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Math Help - Taylor Series

  1. #1
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    Taylor Series

    Compute the Taylor series of (1-x)^(-1/2) by Taylor's formula.


    I used Taylor's formula to get the first few terms but couldn't see a reasonable pattern to generalize...

    Any help at all is appreciated
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  2. #2
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    Quote Originally Posted by cassie00 View Post
    Compute the Taylor series of (1-x)^(-1/2) by Taylor's formula.


    I used Taylor's formula to get the first few terms but couldn't see a reasonable pattern to generalize...

    Any help at all is appreciated
    Since you are concerned with f(0), f'(0), f''(0) etc., keep in mind that 1^n = 1 for all n.

    So you are having trouble generalizing the product \frac{-1}{2}*\frac{-3}{2}*\frac{-5}{2}...?
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  3. #3
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    Quote Originally Posted by icemanfan View Post
    Since you are concerned with f(0), f'(0), f''(0) etc., keep in mind that 1^n = 1 for all n.

    So you are having trouble generalizing the product \frac{-1}{2}*\frac{-3}{2}*\frac{-5}{2}...?

    I thought the summation also involved dividing by n!.
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  4. #4
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    Quote Originally Posted by cassie00 View Post
    Compute the Taylor series of (1-x)^(-1/2) by Taylor's formula.


    I used Taylor's formula to get the first few terms but couldn't see a reasonable pattern to generalize...

    Any help at all is appreciated

    If you know the binomial theorem it would make this problem alot faster?

    f(x)=(1-x)^{-1/2}
    f'(x)=(\frac{1}{2})(1-x)^{-3/2}
    f''(x)=(\frac{1}{2})(\frac{3}{2})(1-x)^{-5/2}
    f'''(x)=(\frac{1}{2})(\frac{3}{2})(\frac{5}{2})(1-x)^{-7/2}


    if we notice the pattern for n > 1 we get

    f^{(n)}(x)=\frac{1\cdot 3\cdot 5 \cdot \cdot \cdot (2n-1)}{2^n}(1-x)^{2n+1}

    This should get you started
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