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A few days ago I asked a question about an interesting property of the partial sums of the series $\sum\sin(nx)/n$. Here's the link: Bound the absolute value of the partial sums of $\sum \frac{\sin(nx)}{n}$

The proof given there left some details for me, in particular I have to prove that $$ \sum_{k = 1}^{n}{1 \over k} \sin\left(k\pi\frac{2m+1}{n+1}\right) > \sum_{k=1}^{n}{1 \over k}\sin\left(k\pi\frac{2m+3}{n+1}\right) $$ for every natural $n$ and $m$ such that $m = 0, 1, \ldots, \left\lfloor\left(n-1\right)/2\right\rfloor$. I don't think this is relatively simple; anyway I tried a few expansions with sum-to-product and viceversa formulas, but everything I tried leads to $$\sum_{k=1}^{n}\frac{\cos(2(m+1)x_k)\sin(x_k)}{k} < 0$$ where $x_k = k\pi/(n+1)$. From here I don't see any further semplification. Also I noted that $\sin(x_k)$ is always positive since $k$ is at most $n$, but we can't say much about $\cos$. Then I tried to pair terms since for $k' = n+1-k$ we have $\sin(x_k) = \sin(x_k')$, which gives $$\sum_{k=1}^{n/2}\left[\frac{\cos(2(m+1)x_k)}{k}+\frac{\cos(2(m+1)x_{k'})}{n+1-k}\right]\sin(x_k) < 0$$ for $n$ even. Then I hoped that the term between square brackets were always negative, but it's not always the case... How can I do?

Felix Marin
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dude
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  • have you tried taking derivative of $f(x) = \sum_{k=1}^n\dfrac{\sin(k\pi\frac{2x+1}{n+1})}{k}$ and see if it is negative on $0\leq x\leq\frac{n-1}{2}$? – dezdichado Dec 06 '20 at 21:27
  • Actually yes, but then I forgot that we can sum the resulting sum, as it is a sum of $\cos(kx)$. I will try again! Thank you in the meantime! – dude Dec 06 '20 at 22:01
  • and yeah so if you just sum that bastard and integrate again, wouldn't you get a closed formula too – dezdichado Dec 06 '20 at 22:20

2 Answers2

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You have already determined the local maxima of a partial sum $S_n(x)$ in $[0,\pi]$ are at $x_{n,m} = \frac{2m+1}{n+1} \pi$ for $m = 0,1, \ldots ,\lfloor\frac{n-1}{2} \rfloor$.

We first show that the vertical distance between successive peaks is diminishing in that

$$S_n\left(\frac{2m+1}{n+1}\pi \right) - S_n\left(\frac{2m+3}{n+1}\pi \right)> S_n\left(\frac{2m+3}{n+1}\pi \right) - S_n\left(\frac{2m+5}{n+1}\pi \right) > \ldots $$

Note that $\frac{2m+3}{n+1}\pi = \frac{2m+1}{n+1}\pi + \frac{2\pi}{n+1}$ and

$$S_n'\left(x\right) - S_n'\left(x+ \frac{2\pi}{n+1} \right) = \frac{\sin \frac{n}{2}x \cos \frac{n+1}{2}x}{\sin \frac{x}{2}} - \frac{\sin \left(\frac{n}{2}x+ \frac{n\pi}{n+1} \right) \cos \left(\frac{n+1}{2}x+\pi\right)}{\sin \left(\frac{x}{2}+ \frac{\pi}{n+1}\right)} \\ =\frac{\sin \frac{n}{2}x \cos \frac{n+1}{2}x}{\sin \frac{x}{2}} - \frac{\sin \left(\frac{n}{2}x- \frac{\pi}{n+1} \right) \cos \frac{n+1}{2}x}{\sin \left(\frac{x}{2}+ \frac{\pi}{n+1}\right)} \\ = \frac{\cos \frac{n+1}{2}x}{\sin \frac{x}{2}\sin \left(\frac{x}{2}+ \frac{\pi}{n+1}\right)}\left[\sin \frac{x}{2}\sin \left(\frac{n}{2}x+ \frac{\pi}{n+1}\right) -\sin \frac{x}{2}\sin \left(\frac{n}{2}x- \frac{\pi}{n+1}\right) \right]$$

Applying angle addition identities and simplifying we eventually get

$$\tag{1}S_n'\left(x\right) - S_n'\left(x+ \frac{2\pi}{n+1} \right) = \frac{\sin \frac{\pi}{n+1}\sin (n+1)x}{\cos \frac{\pi}{n+1} - \cos \left(\frac{\pi}{n+1} +x \right)}$$

Now consider the points

$$\frac{2m+1}{n+1}\pi \leqslant \frac{2m+2}{n+1}\pi-\frac{y}{n+1} \leqslant \frac{2m+2}{n+1} \leqslant \frac{2m+2}{n+1}\pi+\frac{y}{n+1} \leqslant \frac{2m+3}{n+1}\pi, \\ \frac{2m+3}{n+1}\pi \leqslant \frac{2m+4}{n+1}\pi-\frac{y}{n+1} \leqslant \frac{2m+4}{n+1} \leqslant \frac{2m+4}{n+1}\pi+\frac{y}{n+1} \leqslant \frac{2m+5}{n+1}\pi$$

Using (1) with $x_- = \frac{2m+2}{n+1}\pi-\frac{y}{n+1}$, $x'_-= x_- + \frac{2\pi}{n+1} = \frac{2m+4}{n+1}\pi-\frac{y}{n+1}$, $x_+ = \frac{2m+2}{n+1}\pi+\frac{y}{n+1}$, and$x'_+=x_+ + \frac{2\pi}{n+1} = \frac{2m+4}{n+1}\pi+\frac{y}{n+1}$we get

$$G'(y) = \frac{d}{dy} \left[S_n \left(x_-) - S_n \left(x'_-\right) \right] - S_n(x_+) - S_n(x'_+)\right] \\ = \frac{\sin \frac{\pi}{n+1} \sin y}{n+1}\left[\frac{1}{\cos \frac{\pi}{n+1}- \cos \left(\frac{(2m+3)\pi-y}{n+1}\right)} - \frac{1}{\cos \frac{\pi}{n+1}- \cos \left(\frac{(2m+3)\pi+y}{n+1}\right)} \right] $$

Notice that $G(0) = 0$ and $G'(y) > 0$ for $0 < y < \pi$, so $G(\pi) > 0$ which implies

$$S_n\left(\frac{2m+1}{n+1}\pi \right) - S_n\left(\frac{2m+3}{n+1}\pi \right)> S_n\left(\frac{2m+3}{n+1}\pi \right) - S_n\left(\frac{2m+5}{n+1}\pi \right)$$

When $m = \lfloor \frac{n-1}{2}\rfloor -1 $ we have $\frac{2m+3}{n+1} < \pi < \frac{2m+5}{n+1} = \pi + \delta$. Hence,

$$S_n\left(\frac{2m+5}{n+1}\pi \right) = S_n(\pi + \delta) = S_n(\pi + \delta - 2\pi) = S_n(\delta - \pi)= - S_n(\pi - \delta) < 0,$$

Since $\frac{2m+3}{n+1}$ is the last relative extremum point before $\pi$, $S_n(\pi) = 0$, and $\frac{2m+3}{n+1} < \pi - \delta < \pi$, it follows that

$$S_n\left(\frac{2m+3}{n+1}\pi \right) > 0 > S_n\left(\frac{2m+5}{n+1}\pi \right)$$

Thus, for all $m = 0,1,\ldots, \lfloor\frac{n-1}{2}\rfloor -1$ we have

$$S_n\left(\frac{2m+1}{n+1}\pi \right) - S_n\left(\frac{2m+3}{n+1}\pi \right) > 0$$

RRL
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  • Why is $G'(y) > 0$? – dude Dec 07 '20 at 11:38
  • @Peanut: For $0 < y < \pi$ we see immediately that $\sin \frac{\pi}{n+1}\sin y > 0$. Now $\frac{\pi}{n+1} < \frac{(2m+3)\pi - y}{n+1} < \frac{(2m+3) \pi + y}{n+1} < \pi$, and the cosine function is decreasing on the interval $[0,\pi]$. So $0 < \cos \frac{\pi}{n+1}- \cos \left(\frac{(2m+3)\pi-y}{n+1}\right) < \cos \frac{\pi}{n+1}- \cos \left(\frac{(2m+3)\pi+y}{n+1}\right)$ and the term inside $[\ldots]$ must be positive. – RRL Dec 07 '20 at 15:42
  • For this reason I went through a series of steps using trigonometric identities to get the expression $\cos \frac{\pi}{n+1} - \cos \left(\frac{\pi}{n+1} +x \right)$ in the denominator of (1). – RRL Dec 07 '20 at 15:48
  • Thank you! Now I got you! And for the last statement where you say "it can be shown that...", do you think I can prove it myself or is it involved? XD – dude Dec 07 '20 at 19:42
  • This along with the original post is a great series of questions. When I said before that this monotonicity of the maxima was "clear" I meant that I could see from numerical evidence that it was likely and then tried to justify -- which turned out for me to be not so easy. Again to show these differences are all positive follows from another observation so it should be possible to fill in the final gap. I'll work on it some more. – RRL Dec 07 '20 at 20:00
  • Thank you, I appreciate that very much! Hopefully waiting for the final step! – dude Dec 07 '20 at 20:28
  • @Peanut: I made some updates. – RRL Dec 07 '20 at 22:23
  • One last question: everything is clear also from the graph. But why is it $S_n(\pi-\delta) > 0$? Also notice that I think an equality sign should be introduced, since for example when $n=4$, $m$ is at most $0$ and therefore $S_n(\frac{2m+5}{n+1}\pi) = 0$. Apart from this sign, I fail to see why it should be $\ge 0$, though it's super obvious from the graph. Am I missing something? – dude Dec 07 '20 at 23:05
  • There has to be a last maximum point before $\pi$. I called it $\frac{2m+3}{n+1}\pi$ above by shifting the index, but to be more clear it is the last point of the form $\frac{2m+1}{n+1}\pi$ where $m =0,1,\ldots, M$ -- so $x_M =\frac{2M+1}{n+1}\pi < \pi$. SInce $S_n(\pi) =0$ and there are no relative minima between $x_M$ and $\pi$, if $\pi - \delta$ is between these points we must have $S_n(\pi - x_0) > 0$. If it happened to be $0$ then we would have $\pi - \delta = \pi$ which is not the case. Now what do you think the value of $M$ is? – RRL Dec 08 '20 at 00:22
  • If $n$ is odd, $M= \frac{n-1}{2}$ and $\frac{2M+1}{n+1}\pi = \frac{n}{n+1}\pi < \pi$. Also $\frac{2(M+1)+1}{n+1}\pi = \frac{2M+1}{n+1}\pi + \frac{2}{n+1}\pi = \frac{n}{n+1}\pi + \frac{2}{n+1}\pi =\frac{n+2}{n+1}\pi > \pi$, so indeed the index $M$ corresponds to the last maximum before $\pi$. Now work this out for $n$ even. – RRL Dec 08 '20 at 00:26
  • Thank you! Now it's more clear! – dude Dec 08 '20 at 00:37
  • Also if $n$ is even, then the next point beyond the last maximum is $x_{M+1}= \pi$ as you saw but then $S_n(x_{M+1}) = 0$ and we still prove the differences are greater than $0$. – RRL Dec 08 '20 at 00:40
  • Exactly, it doesn't really matter... – dude Dec 08 '20 at 00:41
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Remark: One year ago, I analyzed the monotonicity of the function values at all local minimizers, when I tried to answer Inequality $\sum\limits_{1\le k\le n}\frac{\sin kx}{k}\ge 0$ (Fejer-Jackson). Here, I give the analysis of the monotonicity of the function values at all local maximizers.

Problem: Given positive integer $n\ge 3$, all the local maximizer of $f(x) = \sum_{k=1}^n \frac{\sin kx}{k}$ on $(0, \pi)$ are given by $x_m = \frac{2m+1}{n+1}\pi, m=0, 1, \cdots, \lfloor \frac{n-1}{2}\rfloor$. Prove that $f(x_m) > f(x_{m+1})$ for all $m=0, 1, \cdots, \lfloor \frac{n-1}{2}\rfloor - 1$.

Proof: Since $f(x) = \int_0^x f'(t) \mathrm{d} t + f(0)$, we have \begin{align} f(x_{m+1}) - f(x_m) &= \int_{\frac{2m+1}{n+1}\pi}^{\frac{2m+3}{n+1}\pi} f'(t) \mathrm{d} t = \int_{\frac{2m+1}{n+1}\pi}^{\frac{2m+2}{n+1}\pi} \Big( f'(t) + f'(t + \tfrac{\pi}{n+1})\Big) \mathrm{d} t. \tag{1} \end{align} Since $f'(x) = \sum_{k=1}^n \cos kx = \frac{\sin \frac{nx}{2} \cos \frac{(n+1)x}{2}}{\sin \frac{x}{2}}$, we have, for all $t\in [\frac{2m+1}{n+1}\pi, \frac{2m+2}{n+1}\pi]$, \begin{align*} &f'(t) + f'(t + \tfrac{\pi}{n+1})\\ =\ & \frac{\sin \frac{nt}{2} \cos \frac{(n+1)t}{2}}{\sin \frac{t}{2}} + \frac{\sin (\frac{nt}{2} + \frac{n}{2n+2}\pi) \cos (\frac{(n+1)t}{2} + \frac{\pi}{2})}{\sin (\frac{t}{2} + \frac{\pi}{2n+2})}\\ =\ & - \frac{\sin (\frac{nt}{2} - m\pi)\sin (\frac{(n+1)t}{2} - \frac{(2m+1)\pi}{2})}{\sin \frac{t}{2}} \\ & - \frac{\sin (\frac{nt}{2} + \frac{n}{2n+2}\pi - m\pi)\sin (\frac{(n+1)t}{2} + \frac{\pi}{2} - \frac{(2m+1)\pi}{2})}{\sin (\frac{t}{2} + \frac{\pi}{2n+2})} \tag{2}\\ =\ & - \frac{\sin (\frac{nt}{2} - m\pi)\sin (\frac{(n+1)t}{2} - \frac{(2m+1)\pi}{2})}{\sin \frac{t}{2}} \\ & - \frac{\cos (\frac{nt}{2} + \frac{n}{2n+2}\pi - m\pi - \frac{\pi}{2})\cos (\frac{(n+1)t}{2} - \frac{(2m+1)\pi}{2})}{\sin (\frac{t}{2} + \frac{\pi}{2n+2})}\\ \le\ & - \frac{\sin (\frac{nt}{2} - m\pi)\sin (\frac{(n+1)t}{2} - \frac{(2m+1)\pi}{2})}{\sin (\frac{t}{2} + \frac{\pi}{2n+2})} \\ & - \frac{\cos (\frac{nt}{2} - m\pi )\cos (\frac{(n+1)t}{2} - \frac{(2m+1)\pi}{2})}{\sin (\frac{t}{2} + \frac{\pi}{2n+2})} \tag{3}\\ \le\ & - \frac{\cos \frac{\pi - t}{2}}{\sin (\frac{t}{2} + \frac{\pi}{2n+2})} \\ \le\ & 0 \end{align*} where in (2) we have used $\sin(\alpha - m\pi) \sin (\beta - \frac{(2m+1)\pi}{2}) = -\sin \alpha \cos \beta$, and in (3) we have used \begin{align} &0 < \frac{nt}{2} + \frac{n}{2n+2}\pi - m\pi - \frac{\pi}{2} < \frac{nt}{2} - m\pi < \pi, \\ &0 \le \frac{(n+1)t}{2} - \frac{(2m+1)\pi}{2} \le \frac{\pi}{2},\\ &0 < \frac{t}{2} < \frac{t}{2} + \frac{\pi}{2n+2} < \frac{\pi}{2}. \end{align} Also, note that $f'(\frac{2m+1}{n+1}\pi) + f'(\frac{2m+1}{n+1}\pi + \tfrac{\pi}{n+1}) = -1 < 0$. From (1), we have $f(x_{m+1}) < f(x_m)$. We are done.

River Li
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