Fourier series problem solving for an,bn

[dp182]

Homework Statement

let f(x)={0;-2\leqx\leq0.
x;0\leqx\leq2
find a₀
a_n
b_n
given the period is 4

Homework Equations

a₀=1/L\intf(x)dx
a_n=1/L\intf(x)cos(n\pix/L)
b_n=1/L\intf(x)sin(n\pix/L)

The Attempt at a Solution

so I can get a₀ = 1 but I run into trouble with a_n. so I plug
a_n=1/2\intxcos(n\pix/L) for the interval 0\leqx\leq2 and I get the solution
=(1/n²\pi²)2xn\pisin(n\pix/2)+4cos(n\pix/2) then subbing in for x I get (1/n²\pi²)(4n\pisin(n\pi)+4cos(n\pi)-1) can anyone tell me what I am doing wrong here?

 

[micromass]

Hi dp182!

Homework Statement

let f(x)={0;-2\leqx\leq0.
x;0\leqx\leq2
find a₀
a_n
b_n
given the period is 4

Homework Equations

a₀=1/L\intf(x)dx
a_n=1/L\intf(x)cos(n\pix/L)
b_n=1/L\intf(x)sin(n\pix/L)

The Attempt at a Solution

so I can get a₀ = 1 but I run into trouble with a_n. so I plug
a_n=1/2\intxcos(n\pix/L) for the interval 0\leqx\leq2 and I get the solution
=(1/n²\pi²)2xn\pisin(n\pix/2)+4cos(n\pix/2) then subbing in for x I get (1/n²\pi²)(4n\pisin(n\pi)+4cos(n\pi)-1) can anyone tell me what I am doing wrong here?

You made two mistakes: you forgot the term 1/L, so you still need to divide by 2. Secondly, when you "subbed for x", you made a mistake in "subbing for 0". That is, the correct formula is

\int_0^2{f(x)dx}=F(2)-F(0)

But somehow, you calculated F(0) wrong.

 

[dp182]

but wouldn't F(0) be 4cos(0)/n²pi² which is just 4/n²pi²

 

[micromass]

but wouldn't F(0) be 4cos(0)/n²pi² which is just 4/n²pi²

Indeed, but you wrote -1 instead of -4 in the end.

 

[dp182]

ya but I brought out a 4/n²pi² so its not -1 its -4/n²pi²