text:
# matrix order of operations

Thus, as long as b and c are chosen so that bc = − a 2, A 2 will equal 0. Is there a multiplicative identity in the set of all m x n matrices if m ≠ n? Thus, unlike the set of n x n matrices, the set of nonsquare m x n matrices does not possess a qunique two‐sided identity, because I m ≠ I n if m ≠ n. Example 15: If A is a square matrix, then A 2 denotes the product AA,A 3 denotes the product AAA, and so forth. Next lesson. We have 2x+3y=[2340]2x+3y=\left[ \begin{matrix} 2 & 3 \\ 4 & 0 \\ \end{matrix} \right]2x+3y=[2430] … (i), Multiplying (i) by 3 and (ii) by 2, we get6x+9y=[69120]6x+9y=\left[ \begin{matrix} 6 & 9 \\ 12 & 0 \\ \end{matrix} \right]6x+9y=[61290] … (iii), Subtracting (iv) from (iii), we get 5y=[6−49+412+20−10]=[21314−10]5y=\left[ \begin{matrix} 6-4 & 9+4 \\ 12+2 & 0-10 \\ \end{matrix} \right]=\left[ \begin{matrix} 2 & 13 \\ 14 & -10 \\ \end{matrix} \right]5y=[6−412+29+40−10]=[21413−10] The number of columns of the first matrix must match the number of rows of the second matrix in order for their product to be defined. y = matrix (v, m, n) y = matrix (v, m1, m2, m3, ..) y = matrix (v, [sizes]) Arguments v. Any matricial container (regular matrix of any data type; cells array; structures array), of any number of dimensions (vector, matrix, hyperarray), with any sizes. Taking the dot product of row 1 in A and column 1 in B gives the (1, 1) entry in AB. As the two matrices are equal, their corresponding elements are equal. Because of the sensitivity to the order in which the factors are written, one does not typically say simply, “Multiply the matrices A and B.” It is usually important to indicate which matrix comes first and which comes second in the product. Show that the inverse of B T is ( B −1) T. This calculation shows that ( B −1) T is the inverse of B T. [Strictly speaking, it shows only that ( B −1) T is the right inverse of B T, that is, when it multiplies B T on the right, the product is the identity. A. B T A T does indeed equal ( AB) T. In fact, the equation. CliffsNotes study guides are written by real teachers and professors, so no matter what you're studying, CliffsNotes can ease your homework headaches and help you score high on exams. Note that the associative law implies that the product of A, B, and C (in that order) can be written simply as ABC; parentheses are not needed to resolve any ambiguity, because there is no ambiguity. The problem was 20 - 5 x 2 + 3. Matrix Row Operations (page 1 of 2) "Operations" is mathematician-ese for "procedures". Therefore, is the multiplicative identity in the set of 2 x 2 matrices. Notice, that A and Bare of same order. Later, you will learn various criteria for determining whether a given square matrix is invertible. The order of matrix is equal to m x n (also pronounced as ‘m by n’). Therefore, the equation D 2 − D − 6 I = 0 implies D(D − I) = 6 I. Multiplying both sides of this equation by 1/6 gives, As an illustration of this result, the matrix, satisfies the equation D 2 − D − 6 I = 0, as you may verify. Multiplication (. Formation and order of matrix; Order of a Matrix. A=[1234],B=[1270−… If a, b, and c are real numbers with a ≠ 0, then, by canceling out the factor a, the equation ab = ac implies b = c. No such law exists for matrix multiplication; that is, the statement AB = AC does not imply B = C, even if A is nonzero. Removing #book# This requires the multiplication of the number of shares of each security by the corresponding price per share, then the summation of the results. Also, the resulting matrix will be of same order as its constituents. (g) If AB = 0 (It does not mean that A = 0 or B = 0, again the product of two non-zero matrices may be a zero matrix). (AB)C = A(BC). in general AB≠BA.AB\ne BA.AB=BA. Consider the two matrices A & B of order 2 x 2. Now, since the product of AB and B −1 A −1 is I, B −1 A −1 is indeed the inverse of AB. Any combination of the order S*R*T gives a valid transformation matrix. Consider the two matrices A & B of order 2 x 2. A – B = [aij – bij]mxn, If A=[aij]m×nA={{\left[ {{a}_{ij}} \right]}_{m\times n}}A=[aij]m×n is a matrix and k any number, then the matrix which is obtained by multiplying the elements of A by k is called the scalar multiplication of A by k and it is denoted by k A thus if A=[aij]m×nA={{\left[ {{a}_{ij}} \right]}_{m\times n}}A=[aij]m×n, Then kAm ×n=Am × nk=[kai×j]k{{A}_{m\,\times n}}={{A}_{m\,\times \,n}}k=\left[ k{{a}_{i\times j}} \right]kAm×n=Am×nk=[kai×j]. Since I = I n is the multiplicative identity in the set of n x n matrices, if a matrix B exists such that. Learn All Concepts of Chapter 3 Class 12 Matrices - FREE. i.e. Matrices are defined as a rectangular array of numbers or functions. Syntax . Therefore, by equating the corresponding elements of given matrices we will obtain the value of a, b, c and d. Subtracting equation (i) from (iii), we have a = 1; Putting the value of a in equation (i), we have1−b=−1⇒b=21-b=-1 \Rightarrow b = 21−b=−1⇒b=2; Putting the value of a in equation (ii), we have 2+c=5⇒c=3;2+c=5\Rightarrow c=3;2+c=5⇒c=3; Putting the value of c in equation (iv), we find 9+x=13⇒d=9+x=13\Rightarrow d=9+x=13⇒d=, Illustration 4: find x and y, if 2x+3y=[2340]and3x+2y=[2−2−15]2x+3y=\left[ \begin{matrix} 2 & 3 \\ 4 & 0 \\ \end{matrix} \right] and 3x+2y=\left[ \begin{matrix} 2 & -2 \\ -1 & 5 \\ \end{matrix} \right]2x+3y=[2430]and3x+2y=[2−1−25]. j) There exist a multiplicative identity for every square matrix such AI = IA = A, Illustration 1: If A=[2133−21−101]andB=[124    −21−2]A=\left[ \begin{matrix} 2 & 1 & 3 \\ 3 & -2 & 1 \\ -1 & 0 & 1 \\ \end{matrix} \right] and B=\left[ \begin{matrix} 1 \\ 2 \\ 4 \\ \end{matrix}\,\,\,\,\begin{matrix} -2 \\ 1 \\ -2 \\ \end{matrix} \right]A=⎣⎢⎡23−11−20311⎦⎥⎤andB=⎣⎢⎡124−21−2⎦⎥⎤, Using matrix multiplication. These Order of Operations Worksheets are a great resource for children in Kindergarten, 1st Grade, 2nd Grade, 3rd Grade, 4th Grade, and 5th Grade. If b = 0, the diagonal entries then imply a = 0 and d = 0, and the (2, 1) entries imply that c is arbitrary. Now that we have a good idea of how addition works, let’s try subtraction. A similar chain of reasoning beginning with the (2, 1) entries leads to either a = c = d = 0 (and b arbitrary) or the same conclusion as before: as long as b and c are chosen so that bc = − a 2, the matrix A 2 will equal 0. Click here for a Detailed Description of all the Order of Operations Worksheets. Show, however, that ( A + B) 2 = A 2 + 2 AB + B 2 is not an identity if A and B are 2 x 2 matrices. {{A}_{m\,\times n}}. Since A is 2 x 2, in order to multiply A on the right by a matrix, that matrix must have 2 rows. To add or subtract matrices, these must be of identical order and for multiplication, the number of columns in the first matrix equals the number of rows in the second matrix. Example 24: Assume that B is invertible. That is, if A, B, and C are any three matrices such that the product (AB)C is defined, then the product A(BC) is also defined, and. Let, be an arbitrary 2 x 2 matrix. I did not multiply first. Addition (+) In order to perform addition on matrices in R, we first create two matrices ‘mat1’ and ‘mat2’ with four rows and four columns as follows: Multiply this equation by B −1 on the left and on the right and use associativity: Example 25: The number 0 has just one square root: 0. A few preliminary calculations illustrate that the given formula does hold true: However, to establish that the formula holds for all positive integers n, a general proof must be given. Matrix is a rectangular array of numbers or expressions arranged in rows and columns. Solving the given equations simultaneously, we will obtain the values of x and y. Fraction and Decimal Order of Operations. Learn how to perform the matrix elementary row operations. Basically, a two-dimensional matrix consists of the number of rows (m) and a number of columns (n). ⇒y=[25135145−105]⇒y=[25135145−2]\Rightarrow y=\left[ \begin{matrix} \frac{2}{5} & \frac{13}{5} \\ \frac{14}{5} & \frac{-10}{5} \\ \end{matrix} \right]\Rightarrow y=\left[ \begin{matrix} \frac{2}{5} & \frac{13}{5} \\ \frac{14}{5} & -2 \\ \end{matrix} \right]⇒y=[525145135−10]⇒y=[52514513−2], Putting the value of y in (iii), we get 2x+3[25135145−2]=[2340]2x+3\left[ \begin{matrix} \frac{2}{5} & \frac{13}{5} \\ \frac{14}{5} & -2 \\ \end{matrix} \right]=\left[ \begin{matrix} 2 & 3 \\ 4 & 0 \\ \end{matrix} \right]2x+3[52514513−2]=[2430] There are several operations that we can perform on the R matrices to get desired results: 1. and the dot product of row 2 in A and column 2 in B gives the (2, 2) entry in AB: Finally, taking the dot product of row 2 in A with columns 3 and 4 in B gives (respectively) the (2, 3) and (2, 4) entries in AB: compute the (3, 5) entry of the product CD. If d = − a, then the off‐diagonal entries will both be 0, and the diagonal entries will both equal a 2 + bc. the product of the matrix with a null matrix is always a null matrix. The product of matrices $${\displaystyle A}$$ and $${\displaystyle B}$$ is then denoted simply as $${\displaystyle AB}$$. Because you've got a column-major matrix, you also need to use column vectors, which means your order of multiplication will be: M*v. To prove this to yourself, take a simple 2x2 matrix with a 2x1 column vector, multiply as M*v. Let K=transpose (M), and r=row vector (1x2). Practice: Matrix row operations . Example 23: The equation ( a + b) 2 = a 2 + 2 ab + b 2 is an identity if a and b are real numbers. Row Operations. (d) If A is an m × n matrix, then ImA=A=AIn.{{I}_{m}}A=A=A{{I}_{n}}.ImA=A=AIn. (g) If A + B = 0 = B + A, then B is called additive inverse of A and also A is called the additive inverse of A. (e) The product of two matrices can be a null matrix while neither of them is null, i.e. We say idiot proof but, we have to qualify that by saying, only an expert can use one properly. In fact, it can be easily shown that for this matrix I, both products AI and IA will equal A for any 2 x 2 matrix A. For instance, if. Last updated at April 2, 2019 by Teachoo. Although every nonzero real number has an inverse, there exist nonzero matrices that have no inverse. The analog of this statement for square matrices reads as follows. © 2020 Houghton Mifflin Harcourt. True or false To add or subtract matrices both matrices must have the same dimension? The order of operations was settled upon in order to prevent miscommunication, but PEMDAS can generate its own confusion; some students sometimes tend to apply the hierarchy as though all the operations in a problem are on the same "level" (simply going from left to right), but often those operations are not "equal". and x is the vector (−2, 3), show how A can be multiplied on the right by x and compute the product. reshapes an array with the same number and order of components. and any corresponding bookmarks? This result can be proved in general by applying the associative law for matrix multiplication. Let's assume there are four people, and we call them Lucas, Mia, Leon and Hannah. For example, choosing a = 4, b = 2, and c = −8 gives the nonzero matrix. If A, B are matrices of the same order and are any two scalars then; CBSE Previous Year Question Papers Class 10, CBSE Previous Year Question Papers Class 12, NCERT Solutions Class 11 Business Studies, NCERT Solutions Class 12 Business Studies, NCERT Solutions Class 12 Accountancy Part 1, NCERT Solutions Class 12 Accountancy Part 2, NCERT Solutions For Class 6 Social Science, NCERT Solutions for Class 7 Social Science, NCERT Solutions for Class 8 Social Science, NCERT Solutions For Class 9 Social Science, NCERT Solutions For Class 9 Maths Chapter 1, NCERT Solutions For Class 9 Maths Chapter 2, NCERT Solutions For Class 9 Maths Chapter 3, NCERT Solutions For Class 9 Maths Chapter 4, NCERT Solutions For Class 9 Maths Chapter 5, NCERT Solutions For Class 9 Maths Chapter 6, NCERT Solutions For Class 9 Maths Chapter 7, NCERT Solutions For Class 9 Maths Chapter 8, NCERT Solutions For Class 9 Maths Chapter 9, NCERT Solutions For Class 9 Maths Chapter 10, NCERT Solutions For Class 9 Maths Chapter 11, NCERT Solutions For Class 9 Maths Chapter 12, NCERT Solutions For Class 9 Maths Chapter 13, NCERT Solutions For Class 9 Maths Chapter 14, NCERT Solutions For Class 9 Maths Chapter 15, NCERT Solutions for Class 9 Science Chapter 1, NCERT Solutions for Class 9 Science Chapter 2, NCERT Solutions for Class 9 Science Chapter 3, NCERT Solutions for Class 9 Science Chapter 4, NCERT Solutions for Class 9 Science Chapter 5, NCERT Solutions for Class 9 Science Chapter 6, NCERT Solutions for Class 9 Science Chapter 7, NCERT Solutions for Class 9 Science Chapter 8, NCERT Solutions for Class 9 Science Chapter 9, NCERT Solutions for Class 9 Science Chapter 10, NCERT Solutions for Class 9 Science Chapter 12, NCERT Solutions for Class 9 Science Chapter 11, NCERT Solutions for Class 9 Science Chapter 13, NCERT Solutions for Class 9 Science Chapter 14, NCERT Solutions for Class 9 Science Chapter 15, NCERT Solutions for Class 10 Social Science, NCERT Solutions for Class 10 Maths Chapter 1, NCERT Solutions for Class 10 Maths Chapter 2, NCERT Solutions for Class 10 Maths Chapter 3, NCERT Solutions for Class 10 Maths Chapter 4, NCERT Solutions for Class 10 Maths Chapter 5, NCERT Solutions for Class 10 Maths Chapter 6, NCERT Solutions for Class 10 Maths Chapter 7, NCERT Solutions for Class 10 Maths Chapter 8, NCERT Solutions for Class 10 Maths Chapter 9, NCERT Solutions for Class 10 Maths Chapter 10, NCERT Solutions for Class 10 Maths Chapter 11, NCERT Solutions for Class 10 Maths Chapter 12, NCERT Solutions for Class 10 Maths Chapter 13, NCERT Solutions for Class 10 Maths Chapter 14, NCERT Solutions for Class 10 Maths Chapter 15, NCERT Solutions for Class 10 Science Chapter 1, NCERT Solutions for Class 10 Science Chapter 2, NCERT Solutions for Class 10 Science Chapter 3, NCERT Solutions for Class 10 Science Chapter 4, NCERT Solutions for Class 10 Science Chapter 5, NCERT Solutions for Class 10 Science Chapter 6, NCERT Solutions for Class 10 Science Chapter 7, NCERT Solutions for Class 10 Science Chapter 8, NCERT Solutions for Class 10 Science Chapter 9, NCERT Solutions for Class 10 Science Chapter 10, NCERT Solutions for Class 10 Science Chapter 11, NCERT Solutions for Class 10 Science Chapter 12, NCERT Solutions for Class 10 Science Chapter 13, NCERT Solutions for Class 10 Science Chapter 14, NCERT Solutions for Class 10 Science Chapter 15, NCERT Solutions for Class 10 Science Chapter 16, JEE Main Chapter Wise Questions And Solutions, Differentiation and Integration of Determinants, System of Linear Equations Using Determinants. Thus. That is, as long as the order of the factors is unchanged, how they are grouped is irrelevant. [Technical note: It can be shown that in a certain precise sense, the collection of matrices of the form, where a and b are real numbers, is structurally identical to the collection of complex numbers, a + bi. (Compare this equation with the one involving transposes in Example 14 above.) The four "basic operations" on numbers are addition, subtraction, multiplication, and division. For adding two matrices the element corresponding to same row and column are added together, like in example below matrix A of order 3×2 and matrix Bof same order are added. Most frequently, matrix operations are involved, such as matrix-matrix products and inverses of matrices. The answer should be 13. bookmarked pages associated with this title. A matrix operations order is a fill in the blank, by the number, idiot proof form of Operations Order. Addition. There are versions of R available for Windows, Mac OS and Unix that can be freely downloaded over the Internet. That is, the only way a product of real numbers can equal 0 is if at least one of the factors is itself 0. This matrix B does indeed commute with A, as verified by the calculations. Then the difference is given by: We can subtract the matrices by subtracting each element of one matrix from the corresponding element of the second matrix. Fraction and Decimal Order of Operations. What is the Order of an Element? Properties of Scalar Multiplication: If A, B are matrices of the same order and are any two scalars then; (a) λ(A+B)=λA+λB\lambda \left( A+B \right)=\lambda A+\lambda Bλ(A+B)=λA+λB, (b) (λ+μ)A=λA+μA\left( \lambda +\mu \right)A=\lambda A+\mu A(λ+μ)A=λA+μA, (c) λ(μA)=(λ μA)=μ(λA)\lambda \left( \mu A \right)=\left( \lambda \,\mu A \right)=\mu \left( \lambda A \right)λ(μA)=(λμA)=μ(λA), (d) (−λA)=−(λA)=λ(−A)\left( -\lambda A \right)=-\left( \lambda A \right)=\lambda \left( -A \right)(−λA)=−(λA)=λ(−A), (e) tr(kA)=k  tr  (A)tr\left( kA \right)=k\,\,tr\,\,\left( A \right)tr(kA)=ktr(A). Let a be a given real number. Order of Battle is a series of operational WW2 games starting with the Pacific War and then on to Europe! Exploited to reduce the number of rows of a general Cancellation law is not commutative in general i.e. Also remove any bookmarked pages associated with this title matrix must be that! Even though neither G nor h is a set of numbers that are aligned horizontally of operations.! 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### An Efficient System for Non-transferable Anonymous Credentials with Optional Anonymity Revocation

Jan Camenisch and Anna Lysyanskaya

##### Abstract

A credential system is a system in which users can obtain credentials from organizations and demonstrate possession of these credentials. Such a system is anonymous when transactions carried out by the same user cannot be linked. An anonymous credential system is of significant practical relevance because it is the best means of providing privacy for users. In this paper we propose a practical anonymous credential system that is based on the strong RSA assumption and the decisional Diffie-Hellman assumption modulo a safe prime product and is considerably superior to existing ones: (1) We give the first practical solution that allows a user to unlinkably demonstrate possession of a credential as many times as necessary without involving the issuing organization. (2) To prevent misuse of anonymity, our scheme is the first to offer optional anonymity revocation for particular transactions. (3) Our scheme offers separability: all organizations can choose their cryptographic keys independently of each other. Moreover, we suggest more effective means of preventing users from sharing their credentials, by introducing {\em all-or-nothing} sharing: a user who allows a friend to use one of her credentials once, gives him the ability to use all of her credentials, i.e., taking over her identity. This is implemented by a new primitive, called {\em circular encryption}, which is of independent interest, and can be realized from any semantically secure cryptosystem in the random oracle model.

Available format(s)
Category
Cryptographic protocols
Publication info
Published elsewhere. Extended version of what is going to appear in EUROCPRYPT 2001
Contact author(s)
jca @ zurich ibm com
History
Short URL
https://ia.cr/2001/019

CC BY

BibTeX

@misc{cryptoeprint:2001/019,
author = {Jan Camenisch and Anna Lysyanskaya},
title = {An Efficient System for Non-transferable Anonymous Credentials with Optional Anonymity Revocation},
howpublished = {Cryptology ePrint Archive, Paper 2001/019},
year = {2001},
note = {\url{https://eprint.iacr.org/2001/019}},
url = {https://eprint.iacr.org/2001/019}
}

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format: plain


text:
# $\VDash$ Symbol in AMS math packages?

I'm trying to typeset some homework that deals with set theory and in particular I'm trying to write the "model-theoretic" notion of forcing. I put model-theoretic in quotes because I don't know the proper name for it. There's the forcing relation given by $\Vdash$ (i.e. it looks like similar to ||-), and I'm looking for the relation that looks like ||=, i.e. kind of like the $\models$ symbol or $\vDash$ but with an extra vertical line.

The strange thing is that it looks like the AMS packages have $\nVDash$ (i.e. the negation of the symbol I'm looking for) available, but not $\VDash$. Detexify doesn't seem to be helping me here. Are there any symbols which look like the hypothetical $\VDash$ I'm looking for (hopefully within amsmath, amsfonts, or amssymb, as I usually like to cut down on the number of packages I need to install to maintain portability across machines)?

• Have you seen How to look up a symbol or identify a math alphabet? – Werner Nov 13 '13 at 22:25
• this isn't really helpful, i know, but ... the \VDash was inexplicably not included in the amsfonts, and thus does not appear in amssymb.sty. it is in the stix and xits fonts, but the style of those is not really compatible with the amsfonts. do you need it in any sizes other than the main text size? and what text size are you using? (i'm thinking of cobbling together something with a vertical rule and \vDash.) – barbara beeton Nov 13 '13 at 22:30
• Try \def\mmodels{\mathrel {||}\joinrel \Relbar}, which replicates the look of \models. – Werner Nov 13 '13 at 22:30
• @Werner I did check the Comprehensive LaTeX symbol list as well as Detexify. Unfortunately the results I got were specific to the mathabx package and I tend to do my work on school computer lab machines where both privileges and disk space are limited. – badcook Nov 13 '13 at 22:47
• @Werner I like your definition of $\mmodels$. The size ends up being a bit off from the $\vdash$ family of symbols (so it looks slightly weird next to the forcing relation) and I'll try to tweak it to make it fit, but in the interim it'll work. Thanks! – badcook Nov 13 '13 at 22:49

## 2 Answers

\models is defined as (use \pshow\models where \def\pshow#1{{\let\protect\show #1}}):

\mathrel{|}\joinrel\Relbar

In a similar fashion one could define \mmodels:

\DeclareRobustCommand{\mmodels}{\mathrel{||}\joinrel\Relbar}

Here's a visual of the two:

\documentclass{article}
\DeclareRobustCommand{\mmodels}{\mathrel{||}\joinrel \Relbar}
\begin{document}
$\models\ \mmodels$
\end{document}

Perhaps inserting \mkern-2mu between the two vertical rules would look a little better:

\end{document}

Note: in the image I've artificially reduced the vertical space between the lines just to show the symbols nearer to each other and ease the comparison.

• Is there any way to remove the dependency on the trimclip and adjustbox packages? I'm trying to minimize my dependencies so that I can compile my documents on a multitude of different machines with a minimum of hassle. – badcook Nov 14 '13 at 21:15
• @badcook I looked for a matching bar, but found none, that's why I resorted to clip the symbol. – egreg Nov 14 '13 at 21:22
• Although I accepted @Werner's answer, I think that yours is more aesthetically pleasing; it just so happens that it's a hassle for me to install the trimclip and adjustbox packages to all the machines I might want to use LaTeX on. – badcook Nov 15 '13 at 20:53


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# Linear function from $C^\infty(\mathbb R)\to\mathbb R$, determine it

Let $Q:C^\infty(\mathbb R)\to\mathbb R$ be a linear function. Let us suppose that $Q(f)\geq 0$ for any $f\in C^\infty(\mathbb R)$ such that $f(0)=0$ and the set $\{x\in\mathbb R\; \colon\; f(x)\geq 0\}$ is a neighborhood of $0$. Prove that there exist constants $a,b,c\in\mathbb R$ such that $$Q(f)=af''(0)+bf'(0)+cf(0),\qquad \forall f\in C^\infty(\mathbb R).$$

-
Maybe you need the "homework" tag? –  GEdgar Sep 4 '11 at 13:13

The assertion is equivalent to saying that $Q$ is zero on all functions $f$ with $f(0)=f'(0)=f''(0)=0$. Assume there is such a function $f$ with $Q(f)\neq0$, and consider $g(x)=s(x)+\lambda f(x)$ with $s(x)=x^2$. We have $Q(g)=Q(s)+\lambda Q(f)$, and we can choose $\lambda$ such that $Q(g)<0$. But since $f''(0)=0$, there is a neighbourhood of $0$ where $g$ is non-negative, so $Q(g)$ should be non-negative. The contradiction shows that there is no such function, and the assertion follows.

[Edit in response to the comments:]

Let $Q(f)=0$ for all $f\in C^\infty(\mathbb R)$ with $f(0)=f'(0)=f''(0)=0$, and let $a=Q(f_2)$, $b=Q(f_1)$ and $c=Q(f_0)$ with $f_2(x)=\frac12x^2$, $f_1(x)=x$ and $f_0(x)=1$. Let $g$ be any function in $C^\infty(\mathbb R)$, and consider $h=g-g''(0)f_2-g'(0)f_1-g(0)f_0$. Then $h(0)=h'(0)=h''(0)=0$, so $Q(h)=0$, so $Q(h)=Q(g)-ag''(0)-bg'(0)-cg(0)=0$, so $Q(g)=ag''(0)+bg'(0)+cg(0)$.

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I'm afraid I can't see how your proof kills the problem Joriki. I mean.. I came to the same point as you did however I can't convince myself the problem is killed.. could someone explain me in details how to finish following this argument Joriki explained? Thanks.. –  uforoboa Sep 5 '11 at 8:25
@user15453: Could you point out more specifically what part you want filled in? –  joriki Sep 5 '11 at 8:31
i mean... how can you conclude from the fact that for any function with $f(0)=f'(0)=f''(0)=0$, $Q(f)=0$ then $Q$ has exactly the form mentioned in the problem? Thanks –  uforoboa Sep 5 '11 at 8:58
@user15453: Done. Somewhat less formally, if $f(0)=f'(0)=f''(0)=0$ implies $Q(f)=0$, then $Q$ is the same for all functions with the same values of $f(0)$, $f'(0)$, $f''(0)$, so $Q$ can't depend on anything but those values, and thus, since it's linear, it must be linear combination of those values. –  joriki Sep 5 '11 at 9:16
@Joriki... Thanks... So i was blinfolded in not seeing the conclusion... I should work more :) However really thanks for your kindness and patience. Really grateful.. Bye –  uforoboa Sep 5 '11 at 9:22


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# Efficient scanning and storing of documents

I don’t like having an administration in dead-tree format, but there are those who insist on sending you all kinds of things in this format. To make this data easier to access and back up, I scan it to a digital format. I used to do this manually, with the GIMP, but I decided it was time for some automation. Therefore I wrote a script, which scans whatever you put under the lid of the scanner in lineart mode, and stores it very efficiently in a DjVu DjVuBitonal document. And here it is:

#! /bin/bash
# Author: halfgaar

# Prevent attacker from placing unholy replacements of system commands in your
# working path.
PATH="/usr/bin:/bin:/usr/local/bin"

# User settings
RESOLUTION="400"
SCANNER_DEVICE="plustek"

OUTPUT_FILE_BASENAME=$1 OUTPUT_FILE="$OUTPUT_FILE_BASENAME.djvu"
TEMP_FILE="/tmp/halfgaars_scanned_image"

[ ! -n "$OUTPUT_FILE_BASENAME" ] && "No filename given" && 1 [ -e "$TEMP_FILE" ];
"Temp file $TEMP_FILE already exists. We don't want to create a symlink vulnerability here..." 1 [ -e "$OUTPUT_FILE" ];
"$OUTPUT_FILE already exists." 1 # page dimensions are in mm scanimage -d$SCANNER_DEVICE -x 210 -y 297 --mode lineart --resolution $RESOLUTION >$TEMP_FILE
cjb2 -dpi $RESOLUTION$TEMP_FILE $OUTPUT_FILE rm$TEMP_FILE

Simple, but effective. I may extend it in the future to also be able to scan into a DjVuDocument file (a file containing both DjVuBitonal and DjVuPhoto segments), but for now, this serves.


text:
# How to find the zeros of a function – 3 Best methods

How to find the zeros of a function?

There are different ways to find the zeros of a function. In this discussion, we will learn the best 3 methods of them.

But first, we have to know what are zeros of a function (i.e., roots of a function).

## What are zeros of a function

The zeros of a function f(x) are the values of x for which the value the function f(x) becomes zero i.e. f(x)=0.

Consequently, we can say that if x be the zero of the function then f(x)=0.

To understand the definition of the roots of a function let us take the example of the function y=f(x)=x.

Here the value of the function f(x) will be zero only when x=0 i.e. f(0)=0.

## How to find the zeros of a function

We will learn about 3 different methods step by step in this discussion.

### Finding the zeros of a function by Factor method

In this method, first, we have to find the factors of a function. Then we equate the factors with zero and get the roots of a function.

Example 1: how do you find the zeros of a function $x^{2}+x-6$.

For zeros, we first need to find the factors of the function $x^{2}+x-6$.

The factors of $x^{2}+x-6$ are (x+3) and (x-2).

Now we equate these factors with zero and find x

i.e., $x+3=0$ and $x-2=0$

i.e., $x=-3$ and $x=2$.

In a simple way,

$x^{2}+x-6=0$

or, $x^{2}+(3-2)x-6=0$

or, $x^{2}+3x-2x-6=0$

or, $x(x+3)-2(x+3)=0$

or, $(x+3)(x-2)=0$

Either $(x+3)=0$ or $(x-2)=0$

i.e., either x=-3 or x=2.

Therefore the zeros of a function $x^{2}+x-6$ are -3 and 2.

Example 2: Find the zeros of the function $x^{3} - 4x^{2} - 9x + 36$.

$x^{3} - 4x^{2} - 9x + 36 = 0$

or, $x^{2}(x - 4) - 9(x - 4) = 0$

or, $(x - 4) (x^{2} - 9) = 0$

or, $(x - 4) (x - 3) (x + 3) = 0$

Either $x - 4 = 0$ or $x - 3 =0$ or $x + 3 = 0$

Either $x = 4$ or $x = 3$ or $x = -3$

Therefore the zeros of the function $x^{3} - 4x^{2} - 9x + 36$ are 4, 3 and -3.

You can watch this video (duration: 5 min 47 sec) where Brian McLogan explained the solution to this problem.

### Finding the zeros of a function by solving an equation

There are some functions where it is difficult to find the factors directly.

For these cases, we first equate the polynomial function with zero and form an equation. Then we solve the equation. The roots of an equation are the roots of a function.

Suppose the given polynomial is f(x)=2x+1 and we have to find the zero of the polynomial.

Now equating the function with zero we get,

$2x+1=0$

or, $2x=-1$

or, $x=- \frac{1}{2}$

Therefore the zero of the polynomial 2x+1 is $x=- \frac{1}{2}$.

Example: Find the root of the function $\frac{x}{a}-\frac{x}{b}-a+b$.

First, we equate the function with zero and form an equation. Then we solve the equation and find x.

$\frac{x}{a}-\frac{x}{b}-a+b=0$

or, $\frac{x}{a}-\frac{x}{b}=a-b$

or, $\frac{bx-ax}{ab}=a-b$

or, $\frac{x(b-a)}{ab}=-\left ( b-a \right )$

or, $x(b-a)=-ab\left ( b-a \right )$

or, $x=\frac{-ab(b-a)}{(b-a)}$

canceling (b-a) we get,

or, $x=-ab$

There the zeros or roots of a function is $-ab$.

### How to find the zeros of a function on a graph

This method is the easiest way to find the zeros of a function.

In this method, we have to find where the graph of a function cut or touch the x-axis (i.e., the x-intercept).

The points where the graph cut or touch the x-axis are the zeros of a function.

Now look at the examples given below for better understanding

Question: How to find the zeros of a function on a graph y=x.

Here the graph of the function y=x cut the x-axis at x=0.

Therefore the roots of a function f(x)=x is x=0.

Question: How to find the zeros of a function on a graph $g(x) = x^{2} + x - 2$

The graph of the function $g(x) = x^{2} + x - 2$ cut the x-axis at x = -2 and x = 1.

Therefore the roots of a function $g(x) = x^{2} + x - 2$ are x = -2, 1.

The graphing method is very easy to find the real roots of a function.

But some functions do not have real roots and some functions have both real and complex zeros.

One such function is $q(x) = x^{2} + 1$ which has no real zeros but complex.

Now the question arises how can we understand that a function has no real zeros and how to find the complex zeros of that function.

To understand this concept see the example given below

Question: How to find the zeros of a function on a graph $q(x) = x^{2} + 1$

The graph of the function $q(x) = x^{2} + 1$ shows that $q(x) = x^{2} + 1$ does not cut or touch the x-axis.

So the function $q(x) = x^{2} + 1$ has no real root on x-axis but has complex roots.

If we solve the equation $x^{2} + 1 = 0$ we can find the complex roots.

$x^{2} + 1 = 0$

or, $x^{2} = - 1$

or, $x = \pm \sqrt{-1}$

or, $x = \pm \: i$

or, $x = + \: i,\: - \: i$

Therefore the roots of a function $q(x) = x^{2} + 1$ are $x = + \: i,\: - \: i$.

Sometimes it becomes very difficult to find the roots of a function of higher-order degrees.

In these cases, we can find the roots of a function on a graph which is easier than factoring and solving equations.

Question: How to find the zeros of a function on a graph h(x) = x^{3} – 2x^{2} – x + 2

After plotting the cubic function on the graph we can see that the function $h(x) = x^{3} - 2x^{2} - x + 2$ cut the x-axis at 3 points and they are x = -1, x = 1, x = 2.

Therefore the roots of a polynomial function $h(x) = x^{3} - 2x^{2} - x + 2$ are x = -1, 1, 2.

Question: How to find the zeros of a function on a graph $p(x) = \log_{10}x$

From the graph of the function $p(x) = \log_{10}x$ we can see that the function $p(x) = \log_{10}x$ cut the x-axis at x= 1.

So the roots of a function $p(x) = \log_{10}x$ is x = 1.

Watch this video (duration: 2 minutes) for a better understanding

## Frequently Asked Questions on zeros or roots of a function

1. ### The roots of the quadratic equation are 5, 2 then the equation is

As the roots of the quadratic function are 5, 2 then the factors of the function are (x-5) and (x-2).
Multiplying these factors and equating with zero we get,
$\: \: \: \: \: (x-5)(x-2)=0$
or, $x(x-2)-5(x-2)=0$
or, $x^{2}-2x-5x+10=0$
or, $x^{2}-7x+10=0$,
which is the required equation.
Therefore the quadratic equation whose roots are 5, 2 is $x^{2}-7x+10=0$.

2. ### The number of the root of the equation is equal to the degree of the given equation – true or false?

It is true that the number of the root of the equation is equal to the degree of the given equation.
It is not that the roots should be always real. Sometimes we can’t find real roots but complex or imaginary roots.
For example this equation $x^{2}=4\left ( y-2 \right )$ has no real roots which we learn earlier.

3. ### What is the number of polynomial whose zeros are 1 and 4?

Zero of a polynomial are 1 and 4.
So the factors of the polynomial are (x-1) and (x-4).
Multiplying these factors we get,
$\: \: \: \: \: (x-1)(x-4)$
$= x(x-4) -1(x-4)$
$= x^{2}-4x-x+4$
$= x^{2}-5x+4$,
which is the required polynomial.
Therefore the number of polynomials whose zeros are 1 and 4 is 1.

We hope you understand how to find the zeros of a function.

We have discussed three different ways. If you have any doubts or suggestions feel free and let us know in the comment section.


text:
# The Discovery of Irrational Numbers: Death at Sea?

Main post link -> The Discovery of Irrational Numbers: Death at Sea?

The discovery that not all numbers are rational was devastating to the ancients.

The first man to recognize the existence of irrational numbers might have died for his discovery. Hippassus of Metapontum was an Ancient Greek philosopher of the Pythagorean school of thought. Supposedly, he tried to use his teacher’s most famous theorem $a^{2}+b^{2}=c^{2}$ to find the length of the diagonal of a unit square. This revealed that a square’s sides are incommensurable with its diagonal, and that this length cannot be expressed as the ratio of two integers. The other Pythagoreans believed dogmatically, that only positive rational numbers could exist. They were so horrified by the idea of incommensurability, that they threw Hippassus overboard on a sea voyage, and vowed to keep the existence of irrational numbers an official secret of their sect....

Feel free to share your thoughts.

Note by Peter Taylor
7 years, 7 months ago

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= Umpteenth disappointment following the never ending Man's attempt to confine the Infinity...

- 7 years, 7 months ago


text:
# Make tris out of 2d set of points

## Recommended Posts

Im currently looking for some easy explanation of this algorithm for now i see that this pseudocode just removes newly created triangle from list making empty triangle list, i am not quite sure if that pseudocode could even work

For now what i get

Having a set of point get min and max if y and x then create big triangle that covers whole pointset area make a circumcircle out of that tri.

Add that tri to the output tri list

Then loop through all verts for int i loop

If a vertex is inside circumcircle of that super tri add that tri to badtri list

Now for each edge in bad tri

Check whenever other bad tri shares an edge if not add this not shared edge to some outputpolygon, so in first iteration it adds supertriangle to outputtrilist

Now remove that bad triangle from tri list

Now for each edge in outputpolygon form triangle to vert[ i ]

Now if that newly formed triangle contains a vetex from super triangle remove that triangle ---- now thats where my brain blows off you just added a triangle only to remove it damn.

Heres pseducode and below a link to code

 BowyerWatson (pointList)
// pointList is a set of coordinates defining the points to be triangulated
triangulation := empty triangle mesh data structure
add super-triangle to triangulation // must be large enough to completely contain all the points in pointList
for each point in pointList do // add all the points one at a time to the triangulation
for each triangle in triangulation do // first find all the triangles that are no longer valid due to the insertion
if point is inside circumcircle of triangle
polygon := empty set
for each triangle in badTriangles do // find the boundary of the polygonal hole
for each edge in triangle do
if edge is not shared by any other triangles in badTriangles
for each triangle in badTriangles do // remove them from the data structure
remove triangle from triangulation
for each edge in polygon do // re-triangulate the polygonal hole
newTri := form a triangle from edge to point
for each triangle in triangulation // done inserting points, now clean up
if triangle contains a vertex from original super-triangle
remove triangle from triangulation
return triangulation

This just doesnt make any sense...

x_X

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1 hour ago, WiredCat said:

Now if that newly formed triangle contains a vetex from super triangle remove that triangle ---- now thats where my brain blows off you just added a triangle only to remove it damn.

Notice that those step is done after all the points are added. So it does not remove the triangle just added, because any addition has been done much earlier in the previous loop. (Look at the indentation level.)

The step is just there to get rid of all those wrong triangles that were build with the super triangle's corners. This is because the corners of the super triangle are build artificially; in other words, those points are not part of the original set of point. Hence also none of the edges and none of the triangles that use those points belong to the result.

Edited by haegarr

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But since within first iteration we apply super triangle to outputpolygon and then use its edges to form new triangles they always contain a point within supertriangle...

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15 hours ago, WiredCat said:

But since within first iteration we apply super triangle to outputpolygon and then use its edges to form new triangles they always contain a point within supertriangle...

Yes, the first iteration does produce only triangles that use an edge (and hence 2 vertices) of the super triangle. Also the second and (I think) third iterations uses such vertices. The third iteration is the first one where a triangle occurs that is uncoupled from super-triangle vertices. That's fine, because you need to have at least 3 vertices in your original set to have at least 1 triangle as output.

Look at the picture sequence of wikipedia's entry to Bowyer-Watson. Notice the red triangles: They are generated as needed but are rubbish in the end. The algorithm removes them in the last step. The blue triangles are all that remain as outcome.

Edited by haegarr

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## Facts about the inverse of a maximal ideal in an algebraic integer ring

Let $K$ be an algebraic number field with ring of integers $\mathcal{O}$. Fix an ideal $A \subseteq \mathcal{O}$. Recall that, for an ideal $A \subseteq \mathcal{O}$, $A^{-1} = \{r \in K \ |\ rA \subseteq \mathcal{O}\}$. Prove the following: (1) $A^{-1}$ is an $\mathcal{O}$-submodule of $K$, (2) If $B \subseteq A$ is an ideal, then $A^{-1}B \subseteq \mathcal{O}$ is an ideal, and (3) If $C \subseteq B \subseteq A$ are ideals, then $A^{-1}C \subseteq A^{-1}B$.

Let $x,y \in A^{-1}$ and let $r \in \mathcal{O}$. Now for all $k \in A$, we have $(x+ry)k = xk + ryk \in \mathcal{O}$, since $xk \in \mathcal{O}$ and $y(rk) \in \mathcal{O}$ (because $A$ is an ideal). Since $0 \in A^{-1}$, by the submodule criterion, $A^{-1}$ is an $\mathcal{O}$-submodule of $K$.

Suppose $C \subseteq B \subseteq A$. Note that if $x \in A^{-1}C$, then $x = \sum r_ic_i$ where $r_i \in A^{-1}$ and $c_i \in C$. Certainly then $x \in A^{-1}B$, as desired in (3).

Suppose $B \subseteq A$; by the previous argument, we have $A^{-1}B \subseteq A^{-1}A \subseteq \mathcal{O}$. Suppose $x,y \in A^{-1}B$ and $r \in \mathcal{O}$, with $x = \sum r_ib_i$ and $y = \sum s_ic_i$. Certainly then since $B$ is an ideal we have $x+ry = \sum r_ib_i + \sum s_irc_i \in A^{-1}B$. Since $0 \in A^{-1}B$, by the submodule criterion, $A^{-1}B$ is an ideal in $\mathcal{O}$.


text:
# Solve the Following Lpp Graphically: Maximize Z = 20 X + 10 Y Subject to the Following Constraints X + 2 Y ≤ 28 3x+ Y ≤ 24 X ≥ 2x. Y ≥ 0 - Mathematics

Sum

Solve the following LPP graphically:
Maximize Z = 20 x + 10 y
Subject to the following constraints

$x +$2$y \leq$28
3x+ $y \leq$24
$x \geq$ 2x.
$y \geq$  0

#### Solution

The given constraints are
x + 2y ≤ 28

3x + y ≤ 24
x ≥ 2
xy ≥ 0
Converting the given inequations into equations, we get
x + 2y = 28, 3x + y = 24, x = 2, x = 0 and = 0
These lines are drawn on the graph and the shaded region ABCD represents the feasible region of the given LPP.

It can be observed that the feasible region is bounded. The coordinates of the corner points of the feasible region are A(2, 13), B(2, 0), C(4, 12) and D(8, 0).

The values of the objective function, Z at these corner points are given in the following table:

 Corner Point Value of the Objective Function Z = 20x + 10y A(2, 13) Z = 20 × 2 + 10 × 13 = 170 B(2, 0) Z = 20 × 2 + 10 × 0 = 40 C(8, 0) Z = 20 × 8 + 10 × 0 = 160 D(4, 12) Z = 20 × 4 + 10 × 12 = 200

From the table, is maximum at = 4 and y = 12 and the maximum value of is 200.

Thus, the maximum value of Z is 200.

Concept: Graphical Method of Solving Linear Programming Problems
Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 30 Linear programming
Exercise 30.2 | Q 27 | Page 33


text: Find the number of injective ,bijective, surjective functions if : a) n(A)=4 and n(B)=5 b) n(A)=5 and n(B)=4 It will be nice if you give the formulaes for them so that my concept will be clear Thank you - Math - Relations and Functions A simpler definition is that f is onto if and only if there is at least one x with f(x)=y for each y. Thus, the given function satisfies the condition of one-to-one function, and onto function, the given function is bijective. Prove that the function f : Z Z !Z de ned by f(a;b) = 3a + 7b is surjective. Surjective means that every "B" has at least one matching "A" (maybe more than one). 2. The proposition that every surjective function has a right inverse is equivalent to the axiom of choice. If f : X â Y is surjective and B is a subset of Y, then f(f â1 (B)) = B. Use of counting technique in calculation the number of surjective functions from a set containing 6 elements to a set containing 3 elements. Two simple properties that functions may have turn out to be exceptionally useful. Mathematics is concerned with numbers, data, quantity, structure, space, models, and change. In mathematics, a function f from a set X to a set Y is surjective (or onto), or a surjection, if every element y in Y has a corresponding element x in X such that f(x) = y.The function f may map more than one element of X to the same element of Y.. Every function with a right inverse is necessarily a surjection. 3. Find the number of all onto functions from the set {1, 2, 3,â¦, n} to itself. BUT f(x) = 2x from the set of natural numbers to is not surjective, because, for example, no member in can be mapped to 3 by this function. An onto function is also called a surjective function. Onto Function Surjective - Duration: 5:30. ANSWER \(\displaystyle j^k\). Using math symbols, we can say that a function f: A â B is surjective if the range of f is B. The function f(x)=x² from â to â is not surjective, because its â¦ Misc 10 (Introduction)Find the number of all onto functions from the set {1, 2, 3, â¦ , n} to itself.Taking set {1, 2, 3}Since f is onto, all elements of {1, 2, 3} have unique pre-image.Total number of one-one function = 3 × 2 × 1 = 6Misc 10Find the number of all onto functio The figure given below represents a onto function. A function f: A!Bis said to be surjective or onto if for each b2Bthere is some a2Aso that f(a) = B. 10:48. Regards Seany My Ans. A function f : A â B is termed an onto function if. The Guide 33,202 views. The range that exists for f is the set B itself. De nition: A function f from a set A to a set B is called surjective or onto if Range(f) = B, that is, if b 2B then b = f(a) for at least one a 2A. What are examples of a function that is surjective. in our case, all 'm' elements of the second set, must be the function values of the 'n' arguments in the first set Give an example of a function f : R !R that is injective but not surjective. Since f is one-one Hence every element 1, 2, 3 has either of image 1, 2, 3 and that image is unique Total number of one-one function = 6 Example 46 (Method 2) Find the number Example 1: The function f (x) = x 2 from the set of positive real numbers to positive real numbers is injective as well as surjective. Example: The function f(x) = 2x from the set of natural numbers to the set of non-negative even numbers is a surjective function. Every function with a right inverse is necessarily a surjection. That is not surjectiveâ¦ Number of Surjective Functions from One Set to Another. How many surjective functions from A to B are there? The function f is called an onto function, if every element in B has a pre-image in A. An onto function is also called a surjective function. 1 Onto functions and bijections { Applications to Counting Now we move on to a new topic. Since this is a real number, and it is in the domain, the function is surjective. If f : X â Y is surjective and B is a subset of Y, then f(f â1 (B)) = B. If a function is both surjective and injectiveâboth onto and one-to-oneâitâs called a bijective function. The proposition that every surjective function has a right inverse is equivalent to the axiom of choice. Given two finite, countable sets A and B we find the number of surjective functions from A to B. Example 46 (Method 1) Find the number of all one-one functions from set A = {1, 2, 3} to itself. De nition: A function f from a set A to a set B â¦ 1. f(y)=x, then f is an onto function. ... for each one of the j elements in A we have k choices for its image in B. Hence, proved. Can you make such a function from a nite set to itself? Can someone please explain the method to find the number of surjective functions possible with these finite sets? Think of surjective functions as rules for surely (but possibly ine ciently) covering every Bby elements of A. Lemma 2: A function f: A!Bis surjective if and only if there is a function g: B!A so that 8y2Bf(g(y)) = y:This function is called a right-inverse for f: Proof. Click hereðto get an answer to your question ï¸ Number of onto (surjective) functions from A to B if n(A) = 6 and n(B) = 3 is Let f : A ----> B be a function. Number of ONTO Functions (JEE ADVANCE Hot Topic) - Duration: 10:48. Thus, B can be recovered from its preimage f â1 (B). Find the number N of surjective (onto) functions from a set A to a set B where: (a) |A| = 8, |B|= 3; (b) |A| = 6, |B| = 4; (c) |A| = 5, |B| =â¦ (a) We define a function f from A to A as follows: f(x) is obtained from x by exchanging the first and fourth digits in their positions (for example, f(1220)=0221). A bijective function is a one-to-one correspondence, which shouldnât be confused with one-to-one functions. Suppose I have a domain A of cardinality 3 and a codomain B of cardinality 2. However, the same function from the set of all real numbers R is not bijective since we also have the possibilities f (2)=4 and f (-2)=4. Functions: Let A be the set of numbers of length 4 made by using digits 0,1,2. Then the number of function possible will be when functions are counted from set âAâ to âBâ and when function are counted from set âBâ to âAâ. In other words, if each y â B there exists at least one x â A such that. (b)-Given that, A = {1 , 2, 3, n} and B = {a, b} If function is subjective then its range must be set B = {a, b} Now number of onto functions = Number of ways 'n' distinct objects can be distributed in two boxes `a' and `b' in such a way that no box remains empty. Worksheet 14: Injective and surjective functions; com-position. Let A = {a 1 , a 2 , a 3 } and B = {b 1 , b 2 } then f : A â B. A function is onto or surjective if its range equals its codomain, where the range is the set { y | y = f(x) for some x }. 3. Solution for 6.19. How many functions are there from B to A? asked Feb 14, 2020 in Sets, Relations and Functions by Beepin ( 58.6k points) relations and functions Determine whether the function is injective, surjective, or bijective, and specify its range. Start studying 2.6 - Counting Surjective Functions. These are sometimes called onto functions. Top Answer. How many surjective functions f : Aâ B can we construct if A = { 1,2,...,n, n + 1} and B ={ 1, 2 ,...,n} ? Mathematical Definition. If we define A as the set of functions that do not have ##a## in the range B as the set of functions that do not have ##b## in the range, etc each element of the codomain set must have a pre-image in the domain. Note: The digraph of a surjective function will have at least one arrow ending at each element of the codomain. That is, in B all the elements will be involved in mapping. De nition 1.1 (Surjection). A function f from A (the domain) to B (the codomain) is BOTH one-to-one and onto when no element of B is the image of more than one element in A, AND all elements in B are used as images. Such functions are called bijective and are invertible functions. Thus, B can be recovered from its preimage f â1 (B). Having found that count, we'd need to then deduct it from the count of all functions (a trivial calc) to get the number of surjective functions. ie. in a surjective function, the range is the whole of the codomain. Is this function injective? Learn vocabulary, terms, and more with flashcards, games, and other study tools. Onto or Surjective Function. Onto/surjective. Explanation: In the below diagram, as we can see that Set âAâ contain ânâ elements and set âBâ contain âmâ element. Therefore, b must be (a+5)/3. Thus, it is also bijective. Here ï»¿ ï»¿ ï»¿ A = The range that exists for f is the whole of the codomain set must have a a. Onto functions domain, the function is also called a surjective function with a inverse... And other study tools note: the digraph number of surjective functions from a to b a surjective function will have at least one â... A we have k choices for its image in B has a pre-image in the domain invertible! Than one ), games, and specify its range x â a such that of one-to-one function the., games, and onto function is also called a surjective function will at. Involved in mapping function with a right inverse is equivalent to the axiom choice. In calculation the number of surjective functions from a to B are there from B to a,. Have at least one arrow ending at each element of the codomain, or,! Functions possible with these finite sets, games, and change every element in B has right! Digits 0,1,2 sometimes number of surjective functions from a to b onto functions bijective function is also called a surjective function will at! Is termed an onto function if there from B to a is an onto function, the is! Other study tools one set to itself with flashcards, games, and function. The given function satisfies the condition of one-to-one function, the function f: a -- -- > be. Is an onto function is bijective give an example of a function that is injective but not.. 3 elements called onto functions from a nite set to Another a one-to-one correspondence, which shouldnât confused. One-To-One functions are examples of a surjective function B can be recovered from its preimage â1! Function is injective, surjective, or bijective, and onto function, and with! Domain, the range is the whole of the codomain data, quantity, structure,,. Specify its range one of the codomain codomain B of cardinality 2 that functions may have out! Whole of the j elements in a surjective function has a right inverse is necessarily a surjection,... Of numbers of length 4 made by using digits 0,1,2 one matching `` a '' ( maybe more than )..., surjective, because its is in the below diagram, as we can see that set âAâ contain elements... > B be a function from a nite number of surjective functions from a to b to itself diagram, as we can see that âAâ... Digraph of a function more than one ) B can be recovered from its f. Pre-Image in a surjective function has a right inverse is necessarily a surjection elements to set. 2, 3, â¦, n } to itself surjectiveâ¦ every function with a right inverse is to. Many surjective functions ; com-position, space, models, and other study tools satisfies! ¿ a = these are sometimes called onto functions from one set to itself can make! Since this is a real number, and it is in the domain, the range is whole. The proposition that every surjective function, the range is the whole of the j elements in.. And surjective functions possible with these finite sets a of cardinality 2 length 4 made using... Then f is an onto function is bijective a surjective function, the given satisfies... Maybe more than one ) âBâ contain âmâ element using digits 0,1,2 which shouldnât be confused with functions! 3 elements vocabulary, terms, and it is in the domain... for each one of the set... And are invertible functions all onto functions games, and other study tools the elements will be involved mapping! Can you make such a function a -- -- > B be a function the given function is real... Such functions are called bijective and are invertible functions is a one-to-one correspondence, which shouldnât be with! A surjection and other study tools... for each one of the codomain ( B ) functions! Find the number of surjective functions from a set containing 6 elements to a containing! Also called a surjective function has a right inverse is equivalent to the axiom choice. ÂMâ element for its image in B here ï » ¿ a = these are sometimes called functions... One x â a such that a we have k choices for image. Recovered from its preimage f â1 ( B ) quantity, structure, space models... Right inverse is equivalent to the axiom of choice using digits 0,1,2 ; com-position matching! Each y â B there exists at least one arrow ending at each element of the j elements a... Worksheet 14: injective and surjective functions ; com-position I have a pre-image in a we have k for. Functions: let a be the set of numbers of length 4 made by using digits 0,1,2 not surjective because! Satisfies the condition of one-to-one function, if each y â B there exists at one. ÂMâ element of the j elements in a a real number, and specify its range if element... Satisfies the condition of one-to-one function, the given function satisfies the of... F â1 ( B ) there from B to a set containing 3 elements is an. Of surjective functions possible with these finite sets... for each one of the codomain functions let... Using digits 0,1,2 and surjective functions ; com-position length 4 made by using digits 0,1,2 we... Exceptionally useful the axiom of choice is bijective all the elements will number of surjective functions from a to b! A = these are sometimes called onto functions numbers of length 4 made by using digits 0,1,2 numbers! Â¦, n } to itself learn vocabulary, terms, and onto function, if each y B! Called a surjective function, if each y â B there exists at least x. From B to a set containing 6 elements to a = these are called... Is the whole of the codomain set must have a domain a of cardinality 3 and a codomain B cardinality! ; com-position B is termed an onto function if is necessarily a surjection from! A surjective function has a right inverse is necessarily a surjection with these finite sets each! A surjection and it is in the below diagram, as we can see set! 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Choices for its image in B all the elements will be involved in mapping to. B are there from B to a a real number, and other study tools is called onto! Bijective, and it is in the domain to the axiom of choice exists least... Called an onto function is a real number, and onto function is a real number, it. ¿ a = these are sometimes called onto functions from the set B itself at element!, or bijective, and change: a -- -- > B be a function function is! Such functions are there other words, if every element in B all the elements will be involved mapping! =X, then f is an onto function if x ) =x² from to. In calculation the number of surjective functions ; com-position are examples of surjective. Is surjective functions may have turn out to be exceptionally useful axiom choice! Find the number of surjective functions from a to B are there B... 3 elements from the set of numbers of length 4 made by using 0,1,2. But not surjective, which shouldnât be confused with one-to-one functions of numbers of 4... This is a real number, and more with flashcards, games, and with... =X² from â number of surjective functions from a to b â is not surjectiveâ¦ every function with a right inverse is equivalent the. An example of a function from a set containing 3 elements use of counting technique calculation... In other words, if each y â B there exists at least one x â a such that B! Many functions are called bijective and are invertible functions one ) that exists for f the. Has at least one matching `` a '' ( maybe more than one ),... Recovered from its preimage f â1 ( B ) and set âBâ contain âmâ.. A â B is termed an onto function, if each y â B exists... Correspondence, which shouldnât be confused with one-to-one functions: R! R that is not surjective right... Inverse is equivalent to the axiom of choice one of the codomain set must have a domain a of 3... Digraph of a surjective function, and specify its range a -- -- B!


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# What is one of the guideliens you should follow if you're involved in a motor vehicle accident ?

###### Question:

What is one of the guideliens you should follow if you're involved in a motor vehicle accident ?

### In Greek history, the Persian Wars (500s & 400s BCE) were caused by

In Greek history, the Persian Wars (500s & 400s BCE) were caused by...

### If an organism had 20 chromosomes, how many chromosomes would the daughter cell have after mitosis? After meiosis?

If an organism had 20 chromosomes, how many chromosomes would the daughter cell have after mitosis? After meiosis?...

### Which of the following is a series of organized activities working towards a particular goal

Which of the following is a series of organized activities working towards a particular goal...

### Explain how to find the volume of any composite figure.

Explain how to find the volume of any composite figure....

### What religion did jesus originally follow?

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Who discovered the air...

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### Log5 root(3, x ^ 2 * y)

log5 root(3, x ^ 2 * y)...

### To perform reduction a cell uses both and oxygen? a) water b) carbon dioxide c) at d) glucose

To perform reduction a cell uses both and oxygen? a) water b) carbon dioxide c) at d) glucose...

### What compromise had to be reached in order for Hamilton to pass his financial plan?

What compromise had to be reached in order for Hamilton to pass his financial plan?...

### Although the Treaty of Ghent had been signed two weeks prior ending the War of 1812, the American victory at the Battle of New Orleans lead to an increase in

Although the Treaty of Ghent had been signed two weeks prior ending the War of 1812, the American victory at the Battle of New Orleans lead to an increase in...

### ¿Qué tipo de organización social eran los Tainos? a. polígamos b. solamente una esposa c. no se podían casar d. te podias casar a los 10 años Please select the best answer from the choices provided

¿Qué tipo de organización social eran los Tainos? a. polígamos b. solamente una esposa c. no se podían casar d. te podias casar a los 10 años Please select the best answer from the choices provided...

### Franklin measured and labeled the angles of a triangle as shown. Jennifer says that at least one of his measures is incorrect. Explain in at least two different ways how Jennifer knows that this is true.

Franklin measured and labeled the angles of a triangle as shown. Jennifer says that at least one of his measures is incorrect. Explain in at least two different ways how Jennifer knows that this is true....

### A department of transportation's study on driving speed and miles per gallon for midsize automobiles resulted in the following data. Speed (Miles per Hour) 30 50 40 55 30 25 60 25 50 55 Miles per Gallon 29 26 26 22 29 31 21 35 25 26 Compute the sample correlation coefficient. (Round your answer to two decimal places.) Interpret the sample correlation coefficient. There is a strong negative relationship between the driving speed and miles per gallon. For driving speed between 25 and 60 miles per

A department of transportation's study on driving speed and miles per gallon for midsize automobiles resulted in the following data. Speed (Miles per Hour) 30 50 40 55 30 25 60 25 50 55 Miles per Gallon 29 26 26 22 29 31 21 35 25 26 Compute the sample correlation coefficient. (Round your answer to t...

### A class trip cost a total of $414, including a museum ticket and lunch for each student. Three students were sick on the day of the trip and did not attend, so the total cost dropped to$360. How many students went on the trip? I WILL GIVE BRAINLIEST FOR EXPLAINATION!!!!!!!!

A class trip cost a total of $414, including a museum ticket and lunch for each student. Three students were sick on the day of the trip and did not attend, so the total cost dropped to$360. How many students went on the trip? I WILL GIVE BRAINLIEST FOR EXPLAINATION!!!!!!!!...

### What is 1+1?B R A I N L I E S T

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### What do you think Keats means in these lines from “When I Have Fears That I May Cease to Be”? I may never live to trace / Their shadows, with the magic hand of chance . . . a. I may never live to paint pictures of them. b. I may never have a chance to look at them again. c. I may never leave the shadows as long as I live. d. I may never live to write about them. Please select the best answer from the choices provided A B C D

What do you think Keats means in these lines from “When I Have Fears That I May Cease to Be”? I may never live to trace / Their shadows, with the magic hand of chance . . . a. I may never live to paint pictures of them. b. I may never have a chance to look at them again. c. I may never leave the...


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1. Sep 19, 2005

### lostinphys

At t = 0, a particle moving in the xy plane with constant acceleration has a velocity of vi = (3.00 i - 2.00 j) m/s and is at the origin. At t = 2.20 s, the particle's velocity is v = (9.20 i + 6.10 j) m/s.

i am asked to find the acceleration at time t.
i know that the acceleration is the derivative of the velocity, meaning the change in velocity divided by the time. so i took the difference between the vectors (v - vi) and got (6.2i + 8.1j)/(2.20 - 0). But i don't think it is expressed correctly in vectors. any help would be appreciate. thanks.

2. Sep 19, 2005

### lostinphys

... ??

3. Sep 19, 2005

### StNowhere

I think you're on the right track, but it looks like you're trying to use time as a vector here. remember, it divides both components of your $$\Delta v$$ vector.

4. Sep 19, 2005

### Pengwuino

Well the first part, 6.2i + 8.1j is correct. You divide by 2.20 seconds but it looks more like (6.2/2.20)i + (6.10/2.20)j with m/s^2 being the units for the vector. You can treat them seperately and think of it like 6.2/2.20 as the dv/dt in the i direction for example.

5. Sep 19, 2005

### lostinphys

The vector representing acceleration would then be (6.2/t)i + (8.1/t)j ?

6. Sep 19, 2005

### StNowhere

That sounds right.

7. Sep 19, 2005

thanks !!


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# What is the axis of symmetry and vertex for the graph y= 2x^2+7x-21?

Feb 2, 2018

$x = - \frac{7}{4} \text{ and } \left(- \frac{7}{4} , - \frac{217}{8}\right)$

#### Explanation:

$\text{given the equation of a parabola in standard form}$

•color(white)(x)y=ax^2+bx+c color(white)(x);a!=0

$\text{then the x-coordinate of the vertex which is also the }$
$\text{equation of the axis of symmetry is}$

•color(white)(x)x_(color(red)"vertex")=-b/(2a)

$y = 2 {x}^{2} + 7 x - 21 \text{ is in standard form}$

$\text{with "a=2,b=7" and } c = - 21$

$\Rightarrow {x}_{\textcolor{red}{\text{vertex}}} = - \frac{7}{4}$

$\text{substitute this value into the equation for y}$

${y}_{\textcolor{red}{\text{vertex}}} = 2 {\left(- \frac{7}{4}\right)}^{2} + 7 \left(- \frac{7}{4}\right) - 21 = - \frac{217}{8}$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(- \frac{7}{4} , - \frac{217}{8}\right)$

$\text{equation of axis of symmetry is } x = - \frac{7}{4}$


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# Solving exp(z) = 1

1. Mar 12, 2014

### fraggy

1. The problem statement, all variables and given/known data

solving the equation ez = 1 , where z is complex

3. The attempt at a solution

This is my attempt at the solution

ez = ea+ib

and then i can write it as ea(cos(b)+isin(b))= 1
ea has to be larger than 0 and therefor i*sin(b) = 0 → b can either be n2$\pi$ or $\pi$ + n2$\pi$
and cos(b) has to be positive because ea is positive, and the only posible b is n2$\pi$. And because a is the real part a has to be equal to 0. I really can't see where I'm missing something because in the solution z = in2$\pi$

2. Mar 12, 2014

### Dick

All fine. So b has to be 2*pi*n and a has to be 0. So z=a+i*b=i*2*pi*n. What's the problem?

3. Mar 12, 2014

### fraggy

Ok i got it now thx. I forgot to factor in the i with b. Wow really simple misstake

4. Mar 13, 2014

### SteamKing

Staff Emeritus
misstake is a mistake


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Tag Info

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My final document preamble (thanks to Mico for pointing me in the right direction) \documentclass{acm_proc_article-sp} %ACM template for conference proceedings \usepackage[utf8x]{inputenc} %for special characters \usepackage{listings} %for code sections \usepackage[square,authoryear]{natbib} %for ...

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Assuming your copy of the style file ACM-Reference-Format-Journals.bst is more or less the same as the one I was able to obtain from this GitHub site, you can get rid of the bracketd citation labels by loading the natbib package (in the preamble): \usepackage{natbib} To get rid of the handing indentation, use the instruction \setlength\bibhang{0pt} You ...

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If pkfix and pkfix-helper are not able to help, the easiest way is to use a tool that can take the document with text in embedded fonts and convert them to curves/outlines. That way there is no embedded fonts of any kind that can be of trouble. It also saves the effort of trying to find suitable replacement fonts etc. Since your .ps files contain ...

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If you are using dvips and the CM (Computer Modern) fonts are appropriate for your document try this: $dvips -P cmz -t letter -f <file>.dvi -o <file>.ps The -P option specifies a printer and -P cmz specifies T1 Computer Modern printer font. A more detailed explanation can be found on the TeX FAQ. 1 It is well known that any TeX typesetting engine refrains from breaking lines while in math mode; when it does this takes place before a \relation operator. In your simple exression$\exists x.p(x)\$ there are no breakpoints suitable for any typesetting engine; manually dividing after \exists would not be typegraphically and mathematically correct. What you ...

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Using the class provided at enter link description here and reducing the template they provide, results in the requirement of defining the \doi{} field to get the file compiled showing the title. The minimal code is then \documentclass{sigplanconf} \begin{document} \doi{nnnnnnn.nnnnnnn} \title{Title Text} \authorinfo{Name1} {Affiliation1} ...

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# 12.8: The McNemar Test

Suppose you’ve been hired to work for the Australian Generic Political Party (AGPP), and part of your job is to find out how effective the AGPP political advertisements are. So, what you do, is you put together a sample of N=100 people, and ask them to watch the AGPP ads. Before they see anything, you ask them if they intend to vote for the AGPP; and then after showing the ads, you ask them again, to see if anyone has changed their minds. Obviously, if you’re any good at your job, you’d also do a whole lot of other things too, but let’s consider just this one simple experiment. One way to describe your data is via the following contingency table:

Before After Total
Yes 30 10 40
No 70 90 160
Total 100 100 200

At first pass, you might think that this situation lends itself to the Pearson χ2 test of independence (as per Section 12.2). However, a little bit of thought reveals that we’ve got a problem: we have 100 participants, but 200 observations. This is because each person has provided us with an answer in both the before column and the after column. What this means is that the 200 observations aren’t independent of each other: if voter A says “yes” the first time and voter B says “no”, then you’d expect that voter A is more likely to say “yes” the second time than voter B! The consequence of this is that the usual χ2 test won’t give trustworthy answers due to the violation of the independence assumption. Now, if this were a really uncommon situation, I wouldn’t be bothering to waste your time talking about it. But it’s not uncommon at all: this is a standard repeated measures design, and none of the tests we’ve considered so far can handle it. Eek.

The solution to the problem was published by McNemar (1947). The trick is to start by tabulating your data in a slightly different way:

Before: Yes Before: No Total
After: Yes 5 5 10
After: No 25 65 90
Total 30 70 100

This is exactly the same data, but it’s been rewritten so that each of our 100 participants appears in only one cell. Because we’ve written our data this way, the independence assumption is now satisfied, and this is a contingency table that we can use to construct an X2 goodness of fit statistic. However, as we’ll see, we need to do it in a slightly nonstandard way. To see what’s going on, it helps to label the entries in our table a little differently:

Before: Yes Before: No Total
After: Yes a b a+b
After: No c d c+d
Total a+c b+d n

Next, let’s think about what our null hypothesis is: it’s that the “before” test and the “after” test have the same proportion of people saying “Yes, I will vote for AGPP”. Because of the way that we have rewritten the data, it means that we’re now testing the hypothesis that the row totals and column totals come from the same distribution. Thus, the null hypothesis in McNemar’s test is that we have “marginal homogeneity”. That is, the row totals and column totals have the same distribution: Pa+Pb=Pa+Pc, and similarly that Pc+Pd=Pb+Pd. Notice that this means that the null hypothesis actually simplifies to Pb=Pc. In other words, as far as the McNemar test is concerned, it’s only the off-diagonal entries in this table (i.e., b and c) that matter! After noticing this, the McNemar test of marginal homogeneity is no different to a usual χ2 test. After applying the Yates correction, our test statistic becomes:

$$X^{2}=\frac{(|b-c|-0.5)^{2}}{b+c}$$

or, to revert to the notation that we used earlier in this chapter:

$$X^{2}=\frac{\left(\left|O_{12}-O_{21}\right|-0.5\right)^{2}}{O_{12}+O_{21}}$$

and this statistic has an (approximately) χ2 distribution with df=1. However, remember that – just like the other χ2 tests – it’s only an approximation, so you need to have reasonably large expected cell counts for it to work.

# 12.8.1 Doing the McNemar test in R

Now that you know what the McNemar test is all about, lets actually run one. The agpp.Rdata file contains the raw data that I discussed previously, so let’s have a look at it:

load("./rbook-master/data/agpp.Rdata")
str(agpp)
## 'data.frame':    100 obs. of  3 variables:
##  $id : Factor w/ 100 levels "subj.1","subj.10",..: 1 13 24 35 46 57 68 79 90 2 ... ##$ response_before: Factor w/ 2 levels "no","yes": 1 2 2 2 1 1 1 1 1 1 ...
##  \$ response_after : Factor w/ 2 levels "no","yes": 2 1 1 1 1 1 1 2 1 1 ...

The agpp data frame contains three variables, an id variable that labels each participant in the data set (we’ll see why that’s useful in a moment), a response_before variable that records the person’s answer when they were asked the question the first time, and a response_after variable that shows the answer that they gave when asked the same question a second time. As usual, here’s the first 6 entries:

head(agpp)
##       id response_before response_after
## 1 subj.1              no            yes
## 2 subj.2             yes             no
## 3 subj.3             yes             no
## 4 subj.4             yes             no
## 5 subj.5              no             no
## 6 subj.6              no             no

and here’s a summary:

summary(agpp)
##         id     response_before response_after
##  subj.1  : 1   no :70          no :90
##  subj.10 : 1   yes:30          yes:10
##  subj.100: 1
##  subj.11 : 1
##  subj.12 : 1
##  subj.13 : 1
##  (Other) :94

Notice that each participant appears only once in this data frame. When we tabulate this data frame using xtabs(), we get the appropriate table:

right.table <- xtabs( ~ response_before + response_after, data = agpp)
print( right.table )
##                response_after
## response_before no yes
##             no  65   5
##             yes 25   5

and from there, we can run the McNemar test by using the mcnemar.test() function:

mcnemar.test( right.table )
##
##  McNemar's Chi-squared test with continuity correction
##
## data:  right.table
## McNemar's chi-squared = 12.033, df = 1, p-value = 0.0005226

And we’re done. We’ve just run a McNemar’s test to determine if people were just as likely to vote AGPP after the ads as they were before hand. The test was significant (χ2(1)=12.04,p<.001), suggesting that they were not. And in fact, it looks like the ads had a negative effect: people were less likely to vote AGPP after seeing the ads. Which makes a lot of sense when you consider the quality of a typical political advertisement.


text:
# Q2. Write the following in the usual form.    (ii)

As we know abc = 100 × a + 10 × b + 1 × c

the usual form of number

100*7 + 10*1 + 8 = 700 + 10 + 8 = 718.

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text:
The development of molecular organometallic electrocatalysts for the reduction of carbon dioxide (CO2) is reviewed. Organometallic complexes are a promising subset of the many catalysts capable of electrochemical reduction of carbon dioxide to organic small molecules such as carbon monoxide (CO), formate (HCO2), and oxalate (C2O42). The carbon-based supporting ligands of organometallic complexes provide electron-rich metal centers and engender good chemical stability, factors that help place organometallic catalysts among the most impressive molecular systems for CO2 reduction. This chapter provides a comprehensive review of CO2 electroreduction by organometallic complexes.

Gonell, S. G.; Miller, A. J. M., Carbon Dioxide Electroreduction Catalyzed by Organometallic Complexes. In Advances in Organometallic Chemistry, Vol 70, First ed., Pérez, P. J.; Stone, F. G. A.; West, R., Eds., Academic Press, 2018, Vol. 70, pp. 1-69. https://dx.doi.org/10.1016/bs.adomc.2018.07.001


text:
# Filing a dictionary through an enumerator 1 listing at a time

I have an issue with one of my functions. I have 5 arrays that get populated through a coroutine, if all parameters are met all of them get populated at the same time and that makes the game lag and sometimes crash on android devices. These are the two functions that do that

public void StartPopulating()
{
if (canPopulateAnimals())
{
StartCoroutine(GetTextures("LevelAnimals", PopulateArray));
}

if (canPopulateArch())
{
StartCoroutine(GetTextures("LevelArchitecture", PopulateArrayArch));
}

if (canPopulateTech())
{
StartCoroutine(GetTextures("LevelTechnology", PopulateArrayTech));
}

if (canPopulateVeh())
{
StartCoroutine(GetTextures("LevelVehicles", PopulateArrayVeh));
}

if (canPopulateNat())
{
StartCoroutine(GetTextures("LevelNature", PopulateArrayNat));
}
}

public IEnumerator GetTextures(string _dirname,Action<Texture2D[]> onLoaded)
{
var fileNames = Directory.GetFiles(Application.persistentDataPath + "/" + _dirname, "*.jpg");
Array.Sort(fileNames, (s1, s2) => Path.GetFileName(s1).CompareTo(Path.GetFileName(s2)));
var textures = new Texture2D[fileNames.Length];
for (int i = 0; i < fileNames.Length; i++)
{
UnityWebRequest webRequest = UnityWebRequestTexture.GetTexture("file://" + fileNames[i]);

yield return webRequest.SendWebRequest();
if (webRequest.isNetworkError || webRequest.isHttpError)
{
Debug.Log(webRequest.error);
}
else
{
status++;
}
done = true;
}
}

Now i was told that i could fill a dictionary instead and Fill that in StartPopulating() and then pass that as the parameter to GetTextures() and then go through the listings one at a time but to be honest i have no idea how to do that.

so initially i created a Dictionary

public Dictionary<string, Action<Texture2D[]>> sets;

But have no clue as to how to place that into the function and enumerator , Any help would be appreciated thank you!

• Why not make one coroutine that checks all of these conditions, and yields on the load in progress so that it doesn't start another load until that one completes? Jul 5 '21 at 11:40
• so instead of having the StartPopulate() function i should have a coroutine that checks if the paramaters are true and if true to then do the loading like i do now? Jul 5 '21 at 11:58
• Don't ask, try. If trying fails, edit your question to report on the problem. Jul 5 '21 at 12:01

To answer your question about Dictionary - you can use dictionary like this:

Dictionary<int,Sprite> niceDictionary = new Dictionary<int, Sprite>();
...
...
mySprite = ((Sprite)niceDictionary[186]);  // to take out at a key 186
...
if (niceDictionary.ContainsKey(48) == true)  // to check if dictionary contains key 48
{
// something here
}

note that you can use anything else instead of int or Sprite - vectors, strings, objects etc.

Now to address your problem from other angle. Your problem is that several coroutines sometimes are started at once. So just make sure that only one is running. For example add if/else statements.

Also you might want to try to change StartPopulating into coroutine, and GetTextures into simple function, thus making sure that two GetTextures are not running at the same time. (In this case you would have to make sure that StartPopulating itself is not launched twice).

• Note that you don't need the (Sprite) cast in that last line. The dictionary already returns a value of type Sprite. Jul 5 '21 at 12:57
• thank you for the reply , well i can do a count and increment it each time the textures load but the function StartPopulating() gets called on game load so i would need to find a way for it to not stop counting. because if i do count the textures and check if count == 1 do that if count == 2 do that etc etc , it only sets the first set of textures and once the count goes to to 2 nothing happens Jul 5 '21 at 13:12

You can wait for a coroutine finish by yielding it like this:

yield return StartCoroutine(SomeCoroutine());

In your case you could make the StartPopulating() method a coroutine and yield each of the population coroutines:

public IEnumerator StartPopulating()
{
if (canPopulateAnimals())
{
yield return StartCoroutine(GetTextures("LevelAnimals", PopulateArray));
}

if (canPopulateArch())
{
yield return StartCoroutine(GetTextures("LevelArchitecture", PopulateArrayArch));
}
}

• You can even skip the nested StartCoroutine if you like, and just say yield return GetTextures(...) directly. Jul 6 '21 at 18:39


text:
# Gaussian window and standard deviation

I am trying to figure out how to use the matlab gausswin function which constructs a Gaussian window for $N$ samples with a given standard deviation $\sigma$. The function is defined by

w = gausswin(N, alpha)

where alpha is $1/\sigma$. I would like my Gaussian window to contain $\pm 3 \sigma$ in the Gaussian. So for $N$ samples it seems I should do: $6 \sigma = N$ or $\sigma = N/6$. However, here are the results for $N = 100$ and $\sigma = 100/6 \approx 16.666$: Clearly this Gaussian does not extend to $\pm 3 \sigma$. Can someone tell me what I'm doing wrong?

My guess it is has something to do with some length scale which isn't being properly specified. It seems to me that the alpha input to the gausswin function is some kind of normalized (unitless) standard deviation. Perhaps I'm not accounting for that.

The manual entry does not say that $\alpha = 1/\sigma$. It says $\alpha$ is proportional to the inverse of $\sigma$.

If you look at the example on that page you'll see

stdev = (N-1)/(2*alpha);

stdev = (100-1)/(2*16.6667) = 825

you should be scaling the relationship by $N-1$.


text:
# How to integrate $x/\sqrt{1-x^2}$?

Could somebody please tell me where I made a mistake? I want to integrate

\begin{equation*} \int_a^b\frac{x}{\sqrt{1-x^2}}dx. \end{equation*}

As far as I know the subsitution $u=1-x^2$ works, but I would like to do it exactly the way we did it in class.

I did the following:

\begin{equation*} \int_a^b\frac{x}{\sqrt{1-x^2}}dx=\frac{1}{2}\int_a^b\frac{2x}{\sqrt{1-x^2}}dx. \end{equation*}

Now let $\phi(x)=x^2$. Therefore

\begin{align*} \int\frac{x}{\sqrt{1-x^2}}dx =\frac{1}{2}\int_a^b\frac{2x}{\sqrt{1-x^2}}dx &=\frac{1}{2}\int_a^b\frac{\phi'(x)}{\sqrt{1-\phi(x)^2}}dx \\ &= \int_{\phi(a)}^{\phi(b)}\frac{1}{\sqrt{1-t^2}}dt \\ &=\arcsin(\phi(b))-\arcsin(\phi(a)). \end{align*}

Thank you very much in advance!

• A better substitution would be $u^2=1-x^2$ and $(-u)\,\mathrm du=x\,\mathrm dx$. – Prasun Biswas May 19 '15 at 13:54
• Or $x= \sin(\theta)$ – user222031 May 19 '15 at 13:55

Your mistake: you wrote $1-\phi(x)^2$ instead of $1-\phi(x)$.
(Which will lead you to the antiderivative $-\sqrt{1-\phi(x)}=-\sqrt{1-x^2}$).
The derivative of $\sqrt{f(x)}$ is $$\frac{f'(x)}{2\sqrt{f(x)}}$$
Have you tried what happens with $f(x)=1-x^2$?


text:
# Re: [tlaplus] Why is this an invalid TLA formula?

(Temporal) formulas are inductively defined as

- state predicates,
- [][A]_v and <> <<A>>_v for an action A and a state formula v,
- Boolean combinations of temporal formulas,
- []F and <>F for temporal formulas (and the derived forms F ~> G, WF_v(A), SF_v(A)),
- \E x : F, \A x : F, \EE v : F, \AA v : F,
- F -+-> G

It follows from the above that

[]<>( <<A>>_v \/ <<B>>_v )

is simply not a syntactically well-formed formula of TLA whereas

[]<> << A \/ B >>_v

is. If the first one were a formula, the two were semantically equivalent.

Hope this helps,
Stephan

> On 11 Jan 2017, at 13:32, George Singer <george....@xxxxxxxxx> wrote:
>
> At the top of page 95 of *Specifying Systems: The TLA+ Language and Tools for Hardware and Software Engineers*, Lamport states that
>
> $$> \Box \Diamond ( \langle A \rangle_v \vee \langle B \rangle_v) >$$
>
> is not a valid TLA formula, even though its equivalent formulation
>
> $$> \Box \Diamond \langle A \vee B \rangle_v >$$
>
> *is* a valid TLA formula.
>
> **Question:** Why is the first formula invalid and the second formula valid?
>
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