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

Marks : 05

Years : MAY 2016

$f(-x) =f(x) \Rightarrow f(x) $ is even $F^n$ & bn=0, l=3 $$a_0= \dfrac 1l \int\limits_0^l f(x) dx =\dfrac 13 \int\limits_0^2(9-x^2) dx=\dfrac 13[9x-\dfrac {x^3}3]^3_0 =\dfrac 13 [18]=6 $$

$an=\dfrac 2l \int\limits_0^l f(x) \cos (\dfrac {n\pi x}l)dx =\dfrac 23 \int\limits_0^3 (9-x^2)\cos (\dfrac n\pi x)3 dx \\ =\dfrac 23[(9-x^2) \dfrac {\sin(\dfrac {n\pi x}3)}{ (\dfrac {n\pi}3)} -(-2x) [\dfrac {-\cos(n\pi x/3)}{n^2x^2/9}]+(-2)[\dfrac {-sm(n\pi x/3)}{n^3x^3/27}]]^3_0 \\ =\dfrac 23[-\dfrac {18}{n^2\pi^2}[3(-1)^n-0]] =\dfrac {-36(-1)^n}{n^2\pi^2} \\ f(x) =6-\dfrac {36}{\pi^2} \sum \limits_{n=1}^\infty \dfrac {(-1)^n}{n^2}\cos (\dfrac {n\pi x}3)$