Science:Math Exam Resources/Courses/MATH257/December 2011/Question 03 (b)/Solution 1

In short, the Fourier convergence theorem states the following.

If $f$ is a periodic function such that $f$ and $\displaystyle {}f'$ are piecewise continuous, then the Fouriner series converges to $f(x)$ at all points where $f$ is continuous and converges to ${\frac {1}{2}}[f(x+)+f(x-)]$ at all points where $f$ is discontinuous.

In the above, $f(x+)$ denotes the right hand limit of $f$ at $x$, while $f(x-)$ denotes the left hand limit of $f$ at $x$. In other words, for a real number $a$,

$f(a+)=\lim _{x\to a^{+}}f(x)\;\;{\text{and}}\;\;f(a-)=\lim _{x\to a^{-}}f(x).$

With the Fourier convergence theorem in mind, let's find out the point of convergence of the Fourier cosine series $Cf(x)$ at each point $x$ in the interval $[0,1]$. To do so, we first need to understand the origin of the Fourier cosine series. Recall that the Fourier cosine series is the Fourier series of $f_{\text{even}}$ where $f_{\text{even}}$ is the even extension of $f$. Following the Fourier convergence theorem, to find the point of convergence of the Fourier cosine series, we need to know where the points of continuity and discontinuity of $f_{\text{even}}$ are. We will do so by sketching the graph of $f_{\text{even}}$. To begin, we sketch the graph of $f(x)$.

A sketch of $f(x)$

The graph of $f_{\text{even}}$ is obtained by first reflecting $f$ over the y-axis and hence obtaining a function over the interval $[-1,1]$. We then periodic extend this function to get $f_{\text{even}}$ which will have a period of $2$.

A sketch of the even extension of $f(x)$. The part of the function from $-1$ to $1$ is coloured in red. The red part shows one period of $f_{\text{even}}$.

From the sketch of $f_{\text{even}}$, we see that $f_{\text{even}}$ is continuous everywhere. Hence, by the Fourier convergence theorem, the Fourier cosine series $Cf(x)$ converges to $f_{\text{even}}$ for every real number $x$. Now, since $f$ agrees with $f_{\text{even}}$ at every point in the interval $[0,1]$,

$Cf(x)$ converges to $f(x)=x$ at every point $x$ in the interval $[0,1]$.

Similarly, let $f_{\text{odd}}$ be the odd extension of $f$. Then the Fourier sine series $Sf(x)$ is the Fourier series of $f_{\text{odd}}$. We will sketch $f_{\text{odd}}$ to find its points of continuity and discontinuity.

The graph of $f_{\text{odd}}$ is obtained by first rotating $f$ about the origin by 180 degree and hence obtaining a function over the interval $[-1,1]$. We then periodic extend this function to get $f_{\text{odd}}$ which will have a period of $2$.

A sketch of the odd extension of $f(x)$. The part of the function from $-1$ to $1$ is coloured in red. The red part shows one period of $f_{\text{odd}}$.The values at points of discontinuity is not specified in the sketch.

From the sketch of $f_{\text{odd}}$ is discontinuous at points $x=-1+2k$ with k an integer. Hence, by the Fourier convergence theorem, the Fourier sine series $Sf$ converges to $f_{\text{odd}}$ at every point in the interval $[0,1{\color {red})}$, but at $x=1$, $Sf$ converges to

${\frac {1}{2}}[f_{\text{odd}}(1-)+f_{\text{odd}}(1+)]={\frac {1}{2}}[1+(-1)]=0$.

Since $f_{\text{odd}}$ agrees with $f$ in the interval $[0,1)$,

$Sf(x)$ converges to $f(x)=x$ at every point in the interval $[0,1{\color {red})}$ and at $x=1$, $Sf(1)$ converges to 0.

Finally, the question asks us to verify our conclusions for the Fourier sine series at x = 0 and x = 1. Evaluate the sine series found at part a at $x=0$, we get

$Sf(0)=\sum _{n=1}^{\infty }{\frac {2}{n\pi }}(-1)^{n+1}\underbrace {\sin(n\pi \;0)} _{=0}=\sum _{n=1}^{\infty }0=0$

and evaluating at $x=1$, we get

$Sf(1)=\sum _{n=1}^{\infty }{\frac {2}{n\pi }}(-1)^{n+1}\underbrace {\sin(n\pi )} _{=0}=\sum _{n=1}^{\infty }0=0$

as expected.