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

f(x) = 3 - 2x, -π < x < π

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

.

Here we understand the notation and as

and

.

In this particular case

.

For the limit , with 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,

.

Thus,

.

Note: In the attached figure we only have drawn three repetitions of the 2π-periodic extension of , 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.

add a pic of the graph, then ill edit this "answer" and solve it for you : 3

step-by-step explanation: