Fourier series problem solving for an,bn

[dp182]

Homework Statement

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

Homework Equations

a₀=1/L$\int$f(x)dx
a_n=1/L$\int$f(x)cos(n$\pi$x/L)
b_n=1/L$\int$f(x)sin(n$\pi$x/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$\int$xcos(n$\pi$x/L) for the interval 0$\leq$x$\leq$2 and I get the solution
=(1/n²$\pi$²)2xn$\pi$sin(n$\pi$x/2)+4cos(n$\pi$x/2) then subbing in for x I get (1/n²$\pi$²)(4n$\pi$sin(n$\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$\leq$x$\leq$0.
x;0$\leq$x$\leq$2
find a₀
a_n
b_n
given the period is 4

Homework Equations

a₀=1/L$\int$f(x)dx
a_n=1/L$\int$f(x)cos(n$\pi$x/L)
b_n=1/L$\int$f(x)sin(n$\pi$x/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$\int$xcos(n$\pi$x/L) for the interval 0$\leq$x$\leq$2 and I get the solution
=(1/n²$\pi$²)2xn$\pi$sin(n$\pi$x/2)+4cos(n$\pi$x/2) then subbing in for x I get (1/n²$\pi$²)(4n$\pi$sin(n$\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