Prove that: $\prod_{k=1}^n \sin^2(2^kx)\le (3/4)^{n-1}.$

Asked Jul 14 '20 at 08:23

Active Jul 14 '20 at 15:25

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To prove:$$\prod_{k=1}^n\sin^2(2^kx)\le\left(\frac34\right)^{n-1}.$$

My approach so far:

We have: \begin{align} \prod_{k=1}^n\sin^2(2^kx)&=\left(\prod_{k=1}^n\sin(2^kx)\right)^2\\\\ &=\left[\Im\left\{\prod_{k=1}^ne^{i\cdot2^kx}\right\}\right]^2\\\\ &=\left[\Im\left\{\exp\left(ix\sum_{k=1}^n 2^k\right)\right\}\right]^2\\\\ &=\left[\Im\Big\{\exp\big(ix(2^{n+1}-2^n)\big)\Big\}\right]^2\\\\ &=\sin^2\left[(2^{n+1}-2^n)x\right]. \end{align}

Now how can I prove that (obviously if I haven't done any mistake so far) :

$\sin^2\left[(2^{n+1}-2^n)x\right]\le\left(\frac34\right)^{n-1}\tag*{}\,?$

Please suggest... Thanks in advance.

edited Jul 14 '20 at 15:25

asked Jul 14 '20 at 08:23

Dhrubajyoti Bhattacharjee

1,293

1

This inequality fails for $n=1$ and $x=\frac{\pi}{4}$. In fact, for any $n$ you can choose $x$ to be $\frac{\pi}{2(2^{n+1} -2^n)}$ to make the left expression equal $1$. – Vishu Jul 14 '20 at 08:30
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See here: https://math.stackexchange.com/questions/3749619 – Michael Rozenberg Jul 14 '20 at 08:39
@Tavish, Yeah! So the conditions should be like $n\in\mathbb N\smallsetminus{1}$ and $x\in\mathbb R\wedge x\ne\fracπ{2(2^{n+1}-2^n)}.$ – Dhrubajyoti Bhattacharjee Jul 14 '20 at 08:59
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@Michael Rosenberg, Thank you. – Dhrubajyoti Bhattacharjee Jul 14 '20 at 09:02

Prove that: $\prod_{k=1}^n \sin^2(2^kx)\le (3/4)^{n-1}.$

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