Fourier series - where is f(t) discontinuous


Junior Member
May 14, 2020
The function f is periodic with [imath]T=2\pi[/imath] and is given as f(t)=t where [imath]-\pi \leq t < \pi[/imath]. Is the function odd, even or neither? Also, what points are f(t) discontinuous?

My attempt:

f(-t)=-t, hence f is an odd function. I then set out to find the Fourier series corresponding to f(t) which gives me:

[math]\sum_{n=1}^\infty \dfrac{2}{n}(-1)^{n+1}[/math]
which is correct, but I don't get how f(t) can be discontinuous? I get that if n=0 in the series above we get issues but other than that I have no clue how f is discontinuous at any points since it is a linear function. I probably need to utilize the fact that I know the period of f(t).


Junior Member
Jan 7, 2021
Hi there,
Your class probably teaches the Dirichlet’s conditions for Fourier series. One of those conditions is that the function must be periodic. If you take a look at the function f(x) = x, it isn’t periodic! Just look at the graph, it’s a line splitting the first and third quadrant. In order to use this method, you need to make your function periodic, so the graph will look similar to this: BC72E515-EFC1-4924-86B5-FF23CAD7481A.png
Except that this function is 2-periodic, and is not the function f(x) = x
Now you should see where the points of discontonuity are!