Trouble determining the Fourier Cosine series for a Function

[Arthur Yeh]

Homework Statement

I am only interested in 9 (a)
Determine the Fourier Cosine series of the function g(x) = x(L-x) for 0 < x < L

Homework Equations

The Answer for 9 a.
g(x) = (L^2)/6 - ∑(L^2/(nπ)^2)cos(2nπx/L)

This is the relevant equation given where ω=π/L
f(t) = a₀+∑a_ncos(nωt)
a₀=1/L ∫f(t) dt from 0 to L
a_n=2/L∫f(t)cos(nωt) dt from 0 to L

The Attempt at a Solution

This is my attempt at the solution
g(x) = a₀ - Σa_ncos(nωx)
where
a₀=L^2/6
a_n= -2L^2[(cos(nπ)+1)/(nπ)^2]
I have double checked this answer both manually and through the use of an online integral calculator and i still arrive at this conclusion. As a result I believe my partial integrations are correct but my answer is in the wrong form. Initially I tried changing cos(nπ) to (-1)^n but i didnt get anywhere as i didnt couldn't get rid of the n power. I also tried working backwards from the answer using some double angle identities but did not arrive at any recognizable form.

 

[Gene Naden]

Sure... I believe your integrals are correct. If you write out the first few of the a's you should be able to see that every other one goes to zero because cos(n*pi) is minus one and cos(n*pi)-1 is zero. Only the even n's count. For the even n's, let n=2*i. Then you have a sum from i=1,2,3,4...

 

[Gene Naden]

I meant to say cos(n*pi)+1 is zero

 

[Arthur Yeh]

Thanks alot!
I didnt think to look more closely at its behaviour for odd and even n's

 

[Ray Vickson]

Homework Statement

I am only interested in 9 (a)
Determine the Fourier Cosine series of the function g(x) = x(L-x) for 0 < x < L
View attachment 220239

Homework Equations

The Answer for 9 a.
g(x) = (L^2)/6 - ∑(L^2/(nπ)^2)cos(2nπx/L)
View attachment 220240

This is the relevant equation given where ω=π/L
f(t) = a₀+∑a_ncos(nωt)
a₀=1/L ∫f(t) dt from 0 to L
a_n=2/L∫f(t)cos(nωt) dt from 0 to L
View attachment 220242

The Attempt at a Solution

This is my attempt at the solution
g(x) = a₀ - Σa_ncos(nωx)
where
a₀=L^2/6
a_n= -2L^2[(cos(nπ)+1)/(nπ)^2]
I have double checked this answer both manually and through the use of an online integral calculator and i still arrive at this conclusion. As a result I believe my partial integrations are correct but my answer is in the wrong form. Initially I tried changing cos(nπ) to (-1)^n but i didnt get anywhere as i didnt couldn't get rid of the n power. I also tried working backwards from the answer using some double angle identities but did not arrive at any recognizable form.
View attachment 220238
View attachment 220237

You are developing a very bad habit: posting images instead of typing out your work. Most helpers will not be bothered to look at your images, so will not want to help. (See the post "Guidelines for students and helpers", by Vela, pinned to the start of this forum.)