Thread: Fourier sine and cosine series of x(Pi-x)

f(x) = x(π –x) on [0, π]

Find:
a)Odd completion (sine series)
f(x) ~ ∑bnsin(nx) where bn= 2/π 0∫π f(x)sin(nx)dx

b)Even completion (cosine series)
f(x) ~ a0/2 + ∑ancos(nx) where an= 2/π 0∫π f(x)cos(nx)dx anda0/2 = 1/π 0∫π f(x)dx

c) Use the above results to show ∑1/n^2 = π^2 /6

I get:
Odd completion = ∑(2/n^3)(1 – (-1)^n)sin(nx)
Even completion = (π ^2 /6)+ ∑(-2/n^2)(1+(-1)^n)cos(nx)

But I don’t know how to manipulate these to show that part c) is true

Durham student?

I've been struggling to find the formulas for the odd/even completions and have no just realised i have them under a different name in my notes *d'oh*
i'll have a look at that question again now.

EDIT: where is your factor of 2 for An and Bn coming from?
you seem to have An=2/π 0∫π f(x)cos(nx)dx I don't have the 2 in mine :s

let the fourier series = g(x)

f(x)=g(x) since no discontinuities
x(pi-x)=fourier
RHS=0 when x=0 or pi
putting those x values into both sides gives you 2 things in the sums both over n^2. One gives you the sum for even n the other for odd n. When you add the two together you get pi^2/6

Thanks, I've solved this now, realised I forgot to apply my subsitution to both the top and bottom part of my sum