Question

Find the fourier series of L(x) = 3(x+1) for -2 <= x <= 2.

I got my ao = 3

my an = (3x)/(nPi) * sin(nPix/2)

my bn = (-3x)/(nPi) * cos(nPix/2)

but it seems to be wrong... Help please?

Answer 1

you have to simplify the values of the sin and cos

Answer 2

cos(n pi / 2 ) = 0

Answer 3

sin( n pi/2 ) = (-1)^n

Answer 4

?
My final answer was...

3+ 3x/nPi \[\sum_{n=1}^{infinity} \sin(nPix/2) \cos(nPix/2) - \cos(nPix/2)\sin(nPix/2)\]

and so the cos*sin muinues eachother.. and I was left with only the 3 at the start... and the final answer according to the book is..

3+ \[\[\sum_{n=1}^{infinity} (-12/nPi) * \cos(nPi) * Sin(nPix/2)\]]