# How's Fourier series modified for function f(t)= f(2Pi t)?

1. May 15, 2016

### Poirot

1. The problem statement, all variables and given/known data
How are the coefficients of the Fourier series modified for a function with a period 2πT?

2. Relevant equations
a0 = 1/π ∫π f(x) dx
an = 1/π ∫π f(x) cos(nx) dx
bn = 1/π ∫π f(x) sin(nx) dx

3. The attempt at a solution
I tried letting x= t/T
so dx = dt/T and the limits x = ± π, x = ± πT
and I was going to plug this into the integrals but I don't think it's right.

Any help would be greatly appreciated, thanks!

2. May 15, 2016

Usually the function f(t) is periodic over period T. (not $2 \pi$ T). The integration over t will go from t=0 to t=T. You can readily google the topic to get the precise form of the coefficients, etc.

3. May 15, 2016

### LCKurtz

For general period $2p$ the Fourier coefficients are$$b_n = \frac 1 p \int_{-p}^p f(x) \sin(\frac{n\pi x}{p})~dx$$ $$a_n =\frac 1 p \int_{-p}^p f(x) \cos(\frac{n\pi x}{p})~dx$$ $$a_0 = \frac 1 {2p}\int_{-p}^p f(x)~dx$$
$a_0$ may or may not have the $2$ in the denominator depending on whether you start the FS with $a_0$ or $\frac {a_0} 2$.