Question

Fourier Series Expansion of |sin x| from (-pi, pi)

Answer 1

Can i say that since |sin x| is NOT a single valued function in this interval, FS Expansion cannot be found out ?

Answer 2

it can be found :D

Answer 3

ok, lets start
|sin x| is even function so \(b_n= 0\)
right ?

Answer 4

is it even ?
i thought its odd
lets check

Answer 5

okk its even :D

Answer 6

sin x is odd, so i though |sin x| would be even :P

Answer 7

haha sry :P
ok lets continue
its periodic on 2pi
so ur right b_0 =0

Answer 8

\(\large f(x)=a_0\sum_{n=1}^{\infty} a_n \cos n x + b_n \sin n x \)

Answer 9

ok so \(b_0 =0\)

\(\large f(x)=a_0\sum_{n=1}^{\infty} a_n \cos n x \)

Answer 10

let me show you my work

Answer 11

ohk quick

Answer 12

just check whether i have done any silly mistake

Answer 13

\(\large a_0=\frac{1}{2\pi}\int_{-\pi}^{\pi} \) f(x) dx
since its even then
\(\large a_0=\frac{1}{ \pi}\int_{0}^{\pi} \) f(x) dx
not
\(\large a_0=\frac{2}{ \pi}\int_{0}^{\pi} \) f(x) dx

Answer 14

if f is even

i used that