Give the number to which the Fourier series converges at a point of discontinuity of f.f(x) = 3 - 2x, -π < x < π

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[tex]\frac{f(x_0+)+f(x_0-)}{2}[/tex].Here we understand the notation [tex]f(x+)[/tex] and [tex]f(x-)[/tex] as [tex]f(x_0+) = \lim_{x\rightarrow x_0}f(x), x>x_0[/tex]and[tex]f(x_0-) = \lim_{x\rightarrow x_0}f(x), x<x_0[/tex].In this particular case [tex]f(\pi-) = \lim_{x\rightarrow \pi}(3-2x) = 3-2\pi, x<\pi[/tex].For the limit [tex]f(\pi+) = \lim_{x\rightarrow \pi}(3-2x) [/tex], with [tex]x>\pi[/tex] 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,[tex]f(\pi+) = \lim_{x\rightarrow \pi}(3-2x) = 3+2\pi, x<\pi[/tex].Thus, [tex]\frac{f(\pi+)+f(\pi-)}{2} = \frac{3+2\pi +3-2\pi}{2} = \frac{6}{2} =3[/tex].Note: In the attached figure we only have drawn three repetitions of the 2π-periodic extension of [tex]f[/tex], 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.