Elementary Linear Algebra 7th Edition

Published by Cengage Learning
ISBN 10: 1-13311-087-8
ISBN 13: 978-1-13311-087-3

Chapter 5 - Inner Product Spaces - 5.2 Inner Product Spaces - 5.2 Exercises - Page 245: 41


(a) $\langle f, g\rangle=\frac{2}{e}$ (b) $\langle f, f\rangle=\sqrt \frac{2}{3}$ (c) $\| g \| =\sqrt {\frac{1}{2}(e^2-e^{-2})}$ (d) $d(f,g)=\sqrt{\frac{2}{3}-\frac{4}{e}+\frac{e^2}{2}-\frac{1}{2e^2}}$

Work Step by Step

$f(x)=x, \quad q(x)=e^x.$ (a) $\langle f, g\rangle=\int_{-1}^{1}f(x)g(x) d x=\int_{-1}^{1} xe^x d x=\left[xe^x-e^x\right]_{-1}^{1}=\frac{2}{e}$ (b) $\langle f, f\rangle=\int_{-1}^{1} x^2 d x=\left.\frac{1}{3}x^3\right|_{-1} ^{1}=\frac{2}{3}\Longrightarrow\| f \| =\sqrt{\langle f, f\rangle}=\sqrt \frac{2}{3}$ (c) \begin{aligned} \langle g, g\rangle &=\int_{-1}^{1}e^{2x} d x \\ &=\frac{1}{2}\left.e^{2x} \right|_{-1} ^{1} \\ &=\frac{1}{2}(e^2-e^{-2}). \end{aligned} Hence, we have $\| g \| =\sqrt{\langle g, g\rangle}=\sqrt {\frac{1}{2}(e^2-e^{-2})}$ (d) \begin{aligned}\langle f-g, f-g\rangle &=\int_{-1}^{1}\left(x-e^x\right)^{2} d x \\ &=\int_{-1}^{1}\left(x^{2}+e^{2x}-2xe^x\right) d x \\ &=\left.\frac{1}{3} x^{3}+\frac{1}{2}e^{2x}-xe^x+e^x\right|_{-1} ^{1} \\ &=\frac{2}{3}-\frac{4}{e}+\frac{e^2}{2}-\frac{1}{2e^2}\end{aligned}. Hence, we have $d(f,g)=\| f-g \|= \sqrt{\langle f-g, f-g\rangle}=\sqrt{\frac{2}{3}-\frac{4}{e}+\frac{e^2}{2}-\frac{1}{2e^2}}$
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