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I'd really love your help with this one: I got this Fourier series for $f(x)=x$ in $[-\pi,\pi]$: $$\sum_{1}^{\infty}\frac{2(-1)^{n+1}\sin(nx)}{n}$$ and I need to check if it's (i) pointwise converges,(ii) uniformly converges and if (iii) $$\sum_{1}^{\infty}\left|\frac{2(-1)^{n+1}\sin(nx)}{n}\right|$$ converges.. I tried to do it, and I have couple of specific questions that interrupt me.

(i) Can I claim that the series is pointwise due to Leibniz test? since $\frac{\sin(nx)}{n} \to 0$? I was told that I need to use Lipschitz continuity here. How do I use it?

(ii) I was told that this series is not uniformly converges since $f(x)$ is not continuous in this part. How come? Is it because we always see those functions who we compute their Fourier series as periodic functions with period of $[0,2\pi]$ so in our case the function in $0$ is not continues?

(iii) How come I can't use Dirichlet test for series and claim that the $$\sum_{1}^{\infty}\left|\frac{2(-1)^{n+1}\sin(nx)}n\right|$$ does converges? The test requires a multiple of a series that monotonic converges to $0$ and $\sum_{1}^{N}|\sin(nx)|<M$? Is it wrong? How can you prove that this series does not converge?

Thanks you so much guys!

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For the first question, you can use Abel's transformation. –  Davide Giraudo Dec 29 '11 at 17:52
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For (iii), note that $|\sin(n x)| \ge \sin^2(n x) = (1 - \cos(2nx))/2$ –  Robert Israel Dec 29 '11 at 19:51
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2 Answers 2

up vote 2 down vote accepted

Something that might prove applicable is the Dirichlet Convergence Test. Summation by Parts yields $$ \sum_{k=1}^na_kb_k=A_nb_n+\sum_{k=1}^{n-1}A_k(b_k-b_{k+1})\tag{1} $$ where $\displaystyle A_n=\sum_{k=1}^na_k$. Note that $(-1)^{n+1}\sin(nx)=-\sin(n(\pi+x))$, so we have $$ \begin{align} A_n &=\sum_{k=1}^n(-1)^{n+1}\sin(kx)\\ &=-\sum_{k=1}^n\sin(k(\pi+x))\\ &=\frac{\sin(n(\pi+x)/2)}{\sin((\pi+x)/2)}\sin((n+1)(\pi+x)/2)\tag{2} \end{align} $$

Thus, $|A_n|\le|\sec(x/2)|$. Therefore,

(i) according to $(1)$ and $(2)$, the series converges pointwise except at odd multiples of $\pi$; however, at any multiple of $\pi$, each term is $0$. Thus, the series converges for all $x$.

(ii) according to $(1)$ and $(2)$, the series converges uniformly on compact sets not containing an odd multiple of $\pi$. A sequence of continuous functions cannot converge uniformly to a discontinuous function, so the convergence cannot be uniform in any neighborhood of an odd multiple of $\pi$.

(iii) except at multiples of $\pi$, the series is not absolutely convergent. Here is the outline of a proof.

First, show that if $x$ is not a multiple of $\pi$, at least $1/4$ of the multiples of $x$ have $|\sin(kx)|>1/\sqrt{2}$. Because $|\sin(k(x+n\pi))|=|\sin(kx)|$, we only need to consider $x\in[-\pi/2,\pi/2]$. Since $|\sin(-kx)|=|\sin(kx)|$, we only need to consider $x\in[0,\pi/2]$.

If $x\in[\pi/(n+1),\pi/n]$, then at least $\lfloor n/2\rfloor$ multiples of $x$ are in $[(m+1/4)\pi,(m+3/4)\pi]$ out of at most $n+1$ in $[m\pi,(m+1)\pi]$. Since $(0,\pi/2]=\cup_{n=2}^\infty[\pi/(n+1),\pi/n]$, this means that at least $1/4$ of the multiples of $x$ in $[m\pi,(m+1)\pi]$ are in $[(m+1/4)\pi,(m+3/4)\pi]$, and therefore, have $|\sin(kx)|>1/\sqrt{2}$.

Thus, $1/4$ of each $n$ to $n+1$ consecutive terms of the sum have $|\sin(kx)|>1/\sqrt{2}$. The sum is therefore greater than $$ \frac{1}{4}\sum_{k=n+1}^\infty\frac{2}{k}\frac{1}{\sqrt{2}}\tag{3} $$ which diverges. Thus, if $x$ is not a multiple of $\pi$, the series of absolute values diverges.

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Thanks a lot! you were very helpful. –  Jozef Dec 30 '11 at 15:52
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What do you mean by Leibniz test? And $\sin nx\over n$ converges to 0, not infinity.

Here we regard $f(x)=x$ with the domain $[-\pi,\pi]$.

For i) you could use (if you've had it already) the theorem: If $f$ is continuous and of bounded variation on $[-\pi,\pi]$, then the Fourier series of $f$ converges to $f$ pointwise on $(-\pi,\pi)$ and to the average of the left and right hand limits of the periodic extension of $f$ at the endpoints.

For ii) The above will also tell you that the limit function is not continuous on $[-\pi,\pi]$ (in particular at $x=\pi$ and $x=-\pi$, where the series converges to 0). Since a uniform limit of continuous functions is continuous, it follows that the convergence of the series is not uniform on $[-\pi,\pi]$ (it is uniform on any $[a,b]\subset(-\pi,\pi)\thinspace$).

I'm not sure how you would handle iii); but you might apply this.

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