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Math Help - Taylor Series (1+x)^(1/2)

  1. #1
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    Taylor Series (1+x)^(1/2)

    I'm looking to find an expression for the n^{th} term of the Taylor Series for (1+x)^{\frac{1}{2}} .

    I have found that:

    (1+x)^{\frac{1}{2}} = 1 + \frac{1}{1!} \left( \frac{1}{2} \right) x +\frac{1}{2!} \left( \frac{1}{2} \right) \left( \frac{-1}{2} \right) x^2 + \frac{1}{3!} \left( \frac{1}{2} \right) \left(\frac{-1}{2} \right) \left( \frac{-3}{2} \right) x^3 + \ldots

    The first term (1) is the only term that I can't get to fit the suggested general term:

     \frac{1}{(n - 1)!} \displaystyle\prod_{i = 1}^n \left( \frac{5 - 2i}{2} \right)

    Any suggestions much appreciated, as this is really bugging me!
    Last edited by Harry1W; July 31st 2009 at 06:24 PM.
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  2. #2
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    Quote Originally Posted by Harry1W View Post
    I'm looking to find an expression for the n^{th} term of the Taylor Series for (1+x)^{\frac{1}{2}} .

    I have found that:

    (1+x)^{\frac{1}{2}} = 1 + \frac{1}{1!} \left( \frac{1}{2} \right) x +\frac{1}{2!} \left( \frac{1}{2} \right) \left( \frac{-1}{2} \right) x^2 + \frac{1}{3!} \left( \frac{1}{2} \right) \left(\frac{-1}{2} \right) \left( \frac{-3}{2} \right) x^3 + \ldots

    The first term (1) is the only term that I can't get to fit the suggested general term:

     \frac{1}{(n - 1)!} \displaystyle\prod_{i = 1}^n \left( \frac{5 - 2i}{2} \right)

    Any suggestions much appreciated, as this is really bugging me!
    let f(x)=\sqrt{1+x}. then f^{(n)}(x)=\frac{(-1)^{n+1}(2n)!}{(2n-1)4^n n!} \sqrt{(1+x)^{1-2n}}. thus: \frac{f^{(n)}(0)}{n!}=\frac{(-1)^{n+1}(2n)!}{(2n-1)4^n(n!)^2}. therefore: \sqrt{1+x}=\sum_{n=0}^{\infty} \frac{(-1)^{n+1}(2n)!}{(2n-1)4^n(n!)^2} x^n.
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  3. #3
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    Thanks for that. I'm afraid that I might need to ask for some explanation of how you found that the coeffecient of \sqrt{(1+x)^{1-2n}} is \frac{ (-1)^{n + 1} (2n)! }{ (2n - 1) 4^n n! }? I can see that it works, but I don't see why it works - if that makes sense! In other words, I can see that it generates the desired result - but I'm not sure how I would find it independently. Thanks in advance.
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