4

Conjecture

if $x_{k}>0,k=1,2,\cdots,n$, then $$\sum_{k=1}^{n}x_{k}\cos{k}\cdot\sum_{k=1}^{n}x_{k}\sin{k}\le\dfrac{n+3}{8}\sum_{k=1}^{n}x^2_{k}$$

It use $ab\le\dfrac{(a+b)^2}{4}$.so we have $$\sum_{k=1}^{n}x_{k}\cos{k}\cdot\sum_{k=1}^{n}x_{k}\sin{k}\le\dfrac{(\sum_{k=1}^{n}x_{k}(\cos{k}+\sin{k}))^2}{4}\le\dfrac{\sum_{k=1}^{n}x^2_{k}\sum_{k=1}^{n}(\sin{k}+\cos{k})^2}{4}=\dfrac{n+\sum_{k=1}^{n}\sin{2k}}{4}\sum_{k=1}^{n}x^2_{k}$$

math110
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  • Have you tried using sum of Sine's when angles are in AP? It comes out to be $sin(n)*sin(n+1)/sin(1)$ –  Dec 31 '19 at 04:22
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    If you split into cases depending on whether $\sum x_k(\cos k+\sin k)$ is positive or negative, you only need to bound $\sum \max(0,\cos k+\sin k)^2$ and $\sum \min(0,\cos k+\sin k)^2.$ This is more likely to give the "8". – Dap Dec 31 '19 at 07:51
  • This bound is ridiculously loose! define $x=[x_1,\ldots , x_k]^\intercal, c=[\cos 1,\ldots , \cos k]^\intercal, s=[\sin 1,\ldots , \sin k]^\intercal$, then by Cauchy-Schwarz $(x^\intercal c)(x^\intercal s) \le \Vert x \Vert^2 \sqrt{\Vert c \Vert \Vert s \Vert}$ and the bound you get is one order lower! This is just pathetic, loose Cauchy-Schwarz and we don't even need to go over big guns like Hoeffding, Bernstein and friends! – K.K.McDonald Jan 04 '20 at 20:37
  • @K.K.McDonald As you suggest, use Cauchy-Schwarz. Then we get, doing this correctly: $(x^\intercal c)(x^\intercal s) \le \Vert x \Vert^2 \sqrt{\Vert c \Vert^2 \Vert s \Vert^2} = \Vert x \Vert^2 \Vert c \Vert \Vert s \Vert \propto n \Vert x \Vert^2$. So there is no "one order lower" with Cauchy-Schwarz, but this gives the proportionality with $n \Vert x \Vert^2$. Indeed, it gives numerically $\simeq 0.5 \cdot n \Vert x \Vert^2$. So we need to do a better job to find the proportionality constant which is conjectured (roughly) $\frac{n}{8}$. – Andreas Jan 05 '20 at 16:17

1 Answers1

1

Starting with OP's considerations, and using a trig theorem, $$\sum_{k=1}^{n}x_{k}\cos{k}\cdot\sum_{k=1}^{n}x_{k}\sin{k}\le\dfrac{(\sum_{k=1}^{n}x_{k}(\cos{k}+\sin{k}))^2}{4} = \dfrac{(\sum_{k=1}^{n}x_{k}\sin{(k + \pi/4)})^2}{2} $$

Following Dap's comment: since all $x_k > 0$, we can write $\sum_{k=1}^{n}x_{k}\sin{(k + \pi/4)} \le \sum_{k=1}^{n}x_{k}\max\{0,\sin{(k + \pi/4)} \}$ and also $-\sum_{k=1}^{n}x_{k}\sin{(k + \pi/4)} \le - \sum_{k=1}^{n}x_{k}\min\{0,\sin{(k + \pi/4)} \}$. Note both RHS bounds are positive.

Applied to the square we have $(\sum_{k=1}^{n}x_{k}\sin{(k + \pi/4)})^2 \le (\sum_{k=1}^{n}x_{k}\max\{0,\sin{(k + \pi/4)} \})^2$ and $(\sum_{k=1}^{n}x_{k}\sin{(k + \pi/4)})^2 = (-\sum_{k=1}^{n}x_{k}\sin{(k + \pi/4)})^2 \le (\sum_{k=1}^{n}x_{k}\min\{0,\sin{(k + \pi/4)} \})^2 $.

Either one of these inequalities is used, depending on whether $\sum_{k=1}^{n}x_{k}\sin{(k + \pi/4)} >0 $ or $<0$.

Applying Cauchy-Schwarz gives

$$ \frac{ (\sum_{k=1}^{n}x_{k}\sin{(k + \pi/4)})^2}{\sum_{k=1}^{n}x_{k}^2} \le \sum_{k=1}^{n} (\max\{0,\sin{(k + \pi/4)} \})^2 \\ = \frac12 \sum_{k=1}^{n} (\sin{(k + \pi/4)} )^2 + \frac12 \sum_{k=1}^{n} |\sin{(k + \pi/4)}| \sin{(k + \pi/4)} $$ or $$ \frac{ (\sum_{k=1}^{n}x_{k}\sin{(k + \pi/4)})^2}{\sum_{k=1}^{n}x_{k}^2} \le \sum_{k=1}^{n} (\min\{0,\sin{(k + \pi/4)} \})^2 \\ = \frac12 \sum_{k=1}^{n} (\sin{(k + \pi/4)})^2 - \frac12 \sum_{k=1}^{n} |\sin{(k + \pi/4)}| \sin{(k + \pi/4)} $$ Note that $S_n = \sum_{k=1}^{n} |\sin{(k + \pi/4)}| \sin{(k + \pi/4)} $ has alternatingly positive and negative value intervals in $n$. Since the sum's argument samples all phase values of the $sin$-function, the sum has expectation value zero, and maximum variance is limited to a value independent of $n$. Indeed, it is bounded analytically by $|S_n|< 1.27$ (** see below). This shows that the splitting of the above sum is a good choice, since $S_n$ measures only the full results' variance, which is bounded, with a value independent of $n$.

Therefore, combining both cases above, $$ \frac{2 (\sum_{k=1}^{n}x_{k}\sin{(k + \pi/4)})^2}{\sum_{k=1}^{n}x_{k}^2} \le \pm S_n + \sum_{k=1}^{n} (\sin{(k + \pi/4)})^2 \\ = \pm S_n + \frac{n}{2} + \frac{\sin(2)- 2\sin(1) \cos(2 n + 1) }{4 (1-\cos(2) )} $$ where for the last equation a trig summation was used.

Bounding $\cos(2 n + 1) \ge -1$, and using the above bound $|S_n| < 1.27$, gives $$ \frac{2 (\sum_{k=1}^{n}x_{k}\sin{(k + \pi/4)})^2}{\sum_{k=1}^{n}x_{k}^2} \le 1.27 + \frac{n}{2} + \frac{\sin(2)+ 2\sin(1) }{4 (1-\cos(2) )} \le \frac{n}{2} +1.73 $$

Applying this to the first line gives $$\sum_{k=1}^{n}x_{k}\cos{k}\cdot\sum_{k=1}^{n}x_{k}\sin{k}\le \frac{n + 2 \cdot 1.73}{8} \sum_{k=1}^{n}x_{k}^2= \frac{n + 3.46}{8} \sum_{k=1}^{n}x_{k}^2$$

which is almost what we need.

The required result is obtained when bouding(*) simultaneously $\pm S_n + \frac{\sin(2)- 2\sin(1) \cos(2 n + 1) }{4 (1-\cos(2) )} \le 1.5$, giving $$\sum_{k=1}^{n}x_{k}\cos{k}\cdot\sum_{k=1}^{n}x_{k}\sin{k}\le \frac{n + 2 \cdot 1.5}{8} \sum_{k=1}^{n}x_{k}^2= \frac{n + 3}{8} \sum_{k=1}^{n}x_{k}^2$$

which proves the claim. $\qquad \Box$

Remarks:

The above treatment shows that (and how) a bound $\frac{n + c}{8} \sum_{k=1}^{n}x_{k}^2$ can be obtained with some constant $c$. The value $c=3$ could be obtained by computational evaluation of one limit, whereas the looser bound $c = 3.46$ could be obtained analytically. The leading coefficient $ \frac{n}{8} $ could be established in any case.

(*) numerical values of combined bound obtained by computational evaluation.
(**) $|S_n|< 1.27$ is obtained by an analytic bound which is being derived in this post: $|S_n|< 0.149 + \max_{\pm \rho} |\pm \rho + \frac{4}{15 \pi} (\frac{5 \cos(1/2 + \frac{\pi}{4})}{\sin(\frac12)} - \frac{\cos(3/2 + 3 \frac{\pi}{4})}{\sin(\frac32)} )| = M(\phi= \pi/4)$ with $\rho = \frac{8 (2 + 5 \cos(1))}{15 \pi \sin(\frac32)} \simeq 0.8$. This gives $|S_n|< M(\phi= \pi/4) \simeq 0.149 + \max |\pm 0.8 + 0.3135| \le 1.27$.

Andreas
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  • Did you prove $|S_n|< 1.1$? – River Li Jan 06 '20 at 13:39
  • @RiverLi I proved that $|S_n|$ is bounded with some number which is independent of $n$. The actual value 1.1 comes from computational evaluation. Have you got an idea for an analytic bound, which can be a number $> 1.1$ but independent of $n$? – Andreas Jan 06 '20 at 15:34
  • I have no idea. – River Li Jan 07 '20 at 01:21
  • @RiverLi I have now a proof that $|S_n|$ is bounded in the main text. – Andreas Jan 07 '20 at 23:03
  • Your argument of using Fourier series is impressive. Maybe it relates to Diophantine approximation and irrationality measure of $\pi$. See for example, https://math.stackexchange.com/questions/2998694/does-this-sequence-an-frac1n3-sinn-converge?noredirect=1&lq=1 – River Li Jan 08 '20 at 02:54