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Find the Padé approximation to \( {e}^{-x} \) of degree 5 with \( n = 3 \) and \( m = 2 \) . | Solution To find the Padé approximation we need to choose \( {p}_{0},{p}_{1},{p}_{2},{p}_{3},{q}_{1} \), and \( {q}_{2} \) so that the coefficients of \( {x}^{k} \) for \( k = 0,1,\ldots ,5 \) are 0 in the expression\n\n\[ \left( {1 - x + \frac{{x}^{2}}{2} - \frac{{x}^{3}}{6} + \cdots }\right) \left( {1 + {q}_{1}x + {q... | Yes |
Determine the Chebyshev rational approximation of degree 5 with \( n = 3 \) and \( m = 2 \) . | Solution Finding this approximation requires choosing \( {p}_{0},{p}_{1},{p}_{2},{p}_{3},{q}_{1} \), and \( {q}_{2} \) so that for \( k = 0,1,2,3,4 \), and 5, the coefficients of \( {T}_{k}\left( x\right) \) are 0 in the expansion\n\n\[ \n{\widetilde{P}}_{5}\left( x\right) \left\lbrack {{T}_{0}\left( x\right) + {q}_{1}... | Yes |
Determine the trigonometric polynomial from \( {\mathcal{T}}_{n} \) that approximates \( f\left( x\right) = \left| x\right| ,\;\text{ for } - \pi < x < \pi . | Solution We first need to find the coefficients\n\n\[ {a}_{0} = \frac{1}{\pi }{\int }_{-\pi }^{\pi }\left| x\right| {dx} = - \frac{1}{\pi }{\int }_{-\pi }^{0}{xdx} + \frac{1}{\pi }{\int }_{0}^{\pi }{xdx} = \frac{2}{\pi }{\int }_{0}^{\pi }{xdx} = \pi ,\]\n\n\[ {a}_{k} = \frac{1}{\pi }{\int }_{-\pi }^{\pi }\left| x\right... | Yes |
Lemma 8.12 Suppose that the integer \( r \) is not a multiple of \( {2m} \) . Then\n\n\[ \text{-}\mathop{\sum }\limits_{{j = 0}}^{{{2m} - 1}}\cos r{x}_{j} = 0\text{and}\mathop{\sum }\limits_{{j = 0}}^{{{2m} - 1}}\sin r{x}_{j} = 0\text{.} \]\n\nMoreover, if \( r \) is not a multiple of \( m \), then\n\n\[ \text{-}\matho... | Proof Euler’s Formula states that with \( {i}^{2} = - 1 \), we have, for every real number \( z \) ,\n\n\[ {e}^{iz} = \cos z + i\sin z \]\n\n(8.25)\n\nApplying this result gives\n\n\[ \mathop{\sum }\limits_{{j = 0}}^{{{2m} - 1}}\cos r{x}_{j} + i\mathop{\sum }\limits_{{j = 0}}^{{{2m} - 1}}\sin r{x}_{j} = \mathop{\sum }\... | Yes |
Theorem 8.13 The constants in the summation\n\n\[ \n{S}_{n}\left( x\right) = \frac{{a}_{0}}{2} + {a}_{n}\cos {nx} + \mathop{\sum }\limits_{{k = 1}}^{{n - 1}}\left( {{a}_{k}\cos {kx} + {b}_{k}\sin {kx}}\right) \n\]\n\nthat minimize the least squares sum\n\n\[ \nE\left( {{a}_{0},\ldots ,{a}_{n},{b}_{1},\ldots ,{b}_{n - 1... | The theorem is proved by setting the partial derivatives of \( E \) with respect to the \( {a}_{k} \) ’s and the \( {b}_{k} \) ’s to zero, as was done in Sections 8.1 and 8.2, and applying the orthogonality to simplify the equations. For example,\n\n\[ \n0 = \frac{\partial E}{\partial {b}_{k}} = 2\mathop{\sum }\limits_... | No |
Find \( {S}_{2}\left( x\right) \), the discrete least squares trigonometric polynomial of degree 2 for \( f\left( x\right) = \) \( 2{x}^{2} - 9 \) when \( x \) is in \( \left\lbrack {-\pi ,\pi }\right\rbrack \). | Solution We have \( m = 2\left( 2\right) - 1 = 3 \), so the nodes are\n\n\[ \n{x}_{j} = \pi + \frac{j}{m}\pi \;\text{ and }\;{y}_{j} = f\left( {x}_{j}\right) = 2{x}_{j}^{2} - 9,\;\text{ for }j = 0,1,2,3,4,5.\n\]\n\nThe trigonometric polynomial is\n\n\[ \n{S}_{2}\left( x\right) = \frac{1}{2}{a}_{0} + {a}_{2}\cos {2x} + ... | Yes |
Example 3 Find the discrete least squares approximation \( {S}_{3}\left( x\right) \) for\n\n\[ f\left( x\right) = {x}^{4} - 3{x}^{3} + 2{x}^{2} - \tan x\left( {x - 2}\right) \]\n\nusing the data \( {\left\{ \left( {x}_{j},{y}_{j}\right) \right\} }_{j = 0}^{9} \), where \( {x}_{j} = j/5 \) and \( {y}_{j} = f\left( {x}_{... | Solution We first need the linear transformation from \( \left\lbrack {0,2}\right\rbrack \) to \( \left\lbrack {-\pi ,\pi }\right\rbrack \) given by\n\n\[ {z}_{j} = \pi \left( {{x}_{j} - 1}\right) \]\n\nThen the transformed data have the form\n\n\[ {\left\{ \left( {z}_{j}, f\left( 1 + \frac{{z}_{j}}{\pi }\right) \right... | Yes |
Example 2 Determine the trigonometric interpolating polynomial of degree 4 on \( \\left\\lbrack {0,2}\\right\\rbrack \) for the data \( \\{ \\left( {j/4, f\\left( {j/4}\\right) }\\right) {\\} }_{j = 0}^{7} \), where \( f\\left( x\\right) = {x}^{4} - 3{x}^{3} + 2{x}^{2} - \\tan x\\left( {x - 2}\\right) \) . | Solution We first need to transform the interval \( \\left\\lbrack {0,2}\\right\\rbrack \) to \( \\left\\lbrack {-\\pi ,\\pi }\\right\\rbrack \) . This is given by\n\n\[ \n{z}_{j} = \\pi \\left( {{x}_{j} - 1}\\right) \n\]\n\nso that the input data to Algorithm 8.3 are\n\n\[ \n{\\left\\{ {z}_{j}, f\\left( 1 + \\frac{{z}... | Yes |
Let \( A \) be an \( n \times n \) matrix and \( {R}_{i} \) denote the circle in the complex plane with center \( {a}_{ii} \) and radius \( \mathop{\sum }\limits_{{j = 1, j \neq i}}^{n}\left| {a}_{ij}\right| \) ; that is,\n\n\[ \n{R}_{i} = \left\{ {z \in \mathcal{C}\left| \right| z - {a}_{ii}\left| { \leq \mathop{\sum ... | Proof Suppose that \( \lambda \) is an eigenvalue of \( A \) with associated eigenvector \( \mathbf{x} \), where \( \parallel \mathbf{x}{\parallel }_{\infty } = 1 \) . Since \( A\mathbf{x} = \lambda \mathbf{x} \), the equivalent component representation is\n\n\[ \n\mathop{\sum }\limits_{{j = 1}}^{n}{a}_{ij}{x}_{j} = \l... | Yes |
Determine the Geršgorin circles for the matrix\n\n\[ A = \left\lbrack \begin{array}{rrr} 4 & 1 & 1 \\ 0 & 2 & 1 \\ - 2 & 0 & 9 \end{array}\right\rbrack \]\n\nand use these to find bounds for the spectral radius of \( A \) . | The circles in the Geršgorin Theorem are (see Figure 9.1)\n\n\( {R}_{1} = \{ z \in \mathcal{C}\left| \right| z - 4\left| { \leq 2\} ,\;{R}_{2} = \{ z \in \mathcal{C} \mid }\right| z - 2 \mid \leq 1\} ,\; \) and \( \;{R}_{3} = \{ z \in \mathcal{C}\left| \right| z - 9 \mid \leq 2\} . \)\n\nBecause \( {R}_{1} \) and \( {R... | Yes |
Theorem 9.3 Suppose that \( \left\{ {{\mathbf{v}}^{\left( 1\right) },{\mathbf{v}}^{\left( 2\right) },{\mathbf{v}}^{\left( 3\right) },\ldots ,{\mathbf{v}}^{\left( n\right) }}\right\} \) is a set of \( n \) linearly independent vectors in \( {\mathbb{R}}^{n} \) . Then for any vector \( \mathbf{x} \in {\mathbb{R}}^{n} \) ... | Proof Let \( A \) be the matrix whose columns are the vectors \( {\mathbf{v}}^{\left( 1\right) },{\mathbf{v}}^{\left( 2\right) },\ldots ,{\mathbf{v}}^{\left( n\right) } \) . Then the set \( \left\{ {{\mathbf{v}}^{\left( 1\right) },{\mathbf{v}}^{\left( 2\right) },\ldots ,{\mathbf{v}}^{\left( n\right) }}\right\} \) is li... | Yes |
Show that \( {\mathbf{v}}^{\left( 1\right) } = {\left( 1,0,0\right) }^{t},{\mathbf{v}}^{\left( 2\right) } = {\left( -1,1,1\right) }^{t} \), and \( {\mathbf{v}}^{\left( 3\right) } = {\left( 0,4,2\right) }^{t} \) is a basis for \( {\mathbb{R}}^{3} \) | Let \( {\alpha }_{1},{\alpha }_{2} \), and \( {\alpha }_{3} \) be numbers with \( \;\mathbf{0} = {\alpha }_{1}{\mathbf{v}}^{\left( 1\right) } + {\alpha }_{2}{\mathbf{v}}^{\left( 2\right) } + {\alpha }_{3}{\mathbf{v}}^{\left( 3\right) } \) . Then\n\n\[{\left( 0,0,0\right) }^{t} = {\alpha }_{1}{\left( 1,0,0\right) }^{t} ... | Yes |
Show that a basis can be formed for \( {\mathbb{R}}^{3} \) using the eigenvectors of the \( 3 \times 3 \) matrix\n\n\[ A = \left\lbrack \begin{array}{rrr} 2 & 0 & 0 \\ 1 & 1 & 2 \\ 1 & - 1 & 4 \end{array}\right\rbrack \] | Solution In Example 2 of Section 7.2 we found that \( A \) has the characteristic polynomial\n\n\[ p\left( \lambda \right) = \det \left( {A - {\lambda I}}\right) = \left( {\lambda - 3}\right) {\left( \lambda - 2\right) }^{2}. \]\n\nHence there are two distinct eigenvalues of \( A : {\lambda }_{1} = 3 \) and \( {\lambda... | No |
Show that no collection of eigenvectors of the \( 3 \times 3 \) matrix\n\n\[ B = \left\lbrack \begin{array}{lll} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{array}\right\rbrack \]\n\ncan form a basis for \( {\mathbb{R}}^{3} \) . | Solution This matrix also has the same characteristic polynomial as the matrix \( A \) in Example 3:\n\n\[ p\left( \lambda \right) = \det \left\lbrack \begin{array}{rrr} 2 - \lambda & 1 & 0 \\ 0 & 2 - \lambda & 0 \\ 0 & 0 & 3 - \lambda \end{array}\right\rbrack = \left( {\lambda - 3}\right) {\left( \lambda - 2\right) }^... | Yes |
Example 5 (a) Show that the vectors \( {\mathbf{v}}^{\left( 1\right) } = {\left( 0,4,2\right) }^{t},{\mathbf{v}}^{\left( 2\right) } = {\left( -5, - 1,2\right) }^{t} \), and \( {\mathbf{v}}^{\left( 3\right) } = {\left( 1, - 1,2\right) }^{t} \) form an orthogonal set, and (b) use these to determine a set of orthonormal v... | Solution (a) We have \( \;{\left( {\mathbf{v}}^{\left( 1\right) }\right) }^{t}{\mathbf{v}}^{\left( 2\right) } = 0\left( {-5}\right) + 4\left( {-1}\right) + 2\left( 2\right) = 0 \) ,\n\n\( {\left( {\mathbf{v}}^{\left( 1\right) }\right) }^{t}{\mathbf{v}}^{\left( 3\right) } = 0\left( 1\right) + 4\left( {-1}\right) + 2\lef... | No |
Theorem 9.8 Let \( \\left\\{ {{\\mathbf{x}}_{1},{\\mathbf{x}}_{2},\\ldots ,{\\mathbf{x}}_{k}}\\right\\} \) be a set of \( k \) linearly independent vectors in \( {\\mathbb{R}}^{n} \) . Then \( \\left\\{ {{\\mathbf{v}}_{1},{\\mathbf{v}}_{2},\\ldots ,{\\mathbf{v}}_{k}}\\right\\} \) defined by\n\n\[ \n{\\mathbf{v}}_{1} = ... | The proof of this theorem, discussed in Exercise 16, is a direct verification of the fact that for each \( 1 \\leq i \\leq k \) and \( 1 \\leq j \\leq k \), with \( i \\neq j \), we have \( {\\mathbf{v}}_{i}^{t}{\\mathbf{v}}_{j} = 0 \) . | No |
Example 6 Use the Gram-Schmidt process to determine a set of orthogonal vectors from the linearly independent vectors\n\n\[ \n{\mathbf{x}}^{\left( 1\right) } = {\left( 1,0,0\right) }^{t},\;{\mathbf{x}}^{\left( 2\right) } = {\left( 1,1,0\right) }^{t},\;\text{ and }\;{\mathbf{x}}^{\left( 3\right) } = {\left( 1,1,1\right)... | Solution We have the orthogonal vectors \( {\mathbf{v}}^{\left( 1\right) },{\mathbf{v}}^{\left( 2\right) } \), and \( {\mathbf{v}}^{\left( 3\right) } \), given by\n\n\[ \n{\mathbf{v}}^{\left( 1\right) } = {\mathbf{x}}^{\left( 1\right) } = {\left( 1,0,0\right) }^{t}\n\]\n\n\[ \n{\mathbf{v}}^{\left( 2\right) } = {\left( ... | Yes |
Show that the matrix\n\n\[ \nQ = \left\lbrack {{\mathbf{u}}^{\left( 1\right) },{\mathbf{u}}^{\left( 2\right) },{\mathbf{u}}^{\left( 3\right) }}\right\rbrack = \left\lbrack \begin{matrix} 0 & - \frac{\sqrt{30}}{6} & \frac{\sqrt{6}}{6} \\ \frac{2\sqrt{5}}{5} & - \frac{\sqrt{30}}{30} & - \frac{\sqrt{6}}{6} \\ \frac{\sqrt{... | Solution Note that\n\n\[ \nQ{Q}^{t} = \left\lbrack \begin{matrix} 0 & \frac{-\sqrt{30}}{6} & \frac{\sqrt{6}}{6} \\ \frac{2\sqrt{5}}{5} & - \frac{\sqrt{30}}{30} & - \frac{\sqrt{6}}{6} \\ \frac{\sqrt{5}}{5} & \frac{\sqrt{30}}{15} & \frac{\sqrt{6}}{3} \end{matrix}\right\rbrack \cdot \left\lbrack \begin{matrix} 0 & \frac{2... | Yes |
Theorem 9.12 Suppose \( A \) and \( B \) are similar matrices with \( A = {S}^{-1}{BS} \) and \( \lambda \) is an eigenvalue of \( A \) with associated eigenvector \( \mathbf{x} \) . Then \( \lambda \) is an eigenvalue of \( B \) with associated eigenvector \( S\mathbf{x} \) . | Proof Let \( \mathbf{x} \neq \mathbf{0} \) be such that\n\n\[ \n{S}^{-1}{BS}\mathbf{x} = A\mathbf{x} = \lambda \mathbf{x} \n\]\n\nMultiplying on the left by the matrix \( S \) gives\n\n\[ \n{BS}\mathbf{x} = {\lambda S}\mathbf{x} \n\]\n\nSince \( \mathbf{x} \neq \mathbf{0} \) and \( S \) is nonsingular, \( S\mathbf{x} \... | Yes |
The \( n \times n \) matrix \( A \) is symmetric if and only if there exists a diagonal matrix \( D \) and an orthogonal matrix \( Q \) with \( A = {QD}{Q}^{t} \) . | First suppose that \( A = {QD}{Q}^{t} \), where \( Q \) is orthogonal and \( D \) is diagonal. Then\n\n\[ \n{A}^{t} = {\left( QD{Q}^{t}\right) }^{t} = {\left( {Q}^{t}\right) }^{t}D{Q}^{t} = {QD}{Q}^{t} = A, \n\]\n\nand \( A \) is symmetric.\n\nTo prove that every symmetric matrix \( A \) can be written in the form \( A... | Yes |
Corollary 9.17 Suppose that \( A \) is a symmetric \( n \times n \) matrix. There exist \( n \) eigenvectors of \( A \) that form an orthonormal set, and the eigenvalues of \( A \) are real numbers. | Proof If \( Q = \left( {q}_{ij}\right) \) and \( D = \left( {d}_{ij}\right) \) are the matrices specified in Theorem 9.16, then\n\n\[ D = {Q}^{t}{AQ} = {Q}^{-1}{AQ}\;\text{ implies that }\;{AQ} = {QD}. \]\n\nLet \( 1 \leq i \leq n \) and \( {\mathbf{v}}_{i} = {\left( {q}_{1i},{q}_{2i},\ldots ,{q}_{ni}\right) }^{t} \) b... | Yes |
Theorem 9.18 A symmetric matrix \( A \) is positive definite if and only if all the eigenvalues of \( A \) are positive. | Proof First suppose that \( A \) is positive definite and that \( \lambda \) is an eigenvalue of \( A \) with associated eigenvector \( \mathbf{x} \), with \( \parallel \mathbf{x}{\parallel }_{2} = 1 \) . Then\n\n\[ 0 < {\mathbf{x}}^{t}A\mathbf{x} = \lambda {\mathbf{x}}^{t}\mathbf{x} = \lambda \parallel \mathbf{x}{\par... | Yes |
Use the Power method to approximate the dominant eigenvalue of the matrix\n\n\[ A = \left\lbrack \begin{array}{rrr} - 4 & {14} & 0 \\ - 5 & {13} & 0 \\ - 1 & 0 & 2 \end{array}\right\rbrack \] | Solution This matrix has eigenvalues \( {\lambda }_{1} = 6,{\lambda }_{2} = 3 \), and \( {\lambda }_{3} = 2 \), so the Power method described in Algorithm 9.1 will converge. Let \( {\mathbf{x}}^{\left( 0\right) } = {\left( 1,1,1\right) }^{t} \), then\n\n\[ {\mathbf{y}}^{\left( 1\right) } = A{\mathbf{x}}^{\left( 0\right... | Yes |
Apply both the Power method and the Symmetric Power method to the matrix \[ A = \left\lbrack \begin{array}{rrr} 4 & - 1 & 1 \\ - 1 & 3 & - 2 \\ 1 & - 2 & 3 \end{array}\right\rbrack \] using Aitken’s \( {\Delta }^{2} \) method to accelerate the convergence. | Solution This matrix has eigenvalues \( {\lambda }_{1} = 6,{\lambda }_{2} = 3 \), and \( {\lambda }_{3} = 1 \) . An eigenvector for the eigenvalue 6 is \( {\left( 1, - 1,1\right) }^{t} \) . Applying the Power method to this matrix with initial vector \( {\left( 1,0,0\right) }^{t} \) gives the values in Table 9.2.\n\nTa... | Yes |
Theorem 9.19 9 Suppose that \( A \) is an \( n \times n \) symmetric matrix with eigenvalues \( {\lambda }_{1},{\lambda }_{2},\ldots ,{\lambda }_{n} \) . If we have \( \parallel A\mathbf{x} - \lambda \mathbf{x}{\parallel }_{2} < \varepsilon \) for some real number \( \lambda \) and vector \( \mathbf{x} \) with \( \para... | Proof Suppose that \( {\mathbf{v}}^{\left( 1\right) },{\mathbf{v}}^{\left( 2\right) },\ldots ,{\mathbf{v}}^{\left( n\right) } \) form an orthonormal set of eigenvectors of \( A \) associated, respectively, with the eigenvalues \( {\lambda }_{1},{\lambda }_{2},\ldots ,{\lambda }_{n} \) . By Theorems 9.5 and \( {9.3},\ma... | Yes |
Example 3 Apply the Inverse Power method with \( {\mathbf{x}}^{\left( 0\right) } = {\left( 1,1,1\right) }^{t} \) to the matrix \[ A = \left\lbrack \begin{matrix} - 4 & {14} & 0 \\ - 5 & {13} & 0 \\ - 1 & 0 & 2 \end{matrix}\right\rbrack \;\text{ with }\;q = \frac{{\mathbf{x}}^{\left( 0\right) t}A{\mathbf{x}}^{\left( 0\r... | Solution The Power method was applied to this matrix in Example 1 using the initial vector \( {\mathbf{x}}^{\left( 0\right) } = {\left( 1,1,1\right) }^{t} \) . It gave the approximate eigenvalue \( {\mu }^{\left( {12}\right) } = {6.000837} \) and eigenvector \( {\left( {\mathbf{x}}^{\left( {12}\right) }\right) }^{t} = ... | Yes |
Theorem 9.20 Suppose \( {\lambda }_{1},{\lambda }_{2},\ldots ,{\lambda }_{n} \) are eigenvalues of \( A \) with associated eigenvectors \( {\mathbf{v}}^{\left( 1\right) },{\mathbf{v}}^{\left( 2\right) },\ldots ,{\mathbf{v}}^{\left( n\right) } \) and that \( {\lambda }_{1} \) has multiplicity 1 . Let \( \mathbf{x} \) be... | There are many choices of the vector \( \mathbf{x} \) that could be used in Theorem 9.20. Wielandt deflation proceeds from defining\n\n\[ \mathbf{x} = \frac{1}{{\lambda }_{1}{v}_{i}^{\left( 1\right) }}{\left( {a}_{i1},{a}_{i2},\ldots ,{a}_{in}\right) }^{t}, \]\n\n(9.7)\n\nwhere \( {v}_{i}^{\left( 1\right) } \) is a non... | Yes |
The matrix\n\n\[ A = \left\lbrack \begin{array}{rrr} 4 & - 1 & 1 \\ - 1 & 3 & - 2 \\ 1 & - 2 & 3 \end{array}\right\rbrack \]\n\nhas the dominant eigenvalue \( {\lambda }_{1} = 6 \) with associated unit eigenvector \( {\mathbf{v}}^{\left( 1\right) } = {\left( 1, - 1,1\right) }^{t} \). Assume that this dominant eigenvalu... | Solution The procedure for obtaining a second eigenvalue \( {\lambda }_{2} \) proceeds as follows:\n\n\[ \mathbf{x} = \frac{1}{6}\left\lbrack \begin{array}{r} 4 \\ - 1 \\ 1 \end{array}\right\rbrack = {\left( \frac{2}{3}, - \frac{1}{6},\frac{1}{6}\right) }^{t} \]\n\n\[ {\mathbf{v}}^{\left( 1\right) }{\mathbf{x}}^{t} = \... | Yes |
Apply Householder transformations to the symmetric \( 4 \times 4 \) matrix\n\n\[ A = \left\lbrack \begin{array}{rrrr} 4 & 1 & - 2 & 2 \\ 1 & 2 & 0 & 1 \\ - 2 & 0 & 3 & - 2 \\ 2 & 1 & - 2 & - 1 \end{array}\right\rbrack \]\nto produce a symmetric tridiagonal matrix that is similar to \( A \) . | Solution For the first application of a Householder transformation,\n\n\[ \alpha = - \left( 1\right) {\left( \mathop{\sum }\limits_{{j = 2}}^{4}{a}_{j1}^{2}\right) }^{1/2} = - 3, r = {\left( \frac{1}{2}{\left( -3\right) }^{2} - \frac{1}{2}\left( 1\right) \left( -3\right) \right) }^{1/2} = \sqrt{6}, \]\n\n\[ \mathbf{w} ... | Yes |
Find a rotation matrix \( P \) with the property that \( {PA} \) has a zero entry in the second row and first column, where\n\n\[ A = \left\lbrack \begin{array}{lll} 3 & 1 & 0 \\ 1 & 3 & 1 \\ 0 & 1 & 3 \end{array}\right\rbrack \] | Solution The form of \( P \) is\n\n\[ P = \left\lbrack \begin{matrix} \cos \theta & \sin \theta & 0 \\ - \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{matrix}\right\rbrack \text{ so }{PA} = \left\lbrack \begin{matrix} 3\cos \theta + \sin \theta & \cos \theta + 3\sin \theta & \sin \theta \\ - 3\sin \theta + \cos \thet... | Yes |
Apply one iteration of the QR method to the matrix given in Example 1: | Let \( {A}^{\left( 1\right) } = A \) be the given matrix and \( {P}_{2} \) represent the rotation matrix determined in Example 1. We found, using the notation introduced in the QR method, that\n\n\[ {A}_{2}^{\left( 1\right) } = {P}_{2}{A}^{\left( 1\right) } = \left\lbrack \begin{matrix} \frac{3\sqrt{10}}{10} & \frac{\s... | Yes |
Example 3 Incorporate shifting into the QR method for the matrix\n\n\[ A = \left\lbrack \begin{array}{lll} 3 & 1 & 0 \\ 1 & 3 & 1 \\ 0 & 1 & 3 \end{array}\right\rbrack = \left\lbrack \begin{matrix} {a}_{1}^{\left( 1\right) } & {b}_{2}^{\left( 1\right) } & 0 \\ {b}_{2}^{\left( 1\right) } & {a}_{2}^{\left( 1\right) } & {... | Solution To find the acceleration parameter for shifting requires finding the eigenvalues of\n\n\[ \left\lbrack \begin{array}{ll} {a}_{2}^{\left( 1\right) } & {b}_{3}^{\left( 1\right) } \\ {b}_{3}^{\left( 1\right) } & {a}_{3}^{\left( 1\right) } \end{array}\right\rbrack = \left\lbrack \begin{array}{ll} 3 & 1 \\ 1 & 3 \e... | Yes |
Determine the singular values of the \( 5 \times 3 \) matrix\n\n\[ A = \left\lbrack \begin{array}{lll} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{array}\right\rbrack \] | Solution We have\n\n\[ {A}^{t} = \left\lbrack \begin{array}{lllll} 1 & 0 & 0 & 0 & 1 \\ 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & 1 & 0 & 0 \end{array}\right\rbrack \;\text{ so }\;{A}^{t}A = \left\lbrack \begin{array}{lll} 2 & 1 & 1 \\ 1 & 4 & 1 \\ 1 & 1 & 2 \end{array}\right\rbrack .\n\nThe characteristic polynomial of \( {A}^{t}A... | Yes |
Place the \( 3 \times 3 \) nonlinear system\n\n\[ 3{x}_{1} - \cos \left( {{x}_{2}{x}_{3}}\right) - \frac{1}{2} = 0 \]\n\n\[ {x}_{1}^{2} - {81}{\left( {x}_{2} + {0.1}\right) }^{2} + \sin {x}_{3} + {1.06} = 0, \]\n\n\[ {e}^{-{x}_{1}{x}_{2}} + {20}{x}_{3} + \frac{{10\pi } - 3}{3} = 0 \] | Solution Define the three coordinate functions \( {f}_{1},{f}_{2} \), and \( {f}_{3} \) from \( {\mathbb{R}}^{3} \) to \( \mathbb{R} \) as\n\n\[ {f}_{1}\left( {{x}_{1},{x}_{2},{x}_{3}}\right) = 3{x}_{1} - \cos \left( {{x}_{2}{x}_{3}}\right) - \frac{1}{2}, \]\n\n\[ {f}_{2}\left( {{x}_{1},{x}_{2},{x}_{3}}\right) = {x}_{1... | Yes |
Theorem 10.6 Let \( D = \left\{ {{\left( {x}_{1},{x}_{2},\ldots ,{x}_{n}\right) }^{t} \mid {a}_{i} \leq {x}_{i} \leq {b}_{i}\text{, for each}i = 1,2,\ldots, n}\right\} \) for some collection of constants \( {a}_{1},{a}_{2},\ldots ,{a}_{n} \) and \( {b}_{1},{b}_{2},\ldots ,{b}_{n} \) . Suppose \( \mathbf{G} \) is a cont... | Moreover, suppose that all the component functions of \( \mathbf{G} \) have continuous partial derivatives and a constant \( K < 1 \) exists with\n\n\[ \left| \frac{\partial {g}_{i}\left( \mathbf{x}\right) }{\partial {x}_{j}}\right| \leq \frac{K}{n},\;\text{ whenever }\mathbf{x} \in D, \]\n\nfor each \( j = 1,2,\ldots,... | Yes |
Theorem 10.10 Let \( \mathbf{F}\left( \mathbf{x}\right) \) be continuously differentiable for \( \mathbf{x} \in {\mathbb{R}}^{n} \) . Suppose that the Jacobian matrix \( J\left( \mathbf{x}\right) \) is nonsingular for all \( \mathbf{x} \in {\mathbb{R}}^{n} \) and that a constant \( M \) exists with \( \begin{Vmatrix}{J... | \[ \mathbf{G}\left( {\lambda ,\mathbf{x}\left( \lambda \right) }\right) = \mathbf{0}, \] for all \( \lambda \) in \( \left\lbrack {0,1}\right\rbrack \) . Moreover, \( \mathbf{x}\left( \lambda \right) \) is continuously differentiable and \[ {\mathbf{x}}^{\prime }\left( \lambda \right) = - J{\left( \mathbf{x}\left( \lam... | Yes |
Use the Continuation method with \( \mathbf{x}\left( 0\right) = {\left( 0,0,0\right) }^{t} \) to approximate the solution to\n\n\[ \n{f}_{1}\left( {{x}_{1},{x}_{2},{x}_{3}}\right) = 3{x}_{1} - \cos \left( {{x}_{2}{x}_{3}}\right) - {0.5} = 0, \]\n\n\[ \n{f}_{2}\left( {{x}_{1},{x}_{2},{x}_{3}}\right) = {x}_{1}^{2} - {81}... | Solution The Jacobian matrix is\n\n\[ \nJ\left( \mathbf{x}\right) = \left\lbrack \begin{matrix} 3 & {x}_{3}\sin {x}_{2}{x}_{3} & {x}_{2}\sin {x}_{2}{x}_{3} \\ 2{x}_{1} & - {162}\left( {{x}_{2} + {0.1}}\right) & \cos {x}_{3} \\ - {x}_{2}{e}^{-{x}_{1}{x}_{2}} & - {x}_{1}{e}^{-{x}_{1}{x}_{2}} & {20} \end{matrix}\right\rbr... | Yes |
Use Theorem 11.1 to show that the boundary-value problem\n\n\\[ \n{y}^{\prime \prime } + {e}^{-{xy}} + \sin {y}^{\prime } = 0,\\;\\text{ for }1 \leq x \leq 2\\text{, with }y\\left( 1\\right) = y\\left( 2\\right) = 0, \n\\]\n\nhas a unique solution. | Solution We have\n\n\\[ \nf\\left( {x, y,{y}^{\prime }}\\right) = - {e}^{-{xy}} - \sin {y}^{\prime }. \n\\]\n\nand for all \\( x \\) in \\( \\left\\lbrack {1,2}\\right\\rbrack \\) ,\n\n\\[ \n{f}_{y}\\left( {x, y,{y}^{\prime }}\\right) = x{e}^{-{xy}} > 0\\;\\text{ and }\\;\\left| {{f}_{{y}^{\prime }}\\left( {x, y,{y}^{\... | Yes |
Apply the Linear Shooting technique with \( N = {10} \) to the boundary-value problem \[ {y}^{\prime \prime } = - \frac{2}{x}{y}^{\prime } + \frac{2}{{x}^{2}}y + \frac{\sin \left( {\ln x}\right) }{{x}^{2}},\;\text{ for }1 \leq x \leq 2\text{, with }y\left( 1\right) = 1\text{ and }y\left( 2\right) = 2, \] | Solution Applying Algorithm 11.1 to this problem requires approximating the solutions to the initial-value problems \[ {y}_{1}^{\prime \prime } = - \frac{2}{x}{y}_{1}^{\prime } + \frac{2}{{x}^{2}}{y}_{1} + \frac{\sin \left( {\ln x}\right) }{{x}^{2}},\;\text{ for }1 \leq x \leq 2\text{, with }{y}_{1}\left( 1\right) = 1\... | Yes |
Apply the Shooting method with Newton's Method to the boundary-value problem\n\n\[{y}^{\prime \prime } = \frac{1}{8}\left( {{32} + 2{x}^{3} - y{y}^{\prime }}\right) ,\;\text{ for }1 \leq x \leq 3\text{, with }y\left( 1\right) = {17}\text{ and }y\left( 3\right) = \frac{43}{3}.\]\n\nUse \( N = {20}, M = {10} \), and \( {... | Solution We need approximate solutions to the initial-value problems\n\n\[{y}^{\prime \prime } = \frac{1}{8}\left( {{32} + 2{x}^{3} - y{y}^{\prime }}\right) ,\;\text{ for }1 \leq x \leq 3\text{, with }y\left( 1\right) = {17}\text{ and }{y}^{\prime }\left( 1\right) = {t}_{k},\]\n\nand\n\n\[{z}^{\prime \prime } = \frac{\... | Yes |
Use Algorithm 11.3 with \( N = 9 \) to approximate the solution to the linear boundary-value problem\n\n\[ \n{y}^{\prime \prime } = - \frac{2}{x}{y}^{\prime } + \frac{2}{{x}^{2}}y + \frac{\sin \left( {\ln x}\right) }{{x}^{2}},\;\text{ for }1 \leq x \leq 2\text{, with }y\left( 1\right) = 1\text{ and }y\left( 2\right) = ... | Solution For this example, we will use \( N = 9 \), so \( h = {0.1} \), and we have the same spacing as in Example 2 of Section 11.1. The complete results are listed in Table 11.3.\n\nTable 11.3\n\n<table><thead><tr><th>\( {x}_{i} \)</th><th>\( {w}_{i} \)</th><th>\( y\left( {x}_{i}\right) \)</th><th>\( \left| {{w}_{i} ... | Yes |
Example 2 Apply Richardson's extrapolation to approximate the solution to the boundary-value problem\n\n\\[ \n{y}^{\prime \prime } = - \\frac{2}{x}{y}^{\prime } + \\frac{2}{{x}^{2}}y + \\frac{\\sin \\left( {\\ln x}\\right) }{{x}^{2}},\\;\\text{ for }1 \\leq x \\leq 2,\\text{ with }y\\left( 1\\right) = 1\\text{ and }y\\... | Solution The results are listed in Table 11.4. The first extrapolation is\n\n\\[ \n{\\operatorname{Ext}}_{1i} = \\frac{4{w}_{i}\\left( {h = {0.05}}\\right) - {w}_{i}\\left( {h = {0.1}}\\right) }{3}; \n\\]\n\nthe second extrapolation is\n\n\\[ \n{\\operatorname{Ext}}_{2i} = \\frac{4{w}_{i}\\left( {h = {0.025}}\\right) -... | Yes |
Apply Algorithm 11.4, with \( h = {0.1} \), to the nonlinear boundary-value problem\n\n\[ \n{y}^{\prime \prime } = \frac{1}{8}\left( {{32} + 2{x}^{3} - y{y}^{\prime }}\right) ,\;\text{ for }1 \leq x \leq 3\text{, with }y\left( 1\right) = {17}\text{ and }y\left( 3\right) = \frac{43}{3}, \n\]\n\nand compare the results t... | Solution The stopping procedure used in Algorithm 11.4 was to iterate until values of successive iterates differed by less than \( {10}^{-8} \) . This was accomplished with four iterations. This gives the results in Table 11.5. They are less accurate than those obtained using the nonlinear shooting method, which gave r... | Yes |
Theorem 11.4 Let \( p \in {C}^{1}\left\lbrack {0,1}\right\rbrack, q, f \in C\left\lbrack {0,1}\right\rbrack \), and\n\n\[ p\left( x\right) \geq \delta > 0,\;q\left( x\right) \geq 0,\;\text{ for }0 \leq x \leq 1.\]\n\nThe function \( y \in {C}_{0}^{2}\left\lbrack {0,1}\right\rbrack \) is the unique solution to the diffe... | Details of the proof of this theorem can be found in [Shul], pp. 88-89. It proceeds in three steps. First it is shown that any solution \( y \) to (11.23) also satisfies the equation\n\n\[ {\int }_{0}^{1}f\left( x\right) u\left( x\right) {dx} = {\int }_{0}^{1}p\left( x\right) \frac{dy}{dx}\left( x\right) \frac{du}{dx}\... | Yes |
Example 2 Use the Poisson finite-difference method with \( n = 6, m = 5 \), and a tolerance of \( {10}^{-{10}} \) to approximate the solution to\n\n\[ \n\\frac{{\\partial }^{2}u}{\\partial {x}^{2}}\\left( {x, y}\\right) + \\frac{{\\partial }^{2}u}{\\partial {y}^{2}}\\left( {x, y}\\right) = x{e}^{y},\\;0 < x < 2,\\;0 < ... | Solution Using Algorithm 12.1 with a maximum number of iterations set at \( N = {100} \) gives the results in Table 12.2. The stopping criterion for the Gauss-Seidel method in Step 17 requires that\n\n\[ \n\\left| {{w}_{ij}^{\\left( l\\right) } - {w}_{ij}^{\\left( l - 1\\right) }}\\right| \\leq {10}^{-{10}} \n\]\n\nfor... | Yes |
Use steps sizes (a) \( h = {0.1} \) and \( k = {0.0005} \) and (b) \( h = {0.1} \) and \( k = {0.01} \) to approximate the solution to the heat equation\n\n\[ \n\\frac{\\partial u}{\\partial t}\\left( {x, t}\\right) - \\frac{{\\partial }^{2}u}{\\partial {x}^{2}}\\left( {x, t}\\right) = 0,\\;0 < x < 1,\\;0 \\leq t, \n\]... | Solution (a) Forward-Difference method with \( h = {0.1}, k = {0.0005} \) and \( \\lambda = {\\left( 1\\right) }^{2}({0.0005}/ \) \( {\\left( {0.1}\\right) }^{2}) = {0.05} \) gives the results in the third column of Table 12.3. As can be seem from the fourth column, these results are quite accurate.\n\n(b) Forward-Diff... | Yes |
Approximate the solution to the hyperbolic problem\n\n\\[ \n\\frac{{\\partial }^{2}u}{\\partial {t}^{2}}\\left( {x, t}\\right) - 4\\frac{{\\partial }^{2}u}{\\partial {x}^{2}}\\left( {x, t}\\right) = 0,\\;0 < x < 1,\\;0 < t, \n\\]\n\nwith boundary conditions\n\n\\[ \nu\\left( {0, t}\\right) = u\\left( {1, t}\\right) = 0... | Solution Choosing \\( h = {0.1} \\) and \\( k = {0.05} \\) gives \\( \\lambda = 1, m = {10} \\), and \\( N = {20} \\) . We will choose a maximum time \\( T = 1 \\) and apply the Finite-Difference Algorithm 12.4. This produces the approximations \\( {w}_{i, N} \\) to \\( u\\left( {{0.1i},1}\\right) \\) for \\( i = 0,1,\... | Yes |
Consider the function \( f : {\mathbf{R}}^{2} \rightarrow {\mathbf{R}}^{2} \) given by \[ f\left( {{x}_{1},{x}_{2}}\right) = \left( {\frac{-{x}_{2}}{{x}_{1}^{2} + {x}_{2}^{2}},\frac{{x}_{1}}{{x}_{1}^{2} + {x}_{2}^{2}}}\right) . \] It is easy to show that (2) is satisfied. However, there is no function \( F : {\mathbf{R... | Assume there were; then \[ {\int }_{0}^{2\pi }\frac{d}{d\theta }F\left( {\cos \theta ,\sin \theta }\right) {d\theta } = F\left( {1,0}\right) - F\left( {1,0}\right) = 0. \] On the other hand the chain rule gives \[ \frac{d}{d\theta }F\left( {\cos \theta ,\sin \theta }\right) = \frac{dF}{dx} \cdot \left( {-\sin \theta }\... | Yes |
Theorem 1.4 Let \( U \subset {\mathbf{R}}^{2} \) be an open star-shaped set. For any smooth function \( \left( {{f}_{1},{f}_{2}}\right) : U \rightarrow {\mathbf{R}}^{2} \) that satisfies (2), Question 1.1 has a solution. | Proof. For the sake of simplicity we assume that \( {x}_{0} = 0 \in {\mathbf{R}}^{2} \) . Consider the function \( F : U \rightarrow \mathbf{R} \) ,\n\n\[ F\left( {{x}_{1},{x}_{2}}\right) = {\int }_{0}^{1}\left\lbrack {{x}_{1}{f}_{1}\left( {t{x}_{1}, t{x}_{2}}\right) + {x}_{2}{f}_{2}\left( {t{x}_{1}, t{x}_{2}}\right) }... | Yes |
Theorem 1.5 An open set \( U \subseteq {\mathbf{R}}^{k} \) is connected if and only if \( {H}^{0}\left( U\right) = \mathbf{R} \) . | Proof. Assume that \( \operatorname{grad}\left( f\right) = 0 \) . Then \( f \) is locally constant: each \( {x}_{0} \in U \) has a neighborhood \( V\left( {x}_{0}\right) \) with \( f\left( x\right) = f\left( {x}_{0}\right) \) when \( x \in V\left( {x}_{0}\right) \) . If \( U \) is connected, then every locally constant... | Yes |
Theorem 1.6 For an open star-shaped set in \( {\mathbf{R}}^{3} \) we have that \( {H}^{0}\left( U\right) = \mathbf{R} \) , \( {H}^{1}\left( U\right) = 0 \) and \( {H}^{2}\left( U\right) = 0 \) . | Proof. The values of \( {H}^{0}\left( U\right) \) and \( {H}^{1}\left( U\right) \) are obtained as above, so we shall restrict ourselves to showing that \( {H}^{2}\left( U\right) = 0 \) . It is convenient to assume that \( U \) is star-shaped with respect to 0 . Consider a function \( F : U \rightarrow {\mathbf{R}}^{3}... | Yes |
Example 1.7 Let \( S = \left\{ {\left( {{x}_{1},{x}_{2},{x}_{3}}\right) \in {\mathbf{R}}^{3} \mid {x}_{1}^{2} + {x}_{2}^{2} = 1,{x}_{3} = 0}\right\} \) be the unit circle in the \( \left( {{x}_{1},{x}_{2}}\right) \) -plane. Consider the function \[ f\left( {{x}_{1},{x}_{2},{x}_{3}}\right) = \left( {\frac{-2{x}_{1}{x}_{... | The curve in question is \[ \gamma \left( t\right) = \left( {\sqrt{1 + \cos t},0,\sin t}\right) ,\; - \pi \leq t \leq \pi . \] Assume \( \operatorname{grad}\left( F\right) = f \) as a function on \( U \) . We can determine the integral of \( \frac{d}{dt}F\left( {\gamma \left( t\right) }\right) \) in two ways. On the on... | Yes |
Let \( U \) be an open set in \( {\mathbf{R}}^{k} \) and \( X : U \rightarrow {\mathbf{R}}^{k} \) a smooth function (a smooth vector field). Recall that the energy \( {A}_{\gamma }\left( X\right) \), of \( X \) along a smooth curve \( \gamma : \left\lbrack {a, b}\right\rbrack \rightarrow U \) is defined by the integral... | \[ \left\langle {X \circ \gamma \left( t\right) ,{\gamma }^{\prime }\left( t\right) }\right\rangle = \frac{d}{dt}\Phi \left( {\gamma \left( t\right) }\right) \] by the chain rule; compare Example 1.2. | Yes |
Lemma 2.2 If \( \omega \in {\operatorname{Alt}}^{k}\left( V\right) \) and \( \sigma \in S\left( k\right) \), then\n\n\[ \omega \left( {{\xi }_{\sigma \left( 1\right) },\ldots ,{\xi }_{\sigma \left( k\right) }}\right) = \operatorname{sign}\left( \sigma \right) \omega \left( {{\xi }_{1},\ldots ,{\xi }_{k}}\right) \] | Proof. It is sufficient to prove the formula when \( \sigma = \left( {i, j}\right) \) . Let\n\n\[ {\omega }_{i, j}\left( {\xi ,{\xi }^{\prime }}\right) = \omega \left( {{\xi }_{1},\ldots ,\xi ,\ldots ,{\xi }^{\prime },\ldots ,{\xi }_{k}}\right) \]\n\nwith \( \xi \) and \( {\xi }^{\prime } \) occurring at positions \( i... | Yes |
Lemma 2.6 If \( {\omega }_{1} \in {\operatorname{Alt}}^{p}\left( V\right) \) and \( {\omega }_{2} \in {\operatorname{Alt}}^{q}\left( V\right) \) then \( {\omega }_{1} \land {\omega }_{2} \in {\operatorname{Alt}}^{p + q}\left( V\right) \) . | Proof. We first show that \( \left( {{\omega }_{1} \land {\omega }_{2}}\right) \left( {{\xi }_{1},{\xi }_{2},\ldots ,{\xi }_{p + q}}\right) = 0 \) when \( {\xi }_{1} = {\xi }_{2} \) . We let\n\n(i) \( {S}_{12} = \{ \sigma \in S\left( {p, q}\right) \mid \sigma \left( 1\right) = 1,\sigma \left( {p + 1}\right) = 2\} \)\n\... | Yes |
Lemma 2.7 A k-linear map \( \omega \) is alternating if \( \omega \left( {{\xi }_{1},\ldots ,{\xi }_{k}}\right) = 0 \) for all \( k \) -tuples with \( {\xi }_{i} = {\xi }_{i + 1} \) for some \( 1 \leq i \leq k - 1 \) . | Proof. \( S\left( k\right) \) is generated by the transpositions \( \left( {i, i + 1}\right) \), and by the argument\n\nof Lemma 2.2,\n\n\[ \omega \left( {{\xi }_{1},\ldots ,{\xi }_{i},{\xi }_{i + 1},\ldots ,{\xi }_{k}}\right) = - \omega \left( {{\xi }_{1},\ldots ,{\xi }_{i + 1},{\xi }_{i},\ldots ,{\xi }_{k}}\right) . ... | Yes |
Lemma 2.8 If \( {\omega }_{1} \in {\operatorname{Alt}}^{p}\left( V\right) \) and \( {\omega }_{2} \in {\operatorname{Alt}}^{q}\left( V\right) \) then \( {\omega }_{1} \land {\omega }_{2} = {\left( -1\right) }^{pq}{\omega }_{2} \land {\omega }_{1} \) . | Proof. Let \( \tau \in S\left( {p + q}\right) \) be the element with\n\n\[ \tau \left( 1\right) = p + 1,\tau \left( 2\right) = p + 2,\ldots ,\tau \left( q\right) = p + q \]\n\n\[ \tau \left( {q + 1}\right) = 1,\tau \left( {q + 2}\right) = 2,\ldots ,\tau \left( {p + q}\right) = p. \]\n\nWe have \( \operatorname{sign}\le... | Yes |
Lemma 2.13 For \( 1 \) -forms \( {\omega }_{1},\ldots ,{\omega }_{p} \in {\operatorname{Alt}}^{1}\left( V\right) \) ,\n\n\[ \left( {{\omega }_{1} \land \ldots \land {\omega }_{p}}\right) \left( {{\xi }_{1},\ldots ,{\xi }_{p}}\right) = \det \left( \begin{matrix} {\omega }_{1}\left( {\xi }_{1}\right) & {\omega }_{1}\left... | Proof. The case \( p = 2 \) is obvious. We proceed by induction on \( p \) . According to Definition 2.5,\n\n\[ {\omega }_{1} \land \left( {{\omega }_{2} \land \ldots \land {\omega }_{p}}\right) \vDash \left( {{\xi }_{1},\ldots ,{\xi }_{p}}\right) \]\n\n\[ = \mathop{\sum }\limits_{{j = 1}}^{p}{\left( -1\right) }^{j + 1... | Yes |
Theorem 2.15 Let \( {e}_{1},\ldots ,{e}_{n} \) be a basis of \( V \) and \( {\epsilon }_{1},\ldots ,{\epsilon }_{n} \) the dual basis of \( {\operatorname{Alt}}^{1}\left( V\right) \) . Then \[ {\left\{ {\epsilon }_{\sigma \left( 1\right) } \land {\epsilon }_{\sigma \left( 2\right) } \land \ldots \land {\epsilon }_{\sig... | Proof. Since \( {\epsilon }_{i}\left( {e}_{\jmath }\right) = 0 \) when \( i \neq \jmath \), and \( {\epsilon }_{i}\left( {e}_{\imath }\right) = 1 \), Lemma 2.13 gives \[ {\epsilon }_{{i}_{1}} \land \ldots \land {\epsilon }_{{i}_{p}}\left( {{e}_{{\jmath }_{1}},\ldots ,{e}_{{j}_{p}}}\right) = \left\{ \begin{array}{ll} 0 ... | Yes |
Example 3.3 Let \( {x}_{i} : U \rightarrow \mathbf{R} \) be the \( i \) -th projection. Then \( d{x}_{1} \in {\Omega }^{1}\left( U\right) \) is the constant map \( d{x}_{i} : x \rightarrow {\epsilon }_{i} \) . | This follows from (1). In general, for \( f \in {\Omega }^{0}\left( U\right) \) , (1) shows that\n\n(2)\n\n\[ \n{d}_{x}f\left( \zeta \right) = \frac{\partial f}{\partial {x}_{1}}\left( x\right) {\zeta }^{1} + \cdots + \frac{\partial f}{\partial {x}_{n}}\left( x\right) {\zeta }^{n} \]\n\nwith \( \left( {{\zeta }^{1},\ld... | Yes |
Lemma 3.4 If \( \omega \left( x\right) = f\left( x\right) {\epsilon }_{I} \) then \( {d}_{x}\omega = {d}_{x}f \land {\epsilon }_{I} \) . | Proof. By (1) we have\n\n\[ \n{D}_{x}\omega \left( \zeta \right) = \left( {{D}_{x}f}\right) \left( \zeta \right) {\epsilon }_{I} = \left( {\frac{\partial f}{\partial {x}_{1}}{\zeta }^{1} + \cdots + \frac{\partial f}{\partial {x}_{n}}{\zeta }^{n}}\right) {\epsilon }_{I} = {d}_{x}f\left( \zeta \right) {\epsilon }_{I} \n\... | Yes |
Lemma 3.5 For \( p \geq 0 \) the composition \( {\Omega }^{p}\left( U\right) \rightarrow {\Omega }^{p + 1}\left( U\right) \rightarrow {\Omega }^{p + 2}\left( U\right) \) is identically zero. | Proof. Let \( \omega = f{\epsilon }_{l} \) . Then\n\n\[ \n{d\omega } = {df} \land {\epsilon }_{I} = \frac{\partial f}{\partial {x}_{1}}{\epsilon }_{1} \land {\epsilon }_{I} + \cdots + \frac{\partial f}{\partial {x}_{n}}{\epsilon }_{n} \land {\epsilon }_{I}.\n\]\n\nNow use \( {\epsilon }_{i} \land {\epsilon }_{i} = 0 \)... | Yes |
Lemma 3.6 For \( {\omega }_{1} \in {\Omega }^{p}\left( U\right) \) and \( {\omega }_{2} \in {\Omega }^{q}\left( U\right) \) , \[ d\left( {{\omega }_{1} \land {\omega }_{2}}\right) = d{\omega }_{1} \land {\omega }_{2} + {\left( -1\right) }^{p}{\omega }_{1} \land d{\omega }_{2} \] | Proof. It is sufficient to show the formula when \( {\omega }_{1} = f{\epsilon }_{I} \) and \( {\omega }_{2} = g{\epsilon }_{J} \) . But then \( {\omega }_{1} \land {\omega }_{2} = {fg}{\epsilon }_{I} \land {\epsilon }_{J} \), and \[ d\left( {{\omega }_{1} \land {\omega }_{2}}\right) = d\left( {fg}\right) \land {\epsil... | Yes |
Theorem 3.7 There is precisely one linear operator \( d : {\Omega }^{p}\left( U\right) \rightarrow {\Omega }^{p + 1}\left( U\right), p = \) \( 0,1,\ldots \), such that\n\n(i) \( f \in {\Omega }^{0}\left( U\right) ,{df} = \frac{\partial f}{\partial {x}_{1}}{\epsilon }_{1} + \cdots + \frac{\partial f}{\partial {x}_{n}}{\... | Proof. We have already defined \( d \) with the asserted properties. Conversely assume that \( {d}^{\prime } \) is a linear operator satisfying (i),(ii) and (iii). We will show that \( {d}^{\prime } \) is the exterior differential.\n\nThe first property tells us that \( d = {d}^{\prime } \) on \( {\Omega }^{0}\left( U\... | Yes |
Lemma 3.9 \( {H}^{0}\left( U\right) \) is the vector space of maps \( U \rightarrow \mathbf{R} \) that are constant on each connected component of \( U \) . | Proof. A locally constant function \( f : U \rightarrow \mathbf{R} \) gives a partition of \( U \) into the mutually disjoint open sets \( {f}^{-1}\left( c\right), c \in \mathbf{R} \) . Consequently \( f : U \rightarrow \mathbf{R} \) is locally constant precisely when \( f \) is constant on each connected component of ... | Yes |
For the constant 1 -form \( {\epsilon }_{i} \in {\Omega }^{1}\left( {U}_{2}\right) \) we have that\n\n\[ \n{\phi }^{ * }\left( {\epsilon }_{i}\right) = \mathop{\sum }\limits_{{k = 1}}^{n}\frac{\partial {\phi }_{i}}{\partial {x}_{k}}{\epsilon }_{k} = d{\phi }_{i} \n\]\n\nwith \( {\phi }_{i} \) the \( i \) -th coordinate... | To see this, let \( \zeta \in {\mathbf{R}}^{n} \) . Then\n\n\[ \n{\phi }^{ * }\left( {\epsilon }_{i}\right) \left( \zeta \right) = {\epsilon }_{i}\left( {{D}_{z}\phi \left( \zeta \right) }\right) = {\epsilon }_{i}\left( {\mathop{\sum }\limits_{{k = 1}}^{m}\left( {\mathop{\sum }\limits_{{l = 1}}^{n}\frac{\partial {\phi ... | Yes |
If \( \phi : {\mathbf{R}}^{n} \times \mathbf{R} \rightarrow {\mathbf{R}}^{n} \) is given by \( \phi \left( {x, t}\right) = \psi \left( t\right) x \), where \( \psi \left( t\right) \) is a smooth real valued function. Then | \[ {\phi }^{ * }\left( {d{x}_{i}}\right) = {x}_{i}{\psi }^{\prime }\left( t\right) {dt} + \psi \left( t\right) d{x}_{i} \] | Yes |
Lemma 4.1 Suppose \( 0 \rightarrow A\overset{f}{ \rightarrow }B\overset{g}{ \rightarrow }C \rightarrow 0 \) is a short exact sequence of vector spaces. Then \( B \) is finite-dimensional if both \( A \) and \( C \) are, and \( B \cong A \oplus C \) . | Proof. Choose a basis \( \left\{ {a}_{n}\right\} \) of \( A \) and \( \left\{ {c}_{j}\right\} \) of \( C \) . Since \( g \) is surjective there exist \( {b}_{j} \in B \) with \( g\left( {b}_{j}\right) = {c}_{j} \) . Then \( \left\{ {f\left( {a}_{1}\right) ,{b}_{j}}\right\} \) is a basis of \( B \) : For \( b \in B \) w... | No |
Lemma 4.3 A chain map \( f : {A}^{ * } \rightarrow {B}^{ * } \) induces a linear map\n\n\[ \n{f}^{ * } = {H}^{ * }\left( f\right) : {H}^{p}\left( {A}^{ * }\right) \rightarrow {H}^{p}\left( {B}^{ * }\right) ,\;\text{ for all }p.\n\] | Proof. Let \( a \in {A}^{p} \) be a cycle \( \left( {{d}^{p}a = 0}\right) \) and \( \left\lbrack a\right\rbrack = a + \operatorname{Im}{d}^{p - 1} \) its corresponding cohomology class in \( {H}^{p}\left( {A}^{ * }\right) \) . We define \( {f}^{ * }\left( \left\lbrack a\right\rbrack \right) = \left\lbrack {{f}^{p}\left... | Yes |
For a short exact sequence of chain complexes the sequence\n\n\[ \n{H}^{p}\left( {A}^{ * }\right) \overset{{f}^{ * }}{ \rightarrow }{H}^{p}\left( {B}^{ * }\right) \overset{{g}^{ * }}{ \rightarrow }{H}^{p}\left( {C}^{ * }\right) \n\]\n\nis exact. | Proof. Since \( {g}^{p} \circ {f}^{p} = 0 \) we have\n\n\[ \n{g}^{ * } \circ {f}^{ * }\left( \left\lbrack a\right\rbrack \right) = {g}^{ * }\left( \left\lbrack {{f}^{p}\left( a\right) }\right\rbrack \right) = \left\lbrack {{g}^{p}\left( {{f}^{p}\left( a\right) }\right) }\right\rbrack = 0 \n\]\n\nfor every cohomology cl... | Yes |
Example 4.6 Here is a short exact sequence of chain complexes (the dots indicate that the chain groups are zero) with \( {\partial }^{ * } \neq 0 \) : | One can easily verify that \( {\partial }^{ * }\mathbf{R} \rightarrow \mathbf{R} \) is an isomorphism. | No |
Lemma 4.7 The sequence \( {H}^{p}\left( {B}^{ * }\right) \overset{{g}^{ * }}{ \rightarrow }{H}^{p}\left( {C}^{ * }\right) \overset{{\partial }^{ * }}{ \rightarrow }{H}^{p + 1}\left( {A}^{ * }\right) \) is exact. | Proof. We have \( {\partial }^{ * }{g}^{ * }\left( \left\lbrack b\right\rbrack \right) = {\partial }^{ * }\left\lbrack {{g}^{p}\left( b\right) }\right\rbrack = \left\lbrack {{\left( {f}^{p + 1}\right) }^{-1}\left( {{d}_{B}\left( b\right) }\right) }\right\rbrack = 0 \) . Conversely assume that \( {\partial }^{ \bullet }... | Yes |
Lemma 4.8 The sequence \( {H}^{p}\left( {C}^{ * }\right) \overset{{\partial }^{ * }}{ \rightarrow }{H}^{p + 1}\left( {A}^{ * }\right) \overset{{f}^{ * }}{ \rightarrow }{H}^{p + 1}\left( {B}^{ * }\right) \) is exact. | Proof. We have \( {f}^{ * }{\partial }^{ * }\left( \left\lbrack c\right\rbrack \right) = \left\lbrack {{d}_{B}^{p}\left( b\right) }\right\rbrack = 0 \), where \( {g}^{p}\left( b\right) = c \) . Conversely assume that \( {f}^{ * }\left\lbrack a\right\rbrack = 0 \), i.e. \( {f}^{p + 1}\left( a\right) = {d}_{\beta }^{p}\l... | Yes |
Lemma 4.11 For two chain-homotopic chain maps \( f, g : {A}^{ * } \rightarrow {B}^{ * } \) we have that\n\n\[ \n{f}^{ * } = {g}^{ * } : {H}^{p}\left( {A}^{ * }\right) \rightarrow {H}^{p}\left( {B}^{ * }\right) .\n\] | Proof. If \( \left\lbrack a\right\rbrack \in {H}^{p}\left( {A}^{ * }\right) \) then\n\n\[ \n\left( {{f}^{ * } - {g}^{ * }}\right) \left\lbrack a\right\rbrack = \left\lbrack {{f}^{p}\left( a\right) - {g}^{p}\left( a\right) }\right\rbrack = \left\lbrack {{d}_{B}^{p - 1}s\left( a\right) + s{d}_{A}^{p}\left( a\right) }\rig... | Yes |
Lemma 4.13 If \( {A}^{ * } \) and \( {B}^{ * } \) are chain complexes then\n\n\[ \n{H}^{p}\left( {{A}^{ \bullet } \oplus {B}^{ \bullet }}\right) = {H}^{p}\left( {A}^{ \bullet }\right) \oplus {H}^{p}\left( {B}^{ \bullet }\right) .\n\] | Proof. It is obvious that\n\n\[ \n\operatorname{Ker}\left( {d}_{A \oplus B}^{p}\right) = \operatorname{Ker}{d}_{A}^{p} \oplus \operatorname{Ker}{d}_{B}^{p}\n\]\n\n\[ \n\operatorname{Im}\left( {d}_{A \oplus B}^{p - 1}\right) = \operatorname{Im}{d}_{A}^{p - 1} \oplus \operatorname{Im}{d}_{B}^{p - 1}\n\]\n\nand the lemma ... | Yes |
Theorem 5.2 (Mayer-Vietoris) Let \( {U}_{1} \) and \( {U}_{2} \) be open sets in \( {\mathbf{R}}^{n} \) and \( U = {U}_{1} \cup {U}_{2} \) . There exists an exact sequence of cohomology vector spaces\n\n\[ \cdots \rightarrow {H}^{p}\left( U\right) \overset{{I}^{ * }}{ \rightarrow }{H}^{p}\left( {U}_{1}\right) \oplus {H... | Here \( {I}^{ \bullet }\left( \left\lbrack \omega \right\rbrack \right) = \left( {{i}_{1}^{ \bullet }\left\lbrack \omega \right\rbrack ,{i}_{2}^{ \bullet }\left\lbrack \omega \right\rbrack }\right) \) and \( {J}^{ \bullet }\left( {\left\lbrack {\omega }_{1}\right\rbrack ,\left\lbrack {\omega }_{2}\right\rbrack }\right)... | No |
Corollary 5.3 If \( {U}_{1} \) and \( {U}_{2} \) are disjoint open sets in \( {\mathbf{R}}^{n} \) then\n\n\[ \n{I}^{ * } : {H}^{p}\left( {{U}_{1} \cup {U}_{2}}\right) \rightarrow {H}^{p}\left( {U}_{1}\right) \oplus {H}^{p}\left( {U}_{2}\right)\n\]\n\nis an isomorphism. | Proof. It follows from Theorem 5.1 that\n\n\[ \n{I}^{p} : {\Omega }^{p}\left( {{U}_{1} \cup {U}_{2}}\right) \rightarrow {\Omega }^{p}\left( {U}_{1}\right) \oplus {\Omega }^{p}\left( {U}_{2}\right)\n\]\n\nis an isomorphism, and Lemma 4.13 gives that the corresponding map on cohomology is also an isomorphism. | Yes |
Theorem 5.5 Assume that the open set \( U \) is covered by convex open sets \( {U}_{1},\ldots ,{U}_{r} \) . Then \( {H}^{p}\left( U\right) \) is finitely generated. | Proof. We use induction on the number of open sets. If \( r = 1 \) the assertion follows from the Poincaré lemma. Assume the assertion is proved for \( r - 1 \) and let \( V = {U}_{1} \cup \cdots \cup {U}_{r - 1} \), such that \( U = V \cup {U}_{r} \) . From Theorem 5.2 we have the exact sequence \[ {H}^{p - 1}\left( {... | Yes |
Lemma 6.2 Homotopy is an equivalence relation. | Proof. If \( F \) is a homotopy from \( {f}_{0} \) to \( {f}_{1} \), a homotopy from \( {f}_{1} \) to \( {f}_{0} \) is defined by \( G\left( {x, t}\right) = F\left( {x,1 - t}\right) \) . If \( {f}_{0} \simeq {f}_{1} \) via \( F \) and \( {f}_{1} \simeq {f}_{2} \) via \( G \), then \( {f}_{0} \simeq {f}_{2} \) via\n\n\[... | Yes |
Lemma 6.3 Let \( X, Y \) and \( Z \) be topological spaces and let \( {f}_{\nu } : X \rightarrow Y \) and \( {g}_{\nu } : Y \rightarrow Z \) be continuous maps for \( \nu = 0,1 \) . If \( {f}_{0} \simeq {f}_{1} \) and \( {g}_{0} \simeq {g}_{1} \) then \( {g}_{0} \circ {f}_{0} \simeq {g}_{1} \circ {f}_{1} \) | Proof. Given homotopies \( F \) from \( {f}_{0} \) to \( {f}_{1} \) and \( G \) from \( {g}_{0} \) to \( {g}_{1} \), the homotopy \( H \) from \( {g}_{0} \circ {f}_{0} \) to \( {g}_{1} \circ {f}_{1} \) can be defined by \( H\left( {x, t}\right) = G\left( {F\left( {x, t}\right), t}\right) \) . | Yes |
Example 6.5 Let \( Y \subseteq {\mathbf{R}}^{m} \) have the topology induced by \( {\mathbf{R}}^{m} \) . If, for the continuous maps \( {f}_{\nu } : X \rightarrow Y,\nu = 0,1 \), the line segment in \( {\mathbf{R}}^{m} \) from \( {f}_{0}\left( x\right) \) to \( {f}_{1}\left( x\right) \) is contained in \( Y \) for all ... | \[ F\left( {x, t}\right) = \left( {1 - t}\right) {f}_{0}\left( x\right) + t{f}_{1}\left( x\right) . \] | Yes |
Lemma 6.6 If \( U, V \) are open sets in Euclidean spaces, then\n\n(i) Every continuous map \( h : U \rightarrow V \) is homotopic to a smooth map. | Proof. We use Lemma A. 9 to approximate \( h \) by a smooth map \( f : U \rightarrow V \) . We can choose \( f \) such that \( V \) contains the line segment from \( h\left( x\right) \) to \( f\left( x\right) \) for every \( x \in U \) . Then \( h \simeq f \) by Example 6.5. | Yes |
Theorem 6.7 If \( f, g : U \rightarrow V \) are smooth maps and \( f \simeq g \) then the induced chain maps\n\n\[ \n{f}^{ \bullet },{g}^{ \bullet } : {\Omega }^{ \bullet }\left( V\right) \rightarrow {\Omega }^{ \bullet }\left( U\right)\n\]\n\nare chain-homotopic (see Definition 4.10). | Proof. Recall, from the proof of Theorem 3.15, that every \( p \) -form \( \omega \) on \( U \times \mathbf{R} \) can be written as\n\n\[ \n\omega = \sum {f}_{I}\left( {x, t}\right) d{x}_{I} + \sum {g}_{J}\left( {x, t}\right) {dt} \land d{x}_{J}\n\]\n\nIf \( \phi : U \rightarrow U \times \mathbf{R} \) is the inclusion ... | Yes |
Theorem 6.8 For \( p \in \mathbf{Z} \) and open sets \( U, V, W \) in Euclidean spaces we have\n\n(i) If \( {\phi }_{0},{\phi }_{1} : U \rightarrow V \) are homotopic continuous maps, then\n\n\[{\phi }_{0}^{ \bullet } = {\phi }_{1}^{ \bullet } : {H}^{p}\left( V\right) \rightarrow {H}^{p}\left( U\right)\] | Proof. Choose a smooth map \( f : U \rightarrow V \) with \( {\phi }_{0} \simeq f \) . Lemma 6.2 gives that \( {\phi }_{1} \simeq f \) and (i) immediatcly follows. | No |
Corollary 6.9 (Topological invariance) A homeomorphism \( h.U \rightarrow V \) between open sets in Euclidean spaces induces isomorphisms \( {h}^{ \bullet } : {H}^{p}\left( V\right) \rightarrow {H}^{p}\left( U\right) \) for all \( p \) . | Proof. The corollary follows from Theorem 6.8.(iii), as \( {h}^{-1} : V \rightarrow U \) is a homotopy inverse to \( h \) . | Yes |
Corollary 6.10 If \( U \subseteq {\mathbf{R}}^{n} \) is an open contractible set, then \( {H}^{p}\left( U\right) = 0 \) when \( p > 0 \) and \( {H}^{0}\left( U\right) = \mathbf{R} \) . | Proof. Let \( F : U \times \left\lbrack {0,1}\right\rbrack \rightarrow U \) be a homotopy from \( {f}_{0} = {\mathrm{{id}}}_{U} \) to a constant map \( {f}_{1} \) with value \( {x}_{0} \in U \) . For \( x \in U, F\left( {x, t}\right) \) defines a continuous curve in \( U \) , which connects \( x \) to \( {x}_{0} \) . H... | Yes |
Theorem 6.13 For \( n \geq 2 \) we have the isomorphisms\n\n\[ \n{H}^{p}\left( {{\mathbf{R}}^{n}-\{ 0\} }\right) \cong \left\{ \begin{array}{ll} \mathbf{R} & \text{ if }p = 0, n - 1 \\ 0 & \text{ otherwise. } \end{array}\right.\n\] | Proof. The case \( n = 2 \) was shown in Example 5.4. The general case follows from induction on \( n \), via Proposition 6.11. | No |
Lemma 6.14 For each \( n \geq 2 \), the induced map \( {f}_{A}^{ * } : {H}^{n - 1}\left( {{\mathbf{R}}^{n}-\{ 0\} }\right) \rightarrow \) \( {H}^{n - 1}\left( {{\mathbf{R}}^{n}-\{ 0\} }\right) \) operates by multiplication by \( \det A/\left| {\det A}\right| \in \{ \pm 1\} \) . | Proof. Let \( B \) be obtained from \( A \) by replacing the \( r \) -th row by the sum of the \( r \) -th row and \( c \) times the \( s \) -th row, where \( r \neq s \) and \( c \in \mathbf{R} \), \[ B = \left( {I + c{E}_{r, s}}\right) A \] where \( I \) is the identity matrix and \( {E}_{r, s} \) is the matrix with ... | Yes |
Proposition 6.15 If \( n \neq m \) then \( {\mathbf{R}}^{n} \) and \( {\mathbf{R}}^{m} \) are not homeomorphic. | Proof. A possible homeomorphism \( {\mathbf{R}}^{n} \rightarrow {\mathbf{R}}^{m} \) may be assumed to map 0 to 0 . and would induce a homeomorphism between \( {\mathbf{R}}^{n} - \{ 0\} \) and \( {\mathbf{R}}^{m}\;\{ 0\} \) Hence\n\n\[ \n{H}^{p}\left( {{\mathbf{R}}^{n}-\{ 0\} }\right) \cong {H}^{p}\left( {{\mathbf{R}}^{... | Yes |
Theorem 7.1 (Brouwer's fixed point theorem, 1912) Every continuous map \( f : {D}^{n} \rightarrow {D}^{n} \) has a fixed point. | Proof. Assume that \( f\left( x\right) \neq x \) for all \( x \in {D}^{n} \) . For every \( x \in {D}^{n} \) we can define the point \( g\left( x\right) \in {S}^{n - 1} \) as the point of intersection between \( {S}^{n - 1} \) and the half-line from \( f\left( x\right) \) through \( x \) . We have that \( g\left( x\rig... | Yes |
Lemma 7.2 There is no continuous map \( g : {D}^{n} \rightarrow {S}^{n - 1} \), with \( {g}_{\mid {S}^{n - 1}} = {\mathrm{{id}}}_{{S}^{n - 1}} \) . | Proof. We may assume that \( n \geq 2 \) . For the map \( r : {\mathbf{R}}^{n} - \{ 0\} \rightarrow {\mathbf{R}}^{n} - \{ 0\}, r\left( x\right) = \) \( x/\parallel x\parallel \), we get that \( {\operatorname{id}}_{{\mathbf{R}}^{n}-\{ 0\} } \simeq r \), because \( {\mathbf{R}}^{n} - \{ 0\} \) always contains the line s... | Yes |
Theorem 7.3 The sphere \( {S}^{n} \) has a tangent vector field \( v \) with \( v\left( x\right) \neq 0 \) for all \( x \in {S}^{n} \) if and only if \( n \) is odd. | Proof. Such a vector field \( v \) can be extended to a vector field \( w \) on \( {\mathbf{R}}^{n} - \{ 0\} \) by setting\n\n\[ w\left( x\right) = v\left( \frac{x}{\parallel x\parallel }\right) \]\n\nWe have that \( w\left( x\right) \neq 0 \) and \( w\left( x\right) \cdot x = 0 \) . The expression\n\n\[ F\left( {x, t}... | Yes |
Lemma 7.6 Let \( A \subseteq {\mathbf{R}}^{n} \) and \( B \subseteq {\mathbf{R}}^{m} \) be closed sets and let \( \phi : A \rightarrow B \) be a homeomorphism. There is a homeomorphism \( h \) of \( {\mathbf{R}}^{n + m} \) to itself, such that \[ h\left( {x,{0}_{m}}\right) = \left( {{0}_{n},\phi \left( x\right) }\right... | Proof. By Lemma 7.4 we can extend \( \phi \) to a continuous map \( {f}_{1} : {\mathbf{R}}^{n} \rightarrow {\mathbf{R}}^{m} \) . A homeomorphism \( {h}_{1} : {\mathbf{R}}^{n} \times {\mathbf{R}}^{m} \rightarrow {\mathbf{R}}^{n} \times {\mathbf{R}}^{m} \) is defined by \[ {h}_{1}\left( {x, y}\right) = \left( {x, y + {f}... | Yes |
Corollary 7.7 If \( \phi : A \rightarrow B \) is a homeomorphism between closed subsets \( A \) and \( B \) of \( {\mathbf{R}}^{n} \), then \( \phi \) can be extended to a homeomorphism \( \widetilde{\phi } : {\mathbf{R}}^{2n} \rightarrow {\mathbf{R}}^{2n} \) . | Proof. We merely have to compose the homeomorphism \( h \) from Lemma 7.6 with the homeomorphism of \( {\mathbf{R}}^{2n} = {\mathbf{R}}^{n} \times {\mathbf{R}}^{n} \) to itself that switches the two factors. | Yes |
Theorem 7.8 Assume that \( A \neq {\mathbf{R}}^{n} \) and \( B \neq {\mathbf{R}}^{n} \) are closed subsets of \( {\mathbf{R}}^{n} \) . If \( A \) and \( B \) are homeomorphic, then \[ {H}^{p}\left( {{\mathbf{R}}^{n} - A}\right) \cong {H}^{p}\left( {{\mathbf{R}}^{n} - B}\right) \] | Proof. By induction on \( m \) Proposition 6.11 yields isomorphisms \[ {H}^{p + m}\left( {{\mathbf{R}}^{n + m} - A}\right) \cong {H}^{p}\left( {{\mathbf{R}}^{n} - A}\right) \;\left( {\text{for }p > 0}\right) \] \[ {H}^{m}\left( {{\mathbf{R}}^{n + m} - A}\right) \cong {H}^{0}\left( {{\mathbf{R}}^{n} - A}\right) /\mathbf... | Yes |
Corollary 7.9 If \( A \) and \( B \) are two homeomorphic closed subsets of \( {\mathbf{R}}^{n} \), then \( {\mathbf{R}}^{n} - A \) and \( {\mathbf{R}}^{n} - B \) have the same number of connected components. | Proof. If \( A \neq {\mathbf{R}}^{n} \) and \( B \neq {\mathbf{R}}^{n} \) the assertion follows from Theorem 7.8 and the remarks above. If \( A = {\mathbf{R}}^{n} \) and \( B \neq {\mathbf{R}}^{n} \) then \( {\mathbf{R}}^{n + 1} - A \) has precisely 2 connected components (the open half-spaces), while \( {\mathbf{R}}^{... | Yes |
Theorem 7.10 (Jordan-Brouwer separation theorem) If \( \sum \subseteq {\mathbf{R}}^{n}\left( {n \geq 2}\right) \) is homeomorphic to \( {S}^{n - 1} \) then\n\n(i) \( {\mathbf{R}}^{n} - \sum \) has precisely 2 connected components \( {U}_{1} \) and \( {U}_{2} \), where \( {U}_{1} \) is bounded and \( {U}_{2} \) is unbou... | Proof. Since \( \sum \) is compact, \( \sum \) is closed in \( {\mathbf{R}}^{n} \) . To show (i), it suffices, by Corollary 7.9, to verify it for \( {S}^{n - 1} \subseteq {\mathbf{R}}^{n} \) . The two connected components of \( {\mathbf{R}}^{n} - {S}^{n - 1} \) are\n\n\[ \n{\dot{D}}^{n} = \left\{ {x \in {\mathbf{R}}^{n... | Yes |
Theorem 7.11 If \( A \subseteq {\mathbf{R}}^{n} \) is homeomorphic to \( {D}^{k} \), with \( k \leq n \), then \( {\mathbf{R}}^{n} - A \) is connected. | Proof. Since \( A \) is compact, \( A \) is closed. By Corollary 7.9 it is sufficient to prove the assertion for \( {D}^{k} \subseteq {\mathbf{R}}^{k} \subseteq {\mathbf{R}}^{n} \) . This is left to the reader. | No |
Theorem 7.12 (Brouwer) Let \( U \subseteq {\mathbf{R}}^{n} \) be an arbitrary open set and \( f : U \rightarrow {\mathbf{R}}^{n} \) an injective continuous map. The image \( f\left( U\right) \) is open in \( {\mathbf{R}}^{n} \), and \( f \) maps \( U \) homeomorphically to \( f\left( U\right) \) . | Proof. It is sufficient to prove that \( f\left( U\right) \) is open; the same will then hold for \( f\left( W\right) \), where \( W \subseteq U \) is an arbitrary open subset. This proves continuity of the inverse function from \( f\left( U\right) \) to \( U \) . Consider a closed sphere.\n\n\[ D = \left\{ {x \in {\ma... | Yes |
Corollary 7.13 (Invariance of domain) If \( V \subseteq {\mathbf{R}}^{n} \) has the topology induced by \( {\mathbf{R}}^{n} \) and is homeomorphic to an open subset of \( {\mathbf{R}}^{n} \) then \( V \) is open in \( {\mathbf{R}}^{n} \). | Proof. This follows immediately from Theorem 7.12. | Yes |
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