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Math Help - [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 :

    \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: <br />
\sum\limits_{n=0}^{\infty} {\binom{-\frac{1}{2}}{n}\cdot{(-4)^n}\cdot{x^n}} =\frac{1}{\sqrt{1-4x}}<br />

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

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

    Thus: <br />
\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 } <br />

    <br />
\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} <br />

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

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

    Thus, \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, \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 \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: \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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    <br />
\sum\limits_{n = 0}^\infty  {\binom{2n}{n} \cdot x^n }  = \sum\limits_{n = 0}^\infty  {\tfrac{{\left( {\tfrac{{\left( {2n} \right)!}}<br />
{{n!}}} \right)}}<br />
{{n!}} \cdot x^n } <br />
let <br />
a_n  = \tfrac{{\left( {2n} \right)!}}<br />
{{n!}}<br />

    Note that: <br />
a_{n + 1}  = 2 \cdot \left( {2n + 1} \right) \cdot a_n <br />

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

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