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Math Help - infinite product (sort of)

  1. #1
    Super Member Random Variable's Avatar
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    infinite product (sort of)

    Evaluate  \displaystyle \frac{2}{1} \div \Bigg(\frac{5}{4} \div \bigg( \frac{8}{7} \div \Big(\frac{11}{10} \div \  ... \Big)\bigg) \Bigg)
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  2. #2
    Super Member Random Variable's Avatar
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    Hint

    Spoiler:
    Use the infinite product expression for the sine function.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Random Variable View Post
    Evaluate  \displaystyle \frac{2}{1} \div \Bigg(\frac{5}{4} \div \bigg( \frac{8}{7} \div \Big(\frac{11}{10} \div \  ... \Big)\bigg) \Bigg)
    I'm not certain of the series. Is this in the form:
    \displaystyle \frac{2}{1} \div \left ( \frac{1 + 4}{2 + 2} \div \left ( \frac{(2 + 2)+4}{(1 + 4) + 2} \div \left ( \frac{((1 + 4) + 2) + 4}{((2 + 2) + 4) + 2} \div ~....~ \right ) \right ) \right )

    -Dan
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  4. #4
    Super Member Random Variable's Avatar
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    You're making it too complicated.

    It's equivalent to  \displaystyle \frac{2}{1} * \frac{4}{5} * \frac{8}{7} * \frac{10}{11} * ...

    Now write it as the product of an infinite product for the odd terms and a infinite product for the even terms.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Random Variable View Post
    You're making it too complicated.
    Making things too complicated is my middle name.

    -Dan
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  6. #6
    Super Member Random Variable's Avatar
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    I'm going to post a solution, but there may be other approaches.

     \displaystyle \frac{2}{1} * \frac{4}{5} * \frac{8}{7} * \frac{10}{11} * ... = \prod^{\infty}_{n=0} \frac{6n+2}{6n+1} * \prod^{\infty}_{n=0} \frac{6n+4}{6n+5}

     \displaystyle = \frac{2}{1} * \prod^{\infty}_{n=1} \frac{6n+2}{6n+1} * \prod^{\infty}_{n=1} \frac{6(n-1)+4}{6(n-1)+5}

     \displaystyle = 2 * \prod^{\infty}_{n=1} \frac{6n+2}{6n+1} * \prod^{\infty}_{n=1} \frac{6n-2}{6n-1}

     \displaystyle = 2 \prod^{\infty}_{n=1} \frac{36n^{2}-4}{36n^{2}-1} = 2 \prod^{\infty}_{n=1} \frac{1- (\frac{1}{3n})^{2}}{1-(\frac{1}{6n})^{2}}

     \displaystyle = 2 \ \frac{\prod^{\infty}_{n=1} 1 - (\frac{1}{3n})^{2}}{\prod^{\infty}_{n=1} 1 - (\frac{1}{6n})^{2}}

    Now use the identity  \displaystyle \frac{\sin \pi x}{\pi x} = \prod^{\infty}_{n=1} \Big (1- \frac{x^{2}}{n^{2}} \Big) .

     = \displaystyle 2 \ \frac{\frac{\sin \frac{\pi}{3}}{\frac{\pi}{3}}}{\frac{\sin \frac{\pi}{6}}{\frac{\pi}{6}}} = \sqrt{3}
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