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Math Help - \sum

  1. #1
    Member vernal's Avatar
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    \sum


    Please solve


    thanks...
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  2. #2
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    Re: \sum

    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.
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  3. #3
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    Krizalid's Avatar
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    Re: \sum

    The question isn't clear here, are we looking for convergence or exact value?
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  4. #4
    Member sbhatnagar's Avatar
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    Re: \sum

    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!)
    Last edited by sbhatnagar; December 29th 2011 at 11:54 PM.
    Thanks from vernal
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    Re: \sum

    Quote Originally Posted by sbhatnagar View Post
    It took me 2 hours to figure out this problem!
    Don't worry, two hours added to your mathematical baggage.
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  6. #6
    Member vernal's Avatar
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    Re: \sum

    Quote Originally Posted by sbhatnagar View Post
    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
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  7. #7
    MHF Contributor MarkFL's Avatar
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    Re: \sum

    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}
    Thanks from vernal
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