Math 309B/C Winter 2012, Homework 5


Due February 15

  1. Sketch the following functions and determine whether the following functions are odd or even whenever they are defined (in the given intervals if there are any)
    1. $\tan x$ Answer: This function is odd since $\tan(-x)=\frac{\sin(-x)}{\cos(-x)}=\frac{-\sin x}{\cos x}=-\tan x$.
    2. $x(x^{2}-1)$ for $x\in [-1,1]$ Answer: This function (call it $f(x)$) is odd since \[f(-x)=(-x)((-x)^{2}-1)=-x(x^2-1)=-f(x).\]
    3. $x^{3}$
  2. Find the Fourier series of the following functions and determine their values at the given points.
    1. Let $f$ be the $2$-periodic function defined by $f(x)=\frac{x}{|x|}$ for $x\in [-1,1]$. Find the values of its Fourier series in $[-1,1]$. What is the value at $x=3$? Answer: Note that $|x|=\left\{\begin{array}{cc} x & x\geq 0 \\ -x & x\leq 0\end{array}\right.$. Hence, $\frac{x}{|x|}=\left\{\begin{array}{cc} 1 & 0
    2. Let $f=\left\{\begin{array}{cc} x+\pi & 2x-2\pi\end{array}\right.$ and $f(x+4)=f(x)$. Find the Fourier series and the values in $[-2,2]$.
    3. Let $f$ be the $1$-periodic function $f(x)=(1+\frac{x}{|x|})x^{2}$ for $x\in [-1/2,1/2]$. Find its Fourier series and its values in $[-1/2,1/2]$.
    4. Let $f(x)=\left\{\begin{array}{cc} 0 & 1<|x|<2 \\ x & |x|<1.\end{array}\right.$, and $f(x+4)=f(x)$ (so $f$ is $4$-periodic). Find the values of the Fourier series in $[-2,2]$. Answer: Note that this function is odd. So $a_0=a_n=0$. And, since $f$ is zero outside the interval $-1,1$, \begin{align*} b_{n} & =\frac{1}{2}\int_{-2}^{2}f(x)\sin\frac{n\pi x}{2} =\frac{1}{2}\int_{-1}^{1}f(x)\sin \frac{n\pi x}{2} =\int_{0}^{1}f(x)\sin \frac{n\pi x}{2} =\int_{0}^{1}x\sin \frac{n\pi x}{2} =-\frac{x\cos \frac{n\pi x}{2}}{\frac{n\pi}{2}}|_{0}^{1} +\frac{2}{n\pi}\int_{0}^{1}\cos \frac{n\pi x}{2}\\ & =\frac{-2\cos \frac{n\pi }{2}}{n\pi}+\frac{4}{n^2 \pi^2}\sin \frac{n\pi x}{2}|_{0}^{1} =-\frac{2\cos \frac{n\pi}{2}}{n\pi}+\frac{4}{n^2 \pi^2}\sin \frac{n\pi }{2}.\end{align*} So the Fourier series is \[F(x) =\sum_{n=1}^{\infty}(-\frac{2\cos \frac{n\pi}{2}}{n\pi}+\frac{4}{n^2 \pi^2}\sin \frac{n\pi }{2})\sin \frac{n\pi x}{2}).\] The values of $F$ on $[-2,2]$ are equal to $f(x)$ wherever $f$ is continuous, so that's everywhere except for $-1$ and $1$. At $1$, it is equal to the average of the left and right limits of $f$ at $1$, so \[F(1)=\frac{f(1+)+f(1-)}{2}=\frac{1+0}{2}=\frac{1}{2}.\] Similarly, $F(-1)=\frac{1}{2}$.
    5. Let $f(x)=x^{3}$ for 
  3. Show that the derivative of an even function is odd and the derivative of an odd function is even. Answer: If $f$ is even, then $f(x)=f(-x)$. Differentiating both sides of this equation (and using chain rule) gives $f'(x)=-f'(-x)$, hence $-f'(x)=f'(-x)$, so $f'$ is odd. A similar proof works if $f$ is odd.
  4. Show that if $f$ and $g$ are odd, then $f(g(x))$ is odd, and if $f$ is odd and $g$ is even, then $f(g(x))$ and $g(f(x))$ are even.

    5. The next exercise shows how the Fourier coefficients can tell you how nice or smooth your function is. Basically, if your have a nice enough first derivative, then your Fourier coefficients will be converging to zero at a certain rate (like $\frac{1}{m}$-fast). Hence, if you have a Fourier series whose Fourier coefficients are, say, $\frac{1}{m^{2/3}}$, this means that your function's derivative must either be unbounded, or is not piecewise continuous.
  5. Show that if $f$ is continuous $2L$-periodic function and its derivative is bounded (i.e. for all $x$, $|f'|\leq M$ for some $M>0$), and $a_{m}=\frac{1}{L}\int_{-L}^{L}f(x)\cos \frac{m\pi x}{L}$ are its cosine Fourier coefficients, then $|a_{m}|\leq \frac{2ML}{m\pi}$. Hint: Recall that if $|g|\leq M$, then $\left|\int_{a}^{b}g\right|\leq M(b-a)$. Find a similar bound for $b_{m}$.
    Answer: \begin{align*} |La_{m}| & =\left|\int_{-L}^{L}f(x)\cos \frac{n\pi x}{L}\right| =\left|\frac{Lf(x)\sin \frac{n\pi x}{L}}{n\pi}|_{-L}^{L}-\frac{L}{n\pi}\int_{-L}^{L}f'(x)\sin \frac{n\pi x}{L}\right| \\ & =\left|\frac{L}{n\pi}\int_{-L}^{L}f'(x)\sin \frac{n\pi x}{L}\right| \\ & \leq \frac{L}{n\pi}\int_{-L}^{L}|f'(x)|\cdot|\sin \frac{n\pi x}{L}|\\ & \leq \frac{L}{n\pi}\int_{-L}^{L}M\cdot 1=\frac{2L^{2}M}{n\pi} \end{align*} and hence $|a_{m}|\leq \frac{2LM}{n\pi}$.