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 246: 81


$$\operatorname{proj}_{{g}} {f} =0.$$

Work Step by Step

Let $f(x)=\sin x, \quad g(x)=\cos x$, $\langle f, g\rangle=\int_{-\pi}^{\pi}f(x)g(x) d x$ We have $$\langle f, g\rangle=\int_{-\pi}^{\pi}\sin x\cos x d x=\frac{1}{2}\left[\sin^{2}x\right]_{-\pi}^{\pi}=0,$$ $$\langle g, g\rangle=\int_{-\pi}^{\pi}\cos^2 x d x=\frac{1}{2}\int_{-\pi}^{\pi}(1+\cos 2x) d x=\frac{1}{2}\left[x+\frac{1}{2}\sin 2x\right]_{-\pi}^{\pi}=\pi.$$ Now, the orthogonal projection of $f$ onto $g$ is given by $$\operatorname{proj}_{{g}} {f} =\frac{\langle{f}, {g}\rangle}{\langle{g}, {g}\rangle} {g}=0.$$
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