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Math Help - Binomial Series

  1. #1
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    Binomial Series

    Heres my Question:

    Use the binomial series to expand the function
    f(x)=(3 x)\(sqrt(6+x))
    as a power series

    Find:
    C0=
    C1=
    C2=
    C3=
    C4=
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  2. #2
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    Quote Originally Posted by BKennedy View Post
    Heres my Question:

    Use the binomial series to expand the function
    f(x)=(3 x)\(sqrt(6+x))
    as a power series

    Find:
    C0=
    C1=
    C2=
    C3=
    C4=

    (1+x)^k=1+kx+\frac{k(k-1)}{2!}x^2+\frac{k(k-1)(k-2)}{3!}x^3+...=\sum_{n=0}^{\infty}\binom{k}{n}x^n

    where

    \binom{k}{n}=\frac{k(k-1)(k-2) \cdot\cdot\cdot (k-n+1)}{n!}

    n \ge 1 and \binom{k}{0}=1

    So we need to fix up the function so it can fit the form of the theorem..

    \frac{3x}{\sqrt{6+x}}=\frac{3x}{\sqrt{6(1+\frac{x}  {6})}}=

    =\frac{3x}{\sqrt{6}} \cdot\frac{1}{\sqrt{1+\frac{x}{6}}}=\frac{3x}{\sqr  t{6}}\cdot \left(1+\frac{x}{6} \right)^{-1/2}

    Now that it fits the form needed we can expand the series

    \frac{3x}{\sqrt{6}}\left(1-\frac{1}{2}\left( \frac{x}{6} \right)^1+\frac{\frac{-1}{2} \cdot \frac{-3}{2}}{2!}\left(\frac{x}{6} \right)^2 +\frac{\frac{-1}{2} \cdot \frac{-3}{2} \cdot \frac{-5}{2}}{3!}\left( \frac{x}{6} \right)^3 + \frac{\frac{-1}{2} \cdot \frac{-3}{2} \cdot \frac{-5}{2} \cdot \frac{-7}{2}}{4!}\left( \frac{x}{6} \right)^4+...  \right)

    from here it is just multiplication. Good luck.
    Last edited by TheEmptySet; March 14th 2008 at 02:40 PM.
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