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Expansion of 1/sqrt(1-x)

  1. Apr 16, 2012 #1
    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?
     
  2. jcsd
  3. Apr 17, 2012 #2
    Assuming first of all that you have the correct coefficients for the binomial expansion of 1/√(1+x) as below,
    [tex]\underbrace{\frac{1}{2}}_{n=1}, \underbrace{\frac{1\cdot 3}{2\cdot 4}}_{n=2}, \underbrace{\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}_{n=3}, \underbrace{\frac{1\cdot 3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8}}_{n=4}, ...[/tex]there is a way to get a nice closed formula for the coefficients without double factorials :smile:

    If you look at the denominator of the fourth term for example,
    2·4·6·8 = 2(1)·2(2)·2(3)·2(4) = 24·4!
    then the general form for the denominator is 2nn!


    Now for the numerator of the fourth term,
    1·3·5·7 = (1·2·3·4·5·6·7·8)/(2·4·6·8) = 8!/(24·4!)
    and so the general form for the numerator is (2n)!/(2nn!)

    Combining them together gives you
    [tex]\frac{(2n)!}{(2^nn!)^2}[/tex]
     
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