# Can somebody check my work on this Fourier Series problem?

## The Attempt at a Solution

Since P=2L, L=1 ?

a_o = 1/2 [ ∫(from -1 to 0) -dx + ∫(from 0 to 1) dx ] = 1/2 [ (0-1) + (1-0) ] = 1/2(0) = 0

a_n = - ∫ (from -1 to 0) cosnπx dx + ∫ (from 0 to 1) cosnπx dx = 0

b_n = - ∫ (from -1 to 0) sinnπx dx + (from 0 to 1) sinnπx dx = 2/nπ - 2/nπ*(cosnπ) = 2nπ / (1-cosnπ) = { 4/nπ n is odd ; 0 is even

The problem is my teacher has this as his answer:

Am i doing something incorrectly?

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Dr. Courtney
Gold Member
Why not check by graphing the original function and the Fourier series?

Why not check by graphing the original function and the Fourier series?
I understand you can check through graphing, but I just wanted some verification that my math is correct?

Wait L is suppose to equal 2 correct? Not 1?

Ray Vickson
Homework Helper
Dearly Missed
Wait L is suppose to equal 2 correct? Not 1?
If you mean that one period of the function goes from ##-L## to ##+L## then yes, of course ##L = 2##. That was implied in the question.

Aristotle
If you mean that one period of the function goes from ##-L## to ##+L## then yes, of course ##L = 2##. That was implied in the question.
I figured...
For my b_n I got 2/(n*pi) [1 - cos(n*pi / 2 ).

Is this correct?

vela
Staff Emeritus