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Show that the set {$\cos x, \cos 2x,\cos 3x,- $ } is orthogonal set over $(-\pi,\pi)$, Hence construct orthonormal set

Mumbai university > Electronics and telecommunication engineering, Electronics engineering > Sem 3 > Applied mathematics 3

Marks : 06

Years : MAY 2016

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Let $fm(x) =\cos mx, fn(x) =\cos nx $

  1. For $m\neq n \\ \int\limits_{-\pi}^{\pi}fm (x) fn(x) dx = \int\limits_{-\pi}^{\pi} \cos mx \cos nx dx=\dfrac 12\int\limits_{-\pi}^{\pi}[\cos(m+n)x+\cos(m-n)x]dx \\ =\dfrac 12[\dfrac {\sin (m+n)x}{(m+n)}+\dfrac {\sin(m-n)x}{(m-n)}]_{-\pi}^{\pi} =0 $

  2. For $m=n\\ \int\limits_{-\pi}^{\pi}[fn(x)]^2 dx =\int\limits_{-\pi}^{\pi} \cos^2(nx) dx=2 \int\limits_{0}^{\pi} \dfrac {[1+\cos(2nx)]}2 dx \\ =[x+\dfrac {\sin (2nx)}{2n}]_0^\pi = \pi\neq 0 $

Hence $[\cos x,\cos 2x , -]$ is orthogonal set

For orthonormal set

$\int\limits_{-\pi}^{\pi} \dfrac 1{\sqrt\pi}fn(x). \dfrac 1{\sqrt{\pi}}fn(x) dx=1 \\ \Rightarrow [\dfrac 1{\sqrt\pi}\cos x,\dfrac 1{\sqrt\pi}\cos 2x -]$

is orthonormal set

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