# 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
Homework Helper
I figured...
For my b_n I got 2/(n*pi) [1 - cos(n*pi / 2 ).

Is this correct?
Why don't you do as suggested and graph the resulting series? That'll tell you immediately if you got the right series. If you still want someone to check your work, you need to show it. Just posting the answer is next to useless.