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: 39


(a) $\langle f, g\rangle =0$ (b) $\langle f, f\rangle=\sqrt 2$ (c) $\| g \| = \sqrt \frac{8}{5}$ (d) $d(f,g)= \sqrt\frac{18}{5}$

Work Step by Step

$f(x)=1, \quad q(x)=3x^2-1.$ (a) $\langle f, g\rangle=\int_{-1}^{1}f(x)g(x) d x=\int_{-1}^{1} (3 x^{2}-1) d x=\left[x^{3}-x\right]_{-1}^{1}=0$ (b) $\langle f, f\rangle=\int_{-1}^{1} 1 d x=\left.x\right|_{-1} ^{1}=2 \Longrightarrow\| f \| =\sqrt{\langle f, f\rangle}=\sqrt 2$ (c) \begin{aligned} \langle g, g\rangle &=\int_{-1}^{1}\left(3 x^{2}-1\right)^{2} d x \\ &=\int_{-1}^{1} 9 x^{4}-6 x^{2}+1 d x \\ &=\frac{9}{5} x^{5}-2 x^{3}+\left.x\right|_{-1} ^{1} \\ &=\frac{8}{5}. \end{aligned} Hence, we have $\| g \| =\sqrt{\langle g, g\rangle}=\sqrt \frac{8}{5}$ (d) \begin{aligned}\langle f-g, f-g\rangle &=\int_{-1}^{1}\left(1-\left(3 x^{2}-1\right)\right)^{2} d x \\ &=\int_{-1}^{1}\left(-3 x^{2}+2\right)^{2} d x \\ &=\int_{-1}^{1} 9 x^{4}-12 x^{2}+4 d x \\ &=\frac{9}{5} x^{5}-4 x^{3}+\left.4 x\right|_{-1} ^{1} \\ &=\frac{18}{5} \end{aligned}. Hence, we have $d(f,g)=\| f-g \|= \sqrt{\langle f-g, f-g\rangle}=\sqrt\frac{18}{5}$
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