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Math Help - Nifty series (leads to infinite product of cos)

  1. #1
    Member Pranas's Avatar
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    Nifty series (leads to infinite product of cos)

    Hello.

    I am asked to calculate the number

    \[S = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}} {\sum\limits_{n = 1}^{ + \infty } {\frac{{\tan \left( {\frac{x}{{{2^n}}}} \right)}}{{{2^n}}}} } dx\]

    My approach was

    \[S = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}} {\sum\limits_{n = 1}^{ + \infty } {\frac{{\tan \left( {\frac{x}{{{2^n}}}} \right)}}{{{2^n}}}} } dx = \sum\limits_{n = 1}^{ + \infty } {\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}} {\frac{{\tan \left( {\frac{x}{{{2^n}}}} \right)}}{{{2^n}}}} dx}  = \sum\limits_{n = 1}^{ + \infty } {\ln \left( {\frac{{\cos \frac{\pi }{{3 \cdot {2^{n + 1}}}}}}{{\cos \frac{\pi }{{{2^{n + 1}}}}}}} \right)} \]

    which obviously leads to

    \[S = \ln \left( {\frac{{\prod\limits_{n = 1}^{ + \infty } {\cos \frac{\pi }{{3 \cdot {2^{n + 1}}}}} }}{{\prod\limits_{n = 1}^{ + \infty } {\cos \frac{\pi }{{{2^{n + 1}}}}} }}} \right)\]

    and I am not actually sure how to further calculate it (computer software shows that equations here are correct and I am getting closer to the right answer).

    Thanks in advance.
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  2. #2
    Super Member girdav's Avatar
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    For example we will compute \prod_{j=1}^{+\infty}\cos \frac{\pi}{2^{j+1}}.
    Let P_n := \prod_{j=1}^n\cos \frac{\pi}{2^{j+1}}. What about \sin \frac{\pi}4P_n?
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  3. #3
    Member Pranas's Avatar
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    Quote Originally Posted by girdav View Post
    For example we will compute \prod_{j=1}^{+\infty}\cos \frac{\pi}{2^{j+1}}.
    Let P_n := \prod_{j=1}^n\cos \frac{\pi}{2^{j+1}}. What about \sin \frac{\pi}4P_n?
    Using that logic

    \prod\limits_{j = 1}^{ + \infty } {\cos } \frac{\pi }{{{2^{j + 1}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{\sin \frac{\pi }{{{2^{n + 1}}}} \cdot \prod\limits_{j = 1}^n {\cos } \frac{\pi }{{{2^{j + 1}}}}}}{{\sin \frac{\pi }{{{2^{n + 1}}}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{\frac{1}{{{2^n}}} \cdot \sin \frac{\pi }{2}}}{{\sin \frac{\pi }{{{2^{n + 1}}}}}} = \frac{2}{\pi }

    and I am totally blown away by the elegance of the method

    My humble respect, sir.
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