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In Example 2 we used the data shown in Table 4.3 to approximate the first derivative of \( f\left( x\right) = x{e}^{x} \) at \( x = {2.0} \) . Use the second derivative formula (4.9) to approximate \( {f}^{\prime \prime }\left( {2.0}\right) \) . | The data permits us to determine two approximations for \( {f}^{\prime \prime }\left( {2.0}\right) \) . Using (4.9) with \( h = {0.1} \) gives\n\n\[ \n\frac{1}{0.01}\left\lbrack {f\left( {1.9}\right) - {2f}\left( {2.0}\right) + f\left( {2.1}\right) }\right\rbrack = {100}\left\lbrack {{12.703199} - 2\left( {14.778112}\r... | Yes |
In Example 1 of Section 4.1 we use the forward-difference method with \( h = {0.1} \) and \( h = {0.05} \) to find approximations to \( {f}^{\prime }\left( {1.8}\right) \) for \( f\left( x\right) = \ln \left( x\right) \). Assume that this formula has truncation error \( O\left( h\right) \) and use extrapolation on thes... | In Example 1 of Section 4.1 we found that\n\n\[ \text{with}h = {0.1} : {f}^{\prime }\left( {1.8}\right) \approx {0.5406722},\;\text{and}\;\text{with}h = {0.05} : {f}^{\prime }\left( {1.8}\right) \approx {0.5479795}\text{.} \]\n\nThis implies that\n\n\[ {N}_{1}\left( {0.1}\right) = {0.5406722}\text{ and }{N}_{1}\left( {... | Yes |
Find approximations of order \( O\left( {h}^{2}\right), O\left( {h}^{4}\right) \), and \( O\left( {h}^{6}\right) \) for \( {f}^{\prime }\left( {2.0}\right) \) when \( f\left( x\right) = x{e}^{x} \) and \( h = {0.2} \) . | Solution The constants \( {K}_{1} = - {f}^{\prime \prime \prime }\left( {x}_{0}\right) /6,{K}_{2} = - {f}^{\left( 5\right) }\left( {x}_{0}\right) /{120},\cdots \), are not likely to be known, but this is not important. We only need to know that these constants exist in order to apply extrapolation.\n\nWe have the \( O\... | Yes |
Compare the Trapezoidal rule and Simpson’s rule approximations to \( {\int }_{0}^{2}f\left( x\right) {dx} \) when \( f\left( x\right) \) is (a) \( {x}^{2} \) | Solution On \( \left\lbrack {0,2}\right\rbrack \) the Trapezoidal and Simpson’s rule have the forms\n\n\[ \n\text{Trapezoid:}{\int }_{0}^{2}f\left( x\right) {dx} \approx f\left( 0\right) + f\left( 2\right) \text{and} \n\]\n\n\[ \n\text{Simpson’s:}{\int }_{0}^{2}f\left( x\right) {dx} \approx \frac{1}{3}\left\lbrack {f\l... | Yes |
Theorem 4.2 Suppose that \( \mathop{\sum }\limits_{{i = 0}}^{n}{a}_{i}f\left( {x}_{i}\right) \) denotes the \( \left( {n + 1}\right) \) -point closed Newton-Cotes formula with \( {x}_{0} = a,{x}_{n} = b \), and \( h = \left( {b - a}\right) /n \) . There exists \( \xi \in \left( {a, b}\right) \) for which | \n\[
{\int }_{a}^{b}f\left( x\right) {dx} = \mathop{\sum }\limits_{{i = 0}}^{n}{a}_{i}f\left( {x}_{i}\right) + \frac{{h}^{n + 3}{f}^{\left( n + 2\right) }\left( \xi \right) }{\left( {n + 2}\right) !}{\int }_{0}^{n}{t}^{2}\left( {t - 1}\right) \cdots \left( {t - n}\right) {dt},
\]\n\nif \( n \) is even and \( f \in {C}^... | Yes |
Example 2 Compare the results of the closed and open Newton-Cotes formulas listed as (4.25)-(4.28) and (4.29)-(4.32) when approximating\n\n\[ \n{\int }_{0}^{\pi /4}\sin {xdx} = 1 - \sqrt{2}/2 \approx {0.29289322}. \n\] | Solution For the closed formulas we have\n\n\[ \nn = 1 : \;\frac{\left( \pi /4\right) }{2}\left\lbrack {\sin 0 + \sin \frac{\pi }{4}}\right\rbrack \approx {0.27768018} \n\]\n\n\[ \nn = 2 : \frac{\left( \pi /8\right) }{3}\left\lbrack {\sin 0 + 4\sin \frac{\pi }{8} + \sin \frac{\pi }{4}}\right\rbrack \approx {0.29293264}... | Yes |
Use Simpson’s rule to approximate \( {\int }_{0}^{4}{e}^{x}{dx} \) and compare this to the results obtained by adding the Simpson’s rule approximations for \( {\int }_{0}^{2}{e}^{x}{dx} \) and \( {\int }_{2}^{4}{e}^{x}{dx} \) . Compare these approximations to the sum of Simpson’s rule for \( {\int }_{0}^{1}{e}^{x}{dx},... | Solution Simpson’s rule on \( \left\lbrack {0,4}\right\rbrack \) uses \( h = 2 \) and gives\n\n\[ \n{\int }_{0}^{4}{e}^{x}{dx} \approx \frac{2}{3}\left( {{e}^{0} + 4{e}^{2} + {e}^{4}}\right) = {56.76958}.\n\]\n\nThe exact answer in this case is \( {e}^{4} - {e}^{0} = {53.59815} \), and the error -3.17143 is far larger ... | Yes |
Theorem 4.4 Let \( f \in {C}^{4}\left\lbrack {a, b}\right\rbrack, n \) be even, \( h = \left( {b - a}\right) /n \), and \( {x}_{j} = a + {jh} \), for each \( j = 0,1,\ldots, n \) . There exists a \( \mu \in \left( {a, b}\right) \) for which the Composite Simpson’s rule for \( n \) subintervals can be written with its e... | Notice that the error term for the Composite Simpson’s rule is \( O\left( {h}^{4}\right) \), whereas it was \( O\left( {h}^{5}\right) \) for the standard Simpson’s rule. However, these rates are not comparable because for standard Simpson’s rule we have \( h \) fixed at \( h = \left( {b - a}\right) /2 \), but for Compo... | No |
Example 2 Determine values of \( h \) that will ensure an approximation error of less than 0.00002 when approximating \( {\int }_{0}^{\pi }\sin {xdx} \) and employing (a) Composite Trapezoidal rule and (b) Composite Simpson's rule. | Solution (a) The error form for the Composite Trapezoidal rule for \( f\left( x\right) = \sin x \) on \( \left\lbrack {0,\pi }\right\rbrack \) is \[ \left| {\frac{\pi {h}^{2}}{12}{f}^{\prime \prime }\left( \mu \right) }\right| = \left| {\frac{\pi {h}^{2}}{12}\left( {-\sin \mu }\right) }\right| = \frac{\pi {h}^{2}}{12}\... | Yes |
Example 2 Add an additional extrapolation row to Table 4.10 to approximate \( {\int }_{0}^{\pi }\sin {xdx} \) . | Solution To obtain the additional row we need the trapezoidal approximation\n\n\[ \n{R}_{6,1} = \frac{1}{2}\left\lbrack {{R}_{5,1} + \frac{\pi }{16}\mathop{\sum }\limits_{{k = 1}}^{{2}^{4}}\sin \frac{\left( {{2k} - 1}\right) \pi }{32}}\right\rbrack = {1.99839336}. \n\] \n\nThe values in Table 4.10 give \n\n\[ \n{R}_{6,... | Yes |
Check the accuracy of the error estimate given in (4.38) and (4.39) when applied to the integral\n\n\[ \n{\int }_{0}^{\pi /2}\sin {xdx} = 1 \n\]\n\nby comparing\n\n\[ \n\frac{1}{15}\left| {S\left( {0,\frac{\pi }{2}}\right) - S\left( {0,\frac{\pi }{4}}\right) - S\left( {\frac{\pi }{4},\frac{\pi }{2}}\right) }\right| \;\... | Solution We have\n\n\[ \nS\left( {0,\frac{\pi }{2}}\right) = \frac{\pi /4}{3}\left\lbrack {\sin 0 + 4\sin \frac{\pi }{4} + \sin \frac{\pi }{2}}\right\rbrack = \frac{\pi }{12}\left( {2\sqrt{2} + 1}\right) = {1.002279878} \n\]\n\nand\n\n\[ \nS\left( {0,\frac{\pi }{4}}\right) + S\left( {\frac{\pi }{4},\frac{\pi }{2}}\righ... | Yes |
Approximate \( {\int }_{-1}^{1}{e}^{x}\cos {xdx} \) using Gaussian quadrature with \( n = 3 \) . | The entries in Table 4.12 give us\n\n\[ \n{\int }_{-1}^{1}{e}^{x}\cos {xdx} \approx {0.5}{e}^{0.774596692}\cos {0.774596692} \n\]\n\n\[ \n+ 0.\overline{8}\cos 0 + 0.\overline{5}{e}^{-{0.774596692}}\cos \left( {-{0.774596692}}\right) \n\]\n\n\[ \n= {1.9333904}\text{.} \n\]\n\nIntegration by parts can be used to show tha... | Yes |
Example 2 Consider the integral \( {\int }_{1}^{3}{x}^{6} - {x}^{2}\sin \left( {2x}\right) {dx} = {317.3442466} \) . | Solution (a) Each of the formulas in this part requires 2 evaluations of the function \( f\left( x\right) = \) \( {x}^{6} - {x}^{2}\sin \left( {2x}\right) \) . The Newton-Cotes approximations are\n\n\[ \n\text{Closed}n = 1 : \;\frac{2}{2}\left\lbrack {f\left( 1\right) + f\left( 3\right) }\right\rbrack = {731.6054420}\t... | Yes |
Example 2 Use Gaussian quadrature with \( n = 3 \) in both dimensions to approximate the integral\n\n\[ \n{\int }_{1.4}^{2.0}{\int }_{1.0}^{1.5}\ln \left( {x + {2y}}\right) {dydx} \n\] | Solution Before employing Gaussian quadrature to approximate this integral, we need to transform the region of integration\n\n\[ \nR = \{ \left( {x, y}\right) \mid {1.4} \leq x \leq {2.0},{1.0} \leq y \leq {1.5}\} \n\]\n\ninto\n\n\[ \n\widehat{R} = \{ \left( {u, v}\right) \mid - 1 \leq u \leq 1, - 1 \leq v \leq 1\} . \... | Yes |
Show that the improper integral \( {\int }_{0}^{1}\frac{1}{\sqrt{x}}{dx} \) converges but \( {\int }_{0}^{1}\frac{1}{{x}^{2}}{dx} \) diverges. | For the first integral we have\n\n\[ \n{\int }_{0}^{1}\frac{1}{\sqrt{x}}{dx} = \mathop{\lim }\limits_{{M \rightarrow {0}^{ + }}}{\int }_{M}^{1}{x}^{-1/2}{dx} = {\left. \mathop{\lim }\limits_{{M \rightarrow {0}^{ + }}}2{x}^{1/2}\right| }_{x = M}^{x = 1} = 2 - 0 = 2, \n\] \n\nbut the second integral \n\n\[ \n{\int }_{0}^... | Yes |
Use Composite Simpson’s rule with \( h = {0.25} \) to approximate the value of the improper integral\n\n\[ \n{\int }_{0}^{1}\frac{{e}^{x}}{\sqrt{x}}{dx} \n\] | Solution The fourth Taylor polynomial for \( {e}^{x} \) about \( x = 0 \) is\n\n\[ \n{P}_{4}\left( x\right) = 1 + x + \frac{{x}^{2}}{2} + \frac{{x}^{3}}{6} + \frac{{x}^{4}}{24} \n\]\nso the dominant portion of the approximation to \( {\int }_{0}^{1}\frac{{e}^{x}}{\sqrt{x}}{dx} \) is\n\n\[ \n{\int }_{0}^{1}\frac{{P}_{4}... | Yes |
Approximate the value of the improper integral\n\n\[ I = {\int }_{1}^{\infty }{x}^{-3/2}\sin \frac{1}{x}{dx} \] | Solution We first make the variable change \( t = {x}^{-1} \), which converts the infinite singularity into one with a left endpoint singularity. Then\n\n\[ {dt} = - {x}^{-2}{dx},\;\text{ so }\;{dx} = - {x}^{2}{dt} = - \frac{1}{{t}^{2}}{dt}, \]\n\nand\n\n\[ I = {\int }_{x = 1}^{x = \infty }{x}^{-3/2}\sin \frac{1}{x}{dx... | Yes |
Show that \( f\left( {t, y}\right) = t\left| y\right| \) satisfies a Lipschitz condition on the interval \( D = \{ \left( {t, y}\right) \mid 1 \leq t \leq 2 \) and \( - 3 \leq y \leq 4\} \) . | For each pair of points \( \left( {t,{y}_{1}}\right) \) and \( \left( {t,{y}_{2}}\right) \) in \( D \) we have\n\n\[ \left| {f\left( {t,{y}_{1}}\right) - f\left( {t,{y}_{2}}\right) }\right| = \left| t\right| {y}_{1}\left| {-t}\right| {y}_{2}\parallel = \left| t\right| \begin{Vmatrix}{y}_{1}\end{Vmatrix} - \left| {y}_{2... | Yes |
Theorem 5.3 Suppose \( f\left( {t, y}\right) \) is defined on a convex set \( D \subset {\mathbb{R}}^{2} \) . If a constant \( L > 0 \) exists with\n\nRudolf Lipschitz (1832-1903)\n\[ \n\left| {\frac{\partial f}{\partial y}\left( {t, y}\right) }\right| \leq L,\;\text{ for all }\left( {t, y}\right) \in D \n\]\n\n(5.1)\n... | The proof of Theorem 5.3 is discussed in Exercise 6; it is similar to the proof of the corresponding result for functions of one variable discussed in Exercise 27 of Section 1.1. | No |
Theorem 5.4 Suppose that \( D = \{ \left( {t, y}\right) \mid a \leq t \leq b \) and \( - \infty < y < \infty \} \) and that \( f\left( {t, y}\right) \) is continuous on \( D \) . If \( f \) satisfies a Lipschitz condition on \( D \) in the variable \( y \), then the initial-value problem\n\n\[ \n{y}^{\prime }\left( t\r... | 1ple 2 Use Theorem 5.4 to show that there is a unique solution to the initial-value problem\n\n\[ \n{y}^{\prime } = 1 + t\sin \left( {ty}\right) ,\;0 \leq t \leq 2,\;y\left( 0\right) = 0.\n\]\n\nSolution Holding \( t \) constant and applying the Mean Value Theorem to the function\n\n\[ \nf\left( {t, y}\right) = 1 + t\s... | Yes |
Example 3 Show that the initial-value problem\n\n\\[ \n\\frac{dy}{dt} = y - {t}^{2} + 1,\\;0 \\leq t \\leq 2,\\;y\\left( 0\\right) = {0.5}.\n\\]\n\n(5.4)\n\nis well posed on \\( D = \\{ \\left( {t, y}\\right) \\mid 0 \\leq t \\leq 2 \\) and \\( - \\infty < y < \\infty \\} \\) . | Solution Because\n\n\\[ \n\\left| \\frac{\\partial \\left( {y - {t}^{2} + 1}\\right) }{\\partial y}\\right| = \\left| 1\\right| = 1\n\\]\n\nTheorem 5.3 implies that \\( f\\left( {t, y}\\right) = y - {t}^{2} + 1 \\) satisfies a Lipschitz condition in \\( y \\) on \\( D \\) with Lipschitz constant 1. Since \\( f \\) is c... | Yes |
Use Algorithm 5.1 with \( N = {10} \) to determine approximations, and compare these with the exact values given by \( y\left( t\right) = {\left( t + 1\right) }^{2} - {0.5}{e}^{t} \). | Solution With \( N = {10} \) we have \( h = {0.2},{t}_{i} = {0.2i},{w}_{0} = {0.5} \), and\n\n\[ \n{w}_{i + 1} = {w}_{i} + h\left( {{w}_{i} - {t}_{i}^{2} + 1}\right) = {w}_{i} + {0.2}\left\lbrack {{w}_{i} - {0.04}{i}^{2} + 1}\right\rbrack = {1.2}{w}_{i} - {0.008}{i}^{2} + {0.2}, \n\]\n\nfor \( i = 0,1,\ldots ,9 \) . So... | Yes |
Lemma 5.7 For all \( x \geq - 1 \) and any positive \( m \), we have \( 0 \leq {\left( 1 + x\right) }^{m} \leq {e}^{mx} \) . | Proof Applying Taylor’s Theorem with \( f\left( x\right) = {e}^{x},{x}_{0} = 0 \), and \( n = 1 \) gives\n\n\[ \n{e}^{x} = 1 + x + \frac{1}{2}{x}^{2}{e}^{\xi }\n\]\n\nwhere \( \xi \) is between \( x \) and zero. Thus\n\n\[ \n0 \leq 1 + x \leq 1 + x + \frac{1}{2}{x}^{2}{e}^{\xi } = {e}^{x},\n\]\n\nand, because \( 1 + x ... | Yes |
If \( s \) and \( t \) are positive real numbers, \( {\left\{ {a}_{i}\right\} }_{i = 0}^{k} \) is a sequence satisfying \( {a}_{0} \geq - t/s \), and\n\n\[ \n{a}_{i + 1} \leq \left( {1 + s}\right) {a}_{i} + t,\;\text{ for each }i = 0,1,2,\ldots, k - 1, \n\]\n\nthen\n\n\[ \n{a}_{i + 1} \leq {e}^{\left( {i + 1}\right) s}... | Proof For a fixed integer \( i \), Inequality (5.9) implies that\n\n\[ \n{a}_{i + 1} \leq \left( {1 + s}\right) {a}_{i} + t \n\]\n\n\[ \n\leq \left( {1 + s}\right) \left\lbrack {\left( {1 + s}\right) {a}_{i - 1} + t}\right\rbrack + t = {\left( 1 + s\right) }^{2}{a}_{i - 1} + \left\lbrack {1 + \left( {1 + s}\right) }\ri... | Yes |
Theorem 5.9 Suppose \( f \) is continuous and satisfies a Lipschitz condition with constant \( L \) on\n\n\[ D = \{ \left( {t, y}\right) \mid a \leq t \leq b\text{ and } - \infty < y < \infty \} \]\n\nand that a constant \( M \) exists with\n\n\[ \left| {{y}^{\prime \prime }\left( t\right) }\right| \leq M,\;\text{ for ... | Proof When \( i = 0 \) the result is clearly true, since \( y\left( {t}_{0}\right) = {w}_{0} = \alpha \) .\n\nFrom Eq. (5.7), we have\n\n\[ y\left( {t}_{i + 1}\right) = y\left( {t}_{i}\right) + {hf}\left( {{t}_{i}, y\left( {t}_{i}\right) }\right) + \frac{{h}^{2}}{2}{y}^{\prime \prime }\left( {\xi }_{i}\right) ,\]\n\nfo... | Yes |
The solution to the initial-value problem\n\n\\[ \n{y}^{\prime } = y - {t}^{2} + 1,\;0 \leq t \leq 2,\;y\\left( 0\\right) = {0.5}, \n\\] | Solution Because \\( f\\left( {t, y}\\right) = y - {t}^{2} + 1 \\), we have \\( \\partial f\\left( {t, y}\\right) /\\partial y = 1 \\) for all \\( y \\), so \\( L = 1 \\) . For this problem, the exact solution is \\( y\\left( t\\right) = {\\left( t + 1\\right) }^{2} - {0.5}{e}^{t} \\), so \\( {y}^{\prime \\prime }\\lef... | Yes |
Theorem 5.10 Let \( y\left( t\right) \) denote the unique solution to the initial-value problem\n\n\[ \n{y}^{\prime } = f\left( {t, y}\right) ,\;a \leq t \leq b,\;y\left( a\right) = \alpha \n\]\n\n(5.12)\n\nand \( {u}_{0},{u}_{1},\ldots ,{u}_{N} \) be the approximations obtained using (5.11). If \( \left| {\delta }_{i}... | The error bound (5.13) is no longer linear in \( h \) . In fact, since\n\n\[ \n\mathop{\lim }\limits_{{h \rightarrow 0}}\left( {\frac{hM}{2} + \frac{\delta }{h}}\right) = \infty \n\]\n\nthe error would be expected to become large for sufficiently small values of \( h \) . Calculus can be used to determine a lower bound... | Yes |
Apply Taylor’s method of orders (a) two and (b) four with \( N = {10} \) to the initial-value problem\n\n\[ \n{y}^{\prime } = y - {t}^{2} + 1,\;0 \leq t \leq 2,\;y\left( 0\right) = {0.5}.\n\] | Solution (a) For the method of order two we need the first derivative of \( f\left( {t, y\left( t\right) }\right) = \) \( y\left( t\right) - {t}^{2} + 1 \) with respect to the variable \( t \) . Because \( {y}^{\prime } = y - {t}^{2} + 1 \) we have\n\n\[ \n{f}^{\prime }\left( {t, y\left( t\right) }\right) = \frac{d}{dt... | Yes |
Theorem 5.12 If Taylor’s method of order \( n \) is used to approximate the solution to\n\n\[ \n{y}^{\prime }\left( t\right) = f\left( {t, y\left( t\right) }\right) ,\;a \leq t \leq b,\;y\left( a\right) = \alpha ,\n\]\n\nwith step size \( h \) and if \( y \in {C}^{n + 1}\left\lbrack {a, b}\right\rbrack \), then the loc... | Proof Note that Eq. (5.16) on page 277 can be rewritten\n\n\[ \n{y}_{i + 1} - {y}_{i} - {hf}\left( {{t}_{i},{y}_{i}}\right) - \frac{{h}^{2}}{2}{f}^{\prime }\left( {{t}_{i},{y}_{i}}\right) - \cdots - \frac{{h}^{n}}{n!}{f}^{\left( n - 1\right) }\left( {{t}_{i},{y}_{i}}\right) = \frac{{h}^{n + 1}}{\left( {n + 1}\right) !}... | Yes |
Suppose that \( f\left( {t, y}\right) \) and all its partial derivatives of order less than or equal to \( n + 1 \) are continuous on \( D = \{ \left( {t, y}\right) \mid a \leq t \leq b, c \leq y \leq d\} \), and let \( \left( {{t}_{0},{y}_{0}}\right) \in D \) . For every \( \left( {t, y}\right) \in D \), there exists ... | where\n\n\[ {P}_{n}\left( {t, y}\right) = f\left( {{t}_{0},{y}_{0}}\right) + \left\lbrack {\left( {t - {t}_{0}}\right) \frac{\partial f}{\partial t}\left( {{t}_{0},{y}_{0}}\right) + \left( {y - {y}_{0}}\right) \frac{\partial f}{\partial y}\left( {{t}_{0},{y}_{0}}\right) }\right\rbrack \]\n\n\[ + \left\lbrack {\frac{{\l... | Yes |
Use Maple to determine \( {P}_{2}\left( {t, y}\right) \), the second Taylor polynomial about \( \left( {2,3}\right) \) for the function\n\n\[ f\left( {t, y}\right) = \exp \left\lbrack {-\frac{{\left( t - 2\right) }^{2}}{4} - \frac{{\left( y - 3\right) }^{2}}{4}}\right\rbrack \cos \left( {{2t} + y - 7}\right) \] | Solution To determine \( {P}_{2}\left( {t, y}\right) \) we need the values of \( f \) and its first and second partial derivatives at \( \left( {2,3}\right) \) . The evaluation of the function is easy\n\n\[ f\left( {2,3}\right) = {e}^{\left( -{0}^{2}/4 - {0}^{2}/4\right) }\cos \left( {4 + 3 - 7}\right) = 1, \]\n\nbut t... | Yes |
Example 2 Use the Midpoint method and the Modified Euler method with \( N = {10}, h = {0.2},{t}_{i} = {0.2i} \) , and \( {w}_{0} = {0.5} \) to approximate the solution to our usual example,\n\n\[ \n{y}^{\prime } = y - {t}^{2} + 1,\;0 \leq t \leq 2,\;y\left( 0\right) = {0.5}.\n\] | Solution The difference equations produced from the various formulas are\n\n\[ \n\text{Midpoint method:}\;{w}_{i + 1} = {1.22}{w}_{i} - {0.0088}{i}^{2} - {0.008i} + {0.218}\text{;}\n\]\n\nModified Euler method: \( \;{w}_{i + 1} = {1.22}{w}_{i} - {0.0088}{i}^{2} - {0.008i} + {0.216} \) ,\n\nfor each \( i = 0,1,\ldots ,9... | Yes |
Use the Runge-Kutta method of order four with \( h = {0.2}, N = {10} \), and \( {t}_{i} = {0.2i} \) to obtain approximations to the solution of the initial-value problem\n\n\[ \n{y}^{\prime } = y - {t}^{2} + 1,\;0 \leq t \leq 2,\;y\left( 0\right) = {0.5}. \n\] | Solution The approximation to \( y\left( {0.2}\right) \) is obtained by\n\n\[ \n{w}_{0} = {0.5} \n\]\n\n\[ \n{k}_{1} = {0.2f}\left( {0,{0.5}}\right) = {0.2}\left( {1.5}\right) = {0.3} \n\]\n\n\[ \n{k}_{2} = {0.2f}\left( {{0.1},{0.65}}\right) = {0.328} \n\]\n\n\[ \n{k}_{3} = {0.2f}\left( {{0.1},{0.664}}\right) = {0.3308... | No |
Use these as starting values for the fourth-order Adams-Bashforth method to compute new approximations for \( y\left( {0.8}\right) \) and \( y\left( {1.0}\right) \), and compare these new approximations to those produced by the Runge-Kutta method of order four. | Solution For the fourth-order Adams-Bashforth we have\n\n\[ y\left( {0.8}\right) \approx {w}_{4} = {w}_{3} + \frac{0.2}{24}\left( {{55f}\left( {{0.6},{w}_{3}}\right) - {59f}\left( {{0.4},{w}_{2}}\right) + {37f}\left( {{0.2},{w}_{1}}\right) - {9f}\left( {0,{w}_{0}}\right) }\right) \]\n\n\[ = {1.6489220} + \frac{0.2}{24}... | Yes |
Example 2 Use Eq. (5.31) with \( m = 3 \) to derive the three-step Adams-Bashforth technique. | Solution We have\n\n\[ y\left( {t}_{i + 1}\right) \approx y\left( {t}_{i}\right) + h\left\lbrack {f\left( {{t}_{i}, y\left( {t}_{i}\right) }\right) + \frac{1}{2}\nabla f\left( {{t}_{i}, y\left( {t}_{i}\right) }\right) + \frac{5}{12}{\nabla }^{2}f\left( {{t}_{i}, y\left( {t}_{i}\right) }\right) }\right\rbrack \]\n\n\[ =... | Yes |
Example 3 Determine the local truncation error for the three-step Adams-Bashforth method derived in Example 2. | Solution Considering the form of the error given in Eq. (5.30) and the appropriate entry in Table 5.12 gives\n\n\[ \n{h}^{4}{f}^{\left( 3\right) }\left( {{\mu }_{i}, y\left( {\mu }_{i}\right) }\right) {\left( -1\right) }^{3}{\int }_{0}^{1}\left( \begin{matrix} - s \\ 3 \end{matrix}\right) {ds} = \frac{3{h}^{4}}{8}{f}^{... | Yes |
Consider the initial-value problem\n\n\\[ \n{y}^{\prime } = y - {t}^{2} + 1,\;0 \leq t \leq 2,\;y\left( 0\right) = {0.5}. \n\\] | Use the exact values given from \\( y\\left( t\\right) = {\\left( t + 1\\right) }^{2} - {0.5}{e}^{t} \\) as starting values and \\( h = {0.2} \\) to compare the approximations from (a) by the explicit Adams-Bashforth four-step method and (b) the implicit Adams-Moulton three-step method.\n\nSolution (a) The Adams-Bashfo... | Yes |
Use Adams variable step-size predictor-corrector method with maximum step size \( h{max} = \) 0.2, minimum step size \( {hmin} = {0.01} \), and tolerance \( {TOL} = {10}^{-5} \) to approximate the solution of the initial-value problem\n\n\[ \n{y}^{\prime } = y - {t}^{2} + 1,\;0 \leq t \leq 2,\;y\left( 0\right) = {0.5}.... | Solution We begin with \( h = {hmax} = {0.2} \), and obtain \( {w}_{0},{w}_{1},{w}_{2} \) and \( {w}_{3} \) using Runge-Kutta, then find \( w{p}_{4} \) and \( w{c}_{4} \) by applying the predictor-corrector method. These calculations were done in Example 5 of Section 5.6 where it was determined that the Runge-Kutta app... | Yes |
Transform the second-order initial-value problem\n\n\[ \n{y}^{\prime \prime } - 2{y}^{\prime } + {2y} = {e}^{2t}\sin t,\;\text{ for }0 \leq t \leq 1,\;\text{ with }y\left( 0\right) = - {0.4},{y}^{\prime }\left( 0\right) = - {0.6} \n\]\n\ninto a system of first order initial-value problems, and use the Runge-Kutta metho... | Solution Let \( {u}_{1}\left( t\right) = y\left( t\right) \) and \( {u}_{2}\left( t\right) = {y}^{\prime }\left( t\right) \) . This transforms the second-order equation into the system\n\n\[ \n{u}_{1}^{\prime }\left( t\right) = {u}_{2}\left( t\right) \n\]\n\n\[ \n{u}_{2}^{\prime }\left( t\right) = {e}^{2t}\sin t - 2{u}... | Yes |
Example 1 Show that Euler's method is convergent. | Solution Examining Inequality (5.10) on page 271, in the error-bound formula for Euler's method, we see that under the hypotheses of Theorem 5.9, \[ \mathop{\max }\limits_{{1 \leq i \leq N}}\left| {{w}_{i} - y\left( {t}_{i}\right) }\right| \leq \frac{Mh}{2L}\left| {{e}^{L\left( {b - a}\right) } - 1}\right| . \] However... | Yes |
Verify that the Modified Euler method is stable by showing that it satisfies the hypothesis of Theorem 5.20. | For this method, \n\n\[ \phi \left( {t, w, h}\right) = \frac{1}{2}f\left( {t, w}\right) + \frac{1}{2}f\left( {t + h, w + {hf}\left( {t, w}\right) }\right) . \]\n\nIf \( f \) satisfies a Lipschitz condition on \( \{ \left( {t, w}\right) \mid a \leq t \leq b \) and \( - \infty < w < \infty \} \) in the variable \( w \) w... | Yes |
Theorem 5.21 Suppose the initial-value problem\n\n\\[ \n{y}^{\prime } = f\\left( {t, y}\\right) ,\\;a \\leq t \\leq b,\\;y\\left( a\\right) = \\alpha ,\n\\]\n\nis approximated by an explicit Adams predictor-corrector method with an \\( m \\) -step Adams-Bashforth predictor equation\n\n\\[ \n{w}_{i + 1} = {w}_{i} + h\\l... | Moreover, there exist constants \\( {k}_{1} \\) and \\( {k}_{2} \\) such that\n\n\\[ \n\\left| {{w}_{i} - y\\left( {t}_{i}\\right) }\\right| \\leq \\left\\lbrack {\\mathop{\\max }\\limits_{{0 \\leq j \\leq m - 1}}\\left| {{w}_{j} - y\\left( {t}_{j}\\right) }\\right| + {k}_{1}\\sigma \\left( h\\right) }\\right\\rbrack {... | Yes |
Show that this method is strongly stable. | Solution In this case we have \( m = 4,{a}_{0} = 0,{a}_{1} = 0,{a}_{2} = 0 \), and \( {a}_{3} = 1 \), so the characteristic equation for this Adams-Bashforth method is\n\n\[ 0 = P\left( \lambda \right) = {\lambda }^{4} - {\lambda }^{3} = {\lambda }^{3}\left( {\lambda - 1}\right) .\n\]\n\nThis polynomial has roots \( {\... | Yes |
Example 4 Show that the fourth-order Milne's method, the explicit multistep method given by\n\n\\[ \n{w}_{i + 1} = {w}_{i - 3} + \\frac{4h}{3}\\left\\lbrack {{2f}\\left( {{t}_{i},{w}_{i}}\\right) - f\\left( {{t}_{i - 1},{w}_{i - 1}}\\right) + {2f}\\left( {{t}_{i - 2},{w}_{i - 2}}\\right) }\\right\\rbrack \n\\]\n\nsatis... | Solution The characteristic equation for this method, \\( 0 = P\\left( \\lambda \\right) = {\\lambda }^{4} - 1 \\), has four roots with magnitude one: \\( {\\lambda }_{1} = 1,{\\lambda }_{2} = - 1,{\\lambda }_{3} = i \\), and \\( {\\lambda }_{4} = - i \\) . Because all the roots have magnitude 1 , the method satisfies ... | Yes |
Example 1 Determine the size and respective entries of the matrix\n\n\[ A = \left\lbrack \begin{array}{rrr} 2 & - 1 & 7 \\ 3 & 1 & 0 \end{array}\right\rbrack \] | Solution The matrix has two rows and three columns so it is of size \( 2 \times 3 \) . It entries are described by \( {a}_{11} = 2,{a}_{12} = - 1,{a}_{13} = 7,{a}_{21} = 3,{a}_{22} = 1 \), and \( {a}_{23} = 0 \) . | Yes |
Represent the linear system\n\n\[ \n{E}_{1} : \;{x}_{1} - {x}_{2} + 2{x}_{3} - {x}_{4} = - 8, \]\n\n\[ \n{E}_{2} : \;2{x}_{1} - 2{x}_{2} + 3{x}_{3} - 3{x}_{4} = - {20}, \]\n\n\[ \n{E}_{3} : \;{x}_{1} + {x}_{2} + {x}_{3}\; = - 2, \]\n\n\[ \n{E}_{4} : \;{x}_{1} - {x}_{2} + 4{x}_{3} + 3{x}_{4} = \;4, \]\n\nas an augmented... | Solution The augmented matrix is\n\n\[ \n\widetilde{A} = {\widetilde{A}}^{\left( 1\right) } = \left\lbrack \begin{array}{rrrrrr} 1 & - 1 & 2 & - 1 & \vdots & - 8 \\ 2 & - 2 & 3 & - 3 & \vdots & - {20} \\ 1 & 1 & 1 & 0 & \vdots & - 2 \\ 1 & - 1 & 4 & 3 & \vdots & 4 \end{array}\right\rbrack \]\n\nPerforming the operation... | Yes |
Example 1 Apply Gaussian elimination to the system\n\n\[ \n{E}_{1} : \;{0.003000}{x}_{1} + {59.14}{x}_{2} = {59.17} \n\]\n\n\[ \n{E}_{2} : \;{5.291}{x}_{1} - {6.130}{x}_{2} = {46.78}, \n\]\n\nusing four-digit arithmetic with rounding, and compare the results to the exact solution \( {x}_{1} = {10.00} \) and \( {x}_{2} ... | Solution The first pivot element, \( {a}_{11}^{\left( 1\right) } = {0.003000} \), is small, and its associated multiplier,\n\n\[ \n{m}_{21} = \frac{5.291}{0.003000} = {1763.6}\overline{6} \n\]\n\nrounds to the large number 1764. Performing \( \left( {{E}_{2} - {m}_{21}{E}_{1}}\right) \rightarrow \left( {E}_{2}\right) \... | Yes |
Apply Gaussian elimination to the system\n\n\\[ \n{E}_{1} : \\;{0.003000}{x}_{1} + {59.14}{x}_{2} = {59.17} \n\\]\n\n\\[ \n{E}_{2} : \\;{5.291}{x}_{1} - {6.130}{x}_{2} = {46.78}, \n\\]\n\nusing partial pivoting and four-digit arithmetic with rounding, and compare the results to the exact solution \\( {x}_{1} = {10.00} ... | Solution The partial-pivoting procedure first requires finding\n\n\\[ \n\\max \\left\\{ {\\left| {a}_{11}^{\\left( 1\\right) }\\right| ,\\left| {a}_{21}^{\\left( 1\\right) }\\right| }\\right\\} = \\max \\{ \\left| {0.003000}\\right| ,\\left| {5.291}\\right| \\} = \\left| {5.291}\\right| = \\left| {a}_{21}^{\\left( 1\\r... | Yes |
Determine \( A + B \) and \( {\lambda A} \) when\n\n\[ A = \left\lbrack \begin{array}{rrr} 2 & - 1 & 7 \\ 3 & 1 & 0 \end{array}\right\rbrack ,\;B = \left\lbrack \begin{array}{rrr} 4 & 2 & - 8 \\ 0 & 1 & 6 \end{array}\right\rbrack ,\;\text{ and }\lambda = - 2. \] | Solution We have\n\n\[ A + B = \left\lbrack \begin{array}{rrr} 2 + 4 & - 1 + 2 & 7 - 8 \\ 3 + 0 & 1 + 1 & 0 + 6 \end{array}\right\rbrack = \left\lbrack \begin{array}{rrr} 6 & 1 & - 1 \\ 3 & 2 & 6 \end{array}\right\rbrack ,\]\n\nand\n\n\[ {\lambda A} = \left\lbrack \begin{matrix} - 2\left( 2\right) & - 2\left( {-1}\righ... | Yes |
Theorem 6.5 Let \( A, B \), and \( C \) be \( n \times m \) matrices and \( \lambda \) and \( \mu \) be real numbers. The following properties of addition and scalar multiplication hold:\n\n(i) \( A + B = B + A \) , (ii) \( \left( {A + B}\right) + C = A + \left( {B + C}\right) \) ,\n\n(iii) \( A + O = O + A = A \) , (i... | All these properties follow from similar results concerning the real numbers. | No |
Determine the product \( A\mathbf{b} \) if \( A = \left\lbrack \begin{array}{rr} 3 & 2 \\ - 1 & 1 \\ 6 & 4 \end{array}\right\rbrack \) and \( \mathbf{b} = \left\lbrack \begin{matrix} 3 \\ - 1 \end{matrix}\right\rbrack \) . | Solution Because \( A \) has dimension \( 3 \times 2 \) and \( \mathbf{b} \) has dimension \( 2 \times 1 \), the product is defined and is a vector with three rows. These are\n\n\[ 3\left( 3\right) + 2\left( {-1}\right) = 7,\;\left( {-1}\right) \left( 3\right) + 1\left( {-1}\right) = - 4,\;\text{ and }\;6\left( 3\right... | Yes |
Determine all possible products of the matrices\n\n\[ A = \left\lbrack \begin{array}{rr} 3 & 2 \\ - 1 & 1 \\ 1 & 4 \end{array}\right\rbrack ,\;B = \left\lbrack \begin{array}{rrr} 2 & 1 & - 1 \\ 3 & 1 & 2 \end{array}\right\rbrack \]\n\n\[ C = \left\lbrack \begin{array}{rrrr} 2 & 1 & 0 & 1 \\ - 1 & 3 & 2 & 1 \\ 1 & 1 & 2... | Solution The size of the matrices are\n\n\[ A : 3 \times 2,\;B : 2 \times 3,\;C : 3 \times 4,\;\text{ and }\;D : 2 \times 2. \]\n\nThe products that can be defined, and their dimensions, are:\n\n\[ {AB} : 3 \times 3,\;{BA} : 2 \times 2,\;{AD} : 3 \times 2,\;{BC} : 2 \times 4,\;{DB} : 2 \times 3,\;\text{ and }\;{DD} : 2... | Yes |
Theorem 6.8 Let \( A \) be an \( n \times m \) matrix, \( B \) be an \( m \times k \) matrix, \( C \) be a \( k \times p \) matrix, \( D \) be an \( m \times k \) matrix, and \( \lambda \) be a real number. The following properties hold:\n\n\[ \n\text{(a)}A\left( {BC}\right) = \left( {AB}\right) C;\;\text{(b)}A\left( {... | Proof The verification of the property in part (a) is presented to show the method involved. The other parts can be shown in a similar manner.\n\nTo show that \( A\left( {BC}\right) = \left( {AB}\right) C \), compute the \( {sj} \) -entry of each side of the equation. \( {BC} \) is an \( m \times p \) matrix with \( {s... | No |
Theorem 6.12 For any nonsingular \( n \times n \) matrix \( A \) :\n\n(i) \( {A}^{-1} \) is unique.\n\n(ii) \( {A}^{-1} \) is nonsingular and \( {\left( {A}^{-1}\right) }^{-1} = A \) .\n\n(iii) If \( B \) is also a nonsingular \( n \times n \) matrix, then \( {\left( AB\right) }^{-1} = {B}^{-1}{A}^{-1} \) . | Example 4 Let\n\n\[ A = \left\lbrack \begin{array}{rrr} 1 & 2 & - 1 \\ 2 & 1 & 0 \\ - 1 & 1 & 2 \end{array}\right\rbrack \;\text{ and }\;B = \left\lbrack \begin{array}{rrr} - \frac{2}{9} & \frac{5}{9} & - \frac{1}{9} \\ \frac{4}{9} & - \frac{1}{9} & \frac{2}{9} \\ - \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \end{array}\r... | No |
Find the determinant of the matrix\n\n\[ A = \left\lbrack \begin{array}{rrrr} 2 & - 1 & 3 & 0 \\ 4 & - 2 & 7 & 0 \\ - 3 & - 4 & 1 & 5 \\ 6 & - 6 & 8 & 0 \end{array}\right\rbrack \] using the row or column with the most zero entries. | Solution To compute \( \det A \), it is easiest to use the fourth column:\n\n\[ \det A = {a}_{14}{A}_{14} + {a}_{24}{A}_{24} + {a}_{34}{A}_{34} + {a}_{44}{A}_{44} = 5{A}_{34} = - 5{M}_{34}. \]\n\nEliminating the third row and the fourth column gives\n\n\[ \det A = - 5\det \left\lbrack \begin{array}{lll} 2 & - 1 & 3 \\ ... | Yes |
Compute the determinant of the matrix\n\n\[ A = \left\lbrack \begin{array}{rrrr} 2 & 1 & - 1 & 1 \\ 1 & 1 & 0 & 3 \\ - 1 & 2 & 3 & - 1 \\ 3 & - 1 & - 1 & 2 \end{array}\right\rbrack \]\n\nusing parts (iii), (iv), and (v) of Theorem 6.16, doing the computations in Maple with the LinearAlgebra package. | Solution Matrix \( A \) is defined in Maple by\n\n\( A \mathrel{\text{:=}} \operatorname{Matrix}\left( \left\lbrack {\left\lbrack {2,1, - 1,1}\right\rbrack ,\left\lbrack {1,1,0,3}\right\rbrack ,\left\lbrack {-1,2,3, - 1}\right\rbrack ,\left\lbrack {3, - 1, - 1,2}\right\rbrack }\right\rbrack \right) \)\n\nThe sequence o... | Yes |
Corollary 6.18 Suppose that \( A \) and \( B \) are both \( n \times n \) matrices with either \( {AB} = I \) or \( {BA} = I \) . Then \( B = {A}^{-1} \) (and \( A = {B}^{-1} \) ). | Proof Suppose that \( {AB} = I \) . Then by part (vi) of Theorem 6.16,\n\n\[ 1 = \det \left( I\right) = \det \left( {AB}\right) = \det \left( A\right) \cdot \det \left( B\right) ,\;\text{ so }\;\det \left( A\right) \neq 0\text{ and }\det \left( B\right) \neq 0. \]\n\nThe equivalence of parts (iii) and (iv) of Theorem 6... | Yes |
Determine the \( {LU} \) factorization for matrix \( A \) in the linear system \( A\mathbf{x} = \mathbf{b} \), where\n\n\[ A = \left\lbrack \begin{array}{rrrr} 1 & 1 & 0 & 3 \\ 2 & 1 & - 1 & 1 \\ 3 & - 1 & - 1 & 2 \\ - 1 & 2 & 3 & - 1 \end{array}\right\rbrack \;\text{ and }\;\mathbf{b} = \left\lbrack \begin{array}{r} 1... | The original system was considered in Section 6.1, where we saw that the sequence of operations \( \left( {{E}_{2} - 2{E}_{1}}\right) \rightarrow \left( {E}_{2}\right) ,\left( {{E}_{3} - 3{E}_{1}}\right) \rightarrow \left( {E}_{3}\right) ,\left( {{E}_{4} - \left( {-1}\right) {E}_{1}}\right) \rightarrow \left( {E}_{4}\r... | Yes |
Example 3 3 Determine a factorization in the form \( A = \\left( {{P}^{t}L}\\right) U \) for the matrix\n\n\[ A = \\left\\lbrack \\begin{array}{rrrr} 0 & 0 & - 1 & 1 \\ 1 & 1 & - 1 & 2 \\ - 1 & - 1 & 2 & 0 \\ 1 & 2 & 0 & 2 \\end{array}\\right\\rbrack \] | Solution The matrix \( A \) cannot have an \( {LU} \) factorization because \( {a}_{11} = 0 \) . However, using the row interchange \( \\left( {E}_{1}\\right) \\leftrightarrow \\left( {E}_{2}\\right) \), followed by \( \\left( {{E}_{3} + {E}_{1}}\\right) \\rightarrow \\left( {E}_{3}\\right) \) and \( \\left( {{E}_{4} -... | Yes |
Show that the matrix\n\n\[ A = \left\lbrack \begin{array}{rrr} 2 & - 1 & 0 \\ - 1 & 2 & - 1 \\ 0 & - 1 & 2 \end{array}\right\rbrack \]\n\nis positive definite | Solution Suppose \( \mathbf{x} \) is any three-dimensional column vector. Then\n\n\[ {\mathbf{x}}^{t}A\mathbf{x} = \left\lbrack {{x}_{1},{x}_{2},{x}_{3}}\right\rbrack \left\lbrack \begin{array}{rrr} 2 & - 1 & 0 \\ - 1 & 2 & - 1 \\ 0 & - 1 & 2 \end{array}\right\rbrack \left\lbrack \begin{array}{l} {x}_{1} \\ {x}_{2} \\ ... | Yes |
Theorem 6.23 If \( A \) is an \( n \times n \) positive definite matrix, then\n\n(i) \( A \) has an inverse; (ii) \( {a}_{ii} > 0 \), for each \( i = 1,2,\ldots, n \) ;\n\n(iii) \( \mathop{\max }\limits_{{1 \leq k, j \leq n}}\left| {a}_{kj}\right| \leq \mathop{\max }\limits_{{1 \leq i \leq n}}\left| {a}_{ii}\right| \) ... | Proof\n\n(i) If \( \mathbf{x} \) satisfies \( A\mathbf{x} = \mathbf{0} \), then \( {\mathbf{x}}^{t}A\mathbf{x} = 0 \) . Since \( A \) is positive definite, this implies \( \mathbf{x} = \mathbf{0} \) . Consequently, \( A\mathbf{x} = \mathbf{0} \) has only the zero solution. By Theorem 6.17 on page 398, this is equivalen... | Yes |
Example 2 In Example 1 we used the definition to show that the symmetric matrix\n\n\[ A = \left\lbrack \begin{array}{rrr} 2 & - 1 & 0 \\ - 1 & 2 & - 1 \\ 0 & - 1 & 2 \end{array}\right\rbrack \]\n\nis positive definite. Confirm this using Theorem 6.25. | Solution Note that\n\n\[ \det {A}_{1} = \det \left\lbrack 2\right\rbrack = 2 > 0 \]\n\n\[ \det {A}_{2} = \det \left\lbrack \begin{array}{rr} 2 & - 1 \\ - 1 & 2 \end{array}\right\rbrack = 4 - 1 = 3 > 0, \]\n\nand\n\n\[ \det {A}_{3} = \det \left\lbrack \begin{array}{rrr} 2 & - 1 & 0 \\ - 1 & 2 & - 1 \\ 0 & - 1 & 2 \end{a... | Yes |
Example 3 Determine the \( {LD}{L}^{t} \) factorization of the positive definite matrix\n\n\[ A = \left\lbrack \begin{matrix} 4 & - 1 & 1 \\ - 1 & {4.25} & {2.75} \\ 1 & {2.75} & {3.5} \end{matrix}\right\rbrack \] | Solution The \( {LD}{L}^{t} \) factorization has 1 s on the diagonal of the lower triangular matrix \( L \) so we need to have\n\n\[ A = \left\lbrack \begin{array}{lll} {a}_{11} & {a}_{21} & {a}_{31} \\ {a}_{21} & {a}_{22} & {a}_{32} \\ {a}_{31} & {a}_{32} & {a}_{33} \end{array}\right\rbrack = \left\lbrack \begin{matri... | Yes |
Determine the Crout factorization of the symmetric tridiagonal matrix\n\n\[ \left\lbrack \begin{array}{rrrr} 2 & - 1 & 0 & 0 \\ - 1 & 2 & - 1 & 0 \\ 0 & - 1 & 2 & - 1 \\ 0 & 0 & - 1 & 2 \end{array}\right\rbrack \] | Solution The \( {LU} \) factorization of \( A \) has the form\n\n\[ A = \left\lbrack \begin{matrix} {a}_{11} & {a}_{12} & 0 & 0 \\ {a}_{21} & {a}_{22} & {a}_{23} & 0 \\ 0 & {a}_{32} & {a}_{33} & {a}_{34} \\ 0 & 0 & {a}_{43} & {a}_{44} \end{matrix}\right\rbrack = \left\lbrack \begin{matrix} {l}_{11} & 0 & 0 & 0 \\ {l}_{... | Yes |
Determine the \( {l}_{2} \) norm and the \( {l}_{\infty } \) norm of the vector \( \mathbf{x} = {\left( -1,1, - 2\right) }^{t} \) . | The vector \( \mathbf{x} = {\left( -1,1, - 2\right) }^{t} \) in \( {\mathbb{R}}^{3} \) has norms\n\n\[ \parallel \mathbf{x}{\parallel }_{2} = \sqrt{{\left( -1\right) }^{2} + {\left( 1\right) }^{2} + {\left( -2\right) }^{2}} = \sqrt{6} \]\n\nand\n\n\[ \parallel \mathbf{x}{\parallel }_{\infty } = \max \{ \left| {-1}\righ... | Yes |
Determine the \( {l}_{2} \) and \( {l}_{\infty } \) distances between the exact and approximate solutions. | Solution Measurements of \( \mathbf{x} - \widetilde{\mathbf{x}} \) are given by\n\n\[ \parallel \mathbf{x} - \widetilde{\mathbf{x}}{\parallel }_{\infty } = \max \{ \left| {1 - {1.2001}}\right| ,\left| {1 - {0.99991}}\right| ,\left| {1 - {0.92538}}\right| \} \]\n\n\[ = \max \{ {0.2001},{0.00009},{0.07462}\} = {0.2001} \... | Yes |
Theorem 7.6 The sequence of vectors \( \left\{ {\mathbf{x}}^{\left( k\right) }\right\} \) converges to \( \mathbf{x} \) in \( {\mathbb{R}}^{n} \) with respect to the \( {l}_{\infty } \) norm if and only if \( \mathop{\lim }\limits_{{k \rightarrow \infty }}{x}_{i}^{\left( k\right) } = {x}_{i} \), for each \( i = 1,2,\ld... | Proof Suppose \( \left\{ {\mathbf{x}}^{\left( k\right) }\right\} \) converges to \( \mathbf{x} \) with respect to the \( {l}_{\infty } \) norm. Given any \( \varepsilon > 0 \), there exists an integer \( N\left( \varepsilon \right) \) such that for all \( k \geq N\left( \varepsilon \right) \) ,\n\n\[ \mathop{\max }\lim... | Yes |
Example 3 Show that\n\n\[ \n{\\mathbf{x}}^{\\left( k\\right) } = {\\left( {x}_{1}^{\\left( k\\right) },{x}_{2}^{\\left( k\\right) },{x}_{3}^{\\left( k\\right) },{x}_{4}^{\\left( k\\right) }\\right) }^{t} = {\\left( 1,2 + \\frac{1}{k},\\frac{3}{{k}^{2}},{e}^{-k}\\sin k\\right) }^{t}. \n\]\n\nconverges to \( \\mathbf{x} ... | Solution Because\n\n\[ \n\\mathop{\\lim }\\limits_{{k \\rightarrow \\infty }}1 = 1,\\;\\mathop{\\lim }\\limits_{{k \\rightarrow \\infty }}\\left( {2 + 1/k}\\right) = 2,\\;\\mathop{\\lim }\\limits_{{k \\rightarrow \\infty }}3/{k}^{2} = 0\\;\\text{ and }\\;\\mathop{\\lim }\\limits_{{k \\rightarrow \\infty }}{e}^{-k}\\sin... | Yes |
Theorem 7.7 For each \( \mathbf{x} \in {\mathbb{R}}^{n} \) , \n\n\[ \n\parallel \mathbf{x}{\parallel }_{\infty } \leq \parallel \mathbf{x}{\parallel }_{2} \leq \sqrt{n}\parallel \mathbf{x}{\parallel }_{\infty } \n\] | Proof Let \( {x}_{j} \) be a coordinate of \( \mathbf{x} \) such that \( \parallel \mathbf{x}{\parallel }_{\infty } = \mathop{\max }\limits_{{1 \leq i \leq n}}\left| {x}_{i}\right| = \left| {x}_{j}\right| \) . Then \n\n\[ \n\parallel \mathbf{x}{\parallel }_{\infty }^{2} = {\left| {x}_{j}\right| }^{2} = {x}_{j}^{2} \leq... | Yes |
Example 4 In Example 3, we found that the sequence \( \left\{ {\mathbf{x}}^{\left( k\right) }\right\} \), defined by\n\n\[ \n{\mathbf{x}}^{\left( k\right) } = {\left( 1,2 + \frac{1}{k},\frac{3}{{k}^{2}},{e}^{-k}\sin k\right) }^{t},\n\]\n\nconverges to \( \mathbf{x} = {\left( 1,2,0,0\right) }^{t} \) with respect to the ... | Solution Given any \( \varepsilon > 0 \), there exists an integer \( N\left( {\varepsilon /2}\right) \) with the property that\n\n\[ \n{\begin{Vmatrix}{\mathbf{x}}^{\left( k\right) } - \mathbf{x}\end{Vmatrix}}_{\infty } < \frac{\varepsilon }{2}\n\]\n\nwhenever \( k \geq N\left( {\varepsilon /2}\right) \) . By Theorem 7... | Yes |
If \( \parallel \cdot \parallel \) is a vector norm on \( {\mathbb{R}}^{n} \), then\n\n\[ \parallel A\parallel = \mathop{\max }\limits_{{\parallel \mathbf{x}\parallel = 1}}\parallel A\mathbf{x}\parallel \]\n\n(7.2)\n\nis a matrix norm. | For any \( \mathbf{z} \neq \mathbf{0} \), the vector \( \mathbf{x} = \mathbf{z}/\parallel \mathbf{z}\parallel \) is a unit vector. Hence\n\n\[ \mathop{\max }\limits_{{\parallel \mathbf{x}\parallel = 1}}\parallel A\mathbf{x}\parallel = \mathop{\max }\limits_{{\mathbf{z} \neq \mathbf{0}}}\begin{Vmatrix}{A\left( \frac{\ma... | Yes |
Theorem 7.11 If \( A = \left( {a}_{ij}\right) \) is an \( n \times n \) matrix, then\n\n\[ \parallel A{\parallel }_{\infty } = \mathop{\max }\limits_{{1 \leq i \leq n}}\mathop{\sum }\limits_{{j = 1}}^{n}\left| {a}_{ij}\right| . \]\n | Proof First we show that \( \parallel A{\parallel }_{\infty } \leq \mathop{\max }\limits_{{1 \leq i \leq n}}\mathop{\sum }\limits_{{j = 1}}^{n}\left| {a}_{ij}\right| \) .\n\nLet \( \mathbf{x} \) be an \( n \) -dimensional vector with \( 1 = \parallel \mathbf{x}{\parallel }_{\infty } = \mathop{\max }\limits_{{1 \leq i \... | Yes |
Example 5 Determine \( \parallel A{\parallel }_{\infty } \) for the matrix\n\n\[ A = \left\lbrack \begin{array}{rrr} 1 & 2 & - 1 \\ 0 & 3 & - 1 \\ 5 & - 1 & 1 \end{array}\right\rbrack \] | Solution We have\n\n\[ \mathop{\sum }\limits_{{j = 1}}^{3}\left| {a}_{1j}\right| = \left| 1\right| + \left| 2\right| + \left| {-1}\right| = 4,\;\mathop{\sum }\limits_{{j = 1}}^{3}\left| {a}_{2j}\right| = \left| 0\right| + \left| 3\right| + \left| {-1}\right| = 4, \]\n\nand\n\n\[ \mathop{\sum }\limits_{{j = 1}}^{3}\left... | Yes |
Show that there are no nonzero vectors \( \mathbf{x} \) in \( {\mathbb{R}}^{2} \) with \( A\mathbf{x} \) parallel to \( \mathbf{x} \) if\n\n\[ A = \left\lbrack \begin{array}{rr} 0 & 1 \\ - 1 & 0 \end{array}\right\rbrack \] | Solution The eigenvalues of \( A \) are the solutions to the characteristic polynomial\n\n\[ 0 = \det \left( {A - {\lambda I}}\right) = \det \left\lbrack \begin{array}{rr} - \lambda & 1 \\ - 1 & - \lambda \end{array}\right\rbrack = {\lambda }^{2} + 1, \]\n\nso the eigenvalues of \( A \) are the complex numbers \( {\lam... | Yes |
Determine the eigenvalues and eigenvectors for the matrix\n\n\[ A = \left\lbrack \begin{array}{rrr} 2 & 0 & 0 \\ 1 & 1 & 2 \\ 1 & - 1 & 4 \end{array}\right\rbrack \] | Solution The characteristic polynomial of \( A \) is\n\n\[ p\left( \lambda \right) = \det \left( {A - {\lambda I}}\right) = \det \left\lbrack \begin{array}{rrr} 2 - \lambda & 0 & 0 \\ 1 & 1 - \lambda & 2 \\ 1 & - 1 & 4 - \lambda \end{array}\right\rbrack \]\n\n\[ = - \left( {{\lambda }^{3} - 7{\lambda }^{2} + {16\lambda... | Yes |
Theorem 7.15 If \( A \) is an \( n \times n \) matrix, then\n\n(i) \( \parallel A{\parallel }_{2} = {\left\lbrack \rho \left( {A}^{t}A\right) \right\rbrack }^{1/2} \), \n\n(ii) \( \rho \left( A\right) \leq \parallel A\parallel \), for any natural norm \( \parallel \cdot \parallel \) . | Proof The proof of part (i) requires more information concerning eigenvalues than we presently have available. For the details involved in the proof, see [Or2], p. 21.\n\nTo prove part (ii), suppose \( \lambda \) is an eigenvalue of \( A \) with eigenvector \( \mathbf{x} \) and \( \parallel \mathbf{x}\parallel = 1 \) .... | No |
Example 3 Determine the \( {l}_{2} \) norm of\n\n\[ A = \left\lbrack \begin{array}{rrr} 1 & 1 & 0 \\ 1 & 2 & 1 \\ - 1 & 1 & 2 \end{array}\right\rbrack \] | Solution To apply Theorem 7.15 we need to calculate \( \rho \left( {{A}^{t}A}\right) \), so we first need the eigenvalues of \( {A}^{t}A \) .\n\n\[ {A}^{t}A = \left\lbrack \begin{array}{rrr} 1 & 1 & - 1 \\ 1 & 2 & 1 \\ 0 & 1 & 2 \end{array}\right\rbrack \left\lbrack \begin{array}{rrr} 1 & 1 & 0 \\ 1 & 2 & 1 \\ - 1 & 1 ... | Yes |
Example 4 Show that\n\n\[ A = \left\lbrack \begin{array}{ll} \frac{1}{2} & 0 \\ \frac{1}{4} & \frac{1}{2} \end{array}\right\rbrack \]\n\nis a convergent matrix. | Solution Computing powers of \( A \), we obtain:\n\n\[ {A}^{2} = \left\lbrack \begin{array}{ll} \frac{1}{4} & 0 \\ \frac{1}{4} & \frac{1}{4} \end{array}\right\rbrack ,\;{A}^{3} = \left\lbrack \begin{matrix} \frac{1}{8} & 0 \\ \frac{3}{16} & \frac{1}{8} \end{matrix}\right\rbrack ,\;{A}^{4} = \left\lbrack \begin{matrix} ... | Yes |
Theorem 7.17 The following statements are equivalent.\n\n(i) \( A \) is a convergent matrix.\n\n(ii) \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\begin{Vmatrix}{A}^{n}\end{Vmatrix} = 0 \), for some natural norm.\n\n(iii) \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\begin{Vmatrix}{A}^{n}\end{Vmatrix} = 0 \),... | The proof of this theorem can be found in [IK], p. 14. | No |
The linear system \( A\mathbf{x} = \mathbf{b} \) given by\n\n\[ \n{E}_{1} : \;{10}{x}_{1} - \;{x}_{2} + \;2{x}_{3}\; = 6, \]\n\n\[ \n{E}_{2} : \; - {x}_{1} + {11}{x}_{2} - \;{x}_{3} + 3{x}_{4} = {25}, \]\n\n\[ \n{E}_{3} : \;2{x}_{1} - {x}_{2} + {10}{x}_{3} - {x}_{4} = - {11}, \]\n\n\[ \n{E}_{4} : \;3{x}_{2} - \;{x}_{3}... | Solution We first solve equation \( {E}_{i} \) for \( {x}_{i} \), for each \( i = 1,2,3,4 \), to obtain\n\n\[ \n{x}_{1} = \;\frac{1}{10}{x}_{2} - \frac{1}{5}{x}_{3}\; + \frac{3}{5}, \]\n\n\[ \n{x}_{2} = \frac{1}{11}{x}_{1}\; + \frac{1}{11}{x}_{3} - \frac{3}{11}{x}_{4} + \frac{25}{11}, \]\n\n\[ \n{x}_{3} = - \frac{1}{5}... | Yes |
Example 2 Express the Jacobi iteration method for the linear system \( A\\mathbf{x} = \\mathbf{b} \) given by\n\n\[ \n{E}_{1} : {10}{x}_{1} - {x}_{2} + 2{x}_{3}\\; = 6,\n\]\n\n\[ \n{E}_{2} : \\; - {x}_{1} + {11}{x}_{2} - \\;{x}_{3} + 3{x}_{4} = {25},\n\]\n\n\[ \n{E}_{3} : \\;2{x}_{1} - \\;{x}_{2} + {10}{x}_{3} - \\;{x}... | Solution We saw in Example 1 that the Jacobi method for this system has the form\n\n\[ \n{x}_{1} = \\;\\frac{1}{10}{x}_{2} - \\frac{1}{5}{x}_{3}\\; + \\frac{3}{5},\n\]\n\n\[ \n{x}_{2} = \\frac{1}{11}{x}_{1}\\; + \\frac{1}{11}{x}_{3} - \\frac{3}{11}{x}_{4} + \\frac{25}{11},\n\]\n\n\[ \n{x}_{3} = - \\frac{1}{5}{x}_{1} + ... | Yes |
Use the Gauss-Seidel iterative technique to find approximate solutions to\n\n\\[ \n{10}{x}_{1} - {x}_{2} + 2{x}_{3}\; = 6, \n\\]\n\n\\[ \n- {x}_{1} + {11}{x}_{2} - \;{x}_{3} + 3{x}_{4} = {25} \n\\]\n\n\\[ \n2{x}_{1} - {x}_{2} + {10}{x}_{3} - {x}_{4} = - {11} \n\\]\n\n\\[ \n3{x}_{2} - {x}_{3} + 8{x}_{4} = {15} \n\\]\n\n... | Solution The solution \\( \\mathbf{x} = {\\left( 1,2, - 1,1\\right) }^{t} \\) was approximated by Jacobi’s method in Example 1. For the Gauss-Seidel method we write the system, for each \\( k = 1,2,\\ldots \\) as\n\n\\[ \n{x}_{1}^{\\left( k\\right) } = \\;\\frac{1}{10}{x}_{2}^{\\left( k - 1\\right) } - \\frac{1}{5}{x}_... | Yes |
Lemma 7.18 If the spectral radius satisfies \( \rho \left( T\right) < 1 \), then \( {\left( I - T\right) }^{-1} \) exists, and | Proof Because \( T\mathbf{x} = \lambda \mathbf{x} \) is true precisely when \( \left( {I - T}\right) \mathbf{x} = \left( {1 - \lambda }\right) \mathbf{x} \), we have \( \lambda \) as an eigenvalue of \( T \) precisely when \( 1 - \lambda \) is an eigenvalue of \( I - T \) . But \( \left| \lambda \right| \leq \rho \left... | Yes |
Theorem 7.19 For any \( {\mathbf{x}}^{\left( 0\right) } \in {\mathbb{R}}^{n} \), the sequence \( {\left\{ {\mathbf{x}}^{\left( k\right) }\right\} }_{k = 0}^{\infty } \) defined by\n\n\[ \n{\mathbf{x}}^{\left( k\right) } = T{\mathbf{x}}^{\left( k - 1\right) } + \mathbf{c},\;\text{ for each }k \geq 1, \n\]\n\n(7.11)\n\nc... | Proof First assume that \( \rho \left( T\right) < 1 \) . Then,\n\n\[ \n{\mathbf{x}}^{\left( k\right) } = T{\mathbf{x}}^{\left( k - 1\right) } + \mathbf{c} \n\]\n\n\[ \n= T\left( {T{\mathbf{x}}^{\left( k - 2\right) } + \mathbf{c}}\right) + \mathbf{c} \n\]\n\n\[ \n= {T}^{2}{\mathbf{x}}^{\left( k - 2\right) } + \left( {T ... | Yes |
The linear system \( A\mathbf{x} = \mathbf{b} \) given by\n\n\[ 4{x}_{1} + 3{x}_{2}\; = {24} \]\n\n\[ 3{x}_{1} + 4{x}_{2} - {x}_{3} = {30} \]\n\n\[ \text{-}{x}_{2} + 4{x}_{3} = - {24}\text{,} \]\n\nhas the solution \( {\left( 3,4, - 5\right) }^{t} \) . Compare the iterations from the Gauss-Seidel method and the SOR met... | Solution For each \( k = 1,2,\ldots \), the equations for the Gauss-Seidel method are\n\n\[ {x}_{1}^{\left( k\right) } = - {0.75}{x}_{2}^{\left( k - 1\right) } + 6 \]\n\n\[ {x}_{2}^{\left( k\right) } = - {0.75}{x}_{1}^{\left( k\right) } + {0.25}{x}_{3}^{\left( k - 1\right) } + {7.5}, \]\n\n\[ {x}_{3}^{\left( k\right) }... | Yes |
Theorem 7.26 If \( A \) is positive definite and tridiagonal, then \( \rho \left( {T}_{g}\right) = {\left\lbrack \rho \left( {T}_{j}\right) \right\rbrack }^{2} < 1 \), and the optimal choice of \( \omega \) for the SOR method is\n\n\[ \omega = \frac{2}{1 + \sqrt{1 - {\left\lbrack \rho \left( {T}_{j}\right) \right\rbrac... | With this choice of \( \omega \), we have \( \rho \left( {T}_{\omega }\right) = \omega - 1 \) | No |
Find the optimal choice of \( \omega \) for the SOR method for the matrix\n\n\[ A = \left\lbrack \begin{array}{rrr} 4 & 3 & 0 \\ 3 & 4 & - 1 \\ 0 & - 1 & 4 \end{array}\right\rbrack \] | Solution This matrix is clearly tridiagonal, so we can apply the result in Theorem 7.26 if we can also show that it is positive definite. Because the matrix is symmetric, Theorem 6.24 on page 416 states that it is positive definite if and only if all its leading principle submatrices has a positive determinant. This is... | Yes |
The linear system \( A\mathbf{x} = \mathbf{b} \) given by\n\n\[ \left\lbrack \begin{array}{ll} 1 & 2 \\ {1.0001} & 2 \end{array}\right\rbrack \left\lbrack \begin{array}{l} {x}_{1} \\ {x}_{2} \end{array}\right\rbrack = \left\lbrack \begin{array}{l} 3 \\ {3.0001} \end{array}\right\rbrack \]\n\nhas the unique solution \( ... | Solution We have\n\n\[ \mathbf{r} = \mathbf{b} - A\widetilde{\mathbf{x}} = \left\lbrack \begin{array}{l} 3 \\ {3.0001} \end{array}\right\rbrack - \left\lbrack \begin{array}{ll} 1 & 2 \\ {1.0001} & 2 \end{array}\right\rbrack \left\lbrack \begin{matrix} 3 \\ - {0.0001} \end{matrix}\right\rbrack = \left\lbrack \begin{arra... | Yes |
Theorem 7.27 Suppose that \( \widetilde{\mathbf{x}} \) is an approximation to the solution of \( A\mathbf{x} = \mathbf{b}, A \) is a nonsingular matrix, and \( \mathbf{r} \) is the residual vector for \( \widetilde{\mathbf{x}} \) . Then for any natural norm,\n\n\[ \parallel \mathbf{x} - \widetilde{\mathbf{x}}\parallel ... | Proof Since \( \mathbf{r} = \mathbf{b} - A\widetilde{\mathbf{x}} = A\mathbf{x} - A\widetilde{\mathbf{x}} \) and \( A \) is nonsingular, we have \( \mathbf{x} - \widetilde{\mathbf{x}} = {A}^{-1}\mathbf{r} \) . Theorem 7.11 on page 440 implies that\n\n\[ \parallel \mathbf{x} - \widetilde{\mathbf{x}}\parallel = \begin{Vma... | Yes |
Determine the condition number for the matrix\n\n\[ A = \left\lbrack \begin{array}{ll} 1 & 2 \\ {1.0001} & 2 \end{array}\right\rbrack \] | Solution We saw in Example 1 that the very poor approximation \( {\left( 3, - {0.0001}\right) }^{t} \) to the exact solution \( {\left( 1,1\right) }^{t} \) had a residual vector with small norm, so we should expect the condition number of \( A \) to be large. We have \( \parallel A{\parallel }_{\infty } = \max \{ \left... | Yes |
Use Maple to find the eigenvalues and condition number of the matrix\n\n\[ A = \left\lbrack \begin{matrix} {0.2} & {0.1} & 1 & 1 & 0 \\ {0.1} & 4 & - 1 & 1 & - 1 \\ 1 & - 1 & {60} & 0 & - 2 \\ 1 & 1 & 0 & 8 & 4 \\ 0 & - 1 & - 2 & 4 & {700} \end{matrix}\right\rbrack \]\n\nand compare these with the eigenvalues and condi... | Solution We first need to load the LinearAlgebra package and then enter the matrix.\n\nwith(LinearAlgebra):\n\n\( A \mathrel{\text{:=}} \operatorname{Matrix}(\left\lbrack {\left\lbrack {{0.2},{0.1},1,1,0}\right\rbrack ,\left\lbrack {{0.1},4, - 1,1, - 1}\right\rbrack ,\lbrack 1, - 1,{60},0, - 2}\right\rbrack \),\n\n\( \... | Yes |
Find the least squares line approximating the data in Table 8.1. | Solution We first extend the table to include \( {x}_{i}^{2} \) and \( {x}_{i}{y}_{i} \) and sum the columns. This is shown in Table 8.2.\n\nTable 8.2\n\n<table><thead><tr><th>\( {x}_{i} \)</th><th>\( {y}_{i} \)</th><th>\( {x}_{i}^{2} \)</th><th>\( {x}_{i}{y}_{i} \)</th><th>\( P\left( {x}_{i}\right) = {1.538}{x}_{i} - ... | Yes |
Fit the data in Table 8.3 with the discrete least squares polynomial of degree at most 2. | Solution For this problem, \( n = 2, m = 5 \), and the three normal equations are\n\nTable 8.3\n\n<table><thead><tr><th>\( i \)</th><th>\( {x}_{i} \)</th><th>\( {\mathcal{Y}}_{i} \)</th></tr></thead><tr><td>1</td><td>0</td><td>1.0000</td></tr><tr><td>2</td><td>0.25</td><td>1.2840</td></tr><tr><td>3</td><td>0.50</td><td... | Yes |
Find the least squares approximating polynomial of degree 2 for the function \( f\left( x\right) = \sin {\pi x} \) on the interval \( \left\lbrack {0,1}\right\rbrack \) . | The normal equations for \( {P}_{2}\left( x\right) = {a}_{2}{x}^{2} + {a}_{1}x + {a}_{0} \) are\n\n\[ \n{a}_{0}{\int }_{0}^{1}{1dx} + {a}_{1}{\int }_{0}^{1}{xdx} + {a}_{2}{\int }_{0}^{1}{x}^{2}{dx} = {\int }_{0}^{1}\sin {\pi xdx} \n\]\n\n\[ \n{a}_{0}{\int }_{0}^{1}{xdx} + {a}_{1}{\int }_{0}^{1}{x}^{2}{dx} + {a}_{2}{\in... | Yes |
Theorem 8.2 Suppose that, for each \( j = 0,1,\ldots, n,{\phi }_{j}\left( x\right) \) is a polynomial of degree \( j \) . Then \( \left\{ {{\phi }_{0},\ldots ,{\phi }_{n}}\right\} \) is linearly independent on any interval \( \left\lbrack {a, b}\right\rbrack \) . | Proof Let \( {c}_{0},\ldots ,{c}_{n} \) be real numbers for which\n\n\[ P\left( x\right) = {c}_{0}{\phi }_{0}\left( x\right) + {c}_{1}{\phi }_{1}\left( x\right) + \cdots + {c}_{n}{\phi }_{n}\left( x\right) = 0,\;\text{ for all }x \in \left\lbrack {a, b}\right\rbrack . \]\n\nThe polynomial \( P\left( x\right) \) vanishe... | Yes |
Example 2 Let \( {\phi }_{0}\left( x\right) = 2,{\phi }_{1}\left( x\right) = x - 3 \), and \( {\phi }_{2}\left( x\right) = {x}^{2} + {2x} + 7 \), and \( Q\left( x\right) = {a}_{0} + {a}_{1}x + {a}_{2}{x}^{2} \) . Show that there exist constants \( {c}_{0},{c}_{1} \), and \( {c}_{2} \) such that \( Q\left( x\right) = {c... | Solution By Theorem 8.2, \( \left\{ {{\phi }_{0},{\phi }_{1},{\phi }_{2}}\right\} \) is linearly independent on any interval \( \left\lbrack {a, b}\right\rbrack \) . First note that\n\n\[ 1 = \frac{1}{2}{\phi }_{0}\left( x\right) ,\;x = {\phi }_{1}\left( x\right) + 3 = {\phi }_{1}\left( x\right) + \frac{3}{2}{\phi }_{0... | Yes |
Theorem 8.7 The set of polynomial functions \( \\left\\{ {{\\phi }_{0},{\\phi }_{1},\\ldots ,{\\phi }_{n}}\\right\\} \) defined in the following way is orthogonal on \( \\left\\lbrack {a, b}\\right\\rbrack \) with respect to the weight function \( w \) . | where\n\n---\n\nErhard Schmidt (1876-1959) received his doctorate under the supervision of David Hilbert in 1905 for a problem involving integral equations. Schmidt published a paper in 1907 in which he gave what is now called the Gram-Schmidt process for constructing an orthonormal basis for a set of functions. This g... | Yes |
Theorem 8.9 The Chebyshev polynomial \( {T}_{n}\left( x\right) \) of degree \( n \geq 1 \) has \( n \) simple zeros in \( \left\lbrack {-1,1}\right\rbrack \) at\n\n\[ \n{\bar{x}}_{k} = \cos \left( {\frac{{2k} - 1}{2n}\pi }\right) ,\;\text{ for each }k = 1,2,\ldots, n.\n\]\n\nMoreover, \( {T}_{n}\left( x\right) \) assum... | Proof Let\n\n\[ \n{\bar{x}}_{k} = \cos \left( {\frac{{2k} - 1}{2n}\pi }\right) ,\;\text{ for }k = 1,2,\ldots, n.\n\]\n\nThen\n\n\[ \n{T}_{n}\left( {\bar{x}}_{k}\right) = \cos \left( {n\arccos {\bar{x}}_{k}}\right) = \cos \left( {n\arccos \left( {\cos \left( {\frac{{2k} - 1}{2n}\pi }\right) }\right) }\right) = \cos \lef... | No |
Theorem 8.10 The polynomials of the form \( {\widetilde{T}}_{n}\left( x\right) \), when \( n \geq 1 \), have the property that | Proof Suppose that \( {P}_{n}\left( x\right) \in {\widetilde{\prod }}_{n} \) and that\n\n\[ \mathop{\max }\limits_{{x \in \left\lbrack {-1,1}\right\rbrack }}\left| {{P}_{n}\left( x\right) }\right| \leq \frac{1}{{2}^{n - 1}} = \mathop{\max }\limits_{{x \in \left\lbrack {-1,1}\right\rbrack }}\left| {{\widetilde{T}}_{n}\l... | Yes |
Example 1 Let \( f\left( x\right) = x{e}^{x} \) on \( \left\lbrack {0,{1.5}}\right\rbrack \) . Compare the values given by the Lagrange polynomial with four equally-spaced nodes with those given by the Lagrange polynomial with nodes given by zeros of the fourth Chebyshev polynomial. | Solution The equally-spaced nodes \( {x}_{0} = 0,{x}_{1} = {0.5},{x}_{2} = 1 \), and \( {x}_{3} = {1.5} \) give\n\n\[ \n{L}_{0}\left( x\right) = - {1.3333}{x}^{3} + {4.0000}{x}^{2} - {3.6667x} + 1, \n\] \n\n\[ \n{L}_{1}\left( x\right) = {4.0000}{x}^{3} - {10.000}{x}^{2} + {6.0000x}, \n\] \n\n\[ \n{L}_{2}\left( x\right)... | Yes |
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