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Math Help - binomial series expansion

  1. #1
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    binomial series expansion

    use the binomial series to expand the function as a power series.
    \sqrt{1+x}
    I know how to start the problem but i get stuck.
    using:
    (1+x)^k = \sum_{n=0}^{\infty}\binom{k}{n}x^n = 1 + kx + \frac{k(k-1)}{2!}x^2 + \frac{k(k-1)(k-2)}{3!}x^3 + ...

    so I rewrite \sqrt{1+x} as (1+x)^\frac{1}{2}

    therefore (1+x)^\frac{1}{2} = \sum_{n=0}^{\infty}\binom{\frac{1}{2}}{n}x^n

    after that i dont know what to do
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  2. #2
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    Quote Originally Posted by yoman360 View Post
    use the binomial series to expand the function as a power series.
    \sqrt{1+x}
    I know how to start the problem but i get stuck.
    using:
    (1+x)^k = \sum_{n=0}^{\infty}\binom{k}{n}x^n = 1 + kx + \frac{k(k-1)}{2!}x^2 + \frac{k(k-1)(k-2)}{3!}x^3 + ...

    so I rewrite \sqrt{1+x} as (1+x)^\frac{1}{2}

    therefore (1+x)^\frac{1}{2} = \sum_{n=0}^{\infty}\binom{\frac{1}{2}}{n}x^n

    after that i dont know what to do
    Don't write it using combinations. Use the longhand version.


    (1 + x)^k = 1 + kx + \frac{k(k - 1)}{2!}x^2 + \frac{k(k - 1)(k - 2)}{3!}x^3 + \dots

    So (1 + x)^{\frac{1}{2}} = 1 + \frac{1}{2}x + \frac{\frac{1}{2}\left(\frac{1}{2} - 1\right)}{2!}x^2 + \frac{\frac{1}{2}\left(\frac{1}{2} - 1\right)\left(\frac{1}{2} - 2\right)}{3!}x^3 + \dots

     = 1 + \frac{1}{2}x + \frac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2}x^2 + \frac{\frac{1}{2}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{6}x^3 + \dots

     = 1 + \frac{1}{2}x + \frac{-\frac{1}{4}}{2}x^2 + \frac{\frac{3}{8}}{6}x^3 + \dots

     = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 + \dots
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