Let U(t) be the Fourier series of the 2pi-periodic triangle wave: F(t) = |t|, -pi<t<pi.

[left hull]

Let U(t) be the Fourier series of the 2pi-periodic triangle wave: F(t) = |t|, -pi<t<pi. My value of U(pi) = 0 but it is wrong. Why?

[fallen bloom]

1. Wait patiently for a helper to come along.
2. Once someone helps you, say thank you and close the thread with: ```diff
   +close

[pure trellis]

he 2π-periodic triangle wave F(t) = |t|, -π < t < π is given by:

U(t) = (8/π^2) * ∑[(-1)^(n+1)/(2n-1)^2 * sin((2n-1)t)], where the sum is taken over all odd integers n.

To find U(π), we substitute t = π in the above expression:

U(π) = (8/π^2) * ∑[(-1)^(n+1)/(2n-1)^2 * sin((2n-1)π)]

= (8/π^2) * ∑[(-1)^(n+1)/(2n-1)^2 * sin(nπ)]

Note that sin(nπ) = 0 for all even values of n, and (-1)^(n+1) = -1 for odd values of n. Therefore, the above sum reduces to:

U(π) = (8/π^2) * ∑[(-1)^n/(2n-1)^2]

This is a well-known sum, called the Basel problem, and it is equal to π^2/8. Therefore, we have:

U(π) = (8/π^2) * (π^2/8) = 1

Therefore, the correct value of U(π) is 1, not 0. The mistake may have been due to an error in the calculation or an oversight in the manipulation of the sum.

[left hull]

the correct answer is pi

but it's okay, one mistake shouldn't be too bad

[pure trellis]

sorry

[left hull]

bro, it's okay

UWU...

thank you @pure trellis

[loud schooner]

@left hull has given 1 rep to @pure trellis

[left hull]

+close