Calculate the Fourier Series of f(x)=x?

Answer 1

The function f(x) = x isn't a square integratable funciton, so there isn't a Fourier Series for it. However, for the sawtooth function, the discrete Fourier Series is as follows:

F(x) = Fourier Series of f(x) = (1/2) -(1/π)∑(n:1 to ∞) (1/n) Sin(2nπx)

Answer 2

This does too exist, it just has an infinite number of terms.

Just like any function can be represented by a taylor series, it can be represented by a fourier series.

Laplace of 1/x is 1/s^2

If you can go from laplace to fourier (you can) then you go from complex-fourier to real fourier you get your sum.

matlab gives:
fourier(sym('x'))
=
2*i*pi*Dirac(1,w)

Thats funny, isnt it. Its just dirac(x)*(1+0*i). Its not the full series, just a vector representation on the complex unit circle.

Aha. Thanks jon. That was the fourier transform, not the fourier series. My bad.