1
$\begingroup$

Verify that the fourier series converge uniformly on the interval ${\pi\leq x\leq \pi}$. Also state why this series is differentable in the interval ${\pi\leq x\leq \pi}$, except at the point $x=0$ and describe graphically the function that is represented by the differentiated series for all $x$

$\frac{1}{\pi}+0.5\sin x-\frac{2}{\pi}\sum_{n=1}^\infty \dfrac{\cos 2nx}{4n^{2}-1 }$

What i tried

Using the Weierstrass M test i got $$|\frac{1}{\pi}+0.5\sin x-\frac{2}{\pi}\sum_{n=1}^\infty \dfrac{\cos 2nx}{4n^{2}-1 }|\leq |\frac{1}{\pi}+0.5\sin x+\frac{2}{\pi}\sum_{n=1}^\infty \dfrac{\cos 2nx}{4n^{2}-1 }|$$

$$|\frac{1}{\pi}+0.5\sin x+\frac{2}{\pi}\sum_{n=1}^\infty \dfrac{\cos 2nx}{4n^{2}-1 }| \leq |\frac{1}{\pi}+0.5\ x+\frac{2}{\pi}\sum_{n=1}^\infty \dfrac{\cos 2nx}{4n^{2}-1 }|$$

$$|\frac{1}{\pi}+0.5\ x+\frac{2}{\pi}\sum_{n=1}^\infty \dfrac{\cos 2nx}{4n^{2}-1 }| \leq|\frac{1}{\pi}+0.5\ x+\frac{2x}{\pi}\sum_{n=1}^\infty \dfrac{\ 2n}{4n^{2}-1 }|$$

And since the term

$$\sum_{n=1}^\infty \dfrac{\ 2n}{4n^{2}-1 }$$ converges then by the Weierstrass M test the above fourier seriesM test converges uniformly. Differentating the fourier series term by term i got
$$0.5\cos x+\frac{4n}{\pi}\sum_{n=1}^\infty \dfrac{\sin 2nx}{4n^{2}-1 }$$

I suppose that it is differentiable on the given interval because it is continous on that interval except at $x=0$ but i cant see why is this so and also how to describe the graphically the function that is represented by the differentiated series for all $x$. Could anyone please explain this to me. Thanks

$\endgroup$
  • $\begingroup$ Why in the first passage did you switch the minus sign with a plus? $\endgroup$ – user228113 Aug 31 '16 at 16:12
  • $\begingroup$ The series $\sum_{n\geq 1}\frac{2n}{4n^2-1}$ is not converging. $\endgroup$ – Jack D'Aurizio Aug 31 '16 at 16:13
1
$\begingroup$

The work in the OP has some flaws. Note that we have for all $x$, $\left|\frac{\cos(2nx)}{4n^2-1}\right| \le \frac1{4n^2-1}$, for each $n$. Inasmuch as

$$\begin{align} \sum_{n=1}^\infty \frac{1}{4n^2-1}&<\infty \end{align}$$

the Weierstrass M-Test guarantees that the series $\sum_{n=1}^\infty \frac{\cos(2nx)}{4n^2-1}$ converges uniformly for all $x\in [-\pi,\pi]$.


To analyze whether the series is differentiable, we examine the series of term-by-term derivatives $D(x)$ as given by

$$D(x)=-2\sum_{n=1}^\infty \frac{n\sin(2nx)}{4n^2-1} \tag 1$$

Note that $\sum_{n=1}^N \sin(2nx)=\csc(x)\sin(Nx)\sin((N+1)x)$ is bounded by $|\csc(x)|$ for $x\ne 0,\pi,-\pi$. Furthermore, $\frac{n}{4n^2-1}$ monotonically decreases to $0$ as $n\to \infty$.

Therefore, for any $\delta >0$ and $x\in [-\pi+\delta,-\delta]$ or $x\in [\delta,\pi-\delta]$, Dirichlet's Test guarantees that the series in $(1)$ for $D(x)$ converges uniformly and inasmuch as the original series $\sum_{n=1}^\infty \frac{\cos(2nx)}{4n^2-1}$ also converges on $[-\pi,\pi]$ (actually, we only need that it converge at a single point), we find that

$$D(x)=\frac{d}{dx}\sum_{n=1}^\infty \frac{\cos(2nx)}{4n^2-1}$$

for all $x\ne 0,\pi,-\pi$.


To examine the derivative from the right (left) of $\sum_{n=1}^\infty \frac{\cos(2nx)}{4n^2- 1}$ at $x=-\pi$ ($x=\pi$), we note that

$$\begin{align} \lim_{h\to0^{\pm}}\sum_{n=1}^\infty \frac{\cos(2n(\mp \pi+h))-\cos(2n(\mp\pi))}{h(4n^2- 1)}&=-2\lim_{h\to0^{\pm}}\sum_{n=1}^\infty \frac{\sin^2(nh)}{h(4n^2-1)}\\\\ &\overbrace{=}^{\text{LHR}}\underbrace{-\frac12\lim_{h\to0^{\pm}}\sum_{n=1}^\infty\frac{4n\sin(2nh)}{4n^2-1}}_{=\lim_{x\to \mp\pi}D(x)}\\\\ &=-\frac12\lim_{h\to0^{\pm}}\sum_{n=1}^\infty\left(\frac{\sin(2nh)}{n}+\frac{\sin(2nh)}{n(4n^2-1)}\right)\\\\ &=-\frac12\lim_{h\to0^{\pm}}\sum_{n=1}^\infty\frac{\sin(2nh)}{n}\\\\ &=\mp \frac\pi4 \end{align}$$

It is importatn to observe that $D(\pm\pi)=0\ne \pm\frac\pi4$. This is not inconsistent since $D(x)$ is not the representation of the derivative (from the left or right)) at $x=\pi$ or $x=-\pi$. However, $D(x)$ does have the appropriate limits as $x\to \pm\pi$.

$\endgroup$
  • $\begingroup$ Okay Thanks, how to explain why this series is differentable at the interval except at $x=0$. I know there is a discontinuity at $x=0$ that makes it not differentable but i cant see the disconutity from the equations. Also how do i describe graphically the function that is represented by the differentiated series for all $x$ ? $\endgroup$ – ys wong Aug 31 '16 at 16:27
  • 1
    $\begingroup$ You're welcome. My pleasure. I've edited to add a way forward to proving differentiability. $\endgroup$ – Mark Viola Aug 31 '16 at 16:58
  • $\begingroup$ Weierstrass M does imply that series converges uniformly, but not for that reason. Rather, you want to say that for every $x,$ $$\left | \frac{\cos 2nx}{4n^{2}-1 }\right| \le \frac{1}{4n^{2}-1 }$$ for each $n,$ and $\sum_{n=1}^{\infty}\dfrac{1}{4n^{2}-1 }$ converges. WM then gives the desired uniform convergence. $\endgroup$ – zhw. Jan 26 at 7:15
  • $\begingroup$ @zhw. I've added a new section to discuss the one-sided derivatives at $x=\pm\pi$ and edit the first section as per your comment. $\endgroup$ – Mark Viola Jan 26 at 21:30
  • $\begingroup$ Happy New Year MV! $\endgroup$ – zhw. Jan 26 at 21:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.