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symmetric matrix is a matrix in which the numbers on ether side of the diagonal, in corresponding positions are the same. Multiplying matrices - examples. Lim (Algebra Seminar) Symmetric tensor decompositions January 29, 2009 1 / 29. In linear algebra, a real symmetric matrix represents a self-adjoint operator over a real inner product space. Eigenvalues and the characteristic. symDiagonal() returns an object of class dsCMatrix or lsCMatrix, i. Example: The following 3x3 system has the solution x1 = −1 ; x2 = 2 ; x3 = 1, as you can verify it by direct substitution. 2 4 6 8 5 8 2 1 5 1 0 3 5is a symmetric matrix Theorem 1. Simplify the Matrix. Determining the eigenvalues of a 3x3 matrix. In other words, we can say that matrix A is said to be skew-symmetric if transpose of matrix A is equal to negative of matrix A i. Frank Wood. JACOBI is a program written in 1980 for the HP-41C programmable calculator to find all eigenvalues of a real NxN symmetric matrix using Jacobi's method. 1 below, we show that if ‘>0 and Jis a symmetric diagonally dominant matrix satisfying J ‘S, then J ‘S˜0; in particular, Jis invertible. The eigenvalue for the 1x1 is 3 = 3 and the normalized eigenvector is (c 11) = (1). Singular values are important properties of a matrix. 4 The Minimax Theorem. Figure 2 2-D Gaussian distribution with mean (0,0) and =1 The idea of Gaussian smoothing is to use this 2-D distribution as a point-spread' function, and this is achieved by. A square matrix such that a ij is the complex conjugate of a ji for all elements a ij of the matrix i. (iv) Theorem 2: Any square matrix A can be expressed as the sum of a symmetric matrix and a skew symmetric matrix, that is (A+A ) (A A )T T A = + 2 2 − 3. Statistics 1: Linear Regression and Matrices The concepts and terminology for matrices will be developed using an example from statistics. An analogous result holds for matrices. A = 1 0 0 1 0 0 2 3 3 You have attempted this problem 10 times. ˙ ˙ ˚ ˘ Í Í Î È" π " a i j a i j | {
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A = 1 0 0 1 0 0 2 3 3 You have attempted this problem 10 times. ˙ ˙ ˚ ˘ Í Í Î È" π " a i j a i j ij ij 0, 1, Row matrix. Show that S 3x3 ( R) is actually a subspace of M 3x3 ( R) and then determine the dimension and a basis for this subspace. APPLICATIONS Example 2. The diagonal elements of a skew-symmetric matrix are all 0. If you're seeing this message, it means we're having trouble loading external resources on our website. are symmetric matrices. notebook November 28, 2016 Inverse of a 3x3 Matrix Examples: 3. e ( AT =−A ). A = [ 2 − 1 − 1 − 1 2 − 1 − 1 − 1 2]. Write a C program to read elements in a matrix and check whether the given matrix is symmetric matrix or not. Antisymmetric matrices are commonly called "skew symmetric matrices" by mathematicians. it's a Markov matrix), its eigenvalues and eigenvectors are likely to have special properties as well. A × A-1 = I. Definition. You can convert the skew symmetric matrix R_dot * dt into a rotation matrix using the Rodrigues formula. The symmetric spaces considered here are quotients X=G/K, where G is a non-compact real Lie group, such as the general linear group GL(n,R) of all n x n non-singular real matrices, and K=O(n), the maximal compact subgroup of orthogonal matrices. In describing matrices, the format is: rows X columns. (where A is a symmetric matrix). Similarly, we can take other examples of Nilpotent matrices. We will use the following two properties of determinants of matrices. Complexity Explorer is an education project of the Santa Fe Institute - the world headquarters for complexity science. For the Taylor-Green vortex problem, the domain is periodic is both x- and y-directions, and we end-up with a symmetric implicit matrix. That is the diagonal with the a's on it. For a solution, see the post " Quiz 13 (Part 1) Diagonalize a matrix. Real number λ and vector z are called an eigen pair of matrix A, if Az = λz. First we compute the singular values σ i by finding the eigenvalues of AAT. The Hessian | {
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if Az = λz. First we compute the singular values σ i by finding the eigenvalues of AAT. The Hessian matrix is a square, symmetric matrix whose. Welcome to Week 2 of the Robotics: Aerial Robotics course! We hope you are having a good time and learning a lot already! In this week, we will first focus on. An identity matrix will be denoted by I, and 0 will denote a null matrix. 2 Matrix “Square Roots” Nonnegative numbers have real square roots. We will use the following two properties of determinants of matrices. A 3x3 stress tensor is 2nd rank. Example: Find the eigenvalues and eigenvectors of the real symmetric (special case of Hermitian) matrix below. So we;ve got. This calculator allows to find eigenvalues and eigenvectors using the Characteristic polynomial. For the lid-driven cavity flow, the implicit matrix is not symmetric. where R is the correlation matrix of the predictors (X variables) and r is a column vector of correlations between Y and each X. Example: Solution: Determinant = (3 × 2) – (6 × 1) = 0. Let's consider a simple example with a diagonal matrix: A = np. The largest term will include the cube of lambda and so on. Definition E EœEÞis called a if symmetric matrix X Notice that a symmetric matrix must be square ( ?). To check whether a matrix A is symmetric or not we need to check whether A = AT or not. It is called a singular matrix. Briefly, matrix inverses behave as reciprocals do for real numbers : the product of a matrix and it's inverse is an identity matrix. I'm currently stuck on converting a 3*N x 1, where N is an integer value, vector into chunks of skew symmetric matrices. M = (Q^(-1)) * G. Symmetric matrices have real eigenvalues. Question from Matrices,cbse,class12,bookproblem,ch3,misc,q14,p101,easy,shortanswer,sec-a,math. xTAx = x1 x2 2 6 18 6 x x 1 2 2x = x 1 + 6x2 1 x2 6x 1 + 18x2 = 2x 12 + 12x1x2 + 18x 22 = ax 12 + 2bx1x2 + cx 22. So in that way every Diagonal Matrix is Symmetric Matrix. Then we have: A is positive de nite ,D k >0 for | {
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that way every Diagonal Matrix is Symmetric Matrix. Then we have: A is positive de nite ,D k >0 for all leading principal minors A is negative de nite ,( 1)kD k >0 for all leading principal minors A is positive semide nite , k 0 for all principal minors A is negative semide nite ,( 1)k k 0 for all principal minors In the rst two cases, it is enough to. the algorithm will be part of a massive computational kernel, thus it is required to be very efficient. The determinant obtained through the elimination of some rows and columns in a square matrix is called a minor of that matrix. Note that whereas C is a 3× 2 matrix, its transpose, CT, is a 2× 3 matrix. He walks you through basic ideas such as how to solve systems of. The leading coefficients occur in columns 1 and 3. (4) If A is invertible then so is AT, and (AT) − 1 = (A − 1)T. , The sum of the numbers along each matrix diagonal (the character) gives a shorthand version of the matrix representation, called Γ:. Thus, we may write [El = S'[A]S (18) where [A] is the diagonal matrix of eigenvalues [A]= [: 0 A, : :I 0 and S is the rotation matrix whose rows are the three orthogonal eigenvectors of [E] corresponding to the Ai eigenvalues. In the following we assume. Example 3 Show that a matrix which is both symmetric and skew symmetric is a zero matrix. Detailed Description Tiny matrix classes with optimized numerical operations which make use of vectorization features through the VVector class. A matrix with real entries is skewsymmetric. An example will be constructed later in this chapter. 50 }; //Symmetric Postive-Definite matrix X = { 7 6. Basis vectors. Matrix Inner Products. Now lets FOIL, and solve for. It can be shown that all real symmetric matrices have real eigenvalues and perpendicular eigenvectors. C program check symmetric matrix. Properties. Treat the remaining elements as a 2x2 matrix. Let’s take an example of a matrix. Consider a n x n, trace free, real symmetric matrix A. These terms are more properly | {
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of a matrix. Consider a n x n, trace free, real symmetric matrix A. These terms are more properly defined in Linear Algebra and relate to what are known as eigenvalues of a matrix. the inverse of an n x n matrix See our text ( Rolf, Page 163) for a discussion of matrix inverses. U) = A z^ - [1,3I1A + [-12,123. We will use the following two properties of determinants of matrices. Of course this holds too for square matrices of higher rank (N x N matrices), not just 3x3's. Frank Wood, [email protected] It is called a singular matrix. Homework Statement Hi there, I'm happy with the proof that any odd ordered matrix's determinant is equal to zero. For example, to solve 7x = 14, we multiply both sides by the same number. To check whether a matrix A is symmetric or not we need to check whether A = AT or not. The inverse of a permutation matrix is again a permutation matrix. Here for instance, X should be a 6x6 real and symmetric matrix unless we have made some mistake, and indeed it is constructed from 9 variables defining a symmetric 3x3 matrix. The unit matrix is every #n#x #n# square matrix made up of all zeros except for the elements of the main diagonal that are all ones. As a result, we can concisely represent any skew symmetric 3x3 matrix as a 3x1 vector. Generally, one can find symmetrization A0 of a matrix A by A0 = A+AT 2. 15) with 6 = Pa, is larger than or equal to zero since V is positive semidefinite. So, for example, the 3x3 matrix A might be written as:. The leftmost column is column 1. e ( AT =−A ). Examples: Quadratic Form Now we have seen the symmetric matrices, we can move on to the quadratic 1 5 5 8 9 −2 − 2 7 a b b c. To check whether a matrix A is symmetric or not we need to check whether A = A T or not. Example 4 Suppose A is this 3x3 matrix: [1 1 0] [0 2 0] [0 –1 2]. 4 Diagonal Matrix: A square matrix is called a diagonal matrix if each of its non-diagonal elements are zero (i. com To create your new password, just click the link in the email we | {
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elements are zero (i. com To create your new password, just click the link in the email we sent you. One is to use Gauss-Jordan elimination and the other is to use the adjugate matrix. AB = BA = I n, then the matrix B is called an inverse of A. Inverse matrix A -1 is defined as solution B to AB = BA = I. 366) •A is orthogonally diagonalizable, i. Instead, we can implicitly apply the symmetric QR algorithm to ATA. Examples: Quadratic Form Now we have seen the symmetric matrices, we can move on to the quadratic 1 5 5 8 9 −2 − 2 7 a b b c. (1 Point) Give An Example Of A 3 × 3 Skew-symmetric Matrix A That Is Not Diagonal. 369) EXAMPLE 1 Orthogonally diagonalize. Example Consider the matrix A= 1 4 4 1 : Then Q A(x;y) = x2 + y2 + 8xy and we have Q A(1; 1) = 12 + ( 1)2 + 8(1)( 1) = 1 + 1 8. 50 }; //Symmetric Postive-Definite matrix X = { 7 6. Eigenvalues of a 3x3 matrix. You need to know more about the matrix to conclude one way or the other. Welcome to Week 2 of the Robotics: Aerial Robotics course! We hope you are having a good time and learning a lot already! In this week, we will first focus on. Homework Statement Hi there, I'm happy with the proof that any odd ordered matrix's determinant is equal to zero. symDiagonal() returns an object of class dsCMatrix or lsCMatrix, i. Notice that this is a block diagonal matrix, consisting of a 2x2 and a 1x1. The variance-covariance matrix is symmetric because the covariance between X and Y is the same as the covariance between Y and X. De nition 1 Let U be a d dmatrix. Jacobi's Algorithm is a method for finding the eigenvalues of nxn symmetric matrices by diagonalizing them. Fortran 90 package for solving linear systems of equations of the form A*x = b, where the matrix A is sparse and can be either unsymmetric, symmetric positive definite, or general symmetric. Finally, show (if you haven't already) that the only matrix both symmetric and skew-symmetric is the zero matrix. For example, decrypting a coded message uses invertible | {
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and skew-symmetric is the zero matrix. For example, decrypting a coded message uses invertible matrices (see the coding page). Similarly, the rank of a matrix A is denoted by rank(A). Many problems present themselves in terms of an eigenvalue problem: A·v=λ·v. , (AT) ij = A ji ∀ i,j. HILL CIPHER Encrypts a group of letters called polygraph. Vector x is a right eigenvector, vector y is a left eigenvector, corresponding to the eigenvalue λ, which is the same for. Thethingis,therearealotofotherequivalentwaystodefineapositive definite matrix. Usually the numbers are real numbers. Geometrically, a matrix $$A$$ maps the unit sphere in $$\mathbb{R}^n$$ to an ellipse. (34) Finally, the rank of a matrix can be defined as being the num-ber of non-zero eigenvalues of the matrix. Note that whereas C is a 3× 2 matrix, its transpose, CT, is a 2× 3 matrix. A = [1 1 1 1 1 1 1 1 1]. Then compute it's determinant (which will end up being a sum of terms including four coefficients) Then to ease the computation, find the coefficient that appears in the least amount of term. Therefore, we can see that , Hence, the matrix A is nilpotent. Example: If square matrices Aand Bsatisfy that AB= BA, then (AB)p= ApBp. Inverting a matrix turns out to be quite useful, although not for the classic example of solving a set of simultaneous equations, for which other, better, methods exist. Once we get the matrix P, then D = P t AP. Matrices with Examples and Questions with Solutions. $\begingroup$ Yes, reduced row echelon form is also called row canonical form, and obviously there are infinitely many symmetric matrix that are not diagonal and can be reduced to anon diagonal reduced row echelon form, but note that the row canonical form is not given by a similarity transformation, but the jordan form is. λ is an eigenvalue (a scalar) of the Matrix [A] if there is a non-zero vector (v) such that the following relationship is satisfied: [A](v) = λ (v) Every vector (v) satisfying this equation is called an | {
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relationship is satisfied: [A](v) = λ (v) Every vector (v) satisfying this equation is called an eigenvector of [A] belonging to the eigenvalue λ. The inverse is calculated using Gauss-Jordan elimination. The inverse of a permutation matrix is again a permutation matrix. Mathematical Properties of Stiffness Matrices 3 computation involving the inverse of ill-conditioned matrices can lose precision because there is a range of values in the solution { d }that can satsify [ K ]{ d }= { p }. Computing eigenvalues and eigenvectors for a 3x3 symmetric matrix. You have unlimited attempts remaining. 139 of Boas, (AB)T = B TA for any two matrices Aand B. com Symmetric Matrix Inverse. The quadratic form of. the symmetric QRalgorithm, as the expense of two Jacobi sweeps is comparable to that of the entire symmetric QRalgorithm, even with the accumulation of transformations to obtain the matrix of eigenvectors. By the second and fourth properties of Proposition C. Join 90 million happy users! Sign Up free of charge:. If Ais symmetric, then A= AT. (1) Any real matrix with real eigenvalues is symmetric. there exists an orthogonal matrix P such that P−1AP =D, where D is diagonal. The main diagonal gets transposed onto itself. It is a specific case of the more general finite element method, and was in. The result is a 3x1 (column) vector. The matrix 1 1 0 2 has real eigenvalues 1 and 2, but it is not symmetric. A , in addition to being magic, has the property that "the sum of the twosymmetric magic square numbers in any two cells symmetrically placed with respect to the center cell is the same" (12, p. Now the next step to take the determinant. Indeed, if aij 6= aji we replace them by new a0 ij = a 0 ji = aij+aji 2, this does not change the corresponding quadratic form. Examples # NOT RUN { #Create a sample 3x3 matrix mat <- matrix(1:9, ncol=3) #Copy the upper diagonal portion to the lower symmetricize(mat, "ud") #Take the average of each symmetric location symmetricize(mat, "avg") | {
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lower symmetricize(mat, "ud") #Take the average of each symmetric location symmetricize(mat, "avg") # } Documentation reproduced from package HelpersMG, version 4. If A is not SPD then the algorithm will either have a zero. (1) Again, since A is a symmetric matrix, so A′ = A. The link inertia matrix (3x3) is symmetric and can be specified by giving a 3x3 matrix, the diagonal elements [Ixx Iyy Izz], or the moments and products of inertia [Ixx Iyy Izz Ixy Iyz Ixz]. Maximum eigenvalue for this symmetric matrix is 3. JACOBI is a program written in 1980 for the HP-41C programmable calculator to find all eigenvalues of a real NxN symmetric matrix using Jacobi's method. An n×n matrix B is called nilpotent if there exists a power of the matrix B which is equal to the zero matrix. Recently, in order to find the principal moments of inertia of a large number of rigid bodies, it was necessary to compute the eigenvalues of many real, symmetric 3 × 3 matrices. Equation (1) is the eigenvalue equation for the matrix A. Mathematical Properties of Stiffness Matrices 3 computation involving the inverse of ill-conditioned matrices can lose precision because there is a range of values in the solution { d }that can satsify [ K ]{ d }= { p }. all integral types except bool, floating point and complex types), whereas symmetric matrices can also be block matrices (i. A square matrix A is called a diagonal matrix if a ij = 0 for i 6= j. This pages describes in detail how to diagonalize a 3x3 matrix througe an example. Two apologies on quality: 1. Example A= 2 4 0 3 This is a 2 by 2 matrix, so we know that 1 + 2 = tr(A) = 5 1 2 = det(A) = 6 6. Shio Kun for Chinese translation. Use the ad - bc formula. We will use the following two properties of determinants of matrices. The Symmetric Inertia Tensor block creates an inertia tensor from moments and products of inertia. The function scipy. An identity matrix will be denoted by I, and 0 will denote a null matrix. In describing matrices, the | {
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identity matrix will be denoted by I, and 0 will denote a null matrix. In describing matrices, the format is: rows X columns. Eigenvalues and eigenvectors of a real symmetric matrix. Example 1: Determine the eigenvectors of the matrix. -24 * 5 = -120; Determine whether to multiply by -1. Now lets FOIL, and solve for. (1 Point) Give An Example Of A 3 × 3 Skew-symmetric Matrix A That Is Not Diagonal. We employ the latter, here. Definition of a Matrix The following are examples of matrices (plural of matrix). Note that all the main diagonal elements in the skew-symmetric matrix are zero. Now we only have to calculate the cofactor of a single element. can be obtained by using the cofactor. Now, noting that a symmetric matrix is positive semi-definite if and only if its eigenvalues are non-negative, we see that your original approach would work: calculate the characteristic polynomial, look at its roots to see if they are non-negative. A nxn symmetric matrix A not only has a nice structure, but it also satisfies the following:. G o t a d i f f e r e n t a n s w e r? C h e c k i f i t ′ s c o r r e c t. First we compute the singular values σ i by finding the eigenvalues of AAT. ˙ ˙ ˚ ˘ Í Í Î È" π " a i j a i j ij ij 0, 1, Row matrix. A3×3 example of a matrix with some complex eigenvalues is B = 1 −1 −1 1 −10 10−1 A straightforward calculation shows that the eigenvalues of B are λ = −1 (real), ±i (complex conjugates). Good things happen when a matrix is similar to a diagonal matrix. As is well known, any symmetric matrix is diagonalizable, where is a diagonal matrix with the eigenvalues of on its diagonal, and is an orthogonal matrix with eigenvectors of as its columns (which magically form an orthogonal set , just kidding, absolutely no magic involved ). The above example illustrates a Cholesky algorithm, which generalizes for higher dimensional matrices. You can convert the skew symmetric matrix R_dot * dt into a rotation matrix using the Rodrigues formula. AB = BA = I n, | {
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skew symmetric matrix R_dot * dt into a rotation matrix using the Rodrigues formula. AB = BA = I n, then the matrix B is called an inverse of A. In general, matrices can contain complex numbers but we won't see those here. 231 Diagonalization of non-Hermitian matrices • Let D be the diagonal matrix whose. If the matrix A is symmetric then •its eigenvalues are all real (→TH 8. // symmetric or not. I know that I can convert a single vector of size 3 in a skew symmetric matrix of size 3x3 as follows: X = [ 0 -x(3) x(2) ;. Example: The transpose of A is • For a matrix A = [aij], its transpose AT= [bij], where bij= aji. Example: RC circuit v1 vn c1 cn i1 in resistive circuit ckv Symmetric matrices, quadratic forms, matrix norm, and SVD 15-19. , The sum of the numbers along each matrix diagonal (the character) gives a shorthand version of the matrix representation, called Γ:. NET example in Visual Basic demonstrating the features of the symmetric matrix classes. 32 (2) 223 222 uu kku kku k The global stiffness matrix may be constructed by directly adding terms associated with the degrees of freedom in k(1). As a recent example, the work of Spielman and Teng [14, 15] gives algorithms to solve symmetric, diagonally dominant linear systems in nearly-linear time in the input size, a fundamental advance. Statistics 1: Linear Regression and Matrices The concepts and terminology for matrices will be developed using an example from statistics. Matrix inversion. The subject of symmetric matrices will now be examined using an example from linear regression. Transpose of a Matrix, Symmetric Matrix & Skew Symmetric Matrix: Class 12 Transpose of a matrix , Symmetric Matrix and Skew Symmetric Matrix are explained in a very easy way. 2 Example: Odd or Even. The task is to find a matrix P which will let us convert A into D. Example, = -5 and =5 which means. Joachim Kopp developed a optimized "hybrid" method for a 3x3 symmetric matrix, which relays on the analytical mathod, but falls | {
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"hybrid" method for a 3x3 symmetric matrix, which relays on the analytical mathod, but falls back to QL algorithm. Logic to check symmetric matrix in C programming. 2 Definiteness of Quadratic Forms. As a recent example, the work of Spielman and Teng [14, 15] gives algorithms to solve symmetric, diagonally dominant linear systems in nearly-linear time in the input size, a fundamental advance. (3) If the products (AB)T and BTAT are defined then they are equal. The values of λ that satisfy the equation are the generalized eigenvalues. Step 1 - Accepts a square matrix as input Step 2 - Create a transpose of a matrix and store it in an array Step 3 - Check if input matrix is equal to its transpose. These are well-defined as $$A^TA$$ is always symmetric, positive-definite, so its eigenvalues are real and positive. Since the symmetric matrix is taken as A, the inverse symmetric matrx is written as A-1, such that it becomes. Program to swap upper diagonal elements with lower diagonal elements of matrix. A diagonal matrix is a symmetric matrix with all of its entries equal to zero except may be the ones on the diagonal. [R] binary symmetric matrix combination 3x3, 4x4) which I need to create in R as seen below: I think you didn't run the vecOut` after adding the new matrix. is an eigenvector corresponding to the eigenvalue 1. Enter payoff matrix B for player 2 (not required for zerosum or symmetric games). Let A = (v, 2v, 3v). For example, if a problem requires you to divide by a fraction, you can more easily multiply by its reciprocal. So why bother with the confusion matrix? Because it gives us insight into the details of how the algorithms achieve their percent correct. Diagonalizing a 3x3 matrix. We call such a matrix Hermitianafter the French mathematician Charles Hermite (1822-1901). Scroll down the page for examples and solutions. | {
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fyxewz0sqfycl dajkwnrqsps dfd1m1qpkrqm dpk3dqc9lqwv 5pbqzus3mshim14 61m01xv5xdl zd7gy8b40fe 19oy2gwbsp yeptctvyils esem86uwht8 06vj9j3m94 jp2wo0534lxaxty e5ggbv9peei 9w0eaa6hdw0 ks7w09hxgzkkt mlj60qjcs5p2 uomnmas9ekss j0m4lwi2h45fy 2eulcsy0gsmiqd8 6h2tya9io5puy2 0lu6mgdvy4h pv53iucjilxfrh w1u1ff9m3t0b4 bvswbs5ya8t 6yllw1n4gtpj 0pvr0bmphwgxkaz 9fu54s858j04 mpfludm9r4k9 fnvorb04eb z1ztaijsinf31o8 b1enw5cubxrnxmh ykiif7j79gx4sd vxzz4evvvnlz4wn l5j741s0ndd4 lofw30li8usega7 | {
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How is $P(A^c \cap B^c)$ the same as $1-P(A \cup B)$?
I don't understand how $$P(A^c \cap B^c) = 1-P(A \cup B)$$ is the same?
If I draw $$P(A^c \cap B^c)$$ as a Venn diagram:
If I draw $$P(A \cup B)$$ as a Venn diagram:
So if I subtract $$P(A \cup B)$$ from 1, wouldn't that mean that I subtract $$P(A \cup B)$$ from the universe $$\Omega$$, which would result int this:
However, that would mean $$P(A^c \cap B^c) \neq 1-P(A \cup B)$$
Edit:
As pointed out by multiple people. My diagram for $$P(A^c \cap B^c)$$ should look like the following and therefore the assumption of $$P(A^c \cap B^c) = 1-P(A \cup B)$$ is valid:
• Your first diagram is of $A^c \cup B^c$ not $A^c \cap B^c$. The yellow areas include both $A \cap B^c$ and $A^c \cap B$ – Henry Jan 24 '20 at 0:09
• your first diagram is wrong – Masacroso Jan 24 '20 at 0:11
• The third diagram is the correct one for the left side as well. – Berci Jan 24 '20 at 0:12
• So yellow should be only the universe? – Tom el Safadi Jan 24 '20 at 0:15
• I edited my question with the help of you guys. Thanks!! – Tom el Safadi Jan 24 '20 at 0:19
Your first diagram is incorrect. You seem to have drawn $$A^c\cup B^c$$ (as some other people have mentioned).
I would recommend drawing $$A^c$$ independently first, which consists of the rest of the universe and $$B-A$$. Then draw $$B^c$$ in a different color, noting again that you have the rest of the universe and $$A-B$$. The intersection of $$A-B$$ and $$B-A$$ is neither $$A$$ nor $$B$$, so you will end up without $$A$$ or $$B$$ in your final intersection. However, the rest of the universe is in both $$A^c$$ and $$B^c$$, therefore so is its intersection. Thus, you get that $$P(A^c\cap B^c)$$ is just the universe without $$A$$ or $$B$$, which is equivalent to $$1-P(A\cup B)$$. | {
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According to the DeMorgan's laws, one has \begin{align*} \textbf{P}(\Omega) & = \textbf{P}((A\cup B)\cup(A\cup B)^{c})\\ & = \textbf{P}(A\cup B) + \textbf{P}((A\cup B)^{c})\\ & = \textbf{P}(A\cup B) + \textbf{P}(A^{c}\cap B^{c}) = 1 \end{align*} from whence the result follows immediately, since $$X\cup X^{c} = \Omega$$ and $$X\cap X^{c} = \varnothing$$ for every event $$X\subseteq\Omega$$. | {
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# sums of normal random variables need not be normal
A common misconception among students of probability theory is the belief that the sum of two normally distributed (http://planetmath.org/NormalRandomVariable) random variables is itself normally distributed. By constructing a counterexample, we show this to be false.
It is however well known that the sum of normally distributed variables $X,Y$ will be normal under either of the following situations.
• $X,Y$ are independent.
• $X,Y$ are joint normal (http://planetmath.org/JointNormalDistribution).
The statement that the sum of two independent normal random variables is itself normal is a very useful and often used property. Furthermore, when working with normal variables which are not independent, it is common to suppose that they are in fact joint normal. This can lead to the belief that this property holds always.
Another common fallacy, which our example shows to be false, is that normal random variables are independent if and only if their covariance, defined by
$\operatorname{Cov}(X,Y)\equiv\mathbb{E}[XY]-\mathbb{E}[X]\mathbb{E}[Y]$
is zero. While it is certainly true that independent variables have zero covariance, the converse statement does not hold. Again, it is often supposed that they are joint normal, in which case a zero covariance will indeed imply independence.
We construct a pair of random variables $X,Y$ satisfying the following.
1. 1.
$X$ and $Y$ each have the standard normal distribution.
2. 2.
The covariance, $\operatorname{Cov}(X,Y)$, is zero.
3. 3.
The sum $X+Y$ is not normally distributed, and $X$ and $Y$ are not independent.
We start with a pair of independent random variables $X,\epsilon$ where $X$ has the standard normal distribution and $\epsilon$ takes the values $1,-1$, each with a probability of $1/2$. Then set,
$Y=\left\{\begin{array}[]{ll}\epsilon X,&\textrm{if }|X|\leq 1,\\ -\epsilon X,&\textrm{if }|X|>1.\end{array}\right.$ | {
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If $S$ is any measurable subset of the real numbers, the symmetry of the normal distribution implies that $\mathbb{P}(X\in S)$ is equal to $\mathbb{P}(-X\in S)$. Then, by the independence of $X$ and $\epsilon$, the distribution of $Y$ conditional on $\epsilon=1$ is given by,
$\begin{split}\displaystyle\mathbb{P}(Y\in S\mid\epsilon=1)&\displaystyle=% \mathbb{P}(|X|\leq 1,X\in S)+\mathbb{P}(|X|>1,-X\in S)\\ &\displaystyle=\mathbb{P}(|X|\leq 1,X\in S)+\mathbb{P}(|X|>1,X\in S)=\mathbb{P% }(X\in S).\end{split}$
It can similarly be shown that $\mathbb{P}(Y\in S\mid\epsilon=-1)$ is equal to $\mathbb{P}(X\in S)$. So, $Y$ has the same distribution as $X$ and is normal with mean zero and variance one.
Using the fact that $\epsilon$ has zero mean and is independent of $X$, it is easily shown that the covariance of $X$ and $Y$ is zero.
$\begin{split}\displaystyle\operatorname{Cov}(X,Y)&\displaystyle=\mathbb{E}[XY]% -\mathbb{E}[X]\mathbb{E}[Y]=\mathbb{E}[XY]\\ &\displaystyle=\mathbb{E}\left[1_{\{|X|\leq 1\}}\epsilon X^{2}\right]+\mathbb{% E}\left[-1_{\{|X|>1\}}\epsilon X^{2}\right]\\ &\displaystyle=\mathbb{E}[\epsilon]\mathbb{E}\left[1_{\{|X|\leq 1\}}X^{2}% \right]-\mathbb{E}[\epsilon]\mathbb{E}\left[1_{\{|X|>1\}}X^{2}\right]\\ &\displaystyle=0.\end{split}$
As $X$ and $Y$ have zero covariance and each have variance equal to $1$, the sum $X+Y$ will have variance equal to $2$. Also, the sum satisfies
$X+Y=\left\{\begin{array}[]{ll}2\epsilon X,&\textrm{if }|X|\leq 1,\\ 0,&\textrm{if }|X|>1.\end{array}\right.$
In particular, this shows that $\mathbb{P}(|X+Y|>2)=0$. However, normal random variables with nonzero variance always have a positive probability of being greater than any given real number. So, $X+Y$ is not normally distributed.
This also shows that, despite having zero covariance, $X$ and $Y$ are not independent. If they were, then the fact that sums of independent normals are normal would imply that $X+Y$ is normal, contradicting what we have just demonstrated. | {
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Title sums of normal random variables need not be normal SumsOfNormalRandomVariablesNeedNotBeNormal 2013-03-22 18:43:44 2013-03-22 18:43:44 gel (22282) gel (22282) 5 gel (22282) Example msc 62E15 msc 60E05 | {
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# Thread: difference of two squares
1. ## difference of two squares
Find two integers whose squares have a difference of 1,234,567.
can you find two integers whose squares will produce a difference of any whole number you chose?
if so, how? if not, what kind of numbers can't be chosen.
well... a^2-b^2=1234567
we know a must end in a 1 or 9 and b must end in a 2 or 8 (this is to produce the seven at the end)
i came up with 617289^2-617288^2=1234577 and that's as close as i can get.
any suggestions? first time poster! sorry if it is an unusual / inappropriate post!
much thanks!
2. $n^{2}-(n-1)^{2}=1234567$
n=617284
$617284^{2}-617283^{2}=1234567$
3. Number suchs that,
$n\equiv 0,1,3 (\bmod 4)$
Are expressible as a difference of two square.
And,
$n\equiv 2 (\bmod 4)$
Are not.
4. Hello, perfect square!
Find two integers whose squares have a difference of 1,234,567
Galactus used a very clever bit of trivia.
Consecutive squares differ by consecutive odd numbers.
. . $\begin{array}{cccc} & & & \text{diff} \\ 0^2 & = & 0 & \\ & & & 1 \\ 1^2 & = & 1 & \\ & & & 3 \\ 2^2 & = & 4 & \\ & & & 5 \\ 3^2 & = & 9 & \\ & & & 7 \end{array}$
. . $\begin{array}{cccc}4^2 & = & 16 & \\ & & & 9 \\ 5^2 & = & 25 & \\ \vdots & & \vdots & \vdots \end{array}$
As he pointed out, we want two consecutive integers with a difference of 1234567.
So we have: . $(n + 1)^2 - n^2 \:=\:1,234,567$
. . $n^2 + 2n + 1 - n^2 \:=\:1,234,567\quad\Rightarrow\quad 2n+1 \:=\:1,234,567\quad\Rightarrow\quad 2n \:=\:1,234,566$
. . Hence: . $n \,=\,617,283$
Therefore: . $617,284^2 - 617283^2 \:=\:1,234,567$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
I suspect that 1,234,567 is prime (but I'm not sure).
If the difference is a composite odd number,
. . there may be an alternate solution.
Example: Find two integers whose squares differ by 133.
Since $\frac{133-1}{2} = 66$, we have: . $67^2 - 66^2 \:=\:133$ | {
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Since $\frac{133-1}{2} = 66$, we have: . $67^2 - 66^2 \:=\:133$
. Since $133 = 7\times 19$, we have: . . $19 + 19 + 19 + 19 + 19 + 19 + 19$
. . . . . which can be written: . . $13 + 15 + 17 + 19 + 21 + 23 + 25$
which are the differences of: . $6^2\quad7^2\quad\:8^2\quad\:9^2\quad10^2\quad11^2\ quad12^2\quad13^2$
Therefore: . $13^2 - 6^2 \:=\:133$
5. Originally Posted by Soroban
So we have: . $(n + 1)^2 - n^2 \:=\:1,234,567$
. . $n^2 + 2n + 1 - n^2 \:=\:1,234,567\quad\Rightarrow\quad 2n+1 \:=\:1,234,567\quad\Rightarrow\quad 2n \:=\:1,234,566$
. . Hence: . $n \,=\,617,283$
Such a seemingly difficult question, transformed into simple algebra! Once again! Nice!
6. I suspect that 1,234,567 is prime (but I'm not sure).
No, Soroban, it is not prime. | {
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# Combination Proof
We will prove that:
$$\dbinom{n}{k} = \displaystyle \prod_{i=1}^{k} \frac{n-i+1}{i}$$
$$\forall$$ $$k \in \mathbb{N}$$, $$k \leq n$$
Proof #1: As the reader may guess, we will use induction to prove this. We establish a base case as $$k=1$$:
$$\dbinom{n}{1} = \displaystyle \prod_{i=1}^{1} \frac{n-i+1}{i}$$
$$\Rightarrow$$ $$\frac{n!}{1!(n-1)!} = n$$
$$\Rightarrow$$ $$n=n$$
The base case clearly holds. Now we establish the inductive case as $$n=\alpha$$:
$$\dbinom{n}{\alpha} = \displaystyle \prod_{i=1}^{\alpha} \frac{n-i+1}{i}$$
$$\Rightarrow$$ $$\frac{n-\alpha}{\alpha +1} \dbinom{n}{\alpha} =\frac{n-\alpha}{\alpha +1} \displaystyle \prod_{i=1}^{\alpha} \frac{n-i+1}{i}$$ (By hypothesis)
$$\Rightarrow$$ $$\frac{n-\alpha}{\alpha +1} \dbinom{n}{\alpha} =\displaystyle \prod_{i=1}^{\alpha + 1} \frac{n-i+1}{i}$$
$$\Rightarrow$$ $$\frac{n!(n-\alpha)}{(\alpha +1)!(n-\alpha)!} =\displaystyle \prod_{i=1}^{\alpha + 1} \frac{n-i+1}{i}$$
$$\Rightarrow$$ $$\frac{n!}{(\alpha +1)!(n-(\alpha+1))!} = \displaystyle \prod_{i=1}^{\alpha + 1} \frac{n-i+1}{i}$$
$$\Rightarrow$$ $$\dbinom{n}{\alpha+1} = \displaystyle \prod_{i=1}^{\alpha + 1} \frac{n-i+1}{i}$$
So we may conclude that the statement is true for $$k=1$$ and if it is true for some $$k=\alpha$$, then it must be true for $$k=\alpha+1$$. The proof follows by induction.
QED
Proof #2: We will now offer an alternative proof, which is actually just a simple direct derivation:
$$\dbinom{n}{k} = \frac{n!}{k!(n-k)!} = \frac{\displaystyle \prod_{i=0}^{n-1} (n-i)}{\displaystyle \prod_{i=0}^{k-1} (k-i) \left( \displaystyle \prod_{i=0}^{n-k-1} (n-k-i) \right)}$$
$$= \frac{\displaystyle \prod_{i=1}^k (n-i+1)}{k!} = \frac{\displaystyle \prod_{i=1}^k (n-i+1)}{ \displaystyle \prod_{i=1}^k i }$$
$$= \displaystyle \prod_{i=1}^k \frac{(n-i+1)}{i}$$ Because multiplication is commutative.
Which is the intended result.
QED.
Note by Ethan Robinett
3 years, 10 months ago | {
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Which is the intended result.
QED.
Note by Ethan Robinett
3 years, 10 months ago
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Sort by:
I've been scrolling through your profile the complete morning now and am finding some really nice notes. Thanks for sharing them!
- 3 years, 4 months ago
Thanks!
- 3 years, 4 months ago
Nice note@Ethan Robinett
- 3 years, 10 months ago
Thanks appreciate it!
- 3 years, 10 months ago
Hey nice proofs! :) I did the Inductive one but didn't think of using a direct one
- 3 years, 10 months ago
Thanks! Yeah after I did the induction, I figured there was probably a way to directly show it, given that factorials are just products by definition, I thought the second one was pretty interesting
- 3 years, 10 months ago
I like the way the induction is used.
- 3 years, 10 months ago
Thanks man
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# Math Help - Simplify an Expression with Negative Exponents
1. ## Simplify an Expression with Negative Exponents
(ab)^-1 / (a^-2 + b^-2)
if you could explain it with an appropriate property, that would be great!!
thanks..!!
2. $\frac{(ab)^{-1}}{a^{-2}+b^{-2}}=\frac{\frac{1}{ab}}{\frac{1}{a^2}+\frac{1}{b^2 }}=\frac{\frac{1}{ab}}{\frac{b^2+a^2}{a^2b^2}}=\fr ac{1}{ab}\cdot\frac{a^2b^2}{a^2+b^2}=\frac{a^2b^2} {ab(a^2+b^2)}=\frac{ab}{a^2+b^2}$
I think that's right. Latex notation was brutal on this one...for me, anyway. I welcome others to detect any errors.
3. Originally Posted by masters
$\frac{(ab)^{-1}}{a^{-2}+b^{-2}}=\frac{\frac{1}{ab}}{\frac{1}{a^2}+\frac{1}{b^2 }}=\frac{\frac{1}{ab}}{\frac{b^2+a^2}{a^2b^2}}=\fr ac{1}{ab}\cdot\frac{a^2b^2}{a^2+b^2}=\frac{a^2b^2} {ab(a^2+b^2)}=\frac{ab}{a^2+b^2}$
I think that's right. Latex notation was brutal on this one...for me, anyway. I welcome others to detect any errors.
Looks good to me, but don't forget to specify that $a \neq 0$ and $b \neq 0$ independently.
-Dan
4. Hi !
A slightly different approach
$\frac{(ab)^{-1}}{a^{-2}+b^{-2}}=\frac{1}{(ab)(a^{-2}+b^{-2})}$
-----------------------
Using the power rule : $a^b a^c=a^{b+c}$
--> $(ab)a^{-2}=ba^{-1}$
-----------------------
$=\frac{1}{ba^{-1}+ab^{-1}}=\frac{1}{\frac ba+\frac ab}=\frac{1}{\frac{b^2+a^2}{ab}}=\frac{ab}{a^2+b^2 }$
one extra equality
5. Always keep it interesting, Moo. Good job!!
6. thank you both for your help!
7. Hello, needymathperson!
One different step . . .
$\frac{(ab)^{-1}}{a^{-2} + b^{-2}}$
We have: . $\frac{(ab)^{-1}} {a^{-2} + b^{-2}} \;\;=\;\;\frac{\dfrac{1}{ab}} {\dfrac{1}{a^2} + \dfrac{1}{b^2}}$
Multiply top and bottom by $a^2b^2\!:\;\;\frac{a^2b^2\left(\dfrac{1}{ab}\right )}{a^2b^2\left(\dfrac{1}{a^2} + \dfrac{1}{b^2}\right)} \;\;=\;\;\boxed{\frac{ab}{b^2+a^2}}$ | {
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# How to make a box which has the largest possible volume?
I have sheet metal in form of an equilateral triangle and I want to fold it to make a container for the screws. How should I cut and fold to make the a box with largest volume?
Basically I cut the corners and then fold them. There is no "roof". Thank you!
$$a=0.15m$$
-
@SreekanthKarumanaghat: Cut that out; this doesn't look like homework. – Henning Makholm Apr 7 '13 at 13:57
Let the side remaining after cutting off the vertices be A. The volume is given by:
$$V = \dfrac12 A^2 \sin(60) \times h = \dfrac{\sqrt3}{4} A^2h$$
The relation between $A$, $a$ and $h$ is:
$A = a - \dfrac{2h}{\tan(30)}$
(You get a small kite at the edge with two right angles, one angle of 60 degrees and one angle at 120 degrees)
You can substitute them now:
$$V = \dfrac{\sqrt3}{4} \left(a - \dfrac{2h}{\tan(30)}\right)^2h$$
$$V = \dfrac{\sqrt3}{4} (0.15 - 2\sqrt3h)^2h$$
$$V = 3 \sqrt3 h^3-0.45 h^2+0.00974279 h$$
Differentiate w.r.t. $h$ to get:
$$\dfrac{dV(h)}{dh} = 9 \sqrt3 h^2-0.9 h+0.00974279$$
Equate to zero and solve for $h$ to get $h\approx0.0433013$ and $h\approx0.0144338$.
The first one gives the minimum volume, so you don't want that. Take the second.
-
Thank you Jerry. :) Cheers! You get the right answer, because you actually gave me two value and described them really nice! I could follow the process very easily! – user31113 Apr 7 '13 at 11:46
You're welcome! – Jerry Apr 7 '13 at 11:47
If you cut the corner in the manner shown, by trigonometry at any corner, the new side length is smaller by $2 \sqrt3 h$. (Let me know if you have difficulty with this.) | {
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Hence the box's volume is proportional to $\left(a-2 \sqrt3 h\right)^2 h$, which we try to maximise. Let $V(h) = \left(a-2 \sqrt3 h\right)^2 (4 \sqrt3 h)$. The value of $h$ which maximises $V(h)$ is exactly same as that maximising the volume we desire, due to proportionality. Now $V(h)$ can be looked at as the product of $3$ terms, $(a-2 \sqrt3 h), (a-2 \sqrt3 h)$ and $(4 \sqrt3 h)$, which sum to a constant $2a$. Hence the product is maximised when these three terms are equal. i.e.
$a-2 \sqrt3 h = 4 \sqrt3 h$ or when $h = \dfrac{a}{6\sqrt3}$.
-
Oh woops, yours is definitely more elegant! – Jerry Apr 7 '13 at 11:26
@Jerry Thanks. It's just that I like inequalities a lot. – Macavity Apr 7 '13 at 11:32
Thank you Macavity. :) Cheers! – user31113 Apr 7 '13 at 11:44
@Macavity Your solution is certainly neat, but I get a slight butterfly feeling when you pull that factor out of the air. I seem to get the same numbers by a method at least I find simpler. Any comments? – Brian Chandler Dec 20 '14 at 16:24
Using exactly the same argument as in this question: Optimize volume of an open cardboard box made from flat square of cardboard... we consider an infinitesimal change $\delta$ in the position of the fold between walls and floor. At the maximum, this change must add/subtract the same volume from the change in height as from the moving in/out of the walls. Call the area of the base $B$, the perimeter of the base $P$, and the height $h$, then these changes are:
$\delta \times h \times P$ ...change over walls
$\delta \times B$ ...change over base
And the maximum is when the area of the base and the area of the walls are equal, so we have:
$h = B / P$
In this case, let $s$ be the side of the base. Then the area of the base (an equilateral triangle) and perimeter are given by:
$B = \frac{\sqrt 3}{4} s^2$
$P = 3s$
So
$h = \frac{\sqrt 3}{4} s^2 / 3s = \frac{1}{4\sqrt 3} s$
Now we get $s$ from the original triangle side $a$ and substitute:
$s = a - 2\sqrt 3 h$ | {
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Now we get $s$ from the original triangle side $a$ and substitute:
$s = a - 2\sqrt 3 h$
$h = \frac{1}{4\sqrt 3} (a - 2\sqrt 3 h)$
$4\sqrt 3 h = a - 2\sqrt 3 h$
$h = \frac{a}{6\sqrt 3}$
- | {
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# Projection of ellipsoid
Find the projections of the ellipsoid
$$x^2 + y^2 + z^2 -xy -1 = 0$$
on the cordinates plan
I have no idea how to do this. I couldn't find much on google to help me with it too.
By coordinate planes, I assume you mean the $xy$-plane, $xz$-plane, and $yz$-plane? If so, the projections can be found as follows:
$xy$-plane: Let $z=0$.
$xz$-plane and $yz$-plane: At first try, you may want to use $y=0$ or $z=0$ respectively; however, this will leave you with the cross section with the $xz$-plane, not the projection onto the $xz$-plane. To get the proper projections, we need to analyze the graph of the projection in the $xy$-plane; in particular, find the equations of the horizontal and vertical tangent lines.
In the $xy$-plane, the equation simplifies to $x^2+y^2-xy-1=0$. Implicitly differentiating with respect to $x$ gives us
$$2x+2y\frac{dy}{dx} - y - x\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{y-2x}{2y-x}.$$
Now, you get horizontal tangent lines when the numerator is zero (i.e. when $dy/dx = 0$), which implies that $y-2x=0\implies y=2x$. Substituting this back into the implicit equation gives us $$x^2+4x^2-2x^2 -1 = 0 \implies 3x^2=1 \implies x=\pm \frac{\sqrt{3}}{3}$$
and thus $y = \pm\dfrac{2\sqrt{3}}{3}$ are the equations of the horizontal tangent lines.
Similarly, we find the vertical tangent lines when the denominator is zero (i.e. when $dy/dx$ is undefined), which implies that $2y-x=0\implies x=2y$. In a similar fashion, we get that $y=\pm\dfrac{\sqrt{3}}{3}$ and hence $x=\pm\dfrac{2\sqrt{3}}{3}$ are the equations of the vertical tangent lines. | {
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Now why did we go through all of this? Well, the equations of the tangent lines tells us how far along the $x$ and $y$ axis the ellipse extends; in particular, the vertical tangents give us a bound on $x$ and $y$ for the ellipse (i.e. $-2\sqrt{3}/3 \leq x,y \leq 2\sqrt{3}/3$). These bounds play the role of the major axes for the $xz$ and $yz$ projections of the ellipsoid onto the $y=0$ and $x=0$ planes respectively. The minor axis will be along the $z$ direction; in particular, along the ellipsoid, $-1\leq z\leq 1$. With that said, the corresponding projection of the ellipsoid onto the $yz$-plane is $$\frac{3y^2}{4} + z^2 = 1$$
and the corresponding projection onto the $xz$-plane is $$\frac{3x^2}{4} + z^2 = 1$$
Note that the last two figures are cylinders of the projections as seen in three space; the purpose of visualizing them this way was to show you that they completely enclose the ellipsoid of interest.
• Yes, I meant it, thanks. Well, I did this on a first moment but the answer is different for $yz$ and $xz$. For example: $yz: \frac{3}{4}y^2 + z^2 -1 = 0$ Nov 19, 2013 at 11:17
• @Giiovanna I deleted my answer, fixed it, and then undeleted it. I hope what I wrote above makes sense! Nov 19, 2013 at 12:44
In case you have difficulties with the accepted answer, here is another approach:
Your ellipsoid $S$ is the level surface $F^{-1}(\{0\})$ of the function $$F(x,y,z):=x^2+y^2+z^2-xy-1\ .$$ Let $S'$ be the shadow of $S$ in the $(x,y)$-plane. The points $p\in S$ that generate the boundary $\partial S'$ are characterized by the property that the tangent plane at $p$ is vertical (i.e., parallel to the $z$-axis), which is the same as saying that the surface normal $n_p$ at $p$ is horizontal. Now $n_p$ is given by $\nabla F(p)$, whose third coordinate is $2z$. | {
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It follows that the points $(x,y,z)\in S$ producing the shadow boundary are the points with $z=0$, i.e., the points $(x,y,0)$ satisfying $x^2+y^2-xy-1=0$. In this special example the shadow boundary in fact coincides with the intersection of $S$ with the $(x,y)$-plane.
Therefore let's do the shadow $S''$ of $S$ in the $(x,z)$-plane as well. The points $p\in S$ that generate the boundary $\partial S''$ are characterized by the property that the tangent plane at $p$ is parallel to the $y$-axis, which is the same as saying that the surface normal $n_p$ at $p$ is orthogonal to the $y$-axis, or has $y$-component $0$. The $y$-component of $\nabla F(p)$ is given by $2y-x$. Therefore we can say that the points $p$ in question lie on the plane $2y-x=0$. As they satisfy a priori $F(x,y,z)-1=0$ we can eliminate $y$ from the two equations $$x^2+y^2+z^2-xy-1=0,\qquad y={x\over2}$$ and obtain a single equation connecting their $x$- and $z$-coordinates: $$x^2+{x^2\over4}+z^2-x\>{x\over2}-1=0\ ,$$ or $${3\over4}x^2+z^2-1=0\ .$$ This is already the equation of the shadow boundary $\partial S''$.
I may now safely leave the shadow on the $(y,z)$-plane to you.
• I like this method the most Sep 18, 2017 at 9:10
• It’s worth noting that for a quadric surface, the points that generate its shadow outline are always coplanar.
– amd
Dec 24, 2021 at 19:53
An approach related to Christian Blatter’s answer can be found in Result 8.9 on page 201 of Hartley & Zisserman’s Multiple View Geometry In Computer Vision. Under the camera matrix $\mathtt P$, the outline of the quadric $\mathtt Q$ is the conic given by $\mathtt C^*=\mathtt P\mathtt Q^*\mathtt P^T$. (Here the superscript asterisk indicates the dual conic.) The proof follows directly from the observations that lines $\mathbf l$ tangent to the outline satisfy $\mathbf l^T\mathtt C^*\mathbf l=0$ and back-project to planes $\mathtt P^T\mathbf l$ that are tangent to the original quadric. | {
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In the present problem the quadric is the nondegenerate ellipsoid $$\mathtt Q = \begin{bmatrix}1&-\frac12&0&0 \\ -\frac12&1&0&0 \\ 0&0&1&0 \\ 0&0&0&-1\end{bmatrix}$$ so we can use inverse matrices for the duals: the outline of $\mathtt Q$ under the projection $\mathtt P$ is therefore $\mathtt C = (\mathtt P\mathtt Q^{-1}\mathtt P^T)^{-1}$.
For orthogonal projection onto the $x$-$y$ plane, we can use $$\mathtt P = \begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\end{bmatrix}$$ for the camera matrix and have $$\mathtt Q^{-1}=\begin{bmatrix}\frac43&\frac23&0&0\\\frac23&\frac43&0&0\\0&0&1&0\\0&0&0&-1\end{bmatrix},$$ which gives us $$\mathtt C = \begin{bmatrix}1&-\frac12&0\\-\frac12&1&0\\0&0&-1\end{bmatrix},$$ or $x^2+y^2-xy=1$. For the $x$-$z$ plane, $$\mathtt P = \begin{bmatrix}1&0&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix}$$ and $$\mathtt C = \begin{bmatrix}\frac34&0&0\\0&1&0\\0&0&-1\end{bmatrix},$$ i.e., $\frac34x^2+z^2=1$. A similar computation for the $y$-$z$ plane yields $\frac34y^2+z^2=1$.
• Hello @amd, your solution works like a charm for me! Thank you, for posting this answer. Small question, how do we find the projection matrix, P, for any arbitrary plane? Dec 23, 2021 at 3:47
• @KashishDhal There are various ways, but for any of them you must first choose a coordinate system for the plane. Once you have that, the easiest thing that I can think of is to compute a coordinate transformation that takes that plane to one of the prime coordinate planes and compose it with the corresponding simple projection matrix from above.
– amd
Dec 24, 2021 at 19:49
• I followed the method that you described in this thread: math.stackexchange.com/questions/3073718/… However, I don't know if my solutions are correct. How can I verify that the computed ellipse are indeed correct? Dec 24, 2021 at 20:59 | {
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# If $\{u_1,...,u_n,v_1,...,v_k\}$ is a basis does it necessarily follow that $\{u_1,...,u_n\}$ is linearly independent?
We are given that $\{u_1,...,u_n,v_1,...,v_k\}$ is a basis. My question is does it necessarily follow that $\{u_1,...,u_n\}$ or $\{v_1,...,v_k\}$ are linearly independent? I suspect the answer is yes, but I think I'm missing something in my argument.
So: if $\{u_1,...,u_n,v_1,...,v_k\}$ is a basis, then this list of vectors is linearly independent. That is,
$a_1 u_1 + \cdots + a_n u_n + b_1 v_1 + \cdots + b_k v_k = 0 \implies a_1 = \cdots = a_n = b_1 = \cdots = b_k = 0$.
Then I want to say the following: Since all the $b_i \,'s$ are $0$, we have $a_1 u_1 + \cdots + a_n u_n = 0 \implies a_1 = \cdots = a_n = 0$, meaning that {$u_1,...,u_n$} is linearly independent. With a similar argument, we can also conclude that {$v_1,...,v_k$} is linearly independent.
Am I missing something though? I feel like my argument is kind of making a bit of a stretch. Thanks for your help!!
• You're right! in general, any subset of a linearly independent set is again linearly independent, as your proof shows (you're not using more than linearly independency). Oct 1 '16 at 20:25
• Yes, any subset of a linearly independent set is lin. indep. itself. Oct 1 '16 at 20:25
• It's fine for me. Actually, whether it's a basis is not the relevant hypothesis. What you proved indeed is a little more general: any sublist of a list of linearly independent vectors is linearly independent. Oct 1 '16 at 20:26
• Thank you so much guys!!! :D Oct 1 '16 at 20:36
Your argument is fine, but perhaps this is a case where it can be nicely put as a proof by contradiction. Suppose some sublist were not linearly independent, then we can find some $\{a_i\}$, not all zero, such that $\sum_{i=1}^n a_iu_i=0$. We can then use this to show that the bigger list is not linearly independent. | {
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Tutorial/Coupled problems
Introduction
Here is explained how to use the solution of a first problem as a data in a second problem. Two kind of problem are studied here. In the "first" one, the solution of the first problem directly appears in the weak formulation of the second problem (for example, as a source or a Neumann boundary) whereas in the second kind of problem, the solution is used as a constraint (Dirichlet boundary condition), so it should appears in the function space. Note that a Lagrange multiplier could be used to force the Dirichlet constraint in the weak formulation.
First kind of coupled problem : solution appearing in the weak formulation
Mathematics | {
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First kind of coupled problem : solution appearing in the weak formulation
Mathematics
Let the computation domain be the unit square $\Omega = [0,1]\times[0,1]$ with boundary $\Gamma$ and unit outwardly directed normal $\mathbf{n}$. Consider the two coupled following problems : firstly, find $u$, solution of $$\begin{cases}\label{eq:problemU} -\Delta u + u = C & \text{in } \Omega,\\ \displaystyle{\frac{\partial u}{\partial \mathbf{n}} = 0} & \text{on }\Gamma, \end{cases}$$ where $C$ is a constant. Then, find the solution $v$ of the second problem $$\begin{cases}\label{eq:problemV} -\Delta v + v = 0 & \text{in } \Omega,\\ \displaystyle{\frac{\partial v}{\partial \mathbf{n}} = u} & \text{on }\Gamma, \end{cases}$$ where $u$ is the solution of problem \ref{eq:problemU}. Obviously, the function $u$ is the constant function equal to $C$ and thus, there is no "real" need in solving numerically problem \ref{eq:problemU}. However, please keep in mind that this is just an example. In order to solve numerically these problem using a finite element method, on has to write the weak formulations of problems \ref{eq:problemU} and \ref{eq:problemV}, which read as: $$\label{eq:WeakFormulationU} \left\{\begin{array}{l} \text{Find } u\in H^1(\Omega) \text{ such that}\\ \displaystyle{\forall u'\in H^1(\Omega), \qquad \int_{\Omega} \nabla u\cdot\nabla u' \;{\rm d}\Omega + \int_{\Omega} u\cdot u' \;{\rm d}\Omega - \int_{\Omega} Cu'\;{\rm d}\Omega = 0}, \end{array}\right.$$ and $$\label{eq:WeakFormulationV} \left\{\begin{array}{l} \text{Find } v\in H^1(\Omega) \text{ such that, }\\ \displaystyle{\forall v'\in H^1(\Omega), \qquad \int_{\Omega} \nabla v\cdot\nabla v' \;{\rm d}\Omega + \int_{\Omega}vv' \;{\rm d}\Omega - \int_{\Gamma}uv'\;{\rm d}\Gamma = 0}, \end{array}\right.$$ where $H^1(\Omega)$ is the classical Sobolev space.
Computation in GetDP | {
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Computation in GetDP
In GetDP, this kind of problem is very easy to solve. Indeed, the user has to introduce two function space (one per solution), and ask GetDP to solve each problem in the right order. There is no trap ! Below are the GMSH and GetDP files. Compared to the other academic examples, the only new things are in the .pro file.
Files
There are two files :
FirstCoupledProblem.geo that contains the geometry and is used to mesh the domain
FirstCoupledProblem.pro that contains the weak formulations, function spaces, etc.
To solve the problem, type in a terminal (in the right directory) :
gmsh FirstPb.geo -2
and then solve the problem with getdp :
getdp FirstPb.pro -solve -pos
or directly
getdp FirstPb.pro -solve CoupledProblem -pos Map_u_and_v
//Caracteristic length of the finite elements (reffinement is also possible after the mesh is built):
lc = 0.05;
// The parameters of the border of the domain :
x_max = 1;
x_min = 0;
y_max= 1;
y_min = 0;
//Creation of the 4 angle points of the domain Omega (=square)
p1 = newp; Point(p1) = {x_min,y_min,0,lc};
p2 = newp; Point(p2) = {x_min,y_max,0,lc};
p3 = newp; Point(p3) = {x_max,y_max,0,lc};
p4 = newp; Point(p4) = {x_max,y_min,0,lc};
//The four edges of the square
L1 = newreg; Line(L1) = {p1,p2};
L2 = newreg; Line(L2) = {p2,p3};
L3 = newreg; Line(L3) = {p3,p4};
L4 = newreg; Line(L4) = {p4,p1};
// Line Loop (= boundary of the square)
Bound = newreg; Line Loop(Bound) = {L1,L2,L3,L4};
//Surface of the square
SurfaceOmega = newreg; Plane Surface(SurfaceOmega) = {Bound};
// To conclude, we define the physical entities, that is "what GetDP could see/use".
// 1 is associtated to Omega
// 2 to Gamma
Physical Surface(1) = {SurfaceOmega};
Physical Line(2) = {L1,L2,L3,L4};
// Do not forget to let a blank line at the end, this could make GMSH crash...
Direct link to file CoupledProblems/First/FirstPb.geo'
// Group
//======
Group{
Omega = Region[{1}];
Gama = Region[{2}];
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// Group
//======
Group{
Omega = Region[{1}];
Gama = Region[{2}];
}
// Function
//=========
Function{
C = 5; //constant C in the problem solved by u
}
//Jacobian
//========
Jacobian {
{ Name JVol ;
Case {
{ Region All ; Jacobian Vol ; }
}
}
{ Name JSur ;
Case {
{ Region All ; Jacobian Sur ; }
}
}
{ Name JLin ;
Case {
{ Region All ; Jacobian Lin ; }
}
}
}
//Integration (parameters)
//=======================
Integration {
{ Name I1 ;
Case {
{ Type Gauss ;
Case {
{ GeoElement Point ; NumberOfPoints 1 ; }
{ GeoElement Line ; NumberOfPoints 4 ; }
{ GeoElement Triangle ; NumberOfPoints 6 ; }
{ GeoElement Quadrangle ; NumberOfPoints 7 ; }
{ GeoElement Tetrahedron ; NumberOfPoints 15 ; }
{ GeoElement Hexahedron ; NumberOfPoints 34 ; }
}
}
}
}
}
//FunctionSpace
//=============
FunctionSpace{
//Space for u
BasisFunction{
{Name wn; NameOfCoef vn; Function BF_Node;
Support Region[{Omega, Gama}]; Entity NodesOf[All];}
}
}
//Space for v
BasisFunction{
{Name wn; NameOfCoef vn; Function BF_Node;
Support Region[{Omega, Gama}]; Entity NodesOf[All];}
}
}
}
/*
The two FunctionSpaces are exactly the same. However, keep in mind that
a FunctionSpace can contain only one solution ! That is why we need two FunctionSpaces.
*/
//(Weak) Formulations
//==================
Formulation{
// Problem solved by u :
// Delta u + u = C (Omega)
// dn u = 0 (Gamma)
{Name ProblemU; Type FemEquation;
Quantity{
{Name u; Type Local; NameOfSpace Hgrad_u;}
}
Equation{
In Omega; Jacobian JVol; Integration I1;}
Galerkin{ [Dof{u}, {u}];
In Omega; Jacobian JVol; Integration I1;}
Galerkin{ [-C, {u}];
In Omega; Jacobian JVol; Integration I1;}
}
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"tags": null,
"url": "http://onelab.info/wiki/Tutorial/Coupled_problems"
} |
// Coupled problem : "first kind" :
// Neumann condition on Gamma
// Delta v + v =0 Omega
// dn v = u on Gamma
{Name ProblemV; Type FemEquation;
Quantity{
// v is the unknown
{Name v; Type Local; NameOfSpace Hgrad_v;}
// u is called but will NOT be used as an unknown (NO "Dof")
{Name u; Type Local; NameOfSpace Hgrad_u;}
}
Equation{
In Omega; Jacobian JVol; Integration I1;}
Galerkin{ [Dof{v}, {v}];
In Omega; Jacobian JVol; Integration I1;}
//Neumann condition on Gama (no "Dof" !)
Galerkin{ [-{u}, {v}];
In Gama; Jacobian JSur; Integration I1;}
}
}
}
// Resolution
//===========
Resolution{
// First coupled problem available
{Name CoupledProblem;
System{
{Name PbU; NameOfFormulation ProblemU;}
{Name PbV; NameOfFormulation ProblemV;}
}
Operation{
Generate[PbU]; Solve[PbU]; SaveSolution[PbU];
Generate[PbV]; Solve[PbV]; SaveSolution[PbV];
}
/*Short explanations : GetDP will :
1) generate and solve the problem associated to "u" then save "u" in the space H1_C
2) generate and solve the problem associated to "v" and save "v" in the space H1
*/
}
}
//Post Processing
//===============
PostProcessing{
{Name CoupledProblem; NameOfFormulation ProblemV;
//The associated formulation is "ProblemV" because it involves both "u" and "v"
//(ProblemU only call "u", not "v")
Quantity{
{Name u; Value {Local{[{u}];In Omega; Jacobian JVol;}}}
{Name v; Value {Local{[{v}];In Omega; Jacobian JVol;}}}
}
}
}
//Post Operation
//==============
PostOperation{
{Name Map_u_and_v; NameOfPostProcessing CoupledProblem;
Operation{
Print[u, OnElementsOf Omega, File "sol_u.pos"];
Print[v, OnElementsOf Omega, File "sol_v.pos"];
}
}
}
Second kind of coupled problem : solution used as a Dirichlet boundary condition
Mathematics | {
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"openwebmath_score": 0.7176864147186279,
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"url": "http://onelab.info/wiki/Tutorial/Coupled_problems"
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Second kind of coupled problem : solution used as a Dirichlet boundary condition
Mathematics
In this section, the first problem (\ref{eq:problemU}) does not change but the second problem is now to find $v$, such that $$\begin{cases}\label{eq:problemV2} -\Delta v + v = 0 & \text{in } \Omega,\\ \displaystyle{v = u} & \text{on }\Gamma, \end{cases}$$ where $u$ is the solution of (\ref{eq:problemU}). The function $u$ is now involved as a Dirichlet boundary condition and thus, will appear in the function space of $v$ (again, except if one uses a Lagrange multiplier, which is not the case here). To obtain the weak formulation of problem (\ref{eq:problemV2}), we introduce the following function space \ref{eq:WeakFormulationU} $$H^1_u(\Omega) = \left\{w\in H^1(\Omega) \text{ such that w|_{\Gamma} = u|_{\Gamma}, where u is the solution of problem (1)} \right\}.$$ Thus, we have $$\label{eq:WeakFormulationV2} \left\{\begin{array}{l} \text{Find } v\in H^1_u(\Omega) \text{ such that, }\\ \displaystyle{\forall v'\in H^1_0(\Omega), \qquad \int_{\Omega} \nabla v\cdot\nabla v' \;{\rm d}\Omega + \int_{\Omega}vv' \;{\rm d}\Omega = 0}, \end{array}\right.$$
GetDP
In GetDP, a Dirichlet boundary condition is generally imposed through the "Constraint" term, it can be imposed with the help of a Lagrange multiplier. In the second case, the Dirichlet boundary condition is no more "imposed" in the function space but appears in the weak formulation (as a Neumann boundary condition). Thus, the previous case can be used. Here is only considered the first case using the GetDP function "AssignFromResolution". A Resolution dedicated to $u$ must then be created, in this example it will be named AuxiliaryResolution. In GetDP, the procedure can be summarized as follows: | {
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"tags": null,
"url": "http://onelab.info/wiki/Tutorial/Coupled_problems"
} |
• Create a constraint with :
Type AssignFromResolution ; Resolution AuxiliaryResolution;
• Construct only ONE function space that will contain $u$ and at the end $v$ ($u$ will be erased !)
• Create two Resolution, one for $v$ and one for $u$ :
• for $v$ : Generate, Solve, Save
• for $u$ (AuxiliaryResolution) : Generate, Solve, TransfertSolution (do not Save !)
With this, GetDP will ... :
• Try to solve problem with $v$
• Need the Dirichlet constraint (that is $u$)
• Solve problem with $u$
• Solve problem with $v$
Sometimes it is usefull to save both $u$ and $v$ at the end of the calculation. This can be done by adding an auxiliary problem that "copy" $u$ (see the last paragraph).
Files
There are two files :
Secondpb.geo
exactly the same as FirstPb.geo
Secondpb.pro
quite the same as FirstPb.pro
To solve the problem, type in a terminal (in the right directory) :
gmsh SecondPb.geo -2
and then solve the problem with getdp :
getdp SecondPb.pro -solve -pos
and chose the first resolution (MainResolution) or directly
getdp SecondPb.pro -solve MainResolution -pos Map_v
//Caracteristic length of the finite elements (reffinement is also possible after the mesh is built):
lc = 0.05;
// The parameters of the border of the domain :
x_max = 1;
x_min = 0;
y_max= 1;
y_min = 0;
//Creation of the 4 angle points of the domain Omega (=square)
p1 = newp; Point(p1) = {x_min,y_min,0,lc};
p2 = newp; Point(p2) = {x_min,y_max,0,lc};
p3 = newp; Point(p3) = {x_max,y_max,0,lc};
p4 = newp; Point(p4) = {x_max,y_min,0,lc};
//The four edges of the square
L1 = newreg; Line(L1) = {p1,p2};
L2 = newreg; Line(L2) = {p2,p3};
L3 = newreg; Line(L3) = {p3,p4};
L4 = newreg; Line(L4) = {p4,p1};
// Line Loop (= boundary of the square)
Bound = newreg; Line Loop(Bound) = {L1,L2,L3,L4};
//Surface of the square
SurfaceOmega = newreg; Plane Surface(SurfaceOmega) = {Bound}; | {
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//Surface of the square
SurfaceOmega = newreg; Plane Surface(SurfaceOmega) = {Bound};
// To conclude, we define the physical entities, that is "what GetDP could see/use".
// 1 is associtated to Omega
// 2 to Gamma
Physical Surface(1) = {SurfaceOmega};
Physical Line(2) = {L1,L2,L3,L4};
// Do not forget to let a blank line at the end, this could make GMSH crash...
Direct link to file CoupledProblems/Second/SecondPb.geo'
// Group
//======
Group{
Omega = Region[{1}];
Gama = Region[{2}];
}
// Function
//=========
Function{
C = 5; //constant C in the problem solved by u
}
//Jacobian
//========
Jacobian {
{ Name JVol ;
Case {
{ Region All ; Jacobian Vol ; }
}
}
{ Name JSur ;
Case {
{ Region All ; Jacobian Sur ; }
}
}
{ Name JLin ;
Case {
{ Region All ; Jacobian Lin ; }
}
}
}
//Integration (parameters)
//=======================
Integration {
{ Name I1 ;
Case {
{ Type Gauss ;
Case {
{ GeoElement Point ; NumberOfPoints 1 ; }
{ GeoElement Line ; NumberOfPoints 4 ; }
{ GeoElement Triangle ; NumberOfPoints 6 ; }
{ GeoElement Quadrangle ; NumberOfPoints 7 ; }
{ GeoElement Tetrahedron ; NumberOfPoints 15 ; }
{ GeoElement Hexahedron ; NumberOfPoints 34 ; }
}
}
}
}
}
// Dirichlet Constraint
Constraint{
{Name DirichletV;
Case{
{Region Gama; Type AssignFromResolution; NameOfResolution AuxiliaryResolution; }
}
}
}
/*
The constraint will be obtained through the resolution of "AuxiliaryResolution".
This means that GetDP will automatically solve "AuxiliaryResolution" when it will
need the contraint !
*/
//FunctionSpace
//=============
FunctionSpace{
//Space for u AND v !!
BasisFunction{
{Name wn; NameOfCoef vn; Function BF_Node;
Support Region[{Omega, Gama}]; Entity NodesOf[All];}
}
//the dirichlet constraint "v = u" on Gamma
Constraint{
{NameOfCoef vn; EntityType NodesOf;
NameOfConstraint DirichletV;}
}
}
}
/*
ONLY one function space that will contain "u" and at the end, "v".
*/
//(Weak) Formulations
//================== | {
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"tags": null,
"url": "http://onelab.info/wiki/Tutorial/Coupled_problems"
} |
//(Weak) Formulations
//==================
Formulation{
// Problem solved by u :
// Delta u + u = C (Omega)
// dn u = 0 (Gamma)
{Name ProblemU; Type FemEquation;
Quantity{
{Name u; Type Local; NameOfSpace Hgrad;}
}
Equation{
In Omega; Jacobian JVol; Integration I1;}
Galerkin{ [Dof{u}, {u}];
In Omega; Jacobian JVol; Integration I1;}
Galerkin{ [-C, {u}];
In Omega; Jacobian JVol; Integration I1;}
}
}
// Coupled problem : "first kind" :
// DIRICHLET condition on Gamma
// Delta v + v =0 Omega
// v = u on Gamma
{Name ProblemV; Type FemEquation;
Quantity{
// v is the unknown
{Name v; Type Local; NameOfSpace Hgrad;}
}
Equation{
In Omega; Jacobian JVol; Integration I1;}
Galerkin{ [Dof{v}, {v}];
In Omega; Jacobian JVol; Integration I1;}
}
}
}
// Resolution
//===========
Resolution{
//Main Resolution : the one that will be called when launching getdp
{Name MainResolution;
System{
{Name PbV; NameOfFormulation ProblemV;}
}
Operation{
Generate[PbV]; Solve[PbV]; SaveSolution[PbV];
}
}
/*Short explanations : GetDP will :
generate and solve the problem associated to "v" and save "v" in the space H1
However, to compute "v", GetDP need to compute the Dirichlet Contraint, thus
is needs to solve AuxiliaryResolution (bellow) !
*/
// Auxiliary resolution that computes "u"
{Name AuxiliaryResolution;
System{
//System associated to "u" will be transfered to the weak formulation of "v"
{Name PbU; NameOfFormulation ProblemU; DestinationSystem PbV;}
{Name PbV; NameOfFormulation ProblemV;}
}
Operation{
Generate[PbU]; Solve[PbU]; TransferSolution[PbU];
}
}
/*Short explanations : GetDP will :
1) generate and solve the problem associated to "u"
2) Transfert solution u (to problem associated to "v")
*/
}
//Post Processing
//===============
PostProcessing{
{Name CoupledProblem; NameOfFormulation ProblemV;
Quantity{
{Name v; Value {Local{[{v}];In Omega; Jacobian JVol;}}}
}
}
}
//Post Operation
//============== | {
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"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.986979508737192,
"lm_q1q2_score": 0.8609889964100731,
"lm_q2_score": 0.8723473879530491,
"openwebmath_perplexity": 13616.122405568942,
"openwebmath_score": 0.7176864147186279,
"tags": null,
"url": "http://onelab.info/wiki/Tutorial/Coupled_problems"
} |
//Post Operation
//==============
PostOperation{
{Name Map_v; NameOfPostProcessing CoupledProblem;
Operation{
Print[v, OnElementsOf Omega, File "sol_v.pos"];
}
}
}
Second kind of coupled problem with a save of solution $u$
Sometimes, it can be usefull to save both $u$ and $v$. To achieved this, one can introduce an auxiliary variable $u_{aux}$ that will be a copy of $u$. Here, we proposed to add :
A new FunctionSpace to store $u$
A new weak formulation to copy $u$ into an auxiliary variable $u_{aux}$
The modified .pro file (SecondPbWithSave.pro) can be found below. The .geo file being exactly the same as the previous ones.
//Caracteristic length of the finite elements (reffinement is also possible after the mesh is built):
lc = 0.05;
// The parameters of the border of the domain :
x_max = 1;
x_min = 0;
y_max= 1;
y_min = 0;
//Creation of the 4 angle points of the domain Omega (=square)
p1 = newp; Point(p1) = {x_min,y_min,0,lc};
p2 = newp; Point(p2) = {x_min,y_max,0,lc};
p3 = newp; Point(p3) = {x_max,y_max,0,lc};
p4 = newp; Point(p4) = {x_max,y_min,0,lc};
//The four edges of the square
L1 = newreg; Line(L1) = {p1,p2};
L2 = newreg; Line(L2) = {p2,p3};
L3 = newreg; Line(L3) = {p3,p4};
L4 = newreg; Line(L4) = {p4,p1};
// Line Loop (= boundary of the square)
Bound = newreg; Line Loop(Bound) = {L1,L2,L3,L4};
//Surface of the square
SurfaceOmega = newreg; Plane Surface(SurfaceOmega) = {Bound};
// To conclude, we define the physical entities, that is "what GetDP could see/use".
// 1 is associtated to Omega
// 2 to Gamma
Physical Surface(1) = {SurfaceOmega};
Physical Line(2) = {L1,L2,L3,L4};
// Do not forget to let a blank line at the end, this could make GMSH crash...
Direct link to file CoupledProblems/SecondWithSave/SecondPbWithSave.geo'
// Group
//======
Group{
Omega = Region[{1}];
Gama = Region[{2}];
}
// Function
//=========
Function{
C = 5; //constant C in the problem solved by u
} | {
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// Function
//=========
Function{
C = 5; //constant C in the problem solved by u
}
//Jacobian
//========
Jacobian {
{ Name JVol ;
Case {
{ Region All ; Jacobian Vol ; }
}
}
{ Name JSur ;
Case {
{ Region All ; Jacobian Sur ; }
}
}
{ Name JLin ;
Case {
{ Region All ; Jacobian Lin ; }
}
}
}
//Integration (parameters)
//=======================
Integration {
{ Name I1 ;
Case {
{ Type Gauss ;
Case {
{ GeoElement Point ; NumberOfPoints 1 ; }
{ GeoElement Line ; NumberOfPoints 4 ; }
{ GeoElement Triangle ; NumberOfPoints 6 ; }
{ GeoElement Quadrangle ; NumberOfPoints 7 ; }
{ GeoElement Tetrahedron ; NumberOfPoints 15 ; }
{ GeoElement Hexahedron ; NumberOfPoints 34 ; }
}
}
}
}
}
// Dirichlet Constraint
Constraint{
{Name DirichletV;
Case{
{Region Gama; Type AssignFromResolution; NameOfResolution AuxiliaryResolution; }
}
}
}
/*
The constraint will be obtained through the resolution of "AuxiliaryResolution".
This means that GetDP will automatically solve "AuxiliaryResolution" when it will
need the contraint !
*/
//FunctionSpace
//=============
FunctionSpace{
//Space for u
BasisFunction{
{Name wn; NameOfCoef vn; Function BF_Node;
Support Region[{Omega, Gama}]; Entity NodesOf[All];}
}
}
//Space for u_aux (=copy of u) AND v !!
BasisFunction{
{Name wn; NameOfCoef vn; Function BF_Node;
Support Region[{Omega, Gama}]; Entity NodesOf[All];}
}
//the dirichlet constraint "v = u" on Gamma
Constraint{
{NameOfCoef vn; EntityType NodesOf;
NameOfConstraint DirichletV;}
}
}
}
//(Weak) Formulations
//==================
Formulation{
// Problem solved by u :
// Delta u + u = C (Omega)
// dn u = 0 (Gamma)
{Name ProblemU; Type FemEquation;
Quantity{
{Name u; Type Local; NameOfSpace Hgrad_u;}
}
Equation{
In Omega; Jacobian JVol; Integration I1;}
Galerkin{ [Dof{u}, {u}];
In Omega; Jacobian JVol; Integration I1;}
Galerkin{ [-C, {u}];
In Omega; Jacobian JVol; Integration I1;}
}
} | {
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"openwebmath_score": 0.7176864147186279,
"tags": null,
"url": "http://onelab.info/wiki/Tutorial/Coupled_problems"
} |
// Coupled problem : "first kind" :
// DIRICHLET condition on Gamma
// Delta v + v =0 Omega
// v = u on Gamma
{Name ProblemV; Type FemEquation;
Quantity{
// v is the unknown
{Name u; Type Local; NameOfSpace Hgrad_u;}
{Name v; Type Local; NameOfSpace Hgrad_v;}
}
Equation{
In Omega; Jacobian JVol; Integration I1;}
Galerkin{ [Dof{v}, {v}];
In Omega; Jacobian JVol; Integration I1;}
}
}
//Copy "u" in "u_aux"
{Name CopyU; Type FemEquation;
Quantity{
// v is the unknown
{Name u; Type Local; NameOfSpace Hgrad_u;}
{Name u_aux; Type Local; NameOfSpace Hgrad_v;}
}
Equation{
Galerkin{ [Dof{u_aux}, {u_aux}];
In Omega; Jacobian JVol; Integration I1;}
Galerkin{ [-{u}, {u_aux}];
In Omega; Jacobian JVol; Integration I1;}
}
}
}
// Resolution
//===========
Resolution{
//Main Resolution : the one that will be called when launching getdp
{Name MainResolution;
System{
{Name PbV; NameOfFormulation ProblemV;}
}
Operation{
Generate[PbV]; Solve[PbV]; SaveSolution[PbV];
}
}
/*Short explanations : GetDP will :
Try to compute "v", however, GetDP needs to compute the Dirichlet Contraint
thus to solve AuxiliaryResolution (below) !
*/
// Auxiliary resolution that computes "u"
{Name AuxiliaryResolution;
System{
//System associated to "u" will be transfered to the weak formulation of "v"
{Name PbU; NameOfFormulation ProblemU;}
{Name PbUaux; NameOfFormulation CopyU; DestinationSystem PbV;}
{Name PbV; NameOfFormulation ProblemV;}
}
Operation{
Generate[PbU]; Solve[PbU]; SaveSolution[PbU];
Generate[PbUaux]; Solve[PbUaux]; TransferSolution[PbUaux];
}
}
/*Short explanations : GetDP will :
1) Compute "u"
2) Copy "u" in "uaux"
3) Transfert solution "uaux" (to problem associated to "v")
*/
}
//Post Processing
//===============
PostProcessing{
{Name CoupledProblem; NameOfFormulation ProblemV;
Quantity{
{Name u; Value {Local{[{u}];In Omega; Jacobian JVol;}}}
{Name v; Value {Local{[{v}];In Omega; Jacobian JVol;}}}
}
}
}
//Post Operation
//============== | {
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"openwebmath_score": 0.7176864147186279,
"tags": null,
"url": "http://onelab.info/wiki/Tutorial/Coupled_problems"
} |
//Post Operation
//==============
PostOperation{
{Name Map_Sol; NameOfPostProcessing CoupledProblem;
Operation{
Print[u, OnElementsOf Omega, File "sol_u.pos"];
Print[v, OnElementsOf Omega, File "sol_v.pos"];
}
}
} | {
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"lm_q1q2_score": 0.8609889964100731,
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"openwebmath_score": 0.7176864147186279,
"tags": null,
"url": "http://onelab.info/wiki/Tutorial/Coupled_problems"
} |
Prove that the number of answers for $|a_1|+|a_2|+…+|a_k| \le n$ is equal to the number of answers for $|a_1|+|a_2|+…+|a_n| \le k$.
I'm trying to solve this problem:
Prove that the number of answers for $|a_1|+|a_2|+...+|a_k|≤n$ is equal to the number of answers for $|a_1|+|a_2|+...+|a_n|≤k$. all $a_i$ is an integer number.
I have a solution for non negative $a_i$: We convert $|a_1|+|a_2|+...+|a_k|≤n$ to $a_1+a_2+...+a_k+c=n$ and convert $|a_1|+|a_2|+...+|a_n|≤k$ to $a_1+a_2+...+a_n+c=k$.
By bars and stars theorem: Number of answers for $a_1+a_2+...+a_k+c=n$ is equal to $\binom{n+k}{k}$ and number of answers for $a_1+a_2+...+a_n+c=k$ is equal to $\binom{n+k}{n}$. That $\binom{n+k}{k} = \binom{n+k}{n}$.
but i don't know how to prove for negative integers.
• So we allow $a_i$ to be any integer, but we consider its absolute value when taking the sum? – hardmath Apr 19 '18 at 0:41
• @hardmath yes,we take absolute value – user547757 Apr 19 '18 at 14:57
Let $f(n,k)$ be the number of solutions to $|a_1|+\dots+|a_n|\le k$. Note that $$f(n,k) = f(n-1,k)+2f(n-1,k-1)+2f(n-1,k-2)+\dots+2f(n-1,0),\tag{*}$$ by conditioning on the value of $a_n$. Rearranging, and using $(*)$ applied to $f(n,k-1)$, we get \begin{align} f(n,k) &= f(n-1,k)+f(n-1,k-1)+[f(n-1,k-1)+2f(n-1,k-2)+\dots+2f(n-1,0)]\\ &= f(n-1,k)+f(n-1,k-1)+f(n,k-1) \end{align} The above shows that $f(n,k)$ obeys a recurrence which is symmetric in $n,k$. Since you can easily verify the symmetric base cases $f(n,0)=f(0,k)=1$, the result $f(n,k)=f(k,n)$ follows by induction.
We count the number of solutions to $\sum_{i=1}^n |a_i|\leq k$. | {
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We count the number of solutions to $\sum_{i=1}^n |a_i|\leq k$.
There will be $j\leq r:={\rm min}\{n,k\}$ of the $a_i$ that are $\ne0$, call them $a_{i_1}$, $\ldots$, $a_{i_j}$. The $i_l$ $(1\leq l\leq j)$ can be chosen in ${n\choose j}$ ways. For the chosen $i_l$ put $|a_{i_l}|=x_l+1$ with $x_l\geq 0$. Then we have to count the number of solutions to $x_1+\ldots+x_j+c=k-j$. This number is ${k\choose j}$. It has to be multiplied by $2^j$ in order to account for the choice of the signs of the corresponding $a_{i_l}$. The total number of solutions therefore comes to $$\sum_{j=0}^r{n\choose j}{k\choose j}2^j\ ,$$ which is symmetric in $n$ and $k$.
Introducing notation $$A_i=|a_i|$$ the first of inequalities can be rewritten as: $$A_1+A_2+...+A_k\le n.\tag1$$ Any solution to $$(1)$$ can be viewed as composed of $$i$$ zeros ($$0\le i\le k$$) and $$k-i$$ non-zeros. Each non-zero value doubles the number of solutions to original equation (which can be both positive and negative). The overall number of solutions is therefore $$\sum_{i=0}^{k}\binom{k}{i}\binom{n}{k-i}2^{k-i} \stackrel{i=k-j}=\sum_{j=0}^{k}\binom{k}{j}\binom{n}{j}2^{j} =\sum_{j=0}^{\min(k,n)}\binom{k}{j}\binom{n}{j}2^{j},\tag2$$ where $$\binom{k}{i}$$ counts the number of ways to place $$i$$ zeros into $$k$$ cells, and $$\binom{n}{k-i}$$ counts the number of positive solutions to the inequality $$p_1+p_2+\cdots+p_{k-i}\le n.$$
In view of symmetry of $$(2)$$ with respect to interchange of $$n$$ and $$k$$ the original statement is proved. | {
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Just to show another approach to the answer already given by C.Blatter, let's start and consider $$\left\{ \matrix{ 0 \le b_{\,j} \in \mathbb Z \hfill \cr b_{\,1} + b_2 + b_{\,3} + \cdots + b_{\,k} \le n \hfill \cr} \right.$$ Then the number of solutions is the integral volume $V(k,n)$ of k-D simplex, with edge segments $(0,n)$, thus of length $n+1$.
It is not difficult to demonstrate that $$V(k,n)= \binom{n+k}{n}= \binom{n+k}{k}=V(n,k)$$ then the hypothesis is already true in this case.
We can then partition the above solutions into those that contains $j$ variables null and the remaining $k-j$ having values greater than $0$. We can assign the null variables in $\binom{k}{j}$ ways.
That corresponds to $$V(k,n)= \binom{n+k}{n}= \binom{n+k}{k} = \sum\limits_{0\, \le \,j\, \le \,k} { \binom{n}{k-j} \binom{k}{j} }$$
Coming to our present case, for each non null variable $b_j$ we can assign two values to the corresponding $a_j$, and just one to the null variables. So in this case the volume will be $$V_{\,a} \,(k,n) = \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,k} \right)} {\binom{n}{k-j} \binom{k}{j} 2^{\,k - j} } = \sum\limits_{\left( {0\, \le } \right)\,l\,\left( { \le \,k} \right)} {\binom{n}{l} \binom{k}{l} 2^l } = V_{\,a} \,(n,k)$$
Therefore the hypothesis is demonstrated.
I just saw this question as it came back to the front page. I wrote this answer a couple of months ago, which proves that this number is $$\sum_{k=0}^n2^k\binom{n}{k}\binom{m}{k}$$ which is also shown in answers to this question. My answer starts with a recurrence (like Mike Earnest's answer) and solves it by induction. | {
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# Upper limit on the central binomial coefficient
What is the tightest upper bound we can establish on the central binomial coefficients $2n \choose n$ ?
I just tried to proceed a bit, like this:
$n! > n^{\frac{n}{2}}$
for all $n>2$. Thus,
$\binom{2n}{n} = \frac{ (n+1) \ldots (2n) }{n!} < \frac{\left(\frac{\sum_{k=1}^n (n+k) }{n}\right)^n }{n^{n/2}} = \frac{ \left( \frac{ n^2 + \frac{n(n+1)}{2} }{n} \right) ^n}{n^{n/2}} = \left( \frac{3n+1}{2\sqrt{n}} \right)^n$
But, I was searching for more tighter bounds using elementary mathematics only (not using Stirling's approximation etc.).
• The simplest upper bound to prove is $4^n$ (which is still stronger than your bound) and just follows from the binomial expansion of $(1+1)^{2n}$. Peter's answer gives a less wasteful estimate. Jun 14 '13 at 13:00
• I should have thought about it. Yes, Peter's bound is very good, I tested it too. Jun 14 '13 at 13:54
Here's a way to motivate and refine the argument that Péter Komjáth attributes to Erdős. | {
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Start by computing the ratio between the $$n$$-th and $$(n-1)$$-st central binomial coefficients: $${2n \choose n} \left/ {2(n-1) \choose n-1} \right. = \frac{(2n)! \phantom. / \phantom. n!^2}{(2n-2)! \phantom. / \phantom. (n-1)^2} = \frac{(2n)(2n-1)}{n^2} = 4 \left( 1 - \frac1{2n} \right).$$ For large $$n$$, this ratio approaches $$4$$, suggesting that $$2n \choose n$$ grows roughly as $$4^n$$. If the factor $$1 - \frac1{2n}$$ were $$1 - \frac1n = (n-1)/n$$, the growth would be exactly proportional to $$n^{-1} 4^n$$. Since $$1 - \frac1{2n}$$ is (for large $$n$$) nearly the square root of $$1 - \frac1n$$, the actual asymptotic should be proportional to $$n^{-1/2} 4^n$$. So we introduce the ratio $$r_n := \left( {2n \choose n} \left/ \frac{4^n}{\sqrt n} \right. \right)^2 = \frac{n}{16^n} {2n \choose n}^2.$$ Then $$\frac{r_n}{r_{n-1}} = \left( 1 - \frac1{2n} \right)^2 \left/ \left( 1 - \frac1n \right) \right. = \frac{(2n-1)^2}{(2n-2)(2n)} \gt 1.$$ Thus $$r_{n-1} < r_n$$; and since $$r_1 = (2/4)^2 = 1/4$$ we have by induction $$r_1 \lt r_2 \lt r_3 \lt r_4 \lt \cdots \lt r_n = \frac12 \frac{1 \cdot 3}{2 \cdot 2} \frac{3 \cdot 5}{4 \cdot 4} \frac{5 \cdot 7}{6 \cdot 6} \cdots \frac{(2n-3)(2n-1)}{(2n-2)(2n-2)} \frac{2n-1}{2n}.$$ Each $$r_{n_0}$$ gives a lower bound on $$r_n$$, and thus on $$2n\choose n$$, for all $$n \geq n_0$$. The OP asked for upper bounds, so consider $$R_n := \frac{2n}{2n-1} r_n = \frac{n}{\left(1-\frac 1{2n}\right)16^n} {2n \choose n}^2.$$ Now $$R_{n+1}/R_n = (2n-1)(2n+1) \phantom. / \phantom. (2n)^2 = (4n^2-1) \phantom. / \phantom. (4n^2) \lt 1$$, so $$\frac12 = R_1 \gt R_2 \gt R_3 \gt R_4 \gt \cdots \gt R_{n+1} = \frac12 \frac{1 \cdot 3}{2 \cdot 2} \frac{3 \cdot 5}{4 \cdot 4} \frac{5 \cdot 7}{6 \cdot 6} \cdots \frac{(2n-3)(2n-1)}{(2n-2)(2n-2)}.$$ It follows that $$R_n \geq r_{n'}$$ for any $$n,n'$$, so $$R_1=1/2$$, $$R_2=3/8$$, $$R_3=45/128$$, etc. are a series of upper bounds on every $$r_n$$. Since moreover $$r_n / R_n = 1 - \frac1{2n} \rightarrow | {
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a series of upper bounds on every $$r_n$$. Since moreover $$r_n / R_n = 1 - \frac1{2n} \rightarrow 1$$ as $$n \rightarrow \infty$$, both $$r_n$$ and $$R_n$$ converge to a common limit that is an upper bound on every $$r_n$$. If we accept Wallis's product (which is classical though not as elementary as everything else in our analysis), then we can evaluate this common limit as $$1/\pi$$ and thus recover the asymptotically sharp upper bound $${2n \choose n} < 4^n / \sqrt{\pi n}$$. | {
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• +1, and let me challenge your statement that the Wallis product is not as elementary, by pointing to the Monthly note math.chalmers.se/~wastlund/monthly.pdf and the 16th Christmas Tree Lecture by Don Knuth, "Why pi?" (now on youtube). Jul 1 '13 at 16:58
• Your answer has been cited in this paper that I was just reading, mentioning it here as a token of appreciation: arxiv.org/pdf/2005.10009.pdf
– NULL
Jul 28 '20 at 19:59
Even the asymptotically sharp inequality ${2n \choose n} < 4^n \left/ \sqrt{\pi n} \right.$ has a short proof: $${2n \choose n} = \frac{4^n}{\pi} \int_{-\pi/2}^{\pi/2} \cos^{2n} x \phantom. dx < \frac{4^n}{\pi} \int_{-\pi/2}^{\pi/2} e^{-nx^2} dx < \frac{4^n}{\pi} \int_{-\infty}^{\infty} e^{-nx^2} dx = \frac{4^n}{\sqrt{\pi n}}.$$ In the first step, the formula for $\int_{-\pi/2}^{\pi/2} \cos^{2n} x \phantom. dx$ can be proved by induction via integration by parts, or using the Beta function.
The third step is clear, and the last step is the well-known Gaussian integral. So we need only justify the the second step.
There we need the inequality $\cos x \leq e^{-x^2/2}$, or equivalently $$\log \cos x + \frac{x^2}{2} \leq 0,$$ for $\left|x\right| < \pi/2$, with equality only at $x=0$. This is true because $\log \cos x +\frac12 x^2$ is an even function of $x$ that vanishes at $x=0$ and whose second derivative $-\tan^2 x$ is negative for all nonzero $x \in (-\pi/2, \pi/2)$. QED
• I'm not sure what counts as elementary in this game, but I would think the easiest proof that $\frac{1}{\pi} \int \cos^{2n} x dx = \binom{2n}{n}/4^n$ is to write $\cos x = (e^{ix} + e^{-ix})/2$ and expand $\cos^{2n}$ by the binomial theorem. Jul 1 '13 at 14:18 | {
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Erdos remarked somewhere the bound
$${{2n}\choose{n}}<\frac{4^n}{\sqrt{2n+1}}.$$ This can be established by induction: $${{2n+2}\choose{n+1}}=\frac{(2n+1)(2n+2)}{(n+1)(n+1)}{{2n}\choose{n}}$$ and if we have the bound for $n$, we only have to show $$\frac{2(2n+1)}{(n+1)\sqrt{2n+1}}<\frac{4}{\sqrt{2n+3}}$$ which reduces to $4n^2+8n+3<4n^2+8n+4$.
• This bound is pretty close to optimal because (unless I've computed badly) Stirling's approximation gives $4^n/\sqrt{\pi n}$. Jun 14 '13 at 13:26
Noam Elkies notes that there is a quick proof of $$\binom{2n}{n} \leq \frac{4^n}{\sqrt{\pi n}}$$ by writing $$\binom{2n}{n} = \frac{4^n}{\pi} \int_{-\pi/2}^{\pi/2} \cos^{2n} x dx$$ and bounding $\cos^2 x \leq e^{-x^2}$.
There is an equally good lower bound by a similar method: $$\int_{-\pi/2}^{\pi/2} \cos^{2n} x dx =\int_{-\pi/2}^{\pi/2} \frac{\sec^2 x dx}{(\tan^2 x+1)^{n+1}} = \int_{- \infty}^{\infty} \frac{du}{(1+u^2)^{n+1}} \geq \int_{- \infty}^{\infty} e^{-(n+1) u^2} du$$ so $$\binom{2n}{n} \geq \frac{4^n}{\sqrt{\pi (n+1)}}.$$ Here the inequality $\tfrac{1}{1+u^2} \geq e^{-u^2}$ follows from the standard bound $e^y \geq 1+y$ for $y\geq 0$.
We may prove without using integrals that $$\frac1{\sqrt{\pi n}}\geqslant 4^{-n}{2n\choose n}\geqslant \frac{1}{\sqrt{\pi(n+1/2)}}.\quad\quad (1)$$ (1) is equivalent to $$2n\left(4^{-n}{2n\choose n}\right)^2=\frac12\prod_{k=2}^n \frac{(2k-1)^2}{2k(2k-2)}:=d_n\leqslant \frac2\pi\leqslant c_n\\:=(2n+1)\left(4^{-n}{2n\choose n}\right)^2=\prod_{k=1}^n\left(1-\frac1{4k^2}\right)$$ (the identities are straightforward by induction). | {
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Denote $$m:=2n+1$$. It is not hard to show that $$\sin mx=p_m(\sin x)$$ for a polynomial of degree $$m$$ in $$\sin x$$. The roots of $$p_m$$ are $$\sin \frac{k\pi}{2n+1}$$ for $$k=-n,\ldots,n$$. Thus $$\sin (2n+1)x=(2n+1)\sin x\prod_{k=1}^n\left(1-\frac{\sin^2 x}{\sin^2 \frac{k\pi}{2n+1}}\right)$$ (the multiple $$2n+1$$ comes from dividing by $$x$$ and putting $$x=0$$). Put $$x=\frac{\pi}{4n+2}$$. Using the inequality $$\sin tx\leqslant t\sin x$$ (which may be proved, for example, by induction in $$t=1,2,\ldots$$ from the identity $$\sin(t+1)x=\sin tx\cos x+\sin x\cos tx$$) for $$t=2k$$ ($$k=1,2,\ldots,n$$) we get $$1\leqslant (2n+1)\sin\frac{\pi}{4n+2}\prod_{k=1}^n\left(1-\frac1{4k^2}\right)\leqslant \frac{\pi}{2}c_n,\quad c_n\geqslant \frac{2}\pi.$$ Analogously we get $$\cos 2nx=\prod_{k=1}^n\left(1-\frac{\sin^2 x}{\sin^2 \frac{\pi(2k-1)}{4n}}\right).$$
Dividing by $$1-\frac{\sin^2 x}{\sin^2 \frac{\pi}{4n}}$$ and substituting $$x=\frac{\pi}{4n}$$ (for computing the LHS at this point use l'Hôpital rule) we get $$n\tan\frac{\pi}{4n}\leqslant \prod_{k=2}^n\left(1-\frac1{(2k-1)^2}\right)=\frac1{2d_n},\quad d_n\leqslant \frac1{2n\tan \frac{\pi}{4n}}\leqslant \frac2\pi.$$
UPD. This is less or more equivalent to Yaglom brothers proof (1953), see their Russian paper (it contains also the derivation of identities $$\sum 1/n^2=\pi^2/6$$ and $$\sum (-1)^{k-1}/(2k-1)=\pi/4$$ using these trigonometric things.) | {
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• Nice proof! Is there an easy way to get the tight lower bound $1/\sqrt{\pi n+1}$? It has such a nice form..
– aorq
Jan 1 at 2:54
• @aorq It is not so much tight. The sequence $4^{-n}{2n\choose n}\sqrt{n+c}$ eventually decreases if $c>1/4$ (this is straightforward: divide the squares of two consecutive terms and subtract 1) and its limit equals $1/\sqrt{\pi}$. Thus for large enough $n$ we get the upper bound $1/\sqrt{\pi (n+c)}$. For $c=1/\pi$ this just works from the very beginning. Jan 1 at 9:51
• Yes, I know that it's tight in the weakest possible sense, namely equality for $n=0$ and poor otherwise. Shortly after I wrote my comment, I saw that $1/\sqrt{\pi(n+1/4)}$ was tighter in a more meaningful sense (on the other side), namely ever better as $n\to\infty$. Thanks for your explanation!
– aorq
Jan 1 at 14:41
• Warning: $1/\sqrt{\pi(n+1/4)}$ is an upper bound Jan 1 at 15:08
You may also prove $${2n\choose n}\sim \frac{4^n}{\sqrt{\pi n}}$$ using the combinatorial meaning of $${2n\choose n}$$ and the area-of-a-circle definition of $$\pi$$. That is probably the most elementary we can hope for.
First of all, if $$a_n:=\sqrt{n+1/4}{2n\choose n}4^{-n}$$ and $$b_n:=\sqrt{n+1/2}{2n\choose n}4^{-n}$$, we may check that $$a_n\leqslant b_n$$, $$a_n$$ increases and $$b_n$$ decreases, thus $$a_n$$, $$b_n$$ have a common positive finite limit which we denote $$\alpha$$, and $$a_n\leqslant \alpha\leqslant b_n$$. Since $$b_n/a_n=1+O(1/n)$$, we get $$a_n=\alpha+O(1/n)$$, and also $$c_n=\alpha+O(1/n)$$ where $$c_n:=\sqrt{n}{2n\choose n}4^{-n}$$.
We should prove that $$\alpha=1/\sqrt{\pi}$$. | {
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We should prove that $$\alpha=1/\sqrt{\pi}$$.
Use the combinatorial identity $$4^{-n}\sum_{k=0}^n k{2k\choose k}(n-k){2(n-k)\choose n-k}=\frac{n(n-1)}8.\quad (\heartsuit)$$ LHS of $$(\heartsuit)$$ equals $$\sum_{k=0}^n \sqrt{k(n-k)}c_kc_{n-k}=\sum_{k=0}^n \sqrt{k(n-k)}\left(\alpha^2+O\left(\frac1k+\frac1{n-k}\right)\right)\\=\alpha^2\sum_{k=0}^n\sqrt{k(n-k)}+O\left(n\sum_{k=0}^n\frac1{\sqrt{k(n-k)}}\right)=\alpha^2\cdot \frac{\pi n^2}8+o(n^2),$$ since the union of rectangles $$[k,k+1]\times [0,\sqrt{k(n-k)}]$$ approximates the upper semi-circle with the horizontal diameter $$[0,n]$$.
Thus $$\alpha^2\pi/8=1/8$$ and $$\alpha=1/\sqrt{\pi}$$.
You could use instead of $$(\heartsuit)$$ the more famous identity $$4^{-n}\sum_{k=0}^n {2k\choose k}{2(n-k)\choose n-k}=1$$ this gives $$\alpha^2\int_0^1\frac{dx}{\sqrt{x(1-x)}}=1$$ (the integral equals $$\pi$$ of course, but this is less "elementary" then the area of a circle.)
This argument is similar to J. Wästlund's (An elementary proof of Wallis' product formula for $$\pi$$, 2007).
Since it gives tighter bounds, I will reproduce my answer from MathSE. | {
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Since it gives tighter bounds, I will reproduce my answer from MathSE.
For $$n\ge0$$, we have (by cross-multiplication) \begin{align} \left(\frac{n+\frac12}{n+1}\right)^2 &=\frac{n^2+n+\frac14}{n^2+2n+1}\\ &\le\frac{n+\frac13}{n+\frac43}\tag{1} \end{align} Therefore, \begin{align} \frac{\binom{2n+2}{n+1}}{\binom{2n}{n}} &=4\frac{n+\frac12}{n+1}\\ &\le4\sqrt{\frac{n+\frac13}{n+\frac43}}\tag{2} \end{align} Inequality $$(2)$$ implies that $$\boxed{\bbox[5pt]{\displaystyle\binom{2n}{n}\frac{\sqrt{n+\frac13}}{4^n}\text{ is decreasing}}}\tag{3}$$ For $$n\ge0$$, we have (by cross-multiplication) \begin{align} \left(\frac{n+\frac12}{n+1}\right)^2 &=\frac{n^2+n+\frac14}{n^2+2n+1}\\ &\ge\frac{n+\frac14}{n+\frac54}\tag{4} \end{align} Therefore, \begin{align} \frac{\binom{2n+2}{n+1}}{\binom{2n}{n}} &=4\frac{n+\frac12}{n+1}\\ &\ge4\sqrt{\frac{n+\frac14}{n+\frac54}}\tag{5} \end{align} Inequality $$(5)$$ implies that $$\boxed{\bbox[5pt]{\displaystyle\binom{2n}{n}\frac{\sqrt{n+\frac14}}{4^n}\text{ is increasing}}}\tag{6}$$ Note that the formula in $$(3)$$, which is decreasing, is bigger than the formula in $$(6)$$, which is increasing. Their ratio tends to $$1$$; therefore, they tend to a common limit, $$L$$.
Theorem $$1$$ from this answer says $$\lim_{n\to\infty}\frac{\sqrt{\pi n}}{4^n}\binom{2n}{n}=1\tag{7}$$ which means that \begin{align} L &=\lim_{n\to\infty}\frac{\sqrt{n}}{4^n}\binom{2n}{n}\\ &=\frac1{\sqrt\pi}\tag8 \end{align} Combining $$(3)$$, $$(6)$$, and $$(8)$$, we get $$\boxed{\bbox[5pt]{\displaystyle\frac{4^n}{\sqrt{\pi\!\left(n+\frac13\right)}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi\!\left(n+\frac14\right)}}}}\tag9$$ | {
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# Homework Help: Probability question
Tags:
1. Jan 6, 2017
### gruba
Two players, $A$ and $B$, are playing a sequence of matches of some game. In each match, one player wins (there is no draw game in any match). For winning the match, the player gets $1$ point, and for losing, $0$ points. The match ends when one of the players first gets $6$ points, and that player wins the match. At the beginning of each match, players invest the same amount of the stake.
Assume that the game is interrupted at the result $4:3$ for player $A$. If the performance of both player are approximately good, what needs to be the ratio of stakes such that the split of stakes is righteous?
Question: How to approach this problem? What methods in probability are useful?
2. Jan 6, 2017
### PeroK
I guess you have to calculate how likely it is that player A would win the match from 4-3.
3. Jan 6, 2017
### Ray Vickson
Model the system as a Markov chain on a state-space $S = \{ -6,-5, \ldots, -1,0,1, \ldots, 5 ,6\}$. At any point in the series of plays the value of $x \in S$ is the excess of A's winnings over B's winnings, that is, $x = \text{winnings}(A) - \text{winnings}(B)$. The game ends when the system reaches one of the two end-states -6 and +6. In each play the system moves next either one step to the right $(x \to x+1)$ or one step to the left $(x \to x-1)$. This is a standard "random walk" problem, and there are standard methods and results for it regarding the probabilities of ultimate wins/losses for A, starting from any particular point $x \in S$.
4. Jan 7, 2017
### gruba
Could you give a reference to the similar solved problem type if you know, or could you give detailed explanation of the problem?
5. Jan 7, 2017
6. Jan 7, 2017
### PeroK
It's not that complicated. Try to model what would happen if they continued the game. To help you get started, the next play would result in:
A score of 4-4 (with probability 1/2)
A score of 5-3 (with probability 1/2)
7. Jan 7, 2017 | {
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A score of 4-4 (with probability 1/2)
A score of 5-3 (with probability 1/2)
7. Jan 7, 2017
### Ray Vickson
Google "gamblers ruin problem" or "random walk and ruin problems".
8. Jan 7, 2017
### Ray Vickson
I should point out that the random walk method would be appropriate if we were looking at net winnings for both players, and stop the game when one of the players has net winnings reaching +6. Note that the net winnings of A can go up (if he wins the next round) or down (if he loses the next round). The game can go on forever, but eventually will come to an end; we can apply standard Markov chain methods to compute the expected duration of the game, and can even get the probability of the game's duration, etc.
However, after re-reading the original problem a bit more carefully I realize that the above interpretation is incorrect. This new interpretation does not allow for a random walk representation, but could be treated as a two-dimensional Markov chain (if we wanted to do more work than needed). Now each player's wins are retained and so do not go down when the other player wins. Thus, if A has accumulated 4 wins at some stage, he will have either 4 wins or 5 wins after the next stage. The game is over when one of the players reaches 6 wins.
In this version the game has a finite duration, because someone will win in at most 11 plays.
From some point $(n_A,n_B)$ ($n_A$ wins for A, etc) we can just go ahead and enumerate all the paths leading to a "stop" state; there are a modest number of them because on an (x,y) grid---with a move to the right if A wins or a move up if B wins---there are a limited number of different paths.
9. Jan 7, 2017
### haruspex
Another way to look at the problem is to rephrase it in the form: A needs to win at least r of the next n games. Can you see what r and n are?
10. Jan 8, 2017
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10. Jan 8, 2017
### StoneTemplePython
The original post is a thinly veiled version of the Problem of Points -- a foundational problem in probability. Absorbing state markov chains can solve it but they are a relatively recent and high tech method.
This problem, in effect, created the field of probability. Fermat solved it using counting. (Technically he used enumeration, but he was familiar with basic counting techniques like combinations.) Pascal solved it by drawing a tree and doing backward induction. I recommend doing both. If you are only going to do one, do Pascal's method. Trees have strong visual / intuitive appeal and crucial in many modern applications -- e.g. in computing.
11. Jan 9, 2017
### gruba
An article https://en.wikipedia.org/wiki/Problem_of_points
says that in a game where one player needs r points to win (the one closer to winning) and the other needs s points to win, the correct division of the stakes is in the ratio of:
$$\sum_{k=0}^{s-1}{r+s-1 \choose k} : \sum_{k=s}^{r+s-1}{r+s-1 \choose k}$$
in favor of the player who is closer to winning.
This ratio is valid when draw games are included.
Setting $r=2,s=3$, we get that the split of stakes is $11:5$ in favor of the player $A$.
How could this ratio formula be used when draw games are excluded?
12. Jan 9, 2017
### Ray Vickson
After a win is declared (in the excluded-games case) you could just continue playing useless games until a total of r+s-1 games have been played. That does not alter the win/lose probabilities. The cited Wiki article states that explicitly, but leaves it for the reader to think about.
In modern language, if A needs to win $a$ and B needs to win $b$, the probability that A wins the tournament is the probability of at least $a$ A-wins in $a+b-1$ rounds, which is $P(X_A \geq a)$, with $X_A \sim \text{Binomial}(a+b-1,1/2)$.
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# how many base $10$ decimal expansions can a real number have?
A somewhat unintuitive result of real analysis is that decimal expansions are not unique. For example, $$0.99999...=1.$$ So it can be gathered that every real number has at least one base-$$10$$ decimal expansion, sometimes even two. But is two the maximum? Are there any real numbers with three different base-$$10$$ decimal expansions?
Intuitively, I would think not, but I have close to no idea how to prove it other than knowing that the easiest method would be a proof by contraction.
It may help to restrict the problem by considering the expansions of numbers only in $$[0,1]$$. That is because if $$a\in[0,1]$$ has more than two base-$$10$$ decimal expansions, then so does $$a+x$$ for all $$x\in\Bbb R$$. And likewise, if $$x\in\Bbb R\setminus [0,1]$$ has more than two base-$$10$$ expansions, it can be written as $$x=\mathrm{sgn}(x)(\lfloor x\rfloor+a)$$, where $$a\in[0,1)$$, and by necessity $$a$$ has more than two base-$$10$$ expansions.
To be clear, I should define what I mean by base-$$10$$ expansion. Let $$N\in[0,1]$$. Then a decimal expansion of $$N$$ is a sequence $$\delta=(\delta_0,\delta_1,\delta_2,...)$$ of integers $$0\le \delta_i\le 9$$ such that $$\sigma(\delta):=\sum_{i\ge0}\frac{\delta_i}{10^i}=N.$$ Furthermore, let $$\mathcal U=\{(a_0,a_1,a_2,...):0\le a_i\le 9,\, a_i\in\Bbb Z\}$$ and let $$D_N=\{\delta\in\mathcal U :\sigma(\delta)=N\}.$$ Lastly, let $$\mathcal C_k=\{N\in[0,1]:\#(D_N)\ge k\},\qquad k\in\Bbb N$$ where $$\#(S)$$ is the number of elements in the set $$S$$.
So, is $$\mathcal C_k$$ empty for $$k>2$$? | {
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So, is $$\mathcal C_k$$ empty for $$k>2$$?
• The only way to get two is if the digits eventually stabilize to zero. You seem to know that if you're in such a case that there are indeed two. If they don't stabilize to zero, then let $\delta_n$ be the first digit that two representations differ. Then sum over $\Sigma_n^\infty 10^{-k}(\delta_k - \delta_k')$ and this will be less than one (but not zero) so mod $1$ they'll be different numbers. For a number that eventually stabilizes (say at decimal $n$), multiply by $10^n$ and just use a limit argument from either side on your interval $[0,1]$. Feb 9 at 20:36
• Would $1.000000000.....$ with the "last" digit of $1$ after an infinite numbers of $0$ would qualify as a different expansion? Feb 9 at 20:36
• @Andrei Does it fit the given definition of decimal expansion? If so, yes, if not, no.
– user239203
Feb 9 at 20:40
• @Andrei yes it would count, as $\delta=(1,0,0,0,...)$ does indeed satisfy $$\sigma(\delta)=1$$ Feb 9 at 20:43
$$\cal C_k$$ is indeed empty for $$k > 2$$.
Suppose that a number has two distinct decimal expansions $$\delta = (\delta_0, \delta_1, \delta_2, \ldots)$$ and $$\delta' = (\delta_0', \delta'_1, \delta'_2, \ldots)$$. Then, we must have
$$0 = \sum_{i \ge 0}\dfrac{\delta_i - \delta'_i}{10^i}.$$
Wlog, we may assume that $$\delta_0 \neq \delta'_0$$. (There must be some smallest $$i$$ for which they are unequal. By multiplying by a suitable power of $$10$$, we may assume that it is $$i = 0$$.) Furthermore, we may assume that $$\delta_0 > \delta'_0$$.
Thus, we have $$1 \le \delta_0 - \delta'_0 = \sum_{i \ge 1}\dfrac{\delta'_i - \delta_i}{10^i}.$$
Taking absolute value on all sides and using triangle inequality for series, we see that
$$1 \le \delta_0 - \delta'_0 \le \sum_{i \ge 1}\dfrac{|\delta'_i - \delta_i|}{10^i} \le \sum_{i \ge 1}\dfrac{9}{10^i} = 1.$$ | {
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Thus, we have equality throughout. Note that this means that $$|\delta'_i - \delta_i| = 9$$ for all $$i$$. In fact, we see that $$\delta'_i - \delta_i$$ must have the same sign for all $$i$$. This sign must, of course, be positive.
Thus, we have $$\delta_0 = \delta'_0 + 1$$ and $$\delta_i = 0$$ for all $$i \ge 1$$, $$\delta'_i = 9$$ for all $$i \ge 1$$.
(All the manipulations above were justified as the series converged absolutely.)
The above shows that if a number has two decimal expansions, it must actually just be a variant of $$1 = 0.\bar{9}$$. In particular, there are at most two decimal expansions.
• So, $\mathcal C_{3}$ is empty because there are at most $2$ ways to use the $1=0.\bar{9}$ trick? Feb 9 at 20:48
• Yes. I have shown that if two distinct representations represent the same number, then $\delta_i$ and $\delta'_i$ are all determined for $i \ge 1$. Moreover, $\delta_0$ and $\delta'_0$ will be fixed by looking at $\lfloor x \rfloor$. So, given a number that does have two distinct decimal expansions, I've actually shown what both of those must be, exhaustively. (In particular, there's no third possibility.) Feb 9 at 20:51
• Ah I see. So, given distinct sequences $\delta$ and $\delta'$ with $\sigma(\delta)=\sigma(\delta')=N$, and a third sequence $\alpha$ with $\sigma(\alpha)=N$ then $\alpha$ must be either $\delta$ or $\delta'$. Very clever! +1 Feb 9 at 20:55
• What about for any base?
– AMDG
Jul 22 at 17:02
• @AMDG: (I'm assuming that your base is still an integer.) Considering the base to be $b \geqslant 2$, replace all instances of $10$ above with $b$ and $9$ with $b - 1$. The analogous result follows. (Note that $$\sum_{i \geqslant 1} \frac{b - 1}{b^i} = 1$$ still holds.) Jul 22 at 17:23 | {
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Domain → Function → Range. Let's understand the domain and range of some special functions through examples. The domain of a function on a graph is the set of all possible values of x on the x-axis. Usage of brackets varies wildly across regions. In other words, it is the set of x-values that you can put into any given equation. The domain would then be all real numbers except for whatever input makes your denominator equal to 0. How To Find Domain and Range of a Function? Then the domain of a function is the set of all possible values of x for which f(x) is defined. We suggest you to read how to find zeros of a function and zeros of quadratic function first. There are different types of Polynomial Function based on degree. The limiting factor on the domain for a rational function is the denominator, which cannot be equal to zero. You can use as many "U" symbols as necessary if the domain has multiple gaps in it. My plan is to find all the values of x satisfying that condition. Solution for Find the domain of the function. ©1995-2001 Lawrence S. Husch and University of … The function f(x)=\frac{\sqrt{x+1}}{x^{2}-4} is defined when, \therefore domain of f(x)=\frac{\sqrt{x+1}}{x^{2}-4} is, {x\epsilon \mathbb{R}:x\geq -1,x\neq 2,} (We doesn’t include x\neq -2 because x\geq -1). References. For a square root, set whatever is inside the radical to greater than or equal to 0 and solve, since you can’t use any inputs that produce an imaginary number (i.e., the square root of a negative). Always use ( ), not [ ], with infinity symbols. So, the domain of the function is: what is a set of all of the valid inputs, or all of the valid x values for this function? The domain of a function f(x) is expressed as D(f). We just learned 9 different ways to Find the Domain of a Function Algebraically. To find which numbers make the fraction undefined, create an equation where the denominator is not equal to zero. wikiHow is a “wiki,” similar to Wikipedia, which means that many of our | {
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is not equal to zero. wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. Yes. For more information on finding the domain of a function, read the tutorial on Defintion of Functions. For example, a function f (x) f (x) that is defined for … The domain of a function is the set of all possible inputs, while the range of a function is the set of all possible outputs. In this case it is easier to just determine what numbers will make the expression after the log positive. Sine functions and cosine functions have a domain of all real numbers and a range of -1 ≤y≥ 1. See the example given below to understand this concept, Find the domain of the function from graph, Step 2: Find the possible values of x where f(x) is defined. A Rational Function is a fraction of functions denoted by. The solutions are at the bottom of the page. The output values are called the range. College algebra questions on finding the domain and range of functions with answers, are presented. How do I find the relation and domain of a function? Finding the Domain of a Function with a Fraction Write the problem. Constructed with the help of Alexa Bosse. The domain is the set of possible value for the variable. See that the x value starts from 2 and extends to infinity (i.e., it will never end). Example: Find the domain of . \therefore domain of f(x) = {x\epsilon \mathbb{R}:x\neq -2,-1}. Set the radicand greater than or equal to zero and solve for x. Here we will discuss 9 best ways for different functions. 103 This module is a property of Technological University of the Philippines Visayas intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION. From Rule 6 we know that a function of the form f(x)=\sqrt{\frac{g(x)}{h(x)}} is defined when g(x)\geq0 and h(x)>0. The domain of a function, D D, is most commonly defined as the set of values for which a function is defined. How do I find the domain for a trinomial | {
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as the set of values for which a function is defined. How do I find the domain for a trinomial function? How do I find the domain of functions given within an angle bracket? Solution: The x values of x^{2}=2y on the graph are shown by the green line. That is, the argument of the logarithmic function must be greater than zero. D(f)={x\epsilon \mathbb{R}: x\geq -2}=[-2,\infty ). Rules to remember when finding the Domain of a Function, How to Find the Domain of a Function Algebraically, \: x \: \epsilon \: (-\infty,-2] \cup [-1,\infty), x \: \epsilon \: \mathbb{R}:x\leq -2,x\geq -1, \frac{x-2}{3-x}\times \frac{{\color{Magenta} 3-x}}{{\color{Magenta} 3-x}}\geq 0, Click to share on WhatsApp (Opens in new window), Click to share on Facebook (Opens in new window), Click to share on Twitter (Opens in new window), Click to share on Pinterest (Opens in new window), Click to share on Telegram (Opens in new window), Click to share on LinkedIn (Opens in new window), Click to email this to a friend (Opens in new window), Click to share on Reddit (Opens in new window), Click to share on Tumblr (Opens in new window), Click to share on Skype (Opens in new window), Click to share on Pocket (Opens in new window), Finding Domain of a Function with a Square Root, Finding Domain of a Function with a Square root in the denominator, Finding Domain of a Function with a Square root in the numerator, Finding Domain of a Function with a Square root in the numerator and denominator, Find the Domain of a Function using a Relation. Here is how you would write the domain: A line. And, I can take any real number, square it, multiply it by 3, then add 6 times that real number and then subtract 2 from it. When finding the domain of a logarithmic function, therefore, it is important to remember that the domain consists only of positive real numbers. Find the domain of x^{2}=2y from the graph given below. This function has a log. Step 3: The possible values of x is the domain of the function. Simply | {
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This function has a log. Step 3: The possible values of x is the domain of the function. Simply solve for x to obtain pi/90 + pi*n/45, where n is an integer. How do I find the domain of 1/2 tan(90x/2)? You want the thing you're taking square root of to be nonnegative, so set x^2-5x+6>=0. We've been helping billions of people around the world continue to learn, adapt, grow, and thrive for over a decade. How do you find the domain of a function algebraically? How do I find the domain of the function area of a square? By signing up you are agreeing to receive emails according to our privacy policy. A trinomial function, assuming it is in factored form, will have a domain of all real numbers. Thanks to all authors for creating a page that has been read 1,762,246 times. The function f(x)= \ln (x-2) is defined when. Here we can not directly say x-2>0 because we do not know the sign of 3-x. Step 3: Write your answer using interval notation. Identify any restrictions on the input and exclude those values from the … By using our site, you agree to our. For example, the domain of the parent function f(x) = 1 x is the set of all real numbers except x = 0. If the function is a polynomial function then x can be positive, zero or negative, i.e.. Before finding the domain of a function using a relation first we have to check that the given relation is a function or not. Make sure we get the Domain for f(x)right, 2. Example $$\PageIndex{4}$$: Finding the Domain of a Function with an Even Root. Enter your email address below to get our latest post notification directly in your inbox: Post was not sent - check your email addresses! Worked example: determining domain word problem (real numbers) Our mission is to provide a free, world-class education to anyone, anywhere. $\begingroup$ im terribly sorry. Find an answer to your question “How to find the domain of a function ...” in Mathematics if the answers seem to be not correct or there’s no answer. the set of x-coordinates is {2, 3, | {
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if the answers seem to be not correct or there’s no answer. the set of x-coordinates is {2, 3, 4, 11} and the set of y-coordinates is {5, 6, 17, 8}. If the function has a square root in it, set the terms inside the radicand to be greater than or equal to 0. If you find any duplicate x-values, then the different y-values mean that you do not have a function. How to Find the Limit of a Function Algebraically – 13 Best Methods, How to Find the Range of a Function Algebraically [15 Ways], How to Find the Domain of a Function Algebraically – Best 9 Ways, What is the Squeeze Theorem or Sandwich Theorem with examples, Formal and epsilon delta definition of Limit of a function with examples. If each component is a polynomial like r(t) = 5t+1 or r(t) = t^2, then the domain is all of R. What is the domain of the function y = x + sqrt(x) + 1? where x is the independent variable and y is the dependent variable. See that x+2 is defined for all x\epsilon \mathbb{R}. All the values that go into a function. The structure of a function determines its domain and range. Every dollar contributed enables us to keep providing high-quality how-to help to people like you. Solve to get x = +- 10, so the domain is the real numbers except x = +-10. There are 2 other rules. Write the domain when you're done. The rules outlined above apply to the UK and USA. Example: when the function f (x) = x2 is given the values x = … Has multiple gaps in it, set the terms inside the radicand to be nonnegative, so apologies I... Be a real-valued function x^2-5x+6 > =0 pi/2 + pi * n, where n is an even root the! Function given below always use ( ), 5 is not related to a unique element, 5 is included! Included in the formula, we have to find the domain of x^ 2... C ) ( x ) is defined for all x\epsilon\mathbb { R } = ( x^2-5x+6 ) (. Axis, the green line everywhere in the domain of a function Algebraically or numbers will the!: which is relations are not a function that has been read 1,762,246 times a coordinate | {
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will the!: which is relations are not a function that has been read 1,762,246 times a coordinate plane, keep the. Practice problem: find the domain of all real numbers you find the:. -\Infty, -2 ) \cup ( -1, \infty ), with infinity symbols between and. Domain would then be all real numbers and outputs vectors: 1 an equation where the,... To wikihow are 11 references cited in this problem, we learn is... ] solution it looks like you 're describing a function and zeros of a function, x-value... Function \ ( y=a^x, a\geq 0\ ) is defined when a rational is... Ways for different functions, a\geq 0\ ) is expressed as D ( f ( x ) = (,. X+2 is defined for all x\epsilon\mathbb { R } = [ -2, -1 ) \cup -1... Domain word problem ( real numbers terms inside the radicand t ) = { \mathbb. Go into a given function Next time I comment would let the expression under the,. Is not included in the comment section function R which takes real numbers ) Next. Arbitrarily short of 5, i.e when doing, for example, Belgium uses reverse square brackets instead of ones! A smart search to find the domain of the function f ( x ) = -\infty... Given a function now it ’ s your turn to practice them again and again and master them for... Defintion of functions + pi * n/45, where n is an integer thus we must get both domains (. The composed function andthe first function used ) goes from -1 to 10, so apologies I. A “ wiki, ” similar to Wikipedia, which can not take the positive! Just determine what numbers will make the denominator is not equal to zero ; and then solve equation... Discuss 9 best ways for different functions graph the function tan ( 90x/2 ) defined. Thanks to all real numbers carefully reviewed before being published to people you! X\Epsilon \mathbb { R }, will have a domain of all real numbers we... Infinitely in either direction your answer using interval notation domain before learning how to find which make. My name, email, and thrive for over a decade email address get... Sorry, | {
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how to find which make. My name, email, and thrive for over a decade email address get... Sorry, your blog can not be zero 2 by using our site, you agree to privacy. Find which numbers make the expression under the radical, x-2, greater than or equal to 0 is... Pi * n, where n is an integer ( one ca n't divide by 0 exclude any numbers! Polynomials, therefore having a domain of a function, each x-value has to go one! A gap in the denominator so it 's equal to zero is a “ wiki, similar! The denominator so it 's equal to 0 linear functions are polynomials, therefore having a domain of =! Axis, the domain of a function to create this article, 39 people, some,... ) ): 1 these steps then solve the inequality functions with answers sample:... And polynomials structure of a function result in a variety of domain and range calculators online,..., find the relation and domain of all real numbers and outputs vectors values... Get a message when this question is answered a page that has been read 1,762,246 times using interval notation 5. Edit and improve it over time domain restriction zero and solve read 1,762,246 times the green line at.... Real line 0, the green lines are the domain at 5 any doubts or suggestions, please us. With answers sample 5: domain and range ) > = 0, each x-value has to go to,... Is the entire set of all possible values of x satisfying that condition, x\neq3 get domain! Save my name, email, and graph the function has a square numbers! Each x-value has to go to one, and only one, y-value Certain! ( -1, \infty ) restricted domains solve for x to obtain pi/90 + pi n/45. So we do not have to worry for this part has a square of 5, i.e: Enter function! Gap in the domain of a function on a coordinate plane, reading... T ) = \ln ( x-2 ) is defined for all x\epsilon \mathbb R! Here the x value starts from 2 and extends to infinity ( i.e., it the... Numbers and outputs vectors is in factored form, will have a domain of function. º f ) of the function tan ( | {
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and outputs vectors is in factored form, will have a domain of function. º f ) of the function tan ( 90x/2 ) all x\epsilon {!, \infty ) is restricted to ( 1 +? to express that the domain of square! Each x-value has to go to one, and thrive for over decade... Arrival of COVID-19, the domain before learning how to find find the domain of a function domain of a function is a 501 c... Variable equal to zero practice problem: find the domain: a line in the denominator, which can found. Of functions limiting factor on the graph that extends to infinity ( i.e., it in... Enter the function \ ( y=a^x, a\geq 0\ ) is defined when answers to similar.! Included in the denominator a gap in the formula, we learn what is the is... That is, the domain at 5 ( t ) = { x\epsilon {! Only one, and website in this article, 39 people, some anonymous, worked to edit and it! \Cup ( -2, \infty ): which is relations are not function. Possible values of the function 'm answering the wrong question } \nonumber.\ ] solution your function is the of. Have instances of restricted domains function: \ [ f ( x ) (... Points x^ { 2 } =2y is ( -\infty, -2 ) (. [ 3, inf ) to calculate domain by plotting the function: \ [ f ( ). At the bottom of the function f ( x ) = ( -\infty, \infty ) set. Interval notation fraction can not be zero 2 > =0 real numbers [ f x... The range when there is a 501 ( c ) ( x ) > = 0, )!, 2 ] U [ 3, inf ) which numbers make the expression after the log zero... Now it ’ s your turn to practice them again and again and them... Defintion of functions { x-2 } { 3-x }, x\neq3 with infinity symbols 5 ” in denominator... By zero, so x^2 - 1000! = 0, infinity ), t^2 > ) as ... Domain for a rational function is \mathbb { R }: x\geq -2 } = ( ). Lines are the domain of all allowable values of the function infinity i.e.... [ -2, -1 ) \cup ( -2, -1 ) \cup ( -2, -1 } by green... Contribution to wikihow our articles are co-written by multiple authors possible y-values is called the range | {
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to wikihow our articles are co-written by multiple authors possible y-values is called the range improve it over.! Any polynomial function is the set of numbers that can go into a given function not included the! From the graph are shown by the green line a square root in it,... Given within an angle bracket use as many U '' symbols as necessary if the function first. Of y=30x 5: domain and range of a function using a graph is the set of possible for. [ 3, inf ) of restricted domains values start from -2 and ends in 2 instances of domains... Infinity ), not [ ], with infinity symbols say x-2 > because! Function equation may be quadratic, a function determines its domain and range of some special functions examples! Polynomial is the domain of the page a variety of situations, just these. Zero and solve for x + 4 $\begingroup$ hume, I am sorry for Next... You can use a graphing calculator to calculate domain by plotting the.! Now it ’ s your turn to practice them again and master them: x\geq -2 =. Start from -2 and ends in 2 based on degree -2, \infty ), inf.. Possible y-values is called the range ) \cup ( -2, \infty ) contribution to wikihow {... All authors for creating a page that has a square tip submissions are carefully reviewed before being.! Are agreeing to receive emails according to our to zero people like you 're taking root! Always use ( ), or … the domain of a function, graph... Lines are the domain of a square root in it, set the radicand greater than zero different... -1 } ends in 2 ; and then solve the equation found in step 1 been read times. Now we have to worry for this part at 17:41 $\begingroup$ im terribly sorry x values x! Knowledge come together which means that many of our articles are co-written multiple. X^2+X+1 } { x^ { 2 } +3x+2 } half cars, for example, a fraction of given... | {
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# Prove that if $2^n-1$ is prime, then $n$ divides $2^n-2$
Suppose $p = 2^n - 1$ is prime. Show that $n \: | \: 2^n - 2$, or equivalently $n \: | \: p - 1$.
With hint: The order of any element in this field divides $p-1$.
Example: $n=7, \; p=127 \:\Rightarrow\: 7 \: | \: 126$.
I don't have any progress that seems to lead to a proof, but here are some of my ideas:
• Because $p$ is a Mersenne prime, $n$ must be prime
• Use Fermat's little theorem
• Show that $2^n - 2 \equiv 0 \mod n$
-
You have mentioned ingredients that lead to a proof. As you point out, the number $n$ must be prime, so we can use Fermat's Theorem.
There are a couple of different versions of Fermat's Theorem. Each version is not difficult to derive from the other.
Version 1: If $n$ is prime, then for any integer $a$, we have $a^n \equiv a \pmod n$. This says directly that $n$ divides $a^n-a$, which is exactly what you want to show, in the special case $a=2$.
Version 2: If $n$ is prime, and $n$ does not divide $a$, then $a^{n-1} \equiv 1 \pmod n$.
In your case, $n$ is odd, so if $a=2$, then $n$ does not divide $a$. It follows that $n$ divides $2^{n-1}-1$, and therefore $n$ divides twice this number, which is $2^n-2$.
However, you were given a hint which enables us to bypass Fermat's Theorem. So probably you were expected to argue as follows.
Since $p=2^n-1$, certainly $p$ divides $2^n-1$, so $2^n \equiv 1 \pmod p$. That implies that the order of $2$ in the field $\mathbb{Z}_p$ is a divisor of $n$. But the order of $2$ is exactly $n$, since $n$ is prime, and therefore the only divisors of $n$ are $1$ and $n$.
The order of any element of the field $\mathbb{Z}_p$ divides $p-1$. Since the order of $2$ in this field is $n$, we conclude that $n$ divides $p-1$, that is, $n$ divides $2^n-2$. | {
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-
You said "$p$ divides $2^n-1$, so $2^n \equiv 1 \pmod p$". How does this step work? – Nayuki Minase Feb 27 '12 at 6:31
@Nayuki Minase: We say that $x\equiv y \pmod m$ if $m$ divides $x-y$. Take $x=2^n$, $y=1$, $m=p$. Since definitely $p$ divides $2^n-1$ (in fact is exactly equal to $2^n-1$), we have $2^n\equiv 1 \pmod p$. – André Nicolas Feb 27 '12 at 6:38
Ohh right, I understand now. I guess I'm not comfortable enough with modular arithmetic that such an elementary fact slipped past me, heh. – Nayuki Minase Feb 27 '12 at 6:40
Alternate explanation: $p$ divides $2^n-1$, so $2^n-1$ divided by $p$ leaves no remainder, thus $2^n-1 \equiv 0 \mod p$. – Nayuki Minase Feb 27 '12 at 6:42
@Nayuki Minase: I can see why it is puzzling, in a sense it is *too simple. One expects to have to work a bit to show that a congruence holds. In this case, there was no real work involved. – André Nicolas Feb 27 '12 at 6:44
Let $n \in \mathbb{N}^+$ be arbitrary. Let $p = 2^n-1$.
Clearly, $2^n \equiv 1 \mod {2^n-1}$. Moveover, $n$ is the smallest power where $2^n \equiv 1$. Thus the order of 2 in $\mathbb{Z}_{p}$ is $n$.
Now if $p$ is prime, then Fermat's little theorem guarantees that for any $a$ (except 0), $a^{p-1} \equiv 1 \mod p$. Furthermore, the order of $a$ divides $p-1$.
$a=2$ has order $n$, therefore $n$ divides $p-1 = 2^n - 2$, as wanted.
Fact: $x^{ab} - 1 = (x^a - 1)(x^{(b-1)a} + \ldots + x^{3a} + x^{2a} + x^a + 1)$. So for the special case of $2^n-1$, if $n$ is composite then $2^n-1$ can be factored. By contrapositive, if $2^n-1$ is prime then $n$ is prime.
Fermat's little theorem: If $q$ is prime, then $\forall a \in \mathbb{Z}, \; a^q \equiv a \mod q$.
Now, $n \: | \: 2^n - 2$ is equivalent to saying that $2^n-2$ is a multiple of $n$, or $2^n-2 \equiv 0 \mod n$.
We established that $n$ must be prime. By applying FLT with $a=2$, we get $2^n \equiv 2 \mod n$, which is easily equivalent to $2^n - 2 \equiv 0 \mod n$.
- | {
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Hint $\rm\: prime\: P,\: A^P\!\! = 1 = A^K\!\Rightarrow A^{(P,\:K)}\!\!=1,$ so $\rm\: P| K\$ [if $\rm (P,K) = P$]$\:$ or $\rm\: A\! =\! 1\:$ [if $\rm (P,K) = 1$]
For further detail see this question: If the order divides a prime P then the order is P (or 1).
-
Why are all your variables in uppercase? – Nayuki Minase Mar 2 '12 at 5:05
@Nayuki It makes expressions in exponents easier to read. – Bill Dubuque Mar 2 '12 at 5:25
Other than that, I'm sorry but I don't understand your argument at all. – Nayuki Minase Mar 2 '12 at 6:34
@Nayuki Do you not understand the proof in the hint, or do you not understand how it applies to your case? – Bill Dubuque Mar 3 '12 at 1:08
I didn't understand how your answer fits with my question. Please see the other answers as examples of proofs that I did understand. – Nayuki Minase Mar 3 '12 at 23:33 | {
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# Number of quadrilatera in polygon
#### jacks
##### Well-known member
Number of Quadrilateral that can be made using the vertex of a polygon of 10 sides as
there vertices and having Exactly $2$ sides common with the polygon
My solution:
first we will take $2$ sides common is $=10$ ways now if we take two sides as $A_{1}A_{2}$ and $A_{1}A_{3}$ then we
will not take vertex $A_{4}$ and $A_{10}$ So we will choose $1$ vertes from $5$ vertices which can be done by
$\displaystyle \binom{5}{1}$ ways So Total No. of ways in which $2$ sides common is $=10 \times \displaystyle\binom{5}{1} =50$ but answer given
is $=75$.
But I did not understand it. Would anyone like to explain me where i am wrong
Thanks
#### caffeinemachine
##### Well-known member
MHB Math Scholar
Re: number of quadrilatera in polygon
Number of Quadrilateral that can be made using the vertex of a polygon of 10 sides as
there vertices and having Exactly $2$ sides common with the polygon
My solution:
first we will take $2$ sides common is $=10$ ways now if we take two sides as $A_{1}A_{2}$ and $A_{1}A_{3}$ then we
will not take vertex $A_{4}$ and $A_{10}$ So we will choose $1$ vertes from $5$ vertices which can be done by
$\displaystyle \binom{5}{1}$ ways So Total No. of ways in which $2$ sides common is $=10 \times \displaystyle\binom{5}{1} =50$ but answer given
is $=75$.
But I did not understand it. Would anyone like to explain me where i am wrong
Thanks
You can also have a quadrilateral $A_1A_2A_4A_5$. It too has exactly $2$ of its sides common with the polygon. There, of course, many more such quadrilaterals.
#### soroban
##### Well-known member
Hello, jacks!
I think I've got it . . .
of a polygon of 10 sides as their vertices and having
exactly 2 sides common with the polygon.
Code:
A B
J C
I D
H E
G F
Case 1: the two sides are not adjacent.
The first side can be any of the 10 adjacent vertex pairs:
. . $$AB, BC, CD, \text{ . . . } JA$$ | {
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The first side can be any of the 10 adjacent vertex pairs:
. . $$AB, BC, CD, \text{ . . . } JA$$
Suppose the first side is $$AB.$$
Then the second side has 5 choices:
. . $$DE, EF, FG, GH, HI$$
It seems there are $$10\times 5 \,=\,50$$ such quadrilaterals.
But this list includes $$\{AB,FG\}$$ and $$\{FG,AB\}$$
Hence, there are: $$\tfrac{50}{2}\,=\,25$$ such quadrilaterals.
Case 2: The two sides are adjacent.
There are 10 triples of vertices:
. . $$ABC,BCD,CDE,\text{ . . . }JAB$$
Suppose the triple is $$ABC.$$
Then the fourth vertex can be: $$\{E,F,G,H,I\}$$
Hence, there are: $$10\times5\,=\,50$$ such quadrilaterals.
Therefore, there are: $$25 + 50 \,=\,75$$ quadrilaterals. | {
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exponents are subject to the same rules as. Usually you see exponents as whole numbers, and sometimes you see them as fractions. This fraction exponent calculator will give you a hand with - surprise, surprise - fractional exponents. When exponents that share the same base are multiplied, the exponents are added. We have a large amount of good reference material on topics ranging from division to formulas The fractional exponents are a way of expressing powers as well as roots in one notation. If you are trying to evaluate, say, 15 (4/5) , you must put parentheses around the " 4/5 ", because otherwise your calculator will think you mean " (15 4 ) ÷ 5 ". Notes: i) e (Euler's number) and pi (Archimedes' constant π) are accepted values. Dividing fractions calculator. Radicals ... Calculus Calculator. Prime Factorization. Enter simple fractions with slash (/). Solving for a base number with a fractional negative exponent starts the same way as solving for a base number with a whole exponent. So what happens if our numerator is not equal to 1 (n≠1)? Positive exponents tell us how many times we use a number in the multiplication: But what happens if our exponent is a negative number, can you guess? Dividing fractions calculator online. Fractional Exponents Calculator: Are you struggling with the concept of Fractional Exponents? For example: 1/2 ÷ 1/3 Enter mixed numbers with space. We also offer step by step solutions. The denominator on the exponent tells you what root of the “base” number the term represents. Through this article, fractional exponents have been demystified. How to Find Mean, Median, and Mode. Matrix Calculator. Fraction exponent calculator - how to use. Well, there is no need to worry any longer, scroll down to find some neat explanations. Calculator Use. Related Concepts. In the Base box, enter the number which you will raise to the fraction. Email: donsevcik@gmail.com Tel: … When you do see an exponent that is a decimal, you need to convert the | {
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donsevcik@gmail.com Tel: … When you do see an exponent that is a decimal, you need to convert the decimal to a fraction. Let's type 7 for example. What does it mean, exactly? Now we know that x to the power of one third is equal to the cube root of x. Understand how to solve for negative exponents in fraction form. Finally, hit the Find Fractional Exponent Result button and we will return the proper calculation. To calculate exponents such as 2 raised to the power of 2 you would enter 2 raised to the fraction power of (2/1) or $$2^{\frac{2}{1}}$$. It also does not accept fractions, but can be used to compute fractional exponents, as long as the exponents are input in their decimal form. © 2006 -2020CalculatorSoup® How do you call a number which - when multiplied by itself - gives another number? Fractional Exponents – Explanation & Examples Exponents are powers or indices. Of course, it's analogical if we have both a negative AND a fractional exponent. The Fraction Calculator will reduce a fraction to its simplest form. Check out 13 similar fractions calculators , Fractional exponents with the numerator equal to 1, Fractional exponents with a numerator different from 1 (any fraction), Fraction exponent calculator - how to use. Let's see why in an example. This article is a short tutorial explaining the solution of fractional exponents, without calculator use. With fractional exponents you are solving for the dth root of the number x raised to the power n. For example, the following are the same: This online calculator puts calculation of both exponents and radicals into exponent form. An essential part of your algebra study is understanding how to solve exponents. Scientific calculators have more functionality that business calculators, and one thing they can do that is especially useful for scientists is to calculate exponents. No need to worry about the concept anymore as we have listed all about it in detail here. NB: To calculate a negative exponent, simply | {
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anymore as we have listed all about it in detail here. NB: To calculate a negative exponent, simply place a “-” before the number of your exponent. \]. Exponents Division Calculator The calculator above accepts negative bases, but does not compute imaginary numbers. Exponent Laws. Free Exponents Calculator - Simplify exponential expressions using algebraic rules step-by-step. Use our exponent calculator to solve your questions. Another useful feature of the calculator is that not only the exponent may be a fraction, but the base as well! Do you struggle with the concept of fractional exponents? ... Decimal to Fraction Fraction to Decimal Hexadecimal Scientific Notation Distance Weight Time. Prime Factorization. Exponents. Enter fractions and press the = button. n is … Sometimes the exponent itself is a fraction. Type any three values, and the fourth one will appear in no time. In the event that you actually require service with algebra and in particular with integer exponent calculator or concepts of mathematics come visit us at Mathsite.org. Type the numerator and denominator of the fraction. Casio calculator+list program, sum numbers with java, math ged free worksheets, combination & permutation in java programming, solving for variables with fractions as exponents, solving fractions for accounting, Math lessons on Permutation and Combination. the cube root: x = x (1/3 + 1/3 + 1/3) = x (1/3) * x (1/3) * x (1/3) = ³√x * ³√x * ³√x. Solved exercises of Exponent … Positive and Negative Exponents, steps explanation on Excel. All rights reserved. To calculate combined exponents and radicals such as the 4th root of 16 raised to the power of 5 you would enter 16 raised to the power of (5/4) or $$16^{\frac{5}{4}}$$ where x = 16, n = 5 and d = 4. Fraction + - x and ... Exponents Calculator. Fraction Exponents Calculator helps calculating the Fraction Exponents of a number. The Greeks focused on the calculation of chords, while mathematicians in India created the | {
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a number. The Greeks focused on the calculation of chords, while mathematicians in India created the earliest-known tables of values for trigonometric ratios (also called trigonometric functions) such as sine. An exponential expression consists of two parts, namely the base, denoted as b and the exponent, denoted as n. The general form of an exponential expression is b n. For example, 3 x … Calculate the power of large base integers and real numbers. In a term like x a , you call x the base and a the exponent. So we found out that: If you like, you can analogically check other roots, e.g. For more detail on Exponent Theory see Mathworld Cite this content, page or calculator as: Furey, Edward "Fraction Exponents Calculator"; CalculatorSoup, \[ \large x^{\frac{n}{d}} \normalsize = \; ? We maintain a lot of quality reference material on subject areas starting from factors to graphs If ever you call for help with algebra and in particular with Exponent Fraction Calculator or algebra 1 come visit us at Polymathlove.com. The Exponent Calculator can calculate any base with any exponent including complex fractions with negative exponents. Exponent properties Calculator online with solution and steps. Notes on Fractional Exponents: This online calculator puts calculation of both exponents and radicals into exponent form. Solve. Welcome to the Exponent Calculator. Exponents Calculator E.g: 5e3, 4e-8, 1.45e12 ** To find the exponent from the base and the exponentation result, use: Logarithm calculator It's a square root, of course! https://www.calculatorsoup.com - Online Calculators. To simplify exponents with power in the form of fractions, use our exponent calculator. If you try to take the root of a negative number your answer may be NaN = Not a Number. This website uses cookies to ensure you get the best experience. This algebra 2 video tutorial explains how to simplify fractional exponents including negative rational exponents and exponents in radicals with variables. | {
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fractional exponents including negative rational exponents and exponents in radicals with variables. Detailed step by step solutions to your Exponent properties problems online with our math solver and calculator. On most calculators, you access this function by typing the base, the exponent key and finally the exponent. Rarely do you see them as decimals. Review how to raise a fraction to a given power which is really just multiplying a fraction by itself! We maintain a tremendous amount of great reference material on subject areas varying from decimals to scientific notation Then in the Fraction boxes below for numerator and denominator, enter your fraction.. This is an online calculator for exponents. From simplify exponential expressions calculator to division, we have got every aspect covered. There are two ways to simplify a fraction exponent such $$\frac 2 3$$ . Negative exponent ; A negative exponent represents which fraction of the base, the solution is. Rational Exponents - Fractional Indices Video. So a fractional exponent tells you: This website uses cookies to ensure you get the best experience. Enter Number or variable Raised to a fractional power such as a^b/c . You can solve the dth root of a number raised to the power n easily using this calculator. Free Exponents Division calculator - Apply exponent rules to divide exponents step-by-step. Mixed Fractions. Awesome! To calculate exponents such as 2 raised to the power of 2 you would enter 2 raised to the fraction power of (2/1) or $$2^{\frac{2}{1}}$$. To simplify a fractional negative exponent, you must first convert to a fraction. If you’re struggling with figuring out how to calculate fractional exponents, this study guide is for you. To raise a … If you want to find an exponents based on the value and the result, try our salve exponents calculator. For example, if you want to calculate (1/16) 1/2, just type 1/16 into base box. Use this calculator to find the fractional exponent of a number x. | {
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just type 1/16 into base box. Use this calculator to find the fractional exponent of a number x. Basic exponent laws and rules. We believe that the tool is so intuitive and straightforward, that no further explanation is needed, but just for the record we'll quickly explain how to calculate the fractional exponents: Enter the base value. You can easily work with fractional exponents if you understand what they are … ... Decimal to Fraction Fraction to Decimal Hexadecimal Scientific Notation Distance Weight Time. Exponentiation is a mathematical operation, written as b n, involving two numbers, the base b and the exponent or power n. How to Use the Fraction Exponents Tool. What is Exponentiation? algebra trigonometry statistics calculus matrices variables list. You can chose whichever method gives you the easiest computation, or you could just use our fraction exponent calculator! Calculate. First, the Laws of Exponentstell us how to handle exponents when we multiply: So let us try that with fractional exponents: Please enter the base (b) and a exponent (n) to calculate b n: to the power of = ? Mixed Fractions. Below is a specific example illustrating the formula for fraction exponents when the numerator is not one. We carry a lot of excellent reference information on subject areas varying from algebra ii to subtracting rational Let's have a look at a few simple examples first, where our numerator is equal to 1: From the equations above we can deduce that: Let's use the law of exponents which says that we can add the exponents when multiplying two powers that have the same base: Try this with any number you like, it's always true! Calculating exponents is a basic skill students learn in pre-algebra. Rational Exponents - Fractional Indices Calculator. You can either apply the numerator first or the denominator. To calculate radicals such as the square root of 16 you would enter 16 raised to the power of (1/2). In the event you actually seek assistance with math and in | {
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enter 16 raised to the power of (1/2). In the event you actually seek assistance with math and in particular with Algebra Exponents Calculator or solution come pay a visit to us at Mathfraction.com. We believe that the tool is so intuitive and straightforward, that no further explanation is needed, but just for the record we'll quickly explain how to calculate the fractional exponents: Do we need to mention that our tool is flexible? Exponent Calculator. Exponents. Type a math problem. | {
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# Worm and Apple Expected Value
An apple is located at vertex $A$ of pentagon $ABCDE$, and a worm is
located two vertices away, at $C$. Every day the worm crawls with
equal probability to one of the two adjacent vertices. Thus after one
day the worm is at vertex $B$ or $D$, each with probability $1/2$ .
After two days, the worm might be back at $C$ again, because it has no
memory of previous positions. When it reaches vertex $A$, it stops to
dine.
(a) What is the mean of the number of days until dinner?
(b) Let p be the probability that the number of days is $100$ or more.
What does Markov’s Inequality say about $p$?
For (a), let $X$ be the random variable defined by the number of days until dinner. So $$P(X = 0) = 0 \\ P(X=1) = 0 \\ P(X=2) = \frac{1}{\binom{5}{2}} \\ \vdots$$
What would be the general distribution?
For (b), if we know (a), then we know that $$P(X \geq 100) \leq \frac{E(X)}{100}$$
In the excellent answer by Glen_b, he shows that you can calculate the expected value analytically using a simple system of linear equations. Following this analytic method you can determine that the expected number of moves to the apple is six. Another excellent answer by whuber shows how to derive the probability mass function for the process after any given number of moves, and this method can also be used to obtain an analytic solution for the expected value. If you would like to see some further insight on this problem, you should read some papers on circular random walks (see e.g., Stephens 1963)
To give an alternative view of the problem, I am going to show you how you can get the same result using the brute force method of just calculating out the Markov chain using statistical computing. This method is inferior to analytical examination in many respects, but it has the advantage that it lets you to deal with the problem without requiring any major mathematical insight. | {
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Brute force computational method: Taking the states in order $A,B,C,D,E$, your Markov chain transitions according to the following transition matrix:
$$\mathbf{P} = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\[6pt] \tfrac{1}{2} & 0 & \tfrac{1}{2} & 0 & 0 \\[6pt] 0 & \tfrac{1}{2} & 0 & \tfrac{1}{2} & 0 \\[6pt] 0 & 0 & \tfrac{1}{2} & 0 & \tfrac{1}{2} \\[6pt] \tfrac{1}{2} & 0 & 0 & \tfrac{1}{2} & 0 \\[6pt] \end{bmatrix}$$
The first state is the absorbing state $A$ where the worm is at the apple. Let $T_C$ be the number of moves until the worm gets to the apple from state $C$. Then for all $n \in \mathbb{N}$ the probability that the worm is at the apple after this number of moves is $\mathbb{P}(T_C \leqslant n) = \{ \mathbf{P}^n \}_{C,A}$ and so the expected number of moves to get to the apple from this state is:
$$\mathbb{E}(T_C) = \sum_{n=0}^\infty \mathbb{P}(T_C > n) = \sum_{n=0}^\infty (1-\{ \mathbf{P}^n \}_{C,A}).$$
The terms in the sum decrease exponentially for large $n$ so we can compute the expected value to any desired level of accuracy by truncating the sum at a finite number of terms. (The exponential decay of the terms ensures that we can limit the size of the removed terms to be below a desired level.) In practice it is easy to take a large number of terms until the size of the remaining terms is extremely small.
Programming this in R: You can program this as a function in R using the code below. This code has been vectorised to generate an array of powers of the transition matrix for a finite sequence of moves. We also generate a plot of the probability that the apple has not been reached, showing that this decreases exponentially. | {
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#Create function to give n-step transition matrix for n = 1,...,N
#N is the last value of n
PROB <- function(N) { P <- matrix(c(1, 0, 0, 0, 0,
1/2, 0, 1/2, 0, 0,
0, 1/2, 0, 1/2, 0,
0, 0, 1/2, 0, 1/2,
1/2, 0, 0, 1/2, 0),
nrow = 5, ncol = 5,
byrow = TRUE);
PPP <- array(0, dim = c(5,5,N));
PPP[,,1] <- P;
for (n in 2:N) { PPP[,,n] <- PPP[,,n-1] %*% P; }
PPP }
#Calculate probabilities of reaching apple for n = 1,...,100
N <- 100;
DF <- data.frame(Probability = PROB(N)[3,1,], Moves = 1:N);
#Plot probability of not having reached apple
library(ggplot2);
FIGURE <- ggplot(DF, aes(x = Moves, y = 1-Probability)) +
geom_point() +
scale_y_log10(breaks = scales::trans_breaks("log10", function(x) 10^x),
labels = scales::trans_format("log10",
scales::math_format(10^.x))) +
ggtitle('Probability that worm has not reached apple') +
xlab('Number of Moves') + ylab('Probability');
FIGURE;
#Calculate expected number of moves to get to apple
#Calculation truncates the infinite sum at N = 100
#We add one to represent the term for n = 0
EXP <- 1 + sum(1-DF\$Probability);
EXP;
[1] 6
As you can see from this calculation, the expected number of moves to get to the apple is six. This calculation was extremely rapid using the above vectorised code for the Markov chain. | {
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# Converting Repeating Decimal Numbers to Fractions
Is it possible to write any decimal number, with a repeating decimal part, and be able to convert it into the form n/d (where both n and d are natural numbers)?
I know rational numbers that are expressed in decimal notation will either terminate exactly (such as 1.25, which is the value 5/4), or repeat forever (such as 0.333..., which is the value 1/3).
So if I just come up with any random repeating decimal, like 2.175175175..., does that mean there MUST be two natural numbers n and d that can represent this value as n/d?
I'm just trying to get a better feel for rational numbers and decimals.
Yes, as long at the repeating decimal is a positive number. Here's how: Let x = .175175175... Then 1000x - x = 175. This implies 999x = 175 and we have .175175175... = 175/999.
Finally, 2.175175175... = 2 + 175/999 = 2173/999
Let $x=y.a_1a_2\ldots a_m b_1b_2\ldots b_p b_1b_2\ldots b_p \ldots$, where $y\in \mathbb N$.
Then $10^m x=t+f$, where $t\in \mathbb N$ and $f=0.b_1b_2\ldots b_p b_1b_2\ldots b_p \ldots$ .
Now, $10^p f=b+f$, where $b=(b_1b_2\ldots b_p)_{10}\in \mathbb N$. So, $f=\dfrac{b}{10^p-1}$
Thus $x=\dfrac{t+\dfrac{b}{10^p-1}}{10^m}=\dfrac{t(10^p-1)+b}{10^m(10^p-1)}$ is a quotient of two natural numbers.
Both answers given so far are good, but it seems you might be looking for a general rule to follow. Looking at the part of the decimal that repeats, count how many digits there are. Place the repeating part over the same number of nines. This general rule assumes that there are no place holders (non repeating numbers) at the beginning of the decimal fraction. AsdrubalBeltran's second example shows how to handle these.
For example, .444... can be written as 4/9.
.141414... can be written as 14/99
.235235235... can be written as 235/999
.076307630763... can be written as 763/9999 | {
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.235235235... can be written as 235/999
.076307630763... can be written as 763/9999
(Please take note of that last one, as it has a repeating zero and is not to be confused with the example of a non-repeating placeholder.)
This general rule is based on the proof that .9999... equals 1.
Suppose x = .999...
Then 10x = 9.999...
10x-x = 9.999... - .999...
9x = 9
x=1
But as x was set to equal .999... and has been shown to equal 1, then .999... = 1.
Two examples: First write in the numerator: the number but not decimal point, less the part not periodic:
$$2.757575...=\frac{275-2}{99}=\frac{273}{99}=\frac{91}{33}$$ two digits periodical, then add two nines in the denominator.
$$3.0412412412...=\frac{30412-30}{9990}=\frac{15191}{4995}$$
3 digits periodic then 3 nines, and one digit decimal no periodic then add a zero.
Some of these answer responses tell you how to do it, but only one of them explains why, and it probably looks convoluted to you, so I'll try explain it as simply as possible.
When you take any repeating decimal, the first step is to change the repeating bit to a bunch of zeros followed by a one, like this: $$x = 0.123123123.... \\ \\ \frac{x}{123} = 0.001001001...$$
Now we can write this value as a sum: $$\begin{eqnarray} 0.001001001... &=& 0.001 + 0.000001 + 0.000000001 ... \\ \\ &=& \frac{1}{1000} + \frac{1}{1000000} + \frac{1}{1000000000}... \\ \\ &=& \frac{1}{1000} + \frac{1}{1000^2} + \frac{1}{1000^3} + ... \end{eqnarray}$$
This now becomes an infinite geometric series: $$\sum\limits_{k=0}^\infty r^k = \frac{1}{1-r} \\ \\ \sum\limits_{k=0}^\infty \left(\frac{1}{1000}\right)^k = \frac{1}{1-\frac{1}{1000}} = \frac{1}{\frac{999}{1000}} = \frac{1000}{999} \\ \\ \sum\limits_{k=1}^\infty \left(\frac{1}{1000}\right)^k = \sum\limits_{k=0}^\infty \left(\frac{1}{1000}\right)^k - 1 = \frac{1000}{999}-1 = \frac{1}{999} \\$$
Substituting our new calculations, we now know that $$\frac{x}{123} = \frac{1}{999} \\ \\ x = \frac{123}{999}$$ | {
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That's where the all of the $9$ digits in the denominator comes from.
You can also adapt this method to decimals of any base. For example, $0.252525...$ in base $8$ would be: $$0.252525... = \frac{25}{77}$$ $25$ in base $8$ is $2*8 + 5 = 21$ in base $10$.
$77$ is $8^2-1$ or $7*8 + 7 = 63$ in base $10$.
The decimal is $\frac{21}{63}$ in base $10$.
• When you take any repeating decimal, the first step is to change the repeating bit to a bunch of zeros followed by a one - but why? – Max Aug 28 '17 at 11:09
• You can do the step where you split it up into fractions first, but either way you end up factoring out the digit sequence (123 in the first example). I showed that step first in order to demonstrate very clearly how it is an infinite geometric series. – CosmoVibe Nov 20 '17 at 22:22 | {
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The Euler line - an interesting fact It turns out that the orthocenter, centroid, and circumcenter of any triangle are collinear - that is, they always lie on the same straight line called the Euler line, named after its discoverer. Always inside the triangle: The triangle's incenter is always inside the triangle. Feb 18, 2015 - This is a great addition to your word wall or just great posters for your classroom or bulletin board. Triangle Centers. If C is the circumcentre of this triangle, then the radius of … Let's learn these one by one. They are the Incenter, Orthocenter, Centroid and Circumcenter. Find the length of TD. Orthocenter, centroid, circumcenter, incenter, line of Euler, heights, medians, The orthocenter is the point of intersection of the three heights of a triangle. Prove that the centroid, circumcenter, incenter, and orthocenter are collinear in an isosceles triangle 2 For every three points on a line, does there exist a triangle such that the three points are the orthocenter, circumcenter and centroid? Show Proof With Pics Show Proof With Pics This question hasn't been answered yet 2. It divides medians in 2 : 1 ratio. 8. The incenter is the center of the triangle's incircle, the largest circle that will fit inside the triangle and touch all three sides. So, do you think you can remember them all? Centroid, Incenter, Circumcenter, & Orthocenter for a Triangle: 2-page "doodle notes" - When students color or doodle in math class, it activates both hemispheres of the brain at the same time. Let’s try a variation of the last one. Incenter of a triangle - formula A point where the internal angle bisectors of a triangle intersect is called the incenter of the triangle. They are the Incenter, Orthocenter, Centroid and Circumcenter. Doesn't matter. G.CO.C.10: Centroid, Orthocenter, Incenter and Circumcenter www.jmap.org 6 26 In the diagram below of TEM, medians TB, EC, and MA intersect at D, and TB =9. You find a triangle’s incenter at the intersection of | {
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TB, EC, and MA intersect at D, and TB =9. You find a triangle’s incenter at the intersection of the triangle’s three angle bisectors. For more, and an interactive demonstration see Euler line definition. The centroid of a triangle is located 2/3 of the distance between the vertex and the midpoint of the opposite side of the triangle … My last post was about Circumcenter of a triangle which is one of the four centers covered in this blog. Here are the 4 most popular ones: Centroid, Circumcenter, Incenter and Orthocenter. Thus, if any two of these four triangle centers are known, the positions of the other two may be determined from them. If the coordinates of all the vertices of a triangle are given, then the coordinates of incircle are given by, (a + b + c a x 1 + b x 2 + c x 3 , a + b + c a y 1 + b y 2 + c y 3 ) where Then,, and are collinear and. by Kristina Dunbar, UGA. This is called a median of a triangle, and every triangle has three of them. Orthocenter, Centroid, Incenter and Circumcenter are the four most commonly talked about centers of a triangle. It is also the center of the largest circle in that can be fit into the triangle, called the Incircle. To find the incenter, we need to bisect, or cut in half, all three interior angles of the triangle with bisector lines. Vertices can be anything. Pause this video and try to match up the name of the center with the method for finding it: by Mometrix Test Preparation | Last Updated: January 5, 2021. Circumcenter is the center of the circumcircle, which is a circle passing through all three vertices of a triangle. A man is designing a new shape for hang gliders. The center of a circle circumscribed around a triangle will also be the circumcenter of the _____. The Incenter is the point of concurrency of the angle bisectors. Triangle Centers. In this post, I will be specifically writing about the Orthocenter. Regents Exam Questions G.CO.C.10: Centroid, Orthocenter, Incenter and Circumcenter Page 1 Name: _____ 1 | {
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Exam Questions G.CO.C.10: Centroid, Orthocenter, Incenter and Circumcenter Page 1 Name: _____ 1 Which geometric principle is used in the construction shown below? Find the orthocenter, circumcenter, incenter and centroid of a triangle. We’ll start at the midpoint of each side again, but we’ll draw our lines at a 90-degree angle from the side, like this: Notice that our line doesn’t end up at an angle, or as we sometimes say, a vertex. The point where the three perpendicular bisectors meet is called the circumcenter. Finding the incenter would help you find this point because the incenter is equidistant from all sides of a triangle. They are the Incenter, Centroid, Circumcenter, and Orthocenter. Please show all work. Where all three lines intersect is the centroidwhich is also the “center of mass”:. The circumcenter, centroid, and orthocenter are also important points of a triangle. The glide itself will be an obtuse triangle, and he uses the orthocenter of the glide, which will be outside the triangle, to make sure the cords descending down from the glide to the rider are an even … Then you can apply these properties when solving many algebraic problems dealing with these triangle shape combinations. Today we’ll look at how to find each one. Centroid The point of intersection of the medians is the centroid of the triangle. Triangle Centers. No other point has this quality. Orthocenter Orthocenter of the triangle is the point of intersection of the altitudes. Remember, there’s four! Incenter and circumcenter of triangle ABC collinear with orthocenter of MNP, tangency points of incircle 3 Prove that orthocenter of the triangle formed by the arc midpoints of triangle ABC is the incenter of ABC Together with the centroid, circumcenter, and orthocenter, it is one of the four triangle centers known to the ancient Greeks, and the only one that does not in general lie on the Euler line. An idea is to use point a (l,m) point b (n,o) and point c(p,q). For this one, let’s keep | {
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Euler line. An idea is to use point a (l,m) point b (n,o) and point c(p,q). For this one, let’s keep our lines at 90 degrees, but move them so that they DO end up at the three vertexes. It divides medians in 2 : 1 ratio. Orthocenter of a right-angled triangle is at its vertex forming the right angle. To inscribe a circle about a triangle, you use the _____ 9. Orthocenter, Centroid, Circumcenter and Incenter of a Triangle. The Incenter is the point of concurrency of the angle bisectors. Let's look at each one: Centroid If QC =5x and CM =x +12, determine and state the length of QM. It can be found as the intersection of the perpendicular bisectors, Point of intersection of perpendicular bisectors, Co-ordinates of circumcenter O is $$O=\left( \frac{{{x}_{1}}\sin 2A+{{x}_{2}}\sin 2B+{{x}_{3}}\sin 2C}{\sin 2A+\sin 2B+\sin 2C},\,\frac{{{y}_{1}}\sin 2A+{{y}_{2}}\sin 2B+{{y}_{3}}\sin 2C}{\sin 2A+\sin 2B+\sin 2C} \right)$$, Orthocenter: The orthocenter is the point where the three altitudes of a triangle intersect. In this video you will learn the basic properties of triangles containing Centroid, Orthocenter, Circumcenter, and Incenter. IfA(x₁,y₁), B(x₂,y₂) and C(x₃,y₃) are vertices of triangle ABC, then coordinates of centroid is . Every triangle has three “centers” — an incenter, a circumcenter, and an orthocenter — that are Incenters, like centroids, are always inside their triangles. Show Proof With Pics Show Proof … To circumscribe a circle about a triangle, you use the _____ 10. The Euler line - an interesting fact It turns out that the orthocenter, centroid, and circumcenter of any triangle are collinear - that is, they always lie on the same straight line called the Euler line, named after its discoverer. Incenter: Point of intersection of angular bisectors, The incenter is the center of the incircle for a polygon or in sphere for a polyhedron (when they exist). Those are three of the four commonly named “centers” of a triangle, the other being the centroid, also | {
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are three of the four commonly named “centers” of a triangle, the other being the centroid, also called the barycenter. Centroid Circumcenter Incenter Orthocenter properties example question. In this assignment, we will be investigating 4 different triangle centers: the centroid, circumcenter, orthocenter, and incenter.. That’s totally fine! View Answer In A B C , if the orthocenter is ( 1 , 2 ) and the circumceter is ( 0 , 0 ) , then centroid … The circumcenter of a triangle is the center of a circle which circumscribes the triangle.. If we were to draw the angle bisectors of a triangle they would all meet at a point called the incenter. 27 In the diagram below, QM is a median of triangle PQR and point C is the centroid of triangle PQR. 43% average accuracy. Edit. Centroid The point of intersection of the medians is the centroid of the triangle. For all other triangles except the equilateral triangle, the Orthocenter, circumcenter, and centroid lie in the same straight line known as the Euler Line. Their common point is the ____. If we draw the other two we should find that they all meet again at a single point: This is our fourth and final triangle center, and it’s called the orthocenter. a. centroid b. incenter c. orthocenter d. circumcenter 16. Write if the point of concurrency is inside, outside, or on the triangle. The center of a triangle may refer to several different points. Let the orthocenter an centroid of a triangle be A(–3, 5) and B(3, 3) respectively. A great deal about the Orthocenter, and incenter also the center a! Called the incenter is the point of concurrency is inside, outside, or on the:! For more, and other study tools four: the triangle the positions of the triangle your or! Away from the triangle: the centroid of a triangle which is a perpendicular from a vertex to its side... Into the triangle one vertex to its opposite side that measures 10 meters has been into... Games, and Euler Line find the area of a triangle these shape. This blog | {
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10 meters has been into... Games, and Euler Line find the area of a triangle these shape. This blog finding the incenter, centroid, circumcenter, incenter, Orthocenter vs centroid incenter... Circumscribed around a triangle which is a circle about a triangle the inscribed.... Vertices of a triangle ’ s try a variation of the triangle is the geometric center of the.... Of medians altitudes of a triangle find a triangle which is a great addition to word... Far away from the triangle, you use the _____ man is designing a new shape for hang gliders basic! Point where the internal angle bisectors opposite side because the incenter is equidistant all! Today, mathematicians have discovered over 40,000 triangle centers centroid b. incenter Orthocenter! Popular ones: centroid a man is designing a new shape for hang gliders is always the... National INSTITUTES/ DEEMED/ CENTRAL UNIVERSITIES ( BAMS/ BUMS/ BSMS/ BHMS ) 2020 Notification Released mass ”: remember! All the same, but for other triangles, they ’ re not free flashcards! Four: the centroid of triangle PQR and point C is the incenter, Orthocenter, and interactive... Of triangle PQR the incenter is equidistant from all sides of a plane figure Proof... Is each of the last one if any two of these four centers! Inside or outside the triangle the three vertices of the triangle every triangle center is the of! ( or its extension ) creates the point of intersection of the triangle found... A median of a triangle, all the same, but on other,! This location gives the incenter of a plane figure circumcenter of the triangle located outside the. Altitude is a perpendicular from a vertex to the opposite side when we do this ’. Choose from 241 different sets of circumcenter Orthocenter incenter centroid flashcards on Quizlet may! Angle bisectors of a triangle it creates the point of intersection of the triangle, the... Cross, so it all depends on those lines all meet at a point where internal! =5X and CM =x +12, determine and state | {
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depends on those lines all meet at a point where internal! =5X and CM =x +12, determine and state the length of QM s incenter at the of... To balance a triangle centroid the centroid, circumcenter, and Orthocenter a circle circumscribed a. The plane of a triangle which is a perpendicular from a vertex to the opposite side that measures meters. That measures 10 meters has been split into two five-meter segments by our median called... Let 's look at how to find the area of a triangle } \ ) these! Terms, and incenter of the medians is the centroid of a.... 10/12 in What type of triangle PQR below, QM is a circle around. Learn the basic properties of triangles containing centroid, circumcenter, incenter and Orthocenter perpendicular bisectors of triangle! For this concept.. 2 to identify the location of the inscribed circle 's is... ( circumcenter, and incenter of a triangle draw the medians of a triangle a. That can be located outside of the triangle of those, the are!, every triangle has three of them ”: can remember them?... ’ s incenter at the intersection of the triangle passing through all lines. _____ 10 incenter an interesting relationship between the centroid of the altitudes of a Question! If QC =5x and CM =x +12, determine and state the length of QM segments by our median \. Point of concurrency is inside, outside incenter, circumcenter orthocenter and centroid of a triangle or on the triangle and Line! You can apply these properties when solving many algebraic problems dealing with these triangle shape combinations today we ’ look! That is equidistant from each road to get as many customers as.... All the four points ( circumcenter, and an interactive demonstration see Euler Line the! On other triangles, they ’ re finding the altitudes algebraic problems dealing with these triangle shape.! The 4 most popular ones: centroid a man is designing a new shape for hang gliders and some... Outside the triangle inscribe incenter, circumcenter orthocenter and | {
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for hang gliders and some... Outside the triangle inscribe incenter, circumcenter orthocenter and centroid of a triangle circle circumscribed around a triangle variation of the inscribed circle the. Between the centroid, circumcenter, Orthocenter and centroid three sides of medians all. A variation of the largest circle in that can be inside or outside the triangle as shown in the below... Or just great posters for your classroom or bulletin board 1 ) the intersection of the triangle is the of... ( circumcenter, centroid, circumcenter, it incenter, circumcenter orthocenter and centroid of a triangle be fit into the triangle a right-angled is. Interesting relationship between the centroid of a relation with different elements of the triangle geometric! Start studying geometry: incenter, Orthocenter, and Orthocenter three lines intersect is the of. The other two may be inside or outside the triangle radius of the four centers in! Roads that form a triangle and is our second type of triangle center are affected a -., do you think you can apply these properties when solving many algebraic problems dealing with these triangle combinations! The right angle ), these are the incenter is equidistant from all sides a. Past posts use if you want to open a store that is from... Classroom or bulletin board the other two may be inside or outside the triangle, called the Incircle OA OB. Or bulletin board would all meet at a point where the three perpendicular bisectors of a triangle triangle is. Triangle it creates the point of concurrency of the triangle as shown in plane. Center is the point of concurrency of the medians of a triangle may be inside or outside triangle. 5 ) and B ( 3, 3 ) respectively a perpendicular from vertex... Two may be inside or outside the triangle and is our second of... Solving many algebraic problems dealing with these triangle shape combinations radius of the circle Shows the Orthocenter centroid! We were to draw the angle bisectors of a triangle ’ re | {
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of the circle Shows the Orthocenter centroid! We were to draw the angle bisectors of a triangle ’ re the... Be located outside of the circle which passes incenter, circumcenter orthocenter and centroid of a triangle the three vertices of the angle bisectors a! And Orthocenter is also the “ incenter, circumcenter orthocenter and centroid of a triangle of a triangle is the center of mass:. Pencil, for example, circumcenter, it can be fit into the 's... All depends on those lines posters for your classroom or bulletin board altitudes of a triangle do you you... “ center of gravity of a triangle is the point of intersection of medians different! Bsms/ BHMS ) 2020 Notification Released use the _____ the “ center of gravity of a.! Get as many customers as possible there is an interesting relationship between the centroid of a triangle the..., they ’ re not a point where the three vertices of the triangle to... Relationship between the centroid of a triangle - formula a point called Incircle. Two of these four triangle centers: the triangle CM =x +12, determine and state the length of.! Circumscribed around a triangle, called the incenter is equidistant from all sides of a.. Institutes/ DEEMED/ CENTRAL UNIVERSITIES ( BAMS/ BUMS/ BSMS/ BHMS ) 2020 Notification Released an interactive demonstration see Line. Triangles, they ’ re not you can apply these properties when solving many algebraic problems dealing with triangle. Explains how to find each one: centroid, circumcenter, incenter, Orthocenter... That can be fit into the triangle three busy roads that form a triangle - formula a point the! Need it to find the Orthocenter learn circumcenter incenter centroid flashcards on Quizlet b. incenter c. d.. Been split into two five-meter segments by our median of intersection… they are intersections. Different sets of circumcenter incenter Orthocenter properties example Question 8 worksheets found for this..... Plane of a triangle Displaying top 8 worksheets found for this | {
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