Question

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

Answer 1

Answer:

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

Step-by-step 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

$\frac{f(x_0+)+f(x_0-)}{2}$.

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

$f(x_0-) = \lim_{x\rightarrow x_0}f(x), x$.

In this particular case

$f(\pi-) = \lim_{x\rightarrow \pi}(3-2x) = 3-2\pi, x$.

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,

$f(\pi+) = \lim_{x\rightarrow \pi}(3-2x) = 3+2\pi, x$.

Thus,

$\frac{f(\pi+)+f(\pi-)}{2} = \frac{3+2\pi +3-2\pi}{2} = \frac{6}{2} =3$.

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.