Q2 )Expand f(x) = sin x, -π/2 <x <π/2, in a Fourier series if the period is π.
hope got ppl can help me, thx~~~~~ T_T_T_T_T_T_T
Just like the other one:
The Fourier Series periodic with period 2L, L>0 is defined to be:
(a_0)/2 + sum_{n=1 to infty} [a_n cos(n pi x/L) + b_n sin(n pi x/L)]
where:
a_n = (1/L) integral_{x=-L to L} f(x) cos(n pi x/L) dx
b_n = (1/L) integral_{x=-L to L} f(x) sin(n pi x/L) dx
Now consider f(x) periodic with period pi, where:
f(x) = sin(x), -pi/2<x<pi/2
Now f(x) is odd so all the integrals for the "a" terms are zero. So lets now
consider the "b" terms, here L=pi/2 so:
b_n = (2/pi) integral_{x=-pi/2 to pi/2} sin(x) sin(2 n x) dx
Which I hope I can leave to you.
RonL