Question

(a) Sketch the graph of $f(x)$ over four periods. Find the Fourier series representation for the given function $f(x)$. Use whatever symmetries or other obvious properties the function possesses in order to simplify your calculations. (b) Determine the points at which the Fourier series converges to $f(x)$. At each point $x$ of discontinuity, state the value of $f(x)$ and state the value to which the Fourier series converges. $$f(x)=\left\{\begin{array}{ll}2, & 0 \leq x \leq 1, \ 0, & 1

Short answer

Answer: The Fourier series representation converges to 1 at the points of discontinuity (x=1 and x=2).

Step-by-step solution

Sketch the graph

The function is defined as: $$f(x)=\left\{$$\begin{array}{ll}2,& 0\le x\le 1,\\ 0,& 1<x<2,\end{array}$$ \quad f(x+2)=f(x)\right.$$ The function is a square wave, with period 2. We can sketch the graph by plotting the values of x and their corresponding values of f(x) for one period and then repeating the pattern for three additional periods. The graph will show a square wave with a height of 2.

Find the Fourier series representation

To find the Fourier series representation, we will first find the Fourier coefficients. Since the function is periodic and not odd or even, we will use the general Fourier series formula: $$f(x)\approx \frac{a_0}{2}+\sum _{n=1}^{\mathrm{\infty }}(a_n\cos (nx)+b_n\sin (nx))$$ with $$a_0=\frac{1}{T}{\int }_0^{T}f(x)dx$$ $$a_n=\frac{1}{T}{\int }_0^{T}f(x)\cos (nx)dx$$ and $$b_n=\frac{1}{T}{\int }_0^{T}f(x)\sin (nx)dx$$ where T is the period of the function, in this case 2. First, find $a_0$: $$a_0=\frac{1}{2}{\int }_0^{2}f(x)dx$$ Since the function is defined piecewise, we will split the integral into two parts: $$a_0=\frac{1}{2}({\int }_0^{1}2dx+{\int }_1^{2}0dx)$$ $$a_0=\frac{1}{2}(2(1-0)+0(2-1))=1$$ Next, find $a_n$: $$a_n=\frac{1}{2}{\int }_0^{2}f(x)\cos (nx)dx$$ Split the integral into two parts again: $$a_n=\frac{1}{2}({\int }_0^{1}2\cos (nx)dx+{\int }_1^{2}0dx)$$ $$a_n=\frac{1}{n\pi }(\sin (n\pi )-\sin (0))=0$$ (since $n\pi$ is an integer multiple of $\pi$) Now, find $b_n$: $$b_n=\frac{1}{2}{\int }_0^{2}f(x)\sin (nx)dx$$ Split the integral into two parts again: $$b_n=\frac{1}{2}({\int }_0^{1}2\sin (nx)dx+{\int }_1^{2}0dx)$$ $$b_n=\frac{1}{n\pi }(\cos (n\pi )-1)$$ Finally, substitute the coefficients into the Fourier series formula: $$f(x)\approx \frac{1}{2}+\sum _{n=1}^{\mathrm{\infty }}(0\cdot \cos (nx)+\frac{1}{n\pi }(\cos (n\pi )-1)\sin (nx))$$ Simplify: $$f(x)\approx \frac{1}{2}+\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n\pi }(\cos (n\pi )-1)\sin (nx)$$

Determine points of convergence

By the Dirichlet's theorem, the Fourier series converges to $f(x)$ at all points of continuity. The points of discontinuity are $x=1$ and $x=2$. At these points, the Fourier series converges to the average of the function's values on the right and left of the discontinuity. At $x=1$: $$\frac{f(1^{+})+f(1^{-})}{2}=\frac{0+2}{2}=1$$ At $x=2$: Since the function is periodic with period 2, we have $f(2)=f(0)$: $$\frac{f(2^{+})+f(2^{-})}{2}=\frac{2+0}{2}=1$$ The Fourier series converges to the function except at its points of discontinuity ($x=1$ and $x=2$), where it converges to 1.