Question

Give the number to which the Fourier series converges at a point of discontinuity of f.

f(x) = 3 - 2x, -π < x < π

Answer 1

The Fourier series of f(x) converges to 3 at the points x= π+2kπ, where k is an integer.

Explanation:

First, recall that the function f(x) is extended 2π periodic to the whole real line, in order to obtain a valid Fourier expansion. Remember that a Fourier series is formed by a sines and cosines, which are 2π-periodic.

So, the 2π-periodic expansion of f(x) is discontinuous at the points π+2kπ, in particular π and -π. Check the attached figure to a better understanding.

Now, the Dirichlet theorem on the convergence of a Fourier series tells us that the series converges to the function at the points of continuity, and at points of discontinuity the sum of the series is

`(f(x_0+)+f(x_0-))/(2)`.

Here we understand the notation
`f(x+)` and
`f(x-)` as

`f(x_0+) = \lim_(x\rightarrow x_0)f(x), x>x_0`

and

`f(x_0-) = \lim_(x\rightarrow x_0)f(x), x<x_0`.

In this particular case

`f(\pi-) = \lim_(x\rightarrow \pi)(3-2x) = 3-2\pi, x<\pi`.

For the limit
`f(\pi+) = \lim_(x\rightarrow \pi)(3-2x)`, with
`x>\pi` recall that our function is 2π-periodic, so the values of f near π, with x>π are the same when x is near -π and x>-π. Again, check the attached figure. So,

`f(\pi+) = \lim_(x\rightarrow \pi)(3-2x) = 3+2\pi, x<\pi`.

Thus,

`(f(\pi+)+f(\pi-))/(2) = (3+2\pi +3-2\pi)/(2) = (6)/(2) =3`.

Note: In the attached figure we only have drawn three repetitions of the 2π-periodic extension of
`f`, recall that the extension is ad infinitum. Also, the points drawn in the dotted lines are the sum of the series at the points of discontinuity.