Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
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See our solution for Question 12E from Chapter 6.8 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 12E
Chapter:
Problem:
0
Step-by-Step Solution
Given Information
We have to find the third-order Fourier approximation to $f ( t ) = \cos ^ { 3 } t$ without performing any integration calculations.
Step-1:
We will use the entity $\cos 2 A = 2 \cos ^ { 2 } A - 1$\[\begin{array}{l}{\cos ^3}t = {\cos ^2}t(\cos t)\\ = \left( {\dfrac{{1 + \cos 2t}}{2}} \right)\cos t:\,\,\,\,\,\,{\rm{Since}}\left\{ {\cos 2A = 2{{\cos }^2}A - 1} \right\}\\ = \dfrac{1}{2}(\cos t + \cos 2t\cos t)\\ = \dfrac{1}{2}\cos t + \dfrac{1}{2}\cos 2t\cos t\\ = \dfrac{1}{2}\cos t + \dfrac{1}{4}(2\cos 2t\cos t)\\ = \dfrac{1}{2}\cos t + \dfrac{1}{4}(\cos 3t + \cos t)\\ = \dfrac{2}{4}\cos t + \dfrac{1}{4}\sin 3t\end{array}\]Therefore, the Fourier approximation is
\[{\cos ^3}t = \dfrac{2}{4}\cos t + \dfrac{1}{4}\sin 3t\]
We have to find the third-order Fourier approximation to $f ( t ) = \cos ^ { 3 } t$ without performing any integration calculations.
Step-1:
We will use the entity $\cos 2 A = 2 \cos ^ { 2 } A - 1$\[\begin{array}{l}{\cos ^3}t = {\cos ^2}t(\cos t)\\ = \left( {\dfrac{{1 + \cos 2t}}{2}} \right)\cos t:\,\,\,\,\,\,{\rm{Since}}\left\{ {\cos 2A = 2{{\cos }^2}A - 1} \right\}\\ = \dfrac{1}{2}(\cos t + \cos 2t\cos t)\\ = \dfrac{1}{2}\cos t + \dfrac{1}{2}\cos 2t\cos t\\ = \dfrac{1}{2}\cos t + \dfrac{1}{4}(2\cos 2t\cos t)\\ = \dfrac{1}{2}\cos t + \dfrac{1}{4}(\cos 3t + \cos t)\\ = \dfrac{2}{4}\cos t + \dfrac{1}{4}\sin 3t\end{array}\]Therefore, the Fourier approximation is
\[{\cos ^3}t = \dfrac{2}{4}\cos t + \dfrac{1}{4}\sin 3t\]