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Math Help - MacLaurin Seires

  1. #1
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    MacLaurin Seires

    Hello,
    I am trying to find the Maclaurin series (Taylor series centered about x=0) as well as the radius of converge for
    f(x)=\frac{1}{\sqrt{1+x}}

    Also, assuming the above series represent f(x) on the above interval of convergence, can that series be used to find a Maclaurin series for
    \frac{1}{\sqrt{1-t^2}}? (if it can, how?)

    Am I on the right track with the second part? I am a bit stalled on part a).

    Thank you for any help or direction
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  2. #2
    MHF Contributor chisigma's Avatar
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    Given a real number \alpha , the function f(x) = (1+x)^{\alpha} can be written as McLaurin series for -1 < x < 1 as follows...

    (1+x)^{\alpha}= \sum_{n=0}^{\infty} \binom {\alpha}{n}\cdot x^{n} (1)

    ... where...

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

    Such series is called binomial series. Setting in (1) \alpha= - \frac{1}{2} we have...

    \frac{1}{\sqrt{1+x}}= 1 - \frac{1}{2}\cdot x + \frac{1\cdot 3}{2\cdot 4}\cdot x^{2} - \frac{1\cdot 3 \cdot 5}{2\cdot 4\cdot 6}\cdot x^{3} + \dots + (-1)^{n} \frac{1\cdot 3\cdot 5\dots (2n-1)}{2\cdot 4\cdot 6\dots 2n}\cdot x^{n} +\dots (3)

    If You have to write the McLaurin expansion of f(t)= \frac{1}{\sqrt{1-t^{2}}}, all what you have to do is to write -t^{2} instead of x in (3)...

    Kind regards

    \chi \sigma
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