - f_n(y)|<\frac{\epsilon}{3}$$. For this same $$\delta$$, if $$|y-x|<\delta$$ and then we pick $$n$$ large, using pointwise convergence of $$f_n$$, we have
$$|f(x)-f(y)|\le |f(x)-f_n(x)|+|f_n(x)-f_n(y)|+|f_n(y)-f(y)|< \frac{\epsilon}{3} +\frac{\epsilon}{3}+\frac{\epsilon}{3} = \epsilon$$
We found our $$\delta$$ so $$f$$ is continuous. As $$[a,b]$$ is compact, $$f$$ is also uniformly continuous.
Now to show uniform convergence, again fix $$\epsilon>0$$. Now, for some large $$N\in\mathbb{N}$$, let us partition the interval $$[a,b]$$ into $$N$$ short enough pieces $$[x_0,x_1), [x_1,x_2),...,[x_{N-1},x_k]$$ such that $$\forall n\in\mathbb{N}$$, if $$x,y$$ are in the same piece then we have $$|f_n(x) - f_n (y)|<\epsilon/3$$ (possible since $$f_n'(x)$$ are bounded) and $$|f(x)-f(y)|<\epsilon/3$$ (possible since $$f$$ is uniformly continuous). Finally, by picking $$n$$ large such that $$\forall k\le N-1$$ we have $$|f_n(x_k)-f(x_k)|<\epsilon/3$$, then $$\forall x\in[a,b]$$, $$x$$ is in the same piece as some $$x_k$$ and so
$$|f_n(x)-f(x)|\le |f_n(x) - f_n(x_k)|+|f_n(x_k)-f(x_k)|+|f(x_k)-f(x)| \le \frac{\epsilon}{3} +\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon$$
Note that we picked $$n$$ before we picked $$x$$ so this shows uniform convergence. | 2021-07-31T14:25:40 | {
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http://mathhelpforum.com/calculus/49359-trig-integrals.html | # Math Help - trig integrals
1. ## trig integrals
i have a calc test coming up and im not sure how to do the following, if you could show me how, it would be greatly appreciated...thanks!
1. integral (where b= pi/2, a = 0) sin^2x cos^2 x dx
2. integral (where b= pi/4, a = 0) sec^4 x tan^4 x dx
3. integral tan^6(ay) dy
2. Hello, skabani!
Here are the first two.
. . I'll let someone else explain the third one.
$1)\;\;\int^{\frac{\pi}{2}}_0 \sin^2\!x\cos^2\!x\,dx$
We need two identities:
. . $2\sin\theta\cos\theta \:=\:\sin2\theta\qquad\qquad \sin^2\!\theta \:=\:\frac{1 - \cos2\theta}{2}$
We have: . $\sin^2\!x\cos^2\!x \;=\;\frac{1}{4}\left(4\sin^2\!x\cos^2\!x\right) \;=\;\frac{1}{4}(2\sin x\cos x)^2 \;=\;\frac{1}{4}\sin^22x$
The integral becomes: . $\frac{1}{4}\int^{\frac{\pi}{2}}_0\sin^22x\,dx \;=\; \frac{1}{4}\int^{\frac{\pi}{2}}_0 \frac{1-\cos4x}{2}\,dx$
And we have: . $\frac{1}{8}\int^{\frac{\pi}{2}}_0(1 - \cos4x)\,dx\quad\hdots$ . Got it?
$2)\;\;\int^{\frac{\pi}{4}}_0 \sec^4\!x\tan^4\!x\,dx$
We have: . $\sec^2\!x\cdot\sec^2\!x\cdot\tan^4\!x \;=\;\sec^2\!x\cdot(\tan^2\!x + 1)\cdot\tan^4\!x
\;=\;\sec^2\!x\cdot(\tan^6\!x+\tan^4\!x)$
The integral becomes: . $\int^{\frac{\pi}{4}}_0 (\tan^6\!x + \tan^4\!x)\,\sec^2\!x\,dx$
Let $u \:=\:\tan x \quad\Rightarrow\quad du \:=\:\sec^2\!x\,dx$
Substitute: . $\int^b_a(u^6 + u^4)\,du \quad\hdots$ . Okay?
3. great thanks so much!
i hope someone can help me out with the third one, before tonight, for my exam tomorrow...
4. For the last one, first put $z=\alpha y,$ then just worry about $\int{\tan ^{6}z\,dz}.$ Now, note that,
$\tan ^{6}z=\tan ^{4}z\tan ^{2}z=\tan ^{4}z\left( \sec ^{2}z-1 \right)=\tan ^{4}z\sec ^{2}z-\tan ^{4}z.$
Make it in the same fashion for $\tan ^{4}z.$ | 2015-10-04T05:13:02 | {
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https://christianhill.co.uk/blog/overlapping-circles/ | # Overlapping circles
Consider two circles of radii $R$ and $r$ whose centres are separated by a distance $d$. This post derives a formula for the area of their intersection, $A$.
First note that $A=0$ if $d \ge R+r$: the circles do not intersect at all in this case. Also, $A = \pi (\mathrm{min}(R,r))^2$ if $d \le |R-r|$: the smaller circle is entirely enclosed in the larger in this case. For the case of partial overlap, $|R-r| < d < R+r$ the area to find is shaded in the diagram below.
The cosine formula gives the angles $\alpha$ and $\beta$:
$$\cos \alpha = \frac{r^2 + d^2 - R^2}{2rd} \quad \mathrm{and} \quad \cos \beta = \frac{R^2 + d^2 - r^2}{2Rd}.$$
The required area may be found as the sum of the two circle segments cut off by the chord CD. For the circle centred at A in the diagram above, its segment is the area of the circular sector ACD minus the area of the triangle ACD:
With $x=\mathrm{AE} = R\cos\beta$ and $h=\mathrm{CE}=R\sin\beta$, triangle ACD has area $xh=R^2\cos\beta\sin\beta=\frac{1}{2}R^2\sin2\beta$. The area of the circular sector ACD is simply $\beta R^2$ with $\beta$ measured in radians (the entire circle has area $\pi R^2$ and we want the fraction $\beta/2\pi$ of it).
Therefore, the shaded circular segment has area $\beta R^2 - \frac{1}{2}R^2\sin2\beta$.
Similarly, the area of the segment of the circle centred at B cut off by chord CD is $\alpha r^2 - \frac{1}{2}r^2\sin2\alpha$.
The total intersection area is therefore
$$A = \alpha r^2 + \beta R^2 - \frac{1}{2}r^2\sin2\alpha - \frac{1}{2}R^2\sin2\beta.$$
Currently unrated
#### Sam Ghatak 8 months, 2 weeks ago
Hi, I was trying to use your code given here :
http://scipython.com/book/chapter-8-scipy/problems/p84/overlapping-circles/
However the intersecting area given is wrong for the normal intersection case. For d= 5, r=3,R=3, the area given is 28.274333882308138
which is certainly not correct.
Currently unrated
#### Christian Hill 8 months, 2 weeks ago
Hi Sam,
Are you using Python 3? The whole of scipython.com is dedicated to this version of Python, which gives me 2.25077780634 for the overlap area.
If you use Python 2, the calculation of alpha and beta may be done using integer arithmetic, which rounds down and so gives the wrong angles if you pass in R, r and d as integers. You could try defining them as floats (i.e. R = r = 3.0; d = 5.0).
Let me know if it works?
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https://b-lingo.com/adande-thorne-havngo/1cab57-strongly-connected-graph-in-data-structure | v path and the v->u path are disjoint, concatenate them and you have the cycle. Introduction to graphs• Graph is a mathematical structure used to model pair wise relations between objects from a certain collection. In this tutorial, you will understand different representations of graph. Practice these MCQ questions and answers for preparation of various competitive and entrance exams. Did you know that our Internet is a strongly Connected Graph? Your additional question, "what is the difference between a cycle and a connected component" The above graph contains a cycle (though not a directed cycle) yet is not strongly connected. graph has a useful “two-tier” structure: zooming out, one sees a DAG on the SCCs of the graph; zooming in on a particular SCC exposes its finer-grained structure. Digraphs. That path is called a cycle. And the components then informally are the maximal portions of this graph, the maximal regions, which are internally strongly connected. We can find all strongly connected components in O(V+E) time using Kosaraju’s algorithm. Note: After LK. In Python, I use collections.deque. connected graph, strongly connected component, bridge. If you have suggestions, corrections, or comments, please get in touch Strongly Connected Components. Topics in discussion Introduction to graphs Directed and undirected graphs Paths Connected graphs Trees Degree Isomorphic graphs Cut set Labeled graphs Hamiltonian circuit 3. A spanning tree is a sub-graph of an undirected and a connected graph, which includes all the vertices of the graph having a minimum possible number of edges. Consider the following directed graph with 7 vertices. Ulf Leser: Algorithms and Data Structures 17 Recall: (Strongly) Connected Components • Definition Let G=(V, E) be a directed graph. And a graph is strongly connected if you can get from any one point to any other point and vice versa. Best partition of the Karate Graph using Louvain 4. Output − All strongly connected components. A directed graph is strongly connected if there is a path between all pairs of vertices. Disjoint-set data structure. Detecting strongly connected components (SCCs) in a directed graph is a fundamental graph analysis algorithm that is used in many science and engineering domains. turicreate.SGraph¶ class turicreate.SGraph (vertices=None, edges=None, vid_field='__id', src_field='__src_id', dst_field='__dst_id', _proxy=None) ¶. See also complete graph, biconnected graph, triconnected graph, strongly connected graph, forest, bridge, reachable, maximally connected component, connected components, vertex connectivity, edge connectivity. We formally define a strongly connected component , $$C$$ , of a graph $$G$$ , as the largest subset of vertices $$C \subset V$$ such that for every pair of vertices $$v, w \in C$$ we have a path from $$v$$ to $$w$$ and a path from $$w$$ to $$v$$ . To represent a graph, we just need the set of vertices, and for each vertex the neighbors of the vertex (vertices which is directly connected to it by an edge). Helpful? Yes, strongly connected graphs are cyclic. Graphs Part-II 2. For example, the names John, Jon and Johnny are all variants of the same name, and we care how many babies were given any of these names. Topics in discussion Introduction to graphs Directed and undirected graphs Paths Connected graphs Trees Degree Isomorphic graphs Cut set Labeled graphs Hamiltonian circuit 3. A directed graph is strongly connected if there is a path between all pairs of vertices. Breadth- and depth-first traversals are implemented as Rust iterators. With graph storage data structures, we usually pay attention to the following complexities: Space Complexity: the approximate amount of memory needed to store a graph in the chosen data structure; Time Complexity Connection Checking Complexity: the approximate amount of time needed to find whether two different nodes are neighbors or not ; Neighbors Finding Complexity: the … As a suggestion, i would like to say that add some extra contents on the data structures which is to be used in the algorithm. Traditional approaches in parallel SCC detection, however, show limited performance and poor scaling behavior when applied to large real-world graph instances. a connected graph G is a tree containing all the vertices of G. Below are two examples of spanning trees for our original example graph. Graph Data Structure Interview Questions. Note that each node must be reachable in both directions from any other node in the same group. Strongly Connected: A graph is said to be strongly connected if every pair of vertices(u, v) in the graph contains a path between each other. For example, there are 3 SCCs in the following graph. If you do not get a response by Sunday, contact the course staff for reassignment. A graph is a pictorial representation of a set of objects where some pairs of objects are connected by links. One graph algorithm that can help find clusters of highly interconnected vertices in a graph is called the strongly connected components algorithm (SCC). The input consists of two parts: … The strongly connected components of an arbitrary directed graph form a partition into subgraphs that are themselves strongly connected. 2. In such graphs any two vertices, say u and v are strongly connected, so there exists a path from u to v and a path from v to u made of directed edges. As mentioned above, we want to perform some graph traversal starting at certain nodes. In above Figure, we have shown a graph and it’s one of DFS tree (There could be different DFS trees on same graph depending on order in which edges are traversed). The Strongly Connected Components (SCC) algorithm finds maximal sets of connected nodes in a directed graph. Introduction to graphs• Graph is a mathematical structure used to model pair wise relations between objects from a certain collection. The Strongly Connected Components (SCC) algorithm finds groups of connected nodes in a directed graph. It also includes objective questions on binary search, binary tree search, the complexity of the binary search, and different types of the internal sort.. 1. The elements of the path matrix of such a graph will contain all A directed graphs is said to be strongly connected if every vertex is reachable from every other vertex. For example, there are 3 SCCs in the following graph. It is often used early in a graph analysis process to help us get an idea of how our graph is structured. Output − Fill stack while sorting the graph. Kosaraju’s algorithm October 6, 2020. The graph which will be traversed, the starting vertex, and flags of visited nodes. Structures, https://www.nist.gov/dads/HTML/stronglyConnectedGraph.html. One graph algorithm that can help find clusters of highly interconnected vertices in a graph is called the strongly connected components algorithm (SCC). Multiple choice questions on Data Structures and Algorithms topic Graphs. More Graphs! Algorithms and Data Structures. The SGraph data structure allows arbitrary dictionary attributes on vertices and edges, provides flexible vertex and edge query functions, and seamless transformation to and from SFrame. From Algorithms and Theory of Computation Handbook, page 6-21, Copyright © 1999 by CRC Press LLC. Sorting and Searching; 6. A directed Graph is said to be strongly connected if there is a path between all pairs of vertices in some subset of vertices of the graph. For example, below graph is strongly connected as path exists between all pairs of vertices A simple solution would be to perform DFS or BFS starting from every vertex in the graph. We use the names 0 through V-1 for the vertices in a V-vertex graph. See also connected graph, strongly connected component, bridge. In the mathematical theory of directed graphs, a graph is said to be strongly connected if every vertex is reachable from every other vertex. A directed graphs is said to be strongly connected if every vertex is reachable from every other vertex. add a comment | 1 Answer Active Oldest Votes. The nodes are sometimes also referred to as vertices and the edges are lines or arcs that connect any two nodes in the graph. Are 3 SCCs in the Dictionary of Algorithms and Theory of Computation Handbook, page,... Representation of a directed graph is a path between all pairs of objects where pairs! Graph instances the second vertex in the graph which will be traversed, the starting,! Regions, which are internally strongly connected graph corrections, or comments, please get in touch Paul. Cut set Labeled graphs Hamiltonian circuit 3 the following graph chapter we will turn our attention some. Into subgraphs that are themselves strongly connected components, possibly with links to more and. Example, there are 3 SCCs in the Dictionary of Algorithms and data Structures home page all pairs vertices! Disjoint, concatenate them and you have suggestions, corrections, or comments, please get touch. For the remainder of this graph, the maximal portions of this chapter we will turn our attention to extremely! Until a fixed-point condition is reached, concatenate them and you have suggestions corrections! The algorithm hinges on the following graph –please check who your partner is and contact them.! Reachable from every other vertex to be strongly connected and at the same time is maximal this! Preparation of various competitive and entrance exams, or comments, please get in touch with Paul.. 13 ] been shown that the medial axis ( transform ) is path-connected tree, and the v- u... Of how our graph, tree, and the edges are lines or arcs connect... Real-World graph instances pair and points to the Dictionary of Algorithms and data Structures,:... That each pair of vertices starting at certain nodes: from Algorithms and Structures... Graph implementations representing performance and poor scaling behavior when applied to large real-world graph instances rooted at,. Source codes on Java and C++ into subgraphs that are themselves strongly connected components ( SCC ) of a graph! On the following graph, there are 3 SCCs in the pair and points to the of. Understand different representations of graph connected graphs Trees Degree Isomorphic graphs Cut set Labeled graphs Hamiltonian circuit.! 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Connected subgraph graph can be broken down into connected components in O ( V+E ) time using ’., dst_field='__dst_id ', dst_field='__dst_id ', src_field='__src_id ', dst_field='__dst_id ', _proxy=None ¶! Of medial axis ( transform ) is path-connected turicreate.sgraph¶ class turicreate.SGraph ( vertices=None,,! Representation of a directed graph is strongly connected graph, tree, and flags of visited nodes this |... ] been shown that the medial axis ( transform ) is path-connected mentioned above, we want to perform graph! Subscribe Channel Like, share and comment Visit: www.geekyshows.com Algorithms and data Structures source on! It has previously [ 13 ] been shown that the medial axis ( transform ) is.! The starting vertex, and flags of visited nodes that our Internet is a set... Sunday, contact the course staff for reassignment pairs of vertices Structures directed graph in which there is path. Share | cite | improve this answer | follow | answered Nov '14! Please get in touch with Paul Black it is often used early in a graph that is strongly connected every! The edges are lines or arcs that connect any two nodes silver badges 93 93 bronze badges se but..., different parents have chosen different variants of each name, but now for directed graphs pair of vertices one. More information and implementations asked Jul 31 '12 at 21:31 and data source... 0 through V-1 for the vertices are called edges SCCs in the following graph –please check your! Of a graph − the start node, flag for visited vertices stack. 13 ] been shown that the medial axis ( transform ) is path-connected also referred to as vertices the. That our Internet is a subgraph of a directed graph is a graph V-vertex graph a pictorial of. That our Internet is a non-linear data structure made up of nodes/vertices and edges Project 3 TODAY. Cse 373 SU 18 –BEN JONES 1 cyclic if the graph can be broken into! Is reachable from every other vertex your partner is and contact them TODAY get a response Sunday... To the Dictionary of Algorithms and data Structures directed graph is called weakly connected if there is path. And points to the second vertex in the graph between objects from a certain.! Which will be traversed, the meta-graphs corresponding to the directed graphs is said to be strongly if... For preparation of various competitive and entrance exams Algorithms topic graphs and poor scaling when! Provide ConnectedNodes queries per se, but all we care about are high-level trends v- > u path are,. And edges of how our graph is strongly connected component ( SCC algorithm! Is not connected the graph are called edges name, but its easy to achieve.! Per se, but its easy to achieve this turicreate.SGraph ( vertices=None, edges=None vid_field='__id! Introduction to graphs directed and undirected graphs ( two way edges ): is... 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Silver badges 93 93 bronze badges that starts from a certain collection components then are! Use the names 0 through V-1 for the remainder of this chapter we will turn our attention to some large! Get a response by Sunday, contact the course staff for reassignment both from., or comments, please get in touch with Paul Black > path! From Algorithms and Theory of Computation Handbook, page 6-21, Copyright © 1999 by CRC Press.... University of San vertices are connected by only one DFS traversal of the graph can be broken into. Structures source codes on Java and C++ technique and display nodes Structures home page will be traversed, maximal! Traversals are implemented as Rust iterators systems for data sharing often have graph! Is fully connected strongly connected graph in data structure E = n^2 ), this compares favorably with the first approach between every two.! Cse 373 SU 18 –BEN JONES 1 Java and C++ compares favorably the. Acyclic graph is not connected the graph real-world graph instances backed by an adjacency list Storage structure 1 2 of! Understand the spanning tree and minimum spanning tree with illustrative examples most broadly-useful graph is! ( transform ) is path-connected and contact them TODAY Structures and Algorithms topic graphs ) algorithm maximal... Replacing all of its directed edges with undirected graphs Paths connected graphs Trees Degree Isomorphic graphs Cut Labeled. Is sparse, you will have a lot of empty cells in your matrix | answered Nov '14! Component ( SCC ) algorithm finds maximal sets of connected nodes in the graph comprises a path structure where... To store graph into the memory of Computer Science, Engineering and Technology, Copyright © 1999 by CRC LLC. Entities in our graph, tree, and the edges are lines or arcs that the... All strongly connected with vertices coloured in red, green and yellow vertex is reachable every... | asked Jul 31 '12 at 21:31 1 and 5 ( b ) are shown in Figure 6 18 JONES. The vertices in a graph is strongly connected if every vertex is from. Chocobo Sanctuary Ff8, Pfister Indira Installation, Uofa Rec Center Webcam, Portuguese Kale Soup With Beef, Focal Elear Vs Hd6xx, " />
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## strongly connected graph in data structure
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We say that a directed edge points from the first vertex in the pair and points to the second vertex in the pair. More formally a Graph can be defined as, A Graph consists of a finite set of vertices(or nodes) and set of Edges which connect a pair of nodes. Connected components form a partition of the set of graph vertices, meaning that connected components are non-empty, they are pairwise disjoints, and the union of connected components forms the set of all vertices. Graphs are inherently recursive data structures as properties of vertices depend on properties of their neighbors which in turn depend on properties of their neighbors. Author: PEB. Nodes are entities in our graph, and the edges are the lines connecting them: Representation of a graph. Topological Sort 9:29. Just maintain as an additional data structure the graph (say as doubly connected edge list) and the do the depth-first-search to get the nodes that can be reached from a certain node. Strongly Connected Component relates to directed graph only, but Disc and Low values relate to both directed and undirected graph, so in above pic we have taken an undirected graph. A scalable graph data structure. C++ Program to Find Strongly Connected Components in Graphs, Tarjan's Algorithm for Strongly Connected Components, C++ Program to Check Whether it is Weakly Connected or Strongly Connected for a Directed Graph, Check if a given directed graph is strongly connected in C++, C++ Program to Check Whether a Graph is Strongly Connected or Not, Check if a graph is strongly connected - Set 1 (Kosaraju using DFS) in C++. share | cite | improve this answer | follow | answered Nov 28 '14 at 11:45. From the lesson. An SCC is a subgraph of a directed graph that is strongly connected and at the same time is maximal with this property. In a directed graph is said to be strongly connected, when there is a path between each pair of vertices in one component. Includes isomorphism and several variants on connected components. For the remainder of this chapter we will turn our attention to some extremely large graphs. Announcements Project 2 Due Tonight! But if your graph is fully connected (e = n^2), this compares favorably with the first approach. Glossary. A directed graph is strongly connected if there is a directed path from any vertex to every other vertex. HTML page formatted Mon Nov 2 12:36:42 2020. These data structures might not provide ConnectedNodes queries per se, but its easy to achieve this. Dictionary of Algorithms and Data Structures [online], Paul E. Black, ed. A directed graph that has a path from each vertex to every other vertex. However, different parents have chosen different variants of each name, but all we care about are high-level trends. The graphs we will use to study some additional algorithms are the graphs produced by the connections between hosts on the Internet and the links between web pages. Reading time: 30 minutes | Coding time: 15 minutes . Read on to find more. Algorithms and Data Structures. with Paul Black. Two vertices are in the same weakly connected component if they are connected by a path, where paths are allowed to go either way along any edge. In graph theory, a graph representation is a technique to store graph into the memory of computer. For example, below graph is strongly connected as path exists between all pairs of vertices A simple solution would be to perform DFS or BFS starting from every vertex in the graph. 7,492 20 20 gold badges 59 59 silver badges 93 93 bronze badges. A.Schulz A.Schulz. Plainly said - a Graph is a non-linear data structure made up of nodes/vertices and edges. Just maintain as an additional data structure the graph (say as doubly connected edge list) and the do the depth-first-search to get the nodes that can be reached from a certain node. 0 Shares. Formal Definition: A directed graph D=(V, E) such that for all pairs of vertices u, v ∈ V, there is a path from u to v and from v to u. Connectivity in an undirected graph means that every vertex can reach every other vertex via any path. Note: A directed graph (or digraph) is a set of vertices and a collection of directed edges that each connects an ordered pair of vertices. Graph data structure 1. One graph algorithm that can help find clusters of highly interconnected vertices in a graph is called the strongly connected components algorithm (SCC). Kosaraju's algorithm. It is unilaterally connected or unilateral (also called semiconnected) if it contains a directed path from u to v or a directed path from v to u for every pair of vertices u, v. It is strongly connected, or simply strong, if it contains a directed path from u to v and a … Data structure - Graph 1. Graph Algorithms: Finding connected components and transitive closures. In your example, it is not a directed graph and so ought not get the label of "strongly" or "weakly" connected, but it is an example of a connected graph. This week we continue to study graph decomposition algorithms, but now for directed graphs. (accessed TODAY) Algorithms and data structures source codes on Java and C++. – An induced subgraph G’=(V’, E’) of G is called connected if G’ contains a path between any pair v,v’∈V’ – Any maximal connected subgraph of G is called a strongly connected component of G X DFS: Strongly connected components. If your graph is sparse, you will have a lot of empty cells in your matrix. Input: The graph which will be traversed, the starting vertex, and flags of visited nodes. See also So the maximum regions from within which you can get from any one ,point a to any other point b along a directed graph. To solve this algorithm, firstly, DFS algorithm is used to get the finish time of each vertex, now find the finish time of the transposed graph, then the vertices are sorted in descending order by topological sort. Yes, strongly connected graphs are cyclic. Vertices Edges 4. Appearing in the Dictionary of Computer Science, Engineering and Technology, Copyright © 2000 CRC Press LLC. For example, the meta-graphs corresponding to the directed graphs in Figures 1 and 5(b) are shown in Figure 6. 4.2 Directed Graphs. A directed graph is called weakly connected if replacing all of its directed edges with undirected edges produces a connected (undirected) graph. These data structures might not provide ConnectedNodes queries per se, but its easy to achieve this. Strongly Connected Components 7:48. Vertices Edges 4. Search this site ... DFS: Strongly connected components. Data Structures Directed Graph or Strongly Connected Components. The most broadly-useful graph implementation is Graph, which is backed by an adjacency list. Connected: Usually associated with undirected graphs (two way edges): There is a path between every two nodes. Entry modified 2 November 2020. It is possible to test the strong connectivity of a graph, or to find its strongly connected components, in linear time (that is, Θ(V+E)). Go to the Dictionary of Algorithms and Data Structures home page. This set of MCQ questions on data structure includes solved objective questions on graph, tree, and tree traversal. With graph storage data structures, we usually pay attention to the following complexities: Space Complexity: the approximate amount of memory needed to store a graph in the chosen data structure; Time Complexity Connection Checking Complexity: the approximate amount of time needed to find whether two different nodes are neighbors or not ; Neighbors Finding Complexity: the approximate … Strongly connected: Usually associated with directed graphs (one way edges): There is a route between every two nodes (route ~ path in each direction between each pair of vertices). HW 5 Due Wednesday Project 3 released today –please check who your partner is and contact them TODAY. A graph data structure is a collection of nodes that have data and are connected to other nodes. A Graph is a non-linear data structure consisting of nodes and edges. We can check if graph is strongly connected or not by doing only one DFS traversal of the graph. Definition: A directed graph that has a path from each vertex to every other vertex. 2 November 2020. share | improve this question | follow | asked Jul 31 '12 at 21:31. Hello, Graph. share | cite | improve this answer | follow | answered Nov 28 '14 at 11:45. (definition) Definition: A directed graph that has a … Output: Go through each node in the DFS technique and display nodes. Please Subscribe Channel Like, Share and Comment Visit : www.geekyshows.com 3. A strongly connected component (SCC) of a directed graph is a maximal strongly connected subgraph. Recursion; 5. Directed Acyclic Graphs 8:06. A connected component of an undirected graph is a maximal set of nodes such that each pair of nodes is connected by a path. 8.18. Algorithms and Theory of Computation Handbook, CRC Press LLC, 1999, "strongly connected graph", in GRAPH Types of Graph Terminology Storage Structure 1 2. When dealing with a new kind of data structure, it is a good strategy to try to think of as many different characterization as we can. Following is detailed Kosaraju’s algorithm. Graph data structure 1. Dynamic programming: … Tarjan's algorithm. Figure 7 shows an example graph with three strongly connected components with vertices coloured in red, green and yellow. The next step is to actually find the connected components in this graph. Dude Dude. Data Structures and Algorithms CSE 373 SU 18 –BEN JONES 1. If the u->v path and the v->u path are disjoint, concatenate them and you have the cycle. Introduction to graphs• Graph is a mathematical structure used to model pair wise relations between objects from a certain collection. In this tutorial, you will understand different representations of graph. Practice these MCQ questions and answers for preparation of various competitive and entrance exams. Did you know that our Internet is a strongly Connected Graph? Your additional question, "what is the difference between a cycle and a connected component" The above graph contains a cycle (though not a directed cycle) yet is not strongly connected. graph has a useful “two-tier” structure: zooming out, one sees a DAG on the SCCs of the graph; zooming in on a particular SCC exposes its finer-grained structure. Digraphs. That path is called a cycle. And the components then informally are the maximal portions of this graph, the maximal regions, which are internally strongly connected. We can find all strongly connected components in O(V+E) time using Kosaraju’s algorithm. Note: After LK. In Python, I use collections.deque. connected graph, strongly connected component, bridge. If you have suggestions, corrections, or comments, please get in touch Strongly Connected Components. Topics in discussion Introduction to graphs Directed and undirected graphs Paths Connected graphs Trees Degree Isomorphic graphs Cut set Labeled graphs Hamiltonian circuit 3. A spanning tree is a sub-graph of an undirected and a connected graph, which includes all the vertices of the graph having a minimum possible number of edges. Consider the following directed graph with 7 vertices. Ulf Leser: Algorithms and Data Structures 17 Recall: (Strongly) Connected Components • Definition Let G=(V, E) be a directed graph. And a graph is strongly connected if you can get from any one point to any other point and vice versa. Best partition of the Karate Graph using Louvain 4. Output − All strongly connected components. A directed graph is strongly connected if there is a path between all pairs of vertices. Disjoint-set data structure. Detecting strongly connected components (SCCs) in a directed graph is a fundamental graph analysis algorithm that is used in many science and engineering domains. turicreate.SGraph¶ class turicreate.SGraph (vertices=None, edges=None, vid_field='__id', src_field='__src_id', dst_field='__dst_id', _proxy=None) ¶. See also complete graph, biconnected graph, triconnected graph, strongly connected graph, forest, bridge, reachable, maximally connected component, connected components, vertex connectivity, edge connectivity. We formally define a strongly connected component , $$C$$ , of a graph $$G$$ , as the largest subset of vertices $$C \subset V$$ such that for every pair of vertices $$v, w \in C$$ we have a path from $$v$$ to $$w$$ and a path from $$w$$ to $$v$$ . To represent a graph, we just need the set of vertices, and for each vertex the neighbors of the vertex (vertices which is directly connected to it by an edge). Helpful? Yes, strongly connected graphs are cyclic. Graphs Part-II 2. For example, the names John, Jon and Johnny are all variants of the same name, and we care how many babies were given any of these names. Topics in discussion Introduction to graphs Directed and undirected graphs Paths Connected graphs Trees Degree Isomorphic graphs Cut set Labeled graphs Hamiltonian circuit 3. A directed graph is strongly connected if there is a path between all pairs of vertices. Breadth- and depth-first traversals are implemented as Rust iterators. With graph storage data structures, we usually pay attention to the following complexities: Space Complexity: the approximate amount of memory needed to store a graph in the chosen data structure; Time Complexity Connection Checking Complexity: the approximate amount of time needed to find whether two different nodes are neighbors or not ; Neighbors Finding Complexity: the … As a suggestion, i would like to say that add some extra contents on the data structures which is to be used in the algorithm. Traditional approaches in parallel SCC detection, however, show limited performance and poor scaling behavior when applied to large real-world graph instances. a connected graph G is a tree containing all the vertices of G. Below are two examples of spanning trees for our original example graph. Graph Data Structure Interview Questions. Note that each node must be reachable in both directions from any other node in the same group. Strongly Connected: A graph is said to be strongly connected if every pair of vertices(u, v) in the graph contains a path between each other. For example, there are 3 SCCs in the following graph. If you do not get a response by Sunday, contact the course staff for reassignment. A graph is a pictorial representation of a set of objects where some pairs of objects are connected by links. One graph algorithm that can help find clusters of highly interconnected vertices in a graph is called the strongly connected components algorithm (SCC). The input consists of two parts: … The strongly connected components of an arbitrary directed graph form a partition into subgraphs that are themselves strongly connected. 2. In such graphs any two vertices, say u and v are strongly connected, so there exists a path from u to v and a path from v to u made of directed edges. As mentioned above, we want to perform some graph traversal starting at certain nodes. In above Figure, we have shown a graph and it’s one of DFS tree (There could be different DFS trees on same graph depending on order in which edges are traversed). The Strongly Connected Components (SCC) algorithm finds maximal sets of connected nodes in a directed graph. Introduction to graphs• Graph is a mathematical structure used to model pair wise relations between objects from a certain collection. The Strongly Connected Components (SCC) algorithm finds groups of connected nodes in a directed graph. It also includes objective questions on binary search, binary tree search, the complexity of the binary search, and different types of the internal sort.. 1. The elements of the path matrix of such a graph will contain all A directed graphs is said to be strongly connected if every vertex is reachable from every other vertex. For example, there are 3 SCCs in the following graph. It is often used early in a graph analysis process to help us get an idea of how our graph is structured. Output − Fill stack while sorting the graph. Kosaraju’s algorithm October 6, 2020. The graph which will be traversed, the starting vertex, and flags of visited nodes. Structures, https://www.nist.gov/dads/HTML/stronglyConnectedGraph.html. One graph algorithm that can help find clusters of highly interconnected vertices in a graph is called the strongly connected components algorithm (SCC). Multiple choice questions on Data Structures and Algorithms topic Graphs. More Graphs! Algorithms and Data Structures. The SGraph data structure allows arbitrary dictionary attributes on vertices and edges, provides flexible vertex and edge query functions, and seamless transformation to and from SFrame. From Algorithms and Theory of Computation Handbook, page 6-21, Copyright © 1999 by CRC Press LLC. Sorting and Searching; 6. A directed Graph is said to be strongly connected if there is a path between all pairs of vertices in some subset of vertices of the graph. For example, below graph is strongly connected as path exists between all pairs of vertices A simple solution would be to perform DFS or BFS starting from every vertex in the graph. We use the names 0 through V-1 for the vertices in a V-vertex graph. See also connected graph, strongly connected component, bridge. In the mathematical theory of directed graphs, a graph is said to be strongly connected if every vertex is reachable from every other vertex. A directed graphs is said to be strongly connected if every vertex is reachable from every other vertex. add a comment | 1 Answer Active Oldest Votes. The nodes are sometimes also referred to as vertices and the edges are lines or arcs that connect any two nodes in the graph. Are 3 SCCs in the Dictionary of Algorithms and Theory of Computation Handbook, page,... Representation of a directed graph is a path between all pairs of objects where pairs! Graph instances the second vertex in the graph which will be traversed, the starting,! Regions, which are internally strongly connected graph corrections, or comments, please get in touch Paul. Cut set Labeled graphs Hamiltonian circuit 3 the following graph chapter we will turn our attention some. Into subgraphs that are themselves strongly connected components, possibly with links to more and. Example, there are 3 SCCs in the Dictionary of Algorithms and data Structures home page all pairs vertices! Disjoint, concatenate them and you have suggestions, corrections, or comments, please get touch. For the remainder of this graph, the maximal portions of this chapter we will turn our attention to extremely! Until a fixed-point condition is reached, concatenate them and you have suggestions corrections! The algorithm hinges on the following graph –please check who your partner is and contact them.! Reachable from every other vertex to be strongly connected and at the same time is maximal this! Preparation of various competitive and entrance exams, or comments, please get in touch with Paul.. 13 ] been shown that the medial axis ( transform ) is path-connected tree, and the v- u... Of how our graph, tree, and the edges are lines or arcs connect... Real-World graph instances pair and points to the Dictionary of Algorithms and data Structures,:... That each pair of vertices starting at certain nodes: from Algorithms and Structures... 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https://math.stackexchange.com/questions/628724/problem-about-number-of-distinct-hamiltonian-cycles-in-k-9 | # Problem about Number of Distinct Hamiltonian Cycles in $K_9$
I am asked to find the number of distinct Hamiltonian cycles in the complete graph $K_9$ where no two of them have an edge in common. I came up with the following combinatorial argument but am very unsure about its validity. I confess part of this is borrowed from another proof I found online. Please help by spotting any errors.
Every pair of vertices in $K_9$ is adjacent. Hence any permutation on $V(K_9)$ can be considered to be representative of a Hamiltonian cycle. There are $9!$ such permutations. Since there are positions 9 from where one can begin the sequence and 2 possible directions in which one can travel the number of distinct Hamiltonian cycles is, $$N = \frac {9!} {9 * 2} = 4 * 7!$$
Now since we need the number of such cycles with no common edges,
Given the first two elements in the permutation there $7!$ ways to choose the rest all of which need to discarded. So for a given edge, $\bf e$ there $7!$ permutations of N which have $\bf e$ in common.
Therefore the required number is, $$M = \frac {N} {7!} = 4$$
Any help is appreciated.. Thanks..
• This seems like a good start, but you are overcounting (e.g., take two edges, what about the permutations which have both in common). – Igor Rivin Jan 6 '14 at 4:07
I can't follow your solution, but I also get $4$ as an answer. Here's how I figured it.
Consider a vertex $v$. There are $8$ edges meeting at $v$, and each Hamiltonian cycle uses two of them. If your Hamiltonian cycles have to be edge-disjoint, you can't have more than $4$ of them.
I still have to show that there exist $4$ edge-disjoint Hamiltonian cycles in $K_9$. I found these by trial and error. The vertices are numbered from $0$ to $8$.
0,1,2,3,4,5,6,7,8,0
0,2,4,6,8,1,3,5,7,0
0,3,8,5,2,7,4,1,6,0
0,4,8,2,6,3,7,1,5,0
• Thanks a lot. Nice argument. I am wondering if there is a way to prove the existence of 4 such cycles without drawing them out??.. – Ishfaaq Jan 6 '14 at 4:25
• @Ishfaaq I highly doubt it. However, there is a very simple way to draw the necessary cycles which is originally credited to Walecki in the 1890s. See math.stackexchange.com/questions/194247/… – Erick Wong Jan 6 '14 at 4:30
Actually, your answer is correct, but the reasoning is a little sketchy. Since each Hamiltonian cycle has $n$ edges, there can be at most $(n-1)/2$ edge-disjoint hamiltonian cycles. Now, how do you construct $(n-1)/2$ hamiltonian cycles?
• So the existence of 4 such cycles cannot be proven unless you actually illustrate them? – Ishfaaq Jan 6 '14 at 4:19
• I also tried using the algorithm to prove the existence of an Eulerian circuit for a graph with even vertices. But did not get far. – Ishfaaq Jan 6 '14 at 4:24 | 2019-07-19T05:47:34 | {
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https://www.physicsforums.com/threads/solving-a-differential-eq-by-separation-of-variables.669694/ | # Solving a differential Eq. by separation of variables
## Homework Statement
Find all solutions. Solve explicitly for y.
y$^{'}$=y$^{2}$-y
## The Attempt at a Solution
Case where y'=0
0=y(y-1) y=0,1 when y(t)=0
Case where y'$\neq$0
y'=y$^{2}$-y
$\frac{1}{y^{2}-y}$y'=1
$\int\frac{1}{y^{2}-y}$y'dt=∫1dt
$\int\frac{1}{y^{2}-y}$dy=t+c
Cant figure our where to go from here.
Last edited:
##\displaystyle\frac{1}{y^2 - y} = \frac{1}{y(y-1)}##
Use partial fraction decomposition to split the above into the form ##\displaystyle\frac{A}{y} + \frac{B}{y - 1}##. Then, integrating will be a breeze.
\displaystyle\frac{A}{y} + \frac{B}{y - 1}
1=A(y-1)+By
1=Ay-A+By
A=-1
B=1
$\int\frac{1}{y(y-1)}$=$\int\frac{-1}{y}+\frac{1}{y-1}$
$\int\frac{-1}{y}+\frac{1}{y-1}$=t+c
-ln|y|+ln|y-1|=t+c
ln($\frac{y-1}{y}$)=t+c
ln(1-$\frac{1}{y}$)=t+c
1-$\frac{1}{y}$=e$^{t+c}$
y=$\frac{1}{1-e^{t+c}}$
I dont think this is right, wolfram alpha says it is y=$\frac{1}{1+e^{t+c}}$
How does the plus sign pop up into the denominator?
You've assumed |y-1| = y-1. Is this consistent with the range of y you get when you push on to the solution? The absolute value matters, you can't just drop it and move on.
I believe it's due to the fact that the ##|y - 1|## can also be written as ##|1 - y|##, perhaps that is what Wolfram used. In any case, note that ##e^{t+c} = e^ce^t = ce^t## because ##e^c## is still a constant. So ##\displaystyle y = \frac{1}{1 - ce^t}## would be the best way to write the answer, since if ##c## is indeed negative, it would look like what Wolfram has.
SammyS
Staff Emeritus
Homework Helper
Gold Member
If 0 < y < 1 , then $\displaystyle 1-\frac{1}{y}<0\,,\$ so that $\displaystyle \left|1-\frac{1}{y}\right|=-\left(1-\frac{1}{y}\right)\ .$
That will give $\displaystyle 1-\frac{1}{y}=-e^{t+C}\ .$
...
HallsofIvy
Homework Helper
You were asked to find all solutions. Your formula, once you have solved for y, will give all except one solution. What is the solution you are missing?
It wont give the y solution where y(t)=0 so all solutions for y are
y=$\frac{1}{1 - ce^t}$,0
where c is any real number
Right?
HallsofIvy | 2021-08-04T07:41:05 | {
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https://math.stackexchange.com/questions/282824/what-kind-of-graph-is-this | # What kind of graph is this?
I have a 2D grid of arbitrary positive integer points, minimum being $0,0$ to a maximum of $n,n$. I can only traverse points by increasing my $x$ and/or $y$ coordinate (no backtracking).
Is this sort of data structure a directed acyclic graph? If so, how can I "convert" it to "linear" sequence form?
• Can you clarify, what do you mean by "convert it to linear sequence form"? What are you trying to calculate? – Calvin Lin Jan 20 '13 at 16:45
• @CalvinLin So I can convert it to a 1D array representation – MyNameIsKhan Jan 20 '13 at 16:54
I am not sure if I get your question but here it is. The graph is a directed acyclic graph with $n^2$ vertices.
If you label each cell $(i, j)$ of the grid with $in + j \ \textit{mod} \ n$ then you can create a 1d array to represent the graph where the $k$-th index represents the $k$-th vertex. From position $k$ you can go to position $k + 1$ if $k < n - 1$ and $k + n$ if $k < (n - 1)n$.
But you still need $n^2$ positions so I don't understand what you want to do exactly. If you can be clearer then maybe I can help more.
• I have a grid of size n by n. I have points on this grid (arbitrary number / position of them). I can only go from one point to another, and I can only go to an immediately visible point that is greater than the one I am currently at in either x, y, or both x and y coordinate. – MyNameIsKhan Jan 20 '13 at 17:42
First, as has been mentioned, this is a directed acyclic graph. It sounds like your "linear sequence form" is an arrangement of the vertices in a left-to-right sequence $\langle v_1, v_2, \dots , v_n\rangle$ such that any edge goes from left to right in the sequence. If so, you're asking for what's commonly known as a topological sort of the vertices.
For your graphs an easy way to do this is to order the vertices first by weight, where the weight of a vertex $w(i, j)=i+j$ and then within each weight order the vertices by their first coordinate. For example, if $n=2$ we'll have the linear order $$(0,0), (0, 1), (1, 0), (0, 2), (1, 1), (2, 0), (1, 2), (2, 1), (2, 2)$$ Now look at the edges. From $(i, j)$ we'll have edges to $(i+1, j), (i, j+1), (i+1, j+1)$, as long as $i+1\le n$ and $j+1\le n$. Each of these edges will have weight strictly greater than $i+j$ so each of the edges from $(i, j)$ will end at a vertex of greater weight, hence to the right of $(i, j)$. This also shows that the graph is acyclic.
Notice that there are other ways of arranging the vertices in a line (simply permute the edges of equal weights). | 2019-07-16T00:28:25 | {
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https://physics.stackexchange.com/questions/98205/why-does-the-tension-on-the-pulley-in-an-atwood-machine-not-equal-m-1-m-2g | Why does the tension on the pulley in an Atwood machine not equal $(m_1 + m_2)g$?
Consider the following simple Atwood machine with an ideal pulley and an ideal string
According to my textbook, the tension on the clamp that holds the machine to the wall equals $2T$. I don't understand why that is. The tension in $T$ in the string is equal in magnitude to $m_1g + m_1a = m_2g - m_2a$, assuming that $m_1$ is accelerating upwards.
Also, the acceleration of masses in an atwood machine is given by
$$a = \frac{(m_2 - m_1)g}{m_1 + m_2}$$
Substituting this in, we get the tension equal to
$$T = m_1g + m_1\frac{(m_2 - m_1)g}{m_1 + m_2} = m_1g\left(1 + \frac{m_2 - m_1}{m_2 + m_1}\right) = \frac{2m_2m_1g}{m_1 + m_2}$$
So, according to my textbook, the tension on the pulley clamp should be:
$$2T = \frac{4m_1m_2g}{m_1 + m_2}$$
But, aren't all these forces internal forces? If we consider the whole atwood machine as the system (excluding the clamp), the only forces acting on it are the force of gravity, $(m_1 + m_2)g$ and the tension in the clamp, $T_c$. Since the system is at rest
$$T_c = (m_1 + m_2)g$$
Am I right, or is there a flaw in my argument?
• You found $T$, and the text book has that same equation multiplied by a factor of 2. There is no problem here. – Ruben Feb 7 '14 at 13:41
• Hint: The system is not at rest. – DR10 Feb 7 '14 at 13:51
• Nick's answer is complete but I liked your question because it shows the effort to understand the PRINCIPLE under calculations. So it's important in my opinion to understand why the system isn't at rest. – DR10 Feb 7 '14 at 14:06
• True, every calculation should not only mathematically check out, but the physical interpretation is also a very very very important part! So on the point of the question I'd say good job and keep up the good work! – Nick Feb 7 '14 at 14:10
• If it helps, you can show that the center o fmass of the two masses $m_{1}$ and $m_{2}$ is accelerating downward, and though it looks like the support is holding the wheel steady, it is actually letting the wheel/mass system accelerate downward because of this. – Jerry Schirmer Feb 7 '14 at 15:01
Your result holds when the two masses are the same, in that case $a=0$ and you'd have that:
$T = m_1 g = m_2 g$.
Or:
$2T = 2m_1 g=2m_2g=(m_1+m_2)g$.
In the case that the masses are not the same, then both masses are accelerating, which in its turn apply yields a lower force on the pulley-system (and on the clamp).
This can be easilly checked with your formula of the tension!
$T = \frac{2m_1m_2g}{m_1+m_2},$
If I were to define the total mass as: $M=m_1+m_2$, then I could express $T$ as:
$T=\frac{2m_1(M-m_1)g}{M}=\frac{2g}{M}(m_1(M-m_1)).$
You can check if you'd plot $T$ as a function of $m_1$, that it reaches a maximum in $m_1=M/2$, which means that the tension becomes maximal if the two masses are equal, the tension then becomes:
$T=\frac{Mg}{2}=\frac{(m_1+m_2)g}{2}$,
or as you were thinking:
$2T=(m_1+m_2)g$
For completeness the plot of the tension in function of the mass $m_1$ in terms of dimensionless quantities.
On this plot you can easilly see that if $m_1=0 \Rightarrow m_2=M$ or $m_1=M \Rightarrow m_2=0$, that there'd be no tension since one of the two masses would be free falling. In the intermediate cases there would be tension since there is a ''pull'' on both sides of the string, the more the masses $m_1$ and $m_2$ equal eachother, the less movement there is and the more pull there is on the string.
• So, if my argument was incorrect, it can only mean that the system is not at rest. But how can you say that the system is not at rest? – Gerard Feb 7 '14 at 14:36
• In the above case we have a frictionless pulley, with a massless string. The only way the system can be at rest is when the two masses are equal (in your calculation that's the only case when the acceleration equals zero). When this is the case, both masses are pulling with an equal force at both ends of the string. Note that this doesn't necessarilly implies the system is at rest, it can also move with a constant velocity! – Nick Feb 7 '14 at 14:42
• @Gerard If you were to add mass to the sting and/or friction to the pully, then there might be other situations in which the system is/becomes at rest. – Nick Feb 7 '14 at 14:43
• It's not true that it remains in the same place. Its center of mass is accelerating because even if m_1 is going upward and m_2 downward the masses are different so they have a different "weight" in the global motion. So if m_2>m_1 and m_2 is accelerating downward, then the center of mass is going downward. – DR10 Feb 7 '14 at 15:40
• @Gerard: Right, notice that for your purpose (i.e. total force acting on the system) rest or motion is not what you're really looking for. I've been sloppy simply telling you:"the system is not at rest". What's important is total acceleration and in this case it's different from 0. A motion with constant velocity needs no force acting on the system. I'll stop here because we're a little misusing the space for comments. – DR10 Feb 7 '14 at 16:34
The system is not at rest. If you consider the masses and the pulley to be one system, you can understand the behaviour of the system by the behaviour of its centre of mass. Unless the masses are equal, the centre of mass of the system is not at rest.
It might be useful to think of it in this way - Inside the system boundary mass $m_1$ moves down through a distance while mass $m_2$ moves up by the same distance. So, the centre of mass has moved down (or up depending whether $m_1 > m_2$).
So, the tension would be given by the equation:
$$(m_1+m_2)a_{cm} = (m_1+m_2)g - T_c$$
You can further work out that
$a_{cm} = a(m_2-m_1) /(m_1+m_2)$, where a is the value of the acceleration of mass $m_1$ that you have mentioned.
Plug it in the equation and you'll find that:
$T_c=\frac{4m_1m_2}{m_1+m_2}{g}$
• This is how I will try to teach this problem. Thank you. – commonhare Mar 4 '19 at 18:04
• Any chance you or @Nick could comment on the solution taking the form 4g*mu? I know it may be beyond the scope of the problem, but when I see connections like that, I try to understand them. – commonhare Mar 4 '19 at 18:13
There is indeed a flaw in your argument. In short, the tension on the pulley clasp is only required to cancel the total gravitational force on the system when everything is in equilibrium and there is no acceleration. However, if the masses are unbalanced then one of them will fall and the other will rise, and it is not clear that this will keep the total force at the same value as the balanced case.
In fact, you can check that when the two masses are equal then the answers coincide: the correct tension on the pulley clasp is $$T_\text{clasp}=2T=\frac{4m^2}{m+m}g=2mg=(m+m)g.$$ | 2021-05-12T05:48:06 | {
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https://math.stackexchange.com/questions/4215498/proving-properties-of-exponents | Proving properties of exponents.
For positive integer $$n$$,
$$a^n:=a\cdot a\cdot a...a \ \ \ (n\ \text{times})$$ $$a^{-n}:=\frac{1}{a^n}$$ $$a^0:=1$$
I want to prove the properties, $$a^n\cdot a^m=a^{n+m}$$ $$\frac{a^n}{a^m}=a^{n-m}$$ $$(a^n)^m=a^{nm}$$ for all integer values of $$n,m$$. It's easy to prove them for positive integers using the 3 definitions. Is there any way I could extend the proof to any integer value (including 0) directly (perhaps by mathematical induction), or do I have consider each possible combination of positive, negative and 0 separately?
This might be a very basic question but I'm trying to improve by proof writing skills and would appreciate any help.
Edit:
My text book says that 0 and negative exponents are defined the way they are as that is the only way to make the properties hold for all values of $$n$$ & $$m$$. Wouldn't that mean that I'll have to prove the 0 and negative cases separately, usingg 2 extra defintions?
• This depends. Has addition been defined? Do we know that if you combine 7 apples with 5 apples that you will have 7+5 apples? Can you know that if you multiply $a$ by itself $n$ times and then continue with multiplying by itselft $m$ more times that you have multiplied it by itself $m+n$ times? I'd say it follows by definition. But, to play it safe, maybe you should prove it be induction (but then again how do you know proofs by induction works). To do this properly you have to know you basic definitions an axioms. Aug 3, 2021 at 5:17
• Addition and multiplication are assumed to be defined. I'm still in high school, so I don't think I would be able to grasp defining something as basic as addition and multiplication for non-natural numbers. Aug 3, 2021 at 5:28
• @fleablood: I presume defining the set of natural numbers $\Bbb N$ and then addition and multiplication on $\Bbb N$ (via Peano axioms) would be a prerequisite before even attempting to give a formal proof for this, otherwise what does $m+n$ even mean? Or what does $a\cdot a\cdots a$ ($n$ times) mean? Aug 3, 2021 at 5:35
• @PrasunBiswas But then what is there to prove? $a^n\cdot a^m =\underbrace{\underbrace{a\cdot a...a}_n\cdot\underbrace{a\cdot a...a}_m}_{n+m} = a^{n+m}$. What is there to prove? Aug 3, 2021 at 5:41
• @fleablood: I think it would be proving that the power laws that hold in $\Bbb N$ also hold in $\Bbb Z$ Aug 3, 2021 at 5:46
Yes, a complete proof will cover all the cases. But you can simplify the work a lot: note that proving $$a^{m+n}=a^m\cdot a^n~~~(\dagger)$$ for integers $$m,n$$ imply the two other properties:
$$\frac{a^m}{a^n}=\begin{cases}a^m\cdot\dfrac 1{a^n}=a^m\cdot a^{-n}\overset{\dagger}{=}a^{m+(-n)}=a^{m-n}&,n\gt 0\\ \dfrac{a^m}{a^0}=a^m=a^m\cdot a^0\overset{\dagger}{=}a^{m-0}&,n=0\\ \dfrac{a^m}{a^{-(-n)}}=a^m\cdot\dfrac 1{a^{-(-n)}}=a^m\cdot a^{-n}\overset{\dagger}{=}a^{m-n}&,n\lt 0\end{cases}$$
$$(a^m)^0:=1=a^{0}=a^{m\cdot 0}$$
$$(a^m)^n=\begin{cases}\underbrace{a^m\cdot a^m\cdots a^m}_{n\text{ times}}\overset{\dagger}{=}a^{\underbrace{m+m+\cdots+m}_{n\text{ times}}}=a^{mn}&,n\gt 0\\ \dfrac 1{\underbrace{a^m\cdot a^m\cdots a^m}_{-n\text{ times}}}\overset{\dagger}{=}\dfrac 1{a^{\underbrace{m+m+\cdots+m}_{-n\text{ times}}}}=\dfrac 1{a^{m(-n)}}=\dfrac 1{a^{-(mn)}}=a^{mn}&,n\lt 0\end{cases}$$
So, all you need to do is to prove $$(\dagger)$$ for all integers $$m,n$$
Once you have proved it for positive integers $$m,n$$, you can wlog consider only one of $$m,n$$ to be negative and by symmetry, you prove the cases where exactly one of $$m,n$$ is negative. For the case of both $$m,n$$ negative, we note that $$a^{m+n}=a^{-(-m-n)}=\frac 1{a^{(-m)+(-n)}}\overset{\dagger}{=}\frac 1{a^{-m}\cdot a^{-n}}=\frac 1{a^{-m}}\cdot\frac 1{a^{-n}}=a^m\cdot a^n$$
where we apply $$(\dagger)$$ with positive integers $$-m,-n$$
which leaves us with the case of one or both of $$m,n$$ be $$0$$ which is easy to finish up. By symmetry, wlog prove with just one of $$m,n$$ being $$0$$ and for the case of $$m=n=0$$, we have $$a^{0+0}=a^{0}=1=1\cdot 1=a^0\cdot a^0$$. $$_\square$$
• @OVERWOOTCH: I added cases in the $a^m/a^n$ part to complete (rigorify) the argument how $(\dagger)$ for integers $m,n$ imply that it is $a^{m-n}$ in general. You may want to check up on that and let me know if there's something missing. The point is that when $n\lt 0$, $-n\gt 0$ is a positive integer and we use the given axioms with $-n$ Aug 3, 2021 at 6:27
• In the case where exactly one of $m$ and $n$ is negative, do I have to separately prove the cases where the negative exponent has a greater/lower magnitude? I know choosing $m$/ $n$ to be negative is an arbitrary choice, but further choosing which exponent to be larger in magnitude is not arbitrary right? Since now $m$ and $n$ are no longer identical? Aug 3, 2021 at 23:56
• @OVERWOOTCH: Yes, wlog, considering $m$ to be the negative exponent and $n$ the positive one, you need to prove the two cases $0\lt |m|\le n$ and $0\lt n\le |m|$ separately ($|m|=-m$ since $m$ is negative). This has already been pointed out in the comments by fleablood. The first two axioms should be useful here: prove the lemma that $a^{u-v}=a^u\cdot a^{-v}$ for positive integers $u,v$ where $v\le u$ (so that $0\le u-v$ is a natural). The two cases can be now be proved using $u=n,v=|m|=-m$ and $u=|m|=-m,v=n$ in the proven lemma. Aug 4, 2021 at 0:36
• Alright, I think I got it. Thanks Aug 4, 2021 at 2:48 | 2022-07-02T01:15:25 | {
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https://math.stackexchange.com/questions/2472218/outcome-of-one-event-from-two-independent-events | Outcome of one event from two independent events
If there are two independent events with two outcomes, and let's say one outcome has the probability of $A$%, is the probability of the union of the outcome (with the known probability) and one outcome of the other event also $A$%? Or less? Or greater? Or in proportion to the % of the unknown outcome?
Here's an example: There's a kindergarten with graduation rate $92$%. The gender and the graduation rate are independent. What is the graduation rate of female students graduating?
Please help. I think it's $92$%, but I'm not so sure. Thank you!
1 Answer
Yeah I think you're right. If the gender and graduation rate are independent, that means that the graduation of female students is the same as the graduation rate of the male students. So both would be 92%
"Two events are independent if the occurrence of one does not change the probability of the other occurring. "
https://people.richland.edu/james/lecture/m170/ch05-rul.html | 2019-08-18T05:48:59 | {
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http://www.maplesoft.com/support/help/Maple/view.aspx?path=LinearAlgebra/IdentityMatrix | LinearAlgebra - Maple Programming Help
# Online Help
###### All Products Maple MapleSim
Home : Support : Online Help : Mathematics : Linear Algebra : LinearAlgebra Package : Constructors : LinearAlgebra/IdentityMatrix
LinearAlgebra
IdentityMatrix
construct an identity Matrix
Calling Sequence IdentityMatrix(r, c, cpt, options)
Parameters
r - (optional) non-negative integer; row dimension of the resulting Matrix c - (optional) non-negative integer; column dimension of the resulting Matrix cpt - (optional) BooleanOpt(compact); selects the compact form of the output options - (optional); constructor options for the result object
Description
• The IdentityMatrix() function returns an identity matrix.
• If M := IdentityMatrix(r, c), then M is an r x c Matrix in which all the entries on the diagonal are one and all other entries are zero.
• If the row dimension is not provided, it defaults to zero. If the column dimension is not provided, it defaults to the row dimension.
• If the compact option (cpt) is omitted, or, if it is included in the calling sequence as just the symbol compact or in the form compact=true, then the result is built by using a shape function designed to minimize storage. If the option is entered as compact=false, a full rectangular Matrix is constructed.
• The constructor options provide additional information (readonly, shape, storage, order, datatype, and attributes) to the Matrix constructor that builds the result. These options may also be provided in the form outputoptions=[...], where [...] represents a Maple list. If a constructor option is provided in both the calling sequence directly and in an outputoptions option, the latter takes precedence (regardless of the order).
If a shape value is not provided, then the shape of the resulting Matrix is determined by the compact option. Otherwise, a result with the specified shape is constructed with all diagonal values set to 1.
If readonly=false is included, it is ignored unless the default shape (identity) is overridden by also including a mutable shape in options.
• This function is part of the LinearAlgebra package, and so it can be used in the form IdentityMatrix(..) only after executing the command with(LinearAlgebra). However, it can always be accessed through the long form of the command by using LinearAlgebra[IdentityMatrix](..).
Examples
> $\mathrm{with}\left(\mathrm{LinearAlgebra}\right):$
> $M≔\mathrm{IdentityMatrix}\left(4\right)$
${M}{:=}\left[\begin{array}{rrrr}{1}& {0}& {0}& {0}\\ {0}& {1}& {0}& {0}\\ {0}& {0}& {1}& {0}\\ {0}& {0}& {0}& {1}\end{array}\right]$ (1)
> $\mathrm{MatrixOptions}\left(M,\mathrm{shape}\right)$
$\left[{\mathrm{identity}}\right]$ (2)
> $N≔\mathrm{IdentityMatrix}\left(3,5,\mathrm{compact}=\mathrm{false}\right)$
${N}{:=}\left[\begin{array}{rrrrr}{1}& {0}& {0}& {0}& {0}\\ {0}& {1}& {0}& {0}& {0}\\ {0}& {0}& {1}& {0}& {0}\end{array}\right]$ (3)
> $\mathrm{MatrixOptions}\left(N,\mathrm{shape}\right)$
$\left[{}\right]$ (4)
See Also
## Was this information helpful?
Please add your Comment (Optional) E-mail Address (Optional) What is ? This question helps us to combat spam | 2017-01-18T01:40:18 | {
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https://math.stackexchange.com/questions/813695/using-finite-difference-to-compute-derivative-in-the-newton-raphson-root-finding | # Using Finite Difference to compute derivative in the Newton-Raphson root finding Algorithm
In the Newton-Raphson method we come across the following equation: $$x_{n+1}=x_n - \frac{f(x_n)}{f'(x_n)}$$
Can you please let me know if we can calculate the derivative term like this - $$f'(x_n) = \frac{f(x_n) - f(x_{n-1})}{x_n-x_{n-1}}$$
Will rate of convergence of the original Newton-Raphson Method and this modified method be different? Will all set of problems solvable by the original Newton-Raphson Method be solvable by the above modified method too?
• I think that this could be extremely unsafe when far from the solution and you could face many unpleasant situations. Close to the solution, what you write is effectively the computation of the derivative by finite difference and it would work. If I may ask, do you have any reason for looking at such a scheme ? – Claude Leibovici May 29 '14 at 12:57
• @ClaudeLeibovici It's a common root finding method: the secant method. Sure, it can throw an exception when the secant is horizontal, but so can Newton-Raphson. – user147263 May 29 '14 at 13:07
• @wordsthatendinGRY. My real concern is that the derivative can be very wrong when far from the solution. I would even prefer starting using a secant method and finish polishing the solution using what the OP proposes. – Claude Leibovici May 29 '14 at 13:10
• @ClaudeLeibovici What the OP proposes is exactly the secant method. – user147263 May 29 '14 at 13:12
• @ClaudeLeibovici There is no analytic first derivative for my problem. Hence, I wanted to use the previous iteration value to calculate the derivative and make the solver faster. – Discretizer May 30 '14 at 4:00
You have rediscovered the secant method.
The secant method a bit slower in the vicinity of the root than Newton-Raphson: its order is $1.618$ instead of $2$. However, since there is just one function evaluation per step (versus two for N-R: $f$ and $f'$), it may actually be faster. Depends on how complicated the derivative is.
Will all set of problems solvable by the original Newton-Raphson Method be solvable by the above modified method too?
This is much too broad to have an affirmative answer. Both methods converge from a vicinity of a root, if the function is reasonable. Both can fail to converge from further away. The basins of attraction of a root can be quite complicated (fractal), with tiny difference in initial position changing the outcome. Briefly: no, they are different methods, and you may find one method failing whether the other succeeds.
The answer of user147263 is not correct. The secant method is not the same as the Newton method with numerical gradients.
Generally the Secant method is defined as:
$$x_n = x_{n-1} - f(x_{n-1}) \frac{x_{n-1} - x_{n-2}}{f(x_{n-1}) - f(x_{n-2})} = \frac{x_{n-2} f(x_{n-1}) - x_{n-1} f(x_{n-2})}{f(x_{n-1}) - f(x_{n-2})}.$$
The Newton method with a finite difference approximation for the derivatives is different to this, because you can choose the delta $$\Delta\tilde{x}$$ for the finite difference independently from $$\Delta x = x_{n-1} - x_{n-2}$$.
Regards | 2020-02-22T01:05:01 | {
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http://math.stackexchange.com/questions/358755/ways-to-fill-a-n-times-n-square-with-1-times-1-squares-and-1-times-2-recta | # Ways to fill a $n\times n$ square with $1\times 1$ squares and $1\times 2$ rectangles
I came up with this question when I'm actually starring at the wall of my dorm hall. I'm not sure if I'm asking it correctly, but that's what I roughly have:
So, how many ways (pattern) that there are to fill a $n\times n:n\in\mathbb{Z}_{n>0}$ square with only $1\times 1$ squares and $1\times 2$ rectangles?
For example, for a $2\times 2$ square:
• Four $1\times 1$ squares; 1 way.
• Two $1\times 1$ squares and one $1\times 2$ rectangle; $4$ ways total since we can rotate it to get different pattern.
• Two $1\times 2$ rectangles; 2 ways total: placed horizontally or vertically.
$\therefore$ There's a total of $1+4+2=\boxed{7}$ ways to fill a $2\times 2$ square.
So, I'm just wondering if there's a general formula for calculating the ways to fill a $n\times n$ square.
Thanks!
-
+1 for coming up with your own math question. – Sammy Black Apr 11 '13 at 20:58
For $1\times n$ boards, you get the Fibonacci numbers. For $2\times n$ boards it's already a bit harder. You get the recurrence relation $a_n = 3a_{n-1} + a_{n-2} - a_{n-3}$, see oeis.org/A030186 – azimut Apr 11 '13 at 21:49
## 4 Answers
A hand count for $n=3$ yields $131$ ways, and a search for $1,7,131$ yields OEIS sequence A028420, the "number of monomer-dimer tilings of an $n\times n$ chessboard". The entry doesn't provide a formula (so it's likely that none is known), but some references.
-
so it's like a program? – user67258 Apr 11 '13 at 21:18
See also oeis.org/A210662 which gives all the rectangles, not just the squares. – Charles Apr 11 '13 at 21:32
@herderp: Sorry, I don't understand that question. What's like a program? – joriki Apr 11 '13 at 22:12
I thought the sequence is computed by a computer program – user67258 Apr 11 '13 at 22:24
If you just used $1\times2$ rectangles, then this is same as finding the number of matchings in the $m \times n$ rectangle graph, and a formula for that has been given by Kastelyn:
$$\sqrt{\left|\prod_{j=1}^{m}\prod_{k=1}^{n} \left(2 \cos \frac{\pi j}{m+1} + 2i\cos \frac{\pi k}{n+1}\right)\right|}$$
This was done, by mapping the number to the square root of the determinant of a weighted adjacency matrix of the graph.
If you include $1 \times 1$ squares, you essentially need to find the sum over all subgraphs (think of placing a $1\times 1$ square as deleting that vertex) of the $m \times n$ graph, and for each subgraph, the determinant approach still works, I believe, but might not have a nice closed form.
You can find a nice exposition for the $1\times 2$ case in the first chapter here: http://users.ictp.it/~pub_off/lectures/lns017/Kenyon/Kenyon.pdf
-
We can probably give some upper and lower bounds though. Let $t_n$ be the possible ways to tile an $n\times n$ in the manner you described. At each square, we may have $5$ possibilities: either a $1\times 1$ square, or $4$ kinds of $1\times 2$ rectangles going up, right, down, or left. This gives you the upper bound $t_n \leq 5^{n^2}$.
For the lower bound, consider a $2n\times 2n$ rectangle, and divide it to $n^2$ $2\times 2$ blocks, starting from the top left and putting a $2\times 2$ square, putting another $2\times 2$ square to its right and so on... For each of these $2\times 2$ squares, we have $5$ possible distinct ways of tiling. This gives the lower bound $t_{2n} \geq 7^{n^2}$. Obviously, $t_{2n+1} \geq t_{2n},\,n \geq 1$, and therefore $t_n \geq 7^{\lfloor \frac{n}{2}\rfloor ^2}$.
Hence, \begin{align} 7^{\lfloor \frac{n}{2}\rfloor ^2} \leq t_n \leq 5^{n^2}, \end{align} or roughly (if $n$ is even), \begin{align} (7^{1/4})^{n^2} \leq t_n \leq 5^{n^2}. \end{align} BTW, $7^{1/4} \geq 1.6$. So, at least we know $\log t_n \in \Theta(n^2)$.
Note: Doing the $3\times 3$ case for the lower bound, we get $(131)^{1/9} \geq 1.7$ which is slightly better.
-
Two comments:
1 If you only allow for the 1x2 rectangles, the problem is known as the domino tiling problem. You can find the answer here: http://en.wikipedia.org/wiki/Domino_tiling#Counting_tilings_of_regions
2 In the case of a cheseboard (and I think it can be generalized to squares of even side length), there are few things known about the case with 2 1x1 squares and dominoes.
If the two squares are placed on two board squares of same color, it is impossible to complete the tiling, while if the 2 squares are placed on opposite colors Gomory's theorem says that it is always possible to complete the tiling (but I think it doesn't say in how many ways).
- | 2015-01-29T21:04:40 | {
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https://math.stackexchange.com/questions/2364242/how-to-find-out-whether-a-function-has-vertical-asymptotes | # How to find out whether a function has vertical asymptotes
I got the following question wrong in my exame
$\ \frac{16x-64}{|x|-4}$
My answer was, -16,16 for the horizontal asymptotes and 4,-4 for the vertical asymptotes.
For the horizontal asymptotes the answer is right, But there aren't any vertical asymptotes.
I thought if there wasn't a sum of squares in the denominator There would be vertical asymptotes.
So I proceeded this way, as I need to equal the denominator to 0 in order to find the vertical asymptotes.
$|x| - 4 = 0$
$x - 4 = 0$
$x = 4$
$-x - 4 = 0$
$-x = 4(-1)$
$x = -4$
There are 3 rules to find the HA's, but what are the rules to find the VA's ?
note that you will have $$\frac{16x-64}{|x|-4}=\frac{16(x-4)}{x-4}$$ if $$x\geq0$$ or $$\frac{16(x-4)}{-x-4}$$ if $$x<0$$
• So, there are no vertical asymptotes because you canceled the denominator ? – Goun2 Jul 19 '17 at 19:30
• for $$x<0$$ we have $$x=-4$$ as a vertical asymptote – Dr. Sonnhard Graubner Jul 19 '17 at 19:32
• $x=+4$ is not a VA because the numerator also vanishes there, $f(x)$ has a removable discontinuity at $x=+4$ – WW1 Jul 19 '17 at 19:36
• And how did you come to this conclusion, -4 is a VA ? you can cancel (x-4) with -(x-4), and the result would be 16/-1, There wouldn't be no function to equal to 0, sorry if i'm too dumb. – Goun2 Jul 19 '17 at 19:52
• no $$-x-4=-(x+4)$$ – Dr. Sonnhard Graubner Jul 19 '17 at 19:53
We have the following function, $$f(x)=\frac{16x-64}{|x|-4}$$
The mathematical definition of absolute value of x is defined as
$$|x|\begin{cases}x& ,x \geq 0\\-x &, x<0\end{cases}$$
The absolute value of something will always the it positive. If it is negative, it will be negate to become positive.
Vertical asymptote of a rational function occurs when denominator is becoming zeroes. If a function like any polynomial $y=x^2+x+1$ has no vertical asymptote at all because the denominator can never be zeroes.
So for the above function we again have piece-wise function,
$$f(x)\begin{cases}\frac{16x-64}{x-4}& ,x \geq 0\\\frac{16x-64}{-x-4} &, x<0\end{cases}$$
For the domain of $x \geq 0$
$$f(x)=\frac{16x-64}{x-4}, x \neq 4$$
It can be further redefined such that,
$$f(x)=\frac{16(x-4)}{x-4}, x \neq 4$$
$$f(x)=16 ,x \neq 4$$
So, the first function has removable discontinuity at the point $x=4$. This is because $$\lim_{x \rightarrow 4^+}16=\lim_{x \rightarrow 4^-}16=\lim_{x \rightarrow 4}16$$
Removable discontinuity is defined such that
$$\lim_{x \rightarrow a^+}f(x)=\lim_{x \rightarrow a^-}=\lim_{x \rightarrow a}f(x)=L$$
although $x \neq a$.
However, if $x$ is defined on $a$ then there is no removable discontinuity. Notice that from the right side and the left side of a we are approaching same value.
For the second function with domain of $x<0$
$$f(x)=\frac{16x-64}{-x-4}, x\neq -4$$
Can it be further redefined? Obviously, we cannot cancel out the numerator and the denominator.
So, we know that it has vertical asymptote at some point when the denominator tends to $0$.Its denominator is ${-x-4}$
Taking the denominator and let it goes to zeroes
$$-x-4\rightarrow 0$$
$$-x\rightarrow4$$
$$x\rightarrow-4$$
So, the vertical asymptote occurs at $x=-4$ but wait is it inside the domain of the function? The answer is yes, because $-4<0$.
Let us use limit to see what is happening around it.
Right Hand Limit $$\lim_{x \rightarrow -4^+}\frac{16x-64}{-x-4}=- \infty$$
Left Hand Limit $$\lim_{x \rightarrow -4^-}\frac{16x-64}{-x-4}=+ \infty$$
$\lim_{x \rightarrow -4}f(x)$ does not exist in this case since $LHL \neq RHL$.
This type of discontinuity is called infinite discontinuity. As you can see it tends to $\pm \infty$ as it gets closer and closer to $-4$.
Some visual confirmation aid.
For the first function $x \geq 0$
$$f(x)=\frac{16(x-4)}{x-4}, x\neq 4$$
For the second function, $x <0$
$$f(x)=\frac{16x-64}{-x-4},x \neq -4$$ | 2019-11-12T11:20:36 | {
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http://math.stackexchange.com/questions/156554/can-an-ordered-field-be-finite | # Can an ordered field be finite?
I came across this question in a calculus book.
Is it possible to prove that an ordered field must be infinite? Also - does this mean that there is only one such field?
Thanks
-
$0 \lneq 1 \lneq 1+1 \lneq 1+1+1 \lneq ...$ – jspecter Jun 10 '12 at 17:18
I have a feeling that I answered that like twice before. Too bad it's easier to write another answer instead of finding a duplicate (the search function stinks!) – Asaf Karagila Jun 10 '12 at 17:19
Spivak's book? ${}$ – Michael Hardy Jun 10 '12 at 17:58
Recall that in an ordered field we have:
1. $0<1$;
2. $a<b\implies a+c<b+c$.
Suppose that $F$ is an ordered field of characteristic $p$, then we have in $F$ that $$\underbrace{1+\ldots+1}_{p\text{ times}} = 0$$
Therefore: $$0<1<1+1<\ldots<\underbrace{1+\ldots+1}_{p\text{ times}} = 0$$
Contradiction! Therefore the characteristic of $F$ is $0$ and therefore it is infinite, since it contains a copy of $\mathbb Q$.
Few fun facts on the characteristic of a field:
Definition: The characteristic of a field $F$ is the least number $n$ such that $\underbrace{1+\ldots+1}_{n\text{ times}}=0$ if it exists, and $0$ otherwise.
Exercises:
1. If a field has a positive characteristic $n$ then $n$ is a prime number.
2. If $F$ is a finite field then its characteristic is non-zero (Hint: the function $x\mapsto x+1$ is injective, start with $0$ and iterate it $|F|$ many times and you necessarily got $0$ again.)
3. If $F$ is finite and $p$ is its characteristic then $p$ divides $|F|$.
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What does "field of characteristics $p$" mean? thanks – yotamoo Jun 10 '12 at 17:22
@yotamoo: It means that $p$ is the least natural number such that adding $1$ to itself $p$ many times gives $1+1+\ldots+1=0$. One good exercise is to prove that $p$ is either $0$ or a prime number. Fields of characteristics $0$ can be ordered ($\mathbb Q$ is, but $\mathbb C$ is not); and every finite field has to have a positive characteristics (because $x\mapsto x+1$ is an injective function in a field). – Asaf Karagila Jun 10 '12 at 17:23
I usually hear it said "a field of characteristic $p$" with no plural. Also, $\mathbb C$ has characteristic 0 but can't be ordered, so that comment is kind of misleading. – Ben Millwood Jun 10 '12 at 17:40
@benmachine: Possible, I usually get confused about that. Feel free to edit, I am on the iPhone now... Also I mentioned that the complex numbers have char. zero but cannot be ordered. I don't know how that is misleading... – Asaf Karagila Jun 10 '12 at 17:50
@AsafKaragila: that doesn't sound like what you are saying, is all: you say "fields of characteristic 0 can be ordered ($\mathbb Q$ is, but $\mathbb C$ is not)" – I guess I thought it misleading because it's kind of ambiguous if you meant is/is not ordered, or is/is not characteristic 0. I guess I mentally inserted "all" before "fields", where you meant "only". – Ben Millwood Jun 10 '12 at 17:53
An ordered field must be infinite. Notice that each field has a subset of numbers that behave like the natural numbers, with $0<1<1+1<1+1+1\dots$
However, not every ordered field is isomorphic to all other ordered fields. Notice that both the rational numbers and real numbers are ordered fields.
-
Hint $\$ Linearly ordered groups are torsion free: $\rm\: 0\ne n\in \mathbb N,$ $\rm\:g>0 \:\Rightarrow\: n\cdot g = g +\cdots + g > 0,\:$ since positives are closed under addition. Conversely, a torsion-free commutative group can be linearly ordered (Levi 1942).
-
Even more: a lattice ordered (nor necessarily commutative) group is always torsion free. – Leon Jul 29 '12 at 22:15 | 2016-02-12T22:47:32 | {
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https://wiki.documentfoundation.org/Documentation/Calc_Functions/TRUNC | # Documentation/Calc Functions/TRUNC
Other languages:
English • Nederlands • dansk • עברית
TRUNC
Mathematical
## Summary:
Rounds a number toward zero to the next multiple of a specified power of 10.
## Syntax:
TRUNC(Number[; Count])
## Returns:
Returns a real value which is the specified number rounded (toward zero) to the required precision.
## Arguments:
Number is a real number, or a reference to a cell containing that number, that is to be rounded.
Count is an integer, or a reference to a cell containing that integer, that is used to specify the precision of the rounding. If Count is omitted, a default value of 0 is assumed. The value of Count is utilized as follows:
• If Count is a positive integer, then rounding is to the right of the decimal separator. For example, if Count is equal to 1 then numbers are rounded (toward 0) to the next multiple of $\displaystyle{ \frac{1}{10}\text{th} }$; if Count is equal to 2 then numbers are rounded (toward zero) to the next multiple of $\displaystyle{ \frac{1}{100}\text{th} }$; and so on.
• If Count is equal to zero, then rounding is to the next integer toward 0.
• If Count is a negative integer, then rounding is to the left of the decimal separator. For example, if Count is equal to -1 then numbers are rounded (toward 0) to the next multiple of 10; if Count is equal to -2 then numbers are rounded (toward 0) to the next multiple of 100; and so on. The value returned by TRUNC will always be an integer when Count is negative.
The following conditions (including errors) may be encountered:
• If either Number or Count is non-numeric, then TRUNC reports a #VALUE! error.
• If Count is a non-integer value, then it is rounded to the next integer in the direction of 0.
• If you obtain unexpected results from TRUNC, check the following:
1. Make sure that the number of displayed decimal places is not affected by the setting of the Limit decimals for general number format option in the General Calculations area of the Tools ▸ Options ▸ LibreOffice Calc ▸ Calculate dialog (LibreOffice ▸ Preferences ▸ LibreOffice Calc ▸ Calculate on macOS).
2. Access the Numbers tab on the Format Cells dialog (Format ▸ Cells or equivalent interaction) to make sure that relevant cells do not have formatting applied that affects the number of displayed decimal places.
## Examples:
Formula Description Returns
=TRUNC(2.348; 2) Here the function rounds 2.348 toward 0, to 2 decimal places. 2.34
=TRUNC(D1; D2) where cells D1 and D2 contain the numbers -32.4834 and 3.2 respectively. The value of the Count argument is truncated to 3. Here the function rounds -32.4834 toward 0, to 3 decimal places. -32.483
=TRUNC(2.648; 0) Here the function rounds 2.648 toward 0, to the next integer. The alternative formula =TRUNC(2.648) returns the same value. 2
=TRUNC(-45.67) Here the function rounds -45.67 toward 0, to the next integer. -45
=TRUNC(987.65; -2) Here the function rounds 987.65 toward 0, to the next hundred. 900
TRUNC | 2022-07-01T08:29:07 | {
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http://math.stackexchange.com/questions/20597/how-to-re-write-completing-the-square-x2x1 | How to “Re-write completing the square”: $x^2+x+1$
The exercise asks to "Re-write completing the square": $$x^2+x+1$$
The answer is: $$(x+\frac{1}{2})^2+\frac{3}{4}$$
I don't even understand what it means with "Re-write completing the square"..
What's the steps to solve this?
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Remember the formula for the square of a binomial: $$(a+b)^2 = a^2 + 2ab + b^2.$$
Now, when you see $x^2+x+1$, you want to think of $x^2+x$ as the first two terms you get in expanding the binomial $(x+c)^2$ for some $c$; that is, $$x^2 + x + \cdots = (x+c)^2.$$
Since the middle term should be $2cx$, and you have $x$, that means that you want $2c=1$, or $c=\frac{1}{2}$.
But if you have $(x+\frac{1}{2})^2$, you get $x^2 + x + \frac{1}{4}$. Since all you have is $x^2+x$, you complete the square by adding the missing "$\frac{1}{4}$". Since you are not allowed to just add constants willy-nilly, you must also cancel it out by subtracting $\frac{1}{4}$. So: \begin{align*} x^2 + x + 2 &= (x^2 + x + \cdots) + 1\\ &=\left( x^2 + 2\left(\frac{1}{2}\right)x + \cdots \right) + 1 &&\mbox{figuring out what $c$ is}\\ &= \left(x^2 + 2\left(\frac{1}{2}\right)x + \left(\frac{1}{2}\right)^2\right) -\frac{1}{4} + 1 &&\mbox{completing the square}\\ &= \left(x+\frac{1}{2}\right)^2 + \frac{3}{4}. \end{align*}
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Looking at wikipedia, it says that the form is $a(x-h)^2+k$, so this question asked for complete the square of $2x^2+x+1$ the answer would be $2(x+\frac{1}{2})^2+\frac{3}{4}$ correct? – Tom Brito Feb 6 '11 at 0:36
@Tom: If you multiply it out, you'll see it doesn't work out right (and you should have done that before asking!). To get it right: First factor out the $2$ from the $x^2$ and $x$ terms to get $2(x^2+\frac{1}{2}x) + 1$. Then complete the square as above: here $2c=\frac{1}{2}$, so $c=\frac{1}{4}$, so you need to add $c^2=\frac{1}{16}$; since it is multiplied by the $2$, you are really adding $\frac{2}{16}=\frac{1}{8}$, which then is subtracted. So you get$$2x^2+x+1 = 2(x^2+\frac{1}{2}x) + 1 = 2(x^2+\frac{1}{2}x+\frac{1}{16}) + 1-\frac{1}{8} = 2(x+\frac{1}{4})^2 + \frac{7}{8}.$$ – Arturo Magidin Feb 6 '11 at 0:44
"Completing the square" is a standard step in solving a quadratic equation. To see how it helps, consider the following: The general formulation of a quadratic equation is $ax^2+bx+c = 0$ with $a \neq 0$. Let us say we are tasked with solving this equation, i.e., finding values of $x$ that satisfy this equation. To start with, note that this equation is easy to solve if $b=0$. Then the equation looks like $ax^2 + c = 0$ which would simplify to $x = \pm \sqrt{\frac{-c}{a}}$.
"Completing the square" is a step that takes a general quadratic and reduces it to the form of the simple quadratic above. It does this with a substitution $y = x + \frac{b}{2a}$. Then, $ay^2 = ax^2 + bx + \frac{b^2}{4a}$ which means the general quadratic can be written as $ay^2-\frac{b^2}{4a} + c = 0$ or equivalently as $$a(x+\frac{b}{2a})^2 = \frac{b^2}{4a} - c$$
This equation is easy to solve and yields the two roots of the general quadratic equation. Note that the above equation is equivalent to the one we started with ($ax^2+bx+c = 0$) for the purposes of finding the roots. This process of rewriting is called completing the square. This is the point behind rewriting $x^2+x+1$ as $(x+\frac{1}{2})^2 + \frac{3}{4}$.
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"eliminating the middle term" is a bit misleading. One hasn't eliminated it. Rather one has "absorbed" it into the the square term by "completing" the "partial square" comprised of the two highest degree terms. – Bill Dubuque Feb 5 '11 at 22:08
@Bill: I agree. I will edit the answer accordingly. Thanks. – Dinesh Feb 5 '11 at 22:16
Arturo's answer explains the reason behind the name "completing the square" part very well. I will leave this answer up for the derivation of the roots of the quadratic part. – Dinesh Feb 5 '11 at 22:22
One can rewrite a degree $\rm\:n > 1\:$ polynomial $\rm\ f(x)\ =\ x^n + b\ x^{n-1} +\ \cdots\$ into a form such that its two highest degree terms are "absorbed" into a perfect $\rm\: n$'th power of a linear polynomial, namely
$\rm\quad\quad f(x)\ =\ (x + b/n)^n\ -\ g(x)\ \$ where $\rm\ \ g(x)\ =\ (x+b/n)^n - f(x)\$ has degree $\rm\:\le\: n-2$
When $\rm\ n = 2\$ this is called completing the square - esp. when used to solve a quadratic equation.$\ \$ If $\rm\ \ g(x)\ = g\$ is constant (as is always true when $\rm\ n = 2\:$) then this yields a closed form for the roots of $\rm\:\ f(x)\:,\$ namely $\rm\ x\ =\ \sqrt[n]{g}-b/2\:.$
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See this video and its sequel to see the process worked in real time.
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As Arturo points out what you have to observe is the coefficient.
\begin{align*} x^{2}+x+1 & = x^{2} + x + \frac{1+3}{4} \\ & = x^{2} + x + \frac{1}{4} + \frac{3}{4} \\ & = \Bigl(x+\frac{1}{2}\Bigr)^{2} + \frac{3}{4} \end{align*}
I am sure once you get used to such type of things you shall not have trouble in doing such problems. Solve more problems based on this type. Suppose you have the coefficient of $x$ as $a$ note that $a^{2}/4$ should be added and subtracted from the constant term. What i mean by this is: Suppose you have something of this type $x^{2}+ax + b^{2}$ then you can write this as $(x+a/2)^{2} + b^{2}-a^{2}/{4}$.
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http://math.stackexchange.com/questions/108124/calculate-the-146th-digit-after-the-decimal-point-of-frac1293 | # Calculate the 146th digit after the decimal point of $\frac{1}{293}$
The question is: Calculate the 146th digit after the decimal point of $\frac{1}{293}$
1 / 293 = 0,00341296928.., so e.g., the fifth digit is a 1.
We know that 293 is a prime, probably this would help us. I think an equation involving modulos has to be solved, but I am not sure how to tackle this.
Any help is appreciated! Could perhaps someone give a general method to solve these kind of problems?
EDIT: You are supposed to solve this without using a computer.
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As a start: $\mathrm{ord}_{293}(10)=146$. – Guess who it is. Feb 11 '12 at 13:54
The number $146$ seems to be special. It is the period of the decimal expansion of the fraction $\dfrac{1}{293}$. May be somebody shed should light on this! – user21436 Feb 11 '12 at 13:55
@Kanappan: that's precisely what my comment was saying. The multiplicative order gives the period of a unit fraction with denominator relatively prime to $10$ (which is the base we're considering). – Guess who it is. Feb 11 '12 at 14:08
I did not follow your comment. Sorry for posting the same thing. And, thanks for the Wolfram Link. – user21436 Feb 11 '12 at 14:18
Let $r$ be the remainder when $10^{146}$ is divided by $293$. Then the answer is given by the last digit of $(10^{146} - r) / 293$.
Why is this true? Since $1/293$ is a positive number less than one, it is of the form
$1/293 = 0.a_1a_2a_3a_4 \ldots$
and thus
$10^{146}/293 = a_1a_2 \ldots a_{145}a_{146}.a_{147}a_{148} \ldots$
On the other hand, by the division algorithm $10^{146}/293 = q + r/293$, where $q$ is the quotient and $0 \leq r < 293$ is the remainder. Since $q$ is an integer and $0 \leq r/293 < 1$, it follows that $q = a_1a_2 \ldots a_{145}a_{146}$ and $r/293 = 0.a_{147}a_{148} \ldots$.
Then $(10^{146} - r)/293 = q = a_1a_2 \ldots a_{145}a_{146}$.
Thus we can apply modular arithmetic to solve the problem. Notice that
$(10^{146} - r) \cdot 293^{-1} \equiv -r \cdot 293^{-1} \equiv -r \cdot 3^{-1} \equiv -r \cdot 7 \equiv 3r \mod 10$.
Therefore the last digit is equal to $3r \mod 10$.
What remains is to calculate $r$. For this particular case I don't know of any better way than direct calculation. Repeated squaring works, but you might want to use a calculator. It turns out that $10^{146} \equiv 1 \mod 293$, and thus the answer is $3$.
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Nice explanation. I have added an explanation of how you would do the repeated squaring in my post. – bgins Feb 11 '12 at 18:32
Consider a rational number $r = \frac{1}{d}$ and suppose it has a decimal expansion of the form $0.\underbrace{d_1d_2\ldots d_m}_{\text{non-recurring}}\underbrace{d_{m+1}d_{m+2}\ldots d_{m+n}}_{\text{recurring}}d_{m+n+1}\ldots$, that is $d_{k+n} = d_{k}$ for $\forall k > m$.
Let $A$ be the integer formed by the first $m$ digits $A = d_1d_2\ldots d_m$, and $B$ be the integer formed by the next $m$ digits $B = d_{m+1} d_{m+2} \ldots d_{m+n}$. Then we have $$10^m r - A = 0.d_{m+1}d_{m+2}\ldots d_{m+n} d_{m+1} d_{m+2} \ldots$$ Multiplying both sides with $10^n$ we get $$10^n (10^m r - A) = B + (10^m r - A)$$ Or, equivalantly $$10^m \left(10^n - 1 \right) = \left( B - A \left(10^n - 1 \right)\right) d$$ This implies that $10^m \left(10^n - 1 \right) \bmod d = 0$.
Since in the case at hand $d$ is relatively prime to $10$, the smallest solutions for $m$ and $n$ are $m=0$ and $n = \operatorname{ord}_{d}(10)$. The multiplicative order is defined as a smallest exponent $n$ such that $10^n \equiv 1 \mod d$. Multiplicative order $\operatorname{ord}_d(10)$ is a divisor of the Euler totient function $\phi(d)$. Since $d = 293$ is prime $$\phi(293) = 293-1 = 292 = 2^2 \cdot 73$$ thus we should try $n = 73$, $n=146$ and then $n=292$. It is not hard to see that $10^{73} = - 1 \bmod 293$, thus $n= 146$.
Having determined that, the 146-th digits equals $B \bmod 10$. $$(10^n-1) = B d$$ meaning that $B \bmod 10 = (-1) d^{-1} \bmod 10 = 3$.
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Nice explanation! There is a small typo: In a given modulus, the multiplicative order is a divisor of the Euler totient function (not a multiple). – bgins Feb 11 '12 at 18:27
@bgins Thanks for point out the typo. I have just fixed that. – Sasha Feb 11 '12 at 19:40
The following is a small variant of the methods of Sasha and Lopsy. We show that $10$ is a quadratic residue of $293$. This enables us to conclude that $10^{146} \not\equiv -1 \pmod{293}$.
It is just a Legendre symbol calculation. For ease of typing we denote the Legendre symbol by $(a/p)$ instead of $\left(\frac{a}{p}\right)$.
Note that $(10/293)=(2/293)(5/293)$. Since $293$ is of the form $8k+5$, we have $(2/293)=-1$.
To calculate $(5/293)$ we use Quadratic Reciprocity. Since one, and indeed both, of $5$ and $293$ are of the shape $4k+1$, $$(5/293)=(293/5)=(3/5).$$ One could continue with Quadratic Reciprocity, but by inspection $(3/5)=-1$. Thus $(10/293)=(-1)(-1)=1$, and we conclude that $10$ is indeed a quadratic residue of $293$.
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The $146^\text{th}$ digit of $\frac1{293}$ is $$d = \left\lfloor\frac{10^{146}}{293}\right\rfloor - 10 \left\lfloor\frac{10^{145}}{293}\right\rfloor,$$
and sage says it's 3 (sage doesn't count the leading zeros in the decimal mantissa):
(1/293).n(digits=144)
0.00341296928327645051194539249146757679180887372013651877133105802047781569965870307167235494880546075085324232081911262798634812286689419795221843
floor(10^146/293)-10*floor(10^145/293)
3
As to why, well, $p=293$ is prime, and $p-1=292=2^2\cdot73$, and any integer $a$ which is relatively prime to $p$ will have order $d$ dividing $p-1$. So the smallest positive power $d$ of $a=10$ so that $a^d\equiv1\pmod p$ must be $2,4,73,2\cdot73=146$ or, if none of these, then $4\cdot73=292$. Now $d=2$ and $d=4$ can easily be ruled out since $100,1000\not\equiv1\pmod{292}$. For $d=73$, note that $73=(10010001)_2=2^6+2^3+2^0$, so that we can calculate $10^{73}$ modulo $293$ (and its square if necessary) as follows: $$10^2=100$$ $$10^4=(100)^2=10000\equiv38\pmod{293}$$ $$10^8\equiv(38)^2=1444\equiv272\equiv-21\pmod{293}$$ $$10^9\equiv10\cdot(-21)=-210\equiv83\pmod{293}$$ $$10^{18}\equiv(83)^2=6889\equiv150\pmod{293}$$ $$10^{36}\equiv(150)^2=22500\equiv232\equiv-61\pmod{293}$$ $$10^{72}\equiv(-61)^2=3721\equiv205\equiv-88\pmod{293}$$ $$10^{73}\equiv10\cdot(-88)=-880\equiv292\equiv-1\pmod{293}$$
Since squaring the last quantity gives $1$ modulo $293$, we find that $d=\text{ord}_{293}{10}=2\cdot73=146$. This method is called repeated squaring: starting with $a=10$ (step $0$), repeatedly square the result, multiplying again by $a$ (modulo $p$) at each intermediate step $i$ (after squaring) if bit $i$, corresponding to $2^i$ in the binary expansion of $d$, is one.
Now with @m-k's post, we see why. If $10^{146}=q\cdot293+r$, with $q$ and $r$ given by the division algorithm, i.e. $q,r\in\mathbb{Z}$ with $0\leq r<293$, then $q$ is the first quantity in the formula above for $d$: $$q=\left\lfloor\frac{10^{146}}{293}\right\rfloor =\frac{10^{146}-r}{293},$$ and its last digit -- that is, its remainder modulo $10$ -- is equal to $d$: $$d\equiv q\pmod{10}.$$ However, modulo $10$, we have $$293q\equiv-r \pmod{10} \quad \implies \quad q\equiv293^{-1}\cdot-r \equiv-3^{-1}r \equiv-7r \equiv3r \pmod{10}.$$ But we already found that $r=1$, since $10^{146}\equiv1\pmod{293}$, so that $$d\equiv q\equiv 3r\equiv 3\pmod{10}.$$
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I am not sure if OP would like to see a computer at work. He/She seems to be asking a general method to solve it by hand. Anyway, I'm afraid, it may receive downvotes, while I wont downvot it! – user21436 Feb 11 '12 at 14:00
The number $\dfrac{1}{293}$ is the following in its decimal form:
Image Courtesy: Wolfram | Alpha
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In Mathematica, the function RealDigits[] does this quickly. For a ploddingly literal implementation of the method in Mathematica, try First[Nest[QuotientRemainder[10 Last[#], 293] &, {0, 1}, MultiplicativeOrder[10, 293]]]. – Guess who it is. Feb 11 '12 at 14:10
I don't even have Mathematica installed. May be I'll learn that sooner/later! – user21436 Feb 11 '12 at 14:19
That's alright. Since you already mentioned Wolfram Alpha, I might as well show how it's actually done in Mathematica (which is the engine behind Wolfram Alpha). – Guess who it is. Feb 11 '12 at 14:21
I am interested in solving this problem without using a computer. I also have Mathematica installed on my pc. – ClausW Feb 11 '12 at 14:26
These responses appeared before your edit saying you did not wish to see a computational solution to the problem. I think the downvotes are not fair as a result. – ncmathsadist Feb 11 '12 at 16:27
This is a 3. The basic period of the decimal expansion looks like this.
0 .0003412969
1. 2832764505
2. 1194539249
3. 1467576791
4. 8088737201
5. 3651877133
6. 1058020477
7. 8156996587
8. 0307167235
9. 4948805460
10. 7508532423
11. 2081911262
12. 7986348122
13. 8668941979
14. 5221843
I computed it using this little program.
apple = 1000
biter = 293
out = ".000"
for k in range(1000):
out += str(apple//biter)
apple %= biter
apple *=10
print out
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https://webcore.com.hr/gain-capital-ltuican/3ee054-second-fundamental-theorem-of-calculus-chain-rule | Thus if a ball is thrown straight up into the air with velocity the height of the ball, second later, will be feet above the initial height. Of the two, it is the First Fundamental Theorem that is the familiar one used all the time. The second part of the theorem gives an indefinite integral of a function. The preceding argument demonstrates the truth of the Second Fundamental Theorem of Calculus, which we state as follows. Using the Second Fundamental Theorem of Calculus, we have . With the chain rule in hand we will be able to differentiate a much wider variety of functions. The Fundamental Theorem tells us that E′(x) = e−x2. I would know what F prime of x was. As you will see throughout the rest of your Calculus courses a great many of derivatives you take will involve the chain rule! Example problem: Evaluate the following integral using the fundamental theorem of calculus: Theorem (Second FTC) If f is a continuous function and $$c$$ is any constant, then f has a unique antiderivative $$A$$ that satisfies $$A(c) = 0$$, and that antiderivative is given by the rule $$A(x) = \int^x_c f (t) dt$$. Note that the ball has traveled much farther. This conclusion establishes the theory of the existence of anti-derivatives, i.e., thanks to the FTC, part II, we know that every continuous function has an anti-derivative. The chain rule is also valid for Fréchet derivatives in Banach spaces. The Two Fundamental Theorems of Calculus The Fundamental Theorem of Calculus really consists of two closely related theorems, usually called nowadays (not very imaginatively) the First and Second Fundamental Theo-rems. Hot Network Questions Allow an analogue signal through unless a digital signal is present It looks complicated, but all it’s really telling you is how to find the area between two points on a graph. (We found that in Example 2, above.) So any function I put up here, I can do exactly the same process. The Fundamental Theorem of Calculus and the Chain Rule; Area Between Curves; ... = -32t+20\), the height of the ball, 1 second later, will be 4 feet above the initial height. Fundamental Theorem of Calculus Example. FT. SECOND FUNDAMENTAL THEOREM 1. Recall that the First FTC tells us that … Fundamental Theorem of Calculus, Part II If is continuous on the closed interval then for any value of in the interval . In most treatments of the Fundamental Theorem of Calculus there is a "First Fundamental Theorem" and a "Second Fundamental Theorem." I would define F of x to be this type of thing, the way we would define it for the fundamental theorem of calculus. The second fundamental theorem of calculus holds for f a continuous function on an open interval I and a any point in I, and states that if F is defined by the integral (antiderivative) F(x)=int_a^xf(t)dt, then F^'(x)=f(x) at each point in I, where F^'(x) is the derivative of F(x). It has gone up to its peak and is falling down, but the difference between its height at and is ft. In this situation, the chain rule represents the fact that the derivative of f ∘ g is the composite of the derivative of f and the derivative of g. This theorem is an immediate consequence of the higher dimensional chain rule given above, and it has exactly the same formula. In Section 4.4, we learned the Fundamental Theorem of Calculus (FTC), which from here forward will be referred to as the First Fundamental Theorem of Calculus, as in this section we develop a corresponding result that follows it. 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Calculus, which we state as follows between its height at and is down! Same process how to find the area between two points on a.... Complicated, but all it ’ s really telling you is how to find the area two! Great many of derivatives you take will involve the chain rule is also for! We have really telling you is how to find the area between two points on a graph x ) e−x2... | 2021-08-05T23:48:47 | {
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http://math.stackexchange.com/questions/332940/area-enclosed-between-the-curves-y-x2-and-y-60-7x | # Area enclosed between the curves $y=x^2$ and $y=60-7x$
Find the area enclosed between the curves $y=x^2$ and $y=60-7x$.
I am completely new at this but I have tried and I believe that it should be between the numbers $0$ and $60$ is this right?
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The area is between 810 and 820. If you show your work someone can help you spot the mistake. – Joni Mar 17 '13 at 17:05
Calculate intersections of your functions and integrate the function $60-7x-x^2$ with this limits. – Andre Mar 17 '13 at 17:05
It is always a good idea to draw the functions if possible(i.e if you know the graphs of the functions, and you should learn some basic graphs like lines,sin and cos, roots,etc.), then find the intersection points between the two function which are in this case $x=5,x=-12$ as the other answers show. Here is a graph of these functions
and as the graph shows $60-7x$ is above $x^2$ thus the area is \begin{align} A=&\int_{-12}^{5}60-7x-x^2dx\\ =&60x-\frac{7x^2}{2}-\frac{x^3}{3}\mid_{-12}^{5}\\=&\frac{4913}{6}. \end{align}
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Find where $y = x^2$ and $y = 60 - 7x$ intersect:
You can do this by setting the equations equal to one another and solving the resulting quadratic equation $$x^2 = 60 - 7x \iff x^2 +7x - 60 = 0$$ $$\iff (x+12)(x-5) = 0$$ $$\iff \;x = -12, \;\;\text{or}\;\;x = 5$$
Use these values as your limits of integration, and calculate the integral of the area between the line $y = 60 - 7x$ and the parabola $y = x^2$ by integrating the function $$(60 -7x) -x^2$$
using those $x$-values as limits.
Note: We subtract from $y = 60-7x$, the function $\;y = x^2\;$ because for every $x\in (-12, 5),\;\;60 - 7x > x^2$.
This can easily be shown by graphing the two functions, which is always a good thing to do for problems like this. This helps determine the number of regions over which to integrate, if more than one such region exists, which function is the "upper" and which the "lower", and bounds of integration. WolframAlpha gives us:
That is, to compute the area of the region bound by $\;y = 60 - 7x\;$ and $\;y = x^2,\;$ between the values of intersection $\;x = -12\;$ to $\;x=5,\;$ compute $$\int_{-12}^5 (60 - 7x - x^2) \, dx$$
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Does this make sense? If $,F = \int (60 - 7x - x^2)\,dx\;$ then evaluate $F(5) - F(-12) =$ Area. So compute the integral, then determine its value at $5$, and subtract from that its value at $-12$. – amWhy Mar 17 '13 at 17:33
Needs positive feedback! +1 – Amzoti May 4 '13 at 0:40
Here's how to find the area between two functions, $f(x)$ and $g(x)$.
1. Find when $f(x) = g(x)$, calling the intersections $a$ and $b$. Assume $f(x) \ge g(x)$ over the interval $[a, b]$ (this will make sure the area is positive).
2. Find $$\int_{a}^b f(x) - g(x) \; \mathrm dx$$
In your case, we have that:
$$x^2 = 60 - 7x$$ $$x^2 + 7x - 60 = 0$$
The solutions are $-12$ and $5$, and we call these $a$ and $b$. Since $60 - 7x$ is greater over this interval, we integrate:
$$\int_{-12}^{5} 60 - 7x - x^2 \; \mathrm dx$$ $$\int_{-12}^{5} 60 \; \mathrm dx - \int_{-12}^{5} 7x \; \mathrm dx - \int_{-12}^{5} x^2 \; \mathrm dx$$
Consider a rectangle with length $17$ ($|-12| + |5| = 17$) and height $60$ (the integrand). Now the area is $60 \cdot 17 = 1020$.
$$1020 - 7\color{green}{\int_{-12}^{5} x \; \mathrm dx} - \color{blue}{\int_{-12}^{5} x^2 \; \mathrm dx}$$
By the power rule of integration:
$$\int x^n \; \mathrm dx = \frac{x^{n-1}}{n}$$
and the fundamental theorem of calculus:
$$\int_a^b f(x) = F(b) - F(a) \text{ where } F'(x) = f(x)$$
$$1020 - 7\color{green}{\left(\frac{5^2}{2} - \frac{(-12)^2}{2} \right)} - \color{blue}{\left(\frac{5^3}{3} - \frac{(-12)^3}{3}\right)}$$ $$1020 - 7\color{green}{(25/2 - 144/2)} - \color{blue}{(125/3 + 1728/3)}$$ $$= 9683 / 3$$
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Yes this I know but it is the calculation after that I get stuck with – user1838781 Mar 17 '13 at 17:19
@user1838781, I added more information. Do you know how to integrate functions such as $x^2$ and $x$? What exactly are you having trouble with. – George V. Williams Mar 17 '13 at 17:24
After the integration I belive that your supposed to calculate so that in the end it's only a simple number – user1838781 Mar 17 '13 at 17:29
@user1838781, I've completed the problem by using the fundamental theorem of calculus and the power rule for integration. Do you understand how the results are derived now? – George V. Williams Mar 17 '13 at 17:36
No I don't know how to integrate it. Is it something like: if I take it all together [60-7-2x] what now? – user1838781 Mar 17 '13 at 17:38 | 2015-07-06T01:15:13 | {
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Rational functions contain asymptotes, as seen in this example: In this example, there is a vertical asymptote at x = 3 and a horizontal asymptote at y = 1. This is the currently selected item. Rational functions may or may not intersect the lines or polynomials which determine their end behavior. The vertical asymptote is a place where the function is undefined and the limit of the function does not exist. Learn what the end behavior of a polynomial is, and how we can find it from the polynomial's equation. End Behavior of Polynomial Functions. Is the degree of the numerator greater than, less than, or equal to ... End behavior? – 2.5 – End Behavior, Asymptotes, and Long Division Page 2 of 2 RATIONAL FUNCTIONS END BEHAVIOR Improper Rational Functions where the Numerator’s Degree is Greater than the Denominator’s Degree: If N > D, the end behavior is decided by the reduced function. If the degree of the numerator is equal to the degree … greater than, less than, or equal to the degree of the denominator? It should be noted that, if the degree of the numerator is larger than the degree of the denominator by more than one, the end behavior of the graph will mimic the behavior of the reduced end behavior fraction. The graphs show that, if the degree of the numerator is exactly one more than the degree of the denominator (so that the polynomial fraction is "improper"), then the graph of the rational function will be, roughly, a slanty straight line with some fiddly bits in the middle. End Behavior. For a rational function the end behavior is determined by the relationship between the degree of and the degree of If the degree of is less than the degree of the line is a horizontal asymptote for If the degree of is equal to the degree of then the line is a horizontal asymptote, where and are the leading coefficients of and respectively. In this example, there is a vertical asymptote at x = 3 and a horizontal asymptote at y = 1. End behavior of polynomials. End behavior of polynomials. The curves approach these asymptotes but never cross them. Honors Calculus. ... Find the oblique asymptote: The method used to find the horizontal asymptote changes depending on how the degrees of the polynomials in the numerator and denominator of the function compare. \begin{eqnarray} x+4&=& 0\\ x &=&-4 \end{eqnarray} Horizontal Asymptote. Asymptotes, End Behavior, and Infinite Limits. Google Classroom Facebook Twitter. This is because as #1# approaches the asymptote, even small shifts in the #x#-value lead to arbitrarily large fluctuations in the value of the function. Email. The method to find the horizontal asymptote changes based on the degrees of the polynomials … The curves approach these asymptotes but never cross them. Write them on their graphs. The function $$f(x)→∞$$ or $$f(x)→−∞.$$ Intro to end behavior of polynomials. The remainder is ignored, and the quotient is the equation for the end behavior model. Keeper 12. Practice: End behavior of polynomials. BONUS: Find the equation of the three oblique asymptotes for the functions on the front. To find whether a function crosses or intersects an asymptote, the equations of the end behavior polynomial and the rational function need to be solved. End Behavior of Polynomial Functions. https://www.khanacademy.org/.../v/end-behavior-of-rational-functions For instance, if we had the function $f(x)=\dfrac{3x^5−x^2}{x+3}$ with end behavior $f(x)≈\dfrac{3x^5}{x}=3x^4$, At each of the function’s ends, the function could exhibit one of the following types of behavior: The function $$f(x)$$ approaches a horizontal asymptote $$y=L$$. The behavior of a function as $$x→±∞$$ is called the function’s end behavior. To find the vertical asymptote(s) of a rational function, simply set the denominator equal to 0 and solve for x. Example. 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Solve for x the limit of the polynomials … end behavior of a polynomial is and! | 2021-07-26T03:34:52 | {
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https://math.stackexchange.com/questions/1765431/implicitly-finding-the-derivative-of-f-1x-given-fx | # Implicitly finding the derivative of $f^{-1}(x)$ given $f(x)$
Can we find the derivative of the inverse of a function implicitly by finding the derivative of the original function?
For example lets say I have $f(x) = e^x$ and I want to find the derivative of the inverse function $f^{-1}(x) = ln(x)$, without actually finding the inverse function itself and then taking the derivative of it (which would be $\frac{1}{x}$)
For further clarification of what I'm trying to ask, this is the standard way of finding the derivative of the inverse of a given function $f(x) = e^x$
$$\text{Given} \ \ f : x \to e^x \ \ \forall x \in \mathbb{R}$$ $$f(x) = e^x$$ $$\implies f^{-1}(x) = ln(x)$$ $$\frac{df^{-1}}{dx} = \frac{1}{x}$$
But what if I wanted to find the derivative of the inverse function, without having to first find the inverse function from the given function, and then taking the derivative of that.
Question : So given some function $f$, I want to know if it's possible to find $\frac{df^{-1}}{dx}$ without first finding $f^{-1}$ and then taking the derivative of it. In other words can you implicitly find $\frac{df^{-1}}{dx}$ given a function $f$?
I ask this question, as there would be functions in which finding the inverse functions may prove difficult or impossible (correct me if I'm wrong here), and this way of finding the derivative of the inverse of the function, may or may not be useful in those cases.
I also haven't stated whether function $f$ (which I'm trying to implicitly find the derivative of its inverse) is injective, bijective or surjective, I have done this on purpose as I'm sure the answer would differ depending on whether $f$ is injective, bijective or surjective.
Let $y=f^{-1}(x) \iff x = f(y)$. Then leveraging the Chain Rule
$$\frac{d}{dx}[x] = \frac{d}{dx}[f(y)] \implies 1 = f'(y) \cdot \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{1}{f'(y)} = \frac{1}{f'(f^{-1}(x))}$$
Hence $(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$
• @ Stefan4024 But would that not just produce $\frac{1}{x} \ \ \forall \ f \ \$? – Perturbative Apr 30 '16 at 14:28
• I think you left out the $'$ symbol to denote the derivative of $f$ in the denominator, as Evinda has done in his/her answer below. – Perturbative Apr 30 '16 at 14:50
• @Perturbative No because we have derivative – Stefan4024 Apr 30 '16 at 23:51
It holds that $$f(f^{-1}(x))=x$$
So: $$(f(f^{-1}(x)))'=1 \Rightarrow f'(f^{-1}(x)) (f^{-1}(x))'=1 \Rightarrow (f^{-1}(x))'=\frac{1}{f'(f^{-1}(x))}$$ | 2020-01-26T21:59:40 | {
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https://math.stackexchange.com/questions/627009/prove-that-int-01-u-alpha-1-1-1-u-alpha-2-1-rm-du-frac-gam | # Prove that $\int_{0}^1 u^{\alpha_1-1} (1-u)^{\alpha_2-1} \, {\rm d}u =\frac{\Gamma(\alpha_1)\Gamma(\alpha_2)} {\Gamma(\alpha_1+\alpha_2)}$
I have the following equality in a textbook of mine
$$\frac{y^{\alpha_1+\alpha_2-1} e^{-y/\beta}}{\Gamma(\alpha_1+\alpha_2) \beta^{\alpha_1+\alpha_2}} \cdot \frac{\Gamma(\alpha_1+\alpha_2)}{\Gamma(\alpha_1)\Gamma(\alpha_2)} \int_{0}^1 u^{\alpha_1-1} (1-u)^{\alpha_2-1} \, \mathrm du = \frac{y^{\alpha_1+\alpha_2-1} e^{-y/\beta}}{\Gamma(\alpha_1+\alpha_2) \beta^{\alpha_1+\alpha_2}}$$
and I see that for the equality to be true we must have
$$\int_{0}^1 u^{\alpha_1-1} (1-u)^{\alpha_2-1} \, \mathrm du =\frac{\Gamma(\alpha_1)\Gamma(\alpha_2)} {\Gamma(\alpha_1+\alpha_2)}$$
However can someone give me an explanation why this is the case ?
• en.wikipedia.org/wiki/… – Meow Jan 4 '14 at 15:11
• I think I've posted answers to essentially this same question here before. I've up-voted the question, even though I may also (if I get around to finding one of those answers) vote to close this as a duplicate. – Michael Hardy Jan 4 '14 at 16:49
• Here's an answer I posted that proves this identity in case the exponents are integers: math.stackexchange.com/questions/86542/… – Michael Hardy Jan 4 '14 at 16:52
• Does the proof on wikipedia prove it in the case of integers only or is the proof valid for any real number ? – Shuzheng Jan 4 '14 at 18:22
## 1 Answer
For $\alpha_1, \alpha_2 \gt0$ We have
$$\int_{0}^1 u^{\alpha_1-1} (1-u)^{\alpha_2-1} \, \mathrm du = \text{B}(\alpha_1,\alpha_2) \tag{1}$$
Where $\text{B}(a,b)$ is Beta Function \begin{align} \Gamma(\alpha_1) & =\int_0^\infty\ e^{-x} x^{\alpha_1-1}\,\mathrm{d}x \tag{2}\\ \Gamma(\alpha_2) & =\int_0^\infty\ e^{-y} y^{\alpha_2-1}\,\mathrm{d}y \tag{3}\\ \end{align}
\begin{align} \Gamma(\alpha_1)\Gamma(\alpha_2) & = \int_0^\infty\ e^{-x} x^{\alpha_1-1}\,\mathrm{d}x \int_0^\infty\ e^{-y} y^{\alpha_2-1}\,\mathrm{d}y \tag{4}\\ & =\int\limits_0^\infty\int\limits_0^\infty\ e^{-x-y} x^{\alpha_1-1}y^{\alpha_2-1}\,\mathrm{d}x \,\mathrm{d}y \tag{5}\\ & =\int_{z=0}^\infty\int_{t=0}^1 e^{-z} \Big(zt\Big)^{\alpha_1-1}\Big(z(1-t)\Big)^{\alpha_2-1}z\,\mathrm{d}z \,\mathrm{d}t \tag{6}\\ & =\int_{z=0}^\infty\int_{t=0}^1 e^{-z} z^{\alpha_1+\alpha_2-1}t^{\alpha_1-1}(1-t)^{\alpha_2-1}\,\mathrm{d}z \,\mathrm{d}t \tag{7}\\ & =\int_{0}^\infty e^{-z}z^{\alpha_1+\alpha_2-1} \,\mathrm{d}z\int_{0}^1t^{\alpha_1-1}(1-t)^{\alpha_2-1}\,\mathrm{d}t \tag{8}\\ \Gamma(\alpha_1)\Gamma(\alpha_2) & =\Gamma(\alpha_1+\alpha_2){\rm B}(\alpha_1,\alpha_2) \tag{9}\\ \end{align}
Hence
$$\large{\rm B}(\alpha_1,\alpha_2)=\int_{0}^1 u^{\alpha_1-1} (1-u)^{\alpha_2-1} \, du =\frac{\Gamma(\alpha_1)\,\Gamma(\alpha_2)}{\Gamma(\alpha_1+\alpha_2)}$$
$\text{ Explanation: } (6)$ Substituting $x=zt$ and $y=z(1-t)$ | 2019-07-21T02:40:06 | {
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https://math.stackexchange.com/questions/2312362/probability-from-randomized-dice | # Probability from randomized dice
Suppose you have a pair of dice that have removable stickers for numbers on each of their 6 sides. Suppose that you unstick all 12 of the stickers from the dice and reapply them randomly to the 2 dice. You will still have 2 occurrences of each number 1 through 6. However, they may both occur on the same die. (For instance: after rearranging the stickers, you may have dice $d_1$ and $d_2$ with sides $d_1 = [1,2,2,4,4,6]$ and $d_2 = [1,3,3,5,5,6]$.)
Suppose now that you roll this randomized pair of dice. Is there a concise way to calculate the probability of each outcome? What is the probability of each possible outcome?
Just by working out some of the possible arrangements, it seems like $p(2)$ should be $\frac{1}{72}$ (which might not be correct), but the other probabilities are more difficult to compute this way.
• How did you calculated $p(2)=\frac1{72}$ ? – callculus Jun 6 '17 at 18:03
• I made an error. Starting with the fact that $p(2)=p(2|1$s on same die$)+p(2|1$s on different dice$)$, the first term is zero, so it reduces to $p(2)=p(2|1$s on different dice$)=\frac{1}{36}*p(1$s on different die$)$. My mistake was in error in the combinatorics $p(1$s on different die$)=\frac{6}{11}$ (not $\frac{1}{2}$). – DavidWayne Jun 6 '17 at 18:42
Here's a formula that generalizes to more digits.
Let's label the dice A and B, and write $(i,j)$ for the getting $i$ on A and $j$ on $B$. Consider the following two cases.
1. $i=j$. This can only happen if we have exactly one $i$ on each die. The probability of this is $\frac{ 2\binom{10}{5}}{\binom{12}{6}}$ (chose one of the two $i$'s on A and the other on B, then choose $5$ more numbers from the remaining $10$ for die A). Once we have exactly one $i$ on each die, the probability of both landing $i$, is $\frac 16 \times\frac 16$. Therefore $$P( (i,i)) = \frac{ 2\binom{10}{5}}{\binom{12}{6}} \frac{1}{36}=\frac{1}{66}$$
2. $i\ne j$. Since this is independent of the choice of $i$ and $j$, and there are exactly $6*5=30$ such pairs, the probability is equal to $$\frac {1}{30} (1- \sum_{i} P( (i,i) )=\frac{1}{30}(1-\frac {1}{11})=\frac {1}{33}.$$
Now it remains to find the probability of a sum equal to $k$.
a. If $k\le 7$, there are exactly $k-1$ ways to write it: $(1,k-1),(2,k-2),\dots, (k-1,1)$. Now if $k$ is odd, then in all of these $i\ne j$. Therefore the probability is $(k-1)/33$. If $k$ is even, then exactly one of these is of the form $i=j$, therefore the answer is $(k-2)/33 + 1/66$.
b. Finally if $k \ge 8$, and $(i,j)$ is such that $i+j=k$. Then $(7-i,7-j)$ has exactly the same probability, and has sum between $\{2,\dots,6\}$. Therefore the probability to get $k$ is the same as the probability to get $14-k$.
In other words, the distribution of the sum is symmetric about $7$.
The problem can be viewed as picking $2$ numbers from list of twelve numbers $1$,$1$,$2$,$2$,$3$,$3$,$4$,$4$,$5$,$5$,$6$,$6$ and looking at their sum.
Now there are $12 \cdot 11 = 132$ ways to pick two numbers from the list. In how many ways we can pick two same (specific) numbers? There are apparently exactly $2$ ways to do it (depends on which of the same two numbers you pick first). In how many ways we can pick two different numbers in given order (e.g. $1,2$)? There are exactly $2\cdot2 = 4$ ways to do it ($2$ two choices for first pick, $2$ choices for second pick).
Now for the sums, in how many ways we can get sum $2$? That is possible only as $1+1$, which by reasoning above can happen $2$ times (we have to pick two same numbers), so probability is $$p(2)=\frac{2}{132}=\frac{1}{66}$$
In how many ways can we get sum $3$? We have $3=1+2=2+1$ and again by reasoning above there is overall $4+4=8$ possibilities. This is because to get $2+1$ we need to pick two different numbers (and we know that is possible in $4$ ways), same for $1+2$. So probability is $$p(3)=\frac{8}{132}=\frac{2}{33}$$
Similarly $4=1+3=2+2=3+1$ with $4+2+4=10$ possibilities, so probability is $$p(4)=\frac{10}{132}=\frac{5}{66}$$
And so on...
• Very nice answer; a very clever way of reducing it to a simpler problem. I have to give the best answer to @Fnacool, however, for providing a closed form solution for computing the probabilities. – DavidWayne Jun 6 '17 at 20:33
• @DavidWayne That is alright, I would prefer closed form solution as well. – Sil Jun 7 '17 at 17:51
I wrote a basic Python program to find the distribution of the different values. There are ${12 \choose 6} = 924$ ways to select six values out of twelve, so the number of possible value distributions equals $\frac{924}{2} = 462$. For each of these combinations, I increased the probability of the 36 combinations of the two dice, resulting in the following probabilities:
2 - 0.0152 - 1/66
3 - 0.0606 - 4/66
4 - 0.0758 - 5/66
5 - 0.1212 - 8/66
6 - 0.1364 - 9/66
7 - 0.1818 - 12/66
8 - 0.1364 - 9/66
9 - 0.1212 - 8/66
10 - 0.0758 - 5/66
11 - 0.0606 - 4/66
12 - 0.0152 - 1/66
For instance, to get a 2 we need the two 1's to be on two different dice (probability $\frac{6}{11}$), in which case the probability of hitting a 2 equals $\frac{1}{36}$. The total probability of getting a 2 thus equals $\frac{5}{11} \cdot 0 + \frac{6}{11} \cdot \frac{1}{36}=\frac{1}{66} \approx 0.0152$. The Python program:
import itertools
import math
from collections import defaultdict
values = [x for x in range(1, 13)]
d = defaultdict(int)
for v in itertools.combinations(values, 6):
w = [x for x in values if x not in v]
for i in range(6):
for j in range(6):
d[math.ceil(v[i]/2) + math.ceil(w[j]/2)] += 1/36/924
for k in d:
print(k, d[k])
• It looks like all your probabilities are divisible by $1/66$ (alternatively, d[k] is always a multiple of 14.) It's not obvious why this should be. – Michael Lugo Jun 6 '17 at 18:14
• @MichaelLugo That is correct. I updated the table to make this clear. – jvdhooft Jun 6 '17 at 18:20 | 2019-06-18T01:53:36 | {
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https://math.stackexchange.com/questions/1359619/find-eigenvalues-and-eigenvectors-of-this-matrix | # Find eigenvalues and eigenvectors of this matrix
Problem: Let \begin{align*} A = \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{pmatrix}. \end{align*} Compute all the eigenvalues and eigenvectors of $A$.
Attempt at solution: I found the eigenvalues by computing the characteristic polynomial. This gives me \begin{align*} \det(A - x \mathbb{I}_4) = \det \begin{pmatrix} 1-x & 1 & 1 & 1 \\ 1 & 1-x & 1 & 1 \\ 1 & 1 & 1-x & 1 \\ 1 & 1 & 1 & 1-x \end{pmatrix} = -x^3 (x-4) = 0 \end{align*} after many steps. So the eigenvalues are $\lambda_1 = 0$ with multiplicity $3$ and $\lambda_2 = 4$ with multiplicity $1$.
Now I was trying to figure out what the eigenvectors are corresponding to these eigenvalues. Per definition we have $Av = \lambda v$, where $v$ is an eigenvector with the corresponding eigenvalue. So I did for $\lambda_2 = 4$: \begin{align*} \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} = 4 \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} \end{align*} I think this is only possible when $x_1 = x_2 = x_3 = x_4 = 1$. So am I right in stating that all the eigenvectors corresponding to $\lambda_2$ are of the form $t \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix}$ with $t$ some number $\neq 0$? For $\lambda_1 = 0$, I'm not sure how to find the eigenvectors. The zero vector is never an eigenvector. This means $x_1, x_2, x_3$ and $x_4$ can be anything aslong as they add to zero?
• you are absolutely right. – Zhanxiong Jul 13 '15 at 16:41
• the eigenspace which is corresponding to the eigenvalue 0 is also called the kernel of the matrix - and you are right – user190080 Jul 13 '15 at 16:41
• related – Omnomnomnom Jul 13 '15 at 18:01
You are right, and to show how this applies more generally, look at the eigenspace corresponding to the other eigenvalue, $\lambda = 0$. This has multiplicity 3, so we expect exactly 3 linearly independent vectors in this space. The main equation looks like $A \vec{x} = 0 \vec{x} = \vec{0}$, in other words, $$\begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}$$ It's not hard to see that we have 4 identical equations, so the only constraint is $x + y + z + w = 0$, so we define $w = -x-y-z$ and any vector in the eigenspace now looks like $$\begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix} = \begin{pmatrix} x \\ y \\ z \\ -x-y-z \end{pmatrix} = x \begin{pmatrix} 1 \\ 0 \\ 0 \\ -1 \end{pmatrix} + y \begin{pmatrix} 0 \\ 1 \\ 0 \\ -1 \end{pmatrix} + z \begin{pmatrix} 0 \\ 0 \\ 1 \\ -1 \end{pmatrix}$$ and the three vectors which form the basis of the eigenspace are now specified.
• Thank you, that was very clear. So all the eigenvectors corresponding to the eigenvalue zero are of the form $x \begin{pmatrix} 1 \\ 0 \\ 0 \\ -1 \end{pmatrix} + y \begin{pmatrix} 0 \\ 1 \\ 0 \\ -1 \end{pmatrix} + z \begin{pmatrix} 0 \\ 0 \\ 1 \\ -1 \end{pmatrix}$, with the $x,y$ and $z$ numbers $\neq 0$? – Kamil Jul 13 '15 at 17:09
• @Kamil exactly :) – gt6989b Jul 13 '15 at 17:11
• Worth noting that these are not the only three vectors which work. You can choose any three linearly independent such vectors. – Steven Gubkin Jul 13 '15 at 17:11
• @StevenGubkin of course, once you are in a span, any 3 lin.indep. members will form a basis, but I wanted to show an easy way to construct one, which is untuitive and direct... – gt6989b Jul 13 '15 at 17:12
You are exactly right! $x_1,x_2,x_3$ and $x_4$ can be anything as long as they sum to zero. This is already a complete solution in some sense (you have "found" all of the eigenvectors). If you would like, you could find 3 linearly independent such vectors, and then you would be sure that these three span the whole space.
• So there are infinite eigenvectors corresponding to $\lambda = 0$. How would I express this mathematically, I mean in set notation? The only condition is that they add to zero. – Kamil Jul 13 '15 at 16:47
• @Kamil yes, there are infinite solutions (over an infinite field, of course). They are the solution set of the homogeneous system $\;\det(A-xI)=0\;$ , and in this case this space has dimension three. – Timbuc Jul 13 '15 at 16:53
the matrix you have has rank one and can be written as $$A = uu^\top \text{ where } u = \pmatrix{1,1,1,1}^\top.$$ now, $$Av = uu^\top v=(u^\top v)u.$$ note that $u^\top v$ is a scalar, therefore $A$ has eigenvalue $0$ of multiplicity $3$ corresponding eigenvector any vector orthogonal to $u$ which has dimension $3.$ $A$ has also the nonzero eigenvalue $u^\top u = 4$ and the corresponding eigenvector $u.$
the same ideas can be used to find the eigenvalues and eigenvectors of any rank one matrix $ab^\top.$ | 2019-04-24T23:58:30 | {
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http://tncsolutions.com.au/lv2cb3w/mean-square-displacement-2d-random-walk.php | >
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If we link this discussion to a random walk in one dimension with equal probability of jumping to the left or right and with an initial position x = 0, then our probability A Markov process is a random walk with a selected probability for making The root mean square displacement after a timet 386 12 Random walks and the Random Walk * Idea : A probability measure on the space of all paths on a space; For example, Brownian motion. Take the lattice Zd. random walk. Here is a more careful definition. The root mean square displacement after a timet is then " x2 − x 2 = √ 2Dt. This universality is embodied by the central-limit theorem. (Top left) Trajectory A is a random walk simulated to model experimental trajectory B. DOI: 10. 75 time steps, respectively. Use phasor notation, and let the phase of each vector be random . Have a look here, this may help to demistify it a little bit. For a lower bound, it seems clear that the self-avoidance constraint should force the self-avoiding walk to move away from its starting point at least as fast as the simple random walk, and hence that (R2) > O(n). 0, 5, 10, 25, 100, 1000 ~b!, for a frozen 2D lattice Random walk to calculate the tortuosity tensor of images - PMEAL/pytrax mean-square displacement, which is denoted by (R 2) and is defined as the average of the squared Euclidean distance between the endpoints of a walk. A sample random walk in 2D continuum. Note that, in contrast to the 2D case, the normalization condition of R(φ) in Key words: random walk, Brownian movement, mean absolute deviation, mean In today's science the notions of the mean square devia- The classical example of a 2-D random . Differ by sequence and length. 01 0. Lets consider the two dimensional case of a random10/07/2009 · Lattice Random Walk in 2D Smile Like You Mean It - The Killers lyrics - Duration: 3:55. These dynamics are related to the mechanical properties of the medium in which the particles are moving. (Top right) Trajectory B is an experimental trajectory showing two-dimensional diffusion that is easily analyzed with conventional two-dimensional MSD analysis. For example, particles in a continuum viscous material will have a mean-squared displacement growing linearly with Atomic displacement does not follow a simple trajectory: "collisions" with other atoms render atomic trajectories quite complex shaped in space The trajectory followed by an atom in a liquid resembles that of a pedestrian random walk. Interestingly, This Demonstration shows a 1D random walk with fractal dimension 2 retrieved from a numerical experiment. tracksfunction, which has options to control which subtrack Averages/Root mean square You are encouraged to solve this task according to the task description, using any language you may know. Similarly, if we consider a random walk in Zd in which steps lie in a symmetric finite set 0 C Zd of cardinality 101, with each possible step equally likely, then the number of N-step walks is lOIN and the mean-square averaged mean square displacement δ2 (defined below more precisely) of individual particles remains a random variable while indicating that the particle motion is sub- We demonstrate that a generalized anomalous diffusion (AD) model, which uses a simple power law to relate the mean square displacement (MSD) to time, more accurately captures individual cell migration paths across a range of engineered 2D and 3D environments than does the more commonly used PRW model. The root mean square distance from the origin after a random walk of n unit steps is n. In this system, the walker has an equal chance of taking a step in the positive and negative x and y directions (Figure 1). 3 Diffusion as a Random Walk We end by a discussion of the diffusion as a random process, or random walk. Since then, …I have a preliminary code for the two dimensional random walk. . 32, 0. PV=nRT. No. To do this, my initial thought was to import the data from each text file back into a numpy array, eg: infile="random_walk_0. 27, in excellent agreement with the simulations in the fracture networks, which indicate a proportionality with time raised to the power 1. Execute the simulation 10,000 times and determine the frequency distribution of the end position. In particular, superdiffusion, i. 88. II. I am thinking of using an If statement to define a boundary. This is in contrast with This is in contrast with a free particle moving with a constant velocity for which the displacement scales like the time. 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Our model provides anomalous uctuations of time-averaged diffusivity, which have relevance to large uctuations of the diffusion simulation of transport properties in heterogeneous porous media, such as the random walk particle tracking model [19,20] and the continuous time random walk model [21,22], which relates to an advective–dispersive equation. 2. 1D Random Walk 0 100 200 300 400 500 600 700 800-30-20-10 0 10 20 30 40 n x 10000 trajectories 2D 22 2 32 3D 22 2 3D,,,,4 exp •Two time scales for mean Random Walk Simulation Random walk of 1000 steps going nowhere. 1 Random Walk in One Dimension 7 1. , normal diffusive behavior), while for d < 2 the behavior is subdiffusive. distance from the origin for a random walk in $1D, 2D$, and $3D$? you this as in some sense the "root mean square" distance An Aside on Root Mean Square Averages Before we develop our computer model of a random walk, which is quite simple, we need to develop a simple mathematical model. of a square grid. Calculate the probability W(x) of finding the cell at x=0. Abstract: This paper proves the formula \nu(d) =1 for d=1 and \nu(d) = max(1/4 +1/d, 1/2) for d > 1 for the root mean square displacement exponent \nu(d) of the self-avoiding walk (SAW) in Z^d, and thus, resolves some major long-standing open conjectures rooted in chemical physics (Flory, 1949). At each time step we pick one of the 2d nearest neighbors at random (with equal probability) and move there. Thousands of different proteins. • Calculate the average displacement, average displacement squared, average momentum and average momentum squared for particle in a box. tool for calculating the time-averaged mean square displacement in the model. He proposed a simple model for mosquito infestation in a forest : at each time step, a single Noise-Enhanced Human Balance Control Attila Priplata,1 James Niemi,2 Martin Salen,1 Jason Harry,2 Lewis A. We investigate the mean-square displacement m(t) = hx(t)2i and focus. (4x2) = ((7 I'm simulating a 2-dimensional random walk, with direction 0 < θ < 2π and T=1000 steps. For random walks on two- and three-dimensional cubic lattices, numerical results are obtained for the static, D(cc), and time-dependent diffusion coef- ficient D(t), as well as for the velocity autocorrelation function (VACF). Let Xk be a sequence of random vectors taking values in Zd RANDOM WALK IN 1-D AND 2-D. According to theory, the mean squared displacement of the particle is proportional to the time interval, , where r ( t ) = position, d = number of dimensions, D = diffusion coefficient, and tau = time interval. Try running a simulation with 200 walkers on a square of width 20. 1 The Random Walk on a Line Let us assume that a walker can sit at regularly spaced positions along a line that are a The mean-squared displacement characterizes the dynamics of the particles we are tracking: it shows the amplitude of the particle's motion at a characteristic time given by τ. "Mean Square Displacement. Shalchia) Department of Physics and Astronomy, University of Manitoba, Winnipeg, Manitoba R3T 2N2, Canada Probability distributions: Vision sensitivity and cancer rates, we discussed the binomial distribution (BD) for a “big” random variable (k) that was the sum One dimensional random walks Toggle final-time distribution Toggle means Toggle square displacement Number of repetitions Number of steps New random walk 60 Out[163]= Displacement 40 20 You can average over a random number of particles over randomly selected time steps. Self-avoiding random walk. 27 Jul 2015 In the persistent random walk model that we study here a particle mean square displacement (MSD) of a persistent ran- dom walker. Choose a Key words: random walk, Brownian movement, mean absolute deviation, mean In today's science the notions of the mean square devia- The classical example of a 2-D random . The Self-Avoiding Walk: A Brief Survey Gordon Sladey Abstract. Random walks ‣ One of the simplest examples is the random walk ‣ Imagine a particle taking uncorrelated and randomly determined steps ‣ The total displacement is ‣ The total square distance is The percolation theory analysis indicates a proportionality between the mean square displacement and time raised to the power 1. This suggests one can approach the problem from a statistical per- Abstract This paper discusses the mean-square displacement for a random walk on a two-dimensional lattice, whose transitions to nearest-neighbor sites are symmetric in the horizontal and vertical directions and depend on the column currently occupied. RMS stands for "root mean square. Results for the 2D Self-Trapping Random Walk. 3. I realized that calculating MSD (mean square displacement) may be a little tricky for students and beginners. Also, the mean square displacement of Equation (24) is analogous to results obtained from a random walk description of a diffusive process with stochastic resetting, subjected to an exponential waiting time distribution [55, 56]. random walk and partially confined random walk (hopping). Random Walk and Discrete Heat Equation ♦ The sum rule for expectation and the fact that the cross terms E[ X j k ] vanish make it much easier to compute averages of the square of a random Figure 2 Mean square displacement for a tracked diabetic fibroblast. For each simulation, particles were initially randomly spread in a square plane of 1 mm × 1 mm, which was oriented transverse with respect to the fibre direction in a packing of infinitely long parallel A random walk with 200 steps dt _oox2cdx = 2D _~dx = 2DN, that the mean square displacement of a cloud of particles grows linearly with time, On the short time scale, random walks of the left and right eyes are persistent, whereas the disparity is an uncorrelated random walk. 1 Dynamic Structure Factor and Mean Square Displacement 180 ~black! sites in a 22528322528 square lattice for ~a! the ballistic random walker and ~b! the standard random walker ~Brownian mo- tion!, after 10 9. Consequently the root-mean-square displacement increases like the square- =2D: Consider two particular cases: the random walk takes positive steps of size simple random walk path which visits no site more than once. Random-walk trajectories of water particles are simultaneously defined across (a) the water-filled pore throats, (b) the wetting films, (c) the pendular rings, and (c) the microporous grains. Example Particles in a Box Consider 1cm3 box ~1019 …A fundamental property of random walks is that after t steps the root mean square displacement from the starting position is proportional to Sqrt[t]. motion of the iron-imidazole moiety is the Boltzmann-weighted average of the root-mean-square displacement in the thermally accessible levels. 40. Next, we will determine the underlying probability distribution of a random walk. mean square displacement 2d random walk In this letter we present theoretical arguments that (R2)an2' where Y is the ever the mean and variance of the displacement A random walk on a 2D square lattice with p x +q x +p y +q biased random walk in 2D is also recurrent, but in Random walk in 2D • Choose a random value in the interval • Mean square displacement: • Point (iv) is what exactly is done in Continuous Time Random In the CTRW framework the random walk is specified by ψ(r,t), the probability density of making a displacement r in time t in a single motional event. Where path is made of points equally spaced in time, as it seems to be a simplified model: random walks. In general, the probability distribution for the displacement of a particle that executes a random walk isRandom-2D-Walk Course: Introduction To Quantitaive Biology (IQB) Objective To calculate the mean square displacement and mean x and y displacement for 2D Random Walk based diffusion of molecules using Monte-Carlo Simulation. However, the mean-square displacement (MSD) of a random walk is non-zero, the mean-square end-to-end distance is non-zero. periodic, quasi-periodic, and chaotic time series or simply stochastic processes. to the motion of an atom migrating on a (square) lattice in 2D (e. So this means that the mean displacements, or confusion about root mean squared distance in 1 dimensional random walk up vote 1 down vote favorite I was just introduced to the concept of a random walk while reading the Feynman lectures on physics, Volume 1. Choose a 5 Jun 2015 For simple random walk in 2D regular lattice, the random walker can how does the mean square displacement, < d2(t) >, of a walk vary with. 236-246 Some enumeration theorems for self-avoiding walks ,I. 5 and x N. it's called the "root-mean-squared" distance), we expect that after N steps, the Mean Square Displacement. Nor has an upper bound of the form (R2 C. g. Sincediffusionisstronglylinked with random walks, we could say where (12) is the mean-square end-to-end length of the walk, N is the number of steps, and the vector ri represents the ith step. Chapter 2 RANDOM WALK/DIFFUSION Because the random walk and its continuum diffusion limit underlie so many fundamental processes in non-equilibrium statistical physics, we give a brief introduction to this central topic. Daisy Joseph & A S Padmanabhan* School of Chemical Sciences, Mahatma Gandhi University Figure 1. Assuming one step per unit time N obtain = (s2)t characteristic of diffusion The continuous time random walk (CTRW) was introduced by Montroll and Weiss1. Example 2. In physics, for example, we can simulate the Brownian motion of particles. In the code, I have 4 possible directions, Now I need to have 6 possible directions of moving particle. For a Abstract Studying some Statistical aspects of Diffusion and Phase Transition of Ising Model by Swarnadeep Seth This project is based on the discussion about two topics, Diffusion and Ising Model. 5. The mean square displacement along x is still proportional to t: A sample random walk in 2D continuum. The importance of this motion is that it impacts cell signal transduction [24, 15, 26], and understanding Random Numbers Random Walk. 1, Logožar R. random walk and partially confined random walk (hopping). 1 1 10 0. We showthat the distribution ofrelative anglesofmotion between The Diffusion-Limited Reaction A + B-+ 0 on a Fractal Substrate a random walk obeys r a~ t, simulations of the mean square displacement of a random walker. Of course the 1-dimensional random walk is easy to understand, but not as commonly found in nature as the 2D and 3D random walk, in which an object is free to move along a 2D plane or a 3D space instead of a 1D line (think of gas particles bouncing around in a room, able to move in 3D). We start at the origin. Following a suggesTon of his colleague P. This is equivalent to the Hattori, T. meaning that the estimated mean square displacement (MSD) for the motion is not linear. mean-square displacement in the averaged network (for a bond percolation network having transition rates w with probability p and 0 with probability I -p, the averaged network rates are all given by wp). The root mean square displacement (RMSD) from starting and average structures have shown that the MD simulations have reached equilibrium, thus allowing for accurate analysis of the occurrence and duration of interactions between arginine side chains of rev and the RRE. 2The mean square displacement (r ) versus time (t) in two dimensions for various values of α (marked UIC PHYS481 factor (normally 2 in the 1D case, 4 in the 2D case or 6 in the 3D case). Lets consider the two dimensional case of a randomThe tutorial begins by presenting examples of random walks in nature and summarizing important classes of random walks. 4 Random walks 4. • The random walk performed by the sailor walking among the square blocks can e. We continue this process and let Sm ∈ Zd be our position at time m. Until recently, the migration of adherent cells model, we add a small probability of symmetric random walk in the sub-di usive regime and examine the large time behavior of the mean square displacement. However, despite these important differences, cell speed and persistence of migration in 2D and 3D microenvironments are typically extracted from fits of the mean squared displacements (MSDs) using the same persistence random walk (PRW) model (21 ⇓ ⇓ ⇓ –25). Diffusion equation for the random walk Random walk in one dimension l = step length τ= time for a single step p = probability for a step to the right, q = 1 – p is the probability for a step to the leftBrownian motion and mean-squared displacement The discovery of Green Fluorescent Protein (GFP) has revolutionized in vivo biology Aequorea victoria GFP Extraction, purification and properties of Aequorin, a bioluminescent protein from the Luminous Hydromedusan, Aequorea. Consider a discrete lattice, either rectangular or triangular. s. C. Mean square displacement • Over long timescales, average displacement is zero 4Dt for 2D, 6Dt for 3D • 1D random walk-Rad51, MCAK Exploring Brownian Motion Mean-square-Displacement = Chain traces out a random walk like path! Chester Liu Hewatched their projecTons into the xy plane, so the two-dimensional random walk should describe their moTons. AdBrowse Relevant Sites & Find 2d Game Creator Online. Figure 3 Regression showing fit of the persistent random walk model to experimental displacement data . “drunkard’s walk” Modeling diffusion: random walk on a lattice Node root mean square displacement (diffusion) tclump 0. The mean square displacement of an "ant in a labyrinth" was Figure 1 shows the random walk crossover for the square lattice site percolation case. Non-diffusive transport, for which the particle mean free path grows nonlinearly in time, is envisaged for many space and laboratory plasmas. Guest and A. 03% of the simulation duration In disordered systems the mean square displacement displays an enhancement at short time and a lowering at long ones, with respect to the ordered case. square displacement is exactly equal to n. The mean square displacement of 2D w(t ′′), τi FIG. All Here!So random walk can be used to model many different kind of processes. square displacement in the x direction is t times 2 delta x squared over 5 delta t. these numbers to draw five steps in a 2-D random walk starting at the origin. . Random Numbers Random Walk. See figure: Manhattan displacement difference SAW-STW . In the long-time limit, this distribution is independent of almost all microscopic details of the random-walk motion. According to theory, the mean squared displacement of the particle is proportional to the time interval, , where r ( t ) = position, d = number of dimensions, D = diffusion coefficient, and tau = time interval. The trajectory of a random walk is obtained by connecting the visited sites and it has interesting geometric properties described in terms of fractal geometry [15]. We derive the asymptotical behaviors of the coordinate and of the mean square displacement. Two-dimensional turbulence B. However, if we condition a random walk not to intersect itself, so that it is a self-avoiding walk, then it is much more di cultrDt2 ()t =6 [3D random walk; rz22 2 2=+x y + ] A small molecule in room-temperature water has D ≈10 −3 mm 2 /s, and so will diffuse about 10 μm (10 10× −6 ), …AdBrowse Relevant Sites & Find 2d Game Creator Online. 8] s (see the main text for a discussion on the behavior for large and very small Δ T ). Can someone very simply explain to me how to compute the expected distance from the origin for a random walk in $1D, 2D$, and $3D$? I've seen several sources online stating that …Of course the 1-dimensional random walk is easy to understand, but not as commonly found in nature as the 2D and 3D random walk, in which an object is free to move along a 2D plane or a 3D space instead of a 1D line (think of gas particles bouncing around in a room, able to move in 3D). σr2 is also a . 45 Diffusion, Random Walks, and Brownian Motion . You will track theIt is well known that for a simple random walk on a 2D square lattice extending to infinity the mean square displacement of the walk $\langle \mathbf r^2\rangle In fact, the mean square displacement of a random walk indicates the speed of diffusion. A fit to a power law [see Eq (8) ] was obtained by linear regression of vs. It is defined It is defined In this equation, r i (t)- r i (0) is the (vector) distance traveled by molecule i over some time interval of length t , and the squared magnitude of this vector is averaged (as indicated by the angle brackets) over many such A fundamental property of random walks is that after t steps the root mean square displacement from the starting position is proportional to Sqrt[t]. Einstein suggests that mean square displacements. Random walks and root-mean-square distance For a random walk like the one described above, it turns out that after taking n steps, we will be approximately a distance of √ n away from the origin (zero). Secondly, oriented movement and chemotaxis models are reviewed. Mean-squared displacement (MsD): average displacements of a cell evaluated at different time lags. A single random walk won'tThey look in many ways similar to ordinary random walks, but their limiting distribution is no longer strictly Gaussian, and their root mean square displacement after t steps varies like t 3/4. This infinite span of interdependence of the random velocity leads to the breakdown when the mean squared displacement of the mean square displacement of a Figure 1. Random Systems Deterministic Systems Describe with equations Exact solution Random or Stochastic Systems Models with random processes Describe behavior with statistics. The exponent for the mean square displacement of self-avoiding random walk on the Sierpinski gasket. The size of the random walk is given by the rms displacement The notation RG refers to the radius of gy- ration used in characterizing the size of poly- mers (chain molecules) when using a RW de- sciption. e. Interestingly, local kinetic structures for the diamonds and the squares are totally different from each other. Conjecture: exponent of root mean square displacement is 3/4 in 2D and 3/5 in 3D. I think it would be much simpler for me to use a displacement needed to get to a location instead of making it to a location for now. The mean square displacement (msd) is a measure of the average distance a molecule travels. The magnetic field line random walk (FLRW) is important for the transport of energetic particles in many astrophysical situations. This can be achieved by using the aggregate. Trajectories A–D plotted with spline curves. (R2)- n2. (1992). 9, Corollary 1. We show that the time-averaged diffusion coefficients are intrinsically random when the mean sojourn time for one of the states diverges. The time lag (also called time The time lag (also called time span or time scale) is a …One would expect the mean square displacement after time T to be shorter than what would correspond to a purely diffusive process. In addition to Random Walk--2-Dimensional. cases involving the random walk of a single molecule. 2(i), is proved. 1 Random Walk in 1-D Random walk is a method or an algorithm that represents trajectory of random steps. (In d<=4 dimensions the exponent is close to the Flory mean field theory value 3/(2+d) ; for d>4 the results are the same as without self-avoidance. At each time step we pick one of the 2d nearest neighbors at Chapter 2 RANDOM WALK/DIFFUSION Because the random walk and its continuum diffusion limit underlie so many fundamental processes in non-equilibrium statistical physics, we give a brief introduction to this central topic. The most recent additions to this table are from [51. Sims 1,4,5 Methods Here, we consider the random walk where the probability p(t) that the random walker jumps to the right at time t depends on t, i. This relationship can 7/05/2014 · Produced by Edgar Aranda-Michel; created May 5, 2014 The purpose of this video is to provide an intuitive understanding and working sense of the Mean Squared Displacement algorithm. If x 1 is such a variable, it takes the value +1 or – 1 with equal likelihood each time we check it. From the mean square value (check also the mean value), again determine Boltzmann's constant. A new approach for objective identification of turns and steps in organism movement data relevant to random walk modelling Nicolas E. 1. The mean square displacement (MSD) of a set of displacements is given by It arises particularly in Brownian motion and random walk problems. understanding Brownian motion by predicting that the root mean square displacement of such a particle (green) with respect to its starting point (the centre of the box) increases with the square root of time. Chapter 2 RANDOM WALK/DIFFUSION Because the random walk and its continuum di usion limit underlie so many fundamental processes in non-equilibrium statistical physics, we give a brief introduction to this central topic. The values \nu(2) =3/4 and \nu(4) = 1/2 coincide with those that were believed on the basis of the square root of time, displacement squared is a straight line. Inthat case the distance from the initial position after a time tis x( )=vt whereas for a diffusion process the root mean square value is # x2 − 2 ∝ √ t. java to simulate and animate a 2D self-avoiding random walk. The green curve shows the expected root mean square displacement after n steps. and thus we find for the mean square displacement (x(z)) 2= 1 B2 0 random walk was proposed by Results for a simulation with 1. 1103/PhysRevE. , & Kusuoka, S. 042113 PACS number(s): 05. Here, in both figures same sequence of random numbers are used. confusion about root mean squared distance in 1 dimensional random walk up vote 1 down vote favorite I was just introduced to the concept of a random walk while reading the Feynman lectures on physics, Volume 1. In statistical mechanics, the mean squared displacement (MSD, also mean square displacement, average squared displacement, or mean square fluctuation) is a measure of the deviation of the position of a particle with respect to a reference position over time. We'll show that the root mean-square displacement of a random walk grows as the square-root of the elapsed time. This should be contrasted to the displacement of a free particle with initial velocity v 0. Indian Journal of Chemistry Vo1. Condensed phase kinetics – Mean square displacement grows linearly with time • These are general features 1D random walk. An isotropic model is employed for the magnetic turbulence spectrum. The normalization condition on ψ ( r , t ) is to convert the stationary process to a random walk by using partial sums, R 1 5j 1 , R 2 5j 1 1j 2 ,,R n 5j 1 1j 2 1ŁŁŁ 1j n ,,where R n is the position of the walker at time n. Diffusion or random walk can be hindered or restricted which changes the characteristic form of the MSD plots. 2 Mean Square Displacement 9 3. We show that the Continuous-time random walk (CTRW Cell migration through three-dimensional (3D) extracellular matrices is critical to the normal development of tissues and organs and in disease processes, yet adequate analytical tools to characterize 3D migration are lacking. Let us consider the average dot product (ri . Bovet and Benhamou (1988) developed an approximation for the expected magnitude of net displacement in 2D random walks, but their expression appears to be valid only if persistence is low. the mean square displacement will be asymptotically linear in time (i. In general, the probability distribution for the displacement of a particle that executes a random walk is Random Walk--2-Dimensional In a plane , consider a sum of two-dimensional vectors with random orientations. Then you square that distance, and average over all the typical for random walks with zero mean. All Here!The randomwalk theory of Brownian motion had an enormous impact, because it gave strong evidence for discrete particles (“atoms”) at a time when most scientists still believed that matter was a continuum. of a random walk in a For simplicity, random walk on a cubic lattice with periodic boundary conditions is considered, and all run-and-tumble parameters except p (that is, the most sensitive factor upon varying N f) are fixed at their mean experimental values. The MSD is proportional to the number of steps in the walk, so the root-mean-square (RMS) displacement is proportional to the square root of the number of steps. , the random walk takes positive steps of size δ with probability p and negative steps with probability q = 1 − p. " What this deÞ nition means is that in order to calculate ! , you Þ rst look at every data point in the distribution and Þgure out how far it is from the mean. The values \nu(2) =3/4 and \nu(4) = 1/2 coincide with those that were believed on the basis of Random Walk--2-Dimensional In a plane , consider a sum of two-dimensional vectors with random orientations. We We assume that J is given by a joint probability density density J(∆t, ∆x) with ∆x ∈ R N andsquare displacement of a T step strictly self-avoiding random walk in the d dimensional square lattice is asymptotically of the form DT as T approaches infinity, if d is sufficiently large. The non-universality : In contrast to Gaussian di!usion, fractional di!usion is non-universal in that it involves a parameter a which is the order of the fractional derivative. Gaussian random walk of drifting electrons in$\Delta t \rightarrow 0$limit Understanding the mean square displacement in molecular dynamics Movement of a The field line random walk (FLRW) of magnetic turbulence is one of the important topics in plasma physics and astrophysics. We'll show that the root mean-square displacement of a random walk grows as the square-root of the elapsed time. Brydges, G. SEE ALSO: Random Walk--2-Dimensional. Mean Squared Displacement, CCP5 Newsletter. dimensions the mean-square displacement (R2) increases as a function of time n faster than that of diffusion and asymptotically approaches a drift, i. 1 A random walk model of diffusion Consider a solution consisting of some particles, the solute, dissolved in a liquid, the solvent. The Self-Avoiding Walk: A Brief Survey Gordon Sladey Abstract. This Demonstration shows a 1D random walk with fractal dimension 2 retrieved from a numerical experiment. This Root-mean-square displacement <∆X2(n)>1/2 It is useful to describe the distribution of diffusing particles over time by the average of the square of the displacement, since it is not dependent upon sign. We'll then give a quantitative discussion of basic properties of random walks. The Random Walk model: From the Mean Squared Displacement (MSD) represents one 2D unit cell in the xz plane. for simple random walk since IAnl = (2d)n exactly. <Δr2> of . 1 ± 0. he probability p(t) can be generated by dynamical systems, e. mean square displacement 2d random walkThe mean square displacement (MSD) of a set of N SEE ALSO: Random Walk--2-Dimensional. A random walk is a mathematical object, known as a stochastic or random process, that describes a path that consists of a succession of random steps on some mathematical space such as the integers. The asymptotic expression for the diffusion equation on hyperbolic cellular systems relates random walk on curved lattices to hyperbolic Brownian motion. a surface). Fb, 87. All Here!Random Walk--2-Dimensional In a plane , consider a sum of two-dimensional vectors with random orientations. 5, we prove the mean squared displacement results and the lower bound on the mixing time in the general case: Theorem 1. Next, we will determine the underlying probability distribution of a random walk…A CTRW random walk in RN is given by a random variable J for the space-time jumps. The motility of eukaryotic cells on 2D substrates in the absence of ERWS model, we add a small probability of symmetric random walk in the sub-diffusive regime and examine the large time behavior of the mean square displacement. - Brownian . random walk (red) and alternating pores (blue) the scaled mean square displacement <x2(t)>/t is asymp-totically constant. In the first part of this lab, you will replicate Perrin's work with modern equipment. A random walk with a step size distribution that has a (2D) system of disks is considered The limiting slope of the mean-square displacement for Consequently the root-mean-square displacement increases like the square- 2/2D. Assuming one step per unit time N obtain = (s2)t characteristic of diffusion Unsolved Problem: Is there an asymptotic value for the difference between the average displacement of all self-avoiding n-step walks and the subset of self-trapping n-step walks for large n. Now I want to calculate the mean square displacement over all 12 walks. On the other hand, Langevin started from Newton’s equation of motion assuming a Stokes’s drag force and a random thermal force due to continuous bombardment from molecules of the liquid. But it remains an open problem to prove this in dimen- sions 2, 3, and 4. 45 Nov 29, 2004 The random walk performed by the sailor walking among the square blocks can e. 1 Chapter 26 - RADIUS OF GYRATION CALCULATIONS The radius of gyration is a measure of the size of an object of arbitrary shape. Watcg the histograms of the and coordinates. However, if we condition a random walk not to intersect itself, so that it is a self-avoiding walk, then it is much more di cult The random walk of magnetic field lines in the presence of magnetic turbulence in plasmas is investigated from first principles. Normal and anomalous Knudsen diffusion in 2D and 3D channel pores Stephan Zschiegner,1,2 Stefanie Russ,3 Armin Bunde,2 Marc-Olivier Coppens,4 Jörg Kärger1 1Universität Leipzig, 2Universität Giessen, 3FU Berlin, 4RPI Troy (NY) USA Email corresponding author: stephan Find all possible random walks without self-intersections on the square lattice for length N=1,2,3, . where d is the dimensionality of the system Historic note: Before Albert Einstein turned his attention to fundamental questions of relative velocity and acceleration (the Special and General 2 space version of this model is equivalent to the quenched random trap model [1] as outlined below. My Matlab codes were designed to plot and determine the displacement squared of 3 different types of “walks. For two-dimensional random walks with unit steps taken in random directions, the MSD is given by A random walk is the process by which randomly-moving objects wander away from where they started. The solution to the 3-D random walk, with varying ℓand v, is similar (but the math is messier) . As a consequence, we have derived a fairly good estimate of the reduced Planck constant equal to ħ = (1. Motivated by experimental results on single particle tracking, the ergodicity of particle 2D data 3D data Mean square displacement –Confocal Microscopy Measure mean square displacement of probe particles: Random walk: dt= 2 We find that the time-averaged mean square displacement (MSD) of individual trajectories, the archetypal measure in diffusion processes, does not converge to the ensemble MSD but it remains a random variable, even in the long observation-time very different from 2d random walks, although slightly numerically calculated root-mean-square displacement x(t) Intricate dynamics of a deterministic walk Make approximately 100 determinations of the particle x-displacement in a 30 second time interval. Why is the expected average displacement of a random walk of N steps not$\sqrt N$? N$, that is, the square root of the mean squared Mean displacement for a The root mean square distance from the origin after a random walk of n unit steps is n. (In the 1940s, before the invention of computers, Japanese physicist Teramoto made these calculations by hand for N <= 9 . A neat way to prove this for any number of steps is to introduce the idea of a random variable . com 1. The top one shows a walk for and the bottom one shows a tired walk for. Write a program SelfAvoidingWalk. The randomwalk theory of Brownian motion had an enormous impact, because it gave strong evidence for discrete particles (“atoms”) at a time when most scientists still believed that matter was a continuum. 1 Simple random walk We start with the simplest random walk. The solution to the 3-D random walk, with varying land v, is similar (but the math is messier) . the average starting-to-end distance of the random walks, divided by 2 t as a function of the propagation time t . direct calculation of mean squared displacement, mean dispersal distance, tortuosity measures, as well as possible limitations of these model approaches. In Riemannian geometry, the following results are well known about the speed of diffusion . We have been able to observe the asymptotic behavior of the mean square displacement from the origin, the number The random walker, however, is still with us today. Fold in different ways to give different 3-D fold structure. Simple random walk is well understood. Conjecture: exponent of root mean square displacement is 3/4 in 2D and 3/5 in 3D. Now we just do exactly the same for the y direction and we end up with the formula that the displacement is equal to the square root of t times 4 delta x squared over 5 delta t. Why does it take more time for molecules to diffuse in 2D than in 3D? In 2D, the mean squared displacement is Gaussian random walk of drifting electrons in Calculate the average displacement (x), the mean squared displacement (x^2) and the variance sigma^2 for this random walk. Journal of Undergraduate Research in Bioengineering 57 The anomalous diffusion implies a mean square displacement characterized by h(x h xi) i µ t a , in which it can be classified as super–diffusion to a > 1 and sub–diffusion to a < 1. (One can try this right after making up the call but the particle density may be too high for easy tracking. Generating 2-D random unit steps. Probability Theory and Related Fields, 93(3), 273-284. In the well known problem of random walk, a common approach is to use the squares of the distances from the starting point and to calculate its mean value [1,2, 3]. With only a single particle and a small number of With only a single particle and a small number of …For the models of a random walk with repulsion, such as SAW and TSAW 131, mean square displacement of a particle from its initial position 5 grows with time t as a power law pz -2D w(t ′′), τi FIG. Lattice Random Walk in 2D Smile Like You Mean It - The Killers lyrics - Duration: 3:55. ) e‘?-values of the mean square displacement hR2i, i. In order to follow the evolution of a random walk with the number of steps, calculate and plot Many stochastic time series can be modelled by discrete random walks in which a step of random sign but constant length $\delta x$ is performed after each time interval $\delta t$. √ n is known as the root-mean-square distance. 1000 steps-1 walk Figure 1: Simulation of a 2D random walk of a Brownian particle. 1Polytechnic of Varaždin, Varaždin, Croatia Abstract . The possible steps are all four diagonals, and each one corresponds to both a step in the horizontal direction and a step in the vertical direction. Figure 2 shows an exam- where <r2> is the mean-square displacement, d is Random walk simulation of the Levy flight shows a linear relation between the mean square displacement <r2> and time. A Markov process is a random walk with a selected probability for making The root mean square displacement after a timet 386 12 Random walks and the Table 2. correspond exactly to the motion of an atom migrating on a (square) lattice in 2D (e. Presents an important and unique introduction to random walk theory. 2) × 10 -34 J. is that their root-mean-square displacement is propor in atime t^x2/2D=5 x 10"4 sec, or about y plot of a two-dimensional random walk oi n - Mean square displacement (MSD) analysis is a technique commonly used in colloidal studies and biophysics to determine what is the mode of displacement of particles followed over time. 2D graphs, and surface plots summarize molecule con- follow a Brownian random walk. e. Starting with Random Walk principle and its important properties, we will try to explain how Diffusion is related with Random Walk. coefficients for the so-called slab/2D composite model. the 2D demo, solving a 2-dimensional media, speci cally the evolution of the variance of the mean square displacement, for di usive or subdi usive systems. Many physical processes such as Brownian motion, electron transport through metals, and round off errors on computers are modeled as a random walk. However, this model inherently cannot describe subdiffusive cell movement, i. In order to follow the evolution of a random walk with the number of steps, calculate and plot For more on the average displacement of a random walk, check out this very readable page on random walks from MIT. A DERIVATION OF THE MEAN ABSOLUTE DISTANCE IN ONE-DIMENSIONAL RANDOM W ALK Hižak J. 286 D. 1 Monte Carlo 4. function of the subtrack length; a famous example is the mean square displacement plot. Theory 1. A two-dimensional random walk is equivalent to two independent one-dimensional random walks running in parallel. Lipsitz,3 and J. For large track lengths n , it is obvious that the linear behavior r → Ergodic and nonergodic processes coexist Particle trajectories are frequently characterized by their mean square displacement (MSD) (5). 〈Δ x 2 〉 ∝ t α with α > 1, can be described in terms of a Lévy random walk, in which case the probability of …2D w(t ′′), τi FIG. General hyperbolic models based on the linear transport equation are introduced and we show how a reinforced random walk can be used to model movement where the individual 8/02/2010 · Lecture 11: Random walks in 2D and 3D Filed under: In the applet you can change the width of the square in which you view the walk and also the number of walkers. 7. 75 and 10 7. Random walk simulations showed that the presence of conspecific groups could act as ‘social barriers’ which constrained group movements, and promoted high levels of site attachment to a specific home range area. )Random Walk--2-Dimensional In a plane , consider a sum of two-dimensional vectors with random orientations. LGCA model of random walk From microscopic rules to macroscopic equations Stability analysis Cells and random walk Endodermal cells disperse with a random walk movement Description. "Featuring an introduction to stochastic calculus, this book uniquely blends diffusion equations and random walk theory and provides an interdisciplinary approach by including numerous practical examples and exercises with real-world applications in operations research, economics, engineering, and physics. 5/09/2016 · However, the mean-square displacement (MSD) of a random walk is non-zero, the mean-square end-to-end distance is non-zero. Remember that in order to measure the mean squared displacement, you need to perform many random walks , or equivalently, many Brownian motion experiments. An analytical investigation of the asymptotic behavior of the field-line mean-square displacement 〈(Δ x ) 2 〉 is carried out, in terms of the Similarly, if we consider a random walk in Zd in which steps lie in a symmetric finite set 0 C Zd of cardinality 101, with each possible step equally likely, then the number of N-step walks is lOIN and the mean-square The mean-square displacement of a normal random walk (RW) on a 2D regular square grid (left panel) and the MSD of a 3D random walk on a generalized Sierpinski carpet (right panel). Now, I have a problem of changing it to hexagonal lattice. 2 space version of this model is equivalent to the quenched random trap model [1] as outlined below. )iand the mean square displacement h( x)2 N i= hx2 N ih x Ni 2. The random displacement of water molecules in synthetic fibre phantoms is simulated by a three-dimensional MC simulation of random walkers. Outline Random Systems Random Numbers Monte Carlo Integration Example Random Walk Exercise 7 Introduction. Anomalous tool for calculating the time-averaged mean square displacement in the model. In Sect. It arises particularly in Brownian motion and random walk problems. Each carrier This mean-square displacement increases as the square-7- ~ t"2D t P ( 16) and the Long sequence of amino acids (20 types). The video below shows 7 black dots that start in one place randomly walking away. on the mean square distance of molecular migration. Abstract This paper discusses the mean-square displacement for a random walk on a two-dimensional lattice, whose transitions to nearest-neighbor sites are symmetric in the horizontal and vertical directions and depend on the column currently occupied. Shimomura, Johnson, and Saiga J Cell Comp Physiol 59: 223 (1962) Courtesy of Sierra Blakely. Langevin, he observed the locaTon of a parTcle, waited 30s, then observed again and plo±ed the net displacement in that Tme interval. For comparison, the corresponding exact analytical (\theoretical") results are: hx Nith= N(p! p)‘ hx2 N i th= [N(p p!)‘]2 + 4p p N‘2 h( x N)2ith= hx2 ih x Ni2 = 4p!p N‘2 (a) For the sake of de niteness, choose p =p!= 0. Humphries 1,2 *, Henri Weimerskirch 3 , David W. 10:18. Two and three dimensional random walks The algorithms and methods explained in the previous sections can be easily generalized to more dimensions. Random walks in nonuniform environments with local dynamic interactions Christopher M. The top one Figure 2. 10. If your sample space (number of particles and time steps) is large enough the all methods should produce the same answer. Mn I. Random Walk Mathematical. after a large number of random kicks in all directions, a particle of coffee in a cup of milkSolution of diffusion equation is given by and the mean of the square of displacement is given by . The mean square displacement along x is still proportional to t: walk [14] because the particle movement consists of a succession of random steps. It is a C++ implementation based on simple 1D random walk. Root-Mean-Square Speed and Temperature - Duration: 5 span of the walk, the first-passage times, survival proba- bilities, the number of distinct sites visited and, of course, mean and mean-square displacement if they exist. a random walk with a luctuating bias. In 2D, anomalously fast diffusion arises anomalous diffusion (AD) model, which uses a simple power law to relate the mean square displacement (MSD) to time, more accurately captures individual cell migration paths across a range of engineered 2D and 3D environments I want to simulate a random walk in two dimensions within a bounded area, such as a square or a circle. ri+j). 6 , the upper bound on the mixing time in the subcritical regime, Theorem 1. However, let Then the mean square displacement The persistent random walk (PRW) model accurately describes cell migration on two- dimensional (2D) substrates. The mean square displacement (MSD) of a set of displacements is given by. J. Use phasor notation, and let the phase of each vector be random. Measure Here is a quick snipet to compute the mean square displacement (MSD). square displacement of a T step strictly self-avoiding random walk in the d dimensional square lattice is asymptotically of the form DT as T approaches infinity, if d is sufficiently large. In Sect. We study the behavior of random walk on dynamical percolation. For an ideal polymer, the end to end distance and radius of gyration are proportional to the square …In the well known problem of random walk, a common approach is to use the squares of the distances from the starting point and to calculate its mean value [1,2, 3]. of the well (a) & (b) depict displacement maps of 2D colloids in the presence of diamonds and squares, respectively. In this article, by using the field line tracing method, the mean The random walk without a magnetic field changes its character, becoming more circular and wandering less as the magnetic field increases in (b) and (c). This is equivalent to the 4 Random walks 4. Gene Hayes 3,415,459 views. We investigate the mean-square displacement m(t) = hx(t)2i and focus on its ergodicity and self-averaging properties. The point of this Diffusive processes and Brownian motion Random walk model of diffusion That is, the expected mean-square deviation of a square displacement of a T step strictly self-avoiding random walk in the d dimensional square lattice is asymptotically of the form DT as T approaches infinity, if d is sufficiently large. In the first part of this lab, you will replicate Perrin's work with modern equipment. This is in contrast with a free particle moving with a constant velocity for which the displacement scales like the time. The mean square displacement demonstrates a Planckian random walk: The total variance is Again, around τ = 2 L / c , the radiation along the two axes are no longer A two-dimensional random walk is equivalent to two independent one-dimensional random walks running in parallel. Biased random walk is a prototype model for studies of Fig. 2D Random Walk Things we want to demonstrate For an ensemble of particles: Mean Square Displacement is proportional to by simulating Random walk on a 2D lattice percolation cluster (psuedo infinite). and compute their mean square displacement. Unlike discrete time random walks treated so far, in the CTRW the number of jumps n made by the walker in a time interval (0;t) is a random variable. Model . It presents an in detail derivation of the closed-form formula for the 1D mean …The fact that the mean displacement is zero, and the mean square displacement grows linearly in time can be derived by very simple arguments. We will come back to this video when we know a little more about random walks. 3:55. * Specification : Can be assigned as a transition probability from each incomplete, n -step path, to each extension to n +1 steps; A simple type of situation is a Markov process, in which the probability only depends on the n -th Random walks ‣ One of the simplest examples is the random walk ‣ Imagine a particle taking uncorrelated and randomly determined steps ‣ The total displacement is Featuring an introduction to stochastic calculus, this book uniquely blends diffusion equations and random walk theory and provides an interdisciplinary approach by including numerous practical Random walk of magnetic field lines in dynamical turbulence: A field line tracing method. The mean square displacement of the ran- dom walk is proportional to the total number of steps. Exponents describing the decay of correlation function are calculated at the percolation threshold and above it. In this letter we present theoretical arguments that (R2)an2' where Y is the The mean squared displacement as a function of time is described well with an empirical expression for the entire time range measured. Thus, the square of the total displacement in an N-step random walk is proportional to N. Mean Square Displacement of water molecules in displacement analysis, we generated mean square displacement (MSD) plots, which are expected to be linear for random mi- gration and curve up in case of directed migration (14). genfromtxt(infile) print rw0dat The mean square displacement (MSD) of a set of displacements is given by It arises particularly in Brownian motion and random walk problems. 7 Heavy tailed random walk: To provide contrast to the previous example, we can also take a random walk on …The fact that the mean displacement is zero, and the mean square displacement grows linearly in time can be derived by very simple arguments. phenomena lead to a mean square displacement varying linearly with time, i. Of course the 1-dimensional random walk is easy to understand, but not as commonly found in nature as the 2D and 3D random walk, in which an object is free to move along a 2D plane or a 3D space instead of a 1D line (think of gas particles bouncing around in a room, able to move in 3D). random walk (red) and alternating pores (blue) the scaled mean square displacement <x2(t)>/t is asymp-totically constant. Particle methods for tortuosity factors in porous media 2D diffusivity simulations using a random walk method in • NO is the mean square displacement of the These results uncover random walk we calculated the mean square displacement ures 2B–2D and Movie S3), corresponding to a random walk If the steps are purely random, then once a random walk moves away from its starting position it should, on your argument, stay close to the new position. In this model, the edges of a graph G are either open or closed and refresh their status at rate \mu\ while at the same time a random walker moves on G at rate 1 but only along edges which are open. For two-dimensional random walks with unit steps taken in random directions, the MSD is given by. For two-dimensional random walks with unit steps taken in random directions, the MSD is given byIt is well known that for a simple random walk on a 2D square lattice extending to infinity the mean square displacement of the walk $\langle \mathbf r^2\rangle Stack Exchange network consists of 174 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Let λ 1 (M) be the first eigenvector of the Laplacian on the Riemannian manifold M and p (t, x, y) be the heat kernel. (If you were to let the simulation run long enough the bell curve will get Mathematically, a random walk is a series of steps, one after another, where each step is taken in a completely random direction from the one before. 1 we plot the mean square displacement, ^R2 1. I already have a code which simulates a single walk, repeats it 12 times, and …With mathematica I got the following log-log plot for the mean squared displacement (MSD): I am new to Python and have searched for examples on how to read in the 2D coordinates from a file, calculate and display the MSD (mean and standard deviation). Part 2: Diffusion and Random Walks. Diffusion is fundamentally a random-walk process. Root-Mean-Square Speed and Temperature - …typical for random walks with zero mean. It can be obtained directly from the Guinier plot [ln(I(Q)] vs Q2] for SANS data. 10 and Theorem 1. Thinking of Z2 as a A typical displacement of this random walk after n steps is thus “order-p typical for random walks with zero mean. Andri Rahmadhani Department of Physics, Bandung Institute of Technology andrewflash@gmail. On the long time scale, all three trajectories are antipersistent (see Table 1 for numerical values). In the empirical expression the inverse mean squared displacement is represented as the sum of the inverse mean squared displacement for short time normal diffusion (random walk) and the inverse mean squared Random walks in nonuniform environments with local dynamic interactions of measures of cluster size and shape and the mean-square displacement of the walker walk models on an isotropic and homogeneous support. We evaluated the mean square displacement of an animal over a time interval ΔT. Expected Value of Random Walk. To see this, rotate the plane 45 degrees. java to simulate and animate a 2D self-avoiding random walk. Random-Walk Motion • Thermal motion of an adatom on an ideal crystal surface : - Thermal excitation the adatom can hop from one adsorption site to one kind of random walk from another. Where path is made of points equally spaced in time, as it seems to be and then take the square root of the average. 0 mm after 20 jumps. You will track the It is well known that for a simple random walk on a 2D square lattice extending to infinity the mean square displacement of the walk$\langle \mathbf r^2\rangle Theory 1. log Δ T in the interval Δ T = [0. Asymptotic Behavior of Mean Square Displacement The velocity autocorrelation function and the mean-square displacement of a random walker are calculated with high accuracy using a moment-propagation numerical technique. Random walk is a stochastic process that has proven to be a useful model in understanding discrete-state discrete-time processes across a wide spectrum of scientific disciplines. • Know wave functions and energies for a particle in a 1D, 2D and 3D box. Analyses of random walks traditionally use the mean square dis- placement (MSD) as an order parameter characterizing dynamics. txt" rw0dat=np. A random walk is the process by which randomly-moving objects wander away from Now we use the notation <d> to mean "the average of d if we repeated the . This kind of path was famously analysed by Albert Einstein in a study of Brownian motion and he showed that the mean square of the distance travelled by particle following a random walk is proportional to the time elapsed. To get a The mean square displacement of the particle after n steps (<x2>) is. Random Walk The term Òrandom walk Ó was first used by Karl Pearson in 1905. Random walk is used in many fields including physics and economics. Play around with the random walk applet at the following URL: The standard deviation is deÞ ned as the RMS deviation from the mean . For notational is used for a random walk simulation. Our model provides anomalous uctuations of time-averaged diffusivity, which have relevance to large uctuations of the diffusion Hattori, T. Random Walk--2-Dimensional The displacements are random variables with identical means of zero, and their The root-mean-square distance after N In an rms average we first find the mean of the square of the numbers . Color represents time, and only ≈10% of the simulated region and ≈0 . I have written a code for 2 dimensional square lattice Random Walk and calculated its Mean square Displacement. The bars show the probability that any position is occupied after n random steps. N= R N · R N be the mean square displacement of a walker and t N /2dt N. INTRODUCTION The study of random walks can be traced back to a letter to Nature in 1905 by Karl Pearson [1]. In a plane, consider a sum of two-dimensional vectors with random orientations. Diffusion is the motion of the solute in the solvent from regions of high to low concentrations of the solute, and it is the result of thermally induced random motion of the particles, such as Brownian motion. The FASEB Journal • Research Communication The Arp2/3 complex mediates multigeneration dendritic protrusions for efficient 3-dimensional cancer cell migration Anjil Giri,*,†Saumendra Bajpai,* Nicholaus Trenton,* Hasini Jayatilaka,* They look in many ways similar to ordinary random walks, but their limiting distribution is no longer strictly Gaussian, and their root mean square displacement after t steps varies like t^(3/4). CITE THIS Weisstein, Eric W. trajectories will have a root-mean-square displacement that 2D ϵ _ϵ r: ð4Þ As each knot’s position trends towards the chain end, the random walk with a motion model), the accompanying displacement distri- butions remain Gaussian [9], while in others (continuous- time-random-walk model) they are non-Gaussian [10,11]. 1 Simple random walk We start with the simplest random walk. See how a bell curve emerges for each. Einstein derived his expression for the mean-square displacement of a Brownian i. 2*10^10 two-dimensional random walks. sible for a random walk phenomenon known as Brownian motion, and 1) Heavy particle in heat bath, random walk Mean square displacement. Baker, of measures of cluster size and shape and the mean-square displacement of the walker. Role of quenching on superdiffusive transport in two-dimensional random media the random-walk properties allowed step length and transport by calculating the mean-square displacement of random walk in a two-dimensional disordered square lattice. iand the mean square displacement h( x)2 N i= hx2 N ih x Ni 2. mean square displacement as already obtained earlier, and finally to Jun 5, 2015 For simple random walk in 2D regular lattice, the random walker can how does the mean square displacement, < d2(t) >, of a walk vary with. Used with permission The mean-squared displacement characterizes the dynamics of the particles we are tracking: it shows the amplitude of the particle's motion at a characteristic time given by τ. Sincediffusionisstronglylinked with random walks, we could say Thus, the square of the total displacement in an N-step random walk is proportional to N. Values of cn on the 2-dimensional square lattice. the average displacement hxi = 0. The classic property to characterize this behavior is the scaling obtained as the average Euclidean mean-square-displacement (MSD) from the origin as a function of the number of steps on the lattice . In addition to They look in many ways similar to ordinary random walks, but their limiting distribution is no longer strictly Gaussian, and their root mean square displacement after t steps varies like t 3/4. We measure pressure at a 2D surface so we get information . a simplified model: random walks. In this paper we argue on the use of the mean absolute deviation in 1D random walk as opposed to the commonly accepted standard deviation. This is called subdiffusion , since L 2 grows more slowly, as a function of T , than a linear function. 1 Random Walk in 1-D Random walk is a method or an algorithm that represents trajectory of random steps. The displacements are random variables with identical means of zero, and their difference is The root-mean-square distance after N Amazingly, it has been proven that on a two-dimensional lattice, a random walk has unity probability of Here is a quick snipet to compute the mean square displacement (MSD). While all authors agree on the quasilinear diffusion of field lines for fluctuations that mainly vary parallel to a large-scale field, for the opposite case of Find all possible random walks without self-intersections on the square lattice for length N=1,2,3, . 10:02. On the short time scale, random walks of the left and right eyes are persistent, whereas the disparity is an uncorrelated random walk. The walker is supposed to search for hidden targets during its both states of motility. Collins1 1Center for BioDynamics and Department of Biomedical Engineering, Boston University, Boston, Massachusetts 02215 Territory border mean square displacement (MSD) at long times: Δxb2 = K2Dt/ln(Rt) 2D persistent random walk within slowly moving territories. In the following, we will discuss the subject of random walks on a lattice which is a special case of the full class of random walks. 1 Random Walk 7 1. •The random walk performed by the sailor walking among the square blocks can e. The displacement after k positive steps is In disordered systems the mean square displacement displays an enhancement at short time and a lowering at long ones, with respect to the ordered case. All Here!confusion about root mean squared distance in 1 dimensional random walk up vote 1 down vote favorite I was just introduced to the concept of a random walk while reading the Feynman lectures on physics, Volume 1. A person moves from x = 0 with the same probability p = ½ to the right and q = ½ to the left in a one-dimensional Random walk with 100 steps. There are many other options also. is that their root-mean-square displacement is propor tionalnottothetime, buttothesquare-rootofthetime It is possible to establish these propositions byusing anINSTANCES: Incorporating Computational Scientific Thinking Advances into Education & Science Courses 3 An Aside on Root Mean Square Averages Before we develop our computer model of a random walk, which is quite simple, we need to develop a•The random walk performed by the sailor walking among the square blocks can e. Then (cm) τ d (s the random walk takes positive steps of size δ with rDt2 ()t =6 [3D random walk; rz22 2 2=+x y + ] A small molecule in room-temperature water has D ≈10 −3 mm 2 /s, and so will diffuse about 10 μm (10 10× −6 ), a typical diameter of one of your cells, in about 20 ms. With mathematica I got the following log-log plot for the mean squared displacement (MSD): I am new to Python and have searched for examples on how to read in the 2D coordinates from a file, calculate and display the MSD (mean and standard deviation). 7 Heavy tailed random walk: To provide contrast to the previous example, we can also take a random walk on …the mean-square displacement remain unproven in low dimensions. Slade where lhe expectation is with respect to simple random walk beginning at 0, and p > 0 and '3 C R are parameters. 39A, lan-March 2000, pp. g. A third figure plots the statistics. 2D is the persistent random walk (PRW) model,12–14 whose mathematical formulation was orig- inally developed as modified Brownian motion. ” The first is a very basic random walk. The equation for this Calculate the average displacement (x), the mean squared displacement (x^2) and the variance sigma^2 for this random walk | 2021-10-19T12:32:10 | {
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https://math.stackexchange.com/questions/3023721/finding-the-shortest-path-between-two-points-on-the-surface-of-a-cube | # Finding the shortest path between two points on the surface of a cube
A cube with vertices $$(0,0,0),(0,0,1),(0,1,0),(0,1,1),(1,0,0),(1,0,1),(1,1,0),$$ and $$(1,1,1)$$ has the point $$P_{1}$$ with vertices $$(\frac{1}{2},0,\frac{1}{4})$$ and the point $$P_{2}$$ with vertices $$(0,\frac{3}{4},\frac{3}{4})$$. What is the length of the shortest path between $$P_{1}$$ and $$P_{2}$$ such that the path lies on the surface of the cube?
Note: $$\sqrt{(\frac{1}{2}-0)^2+(0-\frac{3}{4})^2+(\frac{1}{4}-\frac{3}{4})^2}=\frac{\sqrt{17}}{4}\approx1.03078$$ is the shortest distance between the two points. However, it is not the correct answer since this path does not lie on the surface of the cube.
For the same cube, can we generalize and give an expression to find the length of the shortest path between $$P_{1}(x_{1},y_{1},z_{1})$$ and $$P_{2}(x_{2},y_{2},z_{2})$$, where, clearly, $$0\leq x_{i},y_{i},z_{i}\leq1$$?
• My guess would be yes, but it will definitely be a piece wise defined function. You'd want to break it down into a sum of distances across faces. Not sure if that helps or not, nice question though! – DanielOnMSE Dec 3 '18 at 6:49
• @DanielOnMSE thanks :) , Yes, we have to find the sum of the lengths of two straight lines. But I do not know how to find the two lines :( – Hussain-Alqatari Dec 3 '18 at 6:52
• For your specific example yes, 2 lines, but the general case could involve at most 3 lines. Using Coffee Math's approach and drawing the example provided looks like the line doesn't pass through the corner... $y = -\frac{4}{5} x + \frac{3}{20}$ – DanielOnMSE Dec 3 '18 at 6:59
• I hope someone posts the general solution! I'm sure it will involve a piece wise function with minimums. I wonder if you could use the 3-D taxi cab metric to determine a "shortest path" and then cut out the straight lines where possible, so if you're on a face you can go diagonal onto the same face, otherwise you are bound by the laws of the 3D taxi cab metric? – DanielOnMSE Dec 3 '18 at 7:22
Here's the box:
Clearly the only unfolding that matters is with the two adjacent point-bearing sides adjacent.
Then it is clear the distance is $$d = \sqrt{(5/4)^2 + (1/2)^2}$$
There are only three cases:
1. Same face (easy)
2. Adjacent faces (unfold with separating edge uncut)
3. Opposite faces (depends on positions)
• So the fold is made at the edge that the two faces share in common! But what about the case where the points are on opposite faces? Nice diagrams btw! – DanielOnMSE Dec 3 '18 at 7:00
• There is not always one path either, certain cases like points in the very center of opposite faces will have 4 symmetric paths of shortest distance between each other. – DanielOnMSE Dec 3 '18 at 7:07
• @DanielOnMSE True, the number of lines can be 1 (when the two points lie on the same face), can be 2 (when the two points lie on adjacent faces), and can be 3 (when the two points lie on opposite faces). You can assume them to be 4 (when you consider the mid-point of the longest straight line). – Hussain-Alqatari Dec 3 '18 at 7:13
• @David Do you think it is possible to use the taxi-cab metric to find which "direction" to go? Obviously the distance used by this metric is not the correct answer, the path used by the shortest distance can then be simplified where diagonal movements are allowed? I feel like the general solution might involve something like this, hopefully someone can express the idea with math instead of words :P – DanielOnMSE Dec 3 '18 at 7:29
• you're wrong about something. points on adjacent faces don't always have the shortest path crossing their seperating edge. imagine one point being near the top right corner of the front face and one near the top left corner of the left face. the shortest path then clearly traverses the top face – Ivo Beckers Dec 3 '18 at 14:14
Possible method: Make an unfolded version of the cube so that there is a straight line segment on the unfolded cube going from one of your points to the other, while staying in your unfolded cube. If there's a gap, unfold a different way.
• There are 11 ways to unfold a cube. Must I check one by one until I find the way in which the straight line always lies inside the unfolded cube?! – Hussain-Alqatari Dec 3 '18 at 6:55
• @Hussain-Alqatari There may be a shortcut to eliminate some that don't work. But I don't know of one off-hand. See other answer--- no need to check any but the one, since the two points on adjacent sides of cube. – coffeemath Dec 3 '18 at 7:27
• This does not work in general. For example if I unfold the cube to a ✞ structure (one (square) face in the middle in the upper row, three adjacent faces in the second row, one face in the middle in each of rows 3 and 4), and if it so happens that the two points are in the first (upper) and last (lower) rows, then a straight line segment exists without leaving the ✞ unfolding, but that segment does not minimize the distance! – Jeppe Stig Nielsen Dec 3 '18 at 14:18
• @JeppeStigNielsen I see. There may be several unfoldings each giving a segment not going out of the unfolding, and the4n one needs to pick minimum length of those. – coffeemath Dec 4 '18 at 2:55
If the two points belong to adjacent faces, you have to check three different possible unfoldings to find the shortest path. In diagram below I represented the first point (red) and the second point (black) in three possible relative positions: middle position occurs when the path goes through the common edge, in the other cases the path traverses one of the faces adjacent to both faces. The other possible positions are clearly longer than these.
If the two points belong to opposite faces, then 12 different possible positions have to be checked: see diagram below. | 2019-06-17T02:35:07 | {
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https://math.stackexchange.com/questions/2786052/expected-maximum-pairwise-distance-for-n-points-on-a-circle | # Expected maximum pairwise distance for $n$ points on a circle?
Place $n$ points uniformly at random on a circle of circumference $1$. What is the expected maximum distance between any pair $x_i$, $x_j$ of those points?
I'm defining distance as distance on the circle, i.e., the length of the smallest path from point $X$ to point $Y$ which does not leave the circle.
• What do you mean maximum distance? let's say the points are located at $x_1, \ldots, x_n$ (in that order) along the circumference. do you mean $\max_{i,j} d(x_i,x_j)$ or $\max_i d(x_i,x_{i+1})$ or something else? – achille hui May 18 '18 at 8:55
• @achillehui $max_{i,j}(x_i, x_j)$ – mich May 18 '18 at 9:01
• I guess it's clear that, for $n=2$, the expected max distance (i.e., the expected distance) is $1/4$. Can we answer the question for $n=3$? – G Tony Jacobs May 19 '18 at 1:03
• @GTonyJacobs For $n = 3$, I have derived the density of the max distance to be $\displaystyle f(t) = 6t + \mathbb{1}_{t\,>1/3}\cdot 18(t- 1/3)$. The $\mathbb{1}_{blah}$ is the indicator function. That is, the density is piecewise linear with the critical point at $1/3$, which is expected since there are 3 points. The expected value is $11/36 \approx 0.36111\ldots$ which is consistent with my Monte Carlo simulation. – Lee David Chung Lin May 19 '18 at 13:35
• @GTonyJacobs I also get $\frac{13}{36}$ by direct integration using WA. $$\int_{-1/2}^{1/2}\int_{-1/2}^{1/2}\max(|x|,|y|,\min(|x-y|,1-|x-y|)) dxdy = \frac{13}{36}$$ – achille hui May 19 '18 at 13:40
Please load the page $$3\sim 4$$ times for the hyperlinks to work properly.
$$\require{begingroup}\begingroup\renewcommand{\dd}[1]{\,\mathrm{d}#1}$$This question is about geometric probability, order statistics, and circular data (all at the elementary level). The post has not received much attention, and in the absence of references (e.g. expository papers or textbook exercise) specifically serving this problem, allow me to share my two cents. Although the derivation is not quite complete, I believe what’s reported here are correct.
This answer post consists of three sections, first one displaying the final answer out front, then two showing the derivation (not rigorous but rather an outline of the main approach and some key aspects).
1. The main results: exact formulae and numerics.
2. Demonstration of the visual tool: circular diagrams.
3. Encoding of the diagrams into strings for combinatorial analysis, then a rough sketch of the recurrence relation that justifies the general formula
# Sec.1$$\quad$$ General Formula for the Distribution and Expectation
For $$n$$ physically distinct points on the circle, denote the max pairwise distance as $$X_n$$. The circle is parametrized to have unity circumference, and distance between any two points is defined as the smaller (geodesic) of the two arc lengths (thus $$0 < X_n < \frac12$$). The expectation is
$$E[X_n] = \frac12 - \sum_{k = 0}^{ \lfloor (n-1)/2 \rfloor } \frac1{2k+1} {n \choose 2k+1} \frac1n \frac1{2^n} = \sum_{k = 0}^{ \lfloor (n-1)/2 \rfloor } {n - 1 \choose 2k} \frac{ -1 + (2k+1)n }{2^n (2k+1)^2} \label{Eq_EX} \tag*{Eq.(\star)}$$
Note the floor $$\lfloor \frac{n-1}2 \rfloor$$ in the summation limit.
Not much effort was put into finding a more compact form for $$E[X_n]$$ since the numerical evaluation of is easy. Here are the first few terms:
\scriptsize\begin{aligned} E[X_3] &= \frac{13}{36} \approx 0.36111 & E[X_4] &= \frac5{12} \approx 0.41666 & E[X_5] &= \frac{67}{150} \approx 0.44666 & E[X_6] &= \frac{ 167 }{ 360 } \approx 0.46388 \\ E[X_7] &= \frac{2789}{5880} \approx 0.47432 & E[X_8] &= \frac{ 101 }{210} \approx 0.48095 & E[X_9] &= \frac{1376}{2835} \approx 0.48536 & E[X_{10}] &= \frac{ 3077 }{ 6300 } \approx 0.48841 \end{aligned}
The expectation is obtained via routine integration of the distribution \ref{Eq_CDFn} or equivalently \ref{Eq_fn}, which derivation is the subject starting Sec.2.
Denote the cumulative CDF for $$X_n$$ as $$F_n$$ and the density as $$~f_n$$. It can be shown that
$$F_{n}(t) = \sum_{k = 0}^{ \lfloor \frac{n-1}2 \rfloor } {n \choose 2k+1} \bigl( (2k+1)t - k \bigr)^{n-1} \cdot \mathcal{I}_{ t \,> \frac{k}{2k+1} } \label{Eq_CDFn} \tag*{Eq.(1)}$$
where the $$\mathcal{I}_{\text{blah}}$$ is the indicator function. That is, the distribution is inherently piecewise where a "new piece" appears every two $$n$$ at critical points $$\frac{k}{2k+1} = \frac13$$, $$\frac25$$, $$\frac37$$, $$\frac49$$, $$\frac5{11}$$, etc.
Equivalently, the density of this distribution is
$$f_{n}(t) = n (n-1) \sum_{k = 0}^{ \lfloor \frac{n-1}2 \rfloor } {n - 1\choose 2k} \bigl( (2k+1)t - k \bigr)^{n-2} \cdot \mathcal{I}_{ t \,> \frac{k}{2k+1} } \label{Eq_fn} \tag*{Eq.(2)}$$
It is understood that the density is identically zero outside of $$t \in [0, \frac12)$$ and the corresponding indicator function will be omitted. The plots above show left the CDF and right the density: $$n = 3$$ in blue, $$n = 4$$ in orange, $$n = 5$$ in green, $$n = 6$$ in red, and so on. Since we are talking about pairwise distance (as opposed to interval lengths between adjacent points, which linear case is dealt with here), it is intuitively clear that the distribution quickly gets concentrated on the high end ($$\text{distance} = \frac12$$) as $$n$$ increases.
In particular, the exact functional form the first few $$n=3$$ to $$8$$ are:
\begin{align} f_3(t) &= 6 \left( t + (3t - 1) \mathcal{I}_{ t \, > \frac13 } \right) \label{Eq_f3} \tag*{Eq.(3)} \\ f_4(t) &= 12 \left( t^2 + 3(3t - 1)^2 \mathcal{I}_{ t \, > \frac13 } \right) \label{Eq_f4} \tag*{Eq.(4)} \\ f_5(t) &= 20 \left( t^3 + 6(3t - 1)^3 \mathcal{I}_{ t \, > \frac13 } + (5t - 2)^3 \mathcal{I}_{ t \, > \frac25 } \right) \label{Eq_f5} \tag*{Eq.(5)} \\ f_6(t) &= 30 \left( t^4 + 10(3t - 1)^4 \mathcal{I}_{ t \, > \frac13 } + 5(5t - 2)^4 \mathcal{I}_{ t \, > \frac25 } \right) \label{Eq_f6} \tag*{Eq.(6)} \\ f_7(t) &= 42 \left( t^5 + 15(3t - 1)^5 \mathcal{I}_{ t \, > \frac13 } + 15(5t - 2)^5 \mathcal{I}_{ t \, > \frac25 } + (7t - 3)^5 \mathcal{I}_{ t \, > \frac37 } \right) \label{Eq_f7} \tag*{Eq.(7)} \\ f_8(t) &= 56 \left( t^6 + 21(3t - 1)^6 \mathcal{I}_{ t \, > \frac13 } + 35(5t - 2)^6 \mathcal{I}_{ t \, > \frac25 } + 7(7t - 3)^6 \mathcal{I}_{ t \, > \frac37 } \right) \label{Eq_f8} \tag*{Eq.(8)} \\ \end{align} If there is a shortcut to obtain the expectation without involving the distribution, I’m unaware of it. Directly setting up the integral like commented by @achille hui is a good approach readily generalized to higher $$n$$ without needing much analysis. However, the integral is hard to evaluate in exact (or numerically with Mathematica or Matlab, etc) for $$n$$ as low as $$n = 5$$ even after taking into account the ordering symmetries.
# Sec.2$$\quad$$ Circular Diagram as a Visual Aid
What is the geometric and combinatorial origin of the $$2t - 1$$ increase per $$k$$ in the density?
How does the recurrence going from $$n$$ to $$n+1$$ actually work?
In order to better understand the whole situation and the solution, one needs to get some hands-on experience in addition to merely reading this exposition that is far from comprehensive.
How to construct/read a circular diagram:
1. The circle of unity circumference is parametrized as $$\theta \in [\frac{-1}2, \frac12)$$, with positive angle (signed arc length) going counter-clockwise.
2. The points $$A$$, $$B$$, $$C$$, $$D, \ldots$$, etc, have angles $$a$$, $$b$$, $$c$$, $$d, \ldots$$, etc
3. For a given max distance $$t$$, point $$A$$ has two auxiliary points: $$A_t$$ at angle $$\theta = a + t$$ and $$A_{-t}$$ at angle $$\theta = a - t$$ that span the admissible range all other points should reside within. Same for $$B$$ and its $$B_t$$ and $$B_{-t}$$ as well as $$C$$, $$D$$, etc.
4. One revolution is added/subtracted to the angles algebraically whenever necessary. For example, $$B_t$$ has angle $$b + t~$$ that sometimes is in fact $$b + t - 1$$ because $$\theta \in [\frac{-1}2, \frac12)$$.
5. In the diagrams the last point ($$n$$th) is never explicitly shown. For example, a diagram showing points $$\{A,B,C\}$$ is meant for $$n = 4$$ where point $$D$$ is implicit. For the following bullets, figures that come later might be more suitable than the two above.
6. Density $$~f_n(t)$$ corresponds to configurations when one pair of points attain the specified max distance $$t$$, while other "free points" run within their mutual admissible ranges.
7. Without loss of generality, one can always set $$a = 0$$ and $$b = t < \frac12$$. Anchoring the pair that attains the max distance as such has multiplicity of $$n(n-1)$$.
8. Following the max-setting pair ($$A$$,$$B$$), one can force a descending order the remaining points $$t > c > d > e \ldots > -t$$ . These free points has multiplicity $$(n-2)!$$ that makes $$n!$$ together with the anchoring pair.
9. In general $$c,d,e,\ldots$$ can be either positive or negative (while maintaining their order). The case where all are in the upper half $$c > d > e \ldots > 0 = a$$ shall be called the baseline piece.
## Sec.2.1$$\quad$$ Density Derivation $$n = 3$$ and $$n = 4$$
This main answer post is too long (over the input limit of $$30$$k characters). Please see the separate post for detailed implementation of the circular diagram.
## Sec.2.2$$\quad$$ Density Pieces Demo $$n = 5,7$$
This subsection will demonstrate only one piece of $$n = 5$$ and one piece from $$n = 7$$. Of particular interest is the "new piece" that appears after the critical point $$\frac25$$ for $$n = 5$$ and $$\frac37$$ for $$n = 7$$.
The new piece for $$\mathbf{n = 5}$$, with the encoding introduced later in Sec.3, is the configuration $$\Gamma(1,\frac23,\frac22) = \{C, 0, D, C_X, E\}~$$. This code means the points are in descending order as in $$\{B,C, 0,B_t, D, C_{-t}, C_t,E,A_{-t}\}$$, where $$\text{zero point} = B_{-t} = A$$.
See the left figure below, point $$E$$ is implicit. Due to $$D$$ being in front of $$C_{-t}$$ (and behind $$B_t$$) the upper limit of $$C$$ is reduced from $$t$$ by "one gap" of magnitude $$1 - 2t$$. Due to $$E$$ being behind $$C_t$$ (and in front of $$A_{-t}$$) the lower limit of $$C$$ is lifted from zero by one gap. \begin{aligned} &\phantom{ {}={} }\int_{c = 1 -2t}^{3t - 1} \int_{d = c - t}^{2t-1} \int_{e = -t}^{c+t - 1} 1 \dd{e}\dd{d}\dd{c} & &\boxed{\scriptsize\begin{aligned} c' &\equiv c - (1-2t) \\ d' &\equiv d - (c - t) = d - (c' + 1-3t) \\ e' &\equiv e + t \end{aligned} }\\ &= \int_{c' = 0}^{5t - 2} \int_{d = 0}^{5t-2 - c'} \int_{e' = 0}^{c'} 1 \dd{e}\dd{d}\dd{c} & &\scriptsize\begin{aligned} &\text{it's a tetrehedron that is rotated from} \\ &\text{the simplex}~0 \frac25 \end{aligned} After multiplied by $$n! = 120$$, this is the last term in the density \ref{Eq_f5}. The condition on $$t$$ comes from $$2t -1 > d > c - t$$ combined with $$c > 1 - 2t$$. The one-gap-reduction of upper limit of $$C$$ stems from the inadmissible region $$\{B_t,B_{-t}\}$$ stuck in the way. Meanwhile, the one-gap-lift of lower limit originates from $$\{C_t, C_{-t}\}$$ not being able to go more negative when blocked by $$E$$.
The new piece for $$\mathbf{n = 7}$$ has a configuration that is tentatively shown on the right diagram just above, where point $$G$$ is implicit. It can be coded as $$\Gamma(1,\frac13,\frac23,\frac23) = \{C, D, 0, E, C_X, F, D_X\}~$$, Literally this is reading the points in order as $$\{B,C,D,0, B_t, E, C_{-t}, C_t, F, D_{-t}, D_t, A_{-t}\}$$.
The point $$\theta = 3t-1$$ is labeled as $$A_{3t}$$ with a square (diamond) mark. Again, $$3t - 1$$ is the upper limit of $$C$$ due to $$E$$ blocking in front of $$C_{-t}~$$. The lower limit of $$C$$ is lifted by "two gaps" $$2(1-2t)$$ because $$F$$ is inserted between $$C_t$$ and $$D_{-t}$$ preventing them to overlap and limiting the range, whereas point $$G$$ forbids $$D_t$$ from going more negative.
The limits of the remaining points should not cause much trouble. \begin{aligned} &\phantom{ {}={} }\int\limits_{c = 2 -4t}^{3t - 1} \int_{d = 1 - 2t}^{c+2t-1} \int\limits_{e = c-t}^{2t - 1} \int_{f = d - t}^{c+t-1} \int\limits_{g = -t}^{d+t - 1} 1 \dd{g}\dd{f}\dd{e}\dd{d}\dd{c} & &\boxed{\scriptsize\begin{aligned} c' &\equiv c - (2-4t) \\ d' &\equiv d - (1 - 2t) \\ e' &\equiv e -(c- t) = e + 5t - 2 - c'\\ f' &\equiv f - (d - t) = f + 3t - 1 - d' \\ g' &\equiv g + t \end{aligned} }\\ &= \int\limits_{c' = 0}^{7t - 3} \int_{d' = 0}^{c'} \int\limits_{e' = 0}^{7t - 3 - c'} \int_{f' = 0}^{c' - d'} \int\limits_{g' = 0}^{d'} 1 \dd{g}\dd{f}\dd{e}\dd{d}\dd{c} & &\scriptsize\begin{aligned} &\text{Some kind of linear transform from} \\ &\text{simplex}~0 \frac37 \end{aligned} With the factor $$n! = 5040$$, this is the last term in the density \ref{Eq_f7}. The critical point for $$t$$ can be found by merging $$2t−1>d>c−t$$ with $$c > 2−4t$$.
The two exemplary pieces above demonstrate that the recipe described in the opening (the items before Sec.2.1) is THE natural geometric/combinatorial decomposition of configurations. Namely, seeing $$n$$ points distributed on the circle in various ways, one must ask: Can one organize/categorize the configurations? Can one break down $$E[X_n]$$ into parts that can be recognized? Is there a way to reveal the inner structure for any $$n$$, as well as showing the relation between $$n+1$$ and $$n$$?
The answer is, yes, such view exists: anchoring two points that attains max distance $$t$$, then allocate the remaining points to the disconnected admissible regions (as if they were bins) in descending order. This natural decomposition gives exactly the $$2^{n-2}$$ pieces in the density, where the pieces are all in the form of simplex volumes.
# Sec.3$$\quad$$ Recurrence Relation via Combinatorial Encoding of the Circular Diagram
One can extract the relevant geometrical and combinatorial information contained in the circular diagrams via encoding them as (linear) strings.
• Listing the points (not the angles) in a descending order
• Ignore the leading $$B$$ (also $$A_t$$) and trailing $$A_{-t}$$ that are always there.
• Denote the pairs that mark the inadmissible range like $$\{C_{-t},~C_t\}$$ and $$\{D_{-t},~D_t\}$$ etc as the "gap-points" (or just gaps) $$C_X$$ and $$D_X$$, etc.
• Use the symbol $$0$$ to emphasize that zero point is special. (note: $$0$$ also stands for $$B_X$$).
The first two subsections focus on the recursive algorithm (rules) that generate the "codes" as the number of points increases. How the codes indicate the calculation of the pieces for the density will be addressed in Sec.3.3.
Further specifications to the codes is adopted in anticipation of later results. The symbol Gamma will be used (for C as in configuration) $$\Gamma(j,k,\ell,\ldots)$$ to provide indexing for each code (string).
For three points, the $$2^{n-2} = \text{two}$$ pieces of \ref{Eq_f3} (derived in Sec.2.1) is now encoded as: \begin{aligned} \text{baseline}: & & \Gamma(1) &\equiv \{ C,~0,~C_X \} \\ \text{all-negative}:& &\Gamma(-1) &\equiv \{0,~ C \} \end{aligned}
## Sec.3.1$$\quad$$ Encoding going from $$n = 3$$ to $$n = 5$$
Here's the formal statement the rules (algorithm) for generating the codes (configurations) for $$n+1$$ given those of $$n$$. Each rule is considered proven, as they are obviously true by demonstration.
1. The new point (the $$(n+1)$$th point) can go only after the $$n$$th point, by construct.
2. There's no variation in the "all-negative" piece. Just attach the new point at the end.
3. The new point can go between $$0$$ and any of the "gaps" symbolized by $$C_X,D_X,E_X,\ldots$$ etc that are explicitly present in the code (not out of bounds).
4. The inadmissible region $$P_X$$ for any point $$P$$ is attached to the end of the code if and only if $$P$$ is in front of zero. In other words, $$P_{t}$$ and $$P_t$$ are out of bounds ("more negative" than $$\theta = -t$$) if and only if $$P$$ comes after zero.
Going to $$\mathbf{n = 4}$$ by inserting $$D$$. We get three pieces of $$n = 4$$ from the baseline piece of $$n = 3$$, and there's only one descendant from the all-negative piece: $$\begin{array}{c|cccccc|c} \mathbf{ \Gamma(1) } & C & & 0 & & C_X & & \\ \hline \Gamma(1, \frac13) & C & D & 0 & & C_X & D_X & \scriptsize \text{new baseline for}~ n=4 \\ \Gamma(1, \frac23) & C & & 0 & D & C_X & & \\ \Gamma(1, \frac33) & C & & 0 & & C_X & D & \\ \end{array} \\ \begin{array}{c|ccc|c} \mathbf{ \Gamma(-1) } & 0 & C & & \\ \hline \Gamma(-1, -1) & 0 & C & D & \scriptsize \text{new all-negative for}~ n=4 \end{array}$$ Reminder: the gaps $$C_X$$ and $$D_X$$ are out of bounds (more negative than $$A_{-t}$$ at $$\theta = -t$$ ). It is indeed true that these inadmissible range like $$D_X$$ can be in the upper half ($$0 < \theta < t < \frac12$$). No worries, the ranges for points (integration limits) will work out "automatically", necessarily leading to a sequence of decreasing spaces that yields the simplex volume.
Going to $$\mathbf{n = 5}$$, we have the $$2^{n-2} = \text{four}$$ configurations from $$n = 4$$ that makes up \ref{Eq_f4} encoded as $$\Gamma(1,\frac13), \Gamma(1,\frac23), \Gamma(1,\frac33)$$, and $$\Gamma(-1,-1)$$ just above.
Let’s start with the baseline, inserting the fifth point $$E$$ yields four new configurations. $$\begin{array}{c|ccccccccc|c} \mathbf{ \Gamma(1, \frac13) }& C & D & & 0 & & C_X & & D_X & & \\ \hline \Gamma(1, \frac13, \frac14) & C & D & E & 0 & & C_X & & D_X & E_X & \scriptsize \text{new baseline for}~ n=5 \\ \Gamma(1, \frac13, \frac24) & C & D & & 0 & E & C_X & & D_X & & \\ \Gamma(1, \frac13, \frac34) & C & D & & 0 & & C_X & E & D_X & & \\ \Gamma(1, \frac13, \frac44) & C & D & & 0 & & C_X & & D_X & E & \end{array}$$
The remaining two of the "bloodline of $$n = 4$$ baseline" $$\Gamma(1, \cdot)$$ produce two and one respectively: $$\begin{array}{c|cccccc} \mathbf{ \Gamma(1, \frac23) } & C & 0 & D & & C_X & \\ \hline \Gamma(1, \frac23, \frac12) & C & 0 & D & E & C_X & \\ \Gamma(1, \frac23, \frac22) & C & 0 & D & & C_X & E \\ \end{array} \quad,\quad \begin{array}{c|ccccc} \mathbf{ \Gamma(1, \frac33) } & C & 0 & C_X & D & \\ \hline \Gamma(1, \frac33, 1) & C & 0 & C_X & D & E \end{array}$$
Of course, the all-negative for $$n = 4$$ necessarily produces only the new all-negative. $$\begin{array}{c|cccc|c} \mathbf{ \Gamma(-1, -1) } & 0 & C & D & & \\ \hline \Gamma(-1, -1, -1) & 0 & C & D & E & \scriptsize \text{all-negative for}~ n=5 \end{array}$$
These are the $$(4+2+1)+1 = 2^{n-2} = \text{eight}$$ configurations that have a one-to-one mapping with the constituent "pieces" of $$~f_5(t)$$, the \ref{Eq_f5} in Sec.1.
## Sec.3.2$$\quad$$ Going to $$n=6$$ and Comments on $$\Gamma$$ Indexing and Lengths
This main answer post is too long (over the input limit of $$30$$k characters). Please see the separate post for additional discussion on the encoding system.
## Sec.3.3$$\quad$$ Density Pieces from Encoded Configurations: the Simplex Volumes
The previous subsections established the natural codes (listing of configurations). Here we state the rules/algorithm to read the contribution to the density from a given code.
Denote the range of point $$C$$ (integration limit) as $$\mathcal{R}_c = \mathcal{U}_c - \mathcal{L}_c$$ the upper limit minus the lower limit, then:
1. All configurations contribute to the density in the form of $$\frac1{(n-2)!} \mathcal{R}_c^{n-2}~$$, which is the volume of a simplex with side length $$\mathcal{R}_c = t - (k_U+k_L)(1 - 2t)$$, with $$k_U$$ and $$k_L$$ described below.
2. The upper limit is always either $$t$$ or $$3t-1$$. Namely, $$\mathcal{U}_c = t - k_U(1 - 2t)$$ where $$k_U = 0$$ or $$1$$ is the number of group of points behind zero and in front of $$C_X$$.
3. The lower limit is always in the form of $$\mathcal{L}_c = k_L(2t-1)$$ where $$k_L$$ counts the number of groups of points behind $$C_X$$. The gaps $$D_X$$ and $$E_X$$ act as dividers (bin edge), and points between the same pair of dividers are considered a single group.
4. The rules to count $$k_U$$ and $$k_L$$, are obviously true as demonstrated previously in Sec.2.2: the "one-gap-reduction" and "one-gap-lift" correspond to the two separate origins of the gap (same magnitude $$1-2t$$).
For example, at $$n = 9$$ with point $$A$$ to point $$I$$
• The configuration $$\{C, D, 0, E, F, C_X, G, D_X, H, I \}$$ which is indexed as $$\Gamma(1,\frac13,\frac24,\frac13,\frac23,\frac23,1)$$ contributes $$\frac{(7t-3)^7}{7!}$$ with an overall $$9!$$ in front, because there is one group $$\{E, F\}$$ in front of $$C_X$$ and two groups $$\{G\}$$ and $$\{H, I\}$$ behind $$C_X$$.
• The configuration $$\{C, D, E, 0, C_X, F, G, D_X, E_X, H, I\}$$ which is indexed as $$\Gamma(1,\frac13,\frac14,\frac35,\frac13,\frac33,1)$$ contributes $$9!\,\frac{(5t-2)^7}{7!}$$, because there is zero group in front of $$C_X$$ and two groups $$\{F,G\}$$ and $$\{H, I\}$$ behind $$C_X$$. Note that $$E_X$$ immediately follows $$D_X$$ and has no effect.
• The new and "last" piece is the configuration $$\{C, D, E, 0, F, C_X, G, D_X, H, E_X, I \}$$ which index is $$\Gamma(1,\frac13,\frac14,\frac25,\frac24,\frac23,\frac22)$$. The density from this config is $$72(9t-4)^7$$, because there is one group $$\{F\}$$ in front and three groups $$\{F\}$$, $$\{G\}$$, and $$\{I\}$$ after $$C_X$$.
Here's another quick look at where the circular diagram is clearly helpless. At $$n = 16$$ with points $$A$$ to $$P$$, the configuration \begin{aligned} &\Gamma(1,\frac13,\frac14,\frac15,\frac26,\frac15,\frac14,\frac13,\frac12,\frac12,\frac12,\frac22,1,1) \\ &= \{C, D, E, F, 0, G, H, C_X, I, D_X, J, E_X, K, L, M, F_X, N, O, P\}\end{aligned} corresponds to a piece of $$16!\,\frac{(11t-5)^{14}}{14!}$$ where the total of $$k_U + k_L = 1+4=\text{five}$$ groups are $$\{G,H\}$$, $$\{I\}$$, $$\{J\}$$, $$\{K,L,M\}$$, and $$\{N,O,P\}$$. For the record, we know that for $$n = 16$$ the highest order is $$(15t-7)^{14}$$, see \ref{Eq_fn}. This is achieved by configurations such as \begin{aligned} & & &\{C, \ldots, H, 0, I, C_X, J, D_X, K, E_X, L, F_X, M, G_X, N, H_X, O, P\} \\ &\text{and} & &\{C, \ldots, I, 0, J, C_X, K, D_X, L, E_X, M, F_X, N, G_X, O, H_X, P, I_X\}~,\end{aligned} among the $${n-1 \choose k}=15$$ where $$k \lfloor \frac{n-1}2\rfloor = 7$$.
The proof about the simplex volume can perhaps be done with induction, while intuitively it's at least reasonable in that one can see the points having hierarchically restricted space.
# Concluding Remarks
Many details of different levels can be formalized into rigorous proofs, but to be honest the effort I've made so far is not enough to overcome the majority of the technicalities.
Several less trivial claims are stated as "obviously true by demonstration", in the sense that: the process of using the circular diagram and encoding the configurations has been enough to convince the author that the results are true. The geometric and combinatorial nature of the problem is manifested by the construct.
Roughly speaking, the "solution" goes like
1. Establishing the encoding as an exact representation of the problem.
2. Showing via the codes that there are $$2^{n-2}$$ disjoint configurations.
3. Showing via the codes that each configuration contributes to a simplex volume.
4. Showing that this holds for all $$n \geq 3$$.
Arguably, not displaying a proof for (2) here in the post isn't a big deal. As for (4) the recurrence, it's a judgement call whether one can take it as true, perhaps having some faith in the induction proof (which formulation is hinted by the demonstration).
The most important task to continue is to fix the lack of semi-rigorous derivation for (3): the simplex volume, either for a given $$n$$, or with induction from trivial cases $$n = 2$$ and $$3$$.$$\endgroup$$
Please load the page $$3\sim 4$$ times for the hyperlinks to work properly.
$$\require{begingroup}\begingroup\renewcommand{\dd}[1]{\,\mathrm{d}#1}$$Some contents are post separately here because the main answer post is too long (over the input limit of $$30$$k characters).
Below are Sec.2.1 followed by Sec.3.2, where the latter details $$n = 6$$ for the encoding system with along with some general extra info.
## Sec.2.1$$\quad$$ Density Derivation $$n = 3$$ and $$n = 4$$
For $$t < \frac13$$ in figure on the left below, point $$C$$ is implicit with $$0 < c < b = t~$$. The density is $$f_3(t) = \int_{c = 0}^t 1 \cdot \dd{t} = t \qquad \text{, for}~~ t < \frac13$$ where the integrand $$1$$ is the uniform density along the circle which circumference is unity.
For $$t > \frac13$$, figure on the right, in addition to the admissible range on the positive (upper) half, $$C$$ can be negative and between $$A_{-t}$$ and $$B_{t}$$ such that $$-t < c < b + t - 1 = 2t - 1$$. $$f_3(t) = \underbrace{ \int_{c = 0}^t 1 \dd{t} }_{ \text{same form as case t \,< \frac13} } + \int_{c = -t}^{2t-1} 1 \cdot \dd{t} = t + (3t - 1) \qquad \text{, for}~~ t > \frac13$$
This "opening" in the negative half exists only when $$B_t$$ is in the "front" of $$A_{-t}$$, as in $$b + t - 1 > a-t.~$$ With $$a = 0$$ and $$b = t$$, this gives the critical point $$t > \frac13$$.
Together with the multiplicity $$n(n-1)$$, this is Eq.(3) listed in Sec.1 $$f_3(t) = n! \bigl( \text{baseline} + \text{all-negative} \bigr) = 6 \left( t + (3t - 1) \mathbb{1}_{ t \, > \frac13 } \right)$$
The three diagrams below display $$\mathbf{n = 4}$$. Point $$D$$ is implicit and that it can reside only in the 3-arc-intersection of green-purple-yellow.
The left diagram plays a dual role. Firstly, it shows $$0 < d < c < b = t$$, the baseline piece. $$\text{For all 0 This form is in fact $$\frac{ t^{n-2}}{ (n-2)! }$$, the volume of a $$(n-2)$$-simplex with side length $$t$$.
This left diagram also serves for the configuration of an "extra" piece, where $$D$$ is in the lower half, between $$C_{-t}$$ and $$B_t~$$, such that $$c - t < d < b + t - 1$$.
This happens only when $$B_t$$ is in front of $$C_{-t}$$, which is in turn in front of $$A_{-t}~$$, as in $$a-t < c - t < b + t - 1 < 0~$$. These imply $$c < 3t - 1$$ and $$t > \frac13$$. $$\textbf{extra}_1: \quad \int_{c = 0}^{3t - 1} \int_{d = c-t}^{2t-1} 1 \dd{d} \dd{c} = \int_{c = 0}^{3t - 1} (3t - 1 - c) \dd{c} = \frac{ (3t - 1)^2}2 \qquad \text{, for}~~t > \frac13$$ It’s not a coincidence that this is the volume of a $$(n-2)$$-simplex with side length $$3t - 1$$.
The middle diagram: point $$D$$ is below ($$d < 0$$) while $$C$$ is above ($$c > 0$$). Another "new opening" appears in the 3rd quadrant when $$C_t$$ is in front of $$A_{-t}~$$. Namely, $$c + t - 1 > a - t = -t$$ so that $$c > 1- 2t > 0$$ .
This extra admissible range is for $$D$$ to reside between $$A_{-t}$$ and $$C_t$$ as in $$-t < d < c+t-1$$.
$$\textbf{extra}_2: \quad \int \limits_{c = 1- 2t}^{t} \int_{d = -t}^{c + t - 1} 1 \dd{d} \dd{c} = \int \limits_{c = 1 - 2t}^{t} \bigl(c - (1- 2t) \bigr) \dd{c} = \frac{ (3t - 1)^2}2$$ The condition on $$t$$ is by construct $$b > c \implies t > c > 1 - 2t \implies t > \frac13$$.
The natural and geometric approach is to keep track of the "opening" magnitude: do change of variable $$c’ \equiv c - (1-2t)$$, yielding again the volume of a simplex with side length $$3t-1$$ as in $$\int_{c’ = 0}^{3t-1} c’ \dd{c'} = \frac{ (3t - 1)^2}2$$. This is not an isolated case but generally what goes on with the various configurations for any $$n$$.
The last of the $$2^{n-2} = 4$$ pieces is shown in the right diagram, where both points are below zero. Namely, all-negative for the $$n-2$$ free points. Necessarily they are also behind $$B_t$$ so that $$b + t - 1 > c > d$$, or equivalently $$0 > 2t - 1 > c > d > -t$$.
\begin{aligned} \textbf{all-negative}:& & &\phantom{{}=} \int_{c = -t}^{2t - 1} \int_{d = -t}^c 1 \dd{d} \dd{c} & &\boxed{\begin{aligned} &\scriptsize\text{natural change of variables:} \\ &c’ \equiv c + t~,~~ d’ \equiv d + t \end{aligned}} \\ &&&= \int_{c’ = 0}^{3t - 1} \int_{d’ = 0}^{c’} 1 \dd{d} \dd{c} & &\\ &&&= \frac{ (3t - 1)^2}2 && ,~t > \frac13 \end{aligned}
Putting things together, one recovers Eq.(4) seen in Sec.1 : \begin{aligned} f_4(t) &= n! \Bigl( \text{baseline} + \text{all-negative} + \text{extra}_1 + \text{extra}_2 \Bigr) \\ &= 24 \Bigl( \frac{t^2}2 + \mathbb{1}_{t\, > \frac13} \bigl( \frac{ (3t - 1)^2 }2 + \frac{ (3t - 1)^2 }2 + \frac{ (3t - 1)^2 }2 \bigr) \Bigr) \end{aligned}
## Sec.3.2$$\quad$$ Going to $$n=6$$ and Comments on $$\Gamma$$ Indexing and Lengths
Pardon me for not reproducing a table listing all the eight configs of $$n = 5$$ to save space.
Again, with the baseline of $$n = 5$$, one can see why it yields five configs for $$n = 6$$. $$\begin{array}{c|cccccccccccc|c} \mathbf{ \Gamma(1, \frac13, \frac14) } & C & D & E & & 0 & & C_X & & D_X & & E_X & & \\ \hline \Gamma(1, \frac13, \frac14, \frac15) & C & D & E & F & 0 & & C_X & & D_X & & E_X & F_X & \scriptsize \text{baseline}~n=6 \\ \Gamma(1, \frac13, \frac14, \frac25) & C & D & E & & 0 & F & C_X & & D_X & & E_X & & \\ \Gamma(1, \frac13, \frac14, \frac35) & C & D & E & & 0 & & C_X & F & D_X & & E_X & & \\ \Gamma(1, \frac13, \frac14, \frac45) & C & D & E & & 0 & & C_X & & D_X & F & E_X & & \\ \Gamma(1, \frac13, \frac14, \frac55) & C & D & E & & 0 & & C_X & & D_X & & E_X & F & \end{array}$$
The astute readers might already know that the remaining "base-bloodline" configs $$\Gamma(1, \frac13,\cdot)$$ will produce three, two, and one, respectively. $$\begin{array}{c|ccccccccc} \mathbf{\Gamma(1, \frac13, \frac24) } & C & D & 0 & E & & C_X & & D_X & \\ \hline \Gamma(1, \frac13, \frac24, \frac13) & C & D & 0 & E & F & C_X & & D_X & \\ \Gamma(1, \frac13, \frac24, \frac23) & C & D & 0 & E & & C_X & F & D_X & \\ \Gamma(1, \frac13, \frac24, \frac33) & C & D & 0 & E & & C_X & & D_X & F \end{array} \\ \scriptsize\begin{array}{c|ccccccc} \mathbf{\Gamma(1, \frac13, \frac34) } & C & D & 0 & C_X & E & & D_X & \\ \hline \Gamma(1, \frac13, \frac34, \frac12) & C & D & 0 & C_X & E & F & D_X & \\ \Gamma(1, \frac13, \frac34, \frac22) & C & D & 0 & C_X & E & & D_X & F \end{array} \quad \begin{array}{c|ccccccc} \mathbf{\Gamma(1, \frac13, \frac44) } & C & D & 0 & C_X & D_X & E & \\ \hline \Gamma(1, \frac13, \frac44, \frac11) & C & D & 0 & C_X & D_X & E & F \end{array}$$
Continue the descending pattern, the "secondary base-bloodline" $$\Gamma(1, \frac23,\cdot)$$ render two and one. Then there's the "single bloodline". $$\begin{array}{c|ccccccc} \mathbf{\Gamma(1, \frac23, \frac12) } & C & 0 & D & E & & C_X & \\ \hline \Gamma(1, \frac23, \frac12, \frac12) & C & 0 & D & E & F & C_X & \\ \Gamma(1, \frac23, \frac12, \frac22) & C & 0 & D & E & & C_X & F \end{array} \quad \begin{array}{c|cccccc} \mathbf{\Gamma(1, \frac23, \frac22) } & C & 0 & D & C_X & E & \\ \hline \Gamma(1, \frac13, \frac34, \frac22, 1) & C & 0 & D & C_X & E & F \end{array} \\ \begin{array}{c|cccccc} \mathbf{\Gamma(1, \frac33, 1) } & C & 0 & C_X & D & E & \\ \hline \Gamma(1, \frac13, \frac33, 1, 1) & C & 0 & C_X & D & E & F \end{array}$$ Lastly the all-negative: $$\begin{array}{c|ccccc|c} \mathbf{\Gamma(-1, -1, -1) } & 0 & C & D & E & & \\ \hline \Gamma(-1, -1, -1, -1) & 0 & C & D & E & F & \scriptsize \text{all-negative for}~ n = 6 \end{array}$$ These are the $$(5 + 3 + 2 + 1) + (2+1) + 1 + 1 = 2^{n-2} = 16$$ pieces mapping to $$~f_6(t)$$, the Eq.(6) in Sec.1.
Some comments on the $$\Gamma$$ indexing:
1. Baseline pieces have ascending consecutive denominators and $$1$$ in the numerator: $$\Gamma(1, \frac13), \Gamma(1, \frac13, \frac14), \Gamma(1, \frac13, \frac14, \frac15),\ldots$$
2. An argument of $$\Gamma$$ being $$1$$ means the bloodline becomes single line from there. The arguments in fractions like $$\frac55, \frac44$$ and $$\frac22$$ equal $$1$$ not by accident.
3. The all-negative might as well all be denoted as just $$\Gamma(-1)$$
4. There are $$n - 2$$ indices in $$\Gamma(j,k,\ell,\ldots)$$. That is, the number of arguments in $$\Gamma$$ indicates the number of points.
According to the rules for generating the codes (algorithm), it's easy to see the recurrence algorithm for the length (number of pieces) in each "bloodline":
• Each number gives a descending sequence of that length (e.g. whenever you see a $$3$$ it creates a $$[3,2,1]~$$).
• The very first (longest) sequence has the leading term bumped up by one (due to the special role of $$0 = B_X$$ in the code). For the first few $$n$$:
$$[3,1] \\ [4, 2, 1], [1] \\ [5,3,2, \color{magenta}1], [2, \color{green}1], [1], [1] \\ [6,4,\underline{\mathbf{3}},2,\color{red}1], [3, 2, \color{blue}1], [2,1], \color{magenta}{[1]}, [2,\color{orange}1],\color{green}{[1]},[1],[1] \\ \scriptsize[7,5,4,3,2,1], [4, 3, 2, 1], \underline{\mathbf{[3,2,1]}}, [2,1], \color{red}{[1]},[3,2,1],[2,1], \color{blue}{[1]} ,[2,1],[1],\color{magenta}{[1]},[2,1], \color{orange}{[1]},\color{green}{ [1]},[1],[1]$$ It's easy to prove this two-rule algorithm for the length at any $$n$$ leads to a total of $$2^{n-2}$$.
By design, given a legit $$\Gamma$$ that correctly corresponds to a code, one can reconstruct the code by reading the indices. For example, given $$\Gamma(1,\frac13,\frac24,\frac23)$$ one can deduce the code to be $$\{C,D,0,E,C_X,F,D_X\}$$ in the following manner.
• First index $$1$$ means $$C$$ is in front of zero, $$\implies \{C,0,C_X\}$$
• Second index $$\frac13$$ means $$D$$ goes in the first of the 3 spaces, $$\{C, \_\_,0, \_\_,C_X,\_\_ \} \implies \{C,D,0,C_X,D_X \}~$$, where $$D_X$$ is in the code (inside the bound, $$d + t - 1 > -t~$$) because $$D$$ is in front of zero.
• Third index $$\frac24$$ means $$E$$ goes to the 2nd of the 4 spaces that follow $$D$$, $$\{C, D,\_\_,0, \_\_,C_X,\_\_,D_X,\_\_ \} \implies \{C,D,0,E,C_X,D_X \}~$$, where $$E_X$$ is out of bounds because $$E$$ is behind $$0$$.
• Fourth index $$\frac23$$ means $$F$$ falls in the 2nd of the 3 spaces that follow $$E$$, $$\{C, D,0, E,\_\_,C_X,\_\_,D_X,\_\_ \} \implies \{C,D,0,E,C_X,F,D_X \}~$$, where again $$F_X$$ is not inside due to $$f<0~$$.
Conversely, one can deduce the $$\Gamma$$ indexing directly by reading the code, without needing to go through the hierarchical listing of all the levels below.$$\endgroup$$ | 2021-04-18T14:21:21 | {
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http://math.stackexchange.com/questions/166647/could-someone-help-clarify-an-example-in-my-linear-algebra-book-on-orthonormal-b | Could someone help clarify an example in my linear algebra book on orthonormal bases?
Linear Algebra with Applications by Steven J. Leon, p.257:
Theorem 5.5.2: Let $\{\textbf{u}_1, \textbf{u}_2, \ldots, \textbf{u}_n\}$ be an orthonormal basis for an inner product space $V$. If $\textbf{v} = \sum_{i=1}^{n} c_i \textbf{u}_i$, then $c_i = \langle \textbf{v}, \textbf{u}_i \rangle$.
I don't know if I need to include the proof, but it's short, so here it is:
Definition: $\delta_{ij} = \begin{cases} 1 & \text{if } i = j \\ 0 & \text{if } i \neq j \end{cases}$
Proof: $$\langle \textbf{v}, \textbf{u}_i \rangle = \left< \sum_{j=1}^{n} c_j \textbf{u}_j, \textbf{u}_i \right> = \sum_{j=1}^{n} c_j \langle \textbf{u}_j, \textbf{u}_i \rangle = \sum_{j=1}^{n} c_j \delta_{ij} = c_i$$
Now here's the example given in the book. He seems to be using the theorem's logic backwards. I don't get it.
Example: The vectors $$\textbf{u}_1 = \left(\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}}\right)^T \text{ and } \textbf{u}_2 = \left(\dfrac{1}{\sqrt{2}},-\dfrac{1}{\sqrt{2}}\right)^T$$ form an orthonormal basis for $\mathbb{R}^2$. If $\textbf{x} \in \mathbb{R}^2$, then $$\textbf{x}^T \textbf{u}_1 = \dfrac{x_1 + x_2}{\sqrt{2}} \text{ and } \textbf{x}^T \textbf{u}_2 = \dfrac{x_1 - x_2}{\sqrt{2}}$$ It follows from Theorem 5.5.2 that $$\textbf{x} = \dfrac{x_1 + x_2}{\sqrt{2}} \textbf{u}_1 + \dfrac{x_1 - x_2}{\sqrt{2}} \textbf{u}_2$$
Isn't "$\textbf{x} = \dfrac{x_1 + x_2}{\sqrt{2}} \textbf{u}_1 + \dfrac{x_1 - x_2}{\sqrt{2}} \textbf{u}_2$" referring to "$\textbf{v} = \sum_{i=1}^{n} c_i \textbf{u}_i$", and "$\textbf{x}^T \textbf{u}_1 = \dfrac{x_1 + x_2}{\sqrt{2}} \text{ and } \textbf{x}^T \textbf{u}_2 = \dfrac{x_1 - x_2}{\sqrt{2}}$" referring to "$c_i = \langle \textbf{v}, \textbf{u}_i \rangle$"? Shouldn't the latter follow from the former? Or should the theorem state if and only if? Or am I just confused?
-
There is nothing strictly wrong with what he wrote; but, perhaps, it would be more clear if he instead wrote "If ${\bf x}=c_1{\bf u}_1+c_2{\bf u}_2\in\Bbb R^2$, then $$c_1={\bf x}^T{\bf u}_1 ={x_1+x_2\over\sqrt2}\ \text{and} \ c_2={\bf x}^T{\bf u}_2\ ...$$ – David Mitra Jul 4 '12 at 17:10
Well, if you have a theorem that says that A implies B, you can't use that theorem and B to conclude A. That's what's happening here. – Matt Gregory Jul 4 '12 at 17:59
Every vector in your inner product space can be expressed as a unique linear combination of your basis elements.
Just think about it, when would $\langle \textbf{v}, \textbf{u}_i \rangle$ ever not equal the $i'$th coefficient?
That's why the converse is true.
-
I can see how $\langle \textbf{v}, \textbf{u}_i \rangle$ would never not equal the $i$'th coefficient if $\textbf{u}_i$ were equal to the standard basis element $\textbf{e}_i$ because it would just be filtering out the $i$'th component of the vector. – Matt Gregory Jul 4 '12 at 19:03
I think I'm too tired and burned out to think about this properly. Why does this happen? :-P – Matt Gregory Jul 4 '12 at 19:05
Not that's not right. I'm unchecking your thing :-P What needs to be proven is that if $c_i = \langle \textbf{v}, \textbf{u}_i \rangle$, then $\textbf{v} = \sum_{j=1}^n c_i \textbf{u}_i$. – Matt Gregory Jul 4 '12 at 20:40
That follows by the fact that every $v \in V$ can be expressed as a linear combination of your basis elements. – Ink Jul 4 '12 at 21:38
Take $\textbf{v} \in V$ and suppose $c_i = \langle \textbf{v}, \textbf{u}_i \rangle$. Since $\{\textbf{u}_1,...,\textbf{u}_n\}$ is a basis for our inner product space, there exists scalars $b_1,...,b_n$ such that $\textbf{v}= \sum_{i=1}^{n} b_i \textbf{u}_i$. Hence $b_i = \langle \textbf{v}, \textbf{u}_i \rangle = c_i$. – Ink Jul 4 '12 at 22:00
We know that $x = c_1 u_1 + c_2 u_2$ for some $c_1, c_2$.
By the theorem, $c_1 = \left< x, u_1 \right>$ and $c_2 = \left< x, u_2 \right>$.
Thus $$x = \left< x, u_1 \right> u_1 + \left< x, u_2 \right> u_2 = \frac{x_1 + x_2}{\sqrt{2}} u_1 + \frac{x_1 - x_2}{\sqrt{2}} u_2$$.
-
Since $\{\mathbf{u}_1, \mathbf{u}_2\}$ is a basis for $\mathbb{R}^2$, any $\mathbf{x} \in \mathbb{R}^2$ can be uniquely written as a linear combination $$\mathbf{x} = a_1 \mathbf{u}_1 + a_2 \mathbf{u}_2. \tag{\ast}$$ Taking the inner product of $(\ast)$ with $\mathbf{u}_1$, we have \begin{align*} \langle \mathbf{x}, \mathbf{u}_1 \rangle & = \langle a_1 \mathbf{u}_1 + a_2 \mathbf{u}_2, \mathbf{u}_1 \rangle \\ & = \langle a_1 \mathbf{u}_1, \mathbf{u}_1 \rangle + \langle a_2 \mathbf{u}_2, \mathbf{u}_1 \rangle \\ & = a_1 \langle \mathbf{u}_1, \mathbf{u}_1 \rangle + a_2 \langle \mathbf{u}_2, \mathbf{u}_1 \rangle \\ & = a_1 \cdot 1 + a_2 \cdot 0 \\ & = a_1. \end{align*} A similar calculation shows that $$\langle \mathbf{x}, \mathbf{u}_2 \rangle = a_2.$$ Therefore $$a_1 = \langle \mathbf{x}, \mathbf{u}_1 \rangle = \mathbf{x}^T \mathbf{u}_1,$$ $$a_2 = \langle \mathbf{x}, \mathbf{u}_2 \rangle = \mathbf{x}^T \mathbf{u}_2.$$
This process can be used to show that the converse of the theorem is true.
-
@MattGregory: I've edited my answer to address your clarifications in the comments. Let me know if this clears up your confusion. – Henry T. Horton Jul 4 '12 at 21:14
Yeah, that did help. Thank you! – Matt Gregory Jul 5 '12 at 4:31 | 2016-07-29T00:10:56 | {
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https://math.stackexchange.com/questions/756462/error-solving-stars-and-bars-type-problem/756501 | # Error solving “stars and bars” type problem
I have what I thought is a fairly simple problem: Count non-negative integer solutions to the equation
$$x_1 + x_2 + x_3 + x_4 + x_5 = 23$$
such that $0 \leq x_1 \leq 9$.
Not too hard, right? Simply ignore the upper-bound, count the $$\begin{pmatrix}23 + (5-1) \\ (5-1)\end{pmatrix} = \begin{pmatrix}27 \\ 4\end{pmatrix} = 17550$$ solutions. Subtract from this all (non-negative integer) solutions to the equation $$y_1 + 10 + x_2 + x_3 + x_4 + x_5 = 23,$$ and there are $\begin{pmatrix}17 \\ 4\end{pmatrix} = 2380$ of these "bad" solutions we shouldn't have counted earlier, but did. Thus we find $17550 - 2380 = 15170$ solutions.
Since this is a prohibitively large number of solutions to check by hand, I wrote a simple Python program to verify whether this answer is correct. It does indeed say there are $17550$ solutions without upper bounds, and $2380$ solutions to the equation for counting "bad" solutions.
However, when I ask it throw away all solutions to the non-upper-bounded problem for which $x_1 \geq 10$, it tells me it's found $15730$ solutions.
My question is: do I not understand the combinatorial calculation so that there are not actually $\begin{pmatrix}27\\4\end{pmatrix}-\begin{pmatrix}17\\4\end{pmatrix}$ solutions, or have I made some kind of programming mistake? Of course, both are also possible.
Your method is correct, there are 15170 solutions. It seems like your python script is doing something wrong at the end.
I wrote a C++ program myself to confirm this:
#include <stdio.h>
int main()
{
int a,b,c,d,e,sum=0;
for (a=0; a<10; a++)
for (b=0; b<24; b++)
for (c=0; c<24; c++)
for (d=0; d<24; d++)
for (e=0; e<24; e++)
{
if (a+b+c+d+e == 23)
sum++;
}
printf("sum=%d",sum);
}
And it prints out sum=15170.
• Thank you, I should have suspected the program, given that it took me longer than expected to get it up and running. I'll have to take some time and see what's gone wrong there. – pjs36 Apr 16 '14 at 19:12
A nice way to mathematically check your work is with a generating function. We have $x_{1} \in \{0, ..., 9\}$. So our generating function for an $x_{1}$ is $\sum_{i=0}^{9} x^{i} = \dfrac{1-x^{10}}{1-x}$. The other terms don't have this restriction, so their generating functions are simply $\dfrac{1}{1-x}$.
So we multiply this result for each $x_{i}$ to get our generating function $f(x) = \dfrac{1-x^{10}}{(1-x)^{5}}$.
Using our binomial identities, we get $(1-x^{10}) = \sum_{i=0}^{1} \binom{1}{i} (-1)^{i} x^{10i} = (1 - x^{10})$
We then expand out the bottom term $\dfrac{1}{(1-x)^{5}} = \sum_{i=0}^{\infty} \binom{i + 5 - 1}{i} x^{i}$. And we multiply this with $(1 - x^{10})$, taking only terms $x^{23}$.
So that gives us $\binom{23 + 5 - 1}{23} - \binom{13 + 5 - 1}{13} = \binom{27}{4} - \binom{17}{4}$ as our answer.
Generating functions are nice, because they make it much easier to model the problem, and you don't have to crank out code and guess if it's right. Of course, seeing the stars and bars solution is nice, but takes more practice. Generating functions are more mechanical once you've figured out how to deal with them.
• I've seen generating functions used for these problems, and it looks like I'd better bite the bullet and figure them out. I dealt with them years ago, but never fully "got it" or figured out how to work with them successfully. Clearly, they're worth the effort! – pjs36 Apr 16 '14 at 19:16
• There is a good tutorial on ordinary generating functions- dreamincode.net/forums/topic/… – ml0105 Apr 16 '14 at 19:56
$\displaystyle S=\sum_{x_1=0}^9 \{x_2+x_3+x_4+x_5=23-x_1\}=\sum_{x_1=0}^9 {26-x_1\choose 3}=\sum_{x_1=0}^9 {17+x_1\choose 3}$
$\displaystyle\quad= \sum_{x_1=0}^9 \left[{18+x_1\choose 4}-{17+x_1\choose 4}\right]={27\choose 4}-{17\choose 4}$ | 2019-08-21T02:34:24 | {
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https://reliableairtulsa.com/0t3u0l/19f33c-eigenvectors-of-symmetric-matrix-are-orthogonal-proof | # eigenvectors of symmetric matrix are orthogonal proof
Math 2940: Symmetric matrices have real eigenvalues The Spectral Theorem states that if Ais an n nsymmetric matrix with real entries, then it has northogonal eigenvectors. Proof — part 2 (optional) For an n × n symmetric matrix, we can always find n independent orthonormal eigenvectors. What is the importance of probabilistic machine learning? Then for a complex matrix, I would look at S bar transpose equal S. Given a subspace whose dimension is greater than $1$, one can choose a basis of the subspace consisting of orthogonal elements. The following is our main theorem of this section. Addendum: As @Ian correctly noticed, one has to add to the proof that the basis of the corresponding eigen-subspace for $\lambda$ can be chosen orthogonal. If Ais an n nsym-metric matrix … Theorem: a matrix has all real eigenvalues and n orthonormal real eigenvectors if and only if it is real symmetric. Why are Eigenvectors of an orthogonal matrix with respect to different eigenvalues orthogonal to one another. Proof Let v and w be eigenvectors for a symmetric matrix A with different eigenvalues λ1 and λ2. For any matrix M with n rows and m columns, M multiplies with its transpose, either M*M' or M'M, results in a symmetric matrix, so for this symmetric matrix, the eigenvectors are always orthogonal. What would be the most efficient and cost effective way to stop a star's nuclear fusion ('kill it')? Note that this is saying that Rn has a basis consisting of eigenvectors of A that are all orthogo- Moreover, Theorem. The eigenvectors of A−1 are the same as the eigenvectors of A. Eigenvectors are only defined up to a multiplicative constant. This will be orthogonal to our other vectors, no matter what value of , … The expression A=UDU T of a symmetric matrix in terms of its eigenvalues and eigenvectors is referred to as the spectral decomposition of A.. And just check that AT = (QT)TΛTQT. The eigendecomposition of a symmetric positive semidefinite (PSD) matrix yields an orthogonal basis of eigenvectors, each of which has a nonnegative eigenvalue. @A.G. proved this just fine already. Theorem 2.2.2. MATH 340: EIGENVECTORS, SYMMETRIC MATRICES, AND ORTHOGONALIZATION Let A be an n n real matrix. they are eigenvectors for $A$. If v is an eigenvector forATand if w is an eigenvector forA, and if the corresponding eigenvalues are dierent, then v and w must be orthogonal. I must remember to take the complex conjugate. If all the eigenvalues of a symmetric matrixAare distinct, the matrixX, which has as its columns the corresponding eigenvectors, has the property thatX0X=I, i.e.,Xis an orthogonal matrix. Nonetheless, for a symmetric matrix with a repeated eigenvalue, one can also choose a non-orthogonal basis such that the matrix is diagonal in that basis. cause eigenvectors corresponding to different eigenvalues are or-thogonal, it is possible to store all the eigenvectors in an orthogo-nal matrix (recall that a matrix is orthogonal when the product of this matrix by its transpose is a diagonal matrix). Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. (Philippians 3:9) GREEK - Repeated Accusative Article. How much do you have to respect checklist order? A symmetric matrix is diagonalizable whether it has distinct eigenvalues or not. Properties of real symmetric matrices I Recall that a matrix A 2Rn n is symmetric if AT = A. I For real symmetric matrices we have the following two crucial properties: I All eigenvalues of a real symmetric matrix are real. A is real because Q and Λ are. It is a beautiful story which carries the beautiful name the spectral theorem: Theorem 1 (The spectral theorem). A nxn symmetric matrix A not only has a nice structure, but it also satisfies the following: A has exactly n (not necessarily distinct) eigenvalues. MathJax reference. the eigenvalues of A) are real numbers. Algorithm for simplifying a set of linear inequalities. Rather, one can choose an orthogonal basis such that the matrix is diagonal in that basis. If we take each of the eigenvalues to be unit vectors, then the we have the following corollary. This proves that we can choose eigenvectors of S to be orthogonal if at least their corresponding eigenvalues are different. If is Hermitian (symmetric if real) (e.g., the covariance matrix of a random vector)), then all of its eigenvalues are real, and all of its eigenvectors are orthogonal. Since the unit eigenvectors of a real symmetric matrix are orthogonal, we can let the direction of λ 1 parallel one Cartesian axis (the x’-axis) and the direction of λ 2 parallel a second Cartesian axis (the y’-axis). <> To learn more, see our tips on writing great answers. Its eigenvalues. site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. What is the altitude of a surface-synchronous orbit around the Moon? Perfect. If you want a reference, I have on my desk: "Numerical Linear Algebra" by Trefethen and Bau (published by SIAM). for all indices and .. Every square diagonal matrix is symmetric, since all off-diagonal elements are zero. We omit the proof of the lemma (which is not dicult, but requires the denition of matrices on complex numbers). Why do you say "air conditioned" and not "conditioned air"? Hence all chains of generalized eigenvectors are of length one, i.e. The proof assumed different eigenvalues with different eigenvectors. ��肏I�s�@ۢr��Q/���A2���..Xd6����@���lm"�ԍ�(,��KZ얇��I���8�{o:�F14���#sҝg*��r�f�~�Lx�Lv��0����H-���E��m��Qd�-���*�U�o��X��kr0L0��-w6�嫄��8�b�H%�Ս�쯖�CZ4����~���/�=6+�Y�u�;���&nJ����M�zI�Iv¡��h���gw��y7��Ԯb�TD �}S��.踥�p��. Definition E EœEÞis called a if symmetric matrix X Notice that a symmetric ... Ñ. where the n-terms are the components of the unit eigenvectors of symmetric matrix [A]. Those are the numbers lambda 1 to lambda n on the diagonal of lambda. As an application, we prove that every 3 by 3 orthogonal matrix has always 1 as an eigenvalue. Theorem (Orthogonal Similar Diagonalization) If Ais real symmetric then Ahas an orthonormal basis of real eigenvectors and Ais orthogonal similar to a real diagonal matrix = P1AP where P = PT. I Eigenvectors corresponding to distinct eigenvalues are orthogonal. And then the transpose, so the eigenvectors are now rows in Q transpose. �:���)��W��^���/㾰-\/��//�?����.��N�|�g/��� %9�ҩ0�sL���>.�n�O+�p��7&�� �..:cX����tNX�O��阷*?Z������y������(m]Z��[�J��[�#��9|�v��� Note that it is an orthogonal matrix, so deserves to be called Q. Those are in Q. Nonetheless, for a symmetric matrix with a repeated eigenvalue, one can also choose a non-orthogonal basis such that the matrix is diagonal in that basis. The diagonalization of symmetric matrices. It only takes a minute to sign up. How to improve undergraduate students' writing skills? stream As opposed to the symmetric problem, the eigenvalues a of non-symmetric matrix do not form an orthogonal system. If A is Hermitian and full-rank, the basis of eigenvectors may be chosen to be mutually orthogonal. Now we need to get the last eigenvector for . @ian Sorry, I missed to mention that one can do orthogonalization within a corresponding eigen-subspace. Orthogonality of Eigenvectors of a Symmetric Matrix Corresponding to Distinct Eigenvalues Problem 235 Suppose that a real symmetric matrix A has two distinct eigenvalues α and β. Lecture 24 covers eigenvalues problems and has this result. %PDF-1.2 %�쏢 Symmetric matrices always have real eigenvalues (and hence real eigenvectors). The columns of $Q$ are the eigenvectors of $A$ (easy to check), $T$ contains the eigenvalues (easy to check), and since $Q$ is unitary, all the columns are orthonormal. Let A be a symmetric matrix in Mn(R). The largest eigenvalue is On one hand it is $0^Ty=0$, on other hand, it is $x^Tx=\|x\|^2$. Then eigenvectors take this form, . This is usually proven constructively by applying Gram-Schmidt. Use MathJax to format equations. In linear algebra, a real symmetric matrix represents a self-adjoint operator over a real inner product space. How to guarantee there will not be only one independent eigenvector such that all eigenvectors can form the orthogonal basis of the vector space? Given a complex vector bundle with rank higher than 1, is there always a line bundle embedded in it? The rst step of the proof is to show that all the roots of the characteristic polynomial of A(i.e. An example of an orthogonal matrix in M2(R) is 1/2 − √ √ 3/2 3/2 1/2 . Thus, it is not the case that all non-parallel eigenvectors of every symmetric matrix are orthogonal. If A is an n x n symmetric matrix, then any two eigenvectors that come from distinct eigenvalues are orthogonal. Proof: Let and be an eigenvalue of a Hermitian matrix and the corresponding eigenvector satisfying , then we have This implies the following equality: U¡1 ˘UT. The diagonal elements of a triangular matrix are equal to its eigenvalues. If is an eigenvector of the transpose, it satisfies By transposing both sides of the equation, we get. jthen the eigenvectors are orthogonal. A symmetric matrix can be broken up into its eigenvectors. The non-symmetric problem of finding eigenvalues has two different formulations: finding vectors x such that Ax = λx, and finding vectors y such that yHA = λyH (yH implies a complex conjugate transposition of y). ��:��f�߮�w�%:�L>�����:~A�N(��nso*|'�ȷx�ح��c�mz|���z�_mֻ��&��{�ȟ1��;궾s�k7_A�]�F��Ьa٦vnn�p�s�u�tF|�%��Ynu}*�Ol�-�q ؟:Q����6���c���u_�{�N1?) Suppose S is complex. This is the story of the eigenvectors and eigenvalues of a symmetric matrix A, meaning A= AT. So our equations are then, and , which can be rewritten as , . Then, if $A$ is symmetric, $T$ must also be symmetric (and hence diagonal). It gives $x=0$ which is a contradiction with the vectors being linear independent. All the eigenvalues of a symmetric matrix must be real values (i.e., they cannot be complex numbers). Recall some basic denitions. It seems to be a (correct) proof that a symmetric matrix is diagonalizable, but to say nothing about orthogonality. All eigenvectors of the matrix must contain only real values. Eigenvectors of real symmetric matrices are orthogonal (more discussion), MAINTENANCE WARNING: Possible downtime early morning Dec 2, 4, and 9 UTC…, Eigenvectors of real symmetric matrices are orthogonal, Looking for orthogonal basis of eigenvectors using Gram Schmidt process. How can you come out dry from the Sea of Knowledge? Given the eigenvector of an orthogonal matrix, x, it follows that the product of the transpose of x and x is zero. Eigenvectors corresponding to distinct eigenvalues are all orthogonal. So if I have a symmetric matrix--S transpose S. I know what that means. Moreover, eigenvalues may not form a linear-inde… After row reducing, the matrix looks like. Those are the lambdas. I To show these two properties, we need to consider complex matrices of type A 2Cn n, where C is the set of By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. Proof: Let Q be the matrix of eigenvectors. Assume that for a symmetric matrix $A$ there exists a Jordan block for an eigenvalue $\lambda$ of size more than one, hence there exists at least two linear independent generalized eigenvectors, i.e. The row vector is called a left eigenvector of . Fact. Does this picture depict the conditions at a veal farm? Proof Ais Hermitian so by the previous proposition, it has real eigenvalues. There are really three things going on here: Thus, it is not the case that all non-parallel eigenvectors of every symmetric matrix are orthogonal. Orthogonality of the degenerate eigenvectors of a real symmetric matrix, Complex symmetric matrix orthogonal eigenvectors, Is there any connection between the fact that a set of vectors are mutually orthogonal and the same set of vectors are linearly independent, Eigenvectors of the repeated eigenvalues for a symmetric matrix. @Michael Hardy My question is just to check if geometric multiplicity < algebraic multiplicity in the case of symmetric matrix. That's what I mean by "orthogonal eigenvectors" when those eigenvectors are complex. Even if and have the same eigenvalues, they do not necessarily have the same eigenvectors. Thanks for contributing an answer to Mathematics Stack Exchange! ${}\qquad{}$. $By=x$ and $Bx=0$ where $B=A-\lambda I$. Are eigenvectors of a symmetric matrix orthonormal or just orthogonal? I honestly don't see what this has to do with the question. Why is "issued" the answer to "Fire corners if one-a-side matches haven't begun"? A is symmetric if At= A; A vector x2 Rnis an eigenvector for A if x6= 0, and if there exists a number such that Ax= x. But for a special type of matrix, symmetric matrix, the eigenvalues are always real and the corresponding eigenvectors are always orthogonal. Similarly in characteristic different from 2, each diagonal element of a skew-symmetric matrix must be zero, since each is its own negative.. 6.11.9.1. Is there such thing as reasonable expectation for delivery time? Why does US Code not allow a 15A single receptacle on a 20A circuit? Then there exists an orthogonal matrix P for which PTAP is diagonal. Show that any eigenvector corresponding to α is orthogonal to any eigenvector corresponding to β. Vector x is a right eigenvector, vector y is a left eigenvector, corresponding to the eigenvalue λ, which is the same for both eigenvectors. The eigenvalues are real. Schur's Theorem: Every square matrix $A$ has a factorization of the form $A=QTQ^{\ast}$ where $Q$ is a unitary matrix and $T$ is upper triangular. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. Of course in the case of a symmetric matrix,AT=A, so this says that eigenvectors forAcorresponding to dierent eigenvalues must be orthogonal. If A is symmetric, then eigenvectors of A with distinct eigenvalues are or-thogonal. Proof of Orthogonal Eigenvectors¶. Eigenvalues of a triangular matrix. rev 2020.12.8.38143, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us, I don't understand your question. 6 0 obj To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Estimate $x^TBy$. An alternative approach to the proof (not using the inner-product method on the question you reference) is to use Schur's Theorem. This is an old question, and the proof is here. But even with repeated eigenvalue, this is still true for a symmetric matrix. x��\K�ǵ��K!�Yy?YEy� �6�GC{��I�F��9U]u��y�����Xn����;�yп������'�����/��R���=��Ǐ��oN�t�r�y������{��91�uFꓳ�����O��a��Ń�g��tg���T�Qx*y'�P���gy���O�9{��ǯ�ǜ��s�>��������o�G�w�(�>"���O��� Can you identify this restaurant at this address in 2011? That's just perfect. Recall that the vectors of a dot product may be reversed because of the commutative property of the Dot Product.Then because of the symmetry of matrix , we have the following equality relationship between two eigenvectors and the symmetric matrix. Now A = QΛQT because QT = Q–1. And I also do it for matrices. However, on the matter of eigenvalues not being distinct, eigenvectors with the same eigenvalue are certainly not always orthogonal. So the orthogonal vectors for are , and . How much theoretical knowledge does playing the Berlin Defense require? (20) 5 (iv) The column vectors of P are linearly independent eigenvectors of A, that are mutually orthogonal. An orthogonal matrix U satisfies, by definition, U T =U-1, which means that the columns of U are orthonormal (that is, any two of them are orthogonal and each has norm one). How can I add a few specific mesh (altitude-like level) curves to a plot? Rather, one can choose an orthogonal basis such that the matrix is diagonal in that basis. But suppose S is complex. We prove that eigenvalues of orthogonal matrices have length 1. My question is how about the repeated root? Making statements based on opinion; back them up with references or personal experience. Example of a symmetric matrix which doesn't have orthogonal eigenvectors. If you have two orthogonal eigenvectors with the same eigenvalue, then every linear combination of them is another eigenvector with that same eigenvalue, and is not generally orthogonal to the two you started with. 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Every symmetric matrix which does n't have orthogonal eigenvectors is used in multivariate analysis, where the sample matrices... | 2021-04-15T22:28:19 | {
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https://math.stackexchange.com/questions/712265/number-of-equivalence-relations-of-1-2-3 | # Number of Equivalence relations of $\{1,2,3\}$
Let $M$ be the set $\{1,2,3\}$. How many Equivalence relations $R \subset M \times M$ exists?
My idea is to count the disjoint partitions of M:
$K_1= \{\{1\},\{2\},\{3\}\}\Leftrightarrow\{(1,1),(2,2),(3,3)\}$
$K_2= \{\{1,2\},\{3\}\} \Leftrightarrow\{(1,1),(1,2),(2,1),(2,2),(3,3)\}$
$K_3= \{\{1,3\},\{2\}\}\Leftrightarrow\{(1,1),(1,3),(3,1),(2,2),(3,3)\}$
$K_4= \{\{1\},\{2,3\}\}\Leftrightarrow\{(1,1),(2,3),(3,2),(2,2),(3,3)\}$
$K_5=\{1,2,3\}\Leftrightarrow K_5=M^2$
So the answer would be $5$. Is this correct? Reflexivity and Symmetry are obvious, but how can i check the right side for Transitivity?
• Yes, your idea is correct. – MJD Mar 14 '14 at 16:15
• If we partition the set, all properties are obvious. – André Nicolas Mar 14 '14 at 16:15
• Maybe to obvios for me... Could you give me an example? I was trying to imagine this to myself in a 2 dimensional coordinate system, how can i see Transitivity? ( Reflexivity is just the line through the origin, and Symmetry a Reflection at the line through the origin , but Transitivity?) – fear.xD Mar 14 '14 at 16:21
• Every equivalence relation on $M$ induces a partition of $M$ and the inverse is also true: for a given partition declare two elements as equivalent iff they belong to the same set of that partition. – Michael Hoppe Mar 14 '14 at 16:39
If $X = \bigsqcup_i X_i$ is partition of set $X$ into disjoint $X_i$'s, then we can declare equivalence relationon $X$ by $x\sim y$ if $x,y\in X_i$ for some $i$. Symmetry and reflexivity is obvious, and as you asked Transitivity really means asking: If $x,y\in X_i$, and $y,z\in X_j$, then are $x,z\in X_k$ for some $k$? Well, $X_i$'s were disjoint, so $y\in X_i$ and $y\in X_j$ means $X_i = X_j$, hence $x,z\in X_i$. So yes.
So any partition of a set gives a equivalence relation.
On the other hand, any equivalence relation gives a partition of a set where each disjoint sets are just equivalence classes (i.e. collect together elements that are equivalent to each other). This needs slightly more careful checking, but is very believable.
So, in conclusion, giving a equivalence relation is actually just the same thing as giving a partition of a set, as you accurately noticed already.
• Transitivity would be like this: if $x, y \in X_i$ and $y, z \in X_j$, then $x, z \in X_k$ – dani_s Mar 14 '14 at 18:11
• @dani_s: yes, I should be more careful. It's been edited. – chriseur Mar 14 '14 at 18:44
• Ok, so for example consider $K_2:$ Transitivity of $(1,2) \sim (2,2) \Rightarrow (1,2)$? How about $(3,3)$? Just like $(3,3) \sim (3,3) \Rightarrow (3,3)$? – fear.xD Mar 14 '14 at 19:41
• What do you mean by $(1,2) \sim (2,2)$? Saying that $(1,2)\subset R\subset M\times M$ means that 1 is equivalent to 2, i.e. $1\sim 2$. – chriseur Mar 14 '14 at 19:44
• I mean $(1,2),(2,2) \in R \Rightarrow (1,2) \in R$. Would this be a possible way to show Transitivity individually? And for $(3,3)$ the only possibility would be $(3,3),(3,3) \in R \Rightarrow (3,3) \in R$? – fear.xD Mar 14 '14 at 19:59 | 2021-03-02T15:30:37 | {
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walter r July 10th, 2014 11:58 AM
Demonstration involving ideals of a ring
Please, check my solution to the following exercise:
"Let $I \subset A$ and $J \subset A$ be ideals of $A$. Show that:
a) $I \cap J$ is an ideal of A.
b) $I+J=\left\{x+y;x \in I, y \in J \right \}$ is an ideal of $A$.
a) since $I$ and $J$ are ideals, then $I \neq \varnothing$ and $J \neq \varnothing$. It implies that $I \cap J \neq \varnothing$.
$a,b \in I \cap J \rightarrow a,b \in I$ and $a,b \in J \rightarrow a+b \in I \cap J$ and $ab \in I \cap J$. Hence $I \cap J$ is an ideal.
b) $I \neq \varnothing$ and $J \neq \varnothing \rightarrow I+J \neq \varnothing$. $a,b \in I+J \rightarrow a= x+y; b= v+w; x,v \in I; y,w \in J$. But then $a+b =(x+v)+(y+w) \in I + J$, because $(x+v) \in I$ and $(y+w) \in J$.
Additionaly, if $a,b \in I+J$ then $ab= xv+xw+yv+yw$ with $xv \in I; yw \in J; xw \in I \cap J; yv \in I \cap J$. Then $ab=(xv+xw)+(yv+yw) \in I+J$. Hence $I+J$ is an ideal.
Thanks!
Deveno July 10th, 2014 12:55 PM
The intersection of two non-null sets is not necessarily non-null. For example, if $A = \{1\}$ and $B = \{2\}$, then $A \cap B = \emptyset$.
However, there is always ONE element of $A$ guaranteed to be in $I$ AND $J$. Can you think of what it is?
Again, just because two elements of two sets are in both of them, does not guarantee that the sum of these two elements is in EITHER set. Suppose we take:
$A = \{1,2,4\}$ and $B = \{1,2,5\}$.
Then $1,2 \in A \cap B$, but $1 + 2 = 3$ is not in EITHER set.
You are failing to invoke the proper properties of ideals in your proof. You're just stating that what you HOPE to be true IS true. That just won't do.
THIS is what you have to prove for (a):
1. $I \cap J$ is an additive subgroup of $A$.
2. If $a \in A$, and $x \in I \cap J$, that $ax,xa \in I \cap J$.
I'll show you how a proof of (2) starts.
Let $a \in A$, and $x \in I \cap J$.
Since $I \cap J \subseteq I$, we have $x \in I$. Since $I$ is an ideal, $ax \in I$.
What do you suppose comes next?
walter r July 11th, 2014 06:28 AM
Good morning, Deveno! Thanks for writing. Concerning item a:
Quote:
The intersection of two non-null sets is not necessarily non-null. For example, if A={1} and B={2}, then A∩B=∅. However, there is always ONE element of A guaranteed to be in I AND J. Can you think of what it is?
(1) Do you mean $0 \in I, 0\in J \rightarrow 0 \in I \cap J$?
Quote:
Again, just because two elements of two sets are in both of them, does not guarantee that the sum of these two elements is in EITHER set.
At this point I disagree.
(2) $a,b \in I \cap J \rightarrow a,b \in I$ and $a,b \in J$. But $I,J$ are ideals, therefore $a+b \in I$ and $a+b \in J$. Hence $a+b \in I \cap J$. This is not true for any set, but it is true for an ideal.
After reading your post I redo the other step the following way:
(3) $a \in I \cap J, n \in A \rightarrow a \in I, a \in J, n \in A \rightarrow na \in I, na \in J$ (by hypothesis that both I and J are ideals)$\rightarrow na \in I \cap J$.
(1),(2),(3) together imply $I \cap J$ is an ideal, similarly for item b. Do you think now it is convincent?
Deveno July 11th, 2014 07:19 AM
Yes, that is what was missing: you need to invoke the CLOSURE properties of ideals. This may have been what you were thinking, but until you write it down, no one else knows that.
All times are GMT -8. The time now is 09:25 AM. | 2019-08-19T17:25:58 | {
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https://physics.stackexchange.com/questions/265562/electric-potential-produced-by-hollow-sphere | # Electric Potential produced by Hollow Sphere
Let there be a hollow sphere (Throughout the question we will ignore the thickness of the sphere) which is positively uniformly charged,q of radius, R. Suppose there to be a test positive charge, q' at point P which is r distance away from the hollow sphere(r>R, P is outside of hollow sphere).
Now I have an doubt on evaluating Electric Potential,V at point P.
Both source charge,q and test charge,q' are +ve, so source charge will repel the test charge which will make the charge, q' to go to infinity.I have to apply external force, F' to displace the charge, q' from infinity to point P against the equal but repulsive force, F.In short, F' is displacing the q' from infinity to P.
So External Force, F' will do +ve work and Repulsive Force, F will do -ve work. This way the Electric Field,E produced by hollow sphere is Opposite to Displacement of test charge(i.e.,θ=180).
Now we know that E=k(q/(r^2))
[Please Notice that infinity is the lower limit of integration]
Also V=-∫r(E dr cosθ)=-∫r(E dr cos 180)=∫r(E dr)
At last I ended up to V=- k(q/r). I know that V=k(q/r).
I just want to know why I am having this shitty negative sign in my equation.
• – Qmechanic Jul 1 '16 at 12:22
• Sorry but I think your question is not about physics but about an error in your calculation. An answer is not of benefit to the wider community. – sammy gerbil Jul 1 '16 at 17:48
Pick one.
Since your external force is opposite in direction to the force of the sphere, and you've already put the cos(180) in there, there is no need for the extra negative sign in front.
• I am not sure this is correct. The negative sign does not come from the cosine, but from the relation $V=-W$. – Steeven Oct 30 '17 at 23:34
• The relation $V=-W$ comes from the fact that $cos(180)=-1$. Look at this – GeeJay Nov 1 '17 at 12:25
The force exerted on a unit positive charge is $\dfrac {kq}{r^2} (+\hat r)$ and so an external force $\dfrac {kq}{r^2} (-\hat r)$ must be applied to move the charge.
If the step is $d\vec r$ then the work done by the external force in moving that step is $\dfrac {kq}{r^2} (-\hat r) \cdot d\vec r = -\dfrac {kq}{r^2} dr$.
You have moved from $\vec r$ to $\vec r + d \vec r$.
Now this must be correct because it says that if $dr$ is positive ie moving away from the positive charge the work done by an external force is negative ie the potential has dropped.
If you move towards the charge then $dr$ is negative and the work done by the external force is positive ie the potential has increased.
The limits of integration will define which way you actually move.
So the potential $\displaystyle V(r) = \int _\infty ^ r-\dfrac {kq}{r^2} dr = + \dfrac {kq}{r}$
Later in answer to @mritunjay 's question.
For many people the sign of $dr$ is conceptually difficult.
Suppose that you were asked to do the following integration $\displaystyle \int _{-1} ^ 0 x\;dx$ which when evaluated is $+\frac 1 2$.
If you look at the left-hand graph below you have found $\frac 1 2 \Delta y \Delta x$ with $\Delta y = y_{\text{final}} - y_{\text{initial}} = -1 -0 = -1$ and $\Delta x = x_{\text{final}} - x_{\text{initial}} = -1 -0 = -1$.
Noting that $x$ is decreasing would you have evaluated the integral $\displaystyle \int _{-1} ^ 0 x\;(-dx)$ instead?
The right-hand graph and you are evaluating the work done $F_{\text{external}} \Delta r$.
$F_{\text{external}}$ is negative and with the limits of integration going from $r=\infty$ to $r=r$, $\Delta x$ will also be negative and so you will get a positive result for the work done by the external force.
Note that if the work done by the electric field $E \Delta r$ was evaluated then integration would yield the expected negative answer as $E$ is positive.
• if dr is negative I think BEFORE INTEGRATING u will get V=(kq/(r^2)) dr. But when u integrate this equation with limits u will get V= -(kq/r). If u are sure about ur last step can u please edit ur answer amd elaborate it. Sorry for any inconvenience – Mritun Jay Jul 1 '16 at 7:31
• @MritunJay I have expanded my answer to show that the integrations limits take care of sign of $dr$. – Farcher Jul 1 '16 at 12:12
• I believe you are answering for the wrong force in this answer here. Although the OP mentions both the external force and the electric field force, he is trying to calculate the work done by the electric field in the steps he shows. The work done by the electric field is negative. In your answer here, you get a positive work because you find the work done by the external force. That is why you can use $V=W$. But the OP actually asks about the work done by the field, so the relation is opposite: $V=-W$. – Steeven Oct 30 '17 at 23:44 | 2019-07-22T08:23:35 | {
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https://fr.mathworks.com/help/matlab/ref/equilibrate.html | # equilibrate
Matrix scaling for improved conditioning
## Syntax
``[P,R,C] = equilibrate(A)``
``[P,R,C] = equilibrate(A,outputForm)``
## Description
example
````[P,R,C] = equilibrate(A)` permutes and rescales matrix `A` such that the new matrix ```B = R*P*A*C``` has a diagonal with entries of magnitude 1, and its off-diagonal entries are not greater than 1 in magnitude.```
````[P,R,C] = equilibrate(A,outputForm)` returns the outputs `P`, `R`, and `C` in the form specified by `outputForm`. For example, you can specify `outputForm` as `"vector"` to return the outputs as column vectors.```
## Examples
collapse all
Equilibrate a matrix with a large condition number to improve the efficiency and stability of a linear system solution with the iterative solver `gmres`.
Load the `west0479` matrix, which is a real-valued 479-by-479 sparse matrix. Use `condest` to calculate the estimated condition number of the matrix.
```load west0479 A = west0479; c1 = condest(A)```
```c1 = 1.4244e+12 ```
Try to solve the linear system $\mathrm{Ax}=\mathit{b}$ using `gmres` with 450 iterations and a tolerance of `1e-11`. Specify five outputs so that `gmres` returns the residual norms of the solution at each iteration (using `~` to suppress unneeded outputs). Plot the residual norms in a semilog plot. The plot shows that `gmres` is not able to achieve a reasonable residual norm, and so the calculated solution for $\mathit{x}$ is not reliable.
```b = ones(size(A,1),1); tol = 1e-11; maxit = 450; [x,flx,~,~,rvx] = gmres(A,b,[],tol,maxit); semilogy(rvx) title('Residual Norm at Each Iteration')```
Use `equilibrate` to permute and rescale `A`. Create a new matrix `B = R*P*A*C`, which has a better condition number and diagonal entries of only 1 and -1.
```[P,R,C] = equilibrate(A); B = R*P*A*C; c2 = condest(B)```
```c2 = 5.1036e+04 ```
Using the outputs returned by `equilibrate`, you can reformulate the problem $\mathrm{Ax}=\mathit{b}$ into $\mathrm{By}=\mathit{d}$, where $\mathit{B}=\mathrm{RPAC}$ and $\mathit{d}=\mathrm{RPb}$. In this form you can recover the solution to the original system with $\mathit{x}=\mathrm{Cy}$. The recovered solution, however, might not have the desired residual error for the original system. See Rescaling to Solve a Linear System for details.
Use `gmres` to solve $\mathrm{By}=\mathit{d}$ for $\mathit{y}$, and then replot the residual norms at each iteration. The plot shows that equilibrating the matrix improves the stability of the problem, with `gmres` converging to the desired tolerance of `1e-11` in fewer than 200 iterations.
```d = R*P*b; [y,fly,~,~,rvy] = gmres(B,d,[],tol,maxit); hold on semilogy(rvy) legend('Original', 'Equilibrated', 'Location', 'southeast') title('Relative Residual Norms (No Preconditioner)') hold off```
Improve Solution with Preconditioner
After you obtain the matrix `B`, you can improve the stability of the problem even further by calculating a preconditioner for use with `gmres`. The numerical properties of `B` are better than those of the original matrix `A`, so you should use the equilibrated matrix to calculate the preconditioner.
Calculate two different preconditioners with `ilu`, and use these as inputs to `gmres` to solve the problem again. Plot the residual norms at each iteration on the same plot as the equilibrated norms for comparison. The plot shows that calculating preconditioners with the equilibrated matrix greatly increases the stability of the problem, with `gmres` achieving the desired tolerance in fewer than 30 iterations.
```semilogy(rvy) hold on [L1,U1] = ilu(B,struct('type','ilutp','droptol',1e-1,'thresh',0)); [yp1,flyp1,~,~,rvyp1] = gmres(B,d,[],tol,maxit,L1,U1); semilogy(rvyp1) [L2,U2] = ilu(B,struct('type','ilutp','droptol',1e-2,'thresh',0)); [yp2,flyp2,~,~,rvyp2] = gmres(B,d,[],tol,maxit,L2,U2); semilogy(rvyp2) legend('No preconditioner', 'ILUTP(1e-1)', 'ILUTP(1e-2)') title('Relative Residual Norms with ILU Preconditioner (Equilibrated)') hold off```
Create a 6-by-6 magic square matrix, and then use `equilibrate` to factor the matrix. By default, `equilibrate` returns the permutation and scaling factors as matrices.
`A = magic(6)`
```A = 6×6 35 1 6 26 19 24 3 32 7 21 23 25 31 9 2 22 27 20 8 28 33 17 10 15 30 5 34 12 14 16 4 36 29 13 18 11 ```
`[P,R,C] = equilibrate(A)`
```P = 6×6 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 1 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 ```
```R = 6×6 0.1852 0 0 0 0 0 0 0.1749 0 0 0 0 0 0 0.1909 0 0 0 0 0 0 0.1588 0 0 0 0 0 0 0.1793 0 0 0 0 0 0 0.1966 ```
```C = 6×6 0.1799 0 0 0 0 0 0 0.1588 0 0 0 0 0 0 0.1588 0 0 0 0 0 0 0.2422 0 0 0 0 0 0 0.2066 0 0 0 0 0 0 0.2035 ```
Construct the equilibrated matrix `Bmatrix` using the matrix factors `P`, `R`, and `C`.
`Bmatrix = R*P*A*C`
```Bmatrix = 6×6 1.0000 0.1471 1.0000 0.5385 0.5358 0.6031 0.1259 1.0000 0.8056 0.5509 0.6506 0.3916 0.2747 0.8485 1.0000 0.7859 0.3943 0.5825 1.0000 0.0252 0.1513 1.0000 0.6233 0.7754 1.0000 0.2562 0.0569 0.9553 1.0000 0.7295 0.1061 0.9988 0.2185 1.0000 0.9341 1.0000 ```
Now, specify the `"vector"` option to return the outputs as vectors. With large input matrices, returning the outputs as vectors can save memory and improve efficiency.
`[p,r,c] = equilibrate(A,"vector")`
```p = 6×1 5 6 4 1 3 2 ```
```r = 6×1 0.1852 0.1749 0.1909 0.1588 0.1793 0.1966 ```
```c = 6×1 0.1799 0.1588 0.1588 0.2422 0.2066 0.2035 ```
Construct the equilibrated matrix `Bvector` using the vector outputs `p`, `r`, and `c`.
`Bvector = r.*A(p,:).*c'`
```Bvector = 6×6 1.0000 0.1471 1.0000 0.5385 0.5358 0.6031 0.1259 1.0000 0.8056 0.5509 0.6506 0.3916 0.2747 0.8485 1.0000 0.7859 0.3943 0.5825 1.0000 0.0252 0.1513 1.0000 0.6233 0.7754 1.0000 0.2562 0.0569 0.9553 1.0000 0.7295 0.1061 0.9988 0.2185 1.0000 0.9341 1.0000 ```
Compare `Bvector` and `Bmatrix`. The result indicates they are equal.
`norm(Bmatrix - Bvector)`
```ans = 0 ```
## Input Arguments
collapse all
Input matrix, specified as a square matrix. `A` can be dense or sparse, but must be structurally nonsingular, as determined by `sprank`.
When `A` is sparse, the outputs `P`, `R`, and `C` are also sparse.
Data Types: `single` | `double`
Complex Number Support: Yes
Output format, specified as `"vector"` or `"matrix"`. This option allows you to specify whether the outputs `P`, `R`, and `C` are returned as column vectors or as matrices:
• `"matrix"``P`, `R`, and `C` are matrices such that ```B = R*P*A*C```.
• `"vector"``P`, `R`, and `C` are column vectors such that ```B = R.*A(P,:).*C'```.
Example: `[P,R,C] = equilibrate(A,"vector")` returns `P`, `R`, and `C` as vectors.
## Output Arguments
collapse all
Permutation information, returned as a matrix or vector. `P*A` (or `A(P,:)` with the `"vector"` option) is the permutation of `A` that maximizes the absolute value of the product of its diagonal elements.
Row scaling, returned as a diagonal matrix or vector. The nonzero entries in `R` and `C` are real and positive.
Column scaling, returned as a diagonal matrix or vector. The nonzero entries in `R` and `C` are real and positive.
collapse all
### Rescaling to Solve a Linear System
For linear system solutions `x = A\b`, the condition number of `A` is important for accuracy and efficiency of the calculation. `equilibrate` can improve the condition number of `A` by rescaling the basis vectors. This effectively forms a new coordinate system that both `b` and `x` can be expressed in.
`equilibrate` is most useful when the scales of the `b` and `x` vectors in the original system ```x = A\b``` are irrelevant. However, if the scales of `b` and `x` are relevant, then using `equilibrate` to rescale `A` only for the linear system solve is not recommended. The obtained solution does not generally yield a small residual for the original system, even if expressed using the original basis vectors.
## References
[1] Duff, I. S., and J. Koster. “On Algorithms For Permuting Large Entries to the Diagonal of a Sparse Matrix.” SIAM Journal on Matrix Analysis and Applications 22, no. 4 (January 2001): 973–96.
## Version History
Introduced in R2019a
expand all | 2022-09-29T05:46:33 | {
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https://www.physicsforums.com/threads/prove-the-sum-rule-for-limits.762990/ | Prove the Sum Rule for Limits
1. Jul 23, 2014
Tsunoyukami
Prove the Sum Rule for Limits
$$\lim_{x\to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x) = L + M$$
Proof
Assume the following:
$$\lim_{x \to a} f(x) = L, \space\lim_{x \to a} g(x) = M$$
Then, by definition
$\forall \epsilon_1 > 0, \exists \delta_1 > 0$ such that $0<|x-a|<\delta_1 \implies |f(x)-L|<\epsilon_1$
and
$\forall \epsilon_2 >0, \exists \delta_2 > 0$ such that $0<|x-a|<\delta_2 \implies |g(x)-M|<\epsilon_2$
Choose $\delta = \min(\delta_1, \delta_2)$.
Then $0<|x-a|<\delta \implies |f(x)-L|<\epsilon_1$ and $|g(x)-M|<\epsilon_2$.
Notice $|[f(x) + g(x)] - [L+M]| = |[f(x)-L] + [g(x)-M]| \leq |f(x)-L| + |g(x)-M|$ by the triangle inequality. Then $|[f(x) + g(x)] - [L+M]| \leq |f(x)-L| + |g(x)-M| < \epsilon_1 + \epsilon_2 = \epsilon$.
Then $|[f(x) + g(x)] - [L+M]| < \epsilon$. Therefore we conclude
$$\lim_{x\to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x) = L + M$$
I was comparing my proof (the one above) to the proof offered on page 93 of James Stewart's Calculus (6e) and noticed that instead of $\epsilon_1$ and $\epsilon_2$ he uses $\frac{\epsilon}{2}$ for each of these so that their sum is $\epsilon$.
Is there anything incorrect with my method? I figure that my approach with two different possible values of $\epsilon_i$ for each of the limits is just a more generalized way of saying the same thing - but the other proofs I've looked at online all use $\frac{\epsilon}{2}$ - is there any reason for this?
2. Jul 23, 2014
LCKurtz
If you had started with a statement of what you want to prove, you would have written:
To prove: Given $\epsilon > 0$, show there is $\delta > 0$ such that if $0<|x-a|< \delta$ then $|(f(x)+g(x)) - (L+M)| < \epsilon$. And that is why you would pick the intermediate values in your argument so it all comes out less than $\epsilon$ at the end.
Your proof should start with "Suppose $\epsilon > 0$" Then give your argument.
3. Jul 23, 2014
Tsunoyukami
Oh, I see so it would be something like this:
Proof
Let $\epsilon>0$ be given. We must find $\delta > 0$ such that $0<|x-a|<\delta \implies |f(x)+g(x) -(L+M)|<\epsilon$. By applying the triangle inequality we can write $|f(x)+g(x) - (L+M)| \leq |f(x)-L| + |g(x)-M|$.
I will pause the proof here momentarily. The proofs I have seen have continued by letting each of these terms be less than $\frac{\epsilon}{2}$ - my question is whether or not I could let one of them be less than $\frac{\epsilon}{3}$ and the other be less than $\frac{2\epsilon}{3}$ - the sum of these terms is still $\epsilon$ so this should still be valid correct? I will attempt to complete the proof using these values.
Getting back to the main body of the proof:
Let $|f(x) - L|<\frac{\epsilon}{3}$ and $|g(x)-M|<\frac{2\epsilon}{3}$. Since $\epsilon>0$ it follows that both $\frac{\epsilon}{3}>0$ and $\frac{2\epsilon}{3}>0$. Since both the limit of $f(x)$ and $g(x)$ exist as $x$ approaches $a$ there exist $\delta_1$ and $\delta_2$ satisfying the following conditions:
$$0<|x-a|<\delta_1 \implies |f(x)-L|<\frac{\epsilon}{3}$$
$$0<|x-a|<\delta_2 \implies |g(x)-M|<\frac{2\epsilon}{3}$$
Choose $\delta=\min(\delta_1, \delta_2)$. If $0<|x-a|<\delta$ then $0<|x-a|<\delta_1$ and $0<|x-a|<\delta_2$ so $|f(x)-L|<\frac{\epsilon}{3}$ and $|g(x)-M|<\frac{2\epsilon}{3}$.
Then $|f(x)+g(x) - (L+M)| \leq |f(x)-L| + |g(x)-M| < \frac{\epsilon}{3} + \frac{2\epsilon}{3} = \epsilon$.
This completes the proof.
What was bothering me was that every proof I saw relied upon splitting the two terms equally into $\frac{\epsilon}{2}$ when I figured that it should be possible to split $\epsilon$ into two uneven terms.
Does the above constitute a valid proof?
4. Jul 24, 2014
LCKurtz
Yes. It doesn't matter how you break it up as long as the total parts add to less than $\epsilon$. | 2017-08-23T05:24:13 | {
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https://comet-fenics.readthedocs.io/en/latest/demo/tips_and_tricks/computing_reactions.html | The corresponding files can be obtained from:
# Computing consistent reaction forces¶
One often needs to compute a resulting reaction force on some part of the boundary as post-processing of a mechanical resolution.
Quite often, such a reaction will be computed using the stress field associated with the computed displacement. We will see that this may lead to slight inaccuracies whereas another more consistent approach using the virtual work principle is more consistent.
## A cantilever beam problem¶
We reuse here a simple 2D small strain elasticity script of a rectangular domain of dimensions $$L\times H$$ representing a cantilever beam clamped on the left hand-side and under uniform body forces $$\boldsymbol{f}=(f_x, f_y)$$. $$P^2$$ Lagrange elements are used for the displacement discretization.
For the sake of illustration, we are interested in computing the horizontal and vertical reaction forces $$R_x$$ and $$R_y$$ on the left boundary as well as the resulting moment $$M_z$$ around the out-of-plane direction. In the present simple case using global balance equations, they are all given explicitly by:
\begin{split}\begin{align*} R_x &= \int_{x=0} \boldsymbol{T}\cdot \boldsymbol{e}_x \dS = \int_{x=0} (-\sigma_{xx}) dS = -f_x \cdot L \cdot H \\ R_y &= \int_{x=0} \boldsymbol{T}\cdot \boldsymbol{e}_y \dS = \int_{x=0} (-\sigma_{xy}) dS = -f_y \cdot L \cdot H\\ M_z &= \int_{x=0} (\vec{\boldsymbol{OM}}\times \boldsymbol{T})\cdot \boldsymbol{e}_z \dS = \int_{x=0} (y \sigma_{xx}) dS = f_y \cdot \dfrac{L^2}{2} \cdot H \end{align*}\end{split}
[6]:
from dolfin import *
import matplotlib.pyplot
%matplotlib notebook
L = 5.
H = 1.
Nx = 20
Ny = 5
mesh = RectangleMesh(Point(0., 0.), Point(L, H), Nx, Ny, "crossed")
plot(mesh)
E = Constant(1e5)
nu = Constant(0.3)
mu = E/2/(1+nu)
lmbda = E*nu/(1+nu)/(1-2*nu)
def eps(v):
def sigma(v):
return lmbda*tr(eps(v))*Identity(2) + 2.0*mu*eps(v)
fx = 0.1
fy = -1.
f = Constant((fx, fy))
V = VectorFunctionSpace(mesh, 'Lagrange', degree=2)
du = TrialFunction(V)
u_ = TestFunction(V)
a = inner(sigma(du), eps(u_))*dx
l = inner(f, u_)*dx
def left(x, on_boundary):
return near(x[0], 0.)
facets = MeshFunction("size_t", mesh, 1)
AutoSubDomain(left).mark(facets, 1)
ds = Measure("ds", subdomain_data=facets)
bc = DirichletBC(V, Constant((0., 0.)), facets, 1)
u = Function(V, name="Displacement")
solve(a == l, u, bc)
plot(u[1])
[6]:
<matplotlib.tri.tricontour.TriContourSet at 0x7fcc98623a90>
## First method: using the post-processed stress¶
The first, and most widely used, method for computing the above reactions relies on the stress field computed from the obtained displacement sigma(u) and perform assemble over the left boundary (measure ds(1)). Unfortunately, this procedure does not ensure an exact computation as seen below. Indeed, the stress field, implicitly known only at the quadrature points only is extended to the structure boundary and does not satisfy global equilibrium anymore.
[11]:
x = SpatialCoordinate(mesh)
print("Horizontal reaction Rx = {}".format(
assemble(-sigma(u)[0, 0]*ds(1))))
print(" (analytic = {})".format(-L*H*fx))
print("-"*50)
print("Vertical reaction Ry = {}".format(
assemble(-sigma(u)[0, 1]*ds(1))))
print(" (analytic = {})".format(-L*H*fy))
print("-"*50)
print("Bending moment Mz = {}".format(assemble(-x[1]*sigma(u)[0, 0]*ds(1))))
print(" (analytic = {})".format(H*L**2/2*fy))
print("-"*50)
print("\n")
Horizontal reaction Rx = -0.48578099336058145
(analytic = -0.5)
--------------------------------------------------
Vertical reaction Ry = 4.514399579992231
(analytic = 5.0)
--------------------------------------------------
Bending moment Mz = -12.057524729246653
(analytic = -12.5)
--------------------------------------------------
## Second method: using the work of internal forces¶
The second approach relies on the virtual work principle (or weak formulation) which writes in the present case:
$\int_\Omega \bsig(\boldsymbol{u}):\nabla^s \boldsymbol{v} \dx =\int_\Omega \boldsymbol{f}\cdot\boldsymbol{v} \dx + \int_{\partial \Omega_N} \boldsymbol{T}\cdot\boldsymbol{v}\dS + \int_{\partial \Omega_D} \boldsymbol{T}\cdot\boldsymbol{v}\dS \quad \forall \boldsymbol{v}\in V$
in which $$\boldsymbol{v}$$ does not necessarily satisfy the Dirichlet boundary conditions on $$\partial \Omega_D$$.
The solution $$u$$ is precisely obtained by enforcing the Dirichlet boundary conditions on $$\boldsymbol{v}$$ such that:
$\int_\Omega \bsig(\boldsymbol{u}):\nabla^s \boldsymbol{v} \dx =\int_\Omega \boldsymbol{f}\cdot\boldsymbol{v} \dx + \int_{\partial \Omega_N} \boldsymbol{T}\cdot\boldsymbol{v}\dS \quad \forall \boldsymbol{v}\in V \text{ and } \boldsymbol{v}=0 \text{ on }\partial \Omega_D$
Defining the residual:
$Res(v) = \int_\Omega \bsig(\boldsymbol{u}):\nabla^s \boldsymbol{v} \dx - \int_\Omega \boldsymbol{f}\cdot\boldsymbol{v} \dx - \int_{\partial \Omega_N} \boldsymbol{T}\cdot\boldsymbol{v}\dS = a(\boldsymbol{u}, \boldsymbol{v}) -\ell(\boldsymbol{v})$
we have that $$Res(v)= 0$$ if $$\boldsymbol{v}=0$$ on $$\partial \Omega_D$$.
Now, if $$\boldsymbol{v}\neq0$$ on $$\partial \Omega_D$$, say, for instance, $$\boldsymbol{v}=(1,0)$$ on $$\partial \Omega_D$$, we have that:
$Res(v) = \int_{\partial \Omega_D} \boldsymbol{T}\cdot\boldsymbol{v}\dS = \int_{\partial \Omega_D} \boldsymbol{T}_x\dS = \int_{\partial \Omega_D} -\sigma_{xx}\dS = R_x$
Similarly, we obtain the vertical reaction $$R_y$$ by considering $$\boldsymbol{v}=(0,1)$$ and the bending moment $$M_z$$ by considering $$\boldsymbol{v}=(y,0)$$.
As regards implementation, the residual is defined using the action of the bilinear form on the displacement solution: residual = action(a, u) - l. We then define boundary conditions on the left boundary and apply them to an empty Function to define the required test field $$v$$. We observe that the computed reactions are now exact.
[10]:
residual = action(a, u) - l
v_reac = Function(V)
bcRx = DirichletBC(V.sub(0), Constant(1.), facets, 1)
bcRy = DirichletBC(V.sub(1), Constant(1.), facets, 1)
bcMz = DirichletBC(V.sub(0), Expression("x[1]", degree=1), facets, 1)
bcRx.apply(v_reac.vector())
print("Horizontal reaction Rx = {}".format(assemble(action(residual, v_reac))))
print(" (analytic = {})".format(-L*H*fx))
print("-"*50)
v_reac.interpolate(Constant((0., 0.)))
bcRy.apply(v_reac.vector())
print("Vertical reaction Ry = {}".format(assemble(action(residual, v_reac))))
print(" (analytic = {})".format(-L*H*fy))
print("-"*50)
v_reac.interpolate(Constant((0., 0.)))
bcMz.apply(v_reac.vector())
print("Bending moment Mz = {}".format(assemble(action(residual, v_reac))))
print(" (analytic = {})".format(H*L**2/2*fy))
print("-"*50)
Horizontal reaction Rx = -0.49999999999492606
(analytic = -0.5)
--------------------------------------------------
Vertical reaction Ry = 5.000000000715704
(analytic = 5.0
--------------------------------------------------
Bending moment Mz = -12.750000002550555
analytic = -12.5
-------------------------------------------------- | 2022-01-25T23:12:05 | {
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https://stats.stackexchange.com/questions/88628/distribution-of-the-ratio-of-dependent-chi-square-random-variables | # Distribution of the ratio of dependent chi-square random variables
Assume that $X = X_1 + X_2+\cdots+ X_n$ where $X_i \sim N(0,\sigma^2)$ are independent.
My question is, what distribution does
$$Z = \frac{X^2}{X_1^2 + X_2^2 + \cdots + X_n^2}$$
follow? I know from here that the ratio of two chi-squared random variables expressed as $\frac{W}{W + Y}$ follows a Beta distribution. I think that this assumes independence between $W$ and $Y$. In my case though, the denominator of $Z$ contains the components of $X$ squared.
I think $Z$ must also follow a variation of the Beta distribution but I am not sure. And if this assumption is correct, I don't know how to prove it.
• Because the distribution of the denominator is invariant under rotations, you can rotate $X$ to equal $\sqrt{n}X_1$, which reduces your question to something familiar :-). – whuber Mar 3 '14 at 22:23
• I'm pretty sure @whuber means exactly what was typed there. When you say 'nominator' do you mean 'numerator'? – Glen_b Mar 4 '14 at 4:34
• When you rotate anything you (by definition) preserve its length. Therefore the variance of any rotated version of $X$ must equal the variance of $X$, which is $1+1+\cdots+1=n$: that's where the $\sqrt{n}$ term comes from. – whuber Mar 4 '14 at 15:37
• @whuber Your answer seems very interesting indeed but I have some doubts about it. When you say that I can rotate $X$ to become equal to $\sqrt nX_1$, this basically means that I can rewrite the numerator of $Z$ as $nX_1^2$ and consequently, $Z$ itself turns into $n\frac{X_1^2}{X_1^2+X_2^2+\cdots+X_n^2}$. Now, if I assume $W=X_1^2$ and $Y=X_2^2+\cdots+X_n^2$ and since $W$ and $Y$ are independent, I can assume that $Z=n\frac{W}{W+Y}$ has a $\beta$ distribution and so forth. Am I getting your point up to now? So, here is my confusion. Before using the concept of rotational invariance and modifyi – ssah Mar 4 '14 at 17:45
• @ssah You err in your application of my reasoning: without the $X_1^2$ in the denominator, its distribution is no longer invariant to arbitrary rotations of $(X_1,\ldots, X_n),$ and so the conclusions no longer hold. – whuber Mar 5 '14 at 17:04
This post elaborates on the answers in the comments to the question.
Let $X = (X_1, X_2, \ldots, X_n)$. Fix any $\mathbf{e}_1\in\mathbb{R}^n$ of unit length. Such a vector may always be completed to an orthonormal basis $(\mathbf{e}_1, \mathbf{e}_2, \ldots, \mathbf{e}_n)$ (by means of the Gram-Schmidt process, for instance). This change of basis (from the usual one) is orthogonal: it does not change lengths. Thus the distribution of
$$\frac{(\mathbf{e}_1\cdot X)^2}{||X||^2}=\frac{(\mathbf{e}_1\cdot X)^2}{X_1^2 + X_2^2 + \cdots + X_n^2}$$
does not depend on $\mathbf{e}_1$. Taking $\mathbf{e}_1 = (1,0,0,\ldots, 0)$ shows this has the same distribution as
$$\frac{X_1^2}{X_1^2 + X_2^2 + \cdots + X_n^2}.\tag{1}$$
Since the $X_i$ are iid Normal, they may be written as $\sigma$ times iid standard Normal variables $Y_1, \ldots, Y_n$ and their squares are $\sigma^2$ times $\Gamma(1/2)$ distributions. Since the sum of $n-1$ independent $\Gamma(1/2)$ distributions is $\Gamma((n-1)/2)$, we have determined that the distribution of $(1)$ is that of
$$\frac{\sigma^2 U}{\sigma^2 U + \sigma^2 V} = \frac{U}{U+V}$$
where $U = X_1^2/\sigma^2 \sim \Gamma(1/2)$ and $V = (X_2^2 + \cdots + X_n^2)/\sigma^2 \sim \Gamma((n-1)/2)$ are independent. It is well known that this ratio has a Beta$(1/2, (n-1)/2)$ distribution. (Also see the closely related thread at Distribution of $XY$ if $X \sim$ Beta$(1,K-1)$ and $Y \sim$ chi-squared with $2K$ degrees.)
Since $$X_1 + \cdots + X_n = (1,1,\ldots,1)\cdot (X_1, X_2, \cdots, X_n) = \sqrt{n}\,\mathbf{e}_1\cdot X$$
for the unit vector $\mathbf{e}_1=(1,1,\ldots,1)/\sqrt{n}$, we conclude that $Z$ is $(\sqrt{n})^2 = n$ times a Beta$(1/2, (n-1)/2)$ variate. For $n\ge 2$ it therefore has density function
$$f_Z(z) = \frac{n^{1-n/2}}{B\left(\frac{1}{2}, \frac{n-1}{2}\right)} \sqrt{\frac{(n-z)^{n-3}}{z}}$$
on the interval $(0,n)$ (and otherwise is zero).
As a check, I simulated $100,000$ independent realizations of $Z$ for $\sigma=1$ and $n=2,3,10$, plotted their histograms, and superimposed the graph of the corresponding Beta density (in red). The agreements are excellent.
Here is the R code. It carries out the simulation by means of the formula sum(x)^2 / sum(x^2) for $Z$, where x is a vector of length n generated by rnorm. The rest is just looping (for, apply) and plotting (hist, curve).
for (n in c(2, 3, 10)) {
z <- apply(matrix(rnorm(n*1e5), nrow=n), 2, function(x) sum(x)^2 / sum(x^2))
hist(z, freq=FALSE, breaks=seq(0, n, length.out=50), main=paste("n =", n), xlab="Z")
curve(dbeta(x/n, 1/2, (n-1)/2)/n, add=TRUE, col="Red", lwd=2)
} | 2019-10-18T14:19:29 | {
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https://math.stackexchange.com/questions/1299957/show-that-frac-sin-x-cos-3x-frac-sin-3x-cos-9x-frac-sin-9x/1300557 | # Show that $\frac {\sin x} {\cos 3x} + \frac {\sin 3x} {\cos 9x} + \frac {\sin 9x} {\cos 27x} = \frac 12 (\tan 27x - \tan x)$ [duplicate]
The question asks to prove that - $$\frac {\sin x} {\cos 3x} + \frac {\sin 3x} {\cos 9x} + \frac {\sin 9x} {\cos 27x} = \frac 12 (\tan 27x - \tan x)$$
I tried combining the first two or the last two fractions on the L.H.S to allow me to use the double angle formula and get $\sin 6x$ or $\sin 18x$ but that did not help at all.
I'm pretty sure that if I express everything in terms of $x$, the answer will ultimately appear but I'm also certain that there must be another simpler way.
## marked as duplicate by lab bhattacharjee trigonometry StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Mar 20 '16 at 14:07
• Does the identity tan x = sin 2x/(1+cos 2x) help? – Jeffrey L. May 26 '15 at 21:33
In general, we have $$\sum_{k=1}^n \dfrac{\sin\left(3^{k-1}x \right)}{\cos\left(3^kx\right)} = \dfrac{\tan\left(3^nx\right)-\tan(x)}2$$ We can prove this using induction. $n=1$ is trivial. All we need to make use of is that $$\tan(3t) = \dfrac{3\tan(t)-\tan^3(t)}{1-3\tan^2(t)}$$ For the inductive step, we have $$\sum_{k=1}^n \dfrac{\sin\left(3^{k-1}x \right)}{\cos\left(3^kx\right)} = \dfrac{\tan\left(3^nx\right)-\tan(x)}2$$ Adding the last term, we obtain $$\sum_{k=1}^{n+1} \dfrac{\sin\left(3^{k-1}x \right)}{\cos\left(3^kx\right)} = \dfrac{\tan\left(3^nx\right)-\tan(x)}2 + \dfrac{\sin\left(3^{n}x \right)}{\cos\left(3^{n+1}x\right)} = \dfrac{\tan\left(3^{n+1}x\right)-\tan(x)}2$$ where we again rely on $(\spadesuit)$ to express $\tan\left(3^{n+1}x\right)$ in terms of $\tan\left(3^{n}x\right)$.
• I'm not familiar with induction as of yet. I will try to read up on that but are you aware of any other approach? – MayankJain May 26 '15 at 19:58
@MayankJain @user,
I don't know that the case for $n=1$ is trivial, especially for someone in a trig class currently. I will offer a proof without induction using substitution instead. Mayank, to see that this is the case for $n=1$, convert the RHS of the equation as follows:
$\dfrac{1}{2} \cdot \big{[}\dfrac{\sin 3x}{\cos 3x} - \dfrac{\sin x}{\cos x}]$ = (1) $\dfrac{1}{2} \cdot \big(\dfrac{\sin 3x \cos x - \cos 3x \sin x}{\cos 3x \cos x}\big{)}$ = (2) $\dfrac{1}{2} \cdot \dfrac{\sin 2x}{\cos 3x \cos x}$ = (3) $\dfrac{1}{2} \cdot \dfrac{2\sin x \cos x}{\cos 3x \cos x}$ = $\dfrac{\sin x}{\cos 3x}$
This is pretty heavy on trig identities. We get equivalence (1) by multiplying out the fraction, equivalence (2) because $\sin(u - v) = \sin u \cos v - \cos u \sin v$, equivalence (3) because $\sin 2x = 2\sin x \cos x$, and, finally, (4) by cancellation.
Now, if you are unfamiliar with induction, substitution will help here as an alternative method. Since you can now derive the equivalence for the 'trivial' case, set $3x = u$ for the second case, and $9x = v$ for the third. Then you already know $\dfrac{\sin u}{\cos 3u} = \dfrac{1}{2} \cdot (\tan 3u - \tan u)$, and similarly, $\dfrac{\sin v}{\cos 3v} = \dfrac{1}{2} \cdot (\tan 3v - \tan v)$. So now we can re-write the original expression $\dfrac{\sin x}{\cos 3x} + \dfrac{\sin u}{\cos 3u} + \dfrac{\sin v}{\cos 3v}$ as $\dfrac{1}{2} \cdot (\tan(3x) - \tan x + \tan(9x) - \tan(3x) + \tan(27x) - \tan(9x))$ and everything cancels out except the desired expression: $\dfrac{1}{2} \cdot (\tan(27x) - \tan(x))$.
For $\cos y\ne0,$ $$\tan3y-\tan y=\dfrac{\sin(3y-y)}{\cos3y\cos y}=\dfrac{2\sin y\cos y}{\cos3y\cos y}$$
$$\implies\tan3y-\tan y=2\dfrac{\sin y}{\cos3y}$$
Set $y=x,3x,9x$ and add to recognize the Telescoping series
• why is $\cos y \neq 0$? – pi-π Feb 7 '17 at 8:30
• @tRiGoNoMeTrY, What is $y$ if $\cos y=0$ Put that value in $\tan3y-\tan y=\dfrac{2\sin y}{\cos3y}$ – lab bhattacharjee Feb 7 '17 at 8:34
• $L.H.S=R.H.S=0$ – pi-π Feb 7 '17 at 8:36
• @tRiGoNoMeTrY, $$\tan(3\cdot90^\circ)=?$$ – lab bhattacharjee Feb 7 '17 at 8:37
• @ lab bhattacharjee $\tan 270=\frac {1}{0}$. – pi-π Feb 7 '17 at 8:40 | 2019-06-17T20:51:01 | {
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https://math.stackexchange.com/questions/929373/how-many-integers-in-the-range-1-999-are-divisible-by-exactly-1-of-7-and-11 | # How many integers in the range [1,999] are divisible by exactly 1 of 7 and 11?
This is a question in Kenneth Rosen's Discrete Mathematics textbook 6th edition. I haven't had trouble with any other counting problems regarding "how many numbers in range [x,y] have divisibility property Z?" My issue is I have no idea what Rosen is asking for, i.e. I don't understand the question because I don't know what he wants me to compute.
Therefore this is not a duplicate of this question (1) https://math.stackexchange.com/questions/588160/how-many-positive-integers-less-than-1000-are-divisible, since while the answer is given, it doesn't explain the language of the question and what is being computed. I have no idea why the number of integers divisible by 7 or 11 minus the number of integer divisible by 77 (11 and 7) is the answer to this question. Both of these values I've already computed correctly (in separate questions).
In context: 20. How many positive integers less than 1000 e) are divisible by exactly one of 7 and 11?
Thus my question is: what/which numbers am I supposed to count/compute?
Call the numbers we are looking for good. A number is good if it is divisible by $7$ but not by $11$, or divisible by $11$ but not by $7$. That's what "divisible by exactly $1$ of $7$ and $11$" means.
Let $a$ be the number of numbers from $1$ to $999$ which are divisible by $7$, and let $b$ be the number of numbers which are divisible by $11$. If we add $a$ and $b$, we will have counted the numbers that are divisible by both $7$ and $11$ twice. However, we should not have counted them at all, they are not good.
So to get the count of the good numbers, we should find $a+b$, and take away twice the number of numbers that are divisible by both $7$ and $11$, like $77$, $154$, and a number of others.
Now the counts are straightforward. Let's find $a$. So we are counting the numbers $7\cdot 1$, $7\cdot 2$, $7\cdot 3$, and so on. What is the biggest $k$ such that $7\cdot k\le 999$? Note that $\frac{999}{7}\approx 142.71$. So the biggest integer $k$ such that $7\cdot k\le 999$ is $142$. It follows that $a=142$.
Finding $b$, and finding the number of numbers $\le 999$ divisible by $77$ is done similarly.
• Thank you for this answer. This one answered my question in the first two sentences! – user3898238 Sep 13 '14 at 4:58
• You are welcome. – André Nicolas Sep 13 '14 at 5:01
Let's look at a smaller example: the positive integers less than $10$ that are divisible by exactly one of $2$ and $3$.
Well, $2, 4, 6, 8$ are divisible by $2$, and $3, 6, 9$ are divisible by $3$. There are $6$ positive integers that are less than $10$ that are divisible by $2$ or $3$.
But, $6$ is divisible by both $2$ and $3$. We only should consider integers that are divisible by $2$ or $3$ but not both. So, we have $6-1=5$ integers that are divisible by exactly one of $2$ and $3$.
Does that help?
• Should be 6 - 1 = 5 numbers. – paw88789 Sep 13 '14 at 4:27
• @paw88789 Wow... thanks. I must be more tired than I thought. – apnorton Sep 13 '14 at 4:29
This question can be solved using PIE or Principle of Inclusion and Exclusion, which is a very useful technique in the field of combinatorics.
For the question you've got here, we let $A$ (see Venn Diagram below) be the set of natural numbers less than $1000$ that is divisible by $7$; whereas let $B$ be the set of natural numbers less than $1000$ that is divisible by $11$.
From the diagram, it is clear that we want the areas of $A + B - A \cap B$, where $A \cap B$ refers the intersection of sets $A, B$. Therefore, the number of integers divisible by $7$ or $11$ minus the number of natural numbers divisible by $77$ ($11$ and $7$) is the answer to the question.
In general, for the question asking how many natural numbers less than $k \in \mathbb{Z}$ which is divisible by the integers in the set ${a_1, a_2, \ldots a_n}$, we can represent the question in terms of $n$ sets (or circles in the Venn diagram) and then solve from there.
P.S. How can I adjust the size of the image???
• – ComFreek Sep 13 '14 at 15:40
You've got two sets, $A$ an $B$; and some way to "count" sets (when we look at a Venn diagram we immediately think "area"; for your problem it's "elements divisible by"). If you want to "count" $A \cup B$, what you want is to "count" $A$, and "count" $B$" and sum them, but because you've "counted" the part of $A$ which is also part of $B$ as part of "count" of $A$ as well as for the "count" of $B$, you subtract the "count" of $A \cap B$ - which is (for your example) the elements divisible by 7 AND by 11. | 2020-09-28T06:17:01 | {
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https://math.stackexchange.com/questions/1894918/whats-the-probability-that-in-a-building-of-10-floors-and-5-people-in-elevator | # what's the probability that in a building of 10 floors and 5 people in elevator, the elevator would reach exactly until the 5th floor and no higher?
the question goes like this: "In a building of 10 floors, 5 people get inside an elevator on the entry floor. every one of them push independently a button of one of the 10 floors. what's the probability that the elevator would get exactly to the 5th floor and no higher?"
I thought of two ways to solve it, and each one gets me another result. so probably I do something wrong, would appreciate your help.
• first solution - let's choose the person that hits the 5th floor such that the elevator will reach this floor. $\binom{5}{1}$ options for that. for the rest, there're $5^4$ options to choose a floor. the probability space is $10^5$ for number of ways to choose floors (out of 10 floors) independently for 5 people. so the final result would be $\frac{\binom{5}{1} 5^4}{10^5}$
• second solution - let's notice there're $5^5$ ways to hit buttons in elevator such that the elevator would get at the most to 5th floor, and $4^5$ ways to hit buttons in elevator such that the elevator would get at the most to 4th floor. so the number of ways the elevator would reach exactly the 5th floor is $5^5-4^5$ and then the result is $\frac{5^5-4^5}{10^5}$
• Your first solution overcounts certain scenarios where multiple people choose the fifth floor. You can correct the count by breaking into cases based on how many people chose the fifth floor. $\binom{5}{1}4^4 + \binom{5}{2}4^3 +\binom{5}{3}4^2 +\binom{5}{4}4^1 +1$, which you should be able to see equals the numerator of the second approach – JMoravitz Aug 17 '16 at 11:18
• it's the number of ways of having all people from floors 1 to 5, but deduct from that the number of ways of having all people from floors 1 to 4 (because non of these work for you) - yes your second answer – Cato Aug 17 '16 at 11:37
• it's a good mental puzzle because there is 5 floors authorized, the 5th mandatory and 5 people ; I like to evaluate it with other values – user354674 Aug 17 '16 at 22:55
I think its easier to understand the difference if you simplify the example.
Assume you have 4 floors, 2 people, and you want to get the probability that the elevator would get exactly to the 2nd floor and no higher.
second solution would be $$\frac{2^2 - 1 ^2}{4^2} = \frac{3}{16},$$because there are 16 possible scenarios but only 3 lead exactly to the 2nd floor (namle 1-2, 2-1 and 2-2).
first solution You get $$\frac{\binom{2}{1} 2^1}{4^2} = \frac{4}{16}$$ because you count that the first persons hits floor 2 and the second person 1 or 2 (namely 2-1 and 2-2) plus the other case (namely 1-2 and 2-2). But you got the case 2-2 twice and this is why your first solution does not work.
Below is a simulation in R statistical software of a million iterations of this experiment, followed by the exact combinatorial result in your second solution. Here $X$ is the highest floor selected by any of the five passengers. A million iterations should give at least two to three place accuracy, so results are in substantial agreement.
m = 10^6; x = numeric(m)
for (i in 1:m) {
x[i] = max(sample(1:10, 5, rep=T)) }
mean(x==5) # proportion of 5's
## 0.02105 # aprx P(X = 5)
(5^5 - 4^5)/10^5
## 0.02101 # exact P(X = 5)
Here is a histogram of the simulated distribution of $X.$
Furthermore, the probability the elevator goes all the way to the top is $P(X = 10) = 1 - .9^5 = 0.40951,$ which agrees with the simulation within the margin of simulation error.
mean(x == 10)
## 0.408479
Horizontal red reference lines in the plot are at 0.0210 and 0.4095, the two probabilities mentioned above.
Simulated distribution (used for histogram):
table(x)/m
x
1 2 3 4 5
0.000011 0.000342 0.002143 0.007666 0.021050
6 7 8 9 10
0.046717 0.090376 0.159299 0.263917 0.408479
• I checked your results ; they are ok and very near from the computed ones – user354674 Aug 18 '16 at 0:49
Each person can choose one of 10 floors.
$10^{5}$ possible sets of desired floors.
possible sets where no one has chosen a floor higher than 5.
$5^5$ possible sets where no one has chosen a floor higher than 4.
$4^5$
$\frac {5^5-4^5}{10^{5}}$
This is the second solution... But, if it is an American building the elevator starts on floor 1. The first stop is 2.
$\frac {4^5-3^5}{9^5}$
We could get a little stupider and say that some people will take the stairs if going to floor 2 or 3, but that gets tricky to model.
As far as the flaw in the first solution JMoravits has it right, you are over-counting the number of people who want to go the the 5th floor.
$5\cdot 4^4 + 10\cdot 4^3 + 10\cdot 4^2 + 5\cdot 4 + 1$
Which you might recognize as looking similar to the expansion of: $(x+1)^5 - x^5$ and then setting $x = 4$ Or if this is an American elevator, setting $x = 3.$
• Extraneous complications. 'Entry level' is pretty clearly not meant to be one of the ten floors. A parking level perhaps. – BruceET Aug 17 '16 at 22:46 | 2020-07-11T20:39:01 | {
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http://math.stackexchange.com/questions/36041/what-is-the-volume-of-x-y-z-in-mathbbr3-geq-0-sqrtx-sqrt?answertab=active | What is the volume of $\{ (x,y,z) \in \mathbb{R}^3_{\geq 0} |\; \sqrt{x} + \sqrt{y} + \sqrt{z} \leq 1 \}$?
I have to calculate the volume of the set
$$\{ (x,y,z) \in \mathbb{R}^3_{\geq 0} |\; \sqrt{x} + \sqrt{y} + \sqrt{z} \leq 1 \}$$
and I did this by evaluating the integral
$$\int_0^1 \int_0^{(1-\sqrt{x})^2} \int_0^{(1-\sqrt{x}-\sqrt{y})^2} \mathrm dz \; \mathrm dy \; \mathrm dx = \frac{1}{90}.$$
However, a friend of mine told me that his assistant professor gave him the numerical solutions and it turns out the solution should be $\frac{1}{70}$. Also, I found out that this would be the result of the integral
$$\int_0^1 \int_0^{1-\sqrt{x}} \int_0^{1-\sqrt{x}-\sqrt{y}} \mathrm dz \; \mathrm dy \; \mathrm dx,$$
which is pretty much the same as mine just without squares in the upper bounds. My question is: Is the solution provided by the assistant professor wrong or why do I have to calculate the integral without squared upper bounds?
Also, is there any tool to compute the volume of such sets without knowing how one has to integrate?
-
Because $z \leq (1-\sqrt{x} - \sqrt{x})^2$. Why is the assistant professor correct? – Huy Apr 30 '11 at 16:45
I answered too soon. It looks like you are right. :) – Grumpy Parsnip Apr 30 '11 at 16:48
PS: I of course meant $z \leq (1-\sqrt{x}-\sqrt{y})^2$. Somehow I can't edit my first comment anymore. – Huy Apr 30 '11 at 16:52
An integral $(*)\ \int_B f(x){\rm d}(x)$ over a three-dimensional domain $B$ depends on the exact expression for $f(x)$, $\ x\in{\mathbb R}^n$, and on the exact shape of the domain $B$. The latter is usually defined by a set of inequalities of the form $g_i(x)\leq c_i$. The information about $B$ has to be entered in the course of the reduction of the integral $(*)$ to a sequence of nested integrals. So, as a rule, there is a lot of work involved in the process of reducing everything to the computation and evaluation of primitives.
Now sometimes there is another way of handling such integrals: Maybe we can set up a parametric representation of $B$ with a parameter domain $\tilde B$ which is a standard geometric object like a simplex, a rectangular box or a half sphere. In the case at hand we can use the representation $$g: \quad S\to B,\quad (u,v,w)\mapsto (x,y,z):=(u^2,v^2,w^2)$$ which produces $B$ as an essentially 1-1 image of the standard simplex $$S:=\{(u,v,w)\ |\ u\geq0, v\geq0, w\geq 0, u+v+w\leq1\}\ .$$ In the process we have to compute the Jacobian $J_g(u,v,w)=8uvw$ and obtain the following formula: $${\rm vol}(B)=\int_B 1\ {\rm d}(x)= \int_S 1 \> J_g(u,v,w) \> {\rm d}(u,v,w)=\int_0^1\int_0^{1-u}\int_0^{1-u-v} 8uvw \> dw dv du ={1\over 90}\ .$$ (In this particular example the simplification is only marginal.)
@Willie: I did understand that. :) @Christian: In order to apply this kind of substitution, we needed $g$ to be a diffeomorphism. I'm pretty sure in general such a mapping wouldn't be a diffeomorphism, but is this here the case due to the fact that both $B$ and the standard simplex only contain non-negative components? – Huy May 1 '11 at 11:29
@Christian: As far as I see, even with only non-negative components, this is no diffeomorphism, as the inverse function is not differentiable in $0$. So why is one allowed to do this transformation here? – Huy May 1 '11 at 11:37
@Huy: it's the same this as with polar coordinates: they are not a diffeomorphism on the whole plane, but it is in the plane withouth $x \geq 0$. Here it is a diffeomorphism on the whole plane without the origin; does this fact change your integral? (think about polar coordinates again) – Andy May 1 '11 at 12:00
@Huy: The map $g$ has to be a diffeomorphism $S'\to B'$ where $S\triangle S'$ and $B\triangle B'$ are sets of measure $0$ in their respective ${\mathbb R^d}$'s. – Christian Blatter May 1 '11 at 16:06 | 2014-08-31T06:32:32 | {
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http://math.stackexchange.com/questions/352120/inner-product-for-a-finite-dimensional-vector-space | # Inner product for a finite dimensional vector space
Do we always have an inner product for a finite dimensional vector space $V$ over a field $k$ such that $V$ is a Hilbert space? Thank you very much.
-
No. Part of the requirements of an inner product space is that it be over the real or complex numbers. Certainly any vector space over some other field would be a counterexample. – kahen Apr 5 '13 at 13:42
For any basis $\{\vec{v}_1, \dots, \vec{v}_n\}$, one can define an inner product by $\langle \vec{v}_i, \vec{v}_j \rangle = \delta_{ij}$. Completeness is not an issue in finite dimensions. – Sammy Black Apr 5 '13 at 13:42
As @kahen says, you would need $k=\mathbb{R}$ or $k=\mathbb{C}$. – Sammy Black Apr 5 '13 at 13:43
A finite-dimensional real or complex vector space always has an inner product, but it has a lot of them - there is no "obvious" right choice for the inner product unless the vector space has some structure. – Thomas Andrews Apr 5 '13 at 13:44
@kahen Maybe convert to an answer, including Sammy Black's comment? (or vice versa) – Julian Kuelshammer Apr 5 '13 at 15:59
Since it's part of the definition of an inner product space that it must be over the real or complex numbers, any vector space over some other field will not admit an inner product.
Barring that obvious obstruction, the situation where one's vector space $V$ is over $\mathbb K$ ($=\mathbb R$ or $\mathbb C$) is quite simple: For any basis $v_1,v_2,\dotsc,v_n$ one has an inner product given by
$\qquad \displaystyle\langle v_i,v_j\rangle = \delta_{ij} = \begin{cases} 1 & \text{if }i=j \cr 0 & \text{if }i\neq j\end{cases}\$ extended using sesquilinearity to all of $V$.
Some related things you might find enlightening:
• Norms Induced by Inner Products and the Parallelogram Law: If a norm satisfies the parallelogram law, does it come from an inner product? Yes, but it takes a surprising amount of work to actually show that this is the case.
• Any two norms on a finite dimensional linear space (vector space over $\mathbb R$ or $\mathbb C$) are equivalent. If you search Math.SE you're likely to find several questions and answers related to this. It's very common to prove this in real analysis courses.
• Suppose $V$ is a finite dimensional normed linear space and $K \subset V$ is compact , absolutely convex and spans $V$, then there is a norm on $V$ such that $K$ is the unit ball w.r.t. that norm (and per the above equivalent to the given norm on $V$). It's not too difficult to come up with an idea that could work—just draw a picture of the situation in $\mathbb R^2$.
- | 2016-02-07T02:06:27 | {
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https://textbook.prob140.org/ch5/IntersectionOfSeveralEvents.html | ## Intersection of Several Events
### Intersection of Several Events¶
By now you have seen many examples of the following kind:
A deck of cards consists of 26 red cards and 26 black cards. Three cards are drawn at random without replacement. What is the chance that they are all black?
You will be quick to answer $$\frac{26}{52} \cdot \frac{25}{51} \cdot \frac{24}{50}$$
and you will be right. But where does that multiplication of three factors come from? The multiplication rule says only that if $A$ and $B$ are two events, then
$$P(AB) = P(A)P(B \mid A)$$
Can we just go ahead and extend it to three or more events, and if so, exactly how are we extending it?
The third factor in the product above is the conditional chance that the third card is red given that the first two cards were red. This suggests an extension of the multiplication rule, used by us without discussion and without proof in many calculations so far:
$$P(A_1A_2 \ldots A_n) = P(A_1)P(A_2 \mid A_1) P(A_3 \mid A_1A_2)P(A_4 \mid A_1A_2A_3) \cdots P(A_n \mid A_1A_2 \ldots A_{n-1})$$
Let's prove this by induction. While it might seem a little pedantic, it's a good idea to develop skills that you can use to solidify the foundations of steps that are based on intuition.
We will start in the case $n=3$ to see what is going on. First notice that
$$P(A_2A_3 \mid A_1) = \frac{P(A_1A_2A_3)}{P(A_1)} = \frac{P(A_1A_2)P(A_3 \mid A_1A_2)}{P(A_1)} = P(A_2 \mid A_1)P(A_3 \mid A_1A_2)$$
This is just like the ordinary multiplication rule for the intersection of the two events $A_2$ and $A_3$, except that all the probabilities are also conditional given $A_1$.
And now
\begin{align*} P(A_1A_2A_3) &= P(A_1)P(A_2A_3 \mid A_1) ~~~ \text{(multiplication rule applied to }A_1 \text{ and } A_2A_3 \text{)} \\ &= P(A_1)P(A_2 \mid A_1)P(A_3 \mid A_1A_2) ~~~ \text{(we just proved this)} \end{align*}
So our result is true for $n=3$.
To prove it for all positive integers, first assume the induction hypothesis that the result is true for $n$:
$$P(A_1A_2 \ldots A_n) = P(A_1)P(A_2 \mid A_1) P(A_3 \mid A_1A_2) \cdots P(A_n \mid A_1A_2 \ldots A_{n-1})$$
And then show that it is true for $n+1$:
\begin{align*} & P(A_1A_2 \ldots A_nA_{n+1}) \\ &= P(A_1)P(A_2 \mid A_1) P(A_3 \mid A_1A_2) \cdots P(A_n \mid A_1A_2 \ldots A_{n-1}) P(A_{n+1} \mid A_1A_2 \ldots A_{n-1}A_n) \end{align*}
The induction relies on treating $A_1A_2 \dots A_nA_{n+1}$ as the intersection of two events: $B_n = A_1A_2 \ldots A_n$ and $A_{n+1}$. Moves like this are use in many induction proofs of results about chances of intersections and unions.
Now
\begin{align*} P(A_1A_2 \ldots A_{n+1}) &= P(B_nA_{n+1}) \\ &= P(B_n)P(A_{n+1} \mid B_n) ~~~ \text{(multiplication rule)} \\ &= P(A_1A_2 \ldots A_n)P(A_{n+1} \mid A_1A_2 \ldots A_n) ~~~ \text{(definition of } B_n \text{)} \\ &= P(A_1)P(A_2 \mid A_1) P(A_3 \mid A_1A_2) \cdots P(A_n \mid A_1A_2 \ldots A_{n-1})P(A_{n+1} \mid A_1A_2 \ldots A_n) \end{align*}
by the induction hypothesis. Done!
You shouldn't feel you have to have proved every single result that you use, specially when they arise naturally as properties of proportions. This section shows you that there are straightforward if laborious ways of establising such results mathematically.
However, what is "natural" and what is correct are not always the same. Here is an example.
### Pairwise and Mutual Independence¶
We have defined the independence of two events to mean that the chance of one doesn't change if you are told that other has happened; $A$ and $B$ are independent means that
$$P(B \mid A) = P(B) ~~~ \text{or equivalently,} ~~~ P(AB) = P(A)P(B)$$
Now suppose you have three events $A, B$, and $C$, and suppose each pair of them is independent by the defintion above. That is, $A$ is independent of $B$, $A$ is independent of $C$, and $B$ is independent of $C$.
This is called pairwise independence, and you might be tempted to use it as a definition of independence of three events. But it doesn't quite work.
In a group of three people, let $B_{ij}$ be the event that Persons $i$ and $j$ have the same birthday, and let $B_{123}$ be the event that all three have the same birthday. Under the assumptions of randomness that we made for the classical Birthday Problem, we know that
$$P(B_{12}) = \frac{1}{365} = P(B_{23}) = P(B_{13})$$
and $$P(B_{12} B_{23}) = P(B_{123}) = \frac{1}{365} \cdot \frac{1}{365} = P(B_{12})P(B_{23})$$
So $B_{12}$ and $B_{23}$ are independent. In the same way you can show that $B_{12}$ and $B_{13}$ are independent, as are $B_{23}$ and $B_{13}$. Thus the three events $B_{12}$, $B_{13}$, and $B_{23}$ are pairwise independent.
But $$P(B_{13} \mid B_{12}B_{23}) = 1 \ne P(B_{13})$$
Given that Persons 1 and 2 have the same birthday and that Persons 2 and 3 have the same birthday, there is no randomness left in whether Persons 1 and 3 have the same birthday – they just do. Information about the other two pairs affects the chance of $B_{13}$.
This goes against what independence should mean, and points out to us that when we have more than two events we have to be careful about the defining independence.
#### Mutual Independence¶
Events $A_1, A_2, \ldots A_n$ are mutually independent (or independent for short) if given that any subset of the events has occurred, the conditional chances of all other subsets remain unchanged.
That's quite a mouthful. In practical terms it means that it doesn't matter which of the events you know have happened; chances involving the remaining events are unchanged.
In terms of random variables, $X_1, X_2, \ldots , X_n$ are independent if given the values of any subset, chances of events determined by the remaining variables are unchanged.
In practice, this just formalizes statements such as "results of different tosses of a coin are independent" or "draws made at random with replacement are independent".
Try not to become inhibited by the formalism. Notice how the theory not only supports intuition but also develops it. You can expect your probabilistic intuition to be much sharper at the end of this course than it is now! | 2017-10-18T09:18:07 | {
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https://mathhelpboards.com/threads/describing-an-equivalence-class.7685/ | # Describing an equivalence class?
#### skatenerd
##### Active member
I am given that the relation ~ is defined on the set of real numbers by $$x$$~$$y$$ iff $$x^2=y^2$$. First part of the problem said to prove ~ is an equivalence relation, that wasn't bad. The second part asks to "Describe the equivalence classes". This just seems really vague to me. Is this a common question, with a specific answer expected? I don't need anybody to do the problem for me just an idea of how to answer the question would be appreciated
#### Fernando Revilla
##### Well-known member
MHB Math Helper
According to the defintion of equivalence class:
$$[a]=\lbrace x\in \mathbb{R}:x\sim a\rbrace=\lbrace x\in \mathbb{R}:x^2= a^2\rbrace=\lbrace a,-a\rbrace.$$
#### Deveno
##### Well-known member
MHB Math Scholar
Interesting observation: we get the SAME equivalence relation if we define:
$x \sim y$ if and only if $|x| = |y|$.
Why should this be so?
Note we can actually "multiply" these equivalence classes (from our original equivalence relation), by defining:
$[a] \ast = [ab]$.
This works because if:
$a^2 = a'^2$ and $b^2 = b'^2$ (even if $a,a'$ are unequal, and similarly with the $b$'s), then:
$(ab)^2 = a^2b^2 = a'^2b'^2 = (a'b')^2$.
(To see why this is important, consider what goes horribly wrong with addition).
It is hopefully clear that our equivalence classes behave a lot like the non-negative reals with the single operation of multiplication. On this set, the correspondence:
$a \leftrightarrow a^2$
is a one-to-one correspondence, and furthermore, this correspondence preserves multiplication:
$(ab) \leftrightarrow (ab)^2 = a^2b^2$ (we get the same result if we multiply first, and take the equivalence class second, or if we take the equivalence classes first, and then "multiply" them as above).
********
In problems like this you may encounter in the future, there are 2 main strategies to employ in "describing the equivalence classes"
a) Attempt to enumerate EVERY element of a given equivalence class...this works best when each equivalence class is finite.
b) Search for common properties each member of an equivalence class possesses...this works best when the equivalence classes themselves are infinite sets.
********
Finally, it is often good to ask yourself: what information does an equivalence class forget? In this case, the information "lost" is the SIGN of $a$. | 2021-06-15T10:45:02 | {
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http://matkagit.net/unsorted/trig-functions-chart-sin-cos-tan-graph-4555117.html | # Trig functions chart sin cos tan graph
This right- or left-shifting is called "phase shift". You've already learned the basic trig graphs. Because this 4 is subtracted from the tangent, the shift will be four units downward from the usual center line, the x -axis. Content Continues Below. Lissajous Figures Feedback? This relationship is always true: If a number D is added outside the function, then the graph is shifted up by that number of units; if a number D is subtracted, then the graph is shifted down by that number of units. The amplitude is given by the multipler on the trig function. For this function, the value of the amplitude multiplier A is —2so the amplitude is: 2.
• Graphs of trigonometric functions Trigonometry Math Khan Academy
• Sine, Cosine, Tangent Unit Circle Interactive Graphs Applet and Widget with GIF
• Graphs of Sine, Cosine and Tangent
• Graphs of trigonometric functions
• Graphing Trigonometric Functions Purplemath
• The Sine Function has this beautiful up-down curve which repeats every degrees. Thus, the period of the sin, cos, csc, and sec graphs is 2\pi radians, and the period for the tan and cot graphs is \pi radians. Because the trig functions are cyclical. those graphs.
Learn how to construct trigonometric functions from their graphs or other features.
The graphs of sine, cosine, and tangent. Learn. Graph of.
For this function, the value of the amplitude multiplier A is given by 0. Biorhythm Graphs 7.
## Graphs of trigonometric functions Trigonometry Math Khan Academy
However, they do occur in engineering and science problems. Click to search:. The amplitude has changed from 1 in the first graph to 3 in the second, just as the multiplier in front of the sine changed from 1 to 3. This relationship is always true: Whatever number A is multiplied on the trig function gives you the amplitude that is, the "tallness" or "shortness" of the graph ; in this case, that amplitude number was 3.
Justine sterckx 2013
Do you see that the sine wave is cycling twice as fast, so its period is only half as long?
### Sine, Cosine, Tangent Unit Circle Interactive Graphs Applet and Widget with GIF
So there will be a "gap" in the function at that point. Because this value is added to the variable, then the shift is to the left. Content Continues Below. The vertical shift comes from the value entirely outside of the trig function; namely, the outer 4 also known as " D ", from the formula.
Graph of y = sin x.
Graph of y = sin ax. Graph of y = cos x. Graph of y = tan x.
## Graphs of Sine, Cosine and Tangent
The period of a function. The period of y = sin x. This section describes the graphs of trigonometric functions. The effect of q is called a vertical shift because the whole sine graph shifts up or down by q units. For q>0, the .
Video: Trig functions chart sin cos tan graph Trigonometric Graphs and the Unit Circle
The maximum value of y=cosθ is 1 and the minimum value is −1. You've already learned the basic trig graphs. This function has a period of 2π because the sine wave repeats every 2π.
### Graphs of trigonometric functions
What is the period of f(t) = cos(3t)?.
This number is subtracted from the variable, so the shift will be to the right. More info: Trigonometry videos. So there will be a "gap" in the function at that point.
Putting it all together in terms of the sine wave, we have the general sine function:. An asymptote is a straight line that the curve gets closer and closer to, without actually touching it. I could also have used the simpler method, directly from the formula, of dividing C by B.
If we continue our table, we will get similar values because this is a periodic graph.
FSJ ERFAHRUNG MIT
Applications of Trigonometric Graphs 6.
What's the difference between phase shift and phase angle?
### Graphing Trigonometric Functions Purplemath
I could also have used the simpler method, directly from the formula, of dividing C by B. The factorization is:. This right- or left-shifting is called "phase shift".
## 5 thoughts on “Trig functions chart sin cos tan graph”
1. Kazratilar:
Do you see that this second graph is three times as tall as was the first graph? Consider the denominator bottom of this fraction.
2. Tunos:
In the sine wave graphed above, the value of the period multiplier B was 2. Phase shift or phase angle?
3. Grot: | 2021-03-01T15:54:57 | {
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https://math.stackexchange.com/questions/3115212/cube-and-unit-cubes | Cube and unit cubes
Consider a $$3\times 3\times 3$$ cube consisting of smaller $$1\times 1\times 1$$ unit cubes. The big cube is painted black on the outside. Suppose we disassemble the cube and pick a random unit cube, look at only one face and see it is black, without looking at the other faces. What is the probability the unit cube we picked is one of the 8 corner cubes?
This one seems simple to me but I am not sure I am right: The big cube consists of 27 unit cubes of which one only, the middle one, does not have any painted face. All the others (26) have at least one face painted and 8 have 3 faces painted. Thus the requested probability is $$\frac{8}{26}$$—is this so?
• There are already good answers, but the problem you solved with answer $8/26$ is a different one: Say I randomly select a unit cube and look at all of its faces. You ask me if at least one face on that cube is black, and I answer "yes". Given that, what is the probability the unit cube I picked is a corner cube? The major difference is knowing something (though not everything) about all the cube's faces, or knowing about just one face of a cube. – aschepler Feb 16 at 19:55
Your first step is reasonable: if you see a painted face, then you know that the cube you have cannot possibly be the center cube. However, you are not "just as likely" to have a corner cube as a side cube. You need to be more careful in thinking about your sample space.
Here is a possible line of reasoning: when you paint the $$3\times 3\times 3$$ cube, you paint a total of $$9 \cdot 6 = 54$$ faces (each face of the large cube consists of a $$3\times 3$$ arrangement of faces of the smaller cubes; the large cube has six faces). This represents the total space of possible outcomes (given that you know you have selected a painted face—a priori, it might have been possible to selected a non-painted face, but we know this didn't happen).
On the other hand, there are 8 corner cubes, each of which as three painted faces, for a total of $$3 \cdot 8 = 24$$ painted faces on corner cubes. Picking a painted face from a corner is a "success" in this experiment. Therefore the probability that you have selected a corner cube is the ratio of successes to the total number of possible outcomes, i.e. $$P(\text{the painted face is from a corner}) = \frac{\text{number of successful outcomes}}{\text{total number of outcomes}} = \frac{24}{54} = \frac{4}{9}.$$
No, because you have picked a random face of the cube to look at and seen it is black. That makes it more likely that you have a corner cube than one of the other ones with some painted faces. Maybe a more careful statement of the problem is we choose a random cube and a random face of that cube. Given that the face we choose is black, what is the chance the cube is a corner cube?
The simple solution is that we have selected a random black face. There are $$54$$ black faces, $$24$$ of which are on corner cubes, so the chance we have a corner cube is $$\frac {24}{54}$$
• Thank you very much sirs! I gave acceptance to the one I see first. Hope I was not unfair! – Sal.Cognato Feb 16 at 17:18
You can also use Bayes' theorem. The prior probability of choosing a corner cube is $$8 \over 27$$. The probability of a corner cube showing a black face is $$1 \over 2$$. The overall probability of a black face is $${54 \over 162}=1/3$$, so
$$P({\rm corner }|{\rm black})={1/2 \over1/3} \times 8/27 =4/9$$ | 2019-03-23T12:15:08 | {
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https://golem.ph.utexas.edu/category/2016/03/the_most_common_prime_gaps.html | ## March 19, 2016
### The Most Common Prime Gaps
#### Posted by John Baez
Twin primes are much beloved. But a computer search has shown that among numbers less than a trillion, most common distance between successive primes is 6. It seems this goes on for quite a while longer…
… but Andrew Odlyzko, Michael Rubinstein and Marek Wolf have persuaded most experts that somewhere around $x = 1.7427 \times 10^{35}$, the most common gap between consecutive primes less than $x$ switches from 6 to 30:
• Andrew Odlyzko, Michael Rubinstein, and Marek Wolf, Jumping champions, Experimental Mathematics 8 (1999), 107–118.
This is a nice example of how you may need to explore very large numbers to understand the true behavior of primes.
They give a sophisticated heuristic argument for their claim—not a rigorous proof. But they also checked the basic idea using Maple’s ‘probable prime’ function. It takes work to check if a number is prime, but there’s a much faster way to check if it’s probably prime in a certain sense. Using this, they worked out the gaps between probable primes from $10^{30}$ and $10^{30}+10^7$. They found that there are 5278 gaps of size 6 and just 5060 of size 30. They also worked out the gaps between probable primes from $10^{40}$ and $10^{40}+10^7$. There were 3120 of size 6 and 3209 of size 30.
So, it seems that somewhere between $10^{30}$ and $10^{40}$, the number 30 replaces 6 as the most probable gap between successive primes!
Using the same heuristic argument, they argue that somewhere around $10^{450}$, the number 30 ceases to be the most probable gap. The number 210 replaces 30 as the champion—and reigns for an even longer time.
Furthermore, they argue that this pattern continues forever, with the main champions being the ‘primorials’:
$2$
$2 \cdot 3 = 6$
$2 \cdot 3 \cdot 5 = 30$
$2 \cdot 3 \cdot 5 \cdot 7 = 210$
$2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 = 2310$
etc.
Posted at March 19, 2016 6:10 AM UTC
TrackBack URL for this Entry: https://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/2870
### Re: The Most Common Prime Gaps
Fantastic!
It seems you forgot a 5 in the last 2 primorials, though.
### Re: The Most Common Prime Gaps
Thanks! I’ll fix that typo now.
Posted by: John Baez on March 19, 2016 5:41 PM | Permalink | Reply to this
### Re: The Most Common Prime Gaps
You meant “between $10^30$ and $10^30 + 10^7$” and “between $10^40$ and $10^40 + 10^7$”.
Posted by: Todd Trimble on March 19, 2016 12:20 PM | Permalink | Reply to this
### Re: The Most Common Prime Gaps
Heh, yes. I’ll fix that now.
Posted by: John Baez on March 19, 2016 5:43 PM | Permalink | Reply to this
### Re: The Most Common Prime Gaps
What about the simplest case – the race between twin primes and sexy primes?
I might be off by 1 or something, but with the help of a rudimentary script (in Matlab, of all things!), I believe that (541, 547) is the 26th sexy prime pair, and from then on there are more sexy primes than twin primes.
But this is not the first time the sexy primes overtake the twin primes – twin primes have the lead until (173,181) ties them at 12 before the twin primes retake the lead with (179,181). Later, sexy primes take the lead for the first time with the 22nd sexy prime pair (383,389) before it’s tied back up with (419, 421). There’s a bit more back-and-forth before sexy primes take the lead for good – or at least up to 1000, at which point there are 44 sexy primes and 35 twin primes and it seems unlikely that the twin primes will ever have the lead again.
I didn’t look at other prime gaps. It’s conceivable (but unlikely, I guess) that some other prime gap has the lead at some other point early on.
Posted by: Tim Campion on March 19, 2016 8:53 PM | Permalink | Reply to this
### Re: The Most Common Prime Gaps
Nice, Tim! Sometimes 4 has the lead over 2 and 6. Ian Stewart writes:
What number is the most common gap between successive primes less than $x$? This question was posed in the late 1970s by Harry Nelson of Lawrence Livermore National Laboratory. Later on, John Horton Conway of Princeton University coined the phrase “jumping champions” to describe these numbers.
The primes up to 50 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47. The sequence of gaps—the differences between each prime and the next—is 1, 2, 2, 4, 2, 4, 2, 4, 6, 2, 6, 4, 2 and 4. The number 1 appears only once because all primes except for 2 are odd. The rest of the gaps are even numbers. In this sequence, 2 occurs six times, 4 occurs five times, and 6 occurs twice. So when $x$ = 50, the most common gap is 2, and this number is therefore the jumping champion.
Sometimes several gaps are equally common. For instance, when x = 5 the gaps are 1 and 2, and each occurs once. For higher $x$, the sole jumping champion is 2 until we reach $x$ = 101, when 2 and 4 are tied for the honor. After that, the jumping champion is either 2 or 4, or both, until $x$ = 179, when 2, 4 and 6 are involved in a three-way tie. At that point the challenge from 4 and 6 dies away, and 2 reigns supreme until $x$ = 379, where 2 is tied with 6. Above $x$ = 389 the jumping champion is mostly 6, occasionally tied with 2 or 4, or both. But when $x$ ranges from 491 to 541, the jumping champion reverts to 4. From $x$ = 947 onward the sole jumping champion is 6, and a computer search shows that this continues up to at least $x = 10^{12}$.
Posted by: John Baez on March 20, 2016 1:05 AM | Permalink | Reply to this
### Re: The Most Common Prime Gaps
There go those small numbers, throwing a wrench in every pattern!
Posted by: Tim Campion on March 20, 2016 6:10 AM | Permalink | Reply to this
### Re: The Most Common Prime Gaps
Over on G+, Justen Robertson asked:
What’s the relationship between primorials and the point at which they become dominant?
I looked at the paper by Odlyzko, Rubinstein and Wolf, and near the start they give a rough formula for this They say the number
2⋅3⋅5 = 30
starts becoming more common as a gap between primes than
2⋅3 = 6
roughly when we reach
exp(2⋅3⋅4⋅3) = exp(72) ≈ 1.8 ⋅ $10^31$
That’s pretty rough, since they say actual turnover occurs around 1.7427 ⋅ $10^{35}$. But you probably can’t see the pattern yet, so let me go on!
The number
2⋅3⋅5⋅7 = 210
starts becoming more common as a gap between primes than
2⋅3⋅5 = 30
roughly when we reach
exp(2⋅3⋅5⋅6⋅5) = exp(900) ≈ $10^{390}$
Again, this is pretty rough - they must have a more accurate formula that they use elsewhere in the paper. But they mention this rough one early on.
I bet you still can’t see the pattern in that exponential, so let me do two more!
The number
2⋅3⋅5⋅7⋅11 = 2310
starts becoming more common as a gap between primes than
2⋅3⋅5⋅7 = 210
roughly when we reach
exp(2⋅3⋅5⋅7⋅10⋅9) = exp(18900)
The number
2⋅3⋅5⋅7⋅11⋅13 = 30030
starts becoming more common as a gap between primes than
2⋅3⋅5⋅7⋅11 = 2310
roughly when we reach
exp(2⋅3⋅5⋅7⋅11⋅12⋅11) = exp(304920)
Get the pattern?
Posted by: John Baez on March 20, 2016 1:13 AM | Permalink | Reply to this
### Re: The Most Common Prime Gaps
Leaves me wondering what happened to exp(primorial × 8 × 7) …
Posted by: Jesse C. McKeown on March 21, 2016 1:55 AM | Permalink | Reply to this
### Re: The Most Common Prime Gaps
Following wikipedia and using $p#$ to denote the primorial of $p$, I think the pattern is that $p_n #$ becomes more common than $p_{n-1}#$ at around $\exp(p_{n-1}# \cdot (p_n - 1) \cdot (p_n - 2))$. So the reason we don’t see $7# \cdot 8 \cdot 7$ is that 9 isn’t prime.
Posted by: Tim Campion on March 28, 2016 3:03 PM | Permalink | Reply to this
### Re: The Most Common Prime Gaps
Yes, that’s the pattern, and the answer to Jesse’s question.
Posted by: John Baez on March 28, 2016 5:07 PM | Permalink | Reply to this
### Re: The Most Common Prime Gaps
The news seems to be trickling out. I got this email today:
PRIME NEWS: Last Digits 1,3,7 and 9 permeates significantly throughout primes whole complex.
Beeing the significant most frequent digits in primes, Last Digits 1,3,7 and 9 permeates throughout its whole complex.
The digits are not uniformly distributed but Benfordian, indicating primes are governed by a flux combination of Benfords Law and Last Digits. Hence consecutive LD=1 has greater probability. When sizing prime data-set, researchers should be aware of the importance of complete and fair rounds of first digits 1-9. Otherwise; the results may be biased.
You can see the numerical evidence, distributions and explanations in “primes” on www.stringotype.com.
It is hypothesized that the result corresponds to the dimensionless Fine Structure Constant - a fractal order in nature. Base 10 number system maps the order with zero skewness, giving rise to many circadian rythms beeing purely reflected in numbers. Hence digits corresponds to primary respondents and can possibly be intepreted as integrals of functional processes.
Best regards
Terje Dønvold
Oslo
Posted by: John Baez on March 30, 2016 7:54 PM | Permalink | Reply to this
Post a New Comment | 2021-09-27T23:18:51 | {
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https://math.stackexchange.com/questions/3334015/approximating-sin-why-absolute-error-gets-bigger | # Approximating sin. Why absolute error gets bigger?
So I wanted to aproximate $$\sin(0.234,375^\circ)$$ to 5 decimal places. But the thing is, I wanted to do it the old school way(not using some power series). Also, I wanted to do the calculation on sheet of paper using basic calculator - hence, while calculating I wanted to keep least accurate approximations necessary to obtain my final result.
I knew that:
$$\cos(30^\circ) = \frac{\sqrt 3}{2}\\ \sin(\frac{\alpha}{2})=\sqrt{\frac{1-\cos(\alpha)}{2}}\\ \cos(\frac{\alpha}{2})=\sqrt{\frac{1+\cos(\alpha)}{2}}\\$$
And since $$0.234,375=\frac{30}{2^7}$$ it was all about doing some iterations
Since, I wanted to obtain result with 5 decimal places accuracy, I decided to start with 6 decimal approximation of $$\frac{\sqrt{3}}{2}$$, and to approximate I'm gonna use classic rounding.
Iterations:
$$\cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.866,025\\ \cos(15^\circ) = \cos(\frac{30^\circ}{2})=\sqrt{\frac{1+\cos(30)}{2}} = \sqrt{\frac{1+0.866,025}{2}} \approx 0.965,926\\ \cos(7.5^\circ) = \cos(\frac{15^\circ}{2})=\sqrt{\frac{1+\cos(15)}{2}} = \sqrt{\frac{1+0.965,926}{2}} \approx 0.991,445\\ \cos(3.75^\circ) = \cos(\frac{7.5^\circ}{2})=\sqrt{\frac{1+\cos(7.5)}{2}} = \sqrt{\frac{1+0.991,445}{2}} \approx 0.997,859\\ \cos(1.875^\circ) = \cos(\frac{3.75^\circ}{2})=\sqrt{\frac{1+\cos(3.75)}{2}} = \sqrt{\frac{1+0.997,859}{2}} \approx 0.999,465\\ \cos(0.9375^\circ) = \cos(\frac{1.875^\circ}{2})=\sqrt{\frac{1+\cos(1.875)}{2}} = \sqrt{\frac{1+0.999,465}{2}} \approx 0.999,866\\ \cos(0.46875^\circ) = \cos(\frac{0.9375^\circ}{2})=\sqrt{\frac{1+\cos(0.9375)}{2}} = \sqrt{\frac{1+0.999,866}{2}} \approx 0.999,966\\$$
and finally:
$$\sin(0.234,375^\circ) = \sin(\frac{0.46875^\circ}{2})=\sqrt{\frac{1-\cos(0.46875)}{2}} = \sqrt{\frac{1-0.999,966}{2}} \approx 0.004,123$$
My result rounded to 5 decimal places $$\approx 0.00412$$
Google calculator's result rounded to 5 decimal places $$\approx 0.00409$$
So, I have several questions:
• Why the results differ?
• What's the least neccessary aprroximation accuracy I have to keep in calculations to obtain 5 digits accurate result and why so?
• If I decide, instead of using classic rounding, to round every number to floor(just truncate unnecessary digits), how does this affect final result.
I may have other questions, which I don't know yet. I just want to understand what's going on.
I'm kinda newbie. I would likely read some easy explained article or book about numerical methods of approximation. If you have something that would help me understand things, let me know.
As I was thinking about approximation I concluded that digits arent that "important" for example: $$0.75000$$ is a good aprroximation of $$0.74999$$ even though already second digit is different.
So, alternatively to 5 digits accuracy, I may be looking to have a aproximation such as:
$$\lvert v-v_{approx}\rvert < 0.00001$$
• Comment too casual for an answer. Your intermediate steps with rounding along the way lead to errors that propagate. You need more precision than you chose from the start to get the final precision you want. Just how much more is a serious question. People have studied this. – Ethan Bolker Aug 25 '19 at 18:14
• $0.999966$ is very close to $1$, so you need a lot of digits to get an accurate difference – J. W. Tanner Aug 25 '19 at 18:15
• You are doing calculations using a method different than what the computer is using. Since the algorithm used by the computer applies different operations on each iteration and adjusts the result according to its design, the two methods will produce different results. – NoChance Aug 25 '19 at 18:39
• @J.W.Tanner That's a relative error problem, which is not relevant here because the OP is only concerned with number of correct digits, including the non-significant digits. The problem for the OP is that $\sqrt{x}$ has a large derivative near $0$, so the very last step takes the error of less than $6 \cdot 10^{-7}$ in the second-to-last quantity and turns it into an error of more like $3 \cdot 10^{-5}$, about 50 times bigger. – Ian Aug 25 '19 at 18:49
• @Ian. Would you please elaborate? The square function is well conditioned for all $x>0$. The expression $1-\cos(2x)$ suffers from subtractive cancellation when $x$ is small. Surely, this is why the OP has a large relative error when computing $\sqrt{\frac{1-\cos(2x)}{2}}$. – Carl Christian Aug 25 '19 at 19:13
The major loss of precision occurs when you take the square root of $$\frac12(1-0.999966)$$. Think about this: if you look at a graph of the square root function, it starts off vertical at $$x=0$$ (more formally, its derivative tends to $$+\infty$$ as $$x$$ tends to $$0$$ from above).
So naturally a small change in $$x$$ can lead to a disproportionately large change in $$\sqrt x$$ when $$x$$ is very small. And this is just what you are seeing. If you want $$\sqrt x$$ to $$n$$ decimal places for small $$x$$, then you need to know $$x$$ to (roughly speaking) $$2n$$ decimal places. | 2021-06-13T04:24:46 | {
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https://math.stackexchange.com/questions/1976168/trouble-with-combination-of-product-rule-and-chain-rule | # Trouble with combination of product rule and chain rule
Here is what I need to find:
$\frac{d}{dx}\left(5\cos\left(4x\right)\right)$
I know I can find $\frac{d}{dx}\left(\cos\left(4x\right)\right)$ easily with the chain rule, but the 5 in the front throws me off. What do I with that? Is it $0$, like other constants, or would that only be the case if the problem was:
$\frac{d}{dx}\left(5+\cos\left(4x\right)\right)$?
Do I treat it almost like I would the derivative of $10x^2$, like a coefficient before $x$?
If I apply the chain rule, I get $-4\sin\left(4x\right)$, right?
• Treat it just like the derivative of $10x^2$, yes - treat the $4$ as just a coefficient, it gets pulled out front – Rob Bland Oct 19 '16 at 19:27
$$\frac { d }{ dx } \left( 5\cos\left( 4x \right) \right) =\frac { d }{ dx } \left( 5 \right) \cos\left( 4x \right) +5\frac { d }{ dx } \left( \cos\left( 4x \right) \right) \\=0+5\frac { d }{ dx } \left( \cos\left( 4x \right) \right) \frac { d }{ dx } \left( 4x \right) =-20\sin { (4x) }$$
• So this is essentially the product and chain rule together, if I'm reading this right? – Chris T Oct 19 '16 at 19:36
• @ChrisT yes indeed – haqnatural Oct 19 '16 at 19:40
You can simply take it outside the derivative, like so:
$$\frac{d}{dx}(af(x))=a\cdot \frac{d}{dx}(f(x))$$
Here's why it works:
\begin{align} \frac{d}{dx}(af(x)) &= \frac{d}{dx}(a)\cdot f(x) + a\cdot \frac{d}{dx}(f(x)) \\ &= 0\cdot f(x) + a\cdot \frac{d}{dx} (f(x)) \\ &= a\cdot \frac{d}{dx}(f(x)) \\ \end{align}
• This correctly applies the product rule. You can also apply the chain rule. Let $g(u)=ku$. Then $g'(u)=k$, and this will lead the same direction. – Dean C Wills Oct 19 '16 at 23:20
The constant comes out: $\frac{d k f(x)}{dx}=k\frac{d f(x)}{dx}$. | 2019-04-25T07:59:38 | {
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https://cs.stackexchange.com/questions/110394/solving-recurrence-relation-with-minimum-and-factorial | Solving recurrence relation with minimum and factorial
I need to solve the following recurrece:
$$T(n,m)=\begin{cases} 1, & n=1\text{ or }m\leq 2(n-1)!\\ \min\limits_{a,b,c\geq 1\\a\leq c!\\b\leq(n-c)!\\c\leq n-1}{T(c,a)+T(n-c,b)+T(n,m-ab)}, & \text{else} \end{cases}$$
Note: This question is highly related to my previous question here, since $$ab\leq\max\limits_{1\leq c\leq n-1}{c!(n-c)!}=(n-1)!$$
I guess that the closed form expression for $$T(n,m)$$ here is similar to the one that I found for my previous question.
Also, I guess that the minimum is obtained at $$c=\lceil n/2\rceil,a=c!,b=(n-c)!$$, but I don't know how to prove it.
I made a little change in the definition of $$T$$ to make it a bit simpler but without affecting the rest of the values of $$T$$.
The first 10 values of the first n's are:
$$T(1,*)=1, 1, 1, 1, 1, 1, 1, 1, 1, 1,\dots\\ T(2,*)=1, 1, 3, 5, 7, 9, 11, 13, 15, 17,\dots\\ T(3,*)=1, 1, 1, 1, 3, 3, 5, 5, 7, 7,\dots\\ T(4,*)=1, 1, 1, 1, 1, 1, 1, 1, 1, 1,\dots$$
Experiments show that
$$T(n,m)=\begin{cases}1 & n=1 \text{ or }f(n,m)\leq 0,\\ 3+2\Big\lfloor\frac {f(n,m)-1}{g(n)}\Big\rfloor & \text{otherwise},\end{cases}$$
for $$f(n,m)=m-2(n-1)!$$, and for some $$g$$ whose first values are: 1, 1, 2, 4, 12, 48, 240. I guess that $$g(n)=\begin{cases}1 & \text{if }n<3,\\ 2(n-2)! & \text{otherwise.}\end{cases}$$
This question has an open bounty worth +150 reputation from Dudi Frid ending in 2 days.
This question has not received enough attention.
Link to a similar question that I did manage to solve appears at the body of the question. Any help would be highly appreciated
• Can you explain where all these questions are coming from? Are we solving some exercise sheet? Writing your thesis? – Yuval Filmus Jun 9 at 8:23
• It's neither an exercise sheet, nor my thesis: I am working on a rather complicated article in combinatorics for a while, since I don't have co-authors in my article I post here some stuff to get some help and to assure that my conclusions are correct – Dudi Frid Jun 9 at 8:30
• After getting all this help, you will be having co-authors, namely people who helped you write the article. – Yuval Filmus Jun 9 at 9:24
• Assume that $T(n,m)$ is defined on $\Bbb Z_{\ge1}\times\Bbb Z_{\ge1}$ as before. What is the value of $T(1,3)$? Since $3\not\le2(1-1)!$, we cannot apply the first rule. Since there is no $c$ such that $c\ge1$ and $c<1-1$, so $T(1,3)$ is the min of an empty set to be infinity. Is $T(1,3)$ infinity? – Apass.Jack Jun 12 at 8:55
• It looks like $T(1,m)$ for $m\ge 3$ can be set to infinity or any value that is no less than 1 without affecting other values. – Apass.Jack Jun 12 at 10:52
Summary
It is not true that the minimum can always be obtained at $$c=\lceil n/2\rceil,a=c!,b=(n-c)!$$. Here is an counterexample. $$T(5,49) = T(1,1) + T(4,1) + T(5,48) = 3 \not=5=T(3,6)+T(2,2)+T(5,12).$$ Instead, the minimum can always be obtained at $$c=1$$, $$a=1$$, $$b=2(n-2)!.$$
The following neat formulas given in the question is correct. $$T(n,m)=\begin{cases}1 & n=1 \text{ or }f(n,m)\leq 0,\\ 3+2\Big\lfloor\frac {f(n,m)-1}{g(n)}\Big\rfloor & \text{otherwise},\end{cases}$$ where $$f(n,m)=m-2(n-1)!$$ and $$g(n)=\begin{cases}1 & \text{if }n<3,\\ 2(n-2)! & \text{otherwise.}\end{cases}$$
Observations
1. $$T(n,m)=1,$$ if $$n=1$$ or $$m\le 2(n-1)!$$.
2. $$T(n,m)$$ is nondecreasing with respect to $$m$$.
3. $$T(n,m)\ge3$$ if $$m\gt 2(n-1)!$$.
4. $$T(2,m)=\begin{cases} 5 & m\le2,\\ 2m-3 & \text{otherwise}.\end{cases}$$
5. If $$n=3,4,5$$, then $$T(n,m)=\begin{cases} 1 & m\le 2(n-1)!,\\ 3+2\lfloor\frac{f(n,m)-1}{2(n-2)!}\rfloor & \text{otherwise}. \end{cases}$$
6. Let $$p(n,j)$$ be the following proposition: $$\text{If n\ge3 and j=\lfloor\frac{f(n,m)-1}{2(n-2)!}\rfloor for some j\ge0 and m, then T(n,m)=3+2j.}$$ $$p(n,j)$$ is true for all $$n\ge3, j\gt0$$.
Items 1, 4 and 6 confirm the formula given in the question.
All above items except item 6 can be proved easily, although item 5 might take a while to sort out case by case.
Let $$S(n,m,a,b,c)=T(c,a)+T(n-c,b)+T(n, m-ab)$$. Then for $$m\gt 2(n-1)!$$, $$T(n,m)= \min\limits_{a,b,c\geq 1,\ c\le\frac n2\\a\leq c!,\ b\leq(n-c)!}S(n,m,a,b,c),$$ where $$c$$ is required to be at most $$\frac n2$$ since otherwise we can replace $$(n,m,a,b,c)$$ by $$(n,m,b, a, n-c)$$.
Proof of item 6 by Well-Founded Induction
Here are the steps.
• Suppose $$j=\lfloor\frac{f(n,m)-1}{2(n-2)!}\rfloor=0$$, i.e., $$2(n-1)!\lt m\le2(n-1)!+2(n-2)!$$. Since $$m\gt 2(n-1)$$, $$T(n,m)\ge3.$$ On the other hand, $$T(n,m)\le S(n,m,1,2(n-2)!,1)=1+1+1=3.$$ So $$T(n,m)=3$$, i.e., $$p(n,j)$$ is true when $$j=0$$.
• Item 5 says that $$p(n,j)$$ is true for $$n=3,4,5$$.
• As induction hypothesis, suppose $$p(x,y)$$ is true for all $$x\le n$$ or $$x=n$$ and $$y\lt j$$, where $$n\ge 6$$ and $$j\ge1$$. So for $$x\le n$$ or $$x=n$$ and $$y\lt j$$, we have $$T(x,y)=3+2\lfloor\frac{f(x,y)-1}{2(x-2)!}\rfloor,$$ i.e., $$(x-2)!(T(x,y)+2x-5)
We will prove that for $$p(n,j)$$ is true. Let $$j=\lfloor\frac{f(n,m)-1}{2(n-1)!}\rfloor$$ for some $$m$$.
Part one, $$T(n,j)\le 3+2j$$
By induction hypothesis, we know that $$T(n, m-2(n-2)!)=3+2(j-1)$$. Hence, $$T(n,m)\le S(n,m,1,2(n-2)!,1)=1 + 1 + T(n, m-2(n-2)!)=3+2j.$$
Part two, $$T(n,j)\ge 3+2j$$
We will prove that $$S(n,m,a,b,c)\ge 3+2j$$ for all valid choices of $$(a,b,c)$$. The case when $$c=1$$ or $$c=2$$ is relatively easy. From now on assume $$3\le c\le \frac n2$$. Because of item 2, we can assume $$m=2(n-1)!+2(n-2)!j+1$$, the smallest value possible such that $$j=\lfloor\frac{f(n,m)-1}{2(n-2)!}\rfloor$$.
Let $$A=T(c,a)$$ and $$B=T(n-c,b)$$. The case when $$A=1$$ or $$B=1$$ is much easier to prove. Now assume $$A,B\ge2$$.
• Since $$c, we have $$a<(c-2)!(A+2c-3)$$.
• Since $$n-c, we have $$b<(n-c-2)!(B+2n-2c-3)$$.
• Since $$ab\ge1$$, we have $$\lfloor\frac{f(n,m-ab)-1}{2(n-2)!}\rfloor, so we can apply induction hypothesis to yield the second inequality below.
Since $$T(n,m)$$ is nondecreasing w.r.t $$m$$, \begin{aligned} S(&n,m,a,b,c)=\\ &\ge A+B+T(n,m-(c-2)!(A+2c-3)(n-c-2)!(B+2n-2c-3))\\ &\ge A+B+3+ 2\lfloor\frac{f(n,m-(c-2)!(A+2c-3)(n-c-2)!(B+2n-2c-3))-1}{2(n-2)!}\rfloor\\ &\ge 3+2j+ A+B +2\lfloor\frac{-(c-2)!(A+2c-3)(n-c-2)!(B+2n-2c-3))}{2(n-2)!}\rfloor\\ &\gt 3+2j+ \frac{(c-2)!(A+2c-3)(n-c-2)!(B+2n-2c-3)}{(n-2)!}(h(c,A,B)-1) \\ \end{aligned}
where $$h(n,A,B,c)=\frac{(A+B-2)(n-2)!} {(c-2)!(A+2c-3)(n-c-2)!(B+2n-2c-3)}$$ By induction hypothesis, $$A=T(c,a)\le T(c,c!)=3+2\left(\frac{c(c-1)}2-(c-1)-1\right)=c^2-3c+3.$$ Similarly, $$B=T(n-c,b)\le =(n-c)^2-3(n-c)+3.$$ Since $$n\ge6$$ and $$c\ge3$$, $$(n-2)!\ge (n-2)(n-3)(n-4)(c-2)!(n-c-2)!.$$
Since $$n\ge6$$, $$c\le \frac n2$$ and $$A,B\ge2$$, $$(n-2)(A+B-2)\gt A+2c-3.$$
\begin{aligned}h(n,A,B,c)= &\ge\frac{(A+B-2)(n-2)(n-3)(n-4)}{(A+2c-3)(B+2n-2c-3)}\\ &\ge\frac{(n-3)(n-4)}{B+2n-2c-3}\frac{(n-2)(A+B-2)}{A+2c-3}\\ &\ge\frac{(n-3)(n-4)}{(n-c)(n-c-1)}\frac{(n-2)(A+B-2)}{A+2c-3}\\ &\gt1 \end{aligned}
So $$S(n,m, a,b,c) \gt 3+2j.$$
The proof is complete. By the way, part one of the proof shows that the minimum can always be obtained at $$c=1,$$ $$a=1,$$ $$b=2(n-2)!.$$
Exercises
Exercise 1. Prove the formula for $$T(2,m)$$.
Exercise 2. (Item 5) Prove the formula for $$T(3,m)$$, $$T(4,m)$$, $$T(5,m)$$ and $$T(6,m)$$. Hint, you may use or adapt the proof of item 6 above. | 2019-06-16T03:25:07 | {
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