2

I want to show that the following pointwise limit holds: $$\frac{\sum_{n=1}^N\cos(nx)}{N(x^2+1)}\to\begin{cases} 1, & x=0 \\ 0 & x\neq0 \end{cases}$$ The case $x=0$ is easy: we have, for any $\varepsilon>0$: $$\frac{1}{N}\left\vert\sum_{n=1}^N\cos(nx)-N\right\vert=\left\vert\frac{N}{N}-1\right\vert=0<\varepsilon$$ for any $N$. However, I'm more troubled by the case where $x\neq0$. One issue is that we might have that $x$ is a multiple of $2\pi$, in which case we would have: $$\frac{1}{N(x^2+1)}\left\vert\sum_{n=1}^N\cos(nx)\right\vert=\frac{N}{N(x^2+1)}=\frac{1}{x^2+1}$$ for any $N$, in which case we can't make the above arbitrarily small for $x$ fixed.

Graphically, the pointwise limit seems to hold, but I could be mistaken. Any thoughts?

csch2
  • 4,557

2 Answers2

2

It doesn't converge to zero for $x=2\pi$ for example:

$\frac{1}{N((2\pi)^2+1)}\cdot \sum\limits_{n=1}^N\cos(n\cdot 2\pi)= \frac{1}{N((2\pi)^2+1)}\cdot \sum\limits_{n=1}^N1=\frac{1}{(2\pi)^2+1}$

user394334
  • 1,230
  • Looks like my reasoning is right then. Guess this is a lesson not to always trust Desmos! – csch2 Nov 30 '19 at 04:14
  • @csch2 Are you sure you graphed it for fixed $x$, $N$ variable, and looked at what happened when $N$ got big? And not for instance looked at what happened when only $x$ got big? – user394334 Nov 30 '19 at 04:22
  • I plotted $f_N(x)$ as a function of $x$ and scaled $N$ to see how the plot changed with $N$. The issue was that the bumps at multiples of $2\pi$ were outside of my plotting range and also so small as to be unnoticeable. – csch2 Nov 30 '19 at 04:23
  • Question: What about the case where $x$ is not a multiple of $2\pi$? Does $f_n(x)\to0$ whenever $x\neq 2k\pi$ for $k\in\mathbb{Z}$? – csch2 Nov 30 '19 at 04:28
  • @csch2 Good question, I don't really have an answer for that right now. Maybe you can try to ask a new question where you add that constraint. – user394334 Nov 30 '19 at 04:32
2

Half of the Dirichlet Kernel $$\newcommand{\Re}{\operatorname{Re}} \begin{align} \sum_{n=1}^N\cos(nx) &=\Re\left(\sum_{n=1}^Ne^{inx}\right)\tag1\\ &=\Re\left(\frac{e^{ix}-e^{i(N+1)x}}{1-e^{ix}}\right)\tag2\\ &=\frac{\Re\left(e^{ix}-1-e^{i(N+1)x}+e^{iNx}\right)}{2-2\cos(x)}\tag3\\ &=-\frac12+\frac{\cos(Nx)-\cos((N+1)x)}{2-2\cos(x)}\tag4\\ &=\frac{\sin\left(\left(N+\frac12\right)x\right)}{2\sin\left(\frac12x\right)}-\frac12\tag5 \end{align} $$ Explanation:
$(1)$: $\cos(x)=\Re\left(e^{ix}\right)$
$(2)$: sum of a geometric series
$(3)$: multiply by $\frac{1-e^{-ix}}{1-e^{-ix}}$
$(4)$: $\cos(x)=\Re\left(e^{ix}\right)$
$(5)$: $\cos(Nx)-\cos((N+1)x)=2\sin\left(\left(N+\frac12\right)x\right)\sin\left(\frac12x\right)$

Note that $(5)$ also follows from the Dirichlet Kernel: $$ \sum_{n=-N}^N\cos(nx)=\frac{\sin\left(\left(N+\frac12\right)x\right)}{\sin\left(\frac12x\right)}\tag6 $$

Application to the Limit

If $\sin\left(\frac12x\right)\ne0$, then, since $\left|\sin\left(\left(N+\frac12\right)x\right)\right|\le1$, $$ \begin{align} \lim_{N\to\infty}\frac1{N\left(x^2+1\right)}\sum_{n=1}^N\cos(nx) &=\lim_{N\to\infty}\frac1{N\left(x^2+1\right)}\left(\frac{\sin\left(\left(N+\frac12\right)x\right)}{2\sin\left(\frac12x\right)}-\frac12\right)\\[6pt] &=0\tag6 \end{align} $$ If $\sin\left(\frac12x\right)=0$, then $x=2\pi k$ for some $k\in\mathbb{Z}$; and thus, $\cos(nx)=1$. Therefore, $$ \begin{align} \lim_{N\to\infty}\frac1{N\left(x^2+1\right)}\sum_{n=1}^N\cos(nx) &=\lim_{N\to\infty}\frac1{N\left(x^2+1\right)}\,N\\[6pt] &=\frac1{x^2+1}\tag7 \end{align} $$

robjohn
  • 345,667