\sum

**vernal** · December 28th 2011, 07:37 AM

$\sum_{1}^{\infty } \frac{ sin^4(2^n)}{4^n}$
Please solve

thanks...

**Ridley** · December 28th 2011, 09:06 AM

There are probably better solutions, but you can rewrite it like this: $\sin^4 (2^n) = \frac{3-4\cos(2^{n+1})+\cos(2^{n+2})}{8}$

You can then split the original sum into three different sums and use the identity $\cos(2^n) = \Re \left\{e^{i2^n} \right\}$ for the two sums that have cosinus terms in them.

**Krizalid** · December 28th 2011, 05:40 PM

The question isn't clear here, are we looking for convergence or exact value?

**sbhatnagar** · December 29th 2011, 06:53 AM

Let $S_n=\sum_{k=1}^{n}\frac{\sin^4{(2^k)}}{4^k}$

$\begin{align*} S_n &=\frac{\sin^4(2)}{4}+\frac{\sin^4(4)}{16}+ \cdots + \frac{\sin^4(2^n)}{4^n} \\ S_n &= \frac{\sin^2(2)-[\sin^2(2)-\sin^4(2)]}{4}+\frac{\sin^2(4)-[\sin^2(4)-\sin^4(4)]}{16}+\cdots + \frac{\sin^2(2^n)-[\sin^2(2^n)-\sin^4(2^n)]}{4^n} \\ S_n &= \frac{\sin^2(2)}{4}-\frac{\sin^2(2)\cos^2(2)}{4}+\frac{\sin^2(2)\cos^2 (2)}{4}-\frac{\sin^2(4)\cos^2(4)}{16}+\cdots +\frac{\sin^2(2^n)}{4^n}-\frac{\sin^2(2^n) \cos^2(2^n)}{4^n}\\ S_n &= \frac{\sin^2(2)}{4}-\frac{\sin^2( 2^n) \cos^2(2^n)}{4^n} \\ S_n &= \frac{1}{4}\left( \sin^2(2)-\frac{\sin^2(2^{n+1})}{4^n}\right)\end{align*}$

Therefore : $S_{\infty}=\lim_{n \to \infty}\sum_{k=1}^{n}\frac{\sin^4{(2^k)}}{4^k}= \frac{1}{4} \lim_{n \to \infty}\left( \sin^2(2)-\frac{\sin^2(2^{n+1})}{4^n}\right)=\frac{\sin^2(2) }{4}$

Happy New Year!
sbhatnagar( It took me 2 hours to figure out this problem!)

**FernandoRevilla** · December 30th 2011, 03:23 AM

Originally Posted by sbhatnagar

It took me 2 hours to figure out this problem!

Don't worry, two hours added to your mathematical baggage.

**vernal** · December 30th 2011, 05:42 AM

Originally Posted by sbhatnagar

Let $S_n=\sum_{k=1}^{n}\frac{\sin^4{(2^k)}}{4^k}$

$\begin{align*} S_n &=\frac{\sin^4(2)}{4}+\frac{\sin^4(4)}{16}+ \cdots + \frac{\sin^4(2^n)}{4^n} \\ S_n &= \frac{\sin^2(2)-[\sin^2(2)-\sin^4(2)]}{4}+\frac{\sin^2(4)-[\sin^2(4)-\sin^4(4)]}{16}+\cdots + \frac{\sin^2(2^n)-[\sin^2(2^n)-\sin^4(2^n)]}{4^n} \\ S_n &= \frac{\sin^2(2)}{4}-\frac{\sin^2(2)\cos^2(2)}{4}+\frac{\sin^2(2)\cos^2 (2)}{4}-\frac{\sin^2(4)\cos^2(4)}{16}+\cdots +\frac{\sin^2(2^n)}{4^n}-\frac{\sin^2(2^n) \cos^2(2^n)}{4^n}\\ S_n &= \frac{\sin^2(2)}{4}-\frac{\sin^2( 2^n) \cos^2(2^n)}{4^n} \\ S_n &= \frac{1}{4}\left( \sin^2(2)-\frac{\sin^2(2^{n+1})}{4^n}\right)\end{align*}$

Therefore : $S_{\infty}=\lim_{n \to \infty}\sum_{k=1}^{n}\frac{\sin^4{(2^k)}}{4^k}= \frac{1}{4} \lim_{n \to \infty}\left( \sin^2(2)-\frac{\sin^2(2^{n+1})}{4^n}\right)=\frac{\sin^2(2) }{4}$

Happy New Year!
sbhatnagar( It took me 2 hours to figure out this problem!)

thanks for your help .Wish the best

**MarkFL** · December 30th 2011, 12:08 PM

This problem showed up on another forum, and this was my response:

First, we may compute:

$\sin^4\left(2^k\right)=\sin^2\left(2^k\right)\left (1-\cos^2\left(2^k\right)\right)=$

$\sin^2\left(2^k \right)-\sin^2\left(2^k\right)\cos^2\left(2^k\right)=\sin^ 2\left(2^k\right)-\frac{1}{4}\sin^2\left(2^{k+1}\right)$

Then we may write:

$\sum_{k=1}^{\infty}\frac{\sin^4\left(2^k\right)}{4 ^k}=\sum_{k=1}^{\infty}\frac{\sin^2\left(2^k\right )-\frac{1}{4}\sin^2\left(2^{k+1}\right)}{4^k}=$

$\frac{\sin^2(2)}{4}+\sum_{k=2}^{\infty}\frac{\sin^ 2\left(2^k\right)}{4^k}-\sum_{k=1}^{\infty}\frac{\sin^2\left(2^{k+1}\right )}{4^{k+1}}=$

$\frac{\sin^2(2)}{4}+0=\frac{\sin^2(2)}{4}$

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