Finding the uniform convergence of a Fourier series

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

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

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 $$x\to \pm\pi$$.

• 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 $x$ ? – ys wong Aug 31 '16 at 16:27
• You're welcome. My pleasure. I've edited to add a way forward to proving differentiability. – Mark Viola Aug 31 '16 at 16:58
• 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. – zhw. Jan 26 at 7:15
• @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. – Mark Viola Jan 26 at 21:30
• Happy New Year MV! – zhw. Jan 26 at 21:59