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: 84


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

Work Step by Step

Let $f(x)=x, \quad g(x)=\cos 2x$, $\langle f, g\rangle=\int_{-\pi}^{\pi}f(x)g(x) d x$ We have $$\langle f, g\rangle=\int_{-\pi}^{\pi}x \cos 2x d x=\left[\frac{x}{2}\sin 2x+\frac{1}{4} \cos 2x\right]_{-\pi}^{\pi}=0,$$ $$\langle g, g\rangle=\int_{-\pi}^{\pi}\cos^2 2x d x=\frac{1}{2}\int_{-\pi}^{\pi}(1+\cos 4x) d x=\frac{1}{2}\left[x+\frac{1}{4}\sin 4x\right]_{-\pi}^{\pi}=\pi.$$ Now, the orthogonal projection of $f$ onto $g$ is given by $$\operatorname{proj}_{{g}} {f} =\frac{\langle{f}, {g}\rangle}{\langle{g}, {g}\rangle} {g}=0.$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.