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

Marks : 06

Years : MAY 2016

Half range cosine series of f(x) is

$f(x) =a_0 + \sum\limits_{n=1}^\infty a_n \cos(nx) \\ a_0 =\dfrac 1\pi \int\limits_0^{\pi} f(x) dx =\dfrac 1\pi[\int\limits_0^{\pi/2} x\space dx \int\limits^\pi_{\pi/2} (\pi-x)dx] =\dfrac 1\pi [(\dfrac {x^2}2)^{\pi/2}_0 + (\pi x -\dfrac {x^2}2)]^\pi_{\pi/2} \\ =\dfrac 1\pi [\dfrac {\pi^2}8 + \pi^2-\dfrac {\pi^2}2 -\dfrac {\pi^2}2 + \dfrac {\pi^2}8] =\dfrac \pi 4 \\ a_n =\dfrac 2\pi \int\limits_0^{\pi} f(x) \cos nx \space dx =\dfrac 2\pi [\int\limits_0^{\pi/2} x\cos nx \space dx + \int\limits_{\pi/2}^\pi(\pi-x) \cos nx\space dx ] \\ =\dfrac 2\pi[[x\dfrac {\sin nx}n + \dfrac {\cos nx}{n^2}]_0^{\pi/2} + [(\pi-x)\dfrac {\sin nx}n + (-1) \dfrac {\cos nx}{n^2}]^{\pi}_{\pi/2}] \\ =\dfrac 2{\pi n^2}[2\cos(n\pi/2)-(-1)^n -1]\\ f(x) =\dfrac {\pi}4 + \dfrac 2\pi \sum\limits_{n=1}^{\infty} \dfrac {[2\cos(n\pi/2) -(-1)^n -1]}{n^2}\cos(nx)$