Question

Help with Fourier series

Answer 1

Arguments for \[f(x)=\cos(3x)\sin^2(x)\] is a 2π-periodic even function?
Can someone help me?

Answer 2

did you know the formula for Ck and Co ???

Answer 3

sin^2(x) = (1 - cos(2x))/2

Answer 4

cos(2x).cos(3x) = cos(..) + cos( ..) hence you have your Fourier series.

Answer 5

Let said I only have to show the function is 2π-periodic. Can I just do it this way?\[f(x+2π)=\cos(3x+2π)\sin^2(x+2π)=f(x)\]

Answer 6

yes ... you can do it that way.

what is the fourier expansion of sin(x) ?

Answer 7

just sin(x)?

Answer 8

I really have trouble with this fourier....

Answer 9

yes .... just express the above expression as
sin(3x) + cos(x)+ cos(5x) ... etc etc ..

Answer 10

I don't see any problem. in fact this is the easiest you can ever get.

sin^2(x) cos(3x)= (1 - cos(2x))/2 cos(3x) = 1/ 2 cos(3x) - 1/2 (cos(2x) cos(3x)
= ... - 1/4 (cos(5x + cos(x)) = -1/4 cos(x) + 1/2 sin(2x) - 1/4 cos(5x)

Answer 11

If the fourier is 2π-periodic then f(x)=f(x+2π)
If the fourier is even then f(x)=f(-x)
Is this right?

Answer 12

yes ... apparently this is neither even nor odd

Answer 13

Why is it not even?
\[\cos(3x)\sin^2(x)=\cos(-3x)\sin^2(-x)\]

Answer 14

woops!! looks like i did not note
-1/4 cos(x) + 1/2 sin(2x) - 1/4 cos(5x)
^^ this one is cos(3x)

Answer 15

1/ 2 cos(3x) - 1/2 (cos(2x) cos(3x)
= 1/2 cos(3x) - 1/4 (cos(5x + cos(x))

Answer 16

thank you @experimentX...

Answer 17

yw