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Thread: [SOLVED] Finding a power series

  1. #1
    Moo
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    [SOLVED] Finding a power series

    Hello

    This can look like a stupid question, but I just can't see the trick (or maybe because I'm up since 5:30am, I'm too tired to think correctly...)


    Prove :

    $\displaystyle \frac{1}{\sqrt{1-4z}}=\sum_{n \ge 0} {2n \choose n} z^n$


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  2. #2
    Super Member PaulRS's Avatar
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    By the Binomial Theorem: $\displaystyle
    \sum\limits_{n=0}^{\infty} {\binom{-\frac{1}{2}}{n}\cdot{(-4)^n}\cdot{x^n}} =\frac{1}{\sqrt{1-4x}}
    $

    Now prove that: $\displaystyle \binom{-\frac{1}{2}}{n}\cdot{(-4)^n}=\binom{2n}{n}$

    Remember that $\displaystyle \binom{u}{n}=\frac{u\cdot{...}\cdot{(u-n+1)}}{n!}$ $\displaystyle
    u \in \mathbb{R}/n \in \mathbb{N}
    $
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  3. #3
    Super Member PaulRS's Avatar
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    Another way: $\displaystyle
    \int_{ - \pi }^\pi {\cos ^{2n} \left( x \right)dx} = \tfrac{{2\pi }}
    {{4^n }} \cdot \binom{2n}{n}
    $ for all naturals n

    Thus: $\displaystyle
    \sum\limits_{n = 0}^\infty {\left( {\int_{ - \pi }^\pi {\cos ^{2n} \left( x \right)dx} } \right) \cdot \left( {4z} \right)^n } = 2\pi \cdot \sum\limits_{n = 0}^\infty {\binom{2n}{n} \cdot z^n }
    $

    $\displaystyle
    \sum\limits_{n = 0}^\infty {\left( {\int_{ - \pi }^\pi {\cos ^{2n} \left( x \right)dx} } \right) \cdot 4^n \cdot z^n } = \int_{ - \pi }^\pi {\left[ {\sum\limits_{n = 0}^\infty {4^n \cdot z^n \cdot \cos ^{2n} \left( x \right)} } \right]dx}
    $

    Then: $\displaystyle 2\pi \cdot \sum\limits_{n = 0}^\infty {\binom{2n}{n} \cdot z^n } =\int_{ - \pi }^\pi {\tfrac{{dx}}
    {{1 - 4 \cdot z \cdot \cos ^2 \left( x \right)}}}
    $
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  4. #4
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    Another way. ($\displaystyle \Gamma$ is unit circle).

    $\displaystyle {{2n}\choose n} = \frac{1}{2\pi i}\oint_{\Gamma} \frac{(1+z)^{2n}}{z^{n+1}}dz$

    Thus, $\displaystyle \sum_{n=0}^{\infty} {{2n}\choose n} x^n = \frac{1}{2\pi i}\oint_{\Gamma} \sum_{n=0}^{\infty} \frac{(1+z)^{2n}}{z^{n+1}} x^n dz$

    Geometric series, $\displaystyle \frac{1}{2\pi i}\oint_{\Gamma} \frac{z}{z+x(1+z)^2}\cdot \frac{1}{z} dz = \frac{1}{2\pi i}\oint_{\Gamma} \frac{dz}{z+x(1+z)^2} $.

    Now use residue theorem and this integral calculates to $\displaystyle \frac{1}{\sqrt{1-4x}}$.
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  5. #5
    Moo
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    Quote Originally Posted by PaulRS View Post

    Now prove that: $\displaystyle \binom{-\frac{1}{2}}{n}\cdot{(-4)^n}=\binom{2n}{n}$
    Hmm, that's what I was struggling with.. I think I've understood ^^ It's all a matter of denominator ~

    Pooh


    The other methods don't correspond to the general method expected by the paper, but it's quite interesting


    Thanks guys
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  6. #6
    Super Member PaulRS's Avatar
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    $\displaystyle
    \sum\limits_{n = 0}^\infty {\binom{2n}{n} \cdot x^n } = \sum\limits_{n = 0}^\infty {\tfrac{{\left( {\tfrac{{\left( {2n} \right)!}}
    {{n!}}} \right)}}
    {{n!}} \cdot x^n }
    $ let $\displaystyle
    a_n = \tfrac{{\left( {2n} \right)!}}
    {{n!}}
    $

    Note that: $\displaystyle
    a_{n + 1} = 2 \cdot \left( {2n + 1} \right) \cdot a_n
    $

    Thus: $\displaystyle
    \left( {\sum\limits_{n = 0}^\infty {\tfrac{{a_n }}
    {{n!}} \cdot x^n } } \right)^\prime = \sum\limits_{n = 0}^\infty {\tfrac{{2 \cdot \left( {2n + 1} \right) \cdot a_n }}
    {{n!}} \cdot x^n }
    $ let $\displaystyle
    \sum\limits_{n = 0}^\infty {\tfrac{{a_n }}
    {{n!}} \cdot x^n } = y
    $ we have: $\displaystyle
    y^\prime = 2 \cdot \sum\limits_{n = 0}^\infty {\tfrac{{\left( {2n + 1} \right) \cdot a_n }}
    {{n!}} \cdot x^n } = 4 \cdot x \cdot y^\prime + 2y
    $

    Thus our solution satisfies the differential equation: $\displaystyle
    \left( {4 \cdot x - 1} \right) \cdot y^\prime + 2y = 0
    $ which is quite easy to solve (we define $\displaystyle y(0)=\lim_{x\to 0}y(x)=1$)
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