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Deduce taylor series

  1. Apr 1, 2008 #1
    Deduce that the Taylor series about 0 of 1/sqrt(1-4x) is the series summation (2n choose n) x^n.
    From this conclude that summation (2n choose n) x^n converges to 1/sqrt(1-4x) for x in (-1/4,1/4).
    Then show that summation (2n choose n) (-1/4)^n = 1/sqrt(1-4(-1/4)) = 1/sqrt(2)

    What I know:
    Taylor series about 0 of f(x) = (1+x)^r, r is a real number given by summation (r choose n) x^n.
    I know that (r choose n) can be rewritten as r(r-1)(r-2)..(r-n+1)/n!
    and I know from a previous question that the Taylor series converges to f(x) for all x in (-1,1), and summation (2n choose n) x^n converges conditionally at x=-1/4.

    How can I do this question with all this information? I am not sure how to piece it all together I am having a lot of trouble with this course.
     
  2. jcsd
  3. Apr 1, 2008 #2
    Well, you know how to Taylor-expand the function [itex]\frac{1}{\sqrt{1+x}}=(1+x)^{-\frac{1}{2}}[/itex], it just your formula with r = -1/2.

    So then just write -4x instead of x and you have the Taylor expansion for the given function:smile:

    You might want to try and play around with the binomial coefficients ("r chose n") to bring it into the desired form.
     
    Last edited: Apr 1, 2008
  4. Apr 2, 2008 #3
    I have tried to do this, but I am stuck and cannot get them to equal each other.
    For f(x)=(1+(-4x))^(-1/2)
    The taylor series about 0 is:
    sum (-1/2 choose n) (-4x)^n
    expanding binomial coefficients:
    sum -1/2(-1/2-1)(-1/2-2)...(-1/2-n+1) / n! x (-4)^n (x)^n
    sum -1/2(-3/2)(-5/2)...(1/2-n)/n! x (-1)^n (4)^n (x)^n

    While the series I am trying to get is:
    sum (2n choose n) x^n
    expanding binomial coefficients:
    sum 2n(2n-1)(2n-2)..(2n-n+1)/n! (x)^n
    sum 2n(n-1/2)(n-1)(n-3/2)...(n+1)/n! (x)^n

    I am stuck now
     
  5. Apr 3, 2008 #4
    can anyone help me out, i am still stuck in same place
     
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