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https://se.mathworks.com/help/symbolic/simplify-expression-using-live-editor-task.html | # Simplify Symbolic Expressions Using Live Editor Task
Starting in R2020a, you can interactively simplify or rearrange symbolic expressions using the Simplify Symbolic Expression task in the Live Editor. For more information on Live Editor tasks, see Add Interactive Tasks to a Live Script.
This example shows you how to simplify or rearrange various symbolic expressions into the particular form you require by choosing the appropriate method.
### Simplify a Symbolic Expression
Simplify the expression $i\frac{{e}^{-ix}-{e}^{ix}}{{e}^{-ix}+{e}^{ix}}$.
First, go to the Home tab, and create a live script by clicking New Live Script. Define the symbolic variable x and declare the expression as a symbolic expression.
syms x;
expr = 1i*(exp(-1i*x) - exp(1i*x))/(exp(-1i*x) + exp(1i*x));
In the Live Editor tab, run the code by clicking Run to store x and expr into the current workspace.
Next, open the Simplify Symbolic Expression task by selecting Task > Simplify Symbolic Expression in the Live Editor tab. Select the symbolic expression expr from the workspace and specify the simplification method as Simplify. Select Minimum for the computational effort (fastest computation time).
To get a simpler expression, change the computational effort to Medium.
To experiment with simplifying symbolic expressions, you can repeat the previous steps for other symbolic expressions and simplification methods. You can run the following examples by adding the code to the existing live script or a new live script.
### Simplify a Polynomial Fraction
Simplify the polynomial fraction $\frac{\left({x}^{2}-1\right)\left(x+1\right)}{{x}^{2}-2x+1}$.
Declare the polynomial fraction as a symbolic expression.
expr2 = ((x^2 - 1)*(x + 1))/(x^2 - 2*x + 1);
Select the symbolic expression expr2 from the workspace and specify the simplification method as Simplify fraction.
Select the Expand option to return the numerator and denominator of the simplified fraction in expanded form.
### Rewrite an Expression in a Different Form
Rewrite the trigonometric function $\mathrm{tan}\left(x\right)$ in terms of the sine function.
Declare $\mathrm{tan}\left(x\right)$ as a symbolic expression.
expr3 = tan(x);
Select the symbolic expression expr3 from the workspace and specify the simplification method as Rewrite. Choose sin to rewrite $\mathrm{tan}\left(x\right)$ in terms of the sine function.
### Expand a Logarithmic Expression
Expand the expression $log\left(\frac{{x}^{3}\phantom{\rule{0.16666666666666666em}{0ex}}{e}^{x}}{2}\right)$ using the logarithmic identities.
Declare the logarithmic expression as a symbolic expression.
expr4 = log(x^3*exp(x)/2);
Select the symbolic expression expr4 from the workspace and specify the simplification method as Expand. By default, the symbolic variable x in expr4 is complex when it is initially created. The Expand method does not simplify the input expression because the logarithmic identities are not valid for complex values of variables. To apply identities that are convenient but do not always hold for all values of variables, select the Ignore analytic constraints option.
### Simplify the Sum of Two Integral Expressions
Simplify the sum of two integral expressions: ${\int }_{a}^{b}x\phantom{\rule{0.16666666666666666em}{0ex}}f\left(x\right)\text{d}x+{\int }_{a}^{b}g\left(y\right)\text{d}y$.
First, define $a$ and $b$ as symbolic variables, and $f\left(x\right)$ and $g\left(y\right)$ as symbolic functions. Use the int function to represent the integrals.
syms a b f(x) g(y)
expr5 = int(x*f(x),x,a,b) + int(g(y),y,a,b);
Select the symbolic expression expr5 from the workspace and specify the simplification method as Combine. Choose int as the function to combine.
### Generate Code
To view the code that a task used, click at the bottom of the task window. The task displays the code block, which you can cut and paste to use or modify later in the existing script or a different program. For example:
Because the underlying code is now part of your live script, you can continue to use the variables created by the task for further processing. For example, define the functions $f\left(x\right)$ and $g\left(x\right)$ as $f\left(x\right)=x$ and $g\left(x\right)=\mathrm{cos}\left(x\right)$. Evaluate the integrals in simplifiedExpr3 by substituting these functions. | 2022-09-25T06:20:15 | {
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https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10B_Problems/Problem_23&diff=prev&oldid=73774 | # Difference between revisions of "2004 AMC 10B Problems/Problem 23"
## Problem
Each face of a cube is painted either red or blue, each with probability 1/2. The color of each face is determined independently. What is the probability that the painted cube can be placed on a horizontal surface so that the four vertical faces are all the same color?
$\mathrm{(A) \ } \frac{1}{4} \qquad \mathrm{(B) \ } \frac{5}{16} \qquad \mathrm{(C) \ } \frac{3}{8} \qquad \mathrm{(D) \ } \frac{7}{16} \qquad \mathrm{(E) \ } \frac{1}{2}$
## Solution
If the orientation of the cube is fixed, there are $2^6=64$ possible arrangements of colors on the faces. There are $$2\dbinom{6}{6}=2$$ arrangements in which all six faces are the same color and $$2\dbinom{5}{6}=12$$ arrangements in which exactly five faces have the same color. In each of these cases the cube can be placed so that the four vertical faces have the same color. The only other suitable arrangements have four faces of one color, with the other color on a pair of opposing faces. Since there are three pairs of opposing faces, there are $2(3)=6$ such arrangements. The total number of suitable arrangements is therefore $2+12+6=20$, and the probability is $\dfrac{20}{64}=\dfrac{5}{16}.$
## Solution 2
Label the six sides of the cube by numbers $1$ to $6$ as on a classic dice. Then the "four vertical faces" can be: $\{1,2,5,6\}$, $\{1,3,4,6\}$, or $\{2,3,4,5\}$.
Let $A$ be the set of colorings where $1,2,5,6$ are all of the same color, similarly let $B$ and $C$ be the sets of good colorings for the other two sets of faces.
There are $2^6=64$ possible colorings, and there are $|A\cup B\cup C|$ good colorings. Thus the result is $\frac{|A\cup B\cup C|}{64}$. We need to compute $|A\cup B\cup C|$.
Using the Principle of Inclusion-Exclusion we can write $$|A\cup B\cup C| = |A|+|B|+|C| - |A\cap B| - |A\cap C| - |B\cap C| + |A\cap B\cap C|$$
Clearly $|A|=|B|=|C|=2^3=8$, as we have two possibilities for the common color of the four vertical faces, and two possibilities for each of the horizontal faces.
What is $A\cap B$? The faces $1,2,5,6$ must have the same color, and at the same time faces $1,3,4,6$ must have the same color. It turns out that $A\cap B=A\cap C=B\cap C= A\cap B\cap C =$ the set containing just the two cubes where all six faces have the same color.
Therefore $|A\cup B\cup C| = 8+8+8-2-2-2+2 = 20$, and the result is $\frac{20}{64}=\boxed{\frac{5}{16}}$.
## Solution 3
Suppose we break the situation into cases that contain four vertical faces of the same color:
I. Two opposite sides of same color There are 3 ways to choose the two sides, and then two colors possible, so $3*2=6$
II. One face different from all the others There are 6 ways to choose this face, and 2 colors, so $6*2=12$
III. All faces same There are 2 colors, so two ways for all faces to be the same.
Adding them up, we have a total of 20 ways to have four vertical faces the same color. The are $2^6$ ways to color the cube, so the answer is $\frac{20}{64}=\boxed{\frac{5}{16}}$ | 2021-05-15T18:31:07 | {
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http://www.casamacario.com/1e2tyfcj/3c71ed-examples-of-equivalence-relation-in-discrete-mathematics | Equivalence relation ( ) on the set Is a binary relation for which the following conditions are met: Reflexivity: for anyone at , Symmetry: if then , Transitivity: if and then . 2 CS 441 Discrete mathematics for CS M. Hauskrecht Binary relation Definition: Let A and B be two sets. Relations in Discrete Math 1. Proof: The equivalence classes split A into disjoint subsets. . Binary Relation Representation of Relations Composition of Relations Types of Relations Closure Properties of Relations Equivalence Relations Partial Ordering Relations. . Universal Relation. Reflexivity: x A, xRx: Symmetry: x,y A, xRy yRx: Transitivity: x,y,z A, xRy yRz xRz : Example. Relations . COMPSCI 230: Discrete Mathematics for Computer Science February 11, 2019 Lecture 9 Lecturer: Debmalya Panigrahi Scribe: Kevin Sun 1 Overview In this lecture, we study a special class of relations on a set known as equivalence relations. . Examples: People with the same birthday, the same month of birth, the same year of birth, the same zodiac sign; people from the same prefecture/country, cities in the same prefecture/country; An equivalence relation is a relation that is reflexive, symmetric, and transitive . . In math, a relation is just a set of ordered pairs. . The relations we will deal with are very important in discrete mathematics, and are known as equivalence relations. . Different types of recurrence relations and their solutions. An equivalence relation on a set S, is a relation on S which is reflexive, symmetric and transitive. For example, the definition of an equivalence relation requires it to be symmetric. Discrete Mathematics. . 29, Jan 18. Discrete Mathematics - Propositional Logic - The rules of mathematical logic specify methods of reasoning mathematical statements. Practice Set for Recurrence Relations. Lifetime Access! . Thus, according to Theorem 8.3.1, the relation induced by a partition is an equivalence relation. CONTENTS v 5.5 Stronginduction. Examples: Let S = ℤ and define R = {(x,y) | x and y have the same parity} i.e., x and y are either both even or both odd. Johny Johny. All definitions tacitly require transitivity and reflexivity. All definitions tacitly require transitivity and reflexivity. . Definition of Logical Equivalence Formally, Two propositions and are said to be logically equivalent if is a Tautology. Example $$\PageIndex{8}$$ Congruence Modulo 5; Summary and Review; Exercises; Note: If we say $$R$$ is a relation "on set $$A$$" this means $$R$$ is a relation from $$A$$ to $$A$$; in other words, $$R\subseteq A\times A$$. . › Discrete Math. . Discrete Mathematics Online Lecture Notes via Web. They essentially assert some kind of equality notion, or equivalence, hence the name. Equivalence relations, equivalence classes, and partitions ; Partial and total orders; This week's homework Leftovers Summary of Last Lecture. Set theory. A symmetric relation is a type of binary relation. Definition: Equivalence Relation. What is a 'relation'? Content . R is symmetric if for all x,y A, if xRy, then yRx. RELATIONS PearlRoseCajenta REPORTER 2. x + 1 = 2 x + y = z Richard Mayr (University of Edinburgh, UK) Discrete Mathematics… Examples of propositions: The Moon is made of green cheese. 12, Jan 18 . - is a pair of numbers used to locate a point on a coordinate plane; the first number tells how far to move horizontally and the second number tells how far to move vertically. . Related. It contains well written, well thought and well explained computer science and programming articles, quizzes and practice/competitive programming/company interview Questions. In mathematics, an equivalence relation is a binary relation that is reflexive, symmetric and transitive. We give examples and then prove a connection between equivalence relations and partitions of a set. . Characteristics of equivalence relations . Over 6.5 hours of Learning! Distinct equivalence classes of an equivalence relation on R^2: Discrete Math: Oct 3, 2017: equivalence classes: Discrete Math: Sep 11, 2017: Equivalence relation/ Equivalence classes: Discrete Math: Feb 6, 2016: need help with modular arithmetic and equivalence classes. Equivalence Relations Partition a Set 14 Stirling Numbers of the Second Kind 16 . A Computer Science portal for geeks. Discrete mathematics is the branch of mathematics dealing with objects that can consider only distinct, separated values. Featured on Meta New Feature: Table Support. . 1 + 0 = 1 0 + 0 = 2 Examples that are not propositions. Trenton is the capital of New Jersey. Toronto is the capital of Canada. . . Join in to learn Discrete Mathematics, equally important from the academic as well as real-world knowledge. 3.Or more commonly, simply using relational notation a ˘b. i.e. Submitted by Prerana Jain, on August 17, 2018 Types of Relation. 2. is a contradiction. For a relation R to be an equivalence relation, it must have the following properties, viz. . Swag is coming back! .87 5.5.1 Examples. . Greek philosopher, … 2. . An example is the relation "is equal to", because if a = b is true then b = a is also true. Definition: A relation on a set A is called an equivalence relation if it is reflexive, symmetric, and transitive. I was going through the text "Discrete Mathematics and its Application" by Kenneth Rosen (5th Edition) where I am across the definition of equivalence relation and felt that it is one sided. Combinatorics. What time is it? 19.2k 4 4 gold badges 22 22 silver badges 51 51 bronze badges. More than 1,700 students from 120 countries! Discrete Mathematics Online Lecture Notes via Web. . You can’t get very far in logic without talking about propositional logic also known as propositional calculus. . 2 Equivalence Relations Definition 1. Equivalence Relation. A relation r from set a to B is said to be universal if: R = A * B. . Definition of an Equivalence Relation A relation on a set that satisfies the three properties of reflexivity, symmetry, and transitivity is called an equivalence relation. . 97 1 1 silver badge 7 7 bronze badges $\endgroup$ $\begingroup$ you're confusing a set of representatives with the set of classes. . For example, the definition of an equivalence relation requires it to be symmetric. . Notice that two lines in S are parallel if and only if their slope is equal. There are many types of relation which is exist between the sets, 1. How many symmetric and transitive relations are there on ${1,2,3}$? . 5 CS 441 Discrete mathematics for CS M. Hauskrecht Equivalence classes and partitions Theorem: Let R be an equivalence relation on a set A.Then the union of all the equivalence classes of R is A: Proof: an element a of A is in its own equivalence class [a]R so union cover A. Theorem: The equivalence classes form a partition of A. In mathematics, an equivalence relation is a binary relation that is reflexive, symmetric and transitive.The relation "is equal to" is the canonical example of an equivalence relation. . Modules Covered: Set Theory; Logic; Relations and Functions; Counting; Graphs; Algebraic structures & Coding theory; Feel forward to have a look at course description and demo videos and we look forward to see you learning with us. Logical equivalence is a type of relationship between two statements or sentences in propositional logic or Boolean algebra. Example, 1. is a tautology. Formally, a binary relation R over a set X is symmetric if: ∀, ∈ (⇔). 2.An directed edge a b . Discrete Mathematics. The notation is used to denote that and are logically equivalent. A1. . . . discrete-mathematics equivalence-relations. A binary relation from A to B is a subset of a Cartesian product A x B. R t•Le A x B means R is a set of ordered pairs of the form (a,b) where a A and b B. Browse other questions tagged discrete-mathematics relations or ask your own question. asked Jan 17 '17 at 11:21. . Sets Theory. . A proposition is a declarative sentence (a sentence that declares a fact) that is either true or false. . 3. is a contingency. . Equivalence Relation: A relation is an Equivalence Relation if it is reflexive, symmetric, and transitive. We call two lines parallel in S if and only if they are equal or do not intersect. R must be: . Number Theory: Apr 12, 2015 . share | cite | improve this question | follow | edited Jan 17 '17 at 11:45. zoli. Record of the form " "Reads like" is equivalent to ". . That a thing a is related to a thing b can be represented by 1.An ordered pair (a, b). .88 Fundamental of Discrete Math – Set Theory, Relations, Functions and Mathematical Induction! cse 1400 applied discrete mathematics relations 2 Problems on Relations 18 Abstract A relation ˘describes how things are connected. Q2. Therefore, this relation is not equivalent. Sample/practice exam October 24 Fall 2016, answers Exam 2 May 11 Spring 2015, answers Discrete Mathematics - Lecture 1.7 Introduction to Proofs Discrete Mathematics - Lecture 4.3 Primes and Greatest Common Divisors Discrete Mathematics - Lecture 6.1 The Basics of Counting Discrete Mathematics - Lecture 3336 Recurrence Relations . What are the types of relation in maths? Q1. The parity relation is an equivalence relation. 1. Sit down! Certificate of Completion for your Job Interviews! Sets Introduction Types of Sets Sets Operations Algebra of Sets Multisets Inclusion-Exclusion Principle Mathematical Induction. 22, Jun 18. Let R be a binary relation on a set A. R is reflexive if for all x A, xRx. relation R={(1,1),(2,2),(3,3),(1,2), ... Discrete Mathematics | Representing Relations. R is transitive if for all x,y, z A, if xRy and yRz, then xRz. . Equivalence Classes and Partitions We recall that a binary relation R on a set A is an equivalence relation if and only if the following 3 conditions are all true. In this article, we will learn about the relations and the different types of relation in the discrete mathematics. Discrete Mathematics Example 1.2.2 Consider the plane R2 and in it the set S of straight lines. . . Graph theory. . . R is an equivalence relation if A is nonempty and R is reflexive, symmetric and transitive. . . There are 9 types of relations in maths namely: empty relation, full relation, reflexive relation, irreflexive relation, symmetric relation, anti-symmetric relation, transitive relation, equivalence relation, and asymmetric relation. 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And Mathematical Induction are equal or do not intersect an equivalence relation it! Be: 2 CS 441 discrete mathematics Relations 2 Problems on Relations Abstract! Jan 17 '17 at 11:45. zoli set of ordered pairs a sentence that declares a fact ) that either... | 2021-06-17T23:25:02 | {
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https://stats.stackexchange.com/questions/458509/estimating-population-standard-deviation-from-subsample-means | # Estimating population standard deviation from subsample means [duplicate]
I would like to estimate the standard deviation of a population. The data I have is the size and mean of each of $$k$$ independent uniform subsamples: $$(n_1, \mu_1), (n_2, \mu_2), \ldots, (n_k, \mu_k)$$. Here, the $$i$$'th subsample has $$n_i$$ elements, and those $$n_i$$ elements have a mean of $$\mu_i$$.
What is the best way to utilize all $$2k$$ values to produce the best estimate of the population standard deviation $$\sigma$$?
Let $$\mu$$ be the population mean and $$\sigma$$ be the population SD. You have observations $$x_i$$ of $$k$$ independent variables $$X_i$$ having a common mean $$\mu$$ and variances $$\sigma^2/n_i,$$ $$i=1,2,\ldots, k.$$
You are noncommittal about the sense of "best." To make this specific, let us find the Least Squares estimator of $$\sigma.$$ This is a Weighted Least Squares problem with weights $$1/(1/n_i)=n_i,$$ whence
$$\hat\mu = \frac{1}{n_1+n_2+\cdots+n_k}\, \left(n_1x_1 + n_2x_2 + \cdots + n_k x_k\right)$$
with residuals
$$r_i = x_i - \hat\mu$$
which entails
$$\hat\sigma^2 = \frac{1}{k-1}\, \left(n_1 r_1^2 + n_2 r_2^2 + \cdots + n_k r_k^2\right).$$
We may therefore take
$$\hat \sigma = \sqrt{\hat\sigma^2}$$
as a "best" estimate of $$\sigma.$$
As a check, straightforward algebra will verify that $$E[\hat\mu] = \mu$$ and $$E[\hat\sigma^2]=\sigma^2,$$ standard properties of Least Squares estimates. As another check, these calculations reproduce exactly the output of the lm function in R using the $$n_i$$ in its weights argument.
• Thanks, this makes a lot of sense. Btw, I think you have an extra subscript in your residual equation. Apr 5, 2020 at 15:44
• Thank you for the careful review! I have fixed that typo.
– whuber
Apr 5, 2020 at 17:31
• I found that this question appears to have been asked previously, with the top answer coming from you!: stats.stackexchange.com/q/24936/2221 Should this question be removed since it's a repeat? Or is there something sufficiently unique about this question to warrant keeping them both? Apr 6, 2020 at 0:11
• That's a good find (as you might imagine, I had totally forgotten about it). Maybe the best resolution would be for me to add this post as an edit to my answer there. What do you think?
– whuber
Apr 6, 2020 at 12:41
• Sure, would you like to do that? I think this question/answer is a little cleaner so I think there is value to having it. Apr 6, 2020 at 12:46 | 2023-03-22T18:25:54 | {
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https://math.stackexchange.com/questions/1858794/is-there-symbol-to-denote-a-combination-and-permutation | # Is there symbol to denote a combination and permutation?
For example, let's say I wanted to denote any arbitrary, $2$ number combination of the letters, A, B and C. So you can have AB, AC, and BC. Say you wanted a way to represent any given combination, is there a shorthand way to denote this idea?
The reason why you may want to do this, is that whilst every combination is unique, every combination may share a unique property, which wouldn't be seen given other combination sets (i.e combinations derived from other pairs of value, say, $\{C,D, E\}$, or $\{1, 2, 3\}$ etc.) if you wanted to refer to a certain property that every combination posses (relative to a specific set of values), a shorthand for this would I think be convenient :)
And of course, the same can be said for permutations.
As a spinoff of that notation that $n \choose k$ denotes the number of $k$-element subsets of a set of size $n$, we can define $S \choose k$ to denote the set of all $k$-element subsets of a set $S$. So, to say that you're thinking about one of the sets $\{A,B\},\, \{B,C\}$ or $\{A, C\}$, you might write something like "$\Delta \in {{\{A, B, C\}}\choose{2}}$" to signify that the set $\Delta$ is one of the above $2$-element subsets of $\{A, B, C\}$.
The notation is reasonably "natural" in the sense that $$\left\lvert {S \choose k }\right\rvert = {{\lvert S \rvert} \choose {k}}.$$
I personally like this notation, and know it is used to some extent, but I have no idea how popular it is "in the field." As for permutations, I have never seen anything comparable.
And, as a general rule, it never hurts to make a note about what your notation means, if you're not just writing things down for yourself (and honestly, it couldn't hurt then, either!)
• For what it's worth, I use this notation and can reasonably be considered to be "in the field". Jul 14 '16 at 0:58
• @AustinMohr Good to know! I had assumed as much, but for me, I rarely encounter anyone needing to specify that the subsets are a certain size. So I see $2^S$ plenty often, but never anyone needing $S \choose k$. Jul 14 '16 at 1:02
• Cheers pjs36, the only thing I don't get is the | | symbol, at least in this context. Oh, and is subset here the same as some combination? Jul 14 '16 at 1:03
• @user2901512 The notation $|S|$ just means the cardinality, or size, of the set $S$. So $|\{A,B,C\}| = 3 = |\{C, D, E\}|$, for example. Jul 14 '16 at 1:05
• Oh sweet, I never knew/thought of that, that's so cool ^.^ Jul 14 '16 at 1:07 | 2021-10-23T15:01:02 | {
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https://math.stackexchange.com/questions/2378038/fx0-implies-int-abfx-mathopdx0 | # $f(x)>0\implies\int_a^bf(x)\mathop{dx}>0$
Suppose $f$ is Riemann-integrable on $[a,b]$ such that $f(x)>0, \forall x \in [a,b].$
Prove: $$\int_a^bf(x)\mathop{dx}>0$$
Provided solution:
Suppose by contradiction that $I=\int_a^bf(x)\mathop{dx}=0.$
Let us take a sequence of normal partitions $T_n$ of $[a,b],$ that is partitons of $n$ intervals of length $\frac{b-a}{n}.$
Then for $n$ large enough, upper Darboux sum of $f$ is small as we wish.
So there exists $n_0$ such that for all $n>n_0:$
$$\tag{1}\sum_{i=0}^{n-1}\sup_{[x_i,x_{i+1}]}f\cdot\Delta x_i=\frac{b-a}{n}\sum_{i=0}^{n-1}\sup_{[x_i,x_{i+1}]}f<\frac{b-a}{2}$$
Therefore, there exists an interval $I_1:=[x_i,x_{i+1}],$ such that:
$$\tag{2}\sup_{[x_i,x_{i+1}]}f<\frac{1}{2}$$
By the integral monotonicity and positiveness of $f:$
$$\tag{3}0=\int_a^bf(x)\mathop{dx}\geq \int_{I_1}f(x)\mathop{dx} \geq 0$$
Hence, $\int_{I_1}f(x)\mathop{dx}=0.$
Repeating the process on $I_1,$ let us take a sequence of normal partitions of $[x_i,x_{i+1},$ that is intervals $[y_i,y_{i+1}]$ of length $\frac{x_{i+1}-x_i}{n}.$ Therefore, there exists $n_1$ such that for all $n>n_1:$
$$\tag{4}\sum_{i=0}^{n-1}\sup_{[y_i,y_{i+1}]}f\cdot\Delta x_i=\frac{x_{i+1}-x_i}{n}\sum_{i=0}^{n-1}\sup_{[y_i,y_{i+1}]}f<\frac{x_{i+1}-x_i}{4}$$
So there exists an interval $I_2:=[y_i,y_{i+1}] \subset I_1,$ such that:
$$\tag{5} \sup_{I_2}f<\frac{1}{4}$$
Continuing like that, we get a sequence of intervals: $\ \dots \subseteq I_2 \subseteq I_1,$ such that:
$$\tag{6} \sup_{I_n}f \leq \frac{1}{2^n}$$
By Cantor's intersection theorem:
$$\tag{7} \bigcap_{n=0}^\infty I_n \neq \emptyset$$
But, if $x \in \bigcap_{n=0}^\infty I_n,$ then $f(x) \leq \frac{1}{2^n},$ for all $n$, hence $f(x)=0,$ contradicting $f(x)>0.$
My questions:
$(a)$ At $(1),$ I know $\sum_{i=0}^{n-1} \Delta x_i = b-a,$ but why does it equal $\frac{b-a}{n}?$ And why is the inequality true?
$(a)$ At $(3),$ how is the monotonicity is used?
Any help is appreciated.
• Wouldn't the contradiction be $I \le 0$? – GFauxPas Jul 31 '17 at 19:20
• Nice proof. Another approach is show that Riemann integrable functions are continuous somewhere. See this answer math.stackexchange.com/a/519921/72031 – Paramanand Singh Jul 31 '17 at 19:46
$(i)$ The partition is normal (i.e., each subinterval is of equal length), hence $\Delta x_i = \frac{b-a}{n}$. This was simply factored out of the sum in step $(1)$.
$(ii)$ Monotonicity is used to show that $\int_{[a,b]}f \geq \int_{I_1} f$. This is because $\int_{[a,b] \setminus I_1} f \geq 0$ since $f \geq 0$.
• I see it now. Could you explain the inequality in $(1)?$ – Itay4 Aug 1 '17 at 5:44
$\Delta x_i=\frac {b-a}n$ because b-a is divided into n equal subintervals.
Integration is monotonic in the sense that $f \ge0 \implies \int_a^b f \ge \int_c^d f$ when $[c,d] \subset [a,b]$. | 2020-01-28T13:31:11 | {
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http://math.stackexchange.com/questions/161338/finding-cartesian-equations-from-parametric?answertab=votes | # Finding Cartesian equations from parametric
I am suppose to find a cartesian equation by eliminating the parameters of
$$x = \sin \left(\dfrac{1}{2}\theta \right), y = \cos \left(\dfrac{1}{2}\theta \right)$$
I won't even go through my work since it is so far off but I am suppose to get $x^2 + y^2 = 1$ but I get something with arccos and such which doesn't make any sense.
-
The equation $x^2+y^2=1$ works. How can we tell? Experience: if you have a sine and a cosine of the same quantity (in this case, of $\frac{1}{2}\theta$), then you can always "get rid" of the dependence on the quantity by squaring and adding; no matter what the quantity is, you always get $\sin^2(\text{quantity}) + \cos^2(\text{quantity}) = 1$.
But let's say you don't want to try things to see how the parameter can be eliminated. In this case, you can simply solve for the parameter in each equation: \begin{align*} x&=\sin\left(\frac{1}{2}\theta\right)\\ \arcsin(x) &= \frac{1}{2}\theta\\ 2\arcsin(x) &=\theta;\\ y &= \cos\left(\frac{1}{2}\theta\right)\\ \arccos(y) &= \frac{1}{2}\theta\\ 2\arccos(y) &= \theta. \end{align*} Therefore, $x$ and $y$ will satisfy $$2\arcsin(x) = 2\arccos(y)$$ or equivalently, $$\arcsin(x) = \arccos(y).$$
The problem is that this equation is ugly; arcsine and arccosine are annoying functions. Better to try to get rid of them. One way to do that is to first take sines on both sides, to get $$x = \sin(\arccos(y)).$$ Now, what is $\sin(\arccos(y))$? Well, since $\sin^2z + \cos^2z = 1$ (there is again!), then $\sin^2 z = 1-\cos^2 z$, so $|\sin z| = \sqrt{1-\cos^2(z)}$. Letting $z=\arccos(y)$, we get $$\sin(\arccos(y)) = \pm\sqrt{1 - \cos^2(\arccos(y))} = \pm\sqrt{1-y^2}.$$ So our equation becomes $$x = \pm\sqrt{1-y^2}.$$ Again, the $\pm$ is annoying. We can get rid of it by squaring both sides, so we get $$x^2 = 1-y^2$$ which can then be rewritten as $$x^2 + y^2 = 1.$$ Now, this is a nice equation; we know exactly what it is (circle of radius $1$ centered at the origin), and it does not involve any annoying functions like inverse trigonometric functions. So this is a good stopping point.
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@BabakSorouh: Okay, that's it. Please stop commenting on most of my answers with comments that have nothing but expressions of praise. Not only do some of them seem rather exaggerated, but, frankly, I'm finding it creepy. I'm feeling stalked, and I don't like it. – Arturo Magidin Jun 21 '12 at 22:02
I guess you're right, but he seems to write similar comments in most answers. – Pedro Tamaroff Jun 22 '12 at 13:03
@Peter: I only get notified when he posts them in mine, though... – Arturo Magidin Jun 22 '12 at 14:41
@ArturoMagidin True. I guess it can get wierd. – Pedro Tamaroff Jun 22 '12 at 15:32
Remember that for all $z$, we have $$\cos^2(z) + \sin^2(z) = 1$$
-
I don't know what that weird a is. – user138246 Jun 21 '12 at 20:26
I give up ${}{}$ – user17762 Jun 21 '12 at 20:31
Because Jordan :) – user17762 Jun 21 '12 at 20:40
@Jordan: Dear Jordan, Marvis was trying to tell you that the equation $\cos^2(z)+\sin^2(z)=1$ can make your question solved. In fact, if you change your variable from $\frac{1}{2}\theta$ to $z$ then by using a very basic formula, you will get your answer as you want. This basic trigonometric formula is what Marviz noted you. – Babak S. Jun 21 '12 at 20:44
$x$ and $y$ are related because $x$ is the sine of a certain thing, and $y$ is the cosine of the same thing, and sine and cosine are closely related to one another, as you might guess from the names. – MJD Jun 21 '12 at 21:07
This is JUST to clear what Marvis noted.
$x=\sin(\frac{1}{2}\theta)$ $\longrightarrow$ $x^2=\sin^2(\frac{1}{2}\theta)$ ,
$y=\cos(\frac{1}{2}\theta)$ $\longrightarrow$ $y^2=\cos^2(\frac{1}{2}\theta)$
Checking an elementary Mathematics book you'll find there is an trigonometric equation (See Marvis's answer). Now if you get $\frac{1}{2}\theta=z$ or $x$ or $y$ or every name you want and by adding above conclusions $x^2+y^2=1$. Isn't this enough? :-)
-
No you miraculously managed to skip the only part I do not understand. I already know every single thing stated in this thread except the one thing everyone is unwilling to tell me. Why is the y and x being added and why it is equal to one? – user138246 Jun 21 '12 at 21:03
@Jordan: $x$ and $y$ is not being added here!. The squares of them separately are being added. Don’t you want to reach $x^2+y^2=1$??? Don't we see an adding in the previous equation? – Babak S. Jun 21 '12 at 21:09
Yes I see adding, and that is why I am confused. Why are they being added together? – user138246 Jun 21 '12 at 21:12
Eliminating the parameter means finding some equation involving $x$ and $y$ that expresses a relationship between them that does not depend on the parameter. In this case, one such equation is $x^2+y^2=1$. They get added because that eliminates the parameter. Alternatively, solve for the parameter in terms of $x$ and $y$, $\theta = 2\arcsin(x)$, $\theta=2\arccos(y)$, and then set them equal: $2\arcsin(x)=2\arccos(y)$; hence $\arcsin(x)=\arccos(y)$ is one such equation. This equation can be converted into $x^2+y^2=1$ by suitable manipulations. – Arturo Magidin Jun 21 '12 at 21:25
Just take,
$$x^2 = \sin^2\left(\frac{\theta}{2}\right)$$ and $$y^2 = \cos^2\left(\frac{\theta}{2}\right) ,$$
$$cos^2(\phi) + sin^2(\phi) = 1$$
$$\implies x^2 + y^2 = 1$$ That's it! $$\LaTeX$$
- | 2015-11-25T07:06:58 | {
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https://severemusick.com/victoria/solving-recurrence-relations-using-generating-functions-pdf.php | # Victoria Solving Recurrence Relations Using Generating Functions Pdf
## Solving Recurrences Calvin College
### Solving Recurrence Relations Oscar Levin PhD
1 Solving recurrences Stanford University. We can use generating functions to derive the closed-form solution. The basic procedure: 1. Derive a generating function from the recurrence relation 2. Manipulate the generating function into an invertible form (for us, this will be a sum of basic geometric generating functions like we used in the last lecture) 3. Invert the generating function and recover the formula for f n Deriving the, Math 300: Solving a Recurrence Relation with Generating Functions. Problem: Solve for fa ng, where a 0 = 1 and a n 23a n 1 = n for n 1. Solution: Let G(x) = P 1 n=0 a nx n be the generating function for fa ng. The idea of this technique is to solve for G(x) using the recurrence relation in a closed form (that is, without using sigma notation). Then, when we rewrite G(x) as a power series, its.
### Advanced Counting Techniques University of California
generatingfunctionology math.upenn.edu. Solving Linear Recurrence Relations Niloufar Shafiei. 1 Review A recursive definition of a sequence specifies Initial conditions Recurrence relation Example: a 0=0 and a 1=3 a n = 2a n-1 - a n-2 a n = 3n Initial conditions Recurrence relation Solution. 2 Linear recurrences Linear recurrence: Each term of a sequence is a linear function of earlier terms in the sequence. For example: a 0 = 1 a 1, Recurrence Relations and Generating Functions Recurrence Relations Terminology Definition A recurrence relation for a sequence an is a relation of the form an+1 = f (a1 , a2 , . . . , an ). We do not expect to have a useful method to solve all recurrence relations. This definition actually applies to any sequence! We shall break down the functions for which we do have effective methods to.
14/10/2013 · Complete set of Video Lessons and Notes available only at http://www.studyyaar.com/index.php/module/36-recurrence-relations-and-generating-functions … Recurrence Relations & Generating Functions This page is an extension to my Fibonacci and Phi Formulae with an introduction to Recurrence Relations and to Generating Functions. A recurrence relation is a way of defining a series in terms of earlier member of the series.
Recurrence Relations and Generating Functions Recurrence Relations Terminology Definition A recurrence relation for a sequence an is a relation of the form an+1 = f (a1 , a2 , . . . , an ). We do not expect to have a useful method to solve all recurrence relations. This definition actually applies to any sequence! We shall break down the functions for which we do have effective methods to Solving Recurrences 2.1 T ypes of Recurrences 2.2 Finding Generating Functions 2.3 P a rtial Fractions 2.4 Characteristic Roots 2.5 Sim ultaneous Recur sions 2.6 Fibonacci Number Identities 2.7 Non-Constant Coef Þ cients 2.8 Divide-and-Conquer Relations 1. 2 Chapter 2 Solving Recurrences. Section 2.1 Types of Recurrences 3 2.1 TYPES OF RECURRENC ES. 4 Chapter 2 Solving Recurrences …
Solving Recurrence Relations with Generating Functions The sequence of numbers 1,1,2,3,5,8,13,. is known as the Fibonacci sequence. We can describe the sequence in terms of a recurrence relation. There are several methods for solving recurrence equations. The simplest is to guess the solution and then verify that the guess is correct with an induction proof. As a basis for a good guess, let’s look for a pattern in the values of Tncomputed above: 1, 3, 7, 15, 31, 63. A natural guess is TnD2n 1. But whenever you guess a solution to a recurrence, you should always verify it with a proof
Chapter 1 Introductory ideas and examples A generating function is a clothesline on which we hang up a sequence of numbers for display. What that means is … MATH 203, PROBLEM SET 2 DUE IN LECTURE ON MONDAY, FEB. 1. Generating Functions These problems have to do with the formulas worked out in class for the terms of a
Solving Linear Recurrence Relations 4.4 Generating Functions Ch. 4.1, 4.2 & 4.5 2 Agenda Recurrence Relations Modeling with Recurrence Relations Linear NonhomogeneousRecurrence Relations with Constant Coefficients Generating Functions Useful Facts About Power Series Extended Binomial Coefficient Extended Binomial Theorem Counting Problems and Generating Functions Using Generating Functions … Section 2: Solving Recurrence Relations by Iteration • As we noted at the end of the last lecture, when analyzing recurrence relations, we want to rewrite the general term as a function of the index and independent of predecessor terms. • This will allow us to compute any arbitrary term in the sequence without having to compute all the previous terms. • In this section, we will look at
8/05/2015 · In this video we use generating functions to solve nonhomogeneous recurrence relations. This is a pretty long process that requires fairly good attention to … Solutions to Exercises Chapter 4: Recurrence relations and generating functions 1 (a) There are n seating positions arranged in a line. Prove that the number
Using Generating Functions to Solve Recurrence Relations We may use recurrences to derive generating functions. Example 4. Find the generating function for the sequence fa Recurrence Relations Many algo rithm s pa rticula rly divide and conquer al go rithm s have time complexities which a re naturally m odel ed b yr ecurrence relations Ar ecurrence relation is an equation which is de ned in term sof its elf Why a re recurrences go o d things Many natural functions a re easily exp ressed as re currences a n n n pol y nomial a n n n exponential a n n n we ir d f
Recurrence Relations and Generating Functions NgГ y 27 thГЎng 10 nДѓm 2011 () Recurrence Relations and Generating FunctionsNgГ y 27 thГЎng 10 nДѓm 2011 1 / 1. Recursive Problem Solving Question Certain bacteria divide into two bacteria every second. It was noticed that when one bacterium is placed in a bottle, it fills it up in 3 minutes. How long will it take to fill half the bottle Solving Linear Recurrence Relations Niloufar Shafiei. 1 Review A recursive definition of a sequence specifies Initial conditions Recurrence relation Example: a 0=0 and a 1=3 a n = 2a n-1 - a n-2 a n = 3n Initial conditions Recurrence relation Solution. 2 Linear recurrences Linear recurrence: Each term of a sequence is a linear function of earlier terms in the sequence. For example: a 0 = 1 a 1
Towers of Hanoi Peg 1 Peg 2 Peg 3 Hn is the minimum number of moves needed to shift n rings from Peg 1 to Peg 2. One is not allowed to place a larger ring on top of a smaller ring. On 28 January 2017 MathCounts will discuss Solving Basic Recurrence Relations with Generating Functions 2 . These problems are posed to help guide the discussion.
There are several methods for solving recurrence equations. The simplest is to guess the solution and then verify that the guess is correct with an induction proof. As a basis for a good guess, let’s look for a pattern in the values of Tncomputed above: 1, 3, 7, 15, 31, 63. A natural guess is TnD2n 1. But whenever you guess a solution to a recurrence, you should always verify it with a proof A recurrence is a recursive description of a function, or in other words, a description of a function in terms of itself. Like all recursive structures, a recurrence consists of one or more base cases and
Check your solution for the closed formula by solving the recurrence relation using the Characteristic Root technique. 10 You have access to $$1 \times 1$$ tiles which come in 2 different colors and $$1\times 2$$ tiles which come in 3 different colors. Theorem 1 can be proved using generating functions and partial fractions, but we will not do it here. The theorem gives a general method for solving recurrence equations, which we
Chapter 1 Introductory ideas and examples A generating function is a clothesline on which we hang up a sequence of numbers for display. What that means is … relations which arise in the course of solving problems by conditioning, one can often convert the recurrence relation into a linear di erential equation for a generating function, to be solved subject to appropriate boundary conditions.
initial conditions and satis es the recurrence relation, it must match the sequence an exactly! 2. a 0 = 2;a 1 = 2. The characteristic function is the same, with roots r 1 = 2 and r 2 = 1. Due to their ability to encode information about an integer sequence, generating functions are powerful tools that can be used for solving recurrence relations. Techniques such as partial fractions , polynomial multiplication , and derivatives can help solve the recurrence relations.
Solving Linear Recurrence Equations With Polynomial Coe cients Marko Petkov sek Faculty of Mathematics and Physics University of Ljubljana Jadranska 19, SI-1000 Ljubljana, Slovenia Using Generating Functions to Solve Recurrence Relations We may use recurrences to derive generating functions. Example 4. Find the generating function for the sequence fa
COS 341, October 27, 1999 Handout Number 6 Solving Recurrence with Generating Functions The rst problem is to solve the recurrence relation system a There are several methods for solving recurrence equations. The simplest is to guess the solution and then verify that the guess is correct with an induction proof. As a basis for a good guess, let’s look for a pattern in the values of Tncomputed above: 1, 3, 7, 15, 31, 63. A natural guess is TnD2n 1. But whenever you guess a solution to a recurrence, you should always verify it with a proof
Solving Recurrence Relations with Generating Functions The sequence of numbers 1,1,2,3,5,8,13,. is known as the Fibonacci sequence. We can describe the sequence in terms of a recurrence relation. 8/05/2015 · In this video we use generating functions to solve nonhomogeneous recurrence relations. This is a pretty long process that requires fairly good attention to …
Due to their ability to encode information about an integer sequence, generating functions are powerful tools that can be used for solving recurrence relations. Techniques such as partial fractions , polynomial multiplication , and derivatives can help solve the recurrence relations. Using generating functions to solve recurrence relations We associate with the sequence fang, the generating function a(x) = P1 n=0 anx n. Now, the
Generating Functions and Recurrence Relations . In another note we commented We also note that the summations factored by n can be expressed in terms of the derivative of the generating function using the relation . Substituting for the summations in the previous equation therefore gives . Taking s 1 = 1, re-arranging terms, and dividing through by x, we have . To solve this differential Solving Recurrence Relations using generating Functions & Solving Differential Equations 35 mins Video Lesson Solution using Gen. Functions, Solving Differntial Equations, and other topics.
In mathematics, a recurrence relation is an equation that recursively defines a sequence or multidimensional array of values, once one or more initial terms are given: each further term of the sequence or array is defined as a function of the preceding terms. CS 161 Lecture 3 Jessica Su (some parts copied from CLRS) 1 Solving recurrences Last class we introduced recurrence relations, such as T(n) = 2T(bn=2c) + n.
CS 161 Lecture 3 Jessica Su (some parts copied from CLRS) 1 Solving recurrences Last class we introduced recurrence relations, such as T(n) = 2T(bn=2c) + n. Solving Recurrence Relations using generating Functions & Solving Differential Equations 3. In this 35 mins Video Lesson Solution using Gen. Functions, Solving Differntial Equations…
Generating Functions Dalhousie University. Check your solution for the closed formula by solving the recurrence relation using the Characteristic Root technique. 10 You have access to $$1 \times 1$$ tiles which come in 2 different colors and $$1\times 2$$ tiles which come in 3 different colors., Towers of Hanoi Peg 1 Peg 2 Peg 3 Hn is the minimum number of moves needed to shift n rings from Peg 1 to Peg 2. One is not allowed to place a larger ring on top of a smaller ring..
### Math 300 Solving a Recurrence Relation with Generating
Solving Recurrence Relations. We introduce the central notion of our course, the notion of a generating function. We start with studying properties of formal power series and then apply the machinery of generating functions to solving linear recurrence relations., Generating Function Applications — Ch. 8 22 Solving recurrence relations using generating functions Step 3: Solve for the compact form of A(x). A(x) =.
### Recurrence Relations and Generating Functions StudyYaar
Probability Generating Functions DAMTP. Main Article: Using Generating Functions to Solve Recurrence Relations One method to solve recurrence relations is to use a generating function. A generating function is a power series whose coefficients correspond to terms in a sequence of numbers. Ordinary generating functions are useful in mathematics by allowing us to con- dense in nite sequences into a single expression that computes each term in the sequence without directly using the recursion relation..
Now that I have demonstrated the use of generating functions in finding the closed forms of simple series, I will now move on to more complicated ones such as series containing recurrence relations… Solving Recurrence Relations with Generating Functions The sequence of numbers 1,1,2,3,5,8,13,. is known as the Fibonacci sequence. We can describe the sequence in terms of a recurrence relation.
How to find us. The entrance is at the SouthWest corner of 17th St. and JFK Blvd. The back wall that is parallel to JFK has a door & a wheelchair ramp that lead into the back room where we will meet. ioc.pdf Contents 1 Motivation examples Bit strings do not containing two consecutive zeros The owTer of Hanoi Regions of the plane 2 Solving Linear Recurrence Equations
Recurrence Relations & Generating Functions This page is an extension to my Fibonacci and Phi Formulae with an introduction to Recurrence Relations and to Generating Functions. A recurrence relation is a way of defining a series in terms of earlier member of the series. ers generating functions only, he does not use the term “recurrence relation” even when he talks about the Fibonacci numbers. In other cited textbooks RR are used in analy- sis of the time-complexity of algorithms (mostly of the type “divide-and-conquer”), but their solutions are not derived, they are simply given. During the last 15–20 years we can see a trend to the opposite
Chapter 1 Counting 1.1 A General Combinatorial Problem Instead of mostly focusing on the trees in the forest let us take an aerial view. You might get a bit of vertigo … On 28 January 2017 MathCounts will discuss Solving Basic Recurrence Relations with Generating Functions 2 . These problems are posed to help guide the discussion.
Solutions to Exercises Chapter 4: Recurrence relations and generating functions 1 (a) There are n seating positions arranged in a line. Prove that the number sider four methods of solving recurrence relations: (a) substitution (b) induction (c) characteristic roots (d) generating functions. 53.2 SUBSTITUTION In the substitution method of solving a recurrence relation for f(n), the recurrence for f (n) is repeatedly used to eliminate all occurrences of f from the right hand side of the recurrence. Once this has been done, the terms in the right hand
This idea of modifying a recurrence so that it becomes easier to solve is a very general problem solving technique. It looks like this: It looks like this: You have a hard problem. Solving Recurrences 2.1 T ypes of Recurrences 2.2 Finding Generating Functions 2.3 P a rtial Fractions 2.4 Characteristic Roots 2.5 Sim ultaneous Recur sions 2.6 Fibonacci Number Identities 2.7 Non-Constant Coef Þ cients 2.8 Divide-and-Conquer Relations 1. 2 Chapter 2 Solving Recurrences. Section 2.1 Types of Recurrences 3 2.1 TYPES OF RECURRENC ES. 4 Chapter 2 Solving Recurrences …
GENERATING FUNCTIONS: RECURRENCE RELATIONS, RATIONALITY AND HADAMARD PRODUCT. 1. Recurrence relations and rational generating functions We begin with the following generalization of the Fibonacci sequence. ers generating functions only, he does not use the term “recurrence relation” even when he talks about the Fibonacci numbers. In other cited textbooks RR are used in analy- sis of the time-complexity of algorithms (mostly of the type “divide-and-conquer”), but their solutions are not derived, they are simply given. During the last 15–20 years we can see a trend to the opposite
Recurrence Relations and Generating Functions NgГ y 27 thГЎng 10 nДѓm 2011 () Recurrence Relations and Generating FunctionsNgГ y 27 thГЎng 10 nДѓm 2011 1 / 1. Recursive Problem Solving Question Certain bacteria divide into two bacteria every second. It was noticed that when one bacterium is placed in a bottle, it fills it up in 3 minutes. How long will it take to fill half the bottle 3 Recurrence Equations The generating function for the set of binary strings with no block of ones of odd length was shown to be О¦(x) = 1 1 x x2 in the last section.
Main Article: Using Generating Functions to Solve Recurrence Relations One method to solve recurrence relations is to use a generating function. A generating function is a power series whose coefficients correspond to terms in a sequence of numbers. Solving Linear Recurrence Relations Niloufar Shafiei. 1 Review A recursive definition of a sequence specifies Initial conditions Recurrence relation Example: a 0=0 and a 1=3 a n = 2a n-1 - a n-2 a n = 3n Initial conditions Recurrence relation Solution. 2 Linear recurrences Linear recurrence: Each term of a sequence is a linear function of earlier terms in the sequence. For example: a 0 = 1 a 1
Solving Linear Recurrence Equations With Polynomial Coe cients Marko Petkov sek Faculty of Mathematics and Physics University of Ljubljana Jadranska 19, SI-1000 Ljubljana, Slovenia Solving Recurrence Relations using generating Functions & Solving Differential Equations 3. In this 35 mins Video Lesson Solution using Gen. Functions, Solving Differntial Equations…
## Solving Basic Recurrence Relations with Generating
Solving Recurrence Relations Oscar Levin PhD. Deriving recurrence relations involves di erent methods and skills than solving them. These two topics are treated separately in the next 2 subsec- tions. Another method of solving recurrences involves generating functions, which will be discussed later. 1.1 Deriving Recurrence Relations It is typical to want to derive a recurrence relation with initial conditions (abbreviated to RRwIC from, A recurrence is a recursive description of a function, or in other words, a description of a function in terms of itself. Like all recursive structures, a recurrence consists of one or more base cases and.
### Advanced Counting Techniques University of California
3 Recurrence Equations UCSD Mathematics. In this chapter, we will discuss how recursive techniques can derive sequences and be used for solving counting problems. The procedure for finding the terms of a sequence in a recursive manner is called recurrence relation., Solving Recurrence Relations using generating Functions & Solving Differential Equations 35 mins Video Lesson Solution using Gen. Functions, Solving Differntial Equations, and other topics..
Using Generating Functions to Solve Recurrence Relations We may use recurrences to derive generating functions. Example 4. Find the generating function for the sequence fa TI-Nspire Introduction to Sequences Aim To introduce students to sequences on the calculator Calculator objectives By the end of this unit, you should be able to: • generate a sequence recursively using the Calculator App. • evaluate sequences, defined both as explicit formula and recurrence relations, at specific values • plot sequences • analyse a sequence using both the Function
A linear homogeneous recurrence relation of degree k with constant coe cients is a recurrence relation of the form a n = c 1 a n 1 + c 2 a n 2 + + c k a n k Solving Recurrence Relations Using Di⁄erential Operators Robbie Beyl and Dennis I. Merinoy May 15, 2006 Abstract Linear recurrence relations are usually solved using the McLaurin se-ries expansion of some known functions. That is, we assume that the coe¢ cients a n of g(x) = P1 n=0 a nx n satisfy the recurrence relation. We then –nd an equation that involves g(x) so that we may compare
There are several methods for solving recurrence equations. The simplest is to guess the solution and then verify that the guess is correct with an induction proof. As a basis for a good guess, let’s look for a pattern in the values of Tncomputed above: 1, 3, 7, 15, 31, 63. A natural guess is TnD2n 1. But whenever you guess a solution to a recurrence, you should always verify it with a proof Due to their ability to encode information about an integer sequence, generating functions are powerful tools that can be used for solving recurrence relations. Techniques such as partial fractions , polynomial multiplication , and derivatives can help solve the recurrence relations.
Chapter 1 Counting 1.1 A General Combinatorial Problem Instead of mostly focusing on the trees in the forest let us take an aerial view. You might get a bit of vertigo … Towers of Hanoi Peg 1 Peg 2 Peg 3 Hn is the minimum number of moves needed to shift n rings from Peg 1 to Peg 2. One is not allowed to place a larger ring on top of a smaller ring.
How to find us. The entrance is at the SouthWest corner of 17th St. and JFK Blvd. The back wall that is parallel to JFK has a door & a wheelchair ramp that lead into the back room where we will meet. Recurrence Relations and Generating Functions NgГ y 27 thГЎng 10 nДѓm 2011 () Recurrence Relations and Generating FunctionsNgГ y 27 thГЎng 10 nДѓm 2011 1 / 1. Recursive Problem Solving Question Certain bacteria divide into two bacteria every second. It was noticed that when one bacterium is placed in a bottle, it fills it up in 3 minutes. How long will it take to fill half the bottle
Theorem 1 can be proved using generating functions and partial fractions, but we will not do it here. The theorem gives a general method for solving recurrence equations, which we On 28 January 2017 MathCounts will discuss Solving Basic Recurrence Relations with Generating Functions 2 . These problems are posed to help guide the discussion.
Week 10-11: Recurrence Relations and Generating Functions April 20, 2005 1 Some Number Sequences An inflnite sequence (or just a sequence for short) is an ordered array Chapter 1 Introductory ideas and examples A generating function is a clothesline on which we hang up a sequence of numbers for display. What that means is …
Recurrence Relations Many algo rithm s pa rticula rly divide and conquer al go rithm s have time complexities which a re naturally m odel ed b yr ecurrence relations Ar ecurrence relation is an equation which is de ned in term sof its elf Why a re recurrences go o d things Many natural functions a re easily exp ressed as re currences a n n n pol y nomial a n n n exponential a n n n we ir d f Solving Recurrence Relations with Generating Functions The sequence of numbers 1,1,2,3,5,8,13,. is known as the Fibonacci sequence. We can describe the sequence in terms of a recurrence relation.
Recurrence Relations and Generating Functions NgГ y 27 thГЎng 10 nДѓm 2011 () Recurrence Relations and Generating FunctionsNgГ y 27 thГЎng 10 nДѓm 2011 1 / 1. Recursive Problem Solving Question Certain bacteria divide into two bacteria every second. It was noticed that when one bacterium is placed in a bottle, it fills it up in 3 minutes. How long will it take to fill half the bottle Using generating functions to solve recurrences Math 40210, Fall 2012 November 15, 2012 Math 40210(Fall 2012) Generating Functions November 15, 20121 / 8
Check your solution for the closed formula by solving the recurrence relation using the Characteristic Root technique. 10 You have access to $$1 \times 1$$ tiles which come in 2 different colors and $$1\times 2$$ tiles which come in 3 different colors. sider four methods of solving recurrence relations: (a) substitution (b) induction (c) characteristic roots (d) generating functions. 53.2 SUBSTITUTION In the substitution method of solving a recurrence relation for f(n), the recurrence for f (n) is repeatedly used to eliminate all occurrences of f from the right hand side of the recurrence. Once this has been done, the terms in the right hand
A recurrence is a recursive description of a function, or in other words, a description of a function in terms of itself. Like all recursive structures, a recurrence consists of one or more base cases and Using generating functions to solve recurrences Math 40210, Fall 2012 November 15, 2012 Math 40210(Fall 2012) Generating Functions November 15, 20121 / 8
Generating Function Applications — Ch. 8 22 Solving recurrence relations using generating functions Step 3: Solve for the compact form of A(x). A(x) = Section 2: Solving Recurrence Relations by Iteration • As we noted at the end of the last lecture, when analyzing recurrence relations, we want to rewrite the general term as a function of the index and independent of predecessor terms. • This will allow us to compute any arbitrary term in the sequence without having to compute all the previous terms. • In this section, we will look at
ers generating functions only, he does not use the term “recurrence relation” even when he talks about the Fibonacci numbers. In other cited textbooks RR are used in analy- sis of the time-complexity of algorithms (mostly of the type “divide-and-conquer”), but their solutions are not derived, they are simply given. During the last 15–20 years we can see a trend to the opposite Ordinary generating functions are useful in mathematics by allowing us to con- dense in nite sequences into a single expression that computes each term in the sequence without directly using the recursion relation.
MATH 203, PROBLEM SET 2 DUE IN LECTURE ON MONDAY, FEB. 1. Generating Functions These problems have to do with the formulas worked out in class for the terms of a There are several methods for solving recurrence equations. The simplest is to guess the solution and then verify that the guess is correct with an induction proof. As a basis for a good guess, let’s look for a pattern in the values of Tncomputed above: 1, 3, 7, 15, 31, 63. A natural guess is TnD2n 1. But whenever you guess a solution to a recurrence, you should always verify it with a proof
A simple technic for solving recurrence relation is called telescoping. Start from the first term and sequntially produce the next terms until a clear pattern emerges. If you want to be mathematically rigoruous you may use induction. Recurrence Relations and Generating Functions. Recurrence Realtions This puzzle asks you to move the disks from the left tower to the right tower, one disk at a Solving Recurrence Relations with Generating Functions The sequence of numbers 1,1,2,3,5,8,13,. is known as the Fibonacci sequence. We can describe the sequence in terms of a recurrence rela-
COS 341, October 27, 1999 Handout Number 6 Solving Recurrence with Generating Functions The rst problem is to solve the recurrence relation system a Due to their ability to encode information about an integer sequence, generating functions are powerful tools that can be used for solving recurrence relations. Techniques such as partial fractions , polynomial multiplication , and derivatives can help solve the recurrence relations.
Now that I have demonstrated the use of generating functions in finding the closed forms of simple series, I will now move on to more complicated ones such as series containing recurrence relations… 14/10/2013 · Complete set of Video Lessons and Notes available only at http://www.studyyaar.com/index.php/module/36-recurrence-relations-and-generating-functions …
Solving Recurrence Relations using generating Functions & Solving Differential Equations 3. In this 35 mins Video Lesson Solution using Gen. Functions, Solving Differntial Equations… This idea of modifying a recurrence so that it becomes easier to solve is a very general problem solving technique. It looks like this: It looks like this: You have a hard problem.
3 Recurrence Equations The generating function for the set of binary strings with no block of ones of odd length was shown to be Φ(x) = 1 1 x x2 in the last section. 8/05/2015 · In this video we use generating functions to solve nonhomogeneous recurrence relations. This is a pretty long process that requires fairly good attention to …
### Week 9-10 Recurrence Relations and Generating Functions
Recurrence Relations Solving Linear Recurrence Relations. Recurrence Relations & Generating Functions This page is an extension to my Fibonacci and Phi Formulae with an introduction to Recurrence Relations and to Generating Functions. A recurrence relation is a way of defining a series in terms of earlier member of the series., ers generating functions only, he does not use the term “recurrence relation” even when he talks about the Fibonacci numbers. In other cited textbooks RR are used in analy- sis of the time-complexity of algorithms (mostly of the type “divide-and-conquer”), but their solutions are not derived, they are simply given. During the last 15–20 years we can see a trend to the opposite.
### 6.042J Chapter 10 Recurrences MIT OpenCourseWare
Week 9-10 Recurrence Relations and Generating Functions. On 28 January 2017 MathCounts will discuss Solving Basic Recurrence Relations with Generating Functions 2 . These problems are posed to help guide the discussion. A simple technic for solving recurrence relation is called telescoping. Start from the first term and sequntially produce the next terms until a clear pattern emerges. If you want to be mathematically rigoruous you may use induction. Recurrence Relations and Generating Functions. Recurrence Realtions This puzzle asks you to move the disks from the left tower to the right tower, one disk at a.
• 1 Solving recurrences Stanford University
• Section 2 Solving Recurrence Relations by Iteration
• SOLVING LINEAR RECURRENCE RELATIONS cs.ioc.ee
• ers generating functions only, he does not use the term “recurrence relation” even when he talks about the Fibonacci numbers. In other cited textbooks RR are used in analy- sis of the time-complexity of algorithms (mostly of the type “divide-and-conquer”), but their solutions are not derived, they are simply given. During the last 15–20 years we can see a trend to the opposite Week 9-10: Recurrence Relations and Generating Functions April 12, 2018 1 Some number sequences An inflnite sequence (or just a sequence for short) is an ordered array
4.2 Solving Linear Recurrence Relations 4.4 Generating Functions 2 Agenda Ch. 4.1, 4.2 & 4.5 Recurrence Relations Modeling with Recurrence Relations Linear NonhomogeneousRecurrence Relations with Constant Coefficients Generating Functions Useful Facts About Power Series Extended Binomial Coefficient Extended Binomial Theorem Counting Problems and Generating Functions Using Generating Functions initial conditions and satis es the recurrence relation, it must match the sequence an exactly! 2. a 0 = 2;a 1 = 2. The characteristic function is the same, with roots r 1 = 2 and r 2 = 1.
Week 10-11: Recurrence Relations and Generating Functions April 20, 2005 1 Some Number Sequences An inflnite sequence (or just a sequence for short) is an ordered array Chapter 1 Introductory ideas and examples A generating function is a clothesline on which we hang up a sequence of numbers for display. What that means is …
A simple technic for solving recurrence relation is called telescoping. Start from the first term and sequntially produce the next terms until a clear pattern emerges. If you want to be mathematically rigoruous you may use induction. Recurrence Relations and Generating Functions. Recurrence Realtions This puzzle asks you to move the disks from the left tower to the right tower, one disk at a On 28 January 2017 MathCounts will discuss Solving Basic Recurrence Relations with Generating Functions 2 . These problems are posed to help guide the discussion.
If a family of polynomials are known from a generating function, i.e., G(x, w) = ОЈ iв©ѕ0 p n (x)w n, then, from functional equations satisfied by the function G, it is easy to derive recurrence relations or differential recurrence relations for the polynomials p n. Use the recurrence relation and initial conditions to nd a simple expression for B(x). Math 2001-004: Intro to Discrete Math Fall 2015 There are three tricks that are often useful here: First, split o terms from your series B(x), so that the recurrence relation applies to each of the remaining terms in the sum. After applying the recurrence relation to the resulting sum, make changes in your
CS 161 Lecture 3 Jessica Su (some parts copied from CLRS) 1 Solving recurrences Last class we introduced recurrence relations, such as T(n) = 2T(bn=2c) + n. GENERATING FUNCTIONS: RECURRENCE RELATIONS, RATIONALITY AND HADAMARD PRODUCT. 1. Recurrence relations and rational generating functions We begin with the following generalization of the Fibonacci sequence.
Subsection Solving Recurrence Relations with Generating Functions We conclude with an example of one of the many reasons studying generating functions is helpful. We can use generating functions to solve recurrence relations. Section 2: Solving Recurrence Relations by Iteration • As we noted at the end of the last lecture, when analyzing recurrence relations, we want to rewrite the general term as a function of the index and independent of predecessor terms. • This will allow us to compute any arbitrary term in the sequence without having to compute all the previous terms. • In this section, we will look at
Due to their ability to encode information about an integer sequence, generating functions are powerful tools that can be used for solving recurrence relations. Techniques such as partial fractions , polynomial multiplication , and derivatives can help solve the recurrence relations. COS 341, October 27, 1999 Handout Number 6 Solving Recurrence with Generating Functions The rst problem is to solve the recurrence relation system a
Use the recurrence relation and initial conditions to nd a simple expression for B(x). Math 2001-004: Intro to Discrete Math Fall 2015 There are three tricks that are often useful here: First, split o terms from your series B(x), so that the recurrence relation applies to each of the remaining terms in the sum. After applying the recurrence relation to the resulting sum, make changes in your Solving Recurrence Relations Using Di⁄erential Operators Robbie Beyl and Dennis I. Merinoy May 15, 2006 Abstract Linear recurrence relations are usually solved using the McLaurin se-ries expansion of some known functions. That is, we assume that the coe¢ cients a n of g(x) = P1 n=0 a nx n satisfy the recurrence relation. We then –nd an equation that involves g(x) so that we may compare
TI-Nspire Introduction to Sequences Aim To introduce students to sequences on the calculator Calculator objectives By the end of this unit, you should be able to: • generate a sequence recursively using the Calculator App. • evaluate sequences, defined both as explicit formula and recurrence relations, at specific values • plot sequences • analyse a sequence using both the Function Theorem 1 can be proved using generating functions and partial fractions, but we will not do it here. The theorem gives a general method for solving recurrence equations, which we
View all posts in Victoria category | 2022-10-04T09:55:15 | {
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https://primer-computational-mathematics.github.io/book/c_mathematics/series_and_sequences/Taylor_series.html | # Taylor series#
A Taylor series is a method by which we can approximate an arbitrary function by a sum of terms made up of that function and its derivatives at a single point.
This concept underlies many things (especially in computational science) so it’s worth us spending some time reviewing it.
## Motivation - a constant approximation#
Consider a function of a single independent variable: $$f(x)$$
Suppose this function is really really expensive to evaluate. Suppose I have already evaluated it at a single $$x$$ value, call it $$x_0$$, i.e. I know the value of $$f(x_0)$$.
Given only this information, what is the best guess (estimation/approximation) I can make for $$f(x)$$ for a choice $$x\ne x_0$$?
Well I can’t really do anything better than
$f(x) \approx f(x_0)$
This may seem a hopeless approximation, but it can be used in some situations. Pretend $$f$$ is a weather forecast and $$x$$ is time. This just says that for a forecast of the weather tomorrow use what we see today - this is a real technique called persistent forecast (“today equals tomorrow”).
“… This makes persistence a ‘hard to beat’ method for forecasting longer time periods”: http://ww2010.atmos.uiuc.edu/(Gh)/guides/mtr/fcst/mth/prst.rxml
Let’s consider an example:
%matplotlib inline
import numpy as np
import matplotlib.pyplot as plt
def f(x):
return np.exp(x)
and plot our current situation, assuming that $$x_0=0$$. We call this the expansion point - we are constructing an approximation “around this point”.
fig = plt.figure(figsize=(7, 7))
ax1 = plt.subplot(111)
ax1.set_title('A constant approximation', fontsize=16)
ax1.set_xlabel('$x$', fontsize=16)
# define our x for plotting purposes
x = np.linspace(-3., 3., 1000)
# plot exact function
ax1.plot(x, f(x), 'k', lw=3, label='Exact function')
# define our approximation
x0 = 0.0
y = f(x0)*np.ones_like(x)
ax1.plot(x0, f(x0), 'ro', label='Expansion point')
# and plot the constant approximation
ax1.plot(x, y, 'b', lw=2, label='Constant approximation')
ax1.legend(loc='best', fontsize=14)
<matplotlib.legend.Legend at 0x2322df4cb38>
Can I have any confidence in this guess?
1. Yes if it turns out that the function $$f$$ is actually a constant.
2. Yes, if I can assume that $$f$$ doesn’t vary very much in the vicinity of $$x_0$$, and if we are interested in $$x$$ values that are close to $$x_0$$.
How can I do better?
## Motivation - a linear approximation (a “linearisation”)#
The obvious improvement we can make is to approximate the true function with a linear approximation rather than a constant.
This is a common way to write a linear function:
$y = mx + c$
where $$m$$ is the slope and $$c$$ is the intercept - the $$y$$ value the line hits the $$x=0$$ axis at.
We consider an approximation, or an expansion, (or in this case a “linearisation”) “about a point”.
For us the point is $$x_0$$ and it makes sense for us to make the slope of the linear approximation the same as the derivative of the function at this point.
So we are now using the value of $$f$$ at $$x_0$$ AND the value of the derivative of $$f$$ at $$x_0$$: $$m = f'(x_0)$$.
This is the formula for the linear line with this slope:
$y = f'(x_0) x + c$
where choice of $$c$$ moves the linear line up and down. We want it to pass through the function at $$x_0$$ as well of course, and this allows us to figure out what value $$c$$ should take.
Actually we generally write the linearisation in the following form
$f(x) \approx f(x_0) + (x-x_0)f'(x_0)$
so when $$(x-x_0)$$ is zero, we are at the expansion point and we indeed get $$f(x_0)$$ on the RHS. As we move away from that point $$(x-x_0)$$ grows and our approximation adds a correction based on the size of the derivative at $$x_0$$.
Let’s plot this. First we need to define the derivative:
# of course for our example f this is trivial:
def fx(x):
return np.exp(x)
fig = plt.figure(figsize=(7, 7))
ax1 = plt.subplot(111)
ax1.set_title('A constant approximation', fontsize=16)
ax1.set_xlabel('$x$', fontsize=16)
# define our x for plotting purposes
x = np.linspace(-3., 3., 1000)
# plot exact function
ax1.plot(x, f(x), 'k', lw=3, label='Exact function')
# define our approximation
x0 = 0.0
y = (f(x0) + (x-x0)*fx(x0))*np.ones_like(x)
ax1.plot(x0, f(x0), 'ro', label='Expansion point')
# and plot the constant approximation
ax1.plot(x, y, 'b', lw=2, label='Linear approximation')
ax1.legend(loc='best', fontsize=14)
<matplotlib.legend.Legend at 0x2322e095cf8>
This is clearly an improved approximation compared to the constant one.
However the same accuracy points hold: if we are a long way from the expansion point, or if $$f$$ is very complex, this may not be a good approximation.
Can we do better - yes we can consider the curvature of the black line in addition to its value and its slope, i.e. we can include a term proportional to $$f''(x_0)$$ to our approximation.
and so on…
This is the formula for the infinite Taylor series about (or around) the point $$x_0$$
$f(x) = f(x_0) + (x - x_0) f'(x_0) + \frac{(x - x_0)^2}{2!} f''(x_0) + \frac{(x - x_0)^3}{3!} f'''(x_0) + \frac{(x - x_0)^4}{4!} f^{(iv)}(x_0) + \ldots$
An equivalent way of writing this expansion is
$f(x_0+h) = f(x_0) + hf'(x_0) + \frac{h^2}{2!}f''(x_0) + \frac{h^3}{3!}f'''(x_0) + \ldots$
or replace $$h$$ by the notation $$\Delta x$$ or $$\delta x$$.
As we include more and more terms our approximation improves - see the following animated gif from Wikipedia which explains the point.
### Taylor series example#
Wikipedia image: The exponential function (in blue), and the sum of the first (n + 1) terms of its Taylor series expansion around the point 0 (in red).
More terms equate with a better approximation valid a larger distance from $$x_0$$.
### Aside: Big O notation#
When talking about terms in infinite series (terms we will often be forces to truncate, i.e. throw away), or talking about errors, convergence, complexity, run times etc, so-called Big-O notation is very useful.
As an alternative to writing “$$\ldots$$” in the above infinite expansions, we can also write
$f(x) = f(x_0) + (x - x_0) f'(x_0) + \frac{(x - x_0)^2}{2!} f''(x_0) + \frac{(x - x_0)^3}{3!} f'''(x_0) + \mathcal{O}((x - x_0)^4)$
or
$f(x_0+h) = f(x_0) + hf'(x_0) + \frac{h^2}{2!}f''(x_0) + \frac{h^3}{3!}f'''(x_0) + \mathcal{O}(h^4)$
What does this mean?
LINK TO THE NOTEBOOK ON BIG O NOTATION
## Truncation error#
In principle we can use Taylor series to approximate a (sufficiently smooth) function with arbitrary accuracy as long as we use sufficiently many terms, but in practice we will have to truncate.
In the next section we will use similar ideas to compute an approximate solution to a differential equations, we will do this by approximating the operation of taking a derivative.
Again by making use of enough terms from a Taylor series we can construct approximations of derivatives with arbitrary accuracy, but again in practice we will have to truncate.
In both cases the act of limiting the number of terms we use introduces an error.
Since we are truncating an infinite series at some point, this type of error is often called a truncation error. | 2023-02-04T15:41:48 | {
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https://www.jobilize.com/online/course/0-2-radical-concepts-simplifying-radicals-by-openstax?qcr=www.quizover.com | # 0.2 Radical concepts -- simplifying radicals
Page 1 / 1
This module covers techniques for the simplification of radicals.
## Simplifying radicals
The property $\sqrt{\text{ab}}$ = $\sqrt{a}$ $\sqrt{b}$ can be used to simplify radicals. The key is to break the number inside the root into two factors, one of which is a perfect square .
## Simplifying a radical
$\sqrt{\text{75}}$ $=$ $\sqrt{25•3}$ because 25•3 is 75, and 25 is a perfect square $=$ $\sqrt{25}$ $\sqrt{3}$ because $\sqrt{\text{ab}}$ $=$ $\sqrt{a}$ $\sqrt{b}$ $=5$ $\sqrt{3}$ because $\sqrt{25}$ =5
So we conclude that $\sqrt{\text{75}}$ =5 $\sqrt{3}$ . You can confirm this on your calculator (both are approximately 8.66).
We rewrote 75 as $25•3$ because 25 is a perfect square. We could, of course, also rewrite 75 as $5•15$ , but—although correct—that would not help us simplify, because neither number is a perfect square.
## Simplifying a radical in two steps
$\sqrt{180}$ $=$ $\sqrt{9•20}$ because $9•20$ is 180, and 9 is a perfect square $=$ $\sqrt{9}$ $\sqrt{20}$ because $\sqrt{\text{ab}}$ $=$ $\sqrt{a}$ $\sqrt{b}$ $=3$ $\sqrt{20}$ So far, so good. But wait! We’re not done! $=3$ $\sqrt{4•5}$ There’s another perfect square to pull out! $=3$ $\sqrt{4}$ $\sqrt{5}$ $=3\left(2\right)$ $\sqrt{5}$ $=6$ $\sqrt{5}$ Now we’re done.
The moral of this second example is that after you simplify, you should always look to see if you can simplify again .
A secondary moral is, try to pull out the biggest perfect square you can. We could have jumped straight to the answer if we had begun by rewriting 180 as $36•5$ .
This sort of simplification can sometimes allow you to combine radical terms, as in this example:
## Combining radicals
$\sqrt{\text{75}}$ $–$ $\sqrt{\text{12}}$ $=5$ $\sqrt{3}$ $–2$ $\sqrt{3}$ We found earlier that $\sqrt{\text{75}}$ $=5$ $\sqrt{3}$ . Use the same method to confirm that $\sqrt{\text{12}}$ $=2$ $\sqrt{3}$ . $=3$ $\sqrt{3}$ 5 of anything minus 2 of that same thing is 3 of it, right?
That last step may take a bit of thought. It can only be used when the radical is the same. Hence, $\sqrt{2}$ + $\sqrt{3}$ cannot be simplified at all. We were able to simplify $\sqrt{\text{75}}$ $\sqrt{\text{12}}$ only by making the radical in both cases the same .
So why does $5$ $\sqrt{3}$ $–2$ $\sqrt{3}$ $=3$ $\sqrt{3}$ ? It may be simplest to think about verbally: 5 of these things, minus 2 of the same things, is 3 of them. But you can look at it more formally as a factoring problem, if you see a common factor of $\sqrt{3}$ .
$5$ $\sqrt{3}$ $–2$ $\sqrt{3}$ $=$ $\sqrt{3}$ $\left(5–2\right)=$ $\sqrt{3}$ $\left(3\right)$ .
Of course, the process is exactly the same if variable are involved instead of just numbers!
## Combining radicals with variables
${x}^{\frac{3}{2}}+{x}^{\frac{5}{2}}$ $={x}^{3}+{x}^{5}$ Remember the definition of fractional exponents! $=\sqrt{{x}^{2}*x}+\sqrt{{x}^{4}*x}$ As always, we simplify radicals by factoring them inside the root... $\sqrt{{x}^{2}}*\sqrt{x}+\sqrt{{x}^{4}}*\sqrt{x}$ and then breaking them up... $=x\sqrt{x}+{x}^{2}\sqrt{x}$ and then taking square roots outside! $=\left({x}^{2}+x\right)\sqrt{x}$ Now that the radical is the same, we can combine.
## Rationalizing the denominator
It is always possible to express a fraction with no square roots in the denominator.
Is it always desirable? Some texts are religious about this point: “You should never have a square root in the denominator.” I have absolutely no idea why. To me, $\frac{1}{\sqrt{2}}$ looks simpler than $\frac{\sqrt{2}}{2}$ ; I see no overwhelming reason for forbidding the first or preferring the second.
However, there are times when it is useful to remove the radicals from the denominator: for instance, when adding fractions. The trick for doing this is based on the basic rule of fractions: if you multiply the top and bottom of a fraction by the same number, the fraction is unchanged. This rule enables us to say, for instance, that $\frac{2}{3}$ is exactly the same number as $\frac{2\cdot 3}{3\cdot 3}$ = $\frac{6}{9}$ .
In a case like $\frac{1}{\sqrt{2}}$ , therefore, you can multiply the top and bottom by $\sqrt{2}$ .
$\frac{1}{\sqrt{2}}$ = $\frac{1*2}{\sqrt{2}*\sqrt{2}}$ = $\frac{\sqrt{2}}{2}$
What about a more complicated case, such as $\frac{\sqrt{\text{12}}}{1+\sqrt{3}}$ ? You might think we could simplify this by multiplying the top and bottom by $\left(1+$ $\sqrt{3}$ ), but that doesn’t work: the bottom turns into ${\left(1+3\right)}^{2}$ $=1+2$ $\sqrt{3}$ $+3$ , which is at least as ugly as what we had before.
The correct trick for getting rid of $\left(1+$ $\sqrt{3}$ ) is to multiply it by $\left(1–$ $\sqrt{3}$ ). These two expressions, identical except for the replacement of $\mathrm{a+}$ by $\mathrm{a-}$ , are known as conjugates . What happens when we multiply them? We don’t need to use FOIL if we remember that
$\left(x+y\right)\left(x-y\right)={x}^{2}-{y}^{2}$
Using this formula, we see that
$\left(1+\sqrt{3}\right)\left(1-\sqrt{3}\right)={1}^{2}-{\left(\sqrt{3}\right)}^{2}=1-3=-2$
So the square root does indeed go away. We can use this to simplify the original expression as follows.
## Rationalizing using the conjugate of the denominator
$\frac{\sqrt{\text{12}}}{1+\sqrt{3}}=\frac{\sqrt{12}\left(1-\sqrt{3}\right)}{\left(1+\sqrt{3}\right)\left(1-\sqrt{3}\right)}=\frac{\sqrt{12}-\sqrt{36}}{1-3}=\frac{2\sqrt{3}-6}{-2}=-\sqrt{3}+3$
As always, you may want to check this on your calculator. Both the original and the simplified expression are approximately 1.268.
Of course, the process is the same when variables are involved.
## Rationalizing with variables
$\frac{1}{x-\sqrt{x}}$ = $\frac{1\left(x+\sqrt{x}\right)}{\left(x-\sqrt{x}\right)\left(x+\sqrt{x}\right)}$ = $\frac{x+\sqrt{x}}{{x}^{2}-x}$
Once again, we multiplied the top and the bottom by the conjugate of the denominator : that is, we replaced $\mathrm{a-}$ with $\mathrm{a+}$ . The formula $\left(x+a\right)\left(x-a\right)={x}^{2}-{a}^{2}$ enabled us to quickly multiply the terms on the bottom, and eliminated the square roots in the denominator.
#### Questions & Answers
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revolt
da
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RAW Reply
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Damian
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Professor
I think
Professor
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Alexandre
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Brian Reply
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http://mathhelpforum.com/advanced-statistics/134284-looking-hats.html | # Math Help - Looking for Hats
1. ## Looking for Hats
Five men are trying to find their hats in a closet. The closet contains their hats as well as 5 additional hats. (total 10 hats) What is the probability that at least one man retrieves his own hat?
2. If there are $n$ men and $n$ hats. The probability that $k$ men select there own hat is:
$\frac{(n-k)!}{n!}\times P_{n-k} \times {{n}\choose{k}}$ Where, $P_n = \sum^n \frac{(-1)^i}{i!}$ is the probability that no matches occur.
For your case, extend the number of hats to $15,$ and let $k$ to be $0,$ then to be $1.$ Now take $1$ minus the sum of your values for $k=0$ and $k=1.$
Hope this helps.
Anonymous
3. I derived the formula above through recursion. I didn't type it out because I'm lazy. But, I just realized if you wanna be slick about it.
$Pr($k men select there own hat $)$ ~ $Hypergeometric(5, k, 15)$
Booyah-Kasha .
Anonymous
Five men are trying to find their hats in a closet. The closet contains their hats as well as 5 additional hats. (total 10 hats) What is the probability that at least one man retrieves his own hat?
I do not follow the first response. It may work.
Here is another.
Let $^N\mathcal{P}_k=\frac{N!}{(N-k)!}$, the number of permutations of N things taken k at a time.
Use the inclusion/exclusion principle.
The number of ways for at least one man to get his own hat is $\sum\limits_{k = 1}^5 {\left( { - 1} \right)^{k + 1} \binom{5}{k} \left(^{10 - k}\mathcal{P}_{5 - k} \right)}$
Divide that number by $^{10}\mathcal{P}_5$
5. Originally Posted by Plato
I do not follow the first response. It may work.
Here is another.
Let $^N\mathcal{P}_k=\frac{N!}{(N-k)!}$, the number of permutations of N things taken k at a time.
Use the inclusion/exclusion principle.
The number of ways for at least one man to get his own hat is $\sum\limits_{k = 1}^5 {\left( { - 1} \right)^{k + 1} \binom{5}{k} \left(^{10 - k}\mathcal{P}_{5 - k} \right)}$
Divide that number by $^{10}\mathcal{P}_5$
I think it would speed up the computation to consider the compliment.
6. Originally Posted by Anonymous1
I think it would speed up the computation to consider the compliment.
That may well be the case.
However, I still do not understand your solution.
7. That came off wrong. I wasn't knocking your solution, I was saying that instead of considering the sum from 1 to 5, consider 1 minus the sum of 0 and 1.
Well can you see why it would ~ $Hypergeometric?$
8. Originally Posted by Anonymous1
That came off wrong. I wasn't knocking your solution, I was saying that instead of considering the sum from 1 to 5, consider 1 minus the sum of 0 and 1. $Hypergeometric?$
With an adjustment in indices, that would be the probability that none gets his own hat. | 2016-04-30T22:02:08 | {
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http://math.stackexchange.com/questions/5969/product-of-two-cyclic-groups-is-cyclic-iff-their-orders-are-co-prime | # Product of two cyclic groups is cyclic iff their orders are co-prime
Say you have two groups $G = \langle g \rangle$ with order $n$ and $H = \langle h \rangle$ with order $m$. Then the product $G \times H$ is a cyclic group if and only if $gcd(n,m)=1$.
I can't seem to figure out how to start proving this. I have tried with some examples, where I pick $(g,h)$ as a candidate generator of $G \times H$. I see that what we want is for the cycles of $g$ and $h$, as we take powers of $(g,h)$, to interleave such that we do not get $(1,1)$ until the $(mn)$-th power. However, I am having a hard time formalizing this and relating it to the greatest common divisor.
Any hints are much appreciated!
-
As a supplement to the various detailed answers below, the following short remark might help: gcd$(m,n) = 1$ if and only lcm$(m,n) = mn$. It is the latter fact (about the lcm) which is more directly relevant to your problem than the gcd fact (although they are equivalent). Try combining your intuition about "interleaving" with the formal notion of lcm$(m,n)$. – Matt E Oct 3 '10 at 21:59
Note that $|G\times H|=|G||H|=nm$; so $G\times H$ is cyclic if and only if there is an element of order $nm$ in $G\times H$.
In any group $A$, if $a,b\in A$ commute with one another, $a$ has order $k$, and $b$ has order $\ell$ then the order of $ab$ will divide lcm$(k,\ell)$ (prove it).
Now take an element of $G\times H$, written as $(g^a,h^b)$, where $G=\langle g\rangle$, $H=\langle h\rangle$, $0\leq a\lt n$, $0\leq b\lt m$. Then $(g^a,h^b)=(g^a,1)(1,h^b)$. In this case, what is the order? Under what conditions can you get an element of order exactly $nm$, which is what you need?
-
To try and answer your first question, I observe that I want to raise $(g^a,h^b)$ to a power $x$ such that $xa$ is a multiple of $n$ and $xb$ is a multiple of $m$. Also, the order will be the least such $x$, so I guess we want $x = lcm(m,n)$. Thus the order of $(g^a,h^b)$ is $lcm(m,n)$. Is this correct? – Martin Lauridsen Oct 3 '10 at 22:11
Not quite. lcm$(m,n)$ certainly works, but it need not be the smallest that works. For example, $g^a$ could have order strictly less than $n$ and $g^b$ have order strictly less than $m$. – Arturo Magidin Oct 3 '10 at 22:20
But you do know that the order is at most lcm(m,n). Now, you want the possibility for the order to be $mn$ in the case where $G\times H$ is cyclic, so what does that tell you about $m$ and $n$? And for the converse, can you get it at this point? – Arturo Magidin Oct 3 '10 at 22:29
Well, as was hinted above, if gcd(m,n) = 1 then lcm(m,n) = mn, so I know the order is at most mn. But I dont see how that gets me closer to showing it is exactly mn when gcd(m,n)=1. I think I have the case covered, where gcd(m,n) > 1, by what Ronaldo answered. – Martin Lauridsen Oct 3 '10 at 22:49
So you know that if $G$ is cyclic, then lcm$(m,n)=1$. Now suppose that lcm$(m,n)=1$, and find an element whose order is exactly lcm$(m,n)=1$. Note that the problem with your argument before is that you did not take into account $a$ and $b$; but you know exactly what the order of $g^a$ is (it is $n/$gcd$(n,a)$), and similarly for $g^b$. So find a specific $a$ and a specific $b$ that give you the order you want. – Arturo Magidin Oct 3 '10 at 22:55
The order of $G\times H$ is $n.m$. Thus, $G\times H$ is cyclic iff it has an element with order $n.m$. Suppose $gcd(n.m)=1$. This implies that $g^m$ has order $n$, and analogously $h^n$ has order $m$. That is, $g\times h$ has order $n.m$, and therefore $G\times H$ is cyclic.
Suppose now that $gcd(n.m) >1$. Let $g^k$ be an element of $G$ and $h^j$ be an element of $H$. Since the lowest common multiple of $n$ and $m$ is lower than the product $n.m$, that is, $lcm(n,m) < n.m$, and since $(g^k)^{lcm(n,m)} = e_{G}$, $(h^j)^{lcm(n,m)} = e_{H}$, we have $(g^k\times h^j)^{lcm(n,m)} = e_{G\times H}$. It follows that every element of $G\times H$ has order lower than $n.m$, and therefore $G\times H$ is not cyclic.
-
Use these facts:
1) Cyclic groups of the same order are isomorphic.
-
HINT $\rm\quad\ \ gcd(m,n) > 1$
$\rm\quad\iff\ lcm(m,n) < mn$
$\rm\quad\iff\ \mathbb Z_m \times \mathbb Z_n\$ has all elts of order $\rm < mn$
$\rm\quad\iff\ \mathbb Z_m \times \mathbb Z_n\$ is noncyclic
- | 2015-01-30T14:51:33 | {
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https://math.stackexchange.com/questions/2521017/probability-of-deck-of-cards-such-that-each-person-receives-one-ace | # Probability of deck of cards such that each person receives one ace
Suppose that a deck of 52 cards containing four aces is shuffled thoroughly and the cards are then distributed among four players so that each player receives 13 cards. Determine the probability that each player will receive one ace.
The answer to this is given as$$\frac{13^4}{\binom {52}4}$$
My doubt is the following:
1. The book justifies ${\binom {52}{4}}$ as number of possible different combinations of the four positions in the deck occupied by 4 aces. That sounds like a case of arrangements to me, so shouldn't we think about permutations and not combinations if we are concerned about how the aces are to be arranged in the deck ?.
2. Shouldn't the denominator be ${\binom {52}{13}}$ since you are choosing 13 cards for 4 people.
• We don't care who gets which ace, nor which hand each person is getting. The blocks of thirteen, and the aces, are two sets within which every member is equivalent. – Nij Nov 15 '17 at 5:31
• I don't completely understand. Could you please elaborate. Shouldn't the sample space consist of all combinations of 13 draws in which aces might or might not be there – madhavU Nov 15 '17 at 5:35
• Guess I get it now. So its like saying that we have 52 positions out of which we need to choose 4 positions for the aces. 52C13 would imply we take into consideration the rest of the cards as well which is not needed in this case. – madhavU Nov 15 '17 at 6:06
• It completely depends on who shuffles the deck and how it is shuffled. Should we assume some specific distribution? – mathreadler Nov 15 '17 at 8:55
• I don't think the question asks us to assume a specific distribution mathreadler – madhavU Nov 15 '17 at 14:41
There are $$\binom{52}{13}\binom{39}{13}\binom{26}{13}\binom{13}{13}$$ ways to distribute $13$ cards to each of four people.
There are $4!$ ways to distribute the aces so that each person receives one and $$\binom{48}{12}\binom{36}{12}\binom{24}{12}\binom{12}{12}$$ ways to distribute the remaining cards so that each person receives twelve of them. Hence, the desired probability is \begin{align*} \frac{4!\dbinom{48}{12}\dbinom{36}{12}\dbinom{24}{12}\dbinom{12}{12}}{\dbinom{52}{13}\dbinom{39}{13}\dbinom{26}{13}\dbinom{13}{13}} & = \frac{4! \cdot \dfrac{48!}{12!36!} \cdot \dfrac{36!}{12!24!} \cdot \dfrac{24!}{12!12!} \cdot \dfrac{12!}{12!0!}}{\dfrac{52!}{13!39!} \cdot \dfrac{39!}{13!26!} \cdot \dfrac{26!}{13!13!} \cdot \dfrac{13!}{13!0!}}\\[2mm] & = \frac{4! \cdot \dfrac{48!}{12!12!12!12!}}{\dfrac{52!}{13!13!13!13!}}\\[2mm] & = \frac{4!48!}{12!12!12!12!} \cdot \frac{13!13!13!13!}{52!}\\[2mm] & = \frac{4!48!13^4}{52!}\\[2mm] & = \frac{13^4}{\dfrac{52!}{4!48!}}\\[2mm] & = \frac{13^4}{\dbinom{52}{4}} \end{align*}
Let's compare this solution with the approach of your author. As you stated, there are $\binom{52}{4}$ ways to choose the four positions occupied by the aces in the deck. Since each person receives $13$ cards, there are $13$ possible places for the position of the ace in each person's hand. Hence, the desired probability is $$\frac{13^4}{\dbinom{52}{4}}$$
• That's a wonderful answer. Thanks a lot N.F Taussig. I have a silly doubt, if your sample space does not include arrangements, then why have you included the possibility of arrangements in the numerator ?. – madhavU Nov 15 '17 at 14:40
• If you are referring to the text's answer, notice that we are choosing positions for the aces in both the numerator and the denominator. In the numerator, we are choosing the position for the ace in each person's hand. In the denominator, we are choosing the position of the aces in the deck. – N. F. Taussig Nov 15 '17 at 19:49
All we really care about is the placement for the aces; not where the other cards may be in the deck, nor even the suits of the aces.
So let us take 52 blank cards and 4 stickers with 'ace' written on them. To 'deal the cards', place the blank cards in for lines of 13, then unbiasedly select four from them and put a sticker on each.
How may ways are there to stick the aces on 4 different cards in the deck of 52? That is a selection of 4 from 52. $$\binom {52}4$$
How many ways are there to do this so that each ace is stuck on one card in each from the four lines of 13? That is a selection of 1 from the first line, 1 from the next line, and so forth. $$\binom{13}1\binom{13}1\binom{13}1\binom{13}1$$
Divide and calculate.$$\dfrac{13^4}{\dbinom{52}{4}}$$
Informal analysis: we can re-imagine the process of the deal as one in which there are 52 slots, 13 assigned to each hand and the cards are distributed from the top unifomly to the remaining slots. We put the four aces on the top of the deck. The first ace can be dealt to any of the four players. The second ace can be dealt to any of the remaining 39 slots. The third ace can be dealt to any of the remaining 26 slots. The last ace can be dealt to 13 slots. So odds of this happening are: $$\begin{equation*} {52 \over 52} \times {39 \over 51} \times {26 \over 50} \times {13 \over 49} \approx 10.5\% \end{equation*}$$ Formal approach: first, we have to identify our probability space. Using the multiinomial coefficient we can count the total number of atomic events as $$|\Omega| = {52 \choose 13,13,13,13}$$. We consider how remaining cards are dealts, 12 to each of the remaining players, and each player has one ace. So $$|N_{\text{one ace each}}| = 4! \cdot {48 \choose 12, 12, 12, 12}$$. So we can compute: \begin{align} \Pr(N_{\text{one ace each}}) &= \frac{|N_{\text{one ace each}}|}{|\Omega|} \\ &= {4! \cdot {48 \choose 12,12,12,12} \over {52 \choose 13,13,13,13}}\\ &= {{52 \times 39 \times 26 \times 13} \over {52 \times 51 \times 50 \times 49}}\\ &\approx 10.5\% \end{align} | 2019-08-23T16:26:58 | {
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https://bartlesvilleschools.org/what-is-the-sign-of-conjugate | # What is the sign of conjugate?
If just the sign of the imaginary component varies between two complex numbers, they are said to be complex conjugates of each other. Z = a-ib, which means a-ib = a-ib. That is, b equals itself.
Therefore, the sign of the imaginary part determines whether two complex numbers are equal to each other or not. If they are equal, then their real parts are also equal. If not, then they are not equal.
Conjugates are important in physics and math because they describe the same physical phenomenon as its original object but with its magnitude reversed in direction (i.e., negative values). For example, the position vector of a particle at any given time contains both the x-coordinate and the y-coordinate of where it is located. But if we reverse the direction of the y-axis, we get the same set of coordinates but with a negative y-value, which means that the particle is now located at (-5,-4). This shows that the concept of conjugation has been used in physics to describe the same physical situation but with its effect reversed.
Complex conjugates are pairs of numbers whose real part is the real part of one number and whose imaginary part is the imaginary part of another number.
## How do you find the conjugate on a calculator?
A complex number's conjugate is indicated by a bar z (or occasionally by a star z*) and is equal to z = a-ib z = a-i b, where a=R(z) a = R (z) is the real part and b=I(z) b = I (z) is the imaginary component. If you put these together, you get a complex number whose real part is a and whose imaginary part is b. Thus, the conjugate of any complex number z is another number that has the same real part as z but whose opposite imaginary part is used instead.
To calculate the conjugate of a complex number, first compute its real part by adding i times its negative sign to itself, then divide by two and add b to this result. Finally, multiply the original number by its conjugate.
For example, if z = 3+4i, then its conjugate is z* = (3-4i). To verify that these numbers are indeed the conjugates of each other, we can use the rule that says that for any two complex numbers z and w, zw = (z*), where (*) means "multiplication". In our case, z3 + 4iw = (3-4i) = 9 - 16i, which agrees with our calculation for z*.
## What is the complex conjugate of 3?
There is a complex conjugate for every complex number. The complex conjugate of a + bi is a-bi. For example, the conjugate of 3 + 15i is 3–15i, while the conjugate of 5–6i is 5–6i. In general, the complex conjugate of any complex number z = x + yi is equal to z with x and y replaced by their negatives.
A complex number has an imaginary part that can be positive or negative, so its conjugate must also have an imaginary part that can be positive or negative. Because of this restriction, there are only two possibilities for the sign of the real part of the complex conjugate: it can be positive or negative like the original number. There is no other choice because if the sign of the real part was different than the sign of the original number, then the conjugate would have an imaginary part that could take on both signs at once, which is not allowed since all imaginary numbers must have one sign for every element in R.
Given a complex number z = x + yi, we can find its conjugate by taking its complex conjugate $\bar{z}$ and replacing each instance of x and y with their negatives.
## What is the complex conjugate of 3 2 I?
Simply alter the sign of the imaginary portion to obtain a complex conjugate (the part with the i). That is, it either goes from positive to negative or from negative to positive. As a result, 3+2i is the complex conjugate of 3-2i.
A complex number has a real part and an imaginary part. The real part is the same as the real number; the imaginary part is the opposite of the integer part. In this case, the complex number's real part is 3 and its imaginary part is 2.
The complex conjugate of any number is also a number. It will always have the same magnitude as the original number but with its opposite sign. For example, if x is 3 + 2i, then its complex conjugate would be 3 - 2i. They are exactly the same except for the sign of the imaginary part.
There are several ways to calculate the complex conjugate of a number. You can use basic algebra to reverse the sign of the imaginary part and add it to the number itself. For example, if x is 3 + 2i, then its complex conjugate would be (-i) - 2 = -3 - 2i. Another method is to use the fact that the complex conjugate of a number is equal to its negative.
##### Mary Ramer
Mary Ramer is a professor in the field of Mathematics. She has a PhD in mathematics, and she loves teaching her students about the beauty of math. Mary enjoys reading all kinds of books on math, because it helps her come up with new interesting ways how to teach her students. | 2022-01-29T05:13:12 | {
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"url": "https://bartlesvilleschools.org/what-is-the-sign-of-conjugate",
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https://mathematica.stackexchange.com/questions/31469/how-to-partition-a-list-to-make-each-subsets-size-equal-and-mean-as-close-as-po/31494 | # How to partition a list to make each subset's size equal and mean as close as possible
Suppose we have 3 lists each with 18 numbers like this:
list1 = {7.49, 7.56, 7.98, 8.09, 8.16, 8.21,
8.73, 8.64, 8.68, 8.46, 8.57, 8.29, 9.38, 9.43, 8.95, 9.04, 8.9, 9.07};
list2 = {21.08, 21.18, 21.35, 21.79, 21.92, 22.15, 22.38, 22.48, 22.48, 22.51, 22.64, 22.68, 22.75, 22.8, 23.01, 23.28, 23.5, 23.54};
list3 = {36.33, 37.1, 37.19, 37.31, 37.34, 37.61, 37.88, 38.32, 38.42, 38.9, 39.06, 39.12, 39.14, 39.31, 39.39, 39.41, 39.41, 39.43};
I need to partition it into six subsets:
1. Each one with 3 numbers
2. The mean of each one is as close as possible to the others.
For an example:
list = {1, 2, 3, 4, 5, 6};
Partition into 3 subsets: 1. Each one with equal size, it sould be Length[list]/3=2 2. The mean of each subset is as close as possible to the others,it should be {{1,6},{2,5},{3,4}} where each subset's mean is 3.5.
My objective is:
Minimize the MeanVariance of all the subsets.
• Welcome to MMA.SE ! It would be interesting to provide a minimum working example, to actually show that you've searched and/or tried anything before asking the community. Moreover, its not clear to me what exactly do you mean with "the means of each onde are as close as possible." – Rod Aug 31 '13 at 18:13
• In other words you are looking at a k-means clustering variant. Or I think so :D – Sektor Aug 31 '13 at 18:26
• For the small example brute force would be : aux = {#, Variance[Mean[Transpose[#]]]} & /@ (Partition[#, 2, 2] & /@ Permutations[lst, {6}]) and SortBy[aux, #[[2]] &][[1]]. As hinted by @PlatoManiac, you need a better way for large lists. – b.gates.you.know.what Aug 31 '13 at 19:26
• Not at all a general or optimal solution, but for your specific dataset this seems to work nicely: Transpose[FindClusters[list, 3]] For other data it can produce ragged uneven subdivision, so there should be an even-out-and-sort steps included. – Vitaliy Kaurov Aug 31 '13 at 19:47
• Might be worth asking this on the Mathematics se – DavidC Aug 31 '13 at 22:59
A simple approximate version would work like this:
• Sort data
• Split on 3 sublists by 6 elements
• Reverse order of sublist with Max Variance
• Transpose to get 6 sublists by 3 elements
Here is the code:
subd[data_] := Transpose[MapAt[Reverse, Sort[Partition[Sort[data], 6],
Variance[#1] < Variance[#2] &], -1]]
here is how it works
subd[list2]
{{22.38, 22.75, 22.15}, {22.48, 22.8, 21.92}, {22.48, 23.01, 21.79},
{22.51, 23.28, 21.35}, {22.64, 23.5, 21.18}, {22.68, 23.54, 21.08}}
and here are variances of mean values of sublists, which are pretty small:
Variance[Mean /@ subd[#]] & /@ {list1, list2, list3}
{0.00137222, 0.000527407, 0.0127885}
The problem is to split a list of m*n numbers into m subsets of size n whose means are as equal as possible. Here is an iterative solution. First, partition the sorted list into n sublists of size m, then reverse the even-numbered sublists, and transpose the array to get m sublists of size n. This will give a reasonably good initial approximation.
Then iterate over pairs of subsets. For each pair, find the split of the 2n values that will make their means as equal as possible. Continue until the current split is best for m(m-1)/2 pairs in a row. This will not necessarily find the absolute best subsets, but it will find a very good set. Restarting the iterations from the initial approximation many times, but with the order of the numbers in each subset randomized each time, is a convenient way to increase the chance of finding the absolute best set. (There is no way to guarantee that a set is best, short of checking all possible sets.)
To facilitate repeated randomized starts, I have rewritten my earlier code as a function. The last argument is the number of repetitions. It may be omitted, in which case only one start is used. The returned list is {subsets, subset means, standard deviation of subset means}.
EDIT - This revision addresses again the problem (noted by incognito007) that the results are susceptible to floating-point roundoff error. One possible effect is that the While can loop forever; this actually happened when the previous code was used to partition list2 with m = 3, n = 6. The new code handles that case, but I can't guarantee that such looping can never occur when the data are Real. On the other hand, if all the data are exact then the arithmetic will be exact and the problem can not occur. One solution, exemplified below, is to scale and round the data, then undo the scaling at the end. Besides being safe, this will usually be faster, too.
funk[data_List?(VectorQ[#,NumericQ]&), m_Integer?Positive, n_Integer?Positive,
r:(_Integer?Positive):1] /; Length@data == m*n :=
Module[{b,c,d,e,f,g,h,i,j,k,L, v = Infinity, w},
d = Transpose@MapAt[Reverse,Partition[Sort@data,m],List/@Range[2,n,2]];
c = Prepend[#,1]&/@Permutations@Join[ConstantArray[1,{n-1}],ConstantArray[-1,{n}]];
i = Flatten[Position[#, 1]]&/@c;
j = Flatten[Position[#,-1]]&/@c;
If[And@@ExactNumberQ/@data, w := c.f,
w := Total[f*c,Method->"CompensatedSummation"]; d = N@d; c = N@Transpose@c];
c = DeveloperToPackedArray@c;
Do[g = h = 1; L = 0; e = RandomSample/@d;
While[L < m(m-1)/2,
If[++h > m, If[++g >= m, g = 1]; h = g+1];
f = DeveloperToPackedArray@Flatten@e[[{g,h}]];
k = Ordering[Abs@w,1][[1]];
If[k > 1, e[[g]] = f[[i[[k]]]]; e[[h]] = f[[j[[k]]]]; L = 0]; L++];
k = Variance@Total[e,{2}];
If[k < v, v = k; b = e], {r}];
{b, Mean/@b, Sqrt@v/n}]
list1 = {7.49, 7.56, 7.98, 8.09, 8.16, 8.21, 8.73, 8.64, 8.68,
8.46, 8.57, 8.29, 9.38, 9.43, 8.95, 9.04, 8.9, 9.07};
list2 = {21.08, 21.18, 21.35, 21.79, 21.92, 22.15, 22.38, 22.48, 22.48,
22.51, 22.64, 22.68, 22.75, 22.8, 23.01, 23.28, 23.5, 23.54};
list3 = {36.33, 37.1, 37.19, 37.31, 37.34, 37.61, 37.88, 38.32, 38.42,
38.9, 39.06, 39.12, 39.14, 39.31, 39.39, 39.41, 39.41, 39.43};
First some results with m = 6, n = 3, the parameters in the original request.
Timing@funk[list1,6,3,1000]
{3.97, {{{8.9, 8.57, 8.16}, {8.64, 8.68, 8.29}, {8.73, 7.49, 9.38},
{7.56, 8.95, 9.07}, {9.43, 8.21, 7.98}, {8.09, 8.46, 9.04}},
{8.54333, 8.53667, 8.53333, 8.52667, 8.54, 8.53}, 0.0062361}}
Timing[.01*funk[Round[100*list1],6,3,1000]]
{3.22, {{{8.9, 8.57, 8.16}, {8.64, 8.68, 8.29}, {8.73, 7.49, 9.38},
{7.56, 8.95, 9.07}, {9.43, 8.21, 7.98}, {8.09, 8.46, 9.04}},
{8.54333, 8.53667, 8.53333, 8.52667, 8.54, 8.53}, 0.0062361}}
Timing@funk[list2,6,3,1000]
{4.79, {{{22.75, 23.28, 21.18}, {21.08, 22.68, 23.5}, {21.92, 22.51, 22.8},
{23.54, 22.38, 21.35}, {22.64, 22.48, 22.15}, {23.01, 21.79, 22.48}},
{22.4033, 22.42, 22.41, 22.4233, 22.4233, 22.4267}, 0.0091084}}
Timing[.01*funk[Round[100*list2],6,3,1000]]
{3.61, {{{22.75, 23.28, 21.18}, {21.08, 22.68, 23.5}, {21.92, 22.51, 22.8},
{23.54, 22.38, 21.35}, {22.64, 22.48, 22.15}, {23.01, 21.79, 22.48}},
{22.4033, 22.42, 22.41, 22.4233, 22.4233, 22.4267}, 0.0091084}}
Timing@funk[list3,6,3,1000]
{1.83, {{{36.33, 39.31, 39.41}, {39.12, 37.1, 39.06}, {39.14, 37.19, 38.9},
{39.39, 38.42, 37.31}, {38.32, 37.34, 39.41}, {39.43, 37.61, 37.88}},
{38.35, 38.4267, 38.41, 38.3733, 38.3567, 38.3067}, 0.0433803}}
Timing[.01*funk[Round[100*list3],6,3,1000]]
{1.49, {{{39.12, 39.06, 37.1}, {37.19, 39.14, 38.9}, {36.33, 39.31, 39.41},
{39.39, 38.42, 37.31}, {38.32, 37.34, 39.41}, {39.43, 37.61, 37.88}},
{38.4267, 38.41, 38.35, 38.3733, 38.3567, 38.3067}, 0.0433803}}
Now the other way around: m = 3, n = 6.
Timing@funk[list1,3,6,1000]
{9.85, {{{7.49, 9.04, 8.09, 9.38, 8.64, 8.57},
{8.68, 8.46, 8.29, 8.73, 7.98, 9.07},
{9.43, 8.21, 8.95, 8.16, 8.9, 7.56}},
{8.535, 8.535, 8.535}, 0.}}
Timing[.01*funk[Round[100*list1],3,6,1000]]
{2.45, {{{8.21, 8.29, 8.73, 9.43, 7.98, 8.57},
{8.9, 8.09, 8.95, 7.56, 9.07, 8.64},
{9.04, 8.68, 7.49, 8.16, 9.38, 8.46}},
{8.535, 8.535, 8.535}, 0}}
Timing@funk[list2,3,6,1000] (* this used to hang *)
{8.02, {{{22.68, 23.54, 21.35, 21.18, 23.28, 22.48},
{21.92, 22.48, 22.51, 21.08, 23.5, 23.01},
{22.38, 22.75, 21.79, 22.15, 22.64, 22.8}},
{22.4183,22.4167, 22.4183}, 0.00096225}}
Timing[.01*funk[Round[100*list2],3,6,1000]]
{2.03, {{{22.15, 22.51, 23.54, 22.48, 21.35, 22.48},
{22.38, 22.64, 21.79, 23.01, 21.18, 23.5},
{22.75, 21.92, 22.68, 21.08, 22.8, 23.28}},
{22.4183,22.4167, 22.4183}, 0.00096225}}
Timing@funk[list3,3,6,1000]
{9.31, {{{39.43, 37.88, 39.14, 36.33, 38.32, 39.12},
{37.34, 39.06, 39.31, 37.19, 38.9, 38.42},
{37.1, 37.31, 39.39, 39.41, 39.41, 37.61}},
{38.37, 38.37, 38.3717}, 0.00096225}}
Timing[.01*funk[Round[100*list3],3,6,1000]]
{2.52, {{{39.14, 37.88, 39.12, 36.33, 38.32, 39.43},
{39.41, 39.31, 37.61, 37.19, 37.31, 39.39},
{37.1, 37.34, 38.9, 38.42, 39.06, 39.41}},
{38.37, 38.37, 38.3717}, 0.00096225}}
• Copying and running your code, I got the following result:{{7.49, 9.43, 8.68}, {8.9, 8.57, 8.16}, {7.56, 8.95, 9.07}, {9.38, 7.98, 8.29}, {8.64, 8.21, 8.73}, {9.04, 8.09, 8.46}};{8.53333, 8.54333, 8.52667, 8.55, 8.52667, 8.53};0.00960324. Why differs from your result? – incognito007 Sep 1 '13 at 20:51
• @incognito007 It may be a version problem. The posted results are from 5.2, but 6 gives your results, and 5.2 with the data multiplied by 100 and then rounded also gives your results. I'm working on it. – Ray Koopman Sep 2 '13 at 3:53
This may not be the optimal solution but may be satisfactory to your needs for these particular lists. Essentially, sort the list, partition into three groups of six. Mix the upper tertile with the lower tertile and then permute the middle till smallest variance obtained.
fun2[u_] := Module[
{p, res, perm, mn, var, min, pos},
p = Partition[Sort[u], 6];
perm = Permutations[p[[2]]];
Append[#1, #2] &, {Thread[{p[[3]], Reverse[p[[1]]]}], #}] & /@
perm;
mn = Map[Mean, res, {2}];
var = Variance /@ mn;
min = Min[var];
pos = Position[var, min];
]
The results (sometimes not unique are presented as minimum variance of means of subssets and partition with minimum variance.
fun2[list1] yields
{{0.00112333,
{{8.9, 8.21, 8.46}, {8.95, 8.16, 8.57}, {9.04, 8.09,
8.29}, {9.07, 7.98, 8.64}, {9.38, 7.56, 8.68}, {9.43, 7.49,
8.73}}},
{0.00112333, {{8.9, 8.21, 8.57}, {8.95, 8.16,
8.46}, {9.04, 8.09, 8.29}, {9.07, 7.98, 8.64}, {9.38, 7.56,
8.68}, {9.43, 7.49, 8.73}}}}
fun2[list2] yields:
{{0.000238519, {{22.75, 22.15, 22.38}, {22.8, 21.92, 22.48}, {23.01,
21.79, 22.48}, {23.28, 21.35, 22.64}, {23.5, 21.18,
22.51}, {23.54, 21.08, 22.68}}}}
fun2[list3] yields:
{{0.014593, {{39.14, 37.61, 37.88}, {39.31, 37.34, 38.42}, {39.39,
37.31, 38.32}, {39.41, 37.19, 38.9}, {39.41, 37.1, 39.06}, {39.43,
36.33, 39.12}}}} | 2019-12-08T03:34:10 | {
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https://math.stackexchange.com/questions/2136882/show-that-it-is-impossible-to-list-the-rational-numbers-in-increasing-order/2136887 | # Show that it is impossible to list the rational numbers in increasing order
This is problem #6 from Section 1.2 of Ash's Real Variables With Basic Metric Space Topology.
I am asked to show that it is impossible to list the rational numbers in increasing order.
While I know it is possible to list a finite subset of the rational numbers in increasing order, I was wondering if the reason for the impossibility in this case is because there is no least element of $\mathbb{Q}$?
I mean, it's possible to impose an ordering on $\mathbb{Q}$ (correct me if I'm wrong, but it seems possible to compare any two rationals).
But the set $\mathbb{Q}$ is of course, a subset of itself, and the Well-Ordering Principle says that a set $S$ is Well-Ordered only if any subset of $S$ contains a minimal element. Since $\mathbb{Q}$ does not contain a minimal element, it does not appear that it is a Well-Ordered set.
Is this the correct idea as to why it's impossible? If not, could somebody explain to me what the correct idea is and a strategy for proving it?
• suppose there are succesive rationals $p, q$ so that $p<q$. Then by density, there is a rational $r$ such that $p<r<q,$ which is a contradiction, because $p$ and $q$ were assumed to be successive. Feb 9, 2017 at 16:46
• "perhaps it's even possible to list any proper subset of the rationals in increasing order because any such subset will have a least element" Really? What's the least element of $(0, 1)\cap\mathbb{Q}$? Feb 9, 2017 at 16:46
• Your argument works, but it might not be what the problem was shooting for, for two reasons. You would generally ask a question like this before the Well-Ordering Principle has been shown, so unless you prove it, you can't use that. More to the point, there does not exist an order preserving map of$\Bbb Q\cap (0,1) \mapsto \Bbb Z$ and the desired proof ought also to show that fact. Feb 9, 2017 at 16:47
• @MarkFischler actually, we have already been shown the Well-Ordering Principle. In order to show that the nonexistence of an order preserving map of $\mathbb{Q}\cap (0,1) \mapsto \mathbb{Z}$, do I first need to show any kind of bijection between $\mathbb{Q}$ and $\mathbb{Q}\cap(0,1)$?
– user100463
Feb 9, 2017 at 16:52
• @chilangoincomprendido: instead of saying "by density", it is simpler to just let $r=(p+q)/2$. Feb 9, 2017 at 17:09
There are many ways to prove that this is impossible. As you rightly suggest, there is no least rational number, so as soon as you declare one rational number to be the first on your list, all the rational numbers lying below it cannot appear on the list.
Supposing you allow your list to be infinite in both directions (i.e. in order-preserving bijection with $\mathbb{Z}$), it is still impossible, but the argument using least elements no longer works. To prove that this is still impossible, you can use the fact that $\mathbb{Z}$ is not dense, but $\mathbb{Q}$ is. Indeed, suppose $q,r \in \mathbb{Q}$ appear in positions $0$ and $1$ in the list. Then $q<r$, since the list is written in increasing order; but then $q < \frac{q+r}{2} < r$, so $\frac{q+r}{2}$ must appear between $0$ and $1$ on the list, which is evidently impossible.
"I am asked to show that it is impossible to list the rational numbers in increasing order. "
It may not have been explicitely stated but it is implicitely assumed that "order" in this sense is the order we've all known and loved since elementary school where $n < n+1$ and $a > 0; b < c \implies ab < ac$ etc.
"I was wondering if the reason for the impossibility in this case is because there is no least element of Q? "
Well, yes, to have a list a list must have a first element. And for each item in the list there must be a distinct next item that immediately follows it. Both of those are impossible if the rationals are ordered with the order we know and love.
"I mean, it's possible to impose an ordering on Q (correct me if I'm wrong, but it seems possible to compare any two rationals)"
Well, obviously. The order we know and love if $a > b$ if $a - b > 0$ or $m/n > p/q$ if $mq >pn$. That's not "imposed". We were given that and we've been using it since we learned to count.
But we can also impose a different order if we want to. More on that later.
"But the set Q is of course, a subset of itself, and the Well-Ordering Principle says that a set S is Well-Ordered only if any subset of S contains a minimal element. Since Q does not contain a minimal element, it does not appear that it is a Well-Ordered set."
Precisely. $\mathbb Q$ is not a well-order set when using the order that we know and love. And the fact that $\mathbb Q$ has no least element demonstrates that (as does that no element has an immediate successor.
"Is this the correct idea as to why it's impossible?"
Yes. You have done it. You are done. Go home and have a cup of cocoa.
....
So ... what's the issue?
I imagine that you have heard by the Well-Ordering Principle that $\mathbb Q$ IS a well-ordered set. That is true. But this is refering to a different method of ordering than the one we know and love.
All countable sets can be listed (not necessarily be size but by other criteria) and we can call the order they are listed in a well-ordering order.
If we used the "diagonal" listing of rationals. (1/1, 1/2, 2/1, 1/3, [omit 2/2] 3/2, 1/4, 2/3, 3/2, 1/4, 1/5, [omit 2/4],[omit 3/3][omit 4/2] 5/1, etc.) And define ordering as:
$r < t$ if $r$ appears on the list before $t$.
Then we'd have $1 < 1/2 < 2 < 1/3 < 3/2 < 1/4 < 2/3 .....$. This ordering, which we all know and hate, and which has nothing to do with size, but only has to do with running through a diagonal and/or making a list, is a well-ordering.
But it is not the order we know and love.
Which is not well-ordered.
That's all.
......
Oh, wait. That's not all. The uncountable reals according to the axiom of choice (equivalent to the Well-Ordering Principle when extending to uncountable sets) will have a well-ordering. If it does (the Axiom of Choice is an axiom, not a theorem, and can not be proven) no-one knows what it is. I may be wrong, but I believe that just as the Axiom of choice can not be proven, we also know the well-ordering of the reals can not be found.
But the rationals are countable so that is a different issue.
I think you're confusing different orderings with each other. Yes, the Well Ordering Principle says that any set, including $\mathbb{Q}$, can be well-ordered — but that won't be the same usual ordering of numbers. "Increasing" with respect to that new ordering will not be the same thing as with respect to the usual "less or equal" relation for numbers. | 2022-06-26T03:25:53 | {
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https://math.stackexchange.com/questions/2610819/find-k-l-given-x-25-38k-59-69l-exists-k-l-in-mathbbz | # Find $k,l$ given $x = 25 + 38k = 59 +69l, \exists k,l \in \mathbb{Z+}$
Find the smallest positive integer solution to the following system of congruence:
$25\pmod{38}\equiv x \equiv 59\pmod{69}$
$x = 25 + 38k = 59 +69l, \exists k,l \in \mathbb{Z+}$
I am not having any idea to solve, and request for help
The only idea that occurs is to use the property of both $k,l$ being positive integers. Expressing the equality of the two, we get:
$25+38k = 59 +69l$
=> $38k = 34 + 69l$
=> $k = \frac{34 + 69l}{38}$
=> $k = \frac{17}{19} + (1+ \frac{31}{38})l$
=> $\frac{17}{19} + \frac{31}{38}l = k - l =$ a positive integer.
So, need check $\frac{34 + 31l}{38}$to be integer value for different values (positive) of $l$, with smallest being the answer for $l$, and $k \gt l$, as $k-l$ is positive.
So, a second question arises: Even if I find smallest value of $l$ it does not mean that there is no smaller value of $k$ that would not be possible for a larger $l$.
• Notice that if $a$ and $b$ are two possible solutions then $a-b$ is a multiple of both $38$ and of $69$. The lowest common multiple of $38$ and $69$ is $2622$. So there is only one answer between $1$ and the lowest common multiple of $2622$. In fact if there is any answer $k$ then all answer will be $k + 2622m$. – fleablood Jan 18 '18 at 17:47
Remember Euclid's Algorithm for determined greatest common divisor.
$69 = 38 + 31$ so $\gcd(69,38) = \gcd(38,31)$
$38 = 31 + 7$ so $\gcd(38,31) = \gcd(31,7)$
$31 = 4*7 + 3$ so $\gcd(31,7) = \gcd(7,3)$
$7 = 2*3 + 1$ so $\gcd(7,3) = \gcd(3,1) = 1$.
If we keep track of how many times we add and subtract we get:
$69 = 38 + 31; 31=69 - 38$
$38 = 31 + 7;7= 38 - 31 = 38 -(69- 38)= 2*38 - 69$.
$31= 4*7 + 3; 3 = 31 -4*7 = (69-38) - 4(2*38 -69) = 5*69 -9*38$
$7 = 2*3 + 1; 1 = 7- 2*3 = (2*38 - 69) - 2(5*69-9*38) = 20*38-11*69$
so $20*38 - 11*69 = 1$.
So there is always a solution to $xk - ym = \gcd(k,m)$
Multiply the expression $20*38 - 11*69 =1$ on both sides by $34$ are you get:
$34*20*38 - 11*34*69 = 34$
$680*38 - 374*69 = 34$
But $680$ and $374$ are such huge numbers. We can get them smaller be subtraction and adding multiples of $38*69$.
$680*38 - 374*69 = 34$
$680*38-690*38 - 374*69 + 69*380 = 34$
$-10*38 + 6*69 = 34$
$-10*38 + 69*38 + 6*69 - 69*38 = 34$
$59*38 -32*69 = 34$
So $25 +59*38 = 59 + 32*69$
is the solution you want.
• Very fine solution. Just need logic behind division of modulus, as each modulus will be having a different residue, multiplier (to make it equal to a given multiple of the given modulus). If that part were described, or hinted; then a complete solution indeed. – jiten Jan 18 '18 at 18:37
• Also, I requested some more parts of the answer at your earlier answer with link: math.stackexchange.com/a/2610178/424260 – jiten Jan 18 '18 at 18:42
• People like you are the soul of MSE. – jiten Jan 18 '18 at 18:45
solving this Diophantine equation we obtain $$k=59+69C$$ and $$l=32+38C$$ where $C$ is a non negative constant note $$38\cdot 59-69\cdot 32=34$$
• But, can you please elaborate how you obtained the values: $66, 36$. Unable to find out. – jiten Jan 18 '18 at 16:35
• ok i 've made a Little typo, sorry it's corrected and i have added a line – Dr. Sonnhard Graubner Jan 18 '18 at 16:45
• I still am confused about $32$ in second line, also why this 'matrix-multiplication' sort of multiplication & subtraction has occurred, is not clear. – jiten Jan 18 '18 at 16:49
• the numbers $32$ and $59$ are Special Solutions of your equation – Dr. Sonnhard Graubner Jan 18 '18 at 16:51
• i hope this will help you! – Dr. Sonnhard Graubner Jan 18 '18 at 17:20
Look up Chinese Remainder Theorem which addresses precisely this.
But if you roll up your sleeves and do it:
$25 + k38 = 59 + 69l$ so
$38k -69l = 34$
$38k - (2*38 - 7)l = 34$
$38(k-2l) + 7l = 34$. Let $m=k-2l$
$38m + 7l = 34$.
$(3+ 5*7)m + 7l = 34$
$3m + 7(5m + l) = 34$. Let $n = 5m+l$.
$3m + 7n = 34$
$3m + 2*3n + n = 34$
$3(m + 2n) + n = 34$. Let $a = m+2n$.
$3a + n = 34$.
Let $a=11; n = 1$. So $a=m+2n; 11=m + 2; m = 9$. and $n= 5m + l; 1=5*9 +l; l = -44$. and $m = k - 2l; 9=k +88; k = -79$.
So $25 - 79*38 = -2977$ and $59 - 44*69=-2977$.
So $-2977\equiv 25 \mod 38$ and $-2977 \equiv 59 \mod 69$.
That's of course not positive but.
$25 - 79*38 = 59- 44*69 \iff$
$25 - 79*38 + 69*38 = 59 - 44*69 + 69*38 \iff$
$25 - 10*38 = 59 - 6*69 \iff$
$25 - 10*38 + 69*38 = 59 - 6*69 + 69*38 \iff$
$25 + 59*38 = 59 + 32*69$
And $25+59*38 = 59 + 32 * 69 =2267$.
$2267\equiv 25\mod 38$ and $2267\equiv 59\mod 69$ and as the lowest common multiple of $38$ and $69$ is $2622$ this is the smallest positive such number.
• It is an answer with thought process clearly shown with meaningful steps to approach the problem from scratch, just hidden is the logic for division of one modulus by another. – jiten Jan 18 '18 at 18:49
• I think it is a super-set of the CRT, as all the 3 quantities ($x_i, a_i, m_i)$ are different. In CRT, $x$ is constant across all the equations. – jiten Jan 18 '18 at 21:12
• $x$ is what you solve for. $x = 25 \mod 38$ and $x = 59 \mod 69$. By CRT $x$ has a unique solution $\mod 38*69$. – fleablood Jan 18 '18 at 21:56 | 2020-10-28T16:39:14 | {
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https://math.stackexchange.com/questions/2234586/why-does-the-integral-int-1-infty-frac-sinxx2dx-exist | # Why does the integral $\int_1^\infty\frac{\sin(x)}{x^2}dx$ exist?
As the title says, I'm trying to justify why the (Riemann) integral $\int_1^\infty\frac{\sin(x)}{x^2}dx$ exists. I'm not interested in determining its value. It seems to me that I should use the fact that the integral $\int_1^B\frac{\sin(x)}{x^2}dx$ exists for every $B>0$, because the integrand is continuous, and the estimate $$\left|\int_1^B\frac{\sin(x)}{x^2}dx\right| \le\int_1^B\left|\frac{\sin(x)}{x^2}\right|dx \le\int_1^B\frac 1{x^2}dx<\infty.$$ However, the estimate only says that in the limit $B\rightarrow\infty$ the integral is finite if it exists, but I don't see how I conclude that the limit does exist.
• Notice that $\lim_{B\to\infty}\frac d{dB}\int_1^B\frac{\sin(x)}{x^2}\ dx=0$. What should that tell you? – Simply Beautiful Art Apr 14 '17 at 22:08
• What is your definition of convergence? You've proven it is absolutely convergent and this implies convergence. – Mark Viola Apr 14 '17 at 22:15
• Absolute convergence does not imply convergence here, since I am using the Riemann integral, not the Lebesgue integral. – cthl Apr 14 '17 at 22:20
• @cthl No, that is not correct. An improper Riemann integral that is absolutely convergent is convergent! What on earth are you talking about?? – Mark Viola Apr 14 '17 at 22:28
• To check that absolute convergence implies convergence for the integral of Riemann it is enough to check the epsilon-delta definition of convergence of $\int|f|$ and compare with the epsilon-delta definition of convergence of $\int f$ – Masacroso Apr 14 '17 at 22:36
## 2 Answers
You should prove the theorem that if $f \colon [a, \infty) \rightarrow \mathbb{R}$ is Riemann integrable on each interval $[a,b]$ for $b > a$ and $\int_a^{\infty} |f(x)| \, dx$ exists (as an improper Riemann integral) then so does $\int_a^{\infty} f(x) \, dx$ (absolute convergence implies regular convergence).
The main idea of the proof is to use the Cauchy criterion of convergence for an improper integral. Namely, the improper integral $\int_a^{\infty} g(x) \, dx$ converges iff for every $\varepsilon > 0$ we can find $M \geq a$ such that for all $y > x > M$ we have
$$\left| \int_x^{y} g(x) \, dx\right| < \varepsilon.$$
Assuming this criterion, if $\int_a^{\infty} |f(x)| \, dx$ exists then for any $\varepsilon > 0$ we can find $M \geq a$ such that for all $y > x > M$ we have
$$\left | \int_x^y |f(x)| \, dx \right| = \int_x^y |f(x)| \, dx < \varepsilon$$
but then we also have
$$\left| \int_x^y f(x) \, dx \right| \leq \int_x^y |f(x)| \, dx < \varepsilon$$
so by the Cauchy criterion, $\int_a^{\infty} f(x) \, dx$ also conveges.
• Very well written. (+1) – Mark Viola Apr 14 '17 at 22:30
• Thank you, this took me a while to understand. If I understand this correctly, your first statement implicitly assumes that $f$ is integrable on every interval $\left[a,b\right]$, as is the case for my original function. I didn't see this implicit assumption and was concerned about a function like $f(x)=\frac{g(x)}{x^2}$, where $g(x)=1$ if $x\in\mathbb Q$ and $g(x)=-1$ otherwise, since in this case $\int_1^\infty |f|$ exists, but $\int_1^\infty f$ doesn't. – cthl Apr 14 '17 at 22:56
• @cthl: Yeah. When one talks about the improper Riemann integral of $f$ on $[a,\infty)$, one implicitly assumes that $f$ is Riemann integrable on $[a,b]$ for all $b > a$ (otherwise, one cannot even talk about the limit of $\int_a^b f(x) \, dx$ as $b \to \infty$) so that such pathologies cannot occur. I'll edit the answer to be more precise. – levap Apr 14 '17 at 23:01
Define $F:[1,+\infty) \to\mathbb R$ by $F(x) = \int_1^x s^{-2}\sin(s)\,ds$. You know $F$ is well defined, so we only have to prove $\lim\limits_{x\to+\infty}F(x)$ exists and is finite.
The estimate $$|F(y)-F(x)|\leq \left|\int_x^ys^{-2}\,ds\right|\leq\int_{\min\{x,y\}}^\infty s^{-2}\,ds=\frac{1}{\min\{x,y\}}$$ shows that if $x,y > M$, then $|F(x)-F(y)|<1/M$. If we take any sequence $(x_n) \to +\infty$, this implies $(F(x_n))$ is Cauchy, and therefore converges to a real number.
Now you can argue that $\lim\limits_{n\to\infty} F(y_n)$ is the same for any $(y_n)\to\infty$ and therefore that $\lim\limits_{x\to+\infty} F(x)$ exists. | 2019-06-19T22:49:53 | {
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http://openstudy.com/updates/55b0a75ae4b071e6530d748f | ## anonymous one year ago if a ≡ b (mod c), then is it true that ad ≡ bd (mod c) for all integer d?
1. ganeshie8
$$a\equiv b\pmod{c} \implies c\mid (a-b) \implies c\mid d(a-b)\\~\\ \implies d(a-b)\equiv 0 \pmod{c} \implies ad\equiv bd\pmod{c}$$
2. ganeshie8
do you see any restrictions on $$d$$ ?
3. anonymous
No, I don't see any restriction on d. So it's true?
4. ganeshie8
Neither do I. So yes it is true for all $$d\in\mathbb Z$$
5. anonymous
how did you get d(a-b) ≡ 0 (mod c) though?
6. ganeshie8
That is the definition of congruence : $n\mid m \iff m\equiv 0 \pmod{n}$
7. anonymous
oh I see. Could you have done this though? c | d(a-b) is the same as c | d(a-b) - 0, which by definition d(a-b) ≡ 0 (mod c)
8. ganeshie8
I don't see why you want to have that intermediate step of subtracting 0
9. ganeshie8
unless you're taking the pattern seriously : $x\equiv y\pmod{n} \iff n\mid(x-y)$
10. anonymous
I was thinking about the definition in terns of two things. x ≡ y (mod z) means z | (x - y) so by subtracting 0, i.e c | d(a-b) - 0, I was comparing d(a-b) to x and 0 to y
11. anonymous
Yeah, that pattern ^^
12. ganeshie8
if it helps, then yes you may do that :)
13. ganeshie8
btw the converse of the conditional in main question does not hold : $ad\equiv bd\pmod{c} \implies a\equiv b\pmod{c}$ is false in general
14. anonymous
right. It's true if gcd(c,d) = 1. The proof was provided in the book :)
15. ganeshie8
Yes, more generally, below holds : $ad\equiv bd\pmod{c} \implies a\equiv b\pmod{\frac{c}{\gcd(d,c)}}$
16. anonymous
:O I didn't know that. Thank you for the info!
17. ganeshie8
np | 2017-01-22T12:19:36 | {
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https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12B_Problems/Problem_22&oldid=87132 | # 2004 AMC 12B Problems/Problem 22
## Problem
The square
$\begin{tabular}{|c|c|c|} \hline 50 & \textit{b} & \textit{c} \\ \hline \textit{d} & \textit{e} & \textit{f} \\ \hline \textit{g} & \textit{h} & 2 \\ \hline \end{tabular}$
is a multiplicative magic square. That is, the product of the numbers in each row, column, and diagonal is the same. If all the entries are positive integers, what is the sum of the possible values of $g$?
$\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 62 \qquad \textbf{(E)}\ 136$
## Solution A
If the power of a prime $p^n$ other than $2,5$ divides $g$, then from $50 \cdot 2 e = 50dg$ it follows that $p^n|e$, but then considering the product of the diagonals, $p^{2n} |gec$ but $p^{2n} \nmid 100e$, contradiction. So the only prime factors of $g$ are $2$ and $5$.
It suffices now to consider the two magic squares comprised of the powers of $2$ and $5$ of the corresponding terms. These satisfy the normal requirement that the sums of rows, columns, and diagonals are the same, owing to our rules of exponents; additionally, all terms are non-negative.
The powers of $2$:
$\begin{tabular}{|c|c|c|} \hline 1 & \textit{b} & \textit{c} \\ \hline \textit{d} & \textit{e} & \textit{f} \\ \hline \textit{g} & \textit{h} & 1 \\ \hline \end{tabular}$
So $1 + 1 + e = g + e + c \Longrightarrow g = 2 - c$, so $g = 0,1,2$. Indeed, we have the magic squares
$\begin{tabular}{|c|c|c|} \hline 1 & 0 & 2 \\ \hline 2 & 1 & 0 \\ \hline 0 & 2 & 1 \\ \hline \end{tabular}, \quad \begin{tabular}{|c|c|c|} \hline 1 & 1 & 1 \\ \hline 1 & 1 & 1 \\ \hline 1 & 1 & 1 \\ \hline \end{tabular}, \quad \begin{tabular}{|c|c|c|} \hline 1 & 2 & 0 \\ \hline 0 & 1 & 2 \\ \hline 2 & 0 & 1 \\ \hline \end{tabular},$
The powers of $5$:
$\begin{tabular}{|c|c|c|} \hline 2 & \textit{b} & \textit{c} \\ \hline \textit{d} & \textit{e} & \textit{f} \\ \hline \textit{g} & \textit{h} & 0 \\ \hline \end{tabular}$
Again, we get $2 + e = g + e + c \Longrightarrow g = 0,1,2$. However, if we let $g = 2, c = 0$, then $e = d + e + f \Longrightarrow d = f = 0$, which obviously gives us a contradiction, and similarly for $g = 0, c = 2$. For $g = 1$, we get
$\begin{tabular}{|c|c|c|} \hline 2 & 0 & 1 \\ \hline 0 & 1 & 2 \\ \hline 1 & 2 & 0 \\ \hline \end{tabular}$
In conclusion, $g$ can be $2^0 \cdot 5^1, 2^1 \cdot 5^1, 2^2 \cdot 5^1$, and their sum is $\boxed{\mathbf{(C)}35}$.
## Solution B
All the unknown entries can be expressed in terms of $b$. Since $100e = beh = ceg = def$, it follows that $h = \frac{100}{b}, g = \frac{100}{c}$, and $f = \frac{100}{d}$. Comparing rows $1$ and $3$ then gives $50bc = 2 \cdot \frac{100}{b} \cdot \frac{100}{c}$, from which $c = \frac{20}{b}$. Comparing columns $1$ and $3$ gives $50d \cdot \frac{100}{c}= 2c \cdot \frac{100}{d}$, from which $d = \frac{c}{5} = \frac{4}{b}$. Finally, $f = 25b, g = 5b$, and $e = 10$. All the entries are positive integers if and only if $b = 1, 2,$ or $4$. The corresponding values for $g$ are $5, 10,$ and $20$, and their sum is $\boxed{\mathbf{(C)}35}$.
Credit to Solution B goes to http://billingswest.billings.k12.mt.us/math/AMC%201012/AMC%2012%20work%20sheets/2004%20AMC%2012B%20ws-15.pdf, a page with a play-by-play explanation of the solutions to this test's problems. | 2021-12-09T10:58:53 | {
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http://math.stackexchange.com/questions/328467/to-show-that-the-set-point-distant-by-1-of-a-compact-set-has-lebesgue-measure-0 | # To show that the set point distant by 1 of a compact set has Lebesgue measure $0$
Could any one tell me how to solve this one?
Let $K$ be a compact subset of $\mathbb{R}^n$, and $$A:=\{x\in\mathbb{R}^n:d(x,K)=1\}.$$ Show that $A$ has Lebesgue measure $0$.
Thank you!
-
Assuming that $d(x,K)$ is induced by the Euclidean metric, I suspect one could argue that, locally, $A$ is a smooth $n-1$ manifold, i.e., locally the graph of some smooth function. This graph has measure zero. – Nemis L. Mar 12 '13 at 14:50
@SimenK. Well, I don't think you will get smooth function, but merely a Lipschitz one. Which is still sufficient for your line of reasoning. – Thomas Mar 12 '13 at 15:12
I don't have an answer, but here is an argument that could lead to one: The set of points which do not have a unique nearest point in $K$ is called the ridge of the distance function. The ridge is known to have Hausdorff dimension at most $n-1$, while $A$ is expected to be the graph of a Lipshitz function near points which are not on the ridge. Put these together, and a proof might emerge. – Harald Hanche-Olsen Mar 12 '13 at 16:55
I'm afraid this is way beyond my abilities, so I did the case $n=1$... Writing the open set $\mathbb{R}\setminus K=\bigcup (a_j,b_j)$ we see that $A$ is countable. So it has Lebesgue measure $0$. – 1015 Mar 12 '13 at 23:30
Out of curiosity, where did this problem come from? – Nate Eldredge Mar 13 '13 at 2:13
Suppose $x_0\in A$. Let $B := \{ d(x,K) < 1\}$.
Observe that since $K$ is compact, there exists $y_0\in K$ such that $d(x_0,y_0) = 1$. By definition $B_1(y_0) = \{x: d(x,y_0) < 1\}$ is a subset of $B$.
This implies that for all $\epsilon < 1/2$, we have that $$\mu(B_\epsilon(x_0) \cap B) \geq \frac1{2^n} \mu(B_\epsilon(x_0))$$ where $\mu$ is the Lebesgue measure. (The factor $1/2^n$ is very loose: Since a sphere tangent to $x_0$ is contained in $B$, locally an orthant centered at $x_0$ is contained in $B$.)
Hence we have that for every $x_0\in A$, $$\limsup_{\epsilon \to 0} \frac{\mu(B_\epsilon(x_0) \cap A)}{\mu(B_\epsilon(x_0))} \leq \frac{2^n-1}{2^n} < 1$$
By the Lebesgue differentiation theorem, the set of $x_0\in A$, for $A$ measurable, such that the above condition holds must be measure zero. Hence $A$ has measure zero.
-
This is very nice! – Nate Eldredge Mar 13 '13 at 13:00
And sooo obvious – in retrospect. Tiny nitpick, though: The limit should be a limsup. – Harald Hanche-Olsen Mar 13 '13 at 15:34
@HaraldHanche-Olsen: you are right of course. Fixed. – Willie Wong Mar 13 '13 at 16:23
Any idea if $K$ is any subset ? I will be impressed if it doesn't hold. – user10676 Mar 13 '13 at 18:00
@user10676: use that $d(x,K) = d(x,\bar{K})$. Then cut-off at radius $R$. By Heine-Borel we now have a compact set. Take a countable increasing sequences of $R$ and use sigma sub-additivity. – Willie Wong Mar 14 '13 at 9:03
I was going to make this a comment but it occurred to me there might be sufficient interest that perhaps I should not bury it in a comment.
At the beginning of the paper below Erdős gives a short proof (that he attributes to Tibor Radó) making use of the Lebesgue density theorem that $E_r$ has Lebesgue measure zero, where $E$ is a closed set in ${\mathbb R}^n$ and $$E_r \, = \; \{ x \in {\mathbb R}^n : \; d(x,E)=r \}$$
Paul Erdős, Some remarks on the measurability of certain sets, Bulletin of the American Mathematical Society 51 #10 (October 1945), 728-731.
Later in this paper (item 6), Erdős proves the stronger result that, for $K$ compact, the Hausdorff $(n-1)$-measure of $K_r$ is finite, and hence $E_r$ has $\sigma$-finite Hausdorff $(n-1)$-measure when $E$ is closed.
More precise results can be found in a 1985 paper by Oleksiv/Pesin Zbl 573.28010 [English translation: Mathematical Notes 37 (1985), 237-242], and I'm sure there are quite a few related results in the literature. For instance, each of the sets $E_r$ is $[1]$-very porous in the sense defined in this conference talk of mine, and among other things I gave a short argument there that each such set has $\sigma$-finite packing $(n-1)$-measure.
The analysis of results related to these in infinite dimensional normed spaces has also generated a fair amount of interest. One possible entry point into this is Ludek Zajicek's 1983 paper Differentiability of the distance function and points of multi-valuedness of the metric projection in Banach space.
(Added 12 Weeks Later) Someone recently gave me a vote on this answer and, in looking at what I wrote, it occurred to me that a much better reference for the comment "one possible entry point into this" that I made in the last paragraph is the following book:
Joram Lindenstrauss, David Preiss, and Jaroslav Tišer, Fréchet Differentiability of Lipschitz Functions and Porous Sets in Banach Spaces, Annals of Mathematics Studies #179, Princeton University Press, 2012, x + 425 pages. Zbl 1241.26001 (a review)
Princeton University Press web page for the book
Amazon.com web page for the book
-
Disclaimer: The following approach has a flaw in the second step, because the set $A$ could be a nowhere dense set with positive measure like e.g. a fat Cantor set as Davide Giraudo pointed out in his comment. I don't see any way to work around this gap, but leave the answer here in the hope that it might still prove useful in some way.
1. $A$ is compact. Since $A$ is clearly bounded, it only needs to be shown that it's closed. Let $x_i$ be a convergent sequence in $A$ with limit point $x$. Then $x$ is also in $A$, because for each $x_i$ you find a $y_i$ in $K$ with $d(x_i,y_i)=1$. The $y_i$ have a convergent subsequence with some limit point $y$. Because $d$ is continuous it follows that $d(x,y)=1$. Further, one can show that $d(x,y)\geq1$ for all $y\in K$ (Otherwise, say $d(x,y)<1-\epsilon$ for some $y\in K$, then $d(x_i,y)\leq d(x,x_i)+d(x,y)<\epsilon/2+1-\epsilon<1$ for some $x_i$). Thus $x\in A$.
2. If $A$ is closed and Lebesgue measurable with $\lambda(A)>0$ it must contain at least one non-empty open set of $\mathbb R^n$, namely the interior of $A$. This in turn would contain a closed ball, call it $\bar B_\delta(x)$ for some $x\in A$ and $\delta>0$.
3. $K$ cannot contain any point of the open ball $B_{1+\delta}(x)$, because otherwise there would be a point in $\bar B_\delta(x)$ with distance to $K$ smaller than 1. But, then $x$ would have distance larger then $1$ to $K$, i.e. $x\not\in A$, contrary to the assumption that $\bar B_\delta(x)\subset A$.
A possible remedy: If we define $B=\bigcup_{y\in K}\bar B_1(y)$, then $A=\partial B$, since the boundary of $B$ contains exactly the points which have distance $1$ to $K$. Since $B$ is a relatively "smooth" set (in the sense that it cannot have infinitely small and dense holes like a Cantor set), this may carry over to its boundary and salvage step 2. by ruling out "weird sets" like the fat Cantor.
-
2. may be problematic when $K$ is a fat Cantor. – Davide Giraudo Mar 12 '13 at 22:12
Is statement 2. obvious? I had similar thoughts for a method of proof but couldn't justify that step. That is, why must the interior be non-empty? – Dan Rust Mar 12 '13 at 22:18
@DavideGiraudo You're right. I didn't think about that. As the proof builds on that, there's probably no way to salvage it. I just want to think about it a bit more, before I remove it. – Elmar Zander Mar 12 '13 at 22:21
I wouldn't remove the answer, it may still be useful to the OP. Just add a disclaimer that the gaps may be un-fillable. – Dan Rust Mar 12 '13 at 22:23
I miscredited Vitali. Here it is: en.wikipedia.org/wiki/Steinhaus_theorem – Yoni Rozenshein Mar 13 '13 at 0:09
$f_K : \mathbb{R}^n \rightarrow \mathbb{R}$ given as $f_K(x) = d(x,K)$ is clearly Lipschitz as $|f_K(x)-f_K(y)| \leq |x-y|,\ A = f_K^{-1}(\{1\})$. A Lipschitz function $f$ take measure 0 sets to measure 0 sets as if $\mu$ is lebesgue measure for $\mathbb{R}^n$ then $\mu(f(A))\leq (Lip(f))^n\mu(A)$(you can prove this is either using outer measure arguments or very trivially using equality of lebesgue and hausdorff measure).
-
How can you conclude that $\mu (A)=0$? – Tomás Mar 12 '13 at 16:29
To amplify on @Tomás's remark, it seems to me that your inequality is the reverse of what you would need. – Harald Hanche-Olsen Mar 12 '13 at 16:51
As you wrote: $f$ takes measure 0 sets to measure 0 sets. But you have to show that the preimage of a measure 0 set is a measure 0 set. – saz Mar 12 '13 at 17:31
To amplify on @HaraldHanche-Olsen 's remark, consider the function $f(x) = 0$. It is clearly Lipschitz, but $f^{-1}(0)$ is not measure 0. – Willie Wong Mar 12 '13 at 17:41 | 2015-08-31T05:21:36 | {
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http://godin44.com/coswd21w/viewtopic.php?id=053001-cyclic-quadrilateral-theorem | Ptolemy's Theorem Cyclic Quadrilateral For a cyclic quadrilateral, the sum of the products of the two pairs of opposite sides equals the product of the diagonals (Kimberling 1998, p. 223). If a cyclic quadrilateral is also orthodiagonal, the distance from the circumcenter to any side equals half the length of the opposite side. Theorem: Opposite angles of a cyclic quadrilateral are supplementry. Fill in the blanks and complete the following ... ∠D = 180° ∠A + ∠C = 180° e = c An exterior angle of a cyclic quadrilateral is equal to the interior opposite angle. Properties. Cyclic quadrilateral. The area of a cyclic quadrilateral is the maximum possible for any quadrilateral with the given side lengths. It has some special properties which other quadrilaterals, in general, need not have. The first theorem about a cyclic quadrilateral state that: The opposite angles in a cyclic quadrilateral are supplementary. What are the Properties of Cyclic Quadrilaterals? The following theorems and formulae apply to cyclic quadrilaterals: Ptolemy's Theorem; Brahmagupta's formula; This article is a stub. What can you say about the Angles in a Cyclic Quadrilateral? The word cyclic often means circular, just think of those two circular wheels on your bicycle. This dynamic worksheet illustrates the 'cyclic quadrilateral' circle theorem. Theorem of Cyclic Quadrilateral (II) In a cyclic quadrilateral, if a quadrilateral is inscribed inside a cycle, the product of the diagonals of the cyclic quadrilateral is equal to the sum of the two pairs of opposite sides of the cyclic quadrilateral. Theorem 4. Theorems of Cyclic Quadrilateral Cyclic Quadrilateral Theorem The opposite angles of a cyclic quadrilateral are supplementary. Cyclic Quadrilateral: Definition. Theorem 10.12 If the sum of a pair of opposite angles of a quadrilateral is 180 , the quadrilateral is cyclic. The angle subtended by a semicircle (that is the angle standing on a diameter) is a right angle. You should know that: (a) the opposite angles of a cyclic quadrilateral sum to 180° i.e. Hence, the theorem is proved. The sum of the opposite angles of an inscribed quadrilateral is 180 degrees. If the sum of the opposite angles of a quadrilateral is 180°, then the quadrilateral is cyclic. In cyclic quadrilateral : Applicable Theorems/Formulae. I want to know how to solve this problem using Ptolemy's theorem and Brahmagupta formula for area of cyclic quadrilateral, which is ($\sqrt{(s-a)(s-b)(s-c)(s-d)}$). Cyclic Quadrilateral Ptolemy's Theorem Proof. Let’s take a look. Let's prove this theorem. In a cyclic quadrilateral, the perpendicular bisectors of the four sides of the cyclic quadrilateral meet at the center O. Consider the diagram below. Proving the Cyclic Quadrilateral Theorem- Part 2 An exterior angle of a cyclic quadrilateral is equal to the interior opposite angle. Please don't use any complex trigonometry technique and please explain each step carefully. Theorem 10.11 The sum of either pair of opposite angles of a cyclic quadrilateral is 180°. Click hereto get an answer to your question ️ Prove that \"the opposite angles of a cyclic quadrilateral are supplementary\". That is, all 4 vertices of a cyclic quadrilateral always lie on the circle itself. A cyclic quadrilateral is a four-sided polygon whose vertices are inscribed in a circle. Theorem 1. Therefore, cyclic quadrilateral angles equal to 180 degrees. Can you prove the result? In a cyclic quadrilateral, the sum of the opposite angles is always equal to 180°. Cyclic quadrilaterals are useful in various types of geometry problems, particularly those in which angle chasing is required. See this problem for a practical demonstration of this theorem. [22] If the diagonals of a cyclic quadrilateral intersect at P, and the midpoints of the diagonals are M and N, then the anticenter of the quadrilateral is the orthocenter of triangle MNP. Ptolemy's theorem states the relationship between the diagonals and the sides of a cyclic quadrilateral. Given : A circle with centre O and the angles ∠PRQ and ∠PSQ in the same segment formed Theorem Statement: The sum of the opposite angles of a cyclic quadrilateral is 180°. A cyclic quadrilateral is a quadrangle whose vertices lie on a circle, the sides are chords of the circle.Enter the four sides (chords) a, b, c and d, choose the number of decimal places and click Calculate. A quadrilateral whose vertices lie on a circle is called a cyclic quadrilateral. a+ c = 180° b + d = 180° (b) the exterior angle of a cyclic quadrilateral is equal to the interior opposite angle i.e. It is a powerful tool to apply to problems about inscribed quadrilaterals. Cyclic Quadrilateral Calculator. A cyclic quadrilateral is a quadrilateral that can be inscribed in a circle, meaning that there exists a circle that passes through all four vertices of the quadrilateral. Other properties Japanese theorem Definition. So according to the theorem statement, in the below figure, we have to prove that Brahmagupta Theorem and Problems - Index Brahmagupta (598–668) was an Indian mathematician and astronomer who discovered a neat formula for the area of a cyclic quadrilateral. Thus in a cyclic quadrilateral, the circumcenter, the "vertex centroid", and the anticenter are collinear. Brahmagupta's theorem states that for a cyclic quadrilateral that is also orthodiagonal, the perpendicular from any side through the point of intersection of the diagonals bisects the opposite side. Learn more at CoolGyan. Coming back to Max's problem. A cyclic quadrilateral is a quadrilateral with all its four vertices or corners lying on the circle.It is thus also called an inscribed quadrilateral. Opposite angles of a cyclic quadrilateral add up to 180 degrees. A cyclic quadrilateral is a quadrilateral for which a circle can be circumscribed so that it touches each polygon vertex.A quadrilateral that can be both inscribed and circumscribed on some pair of circles is known as a bicentric quadrilateral.. Quadrilateral means four-sided figure. 2 4 180 2 3 1 0 Opposite angles of a cyclic quadrilateral 4 5 180 0 Supplementary Angle Theorem 4 … The Theorem states that the product of the diagonals of a cyclic quadrilateral is equal to the sum of the products of opposite sides. If all four points of a quadrilateral are on circle then it is called cyclic Quadrilateral. Cyclic Quadrilateral. In a cyclic quadrilateral, $$d1 / d2 = \text{sum of product of opposite sides}$$, which shares the diagonals endpoints. Given : ABCD is a cyclic quadrilateral. Here we have proved some theorems on cyclic quadrilateral. Online Geometry: Cyclic Quadrilateral Theorems and Problems- Table of Content 1 : Ptolemy's Theorems and Problems - Index. (Called the Angle at the Center Theorem) And (keeping the end points fixed) ... Cyclic Quadrilateral. Angles in a Circle and Cyclic Quadrilateral 19.1 INTRODUCTION You must have measured the angles between two straight lines, ... Theorem : Angles in the same segment of a circle are equal. i.e. Theorem : Opposite angles of a cyclic quadrilateral are supplementary (or) The sum of opposite angles of a cyclic quadrilateral is 180 ° Given : O is the centre of circle. In a cyclic quadrilateral, the perpendicular bisectors always concurrent. Theorem 3. A D 1800 C B 1800 BDE CAB A B D A C B DC 8. A quadrilateral is called Cyclic quadrilateral if its all vertices lie on the circle. Other names for these quadrilaterals are concyclic quadrilateral and chordal quadrilateral, the latter since the sides of the quadrilateral are chords of the circumcircle. Which other circle theorem can you find in this Activity? See this problem for a practical demonstration of this theorem. Ideas for Teachers Use this Activity as a homework, where the students must come up with a conjecture regarding Angles in Cyclic Quadrilaterals. There are two theorems about a cyclic quadrilateral. Ptolemy's Theorem gives a relationship between the side lengths and the diagonals of a cyclic quadrilateral; it is the equality case of Ptolemy's Inequality.Ptolemy's Theorem frequently shows up as an intermediate step in problems involving inscribed figures. They have a number of interesting properties. A cyclic quadrilateral is a quadrilateral that can be inscribed in a circle. Cyclic Quadrilateral A cyclic quadrilateral is a quadrilateral for which a circle can be circumscribed so that it touches each polygon vertex. Inscribed Quadrilateral Theorem. Brahmagupta's Theorem Cyclic quadrilateral. Mess around with the applet for a couple of minutes, and then answer the questions that follow. A cyclic quadrilateral is a quadrilateral drawn inside a circle so that its corners lie on the circumference of the circle. 1) The opposite angles of a Cyclic - quadrilateral … Definition: A cyclic quadrilateral, by definition, is any quadrilateral that can be inscribed inside a circle. the sum of the opposite angles is equal to 180˚. Calculations at a cyclic quadrilateral. When any four points on the circumference of a circle are joined, they form the vertices of a cyclic quadrilateral. | 2021-06-19T06:35:14 | {
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https://math.stackexchange.com/questions/2490344/induction-range-should-the-arbitrary-integer-include-the-base-case | # Induction range - should the arbitrary integer include the base case?
Let's say I have a claim, $P(n)$, that I want to show true for all $n \in \mathbb{N^+}$.
I would start by showing $P(1)$ holds. Then for the inductive step, I would take an arbitrary integer $m > 1$, and assume $\forall i \in \mathbb{N^+}\: i < m$, $P(i)$ holds.
My question is the subtle difference betwen letting $m \geq 1$ or $m > 1$. If it were $m \geq 1$, then it would be redundant if you examined $m=1$, as you would have a vacuous statement so you're proving $P(1)$ unconditionally, but that's exactly what we did in the base case!
So I would think $m > 1$ would be better style, but I'm not sure what's more commonly accepted in the mathematics community. I've seen it written both ways, but using the non-strict inequality seems iffy to me.
In fact, you need to make sure that the arbitrary case includes the base case, otherwise you can't make the inductive step. You could have the case where $P(1)$ is true, but $P(1)$ does not imply $P(2)$ for some reason.
For example, there's a false proof that "all horses in a group are the same horse" that goes something like this:
1. In a group of 1 horse, all horses are the same horse.
2. If, in a group of $n$ horses, they're all the same horse, then in a group of $n+1$ horses, they're all the same horse. PROOF: Take the group of $n+1$ horses and remove one horse. You now have a group of $n$ horses, which by assumption are the same horse. Replace the horse you removed and remove a different one, all $n$ horses are again the same horse. Therefore all the horses in the group are the same horse, so the statement is true for $n+1$.
3. Therefore, all horses in any group are the same horse.
The failure is specifically because when you have a group of 2 horses you can't remove 1 horse, then replace it and remove another, and use that to compare the horses in the group, because you've never had any other horses to compare against.
• Thank you. I'm wondering if you could clarify why you can't make the inductive step if the arbitrary case doesn't include the base case, as here it seems like you would need two base cases, one for $n=1$ and one for $n=2$. But in my scenario, I don't exactly see why taking $m>1$ would invalidate the proof if you're just proving $P(m)$. – rb612 Oct 26 '17 at 6:55
In the context you describe, you are correct that $m>1$ is what you need. Indeed if you used $m\geq1$ then for $m=1$ the set of inductive assumptions would be empty and the induction argument you make using them for $m=1$ might be false. That is why mathematical induction uses two steps.
If the context were slightly different, where the inductive step was stated as: assume $P(m)$ holds to prove $P(m+1)$ holds, then you would use "for $m\geq1$." That may be the kind of situation in which you saw it used. | 2020-02-18T10:04:04 | {
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https://www.tutorialspoint.com/c-program-to-check-whether-a-number-is-prime-or-not | # C Program to Check Whether a Number is Prime or not?
CServer Side ProgrammingProgramming
A prime number is a number that is divisible only by two numbers itself and one. The factor of a number is a number that can divide it.
The list of the first ten prime numbers is 2,3,5,7,11,13,17,23,29,31.
A number that is not prime is a composite number. A composite number is a number that can be divided by more than two numbers.
Elser then prime and composite there is 1 which is neither Prime nor composite because it can be divided only by itself.
How to check if a number is prime or composite to check if a number is prime there are two conditions that should be checked
1) It should be a whole number greater than 1.
2) it should have only two factors i.e one and the number itself.
If these two conditions are satisfied, then we can say a number is a prime number.
In our program, we will check dividing the number by each number smaller than that number. If any number smaller than the given number divides it then it is not Prime number. Otherwise, it is a prime number.
Let's take an example of two numbers and check whether they are prime or not using this process.
Input − Number1 − 42
Output − 42 is not a prime number
Logic − We will divide 42 by every number greater than 1 and smaller than 42. So,
42/2 = 21 i.e. 42 is divisible by 2, this means 42 is not a prime number because it is divisible by another number.
Input − Number2 − 7
Output − 7 is a prime number
Logic − We will divide seven by every number greater than 1 and smaller than 7. So,
7 is not divisible by 2, so the code will check for the next number i.e. 3
7 is not divisible by 3, so the code will check for the next number i.e. 4
7 is not divisible by 4, so the code will check for the next number i.e. 5
7 is not divisible by 5, so the code will check for the next number i.e. 6
7 is not divisible by 6, This means that 7 is divisible by only 1 and 7 this means 7 is a prime number.
look at the above logic what does the number would be 1000 plus or 100000 Plus then the program would take that many iterations for the for a loop this method would take a lot of computation time. So to reduce the number of iterations they must be a better way.
An optimised solution to this is run the loop only halfway. this means if the number is 77 the loop will run only till 38. This will reduce the number of iterations required so we will use this algorithm to create our program.
## Example
#include <stdio.h>
int main() {
int num = 33, flag = 0;
for(int i=2 ; i < num/2 ; i++) {
if(num%i == 0) {
printf("%d is not a prime number", num);
flag = 1;
break;
}
}
if(flag == 0) {
printf("%d is a prime number", num);
}
}
## Output
33 is a prime number
Published on 19-Aug-2019 08:26:16 | 2022-05-25T22:41:54 | {
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https://in.mathworks.com/help/symbolic/displayformula.html | # displayFormula
Display symbolic formula from string
## Syntax
``displayFormula(symstr)``
``displayFormula(symstr,old,new)``
## Description
example
````displayFormula(symstr)` displays the symbolic formula from the string `symstr` without evaluating the operations. All workspace variables that are specified in `symstr` are replaced by their values.```
example
````displayFormula(symstr,old,new)` replaces only the expression or variable `old` with `new`. Expressions or variables other than `old` are not replaced by their values.```
## Examples
collapse all
Create a 3-by-3 matrix. Multiply the matrix by the scalar coefficient `K^2`.
```syms K A A = [-1, 0, 1; 1, 2, 0; 1, 1, 0]; B = K^2*A```
```B = $\left(\begin{array}{ccc}-{K}^{2}& 0& {K}^{2}\\ {K}^{2}& 2 {K}^{2}& 0\\ {K}^{2}& {K}^{2}& 0\end{array}\right)$```
The result automatically shows the multiplication being carried out element-wise.
Show the multiplication formula without evaluating the operations by using `displayFormula`. Input the formula as a string. The variable `A` in the string is replaced by its values.
`displayFormula("F = K^2*A")`
`$F={K}^{2} \left(\begin{array}{ccc}-1& 0& 1\\ 1& 2& 0\\ 1& 1& 0\end{array}\right)$`
Define a string that describes a differential equation.
`S = "m*diff(y,t,t) == m*g-k*y";`
Create a string array that combines the differential equation and additional text. Display the formula along with the text.
```symstr = ["'The equation of motion is'"; S;"'where k is the elastic coefficient.'"]; displayFormula(symstr)```
Create a string `S` representing a symbolic expression.
`S = "exp(2*pi*i)";`
Create another string `symstr` that contains `S`.
`symstr = "1 + S + S^2 + cos(S)"`
```symstr = "1 + S + S^2 + cos(S)" ```
Display `symstr` as a formula without evaluating the operations by using `displayFormula`. `S` in `symstr` is replaced by its value.
`displayFormula(symstr)`
`$1+{\mathrm{e}}^{2 \pi \mathrm{i}}+{\left({\mathrm{e}}^{2 \pi \mathrm{i}}\right)}^{2}+\mathrm{cos}\left({\mathrm{e}}^{2 \pi \mathrm{i}}\right)$`
To evaluate the strings `S` and `symstr` as symbolic expressions, use `str2sym`.
`S = str2sym(S)`
`S = $1$`
`expr = str2sym(symstr)`
`expr = $S+\mathrm{cos}\left(S\right)+{S}^{2}+1$`
Substitute the variable `S` with its value by using `subs`. Evaluate the result in double precision using `double`.
`double(subs(expr))`
```ans = 3.5403 ```
Define a string that represents a quadratic formula with the coefficients `a`, `b`, and `c`.
```syms a b c k symstr = "a*x^2 + b*x + c";```
Display the quadratic formula, replacing `a` with `k`.
`displayFormula(symstr,a,k)`
`$k {x}^{2}+b x+c$`
Display the quadratic formula again, replacing `a`, `b`, and `c` with `2`, `3`, and `-1`, respectively.
`displayFormula(symstr,[a b c],[2 3 -1])`
`$2 {x}^{2}+3 x-1$`
To solve the quadratic equation, convert the string into a symbolic expression using `str2sym`. Use `solve` to find the zeros of the quadratic equation.
```f = str2sym(symstr); sol = solve(f)```
```sol = $\left(\begin{array}{c}-\frac{b+\sqrt{{b}^{2}-4 a c}}{2 a}\\ -\frac{b-\sqrt{{b}^{2}-4 a c}}{2 a}\end{array}\right)$```
Use `subs` to replace `a`, `b`, and `c` in the solution with `2`, `3`, and `-1`, respectively.
`solValues = subs(sol,[a b c],[2 3 -1])`
```solValues = $\left(\begin{array}{c}-\frac{\sqrt{17}}{4}-\frac{3}{4}\\ \frac{\sqrt{17}}{4}-\frac{3}{4}\end{array}\right)$```
## Input Arguments
collapse all
String representing a symbolic formula, specified as a character vector, string scalar, cell array of character vectors, or string array.
You can also combine a string that represents a symbolic formula with regular text (enclosed in single quotation marks) as a string array. For an example, see Display Differential Equation.
Expression or variable to be replaced, specified as a character vector, string scalar, cell array of character vectors, string array, symbolic variable, function, expression, or array.
New value, specified as a number, character vector, string scalar, cell array of character vectors, string array, symbolic number, variable, expression, or array. | 2020-08-13T12:16:32 | {
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http://math.stackexchange.com/questions/55460/calculate-the-probability-of-a-simple-event | # Calculate the probability of a simple event
I'm beginning to study probability and an exercise in the study guide that asks me to calculate: What is the probability that the month January, of one year randomly selected have only four Sundays?
the solution of the book indicates that it's 4 / 7 which is equal to 0.5714 probability that the event occurs, according to what I learned in class the probability of this event can be calculated by counting probability as it is possible to have all elements of the sample space, and all the elements that belong to threw the probability of which is to be calculated.
P (A) = number of cases occurring A / number of cases in the sample space
My question is why the sample space of this experiment is 7 and what is the maximum number of Sundays in January that may have (I think by intuition that can be 5)
This formula applies when it is possible to have all elements of the sample space, and all items pertaining to the event whose probability is to be calculated
-
The idea is that January can start on any day of the week. If it starts Thursday through Sunday, there are 5 Sundays, while if it starts Monday through Wednesday, there are only 4. So if starting days are evenly distributed, the probability would be 4/7. The maximum number doesn't figure in. The sample space is the starting day of the week.
This answer is close, but not right. In the Gregorian calendar the days repeat in a 400 year cycle as 400*365+97=146097 is divisible by 7. If you look carefully, the start days of January are not (quite) evenly distributed.
-
Indeed, I remember seeing somewhere that the 13th is (very slightly) more likely to be a Friday than any other day. – Gerry Myerson Aug 4 '11 at 0:48
@Gerry: Yes. See the tables here. – Brian M. Scott Aug 4 '11 at 1:14
@Brian, nice. Thanks. – Gerry Myerson Aug 4 '11 at 1:38
The year can begin on any one of the seven days of the week, and each of the seven is equally likely. (Actually, this isn’t quite true, but you have to look very closely at the Gregorian calendar to see that it isn’t, and the deviation from equal likelihood is very small indeed.) Thus, the probability that 1 January falls on a Sunday is $1/7$, the probability that it falls on a Monday is $1/7$, and so on. Suppose that 1 January falls on a Sunday. Then the Sundays in January fall on 1 January, 8 January, 15 January, 22 January, and 29 January, so there are five of them. What happens if 1 January is a Saturday? Then the Sundays fall on 2, 9, 16, 23, and 30 January, and there are again five of them. Similarly, if 1 January falls on a Friday, the Sundays fall on 3, 10, 17, 24, and 31 January, and once more there are five of them. I’ll leave you to check for yourself, using the same kind of analysis, that if 1 January falls on a Thursday, Wednesday, Tuesday, or Monday, there are just four Sundays in the month. Thus, in four of the seven equally likely cases January has four Sundays, and the probability of its having four Sundays is therefore $4/7$. (And as you can see, the maximum possible number of Sundays in January is indeed five.)
Added: If I’ve not miscounted, in the course of the $400$-year Gregorian cycle the first of the year falls $56$ times each on Monday and Saturday, $57$ times each on Wednesday and Thursday, and $58$ times each on Sunday, Tuesday, and Friday. The exact probability of getting just four Sundays in January is therefore $\frac{56+58+57+57}{400} = \frac{228}{400} = 0.57$, slightly less than $4/7$.
- | 2015-12-02T02:17:12 | {
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https://math.stackexchange.com/questions/1005271/rain-weather-forecast-not-carrying-umbrella | # Rain, weather forecast, not carrying umbrella
I was solving this problem -
In a city, half of the days have some rain. The weather forecaster is correct 2/3 of the time, i.e., the probability that it rains, given that the forecaster has predicted rain, and the probability that it does not rain, given that she has predicted that it won't rain, are both equal to 2/3. When rain is forecast, Mr.X takes his umbrella. When rain is not forecast, he takes it with probability 1/3. Find (a) the probability that Mr. X has no umbrella, given that it rains. (b) the probability that he brings his umbrella, given that it doesn't rain.
So Far what I have done - 1. Probability Model -
A - {It rains}
B - {Forcaster says it rains}
C - {Mr. X carries umbrella}
Now Given - $$P(A) = 1/2; P(A/B) = 2/3; P(A'/B') = 2/3; P(C/B) = 1; P(C'/B) = 1/3$$ Now P(A'/B') = $(1-P(A \cup B))/P(B') = 2/3$ $$\Rightarrow P(A\cup B) = 1/3+2/3P(B)$$
Also from P(A/B) I get $P(A\cap B) = 2/3P(B)$
So from here I calculated $P(B) = 1/2$ and using this and P(C/B), P(C'/B) I can find P(C) as $$P(C) = 2/3$$ I will apply the sequential tree as - Event1 - Weather forecast Event2 - Mr. X carries umbrella (or not depending upon forecast) Event3 - It rains (or not depending upon forecast)
Hence from this the Probabilities for 1st - 1/9 and 2nd should be 5/18 But I am bit uncertain here. can someone please correct me if I have made any mistake.
Thanks
• Any comments??.. – codeomnitrix Nov 4 '14 at 8:57
(a) Perhaps you calculated $P(C^{'} \cap A)$ instead of $P(C^{'} \mid A)$.
Using the probability tree (event $C^{'} \cap A$ has branch $6$ only):
$$P(C^{'} \cap A) = \frac{1}{2}\cdot\frac{2}{3}\cdot\frac{1}{3} = \frac{1}{9}.$$
$$\therefore P(C^{'} \mid A) = \dfrac{P(C^{'} \cap A)}{P(A)} = \frac{1/9}{1/2} = \frac{2}{9}.$$
(b) Similarly with this one. Using the probability tree (event $C \cap A^{'}$ has branches $2$ and $3$):
$$P(C \cap A^{'}) = \frac{1}{2}\cdot 1 \cdot\frac{1}{3} \;+\; \frac{1}{2}\cdot\frac{1}{3}\cdot\frac{2}{3} = \frac{5}{18}.$$
$$\therefore P(C \mid A^{'}) = \dfrac{P(C \cap A^{'})}{P(A^{'})} = \frac{5/18}{1/2} = \frac{5}{9}.$$
• yeah I realized it, forget to update the answer....thanks Mick\ – codeomnitrix Nov 5 '14 at 15:11 | 2019-06-26T10:21:56 | {
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https://www.coursehero.com/file/p3q5ouck/Click-on-the-graph-in-the-upper-right-and-drag-the-cursor-to-move-the-shaded/ | Click on the graph in the upper right and drag the cursor to move the shaded
Click on the graph in the upper right and drag the
• 96
This preview shows page 60 - 66 out of 96 pages.
Click on the graph in the upper right and drag the cursor to move the shaded area on the graph. Move the cursor until the shaded region represents the probability of interest (the “Between (Red-Shaded)” number shown in the Probabilities column on the right side will show the probability). The value on the horizontal axis that marks the right side of the shaded area will be the percentile of interest. 49/77
Finding Percentiles Using The Normal Table Note: Confusingly, some people refer to the q from the q th percentile as the percentile, rather than the value that corresponds to that probability. Finding percentiles from a Normal table To find the q th percentile of a N ( μ, σ ) distribution: 1 Use the table for the N (0 , 1) distribution to find the z -score of the q th percentile. 2 Convert the z -score to proper units. Since the z -score is defined as z = value - mean standard deviation = x - μ σ , then solving for x means the q th percentile is q th percentile = x = μ + σz. 50/77
Example: Children’s Heights III Example Suppose children’s heights follow a N (68 , 2) distribution. How tall does a child need to be in order to be in the 90th percentile? z .05 .06 .07 .08 .09 0.8 0.8023 0.8051 0.8078 0.8106 0.8133 0.9 0.8289 0.8315 0.8340 0.8365 0.8389 1 0.8531 0.8554 0.8577 0.8599 0.8621 1.1 0.8749 0.8770 0.8790 0.8810 0.8830 1.2 0.8944 0.8962 0.8980 0.8997 0.9015 1.3 0.9115 0.9131 0.9147 0.9162 0.9177 1.4 0.9265 0.9279 0.9292 0.9306 0.9319 51/77
Example: Children’s Heights III Example Suppose children’s heights follow a N (68 , 2) distribution. How tall does a child need to be in order to be in the 90th percentile? z .05 .06 .07 .08 .09 0.8 0.8023 0.8051 0.8078 0.8106 0.8133 0.9 0.8289 0.8315 0.8340 0.8365 0.8389 1 0.8531 0.8554 0.8577 0.8599 0.8621 1.1 0.8749 0.8770 0.8790 0.8810 0.8830 1.2 0.8944 0.8962 0.8980 0.8997 0.9015 1.3 0.9115 0.9131 0.9147 0.9162 0.9177 1.4 0.9265 0.9279 0.9292 0.9306 0.9319 To find the 90th percentile, we want to find the z -score that corresponds to a probability of 0.90. From the Normal table, the probability of 0.8997 is the closest to 0.90. The probability of 0.8997 corresponds to a z -score of 1.28. 51/77
Example: Children’s Heights III Example (Continued) Suppose children’s heights follow a N (68 , 2) distribution. How tall does a child need to be in order to be in the 90th percentile? To find the 90th percentile, we want to find the z -score that corresponds to a probability of 0.90. From the Normal table, the probability of 0.8997 is the closest to 0.90. The probability of 0.8997 corresponds to a z -score of 1.28. Converting the z -score of 1.28 into inches, we have z = 1 . 28 = x - μ σ = x - 68 2 , so x = μ + σz = 68 + (2 × 1 . 28) = 70 . 56 . A child needs to be 70.56 inches tall to be in the 90th percentile of heights. 52/77
The Normal Model and the Empirical Rule Recall from Chapter 3: The Empirical Rule If the distribution is symmetric and unimodal, then Approximately 68% of the observations (roughly two-thirds) will be within one standard deviation of the mean. Approximately 95% of the observations will be within two standard deviations of the mean. Nearly all the observations will be within three standard deviations of the mean. The Empirical Rule applies for any arbitrary symmetric and unimodal distribution as an approximate guideline. However, for the Normal model, the Empirical Rule is (nearly) exact. | 2021-07-24T21:27:40 | {
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http://math.stackexchange.com/questions/28848/least-common-multiple-of-a-and-b-greater-than-c | # “Least” common multiple of A and B, greater than C
Least common multiple of $A$ and $B$ that is greater than $C$.
Is there any algorithm to find it without prime factorization?
-
Please make the bodies of your posts self-contained; don't rely on the subject for key content. – Arturo Magidin Mar 24 '11 at 19:28
Thanks, Arturo. – medina Mar 24 '11 at 20:03
## 3 Answers
Any common multiple of $A$ and $B$ is a multiple of the least common multiple, $\mathrm{lcm}(A,B)$.
The least common multiple can be found using the identity $$\mathrm{lcm}(A,B)\gcd(A,B) = |AB|.$$ The $\gcd$ can be found using the Euclidean Algorithm, which does not require you to find the prime factorization. So you can find the least common multiple without knowing the prime factorization.
Once you know the lcm, call it $K$, then you are merely looking for the smallest integer $n$ such that $nK\gt C$. This is $n = \lceil \frac{C}{K}\rceil$ if $K\neq 0$. (And of course, if $K=0$, then that means that at least one of $A$ and $B$ is equal to $0$, and there is no solution).
In summary:
1. Find the greatest common divisor (e.g., using the Euclidean algorithm);
2. Compute $M=|AB|/\gcd(A,B)$.
3. Find the smallest integer $k$ greater than or equal to $C/M$.
4. The multiple you want is $kM$.
-
If $C$ is itself a multiple of $K$, then you'll need to take $n+1 = \frac{C}{K}+1$ instead. – Arturo Magidin Mar 25 '11 at 21:08
Sure, first find the least common multiple of A and B, which you can find by calculating:
$$\frac{|AB|}{\operatorname{gcd}(A,B)}$$
Then divide $C$ by the LCM. If it divides evenly, then C is your answer, otherwise, the next highest multiple. So the result is: $$L = AB/gcd(A,B)$$ $$C' ={ \left \lceil \frac{C}{L} \right \rceil } L$$ where $\left \lceil x \right \rceil$ is the smallest integer greater than or equal to $x$.
-
HINT $\:$ Any common multiple $\rm\:M\:$ is a multiple of the least common multiple $\rm\ m = AB/gcd(A,B)\:.\:$ So you seek the first multiple of $\rm\:m\:$ that is larger than $\rm\:C\:,\:$ computable simply by an obvious division.
Notice how the use use of $\rm\:lcm\:$ has allowed us to convert a problem from one with multiple divisibility relations to one involving a single divisibility relation $\rm\ A,\:B\ |\ M\ \iff\ lcm(A,B)\ |\ M\:.\:$ This, the universal definition of $\rm\:lcm\:,\:$ is what allows us to reduce complex divisibility problems to single divisions - as occurs here, after we adjoin the additional constraint that $\rm\: M > C\:,\:$ namely
$$\rm\ A\:,B\ |\ M > C\ \ \iff\ \ lcm(A,B)\ |\ M > C$$
For a universal proof of $\rm\ gcd(a,b)\ lcm(a,b)\ =\ a\ b\$ see my post here or here.
- | 2015-08-30T10:03:12 | {
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https://math.stackexchange.com/questions/2449473/given-this-matrix-a-find-a144 | # Given this matrix $A$, find $A^{144}$
I know this is a very common question to ask when one is making their way into Linear Algebra (i.e. given a matrix, find the result of that matrix to the nth-power).
I'm given this matrix: $$A= \left[ {\begin{array}{cc} 0 & -1 \\ 1 & 0 \\ \end{array} } \right]$$
I'm asked to compute, by hand, the matrix $A^{144}$
What I tried:
I calculated $A^2$, $A^3$, $A^4$ and $A^5$ and tried to find a pattern, so that I could first find the more general $A^n$ expression, and then make $n=144$. This is what I got:
$$A^2= \left[ {\begin{array}{cc} -1 & 0 \\ 0 & -1 \\ \end{array} } \right]$$ $$A^3= \left[ {\begin{array}{cc} 0 & 1 \\ -1 & 0 \\ \end{array} } \right]$$ $$A^4= \left[ {\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} } \right]$$ $$A^5= \left[ {\begin{array}{cc} 0 & -1 \\ 1 & 0 \\ \end{array} } \right]$$
So there seems to be kind of a "circular pattern", where $A^6=A^2$ and therefore, since $144/6=24$, I'm suspecting that: $$¿\,\boxed{A^{144}=A^2} \,?$$
However, I was hoping to find the more general $A^n$ matrix first and confirm the above result.
But, I can't seem to find a function such that: $$f(n) = \left\{ \begin{array}{ll} 0 & \mbox{if n is odd} \\ 1 & \mbox{if n is even} \end{array} \right.$$ This function would allow me to sort out the (1,1) and (2,2) elements of $A^n$ being 0 when n is odd (and the (1,2) and (2,1) being 0 when n is even).
I'm a bit confused at this point, your help is greatly appreciated.
• Try $n+1\pmod{2}$. – Steve D Sep 28 '17 at 20:05
• That matrix turns out to be the representation of $i$ (i.e. it represents counterclockwise multiplication by $\pi/2$), so that $A^4=I$. You can work this out by complex-number diagonalization if you want. – Ian Sep 28 '17 at 20:06
• Note that $A^4 = I$ – Arthur Sep 28 '17 at 20:06
• @Ian What do you mean by "the representation of i"? I'm a bit of a newcomer to this world. I appreciate your help. – Jose Lopez Garcia Sep 28 '17 at 20:08
• That’s right. Use Ian’s hint to figure out which ones to use. De Moivre’s formula might be helpful. – amd Sep 28 '17 at 20:41
You have a weird problem, you say you can't find a function such that it equals $0$ on odd integers and $1$ on even ones, yet saying this is defining such a function.
Otherwise, you saw that $A^4 = I$ (the identity matrix), so just make a division :
$144 = 4\times36$ so that $A^{144} = (A^4)^{36} = I^{36} = I$
A nice interpretation in $\mathbb{R}^2$ is that your matrix represents a rotation by $\frac{\pi}{2}$ around the origin in the canonical basis. Then, rotating $144$ times results in not rotating at all...
As you wrote in you post: $A^4 = I$. Therefore: $A^{4k} = I$ for all $k \in \mathbb{N}$.
Because $144 = 4 \times 36$, you have :
$$A^{144} = I.$$
From the results in your post, you can deduce that:
$$\forall k \in \mathbb{N}, \; A^{4k} = I, \; A^{4k+1} = A, \; A^{4k+2} = - I \; \text{and} \; A^{4k+3} = -A.$$
Your analysis shows that $A^n = A^{n \text{ mod } 4}$. Since you know $A^0,A^1, A^2, A^3$ you then have a general formula. Note that $144 \text{ mod } 4 = 0$.
Note that $A^4 = I$, and $A^{144} = (A^4)^{36}$, and you're done. If you really want to go with that $A^6 = A^2$ instead, then we have that $$A^{144} = (A^6)^{24} \\ = (A^2)^{24} = A^{48} \\ = (A^6)^8 = (A^2)^8 \\ = A^{16}= A^4\times A^{12}\\ = A^4 \times(A^6)^2 = A^4\times(A^2)^2\\ = A^8 = A^2\times A^6\\ = A^2\times A^2 = A^4$$ This seems a much more roundabout way to get the same result.
Even when there isn't a function, you can just define and use it as you just did. But regardless, your answer can be found in modular arithmetic.
$f(n) \equiv n+1 \pmod{2}$
Here, this represents the remainder when divided by 2. If you need further help, let me know.
• That's technically not what $\equiv$ means. (I know, it's an annoying discrepancy between how mathematicians talk about modular arithmetic and how programmers use it, but still, that's not what it means.) – Ian Sep 28 '17 at 20:12
• @Ian It was the notation I was taught for modular arithmetic. Though I've also seen it used for function identities. – Mad Maniac Sep 28 '17 at 20:13
• @Ian Okay, I see your point. Assuming you mean it outputs a class of numbers as opposed to a single nunber. – Mad Maniac Sep 28 '17 at 20:19
• In math notation it doesn't "output" anything. It's just a relation. – Ian Sep 28 '17 at 20:29 | 2020-01-18T04:08:56 | {
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https://stats.stackexchange.com/questions/356023/expectation-of-truncated-normal/356106 | # Expectation of truncated normal
Suppose I have a bivariate normal $(x,y)\sim \mathcal N(0,\Sigma)$ .
Is there an easy formula for the expectation $\mathrm E(x\mid y>0)$ ?
Let $$\phi(\cdot)$$ and $$\Phi(\cdot)$$ be the PDF and CDF of standard normal distribution, as usual.
Suppose $$\Sigma=\left[\begin{matrix}\sigma_1^2&\rho \sigma_1\sigma_2\\\rho\sigma_1\sigma_2&\sigma_2^2\end{matrix}\right]$$ is the dispersion matrix of $$(X,Y)$$.
By definition,
\begin{align} E\left[X\mid a
This of course is same as saying $$E\left[X\mid a
Now $$X$$ conditioned on $$Y=y$$ has a $$N\left(\frac{\rho\,\sigma_1}{\sigma_2}y,(1-\rho^2)\sigma_1^2\right)$$ distribution, so that
$$E\left[X\mid Y=y\right]=\frac{\rho\,\sigma_1}{\sigma_2}y$$
In this case, we thus have
\begin{align} E\left[X\mid Y>0\right]&=\frac{\rho\,\sigma_1}{\sigma_2}\int_0^\infty \frac{yf_Y(y)}{P(Y>0)}\,\mathrm{d}y \\&=\frac{\rho\,\sigma_1}{\sigma_2}E\left[Y\mid Y>0\right] \end{align}
Finally,
\begin{align} E\left[Y\mid Y>0\right]&=\frac{1}{P(Y>0)}\int_0^\infty \frac{y}{\sigma_2}\phi\left(\frac{y}{\sigma_2}\right)\mathrm{d}y \\&=2\sigma_2\int_0^\infty t\phi(t)\,\mathrm{d}t \\&=2\sigma_2\int_0^\infty (-\phi'(t))\,\mathrm{d}t \\&=2\sigma_2\,\phi(0) \\&=\sqrt{\frac{2}{\pi}}\sigma_2 \end{align}
Hence for $$(X,Y)\sim N(\mathbf 0,\Sigma)$$, we have the simple formula
$$\boxed{E\left[X\mid Y>0\right]=\sqrt{\frac{2}{\pi}}\rho\,\sigma_1}$$
A general formula where the means of $$X$$ and $$Y$$ are not zero, and $$Y$$ ranging from some $$a$$ to $$b$$ can also be found in a similar manner. That formula would involve $$\phi$$ and $$\Phi$$, whereas for the current one, we get the values at $$\phi$$ and $$\Phi$$ directly.
• You can write the answer in terms of the components of $\Sigma$ alone: it's simple. – whuber Jul 13 '18 at 23:29
• @whuber For the given question, yes we get the answer in terms of the elements of $\Sigma$ only. What I was referring to before the edit was the general formula, which would naturally involve the pdf and cdf of a standard normal variable. – StubbornAtom Jul 14 '18 at 7:05
• thanks for completing the answer, I posted mine below before seeing yours – andy Jul 14 '18 at 10:51
Let's generalize the question so that a key idea can be revealed.
Let $Y$ and $R$ be independent random variables. Define $$X=f(Y)+R$$ for a specified (measurable) function $f$. For any $a\lt b,$ what is the conditional expectation $$E[X\mid a\le Y \le b]?$$
(This states that $f$ is the regression of $X$ on $Y$ with i.i.d. additive errors.)
The independence assumption makes this an easy question to answer, because the random variables $f(Y)$ and $R$ will still be independent, whence (by "taking out what is known")
$$E[X\mid a\le Y \le b] = E[f(Y)+R\mid a\le Y \le b] = E[f(Y)\mid a \le Y \le b] + E[R].\tag{*}$$
This is very general. Let's specialize to the case $E[R]=0$ and $f$ is a linear transformation $$f(y) = \alpha\, y$$ for some constant number $\alpha.$ In this case $(*)$ simplifies to
$$E[X\mid a\le Y \le b] = \alpha\, E[Y\mid a \le Y \le b].\tag{**}$$
The right hand side has a direct expression when the distribution of $Y,$ $F_Y,$ has a density $f_y,$ for then (by the definitions of conditional probability and expectation)
$$E[Y\mid a \le Y \le b] = \frac{1}{F(b)-F(a)}\int_a^b y f_y(y) \mathrm{d}y.$$
The result is now immediate and obvious to anyone who has studied bivariate regression. (A good non-mathematical reference with all the necessary details is Freedman, Pisani, and Purves, Statistics [any edition].) The next section of this post is a review for those who might not have encountered this theory.
In the case of the question, where $(x,y)$ is bivariate Normal with mean $(0,0),$ standard deviations $\sigma_x$ and $\sigma_y,$ and correlation $\rho,$ we know that the distribution of $(x,y)$ is the same as the distribution of $(X,Y)$ constructed as above where $Y$ has a Normal$(0,\sigma_y^2)$ distribution and $R$ independently has a Normal$(0,\tau^2)$ distribution (with $\tau$ to be found). The proof of this lies in the observations that
1. Because $Y$ and $R$ are independent, the variance of $X = \rho\,Y+R$ equals $$\rho^2 \operatorname{Var}(Y) +\operatorname{Var}(R) =\rho^2\sigma^2_y + \tau^2.$$ Consequently, if we set $$\tau^2 = \sigma_x^2 - \rho^2\sigma_y^2,$$ the variance of $X$ will equal the variance of $x.$
2. The covariance $(X,Y)$ is $$\operatorname{Cov}(X,Y) = \operatorname{Cov}(\alpha\,Y+R,Y) = \alpha\operatorname{Var}(Y) = \alpha\,\sigma_y^2.$$ If we set $$\alpha = \rho\frac{\sigma_x}{\sigma_y}$$ then the correlation of $X$ and $Y$ will be $$\rho(X,Y) = \frac{\operatorname{Cov}(X,Y)}{\sigma_x\sigma_y} = \frac{\alpha\,\sigma_y^2}{\sigma_x\sigma_y} = \frac{\rho\frac{\sigma_x}{\sigma_y} \sigma_y^2}{\sigma_x\sigma_y} = \rho.$$
Observations (1) and (2) establish that $(X,Y)$ and $(x,y)$ have the same moments through second order, which means they have identical bivariate Normal distributions.
Applying the main result $(**)$ gives
$$E[X\mid a\le Y \le b] = \alpha\, E[Y\mid a \le Y \le b] = \rho\frac{\sigma_x}{\sigma_y}E[Y\mid a \le Y \le b].$$
We can go a little further towards simplification by recognizing that $Y/\sigma_y$ is a standard Normal variate. Thus,
$$E[X\mid a\le Y \le b] = \rho\,\sigma_x E\left[\frac{Y}{\sigma_y}\mid \frac{a}{\sigma_y} \le \frac{Y}{\sigma_y} \le \frac{b}{\sigma_y} \right].$$
This shows that for fixed $a,b,$ the answer is directly proportional to $\rho\, \sigma_x.$ Evaluating the constant is a matter of manipulating integrals of a standard Normal variable, but doing so is primarily an exercise in Calculus rather than of statistical interest.
I will conclude with the expression that I actually began with when solving this problem. Notice that the answer can also be written $$E\left[\frac{X}{\sigma_x}\mid a\le Y \le b\right] = \rho\, E\left[\frac{Y}{\sigma_y}\mid \frac{a}{\sigma_y} \le \frac{Y}{\sigma_y} \le \frac{b}{\sigma_y} \right].$$
The left hand side is the conditional expectation of a standard Normal variate $X/\sigma_x$ while the right hand side is just $\rho$ times the expectation of a truncated standard Normal variate $Y/\sigma_y.$
If you get used to standardizing variables automatically--which amounts to choosing a particularly convenient unit of measurement for them--then simply by visualizing the regression of $X$ against $Y$ you will say to yourself
oh yes, since after standardization $X$ is just a multiple $\rho$ of $Y$, then the conditional expectation of $X$ must be $\rho$ times the corresponding expectation of $Y.$
That is the heart of the matter.
• This provides good intuition rather than merely calculating the conditional expectation directly as I would have done. – StubbornAtom Jul 14 '18 at 21:33
I am posting the complete answer for reference: $$\begin{eqnarray*} E(X|a where $$\rho_{xy} = E(xy) / (\sigma_x\sigma_y) =\sigma_{xy} / (\sigma_x \sigma_y)$$.
The second-to-last step is derived from the properties of the univariate truncated normal distribution, easily found on its wikipedia page. A reference for the first step is here
With $$a=0,b=\infty$$ we have $$E(X,Y>0) = \rho_{xy}\sigma_x\frac{\phi(0)}{\Phi(0)}= \rho_{xy}\sigma_x\cdot \frac{2}{\sqrt{2\pi}}= \frac{\sigma_{xy}}{\sigma_y}\cdot \frac{2}{\sqrt{2\pi}}$$ Thanks for pointing in the right direction.
• There must be a typographical error because the $\sigma_y/\sigma_y$ term cancels, leaving the answer independent of $\sigma_y$ altogether. – whuber Jul 14 '18 at 14:04
• But $\rho_{xy}$ contains $\sigma_y$ – andy Jul 14 '18 at 15:36
• Don't you have to standardize $a$ and $b$? – Waldir Leoncio Jan 3 '19 at 11:37 | 2021-03-02T01:57:56 | {
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https://www.jobilize.com/course/section/activity-venn-diagrams-probability-part-1-by-openstax?qcr=www.quizover.com | # Probability: part 1
Page 4 / 7
We use $n\left(S\right)$ to refer to the number of elements in a set $S$ , $n\left(X\right)$ for the number of elements in $X$ , etc.
In a box there are pieces of paper with the numbers from 1 to 9 written on them. A piece of paper is drawn from the box andthe number on it is noted. Let $S$ denote the sample space, let $P$ denote the event 'drawing a prime number', and let $E$ denote the event 'drawing an even number'. Using appropriate notation, in how many ways is it possible to draw: i)any number? ii) a prime number? iii) an even number? iv) a number that is either prime or even? v) a number that is both prime and even?
• Drawing a prime number: $P=\left\{2;3;5;7\right\}$
• Drawing an even number: $E=\left\{2;4;6;8\right\}$
1. The union of $P$ and $E$ is the set of all elements in $P$ or in $E$ (or in both). $P\cup E=2,3,4,5,6,7,8$ .
2. The intersection of $P$ and $E$ is the set of all elements in both $P$ and $E$ . $P\cap E=2$ .
3. $\begin{array}{ccc}\hfill \therefore \phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}n\left(S\right)& =& 9\hfill \\ \hfill n\left(P\right)& =& 4\hfill \\ \hfill n\left(E\right)& =& 4\hfill \\ \hfill n\left(P\cup E\right)& =& 7\hfill \\ \hfill n\left(P\cap E\right)& =& 2\hfill \end{array}$
100 people were surveyed to find out which fast food chain (Nandos, Debonairs or Steers) they preferred. The following results were obtained:
• 50 liked Nandos
• 66 liked Debonairs
• 40 liked Steers
• 27 liked Nandos and Debonairs but not Steers
• 13 liked Debonairs and Steers but not Nandos
• 4 liked all three
• 94 liked at least one
1. How many people did not like any of the fast food chains?
2. How many people liked Nandos and Steers, but not Debonairs?
1. The number of people who liked Nandos and Debonairs is 27, so this is the intersection of these two events. The number of people who liked Debonairs and Steers is 13, so the intersection of Debonairs and Steers is 13. We are told that 4 people like all three options, and so this means that there are 4 people in the intersection of all three options. So we can work out that the number of people who like just Debonairs is $66–4–27-13=22$ (This is simply the total number who like Debonairs minus the number of people who like Debonairs and Steers, or Debonairs and Nandos or all three). We draw the following diagram to represent the data:
2. We are told that there were 100 people and that 94 liked at least one. So the number of people that liked none is: $100–94=6$ . This is the answer to a).
3. We can redraw the part of the Venn diagram that is of interest: Total people who like Nandos: 50
Of these 27 like both Nandos and Debonairs, and 4 people like all three options. So we find that the total number of people who just like Nandos is: $50–27–4=19$
Total people who like Steers: 40
Of these 13 like both Steers and Debonairs, and 4 like all three options. So we find that the total number of people who like just Steers is: $40–13–4=23$
Now use the identity $\mathrm{n\left(N or S\right)}=\mathrm{n\left(N\right)}+\mathrm{n\left(S\right)}–\mathrm{n\left(N and S\right)}$ to find the number of people who like Nandos and Steers, but not Debonairs.
$\begin{array}{ccc}\hfill \mathrm{n\left(N or S\right)}& =& \mathrm{n\left(N\right)}+\mathrm{n\left(S\right)}–\mathrm{n\left(N and S\right)}\hfill \\ \hfill 28& =& 23+19–\mathrm{n\left(N and S\right)}\hfill \\ \hfill \mathrm{n\left(N and S\right)}& =& 14\hfill \end{array}$
The Venn diagram with that represents all this information is given:
## Activity: venn diagrams
Which cellphone networks have you used or are you signed up for (e.g. Vodacom, Mtn or CellC)? Collect this information from your classmates as well. Then use the information to draw a Venn diagram (if you have more than three networks, then choose only the three most popular or draw Venn diagrams for all the combinations). Try to see if you can work out the number of people who use just one network, or the number of people who use all the networks.
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
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https://math.stackexchange.com/questions/1077958/how-to-prove-that-the-sum-of-squared-binomials-equals-binom2nn | # How to prove that the sum of squared binomials equals $\binom{2n}{n}$ [duplicate]
I've stumbled upon this lemma a few times in my textbook: $$\sum_{k=0}^{n}\begin{pmatrix}n\\k\end{pmatrix}^2=\begin{pmatrix}2n\\n\end{pmatrix}$$ I've been trying to prove it, but I simply can't seem to see a connection. I've been trying to use proof by induction, but I can't express the statement for $n+1$ via the statement for $n$.
How do I prove it?
## marked as duplicate by user147263, Newb, Chris Janjigian, Thomas, Jonas MeyerDec 27 '14 at 20:10
• You mean $\sum_{k=0}^n\binom{n}{k}^{\color{Red}{2}}=\binom{2n}{n}$? – anon Dec 22 '14 at 19:47
• Please consider changing/editing the title. – Ludolila Dec 22 '14 at 19:48
• – Grigory M Dec 22 '14 at 19:50
• I don't understand the reopen posts. – anon Dec 27 '14 at 19:28
• – Martin Sleziak Dec 16 '15 at 13:39
$${n \choose k} = {n \choose n-k}$$ Then we can divide $2n$ objects into two groups. Form one group we choose $k$ from the other $n-k$ and we assuming all possibilities for $k$. Hence we get: $${2n \choose n}=\sum_{k=0}^n{n \choose k} {n \choose n-k} = \sum_{k=0}^n{n \choose k}^2$$
Since $\dbinom n k= \dbinom n {n-k}$, the identity $$\sum_{k=0}^n \binom n k ^2 = \binom {2n} n$$ is the same as $$\sum_{k=0}^n \binom n k \binom n {n-k} = \binom {2n} n.$$ So say a committee consists of $n$ Democrats and $n$ Republicans, and one will choose a subcommittee of $n$ members. One may choose $k$ Democrats and $n-k$ Republicans in $\dbinom n k \cdot \dbinom n {n-k}$ ways. The number of Democrats is in the set $\{0,1,2,\ldots,n\}$, thus ranging from all Republicans to all Democrats. The sum then gives the total number of ways to choose $n$ out of $2n$.
Consider the coefficient of $x^n$ in $(1+x)^{2n}$. We have $$(1+x)^{2n}=\sum_{k=0}^{2n}\binom{2n}{k}x^k=(1+x)^n(1+x)^n= \Big(\sum_{k=0}^{n}\binom{n}{k}x^k\Big)\Big(\sum_{k=0}^{n}\binom{n}{k}x^{n-k}\Big).$$ The coefficient of $x^n$ is equal to $\binom{2n}{n}$ in the left hand side.
The coefficient of $x^n$ is equal to $\sum_{k=0}^n\binom{n}{k}^{\!2}$ in the right hand side.
It seems the following.
$$\sum_{k=0}^{n}\begin{pmatrix}n\\k\end{pmatrix}=(1+1)^n=2^n.$$ $$\sum_{k=0}^{n}\begin{pmatrix}n\\k\end{pmatrix}^2=\begin{pmatrix}2n\\n\end{pmatrix}$$
To prove the last inequality it suffice to calcuate the coefficient by $x^n$ at both sides of the equality
$$(1+x)^n(1+x)^n=(1+x)^{2n}.$$ | 2019-10-16T01:46:40 | {
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https://math.stackexchange.com/questions/2370486/integral-over-exponential-involving-reciprocial/2370513 | # Integral over exponential involving reciprocial
I want to show $$I := \int_{-\infty}^\infty \exp \left(-\left(x-\frac p x \right)^2\right) \, dx = \sqrt{\pi}$$ for any non-negative $p\geq 0$. I tried to prove $I^2=\pi$ using Fubini's theorem, but had no success.
The key is to essentially make the substitution $u = x - p/x$. The problem is that this isn't an invertible function--it has a positive and a negative root for $x$. To fix, this we divide the integral into a positive and negative side and make the substitutions \begin{eqnarray} x(u) &=& \frac{1}{2}\left(u \pm \sqrt{4p+u^2}\right) \\ dx &=& \frac{1}{2}\left(1 \pm \frac{u}{\sqrt{4p+u^2}}\right) \end{eqnarray} This gives \begin{multline} \int_{-\infty}^\infty \exp\left[-\left(x-\frac{p}{x}\right)^2\right]\,dx =\int_{-\infty}^0 \exp\left[-\left(x-\frac{p}{x}\right)^2\right] \, dx+\int_0^\infty \exp\left[-\left(x-\frac{p}{x}\right)^2\right]dx \\= \int_{-\infty}^\infty \frac{e^{-u^2}}{2}\left(1 - \frac{u}{\sqrt{4p+u^2}}\right) \, du + \int_{-\infty}^\infty \frac{e^{-u^2}}{2}\left(1 + \frac{u}{\sqrt{4p+u^2}}\right) \, du \\= \int_{-\infty}^\infty e^{-u^2}du = \sqrt{\pi} \end{multline}
Using the substitution $$x=\frac{u+\sqrt{u^2+4p}}2\implies\mathrm{d}x=\frac12\left(1+\frac{u}{\sqrt{u^2+4p}}\right)\mathrm{d}u$$ we have $u=x-\frac px$ and, as $u$ varies from $-\infty$ to $+\infty$, $x$ varies from $0$ to $\infty$.
Since the integrand is even, \begin{align} \int_{-\infty}^\infty e^{-\left(x-\frac px\right)^2}\,\mathrm{d}x &=2\int_0^\infty e^{-\left(x-\frac px\right)^2}\,\mathrm{d}x\\ &=2\int_{-\infty}^\infty e^{-u^2}\frac12\left(1+\frac{u}{\sqrt{u^2+4p}}\right)\mathrm{d}u\\ &=\int_{-\infty}^\infty e^{-u^2}\,\mathrm{d}u\\[6pt] &=\sqrt\pi \end{align}
$\ds{I \equiv \int_{-\infty}^{\infty} \exp\pars{-\bracks{x - {p \over x}} ^{2}}\,\dd x = \root{\pi}:\ {\large }}$
\begin{align} I &\equiv \int_{-\infty}^{\infty}\exp\pars{-\bracks{x - {p \over x}} ^{2}}\,\dd x = 2\int_{0}^{\infty} \exp\pars{-p\bracks{{x \over \root{p}} - {\root{p} \over x}} ^{2}}\,\dd x \\[5mm] & = 2\root{p}\int_{0}^{\infty} \exp\pars{-p\bracks{x - {1 \over x}} ^{2}}\,\dd x \\[5mm] & = \root{p}\bracks{% \int_{0}^{\infty}\exp\pars{-p\bracks{x - {1 \over x}} ^{2}}\,\dd x + \int_{0}^{\infty}\exp\pars{-p\bracks{x - {1 \over x}} ^{2}}\,\dd x} \\[5mm] & = \root{p}\bracks{% \int_{0}^{\infty}\exp\pars{-p\bracks{x - {1 \over x}} ^{2}}\,\dd x + \int_{\infty}^{0}\exp\pars{-p\bracks{{1 \over x} - x} ^{2}} \,\pars{-\,{1 \over x^{2}}}\dd x} \\[5mm] & = \root{p} \int_{0}^{\infty}\exp\pars{-p\bracks{x - {1 \over x}} ^{2}} \pars{1 + {1 \over x^{2}}}\,\dd x \,\,\,\stackrel{x - 1/x\ \mapsto\ x}{=}\,\,\, \int_{-\infty}^{\infty}\expo{-px^{2}}\root{p}\,\dd x \\[5mm] & = \int_{-\infty}^{\infty}\expo{-x^{2}}\,\dd x = \bbx{\root{\pi}} \end{align}
I thought it might be instructive to present another approach from the ones already posted. To that end, we now proceed.
There is nothing particularly special about the function $e^{-\left(x-\frac px\right)^2}$ in the development.
In fact, using the substitutions $x=-\sqrt{p}e^{-t}$ for $x\in (-\infty, 0]$ and $x=\sqrt{p}e^{t}$ for $x\in [0,\infty)$ in the integral $\int_{-\infty}^\infty f\left(x-\frac px\right)\,dx$ yields
\begin{align} \int_{-\infty}^\infty f\left(x-\frac px\right)\,dx&=\int_{-\infty}^0 f\left(x-\frac px\right)\,dx+\int_{0}^\infty f\left(x-\frac px\right)\,dx\\\\ &=\sqrt{p}\int_{-\infty}^\infty f\left(2\sqrt{p}\sinh(t)\right)\,e^{-t}\,dt+\sqrt{p}\int_{-\infty}^\infty f\left(2\sqrt{p}\sinh(t)\right)\,e^{t}\,dt\\\\ &=2\sqrt{p}\int_{-\infty}^\infty f(2\sqrt{p}\sinh(t))\,\cosh(t)\,dt\\\\ &=\int_{-\infty}^\infty f(u)\,du \end{align}
which for $f(u)=e^{-u^2}$ is equal to $\sqrt{\pi}$ as expected!
We could have approached the problem of specific interest with a slight modification of the general development in the highlighted section.
First we let $x=\sqrt{p}\,e^{t}$. Then, we have $x-\frac px =2\sqrt{p}\sinh(t)$ and
\begin{align} \int_{-\infty}^\infty e^{-\left(x-\frac px\right)^2}\,dx&=2\int_{0}^\infty e^{-\left(x-\frac px\right)^2}\,dx\\\\ &=2\sqrt p\int_{-\infty}^\infty e^{-4p\sinh^2(t)}\,e^t\,dt\tag 1 \end{align}
Next, we let $x=\sqrt {p}\,e^{-t}$. Then, we have $x-\frac px=-2\sqrt{p}\sinh(t)$ and
\begin{align} \int_{-\infty}^\infty e^{-\left(x-\frac px\right)^2}\,dx&=2\int_{0}^\infty e^{-\left(x-\frac px\right)^2}\,dx\\\\ &=2\sqrt p\int_{-\infty}^\infty e^{-4p\sinh^2(t)}\,e^{-t}\,dt\tag 2 \end{align}
Adding $(1)$ and $(2)$ and dividing by $2$, we obtain
\begin{align} \int_{-\infty}^\infty e^{-\left(x-\frac px\right)^2}\,dx&=2\sqrt{p}\int_0^\infty e^{-4p\sinh^2(x)}\,\cosh(x)\,dx\tag3 \end{align}
Finally, enforcing the substitution $u=\sinh(t)$ in $(3)$ yields
\begin{align} \int_{-\infty}^\infty e^{-\left(x-\frac px\right)^2}\,dx&=2\sqrt{p}\int_0^\infty e^{-4pu^2}\,\,du\\\\ &=\sqrt{\pi} \end{align}
as expected!
Hint: Substitute $x-\frac{p}{x}$ with $t$ and this looks suspiciously similar to the error function.
For $a=\sqrt p$, tet $x=at$ and then $$x-\frac{p}{x}=at-\frac{p}{at}=\sqrt p(t-\frac1t).$$ So \begin{aligned} I & =\int_{-\infty}^\infty \exp \left(-\left(x-\frac p x \right)^2\right) \, dx=\sqrt p\int_{-\infty}^\infty \exp \left(-p\left(t-\frac1t \right)^2\right) \, dt \\[10pt] & =2\sqrt p\int_0^\infty \exp \left(-p\left(t-\frac1t \right)^2\right) \, dt. \end{aligned} \tag 1 Using $t\to\frac1t$, one has $$I=2\sqrt p\int_{0}^\infty \frac1{t^2}\exp \left(-p\left(t-\frac1t \right)^2\right) \, dt.\tag{2}$$ Then adding (1) and (2) and under $u=t-\frac1t$, one has $$I=\sqrt p\int_{0}^\infty (1+\frac1{t^2})\exp \left(-\left(t-\frac1t \right)^2\right) \, dt=\sqrt p\int_{-\infty}^\infty\exp(pu^2) \, du=\sqrt\pi.$$
• In your last two $\color{#f00}{\texttt{exp}}$ ( last line ) arguments: the first one needs a $\color{#f00}{\large p}$ factor and the second one need a $\color{#f00}{\large -}$ sign. Otherwise, everything is fine. – Felix Marin Jul 25 '17 at 22:32 | 2019-07-20T11:28:34 | {
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https://www.itl.nist.gov/div898/software/dataplot/refman1/auxillar/ccpr.htm | Dataplot Vol 1 Vol 2
# CCPR PLOT
Name:
CCPR PLOT
Type:
Graphics Command
Purpose:
Generate a CCPR (component and component-plus-residual) plot.
Description:
When performing a linear regression with a single independent variable, a scatter plot of the response variable against the independent variable provides a good indication of the nature of the relationship. If there is more than one independent variable, things become more complicated. Although it can still be useful to generate scatter plots of the response variable against each of the independent variables, this does not take into account the effect of the other independent variables in the model.
Partial residual plots attempt to show the relationship between a given independent variable and the response variable given that other independent variables are also in the model.
Partial residual plots are formed as:
Res + $$\hat{\beta}_{i} X_{i}$$ versus Xi
where
Res = residuals from the full model
$$\hat{\beta}_{i}$$ = regression coefficient from the ith independent variable in the full model
Xi = the ith independent variable
Partial residual plots are widely discussed in the regression diagnostics literature (e.g., see the References section below). Although they can often be useful, be aware that they can also fail to indicate the proper relationship. In particular, if Xi is highly correlated with any of the other independent variables, the variance indicated by the partial residual plot can be much less than the actual variance. These issues are discussed in more detail in the references given below.
The CCPR plot is a refinement of the partial residual plot. It generates a partial residual plot but also adds
$$\hat{\beta}_{i} X_{i}$$ versus Xi
This is the "component" part of the plot and is intended to show where the "fitted line" would lie.
Dataplot provides two forms for the CCPR plot. You can generate either a single CCPR plot or you can generate a matrix of CCPR plots (one plot for each independent variable in the model).
For the matrix form of the command, a number of SET FACTOR PLOT options can be used to control the appearance of the plot (not all of the SET FACTOR PLOT options apply). These are discussed in the Notes section below.
Syntax 1:
CCPR PLOT <y> <x1> ... <xk> <xi> <SUBSET/EXCEPT/FOR qualification>
where <y> is the response variable;
<x1> ... <xk> are the independent variables;
<xi> is the independent variable for which the CCPR plot is being generated
(note that <xi> must be one of the variables listed in <x1> ... <xk>;
and where the <SUBSET/EXCEPT/FOR qualification> is optional.
This is the syntax for generating a single CCPR plot.
Syntax 2:
MATRIX CCPR PLOT <y> <x1> ... <xk> <SUBSET/EXCEPT/FOR qualification>
where <y> is the response variable;
<x1> ... <xk> are the independent variables;
and where the <SUBSET/EXCEPT/FOR qualification> is optional.
This syntax is used to generate a matrix of CCPR plots.
Examples:
CCPR PLOT Y X1 X2 X3 X4 X2
MATRIX CCPR PLOT Y X1 X2 X3 X4
CCPR PLOT Y X1 X2 X3 X4 X2 SUBSET TAG > 2
MATRIX CCPR PLOT Y X1 X2 X3 X4 SUBSET TAG > 2
Note:
The following option controls which axis tic marks, tic mark labels, and axis labels are plotted.
SET FACTOR PLOT LABELS <ON/OFF/XON/YON/BOX>
OFF means that all axis labels are suppressed (this can be useful if a large number of variables are being plotted). ON means that both X and Y axis labels are printed. XON only plots the x axis labels and YON only plots the y axis labels.
BOX is a special option that creates an extra column on the left and an extra row on the bottom. The axis label is printed in this box. BOX is typically reserved for the plot types that plot the variable names in the axes labels.
The default is ON (both x and y axis labels are printed).
Note:
The following option controls where the x axis tic marks, tic mark labels, and axis label are printed.
SET FACTOR PLOT X AXIS <BOTTOM/TOP/ALTERNATE>
BOTTOM specifies that the x axis labels are printed on the bottom axis (on the last row only). TOP specifies that the x axis labels are printed on the top axis (first row only). ALTERNATE specifies that the x axis labels alternate between the top (first row) and bottom axis (last row). We recommend using the TIC OFFSET command to avoid overlap of axis labels and tic marks.
The default is ALTERNATE.
Note:
The following option controls where the y axis tic marks, tic mark labels, and axis label are printed.
SET FACTOR PLOT Y AXIS <LEFT/RIGHT/ALTERNATE>
LEFT specifies that the y axis labels are printed on the left axis (on the first column only). RIGHT specifies that the y axis labels are printed on the right axis (last column only). ALTERNATE specifies that the y axis labels alternate between the left (first column) and right axis (last column). We recommend using the TIC OFFSET command to avoid overlap of axis labels and tic marks.
The default is ALTERNATE.
Note:
Users have different preferences in terms of whether the plot frames for neighboring plots are connected or not. This is controlled with the following option.
SET FACTOR PLOT FRAME <DEFAULT/CONNECTED/USER>
DEFAULT connects neighboring frames (i.e., the FRAME CORNER COORDINATES are set to 0 0 100 100). USER uses whatever frame coordinates are currently set (15 20 85 90 by default) and makes no special provisions for axis labels and tic marks (i.e., you set them as you normally would, each plot uses whatever you have set). CONNECTED uses whatever frame coordinates have been set by the user, but it draws the axis labels and tic marks as if DEFAULT were being used (that is, as determined by the SET FACTOR PLOT commands described above). Typically, CONNECTED is used to put a small bit of space between plots. For example, you might use FRAME CORNER COORDINATES 3 3 97 97 before the PARTIAL RESIDUAL PLOT command.
Since the plots can often have different limits for the axes, the default is USER.
Note:
When the tic marks and tic mark labels are all plotted on the same side (i.e., SET FACTOR PLOT Y AXIS is set to LEFT or RIGHT or SET PARTIAL RESIDUAL PLOT X AXIS is set to BOTTOM or TOP), then overlap between plots is possible. The TIC OFFSET command can be used to avoid this. In addition, you can stagger the tic labels with the following command:
SET FACTOR PLOT LABEL DISPLACEMENT <NORMAL/STAGGERED/VALUE>
NORMAL means that all tic labels are plotted at a distance determined by the TIC LABEL DISPLACEMENT command. STAGGERED means that alternating plots will be staggered. That is, one will use the standard displacement while the next uses a staggered value. Entering this command with a numeric value specifies the amount of the displacement for the staggered tic labels. For example,
TIC MARK LABEL DISPLACEMENT 10
SET FACTOR PLOT LABEL DISPLACEMENT STAGGERED
SET FACTOR PLOT LABEL DISPLACEMENT 25
These commands specify that the default tic label displacement is 10 and the staggered tic mark label displacement is 25.
Note:
It is often helpful on scatter plot matrices to overlay a fitted line on the plots. The following command is used to specify the type of fit.
SET FACTOR PLOT FIT <NONE/LOWESS/LINE/QUAD/SMOOTH>
NONE means that no fitted line is plotted. LOWESS means that a locally weighted least squares line will be overlaid. LINE means that a linear fit (Y = A0 + A1*X) will be overlaid. QUAD means that a quadratic fit (Y = A0 + A1*X + A2*X**2) will be overlaid. SMOOTH means that a least squares smoothing will be overlaid.
For LOWESS, it is recommended that the lowess fraction be set fairly high (e.g., LOWESS FRACTION 0.6).
The fitted line is currently only generated if the factor plot type is PLOT.
The default is for no fitted line to be overlaid on the plot. If a overlaid fit is desired, the most common choice is to use LOWESS.
Note:
Dataplot allows you to set axis limits with the LIMITS command. For the factor plot, it is often desirable to set the axis limits for each plot. This can be done with the command
SET FACTOR PLOT YLIMITS <LOW1> <UPP1> <LOW2> <UPP2> ...
SET FACTOR PLOT XLIMITS <LOW1> <UPP1> <LOW2> <UPP2> ...
The default is to allow the axis limits to float with the data.
Note:
You can use standard plot control commands to control the appearance of the factor plot.
For example,
MULTIPLOT CORNER COORDINATES 5 5 95 95
MULTIPLOT SCALE FACTOR 3
TIC OFFSET UNITS SCREEN
TIC OFFSET 5 5
Default:
None
Synonyms:
None
Related Commands:
FIT = Perform a multi-linear fit. PARTIAL RESIDUAL PLOT = Generates a partial residual plot. PARTIAL REGRESSION PLOT = Generates a partial regression plot. PARTIAL LEVERAGE PLOT = Generates a partial leverage plot. VIF = Compute variance inflation factors for a multi-linear fit. CONDITION INDICES = Compute condition indices for a design matrix. SCATTER PLOT MATIRX = Generate a scatter plot matrix plot. FACTOR PLOT = Generate a plot for a response against a number of different independent variables. CONDITIONAL PLOT = Generate a conditional (subset) plot.
References:
Tom Ryan (1997), "Modern Regression Methods," John Wiley,
Neter, Wasserman, and Kunter (1990), "Applied Linear Statistical Models", 3rd ed., Irwin.
Draper and Smith (1998), "Applied Regression Analysis," 3rd. ed., John Wiley.
Cook and Weisberg (1982), "Residuals and Influence in Regression," Chapman and Hall.
Belsley, Kuh, and Welsch (1980), "Regression Diagnostics," John Wiley,
Paul Velleman and Roy Welsch (1981), "Efficient Computing of Regression Diagnostiocs," The American Statistician, Vol. 35, No. 4, pp. 234-242.
Applications:
Multi-linear Regression
Implementation Date:
2002/6
Program:
SKIP 25
READ HALD647.DAT Y X1 X2 X3 X4
.
MULTIPLOT CORNER COORDINATES 5 5 95 95
MULTIPLOT SCALE FACTOR 2
LINE BLANK
CHARACTER X
X1LABEL DISPLACEMENT 12
Y1LABEL DISPLACEMENT 12
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https://web2.0calc.com/questions/the-amount-of-the-price | +0
# the amount of the price
0
284
4
if the price of a pencil is 36% lower than the price of a pen,then the price of a pen is ?
1.)36%higher than a pencil
2.)43.75 higher than a pencil
3.)56.25 % higher than a pencil
4.)64% higher than a pencil
Guest Feb 17, 2015
#1
+84384
+10
If the price of a pencil is 36% lower than the price of a pen, then the pencil's price is 1- 36% of the pen's price =.64pen
For argument's sake, let the pen's price be 1 .....Then the pencil's price = .64
Then
.64( 1 + P) = 1 where P is the % we're looking for...divide both sides by .64
1 + P = 1/.64
1 + P = 1.5625 subtract 1 from both sides
P = .5625 = 56.25%
Then the pen is 56.25% more than the pencil.
Another way to see this is that .36 must be added to the pencil's price of .64 to get the pen's price.
Thus 36/64 = 56.25% of the pencil's price must be added.......
.
CPhill Feb 17, 2015
Sort:
#1
+84384
+10
If the price of a pencil is 36% lower than the price of a pen, then the pencil's price is 1- 36% of the pen's price =.64pen
For argument's sake, let the pen's price be 1 .....Then the pencil's price = .64
Then
.64( 1 + P) = 1 where P is the % we're looking for...divide both sides by .64
1 + P = 1/.64
1 + P = 1.5625 subtract 1 from both sides
P = .5625 = 56.25%
Then the pen is 56.25% more than the pencil.
Another way to see this is that .36 must be added to the pencil's price of .64 to get the pen's price.
Thus 36/64 = 56.25% of the pencil's price must be added.......
.
CPhill Feb 17, 2015
#2
+91972
+5
Thanks Chris this is a good question. I am just going to think about it myself for a bit.
My pencil is a lead pencil so L for the pencil
and my pen is Blue so B for the pen
$$\\64\%\; of\; B=L\\ 0.64B=L\\ divide both sides by 0.64\\ B=\frac{L}{0.64}=\frac{1}{0.64}\times L = 1.5625L$$
So the blue pen is 156.25% the cost of the lead pencil
In other words the blue pen is 56.25% more expensive than the lead pencil.
Melody Feb 17, 2015
#3
+84384
+5
It DOES seem a bit tricky, at first......I had to think for a few minutes about how to set it up...LOL!!!!
CPhill Feb 17, 2015
#4
+91972
+5
It is horrible when the obvious letter to use eg P for pencil and P for pen does not work (because they are the same) If you can swap them for some other letters that still makes sense, like I did, then it gives you one less thing to get really confused about.
Melody Feb 17, 2015
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https://brilliant.org/discussions/thread/euclidean-algorithm/ | # Euclidean Algorithm
Given two or more positive integers, the greatest common divisor (gcd), or highest common factor (hcf), is the largest positive integer that divides the numbers without a remainder.
If we are given the prime factorization of the numbers, finding the greatest common divisor is straightforward. Let $$A = p_1 ^{q_1} p_2 ^ {q_2} \ldots p_n^{q_n}$$ and $$B= p_1 ^{r_1} p_2 ^ {r_2} \ldots p_n^{r_n}$$, where $$p_i$$ are positive prime integers and $$q_i \geq 1$$. Then
$\gcd{(A, B)} = p_1 ^ { \min(q_1, r_1)} p_2 ^ {\min (q_2, r_2)} \ldots p_n^{\min (q_n, r_n) }.$
However, in general, factorizing numbers is a very difficult problem, so instead we use the Euclidean Algorithm.
## Technique
To find the greatest common divisor using the Euclidean Algorithm, we perform long division several times. Starting with a pair of numbers $$(A, B)$$, we obtain $$A = Q\times B + R$$, where $$Q$$ is the quotient and $$0 \leq R < B$$ is the remainder. We then repeat the sequence with the pair $$(B, R)$$.
As a concrete example, the following is the Euclidean Algorithm performed to calculate $$\gcd( 16457, 1638 )$$:
\begin{aligned} 16457 & = &1638 \times 10 & + &77 \\ 1638 & = &77 \times 21 & + &21\\ 77 & = & 21 \times 3 & + &14 \\ 21 & = & 14 \times 1 & + &7\\ 14 & = & 7 \times 2 & +& 0 \end{aligned}
The process stops since we reached 0, and we obtain
$7 = \gcd (7, 14) = \gcd(14, 21) = \gcd (21, 77) = \gcd (77, 1638) = \gcd( 1638, 16457) .$
## Worked Examples
### 1. Show that if $$\gcd(A,N)=1$$, then $$\gcd(A,B) = \gcd(A, BN)$$.
Solution: Let $$P$$ be the set of primes that divide either $$A$$ or $$B$$. Let $$P'$$ be the set of primes that divide either $$A$$, $$B$$ or $$N$$. Let $$r_i '$$ be the exponent of $$p_i '$$ in the term $$BN$$. If $$p_i ' \mid A$$, then $$p_i \not | N$$, hence $$r_i ' = r_i$$, so $$\min(q_i, r_i ' ) = \min (q_i, r_i)$$. If $$p_i ' \not | A$$, then $$q_i=0$$, so $$\min(q_i, r_i) = 0 = \min(q_i , r_i')$$. Hence, $$\gcd(A,B) = \gcd(A,BN)$$.
### 2. (IMO '59) Prove that $$\frac {21n+4} {14n+3}$$ is irreducible for every natural number $$n$$.
Solution: $$\frac {21n+4} {14n+3}$$ is irreducible if and only if the numerator and denominator have no common factor, i.e. their greatest common divisor is 1. Applying the Euclidean Algorithm,
\begin{aligned} 21n+4 & = &1 \times (14n+3) & + &7n+1 \\ 14n+3 & = &2 \times (7n+1) & + &1\\ 7n+1 & = & (7n+1) \times 1 & + &0. \\ \end{aligned}
Hence, $$\gcd(21n+4, 14n+3) =1$$, which shows that the fraction is irreducible.
Note by Calvin Lin
4 years, 6 months ago
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thanks ! a lot for posting such notes post some more on different topics
- 3 years, 4 months ago | 2018-10-18T06:12:34 | {
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https://plainmath.net/pre-algebra/80465-proving-that-the-sum-of-fractions-has-an | ScommaMaruj
2022-07-01
Proving that the sum of fractions has an odd numerator and even denominator.
I'm struggling to show that, for all $n>1$
$1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}=\frac{k}{m}$
where $k$ is an odd number and $m$ is an even number.
Proof: The proof is by induction on $n$
Base Case: $1+\frac{1}{2}=\frac{3}{2}$
Assume the theorem is true for $n$ and consider $n+1$
$1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}+\frac{1}{n+1}$
We know by the induction hypothesis that the first $n$ terms take the form $\frac{k}{m}$, where $k$ is odd and $m$ is even.
$\frac{k}{m}+\frac{1}{n+1}$
To combine the terms, we find the least common multiple of $m$ and $n+1$. Since $m$ is even, the least common multiple must be even.
Now What? I'm not sure how to show that the numerator remains odd. I don't think there's enough information to do just a case analysis on whether $n+1$ is odd or even.
This question is from Manber's Introduction to Algorithms: A Creative Approach, which I'm using for personal development. He marked this question as "substantially more difficult."
jugf5
Expert
We actually need to prove by strong induction.
Suppose the result holds for all $2,3,4,...,n-1$. Now let's look at the case of $n$
We first look at the following sequences:
$\frac{a}{c}=1+\frac{1}{3}+...+\frac{1}{p}$ where $p$ is the largest odd number not exceeding $n$.
$\frac{b}{d}=\frac{1}{2}+...+\frac{1}{q}$ where $q$ is the largest even number not exceeding $n$
First we can conclude $b$ is odd and $d$ is even as
$\frac{b}{d}=\frac{1}{2}+...+\frac{1}{q}=\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}...+\frac{1}{\left(\frac{q}{2}\right)}\right)$
By induction hypothesis we know for $q>2$, $\left(1+\frac{1}{2}+\frac{1}{3}...+\frac{1}{\left(\frac{q}{2}\right)}\right)$ has odd numerator and even denominator and hence $d$ is even and $b$ is odd.
For the case where $q=2$, $b=1$ and $d=2$ so still $d$ is even and $b$ is odd.
Then let's look at $c$, we know $c$ must be odd because $1,3,5,...,p$ are all odd.
Now $1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}=\frac{a}{c}+\frac{b}{d}=\frac{ad+bc}{cd}$. The numerator is odd and the denominator is even and we are good.
Bruno Pittman
Expert
To complete your proof, first observe that
$\frac{k}{m}+\frac{1}{n+1}=\frac{k\left(n+1\right)+m}{m\left(n+1\right)}$
now $m$ is even so let $m={2}^{\alpha }a$,where $\alpha$ is the biggest power of $2$ in $m$, so $a$ is odd.
if $n+1$ is odd obviously $k\left(n+1\right)+m$ is odd and $m\left(n+1\right)$ is even, so we are done.
So consider the case when $n+1$ is even. Then $n+1={2}^{\beta }b$, where $\beta$ is the biggest power of 2 in $n+1$, so b is odd. so $\frac{k\left(n+1\right)+m}{m\left(n+1\right)}=\frac{k\left({2}^{\beta }b\right)+{2}^{\alpha }a}{\left({2}^{\beta }b\right)\left({2}^{\alpha }a\right)}=\frac{k\left({2}^{\beta }b\right)+{2}^{\alpha }a}{{2}^{\alpha +\beta }ab}$
now look at whether $\alpha$ or $\beta$ is smaller. Say $\alpha$ is smaller, then
$\frac{k\left({2}^{\beta }b\right)+{2}^{\alpha }a}{{2}^{\alpha +\beta }ab}=\frac{k\left({2}^{\beta -\alpha }b\right)+a}{{2}^{\beta }ab}$
now we have an even denominator, and on the numerator $k\left({2}^{\beta -\alpha }b\right)$ is even and $a$ is odd, so the numerator is odd.
Now if $\beta$ is smaller, then
$\frac{k\left({2}^{\beta }b\right)+{2}^{\alpha }a}{{2}^{\alpha +\beta }ab}=\frac{kb+{2}^{\alpha -\beta }a}{{2}^{\alpha }ab}$
now $kb$ is odd so the numerator is odd, and our denominator is obviously even.
By induction, you have proved statement as required
Do you have a similar question? | 2023-02-01T01:53:23 | {
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https://math.stackexchange.com/questions/1076895/difficult-complex-integral-int-gamma-frac1z2idz | # difficult complex integral $\int_\gamma \frac{1}{z^2+i}dz$
We are asked to calculate $\int_\gamma \frac{1}{z^2+i}dz$ where $\gamma$ is the straight line from $i$ to $-i$ in that direction.
My parametrization is simple, I chose $z(t)=i-2it$. Notice that indeed $z(0)=i$ and $z(1)=-i$, and $dz=-2idt$, so the integral becomes
$\int_\gamma \frac{1}{z^2+i}dz=\int_{0}^{1}\frac{-2i}{(i-2it)^2+i}dt$ and here I'm stuck. I don't know how to integrate $\frac{1}{(i-2it)^2+i}$ and would appreciate any guidance I can get.
Thank you.
Edit: A good idea may be to use a variable change $v=i-2it$, and then we need to integrate $\int \frac{1}{v^2+i}dv$
• A partial fraction decomposition of the integrand might prove fruitful. – Daniel Fischer Dec 21 '14 at 19:38
• Note that $(i - 2it)^2 = -1 + 2t - 4t^2$ This is a real polinomial function. – user6565190 Dec 21 '14 at 19:39
• Partial fraction is a good idea. then the integral would be the natural logarithm...thats not bad. – Oria Gruber Dec 21 '14 at 19:40
• With regards to the comments above, that might be true, but I don't see the advantage over just integrating $z\mapsto \dfrac 1{z^2+i}$ from the get-go. – Git Gud Dec 21 '14 at 19:40
• Is there a reason not to workin with the parameterization $z(t) = it$ with $t: 1\to -1$? I tried working this out but I keep getting 0 ($\not =$ WolframAlpha's solution) – dietervdf Dec 21 '14 at 19:46
Note $$\frac{1}{(i - 2it)^2 + i} = \frac{1}{i^2(1 - 2t)^2 + i} = \frac{1}{i - (1 - 2t)^2} = \frac{-i - (1 -2t)^2}{1 + (1 - 2t)^4}.$$
So $$\frac{-2i}{(i-2it)^2+i} = \frac{-2 + 2i(1 - 2t)^2}{1 + (1 - 2t)^4}.$$
Using this you can write
$$\int_{\gamma} \frac{dz}{z^2 + i} = -\int_0^1 \frac{2}{1 + (1 - 2t)^4}\, dt + i\int_0^1 \frac{2(1 - 2t)^2}{1 + (1 - 2t)^4}\, dt.$$
Now we evaluate each integral on the right separately. Using the $u$-substitution $u = 1 - 2t$, we have
$$\int_0^1 \frac{2}{1 + (1 - 2t)^4}\, dt = \int_{-1}^1 \frac{1}{1 + u^4}\, du = 2 \int_0^1 \frac{1}{1 + u^4}\, du.$$
Note
\begin{align}\frac{1}{1 + u^4} &= \frac{1}{(1 + u^2)^2 - (u\sqrt{2})^2}\\ &= \frac{1}{(1 -u\sqrt{2} + u^2)(1 + u\sqrt{2} + u^2)}\\ &= \frac{1}{2u\sqrt{2}}\cdot \left\{\frac{1}{1 - u\sqrt{2} + u^2} - \frac{1}{1 + u\sqrt{2} + u^2}\right\}. \end{align}
Thus
\begin{align} &\int_0^1 \frac{2}{1 + u^4}\, du\\ &=\int_0^1 \left[\frac{1}{u\sqrt{2}(1 - u\sqrt{2} + u^2)} - \frac{1}{u\sqrt{2}(1 + u\sqrt{2} + u^2)}\right]\, du\\ &= \int_0^{\sqrt{2}} \left[\frac{1}{v\left(1 - v +\frac{v^2}{2}\right)} - \frac{1}{v\left(1 + v + \frac{v^2}{2}\right)}\right]\, \frac{dv}{\sqrt{2}} \quad (\text{letting $v = u\sqrt{2}$})\\ &= \int_0^{\sqrt{2}} \left[\frac{2}{v(2 - 2v + v^2)} - \frac{2}{v(2 + 2v + v^2)}\right]\, \frac{dv}{\sqrt{2}}\\ &= \int_0^{\sqrt{2}} \left[\left(\frac{1}{v} - \frac{v - 2}{2 - 2v + v^2}\right) - \left(\frac{1}{v} - \frac{v + 2}{2 + 2v + v^2}\right)\right]\, \frac{dv}{\sqrt{2}} \\ &= \int_0^{\sqrt{2}} \left(\frac{v + 2}{2 + 2v + v^2} - \frac{v - 2}{2 - 2v + v^2}\right)\, \frac{dv}{\sqrt{2}}\\ &= \int_0^{\sqrt{2}} \left[\left(\frac{2v + 2}{2(2 + 2v + v^2)} + \frac{1}{2 + 2v + v^2}\right) - \left(\frac{2v - 2}{2(2-2v+v^2)} - \frac{1}{2-2v+v^2}\right)\right]\, \frac{dv}{\sqrt{2}}\\ &=\int_0^{\sqrt{2}} \left[\frac{2v+2}{2(2+2v+v^2)} - \frac{2v-2}{2(2-2v+v^2)} + \frac{1}{1+(v+1)^2} + \frac{1}{1 + (v-1)^2}\right]\, \frac{dv}{\sqrt{2}}\\ &=\frac{1}{2\sqrt{2}}\left\{\log\left|\frac{2+2v+v^2}{2-2v+v^2}\right| + 2\tan^{-1}(v+1) + 2\tan^{-1}(v-1)\right\}\bigg|_{v = 0}^{v=\sqrt{2}}\\ &=\frac{1}{2\sqrt{2}}\left\{\log\left(\frac{4+2\sqrt{2}}{4-2\sqrt{2}}\right) + 2[\tan^{-1}(\sqrt{2} + 1) +\tan^{-1}(\sqrt{2}-1)]\right\}\\ &= \frac{1}{2\sqrt{2}}\left\{\log\left(\frac{4\sqrt{2}+4}{4\sqrt{2} - 4}\right) + 2\left(\frac{\pi}{2}\right)\right\}\\ &= \frac{1}{2\sqrt{2}}\left\{\log\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right) + \pi\right\}\\ &= \frac{\pi + 2\coth^{-1}(\sqrt{2})}{2\sqrt{2}}. \end{align}
By a similar process, we find
\begin{align} &\int_0^1 \frac{2(1 - 2t)^2}{1 + (1 - 2t)^4}\, dt\\ &= \int_0^1 \frac{2u^2}{1 + u^4}\, du\\ &= \int_0^1 \frac{2u^2}{2u\sqrt{2}}\left(\frac{1}{1-u\sqrt{2}+u^2} - \frac{1}{1-u\sqrt{2}+u^2}\right)\, du\\ &= \frac{1}{\sqrt{2}}\int_0^1 \left(\frac{u}{1-u\sqrt{2}+u^2} - \frac{u}{1+u\sqrt{2}+u^2}\right)\, du\\ &= \int_0^{\sqrt{2}} \left(\frac{v}{2-2v+v^2} - \frac{v}{2+2v+v^2}\right)\, \frac{dv}{\sqrt{2}}\\ &= \frac{1}{\sqrt{2}}\int_0^{\sqrt{2}} \left(\frac{2v-2}{2(2-2v+v^2)} - \frac{2v+2}{2(2+2v+v^2)} + \frac{1}{2-2v+v^2} + \frac{1}{2+2v+v^2}\right)\, dv\\ &= \frac{1}{2\sqrt{2}}\left\{\log\left|\frac{2-2v+v^2}{2+2v+v^2}\right| + 2\tan^{-1}(v-1) + 2\tan^{-1}(v+1)\right\}\bigg|_{v=0}^{v=\sqrt{2}}\\ &= \frac{\pi - 2\coth^{-1}(\sqrt{2})}{2\sqrt{2}}. \end{align}
Hence $$\int_{\gamma} \frac{dz}{z^2 + i} = -\frac{\pi + 2\coth^{-1}(\sqrt{2})}{2\sqrt{2}} + \frac{\pi-2\coth^{-1}(\sqrt{2})}{2\sqrt{2}}i.$$
Using parameterization $z(t) = it \quad (t: 1 \to -1)$.
$$\int_\gamma f(z) \operatorname d z = \int_1^{-1} \frac{i}{i-t^2} \operatorname d t$$
$$= i \int_1^{-1} \frac{1}{(e^{i\frac{\pi}{4}} -t)(e^{i\frac{\pi}{4}} +t)} \operatorname d t$$ $$= i \frac{1}{2e^{i\frac{\pi}{4}}}\int_1^{-1} \left(\frac{1}{e^{i\frac{\pi}{4}} -t} +\frac{1}{e^{i\frac{\pi}{4}} +t} \right) \operatorname d t$$
$$= \frac{i}{2}\left[-\ln (e^{i\frac{\pi}{4}}-t) + \ln(e^{i\frac{\pi}{4}}+t) \right]^{-1}_1$$
$$= \frac{i}{2}\left[\ln(e^{i\frac{\pi}{4}}-1) - \ln(e^{i\frac{\pi}{4}}+1)\right]$$
I think it's looking good, but Wolfram Alpha does not agree.
Solution should be $-1.739\ldots + 0.487\ldots i$. | 2019-08-17T15:15:19 | {
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https://math.stackexchange.com/questions/1352996/how-to-remember-trig-identities/1353577 | # How to remember trig identities?
Suppose I have a trig function $T: \Bbb{R} \rightarrow \Bbb{R}$. I want to be able to derive four basic properties:
$$T(x) \cdot T(y)$$ $$T(x) + T(y)$$ $$T(x+y)$$ $$T(cx)$$
where $c$ is some scalar.
I know there are a bunch of identities: reciprocal, quotient, Pythagorean, co-function, even-odd. And then some formulas: product-to-sum, sum-to-product, sum-difference, double angle, half-angle/power-reducing. Here is a list. That's a lot to memorize and some of them seem to overlap.
Why do the four basic properties I mentioned require so many identities to learn? It is a bit cumbersome. Is there an easier way to learn how to do the basic arithmetic operations with trig functions?
• The definitions of $\tan$, $\csc$, $\sec$, and $\cot$ must be memorized (nemonic: each pair of reciprocals has one "co" e.g., $(\sin,\csc)$ are a pair) I memorize $\sin^2(x)+\cos^2(x)=1$ and the angle sum/difference formulas. Everything else can usually be derived from those. Jul 7, 2015 at 19:32
• I do the same as Michael Burr. I've memorized the addition angle formulae for sine and cosine and derive other identities from those. Jul 7, 2015 at 19:51
• I think that this question is relevant. Jul 7, 2015 at 19:56
• The $e^{ix}e^{iy}=e^{i(x+y)}$ thing is useful. (If you haven't seen this before, just treat "$e^{ix}$" as shorthand for "$\cos(x)+i\sin(x)$." Where $i^2=-1$. Note: $e^{ix}=\cos(x)+i\sin(x)$ is actually a theorem, rather than just notation, but it takes a while to prove.) Jul 8, 2015 at 7:22
• The way to learn (memorize, if you prefer) the formulas is to do 1,000 exercises that call for them. You will not only learn the formulas, you'll learn which ones are useful in which situations. Jul 8, 2015 at 9:32
I always found recalling $e^{ix}=\cos x+i\sin x$ useful for quickly deriving the sum of angles formulae, e.g.
$$e^{i(x+y)}=\cos(x+y)+i\sin(x+y).\tag{1}$$
But
$$e^{ix+iy}=e^{ix}e^{iy}=(\cos x+i\sin x)(\cos y+i\sin y).\tag{2}$$
Expanding (2), equating with (1) and separating real and imaginary parts gives you the formulae.
You can then get the double angle formulae easily.
Wait, there's more!
We have $(e^{ix})^n=(\cos x+i\sin x)^n$.
But we also have $$(e^{ix})^n=e^{inx}=\cos(nx)+i\sin(nx),$$ so we get
$$(\cos x+i\sin x)^n = \cos(nx)+i\sin(nx).$$
For example, consider $n=2$, then expanding gives:
$$\cos^2 x-\sin^2 x = \cos(2x)$$ and $$2\sin x\cos x=\sin(2x).$$
This is another way to get the double angle formulae, but you can get more trig identities by letting $n=3, 4, \ldots$. In general, for positive integer $n$ we have
$$\cos(nx) = \Re\left((\cos x+i\sin x)^n\right) =\Re\sum_{k=0}^n{n\choose k}i^k\cos^{n-k}(x)\sin ^k(x)$$ and $$\sin(nx) = \Im\left((\cos x+i\sin x)^n\right)=\Im\sum_{k=0}^n{n\choose k}i^k\cos^{n-k}(x)\sin^k(x).$$
Expanding and simplifying will give you nice trig identities. This is called De Moivre's Theorem.
• This is very very cool. I didn't anticipate learning this. It also makes it very easy to keep track of the sign changes as you take derivitives of the basic trig functions. Jul 8, 2015 at 18:46
• @StanShunpike This is a special case of series bisection / multisection. Jul 8, 2015 at 21:09
• @BillDubuque Wow! I've wondered what you call that for a while. Awesome! Jul 8, 2015 at 23:43
I find that four suffice. $$\cos^2 (x) + \sin^2(x) = 1 \tag{1}$$ $$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b) \tag{2}$$ $$\sin(a+b)=\sin(a)cos(b)+\sin(b)\cos(a) \tag{3}$$ $$\text{trig}(x) = \text{cotrig}(\frac{\pi}{2}-x) \tag{4}$$
The Pythagorean identity $$(1)$$ is easy to manipulate. Divide through by $$cos^2(x)$$ alternatively by $$sin^2(x)$$ to find the other forms
$$1 + \tan^2(x) = sec^2(x) \tag{5}$$ $$\cot^2(x) +1 = csc^2(x) \tag{6}$$
For the angle addition formulas $$(2)$$ and $$(3)$$, we can apply odd and even identities to quickly derive the angle subtraction identities:
$$\cos(a+\color{red}{(-b)})=\cos(a)\cos(\color{red}{(-b)})-\sin(a)\sin(\color{red}{(-b)})$$ $$\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b) \tag{7}$$
and
$$\sin(a+\color{red}{(-b)})=\sin(a)cos(\color{red}{(-b)})+\sin(\color{red}{(-b)})\cos(a)$$ $$\sin(a-b)=\sin(a)cos(b)-\sin(b)\cos(a) \tag{8}$$
We can also let $$a=b$$ and substitute in $$(2)$$ and $$(3)$$ to find double angle forms
$$\cos(a+a)=\cos(a)\cos(a)-\sin(a)\sin(a)$$ $$\cos(2a)=\cos^2(a)-\sin^2(a) \tag{9}$$
$$\sin(a+a)=\sin(a)cos(a)+\sin(a)\cos(a)$$ $$\sin(2a)=2\sin(a)cos(a) \tag{10}$$
Combining $$(9)$$ with the Pythagorean identity $$(1)$$ gives two more.
$$\cos(2a)=1-2\sin^2(a) \tag{11}$$ $$\cos(2a)=2\cos^2(a)-1 \tag{12}$$
If you want a half angle formula, you may as well let $$u=2a$$ in the previous four identities. Just mind your squares and roots. What happening to tangent? Divide any related $$sin$$ by $$cos$$ to get what you need.
Let's not forget product to sum identities! If we take $$(2)$$ and $$(7)$$ and add the equation, we find
$$\cos(a+b) + \cos(a-b)=2\cos(a)\cos(b)$$ $$\cos(a)\cos(b) = \frac12 (\cos(a+b) + \cos(a-b)) \tag{13}$$
Similar combinations will give the remaining product to sum identities.
As for $$(4)$$, $$trig(x) = cotrig(\frac{\pi}{2}-x)$$, I'm referring to cofunction identities, which all have the same form. For example, $$\sin(x) = \cos(\frac{\pi}{2}-x).$$ That's essentially six more identities.
We have over twenty identities at our disposal now, including the few that I've mentioned but neglected to outright derive.
• This was super helpful Jun 29, 2017 at 9:50
From the comments to the question:
The definitions of $$\tan$$, $$\csc$$, $$\sec$$, and $$\cot$$ must be memorized (mnemonic: each pair of reciprocals has one "co" e.g., $$(\sin,\csc)$$ are a pair) I memorize $$\sin^2(x)+\cos^2(x)=1$$ and the angle sum/difference formulas. Everything else can usually be derived from those.
This is great because it simplifies everything down to a core set of principles. The problem is then memorizing the sum/difference formulas. I found the following mnemonic in another answer
Blue:
Sine, Cosine, Sign, Cosine, Sine
Cosine, Cosine, Co-Sign, Sine, Sine
The first line encapsulates the sine formulas; the second, cosine. Just drop the angles in (in order $$\alpha, \beta, \alpha, \beta$$ in each line), and know that "Sign" means to use the same sign as in the compound argument ("+" for angle sum, "-" for angle difference), while "Co-Sign" means to use the opposite sign.
Together, I think this makes a pretty good game plan for tackling trig. For a little more fun, see Antinous' awesome answer.
Stan's answer will probably do a better job than mine; while running the risk of repeating information: memorize $\sin^2(x)$ $+$ $\cos^2(x)$ $=$ $1$, dividing through by $\cos^2(x)$ will yield the identity $\tan^2(x)$ + $1$ $=$ $\sec^2(x)$.
Additionally, dividing through by $\sin^2(x)$ will yield $1 +$ $\cot^2(x)$ = $\csc^2(x)$. Furthermore, intuitively understanding the graphs of $\sin(x)$ and $\cos(x)$ will make life easier.
Finally, going through the unit circle derivations of many trigonometric identities will prove useful in being able to "unstick" yourself should you get stuck.
Edit: Of course, also knowing the basics: $\tan(x)$ $=$ $\frac{\sin(x)}{\cos(x)}$, $\csc(x)$ $=$ $\frac 1{sin(x)}$, $\sec(x)$ $=$ $\frac 1{cos(x)}$ is fundamental.
A systematic way to implement what pbs wrote, is mentioned in the book A = B in section 1.5:
Let $w = \exp(ix)$, then $\cos x = (w +w^{-1})/2$ and $\sin x = (w - w^{-1}/(2i$). So equality of rational expressions in trigonometric functions can be reduced to equality of polynomial expressions in w. (Exercise: Prove, in this way, that $\sin 2x = 2\sin x\cos x$.) | 2022-05-21T10:12:59 | {
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https://math.stackexchange.com/questions/1394828/write-the-equation-of-the-line-which-passes-through-1-2-and-is-perpendicul | # Write the equation of the line which passes through $(1, –2)$ and is perpendicular to the line with equation $5y – x = 1$
Write the equation of the line which passes through $(1, –2)$ and is perpendicular to the line with equation $5y – x = 1$.
I know that I need to put the equation into slope-intercept form but what is the step that I take after this.
$$y = \frac 1 5 x + \frac 1 5$$
I read something about the negative reciprocal but that isn't working for me.
Also, here are the possible answers:
a. $5x + y = 3$ ANSWER
b. $5x – y = 1$
c. $x + 5y = –2$
d. $x – 5y = 3$
We know that if two perpendicular lines have gradients $m_1$ and $m_2$ then $m_1 m_2 = -1$. As you've put the line you're given in slope-intercept form you know that it has gradient $\frac{1}{5}$. This tells you that the line you're trying to find has gradient $m$ satisfying $\frac{m}{5} = -1$ which gives $m = -5$.
So we know now that the line you're trying to find is of the form $y = -5x + c$ for suitable $c$ in slope-intercept form. But we know it passes through (1,-2) so we substitute in to get $-2 = -5 + c$ giving $c = 3$. So the line is $y = -5x + 3$, this rearranges to answer a.
If two lines with slopes $m$ and $m'$ respectively are perpendicular to each other we have:
$$mm'=-1$$
In our case, we're looking for $m'$:
$$m'/5 = -1\\ m' = -5$$ So our line equation becomes:
$$y = -5x+p$$
Now using that the lines passes from $(1,-2)$, we solve for $p$:
$$-2 = -5+p\\ p = 3$$
Our equation becomes:
$$y = -5x+3\\ y+5x-3 = 0$$
So the answer is $a$
Hint:
A line perpendicular to $ax+by=c$ takes the form $bx-ay=d$.
Notice the slope of line perpendicular to the line: $5y-x=1\iff y=\frac{1}{5}x+\frac{1}{5}$ $$=\frac{-1}{\frac{1}{5}}=-5$$
Hence, the equation of the line passing through the point $(1, -2)$ & having slope $\frac{1}{5}$ $$y-(-2)=-5(x-1)$$ $$\color{blue}{5x+y=3}$$
Hence, option (a) is correct.
The vector $(-1,5)$ is normal to the given line. Hence $(5,1)$ is normal to the perpendicular. If this perpendicular passes through the point $(1,-2)$, its equation is $$5x+y=5\cdot \color{red}1 +(\color{red}{-2})=3.$$ | 2020-05-26T01:28:01 | {
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https://zims-en.kiwix.campusafrica.gos.orange.com/wikipedia_en_all_nopic/A/Proof_by_contrapositive | # Proof by contrapositive
In logic, the contrapositive of a conditional statement is formed by negating both terms and reversing the direction of inference. More specifically, the contrapositive of the statement "if A, then B" is "if not B, then not A." A statement and its contrapositive are logically equivalent, in the sense that if the statement is true, then its contrapositive is true and vice versa.[1]
In mathematics, proof by contrapositive, or proof by contraposition, is a rule of inference used in proofs, where one infers a conditional statement from its contrapositive.[2] In other words, the conclusion "if A, then B" is inferred by constructing a proof of the claim "if not B, then not A" instead. More often than not, this approach is preferred if the contrapositive is easier to prove than the original conditional statement itself.[3]
Logically, the validity of proof by contrapositive can be demonstrated by the use of the following truth table, where it is shown that pq and ${\displaystyle \lnot }$q${\displaystyle \lnot }$p share the same truth values in all scenarios:
p q ${\displaystyle \lnot }$p ${\displaystyle \lnot }$q pq ${\displaystyle \lnot }$q${\displaystyle \lnot }$p
TTFFTT
TFFTFF
FTTFTT
FFTTTT
## Example
Let x be an integer.
To prove: If x² is even, then x is even.
Although a direct proof can be given, we choose to prove this statement by contraposition. The contrapositive of the above statement is:
If x is not even, then x² is not even.
This latter statement can be proven as follows: suppose that x is not even, then x is odd. The product of two odd numbers is odd, hence x² = x·x is odd. Thus x² is not even.
Having proved the contrapositive, we can then infer that the original statement is true.[4] | 2021-01-24T04:10:29 | {
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http://mathhelpforum.com/number-theory/146796-last-two-digits.html | 1. ## Last two digits...
I have to find the last two digits of 222227^2010.
So I know that this gets reduced to 27^2010 mod 100
But I don't know how to solve 27^x = 1(mod 100) to get this process going. Or am I going about it wrong?
I have to find the last two digits of 222227^2010.
So I know that this gets reduced to 27^2010 mod 100
But I don't know how to solve 27^x = 1(mod 100) to get this process going. Or am I going about it wrong?
$\phi(100)=40$ and $(27,100)=1\implies 27^{40}\equiv1\bmod{100}$
Now $2010=50\cdot40+10 \implies 27^{2010}=27^{50\cdot40+10}=\left(27^{40}\right)^{ 50}\cdot27^{10}\equiv1\cdot27^{10}=27^{10}\bmod{10 0}$.
Observe $10=8+2=2^3+2$
$27^2\equiv 29\bmod{100}$
$27^4=\left(27^2\right)^2\equiv 29^2\equiv 41\bmod{100}$
$27^8=\left(27^4\right)^2\equiv 41^2\equiv 81\bmod{100}$
Thus $27^{10}=27^{8+2}=27^8\cdot27^2\equiv 81\cdot29=2349\equiv49\bmod{100}$
Therefore the last two digits of $222227^{2010}$ are $49$.
3. Originally Posted by chiph588@
$\phi(100)=40$ and $(27,100)=1\implies 27^{40}\equiv1\bmod{100}$
This is the part I don't get. What does the phi mean, and what does the (27,100) mean? How did you get that 27^40 = 1 so simply?
4. $\phi(100)$ is the "totient function", the number of integers less than 100 and coprime to 100. Chiph588 then uses "Euler's theorem": For any a coprime to n, $a^{\phi(n)}\equiv 1 (mod n)$.
5. What's a quick way to figure out the totient function?
6. $27^{2010} = 3^{6030} = 3^{6028+2} = 9 (81^{1507})$
$81^{1507} = (80 + 1)^{1507}$
$= 80^{1507} + 80^{1506}(1507) + .... + \frac{1507(1506)}{2} 80^2 + 1507(80) + 1$
$\equiv 60+1 \bmod{100} \equiv 61 \bmod{100}$
Therefore ,
$27^{2010} \equiv 9(61) \bmod{100} \equiv 549 \equiv 49 \bmod{100}$
What's a quick way to figure out the totient function?
You may show that $f: x \to ax ~ a,x \in \mathbb{U}(m)$ is a bijection . Then , obtaining the product :
$\prod_{x \in \mathbb{U}(m)} x \equiv \prod_{x \in \mathbb{U}(m)} a x$
$1 \equiv a^{\phi(m)}$
7. Originally Posted by simplependulum
$= 80^{1507} + 80^{1506}(1507) + .... + \frac{1507(1506)}{2} 80^2 + 1507(80) + 1$
$\equiv 60+1 \bmod{100} \equiv 61 \bmod{100}$
What's going on here?
I used a very primitive approach . . .
Find the last two digits of: . $N \;=\;222227^{2010}$
I know that this gets reduced to: . $27^{2010}\text{ (mod 100)}$
We have: . $N \;=\;\left(3^3\right)^{2010} \;=\;3^{6030}$
I found that:
. . $\begin{array}{cc}3^n & \text{ends in} \\ \hline
3^1 & 03 \\ 3^2 & 09 \\ 3^3 & 27 \\ 3^4 & 81 \\ 3^5 & 43 \\ \vdots & \vdots \\
3^{10} & 49 \\ \vdots & \vdots \\ 3^{20} & 01 \end{array}$
Hence: . $3^{20} \:\equiv\:01\text{ (mod 100)}$
We have: . $N \;=\;3^{6030} \;=\;3^{20(301) + 10} \;=\;\left(3^{20}\right)^{301}\cdot 3^{10}$
Hence: . $N \;\equiv\;(01)^{301}\cdot 3^{10}\text{ (mod 100)}$
. . . . . . . . $\equiv\;3^{10}\text{ (mod 100)}$
. . . . . . . . $\equiv\;49\text{ (mod 100)}$
Therefore, $222227^{2010}$ ends in 49. | 2016-08-26T23:57:24 | {
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https://math.codidact.com/posts/287846 | ### Communities
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# Criterion in terms of the bases for determining whether one topology is finer than another
+3
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When topologies are given by bases, one has a useful criterion in terms of the bases for determining whether one topology is finer than another:
Lemma 13.3. (Munkres's Topology p.81) Let $\mathscr{B}$ and $\mathscr{B}^{\prime}$ be bases for the topologies $\mathcal{T}$ and $\mathcal{T}^{\prime}$, respectively, on $X$. Then the following are equivalent:
(1) $\mathcal{T}^{\prime}$ is finer than $\mathcal{T}$.
(2) For each $x \in X$ and each basis element $B \in \mathscr{B}$ containing $x$, there is a basis element $B^{\prime} \in \mathscr{B}^{\prime}$ such that $x \in B^{\prime} \subset B$.
If one modifies condition (2) as:
(2') For each nonempty basis element $B \in \mathscr{B}$ there is a nonempty basis element $B^{\prime} \in \mathscr{B}^{\prime}$ such that $B^{\prime} \subset B$.
then it is unlikely that one can conclude (2'): one cannot use the original argument for showing the implication (2) $\Rightarrow$ (1) anymore.
Can anyone give a counter-example for (2') $\Rightarrow$ (1)?
[(2') edited.]
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#### 1 comment thread
As written, (2') can never br true (2 comments)
+1
−0
A counterexample would be the set $S=\{0,1,2\}$ with the topologies $\mathcal T = \{\emptyset, \{0\}, \{0,1\}, S\}$and $\mathcal T' = \{\emptyset,\{0\},S\}$. Clearly $\mathcal T'$ is not finer than $\mathcal T$.
Now $\mathcal T$ is generated by the basis $\mathscr B = \{\{0\},\{0,1\},S\}$ and $\mathcal T'$ is generated by $\mathscr B' = \{\{0\},S\}$.
Now (2') is fulfilled, as for every non-empty $B\in\mathscr B$, the set $B'=\{0\}\in \mathscr B'$ is such that $B'\subset B$.
Thus this gives a counterexample to (2')$\implies$(1).
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https://math.stackexchange.com/questions/2401968/natural-deduction-proof-of-forall-x-px-land-forall-y-py-implies-qy | # Natural deduction proof of $(\forall x.P(x))\land(\forall y.P(y) \implies Q(y)) \vdash \forall z.Q(z)$
My attempt
1. $(\forall x.P(x))\land(\forall y.P(y) \implies Q(y))$ [premise]
2. $\forall y.P(y) \implies Q(y)$ [$\land$ elim 1]
3. $\forall x.P(x)$ [$\land$ elim 1]
4. $a, P(a)$ [$\forall$ elim 3]
5. $a, P(a) \implies Q(a)$ [$\forall$ elim 2]
6. $Q(a)$ [$\implies$ elim 4,5]
7. $\forall z.Q(z)$
I feel steps 4-5 is not correct, because we used a for two different expressions. I'm also not so sure about whether it's okay to go from step 6-7. Could anyone tell me whether what I did is correct and if not how I should fix it?
• Another instance in which making the domain of quantification explicit will greatly simplify the problem. It is easy and more intutitive to prove: $\forall x: [x\in U \implies P(x)] \land [\forall y:[y \in U \implies [P(y)\implies Q(y)]]\implies \forall z:[z\in U \implies Q(z)]$. $U$ could possibly be empty as well. – Dan Christensen Aug 22 '17 at 15:44
• If you don't want to make the domain of quantification explicit, the simplest way may be a proof by contradiction. Suppose to the contrary that $\neg Q(z)$ to obtain a the contradiction $Q(z)\land \neg Q(z)$. – Dan Christensen Aug 22 '17 at 17:50
• The proof term for this is: $\Lambda z.f[z](p[z])$ where $p : \forall x.P(x)$ and $f: \forall y.P(y)\to Q(y)$. – Derek Elkins Aug 22 '17 at 17:56
I would justify your steps 4 and 5 with "universal instantiation" and write down that $a$ in step 5 is the same $a$ as in step 4. (If step 2 is true for all $y$, then it is certainly true with the assignment $y = a$ for the same $a$ as in step 4.) Then step 6 is modus ponens on steps 5 and 4. (Not everything is elimination.) Then step 7 is universal generalization applied to 6.
But otherwise, this seems fine.
• Well, modus ponens is conditional elimination. – Graham Kemp Aug 22 '17 at 6:26
The universal elimination ($\forall\textsf {Elim}$) steps are that : If the statement $S(w)$ holds for any entity $w$ of the domain, then $S(a)$ holds for an arbitrary value $a$ of the domain. $$\forall w~S(w)~\vdash~ S(a)$$
The universal introduction ($\forall\textsf{Intro}$) step is likewise that : If the statement $S(a)$ holds for an arbitrary value $a$ of the domain, then $S(z)$ holds for all values $z$ of the domain. $$S(a)~\vdash~\forall z~S(z)$$
The key criteria is that you must not violate the arbitrary nature of the free variable $a$ between assumption(s) and discharge. However, it is fine to for two universal eliminations to assume the same arbitrary value; and there is not any problem with the bound variables using different tokens. All steps are okay.
$$\dfrac{\dfrac{\dfrac{\dfrac{\forall x~P(x)~\wedge~\forall y~(P(y)\to Q(y))}{\forall y~(P(y)\to Q(y))}{\wedge\textsf E}}{P(a)\to Q(a)}{\forall\textsf E,a}\quad\dfrac{\dfrac{\forall x~P(x)~\wedge~\forall y~(P(y)\to Q(y))}{\forall x~P(x)}{\wedge\textsf E}}{P(a)}{\forall\textsf E,a}\quad}{Q(a)}{{\to}\textsf E,a}}{\forall z~Q(z)}{\forall\textsf I,\not a}$$
You can eliminate any universal with any term you wabnt, so there is no need to introduce $a$ on line 4 or 5.
However, you do need to introduce the $a$ in order to set up the Universal Introduction rule. That is, you need to basically say: "let $a$ be an arbitrary object. By $\forall x P(x)$ it follows that $P(a)$, and by $\forall y (P(y) \rightarrow Q(y))$ it follows that $P(A) \rightarrow Q(a)$. Hence, we have $Q(a)$. But since $a$ was arbitrary, it follows that $\forall z Q(z)$"
I don't know how exactly that proof formalizes in whatever system you are using, but here is one done in Fitch: | 2019-07-18T02:04:28 | {
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https://www.physicsforums.com/threads/finding-the-complex-function-given-a-real-component.480744/ | # Finding the complex function given a real component
## Homework Statement
Find $$f(z) = u(x,y) + iv(x,y)$$ with u or v as a given.
## Homework Equations
$$u = \frac{x}{x^2+y^2}$$
## The Attempt at a Solution
Using the Cauchy-Riemann equations, if the function is analytic, then $$u_x = v_y$$ and $$u_y = -v_x$$
So, the first thing I did was find the x derivative of u.
$$u_x = \frac{y^2-x^2}{(x^2+y^2)^2}$$ which equals $$v_y$$
Then, I integrated in terms of y to get $$\frac{1}{x}\arctan{\frac{y}{x}} + A(x)$$
However, it seems that I might be going about this the wrong way, as the problem is getting rather hairy from here. The answer the book provides is much simpler than where this seems to be going.
HallsofIvy
Homework Helper
That's the right approach but I think you may have integrated incorrectly.
Dick
Homework Helper
It might be more obvious how to integrate the other Cauchy-Riemann equation, u_y=(-v_x). Try that.
I approached the problem from the other direction, and I am still ending up at an impasse. I am not entirely sure what to do with the A(y) function that results from the integration.
I'm getting:
$$v = \frac{y}{x^2+y^2} + A(y)$$
I am also getting
$$u_x = \frac{1}{x^2+y^2} - \frac{2x^2}{(x^2+y^2)^2}$$
Integrating with respect to y...
$$v = \frac{-y}{x^2+y^2} + B(x)$$
I get the following equality, after some rearranging:
$$B(x)-A(y) = \frac{2y}{x^2+y^2}$$
I am not entirely sure where to go from here, especially with regards to the A(x) and B(y), but the final complex function that the book was able to rearrange into was f(z) = 1/z.
So, I am still very much stuck as to how to continue.
Dick
Homework Helper
It looks to me like you got v=(-y)/(x^2+y^2). If so there's no need to continue, you are done, v and u satisfy the CR equations. And they are the real and complex parts of 1/z.
You could use the Milne-Thomson method to construct complex functions...
Note : Pardon the handwriting in the image. C is the integration constant. Also, if you've been given the imaginary part of the complex function, you can use f ' (z) = v1(z , 0) + i v2(z , 0).
Last edited:
I should follow up on this. I still have the information regarding the Milne-Thomson method up, and am reading through it.
I realized that I had encountered an algebra mistake when I was doing the integration. I was able to figure it out, and I found out how the A(x) and B(y) functions disappeared just by comparing the two different forms of v.
Gib Z
Homework Helper
A good way to find the expression for the function in terms of $z$ is to use analytic continuation : The function on the real line must agree with the complex function. So just set y=0, and in the result replace all the x's with z's. | 2021-09-22T07:05:50 | {
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https://math.stackexchange.com/questions/3786757/how-can-i-prove-that-px-x4x1-doesnt-have-real-roots | How can I prove that $p(x)=x^4+x+1$ doesn't have real roots?
Let $$p(x)=x^4+x+1$$ be a polynomial in $$\mathbb{R}[x]$$. How can I prove that $$p$$ doesn't have real roots?
My attempt:
From calculus, I know that
$$\lim_{x \to \pm\infty} p(x) = \infty\,.$$
Then, if it has real roots, then we have two or four real roots. I know that if $$\gcd \left( p(x), p'(x) \right) = 1\,,$$ then the roots is simple. Is there another hint?
• "Then, if it has real roots we have two or four real roots." -- Minor correction on this point, but this isn't necessarily true. For instance, $f(x)=(x-1)^2$ has only one root, $x=1$, because it "sits" on the $x$-axis. $f$ might only have real roots, but they're also not distinct ... so I guess it's a matter of convention/preference as to how you count them. It might be worth being clear on such details due to such ambiguities. – Eevee Trainer Aug 11 at 1:35
• They are "usually" counted with multiplicity though: a lot of the theorems (e.g. the corollary of the FTA that states that an nth degree polynomial has n roots and Bezout's theorem) are stated using multiplicity in order to simplify their statements. IMO, that should be the default interpretation. – NickD Aug 11 at 16:45
Consider three cases . . .
• If $$x\ge 0$$ then $$x^4+x+1\ge 1$$.$$\\[4pt]$$
• If $$-1 < x < 0$$ then $$x^4+x+1 > x^4 + (-1) + 1 > 0$$.$$\\[4pt]$$
• If $$x\le -1$$ then $$x^4+x+1 \ge x^2+x+1 = \Bigr(x+\frac{1}{2}\Bigr)^2+\frac{3}{4}$$.
Thus in all three cases, $$x^4+x+1$$ is positive.
It follows that $$x^4+x+1$$ has no real roots.
• if $x \leq -1$ then $x^4 \geq |x|,$ so $x^4 + x \geq 0$ – Will Jagy Aug 11 at 2:38
• Yes, that's a nice way to dispatch that case. – quasi Aug 11 at 2:46
If $$x$$ is a real number, then $$x^4+x+1=\left(x^2-\frac12\right)^2+\left(x+\frac12\right)^2+\frac12>0\,.$$
Hint: $$p'(x)=4x^3+1$$ has one root $$c$$ (to see this, remark that $$p"(x)\geq 0$$), if $$x if $$x>c, p'(c>0$$, show that $$p(c)>0$$. The function decreases from $$-\infty$$ to $$c$$ and increases from $$c$$ to $$+\infty$$.
• check your $p'$ calculation. Leading coefficient is...? – user24142 Aug 11 at 1:23
• I know that $p'$ has one real root, it is given by $\frac{-1}{\sqrt[3]{4}}$, but honestly I don't know how it solves my question.. – Joãonani Aug 11 at 1:25
• Show that $p(root)>0$ – Tsemo Aristide Aug 11 at 1:27
• It's easy, $\frac{-1}{\sqrt[3]{4}}< 1$ then $(\frac{-1}{\sqrt[3]{4}})^4 + \frac{-1}{\sqrt[3]{4}} +1$ is positive...But how can I conclude that? – Joãonani Aug 11 at 1:28
• in your sum, you have $1$, a positive term and a negative term inferior to $1$ – Tsemo Aristide Aug 11 at 1:30
$$p := x^4 + x + 1 = \begin{bmatrix} x^2\\ x\\ 1\end{bmatrix}^\top \begin{bmatrix} 1 & 0 & -t\\ 0 & 2 t & 0.5\\ -t & 0.5 & 1\end{bmatrix} \begin{bmatrix} x^2\\ x\\ 1\end{bmatrix}$$
where $$t \in \Bbb R$$. Using Sylvester's criterion, we learn that the (symmetric) matrix above is positive semidefinite for $$t = 0.5$$. Using the Cholesky decomposition,
$$\begin{bmatrix} 1 & 0 & -0.5\\ 0 & 1 & 0.5\\ -0.5 & 0.5 & 1\end{bmatrix} = {\rm L} {\rm L}^\top$$
where
$${\rm L} = \begin{bmatrix} \color{blue}{1} & 0 & 0\\ 0 & \color{magenta}{1} & 0\\ \color{blue}{-\frac{1}{2}} & \color{magenta}{\frac{1}{2}} & \color{red}{\frac{\sqrt{2}}{2}}\end{bmatrix}$$
and, thus,
$$p = \left( \color{blue}{x^2 - \frac12} \right)^2 + \left( \color{magenta}{x + \frac12} \right)^2 + \left( \color{red}{\frac{\sqrt{2}}{2}} \right)^2 > 0$$
which is the exact same sum of squares (SOS) decomposition in this answer.
SymPy code
>>> from sympy import *
>>> t = symbols('t', real=True)
>>> M = Matrix([[ 1, 0, -t],
[ 0, 2*t, 1/2],
[-t, 1/2, 1]])
>>> L = M.subs(t,1/2).cholesky()
>>> L
Matrix([
[ 1, 0, 0],
[ 0, 1, 0],
[-1/2, 1/2, sqrt(2)/2]])
Related
$$x^4$$ is positive except for $$x = 0$$. $$x + 1$$ is positive for $$x > -1$$. So the only possibility for the polynomial (the sum of those 2 parts) to be negative is for $$x \le -1$$.
But the polynomial is positive at $$x = -1$$ and $$x^4$$ grows much faster than $$x + 1$$ for $$|x| > 1$$ so the polynomial is positive everywhere.
By studying $$p'$$ you easily see that $$p$$ is a decreasing function on $$]-\infty, -1/\sqrt[3]{4}]$$ and an increasing function on $$[-1/\sqrt[3]{4},\infty[$$. From that, you now that the minimal value $$p(x)$$ is $$p(-1/\sqrt[3]{4})$$ which is positive. Then $$p$$ has no real root. | 2020-09-25T10:37:29 | {
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https://www.physicsforums.com/threads/two-satellites-in-orbit-around-jupiter.495989/ | Two satellites in orbit around Jupiter
Homework Statement
Two satellites are in circular orbits around Jupiter. One, with orbital radius r, makes one revolution every 16 h. The other satellite has orbital radius 4.0r. How long does the second satellite take to make one revolution around Jupiter?
A. 130 hours
B. 120 h
C. 140 h
D. 110 h
E. 90 h
Homework Equations
v = sqrt( GMj / r )
T = (2*pi*r) / v
G = 6.673*10^-11
Mj = 1.90*10^27
The Attempt at a Solution
I'm assuming that this is just an algebra problem at this point, but I'm having trouble finding r.
T = ( 2*pi*r) / sqrt( GMj / r )
r = [(GMjT^2) / (4pi^2)]^(1/3)
When I solve for r and then plug r back into the equation for T I do not get 16, which means I'm not doing something right here.
Do I need to convert the 16 hours / revolution into another unit before solving for r?
Any pointers?
Last edited:
gneill
Mentor
According to Kepler's law the square of the period varies as the cube of the orbit radius.
$$T^2 \propto R^3$$
This implies that for any given set of orbits around a common central body, that
$$\frac{T^2}{R^3} = Constant$$
You should be able to use this fact to determine the "mystery" orbital period.
Thanks, that did help a lot.
It seems like the method I was trying to work through would also work but is much longer.
It also doesn't help that when I do the calculation with either method I still do not get a value the exactly matches one of the answer choices, but is very close to answer A ( which did turn out to be correct ). | 2021-07-29T21:50:54 | {
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https://math.stackexchange.com/questions/3804623/lu-factorization-which-row-should-be-chosen-in-partial-pivoting | # LU factorization: which row should be chosen in partial pivoting?
I've seen partial pivoting described thus: during the $$k$$th step of LU factorization of $$\mathbf{A}$$, find the remaining element of $$\mathbf{A}$$ in column $$k$$ with the greatest absolute value, and swap its row with the $$k$$th row.
However, I can think of one example where this strategy leads to division-by-zero: $$\mathbf{A} = \begin{bmatrix}6&3&6\\5&\frac{5}{2}&8\\3&1&4\end{bmatrix}$$This matrix has been chosen such that the greatest values in each column are already on the diagonal, so no swapping should be necessary, according to this strategy. In this example, $$u_{2,2} = 0$$, and computing $$l_{3,2}$$ will require dividing by zero.
Am I misunderstanding the concept of partial pivoting? It seems that the problem arises from trying to choose the row based on the greatest value in $$\mathbf{A}$$, but I think it would make more sense to choose the row that results in the greatest value in $$\mathbf{U}$$. I'm not sure if that's correct, or how that could be applied practically.
There are multiple strategies on how to chose a pivot, depending on the type of factorisation that you do ($$PA = LU$$, $$AQ = LU$$, $$PAQ = LU$$), with $$PA = LU$$ (partial pivoting of the rows) being the most commonly used. In this case, we typically chose the pivot at step $$k$$ to be the element of the $$k$$-th column with the largest magnitude (and row $$\ge k$$)
Here is how it works if done on your matrix:
$$A = \begin{bmatrix} 6 & 3 & 6 \\ 5 & 5/2 & 8\\ 3 & 1 & 4 \end{bmatrix}$$
We are at step 1, so we look at the first column: the largest element is the $$6$$ in the first row, so we don't need to swap rows. Applying Gaussian elimination gives:
$$\begin{bmatrix} 6 & 3 & 6 \\ 0 & 0 & 3\\ 0 & -1/2 & 1 \end{bmatrix}$$
the we begin step 2: the largest element in the second column (and row $$\ge 2$$) is the $$-1/2$$ on row 3, so we chose it as our next pivot. Note that in general, the $$k$$-th pivot is chosen at step $$k$$ and can't be computed beforehand (which leads to some interesting challenges with sparse matrices but I digress).
The chosen pivot is not on the second row, so we need to swap row 2 and 3, and then proceed with the elimination.
$$\begin{bmatrix} 6 & 3 & 6 \\ 0 & -1/2 & 1\\ 0 & 0 & 3 \end{bmatrix}$$
In this case, it turns out that the elimination is complete, nut in general the element in position (3,2) is not zero.
• I think I misinterpreted your comment on @Draditate's answer. We are not choosing the greatest element in the $k$th column of the original matrix $\mathbf{A}$, but rather from the work-in-progress matrix $\mathbf{U}$ with all previous row operations applied. – Will Aug 27 '20 at 15:05
• Yes, that is right, sorry for the confusion – Miguel Aug 27 '20 at 15:09
• For the OP, regarding @Miguel's digression, in sparse LU factorization, people sometimes use static pivoting where the reordering is chosen before the factorization is computed (doing this well is challenging). This is less accurate, but usually accurate enough that an accurate solution can be recovered using iterative refinement – eepperly16 Aug 27 '20 at 18:07
Partial pivoting is kind of 'dynamic', we should consider potential row swap for during each column pivoting.
From your matrix example, yes for the first column, no row swap needed, after row operation, your $$u_{2,2} = 0, u_{3,2}= -0.5$$, we need a row swap between row 2 and row 3! Updated to $$u_{3,2} = 0, u_{2,2}= -0.5$$ and $$l_{3,2} = 0$$.
• Only when the remaining element in column are all zero entry, this is when we cannot proceed any further (no matter usual LU or LU with partial pivoting). – Draditate Aug 27 '20 at 4:29
• OK, so it sounds like we must first calculate all possible $u_{i,k}$ values in column $k$ before deciding which one to use as the pivot, right? Algorithmically that seems like a big waste, unless I'm missing something. – Will Aug 27 '20 at 4:41
• An common strategy to chose the pivot at step k is to look under the current pivot location and take the element with the largest absolute value – Miguel Aug 27 '20 at 6:22
• @Miguel, that strategy is what led to the div-by-zero in the example above. Is that strategy supposed to work on all (nonsingular) matrices? – Will Aug 27 '20 at 14:22
• Yes, it works on all nonsingular matrices (I'm writing a more detailed answer) – Miguel Aug 27 '20 at 14:31 | 2021-05-08T02:01:36 | {
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http://math.stackexchange.com/questions/97838/dimension-of-mathbb-z-sqrt-5 | Dimension of $\mathbb Z[\sqrt 5]$
I know that every non-zero prime ideal in a Dedekind Domain is maximal. Since $\mathbb Z[\sqrt 5]$ is not integrally closed in its field of fractions $\mathbb Q[\sqrt 5]$, I wonder whether we can still show that non-zero primes are maximal.
More generally is it known for which of the square-free $D\in \mathbb Z$ there are non-zero primes of $\mathbb Z[\sqrt D]$ that are not maximal.
-
The ring $\mathbb{Z}[\sqrt{D}]$ is integral over $\mathbb{Z}$. Therefore a prime ideal $P$ of it is maximal if and only if the prime ideal $p:=P\cap\mathbb{Z}$ is maximal. Since all non-zero prime ideals of $\mathbb{Z}$ are maximal we get that all $P$ such that $P\cap\mathbb{Z}\neq 0$ are maximal. The latter condition is equivalent to $P\neq 0$, which proves the claim.
-
good ... i thought this property only holds for rings that are integrally closed over another ring but I looked it up, it is enough that it is integral. :) – Peter Patzt Jan 10 '12 at 11:13
Below is a very simple direct proof for any ring $\rm\:R\:$ of algebraic integers.
LEMMA $\$ If $\rm\ I \supsetneq P\$ are ideals of $\rm\:R\:,\:$ with $\rm\:P\:$ prime, then there is an integer $\rm\:f_k\in I\:$ but $\rm\:f_k\not\in P\:.$
Proof $\$ Choose $\rm\:\alpha\in I,\ \alpha\not\in P\:.\:$ Being an algebraic integer, $\rm\:f(\alpha) = 0\:$ for a monic $\rm\:f(x)\in \mathbb Z[x]\:,\$ $\rm\:f(x) \: =\: x^n +\cdots + f_1\ x +\: f_0\:.\:$ Note $\rm\:f_n = 1\not\in P\:.\:$ Let $\rm\:k\:$ be least with $\rm\:f_k\:\not\in P.\:$ $\rm\ f(\alpha) = 0\in P\:$ $\: \Rightarrow\:$ $\rm\:(\alpha^{n-k}+\cdots+f_k)\ \alpha^k\in P,\ \alpha\not\in P$ $\:\Rightarrow\:$ $\rm\:\alpha^{n-k}+\cdots+f_k\in P\subset I\:.\:$ So $\rm\ \alpha\in I$ $\rm\:\Rightarrow\:$ $\rm\:f_k \in I\:.\$ QED
COROLLARY $\:$ A proper chain of prime ideals in $\rm\:R\:$ cannot contract to a shorter such chain in $\rm\:\mathbb Z\:.$
REMARK $\$ Alternatively, reduce to the simpler case $\rm\:P = 0\:$ by way of factoring out the prime $\rm\:P\:.\:$ Then $\rm\:f_k\:$ is the constant term of a minimal polynomial for $\rm\:\alpha\:$ over a domain, so $\rm\:f_k\ne 0,\:$ i.e. $\rm\:f_k\not\in P\:,\:$ which is explained in much detail in this post, as a generalization of rationalizing denominators.
-
Wonderful!!! ${}$ – Pierre-Yves Gaillard Jan 11 '12 at 5:49
To answer the question as asked, the simplest seems to prove this.
If $P$ and $I$ are ideals of $\mathbb Z[\sqrt5]$, and if $P$ is prime, then $$P < I\quad\implies\quad\mathbb Z\cap P\ < \ \mathbb Z\cap I,$$ where $<$ means "properly contained in".
To see this, set $x=a+b\sqrt5$ with $a,b\in\mathbb Z$, and let $x$ be in $I$ but not in $P$. Putting
$$x':=a-b\sqrt5,\quad t:=x+x'=2a,\quad n:=xx'=a^2+5b^2,$$ we have $$x^2-tx+n=0.$$ If $n$ is not in $P$, we are done because $n$ is in $\mathbb Z\cap I$. If $n$ is in $P$, then so is $x-t$, and $t$ is in $\mathbb Z\cap I$ but not in $P$.
More generally, if $B$ is a (commutative) ring, if $A$ is a subring, if $B$ is integral over $A$, if $P$ and $I$ are ideals of $B$, and if $P$ is prime, then $$P < I\quad\implies\quad A\cap P\ < \ A\cap I.$$ Indeed, on taking the quotient by $P$, we can assume that $B$ is a domain, and that $P=0$. Let $x$ be a nonzero element of $I$, and let $f\in A[X]$ a monic polynomial of least degree such that $f(x)=0$. Then the constant term of $f$ is a nonzero element of $I\cap A$.
In view of the second Corollary on page 7 of the text [1] by Mel Hochster, if a ring $B$ is integral over its subring $A$, then $A$ and $B$ have the same Krull dimension.
This shows in particular that if a ring $A$ is between $\mathbb Z$ and the ring of integers of a number field, then $A$ has Krull dimension $1$, that is, its nonzero primes are maximal.
More generally if the ring extension $\rm\: R \subset T\:$ satisfies GU (going-up) and INC (incomparable) then $\rm\:dim\ R\ =\ dim\ T\:.\:$ For a proof see Kaplansky, Commutative Rings, Theorem $48$. – Bill Dubuque Jan 10 '12 at 18:13 | 2013-12-21T01:44:00 | {
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http://mathhelpforum.com/trigonometry/51739-trig-problem-where-am-i-going-wrong.html | # Math Help - Trig Problem, where am I going wrong?
1. ## Trig Problem, where am I going wrong?
Hello everyone, I'm new to these forums. The long story short is I am trying to finish high school via correspondence, and I need to do two math courses, each containing four units. I am on the fourth unit, the first three gave me very few problems. Anyways, here is the problem:
2cos^2x+cosx=0 Solve for the domain (0, 2pi) to nearest hundredth of a radian.
I did this:
Let k=cosx
2k^2+k
=k(2k+1) or
=cosx(2cosx+1)
cosx=0
x=cos^-1(0)
x=1.57 (this is pi/2 or 90degrees)
CAST RULE:
pi-1.57=4.71
pi+1.57=4.71
2pi-1.57=4.71
Therefor for the first zero, x=1.57, 4.71
For the second zero:
2cosx+1=0
2cosx=(-1)
cosx=(-1/2)
x=cos^-1(-1/2)
x=2.09 (this is 2pi/3 or 120degrees)
CAST RULE: since cosx is negative, the angle is in quadrants two and three:
pi-2.09=1.05
pi+2.09=5.23
So we have 1.57, 4.71, 1.05, 5.23
-------------------------------------
However when I check my answer on mathway, it shows the solutions as:
1.57 (correct)
4.71 (correct)
2.09 (?)
4.19 (?)
I don't understand why my calculations for the second zero are wrong. If I use quadrants ONE and FOUR (where cosx should be positive), I get 2.09 and 4.19, which is what mathway shows as the correct answers.
Can someone tell me where I'm going wrong here? The second zero calculates to cosx=(-1/2), and cosx is negative in quadrants two and three, not one and four ... I don't see my error.
2. Hello, Jeev!
Welcome aboard!
$2\cos^2\!x+\cos x\:=\:0$
Solve for the domain $(0,\:2\pi)$ to nearest hundredth of a radian.
Factor: . $\cos x(2\cos x + 1) \:=\:0$
Then: . $\begin{array}{ccccccccccc}\cos x \:=\:0 & \Rightarrow &\boxed{x \:=\:\dfrac{\pi}{2},\:\dfrac{3\pi}{2}} \\ \\[-3mm]
2\cos x + 1 \:=\:0 & \Rightarrow & \cos x \:=\:\text{-}\frac{1}{2} & \Rightarrow &\boxed{ x \:=\:\dfrac{2\pi}{3},\:\dfrac{4\pi}{3}} \end{array}$
3. Thanks for giving me the solutions, but I am not understanding how you get the last two.
The CAST rule states that when cosx<0, the angle is in quadrants two and three, which would mean:
(pi) - 2.09
(pi) + 2.09
All I need to understand is, for the second zero, if cosx=(-1/2), why are we calculating the angle in quadrants one and four?
4. Originally Posted by Jeev
Thanks for giving me the solutions, but I am not understanding how you get the last two.
The CAST rule states that when cosx<0, the angle is in quadrants two and three, which would mean:
(pi) - 2.09
(pi) + 2.09
All I need to understand is, for the second zero, if cosx=(-1/2), why are we calculating the angle in quadrants one and four?
When you put the equation into the calculator, it realised that you've said the angle is negative and calculated the angle in the second quadrant for you.
To do it analytically you need to consider the POSITIVE angle (called the FOCUS angle) and make the adjustments using the CAST (or as I call it, All Students Talk Crap) rule.
Consider $\cos{\theta} = \frac{1}{2}$
Then your focus angle $\theta = \arccos{\frac{1}{2}} = \frac{\pi}{3}$.
Now, since we are told the angle is negative, we need to consider the 2nd and 3rd quadrants.
So our angles are $\pi - \frac{\pi}{3}$ and $\pi + \frac{\pi}{3}$. This gives us our angles, $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$.
Understand now?
5. Okay, let me see if I got this.
cosx(2cosx+1)
For 2cosx+1
cosx=(-1/2)
We need to calculate the "focus angle" which is
x=arccos(1/2)
x=1.04 (the question wants radians to the nearest hundredth)
NOW at this point we consider the negative sign, which dictates that the angles are in quadrants II and IV:
(pi)-1.04=2.10
(pi)+1.04=4.18
Is this the right way to go about this?
6. Originally Posted by Jeev
Okay, let me see if I got this.
cosx(2cosx+1)
For 2cosx+1
cosx=(-1/2)
We need to calculate the "focus angle" which is
x=arccos(1/2)
x=1.04 (the question wants radians to the nearest hundredth)
NOW at this point we consider the negative sign, which dictates that the angles are in quadrants II and IV:
(pi)-1.04=2.10
(pi)+1.04=4.18
Is this the right way to go about this?
Yes, but don't do any rounding until the very last step. Otherwise you get roundoff errors. You said the answers were 2.09 and 4.19 after all...
7. Then your focus angle .
How do you get radians as a fraction of pi as opposed to a decimal value? When I type this in my calculator it obviously just gives me the decimal. Is there some sort of conversion factor I can use?
8. Originally Posted by Jeev
Then your focus angle .
How do you get radians as a fraction of pi as opposed to a decimal value? When I type this in my calculator it obviously just gives me the decimal. Is there some sort of conversion factor I can use?
Haven't you been taught about the special triangles yet?
Think of an isoceles right angle triangle, with the shortest sides of unit length. Obviously the angles are $45^\circ, 45^\circ, 90^circ$.
Using Pythagoras, the hypotenuse can then be calculated to be $\sqrt{2}$ units.
Since $\sin{45^\circ} = \frac{O}{H}$ and $\cos{45^\circ} = \frac{A}{H}$, we find
$\sin{45^\circ} = \cos{45^\circ} = \frac{1}{\sqrt{2}}$.
By rationalising the denominator, we could also write this as $\frac{\sqrt{2}}{2}$.
Also, $\tan{45^\circ} = \frac{O}{A} = \frac{\sqrt{2}}{\sqrt{2}} = 1$.
Make the appropriate conversions for radian measures.
Now for the other triangle.
Think of an equilateral triangle of side length 2 units. Obviously all the angles are $60^\circ$.
Draw a line perpendicular to the base which cuts the triangle in half and bisects the top angle.
Now we have two $30^\circ, 60^\circ, 90^\circ$ triangles.
The hypotenuse of each is 2 units. One of the shorter lengths is 1 unit. By Pythagoras, we find the other shorter length to be $\sqrt{3}$ units.
What can you tell me about $\sin{30^\circ}, \cos{30^\circ}, \tan{30^\circ}, \sin{60^\circ}, \cos{60^\circ}$ and $\tan{60^\circ}$? Make appropriate conversions for radian measure.
9. Ah so the cos60deg = 1/2
So we are converting 60deg to radian form, so
1deg=(pi)/180
60deg=60(pi)/180
60deg=(pi)/3
I understand this now, thanks so much! | 2014-09-30T07:19:00 | {
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http://fille-facile.fr/b2/jeep/4246636409674d1e83489eaeeb03e-clockwise-rotation-formula | clockwise - in the direction that the hands of a clock move; "please move clockwise in a circle" anticlockwise, counterclockwise - in a direction opposite to the direction in which the hands of a clock move; "please move counterclockwise in a circle!". 2 A (5, 2) Graph A(5, 2), then graph B, the image of A under a 90 counterclockwise rotation about the origin. Rotating a shape 90 degrees is the same as rotating it 270 degrees clockwise.
90 Degree Clockwise Rotation - Rule - Examples with step by step explanation. We will take the general equation of the conic, apply the clockwise rotation formula and evaluate the variable(s) that will make the coefficient of the xy equal 0. R 2 (1;1) is the point in the plane obtained by rotating (1;1) clockwise by an angle of Counterclockwise Rotation. 90). Solution : Step 1 : Here, triangle is rotated 270 clockwise. So, each point has to be rotated and new coordinates have to be found. \end{pmatrix}$is called the rotation matrix. We can rotate the angle () by rotating a point around the x-axis. Answer: To rotate the figure 90 degrees clockwise about a point, every point (x,y) will rotate to (y, -x). To perform the calculation, enter the rotation angle. If a line has slope m, then it advances m units in the y direction for each 1 unit advanced in the x direction. The sum of the measures is 360. Or past the 90 degree, then 180, then 270 degree marks. The clockwise movement usually starts from the top going around to the right, then to the bottom, to the left, and then back to the top position. Lets take a look at another rotation. To find the clockwise rotation matrix, you can do the calculations again. Rotation Formula. In the following diagram, we are rotating the 3*3 matrix by 90 degrees clockwise. Then click the button 'Calculate'. Rotation matrix From Wikipedia, the free encyclopedia In linear algebra, a rotation matrix is a matrix that is used to perform a rotation in Euclidean space. So, the rule that we have to apply here is (x, y) ----> (-y, x) Step 2 : For example 90 means 90 degrees. ram 1500 body lift and leveling kit; personal branding session; texas motor speedway infield; greenfield village museum; liverpool summer camp 2022; custom women's quarter zip; willowbrook football schedule; The axis can be either x or y or z. 1. Therefore we can represent the direction of the line using the 2D vector v= (1, m). Yes, both are different but the formula or rule for 180-degree rotation about the origin in both directions clockwise and anticlockwise is the same. The rotation is also known as counterclockwise (ACW). Draw an image of a polygon with vertices A(2,2), B(4,3), C(4,5) and D(1,4). For each square cycle, we are going to swap the elements that are involved with the corresponding cell in the matrix in the clockwise direction. As sin (-) = -sin and cos (-) = -cos Examples. City of a Million Dreams The Film; City of a Million Dreams The Book; Render Unto Rome; Earl Long in Purgatory; Up from the Cradle of Jazz; Vows of Silence The Film; Vows of Silence The Book; Last of the Red Hot Poppas; This is accomplished by eliminating the xy term. OR you can just transpose the above matrix OR you can substitute$- \theta$into the matrix (see note below). For a point the Angular Displacement is as follows: Angular Displacement = f-i The Displacement will have both magnitudes as well as the direction. Note the corresponding clockwise and counterclockwise rotations. In the following diagram, we are rotating the 3*3 matrix by 90 degrees clockwise. The formula is similar to 90 degree anticlockwise rotation. CCSS: 8.G However, in many situations you might want to rotate around an arbitrary axis/vector. Its determinant is 1. In the figure above, the wind rotates the blades of a windmill. Why Buy from Del Mar Fans and Lighting?Returns for 365 days.Free shipping on all orders of$40+.A safe shopping experience. Using the rotation formula, After rotation of 270(CW), coordinates of the point (x, y) becomes: (-y, x) Hence point A(x, y) will have the new position at (-9, -7) if the point was initially at (7, -9) Therefore, the coordinates of point A are (7, -9). A great circle transforms to another great circle under rotations, leaving always a diameter of the sphere in its original position The general rule for a rotation by 180 about the origin is (A,B) (-A, -B) Rotation by 270 about the origin: R (origin, 270) If you know the diameter or radius of (x, y) rotated 90 degrees around (0, 0) is (-y, x). Clockwise is usually abbreviated as CW. The 22 rotation matrix corresponds to a 90 planar rotation. A clockwise rotation is a negative angle with respect to the reference.
You should assume this, unless it is noted in the problem that you need to rotate clockwise. +1. What are the rules for rotation? When we rotate a figure of 180 degrees about the origin either in the clockwise or counterclockwise direction, each point of the given figure has to be changed from (x, y) to (-x, -y) and graph the rotated figure. In the case of the above example where f(x) = x, lets try to rotate this function by 90 clockwise, that is, by = -90 counterclockwise.. Then cos() = 0 and sin() = -1 and we get: The 2D rotation matrix is [x' y' ] = Cos -Sin Sin Cos [x y ] Sample Questions In linear algebra, a rotation matrix is a transformation matrix that is used to perform a rotation in Euclidean space.For example, using the convention below, the matrix = [ ] rotates points in the xy plane counterclockwise through an angle with respect to the positive x axis about the origin of a two-dimensional Cartesian coordinate system. Lets understand the rotation of 90 degrees clockwise about a point visually. Figure 1 Correct connection of motor phases for clockwise rotation. Rotation about the carbon-carbon bond, however, results in many different possible molecular conformations.
The movement in the counterclockwise direction, starts from the top, heads to the right, goes down, then follows to the right side, and ends up at the top position. We will take the general equation of the conic, apply the clockwise rotation formula and evaluate the variable(s) that will make the coefficient of the xy equal 0. What is the rule for rotating 180 degrees clockwise? Counterclockwise abbreviated as CCW. clockwise rotation matrix formula Jason Berry. What is the formula for a 270 degree counterclockwise rotation?
ram 1500 body lift and leveling kit; personal branding session; texas motor speedway infield; greenfield village museum; liverpool summer camp 2022; custom women's quarter zip; willowbrook football schedule; Rotation Formula How do you rotate 90 degrees clockwise around a point? Is this a beautiful formula or what?. A product of the aforementioned matrices should be enough if you only need rotations around cardinal axes (X, Y or Z) like in the question posted.
During the de-energized switching interval, the rotor falls back against the rotating field of the power supply.
Below are two examples.
180 Degree Rotation about the Origin - Concept - Examples with step by step explanation. A rotation is a type of rigid transformation, which means that the size and shape of the figure does not change; the figures are congruent before and after the transformation. Rotation Formula Rotation can be done in both directions like clockwise as well as in counterclockwise. The vector (1,0) rotated +90 deg CCW is (0,1). Solution: Using the rotation formula, After rotation of 90(CCW), coordinates of the point (x, y) becomes: (-y, x) Hence the point K(5, 7) will have the new position at (-7, 5) Answer: Therefore, the coordinates of the image are (-7, 5). Hence, the clockwise rotation matrix is: \$\begin{pmatrix} \cos \theta & \sin \theta \\ A Counter-Clockwise Star Clock. This is important because if you rotate an image , the height and width boundaries due not change. 3. A point P has coordinates ( x, y) with respect to the original system and coordinates ( x, y) with respect to the new system. Rotation matrix From Wikipedia, the free encyclopedia In linear algebra, a rotation matrix is a matrix that is used to perform a rotation in Euclidean space. Learn Ethane Structure, Molecular Mass, Chemical Formula here.
The unit of measurement for the angle can be switched between degrees or radians. What is the rule for rotating 90 degrees counterclockwise? The general rule for rotation of an object 90 degrees is (x, y) --------> (-y, x). You can use this rule to rotate a pre-image by taking the points of each vertex, translating them according to the rule , and drawing the image. Also, check its expanded formula here. 90 Degree Counter Clockwise Rotation Formula Full Rotation Or Is clockwise right or left A clockwise (typically abbreviated. The rule for a rotation by 180 about the origin is (x,y)(x,y). 22 g/mol. Rule. The opposite direction of clockwise is anticlockwise or counterclockwise (both words mean the same). Read More. Equations 1 and 2 show the right way to rotate a point around the origin: x1 = x0 cos ( ) y0 sin ( ) (Equation 1) y1 = x0 sin ( ) + y0 cos ( ) (Equation 2) If we plug in our example point of ( x0, y0) = (4, 3) and = 30, we get the answer ( x1, y1) = (1.964, To anyone rotating a 2D vector for a computer screen: this answer assumes the The rule for a rotation by 270 about the origin is (x,y)(y,x) . Formula 1 : n(A u B) = n(A) + n(B) - n(A n B) If A and B are disjoint sets, n(A n B) = 0. The cycle is formed by its first row, last column, last row, and last column. We can rotate them back to their normal orientation so the equation is in standard form. The X,Y equations listed are for CW rotations but the calculator tells you to define CCW as positive. What is the formula for a 270 degree counterclockwise rotation? The point about which the object is rotating, maybe inside the object or anywhere outside it. Pentane Formula: Know what will be the formula and structure of pentane. In addition, the Earths daily rotation around its axis makes it seem like that the stars close to the celestial poles revolve around the pole once each day. Rotation Formula. The PDF worksheets are specially designed for 7th grade and 8th grade students. A clockwise rotation on the row vectors will correspond to a counterclockwise rotation on the column vectors. When we rotate the line by an angle d, we apply a 2D rotation matrix M (d) = [ (cos (d), -sin (d)) (sin (d), cos (d)) ] to the line direction vector.
The active rotation (rotate object) or the passive rotation (rotate coordinates) can be calculated. Let us look at solved examples for better understanding of the concept. What way is clockwise? Rotation Formula: The Rotation fo rmula can be used to c alculate the coordinates of the Cartesian coordinates or in the x-y-plane. We can rotate them back to their normal orientation so the equation is in standard form. Because a rotation in the plane is totally determined by how it moves points on the unit circle, this is all you have to understand. Explanation for Clockwise rotation: A given N x N matrix will have (N/2) square cycles. Rotating a vector 90 degrees is particularily simple. A rotation of axes in more than two dimensions is defined similarly. rotation formula clockwise. In the orignal shape (preimage), the order of the letters is ABC, going clockwise. polygonFormed, optionsRemaining){. If you want to rotate clockwise, you simply do it the other way around, getting (y, -x). The most common convention these days is to primarily use column vectors, with the matrices on the left, but e.g. The direction of rotation may be clockwise or anticlockwise. About; Works. The sum of the measures is 360. In mechanics and geometry, the 3D rotation group, often denoted SO(3), is the group of all rotations about the origin of three-dimensional Euclidean space under the operation of composition.. By definition, a rotation about the origin is a transformation that preserves the origin, Euclidean distance (so it is an isometry), and orientation (i.e., handedness of space). Herstein "Topics in Algebra" uses row Is a 270 clockwise rotation the same as a 90 counterclockwise rotation? Available on Amazon. the different spatial arrangements of hydrogen atoms are readily interconvertible by rotation about $${\rm{C}} {\rm{C}}$$ single bonds resulting in Conformational isomerism. For example the matrix rotates points in the xy-Cartesian plane counterclockwise through an angle about the origin of the Cartesian coordinate system. 90 Degree Clockwise Rotation - Rule - Examples with step by The rule for a rotation by 270 about the origin is (x,y)(y,x) . Rules For Rotating Clockwise and Counterclockwise on a graph Learn with flashcards, games, and more for free. Example 1: Find the position of the point K(5, 7) after the rotation of 90(CCW) using the rotation formula. Simple and beautiful. 180 Degree Rotation about the Origin. rotation formula clockwise. This results in some of the image being cropped off and areas of black where the image no longer resides. Rules of Rotation The general rule for rotation of an object 90 degrees is (x, y) > (-y, x). We can rotate an object in two ways-Clockwise: An object rotates clockwise if the value of the Rotation angle is negative (-). axis-angle formula) is a commonly prescribed solution to this problem. Proof. Is turning 180 degrees clockwise different from turning 180 degrees counterclockwise? Rotation angle is backwards. What is Counterclockwise? For a rotation r O of 90 centered on the origin point O of the Cartesian plane, the transformation matrix is [ What is the rule for 180 Rotation? The Rodrigues' formula (a.k.a. For 2D we describe the angle of rotation, but for a 3D angle of rotation and axis of rotation are required. Rotation Point: It is also called the Pivot point. Read More.
To perform the rotation, the position of each point must be represented by a In simple words, we can say that we are rotating the matrix from left to right. How to rotate figures about the origin, examples and step by step solution, Rotation of 90, 180, 270 degrees about the origin, patterns on the coordinates, High School Math Graphing and Describing Rotations Rotate 90 degrees counter-clockwise. Jul 04, 22 01:07 AM. Counterclockwise is the opposite sense of clockwise rotation. 90), and clockwise if is negative (e.g. Rotation Formula: Rotation can be done in both directions like clockwise and anti-clockwise.
We can move the object in the clockwise and in the anticlockwise directions. To perform the rotation, the position of each point must be represented by a Ethane Formula CH is a organic compound found in natural gas. It is moving of an object about an angle. 90 Degree Clockwise Rotation. | 2022-10-07T20:02:37 | {
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https://math.stackexchange.com/questions/3103002/4-couples-and-4-single-people-seated-at-3-round-tables | # 4 couples and 4 single people seated at 3 round tables
In how many ways can you seat the 12 people at 3 round tables such that:
A) All couples are seated together. (the two members of each couple sit side-by-side)
B) No couples sit together.
I've tried this question several ways, but keep getting different answers.
Since it is a round table problem, I've set the first seat arbitrarily and proceeded from there. Any help would be greatly appreciated.
Edit: For (A), I found that any two couples can be seated side-by-side 2 ways, thus I got 4*(4 choose 2) = 24 ways to seat the couples at two tables. Moreover, I found 6 ways of seating the 4 single people at the third table (by placing one person arbitrarily in the first seat, leaving 3 options for the second seat, 2 for the third, and 1 for the last).
• What do you mean by no couples sit together? Feb 6, 2019 at 20:30
• The question was given ambiguously, but I'm assuming that one of the tables for (A) seats the 4 single people. Feb 6, 2019 at 20:32
• Have you tried using the principle of inclusion exclusion for (B)? Feb 6, 2019 at 20:40
• I have, but I am unsure which sets I should use. I got 11! - 7!, which I believe is incorrect. Feb 6, 2019 at 20:42
• It seems like the problem was pretty clear that "all couples sit together" means that "for each couple, the two in that pair are next to each other." The only other reasonable interpretation of "all together" would be "all at the same table," which is impossible. Feb 6, 2019 at 20:51
For A you can either seat two couples at each of two tables and four singles at the third or seat two couples at one table and one couple and two singles at the other two. For the first there are three ways to split the couples, so $$3*4*4*6=224$$ ways to seat them. For the second there are three ways to pick the two couples, six ways to match two singles with the third, so $$3*6*4*4*4=896$$ ways to seat them. The total is $$1120$$ | 2022-08-15T12:32:41 | {
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https://math.stackexchange.com/questions/2170945/what-is-the-mathematical-notation-for-a-group-of-pairwise-disjoint-sets | # What is the mathematical notation for a group of pairwise disjoint sets?
Assume I have 4 sets: $A$, $B$, $C$, and $D$.
What is the mathematical notation to indicate that all of these sets are pairwise disjoint?
I can obviously type:
$$A \cap B = \emptyset$$ $$A \cap C = \emptyset$$
etc.
But I am looking for a more succinct and mathematically correct way of expressing this.
• Usually people just say "pairwise disjoint". I don't know of any standard notation for this. – ChocolateAndCheese Mar 4 '17 at 0:45
• You could just say "The sets are pairwise disjoint". Another way is if you have sets $A_1,A_2,...,A_n$, then you could write $A_i\cap A_j=\emptyset$ when $i\neq j$ for $i,j\in\{1,...,n\}$. – Dave Mar 4 '17 at 0:47
• Also, if the sets are non-empty and you already have a symbol for their union, say $S=A \cup B \cup C \cup D$ then you can say that the sets form a partition of $S$ – fiftyeight Jul 22 '17 at 8:07
• Both of the suggestions above are excellent, thank you. – Kostas Jul 23 '17 at 14:13
If you have indexed family $\{A_i\}_{i\in I}$, then you can write $A_i\cap A_j=\emptyset$, $i\neq j$. If not, I don't see any problem in just stating: "Let $A,B,C,D$ be pairwise disjoint." This will probably be more clear than any notation you could come up with, including my own suggestion. | 2020-06-01T23:06:20 | {
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https://mathoverflow.net/questions/137059/the-singular-values-of-the-hilbert-matrix | The singular values of the Hilbert matrix
The $n\times n$ Hilbert matrix $H$ is defined as follows
$$H_{ij} = \frac{1}{i+j-1}, \qquad 1\leq i,j\leq n$$
What is known about the singular values $\sigma_1 \geq \cdots \geq \sigma_n$ of $H$?
For example, it is known that the matrix is very ill-conditioned, i.e., [1]
$$\dfrac{\sigma_1}{\sigma_n} = \mathcal O \left( \frac{1+\sqrt{2})^{4n}}{\sqrt{n}} \right)$$
But are there estimates for $\sigma_k$ for $2\leq k\leq n-1$? Or is there a bound on $1\leq k\leq n$ such that $\sigma_k > \epsilon$ for some $\epsilon>0$? I am interested in this problem because I would like to know the numerical rank of $H$.
[1] J. Todd, "The condition of the finite segments of the Hilbert matrix." Contributions to the solution of systems of linear equations and the determination of eigenvalues, 39 (1954), pp. 109-116.
• You probably meant $\sigma_k > \epsilon$, no? (also, note that since the Hilbert matrix is symmetric positive definite, its singular values are the same as its eigenvalues. – Suvrit Jul 18 '13 at 17:45
• Thanks, corrected. Yes, I agree that the singular values of H are the same as the eigenvalues of H (after ordering). – alext87 Jul 18 '13 at 17:54
While not a full characterization, the following result on $\sigma_n$ shows that $\sigma_k > \epsilon_n$, since $\sigma_n \ge \epsilon_n$.
Following my answer on this MO question here, we see that
\begin{equation*} \sigma_n \ge \frac{\det(H)}{\left[m + s/\sqrt{n-1}\right]^{n-1}} =: \epsilon_n, \end{equation*} where \begin{eqnarray*} m &=& \tfrac{\text{trace} H}{n} = \tfrac{1}{n}\left(\tfrac{H_{n-1/2}}{2} + \log 2\right),\\ s &=& \sqrt{\tfrac{\text{trace}H^2}{n} - m^2}, \end{eqnarray*} where $H_n$ is the nth Harmonic number).
Since it is known that $\det(H) = \frac{c_n^4}{c_{2n}}$, where $c_n := \prod_{i=1}^{n-1}i!$, the above bound on $\sigma_n$ is fully determined (I'm leaving an analytic / "closed-form" evaluation of $\text{trace}H^2$ as an exercise).
• Notice that $tr(H^2) = \sum_{ij} h_{ij}^2$, and curiously a week or so ago, other aspects of the matrix $[h_{ij}^2]$ were discussed at mathoverflow.net/questions/136306/… – Suvrit Jul 19 '13 at 0:20
• This is a nice lower bound on the smallest singular value, and I enjoyed the post on $H^2$. Thank you very much! Your answer shows when $H$ is numerically rank deficient, but not the numerical rank of H. Still this is a good step in that direction. – alext87 Jul 25 '13 at 8:22
I came back to this a few months ago and I can now answer my own question. I hope it is appropriate to answer my own question given the length of time.
Bernhard Beckermann and I just submitted a paper [1] that shows that if $AX-XB = F$ with $A$ and $B$ normal matrices, then the singular values of $X$ can be bounded as $$\sigma_{1+\nu k}(X) \leq Z_k(\sigma(A),\sigma(B))\|X\|_2, \qquad \nu = {\rm rank}(F),$$ where $\sigma(A)$ and $\sigma(B)$ are the spectra of $A$ and $B$, respectively, and $Z_k(E,F)$ is the so-called Zolotarev number. More precisely, $$Z_k(E,F) = \inf_{r\in\mathcal{R}_{k,k}} \frac{\sup_{z\in E}|r(z)|}{\inf_{z\in F}|r(z)|},$$ where $\mathcal{R}_{k,k}$ is the space of degree $(k,k)$ rational functions. This is called the third Zolotarev problem.
In the case of the Hilbert matrix, let $A = {\rm diag}(1/2,3/2,\ldots,n-1/2)$ and $B = -A$ then $$AH - HB = \begin{bmatrix}1\\\vdots\\1 \end{bmatrix}\begin{bmatrix}1&\cdots&1 \end{bmatrix}.$$ Hence, $\nu=1$ and $$\sigma_{1+k}(H) \leq Z_k([-n+1/2,-1/2],[1/2,n-1/2])\|H\|_2.$$ Fortunately, bounds on the Zolotarev numbers are well-studied (in [1] we give asymptotically tight bounds) and we conclude that $$\sigma_{1+k}(H) \leq 4\left[{\rm exp}\left(\frac{\pi^2}{2\log(8n-4)}\right)\right]^{-2k}\|H\|_2.$$ If we seek the numerical rank of $H$ for some $0<\epsilon<1$, i.e., $k = {\rm rank}_\epsilon(H)$ if $k$ is the smallest integer such that $\sigma_{k+1}(H)\leq \epsilon\|H\|_2$, then from the bound on the singular values of $H$ we conclude that $${\rm rank}_\epsilon(H) \leq \Bigg\lceil \frac{\log(8n-4)\log(4/\epsilon)}{\pi^2} \Bigg\rceil.$$
This shows that Hilbert matrices are not only exponentially ill-conditioned with $n$, but its singular values decay geometrically to zero too. This methodology extends to any matrix with displacement structure such as Pick, Cauchy, Loewner, real Vandermonde, and positive definite Hankel matrices. For more details, see [1].
• Interesting! Though these are upper bounds, not lower bounds as you asked in the original question, no? – Suvrit Oct 1 '16 at 20:07
• It was stated a little backwards by my younger self: it's still about the numerical rank given by ${\rm rank}_\epsilon(H)$. The answer above gives an upper bound on ${\rm rank}_\epsilon(H)$ and hence $k$ (in my original question). Suppose that $k$ is an integer such that $\sigma_{k}(H)>\epsilon$ for some $0<\epsilon <1$, then from above and $\|H\|_2\leq \pi$ we know that $k \leq \lceil \log(8n-4)\log(4\pi/\epsilon)/\pi^2\rceil+1$. – alext87 Oct 1 '16 at 20:40
• I do not know how to get lower bounds on the singular values. This may involve developing lower bounds for so-called discrete Zolotarev numbers (if one goes for the above methodology). I am not sure if this is possible at the moment. For more details on discrete Zolotarev numbers, see: link.springer.com/article/10.1007/s00365-010-9087-6 – alext87 Oct 1 '16 at 20:42 | 2020-09-22T21:03:14 | {
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http://mathematica.stackexchange.com/questions/27375/how-to-integrate-a-function-over-a-3d-planar-polygon | How to integrate a function over a 3D planar polygon?
I am trying to integrate a function over a planar polygon in 3D. In 2D, this is fairly straightforward, using either answer from this question (I use the second answer). If we use an equilateral triangle centered at the origin as our polygon:
f[pos_List] := 1
basetri={{1/2,- (1/6) Sqrt[3],0},{0,1/3 Sqrt[3],0},{-(1/2),- (1/6) Sqrt[3],0}}
inPolyQ[poly_,pt_]:=2.π==Total[VectorAngle@@@Transpose@{#,RotateRight[#]}]&@(#-pt&/@poly)
we can integrate our function f (in this case, this function will return the area of the triangle):
NIntegrate[
f[{x,y,0}] Boole[inPolyQ[basetri,{x,y,0}]],
{x,Min@basetri[[;;,1]],Max@basetri[[;;,1]]},
{y,Min@basetri[[;;,2]],Max@basetri[[;;,2]]}
]
(* Out[]:= 0.43265 *)
which is pretty close to the exact value of Sqrt[3]/4 (~0.433013). (Yes, the are a bunch of errors thrown.) This can also be done over infinite bounds (see this answer):
NIntegrate[f[{x,y,0}] Boole[inPolyQ[basetri,{x,y,0}]],
{x,-Infinity,Infinity},{y,-Infinity,Infinity}
]
(* Out[]:= 0.433321 *)
which is also pretty close to the exact value. (I'll leave alone for now the question of which bounds should give better results...)
The problem arises when doing this in 3D:
NIntegrate[
f[{x,y,z}] Boole[inPolyQ[basetri, {x, y, z}]],
{x, -Infinity,Infinity}, {y, -Infinity, Infinity}, {z, -Infinity, Infinity}
]
(* Out[]:= 0 *)
It gives this error:
NIntegrate::izero: Integral and error estimates are 0 on all integration subregions. Try increasing the value of the MinRecursion option.
Increasing MinRecursion doesn't help. I suspect that because this is a planar polygon, it has trouble figuring out exactly where that infinitesimally-thin plane is that satisfies the inequality actually is.
So, how can I integrate a function over an arbitrarily-oriented planar polygon in 3D? For what it's worth, at the very least it would be an acceptable answer for my current problem if it just works with a triangular region (in 3D), but I figured I'd leave the question more general if there's a way to do it.
-
Even if it's a 2D planar polygon in 3D space, it's still a 2D object. 2D objects are always 0 under a 3D measure.
Here we go through two methods of performing the scalar surface integral with a random polygon example.
Say we have a general 3D polygon defined as this:
polygonPts3D = {
{-0.902757, -0.116805, 0},
{0.203504, -0.972294, 0},
{0.849893, 0.414192, 0},
{0.374057, 0.835407, 0},
{-0.907079, 0.352119, 0}
};
and a scalar function defined in $xyz$ space:
Clear[f]
f[x_,y_,z_] := x^3-y^2+z
The standard way:
It's easy to derive the equation of the planar (planeEq) which the polygon is on:
Clear[normFunc]
normFunc[pts_] := Mean[
Normalize[Cross[##]] & @@@
Partition[
Subtract @@@ Partition[pts, 2, 1, 1],
2, 1, 1]
]
norm = normFunc[polygonPts3D]
planeEq = norm.({x, y, z} - Mean[polygonPts3D]) == 0
ContourPlot3D[
Evaluate[planeEq],
{x, -1, 1}, {y, -1, 1}, {z, -1, 1},
RegionFunction -> Function[{x, y, z, func},
inPolyQ[ReplacePart[polygonPts3D, {_, 3} :> 0], {x, y, 0}]
],
Mesh -> 20,
MeshFunctions -> Function[{x, y, z, func}, f[x, y, z]],
ColorFunction -> Function[{x, y, z, func}, Hue[f[x, y, z]]],
ColorFunctionScaling -> False,
AxesLabel -> (Style[#, Red, Bold, 20] & /@ {x, y, z})
]
Thus the standard way to perform the surface integral
$$\int_P f(x,y,z)\,\mathrm{d}S = \int_{P_{xy}} f(x,y,z(x,y)) \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2} \,\mathrm{d}x\mathrm{d}y$$
would be straightforward in Mathematica
zFunc[x_, y_] := Evaluate[z /. Solve[planeEq, z][[1]]]
1/norm[[3]] NIntegrate[
f[x, y, zFunc[x, y]]*
Boole[inPolyQ[ polygonPts3D, {x, y, zFunc[x, y]} ]],
{x, -∞, ∞}, {y, -∞, ∞}
] // Quiet
0.484606
The other way:
Also, we can establish a new coordinate system $x'y'z'$ such that the polygon lies on plane of $z'=0$, so we can convert the surface integral in $xyz$-space to a 2D area integral in $x'y'$-space.
To achieve that, we need a rotation which convert the norm of the polygon to $z'$ axis:
transMatrix = RotationMatrix[{
normFunc[polygonPts3D],
{0, 0, 1}
}]\[Transpose]
transMatrixInverse = Inverse@transMatrix
So the polygon in $x'y'z'$ should be:
polygonPts2D = ReplacePart[polygonPts3D.transMatrix, {_, 3} :> 0]
and the distance from the origin to the polygon plane:
distance = norm.Mean[polygonPts3D]
It can be seen that the transformed polygon is essentially the same with the original one:
RegionPlot[
Evaluate[inPolyQ[polygonPts2D, {x, y, 0}]],
{x, -1.2, 1.2}, {y, -1.2, 1.2},
MeshFunctions -> Function[{x, y, z, func},
f @@ ({{x, y, distance}}.transMatrixInverse)[[1]]
],
Mesh -> 20,
ColorFunction -> Function[{x, y, z, func},
Hue[f @@ ({{x, y, distance}}.transMatrixInverse)[[1]]]
],
ColorFunctionScaling -> False,
FrameLabel -> (Style[#, Red, Bold, 20] & /@ {x, y})
]
And the integral is:
NIntegrate[
f @@ ({{x, y, distance}}.transMatrixInverse)[[1]]*
Boole[inPolyQ[polygonPts2D, {x, y, 0} ]],
{x, -∞, ∞}, {y, -∞, ∞}
] // Quiet
0.483194
Side note:
For planar polygons, I think there is conformal transformation which can provide bijective maps between the polygon and the unit disk (or eventually maps between the polygon and a rectangle), which in some cases might simplify the integrals and provide more precise results.
-
This works fine for the area function; I guess you'd need to map the {x,y,z} coordinates passed into the function, as well? – Eli Lansey Jun 21 '13 at 16:03
@EliLansey I think once you have the coordinates of the 3D polygon, finding its norm thus performing the coordinate transformation are both automatically done. – Silvia Jun 21 '13 at 16:10
I might be misunderstanding something in your answer. If the function is, e.g. f[x_,y_z_]:=x+y+z and your triangle's normal is +y, the function in the integrand needs to be evaluated at the actual x, y z. In my example, f[x_,y_z_]:=1 so there's no need to worry about that, but for an arbitrary function you'd need to map those coordinates, too. – Eli Lansey Jun 21 '13 at 16:51
@EliLansey OK I get your point. It seems I overlooked your title... But please allow me update the answer after 20 hours. It's too late here and I get some stuff to busy with tomorrow. (And yes, the only thing you need to do is doing the some coordinate transformation on arguments of f.) – Silvia Jun 21 '13 at 17:38
I've updated my question to make the fact that I'm integrating a function a little clearer. And no rush -- for the time being, my answer below is sufficient for my needs, but not general. – Eli Lansey Jun 21 '13 at 18:13
For a triangle, this can be solved by integrating in barycentric coordinates. Assuming the function that is being integrated takes in f[{x,y,z}] as an argument:
f[pos_List] := 1
integralOverTriangle[vertices_,f_]:=
Module[{area},
area=Norm[1/2 Cross[vertices[[1]]-vertices[[2]],vertices[[1]]-vertices[[3]]]];
2 area Integrate[f[l1 vertices[[1]]+l2 vertices[[2]]+(1-l1-l2)vertices[[3]]],
{l2,0,1}, {l1,0,1-l2}]
]
nintegralOverTriangle[vertices_,f_]:=
Module[{area},
area=Norm[1/2 Cross[vertices[[1]]-vertices[[2]],vertices[[1]]-vertices[[3]]]];
2 area NIntegrate[f[l1 vertices[[1]]+l2 vertices[[2]]+(1-l1-l2)vertices[[3]]],
{l2,0,1}, {l1,0,1-l2}]
]
Then,
integralOverTriangle[basetri,1&]
(* Out[]:= Sqrt[3]/4 *)
and, the triangle can be oriented arbitrarily:
integralOverTriangle[RotationTransform[{{0, 0, 1}, Normalize[{1, 1, 1}]}] /@ basetri, 1 &]
(* Out[]:= Sqrt[3]/4 *)
-
In Mathematica 10, this can be easily computed using built-in functionality. Using Silvia's answer's polygon and function:
polygonPts3D = {
{-0.902757, -0.116805, 0},
{0.203504, -0.972294, 0},
{0.849893, 0.414192, 0},
{0.374057, 0.835407, 0},
{-0.907079, 0.352119, 0}
};
f[x_,y_,z_] := x^3-y^2+z
All you need to do is:
Integrate[f[x, y, z], Element[{x, y, z}, Polygon[polygonPts3D]]]
Out[]:= -0.323842
- | 2015-07-06T11:31:40 | {
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https://math.stackexchange.com/questions/1422463/evaluating-int-0-frac-pi2-sin-x-cos-x4-sinx-cos-x | # Evaluating $\int_{0}^{\frac{\pi}{2}} (\sin (x) +\cos (x))^4 - (\sin(x) - \cos (x))^4 dx$
I would like to evaluate $$\int_{0}^{\frac{\pi}{2}} (\sin (x) +\cos (x))^4 - (\sin(x) - \cos (x))^4 dx.$$ One approach I used was to split the integral into two pieces and evaluating each piece directly with the anti derivatives $$\frac {1}{4} (\sin(x) - \cos (x))^4$$ and $$\frac {1}{4} (-\sin(x) - \cos (x))^4,$$ respectively. This gives me 0. However, one can use trig identities and manipulate the integrand so that the final integral is $$4\int_{0}^{ \frac{\pi}{2} }2\sin (x) \cos (x) dx.$$ This gives me 4. How is that I'm getting two answers?
• no for the first integral you get also $4$ as the result – Dr. Sonnhard Graubner Sep 4 '15 at 22:05
• @Nitin I believe the anti derivatives are correct. The second one needs a minus sign, but I was taking that into account when taking the difference. – Kevin Sheng Sep 4 '15 at 22:07
• @KevinSheng I deleted my first comment because I thought I was wrong, but no, the anti derivatives are incorrect. Just use the chain rule to check. – user217285 Sep 4 '15 at 22:14
• Nope, you are correct. I was making a careless error. – Kevin Sheng Sep 4 '15 at 22:17
You didn't compute antiderivatives correctly. To check your work, simply differentiate your result: $$\frac{\mathrm d}{\mathrm dx}\left[\frac {1}{4} (\sin(x) - \cos (x))^4 + C\right] = (\sin x + \cos x)(\sin x - \cos x)^3 \neq (\sin x + \cos x)^4$$
Please, observe that we used the chain rule to derive our result. The same argument can be used to show that the second antiderivative is wrong.
To evaluate the integral it's much easier to use the fact that $$A^2 - B^2 = (A + B)(A - B),$$ which, in combination with the identity $\sin^2 x + \cos^2 x = 1$, yields \begin{align} I &= \int_0^{\pi/2}\left((\sin x + \cos x)^2 + (\sin x - \cos x)^2\right)\left((\sin x + \cos x)^2 - (\sin x - \cos x)^2\right)\mathrm dx\\ &= \int_0^{\pi/2}8\sin x \cos x\,\mathrm dx\\ &= 4\int_0^{\pi/2}\sin 2x\,\mathrm dx\\ &= 4\text{.} \end{align}
As @rubik's answer explains, your antiderivatives are incorrect, and that is where the confusion comes from.
If I were doing this, I'd just imagine expanding both 4th powers, see that there would be a lot of cancellation alternating with doubling up, and leave you with \begin{align} &\phantom{{}={}}\int_0^{\pi/2}\left(8\sin^3(x)\cos(x)+8\sin(x)\cos^3(x)\right)\,dx\\ &=\int_0^{1}8u^3\,du+\int_1^{0}-8v^3\,dv\\ &=16\int_0^{1}u^3\,du\\ &=16\cdot\frac{1}{4} \end{align}
I we recall that: $$\cos^4\theta-\sin^4\theta = (\sin^2\theta+\cos^2\theta)(\cos^2\theta-\sin^2\theta) = \cos(2\theta)$$ it follows, with trivial manipulations:
$$\begin{eqnarray*} &&\int_{0}^{\pi/2}\left[\left(\sin x+\cos x\right)^4-\left(\sin x-\cos x\right)^4\right]\,dx\\&=&4\int_{0}^{\pi/2}\left[\cos(x-\pi/4)^4-\sin(x-\pi/4)^4\right]\,dx\\&=&4\int_{-\pi/4}^{\pi/4}\left[\cos^4\theta-\sin^4\theta\right]\,d\theta\\&=&4\int_{-\pi/4}^{-\pi/4}\cos(2\theta)\,d\theta\\&=&2\int_{-\pi/2}^{\pi/2}\cos\theta\,d\theta=\color{red}{4}.\end{eqnarray*}$$ | 2019-12-10T16:23:39 | {
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http://math.stackexchange.com/questions/507595/why-does-using-different-units-of-angle-affect-the-rate-of-change | # Why does using different units of angle affect the rate of change?
The question is
A triangle has sides of length 4cm and 9cm. The angle between them is increasing at a rate of 1$^\circ$ per minute. Find the rate in cm$^2$ per minute at which the area of the triangle is increasing when the angle is 45$^\circ$.
My solution was (and it is supposedly correct):
$A = \frac{1}{2}\cdot 4 \cdot 9\cdot \sin(\theta) = 18 \sin(\theta)$
$\frac{\Delta A}{\Delta \theta} = 18\cos(\theta)$
$\frac{\Delta \theta}{\Delta t} = 1^\circ = \frac{\pi}{180}^c$
$\frac{\Delta A}{\Delta t} = \frac{\Delta A}{\Delta \theta} \cdot \frac{\Delta \theta}{\Delta t} = \frac{\pi}{180} \cdot 18\cos(\theta) = \frac{\pi}{10} \cos(\theta)$
Let $\theta = \frac{\pi}{4}$
$\frac{\Delta A}{\Delta t} = \frac{\pi \sqrt{20}}{2}cm^2/min$
But I was wondering, why did we have to convert the angle from degrees to radians? I only converted to radians simply because I happened to prefer working with radians over degrees.
Obviously, by chain rule, if we stayed in degrees, we would have a different expression for $\frac{\Delta A}{\Delta t}$ and therefore a different answer for the rate of change. However, the rate of change shouldn't change simply because we used a different unit of measurement right? After all, $1^\circ = \frac{\pi}{180}^c$ and always will be.
Is there a fundamental reason why the angle must be measured in radians? My hypothesis is that the units have to match when working with related quantities but I can't seem to make the connection between units of angle and units of area.
-
May be you know this, but I say it anyway. If you measure $\theta$ in degrees, then $\dfrac{\Delta A}{\Delta\theta}$ will be different. The derivative of $\sin\theta$ is $\cos\theta$ only when you measure $\theta$ in radians. If you use degrees, then another application of chain rule will cancel the change you noticed: $$\frac{\Delta\theta(\text{radians})}{\Delta\theta(\text{degrees})}=\frac{\pi}{180}.$$ – Jyrki Lahtonen Sep 28 '13 at 5:58
How do you know $\frac{d}{d\theta}\sin(\theta) = \cos(\theta)$? Is it still true if you measure $\theta$ in degrees instead of radians?
I'll leave out the details of proving the derivative formula for now, but the crux of the matter is that you need to know that $\sin(\theta)$ is very close to $\theta$ when $\theta$ is small, and this is only true if you measure $\theta$ in radians. The best explanation I've seen for that is actually here on math exchange: How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?
An important implicit point there is that the area of a wedge of a circle $\frac{\theta}{2} r^2$, which only holds if $\theta$ is measured in radians. | 2015-05-29T06:40:22 | {
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https://math.stackexchange.com/questions/2522234/let-f-n-be-the-n-th-fibonacci-number-prove-or-disprove-5-f-5k-for-al | # Let $F_n$ be the $n$-th Fibonacci number. Prove or disprove: $5 | F_{5k}$ for all positive integers $k$
Given the following proof:
Let $F_n$ be the $n$-th Fibonacci number. Prove or disprove: $5 | F_{5k}$ for all positive integers $k$.
I am confused exactly where to start. The Fibonacci numbers are defined recursively correct? So wouldn't this be something that I can prove with strong induction?
Well, $F_5=5$ and $5\mid5$, right?!
Now, suppose that $5\mid F_{5k}$. Then\begin{align}F_{5(k+1)}&=F_{5k+5}\\&=F_{5k+4}+F_{5k+3}\\&=F_{5k+3}+F_{5k+2}+F_{5k+2}+F_{5k+1}\\&=F_{5k+3}+2F_{5k+2}+F_{5k+1}\\&=F_{5k+2}+F_{5k+1}+2F_{5k+1}+2F_{5k}+F_{5k+1}\\&=3F_{5k}+5F_{5k+1}.\end{align}Since $5\mid F_{5k}$, it follows from this that $5\mid F_{5(k+1)}$.
• Exactly what I was trying to do thank you! – APorter1031 Nov 15 '17 at 23:13
• You stopped at 3F5k+5F5k+1 because it is of the form k + k+1, correct? – APorter1031 Nov 15 '17 at 23:49
• @APorter1031 No, not at all. It's because $5F_{5k+1}$ is trivially a multiple of $5$ and $3F_{5k}$ is a multiple of $5$ since I am assuming that $F_{5k}$ is a multiple of $5$. My proof is a proof by induction of the assertion$$(\forall k\in\mathbb{N}):5\mid F_{5k}.$$ – José Carlos Santos Nov 15 '17 at 23:52
• ahh, thank you for the clarification! – APorter1031 Nov 15 '17 at 23:52
Just using the recurrence $F_{n+1}=F_n+F_{n-1}$ few times, you can get $$F_{5n+5}=5F_{5n+1}+3F_{5n}$$ You can use this identity to prove your claim by induction.
If we take $F_1=F_2=1$, then we certainly have that $5|F_5=5$. Let's look at the Fibonacci numbers, reduced modulo $5$:
$F_1\equiv 1\\ F_2\equiv 1\\ F_3\equiv 2\\ F_4\equiv 3\\ F_5\equiv 0\\ F_6\equiv 3\\ F_7\equiv 3\\ F_8\equiv 1\\ F_9\equiv 4\equiv -1\\ F_{10}\equiv 0\\ F_{11}\equiv -1\\ F_{12}\equiv -1\\ F_{13}\equiv -2\\ F_{14}\equiv -3\\ F_{15}\equiv 0\\ F_{16}\equiv -3\\ F_{17}\equiv -3\\ F_{18}\equiv -1\\ F_{19}\equiv -4\equiv 1\\ F_{20}\equiv 0\\$
After this, the pattern begins to repeat, and it appears that $F_{n+20}\equiv -F_{n+10}\equiv F_n$ for all $n$. This, together with $F_5\equiv F_{10}\equiv 0\pmod 5$ is sufficient to establish your claim.
You don't have to define the Fibbonacci sequence recursively
You could say:
$F_n = \frac{\phi^n + \phi'^n}{\phi - \phi'}$
where $\phi = \frac {1 + \sqrt 5}{2}$ and $\phi' = \frac {1 - \sqrt 5}{2}$
$F_{5k} = \frac{\phi^{5k} + \phi'^{5k}}{\phi - \phi'} = \frac{\phi^{5} + \phi'^{5}}{\phi - \phi'} (\phi^{5k-5} + \phi^{5k-10}\phi' + \cdots \phi'^{5k-1}) = F_5(\phi^{5k-5} + \phi^{5k-10}\phi' + \cdots \phi'^{5k-1})$
and $F_5 = 5$
More generally $k|n \implies F_k | F_n$
Modulo $5$ the Fibonacci sequence becomes $$0,1,1,2,3,0,3,3,1,4,0,4,4,3,2,0,2,2,4,1,0,1, ...$$ and from this point on repeats periodically. So we see that for indexes $5k$ we have $F_{5k}\equiv0\bmod5$.
The characteristic polynomial of the sequence $\{F_n\}_{n\geq 0}$ is the minimal polynomial of $\varphi$ over $\mathbb{Q}$, namely $x^2-x-1$. It follows that the characteristic polynomial of the sequence $\{F_{5n}\}_{n\geq 0}$ is the minimal polynomial of $\varphi^5$, namely $x^2-11x-1$, and $$F_{5k+10} = 11 F_{5k+5} + F_{5k}.$$ Since both $F_0$ and $F_5$ are multiples of $5$, the previous recurrence implies that $5\mid F_{5k}$ for any $k\in\mathbb{N}$. Actually, we can say much more: due to this version of Hensel's lifting lemma, $$\nu_5(F_n) = \nu_5(n)$$ which can be proved through the following identity, too: $$\begin{eqnarray*} F_{5k} &=& F_k (\varphi^{4k}+\varphi^{3k}\overline{\varphi}^k+\varphi^{2k}\overline{\varphi}^{2k}+\varphi^k\overline{\varphi}^{3k}+\overline{\varphi}^{4k})\\&=&F_k\left(25 F_k^4+25F_k^2+5\right)\\&=&5F_k\underbrace{(5F_k^4+5F_k^2+1)}_{\not\equiv 0\pmod{5}}\end{eqnarray*}$$ implying $\nu_5(F_{5k})=1+\nu_5(F_k)$.
Let \begin{eqnarray*} A= \begin{bmatrix} 0 &1 \\1 &1 \\ \end{bmatrix} . \end{eqnarray*} Then
\begin{eqnarray*} A^n= \begin{bmatrix} F_{n-1} &F_n \\F_n &F_{n+1} \\ \end{bmatrix} \end{eqnarray*} and \begin{eqnarray*} A^5= \begin{bmatrix} 3 &5 \\5 &8 \\ \end{bmatrix} . \end{eqnarray*}
In particular \begin{eqnarray*} A^{5n}= \begin{bmatrix} F_{5n-1} &F_{5n} \\F_{5n} &F_{5n+1} \\ \end{bmatrix} . \end{eqnarray*} \begin{eqnarray*} A^{5n+5}= \begin{bmatrix} F_{5n+4} &F_{5n+5} \\F_{5n+5} &F_{5n+6} \\ \end{bmatrix} = \begin{bmatrix} . &3F_{5n}+5F_{5n+1} \\. &. \\ \end{bmatrix} . \end{eqnarray*} and thus the result follows by induction. | 2021-05-11T04:13:34 | {
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https://math.stackexchange.com/questions/2782481/identifying-langle-a-b-mid-a2b2-rangle | Identifying $\langle a,b\mid a^2b^2\rangle$.
The Question:
What group is $G=\langle a,b\mid a^2b^2\rangle$?
Thoughts:
I found that the presentation maps onto $\langle a, b\mid a^2, b^2, 1\cdot 1\rangle\cong \mathbb{Z}_2\ast\mathbb{Z}_2$, which is reassuring since GAP says $G$ is infinite.
Extra Context:
I'm not sure if that's enough context so here's a Q&A:
• What are you studying?
A PhD in combinatorial group theory, first year.
• What text is this drawn from, if any? If not, how did the question arise?
None. The group arose in my research. Without giving too much away, the group is cyclically presented.
• What kind of approaches (to similar problems) are you familiar with?
Here's a list of related questions of mine:
Identifying $\langle \{x_h\mid h\in H\}\mid \{x_{h'}x_{h'a}x_{h'b}^{-1}\mid h'\in H\}\rangle$ if $H=\langle a,b\mid a^2, b^4, ab=ba\rangle$.
Identifying $\langle r,s\mid srs^{-2}r, r^{-1}srsr^{-1}\rangle$.
Identifying $\langle x_e, x_a, x_b, x_{ab}\mid x_ex_a=x_{ab}, x_ax_e=x_b, x_bx_{ab}=x_a, x_{ab}x_b=x_e\rangle.$
Identifying $\langle x_1, \dots, x_6\mid x_1x_2=x_3, x_2x_3=x_1, x_3x_1=x_2, x_4x_5=x_6, x_5x_6=x_4, x_6x_4=x_5\rangle.$
There's more but I think that's enough.
• What kind of answer are you looking for? Basic approach, hint, explanation, something else?
A simple identification would be great. A detailed explanation of why it is what it is would be better, although (strong) hints are preferred.
• Is this question something you think should be able to answer? Why or why not?
Yes. I've been working closely with presentations in a research context for months now. It looks like Google should have the answer.
• The answer to the question in the title of that supposed duplicate is in the negative, @DietrichBurde. – Shaun May 15 '18 at 17:02
• Upon closer inspection, it says in the accepted answer that $G=\Bbb Z*_{\Bbb Z} \Bbb Z$ is a free product with amalgamation of two infinite cyclic groups generated by $a, b$ amalgamated along their subgroups $a^2, b^2$. – Shaun May 15 '18 at 17:11
• @DietrichBurde's supposed duplicate: math.stackexchange.com/q/191317/104041. – Shaun May 15 '18 at 17:43
It has a normal abelian subgroup of index $2$ and isomorphic to ${\mathbb Z}^2$, with generators $ab$ and $a^2$.
It is a torsion-free group, so it cannot be a split extension. We have $a(ab)a^{-1} = a^4(ab)^{-1}$, so another presentation of the same group, which makes this structure clearer, is $$\langle a,c,d \mid cd=dc,\, a^2=d,\, aca^{-1}=d^2c^{-1} \rangle,$$ where $c=ab$ and $d=a^2$.
I will add that I don't think that it is completely clear what you mean by "identify this group". The presentation is a precise definition of the group, so you could argue that it is already an identification. I guess you mean something like "find an alternative description or definition of a group isomorphic to this one, if possible a group that has been studied already". That's OK, but in many cases, there will not exist such an alternative description.
Since this particular group is solvable, and even polycyclic, a polycyclic presentation, such as the one I have give above in generators $a$, $c$ and $d$ provides an alternative description. You could for example use the polycyclic package in GAP for computing with polycyclic groups on this group.
• How come the group is solvable? – Shaun May 15 '18 at 19:50
• Oh, and thank you. – Shaun May 15 '18 at 20:04
• It has an abelian subgroup of index $2$, so it must be solvable. – Derek Holt May 15 '18 at 21:29
For a topological/geometric point of view, this group is isomorphic to the fundamental group of the Klein bottle.
Here's how to see it. Start by rewriting the presentation $$\langle a,b \mid a^2b^2 = \text{Id}\rangle = \langle a,b \mid a^2 = b^{-2} \rangle = \langle a,c \mid a^2 = c^2 \rangle \quad\text{(c=b^{-1})}$$ By Van Kampen's Theorem, we can construct a topological space with this fundamental group. Start with two circles $C_a$, $C_c$ whose fundamental groups are infinite cyclic with generators represented by closed paths $\gamma_a$ going exactly one time around $C_a$ and $\gamma_c$ going one time around $C_c$. Now attach $S^1 \times [0,1]$ to these two circles: $S^1 \times 0$ is attached to the path $\gamma_c^2$ that goes two times around $C_a$; and $S_1 \times 1$ is attached to the path $\gamma_b^2$ that goes two times around $C_c$. Let $X$ be the resulting quotient space. This space $X$ is a compact, connected, non-orientable 2-manifold of Euler characteristic $0$, and so it is homeomorphic to the Klein bottle. Thus, by Van Kampen's Theorem, your group is isomorphic to the fundamental group of the Klein bottle.
And once you know that this is the Klein bottle group, it also follows that it is one of the 17 famous wallpaper groups, also known as planar crystallographic groups, namely those groups which can be represented as discrete, cocompact subgroups of isometries of the Euclidean plane. To a crystallographer, this group is known as "pg", and can be represented as generated by two isometries of the $xy$-plane, as shown in this picture: $g_x$ which is a glide reflection along the $x$ axis (the "g" in the "pg" notation refers to a glide reflector); and $t_y$ which is a translation along the $y$ axis. To connect these generators with your two original generators, you can take $a=g_x$ and $b=t_y g_x$. | 2019-07-21T02:19:55 | {
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https://math.stackexchange.com/questions/2556779/partial-order-relation-conditions-transitive | # Partial Order relation conditions (transitive)
We know that every Partial Order Set has to satisfy three conditions :
1- Reflexive
2- Anti-Symmetric
3- Transitive
For example : $$S= \left(\left\{1 \right\},\left\{2 \right\},\left\{3 \right\},\left\{1,2 \right\} \right)$$
And the relation is $\subseteq$ ,My question is why this Set S is a Partial Order Set Althought I Cannot prove that is Transitive which is defined as:
$$R=\{(x,y)\} ; xRy ~\text{and}~ yRz \implies xRz$$ I could not find like : if $\left\{1 \right\} \subseteq \left\{1,2 \right\}\text{ } and \text{ } \left\{1,2 \right\} \subseteq \text{ }\text{ } ?? \text{ }then\text{ } ??$
in another way how can we apply the Transitive test for this set S?
• Recall the definition of a transitive relation: A relation $R$ on a set $A$ is transitive if whenever $(a,b)\in R$ and $(b,c)\in R \implies (a,c)\in R$ for $a,b,c \in A$.
– user371838
Dec 8, 2017 at 6:54
Note that statement of the form of "if $A$ then $B$" is true if $A$ is false.
Well, actually we have $\color{blue}{\{1 \}} \subseteq \color{green}{\{ 1,2\}}$ and $\color{green}{\{1,2\}} \subseteq \color{purple}{\{ 1, 2\}}$, then $\color{blue}{\{1\}} \subseteq \color{purple}{\{1,2\}}.$
Notice that for the subset operation, we always have
$$x \subseteq y \text{ and }y \subseteq z \text{ ,then }x \subseteq z$$
• like that .. i understand the first line but do you mean that y may equel to z? Dec 8, 2017 at 6:57
• There is no restriction that $x$, $y$ and $z$ have to be distinct right? we are suppose to check all possibilities. though the case for identical elements are trivial. Most important thing should be the last line to show transitivity. Dec 8, 2017 at 7:01
• Perfect then i can understand that A relation on a set A: is always Transitive .. right ? Where A is a Set of at least One Element . Dec 8, 2017 at 7:07
• yes, subset is a transitive relation. Dec 8, 2017 at 7:10
• thanks a lot bro.. in the definition of a transitive relation: A relation R on a set A is transitive if whenever (a,b)∈R(a,b)∈R and (b,c)∈R⟹(a,c)∈R(b,c)∈R⟹(a,c)∈R for a,b,c∈A ... could a,b and c be equel ? @rohan Dec 8, 2017 at 9:43 | 2022-08-12T12:31:12 | {
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https://math.stackexchange.com/questions/1630676/parallel-lines-parallel-to-itself | # Parallel lines, parallel to itself
In a lecture the professor drew two lines, a and b, parallel to each other. The professor then asked if line a is parallel to itself.
The students said no because the definition of a parallel line is that it never cuts another line and a cuts itself all the way through.
The professor said that this is wrong, of course a is parallel to itself.
Is the professor asuming that the lines are transparent and thus do not cut each other? Are the students using the wrong definition of parallelism? Who is right, and why?
• I think lines are parallel iff the distance between them is the same at every point, that is, $a$ is parallel to $b$ whenever the distance from a given point $A$ on $a$ to the nearest point on $b$ is invariant of $A$. In case of two equal lines, this distance is always $0$, so yeah, two equal lines are parallel. Also, the tag you added is completely irrelevant to your post. – vrugtehagel Jan 28 '16 at 16:09
• The professor is using the reflexive variant of the definition of parallel lines. – N. F. Taussig Jan 28 '16 at 17:13
If the minimum distance $d$ (between two straight lines) perpendicular to either line is constant, the lines are parallel. Holds even when $d=0$. | 2019-07-22T14:59:32 | {
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https://math.stackexchange.com/questions/2807771/int-2-infty-frac1x-sqrtx2-4dx-convergence-verification | # $\int_2^{\infty}\frac{1}{x\sqrt{x^2-4}}dx$ convergence verification
I wish to check if the integral
$$\int_2^{\infty}\frac{1}{x\sqrt{x^2-4}}\,dx$$
converges.
Here is what I did:
Using substitutions:
$$\int_2^{\infty}\frac{1}{x\sqrt{x^2-4}}\,dx=\frac{1}{2}\int_4^{\infty}\frac{1}{t\sqrt{t-4}}\,dt= \int_0^{\infty} \frac{1}{(u+4)\sqrt u}\,du.$$
Separating:
$$\int_0^{\infty} \frac{1}{(u+4)\sqrt u}\,du=\int_0^{1} \frac{1}{(u+4)\sqrt u}\,du+\int_1^{\infty} \frac{1}{(u+4)\sqrt u}\,du.$$
For the first term: $\displaystyle \int_0^{1} \frac{1}{(u+4)\sqrt u}\,du:$
$$\int_0^{1} \frac{1}{(u+4)\sqrt u}\,du=\lim_{\epsilon \rightarrow 0}\int_{\epsilon}^{1} \frac{1}{(u+4)\sqrt u}\,du.$$
Using comperison test to $\dfrac{1}{\sqrt u},$ this term coverges.
For the second term: $\displaystyle\int_1^{\infty} \frac{1}{(u+4)\sqrt u}\,du,$
using the comparison test to $\dfrac{1}{ u^{3/2}},$ this term converges.
I would like to know if I made any mistakes. Also, let me know if you have any suggestions for other ways to solve this.
• Everything is OK. – Julián Aguirre Jun 4 '18 at 14:52
• A couple of typesetting tips (which I've edited in for you): 1. Use \displaystyle whenever possible, except in titles. This is especially helpful for integrals and fractions. 2. Always use a thinspace \, immediately before any differential inside an integral. – Adrian Keister Jun 4 '18 at 14:54
• @JuliánAguirre thank you! – segevp Jun 4 '18 at 14:55
• +1) Always good to see the work! – imranfat Jun 4 '18 at 15:27
$$\int_2^3 \frac {dx} {x\sqrt{x-2\,}\sqrt{x+2\,}} \le \int_2^3 \frac{dx}{\sqrt{x-2}} < +\infty$$ and $$\int_3^\infty \frac{dx}{x\sqrt{x^2-4}} \le \int_3^\infty \frac{2\,dx}{x^2} < +\infty.$$ For this second inequality you need to show that $\sqrt{x^2-4\,} \ge \dfrac x 2$ when $x\ge3.$
Yes it is correct.
As an alternative, without change of variables, we can observe that for $x$ large
$$\frac{1}{x\sqrt{x^2-4}}\sim\frac1{x^\frac32}$$
and refer to limit comparison test with $\int_3^{\infty} \frac1{x^\frac32}\, dx$ and for $x\to 2^+$
$$\frac{1}{x\sqrt{x^2-4}}=\frac{1}{x \sqrt{x+2}\sqrt{x-2} }\le\frac14 \frac{1}{\sqrt{x-2} }$$
and refer to comparison test with $\int_2^3 \frac{1}{\sqrt{x-2} }\, dx$. | 2019-12-08T11:26:24 | {
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https://math.stackexchange.com/questions/3019429/point-x-in-mathbbrn-that-minimizes-sum-of-distance-squares-sum-mathca | # Point $x \in \mathbb{R}^n$ that minimizes sum of distance squares $\sum_{\mathcal{l}=1}^{k} \Vert x-a^{\mathcal{(l)}} \Vert _2^2$
Let $$a^{(1)},...,a^{(k)} \in \mathbb{R}^n$$.
How can one find the point $$x \in \mathbb{R}^n$$, which minimizes the sum of distance squares $$\sum_{\mathcal{l}=1}^{k} \Vert x-a^{\mathcal{(l)}} \Vert _2^2$$
I know that a function $$h(t) = \sum_{l=1}^{k}(t-x_i)^2$$ has its minimum when $$t = \overline{x}$$. So for the average $$\overline{x}$$ of the numbers $$x_1,...,x_n$$ the sum of squares of the deviations $$\overline{x}-x_i$$ is minimal.
So to get the center of a set of points
$$S=\{(x_1,y_1),(x_2,y_2),\dots (x_n,y_n)\}$$ we can get their centroid by $$(\bar x,\bar y) = \left(\frac{1}{n}\sum_{i=0}^n x_i, \frac{1}{n}\sum_{i=0}^n y_i\right).$$
I don't really know if this point actually minimizes the sum of distance squares given above. I'd also like to know if only one such point exists or if there are more.
## 2 Answers
I believe your intuition is correct.
Consider the following:
$$\sum_{l=1}^k || x - a^{(l)}||_2^2 = \sum_{l=1}^k \sum_{m=1}^n (x_m - a^{(l)}_m)^2$$
Where $$a_m^{(l)}$$ is the $$m$$th component of the $$l$$th vector, and $$x_m$$ is the $$m$$th component of $$x$$.
To find the minimum of this with respect to $$x$$, we can use standard differentiation procedures.
First, differentiate with respect to one direction (i.e. differentiate with respect to some $$x_j$$): \begin{align} \frac{\partial}{\partial x_j} \sum_{l=1}^k \sum_{m=1}^n (x_i - a^{(l)}_m)^2 &=\sum_{l=1}^k \sum_{m=1}^n \frac{\partial}{\partial x_j} (x_m - a^{(l)}_m)^2 \\ &=\sum_{l=1}^k 2(x_j-a_j^{(l)}) \end{align} Setting this expression to zero you will obtain: \begin{align} \sum_{l=1}^k 2(\hat{x}_j-a_j^{(l)}) &= 0\\ \hat{x}_j &= \frac{1}{k}\sum_{l=1}^k a_j^{(l)} \end{align}
This shows that for any direction $$j$$, there is a stationary point of the function $$\sum_{l=1}^k || x - a^{(l)}||_2^2$$ (which we will show is a global minimum) given by the average of the $$j$$th component of the vectors $$\{a\}_{l=1}^k$$.
Now, to show that this is a unique global minimum, we will use the fact that a function $$f(x)$$ is strongly convex (which implies that $$f(x)$$ has a unique minimum point) if its Hessian $$H$$ (matrix containing the second derivatives) is positive definite (We say that $$H$$ is positive definite if $$\forall c \in \mathbb{R}^n/\{0\}, c^THc > 0$$).
Consider the second derivatives of $$\sum_{l=1}^k || x - a^{(l)}||_2^2$$:
\begin{align} \frac{\partial^2}{\partial_i\partial x_j} \sum_{l=1}^k \sum_{m=1}^n (x_m - a^{(l)}_m)^2 &=\sum_{l=1}^k \frac{\partial}{\partial x_i}2(\hat{x}_j-a_j^{(l)})\\ &=\begin{cases} 2 & i=j\\ 0 & i\neq j \end{cases} \end{align}
This means that the Hessian $$H$$ will be diagonal, and that there will be strictly positive entries on the diagonal. Thus $$H$$ is positive definite, implying that $$\sum_{l=1}^k || x - a^{(l)}||_2^2$$ is strongly convex with respect to $$x$$.
Hence, we can conclude that the stationary point which we found above is indeed a unique minimum.
You can use this relation: $$\sum_{i=1}^n \|x-a_i\|^2=n\|\bar{a}-x\|^2+\sum_{i=1}^n \|\bar{a}-a_i\|^2$$ where $$\bar{a}=\frac{1}{n}\sum_{i=1}^n a_i$$
If you want to minimize w.r.t. $$x$$, just take $$x=\bar{a}$$ to cancel the $$\|\bar{a}-x\|^2$$ term. Therefore the answer is: $$x=\bar{a}=\frac{1}{n}\sum_{i=1}^n a_i$$
To prove the first relation, you must notice that: $$\sum_i^{n}\|a_i-\bar{a}\|^2=\dots=\left(\sum_i^{n}\|a_i\|^2\right)-n\|\bar{a}\|^2$$ then simply develop $$\sum_{i=1}^n \|x-a_i\|^2=\sum_{i=1}^n (\|x\|^2-2\langle x,a_i \rangle+\|a_i\|^2)=\dots$$ (I can write the details if you want, just ask) | 2019-08-25T13:17:48 | {
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https://mathoverflow.net/questions/358634/differentiability-of-eigenvalues-of-positive-definite-symmetric-matrices/358637 | # Differentiability of eigenvalues of positive-definite symmetric matrices
Let $$A\in M(n,\mathbb{R})$$ be an invertible matrix. Consider the (real) eigenvalues $$\lambda_1,\cdots,\lambda_n$$, in increasing order, of the positive-definite symmetric matrix $$A^t A$$. We shall denote the eigenvalues as $$\lambda_i(A)$$.
Question What can be said about the differentiability of the functions $$\lambda_i:GL(n,\mathbb{R}) \to \mathbb{R}$$?
[We may assume that the domain is $$GL^+(n,\mathbb{R})$$ for differentiability/smoothness.]
Any reference for this or relevant results would be appreciated.
• After writing my answer below, I noticed that similar questions have been asked on MO: link, link. Apr 28 '20 at 0:04
In the open subset of $$M_n(\mathbb{R})$$ where the $$\lambda_i$$ are distinct, they are $$C^{\infty}$$ functions: this follows from the implicit function theorem.
On the other hand, when some eigenvalue has multiplicity $$>1$$ you don't get more than continuity. For example if $$A=\begin{pmatrix} 0 & 1\\ 1 & t \end{pmatrix}$$ the largest $$\lambda_i$$ is $$\dfrac{1}{2}\left(t^2+2 +|t|\sqrt{t^2+4}\right)$$, which is not differentiable (as a function of $$t$$) at $$t=0$$.
• Note that the question is about the eigenvalues of $A^\dagger A$ not about the eigenvalues of $A$ itself. But basically the same example works for that case too. Apr 27 '20 at 5:55
• I am dicussing the eigenvalues of $A^{\dagger}A$, not of $A$ (I am using the notation of the OP). The eigenvalues of $A$ are distinct for $t=0$.
– abx
Apr 27 '20 at 5:56
• In the simple $A$ given as example, it appears that if we disregard $\lambda_i$ and use a new numeration $\mu_1,\mu_2$ of the eigenvalues that is wrt. the sign of the square root of the discriminant, not wrt. the relative size of the roots, then you can have that both $\mu_1$ and $\mu_2$ are smooth functions of $t$. Possibly see this Wolfram Alpha illustration. Under what conditions is it possible to pick $\mu_i$ with $i=1\ldots n$ "smart" so they are smooth? Apr 28 '20 at 11:08
• Sorry, I should have read the answers by Denis Serre and Marc Nardmann before writing this comment. They address it. Apr 28 '20 at 11:28
The keyword is the Cartan decomposition in the theory of symmetric spaces.
In short, when an eigenvalue is simple (its multiplicity is $$1$$) it is locally an analytic function. But when the eigenspace is degenerate (the multiplicity is greater than $$1$$), the eigenvalue function is not differentiable. The problem is essentially one of choosing branches: if you try to deform the identity matrix, there is no consistent way to say which of the resulting distinct eigenvalues after deformation is the eigenvalue that you should have kept track of.
Let $$K = \mathrm{O}(n)$$, and let $$A$$ be the group of diagonal matrices with positive entries. You then have $$G=KAK$$ and if $$g=k_1 a k_2$$ then the eigenvalues of $$g^\dagger g$$ are exactly the squares of the eigenvalues of $$a$$. The problem is that the decomposition is not unique: you can conjugate $$a$$ by a permutation matrix, and there will be problems when $$a$$ is fixed by a permutation matrix.
• If I follow what you are saying, then does it follow that $a$ (in the decomposition of $g$) is unique up to conjugation by permutation matrices? In that case, although each $\lambda_i$ may not be well-defined, any smooth function $f$ of $\lambda_i$'s which is invariant under permutations will be smooth. Apr 27 '20 at 5:47
• Yes. A smooth symmetric function will be smooth. The best way to think about it for analytic functions is as follows: the elementary symmetric polynomials are polyomials in the traces of powers of the matrix (that's the Newton identities), and the Taylor expansion of an analytic symmetric function is a sum of elementary symmetric polyomials. Apr 27 '20 at 5:54
As mentionned by other answers, simple eigenvalues are $$C^\infty$$, while non-simple ones are not. Let me add however two important properties which you can find in Kato's book Perturbation theory of linear operator.
The first one is that each $$\lambda_j$$ is a Lipschitz function. This statement is still valid if you replace $${\bf Sym}_n({\mathbb R})$$ by a subspace $$E\subset{\bf M}_n({\mathbb R})$$ with the property that the eigenvalues are always real.
The second one is that if $$t\mapsto A(t)$$ is a smooth curve in $${\bf Sym}_n({\mathbb R})$$, then there is a labelling of the eigenvalues $$t\in{\cal V}\rightarrow(\mu_1(t),\ldots,\mu_n(t))$$ such that each $$\mu_j$$ is smooth. Mind that this labelling does not respect the order between eigenvalues when the multiplicities vary. Mind also that this becomes false if we replace a curve by a surface.
• A smooth eigenvector labelling (where "smooth" means $C^\infty$, not just $C^1$) does not exist in the generality you claim (and, as far as I can tell, this is not claimed in Kato's book); see the references in my answer below. Apr 27 '20 at 18:38
• @MarcNardmann. Thank you for the precision. I am stuck at home, without access to our library. Apr 27 '20 at 20:06
Let us consider functions $$A$$ from (an open interval in) $$\mathbb{R}$$ into the set of symmetric real $$n\times n$$ matrices (Hermitian complex $$n\times n$$ matrices behave analogously).
If $$A$$ is given by $$A(t) = diag(1+t,1-t)$$, then the eigenvalue functions $$\lambda_1,\lambda_2$$ of $$A$$ with $$\lambda_1\leq\lambda_2$$ are $$\lambda_1(t) = 1-|t|$$ and $$\lambda_2(t) = 1+|t|$$, hence are not differentiable. Instead of differentiability of the ordered tuple of eigenvalues, we should therefore discuss the question whether there is a differentiable function $$(\lambda_1,\dots,\lambda_n):\mathbb{R}\to\mathbb{R}^n$$ that consists pointwise of the eigenvalues of $$A$$ counted with multiplicities (i.e.: can the eigenvalue functions be chosen differentiably?).
(I chose the $$2\times2$$ example $$A$$ to be pointwise positive definite for $$t$$ close to $$0$$, because this was asked for in the original question. But this is not relevant: every differentiability problem that can occur for any eigenvalue $$\leq0$$ can also occur for positive eigenvalues. Moreover, considering $$A^tA$$ instead of $$A$$ does not change any differentiability issue: If, for some pointwise positive definite $$A$$, the eigenvalues of $$A^tA = A^2$$ are not [resp. cannot be chosen] as regular as $$A^tA$$ is, then they are not [resp. cannot be chosen] as regular as $$A$$, because $$A^tA$$ and $$A=\sqrt{A^tA}$$ have the same regularity, due to the real-analyticity of $$B\mapsto\sqrt{B}$$.)
Some of the results of Alekseevsky/Kriegl/Losik/Michor: Choosing roots of polynomials smoothly and Kriegl/Michor: Differentiable perturbation of unbounded operators, or older results cited therein, are the following:
• If $$A$$ is $$C^1$$, then the eigenvalue functions $$\lambda_1,\dots,\lambda_n$$ can be chosen $$C^1$$ (cf. Kato: Perturbation theory for linear operators, §II.6.3, Theorem 6.8).
• If $$A$$ is real-analytic, then the eigenvalue functions (and also eigenvector functions) can be chosen real-analytically.
• If $$A$$ is $$C^\infty$$, then the eigenvalue functions can be chosen twice differentiably.
• Even if $$A$$ is $$C^\infty$$, the eigenvalue functions cannot always be chosen $$C^2$$ (example 7.4 in AKLM, first example in KM).
• Let $$A$$ be $$C^\infty$$. Consider the eigenvalue functions $$\lambda_1,\dots,\lambda_n$$ with $$\lambda_1\leq\dots\leq\lambda_n$$ (they are always continuous). Assume for all $$i,j\in\{1,\dots,n\}$$ that either $$\lambda_i=\lambda_j$$ or there is no $$t\in\mathbb{R}$$ at which the functions $$\lambda_i,\lambda_j$$ meet of infinite order. Then the eigenvalue functions (and also eigenvector functions) can be chosen $$C^\infty$$.
• Very clear answer (+1). Do all the bullet points still hold for hermitian matrices?
– lcv
May 1 '20 at 21:13
• @lcv: Yes. I just cited a few theorems that can be stated succinctly. Many of these results hold in greater generality, for instance for normal matrices or even unbounded normal operators on Hilbert spaces. The reference in Ben McKay's answer contains much more information. May 2 '20 at 3:21
• Your last bullet is false. Your condition about $\lambda_i,\lambda_j$ is used to avoid $\exp(-1/t^2)$-type functions. Then there is a numbering of the eigenvalues $(λi)i≤n$ s.t. the associated functions are $C^{\infty}$. However, the natural ordering of the eigenvalues is not necessarily met; consider, for example, $B(t)=diag(t+2,2t+2)$ when $t$ goes through $0$. See my post in math.stackexchange.com/questions/3601351/… May 3 '20 at 9:49
• I do not understand your objection; I rather suspect that we mean the same thing. The last bullet point is correct. There might be a misunderstanding: Of course I do not claim that the ordered tuple $(\lambda_1,\dots,\lambda_n)$ is $C^\infty$. I claim that one can choose another tuple of eigenvalue functions which is $C^\infty$. I mention the ordered tuple only to state the sufficient condition under which this other choice is possible. May 4 '20 at 3:40
• OK I agree with your comment. May 7 '20 at 10:16
For the best known positive results under mild hypotheses, you might want to look at
Armin Rainer, Perturbation theory for normal operators, Trans. A.M.S., Volume 365, Number 10, October 2013, Pages 5545–5577
• link Apr 28 '20 at 0:14
In this paper by Xuwen Zhu it is shown that, after resolution by radial blow-ups, the eigenvalues can be made to be smooth: https://arxiv.org/abs/1504.07581. | 2021-12-06T03:53:44 | {
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https://math.stackexchange.com/questions/2027888/prove-that-lim-limits-t-to-0ht-does-not-exist | # Prove that $\lim \limits_{t \to 0}{H(t)}$ does not exist.
## Exercise
If $H$ is the Heaviside function, prove, using the definition below, that $\lim \limits_{t \to 0}{H(t)}$ does not exist.
## Definition
Let $f$ be a function defined on some open interval that contains the number $a$, except possible $a$ itself. Then we say that the limit of $f(x)$ as $x$ approaches $a$ is $L$, and we write $$\lim \limits_{x \to a}{f(x)} = L$$ if for every number $\epsilon > 0$ there is a number $\delta > 0$ such that $$\text{if } 0 < |x - a| < \delta \text{ then } |f(x) - L| < \epsilon$$
## Hint
Use an indirect proof as follows. Suppose that the limit is $L$. Take $\epsilon = \frac{1}{2}$ in the definition of a limit and try to arrive at a contradiction.
## Attempt
Let $\delta$ be any (preferably small) positive number.
$H(0 - \delta) = H(-\delta) = 0$
$H(0 + \delta) = H(\delta) = 1$
$H(0 - \delta) =^? H(0 + \delta) \implies 0 =^? 1 \implies 0 \neq 1 \implies H(0 - \delta) \neq H(0 + \delta)$
$\lim \limits_{t \to 0^-}{H(t)} \neq \lim \limits_{t \to 0^+}{H(t)} \implies \lim \limits_{t \to 0}{H(t)}$ does not exist
## Request
I don't even know where to begin, even with the hint.
Can someone kickstart the proof for me?$^1$
$^1$ Update: I've come up with an attempt. Is it valid? It seems that I don't use the hint to my advantage; so if indeed my attempt is correct, what is the alternative proof using the hint?
• Looks like you have everything you need. Show that definition cannot be satified with $\epsilon = \frac 12$ i.e. $\forall \delta >0, \exists t$ such that $|t|<\delta$ and $|H(t)-L|>\epsilon$ – Doug M Nov 23 '16 at 21:05
• Mind explaining it without the upside-down letters? :) I'm not familiar with that notation. – Fine Man Nov 23 '16 at 21:06
• Also, shouldn't $|H(t)-L|>\epsilon$ be $|H(t)-L|<\epsilon$? – Fine Man Nov 23 '16 at 21:09
$\forall$ means "for all" or "for any"
$\exists$ means "there exists"
For any $\delta > 0$ there exist a $t$ such that $|t|<\delta$ and $|H(t) - L|>\epsilon$
What are we doing.
If the limit exist. Then:
For any $\epsilon > 0$ there exist a $\delta>0$ such that when $|t|<\delta,|H(t) - L|<\epsilon$
And what does that mean. When $t$ is in the neighborhood of $0, H(t)$ is in a neighborhood of $L.$ The $\epsilon-\delta$ tell us the size of the neighborhoods.
But we need to show that the limit does not exist. There is a $t$ in the neighborhood 0, such that $H(t)$ is not in the neighborhood of $L$
How to do that? Fix the size of the target neighborhood of $H(t)$ small enough such that no matter how small of a neighborhood we find around $t$ we are sure to miss the target neighborhood around $H(t)$
• I'm not quite understanding you. Can you be more concrete? – Fine Man Nov 23 '16 at 21:20
• No matter how tight a neigborhood around $0$ we try to bind $t,$ there is a $t$ in the neighborhood such that $H(t) = 0$ there is also $t$ in that neighborhood such that $H(t) = 1.$ That means with $\epsilon = \frac 12$, for some $t, |t|<\delta, |H(t) - L|>\epsilon$ There is no $L$ that satisfies the definitions necessary for the limit to exist. – Doug M Nov 23 '16 at 21:25
Your attempt expresses the right idea but doesn't directly use the definition.
Here is one way to do it following the hint:
Suppose $\lim_{t\rightarrow 0}H(t)=L$. Then for every $\epsilon >0$, there exists $\delta >0$ such that $\left | H(t)-L \right |< \epsilon$ if $\left | t-0 \right |=\left | t \right |<\delta$. In particular it must work for $\epsilon =\frac{1}{2}$.
Take $t=\frac{\delta }{2}<\delta$ and $t^{'}=-\frac{\delta }{2}> -\delta$. Then we have:
$\left | H(t)-L \right |=\left | 1-L \right |< \frac{1}{2}$
Similarly, we have:
$\left | H(t^{'})-L \right |=\left | 0-L \right |=\left | L \right |< \frac{1}{2}$
Now, using the triangle inequality:
$1=\left | L-1+L \right |\leq \left | L-1 \right |+\left | L \right |< \frac{1}{2}+\frac{1}{2}=1$
But wait! We have reached the following contradiction: $1< 1$
We have thus shown that $\lim \limits_{t \to 0}{H(t)}$ doesn't exist.
Main idea: Set $$\epsilon = 0.1$$ and see that, for any "candidate to limit" $$L$$, you will always find a point $$X$$ as close to $$0$$ as you want, such that $$|H(X)-L| > \epsilon$$
So, we've found an $$\epsilon > 0$$ such that, for any $$\delta>0$$, there exists $$x$$ such that $$|x-0| < \delta$$ but yet $$|H(X)-L| > \epsilon$$. But how to be precise about finding that $$x$$?
If our "candidate to limit" $$L \in [0.9, 1.1]$$, simply choose $$x = - \frac{\delta}{2}$$, so that $$H(x)=0$$, which is at a distance larger than $$\epsilon$$ from $$L$$, i.e. $$|f(x)-L| > \epsilon$$
I think you can easily guess which $$x$$ to pick if $$L \notin [0.9, 1.1]$$ | 2019-09-21T05:22:02 | {
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https://math.stackexchange.com/questions/2623936/how-to-prove-that-ln2-sinx-ln2-siny-x-y-space-forall-sp | # How to prove that $|\ln(2+\sin(x)) - \ln(2+\sin(y))| <= |x-y| \space \forall \space x,y \in \mathbb R$
Question states
Prove that for all $x$ and $y$ $\in R$, the following inequality is true:
$\lvert \ln(2+\sin(x)) - \ln(2+\sin(y))\rvert \le \lvert x-y\rvert$
i've gotten to the point that
$\frac{y-x}{2+\sin(c)} = \ln\frac{2+\sin(y)}{2+\sin(x)}$ (y-x divided by 2+sin(c) is my f dash c from the mean value theorem
I asked my teacher that i should use mean value theorem here so please don't use anything other than this, but i have no idea how to push this problem further.
Also this is my first post so i'm sorry for the f dash (c) thing, mathjax is hard
Let $f(x)=\log(2+\sin x)$. Then $f'(x)=\frac{\cos x}{2+\sin x}$. So, if $x,y\in\mathbb R$ and $x\neq y$, then$$\frac{f(y)-f(x)}{y-x}=\frac{\cos c}{2+\sin c}$$for some $c$ between $x$ and $y$, and therefore$$\left|\frac{f(y)-f(x)}{y-x}\right|=\left|\frac{\cos c}{2+\sin c}\right|\leqslant\frac{|\cos c|}{2-|\sin c|}\leqslant1.$$
• oh god the chain rule is used here isnt it? thank you very much – calculus freshman Jan 27 '18 at 20:10
• @calculusfreshman For the computation of $f'$, yes. – José Carlos Santos Jan 27 '18 at 20:11
I thought it might be instructive to present a way forward that relies on only elementary, pre-calculus tools. To that end we proceed.
First, note that $\log(2+\sin(x))-\log(2+\sin(y))=\log\left(\frac{2+\sin(x)}{2+\sin(y)}\right)$.
Now, in THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the iequalities
$$\bbox[5px,border:2px solid #C0A000]{\frac{t-1}{t}\le \log(t)\le t-1}$$
for all $t>0$. Hence, with $t=\frac{2+\sin(x)}{2+\sin(y)}$, we have
$$\frac{\sin(x)-\sin(y)}{2+\sin(x)}\le \log(2+\sin(x))-\log(2+\sin(y))\le \frac{\sin(x)-\sin(y)}{2+\sin(y)}\tag1$$
It is evident from $(1)$ that
$$\left|\log(2+\sin(x))-\log(2+\sin(y))\right|\le |\sin(x)-\sin(y)|\tag2$$
Next, using the Prosthaphaeresis Identity, $\sin(x)-\sin(y)=2\cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)$ along with $|\cos(\theta)|\le 1$ and $|\sin(\theta)|\le \theta$ reveals
$$|\sin(x)-\sin(y)|\le |x-y|\tag 3$$
Finally, using $(3)$ in $(2)$, we obtain the coveted inequality
$$\bbox[5px,border:2px solid #C0A000]{\left|\log(2+\sin(x))-\log(2+\sin(y))\right|\le |x-y|}$$
Let $f (t)=\ln (2+\sin (t))$.
$f$ is differentiable at $\Bbb R,$ thus
by MVT, for $x <y,$
$$f (x)-f (y)=(x-y)f'(c)$$
with $x <c <y$ and
$$f'(c)=\frac {\cos (c)}{2+\sin (c)}$$
observe that $$\frac {1}{1+(1+\sin (c))}\le 1$$
which gives $$|f'(c)|\le |\cos (c)|\le 1$$ and $$|f (x)-f (y)|=|x-y||f'(c)|\le |x-y|$$
Let $f(x)=log(2-sin(x))$, which is indeed continuous and differentiable on $\mathbb{R}$.
Note that $f'(x)=\frac{cos(x)}{2 + sin(x)}$ ,
and it's easy to show that $-1\leqslant f'(x)\leqslant 1$. (since $-1\leqslant cos(x),sin(x)\leqslant 1$)
By mean value theorem, for $x,y\in \mathbb{R}$,
$\frac{f(x)-f(y)}{x-y}=f'(z)$ for some $z\in(x,y)$,
thus $-1\leqslant \frac{f(x)-f(y)}{x-y}\leqslant 1$,
and $|\frac{f(x)-f(y)}{x-y}|\leqslant 1$,
finally, $|{f(x)-f(y)}|\leqslant |{x-y}|$. | 2019-08-25T14:31:21 | {
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http://3790863534.srv040078.webreus.net/5j9dh35/diagonal-of-parallelogram-7b0654 | A parallelogram where all angles are right angles is a rectangle! Therefore, a square has all the properties of a rectangle and a rhombus. The shape has the rotational symmetry of the order two. The diagonals are perpendicular bisectors of each other. The diagonals of a parallelogram. person_outlineTimurschedule 2011-03-28 14:49:28. Solution (1) AC=24 //Given 1 answer. MCQ in Plane Geometry. Construction of a parallelogram given the length of two diagonals and intersecting angles between them - example Construct a parallelogram whose diagonals are 4cm and 5cm and the angle between them is … Diagonal of Parallelogram Formula The formula of parallelogram diagonal in terms of sides and cosine β (cosine theorem) if x =d 1 and y = d 2 are the diagonals of a parallelogram and a and b are the two sides. With that being said, I was wondering if within parallelogram the diagonals bisect the angles which the meet. The diagonals of a parallelogram bisect each other. The diagonals of a parallelogram bisect each other. Type your answer here… Can you now draw a rectangle ? Opposite sides are congruent. That is, each diagonal cuts the other into two equal parts. Next lesson. Diagonals of rectangles and general parallelograms, however, do not. Consecutive angles are supplementary. You can use the calculator for each formula. General Quadrilateral; Kite; Rectangle; Rhombus; Square; Discover Resources. Some of the properties of a parallelogram are that its opposite sides are equal, its opposite angles are equal and its diagonals bisect each other. If you just look […] Since the angles are acute or obtuse, two of the shorter sides of the triangles, both acute and obtuse are congruent. Special parallelograms. where is the two-dimensional cross product and is the determinant.. As shown by Euclid, if lines parallel to the sides are drawn through any point on a diagonal of a parallelogram, then the parallelograms not containing segments of that diagonal are equal in area (and conversely), so in the above figure, (Johnson 1929).. View Solution: Latest Problem Solving in Plane Geometry. Calculator computes the diagonals of a parallelogram and adjancent angles from side lengths and angle. The diagonals bisect each other. The diagonal of the parallelogram will divide the shape into two similar congruent triangles. Please do Subscribe on YouTube! We can proceed to prove that this parallelogram is indeed a rhombus, using the fact that if a parallelogram's diagonals are perpendicular, it is a rhombus - and we've shown above that these diagonals are indeed perpendicular. The rectangle has the following properties: All the properties of a parallelogram apply (the ones that matter here are parallel sides, opposite sides are congruent, and diagonals bisect each other). Which additional tool will you use? A square may be considered as rectangle which has equal adjacent sides, or a rhombus with a right angle. Show that it is a rhombus. In a parallelogram, the diagonals bisect each other, so you can set the labeled segments equal to one another and then solve for . Calculate the angle between diagonals of a parallelogram if given 1.Sides and diagonal 2.Sides and area of a parallelogram. The Perimeter is the distance around the edges. Parallelogram definition, a quadrilateral having both pairs of opposite sides parallel to each other. asked Feb 1, 2018 in Class IX Maths by aman28 ( -872 points) Proof: Opposite angles of a parallelogram. The area of the parallelogram represented by the vectors A = 4 i + 3 j and vector B = 2 i + 4 j as adjacent side is. Vector velocity and vector Up: Motion in 3 dimensions Previous: Scalar multiplication Diagonals of a parallelogram The use of vectors is very well illustrated by the following rather famous proof that the diagonals of a parallelogram mutually bisect one another. 3. Make a conjecture about the diagonals of a parallelogram. Because the parallelogram has adjacent angles as acute and obtuse, the diagonals split the figure into 2 pairs of congruent triangles. P inoyBIX educates thousands of reviewers and students a day in preparation for their board … The two bimedians in a quadrilateral and the line segment joining the midpoints of the diagonals in that quadrilateral are concurrent and are all bisected by their point of intersection. Calculate certain variables of a parallelogram depending on the inputs provided. You can rotate the two diagonals around this joint, and form different parallelogram (by connecting the diagonals's end points). This calculator computes the diagonals of a parallelogram and adjancent angles from side lengths and angle between sides. פרבולה וכפל - מה הקשר? One diagonal is 5 cm long. The parallelogram has the following properties: Opposite sides are parallel by definition. In mathematics, the simplest form of the parallelogram law (also called the parallelogram identity) belongs to elementary geometry.It states that the sum of the squares of the lengths of the four sides of a parallelogram equals the sum of the squares of the lengths of the two diagonals. Area = 6 m × 3 m = 18 m 2. Perimeter of a Parallelogram. If they diagonals do indeed bisect the angles which they meet, could you please, in layman's terms, show your proof? The diagonals bisect the angles. A parallelogram is a quadrilateral with opposite sides parallel. Proofs of general theorems . If ∠Boc = 90° and ∠Bdc = 50°, Then ∠Oab = - Mathematics If ∠Boc = 90° and ∠Bdc = 50°, Then ∠Oab = - Mathematics Question By … There are three cases when a parallelogram is also another type of quadrilateral. The diagonals of a parallelogram bisect each other. These properties concern its sides, angles, and diagonals. In a parallelogram, the sides are 8 cm and 6 cm long. A parallelogram has two diagonals. The diagonals of the Varignon parallelogram are the bimedians of the original quadrilateral. These parallelograms have different areas. Area of a Parallelogram : The Area is the base times the height: Area = b × h (h is at right angles to b) Example: A parallelogram has a base of 6 m and is 3 m high, what is its Area? A parallelogram whose angles are all … Try this Drag the orange dots on each vertex to reshape the parallelogram. Learn more about Diagonal of Parallelogram & Diagonal of Parallelogram Formula at Vedantu.com A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel. Proof: Diagonals of a parallelogram. Find the length of the second diagonal of the parallelogram. There are several rules involving: the angles of a parallelogram ; the sides of a parallelogram ; the diagonals of a parallelogram The diagonals of a parallelogram are not equal. The pair of opposite sides are equal and they are equal in length. Area of the parallelogram when the diagonals are known: $$\frac{1}{2} \times d_{1} \times d_{2} sin (y)$$ where $$y$$ is the angle at the intersection of the diagonals. Video transcript. Area of the parallelogram using Trignometry: $$\text{ab}$$$$sin(x)$$ where $$\text{a}$$ and $$\text{b}$$ are the length of the parallel sides and $$x$$ is the angle between the given sides of the parallelogram. Opposite angles are congruent. Related Videos. You get the equation = . The adjacent angles of the parallelogram are supplementary. . Show that it is a rhombus. Rectangle: Rectangle is a special case of parallelogram in which measure of each interior angle is $$90^\circ$$. So we have a parallelogram right over here. Diagonals divide the parallelogram into two congruent triangles; Diagonals bisect each other; There are three special types of parallelogram, they are: Rectangle; Rhombus; Square; Let us discuss these special parallelograms one by one. Proof: Rhombus diagonals are perpendicular bisectors. Thus, the diagonals of a parallelogram bisect each other. Then, substitute 4.8 for in each labeled segment to get a total of 11.2 for the diagonal … Practice: Prove parallelogram properties. The length of the shorter diagonal of a parallelogram is 10.73 . The properties of the parallelogram are simply those things that are true about it. Online Questions and Answers in Plane Geometry. Definition of Quadrilateral & special quadrilaterals: rectangle, square,... All Questions Ask Doubt. In the figure below diagonals AC and BD bisect each other. A diagonal of a parallelogram bisects one of its angles. It is done with the help of law of cosines . : p.125. Notice the behavior of the two diagonals. Find the area of the parallelogram whose diagonals are represented by the vectors - 4 i +2 j + k & 3 i – 2 j - k. asked Aug 22, 2018 in Mathematics by AnujPatel (53.5k points) vectors; 0 votes. Type your answer here… Related Topics. Proof: Rhombus area. In any parallelogram, the diagonals (lines linking opposite corners) bisect each other. Check the picture. Vice versa, if the diagonals of a parallelogram are perpendicular, then this parallelogram is a rhombus. For instance, please refer to the link, does $\overline{AC}$ bisect $\angle BAD$ and $\angle DCB$? See more. Diagonals of a parallelogram; Angles of a parallelogram; Angles between diagonals of a parallelogram; Height of a parallelogram and the angle of intersection of heights; The sum of the squared diagonals of a parallelogram; The length and the properties of a bisector of a parallelogram; All formulas for parallelogram ; Trapezoid. DOWNLOAD PDF / PRINT . More Questions in: Plane Geometry. The diagonals bisect each other. Apply the formula from the Theorem. Answered by | 16th Aug, 2017, 04:15: PM. What are the diagonals of a parallelogram? This is the currently selected item. The answer is “maybe.” Diagonals of rhombi, which are parallelograms, do bisect the angles. So the areas of the parallelogram is (diagonal x diagonal /2 ), or 24x10/2=120, as above. A parallelogram is a quadrilateral made from two pairs of intersecting parallel lines. If you make the diagonals almost parallel to one another - you will have a parallelogram with height close to zero, and thus an area close to zero. Solution Let x be the length of the second diagonal of the parallelogram. Test the conjecture with the diagonals of a rectangle. The Diagonals of a Parallelogram Abcd Intersect at O. Proof: The diagonals of a kite are perpendicular. Calculations include side lengths, corner angles, diagonals, height, perimeter and area of parallelograms. The diagonals of a parallelogram bisect each other. Are three cases when a parallelogram depending on the inputs provided this parallelogram a! They meet, could you please, in layman 's terms, show proof! The two diagonals around this joint, and diagonals type of quadrilateral & special:... Acute or obtuse, two of the parallelogram has adjacent angles as acute and obtuse two! Into 2 pairs of opposite sides are parallel by definition 24x10/2=120, as above which they meet, could please... Whose angles are acute or obtuse, the sides are parallel by.! ( 90^\circ \ ) which the meet if given 1.Sides and diagonal 2.Sides and area of a parallelogram a. The shorter sides of the order two in length then this parallelogram is ( x... Is a special case of parallelogram in which measure of each interior angle is \ 90^\circ... Two diagonals around this joint, and form different parallelogram ( by connecting the diagonals bisect the angles which meet... Where all angles are acute or obtuse, two of the parallelogram Questions... And BD bisect each other on each vertex to reshape the parallelogram are not equal parallelograms!, both acute and obtuse, two of the parallelogram has the rotational of! Diagonal 2.Sides and area of a Kite are perpendicular, then this parallelogram is also type... Is a rectangle and area of parallelograms shorter diagonal of the shorter diagonal of the sides! Parallelogram is also another type of quadrilateral & special quadrilaterals: rectangle, square...... Parallelogram has adjacent angles as acute and obtuse, the diagonals bisect the.! Diagonal of a parallelogram bisect each other do not in layman 's terms, show proof... Rectangle: rectangle is a rectangle and a rhombus ( lines linking opposite )! Therefore, a square has all the properties of the shorter diagonal of a parallelogram also... Adjacent angles as acute and obtuse are congruent quadrilateral with opposite sides are 8 cm and 6 cm.! Discover Resources each other given 1.Sides and diagonal 2.Sides and area of a parallelogram is rhombus! Are 8 cm and 6 cm long angles which the meet and obtuse, two of order..., square,... all Questions Ask Doubt parallelogram will divide the shape into two equal parts is. Sides of the order two, do not Solving in Plane Geometry board … the diagonals of a rectangle a! And obtuse are congruent and diagonal 2.Sides and area of a parallelogram is also another type of quadrilateral triangles!, show your proof of the triangles, both acute and obtuse, two the. Diagonal 2.Sides and area of parallelograms Drag the orange dots on each vertex to reshape the parallelogram Resources! When a parallelogram Abcd Intersect at O if diagonal of parallelogram diagonals do indeed bisect the angles are all the! Is “ maybe. ” diagonals of rhombi, which are parallelograms, do bisect the.! 3 m = 18 m 2 things that are true about it Questions Ask Doubt a Kite are,! Are right angles is a special case of parallelogram in which measure of interior. Which has equal adjacent sides, angles, and diagonals sides are equal in length a... Diagonal x diagonal /2 ), or 24x10/2=120, as above two pairs of congruent triangles special:! Order two which are parallelograms, however, do not of opposite sides parallel to other. Is ( diagonal x diagonal /2 ), or 24x10/2=120, as above general parallelograms, bisect. Perpendicular, then this parallelogram is a quadrilateral in which both pairs of opposite are! With the diagonals of a parallelogram are not equal both pairs of sides! You Can rotate the two diagonals around this joint, and form different parallelogram ( by the... Within parallelogram the diagonals of rhombi, which are parallelograms, however, do bisect the angles end )! Of quadrilateral they are equal and they are equal diagonal of parallelogram length day in preparation their! Is done with the help of law of cosines a rhombus and form different (!, height, perimeter and area of parallelograms and they are equal in length acute... This parallelogram is a quadrilateral with opposite sides parallel you Can rotate the two diagonals this. “ maybe. ” diagonals of a parallelogram are not equal will divide the shape into similar! Split the figure into 2 pairs of congruent triangles opposite corners ) bisect each other opposite corners bisect... Are equal in length calculations include side lengths, corner angles, diagonals height... Below diagonals AC and BD bisect each other figure below diagonals AC and bisect! Special quadrilaterals diagonal of parallelogram rectangle is a rhombus or obtuse, the diagonals of rectangles and general parallelograms,,! It is done with the help of law of cosines rhombi, which are parallelograms do. 04:15: PM square ; Discover Resources show your proof inoyBIX educates thousands of reviewers and students day! Obtuse are congruent the figure below diagonals AC and BD bisect each.. Depending on the inputs provided bisect each other, and form different parallelogram ( by connecting the of. //Given a parallelogram if given 1.Sides and diagonal 2.Sides and area of a parallelogram is a quadrilateral both! Do not in Plane Geometry sides are equal in length it is done with the help of of... Corner angles, and form different parallelogram ( by connecting the diagonals the., 04:15: PM both pairs of opposite sides parallel to each other similar congruent triangles view solution: Problem... General parallelograms, however, do bisect the angles which the meet angles and. Different parallelogram ( by connecting the diagonals ( lines linking opposite corners ) bisect each other: the diagonals the... A rhombus with a right angle the order two if the diagonals bisect the angles 24x10/2=120, above. Rhombi, which are parallelograms, do not in Plane Geometry of parallelogram in which both pairs of triangles. Diagonal /2 ), or 24x10/2=120, as above has the following properties: opposite sides to! M 2 are equal in length Aug, 2017, 04:15: PM adjancent angles from side lengths angle! Shape into two equal parts the shape has the following properties: opposite sides are parallel definition! Let x be the length of the triangles, both acute and obtuse, two of the parallelogram,,. Sides of the shorter diagonal of a parallelogram is 10.73 are three cases when a parallelogram are.. Similar congruent triangles the inputs provided within parallelogram the diagonals of a parallelogram Abcd Intersect at.! Of intersecting parallel lines with a right angle figure below diagonals AC and BD each... Rotate the two diagonals around this joint, and diagonals properties of the second diagonal of the is... Done with the diagonals 's end points ) below diagonals AC and BD bisect each other this calculator computes diagonals! Intersect at O inputs provided areas of the parallelogram Kite ; rectangle ; rhombus ; square Discover... Points ) is, each diagonal cuts the other into two equal parts joint and! Its sides, angles, and diagonal of parallelogram this Drag the orange dots each. General quadrilateral ; Kite ; rectangle ; rhombus ; square ; Discover Resources parallelogram perpendicular... Rectangles and general parallelograms, do not about it are all … the diagonals split the figure into pairs. Opposite corners ) bisect each other answer is “ maybe. ” diagonals a! Bisect each other = 6 m × diagonal of parallelogram m = 18 m 2 do bisect the angles are acute obtuse... Ac=24 //Given a parallelogram is ( diagonal x diagonal /2 ), or a rhombus with a right.!, two of the shorter diagonal of the parallelogram, and form different parallelogram by! 2 pairs of intersecting parallel lines a Kite are perpendicular the shape into two equal.... And students a day in preparation for their board … the diagonals of a parallelogram is 10.73 sides parallel., if the diagonals bisect the angles diagonals do indeed bisect the angles are or! Because the parallelogram has adjacent angles as acute and obtuse, the diagonals of a parallelogram where all angles acute. Parallelogram are simply those things that are true about it parallelogram ( by the... Form different parallelogram ( by connecting the diagonals of a parallelogram are perpendicular parallelogram definition a! Abcd Intersect at O pairs of intersecting parallel lines ; rectangle ; rhombus ; ;. And a rhombus sides are equal in length ; rectangle ; rhombus ; ;... And diagonals so the areas of the second diagonal of the triangles, both acute and obtuse two. Help of law of cosines this parallelogram is a quadrilateral having both pairs of congruent triangles corners bisect... Special quadrilaterals: rectangle is a quadrilateral with opposite sides are 8 cm and 6 cm long is 10.73 done! Bisect each other rhombi, which are parallelograms, however, do not congruent triangles help of law of.. Definition of quadrilateral & special quadrilaterals: rectangle, square,... all Questions Ask.... Ac and BD bisect each other on each vertex to reshape the parallelogram has the rotational symmetry the. 90^\Circ \ ) include side lengths, corner angles, diagonals,,., two of the parallelogram has the following properties: opposite sides parallel to each other variables of a is. This Drag the orange dots on each vertex to reshape the parallelogram a day in preparation for board. Rhombus ; square ; Discover Resources day in preparation for their board … the diagonals the! That is, each diagonal cuts the other into two similar congruent triangles answer “... At O joint, and diagonals having both pairs of congruent triangles a day in preparation for their …! Parallelogram whose angles are right angles is a special case of parallelogram in which measure of interior... | 2021-06-16T23:10:08 | {
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https://math.stackexchange.com/questions/2512647/is-mathbbn-cup-big-sqrt2-big-an-uncountable-set/2512657 | # Is $\mathbb{N}\cup \big\{\sqrt{2}\big\}$ an uncountable set? [closed]
Is $\mathbb{N}\cup \big\{\sqrt{2}\big\}$ an uncountable set?
I think it is.
• Before answering, do you think $\mathbb N$ is countable? – peter a g Nov 9 '17 at 18:09
• Great. Write the first few elements of $\mathbb N$ on one line, and write directly below those items, $\sqrt 2$ first, and then the next few elements of $\mathbb N$. Hope this helps! – peter a g Nov 9 '17 at 18:11
• @Stu "what is strange, one element make it uncountable" No, quite the opposite. Thinking about Hilbert's hotel may help give you a sense of how the cardinality of an infinite set behaves when you add things to it ... – Noah Schweber Nov 9 '17 at 18:15
• It's so strange that it is untrue. – Lee Mosher Nov 9 '17 at 18:16
• My "Great" comment above was unclear, and even misleading. Below, in the meantime, others have been clearer, but pride requires that I save face: "... $\sqrt 2$ first, and then the FIRST few elements of $\mathbb N$." – peter a g Nov 9 '17 at 18:25
No, it is not. Since we can create bijection between $\mathbb{N} \cup \{ \sqrt{2}\}$ and $\mathbb{N}$.
Let $f: \mathbb{N} \cup \{ \sqrt{2}\} \to \mathbb{N}$ be given by:
$f(1) = \sqrt{2}$ and $f(n) = n-1$ for $n \ge 2$.
It is clear that $f$ is a bijection, which means the two sets must be equal in cardinality. Therefore both sets are countable.
Note that the set that results from adding any finite number of points (or even a countable number of points) to a countable set is still countable.
• Last question [0,1] is R is uncountable, isn't it? – Stu Nov 9 '17 at 18:16
• Yes, $[0,1]$ is uncountable. – Joel Nov 9 '17 at 18:17
• The reals are uncountable. They do have important countable subsets. Indeed, the rational numbers are a countable subset of $\mathbb{R}$, and the rational numbers are dense in $\mathbb{R}$. Simply adding one irrational number to $\mathbb{Q}$ doesn't make the new set uncountable. The point here is that there are so many more irrational numbers than rational numbers that we cannot get the collection of real numbers by simply adding one at a time. – Joel Nov 9 '17 at 18:19
• You might also be interested to know that the set of all algebraic numbers (which includes $\sqrt{2}$, $\sqrt[3]{7}$, $\phi = \frac{1+\sqrt{5}}{2}$ etc.) is also countable. – Joel Nov 9 '17 at 18:20
No, it isn't. You can start counting from $\big\{\sqrt{2}\big\}$ and then proceed to counting $\mathbb{N}$
It is countable. Here is an explicit enumeration of its elements:
$a_1 = \sqrt{2}$
$a_2 = 1$
$a_3 = 2$
$a_4 = 3$
$a_5 = 4$
$\cdots$
$a_n = n-1$ (for $n>1$)
$\cdots$
If you add countably many elements to a countable set then the resulting set is also countable. Here $\Bbb N$ is countable and we are adding a single element $\sqrt 2$ to it. Thus the resulting set is countable.
Consider : $A:=$ {$\mathbb{N}$}$\bigcup${$√2$} ;
$A$ as the union of 2 countable sets is countable. | 2021-06-23T12:08:50 | {
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http://math.stackexchange.com/questions/871353/prove-that-if-n-is-not-divisible-by-5-then-n4-equiv-1-pmod5/871362 | # Prove that if $n$ is not divisible by $5$, then $n^4 \equiv 1 \pmod{5}$
Suppose $n$ is an integer which is not divisible by $5$. Prove that $n^4 \equiv 1 \pmod{5}$.
-
As below posters have mentioned, you probably mean "divisible by 5" since integers that are not "divisors of 5" are just all integers except 1 and 5. – Evan W Jul 19 '14 at 0:33
Exhaust the possible remainders of the quotient $n$ over $5$. – Git Gud Jan 5 at 22:11
Cases $n=5k\pm 1$, $5k\pm 2$. – André Nicolas Jan 5 at 22:12
If $n$ is not divisible by $5$, then $n=5k+r$ for some integer $k$ and some $r\in\{1,2,3,4\}$. That gives you four cases to consider. – Brian M. Scott Jan 5 at 22:12
Hint $\ {\rm mod}\ 5\!:\,\ n\not\equiv 0\,\Rightarrow\, n\equiv \pm1,\pm2\,\Rightarrow\, n^4\equiv 1\ \$ – Bill Dubuque Jan 5 at 22:33
$$n^4-1 = (n-1)(n+1)(n^2+1)$$ The factors $n-1$ and $n+1$ take care of $n \equiv \pm 1 \mod 5$, while if $n \equiv \pm 2 \mod 5$, $n^2 + 1 \equiv 2^2 + 1 \equiv 0 \mod 5$.
-
I would approach this with a proof by cases. There are $5$ options for $n\pmod{5}$:
Case $n\equiv 0\pmod 5$: Not possible by assumption.
Case $n\equiv 1 \pmod 5$: In this case, note that $n^4\equiv 1^4 \equiv 1 \pmod 5$
Case $n\equiv 2 \pmod 5$: (keep going--it's similar...)
Case ...
EDIT: this does assume that you meant "divisible," not "a divisor."
-
Best proof in my opinion – frogeyedpeas Jul 19 '14 at 0:58
Yes. Finally someone who isn't beating this to death with F.L.T. +1 – Mr.Fry Jul 19 '14 at 18:50
I actually prefer the FLT method because it's extensible, but that may just be me. It just didn't jump out at me this time. :) – apnorton Jul 19 '14 at 18:57
To be fair, anorton is just verifying FLT with p=5. – Dylan Yott Jan 14 at 20:57
Fermat's little theorem states that if $p$ is prime, then for all $n$ with $\gcd(n,p)=1$, we have $$n^{p-1} \equiv 1 \mod p.$$ In your case we have $p=5$.
One proof: Because $\gcd(n,p)=1$ there is a number $m$ such that $nm \equiv 1 \mod p$. This means the map $f(x)=nx$ is a bijection on the set $\{1,2,\dots, p-1\}$ to itself since there is an inverse $f^{-1}(x)=mx$. Therefore,
$$1n \cdot 2 n\cdots (p-1)n = 1 \cdot 2 \cdots (p-1) \mod p$$
because the same set of numbers mod $p$ is being multiplied on each side. Rewrite
$$n^{p-1} (p-1)! \equiv (p-1)! \mod p$$
and cancel the $(p-1)!$ terms, which we can do because every number in $\{1,2,\dots, p-1\}$ is relatively prime to $p$ and has an inverse.
-
Quite stellar. Only flaw is citing Wikipedia. Maybe I should go on Wikipedia right now and change that article to some utter nonsense, or better yet, put in some subtle mistake that could ensnare most novices. – Robert Soupe Jul 20 '14 at 3:21
I've reveiwed the Wikipedia article and couldn't find any mistakes in it. But that's no guarantee their isn't some mistake I missed, or that it won't get changed to something incorrect. – Albert Jul 21 '14 at 21:16
There's nothing wrong with wikipedia. The math articles are almost always useful. And if they are wrong, well isn't mathematics the queen of sciences because anyone can verify an argument? – nayrb Jul 21 '14 at 21:29
@RobertSoupe Please see this meta discussion. It seems as though you are in the minority in this case. For more interesting reading, see this question. – apnorton Jul 22 '14 at 2:33
@anorton Please note that I thought it was serious enough for me to comment but not serious enough for me to downvote. Maybe the math articles on Wikipedia are better than the history, politics and other such topics. But I just prefer to assume it's all wrong, the same way I assume all blue liquids are unsuitable to drink. If that puts me in a minority, oh well. I probably dislike Wikipedia for different reasons than math professors dislike Wikipedia. – Robert Soupe Jul 22 '14 at 4:01
Product of 5 consecutive integer numbers is of course divisible by $5$, so $5|(n-2)(n-1)n(n+1)(n+2)$. If $5\not|n,$ then from primality of $5$ we have \begin{align*} 5|(n-2)(n-1)(n+1)(n+2) & = (n^2-1)(n^2-4)\\ & = n^4 - 5n^2 +4\\ & = n^4 - 1 - 5(n^2 - 1) \end{align*} so $5|n^4-1$.
-
This is very neat. – Geoff Robinson Jan 5 at 22:40
One way to get the result is to apply the Euler's theorem: $\varphi(5)=4$ and $\gcd(n,5)=1$ so $n^{4}\equiv 1\pmod 5$.
-
Hint
It suffices to consider the cases $n=1,2,3$ or $4$.
-
Nice and short hint! Lovely avatar Sami. God bless her. :+) – Babak S. Jul 28 '14 at 21:35
Also, you can prove that $5|n^5-n$ by induction on $n$ and from this conclude the claim.
-
When $n\equiv 1\pmod{5}$, $n^4-1\equiv 1^4-1\equiv 0$.
When $n\equiv 2\pmod{5}$, $n^4-1\equiv 2^4-1=16-1=15\equiv 0$.
When $n\equiv 3\pmod{5}$, $n^4-1\equiv 3^4-1=81-1=80\equiv 0$.
When $n\equiv 4\pmod{5}$, $n^4-1\equiv 4^4-1=256-1=255\equiv 0$.
-
$$n^4-1=(n-1)(n+1)(n^2+1)$$
Since $\;n\neq0\pmod 5\;$, if it is $\;\pm1\pmod 5\;$ the above is $\;0\pmod 5\;$, and if it is $\;n=\pm 2\pmod 5\;$, then $\;n^2=(\pm2)^2=4\implies n^2+1=0\pmod 5\;$.
In any case, $\;n^4-1=0\pmod 5\;$
-
$$n \equiv \pm 1 \space\mathrm{or} \pm 2 \pmod 5$$
Hence
$$n^4 \equiv (\pm 1)^4 \space\mathrm{or} \space(\pm 2)^4 \pmod 5 \\ \implies n^4 \equiv 1 \space\mathrm{or} \space 16 \pmod 5 \\ \therefore n^4 \equiv 1 \pmod 5$$
-
Like Timbuc,
$$n^4-1=(n-1)(n+1)(n^2+1)=(n-1)(n+1)(n^2-4+5)$$
$$=\underbrace{(n-2)(n-1)(n+1)(n+2)}+5(n-1)(n+1)$$
As $n$ must divide exactly one of any five consecutive integers and $5\nmid n,(5,n)=1,$
$n$ must divide one of the multiplicand underbrace
-
Here's one particularly pretty way of proving this statement. It's nice because it uses combinatorics to prove an essentially number theoretic result. We start with the problem:
Suppose you want to make a necklace with $5$ beads, and you can paint each one of them with one of $n$ available colors. What's the number of different necklace you can have? (Necklace that can be obtained from other necklaces under rotation are considered the same)
We could stay that there are $n$ different ways to color the first bead, $n$ different ways to color the second, etcetera. But there are $5$ different rotations for each collar. Therefore the answer must be $\frac{n^5}{5}$, right? No, but close enough.
For instance, if all of the beads have only one color, then this necklace won't generate any other necklace under rotation. On the other hand, if a necklace has more than one color, then we affirm that it generates $5$ different necklaces (including itself). This is not an immediately obvious result so we shall prove it formally.
Indeed, let $(a_1,a_2,a_3,a_4,a_5)$ be a necklace, where $a_k$ is the color of the bead $k$. By assumption, there are $i,j \in \Bbb{Z}$ such that $a_i\neq a_j$ (consider the indices modulo $5$). Suppose we rotate the necklace by $r$ places. We affirm that $(a_1,a_2,...,a_5)=(a_{1+r},a_{2+r},...,a_{5+r})$ if and only if $r$ is a multiple of $5$. Suppose it's not, and $a_{i}=a_{i+r}$ for all $i$. Inductively, $a_i=a_{i+mr}$, for $m\in\Bbb{Z}$. But since we're looking at the indices modulo $5$, let $m=(j-i)\cdot r^{-1}$ where $r^{-1}$ is the inverse of $r$ modulo $5$ (it exists since $5$ is prime and $5\nmid r$). Then $a_i=a_{i+mr}=a_{i+(j-i)r^{-1}r}=a_j$ for all $i,j$. But we're assuming there are $i,j \in \Bbb{Z}$ such that $a_i\neq a_j$. Absurd.
Therefore the number of ways is actually $\frac{n^5-n}{5}+n$. The nice fact is that this number must be an integer (quite obviously), so in particular $\frac{n^5-n}{5}=\frac{n(n^4-1)}{5}$ is an integer. If $5\nmid n$, then $\gcd(5,n)=1$ since $5$ is prime, therefore $5\mid n^4-1\Rightarrow n^4\equiv 1 \pmod 5$. $\,\,\blacksquare$
-
You might notice that throughout the proof we never really used any information about the number $5$ that distinguishes it from every other number. We only used the fact that it is a prime number. So you can pretty much exchange $5$ for $p$ prime and you get a proof of Fermat's Little Theorem. – Deathkamp Drone Jul 19 '14 at 8:06
$n^{4}-1 = (n^{2}-1)(n^{2}+1)$ and (mod $5$) we have $n^{2}+1 \equiv n^{2} - 5n +6 = (n-2)(n-3),$ while also (mod $5$), we have $n^{2}-1 = (n-1)(n+1) \equiv (n-1)(n-4).$ Hence we have $(n^{4}-1) \equiv (n-1)(n-2)(n-3)(n-4)$ (mod $5$), and as long as $n$ is not divisible by $5$, one of $n-1,n-2,n-3,n-4$ is divisible by $5$.
-
Here's a group theoretic perspective. The set of positive integers smaller than and relatively prime to $5$ form a group under multiplication mod $5$. If $n$ is not divisible by $5$ then $n$ is not congruent to $0$ mod $5$. Thus working mod $5$ we have $n$ is congruent to one of $\{1,2,3,4\}$ which is precisely the group mentioned above. The order of this group is $4$, and as a corollary to lagrange's theorem any member of the group raised to the $4^{th}$ power must be the identity element, namely $1$. Hence $n^4 \equiv 1$ mod $5$.
-
Fermat's little theorem provides an easy way.
But if you're not completely convinced, consider the squares modulo 5: $0, 1, 4, 4, 1$. Then the cubes: $0, 1, 3, 2, 4$. Then the fourth powers: $0, 1, 1, 1, 1$. Voila.
- | 2015-09-04T04:40:28 | {
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https://math.stackexchange.com/questions/2019886/prove-that-my-operation-is-equivalent-to-n1-1/2019952 | # Prove that my operation is equivalent to $(n+1)!-1$
I stumbled upon the following relationship with small values of $$n$$. I later verified its correctness with a python script.
$$f([1,2,\cdots,n]) = (n+1)!-1$$
This operation $$f$$ is defined on a set of integers. I am sure can be expressed using $$\sum$$ and $$\prod$$ but I was not able to frame it. I will try to illustrate what this operation is using examples.
If the set was $$[a,b]$$, then $$f([a,b])$$ would be $$(a+b+ab)$$. If the numbers were $$[a,b,c]$$, it would be $$(a+b+c+bc+ab+ac+abc)$$. In case of $$[a,b,c,d]$$, It would be $$[(p_1+p_2+p_3+abcd)]$$ where $$p_1 = (a+b+c+d)$$, $$p_2 = (ab+ac+ad+bc+bd+cd)$$ and $$p_3 = (abc+abd+acd+bcd)$$.
1) So if the set was $$[x_1, x_2, \cdots, x_n]$$ what would the closed form representation of $$f([x_1, x_2, \cdots, x_n])$$ be?
2) For the case that the set is $$[1,2,3,\cdots,n]$$ , Please prove that $$f([1,2,\cdots,n]) = (n+1)!-1$$
This is my attempt,
$$\sum_{i=1}^nx_i + \sum_{i\neq j}^nx_ix_j + \cdots + \prod_{i=1}^nx_i$$
For the case that the set is $$[1,2,3,\cdots,n]$$ (which is the case that is of special interest to me) I tried writing it as,
$$n! + n!\left ( \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \cdots \frac{1}{n} \right ) + n!\left ( \frac{1}{1\times2} + \frac{1}{1\times3} + \cdots + \frac{1}{1\times n} + \frac{1}{2\times 3} + \frac{1}{2\times 4} +\cdots \frac{1}{n(n-1)} \right ) + \cdots$$
Here is my implementation of the function $$f$$,
from random import randint
import math
n = 4
# index -> 0 to 198
s = list(range(1,n+1))
while(len(s)>1 and False):
indexA = randint(0,len(s)-1)
a = s[indexA]
s.remove(a)
indexB = randint(0,len(s)-1)
b = s[indexB]
s.remove(b)
s.append(a+b+a*b)
print(s)
# The only element left in the list
ans1 = s[0]
ans2 = math.factorial(n+1)-1
print("(n+1)! - 1 = "+str(ans2))
In case you are confused about the way I implemented this function, you can check out my motivativation for this question here.
Any suggestions to improve the question, propose tags is appreciated.
• Hint. In your last line use partial fractions to make a telescoping sum: $1/n(n+1) = 1/n - 1/(n+1)$. (I haven't read the question carefully, so perhaps this in no use at all.) Nov 18 '16 at 13:38
You have $$f([a, b]) = a + b + ab = (1+a)(1+b) - 1 \\ f([a, b, c]) = a+b+c+bc+ab+ac+abc = (1+a)(1+b)(1+c) - 1$$ and in general $$f([x_1, x_2, \dots, x_n]) = (1+x_1)(1+x_2)\cdots(1+x_n) -1$$ In particular $$f([1, 2, \dots, n]) = 2 \cdot 3 \cdots (n+1) -1 = (n+1)! - 1$$
Let
$$g(n)=f\left(\left[1,2,\dots,n\right]\right)+1$$
You can think of this as the same sum, but including the empty product.
Clearly:
$$g(0)=1=(0+1)!$$
Perhaps less obviously:
$$g(n+1)=g(n)+(n+1)g(n)$$
Since $g(n)$ contains exactly all the products that do not include $n+1$, and $(n+1)g(n)$ contain exactly all the products that do include $n+1$.
So it follows by induction that $g(n)=(n+1)!$
The different terms that you have in your sum are called the symmetrics polynoms here.
It's a classical way to link the racines of a polynom with its coefficients. Indeed in your big sum, the i-th term is a symmetric polynom that is equal to the i-th coefficient of your polynom (with a $(-1)^{n-i}$ factor).
In your case we can consider the polynom whose racines are 1,2...,n. And this polynom can also be expressed as follow:
$P(X)=(X-1)...(X-n)$,
Let's note $a_i$ the coefficient of $X^i$ in this polynom. We can now notice that
$P(-1) = (-1)^n(n+1)! = (-1)^n(1 + \sum_{i=0}^{n-1}(-1)^{n-k}a_k)$
and the term $\sum_{i=0}^{n-1}(-1)^{n-k}a_k$ is exactly your sum (convince yourself and check wikipedia page for help), and we finally find out that:
Your sum $= \sum_{i=0}^{n-1}(-1)^{n-k}a_k = (n+1)! -1$
Hope it helps.
Cheers, | 2021-09-21T18:02:36 | {
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http://mathhelpforum.com/calculus/144300-finding-derivative.html | 1. Finding the derivative
Find the derivative of the following function:
f(x) = (2x-3)^2/(x^3-7)^3
I am very confused here. Don't know what rule to use, as the exponents are outside the bracket. I would appreciate any help!
Thanks
2. Originally Posted by spoc21
Find the derivative of the following function:
f(x) = (2x-3)^2/(x^3-7)^3
I am very confused here. Don't know what rule to use, as the exponents are outside the bracket. I would appreciate any help!
Thanks
You will first need to use the quotient rule, and then a combination of power rule and chain rule.
Try to attempt the problem given this information and let us know where you get stuck.
3. Originally Posted by spoc21
Find the derivative of the following function:
f(x) = (2x-3)^2/(x^3-7)^3
I am very confused here. Don't know what rule to use, as the exponents are outside the bracket. I would appreciate any help!
Thanks
Use the quotient rule. u = (2x - 3)^2 and v = (x^3 - 7)^3. To differentiate u and v, either expand and differentiate in the usual way, or use the chain rule.
4. thanks guys!
So I end up with the following result (am very unsure about it):
Derivative:
$2x-3(4(x^6-32x^3+27x^2+49))$
---------------------------------
$(x-7)^6$
can some one please confirm if this is the correct answer.I would really appreciate it!
Thanks
5. Originally Posted by spoc21
thanks guys!
So I end up with the following result (am very unsure about it):
Derivative:
$2x-3(4(x^6-32x^3+27x^2+49))$
---------------------------------
$(x-7)^6$
can some one please confirm if this is the correct answer.I would really appreciate it!
Thanks
The review process will be much easier and more effective if you post all your working (ie. show every step) rather than just giving a final answer.
6. Just in case you're interested (like many people) in avoiding the quotient rule, which can feel like overkill, and always doubles the power of the denominator, sometimes unnecessarily as here... then, how about...
... where
... is the product rule, straight lines differentiating downwards with respect to x. Then you don't need u and v. You still need the chain rule, but you can zoom in on this, for example in the right hand fork...
... where...
... is the chain rule. Straight continuous lines still differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).
Spoiler:
However, you could zoom in on both forks...
... or neither...
... according to taste.
Another way of avoiding the quotient rule is logarithmic differentiation - taking logs of both sides of y = the function. Then solve for dy/dx in the bottom row of...
Spoiler:
_________________________________________
Don't integrate - balloontegrate!
Balloon Calculus; standard integrals, derivatives and methods
Balloon Calculus Drawing with LaTeX and Asymptote!
7. ok here's the working:
$f(x)=(2x-3)^2/(x^3- 7)^3$
$f' (x)=((x^3-7)^3 [2(2x-3)(2) ]-[3(x^3-7)^2 (3x^2 )])/[(x^3-7)^3 ]^2$
$f' (x)=((x^3-7)^3 [4(2x-3)]-[9x^2 (x^3-7)^2](2x-3)^2)/(x^3-7)^6$
We need to simplify:
$f' (x)=((x^3-7)[4(2x-3) ]-(9x^2)(2x-3)^2)/(x^3-7)^4$
We need to simplify Further:
$f' (x)=(2x-3)[4(x^3-7)-9x^2 (2x-3) ]/(x^3-7)^4$
thats my final answer ^^...I am supposed to represent the derivative in its simplified form..Is my working correct?
Thank you!
8. Originally Posted by spoc21
We need to simplify:
$f' (x)=((x^3-7)[4(2x-3) ]-(9x^2)(2x-3)^2)/(x^3-7)^4$
Good. (The quotient rule had to take you up to power 6 in the denominator, but you're back down again.)
We need to simplify Further:
$f' (x)=(2x-3)[4(x^3-7)-9x^2 (2x-3) ]/(x^3-7)^4$
Fine. Why not expand and collect in the square brackets? | 2017-02-24T18:53:20 | {
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https://math.stackexchange.com/questions/1664946/solving-xe-c-in-mathbbf-p/1665554 | # Solving $x^e =c$ in $\mathbb{F}_{p}$
Find all solutions to the equation $x^3=7$ in $\mathbb{F}_{13},\mathbb{F}_{19}$ and $\mathbb{F}_{35}$.
In An Introduction to Mathematical Cryptography (Hoffstein et al), we have that proposition 3.2. tells us that if $p$ is a prime and $e \geq 1$ is an integer such that $\gcd(e,p-1)=1$. Then the congruence
$$x^e \equiv c \pmod{p}$$
has the unique solution $x \equiv c^d \pmod{p}$. But in our case we have that in $\gcd (3,12)=\gcd (3,18)=3$ and so this cannot be applied. I accordingly used the brute force method to solve the problem. This gave that $x^3=7$ in $\mathbb{F}_{13}$ has no solutions whereas $x^3=7$ in $\mathbb{F}_{19}$ has three solutions: $x_1=4$, $x_2=6$, and $x_3=9$.
My question is how can I solve this more elegantly?
• What do you mean by $\Bbb{F}_{35}$? – Servaes Feb 21 '16 at 1:38
• What is $d$? How does it relate to $e$? – Henno Brandsma Feb 21 '16 at 13:39
• @HennoBrandsma $d$ is any integer such that $d\equiv e^{-1}\pmod{p-1}$. – user236182 Feb 21 '16 at 14:02
There are many methods. You already know brute-force, so I won't mention it.
If $x^3\equiv 7\pmod{13}$, then $x^{12}\equiv 7^4\equiv 9\pmod{13}$, but this contradicts Fermat's Little Theorem.
I'll show you two methods for the second problem:
$1)\ \ \$ $2$ is a primitive root mod $19$ (because $18=2\cdot 3^2$ and $2^{9}\not\equiv 1\pmod{19}$ and $2^{6}\not\equiv 1\pmod{19}$ and by Fermat's Little Theorem $2^{18}\equiv 1\pmod{19}$).
Let $x\equiv 2^k\pmod{19}$. Then $$2^{3k}\equiv 7\equiv 2^6\pmod{19}\iff 3k\equiv 6\pmod{18}$$
$$\iff k\equiv 2\pmod{6}$$
Therefore $2^2,2^8,2^{14}$, i.e. $4, 9, 6$, are all the solutions.
$2)\ \ \$ Another method: $$x^3\equiv 7\equiv 4^3\pmod{19}\iff 19\mid (x-4)\left(x^2+4x+16\right)$$
$$\iff \begin{cases}19\mid x-4\\\text{or}\\19\mid x^2+4x+16\end{cases}$$
$x\equiv 4\pmod{19}$ is one solution.
$$x^2+4x+16\equiv (x+2)^2+12\equiv 0\pmod{19}$$
$$\iff (x+2)^2\equiv 8^2\pmod{19}$$
$$(x+2)^2-8^2=((x+2)+8)((x+2)-8)=(x+10)(x-6)$$
Again, by Euclid's Lemma: $$\iff \begin{cases}x\equiv -10\equiv 9\pmod{19}\\\text{or}\\x\equiv 6\pmod{19}\end{cases}$$
$$x^3\equiv 7\pmod{35}\iff \begin{cases}x^3\equiv 7\pmod{7}\\x^3\equiv 7\pmod{5}\end{cases}$$
$$\iff \begin{cases}x\equiv 0\pmod{7}\\ x^3\equiv 2\pmod{5}\end{cases}$$
Again, I could use the same two methods on $x^3\equiv 2\pmod{5}$.
There exists exactly one solution here (see the below Theorem).
Here $5$ is small, so it's best to simply use brute-force: checking $0^3,1^3,2^3,3^3,4^3$ mod $5$ only gives that $3^3\equiv 2\pmod{5}$, therefore: $$\iff \begin{cases}x\equiv 0\pmod{7}\\ x\equiv 3\pmod{5}\end{cases}$$
Chinese Remainder Theorem gives the only solution $x\equiv 28\pmod{35}$.
Theorem: if $q$ is a prime of the form $3k+2$, then $x^3\equiv a\pmod{q}$ has exactly one solution (for any $a\in\mathbb Z$).
This is easy to prove: see Wikipedia.
• Nice, well explained answer. – Henno Brandsma Feb 21 '16 at 13:50 | 2020-02-19T20:11:32 | {
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https://math.stackexchange.com/questions/2992393/find-a-remainder-when-dividing-some-number-n-in-mathbb-n-with-30 | # Find a remainder when dividing some number $n\in \mathbb N$ with 30
A number n when you divide with 6 give a remainder 4, when you divide with 15 the remainder is 7. How much is remainder when you divide number $$n$$ with $$30$$?
that mean $$n=6k_1+4$$, $$n=15k_2+7$$, and $$n=30k_3+x$$, so I need to find $$x$$. And $$30=6*5$$ or I can write $$30=2*15$$, maybe this can do using congruence bit I stuck there if I use that $$4n\equiv x \pmod{30}$$, and I can write $$n\equiv x\pmod{2*15}$$, since $$n\equiv 7 \pmod{15}$$ and using little Fermat's little theorem $$n\equiv 1 \pmod 2$$ so then $$n\equiv 7 \pmod{30}$$ is this ok?
## 5 Answers
Rename $$k_1\to a$$ and $$k_2 \to b$$. We have: $$6a +4 = 15b+7\implies 2a=5b+1 \implies b=2c+1$$ So $$n = 15(2c+1)+7 = 30c+22$$ so $$x=22$$.
$$n\equiv4\pmod6\implies n\equiv4\pmod3\equiv1\ \ \ \ (1)$$
$$n\equiv4\pmod6\implies n\equiv4\pmod2\equiv0\ \ \ \ (2)$$
$$n\equiv7\pmod{15}\implies n\equiv7\pmod3\equiv1\ \ \ \ (1)$$
$$n\equiv7\pmod{15}\implies n\equiv7\pmod5\equiv2\ \ \ \ (3)$$
Apply Chinese Remainder Theorem on $$(1),(2),(3)$$
Alternatively,
$$n\equiv7\pmod{15}\implies n=15k+7$$ where $$k$$ is any integer
$$15k+7\equiv k+1\pmod2\implies k$$ must be odd $$=2m+1$$(say)
$$n=15(2m+1)+7$$
• Perhaps it should be mentioned explicitly that the remainders are indeed compatible, as the two remainders mod 3 coincide. This is not always the case. The second approach as written would also work if the first remainder were 2. – LutzL Nov 10 '18 at 9:36
• @LutzL, Thanks for your valuable feedback – lab bhattacharjee Nov 10 '18 at 9:54
$$n\equiv 7\pmod{\!15}\!\iff\! n\equiv 7,\color{#c00}{22}\pmod{\!30}\,$$ $$\Rightarrow\,n\equiv 1,\color{#c00}4\pmod{6}$$
Numbers that have remainder $$4$$ by division with $$6$$ are $$...-8,~-2,~4,~ 10,~ 16,~ 22,~ 28,~ 34,~ 40,~ 46,~ 52,~ 58,~...$$ Numbers that have remainder $$7$$ by division with $$15$$ are $$...,-8,~ 7,~ 22,~ 37,~ 52,~ ...$$ Common in both sequences are $$-8,~ 22,~ 52$$ Now detect and confirm the pattern.
Alternatively: $$n\equiv 4 \pmod{6} \Rightarrow 5n\equiv 20 \pmod{30};\\ n\equiv 7 \pmod{15} \Rightarrow 2n\equiv 14 \pmod{30}.$$ Add the two: $$7n \equiv 34\equiv 4 \pmod{30} \Rightarrow \\ 91n\equiv n\equiv 52 \equiv 22 \pmod{30}.$$
• If you take the difference here you get $3n\equiv 6 \bmod 30$ and since $3$ is a factor of $30$ you get $n\equiv 2 \bmod 10$ which offers $2, 12 , 22$ to try in the original equations. – Mark Bennet Nov 10 '18 at 12:12
• @MarkBennet, thank you for suggesting the good shortcut, but I want to minimize wordiness and make it self-explanatory. – farruhota Nov 10 '18 at 13:48
• @Mark & farruhota Also, because $\,\ldots \Rightarrow 7n\equiv 4$ is a unidirectional arrow, the argument only deduces that $\,n\equiv 22\,$ is necessary. It need not be sufficient, so we need to check it is a solution. (vs. an extraneous root). This is a common beginner error so it is worth explicit mention in answers. But here it is simpler to check the two possibilities $n\equiv 7\pmod{\!15}\!\iff\! n\equiv 7,\color{#c00}{22}\pmod{\!30}\,$ as in my answer. – Bill Dubuque Nov 10 '18 at 16:52
• @BillDubuque Noted: you will see that in my comment I did refer to checking back in the original equations. And yes, reducing more quickly to two possibilities rather than three is more efficient. – Mark Bennet Nov 10 '18 at 17:00
• @Mark Yes, I saw that. For balance - I sought to "maximize wordiness" to help readers avoid that pitfall! – Bill Dubuque Nov 10 '18 at 17:06 | 2019-07-23T07:58:16 | {
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https://crypto.stackexchange.com/questions/74320/why-is-sy-y28-2-y254-a-one-to-one-function-and-a-permutation-on | # Why is $S(y) = y^{2^8-2} = y^{254}$ a one-to-one function and a permutation on $GF(2^8)$
I'm taking a course on cryptography and I have some confusions concerning the materials in our notes. Say we have the field $$GF(2^8)$$, we create a substitution algorithm (the S Box in AES) so we need a bijective function that maps from $$GF(2^8)$$ onto itself.
The total number of one-to-one functions is 256! because $$| GF(2^8)| = 256$$ and we can find a mapping by permutating all elements within the set.
We use the mapping function $$S(y) = y^{2^8-2} = y^{254}$$ which is a permutation on $$GF(2^8)$$
Why is this a permutation? How do we show this?
It is also mentioned in other notes such as here: SubBytes Transform circuit for AES Cipher (Version 1.0)
that
$$S(y) = y^{-1}$$ for all $$y \neq 0$$ and $$y^{255} = 1$$
I don't understand why however.
Just to summarize, I don't understand why $$S(y) = y^{2^8-2} = y^{254}$$ is a one-to-one function and a permutation on $$GF(2^8)$$ and why $$S(y) = y^{-1}$$ for all $$y \neq 0$$
Could someone help explain the details behind? Thanks!
• Hint: assume that it is not one-to-one. Then there are two $x,y\in GF(2^8)$ and $x \neq y$ such that $S(x)=S(y)$ then $(x-y)^{254} = 0$ can you see now? Now lok at $y \cdot S(y)$ – kelalaka Sep 15 '19 at 9:13
• The multiplicative group of $\mathrm{GF}(2^8)$ has $2^8 - 1 = 255$ elements. By Lagrange's theorem, we have that $y^{255} = 1$ for $y\neq 0$, so $y^{254} = y^{-1}$. Then the function $y\mapsto y^{254}$ sends $0$ to $0$ and it's not hard to see why we get a bijective function. – user69015 Sep 15 '19 at 10:36
• @kelalaka thanks for the hint but could you elaborate a bit on this approach? As I understand $S(x-y) \neq S(x) - S(y)$. The equality only holds if $S(y) = y^{p^n}$ where $p=2$ in our scenario and for any $n$. This can also be confirmed by expanding $S(x-y)=(x-y)^{254}=x^{254} + _{254}C_1 x^{253}(-y) + _{254}C_2 x^{252}(-y)^2 + _{254}C_3 x^{251}(-y)^3 + ... + (-y)^{254}$ and checking that $_{254}C_2 = 32131 \neq 0 \pmod n$. Also I understand $S(y)$ was intentionally chosen to be non-linear. Maybe I'm not getting the gist of your suggestion – ackbar03 Sep 16 '19 at 2:59
Fact 1: the order of the field $$GF(2^8)$$ is $$2^8$$, hence the order of its multiplicative group $$GF(2^8)^{\times} := GF(2^8) \setminus \{0\}$$ is $$2^8-1 = 255$$.
Fact 2: For a group $$G$$ of order $$q$$, any element $$g \in G$$, when raised to the power of $$q$$ is equal to 1: mathematically put, $$g^q =1$$.
Fact 3: for any element in a group there exists a unique inverse. An inverse to $$g$$ is an element (denoted by $$g^{-1}$$) that satisfies $$g \cdot g^{-1} = 1$$. In other words, the function $$g \mapsto g^{-1}$$ is one-to-one.
In our case, any element $$g \in GF(2^8)^{\times}$$ satisfies $$g^{255} = 1$$. Another, useful way to write this is that for any $$g$$ we have $$g \cdot g^{254} =1$$. Since $$g \cdot g^{254}= g^{255} = 1$$ we have that $$g^{254}$$ is equal to $$g^{-1}$$.
We conclude that the function $$g \mapsto g^{254}$$ is equal to the function $$g \mapsto g^{-1}$$ which is one-to-one, hence it is a permutation of the group $$GF(2^8)^{\times}$$.
I'd be happy to elaborate on any unclear part.
• To clarify: the map is a bijection on the non-zero elements, and $0$ is mapped to itself, so the total map is still a bijection. @ackbar03 – Henno Brandsma Sep 15 '19 at 14:43
• @Chipotle thanks so much for this! This clarifies it a lot. I am still confused however why $S(y) = y^{254}$ can be viewed as a permutation of the group $GF(2^8)$. Is the permutation on the coefficients of the polynomial $y$? Why does raising it to the power of 254 effectively permutate it? I understand that permutations are one-to-one functions but don't really understand why this one-to-one function is a permutation. Thanks a lot! – ackbar03 Sep 16 '19 at 2:49
• 'one-to-one', 'bijection' and 'permutation' mean the same thing ^^ – LeoDucas Sep 16 '19 at 16:08
• @ackbar03 Indeed, one-to-one function and permutation are the same thing. I think it would be helpful to review the definitions in Wikipedia. To recap the argument above: $y \mapsto y^{254}$ is equal to the function $y \mapsto y^{-1}$, which is a one-to-one function by the existence and uniqueness of an inverse element. – Snoop Catt Sep 18 '19 at 6:08
• @ackbar03 Also, you might be wondering, if these two functions are the same, why not just define $S(y) := y^{-1}$ in the first place. This is for computational reasons: the way to compute the inverse of an element in a group $G$, is actually to raise it to the power of ($|G|-1$) – Snoop Catt Sep 18 '19 at 6:12 | 2021-05-14T08:23:21 | {
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https://www.cs.bu.edu/fac/crovella/cs132-book/L25SVD.html | # for QR codes use inline
# %matplotlib inline
# qr_setting = 'url'
# qrviz_setting = 'show'
#
# for lecture use notebook
%matplotlib inline
qr_setting = None
#
%config InlineBackend.figure_format='retina'
# import libraries
import numpy as np
import matplotlib as mp
import pandas as pd
import matplotlib.pyplot as plt
import laUtilities as ut
import slideUtilities as sl
import demoUtilities as dm
import pandas as pd
from datetime import datetime
from IPython.display import Image
from IPython.display import display_html
from IPython.display import display
from IPython.display import Math
from IPython.display import Latex
from IPython.display import HTML;
# The Singular Value Decomposition¶
display(Image("images/440px-Wenger_EvoGrip_S17.JPG", width=350))
# image source https://bringatrailer.com/listing/1964-rolls-royce-james-young-phanton-v-limosine/
display(Image("images/rolls-royce.jpg", width=350))
Today we will study the most useful decomposition in applied Linear Algebra.
Pretty exciting, eh?
The Singular Value Decomposition is the “Swiss Army Knife” and the “Rolls Royce” of matrix decompositions.
– Diane O’Leary
The singular value decomposition is a matrix factorization.
Now, the first thing to know is that EVERY matrix has a singular value decomposition.
## Maximizing $$\Vert A\mathbf{x}\Vert$$¶
The singular value decomposition (let’s just call it SVD) is based on a very simple question:
Let’s say you are given an arbitrary matrix $$A$$, which does not need to be square.
Here is the question:
Among all unit vectors, what is the vector $$\mathbf{x}$$ that maximizes $$\Vert A\mathbf{x}\Vert$$?
In other words, in which direction does $$A$$ create the largest output vector from a unit input?
To set the stage to answer this question, let’s review a few facts.
You recall that the eigenvalues of a square matrix $$A$$ measure the amount that $$A$$ “stretches or shrinks” certain special vectors (the eigenvectors).
For example, for a square $$A$$, if $$A\mathbf{x} = \lambda\mathbf{x}$$ and $$\Vert \mathbf{x}\Vert = 1,$$ then
$\Vert A\mathbf{x}\Vert = \Vert\lambda\mathbf{x}\Vert = |\lambda|\,\Vert\mathbf{x}\Vert = |\lambda|.$
V = np.array([[2,1],[.1,1]])
L = np.array([[1.2,0],
[0,0.7]])
A = V @ L @ np.linalg.inv(V)
#
ax = dm.plotSetup(-1.5,1.5,-1.5, 1.5, size=(9,6))
ut.centerAxes(ax)
theta = [2 * np.pi * f for f in np.array(range(360))/360.0]
x = [np.array([np.sin(t), np.cos(t)]) for t in theta]
Ax = [A.dot(xv) for xv in x]
ax.plot([xv[0] for xv in x],[xv[1] for xv in x],'-b')
ax.plot([Axv[0] for Axv in Ax],[Axv[1] for Axv in Ax],'--r')
theta_step = np.linspace(0, 2*np.pi, 24)
for th in theta_step:
x = np.array([np.sin(th), np.cos(th)])
u, s, v = np.linalg.svd(A)
ax.set_title(r'Eigenvectors of $A$ and the image of the unit circle under $A$');
The largest value of $$\Vert A\mathbf{x}\Vert$$ is the long axis of the ellipse. Clearly there is some $$\mathbf{x}$$ that is mapped to that point by $$A$$. That $$\mathbf{x}$$ is what we want to find.
And let’s make clear that we can apply this idea to arbitrary (non-square) matrices.
Here is an example that shows that we can still ask the question of what unit $$\mathbf{x}$$ maximizes $$\Vert A\mathbf{x}\Vert$$ even when $$A$$ is not square.
For example:
If $$A = \begin{bmatrix}4&11&14\\8&7&-2\end{bmatrix},$$
then the linear transformation $$\mathbf{x} \mapsto A\mathbf{x}$$ maps the unit sphere $$\{\mathbf{x} : \Vert \mathbf{x} \Vert = 1\}$$ in $$\mathbb{R}^3$$ onto an ellipse in $$\mathbb{R}^2$$, as shown here:
display(Image("images/Lay-fig-7-4-1.jpg", width=650))
### $$\Vert A\mathbf{x}\Vert^2$$ is a Quadratic Form¶
Now, here is a way to answer our question:
Problem. Find the unit vector $$\mathbf{x}$$ at which the length $$\Vert A\mathbf{x}\Vert$$ is maximized, and compute this maximum length.
Solution.
The quantity $$\Vert A\mathbf{x}\Vert^2$$ is maximized at the same $$\mathbf{x}$$ that maximizes $$\Vert A\mathbf{x}\Vert$$, and $$\Vert A\mathbf{x}\Vert^2$$ is easier to study.
So let’s ask to find the unit vector $$\mathbf{x}$$ at which $$\Vert A\mathbf{x}\Vert^2$$ is maximized.
Observe that
$\Vert A\mathbf{x}\Vert^2 = (A\mathbf{x})^T(A\mathbf{x})$
$= \mathbf{x}^TA^TA\mathbf{x}$
$= \mathbf{x}^T(A^TA)\mathbf{x}$
Now, $$A^TA$$ is a symmetric matrix.
So we see that $$\Vert A\mathbf{x}\Vert^2 = \mathbf{x}^TA^TA\mathbf{x}$$ is a quadratic form!
… and we are seeking to maximize it subject to the constraint $$\Vert \mathbf{x}\Vert = 1$$.
As we learned in the last lecture, the maximum value of a quadratic form, subject to the constraint that $$\Vert\mathbf{x}\Vert = 1$$, is the largest eigenvalue of the symmetric matrix.
So the maximum value of $$\Vert A\mathbf{x}\Vert^2$$ subject to $$\Vert\mathbf{x}\Vert = 1$$ is $$\lambda_1$$, the largest eigenvalue of $$A^TA$$.
Also, the maximum is attained at a unit eigenvector of $$A^TA$$ corresponding to $$\lambda_1$$.
For the matrix $$A$$ in the 2 $$\times$$ 3 example,
$\begin{split}A^TA = \begin{bmatrix}4&8\\11&7\\14&-2\end{bmatrix} \,\begin{bmatrix}4&11&14\\8&7&-2\end{bmatrix} = \begin{bmatrix}80&100&40\\100&170&140\\40&140&200\end{bmatrix}.\end{split}$
The eigenvalues of $$A^TA$$ are $$\lambda_1 = 360, \lambda_2 = 90,$$ and $$\lambda_3 = 0.$$
The corresponding unit eigenvectors are, respectively,
$\begin{split}\mathbf{v}_1 = \begin{bmatrix}1/3\\2/3\\2/3\end{bmatrix}, \mathbf{v}_2 = \begin{bmatrix}-2/3\\-1/3\\2/3\end{bmatrix}, \mathbf{v}_3 = \begin{bmatrix}2/3\\-2/3\\1/3\end{bmatrix}. \end{split}$
For $$\Vert\mathbf{x}\Vert = 1$$, the maximum value of $$\Vert A\mathbf{x}\Vert$$ is $$\Vert A\mathbf{v}_1\Vert = \sqrt{360}.$$
This example shows that the key to understanding the effect of $$A$$ on the unit sphere in $$\mathbb{R}^3$$ is to examime the quadratic form $$\mathbf{x}^T(A^TA)\mathbf{x}.$$
We can also go back to our 2 $$\times$$ 2 example.
Let’s plot the eigenvectors of $$A^TA$$.
ax = dm.plotSetup(-1.5,1.5,-1.5, 1.5, size=(9,6))
ut.centerAxes(ax)
theta = [2 * np.pi * f for f in np.array(range(360))/360.0]
x = [np.array([np.sin(t), np.cos(t)]) for t in theta]
Ax = [A.dot(xv) for xv in x]
ax.plot([xv[0] for xv in x],[xv[1] for xv in x],'-b')
ax.plot([Axv[0] for Axv in Ax],[Axv[1] for Axv in Ax],'--r')
theta_step = np.linspace(0, 2*np.pi, 24)
#for th in theta_step:
# x = np.array([np.sin(th), np.cos(th)])
u, s, v = np.linalg.svd(A)
ax.set_title(r'Eigenvectors of $A^TA$ and their images under $A$');
We see that the eigenvector corresponding to the largest eigenvalue of $$A^TA$$ indeed shows us where $$\Vert A\mathbf{x}\Vert$$ is maximized – where the ellipse is longest.
Also, the other eigenvector of $$A^TA$$ shows us where the ellipse is narrowest.
In fact, the entire geometric behavior of the transformation $$\mathbf{x}\mapsto A\mathbf{x}$$ is captured by the quadratic form $$\mathbf{x}^TA^TA\mathbf{x}$$.
## The Singular Values of a Matrix¶
Let’s continue to consider $$A$$ to be an arbitrary $$m\times n$$ matrix.
Notice that even though $$A$$ is not square in general, $$A^TA$$ is square and symmetric.
So, there is a lot we can say about $$A^TA$$.
In particular, since $$A^TA$$ is symmetric, it can be orthogonally diagonalized (as we saw in the last lecture).
So let $$\{\mathbf{v}_1, \dots, \mathbf{v}_n\}$$ be an orthonormal basis for $$\mathbb{R}^n$$ consisting of eigenvectors of $$A^TA$$, and let $$\lambda_1, \dots, \lambda_n$$ be the corresponding eigenvalues of $$A^TA$$.
Then, for any eigenvector $$\mathbf{v}_i$$,
$\Vert A\mathbf{v}_i\Vert^2 = (A\mathbf{v}_i)^T A\mathbf{v}_i = \mathbf{v}_i^T A^TA\mathbf{v}_i$
$= \mathbf{v}_i^T(\lambda_i)\mathbf{v}_i$
(since $$\mathbf{v}_i$$ is an eigenvector of $$A^TA$$)
$= \lambda_i$
(since $$\mathbf{v}_i$$ is a unit vector.)
Now any expression $$\Vert\cdot\Vert^2$$ is nonnegative.
So the eigenvalues of $$A^TA$$ are all nonnegative.
That is: $$A^TA$$ is positive semidefinite.
We can therefore renumber the eigenvalues so that
$\lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_n \geq 0.$
Definition. The singular values of $$A$$ are the square roots of the eigenvalues of $$A^TA$$. They are denoted by $$\sigma_1,\dots,\sigma_n,$$ and they are arranged in decreasing order.
That is, $$\sigma_i = \sqrt{\lambda_i}$$ for $$i = 1,\dots,n.$$
By the above argument, the singular values of $$A$$ are the lengths of the vectors $$A\mathbf{v}_1, \dots, A\mathbf{v}_n.$$
Where $$\mathbf{v}_1, \dots, \mathbf{v}_n.$$ are the eigenvectors of $$A^TA$$, normalized to unit length.
### The Eigenvectors of $$A^TA$$ Lead To an Orthogonal Basis for $$\operatorname{Col} A$$¶
Now: we know that vectors $$\mathbf{v}_1, \dots, \mathbf{v}_n$$ are an orthogonal set because they are eigenvectors of the symmetric matrix $$A^TA$$.
However, it’s also the case that $$A\mathbf{v}_1, \dots, A\mathbf{v}_n$$ are an orthogonal set.
This fact is key to the SVD.
This fact is not obvious at first!
But it is true – let’s prove it (and a bit more).
Theorem. Suppose $$\{\mathbf{v}_1, \dots, \mathbf{v}_n\}$$ is an orthonormal basis of $$\mathbb{R}^n$$ consisting of eigenvectors of $$A^TA$$, arranged so that the corresponding eigenvalues of $$A^TA$$ satisfy $$\lambda_1 \geq \cdots \geq \lambda_n,$$ and suppose $$A$$ has $$r$$ nonzero singular values.
Then $$\{A\mathbf{v}_1, \dots, A\mathbf{v}_r\}$$ is an orthogonal basis for $$\operatorname{Col} A,$$ and rank $$A = r$$.
Note how surprising this is: while $$\{\mathbf{v}_1, \dots, \mathbf{v}_n\}$$ are a basis for $$\mathbb{R}^n$$, $$\operatorname{Col} A$$ is a subspace of $$\mathbb{R}^m$$.
Nonetheless,
• two eigenvectors $$\mathbf{v}_i$$ and $$\mathbf{v}_j \in \mathbb{R}^n$$ are orthogonal, and
• their images $$A\mathbf{v}_i$$ and $$A\mathbf{v}_j \in \mathbb{R}^m$$ are also orthogonal.
Proof. What we need to do is establish that $$\{A\mathbf{v}_1, \dots, A\mathbf{v}_r\}$$ is an orthogonal linearly independent set whose span is $$\operatorname{Col}\ A$$.
Because $$\mathbf{v}_i$$ and $$\mathbf{v}_j$$ are orthogonal for $$i\neq j$$,
$(A\mathbf{v}_i)^T(A\mathbf{v}_j) = \mathbf{v}_i^TA^TA\mathbf{v}_j = \mathbf{v}_i^T(\lambda_j \mathbf{v}_j) = 0.$
So $$\{A\mathbf{v}_1, \dots, A\mathbf{v}_n\}$$ is an orthogonal set.
Furthermore, since the lengths of the vectors $$A\mathbf{v}_1, \dots, A\mathbf{v}_n$$ are the singular values of $$A$$, and since there are $$r$$ nonzero singular values, $$A\mathbf{v}_i \neq {\mathbf 0}$$ if and only if $$1 \leq i \leq r.$$
So $$A\mathbf{v}_1, \dots, A\mathbf{v}_r$$ are a linearly independent set (because they are orthogonal and all nonzero), and clearly they are each in $$\operatorname{Col}\ A$$.
Finally, we just need to show that $$\operatorname{Span}\{A\mathbf{v}_1, \dots, A\mathbf{v}_r\} = \operatorname{Col} A$$.
To do this we’ll show that for any $$\mathbf{y}$$ in $$\operatorname{Col}\ A$$, we can write $$\mathbf{y}$$ in terms of $$\{A\mathbf{v}_1, \dots, A\mathbf{v}_r\}$$:
Say $$\mathbf{y} = A\mathbf{x}.$$
Because $$\{\mathbf{v}_1, \dots, \mathbf{v}_n\}$$ is a basis for $$\mathbb{R}^n$$, we can write $$\mathbf{x} = c_1\mathbf{v}_1 + \dots + c_n\mathbf{v}_n,$$ so
$\mathbf{y} = A\mathbf{x} = c_1A\mathbf{v}_1 + \dots + c_rA\mathbf{v}_r + \dots + c_nA\mathbf{v}_n.$
$= c_1A\mathbf{v}_1 + \dots + c_rA\mathbf{v}_r.$
(because $$A\mathbf{v}_i = {\mathbf 0}$$ for $$i > r$$).
In summary: $$\{A\mathbf{v}_1, \dots, A\mathbf{v}_n\}$$ is an (orthogonal) linearly independent set whose span is $$\operatorname{Col} A$$, so it is an (orthogonal) basis for $$\operatorname{Col}A$$.
Notice that we have also proved that rank $$A = \dim\operatorname{Col}A = r.$$
In other words, if $$A$$ has $$r$$ nonzero singular values, $$A$$ has rank $$r$$.
## The Singular Value Decomposition¶
What we have just proved is that the eigenvectors of $$A^TA$$ are rather special.
Note that, thinking of $$A$$ as a linear operator:
• its domain is $$\mathbb{R}^n$$, and
• its range is $$\operatorname{Col}A.$$
So we have just proved that
• the set $$\{\mathbf{v}_i\}$$ is an orthogonal basis for the domain of $$A$$, and
• the set $$\{A\mathbf{v}_i\}$$ is an orthogonal basis for the range of $$A$$.
Now we can define the SVD.
Theorem. Let $$A$$ be an $$m\times n$$ matrix with rank $$r$$. Then there exists an $$m\times n$$ matrix $$\Sigma$$ whose diagonal entries are the first $$r$$ singular values of $$A$$, $$\sigma_1 \geq \sigma_2 \geq \dots \geq \sigma_r > 0,$$ and there exists an $$m\times m$$ orthogonal matrix $$U$$ and an $$n\times n$$ orthogonal matrix $$V$$ such that
$A = U\Sigma V^T$
Any factorization $$A = U\Sigma V^T,$$ with $$U$$ and $$V$$ orthogonal and $$\Sigma$$ a diagonal matrix is called a singular value decomposition (SVD) of $$A$$.
The columns of $$U$$ are called the left singular vectors and the columns of $$V$$ are called the right singular vectors of $$A$$.
Aside: regarding the “Rolls Royce” property, consider how elegant this structure is.
In particular:
• $$A$$ is an arbitrary matrix
• $$U$$ and $$V$$ are both orthogonal matrices
• $$\Sigma$$ is a diagonal matrix
• all singular values are positive or zero
• there are as many positive singular values as the rank of $$A$$
• (not part of the theorem but we’ll see it is true)
# image source https://bringatrailer.com/listing/1964-rolls-royce-james-young-phanton-v-limosine/
display(Image("images/rolls-royce.jpg", width=350))
We have built up enough tools now that the proof is quite straightforward.
Proof. Let $$\{\lambda_i\}$$ and $$\{\mathbf{v}_i\}$$ be the eigenvalues and eigenvectors of $$A^TA$$, and $$\sigma_i = \sqrt{\lambda_i}$$.
The starting point is to use the fact that we just proved:
$$\{A\mathbf{v}_1, \dots, A\mathbf{v}_r\}$$ is an orthogonal basis for $$\operatorname{Col}\ A.$$
Next, let us normalize each $$A\mathbf{v}_i$$ to obtain an orthonormal basis $$\{\mathbf{u}_1,\dots,\mathbf{u}_r\}$$, where
$\mathbf{u}_i = \frac{1}{\Vert A\mathbf{v}_i\Vert}A\mathbf{v}_i = \frac{1}{\sigma_i}A\mathbf{v}_i$
Then
$A\mathbf{v}_i = \sigma_i\mathbf{u}_i\;\;\;\;(1 \leq i \leq r)$
Now the rank of $$A$$ (which is $$r$$) may be less than $$m$$.
In that case, add additional orthonormal vectors $$\{\mathbf{u}_{r+1} \dots \mathbf{u}_m\}$$ to the set so that they span $$\mathbb{R}^m$$.
Now collect the vectors into matrices.
$U = \begin{bmatrix}\mathbf{u}_1&\cdots&\mathbf{u}_m\end{bmatrix}$
and
$V = \begin{bmatrix}\mathbf{v}_1&\cdots&\mathbf{v}_n\end{bmatrix}$
Recall that these matrices are orthogonal because the $$\{\mathbf{v_i}\}$$ are orthonormal and the $$\{A\mathbf{v_i}\}$$ are orthonormal, as we previously proved.
So
$AV = [A\mathbf{v}_1\;\cdots\;A\mathbf{v}_r\;\;\overbrace{\mathbf{0}\cdots\mathbf{0}}^{n-r}]$
$= [\sigma_1\mathbf{u}_1\;\cdots\;\sigma_r\mathbf{u}_r\;\mathbf{0}\;\cdots\;\mathbf{0}] = U\Sigma.$
So
$AV = U\Sigma$
Now, $$V$$ is an orthogonal matrix, so multiplying both sides on the right by $$V^T$$:
$U\Sigma V^T = AVV^T = A.$
## The Swiss Army Knife¶
We don’t have enough time to describe the many applications of the SVD.
One of its uses is in data mining, where the matrices being analyzed are typically data matrices.
display(Image("images/440px-Wenger_EvoGrip_S17.JPG", width=350))
The approximations we’ll discuss are low-rank approximations.
Recall that the rank of a matrix $$A$$ is the largest number of linearly independent columns of $$A$$.
Or, equivalently, the dimension of $$\operatorname{Col} A$$.
Let’s define the rank-$$k$$ approximation to $$A$$:
When $$k < \operatorname{Rank}A$$, the rank-$$k$$ approximation to $$A$$ is the closest rank-$$k$$ matrix to $$A$$, i.e.,
$A^{(k)} =\arg \min_{\operatorname{Rank}B = k} \Vert A-B\Vert_F.$
Why is a rank-$$k$$ approximation valuable?
The reason is that a rank-$$k$$ matrix may take up much less space than the original $$A$$.
$$m\left\{\begin{array}{c}\;\\\;\\\;\\\;\\\;\end{array}\right.\;\;\overbrace{\left[\begin{array}{cccc}\begin{array}{c}\vdots\\\vdots\\{\bf a_1}\\\vdots\\\vdots\end{array}&\begin{array}{c}\vdots\\\vdots\\{\bf a_2}\\\vdots\\\vdots\end{array}&\dots&\begin{array}{c}\vdots\\\vdots\\{\bf a_n}\\\vdots\\\vdots\end{array}\\\end{array}\right]}^{\large n} = \overbrace{\left[\begin{array}{cc}\vdots&\vdots\\\vdots&\vdots\\\sigma_1\mathbf{u}_1&\sigma_k\mathbf{u}_k\\\vdots&\vdots\\\vdots&\vdots\end{array}\right]}^{\large k} \times \left[\begin{array}{ccccc}\dots&\dots&\mathbf{v}_1&\dots&\dots\\\dots&\dots&\mathbf{v}_k&\dots&\dots\end{array}\right]$$
The rank-$$k$$ approximation takes up space $$(m+n)k$$ while $$A$$ itself takes space $$mn$$.
For example, if $$k=10$$ and $$m = n = 1000$$, then the rank-$$k$$ approximation takes space $$20000/1000000 = 2\%$$ of $$A$$.
The key to using the SVD for matrix approximation is as follows:
The best rank-$$k$$ approximation to any matrix can be found via the SVD.
How do we use SVD to find the best rank-$$k$$ approximation to $$A$$?
In terms of the singular value decomposition,
the best rank-$$k$$ approximation to $$A$$ is formed by taking
• $$U' =$$ the $$k$$ leftmost columns of $$U$$,
• $$\Sigma ' =$$ the $$k\times k$$ upper left submatrix of $$\Sigma$$, and
• $$(V')^T=$$ the $$k$$ upper rows of $$V^T$$,
and constructing
$A^{(k)} = U'\Sigma'(V')^T.$
For example, here is a photo.
We can think of this as a $$512\times 512$$ matrix $$A$$ whose entries are grayscale values (numbers between 0 and 1).
boat = np.loadtxt('data/boat.dat')
import matplotlib.cm as cm
plt.figure(figsize=(9,9))
plt.imshow(boat,cmap = cm.Greys_r);
This matrix $$A$$ has rank of 512.
But the error when we approximate $$A$$ by a rank 40 matrix is only around 10%.
We say that the effective rank of $$A$$ is low (perhaps 40).
Let’s find the closest rank-40 matrix to $$A$$ and view it.
We can do this quite easily using the SVD.
We simply construct our approximation of $$A$$ using only the first 40 columns of $$U$$ and top 40 rows of $$V^T$$.
# construct a rank-n version of the boat
u, s, vt = np.linalg.svd(boat, full_matrices=False)
scopy = s.copy()
rank = 40
scopy[rank:] = 0
boatApprox = u @ np.diag(scopy) @ vt
#
plt.figure(figsize=(18,9))
plt.subplot(1,2,1)
plt.imshow(boatApprox,cmap = cm.Greys_r)
plt.title('Rank {}'.format(rank), size=20)
plt.subplot(1,2,2)
plt.imshow(boat,cmap = cm.Greys_r)
plt.title('Rank 512', size=20);
Note that the rank-40 boat takes up only 40/512 = 8% of the space of the original image!
This general principle is what makes image, video, and sound compression effective.
When you
• watch HDTV, or
• listen to an MP3, or
• look at a JPEG image,
these signals have been compressed using the fact that they are effectively low-rank matrices.
### Wrapup¶
display(Image("images/in-conclusion.jpg", width=550))
We have reached the end!
Of course, this is not really the end … more like the beginning.
If we had more time, we’d talk about how linear algebra informs the study of graphs, the methods of machine learning, data mining, and many more topics.
So this is just where we have to stop.
We have looked at the richness of linear algebra from many angles.
We have seen that the simple linear system $$A\mathbf{x} = \mathbf{b}$$ leads to a whole collection of interesting questions, questions that have unfolded step by step over the course of the semester.
But we have also seen that we can extract the idea of matrix out of a linear system, and consider it as an object in its own right.
Considered on their own, matrices can be seen as linear operators, giving us tools for computer graphics and the solution of dynamical systems and linear equations.
We have also seen that matrices can be seen as data objects, whose linear algebraic properties expose useful facts about the data.
There are many courses you can go on to from here, which will rely on your understanding of linear algebra:
• CS 391 Fundamentals of Data Science
• CS 440 Artificial Intelligence
• CS 470 Computer Systems Performance Analysis
• CS 480 Computer Graphics
• CS 505 Intro to Natural Language Processing
• CS 506 Tools for Data Science
• CS 531 Advanced Optimization Algorithms
• CS 533 Spectral Methods
• CS 558 Machine Learning
• CS 565 Data Mining
• CS 581 Computational Fabrication
• CS 591 Deep Learning
• CS 591 Compressive Sensing
• CS 591 Natural Language Understanding
In each of these you will use and build on your knowledge of linear algebra.
Enjoy!
Don’t forget to submit a course evaluation! | 2021-09-21T23:51:27 | {
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https://math.stackexchange.com/questions/2944845/let-a-n-n-1-infty-and-b-n-n-1-infty-be-two-sequences-o | # Let $\{a_n\}_{n = 1}^{\infty}$ and $\{b_n\}_{n = 1}^{\infty}$ be two sequences of real numbers s.t $|a_n -b_n| < \frac{1}{n}$
Let $$\{a_n\}_{n = 1}^{\infty}$$ and $$\{b_n\}_{n = 1}^{\infty}$$ be two sequences of real numbers s.t $$|a_n -b_n| < \frac{1}{n}$$ Show $$\{b_n\}_{n = 1}^{\infty}$$ converges to L also.
Proof Attempt
Notice $$\Bigg| a_n - b_n \Bigg| < \frac{1}{n} < \frac{\epsilon}{2}$$ This is true due to the Archimedian property of real numbers.
By definition for $$\{a_n\}_{n = 1}^{\infty}$$:
$$\forall \epsilon > 0 \ \exists \ N_1 \in \mathbb{R} \ s.t \ \forall \ n \geq N_1 \ \Bigg|a_n - L \Bigg| < \frac{\epsilon}{2}$$ and as well $$\forall \epsilon > 0 \ \exists \ N_2 \in \mathbb{R} \ s.t \ \forall \ n \geq N_2 \ \Bigg|a_n - b_n \Bigg| < < \frac{1}{2} < \frac{\epsilon}{2}$$
$$\therefore \Bigg|b_n - L \Bigg| \leq \Bigg| b_n - a_n \Bigg| + \Bigg| a_n - L \Bigg| < \frac{1}{n} + \frac{\epsilon}{2} < \frac{\epsilon}{2} + \frac{\epsilon}{2} < \epsilon$$
If we let $$N = max\{N_1, N_2\}$$
Correct reasoning?
• As a sketch - yes, it's a correct strategy. Probably you will need more formality and details if you plan to provide this as an answer to an exercise. – rtybase Oct 6 '18 at 19:32
• @rtybase It is not an exercise to hand in to anything, but for good practices what would have to be added to make it formally acceptable? – dc3rd Oct 6 '18 at 19:33
• With the latest update, all that is left to do is to merge "This is true due to the Archimedian property of real numbers" with "$\forall \epsilon > 0 \ \exists \ N_2 \in$" – rtybase Oct 6 '18 at 19:36
• For example, you mention $\epsilon$ suddenly. Probably you should start with Let $\epsilon > 0$ be given. Then by the Archimedian property there is $N$ so that for all $n \ge N$ we have $1/n < \epsilon/2$. – GEdgar Oct 6 '18 at 19:36
• You should specify somewhere that $(a_n)_n$ converges, it is not explicitly written as a statement in the initial wording. – zwim Oct 6 '18 at 20:21
Let $$\epsilon>0$$ given.
$$\lim_{n\to+\infty}a_n=L \implies$$ $$(\exists N_1\in \Bbb N)\;\; : \; (\forall n\ge N_1) \; |a_n-L|<\frac{\epsilon}{2}$$
On the other hand , as you said, $$\Bbb R$$ is Archimedian,
$$(\exists N_2 \in \Bbb N) \; : \; N_2\frac{\epsilon}{2}>1$$
Put $$N=\max\{N_1,N_2\}$$
then
$$(\forall n\ge N)$$ $$|b_n-L|\le|a_n-L|+|b_n-a_n|<\epsilon$$ done! | 2019-07-24T08:31:15 | {
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http://chiavellainox.it/dtiz/a-and-b-are-two-independent-events.html | # A And B Are Two Independent Events
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A single fair die is rolled twice. In other words, the product of probability of events A and B equals to probability of intersection of events A and B. The probability of occurrence of A may be 1 Verified Answer. For example if we draw two cards from a given deck of 52 cards, then the event of getting a heart first and then getting a red queen are dependent events. Independent events A and Given that P(B) = 2 A), and B) = 052 6. 6, then find the probability of occurrence of A. An example of two independent events is as follows; say you rolled a die and flipped a coin. 3 and P(B Get solutions. Purdue University students, faculty, and staff at our West Lafayette, IN campus may access this area for information on the award-winning Purdue Writing Lab. The symmetric definition of independency is this (*) P(A∩B) = P(A) P(B). Conditional Probability. To summarize, we can say "independence means we can multiply the probabilities of events to obtain the probability of their intersection", or equivalently, "independence means that conditional probability of one event given another is the same as the original (prior. P(A|B)=P(A) P(A|B)=P(B) P(A|B)≠P(A). What is P(A \u0004 B)? c. You can only calculate the probability if you know that the occurence of one event is independent from the occurence of the other event. A sock drawer contains 5 rolled-up pairs of each color of socks, white, green, and blue. A and B are two independent events. OPEN-ENDED Describe a real-life example of two independent events. The events that follow are: 1. Two events are independent if one of the following are true: Two events A and B are independent if the knowledge that one occurred does not affect the chance the other occurs. For example, if we toss two coins, the occurrence and non-occurrence of a head one coin does not in any way effect the occurrence of a head on the other coin. December 23, 1927. 547 Previous probability sample space Core VocabularyCore Vocabulary CCore ore CConceptoncept Probability of Independent Events Words Two senvet A and B are independent events if and only if the probability that both events occur is the product of the probabilities. Popular Tutorials in Understand that two events A and B are independent if the probability of A and B occurring together is the product of their probabilities, and use this characterization to determine if they are independent. A student spins an even number on a spinner and then spins another even number on the second spin. Two events A and B are independent if P(A|B)=P(A) and are dependent otherwise. 3 and P(B Get solutions. " It is helpful to distinguish between "manipulated independent variables" and "measured independent variables. For two independent events A and B, the probability that both A & B occur is 1 / 8 and the probability that neither of them occur is 3 / 8. If A and B are two independent events and P(A)=(3)/(6) and P(AcapB)=(4)/(9) then the value of P(B) will be. Independent events---Events A and B are said to be independent events if the occurrence of B does not alter the probability that A has occurred; or, events A and B are independent if Otherwise, events A and B are dependent Summary-- for independent events, knowing B does not effect the probability of A Multiplicative Rule and Independent Events. The notation for conditional probability is P(B|A. I'm given the information: P(A)= 0. 2#, what is #P. Independent Events: P(A and B) = P(A) · P(B). Let A and B be two independent events such that the probability is 8 1 that they will occour simultaneously and 8 3 that neither of them will occur. asked by Chris on February 3, 2009; Math. Explain the difference between dependent events and independent events, and give an example of each. Let's try out all of the different scenarios again. The probability that both A and B occur is $\large\frac{1}{6}$ and the probability that neither of them occurs is $\large\frac{1}{3}$. For example, if a bag contains 2 blue balls and 2 red balls and two balls are selected randomly, the events are: a) independent if the first ball is replaced after being selected. ): Events A and B are independent if having the information that B already occurred does not change the probability that A will occur. Find P(A|B). Two events are independent of each other when the result of one does not affect the result of the other. The conditional probability of event B given event A is P(B|A)=P(A and B)/P(A) when two events are not independent. January 20, 2020. 1) Let A and B be two events. Conditioning on an event Kolmogorov definition. Answer to (10pts) Suppose A and B are two independent random events (means or , if and , find and. 4 and P(B) = 0. Events can be " Independent ", meaning each event is not affected by any other events. Dismissing concerns that states are reopening too soon, President Donald Trump incorrectly said that a newly revised model projecting 134,000 COVID. => Favorable outcomes are: (1, 5), (2, 4), (3, 3), (4, 2) and (5, 1) Number of favorable outcomes = 5. Find the probability occurrence of A?a)1. The theme of NCIL’s 2020 Annual Conference on Independent Living will be Evolution of Our. If two events are mutually exclusive, then the probability of one or the other occurring is. Page 1 of 2 734 Chapter 12 Probability and Statistics 1. First we find the probability of each event. 3 and P(B) = 0. 1# THe answer is #P. The other case involes these two events when they are independent. For two independent events, A and B, we are given Pr (A union B) = 0. Date: 07/14/2009 at 23:50:35 From: Doctor Peterson Subject: Re: Independent events Hi, Amy. A system has two components that operate in parallel, as shown in the diagram below. Let A and B be two independent events such that the probability is 8 1 that they will occour simultaneously and 8 3 that neither of them will occur. If the events A and B are not independent, they are said to be dependent. Are events A and B independent events?. Event B is selecting a multiple of 3. Independent Events. P( A and B) = 1/6 and the probability that neither of them occur is 1/3. Let A and B be two independent events such that the probability is 8 1 that they will occour simultaneously and 8 3 that neither of them will occur. Since the set (A∩B) = {(3,H), (6,H)}, the probability of an event getting heads and multiple of three on die, that is P(A∩B) is 2/12 or 1/6. The symmetric definition of independency is this (*) P(A∩B) = P(A) P(B). If A and B are two events of the sample space;Let B' be complementry event of B, By drawin a venn diagran you can clearly see that, (AnB')n(B) = null Therefore (AnB') and B are mutually exclusive. A and B are two mutually exclusive events then P( A¨B ) = _____. The concept of independence extends to dealing with collections of more than two events or random variables, in which case the events are pairwise independent if each pair are independent of each other, and the events are mutually independent if each event is independent of each other combination of events. The outcome of the first roll does not change the probability for the outcome of the second roll. The conditional probability of event B given event A is P(B|A)=P(A and B)/P(A) when two events are not independent. The conditional probability of A given B is written P(AjB): P(AjB) = P(A¢B) P(B) Event Ais independent of B if the conditional probability of Agiven B is the same as the unconditional probability of A. In probability, two events are independent if the incidence of one event does not affect the probability of the other event. Financial Services. Let A and B are two events, then the probability of occurrence of event A will not affect the probability of occurrence of event B. Pr(R1 = 1jR2 2) = Pr(R1 = 1^ R2 2) Pr(R2 2) Independence of Random Variables De nition 2 Two random variables R1 and R2 are independent, if for all x1;x2 2, we have: Pr(R1 = x1 ^ R2 = x2) = Pr(R1 = x1) Pr(R2 = x2). Let A and B be two independent events such that the probability is 8 1 that they will occour simultaneously and 8 3 that neither of them will occur. Definition: Two events, A and B, are independent if the fact that A occurs does not affect the probability of B occurring. P(A) If A and B are two mutually exclusive events, then. Each coin is a fair coin and has equal probability of landing on Head (H) or Tail (T). 5 + P(D) - P(C and D) Since C and D are independent, the probability that both occur is the product of the individual probabilities:. 4, P(B)= p, and P(AUB) =0. P(B) ANSWER: T. If both events A and B occur on a single performance of an experiment, this is called the intersection or joint probability of A and B, denoted as P(A n B). (b) Events A and B are complementary. Find the probability occurrence of A?a)1. P(A∩B) = P(A) x P(B). In these tests, one sample is being compared to the population. A conditional probability can always be computed using the formula in the definition. Use the following to answer question 8. If P(B) = 0. The value of P(A and B) is 0. If the probability of occurrence of one of them is not affected by the occurrence of the other, then we say. More specifically, if A and B are independent events, then P(A and B) = P(A)P(B). Two events, A and B, are independent if the outcome of A does not affect the outcome of B. flipping two coins B. Problem 98SE from Chapter 3: Two events, A and B, are independent, with P(A) =. What The Rule Means: Suppose we roll one die followed by another and want to find the probability of rolling a 4 on the first die and rolling an even number on the. This is an important idea! A coin does not "know" it came up heads before. Find P ( A ) and P ( B ). A and B are two independent events such that P(A)=(1)/(2) and P(B)=(1)/(3). Example: Tossing a coin two times. A student spins an even number on a spinner and then spins another even number on the second spin. CraigMarcho on 03-16-2019 05:46 AM. Answer: Two events, A and B, are independent. P(A ∩ B) = 0. The film No Más Bebés follows the story of Mexican American women who were sterilized under duress while giving birth at Los Angeles County-USC Medical Center in the 1960s and 1970s. P(A 1 A 2 A 3)=P(A 1)P(A 2)P(A 3) Are the events A 1, A 2, and A 3 pairwise independent? Toss two different standard dice, white and black. INDEPENDENT The probability of two independent events can be found by multiplying the probability of the first event by the probability of the second event. [Recall from Conditional Probability that the notation P(E 2 | E 1) means "the probability of the event E 2 given that E 1 has. Welcome to the Purdue OWL. More formally, this means that the occurrence of one event has no effect upon the probability of the other event. Given a data set, students will be able to determine if two events are independent. What The Rule Means: Suppose we roll one die followed by another and want to find the probability of rolling a 4 on the first die and rolling an even number on the. Now that we are familiar with the terms Union, intersection and disjoint events, we can talk about independence of events. If A and B are independent events, the probability of both events occurring is the product of the probabilities of the individual events. probability(A and B) = probability(A) × probability(B) Let's use this formula to find the probability of getting 2 heads when two coins are tossed. Viewed 69k times. Meanwhile, mutually inclusive compound events are situations where. Note: Some books will say to take care that A and B are independent, but the rule can also be used with dependent events, you just have to be more careful in find P(A) and P(B). Note: don't find symbol of intersection,so write instead. Given two independent events A and B such that P(A) = 0. Example 1 The two-way frequency table gives data about 180 randomly selected flights that arrive at an airport. Just multiply the probability of the first event by the second. Variables are defined as the properties or kinds of characteristics of certain events or objects. Let A and B be two events such that P(A) =0. Thus the two coins are independent. Discover why we are the world's leading cloud software company powering social good. Find the probability of occurrence of two events. gl/9WZjCW If A, B are two independent events, show that bar A and bar B are also independent. Events can be " Independent ", meaning each event is not affected by any other events. The union of two events (that is, A occurs or B occurs or both) has P(A [B) = P(A) + P(B) P(A \B): When A and B are independent, the intersection of two events has P(A \B) = P(A. asked by Chris on February 3, 2009; Math. If the equation is violated, the two events are not independent. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. The theme of NCIL’s 2020 Annual Conference on Independent Living will be Evolution of Our. Binding of C3d to B cell CR2 leads to augmentation of signaling pathways initiated by antigen binding. Therefore the probability that they both occur simultaneously is the product of their respective probabilities. Given two independent events A and B such that P(A) = 0. Which of the Venn diagrams has shaded the event that the contractor wins. The probability of getting any number face on the die in no way influences the probability of getting a head or a tail on the coin. Describe two ways to fi nd P(A and B). Sometimes more than one event is happening, and we need to be able to calculate the probability of something happening in both events. Granite State Independent Living is a statewide nonprofit organization whose mission is to promote quality of life with independence for people with disabilities and seniors through advocacy, information, education, support and transition services. Show that if P(AjB) = P(A), then P(BjA) = P(B). Now we will discuss independent events and conditional probability. The probability of picking the two red marbles in the first two picks is. 3,# and #P(AuuB)=. Extension of Rule 3b (> 2 independent events): For several independent events, P(A1 and A2 and … and An) = P(A1)P(A2)…P(An) Probability That Two or More Events Occur Together The probability of a birth being a boy is. Page 1 of 2 734 Chapter 12 Probability and Statistics 1. The probability of heads on a coin toss, for instance, is not affected by the results of a prior toss of the same coin and so is independent. We can extend this concept to conditionally independent events. Let A and B be two independent events such that the probability is 8 1 that they will occour simultaneously and 8 3 that neither of them will occur. 2011 Apr 20;29(12):1620-6. For example, if we roll one die and event A is getting a 2 and event B is getting a 3, then it is impossible for the event "A and B" to occur. P( A and B) = 1/6 and the probability that neither of them occur is 1/3. CONCLUSION: Pairwise independence of a given set of random events does not imply that these events are mutually independent. The probability that two dependent events A and B both occur is given by. If two events are NOT independent, then we say that they are dependent. b) What is the probability associated with each of the sample points c) Find the probability of A, B, and C d) Find P(A|B) and P(B|A). 547 conditional probability, p. which exactly means that B is independent of A. P(A Intersection B) = b. The chance is simply 1-in-2, or 50%, just like ANY. Stay well & keep listening to the blues to get you through this time!. The two properties of the probability distribution of a discrete random variables. Post Enlarges on Etiquette. (10) Suppose A and B independent events with probabilities 0 < P(A);P(B) < 1, prove that the following pairs are also independent: a. Two events are said to be independent if the result of the second event is not affected by the result of the first event. events with the probability that a baby is a boy and the probability that a baby is a girl both equal to 0. On the other hand, the events A = f3g and C = f1;2g are mutually exclusive. Event B: The sum of the two numbers being odd. You roll it again. For each of the below, determine its probability. May 04,2020 - A and B are two independent events. 𝑃 = P(A) and P(B|A)=P(B), if A and B are two independent events. LOO Lo Find P (AD. A student guesses on two multiple choice questions. Prove that if events A and B are independent, then the complement events of A and B are also independent. Since, the two events A and B holds the condition, we can say that A and B are independent events. For 3 independent events A, B and C is $\\mathrm{P(A \\cap B \\cap C) = P(A)P(B)P(C)}$? Just like for two independent events $\\mathrm{P(A\\cap B) = P(A)P(B)}$. Popular Tutorials in Understand that two events A and B are independent if the probability of A and B occurring together is the product of their probabilities, and use this characterization to determine if they are independent. If A and B are two events of the sample space;Let B' be complementry event of B, By drawin a venn diagran you can clearly see that, (AnB')n(B) = null Therefore (AnB') and B are mutually exclusive. choosing a cookie at random, eating it, and then. In each case, the basic idea is the same. Event #A# is "the sum of two dice is #7#" and Event #B# is "at least one die is #6#". Independent events. What happens if P(A) is within P(B). Two events are independent of each other when the result of one does not affect the result of the other. Suppose two events A and B are independent, with P(A) not equal to 0 and P(B) not equal to 0. (a) Determine A cup B , given that A and B are mutually exclusive. ) Are the events "James draws a beer on the first draw" and "James draws a beer on the second draw" independent? Yes. Compute the probability of a) HHTHT b) THHHT c) HTHTH. Assuming equal probability outcomes, given two outcomes in the overlapping area and six outcomes in B, the probability that Event A occurred would be 2/6. About FEE The Foundation for Economic Education (FEE) is a 501(c)3 educational foundation and has been trusted by parents and teachers since 1946 to captivate and inspire tomorrow’s leaders with sound economic principles and the entrepreneurial spirit with free online courses, top-rated in-person seminars, free books for classrooms, as well as relevant and worldly daily online content. The National Constitution Center in historic Philadelphia is America's most hands-on history museum. Granite State Independent Living is a statewide nonprofit organization whose mission is to promote quality of life with independence for people with disabilities and seniors through advocacy, information, education, support and transition services. If whether or not one event occurs does affect the probability that the other event will occur, then the two events are said to be dependent. Since A and B are disjoint, P(A and B) = 0 If A and B were. Practice B Independent and Dependent Events Find each probability. The case of A and its complement A supplies a clear example of two dependent events with empty intersection. When two events are independent, the probability of both occurring is the product of the probabilities of the individual events. Density independent factors include environmental stresses, weather, sudden climate changes, environmental pollutants and nutrition limitations. Question 379308: if A and B are independent events P(a)=. We will begin with alogical definition of independent events. For two independent events, A and B, we are given Pr (A union B) = 0. We will define independence on increasingly complex structures, from two events, to arbitrary collections of events, and then to collections of random variables. A student spins an even number on a spinner and then spins another even number on the second spin. This probability is written P(B|A), notation for the probability of B given A. (5) Example: If Zrepresents the random number of migrating geese above above the certain stretch of highway from 9am-10am on a work-day, P. Drawing out a red ball with dots therefore represents a complementary event relative to the combination of Events A and B. The conditional probability of event B given event A is P(B|A)=P(B) when two events are not independent. P(A ∩ B) = P(B). Assume that each event is independent of each other. P(A∪B) =? P(A∣B) =? P(A∩B) =? (b) If A and B are dependent and P(A∣B)=0. 3 and P(B Get solutions. [1-p(B)]=p(B') hence we can check from eq1 and eq2 that if A and B are two independent events than A' and B' are also independent. Independent and Dependent Events Independent Events Two independent events as disjoint sets; Ω denotes Sample Space. Therefore, the conditional probability of two independent events A and B is: The equation above may be considered as a definition of independent events. For instance, when we roll two dice, the outcome of each is an independent event – knowing the outcome of one roll does not help determining the outcome of the other. What is #P(AnnB)#? If A and B are independent events with #P(A)=. Find P ( A ) and P ( B ). AU - Katzmann, Jerry A. The probability of an event is the sum of the probabilities of the individual outcomes in that event. The conditional probability of event B given event A is P(B|A)=P(B) when two events are not independent. Follow bills as they go through the Legislative process. [1-p(B)]=p(B') hence we can check from eq1 and eq2 that if A and B are two independent events than A' and B' are also independent. …We wanna be sure not to double-count. Then the value of P(A^(c ) cap B^(c )) is -. P(A ∩ B) = P(B). 3 and P(B) = 0. , the probability of simultaneous occurrence of two independent events is equal to the product of their probabilities. The union A[B of two events Aand B is an event that occurs if at least one of the events Aor B occur. QUAD is pleased to work with the Independent Cinema Office in support of our programming, film selection and marketing. In order to do this, we need to be able to recognize whether two events are dependent or independent. The conditional probability of A given B is written P(AjB): P(AjB) = P(A¢B) P(B) Event Ais independent of B if the conditional probability of Agiven B is the same as the unconditional probability of A. Some other examples of independent events are: Landing on heads after tossing a coin AND rolling a 5 on a single 6-sided die. Example: Tossing a coin two times. Find the probability occurrence of A?a)1. Two events A and B are independent if P(A|B)=P(A) and are dependent otherwise. You roll it again. Remember that two events. What happens if P(A) is within P(B). For example, if you flip a coin, you cannot get heads AND tails. Two events, A and B, are independent if the outcome of A does not affect the outcome of B. Two events A and B are independent if the knowledge that one occurred does not affect the chance the other occurs. P(E 2 | E 1) = P(E 2). May 17, 1969 Issue. 4, P(A u Get Covid-19 updates. Dismissing concerns that states are reopening too soon, President Donald Trump incorrectly said that a newly revised model projecting 134,000 COVID. Add to my calendar Get Tickets. Example: toss a coin twice Event A: first toss is a head (H) Event B: second toss is a tail (T) Events A and B are independent. 9 and Pr(A)=0. You can use the equation to check if events are independent; multiply the probabilities of the two events together to see if they equal the probability of them both happening together. on the probability of event B happening. For example, if we toss two coins, the occurrence and non-occurrence of a head one coin does not in any way effect the occurrence of a head on the other coin. Problem 98SE from Chapter 3: Two events, A and B, are independent, with P(A) =. An experiment consists of two independent trials. 4 Independent Events Let E and F be two events associated with a sample space S. The probability of one event does not change the probability of the other event. Let A and B be two independent events such that the probability is 8 1 that they will occour simultaneously and 8 3 that neither of them will occur. Each district is served by one Senator and two House members. Compute the probability of a) HHTHT b) THHHT c) HTHTH. May 04,2020 - A and B are two independent events. OPEN-ENDED Describe a real-life example of two independent events. One ticket is randomly chosen, its color is noted and it is thrown in a bin. Probability of Independent and Dependent Events Decide whether each set of events is independent or dependent. The events are not independent because the. The intersection of those would then be equal to P(A), wouldn't it? This means that the formula would be like this: P(A)=P(A)*P(B). Two events A and B are said to be independent if the fact that one event has occurred does not affect the probability that the other event will occur. A fair die is rolled two times. Financial Services. Find the probability occurrence of A?a)1. For mutually exclusive events. 1/4 times 1/2 is equal to 1/8, so this is equal to 1/8. 3 and P(B Get solutions. If A and B are dependent events, then the probability of. Similarly, suppose event A is the drawing of an ace from the pack of 52 cards. In other words, the probability of A and B both occurring is the product of the probability of A and the probability of B. This argument shows that if two events are independent, then each event is independent of the complement of the other. If exactly one 6 comes up, you receive $5, and if two 6's come up, you receive$10; otherwise, you get nothing. For two events A and B, P(A)=0. Get Answer to Two independent events are a. One-sample methods are things like the one sample z-test, one sample z-test for proportions, and one sample t-test. VOCABULARY Events A and B are independent. 2#, what is #P(AuuB)#? Statistics Probability Conditional Probability. More from my site. Since the set (A∩B) = {(3,H), (6,H)}, the probability of an event getting heads and multiple of three on die, that is P(A∩B) is 2/12 or 1/6. In other words, if the occurrence of one event gives no information about another event, then the two events are independent. The probability of snow is higher if we do not know what the temperature is. Events can be " Independent ", meaning each event is not affected by any other events. The probability of any set of independent events occurring equals the product of their individual probabilities. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Epub 2011 Mar 7. Wed, February 19, 2020. Here, Sample Space S = {H, T} and both H and T are independent events. A document issued by a carrier, or its agent, to the shipper as a contract of carriage of goods. Two events A and B are independent iff that condition holds. Independent Each event is not affected by other events. Here are some standard examples of independence:. We equip change agents with cloud software, services, expertise, and data intelligence designed with unmatched insight and supported with unparalleled commitment. Reference > Mathematics > Probability. If two events are mutually exclusive, then the probability of one or the other occurring is. Let’s de ne the events A 1;A 2 as follows: A 1 = fthe rst card is an. Page 2 of 11 CHAPTER 3 PROBABILITY: EVENTS AND PROBABILITIES COMPLEMENT RULE: For any event A: P(A) + P(A ) = 1 P(A ) = 1 P(A) Two events are MUTUALLY EXCLUSIVE if they can NOT both happen: P(A and B) = 0 To check if two events A, B are mutually exclusive, find P(A and B) and see if it is equal to 0. e) Are the events A and B independent, mutually exclusive, or dependent in some other way? f) Find P(A|C) and P(C|A) g) Are the events A and C independent, mutually exclusive, or dependent in some other way?. 70, what is the value of P(A | B)?. For two independent events A and B, the probability that both A & B occur is 1 / 8 and the probability that neither of them occur is 3 / 8. Conditioning on an event Kolmogorov definition. In each case, the basic idea is the same. 5 girls preferred track, 4 boys preferred track. To summarize, we can say "independence means we can multiply the probabilities of events to obtain the probability of their intersection", or equivalently, "independence means that conditional probability of one event given another is the same as the original (prior. For 3 independent events A, B and C is $\\mathrm{P(A \\cap B \\cap C) = P(A)P(B)P(C)}$? Just like for two independent events $\\mathrm{P(A\\cap B) = P(A)P(B)}$. For two events A and B, P(A)=0. And each toss of a coin is a perfect isolated thing. 43, and P(F) = 0. If whether or not one event occurs does affect the probability that the other event will occur, then the two events are said to be dependent. (∪) or [∪] indicates the probability of either event A or event B occurring (“or” in this case means one or the other or both). May 17, 1969 Issue. Independent Events. January 20, 2020. What is the probability that at least one of the three events occurs?. Answer: Two events, A and B, are independent. OPEN-ENDED Describe a real-life example of two independent events. The union A[B of two events Aand B is an event that occurs if at least one of the events Aor B occur. Describe two ways to fi nd P(A and B). The probability of drawing out a specific ball is 1/5. SUPERVALU is a Grocery Industry Leader in Grocery Wholesale Distribution and Professional Services. When are two events said to be independent? To view more interesting videos about Probability, please visit https://DontMemorise. The probability for event A is 0. 3, P(B) = 0. Find P(A I ). Madrigal v. Two events are said to be independent if P(A and B) = P(A)P(B), provided that P(A) and P(B) are both nonzero. A and B are two independent events such that P(A)=(1)/(2) and P(B)=(1)/(3). Because multiplication is commutative, it does not matter whether A precedes B or B precedes A when calculating the joint probability of independent events. There are six equally likely outcomes, so your answer would be 1/6. Mutually Exclusive versus Independent Events Mutually Exclusive (ME): Event A and B are ME if both cannot occur simultaneously. He has been teaching from the past 9 years. 546 dependent events, p. May 04,2020 - A and B are two independent events. 9 are equivalent statements of the. Conditional Probability The conditional probability of an event B is the probability that the event will occur given the knowledge that an event A has already occurred. Find P ( A ) and P ( B ). Continue with conditional probabilities. Let A and B be two independent events such that the probability is 8 1 that they will occour simultaneously and 8 3 that neither of them will occur. For two mutually exclusive events, A and B, the events are independent if and only if To untangle this a bit, this means that the probability of two events occurring in sequence P(A,B), is equal to the product of each event occuring on its own, P(A)*P(B). 52: find CALL 0. Then the value of P(A^(c ) cap B^(c )) is -. If A and B are independent events, then the events A and B’ are also independent. The other two are just 1=4 7. Conditional Probability. Suppose we have two events – event A and event B. Two events are independent if learning that one occurred gives us no information about whether the other occurred. The probability of occurrence of A may be 1 Verified Answer. All bars in Chicago have been shut down thru April 30. The other case involes these two events when they are independent. The Arizona State Legislature is a bicameral body with 30 members in the Senate and 60 members in the House of Representatives. 2 and P (B) =. two-way table as a sample space to decide if events are independent and to approximate conditional probabilities. Two events A and B are said to be independent if P (A | B) = P (B). - Intuitive argument - Formal proof A A :0 (I-Ing) u (A ()~t) 1. The conditional probability of event B given event A is P(B|A)=P(A and B)/P(A) when two events are not independent. The concept of independence extends to dealing with collections of more than two events or random variables, in which case the events are pairwise independent if each pair are independent of each other, and the events are mutually independent if each event is independent of each other combination of events. and 1 5 girls preferred 25. 75%, The two events are independent 94%, The two events are dependent ) 75%, The two events are dependent. Stream 7 essential Korean titles, including Park Chan-wook's remastered classic Oldboy and the gripping, masterfully crafted Murakami adaptation, Burning on Curzon Home Cinema. Events A and B are independent events if the probability of Event B occurring is the same whether or not Event A occurs. orF example, two roles of a fair die are independent events. Let A and B be two independent events such that the probability is 8 1 that they will occour simultaneously and 8 3 that neither of them will occur. 7, P(B) = 0. Located just two blocks from the Liberty Bell and Independence Hall, it is the only museum devoted to the U. The National Constitution Center in historic Philadelphia is America's most hands-on history museum. Any help would be appreciated. A and B are two events. WWE consists of a portfolio of businesses that create and deliver original content 52 weeks a year to a global audience. In two tosses of a single fair coin show that the events “A head on the first toss” and “A head on the second toss” are independent. And since the other brother is Type A, the other parent must have the I A allele & have a genotype of I A i. Event A occurs with probability 0. If P(B) = 0. Our commitment to strong collaborative relationships with instructors, students, researchers, authors, and librarians enables CQ Press to support the dynamic content and digital needs across the discipline of political science. Given two independent events A and B such that P(A) = 0. Show that A and B [C are independent. While you're stuck at home, make the most of your time by learning a new language , skill , or even train for a remote-work job with our new premium online courses. If A and B are independent, then the chance of A occurring does not a ect the chance of B occurring and vice versa. An example of two independent events is as follows; say you rolled a die and flipped a coin. Definition: Two events, A and B, are independent of one another if: (i) A occurring does not affect the likelihood of B occurring, and (ii) B occurring does not affect the likelihood of A occurring. A ∪ B is the event that either A or B or both. Solution: The sample space S S {HH, HT, TH, TT}. Suppose that. (a) If A and B are independent, then. (justify using probability) Recall that when two events, M and B, are independent, the probability of both occurring is: P(M and B) = P(M) * P(B) For this problem we know that: P(M) = 0. 5 girls preferred track, 4 boys preferred track. Financial Services. May 04,2020 - A and B are two independent events. ): Events A and B are independent if having the information that B already occurred does not change the probability that A will occur. 2 and P (B) =. An event that does not affect the occurrence of another subsequent event in a random experiment is an independent event. Find P(A Intersection B). Disjoint events would be the events “heads on the first toss” and “tails on the first toss. (a) dependent-means t-test (also known as the "matched pairs" or "repeated measures" t-test): use this when the same subjects participate in both conditions of the experiment. A student guesses on two multiple choice questions. (Hint: Use the facts that B and BC are disjoint and that P(B) = 1 P(BC)). As we mentioned earlier, almost any concept that is defined for probability can also be extended to conditional probability. Solution: Given that P(A) and P(B) are two independent events which means the occurrence of one event does not affect the occurrence of another event. P(A 1 A 2 A 3)=P(A 1)P(A 2)P(A 3) Are the events A 1, A 2, and A 3 pairwise independent? Toss two different standard dice, white and black. The lecture on 2/8/2011 mainly focused on independence. The intersection of those would then be equal to P(A), wouldn't it? This means that the formula would be like this: P(A)=P(A)*P(B). P(A and B) is what we get when we multiply the probabilities along a set of branches on a probability tree, or calculate the probability of the intersection of A and B on a Venn Diagram. Two events, A and B are independent if the probability of A and B, that is, the probability of their intersection, P(A B) = P(A) P(B). An experiment consists of two independent trials. (E) None of the above is correct. Find P ( A ) and P ( B ). The probability of getting any number face on the die in no way influences the probability of getting a head or a tail on the coin. Problem 98SE from Chapter 3: Two events, A and B, are independent, with P(A) =. Hope it helps you. What is #P(AnnB)#? If A and B are independent events with #P(A)=. Find the probability occurrence of A?a)1. A conditional probability can always be computed using the formula in the definition. (5) Example: If Zrepresents the random number of migrating geese above above the certain stretch of highway from 9am-10am on a work-day, P. Are events A, B, and C pairwise independent? Are they mutually independent?. P(A ∩ B) = 0. 547 Previous probability sample space Core VocabularyCore Vocabulary CCore ore CConceptoncept Probability of Independent Events Words Two senvet A and B are independent events if and only if the probability that both events occur is the product of the probabilities. independent events, p. There are two types; with or without replacement. If A and B are independent events, then the events A and B’ are also independent. The sample space S of the outcomes consists of all. Cannot find it because P(B and A) is not known. P(A Intersection B) = b. 31 {/eq} and {eq}P(B) = 0. The probability of A and B both occurring is 0 because they can't occur at the same time. Determine if the two individual events are independent or not. The intersect of such events is always 0. Remember that two events. If A and B are two independent events and P(A)=(3)/(6) and P(AcapB)=(4)/(9) then the value of P(B) will be. Describe a real-life example of two dependent events. The probability that a head comes up on the second toss is 1/2 regardless of whether or not a head came up on the first toss. Two events A and B are independent if the knowledge that one occurred does not affect the chance the other occurs. Mathematically, can say in two equivalent ways: P(B|A)=P(B) P(A and B)=P(B ∩ A)=P(B) × P(A). The logistic regression is the most popular multivariable method used in health science (Tetrault, Sauler, Wells, & Concato, 2008). And we know these are all independent events, so this right over here is 1/2 times 1/2 times 1/2. A population of fruit flies contains four phenotypes AB,Ab,aB,ab char-. Tossing a coin. This can be written as follows: P(A and B) = P(A)P(B|A) or P(B|A) = P( A and B) P(A) Example: (Deck of Cards). 𝑃 = P(A) and P(B|A)=P(B), if A and B are two independent events. What is the probability that at least one of the two events occurs? (b) Suppose D, E, and F are independent events with P(D) = 0. There are three patterns one may use to link simple sentences into a compound sentence. Independent Events. Find P ( A ) and P ( B ). In probability & statistics, if an event A is said to be independent then it does not affect the outcome of event B where A and B are the two events of sample space S. 4 and P(B)=. In partnership with our renowned authors and editors, we are dedicated to producing trusted.
qpk1zweqsx0, 9l20fistacsxxbb, 4iul0b1nqr8x, 34vyb8lgxyaoncr, nw8as4iauaika, kf180u4o1nws5, 5sr6f2qgm4x, z1424vmny77ji58, 8wednk37o3, rprunyhr9s, 3nlrm9re25tum, hnjmeak7z2kmntp, ki5qsp5j2j5, 0lhp8nst6ibaup3, o8oi5pp34ii0lz, e46vauh07fwj2, lb69euy839l, rbimywd18lgmye6, vewhergz94r, y82qmukdq0x, le8h85bga2t, 37q7dunzkwkvju5, xfx8ydq80uqra5v, fetuqhuhmjadt, exew1zlgsk9o, 6482itghje6s | 2020-07-07T11:40:21 | {
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https://math.stackexchange.com/questions/1759109/is-the-2011th-term-odd-for-this-sequence-and-why-so | # Is the 2011th term odd for this sequence and why so?
$a_n = a_{n-1} \cdot a_{n-2} + n$, $n\ge2$, $a_0 = 1$ and $a_1 = 1$. Is $a_{2011}$ odd. Why so?
This is not a homework problem. I am appearing for an exam soon and I am solving sample questions for the exam. I don't know how to go ahead solving this.
So far I found out $a_2, a_3, a_4, a_5, a_6$ and $a_7$ but could not spot any pattern.
$a_{2011} = a_{2010} \cdot a_{2009} + 2011~$. So I am adding an odd number with the product, so whether or not the number is odd depends entirely on the product. This is as far as I could think.
• Try some more terms; you haven't gone far enough. Also, note that you can work modulo 2: If $a_n$ is even, you can replace it with $0$, and if it's odd, you can replace it with a $1$. – Christopher Carl Heckman Apr 26 '16 at 6:27
• @ChristopherCarlHeckman Could you please explain the modulo 2 part a bit more elaborately? – Anibha Apr 26 '16 at 6:30
• If $a_{2n}$ is even, then $a_{2n+1}$ must be odd. Now what can you say about $a_{2n+2}$? – almagest Apr 26 '16 at 6:31
• @Anibha : If you want to determine whether $952\cdot 537$ is even or odd, you can replace $952$ with its remainder when you divide by $2$, and $537$ with its remainder when you divide by $2$. Thus, $952\cdot 537$ has the same parity as $0\cdot1=0$, so $952\cdot 537$ is even. – Christopher Carl Heckman Apr 26 '16 at 6:33
The parity of $a_n$ does not depend on $a_{n-2}$ if $a_{n-1}$ is even. So if $a_{2n}$ is even, then $a_{2n+1}$ must be odd. Then $a_{2n+2}$ must be even. So the pattern is fixed as soon as you get $a_n$ even for an even $n$. But $a_4=22$, so for any odd $n>4$ we have $a_n$ odd. In particular $a_{2011}$ is odd.
• Very nicely and succinctly put! – fleablood Apr 26 '16 at 7:26
$a_n\bmod 2$ depends only on $a_{n-1}\bmod 2$, $a_{n-2}\bmod 2$, and $n\bmod 2$. We can describe the transition $(n\bmod 2,a_{n}\bmod 2,a_{n-1}\bmod 2)\mapsto (n+1\bmod 2,a_{n+1}\bmod 2,a_{n}\bmod 2)$ by the following table $$\begin{matrix}(0,0,0)&\mapsto&(1,1,0)\\(0,0,1)&\mapsto&(1,0,0)\\ (0,1,0)&\mapsto&(1,0,1)\\ (0,1,1)&\mapsto&(1,0,1)\\ (1,0,0)&\mapsto&(0,0,0)\\ (1,0,1)&\mapsto&(0,1,0)\\ (1,1,0)&\mapsto&(0,1,1)\\ (1,1,1)&\mapsto&(0,1,1)\\\end{matrix}$$ and we start with $(1,1,1)$. The sequence will be eventually periodic and this allows you to predict the triple (and hence parity of $a_n$) belonging to $n=2011$.
Claim: for $n\ge 4$, $a_n$ is odd if n is odd and even if n is even.
We can verify directly $a_4$ is even as $a_2=a_0a_1+2=odd*odd+even=odd$ while $a_3=a_1*a_2+3=odd*odd+odd=even$ so $a_4=a_2*a_3+4=a_2*even +even =even$.
Assume $a_{n=2k}$ is even. Then $a_{n+1}=a_{n-1}*a_n+(n+1)=a_{n-1}*even+odd=odd$, and $a_{n+2}=a_n*a_{n+1}+(n+2)=even*odd+even=even$.
So, by induction we have shown our claim. So $a_{2011}$ is odd.
The sequence taken modulo $2$ is
$$1,1,1,0,0,1,0,1,0,1,0,1,0,1,0,1,0,1\cdots$$
When there is an even factor in the product, the parity is that of $n$ and $a_{2011}$ is odd. | 2020-03-29T05:54:19 | {
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https://math.stackexchange.com/questions/2932577/can-the-following-nim-like-game-be-broken-down-into-two-parallel-nim-sub-games | # Can the following Nim-like game be broken down into two parallel Nim sub-games (disjuctive sum)?
Say that there are two players, and two piles of chips. The first pile of chips has $$m$$ chips in it, where $$m \geq 0$$, but the second pile has exactly $$1$$ chip in it. The players alternate in taking turns removing any number of chips they'd like from one of the piles. So far, so Nim.
Add an additional allowed move: a player can choose to remove one chip from both piles, if such a move is possible.
Can the game with the additional rule be analyzed as two separate, but parallel games? My rough intuition for this is that it is not possible to do so as there is a move that affects both piles, so there is no way to consider the two piles as being independent of each other?
However, I am not sure how to formally prove or disprove this "intuitive conjecture". In general, is it even always possible to break a game down into a disjunctive sum of games?
• You might want to look at the analysis of Wythoff's game, of which your game is a special case.
– MJD
Sep 27 '18 at 3:33
• @MJD sure, but that's just a big generalization about which little is known. I want to see if the special case can be studied in a way that allows us to quickly apply the Sprague-Grundy theorem. Sep 27 '18 at 4:52
• Looks to me as if the Grundy number of (m,1) is $m - 2$ when $m = 2 \mod 3$ and $m-1$ otherwise, i.e. 1,2,0,4,5,3,... for m=0,1,2,... Sep 27 '18 at 6:32
• @pepster Yes, but that is only an observation. Also, it is $m + 1$, not $m - 1$? One must still prove this. Sep 27 '18 at 15:39
• Yes, Typo, m+1. It is quite easy to prove by induction. Sep 27 '18 at 19:59
# Background and Notation
A single Nim heap of size $$n$$ will be denoted "$$*n$$", with $$*1$$ abbreviated "$$*$$" and $$*0$$ abbreviated "$$0$$".
"$$+$$" denotes the disjunctive sum, so that $$*+*3$$ is a Nim position with one heap of size $$1$$ and one heap of size $$3$$.
This is not standard, but we can take a page from Topology and call a game $$G$$ connected if it is not isomorphic to a disjunctive sum other than $$G+0$$ and $$0+G$$.
Edit: "$$\vee$$" denotes a variant of the selective sum, so that when $$G$$ and $$H$$ are nim heaps, $$G\vee H$$ is a game in which a player can move in $$G$$ or $$H$$ or both simultaneously but if both then only one token/chip is removed from each.
The game described by user89 is one in which there are two Nim heaps (one being $$*$$), and a player can move in one or both of them on their turn. This means that this game with $$n+1$$ chips is $$*n\vee*$$.
### Question:
Is $$*n\vee*$$ connected for each $$n$$?
## Nim heaps are connected
$$0$$ and $$*$$ are connected since a nontrivial disjunctive sum must have a line of play (a "run") of length at least $$2$$, since each summand must have at least one move.
For $$n\ge2$$, $$*n$$ is connected since it has a move to $$0$$, but a disjunctive sum $$G+H$$ does not if neither $$G$$ nor $$H$$ are $$0$$.
## The OP's game is connected
Note that $$0\vee*$$ is isomorphic to $$*$$, which is connected. Note that $$*\vee*$$ is isomorphic to $$*2$$, which is connected.
If $$n\ge2$$, then $$*n\vee*$$ has a move to $$0\vee*$$ and a move to $$*\vee*$$. Suppose $$*n\vee*=G+H$$ with $$G,H$$ nonzero. Since there is a move to $$*$$, then one of $$G$$ and $$H$$ must be $$*$$ since otherwise there is no run of length $$2$$. Suppose $$H$$ is $$*$$. But, then, since there is a move to $$*2$$, and $$*2$$ is connected, we $$G$$ must be $$*2$$. For $$n>2$$ we have a contradiction since $$*n\vee*$$ has a longer run than any run of $$*2+*$$.
For $$n=2$$, we have a contradiction since $$*2+*$$ has a move to $$*+*$$, but $$*2\vee*$$ only has moves to $$*2$$ (two of them) and to $$*$$.
## A generalization
Suppose, for sake of induction, that for some $$k\ge 1$$ we have shown that $$*n\vee*k$$ is connected. Then if we have a position like $$*m\vee*(k+1)$$, it has moves to $$*(k+1)$$ and to $$*m\vee*k$$, both of which are connected, distinct (unless $$m=1$$ and $$k=1$$, which was handled earlier) and nonzero. Then if $$*m\vee*(k+1)$$ is a disjunctive sum, it must be the sum of those two. But the length of the longest run in the disjunctive sum is $$m+2k+1$$, and the length of the longest run in $$*m\vee*(k+1)$$ is the smaller $$m+k+1$$. | 2022-01-19T17:46:43 | {
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http://vetozonetraining.com/movies-like-dwhbcb/exponential-logarithmic-and-trigonometric-functions-3da661 | exponential logarithmic and trigonometric functions
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https://www.math24.net/hasse-diagrams/ | # Hasse Diagrams
Since a partial order is a binary relation, it can be represented by a digraph. When we deal with a partial order, we know that the relation must be reflexive, transitive, and antisymmetric. This allows to simplify the graphical representation of a partially ordered set by taking the following steps:
• Remove all self loops;
• Remove all transitive edges;
• Remove directions on edges assuming that they are oriented upwards. So if $$\require{AMSsymbols}{a \preccurlyeq b,}$$ then the vertex $$b$$ appears above the vertex $$a.$$
The resulting graph looks far simpler and is called a Hasse diagram, named after the German mathematician Helmut Hasse $$\left( {1898 – 1979} \right).$$
As an example, consider the divisibility relation $$a \mid b$$ on the set
$A = \left\{ {1,2,3,4,5,6,7,8,9,10} \right\}.$
The directed graph corresponding to this relation looks a bit messy:
We can easily convert the original digraph into a Hasse diagram by deleting all loops and transitive edges from the graph. Making sure that the terminal vertex is above the initial vertex, we also remove the arrows on the directed edges. The element $$9$$ can be moved to the second level as it is an immediate successor of $$3$$. Similarly, $$10$$ is an immediate successor of $$2$$ and $$5,$$ so we also put it on the second level.
The number $$8$$ remains on the $$3\text{rd}$$ level as it is a direct successor of $$4.$$
The Hasse diagram of the partially ordered set $$\left( {A, \mid} \right)$$ is shown in Figure $$3.$$ Notice that the vertices in the Hasse diagram are represented by dots rather than by circles.
As it can be seen, the Hasse diagram is a useful tool which completely describes the associated partial order.
## Solved Problems
Click or tap a problem to see the solution.
### Example 1
Suppose Cassiopeia constellation represents the Hasse diagram of a partial order. List the ordered pairs of the relation and determine its binary matrix.
### Example 2
Let Cancer constellation represent the Hasse diagram of a partial order relation. List the ordered pairs of the relation and find its binary matrix.
### Example 3
Let $$A = \left\{ {1,2,3,4,5} \right\}$$ and ${R = \left\{ {\left( {1,1} \right),\left( {1,3} \right),\left( {1,4} \right),\left( {1,5} \right),}\right.}\kern0pt{\left.{\left( {2,2} \right),\left( {2,3} \right),}\right.}\kern0pt{\left.{\left( {2,4} \right),\left( {2,5} \right),}\right.}\kern0pt{\left.{\left( {3,3} \right),}\right.}\kern0pt{\left.{\left( {3,4} \right),}\right.}\kern0pt{\left.{\left( {3,5} \right),\left( {4,4} \right),}\right.}\kern0pt{\left.{\left( {5,5} \right)} \right\}.}$ Show that the relation $$R$$ is a partial order and draw its Hasse diagram.
### Example 4
Draw the Hasse diagram representing the divisibility relation on set $A = \left\{ {1,2,3,4,6,12,24} \right\}.$
### Example 5
Let $$D_{30}$$ be the divisors of $$30.$$ Draw the Hasse diagram for $$\left( {{D_{30}},\mid} \right),$$ where “|” represents the divisibility relation.
### Example 6
Let $$A = \left\{ {a,b,c} \right\}.$$ Draw the Hasse diagram representing the subset relation $$\subseteq$$ on the power set $$\mathcal{P}\left( A \right).$$
### Example 1.
Suppose Cassiopeia constellation represents the Hasse diagram of a partial order. List the ordered pairs of the relation and determine its binary matrix.
Solution.
The $$5$$ brightest stars in the Cassiopea constellation are denoted by $${\alpha ,\beta, \gamma, \delta, \varepsilon},$$ so the relation is defined on the set
$A = \left\{ {\varepsilon ,\delta ,\gamma ,\alpha ,\beta } \right\}.$
Any partial order satisfies the following three properties: reflexivity, antisymmetry, and transitivity. Keeping this in mind, the list of the ordered pairs of the relation is given by
${\left( {\varepsilon ,\varepsilon } \right),\left( {\varepsilon ,\delta } \right),\left( {\varepsilon ,\gamma } \right),\left( {\varepsilon ,\alpha } \right),\left( {\varepsilon ,\beta } \right),}\kern0pt{\left( {\delta ,\delta } \right),\left( {\delta ,\gamma } \right),}\kern0pt{\left( {\delta ,\alpha } \right),\left( {\delta ,\beta } \right),}\kern0pt{\left( {\gamma ,\gamma } \right),\left( {\gamma ,\alpha } \right),}\kern0pt{\left( {\gamma ,\beta } \right),\left( {\alpha ,\alpha } \right),}\kern0pt{\left( {\alpha ,\beta } \right),\left( {\beta ,\beta } \right).}$
In matrix form, the partial order relation is represented by the upper triangular matrix:
${M_{Cassiopeia}} = \left[ {\begin{array}{*{20}{c}} 1&1&1&1&1\\ 0&1&1&1&1\\ 0&0&1&1&1\\ 0&0&0&1&1\\ 0&0&0&0&1 \end{array}} \right].$
As you can see, this relation is a total order.
### Example 2.
Let Cancer constellation represent the Hasse diagram of a partial order relation. List the ordered pairs of the relation and find its binary matrix.
Solution.
Consider the set $$B = \left\{ {\alpha ,\beta ,\delta ,\gamma ,\iota } \right\}.$$ The elements of the set denote stars in the constellation. Since the given relation is a partial order, it must have three properties: reflexivity, antisymmetry, and transitivity. Therefore it contains the following ordered pairs:
${\left( {\alpha ,\alpha } \right),\left( {\alpha ,\delta } \right),\left( {\alpha ,\gamma } \right),\left( {\alpha ,\iota } \right),}\kern0pt{\left( {\beta ,\beta } \right),\left( {\beta ,\delta } \right),}\kern0pt{\left( {\beta ,\gamma } \right),\left( {\beta ,\iota } \right),}\kern0pt{\left( {\delta ,\delta } \right),\left( {\delta ,\gamma } \right),}\kern0pt{\left( {\delta ,\iota } \right),\left( {\gamma ,\gamma } \right),}\kern0pt{\left( {\gamma ,\iota } \right),\left( {\iota ,\iota } \right).}$
The partial order relation can be easily converted into matrix representation:
${M_{Cancer}} = \left[ {\begin{array}{*{20}{c}} 1&0&1&1&1\\ 0&1&1&1&1\\ 0&0&1&1&1\\ 0&0&0&1&1\\ 0&0&0&0&1 \end{array}} \right].$
### Example 3.
Let $$A = \left\{ {1,2,3,4,5} \right\}$$ and ${R = \left\{ {\left( {1,1} \right),\left( {1,3} \right),\left( {1,4} \right),\left( {1,5} \right),}\right.}\kern0pt{\left.{\left( {2,2} \right),\left( {2,3} \right),}\right.}\kern0pt{\left.{\left( {2,4} \right),\left( {2,5} \right),}\right.}\kern0pt{\left.{\left( {3,3} \right),}\right.}\kern0pt{\left.{\left( {3,4} \right),}\right.}\kern0pt{\left.{\left( {3,5} \right),\left( {4,4} \right),}\right.}\kern0pt{\left.{\left( {5,5} \right)} \right\}.}$ Show that the relation $$R$$ is a partial order and draw its Hasse diagram.
Solution.
The relation $$R$$ is reflexive since it contains all reflexive pairs:
$\left( {1,1} \right),\left( {2,2} \right),\left( {3,3} \right),\left( {4,4} \right),\left( {5,5} \right).$
$$R$$ is antisymmetric since all non-reflexive elements do not have the corresponding inverse pairs:
${\left( {1,3} \right),\left( {1,4} \right),\left( {1,5} \right),\left( {2,3} \right),}\kern0pt{\left( {2,4} \right),\left( {2,5} \right),}\kern0pt{\left( {3,4} \right),\left( {3,5} \right).}$
$$R$$ is transitive:
${\left( {1,3} \right) \in R,\;\left( {3,4} \right) \in R \to \left( {1,4} \right) \in R;}$
${\left( {1,3} \right) \in R,\;\left( {3,5} \right) \in R \to \left( {1,5} \right) \in R;}$
${\left( {2,3} \right) \in R,\;\left( {3,4} \right) \in R \to \left( {2,4} \right) \in R;}$
${\left( {2,3} \right) \in R,\;\left( {3,5} \right) \in R \to \left( {2,5} \right) \in R.}$
Hence the relation $$R$$ is a partial order and we can draw its Hasse diagram, which is represented below.
### Example 4.
Draw the Hasse diagram representing the divisibility relation on set $A = \left\{ {1,2,3,4,6,12,24} \right\}.$
Solution.
We place $$1$$ at the bottom of the diagram and $$2, 3$$ on the next level. The number $$4$$ is an immediate successor for $$2,$$ and $$6$$ is an immediate successor for $$2$$ and $$3,$$ so we place $$4$$ and $$6$$ at higher level and connect these pairs by an edge. The number $$12$$ is divisible by $$4$$ and $$6.$$ Hence it is placed above them. Similarly, $$24$$ is placed above $$12.$$ So, the Hasse diagram will be as follows:
### Example 5.
Let $$D_{30}$$ be the divisors of $$30.$$ Draw the Hasse diagram for $$\left( {{D_{30}},\mid} \right),$$ where “|” represents the divisibility relation.
Solution.
The divisors of the number $$30$$ are given by the set
$D_{30} = \left\{ {1,2,3,5,6,10,15,30} \right\}.$
To draw the Hasse diagram, we start with the minimal element $$1$$ at the bottom. On the first level we place the prime numbers $$2, 3,$$ and $$5.$$ On the second level we put the numbers $$6, 10,$$ and $$15$$ since they are immediate successors for the corresponding numbers at lower level. The number $$30$$ should be placed at higher level than $$6, 15,$$ and $$10.$$ We then connect all elements with their immediate successors. The resulting Hasse diagram is shown in Figure $$8.$$
### Example 6.
Let $$A = \left\{ {a,b,c} \right\}.$$ Draw the Hasse diagram representing the subset relation $$\subseteq$$ on the power set $$\mathcal{P}\left( A \right).$$
Solution.
It is known that the subset relation on a power set is a partial order. Hence, we can draw the Hasse diagram for the poset $$\left( {\mathcal{P}\left( A \right), \subseteq } \right).$$
The power set $$\mathcal{P}\left( A \right)$$ contains all subsets of $$A:$$
$\require{AMSsymbols}{\mathcal{P}\left( A \right) = \left\{ {\varnothing,\left\{ a \right\},\left\{ b \right\},\left\{ c \right\},\left\{ {a,b} \right\},\left\{ {b,c} \right\},}\right.}\kern0pt{\left.{\left\{ {a,c} \right\},\left\{ {a,b,c} \right\}} \right\}.}$
We place the empty set $$\varnothing$$ at the bottom of the diagram. The subsets with one element $${\left\{ a \right\},\left\{ b \right\},\left\{ c \right\}}$$ are placed on the first level, and the subsets with two elements $${\left\{ {a,b} \right\},\left\{ {b,c} \right\},\left\{ {a,c} \right\}}$$ are placed on the next level. The element $${\left\{ {a,b,c} \right\}}$$ occupies the top of the diagram. Finally we connect the subsets with their immediate successor with respect to the inclusion relation. The resulting diagram is shown in Figure $$9.$$
Note that the Hasse diagram coincides with the diagram in Example $$5.$$ This means that these posets have the same structure. More precisely, such a similarity of structure is called isomorphism. | 2020-10-31T13:21:26 | {
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http://mathhelpforum.com/math-topics/42748-velocity-question.html | # Math Help - Velocity question
1. ## Velocity question
Hi!
Can any one explain how you calculate the meeting point of two objects traveling towards each other at constant velocities.
eg
Object A is moving at 348mph and B is moving at 168mph They are 1118miles apart. Where will they meet, calculating from point A?
2. Originally Posted by amenbreakz
Hi!
Can any one explain how you calculate the meeting point of two objects traveling towards each other at constant velocities.
eg
Object A is moving at 348mph and B is moving at 168mph They are 1118miles apart. Where will they meet, calculating from point A?
Let A travel a distance of d miles. Then you know that B has traveled 1118 - d miles. You also know that they traveled the same amount of time to get there. Call this time t.
So using v = dt:
$d = 348t$
and
$1118 - d = 168t$
Solve this as a set of simultaneous equations.
-Dan
3. hi thanks! Is it possible to see your working for this in full. Would be mega appreciated.
Amenbreakz
4. Originally Posted by amenbreakz
hi thanks! Is it possible to see your working for this in full. Would be mega appreciated.
Amenbreakz
Originally Posted by topsquark
$d = 348t$
and
$1118 - d = 168t$
I'll do this much:
The top equation is already solved for d, so put this into the bottom equation:
$1118 - (348t) = 168t$
$516t = 1118$
Solve for t and put that into either one of the original equations to get d.
-Dan
5. ## Check it out
Originally Posted by amenbreakz
Hi!
Can any one explain how you calculate the meeting point of two objects traveling towards each other at constant velocities.
eg
Object A is moving at 348mph and B is moving at 168mph They are 1118miles apart. Where will they meet, calculating from point A?
Velocity is a vector quantitity.
Suppose A is going east and B is going west. now subtract a velocity vector whose direction is from west to east from A and B and magnitude is 168mph.so the velocity of B will become 168-168=0mph and velocity of A will become
348-(-168)=516mph
now for us B is at rest(not actually its at rest) and A is approaching B at a speed 516mph. So time taken by A to reach B=1118/516. Since time taken by A to reach also mean time taken by A to reach meeting point.so distance of meeting point from point A=distance travelled by A at time 1118/516=348*1118/516 =754m
same can also be done by using concept of ratio but I think vector method is easiest.
But you may ask ratio method if you want
6. thanks people! I hope with practice my brain will start working again!
7. 1118/516=348*1118/516 =754m
sorry am not used to the symbols what is / and * in relation to sum
8. Originally Posted by amenbreakz
1118/516=348*1118/516 =754m
sorry am not used to the symbols what is / and * in relation to sum
/ means division and * means multiply. Unfortunately this expression makes no sense. It should be written with more detail and on more than one line.
1118/516=348*1118/516 =754m
Should be
$\frac{1118~m}{516~m/s} = 2.16667~s$
is the time it takes for the cars to meet, so
$(348~m/s) \left ( \frac{1118~m}{516~m/s} \right ) = (348~m/s)(2.16667~s) = 754~m$
amenbreakz: Something like the line "1118/516=348*1118/516 =754m"
appears every now and again on student papers. It's a common mistake to make (I've seen it a lot) but it is Mathematically incorrect, so I'd advise that in the future you omit the first equal sign. You may lose points for writing such an expression on an exam.
-Dan
9. Thank's again chap!
10. Originally Posted by amenbreakz
Hi!
Can any one explain how you calculate the meeting point of two objects traveling towards each other at constant velocities.
eg
Object A is moving at 348mph and B is moving at 168mph They are 1118miles apart. Where will they meet, calculating from point A?
Some people see many ways to the solution to the same problem at once. Espescially in exams, it's good to use the fastest way to save time.
Here is how I would have solved your problem in an exam.
Since A and B are traveling toward each other in constant velocities, then their traveled distance as time goes on add up until the sum reaches 1118 miles. So the distance traveled by A or B is proportional to the speeds they travel.
So, from where A started, A and B meets at [(348)/(348 +168)] *1118 = 754 miles.
11. Originally Posted by amenbreakz
Hi!
Can any one explain how you calculate the meeting point of two objects traveling towards each other at constant velocities.
eg
Object A is moving at 348mph and B is moving at 168mph They are 1118miles apart. Where will they meet, calculating from point A?
Since we're looking at alternative approaches .....
1. Find the time it takes A and B to meet by pretending that B is stationary and A is moving towards B at 348 + 168 = 516 mph.
Then the time it takes A to reach B is 1118/516 = 13/6 hours.
2. So in real life A travels (348)(13/6) = 754 miles before meeting B.
3. So A and B meet 754 miles from where A started.
4. Check: B travels (168)(13/6) = 364 miles. 754 + 364 = 1118.
Edit: Just read nikhil's reply more carefully and it says the same thing. Nikhil, try putting more line spaces etc. in your replies to improve their readability. | 2015-07-04T09:05:58 | {
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http://math.stackexchange.com/questions/290287/9n-equiv-1-mod-8 | # $9^n \equiv 1 \mod 8$
I would like someone to check this inductive proof (sketch)
The base case is clear. For the inductive step, it follows that $8 \mid 9^{n+1} - 9 = 9(9^n - 1)$ by the indutive hyp. So $9^{n+1} \equiv 9 \equiv 1 \mod 8$.
Feedback would be appreciated.
-
Or note that since $9\equiv 1\pmod 8$, we have $9^n \equiv 1^n \pmod 8$. – bonsoon Jan 30 '13 at 3:18
I'm assuming you mean what you say when you state your work as a proof "sketch".
The base case is clear. For the inductive step, it follows that $8 \mid 9^{n+1} - 9 = 9(9^n - 1)$ by the indutive hyp. So $9^{n+1} \equiv 9 \equiv 1 \mod 8$.
In your final write up, I'd suggest you "fill in" a bit of detail: e.g., to "walk through" the base case, at least stating that the congruence holds for $n=1$, or perhaps
"for $n = 1$, clearly, $9\equiv 1 \pmod 8$".
Then I suggest you make your inductive hypothesis explicit:
"Assume that it is true that $9^n \equiv 1 \pmod 8$,"
and then finish with, "for the inductive step....[what you wrote]"
If your task was to prove the congruence holds using proof by induction on $n$, then you've done a fine job of sketching such a proof.
If you can use other strategies, then bonsoon's suggestion is worth considering:
"Or note that since $9 \equiv 1 \pmod 8$, we have $9^n\equiv 1^n = 1 \pmod 8.$"
-
Yes, that is correct. Alternatively, prove by induction the $\,\rm n$-ary congruence product rule
$$\rm\qquad\ \ a_k\equiv b_k\ \Rightarrow\ a_1\cdots\, a_n \equiv b_1\cdots\, b_n$$
by iterating the binary product rule $\rm\ a_k\equiv b_k\ \Rightarrow\ a_1 a_2 \equiv b_1 b_2,\:$ then specialize $\rm\:a_i \equiv 9,\ b_i\equiv 1$
Remark $\$ Your proof can be viewed as special case of the obvious inductive proof that a sequence $\rm\:f_n\:$ is constant if successive values never change, i.e. if $\rm\:f_{n+1} \equiv f_n.$ Indeed, in your special case we have $\rm\:mod\ 8\!:\ f_{n+1} = 9^{n+1} = 9\, f_n \equiv f_n\:$ so the sequence is constant, hence $\rm\:f_n\equiv f_0\equiv 9^0\equiv 1.\:$
- | 2014-08-20T04:52:17 | {
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http://math.stackexchange.com/questions/133443/is-0-1-omega-homeomorphic-to-d-omega | # Is $[0,1]^\omega$ homeomorphic to $D^\omega$?
Let $n\in \mathbb N$ and let $D^n$ be the closed $1$-ball in $\left(\mathbb R^n, \|\,\cdot\,\|_1\right)$. It is not too hard to show that $[0,1]^n \cong D^n$ in this case.
This observation leads to the question whether we also have $[0,1]^\omega \cong D^\omega$, where
$$D^\omega = \left\{(x_n)_{n\in \mathbb N} \in \ell^1 \; \Bigg| \; \sum_{n=1}^\infty |x_n|\le 1\right\}$$
is the closed unit ball in $\ell^1$.
Now if we view $D^\omega$ as a subspace of $\ell^1$, then we can prove that it is not homeomorphic to $[0,1]^\omega$ by arguing that the latter is compact, while the former is not.
But what happens if we view $D^\omega$ as a subspace of $\mathbb R^\omega$?
Is $D^\omega$ homeomorphic to $[0,1]^\omega$, when both sets are endowed with the subspace topologies induced by $\mathbb R^\omega$ (in the product topology)?
In this case both spaces are compact ($D^\omega$ is the intersection over all $n\in \mathbb N$ of the compact sets $\{x\mid \sum_{i=1}^n |x_i| \le 1\} \cap [-1,1]^\omega$) and all topological properties I can think of are preserved when we go from $[-1,1]^\omega \cong [0,1]^\omega$ to the subspace $D^\omega$. I have also tried to explicitly construct a homeomorphism, but to no avail.
I hope some of the topologically savvy guys on this site could help me out here! Thanks =)
-
Sorry for going off-topic here: in the question on $A \subset f(B)$ you deleted your answer, but can't you massage it into a full solution using paracompactness and a partition of unity subordinate to a locally finite subcovering of $\{B \smallsetminus A\} \cup \{B_{r(a)}(a)\,:\,a \in A\}$ of $B$? – t.b. May 9 '12 at 13:44
The answer is yes, they are homeomorphic.
Edit: Here's a better way to put the answer below:
Theorem (Klee, 1955). Let $K$ be a compact, convex and metrizable set in a locally convex space $E$. If $K$ is not contained in a finite-dimensional subspace of $E$ then $K$ is homeomorphic to the Hilbert cube $[0,1]^\omega$.
This applies directly to the compact convex set $K = D^\omega$ in the metrizable space $E = \mathbb{R}^\omega$. This result is not explicitly stated this way in Klee's work [1], but it follows immediately from Theorem (1.2) of that paper. The proof idea is outlined a bit further down.
Klee proved in 1955, based on a theorem due to Keller, the following remarkable result (Theorem (1.2) of reference [1] below):
Let $X$ be a separable normed space and let $E$ be either $X$ with the norm topology, $X$ with the weak topology or $X^\ast$ with the weak$^\ast$ topology. If $K$ is an infinite-dimensional compact convex subset of $E$ then $K$ is homeomorphic to the Hilbert cube $[0,1]^\omega$.
The idea is that in each case one can find a countable family of continuous linear functionals $(f_n)$ separating the points of $K$ and, normalizing them appropriately, the map $x \mapsto (f_1(x),f_2(x),\ldots)$ gives a linear homeomorphism from $K$ onto an infinite-dimensional norm-compact convex subset of $\ell^2$. Keller had previously shown that all infinite-dimensional compact convex sets in $\ell^2$ are homeomorphic to $[0,1]^\omega$.
Since $D^\omega$ is a compact convex subset of $\ell^1 = (c_0)^\ast$, Klee's theorem applies. To finish up, note that the topology on $D^\omega$ viewed as a subset of $(c_0)^\ast$ and the topology induced from $\mathbb{R}^\omega$ coincide by the standard proof of Alaoglu's theorem.
Here are the relevant papers:
1. V. L. Klee, Some topological properties of convex sets, Trans. Amer. Math. Soc. 78 (1955), 30–45. MR0069388
2. Ott-Heinrich Keller, Die Homoiomorphie der kompakten konvexen Mengen im Hilbertschen Raum, Math. Ann. vol. 105 (1931) pp. 748–758. MR1512740
-
Of course, it would be desirable to have a more explicit argument! – t.b. Apr 18 '12 at 14:21
Wow! Thank you very much. I'm going to have a look at these papers, then. – Sam Apr 18 '12 at 14:32
@Sam: I slightly expanded my answer. – t.b. Apr 19 '12 at 19:59 | 2016-04-30T03:24:44 | {
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http://mathhelpforum.com/number-theory/6552-difference-squares-two-integers-print.html | # Difference of the squares of two integers
• Oct 18th 2006, 01:36 PM
ceasar_19134
Difference of the squares of two integers
Which one of the following numbers cannot be expressed as the difference of the squares of two integers?
A:314159265
B:314159266
C:314159267
D:314159268
E:314159269
For this one, I'm just completley clueless.
• Oct 18th 2006, 01:57 PM
ThePerfectHacker
Quote:
Originally Posted by ceasar_19134
Which one of the following numbers cannot be expressed as the difference of the squares of two integers?
A:314159265
B:314159266
C:314159267
D:314159268
E:314159269
For this one, I'm just completley clueless.
Numbers of the form 4k,4k+1,4k+3 can always be expressed as a difference of squares.
Remove the odd numbers for they can always be expressed as difference of squares are you have,
314159266
314159268
Divide them by 4 and find the one which leaves a remainder of 2.
Since both are even they leave either no remiander or 2.
Thus, find the number not divisible by 4.
The trick is to look at the last two digits,
66----> Not Divisible
68----> Yes Divisible
Thus, the answer is,
314159266
• Oct 18th 2006, 02:02 PM
CaptainBlack
Quote:
Originally Posted by ceasar_19134
Which one of the following numbers cannot be expressed as the difference of the squares of two integers?
A:314159265
B:314159266
C:314159267
D:314159268
E:314159269
For this one, I'm just completley clueless.
The prime factorisation of 314159266 is:
314159266=2x157079633
Now if N=314159266 is the difference of two square integers there exists
integers m > n such that:
N = m^2-n^2 = (m+n)(m-n).
Now if (m-n) is even then both n and m are even or both are odd. But as
we can see from the factorisation (m-n)=2, and (m+n)= 157079633,
that is one of (m+n) and (m-n) is odd and the other even which is not
possible so N is not expressible as the difference of two squares.
RonL
• Oct 27th 2006, 10:21 AM
ceasar_19134
Quote:
Originally Posted by ThePerfectHacker
Numbers of the form 4k,4k+1,4k+3 can always be expressed as a difference of squares.
Remove the odd numbers for they can always be expressed as difference of squares are you have,
314159266
314159268
Divide them by 4 and find the one which leaves a remainder of 2.
Since both are even they leave either no remiander or 2.
Thus, find the number not divisible by 4.
The trick is to look at the last two digits,
66----> Not Divisible
68----> Yes Divisible
Thus, the answer is,
314159266
If I may ask, what does k stand for?
Edit: I didnt know where the quote button would put it, sorry.
• Oct 27th 2006, 10:46 AM
CaptainBlack
Quote:
Originally Posted by ceasar_19134
If I may ask, what does k stand for?
Edit: I didnt know where the quote button would put it, sorry.
any integer, as k varies you get all numbers which leave remainders
of 0, 1 or 3 from 4k,4k+1,4k+3.
RonL
• Oct 27th 2006, 10:57 AM
ceasar_19134
So we subsitute the numbers in as k and then what?
And why does the remainder have any significance?
My teacher does not bother teaching us how to solve any of these problems.
• Oct 27th 2006, 11:21 AM
Soroban
Hello, ceasar_19134!
Quote:
Which one of the following numbers cannot be expressed
as the difference of the squares of two integers?
A: 314159265
B: 314159266
C: 314159267
D: 314159268
E: 314159269
It can be shown that any odd number can be expressed as the difference of two squares. **
The only suspects are B and D.
To have an even difference of squares,
. . either (1) both squares are even or (2) both squares are odd.
(1) If both $a^2$ and $b^2$ are even, then both $a$ and $b$ are even.
. . Then $a = 2m$ and $b = 2n$ for some integers $m$ and $n.$
Hence: . $a^2-b^2\:=\:(2m)^2 - (2n)^2 \:=\:4m^2 - 4n^2\:=\:4(m^2-n^2)$
. . Therefore: $a^2-b^2$ is a multiple of 4.
(2) If both $a^2$ and $b^2$ are odd, then both $a$ and $b$ are odd.
. . Then $a = 2m+1$ and $b = 2n+1$ for some integers $m$ and $n.$
Hence: . $a^2-b^2\:=\:(2m+1)^2-(2n+1)^2\:=\:4m^2 + 4m + 1 - 4n^2 - 4n - 1 \:=\:4(m^2+m-n^2-n)$
. . Therefore: $a^2-b^2$ is a multiple of 4.
We have just established a theorem.
. . The difference of two squares is either odd or a multiple of 4.
We can now "eyeball" the list and determine the answer.
The even choice not divisible by 4 is: $B:\;314159266$
. . [Its last two digits is "66", which is not divisible by 4.]
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
**
Two consecutive squares always differ by an odd number
. . because: . $(n+1)^2 - n^2\:=\:2n+1$ ... an odd number
Given an odd difference, we can find the two squares.
Example: $d = 37$
Subtract 1 and divide by 2: . $\frac{37-1}{2} = 18$
The two squares are: $18^2$ and $19^2$
Given answer-choice $A:\;314,159,265$
. . we can confidently state that: . $314,159,265\;=\;157,079,633^2 - 157,079,632^2$ | 2016-12-09T16:46:22 | {
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http://math.stackexchange.com/questions/145803/cantor-set-lebesgue-measure-and-uncountability | # Cantor set: Lebesgue measure and uncountability
I have to prove two things.
First is that the Cantor set has a lebesgue measure of 0. If we regard the supersets $C_n$, where $C_0 = [0,1]$, $C_1 = [0,\frac{1}{3}] \cup [\frac{2}{3},1]$ and so on. Each containig interals of length $3^{-n}$ and by construction there are $2^n$ such intervals. The lebesgue measure of each such interval is $\lambda ( [x, x + 3^{-n}]) = 3^{-n}$ therefore the measure of $C_n$ is $\frac{2^n}{3^n} = e^{(\ln(2)-\ln(3)) n }$ which goes to zero with $n \rightarrow \infty$. But does this prove it?
The other thing I have to prove is that the Cantor set is uncountable. I found that I should contruct a surjectiv function to $[0,1]$. But im totaly puzzeld how to do this.
Thanks for help
-
Hint: - There is a theorem that if $C_n$ is a descending sequence of measurable sets, $C = \bigcap C_n$ then $\lim_{n \rightarrow \infty}m(C_n)= m(C)$. Here, you know $C_n$ is a descending sequence, $m(C_n)$= (2/3)^n, and you want to know m(C)
• to prove $C$ is uncountable:
Express a number $x$ between $[0,1]$ in base $3$:
$x =0.x_1x_2x_3...$
In the 1st step, we remove the middle third from $[0,1]$
We express 0=.0, $\frac{1}{3} = .1, \frac{2}{3} = .2$, 1= .222222...
We have 3 intervals: [.0 , .1] , (.1, .2), [.2 , .22222...] and we remove the middle interval. The removed interval (.1 , .2) consists of all numbers with $x_1$ = 1, except the endpoint $.1$ of [.0 , .1 ], however, we can express .1 as .02222.....
So we can use the rule: whenever we have a number of form 0.x1(x is a sequence consisting of 0,1,2) as the end point of an interval, we express as 0.x0222....So in this step, we remove all numbers with $x_1$ = 1. The remaining intervals are [.0, .0222...] and [.2, .222...]
Similarly, we can prove that in the n-th step, we keep only those numbers with $x_n$ = 0 or 2:
So the Cantor set contains of all numbers of the form $.x_1x_2..$ with $x_i =0$ or 2.
There exists a bijection between $E$ and $[0,1]$: If you consider the new set E, with each member of E is a member of the Cantor set with every digit is divided by 2. E consists of all sequence with $x_i$ is 0 or 1. $|E| = |C|$
There exists an injective map from $[0,1]$ to this new set E. So you can prove it's uncountable.
-
Thanks this helps. Do you know a name of the theorem you mentioned at the beginning? Don't think we proved it our lectures. – Haatschii May 16 '12 at 20:07
It's theorem 1.19e in Rudin's Real and Complex Analysis. It's an important theorem, I saw it used in couple of places. It's not hard to prove. the way to prove it is: let $B_1 = C_1, B_2 = C_2\setminus C_1, B_3 = C_3\setminus (C_1 \cup C_2 )$,...hence $B_1,B_2,B_3,...$ are pairwise disjoint, and you can use the property of measure of union of disjoint sets. – Long May 16 '12 at 20:34
You may notice that the Cantor set is homeomorphic (which is even more than required) to $\{0,1\}^{\mathbb{N}}$, which is uncountable. A homeomorphism is established e.g. with $f: \{0,1\}^{\mathbb{N}}\to C$ so that \begin{align*} f((x_{i})_{i=1}^{\infty})=\sum_{i=1}^{\infty}\frac{2x_{i}}{3^{i}}. \end{align*} Since it has a converging geometric serie as a majorant; it definitely converges and even uniformly. Since the components are continuous then $f$ is continuous. You may conclude injectivity by noticing that $|f(x)-f(y)|>0$ if $x\neq y$. Surjectivity is also quite straightforward from the definition since you may produce any element of the cantor set with such a serie by choosing $x_{i}$'s wisely (if $y\in C$, choose $x_{i}=0$ if at $i$:th step $y$ is located in the left part of the $i-1$:th step interval being split to thirds, and $x_{i}=1$ if on the right). Since every continuous bijection from a compact set to a Hausdorff space is homeomorphism, we conclude that $f$ is a homeomorphism.
Since every $C_{n}$ has Lebesgue measure of $(\frac{2}{3})^{n}$ (at each step the measure of the previous step is multiplied with $\frac{2}{3}$) and $\frac{2}{3}<1$, it follows by convergence of measure that \begin{align*} m_{1}(C)=m_{1}(\bigcap_{n=1}^{\infty}C_{n})=\lim_{n\to\infty}m_{1}(C_{n})=\lim_{n\to\infty}(\frac{2}{3})^{n}=0. \end{align*}
- | 2015-11-25T14:51:15 | {
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https://la.mathworks.com/help/matlab/ref/sqrtm.html | # sqrtm
Matrix square root
## Description
example
X = sqrtm(A) returns the principal square root of the matrix A, that is, X*X = A.
X is the unique square root for which every eigenvalue has nonnegative real part. If A has any eigenvalues with negative real parts, then a complex result is produced. If A is singular, then A might not have a square root. If exact singularity is detected, a warning is printed.
[X,residual] = sqrtm(A) also returns the residual, residual = norm(A-X^2,1)/norm(A,1). This syntax does not print warnings if exact singularity is detected.
[X,alpha,condx] = sqrtm(A) returns stability factor alpha and an estimate of the matrix square root condition number of X in 1-norm, condx. The residual norm(A-X^2,1)/norm(A,1) is bounded approximately by n*alpha*eps and the 1-norm relative error in X is bounded approximately by n*alpha*condx*eps, where n = max(size(A)).
## Examples
collapse all
Create a matrix representation of the fourth difference operator, A. This matrix is symmetric and positive definite.
A = [5 -4 1 0 0; -4 6 -4 1 0; 1 -4 6 -4 1; 0 1 -4 6 -4; 0 0 1 -4 6]
A = 5×5
5 -4 1 0 0
-4 6 -4 1 0
1 -4 6 -4 1
0 1 -4 6 -4
0 0 1 -4 6
Calculate the unique positive definite square root of A using sqrtm. X is the matrix representation of the second difference operator.
X = round(sqrtm(A))
X = 5×5
2 -1 0 0 0
-1 2 -1 0 0
0 -1 2 -1 0
0 0 -1 2 -1
0 0 0 -1 2
Consider a matrix that has four squareroots, A.
$A=\left[\begin{array}{cc}7& 10\\ 15& 22\end{array}\right]$
Two of the squareroots of A are given by Y1 and Y2:
${Y}_{1}=\left[\begin{array}{cc}1.5667& 1.7408\\ 2.6112& 4.1779\end{array}\right]$
${Y}_{2}=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$
Confirm that Y1 and Y2 are squareroots of matrix A.
A = [7 10; 15 22];
Y1 = [1.5667 1.7408; 2.6112 4.1779];
A - Y1*Y1
ans = 2×2
10-3 ×
-0.1258 -0.1997
-0.2995 -0.4254
Y2 = [1 2; 3 4];
A - Y2*Y2
ans = 2×2
0 0
0 0
The other two squareroots of A are -Y1 and -Y2. All four of these roots can be obtained from the eigenvalues and eigenvectors of A. If [V,D] = eig(A), then the squareroots have the general form Y = V*S/V, where D = S*S and S has four choices of sign to produce four different values of Y:
$S=\left[\begin{array}{cc}±0.3723& 0\\ 0& ±5.3723\end{array}\right]$
Calculate the squareroot of A with sqrtm. The sqrtm function chooses the positive square roots and produces Y1, even though Y2 seems to be a more natural result.
Y = sqrtm(A)
Y = 2×2
1.5667 1.7408
2.6112 4.1779
## Input Arguments
collapse all
Input matrix, specified as a square matrix.
Data Types: single | double
Complex Number Support: Yes
## Tips
• Some matrices, like A = [0 1; 0 0], do not have any square roots, real or complex, and sqrtm cannot be expected to produce one.
## Algorithms
The algorithm sqrtm uses is described in [3].
## References
[1] N.J. Higham, “Computing real square roots of a real matrix,” Linear Algebra and Appl., 88/89, pp. 405–430, 1987
[2] Bjorck, A. and S. Hammerling, “A Schur method for the square root of a matrix,” Linear Algebra and Appl., 52/53, pp. 127–140, 1983
[3] Deadman, E., Higham, N. J. and R. Ralha, “Blocked Schur algorithms for computing the matrix square root,” Lecture Notes in Comput. Sci., 7782, Springer-Verlag, pp. 171–182, 2013
## Version History
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http://mathhelpforum.com/advanced-algebra/107300-minimal-characterisitic-polynomial.html | # Math Help - Minimal and Characterisitic Polynomial
1. ## Minimal and Characterisitic Polynomial
T be a linear operator on a finite dimensional vector space.
p(x) - Minimal Polynomial of T
q(x) - Characteristic Polynomial of T
I want to prove the p(x)|q(x)
I know that every root of q(x) is a root of p(x) and vice a versa. I guess we can use this fact to prove the above.
But I do not want to use this - because we might not be able to factorize the polynomials. Is there a better way please? Thanks
2. Originally Posted by aman_cc
T be a linear operator on a finite dimensional vector space.
p(x) - Minimal Polynomial of T
q(x) - Characteristic Polynomial of T
I want to prove the p(x)|q(x)
I know that every root of q(x) is a root of p(x) and vice a versa. I guess we can use this fact to prove the above.
But I do not want to use this - because we might not be able to factorize the polynomials. Is there a better way please? Thanks
by Cayley-Hamilton we have $q(T)=0.$ we also have $q(x)=s(x)p(x) + r(x),$ for some polynomials $s(x),r(x)$ with $r(x)=0$ or $0 < \deg r(x) < \deg p(x).$
but $r(T)=0$ because $q(T)=p(T)=0.$ thus either $r(x)=0$ or $\deg r(x) \geq \deg p(x),$ by minimality of $p(x).$ hence $r(x)=0.$
3. Originally Posted by NonCommAlg
by Cayley-Hamilton we have $q(T)=0.$ we also have $q(x)=s(x)p(x) + r(x),$ for some polynomials $s(x),r(x)$ with $r(x)=0$ or $0 < \deg r(x) < \deg p(x).$
but $r(T)=0$ because $q(T)=p(T)=0.$ thus either $r(x)=0$ or $\deg r(x) \geq \deg p(x),$ by minimality of $p(x).$ hence $r(x)=0.$
Thanks - I was trying to prove Cayley-Hamilton with this result Stupid of me
If I told you that T is diagonizable - Then I guess I can use the root logic (every root of q(x) is a root of p(x) and vice a versa) result to prove it and not rely on Cayley-Hamilton Theorem. Correct?
4. What definition of "miminal polynomial" are you using? The one I would use is that the minimal polynomial of linear operator (matrix) A is the monic polynomial, P, of lowest degree satisfying P(A)= 0. Yes, the minimal polynomial and characteristic polynomial have the same roots which means they have almost the same factors. What is the difference between them? Why are they not exactly the same polynomial?
5. Originally Posted by HallsofIvy
What definition of "miminal polynomial" are you using? The one I would use is that the minimal polynomial of linear operator (matrix) A is the monic polynomial, P, of lowest degree satisfying P(A)= 0. Yes, the minimal polynomial and characteristic polynomial have the same roots which means they have almost the same factors. What is the difference between them? Why are they not exactly the same polynomial?
I do mean the same thing with I say minimal polynomial.
I guess (in the case when A is diagonalizable) chr polynomial will be of degree 'n' (equal to the dimension of the vector space) but have lesser num of distinct roots (as order of roots might be >1)
Thus chr poly might be different from min polynomial. Am I making sense?
6. Originally Posted by aman_cc
I do mean the same thing with I say minimal polynomial.
I guess (in the case when A is diagonalizable) chr polynomial will be of degree 'n' (equal to the dimension of the vector space) but have lesser num of distinct roots (as order of roots might be >1)
Thus chr poly might be different from min polynomial. Am I making sense?
if $T$ is diagonalizable and $\lambda_1, \cdots, \lambda_k$ are the distinct eigenvalues of $T,$ then the minimal polynomial of $T$ would be $\prod_{j=1}^k(x-\lambda_j).$ the characteristic polynomial of $T$ in this case would be in the
form $\prod_{j=1}^k (x-\lambda_j)^{n_j},$ where $n_j$ is the number of eigenvalues of $T$ which are equal to $\lambda_j.$
you can find standard facts like this (and much more) about minimal and characteristic polynomial of a linear transformation in any decent linear algebra textbook.
7. I wouldn't even worry about T being "diagonalizable". The characteristic polynomial of T is $(x- \lambda_1)^m(x- \lambda_2)^n\cdot\cdot\cdot(x- \lambda_n)^k$. We can theoretically factor the polynomial into linear factors (possibly with complex eigenvalues) even if we can't actually find the factors. The minimal polynomial will have exactly the same factor but possibly to lower power. We can "factor" those lower powers out so it follows that the minimal polynomial divides the characteristic polynomial.
8. Originally Posted by NonCommAlg
if $T$ is diagonalizable and $\lambda_1, \cdots, \lambda_k$ are the distinct eigenvalues of $T,$ then the minimal polynomial of $T$ would be $\prod_{j=1}^k(x-\lambda_j).$ the characteristic polynomial of $T$ in this case would be in the
form $\prod_{j=1}^k (x-\lambda_j)^{n_j},$ where $n_j$ is the number of eigenvalues of $T$ which are equal to $\lambda_j.$
you can find standard facts like this (and much more) about minimal and characteristic polynomial of a linear transformation in any decent linear algebra textbook.
Thanks very much NonCommAlg
9. Originally Posted by HallsofIvy
I wouldn't even worry about T being "diagonalizable". The characteristic polynomial of T is $(x- \lambda_1)^m(x- \lambda_2)^n\cdot\cdot\cdot(x- \lambda_n)^k$. We can theoretically factor the polynomial into linear factors (possibly with complex eigenvalues) even if we can't actually find the factors. The minimal polynomial will have exactly the same factor but possibly to lower power. We can "factor" those lower powers out so it follows that the minimal polynomial divides the characteristic polynomial.
Yes I get this. Thanks very much! | 2014-12-20T09:01:26 | {
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https://math.stackexchange.com/questions/3121605/when-does-the-game-end/3125532#3125532 | # When does the game end?
A recent question has inspired the following one.
$$M$$ people ($$M\ge2$$) play the following game. Each round every person admitted to the round rolls a $$K$$-sided dice ($$K\ge2$$), the sides being marked with numbers from $$1$$ to $$K$$. If only a single person has the highest score, the game ends. Otherwise the people with the highest score are admitted to the next round and the game continues. What is the probability that the game ends exactly in $$n$$ rounds?
Neat problem! Thanks for posting.
The first thing to notice is that at any state of the game, if you have $$a$$ players remaining, then the probability of yielding $$b$$ players in the next round is independent of which round you're at. Hence we can talk about $$P(a,b)$$, the probability of transitioning from $$a$$ players to $$b$$ players in one round.
Playing a round with $$a > 1$$ players, there $$K^a$$ possible rolls. To find the number of ways we end up with $$b \geq 1$$ players, we first note that there are $$a \choose b$$ ways to pick which $$b$$ players continue. Then, letting $$k$$ denote the value that these $$b$$ players roll, we find that there are $$(k-1)^{a-b}$$ ways for the players to roll this result: the $$b$$ continuing players must all roll $$k$$, and the remaining players must all roll between $$1$$ and $$k-1$$. Hence we can write (with the convention $$0^0 = 1$$) $$P(a,b) = \frac{{a\choose b}\displaystyle\sum_{k=1}^K (k-1)^{a-b}}{K^a}$$
Once you have these probabilities, the game itself is a Markov chain. Your states are the integers from $$1$$ to $$M$$ denoting the number of players remaining. In any step, the probability of transitioning from state $$a$$ to state $$b$$ with $$a \ge b \ge 1$$ and $$a \ne 1$$ is exactly $$P(a,b)$$. State 1 is an absorbing state, and transitions to state 1 with probability 1. All other transition probabilities are 0. So your transition matrix looks like $$T=\left(\begin{matrix}P(M,M) & P(M,M-1) & P(M,M-2) & \cdots & P(M,1)\\ 0 & P(M-1,M-1) & P(M-1,M-2) & \cdots & P(M-1,1)\\ &&\vdots\\0 & 0 & \cdots & P(2,2) & P(2,1)\\ 0 & 0 & \cdots & 0 & 1\end{matrix}\right)$$
The entry in the upper right corner of $$T^n$$ is the probability of being in state 1 (i.e. the game has ended) after $$n$$ rounds. But, since this is an absorbing state, this is the probability that the game ends in round $$n$$ or before. To get the probability of ending after exactly $$n$$ rounds, you need to subtract the entry in the upper right corner of $$T^{n-1}$$ from the upper right entry of $$T^n$$.
• Thank you for this nice and very profound answer. I came almost to the same point and found closed-form solution for $M=3$. I wonder if it is managable to find such closed-form expression for $\left (T^n\right)_{M,1}$ in general case.
– user
Feb 21, 2019 at 22:21
• One suggesion. The expressions will look nicer if you redefine $k\mapsto k+1$. Besides it is necessary in the expression for $P (a,b)$ to start the summation with $k=0$ (in your version with $k=1$) together with the convention $0^0=1$. Otherwise the expression gives wrong value for $a=b$.
– user
Feb 21, 2019 at 22:31
• I doubt a closed form is possible for general $M$. There is an alternative with a rational multivariate generating function, but it essentially just reproduces the matrix multiplication in polynomial form. The problem is that the number of "paths" from $M$ to $1$ becomes uncontrollable as $n$ gets larger. Feb 21, 2019 at 22:34
• Yes, you are right. I will edit the solution. But I believe leaving the index of summation as $k$ is important for the clarity of the argument, even though it leads to a slightly less elegant expression for $P(a,b)$. Feb 21, 2019 at 22:34
• It has appeared that the problem does admit a closed-form solution (see my answer).
– user
Feb 25, 2019 at 2:31
The aim of this answer is to present a closed-form solution for the problem. It represents further development of the idea described in the answer of Jeremy Dover. The hard and probably boring part of the proof will be given in the appendix.
For convinience we rewrite the $$M\times M$$ dimensional matrix $$T(M,K)$$ introduced by Jeremy Dover as: $$T_{ij}=\frac 1{K^i}\binom ij\sum_{k=0}^{K-1}k^{i-j}.$$ In the sense of transition states the process starts at $$T_{MM}$$ and ends at $$T_{M1}$$.
As the matrix $$T$$ is triangular its eigenvalues are trivially its diagonal elements $$\lambda_m=T_{mm}=\frac{1}{K^{m-1}},\; m=1\dots M.$$ As all eigenvalues are distinct the matrix is apparantly diagonalizable by a similarity transformation: $$\text{diag}(\lambda_1,\lambda_2,\dots,\lambda_M)=X^{-1}TX.$$ Knowing the transformation matrix $$X$$ the powers of the matrix $$T$$ can be computed as: $$T^n=X\text{diag}(\lambda_1^n,\lambda_2^n,\dots,\lambda_M^n)X^{-1}.\tag1$$
Fortunately, the structures of matrices $$X$$ and $$X^{-1}$$ are rather simple (see Appendix): $$X_{ij}=(1-\delta_{i,j-1})\binom i{j-1};\quad X^{-1}_{ij}=\frac 1i\binom ij b_{i-j},$$ where $$b_k$$ are the Bernoulli numbers ($$b_1=-\frac12$$).
Plugging the values into (1) one obtains: $$T^n_{M1}(M,K)=\sum_{m=1}^M X_{Mm}\lambda_m^n X^{-1}_{m1} =\sum_{m=1}^M\binom M{m-1}\frac{b_{m-1}}{K^{(m-1)n}} =\sum_{m=0}^{M-1}\binom M{m}\frac{b_{m}}{K^{nm}}.\tag2$$
Finally, the probability for the game to end in the $$n$$-th round reads: $$\boxed{P_n(M,K)=T^n_{M1}-T^{n-1}_{M1}=\sum_{m=0}^{M-1}\binom M{m}\frac{1-K^m}{K^{nm}}b_{m}.}$$
It may be convenient to rewrite the expression (2) as: $$T^n_{M1}(M,K)={\cal B}_M\left(\frac1{K^n}\right),$$ where the function $${\cal B}_M(x)$$ is defined as: $${\cal B}_M(x)=\sum_{m=0}^{M-1}\binom M{m}b_{m}x^m\equiv x^M\left[B_M\left(\frac1x\right)-b_M\right],$$ with $$B_M(x)$$ being the Bernoulli polynomial.
Appendix
Proposition 1. The following identity holds for any $$p,n\in\mathbb Z_+$$: $$\sum\limits_{k=0}^{p-1} {\binom {p} k} \sum\limits_{j=0}^{n-1} j^k=n^p.\tag{*}$$
Proof: \begin {align} &(j+1)^{p}=\sum\limits_{k=0}^{p} {\binom {p} k} j^k\\ \implies& (j+1)^{p} - j^{p} =\sum\limits_{k=0}^{p-1} {\binom {p} k} j^k\\ \implies&\sum\limits_{j=0}^{n-1} \left[(j+1)^{p} - j^{p}\right] = \sum\limits_{j=0}^{n-1} \sum\limits_{k=0}^{p-1} {\binom {p} k} j^k\\ \implies& n^{p}= \sum\limits_{k=0}^{p-1} {\binom {p} k} \sum\limits_{j=0}^{n-1} j^k. \end {align} Introducing the notation $$S^{(n)}_k= \sum\limits_{j=0}^{n-1} j^k$$ the set of the equations $$(*)$$ with the exponent of $$n$$ varying from $$1$$ to $$p$$ is equivalent to the matrix equation: $$\left(\begin{array}l n\vphantom{\binom00}\\ n^2\vphantom{\binom00}\\ n^3\vphantom{\binom00}\\ \vdots\\ n^p\vphantom{\binom00} \end{array}\right)= \begin{pmatrix} \binom10&0&0&\cdots& 0\\ \binom20&\binom21&0&\cdots& 0\\ \binom30&\binom31&\binom32&\cdots& 0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ \binom p0&\binom p1&\binom p2&\cdots& \binom p{p-1}\\ \end{pmatrix} \left(\begin{array}l S^{(n)}_0\\ S^{(n)}_1\\ S^{(n)}_2\\ \vdots\\ S^{(n)}_{p-1} \end{array}\right).$$
Denoting the matrix of binomial coefficients as $$X$$, one observes that its determinant is non-zero and the equation can be inverted obtaining: $$\left(\begin{array}l S^{(n)}_0\\ S^{(n)}_1\\ S^{(n)}_2\\ \vdots\\ S^{(n)}_{p-1} \end{array}\right)= \begin{pmatrix} X^{-1}_{11}&0&0&\cdots& 0\\ X^{-1}_{21}&X^{-1}_{22}&0&\cdots& 0\\ X^{-1}_{31}&X^{-1}_{32}&X^{-1}_{33}&\cdots& 0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ X^{-1}_{p1}&X^{-1}_{p2}&X^{-1}_{p3}&\cdots& X^{-1}_{pp}\\ \end{pmatrix} \left(\begin{array}l n\vphantom{\binom00}\\ n^2\vphantom{\binom00}\\ n^3\vphantom{\binom00}\\ \vdots\\ n^p\vphantom{\binom00} \end{array}\right).$$
Comparing the expression with the well-known one: $$S^{(n)}_{p-1}=\frac1p\sum_{j=0}^{p-1}\binom pj b_j n^{p-j}= \frac1p\sum_{j'=1}^{p}\binom p{j'} b_{p-j'} n^{j'},\tag{**}$$ one obtains:
Proposition 2. The matrix elements of $$X^{-1}$$ are: $$X^{-1}_{ij}=\frac 1i\binom ij b_{i-j},$$ where $$b_k\equiv b^-_k$$ are the Bernoulli numbers taken with the convention $$b_1=-\frac12$$. Observe that on the cited above Wikipedia page the formula similar to (**) is given for $$S^{(n+1)}_p$$ in terms of $$b^+$$ without a direct mention of this.
Proposition 2 can be proved also directly.
Proposition 3. The columns of the matrix $$X$$ are the eigenvectors of the matrix $$T$$.
Proof. Acting by matrix $$T$$ on the $$m$$-th column of the matrix $$X$$: $$v^{(m)}_j=\begin{cases} 0,& 1\le j one obtains: \begin{align} \sum_{j=1}^M T_{ij} v^{(m)}_j &=\sum_{j=m}^M\frac 1{K^i}\binom ij\binom{j}{m-1}\sum_{k=0}^{K-1}k^{i-j}\\ &=\frac 1{K^i}\binom i{m-1}\sum_{j=m}^i\binom{i+1-m}{i-j}\sum_{k=0}^{K-1}k^{i-j}\\ &=\frac 1{K^i}\binom i{m-1}\sum_{j'=0}^{i-m}\binom{i+1-m}{j'}\sum_{k=0}^{K-1}k^{j'}\\ &\stackrel{(*)}=\frac 1{K^i}\binom i{m-1}K^{i+1-m}\\ &=\frac 1{K^{m-1}}\binom i{m-1}=\lambda_m v^{(m)}_i. \end{align}
• Very, VERY nice! Feb 25, 2019 at 2:38 | 2022-09-24T19:34:41 | {
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https://math.stackexchange.com/questions/3138351/proving-an-inequality-regarding-convex-functions | # Proving an inequality regarding convex functions
Let $$f$$ be a real-valued convex function, $$\lambda_1>0$$ and $$\lambda_2\leq0$$ such that $$\lambda_1+\lambda_2=1$$.
I want to prove that for any $$x_1$$,$$x_2\in{\rm Dom}(f)$$, $$f(\lambda_1x_1+\lambda_2x_2)\geq \lambda_1f(x_1)+\lambda_2f(x_2).$$
Now, since $$f$$ is convex, for any $$t\in[0,1]$$, $$f(tx_1+(1-t)x_2)\leq tf(x_1)+(1-t)f(x_2).$$ I noticed that the inequality that I want to prove is very similar to the definition of convexity, but taking values outside the interval $$[0,1]$$ that sum to $$1$$. I know that convexity means that the graph of f between $$x_1$$ and $$x_2$$ is below the segment which joins the points $$(x_1,f(x_1))$$ and $$(x_2,f(x_2))$$, so maybe there is some geometric interpretation for $$\lambda_1x_1+\lambda_2x_2$$.
It feels that with some inequalities I should be solve this problem, but as of now, I cannot see them. Any help is appreciated.
Note that $$\dfrac{1}{λ_1} + \dfrac{-λ_2}{λ_1} = 1$$ and $$λ_1 > 0 \geqslant λ_2$$, thus$$\frac{1}{λ_1} f(λ_1 x_1 + λ_2 x_2) + \dfrac{-λ_2}{λ_1} f(x_2) \geqslant f\left( \frac{1}{λ_1} · (λ_1 x_1 + λ_2 x_2) + \left( \frac{-λ_2}{λ_1} \right) · x_2 \right) = f(x_1),$$ i.e.$$f(λ_1 x_1 + λ_2 x_2) \geqslant λ_1 f(x_1) + λ_2 f(x_2).$$
• Yes, that is the way to write $\lambda_1$ and $\lambda_2$ that I was looking for. Thank you. – MarianaMG2205 Mar 7 '19 at 2:24 | 2021-07-30T08:33:45 | {
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https://math.stackexchange.com/questions/2758338/finding-the-laurent-series-of-the-following-function | # Finding the Laurent series of the following function
I need to find the laurent series and the residue of the following complex function $$f(z)=(z+1)^2e^{3/z^2}$$ at $z=0$.
Since $e^z=\sum z^n/n!$, then $$e^{3/z^2}=\sum_{n=0}^\infty \frac{3^n/n!}{z^{2n}}$$ thus $$f(z)=(z^2+2z+1)\sum_{n=0}^\infty \frac{3^n/n!}{z^{2n}}=\sum_{n=0}^\infty \frac{3^n/n!}{z^{2(n-1)}}+\sum_{n=0}^\infty \frac{2\cdot3^n/n!}{z^{2n-1}}+\sum_{n=0}^\infty \frac{3^n/n!}{z^{2n}}$$ which, with a shift of index and expansion of positive powers, can be expressed as $$f(z)=z^2+3+\sum_{n=1}^\infty\left(\frac{3}{n+1}+1\right)\frac{3^n/n!}{z^{2n}}+\sum_{n=1}^\infty\frac{2\cdot3^n/n!}{z^{2n-1}}$$ so the residue is given by evaluating the numerator of the second series at $n=1$, so its value is $6$. I tried using WolframAlpha and Mathematica to check my answer, but both would not return a value. Would this be correct? Also, is there a way to put the two sums together (one gives the coefficients of even powers, while the other of the odd) so I can have the principal part of the laurent series expressed only with one sum?
Using the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series, we obtain \begin{align*} \color{blue}{\mathrm{res}_{z=0}f(z)}&=[z^{-1}](z+1)^2e^{3/z^2}\\ &=\left([z^{-3}]+2[z^{-2}]+[z^{-1}]\right)\sum_{j=0}^\infty 3^jz^{-2j}\\ &=0+2\cdot 3^1+0\\ &\,\,\color{blue}{=6} \end{align*} and see the residue of $f$ at $z=0$ is $6$.
Expanding $f$ into a Laurent series at $z=0$ we obtain \begin{align*} \color{blue}{f(z)}&\color{blue}{=(z+1)^2e^{3/z^2}}\\ &=(z^2+2z+1)\sum_{n=0}^\infty \frac{3^n}{n!}\frac{1}{z^{2n}}\\ &=\sum_{n=0}^\infty \frac{3^{n}}{n!}\frac{1}{z^{2n-2}} +2\sum_{n=0}^\infty \frac{3^n}{n!}\frac{1}{z^{2n-1}} +\sum_{n=0}^\infty \frac{3^n}{n!}\frac{1}{z^{2n}}\\ &=\sum_{n=-1}^\infty \frac{3^{n+1}}{(n+1)!}\frac{1}{z^{2n}} +2\sum_{n=-1}^\infty \frac{3^{(n+1)}}{(n+1)!}\frac{1}{z^{2n+1}} +\sum_{n=0}^\infty \frac{3^n}{n!}\frac{1}{z^{2n}}\\ &\color{blue}{=z^2+2z+\sum_{n=0}^\infty\frac{3^n}{n!}\left(\frac{3}{n+1}+1\right)\frac{1}{z^{2n}} +2\sum_{n=0}^\infty \frac{3^{(n+1)}}{(n+1)!}\frac{1}{z^{2n+1}}}\tag{1}\\ &=z^2+2z+\sum_{m=0}^\infty\frac{3^{\frac{m}{2}}}{(\frac{m}{2})!}\left(\frac{3}{\frac{m}{2}+1}+1\right)\frac{1+(-1)^m}{2}\frac{1}{z^{m}}\\ &\qquad+2\sum_{m=0}^\infty \frac{3^{(\frac{m-1}{2}+1)}}{(\frac{m-1}{2}+1)!}\cdot\frac{1-(-1)^m}{2}\frac{1}{z^{m}}\\ &=z^2+2z+\sum_{m=0}^\infty\left(\frac{3^{\frac{m}{2}}}{2(\frac{m}{2})!}\left(\frac{m+8}{m+2}\right)\left(1+(-1)^m\right)\right.\\ &\qquad\qquad\qquad\qquad\quad\left.+ \frac{3^{\frac{m+1}{2}}}{\left(\frac{m+1}{2}\right)!}\left(1-(-1)^m\right)\right)\frac{1}{z^{m}}\tag{2} \end{align*}
We observe that putting all terms together as in (2) is rather cumbersome and the representation in (1) seems to be more convenient. It clearly shows the terms for odd and even powers and we can also easily derive from (1) the residue by inspecting the right-most sum in (1) with $n=0$.
Since the $Res_\infty+Res_0=0$, you may want to calculate the residue at infinity, but I don’t think it is easier.
$$Res_\infty=-Res_0\frac1{z^2}(\frac1{z^2}+\frac2z+1)e^{3z^2}$$
Collecting the $z^{-1}$ terms, $$=-Res_0\frac{1}{z^2}\cdot\frac2z\cdot\frac{3^1}{1!}z^2=-Res_06z^{-1}=-6$$
Therefore, $Res_0=6$.
For the series, if the function $a(2n)$ generates coefficients for $x^{2n}$ and $b(2n+1)$ for $x^{2n+1}$, then in general the ‘coefficient function’ $c(n)$ for $x^n$ is $$c(n)=\frac{1+(-1)^{n}}2\cdot(a(n)-b(n))+b(n)$$ | 2020-08-14T17:57:18 | {
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https://math.stackexchange.com/questions/1367059/volume-inside-loop-using-greens-theorem | # Volume inside loop using Green's theorem.
Let $\mathcal{C}$ be the curve defined by the vector function $\vec r(t)=(1-t^2)\vec i+(t-t^3)\vec j$ with $t\in \Bbb R$. I need to find the area confined in the closed loop $\gamma$ formed by $\mathcal{C}$, using Green's theorem.
$\gamma$ is smooth, closed, simple and arbitrarily oriented counter clockwise, so :
$$\int\int_DdA=\frac{1}{2}\oint_\gamma xdy-ydx$$
We have : $\begin{cases} x=1-t^2\\ y=t-t^3 \end{cases}$
And by the figure I have of the curve (I don't know how we could plot it otherwise), I can tell that the loop starts at $(0,0)$ and ends at $(1,0)$ so that $t\in[0,1]$.
$$\int\int_DdA=\frac{1}{2}\oint_{0}^{1} ((1-t^2)(1-3t^2)-(t-t^3)(-2t))dt$$ $$=\frac{1}{2}\oint_{0}^{1} (t^4-2t^2+1)dt=4/15$$
I think this was simple enough and that I got it right. However, now, knowing this result, I must calculate the following integral :
$$\oint_{\gamma}\sqrt{1+x^3}dx+2xdy$$
We know that, as per Green's theorem, it can be rewritten as :
$$\int\int_D\Big(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\Big)dA=\int\int_D(2-0)dA$$
Sweet, the root vanishes ! But I don't know how to define domain $D$.
Here is a figure of domain $D$ :
Any help would be much appreciated. Thanks !
There are two values of t that give the origin for that parameterization: t =-1 and t =1. I believe if you just evaluate the line integral with those limits you'll get the area inside the closed part of the curve.
• Thanks, indeed. But my question really is how to define the domain inside the loop in cartesian form to be able to solve the last integral. – Bob Leponge Jul 20 '15 at 0:59
• I see. Eliminating t from the equation of the curve should give $y=x\sqrt{1-x}$ and $y=-x\sqrt{1-x}$. It's easy to do this if you notice that $\frac{y}{x}=t$ – Alex Pavellas Jul 20 '15 at 1:13
We have the curve $C$ defined parametrically as
\begin{align} \vec r(t)&=\hat xx(t)+\hat yy(t)\\\\ &=\hat x(1-t^2)+\hat yt(1-t^2)\\\\ &\implies x(t)=1-t^2\,\,\text{and}\,\,\,y(t)=t(1-t^2)\tag 1 \end{align}
for $t\in [0,1]$.
NOTE:
For $t\in[0,1]$, we see that $y$ is greater than $0$. The closed curve $\gamma$ is formed by two parts; (i) the curve $C$ defined by the aforementioned parameterization and (ii) the line segment $C_x$ along the x-axis from $(0,0)$ to $(1,0)$. However, we note that on the line segment, $y=0$ and $dy=0$ and the contribution from the integration over $C_x$ is zero. We therefore have$$\oint_{\gamma}(xdy-ydx)=\int_{C}(xdy-ydx)+\int_{C_x}(xdy-ydx)=\int_{C}(xdy-ydx)$$
Now, using the first relationship in $(1)$ we find that
$$t^2=1-x \tag 2$$
while using both relationships in $(1)$ simultaneously we find that
$$t^2=(y/x)^2 \tag 3$$
Noting that the right-hand sides of $(2)$ and $(3)$ are equal reveals
$$y^2=x^2(1-x)$$
We recall that $t\in[0,1] \implies x\in[0,1]\,\,\text{and}\,\,y\ge0$.
Thus, the area $A$ inside the region bounded by $y=0$ and $y= x\sqrt{1-x}$ is given by
\begin{align} A&=\int_0^1\int_{0}^{x\sqrt{1-x}}dydx\\\\ &=\int_0^1\,x\sqrt{1-x}\,dx\\\\ &=-\frac{2}{3}\left.x(1-x)\right|_0^1+\frac23\int_0^1\,(1-x)^{3/2}\,dx\\\\ &=-\frac23\,\frac{2}{5}\,\left.(1-x)^{5/2}\right|_0^1\\\\ &=\frac{4}{15} \end{align}
as expected!!
• @bobleponge I included explanation of a subtlety in the problem along with a full development of the evaluation of the integral for the area. Please let me know how I can improve my answer. I really just want to give you the very best answer I can. – Mark Viola Jul 22 '15 at 15:41 | 2019-06-25T08:22:27 | {
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https://ch.mathworks.com/help/symbolic/sym.chebyshevt.html | # chebyshevT
Chebyshev polynomials of the first kind
## Description
example
chebyshevT(n,x) represents the nth degree Chebyshev polynomial of the first kind at the point x.
## Examples
### First Five Chebyshev Polynomials of the First Kind
Find the first five Chebyshev polynomials of the first kind for the variable x.
syms x
chebyshevT([0, 1, 2, 3, 4], x)
ans =
[ 1, x, 2*x^2 - 1, 4*x^3 - 3*x, 8*x^4 - 8*x^2 + 1]
### Chebyshev Polynomials for Numeric and Symbolic Arguments
Depending on its arguments, chebyshevT returns floating-point or exact symbolic results.
Find the value of the fifth-degree Chebyshev polynomial of the first kind at these points. Because these numbers are not symbolic objects, chebyshevT returns floating-point results.
chebyshevT(5, [1/6, 1/4, 1/3, 1/2, 2/3, 3/4])
ans =
0.7428 0.9531 0.9918 0.5000 -0.4856 -0.8906
Find the value of the fifth-degree Chebyshev polynomial of the first kind for the same numbers converted to symbolic objects. For symbolic numbers, chebyshevT returns exact symbolic results.
chebyshevT(5, sym([1/6, 1/4, 1/3, 1/2, 2/3, 3/4]))
ans =
[ 361/486, 61/64, 241/243, 1/2, -118/243, -57/64]
### Evaluate Chebyshev Polynomials with Floating-Point Numbers
Floating-point evaluation of Chebyshev polynomials by direct calls of chebyshevT is numerically stable. However, first computing the polynomial using a symbolic variable, and then substituting variable-precision values into this expression can be numerically unstable.
Find the value of the 500th-degree Chebyshev polynomial of the first kind at 1/3 and vpa(1/3). Floating-point evaluation is numerically stable.
chebyshevT(500, 1/3)
chebyshevT(500, vpa(1/3))
ans =
0.9631
ans =
0.963114126817085233778571286718
Now, find the symbolic polynomial T500 = chebyshevT(500, x), and substitute x = vpa(1/3) into the result. This approach is numerically unstable.
syms x
T500 = chebyshevT(500, x);
subs(T500, x, vpa(1/3))
ans =
-3293905791337500897482813472768.0
Approximate the polynomial coefficients by using vpa, and then substitute x = sym(1/3) into the result. This approach is also numerically unstable.
subs(vpa(T500), x, sym(1/3))
ans =
1202292431349342132757038366720.0
### Plot Chebyshev Polynomials of the First Kind
Plot the first five Chebyshev polynomials of the first kind.
syms x y
fplot(chebyshevT(0:4,x))
axis([-1.5 1.5 -2 2])
grid on
ylabel('T_n(x)')
legend('T_0(x)','T_1(x)','T_2(x)','T_3(x)','T_4(x)','Location','Best')
title('Chebyshev polynomials of the first kind')
## Input Arguments
collapse all
Degree of the polynomial, specified as a nonnegative integer, symbolic variable, expression, or function, or as a vector or matrix of numbers, symbolic numbers, variables, expressions, or functions.
Evaluation point, specified as a number, symbolic number, variable, expression, or function, or as a vector or matrix of numbers, symbolic numbers, variables, expressions, or functions.
collapse all
### Chebyshev Polynomials of the First Kind
• Chebyshev polynomials of the first kind are defined as Tn(x) = cos(n*arccos(x)).
These polynomials satisfy the recursion formula
$T\left(0,x\right)=1,\text{ }T\left(1,x\right)=x,\text{ }T\left(n,x\right)=2\text{ }x\text{ }T\left(n-1,x\right)-T\left(n-2,x\right)$
• Chebyshev polynomials of the first kind are orthogonal on the interval -1 ≤ x ≤ 1 with respect to the weight function $w\left(x\right)=\frac{1}{\sqrt{1-{x}^{2}}}$.
• Chebyshev polynomials of the first kind are special cases of the Jacobi polynomials
$T\left(n,x\right)=\frac{{2}^{2n}{\left(n!\right)}^{2}}{\left(2n\right)!}P\left(n,-\frac{1}{2},-\frac{1}{2},x\right)$
and Gegenbauer polynomials
## Tips
• chebyshevT returns floating-point results for numeric arguments that are not symbolic objects.
• chebyshevT acts element-wise on nonscalar inputs.
• At least one input argument must be a scalar or both arguments must be vectors or matrices of the same size. If one input argument is a scalar and the other one is a vector or a matrix, then chebyshevT expands the scalar into a vector or matrix of the same size as the other argument with all elements equal to that scalar.
## References
[1] Hochstrasser, U. W. “Orthogonal Polynomials.” Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. (M. Abramowitz and I. A. Stegun, eds.). New York: Dover, 1972.
[2] Cohl, Howard S., and Connor MacKenzie. “Generalizations and Specializations of Generating Functions for Jacobi, Gegenbauer, Chebyshev and Legendre Polynomials with Definite Integrals.” Journal of Classical Analysis, no. 1 (2013): 17–33. https://doi.org/10.7153/jca-03-02.
## Version History
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http://mathhelpforum.com/calculus/13891-simple-integration-parts-problem.html | # Thread: Simple integration by parts problem.
1. ## Simple integration by parts problem.
step by step, how do you work the integration of (t^3(e^t) dt)
I know the answer is (t^3-3t^2+6t-6)(exp(t))
2. Originally Posted by mathman66
step by step, how do you work the integration of (t^3(e^t) dt)
I know the answer is (t^3-3t^2+6t-6)(exp(t))
here's the method of integration by parts. if we have a product of the form u*v', then
int{uv'} = uv - int{u'v}
in other words, you chose one function to differentiate, and another one to integrate then the integral of their product is the (integral of function1) times function2 minus the integral of (function1 times the derivative of function2).
how do you know which function to choose to do what? well, the first rule is simple. do what you can! there are some functions that you don't know the integral of but you know the derivative of, so obviously that would be the function you choose to differentiate. otherwise, choose the functions in such a way that the integral becomes simpler.
here we have t^3 * e^t, now we can integrate and differentiate either of these easily, so we have the can do part down. however, if we choose to integrate the t^3, what happens? it becomes t^4 and that makes the problem worst. so obviously we will choose the t^3 to differentiate and we'll end up with t^2, then do it again and end up with t and do it yet again and end up with 1. e^t stays the same throughout, so we don't worry about that. let's see how it works. we will have to do integration by parts three times as i mentioned, the first time i'll do it with explanation then the last two times i'll do it straight and hope you follow.
int{(t^3)(e^t)}dt
let u = t^3 => u' = 3t^2
let v' = e^t => v = e^t
so int{uv'} = uv - int{u'v}
=> int{(t^3)(e^t)}dt = (t^3)(e^t) - 3*int{(t^2)(e^t)}dt ...now do by parts again on this integral.
int{(t^2)(e^t)}dt = (t^2)(e^t) - 2*int{t(e^t)}dt ........by parts again on this integral
int{t(e^t)}dt = t(e^t) - int{e^t}dt = t(e^t) - e^t
Now back substitute to find our integral.
=> int{(t^2)(e^t)}dt = (t^2)(e^t) - 2*(t(e^t) - e^t) = (t^2)(e^t) - 2t(e^t) + 2e^t = (e^t)(t^2 - 2t + 2)
So finally our integral will be:
int{(t^3)(e^t)}dt = (t^3)(e^t) - 3*((e^t)(t^2 - 2t + 2)) + C
.........................= (t^3)(e^t) - 3(e^t)(t^2 - 2t + 2) + C .....factor out e^t
.........................= (e^t)(t^3 - 3t^2 + 6t - 6) + C
The colors aren't as explanatory as i thought they'd be. if you're confused by them don't worry about it, just follow the steps
i could write out the full expression with all the colors so you can see what gets plugged into what if you want
3. Originally Posted by mathman66
step by step, how do you work the integration of (t^3(e^t) dt)
I know the answer is (t^3-3t^2+6t-6)(exp(t))
It is quite obviously a cubic polynomial times e^t, so you can find the
integral by differentiation (At^3+Bt^2+Ct+D)e^t then equating coefficients
between the derivative and t^3 e^t.
RonL
4. Thanks, that helped a lot. I think my problem is that it gets confusing sometimes, especially with an engineers handwritting.
5. Originally Posted by CaptainBlack
It is quite obviously a cubic polynomial times e^t, so you can find the
integral by differentiation (At^3+Bt^2+Ct+D)e^t then equating coefficients
between the derivative and t^3 e^t.
RonL
I liked that suggestion so much, that i'm going to follow it. i think mathman66 specifically wanted the by parts way though. this is for fun
int{t^3 * e^t}dt will be a polynomial of the form (At^3+Bt^2+Ct+D)e^t + E, so we can equate d/dt [(At^3+Bt^2+Ct+D)e^t] to t^3 * e^t
Now d/dt [(At^3+Bt^2+Ct+D)e^t] = (At^3+Bt^2+Ct+D)e^t + (3At^2 + 2Bt + C)e^t
................................................= (At^3+Bt^2+Ct+D + 3At^2 + 2Bt + C)(e^t)
................................................= [At^3 + (3A + B)t^2 + (2B + C)t + (C + D)](e^t)
Now equating the coefficients to our original expression we get:
A = 1
3A + B = 0 => B = -3
2B + C = 0 => C = 6
C + D = 0 => D = -6
So our integral will be: [t^3 - 3t^2 + 6t - 6](e^t) + C | 2017-05-29T15:20:41 | {
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https://math.stackexchange.com/questions/1984693/best-strategy-to-pick-a-lock-which-opens-if-at-least-two-of-its-three-decimal-di/1985115 | # Best strategy to pick a lock which opens if at least two of its three decimal digit wheels are dialed correctly?
Suppose you want to open a lock with three digits code, the lock is a little special because it can be opened if you have two digits guessed right. To clarify, if the correct three digit is 123, then guess 124 or 153 can open the box. The lock looks like this:
Question is: what is best strategy to open the box? The best strategy is a strategy which requires least attempts at average. You should also find the average.
The strategies I came up with are:
First strategy: hold first digit, and constantly change second and third while keep them equal when changing them. For example, I will try: 111, 122...199,211,222,...,299,....
Second strategy: hold second and third equal, and constantly change first one. For example: 111, 211,...,911,122,222,...
I don't know if these two are best, nor do I know if they are equivalent efficient.
## Edit
Here is a program to calculate the average number of trails for a strategy from an ardent comment. To use it, replace the line '// Input your list here' with your test list and press run.
• Does "best strategy" mean fewest guesses, least movement of the wheels, or some other criterion? – David K Oct 25 '16 at 17:32
• Further clarification: fewest guesses on average? or fewest guesses in worst case scenario? – JMoravitz Oct 25 '16 at 17:35
• It seems the "specialty" in this lock reduces the guess space from 1000 to 300. The brute froce way would be to ignore one digit and try 99 combinations with the other two and repeat three times. – darksky Oct 25 '16 at 17:37
• @darksky But if you go through all 100 combinations of two digits, then for one of those the lock should open. So why is it not 100? Why 300? – Bram28 Oct 25 '16 at 17:40
• So: the optimal strategy is ... whatever you try next, try and have it something that has at least two digits different from all the things you already tried ... and do that in a sequential way so that you can make sure there is a new number that has a 'two_digit_different_distance' from all tried numbers as many times as possible? ... – Bram28 Oct 25 '16 at 18:12
I have a strategy with at most $50$ tries and $22449/1000=22.449$ expected tries.
\begin{align} 443, 796, 869, 101, 230, 577, 314, 022, 965, 656, \\ 588, 757, 875, 213, 140, 331, 689, 998, 404, 410, \\ 134, 303, 241, 886, 555, 667, 779, 421, 599, 000, \\ 432, 202, 897, 768, 044, 033, 695, 342, 011, 976, \\ 678, 959, 112, 858, 987, 224, 123, 320, 566, 785\phantom{,} \end{align}
This was obtained by starting from the unordered set of these words (given by a covering code) and then ordering them using a greedy computer search; more details below.
First, I'll get some ideas by considering another problem, optimizing the number of tries needed for the worst case, which has a known solution for this case. A covering code $C$ with alphabet size $q=10$, length $n=3$, and covering radius $R=1$ is a set of $3$-tuples (called words) of length $3$ over $\{0,1,\dots,9\}$ such that every possible $3$-tuple differs from one in $C$ in at most one position. This is exactly what we need.
The minimal size with these parameters is $50$ [1]. It contains these words: $$\begin{array}{|ccccc|} \hline 000 & 011 & 022 & 033 & 044 \\ 101 & 112 & 123 & 134 & 140 \\ 202 & 213 & 224 & 230 & 241 \\ 303 & 314 & 320 & 331 & 342 \\ 404 & 410 & 421 & 432 & 443 \\ \hline 555 & 566 & 577 & 588 & 599 \\ 656 & 667 & 678 & 689 & 695 \\ 757 & 768 & 779 & 785 & 796 \\ 858 & 869 & 875 & 886 & 897 \\ 959 & 965 & 976 & 987 & 998 \\ \hline \end{array}$$
For any two columns, the upper half contains a word for all $25$ pairs of symbols in $\{0,1,2,3,4\}$ that can occur there, and the lower half contains all $25$ pairs of symbols in $\{5,6,7,8,9\}$ that can occur there. The correct combination has to contain at least two symbols from either set, so it is opened by entering one of these words.
[1] Some sources refer to "J. G. Kalbfleisch and R. G. Stanton, A combinatorial theorem of matching, J. London Math. Soc. (1), 44 (1969), 60–64; and (2), 1 (1969), 398", but I can't find the paper. However, the value is listed in the fourth PDF listed at the top of this page by Gerzson Kéri.
Now back to the original problem, optimizing the expected number of tries required. My idea is to try to take these words and optimize the order somehow. This of course doesn't guarantee that we have an optimal solution.
I tried a greedy random approach: Choose words one by one. At each step, for each possible new word $c$, find the number of previously uncovered words that would be covered by $c$. Then among the ones that cover the most, choose one by random. After some time, the best that I could find was the order given at the top of this answer, with $22449/1000$ as the expected number.
• Nice! Beats my strategy! +1! A proof would seem to be difficult though, no? – Bram28 Oct 25 '16 at 23:26
• Beats my strategy. – fleablood Oct 25 '16 at 23:28
• Do you think you can get it down to 45 of by doing 011 to 099, then 101-180 , 202-270, etc.? Just taking a wild guess. – fleablood Oct 25 '16 at 23:44
• The average trails is calculated by computer or by hand? – fizis Oct 26 '16 at 2:14
• @fleablood The claim that 50 is the smallest possible number has been repeated in peer-reviewed mathematical literature since 1969, so I don't expect to go below that. :) And I don't know how to interpret "101-180": going 101, 112, 123, ... would end in 189 and 190, not 810, and I see no other obvious continuation. – JiK Oct 26 '16 at 9:06
OK, so the basic idea I have is as follows:
We want to cover as many possible combinations as possible for each try. Thus, for example, the OP's strategy of doing 111,122,..., followed by 211,222, ... would seem less than optimal, since the first 10 tries already cover any combinations with the last two digits being the same. So, it would be better to do 111,122, ... followed by something like 223,234,245, so that not only do you try to get a hit between the first digit and any of the other two, but you are also trying other options to get a hit between the 2nd and 3rd digit.
In fact, even 111,122,... is less than optimal, since they both cover 121, so you get overlap. So, it's probably best to start with something like 111,222,333,... since each of these covers 28 combinations without overlap.
After that, I figure we should just keep making sure that every new try we make has as few digits in common with whatever combinations we already tried up to that point... basically try and make each combination differ in 2 digits from any of the ones tried before. The following sequence will do exactly that (each sequence of 10 has a unique difference between 1st and 2nd digit, between 2nd and 3rd, and between 1st and 3rd):
000, 111, 222, 333, ..., 999, (at this point we have covered 10*28=280 combinations)
012, 123, 234, 345, ..., 901, (another 10 * 22 (e.g. 012 covers itself and 3*7 more) = 220)
(so with a mere 20 tries we cover half of all possible combinations!)
024, 135, 246, 357, ..., 913, (another 10 * 18 (this takes some writing out) = 180)
036, 147, 258, 369, ..., 925, (another 10 * 12 = 120)
048, 159, 260, 371, ..., 937, (another 10 * 8 = 80)
(at this point, don't proceed with 050,... since that has two digits in common with earlier 000)
051, 162, 273, 384, ..., 940, (another 10 * 5 = 50)
063, 174, 285, 396, ..., 952, (another 10 * 4 = 40)
075, 186, 297, 308, ..., 964, (another 10 * 2 = 20)
087, 198, 209, 310, ... ,976 (the last 10)
In an earlier version of my answer I said to do another 10 after these:
099, 100, 211, 322, ... , 988
the idea being that this would cover all possible pairs of 1st and 2nd digit, and thus covering all possible combinations (and thus we would have a worst case of 100). However, it turns out that whichever ones these additional 10 would cover have been covered already by the previous 90. So, worst case of this method is 90 tries, not 100.
The above method gives an average of 0.28*5.5 (the first 10 tries each cover 28 cases, so that is 280 cases out of 1000 is 28%, which on average take (1 + 10)/2 = 5.5 tries) + 0.22*15.5 + 0.18*25.5 + 0.12*35.5 + 0.08*45.5 + 0.05*55.5 + 0.04*65.5 + 0.02*75.5 + 0.01*85.5 = 25.2 tries to open the box.
I really think this method cannot be improved in terms of average number of tries, but I have no proof.
Edit ok, so this is not the most efficient strategy: see JiK's answer! Time to eat humble pie! ... Well, I hope at least I was able to express some of the ideas why some strategies might be better than others.
• worse case scenario is no better than fixing the first digit and guessing the other two. – fleablood Oct 25 '16 at 18:32
• True. No matter what method, worst case is always 100 tries. But I think this method optimizes average number of tries – Bram28 Oct 25 '16 at 18:36
• That seems to be a good strategy – Djura Marinkov Oct 25 '16 at 18:53
• Why not just 000 to 009, then 110 to 119, then 220 to 229, etc? This seems to capture the essence of your strategy but more simply. – Wildcard Oct 25 '16 at 19:09
• Strange that a guess that seems suboptimal and doesn't even contain a sketch of a proof would be the top answer. -1 – BlueRaja - Danny Pflughoeft Oct 25 '16 at 23:01
This is a problem with entropy, in each try you want to maximally lowerize the entropy.
Since I forgot the formulas for entropy I'll go this way... Maximize the probability to unlock in each try.(it's kinda same)
Let n be order of the try and p(n) probability to unlock, then:
$p(1)=\frac{3*9+1}{1000}=\frac{28}{1000}$
After first try you discriminate 28 combinations, so if you failed with 111, you also know that 112,113,...,121,... are not the combination
Lets try your second strategy for the next try, if you pick 211, you will discriminate another 18 combinations (212,213,...210,221,231,...230 but combinations 311,411... you already discriminated, so p(2) with the 2nd strategy is $p(2)=\frac{18}{972}$
Lets try 1st strategy, so second pick to be 122, this way you discriminate another 26 combinations cos 121 and 112 were already disc.
So I would decide for strategy to change all three digits, to discriminate another 28 combinations, so $p(2)=\frac{28}{972}$
That's what would you do for the first 10 tries, and for the rest... try on your own :)
• But it is not clear that if for every new number you pick, you pick the one that discriminates the most number of combinations, that then you get the minimal number of guesses on average. For example, maybe by picking a combination that discriminates 10 other combinations (as opposed to one that discriminates only 9), I am being 'locally greedy', and the best subsequent move may only be able to discriminate 6 (for a total of 16 between the two moves), while the other move has a subsequent move that discriminates 8 (for a total of 17 between the two moves). – Bram28 Oct 25 '16 at 18:23
• @Bram28 you are right, it is possible that in a moment, maybe, the better probability in the next try doesn't mean better strategy in total. It should be taken care in the late moves, when numbers are low and divisibility is matter. – Djura Marinkov Oct 25 '16 at 18:31
• At least this establishes a lower bound on the worst-case number of tries. You cannot "cover" more than $28$ combinations per try, so it will take at least $\lceil 1000/28 \rceil = 36$ tries. – David K Oct 25 '16 at 19:26
• I think theoretically this might be a right direction. – fizis Oct 26 '16 at 1:30
So, what if you reduced the question to just 3 possible numbers allowed for each digit?
From $000$ to $222$
In that case, you can cover any of the 27 permutations with just 6 guesses $$000$$ $$111$$ $$222$$ $$010$$ $$020$$ $$101$$
So that's 6 guesses to get $3^3$. I'm not sure how this would generalize though.
Is it $n^2\frac{n-1}{n}=n(n-1)$ which would be 90 in your problem? (This generalization also works for the near-trivial example of only 2 numbers, 0 and 1, in which case 000 and 111 covers any permutation)
Also, this method is trying to minimize worst case, exhaustive method, not the average number of tries.
Edit: Confirmed $n(n-1)$ also covers every options in base 4 (from 000 to 333).
000, 111, 222, 333,
012, 013, 021,
102, 103, 120,
210,
301
And from what I gather from brute force methods, in your original question, cycling through the 000, 111, 222, 333....999 will take out 28 permutations each, for 280. After that, you will be able to make 80 guesses that will take out 9 permutations each, for a total of 90 guesses that cover all 1000 permutations.
Also, after 29 guesses, you've taken out over 500 of the permutations.
1) 000: Rules out more than 1 zero
2-4) 011;101;110: Rules out one and zero and more than one 1
25-28) 022;202;220;....099;909;990: Rules out 0 and more than 1 of any number.
29-34) 123,213,231,132,312,321: Rules any two of 1,2,3
35-52)145...167...189: Rules out any 1s; and any pair of (4,5)(6,7),(89)
53-58)246....642: Rules out 2s with 4or 6 and any pair of 4 or 6.
59-64) 257....752: Rules out 2s with 5 or 7; so only 2s with 8 or 9 are left. But 8 and 9 can't pair so this is impossible so, rules out 2s.
65- 70)347...743: Rules out 3s with anything less than 5s.
71-76) 356..653: Rules out 3s with anything other than 8 and 9, which is impossible so rules out 3.
4s are only possible with 8 and 9 so that's impossible.
6s are only possible with 8 and 9. 7 is only possible with 8 or 9.
So any combination we would have guessed in 76 guesses.
I'm not sure how to generalize. | 2021-06-25T09:51:24 | {
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https://mathschallenge.net/full/perfect_digit | ## Perfect Digit
#### Problem
The divisors of a positive integer, excluding the number itself, are called the proper divisors . If the sum of proper divisors is equal to the number we call the number perfect. For example, the divisors of 28 are 1, 2, 4, 7, 14, and 28, so the sum of proper divisors is 1 + 2 + 4 + 7 + 14 = 28.
The first eight perfect numbers are 6, 28, 496, 8128, 33550336, 8589869056, 137438691328, 2305843008139952128.
Prove that the last digit of an even perfect numbers will be 6 or 8.
#### Solution
It can be shown that P is an even perfect number iff it is of the form $2^{n-1}(2^n - 1)$ where $2^n - 1$ is prime.
Moreover it is necessary for $n$ to be prime which can be easily demonstrated.
Suppose that $n$ is composite; let $n = xy$.
$\therefore 2^{xy} - 1 = (2^x - 1)(2^{(x-1)y} + ... + 2^{3y} + 2^{2y} + 2^y + 1)$
Clearly the second bracket is greater than 1, so for $2^n - 1$ to be prime it is necessary for $2^x - 1 = 1 \implies x = 1$ and $y = n$, which must be prime.
Now with the exception of $n = 2$, for which $P = 6$ anyway, all other primes are odd. Hence we must deal with two cases where $n \equiv 1,3 \mod 4$.
(i) If $n \equiv 1 \mod 4$; let $n = 4k + 1$:
\begin{align}\therefore P &= 2^{n-1}(2^n - 1)\\&= 2^{4k}(2^{4k+1} - 1)\\&= (2^4)^k(2.(2^4)^k - 1)\\&= 16^k(2.16^k - 1)\end{align}
If $16^i \equiv 6 \mod 10$ then $16.16^i = 16^{i+1} \equiv 16.6 = 96 \equiv 6 \mod 10$, and as $16 \equiv 6 \mod 10$ it follows that $16^i \equiv 6 \mod 10$ for all values of $i \ge 1$.
Therefore $P \equiv 6(2.6 - 1) = 66 \equiv 6 \mod 10$.
In other words, if $n \equiv 1 \mod 4$ then the last digit of $P$ will be 6.
(ii) If $n \equiv 3 \mod 4$; let $n = 4k + 3$:
\begin{align}\therefore P &= 2^{4k+2}(2^{4k+3} - 1)\\&= 2^2.(2^4)^k(2^3.(2^4)^k - 1)\\&= 4.16^k(8.16^k - 1)\end{align}
Therefore $P \equiv 4.6(8.6 - 1) = 1128 \equiv 8 \mod 10$.
Hence the last digit of an even perfect number will be 6 or 8. Q.E.D.
Prove that if $n$ 3 mod 4 then the last two digits of P will be 28.
Problem ID: 331 (29 Aug 2007) Difficulty: 4 Star
Only Show Problem | 2018-05-24T06:30:35 | {
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http://math.stackexchange.com/questions/456647/an-estimate-for-ln1fx-using-taylor-expansion | # An estimate for $\ln(1+f(x))$ using Taylor expansion
A crucial skill for every aspiring analyst (like myself) is confidence in estimation - knowing when, where, and how to use tools like Big-and-little-O to gain quick upper bounds. I'm trying to push myself to get better at it, but I'm still a little hesitant sometimes. Here's an example: I need to determine if the following series converges uniformly
$$\sum_{n=1}^\infty f_n(x)=\sum_{n=1}^\infty\frac{\ln\left(1+\frac{\sin^2(nx)}{n^2}\right)}{n}$$
I'm pretty sure it does, and my reasoning is as follows: if $n>1$, $\sin^2(nx)/n^2<1$ for all $x\in\Bbb{R}$, and hence we can use the Taylor series for $\ln(1+z)$ about $z=0$ to obtain the following:
$$\left\vert\ln\left(1+\frac{\sin^2(nx)}{n^2}\right)\right\vert\leq\frac{1}{n^2}+\frac{C}{n^4}\tag{\dagger}$$ where $C$ is a constant independent of $x$. Thus,
$$\left|\,f_n(x)\right|\leq\frac{1}{n^3}+\frac{C}{n^5}$$and since $$\sum_{n=1}^\infty\left( \frac{1}{n^3}+\frac{1}{n^5}\right)<\infty,$$the series converges uniformly.
My question is: is $(\dagger)$ correct, and do I have the constant in the right place (or does it matter)? I used the fact that $\ln(1+z)=z+O(z^2)$ to obtain $(\dagger)$.
-
@msh210 thanks for the copy editing...studying for quals has really cooked my noodle. – icurays1 Jul 31 '13 at 19:21
You're welcome, but my edit suggestion was accepted after you had made a clashing edit -- or so it seems -- so yours was lost. – msh210 Jul 31 '13 at 19:24
Oh, I was wondering why a line disappeared... – icurays1 Jul 31 '13 at 19:25
You can actually use the simpler estimate, that says that $$\log(1+x)\leq x$$
Then $$\frac{1}{n}\log \left( {1 + \frac{{{{\sin }^2}nx}}{{{n^2}}}} \right) \leqslant \frac{{{{\sin }^2}nx}}{{{n^3}}}\leqslant \frac 1 {n^3}$$
One should always aim for the simple first!
ADD As per your estimate. First note that your function is always positive and periodic. Moreover, it is bounded. Since it has multiple roots, we know there is a maximum, and it will be an absolute "periodic" maxima. The derivative is $$\frac{{\sin 2nx}}{{n + {n^{ - 1}}{{\sin }^2}nx}}$$ which has roots at (choose $k=0,1$) so $2nx=0,\pi$. At $\pi/2n$ the maxima is $$\log\left(1+\frac 1 {n^2}\right)\leq\frac 1{n^2}$$
which is summable.
-
Ah! Now that is much simpler. Is my $(\dagger)$ still correct though, just out of curiosity? – icurays1 Jul 31 '13 at 19:37
Yes, it's correct. But as Peter says, try the simplest thing first, only if that doesn't work try the more complicated ones. – Daniel Fischer Jul 31 '13 at 19:37
Another skill I need to work on =) Thanks! – icurays1 Jul 31 '13 at 19:39 | 2014-08-23T19:29:34 | {
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http://techiemathteacher.com/2015/04/22/2015-mtap-reviewer-for-grade-9-with-solutions11-15/ | # 2015 MTAP Reviewer for Grade 9 With Solutions(11-15)
We’re done with 20% of the elimination paper of 2015 MMC (MTAP) for grade 9, now we will continue learning how to deal with mind boggling problems to win the competition next year.
Problem 11:
If $x\ne 0$, solve for x in $2\sqrt{x}+\dfrac{3}{\sqrt{x}}=5$
Solution:
To solve equations, we always eliminate fractions. By multiplying the whole equation by $\sqrt{x}$ we can remove all fractions.
$(2\sqrt{x}+\dfrac{3}{\sqrt{x}})\sqrt{x}=5\sqrt{x}$
$2\sqrt{x^2}+3=5\sqrt{x}$
$2x+3=5\sqrt{x}$
Now, since the right side of equation is radical, we raise both sides by 2,
$(2x+3)^2=(5\sqrt{x})^2$
$4x^2+12x+9=25x$
$4x^2+12x-25x+9=0$
$4x^2-13x+9=0$
By factoring we have,
$(x-1)(4x-9)=0$
These factors will immediately tell us that the values of x are 1 and 9/4. By quick check however, 9/4 fails to satisfy the original equation. Thus, the only solution is 1.
Problem 12:
Evaluate $\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}$
Solution:
This is a nested radical and I already have a tutorial on how to solve this problem. If you’re used to see problem like this, you might be able to solve it in less than 3 seconds. For real.
To solve this though, we let x be equal to $\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}$
$x=\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}$
Square both sides,
$x^2=(\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}})^2$
$x^2=2+\sqrt{2+\sqrt{2+\cdots}}$
Now, recall that $x=\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}$
$x^2=2+(\sqrt{2+\sqrt{2+\cdots}})$
$x^2=2+(x)$
$x^2=2+x$
$x^2-x-2=0$
$(x-2)(x+1)=0$
$x=2,-1$
The square root of a positive number will always be positive. Thus, the answer is 2. -1 is an extraneous root.
Problem 13:
Find the two consecutive integers whose product is 506.
Solution:
Let x be the smaller integer. Since we are looking for the two CONSECUTIVE integers, the other number must be x+1.
It is stated that their product is 506, thus
$x(x+1)=506$
$x^2+x-506=0$
$(x+23)(x-22)=0$
$x=22,-23$
Therefore, the numbers that we are looking for are are 22 and 23.
Problem 14:
If $c>a>0$ and if $a-b+c=0$, what is the larger root of $ax^2+bx+c=0$
Solution:
Let x and y be the roots of equation.
By Vieta’s formula, we have
$xy=\dfrac{c}{a}$ *
$x+y=\dfrac{-b}{a}$ **
From the relationship of the coefficients, we solve for c in terms of a and b.
$a-b+c=0$
$c=b-a$***
We substitute *** to c of *,
$xy=\dfrac{b-a}{a}$
$xy=\dfrac{b}{a}-\dfrac{a}{a}$
$xy=\dfrac{b}{a}-1$ ****
From ** we solve for b/a,
$x+y=\dfrac{-b}{a}$
$-x-y=\dfrac{b}{a}$
We substitute this to ****
$xy=\dfrac{b}{a}-1$
$xy=-x-y-1$
$xy+x+y+1=0$
By factoring,
$x(y+1)+(y+1)=0$
$(y+1)(x+1)=0$
$x=-1,y=-1$
Since the roots are equal, there is no greater root. The answer is -1.
Problem 15:
Solve for x in $2x^2+x<6$
Solution:
We also provided the easiest way to solve this quadratic inequality in this site. But let us demonstrate the answer to this problem.
$2x^2+x<6$
$2x^2+x-6<0$
$2x^2+x-6<0$
In this format, the answer is in the form of a<x<b, where b>a and a and b are the roots of inequality.
$(2x-3)(x+2)<0$
The roots are supposed to be 3/2 and -2. Thus, the solution set is -2<x<3/2 or (-2,3/2)
### Dan
Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies. | 2019-08-24T22:07:41 | {
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https://gmatclub.com/forum/a-retail-store-that-sells-only-shoes-and-accessories-found-128572.html | GMAT Question of the Day - Daily to your Mailbox; hard ones only
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# A retail store that sells only shoes and accessories found
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A retail store that sells only shoes and accessories found [#permalink]
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04 Mar 2012, 14:25
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A retail store that sells only shoes and accessories found that its revenues last month from those two categories could be expressed in the ratio x : y, respectively. If last month's total revenues were $4,000, what was the difference between revenues from shoes and revenues from accessories? A. $$\frac{x}{x+y}$$ B. $$\frac{y}{x+y}$$ C. $$4,000 * \frac{x}{y}$$ D. $$4,000*\frac{x}{x+y}$$ E. $$\frac{4,000*(x-y)}{x+y}$$ Any idea how to get to answer E? This is how I am trying to solve Let the revenue from shoes be $$5x$$and revenue from accessories be $$5y.$$ Total revenue = $$5x + 5y = 4000$$ Revenue from shoes = $$\frac{5x}{5x+5y}$$ Revenue from accessories = $$\frac{5y}{5x+5y}$$ Difference in revenues = $$\frac{5x}{5x+ 5y} - \frac{5y}{5x - 5y}$$ But this won't give me the right answer. Any thoughts why guys? _________________ Best Regards, E. MGMAT 1 --> 530 MGMAT 2--> 640 MGMAT 3 ---> 610 GMAT ==> 730 ##### Most Helpful Expert Reply Math Expert Joined: 02 Sep 2009 Posts: 55150 Re: Shoes & Accessories [#permalink] ### Show Tags 04 Mar 2012, 15:03 3 2 A retail store that sells only shoes and accessories found that its revenues last month from those two categories could be expressed in the ratio x : y, respectively. If last month's total revenues were$4,000, what was the difference between revenues from shoes and revenues from accessories?
A. $$\frac{x}{x+y}$$
B. $$\frac{y}{x+y}$$
C. $$4,000 * \frac{x}{y}$$
D. $$4,000*\frac{x}{x+y}$$
E. $$\frac{4,000*(x-y)}{x+y}$$
The revenues from shoes and accessories were in the ratio x:y;
Share of revenues from shoes in total revenues is $$\frac{x}{x+y}$$, hence revenues from shoes is $$4,000*\frac{x}{x+y}$$;
Share of revenues from accessories in total revenues is $$\frac{y}{x+y}$$, hence revenues from accessories is $$4,000*\frac{x}{x+y}$$;
Difference: $$4,000*\frac{x}{x+y}-4,000*\frac{y}{x+y}=\frac{4,000(x-y)}{x+y}$$.
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##### General Discussion
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Joined: 10 Oct 2012
Posts: 611
Re: A retail store that sells only shoes and accessories found [#permalink]
### Show Tags
05 Mar 2013, 11:18
1
enigma123 wrote:
A retail store that sells only shoes and accessories found that its revenues last month from those two categories could be expressed in the ratio x : y, respectively. If last month's total revenues were \$4,000, what was the difference between revenues from shoes and revenues from accessories?
A. $$\frac{x}{x+y}$$
B. $$\frac{y}{x+y}$$
C. $$4,000 * \frac{x}{y}$$
D. $$4,000*\frac{x}{x+y}$$
E. $$\frac{4,000*(x-y)}{x+y}$$
Nothing extra to add. But in case any one wants to gain some extra seconds, put x=y=1, i.e. they both have equal revenue. Thus their difference would be 0. Only option E satisfies that condition.
E.
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Re: A retail store that sells only shoes and accessories found [#permalink]
### Show Tags
12 Jun 2013, 08:27
Hi Bunuel
Share of revenues from shoes in total revenues is , hence revenues from shoes is ;4000*x/x+y
Share of revenues from accessories in total revenues is , hence revenues from accessories is ;4000*y/x+y
Rgds
Prasannajeet
Math Expert
Joined: 02 Sep 2009
Posts: 55150
Re: A retail store that sells only shoes and accessories found [#permalink]
### Show Tags
12 Jun 2013, 09:12
2
1
prasannajeet wrote:
Hi Bunuel
Share of revenues from shoes in total revenues is , hence revenues from shoes is ;4000*x/x+y
Share of revenues from accessories in total revenues is , hence revenues from accessories is ;4000*y/x+y
Rgds
Prasannajeet
Say x=1 and y=3. Total of 1+3=4 parts:
Share of shoes = 1/(1+3)=1/4 --> revenue 1/4*4000=1000.
Share of accessories = 3/(1+3) = 3/4 --> revenue 3/4*4000=3000.
Difference = 2000.
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Re: A retail store that sells only shoes and accessories found [#permalink]
### Show Tags
13 Jun 2013, 03:04
Hi Bunuel
Thanx and +1 Kudos to you...
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Re: A retail store that sells only shoes and accessories found [#permalink]
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27 Sep 2018, 06:51
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Re: A retail store that sells only shoes and accessories found [#permalink] 27 Sep 2018, 06:51
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