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Math Help - Expansion of 1/sqrt(1-x)

  1. #1
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    Expansion of 1/sqrt(1-x)

    How do you find the binomial expansion of 1/sqrt(1-x) in series form?


    I know what the term by term expansion is but i'm trying to find the series representation,


    The closest i have found involved double factorials and i'm sure there's an easier representation,


    I've been trying to use the binomial theorem but i get fractional factorials which just give ∞.


    Is there some formula that i haven't been able to find to apply to this?
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  2. #2
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    Re: Expansion of 1/sqrt(1-x)

    Quote Originally Posted by Daniiel View Post
    How do you find the binomial expansion of 1/sqrt(1-x) in series form?


    I know what the term by term expansion is but i'm trying to find the series representation,


    The closest i have found involved double factorials and i'm sure there's an easier representation,


    I've been trying to use the binomial theorem but i get fractional factorials which just give ∞.


    Is there some formula that i haven't been able to find to apply to this?
    \displaystyle \begin{align*} \frac{1}{\sqrt{1 - x}} = (1 - x)^{-\frac{1}{2}} \end{align*}

    Now expand using the Binomial Series.
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    Re: Expansion of 1/sqrt(1-x)

    but alpha = -1/2

    using the binomial coefficient formula don't you get

    (-1/2)!/k!(-1/2-k)!

    the term by term part works, but to find it in terms of the sum is not going so well for me
    Last edited by Daniiel; April 15th 2012 at 10:08 PM.
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    Re: Expansion of 1/sqrt(1-x)

    Quote Originally Posted by Daniiel View Post
    but alpha = -1/2

    using the binomial coefficient formula don't you get

    (-1/2)!/k!(-1/2-k)!

    the term by term part works, but to find it in terms of the sum is not going so well for me
    Did you read what it said about the generalised binomial coefficients?

    \displaystyle \begin{align*} {\alpha\choose{k}} = \frac{\alpha \cdot (\alpha - 1) \cdot (\alpha - 2) \cdot \dots \cdot (\alpha - k + 1)}{k!} \end{align*}

    What will you do in this case where \displaystyle \begin{align*} \alpha = -\frac{1}{2} \end{align*}?
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    Re: Expansion of 1/sqrt(1-x)

    So it has to be

    \frac{1}{\sqrt{1-x}}=\sum\limits_{k=0}^{\infty }{\frac{\left( -\frac{1}{2} \right)\left( -\frac{3}{2} \right)....\left( \frac{1}{2}-k \right)}{k!}}{{(-x)}^{k}}
    Last edited by Daniiel; April 16th 2012 at 05:55 AM. Reason: fixed
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    Re: Expansion of 1/sqrt(1-x)

    Not quite, remember that \displaystyle \begin{align*} (1 - x)^{-\frac{1}{2}} = [1 + (-x)]^{-\frac{1}{2}} \end{align*}, so you should actually have \displaystyle \begin{align*} (-x)^k \end{align*} inside your sum.
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    Re: Expansion of 1/sqrt(1-x)

    ahh right,

    ahah I actually had that but my edit was editing the negative out, one of my friends said it shouldn't be negative, guess he was wrong!

    haha thanks!
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    Re: Expansion of 1/sqrt(1-x)

    i still cant understand how did you exactly solve it?
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    Re: Expansion of 1/sqrt(1-x)

    what do you mean by solve it?

    you pretty much put -1/2 = alpha into that forumla,

    when I was first trying to do it, I was trying to find a way to do it without having to write it in that form, (a+b)(a+2b)....(a+nb) (just an example of what i was trying to avoid, the ... product), I was hoping to have it in terms of factorials or something,
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    Re: Expansion of 1/sqrt(1-x)

    Quote Originally Posted by Daniiel View Post
    what do you mean by solve it?

    you pretty much put -1/2 = alpha into that forumla,

    when I was first trying to do it, I was trying to find a way to do it without having to write it in that form, (a+b)(a+2b)....(a+nb) (just an example of what i was trying to avoid, the ... product), I was hoping to have it in terms of factorials or something,
    You could always write it in terms of the generalised binomial coefficients...
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