6

I would like to prove that for every $x \in \mathbb{R}, n$ natural, we have $$\!\left|\sin(x)+\frac{\sin(2x)}{2}+\ldots+\frac{\sin{(nx)}}{n}\right| \le \int_{0}^{\pi}\frac{\sin x}{x}\,dx$$

We can of course restrict our attention for $x \in [0, 2\pi]$. So far I tried a few things: letting $S_n(x)$ denote the quantity inside the absolute value, we have$$S_n'(x) = \sum_{k=1}^{n}\cos(kx) = -\frac1{2}+\frac{\sin(n+\frac12)x}{2\sin\frac{x}{2}}$$

and hence $S_n(x)$ has critical points at $x = \frac{2k\pi}{n}$ or $x = \frac{(2k+1)\pi}{n+1}$ where $k$ is an integer. Moreover:$$S_n(x) = -\frac{x}{2}+\int_{0}^{x}\frac{\sin(n+\frac12)t}{2\sin\frac{t}{2}}\,dt$$ I also know that the function $F(x) = \int_{0}^{x}\frac{\sin(t)}{t}\,dt$ has an absolute maximum for $x = \pi$. I struggle to bound the integral in the expression for $S_n$. Can you help me? Or maybe there's a different approach to solve this?

Peanut
  • 1,664

1 Answers1

3

This is the partial sum of the Fourier series for the sawtooth wave where $S(x) = \sum_{n=1}^\infty \frac{\sin nx}{n} = \frac{\pi-x}{2}$ on $(0,2\pi)$. The waveform has a discontinuity at $x=0$ and $S(x) \to \frac{\pi}{2}$ as $x \to 0+$.

Note that

$$\int_0^\pi \frac{\sin x}{x} \, dx = 1.85194\ldots > \frac{\pi}{2}$$

However, the partial sums overshoot the waveform (Gibbs phenomena) and it is proved here that

$$\limsup_{n \to \infty}\sup_{x \in [0,\pi]} S_n(x) \geqslant \int_0^\pi \frac{\sin x}{x} \, dx $$

We can prove that $\displaystyle|S_n(x)| \leqslant \int_0^\pi \frac{\sin x}{x} \, dx$ for all $n \in \mathbb{N}$ and $x \in [0,2\pi]$.

By periodicity and anti-symmetry it is enough to consider the interval $[0,\pi]$. Here the relative maxima of $S_n (x)$ occur at $x_{n,m} = \frac{2m+1}{n+1}\pi$ for $m = 0,1, \ldots, \lfloor\frac{n-1}{2} \rfloor$.

It can be shown that $S_n(x_{n,m}) > S_n(x_{n,m+1})$ so that there is an absolute maximum of $S_n(x)$ on $[0,\pi]$ at $x_{n,0} = \frac{\pi}{n+1}$. It can also be shown that $S_n(x_{n,0})$ is an increasing sequence, such that

$$S_n(x_{n,0}) \nearrow \lim_{n \to \infty}\sum_{k=1}^n \frac{\sin k x_{n,0}}{k} = \lim_{n \to \infty}\frac{\pi}{n+1}\sum_{k=1}^n \frac{\sin \frac{k\pi}{n+1}}{\frac{k\pi}{n+1}} \\ = \int_0^\pi \frac{\sin x}{x} \, dx$$

RRL
  • 90,707
  • Yes! I was looking at that answer, but unfortunetely you bound the reminder for large $n$; what I'm looking for is a proof for every $n$. – Peanut Dec 02 '20 at 17:18
  • So if it isn't a uniform upper bound, is my statement wrong? – Peanut Dec 02 '20 at 17:19
  • Another thing I'm not very convinced of is the $>$ sign. Shouldn't it be $\ge$? – Peanut Dec 02 '20 at 17:23
  • 1
    I’ll take another look. – RRL Dec 02 '20 at 17:26
  • It should be $\geqslant$. It is still possible that the integral is the upper bound at this point. – RRL Dec 02 '20 at 17:59
  • We also have $S_n\left(\frac{\pi}{n}\right) = \frac{\pi}{n}\sum_{k=1}^n \frac{\sin \frac{k\pi}{n}}{\frac{k\pi}{n}} \to \int_0^\pi \frac{\sin x}{x} , dx$ as a convergent sequence of Riemann sums. – RRL Dec 02 '20 at 18:25
  • Exactly! Anyway, let me know if you can adapt your previous proof; I'm also working on it, so I'll edit the question in case I get something! – Peanut Dec 02 '20 at 19:21
  • @Peanut: The integral is the upper bound. See addition to answer, leaving some details to you. – RRL Dec 05 '20 at 00:13
  • Hi RRL, do you have any hint on how to prove that: $$\sum_{k = 1}^{n}\frac{\sin(k\pi\frac{2m+1}{n+1})}{k} > \sum_{k = 1}^{n}\frac{\sin(k\pi\frac{2m+3}{n+1})}{k}$$ – Peanut Dec 05 '20 at 14:10
  • It is clearly true or do you disagree? The proof is a bit involved. – RRL Dec 05 '20 at 19:51
  • For me it's not very clear why :( ... maybe you can tell me how you would proceed to prove it? For example I used the sum-to-product formula and wrote it like a sum of $\sin(\ldots) \cdot \cos(\ldots) / k > 0$ but it's still cumbersome... – Peanut Dec 05 '20 at 22:37
  • So since there's one more proof left, i.e. $S_n(x_{n,0})$ is increasing, I tried to do it, but it's still very complicated...do you think I should ask another question? It looks increasing from the graph, but it's hard to prove it, because we cannot even employ functions and derivatives here. – Peanut Dec 08 '20 at 16:18
  • 1
    Yes I think another question is warranted. It is another non-trivial exercise. I have a few ideas but nothing complete. It is interesting that the key step in the proof is very simple -- that the partial sums evaluated at the absolute maximum form a sequence of Riemann sums which converges to $\int_0^\pi \frac{\sin x}{x} , dx$. It is all the other details that are very hard to prove and I think it is too much for one question here. Also maybe someone finds a simple solution we are not seeing, but I doubt it. – RRL Dec 08 '20 at 16:40
  • By the way what was the motivation for this problem? – RRL Dec 08 '20 at 16:41
  • Just an exercise which is torturing me XD – Peanut Dec 08 '20 at 16:57
  • Hi RRL, I opened another question here, in case you are still working on the problem https://math.stackexchange.com/questions/3940360/proving-a-n-sum-k-1n-frac1k-sin-frack-pin1-is-increasing. Cheers! – Peanut Dec 08 '20 at 20:32
  • I posted an answer. The observation that this is a Riemann sum is not relevant for what you want to show here. – RRL Dec 08 '20 at 21:05