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Math Help - Taylor Series 1/x^2 at x=1

  1. #1
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    Taylor Series 1/x^2 at x=1

    Can i make 1/(x^2) into -1/(((1- ((x-1)^2)+2x)))
    and use f(x) = 1/(1-u) where u = ((x-1)^2)+2x))
    and get -sigma : ((x-1)^2 + 2x))^n
    ??
    Or is that illegal
    How about 1/(x^2) = d/dx [-1/x]
    but what is the taylor series for 1/x
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  2. #2
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    Quote Originally Posted by khuezy View Post
    Can i make 1/(x^2) into -1/(((1- ((x-1)^2)+2x)))
    and use f(x) = 1/(1-u) where u = ((x-1)^2)+2x))
    and get -sigma : ((x-1)^2 + 2x))^n
    ??
    Or is that illegal
    How about 1/(x^2) = d/dx [-1/x]
    but what is the taylor series for 1/x
    Does -\Sigma_{n=0}^{\infty} (x^2 + 1)^n (which is what your exprssion reduces to) have the form of a Taylor Series expanded around x = 1 (your post title suggests that you want it to .....)?
    Last edited by mr fantastic; October 8th 2008 at 08:19 PM. Reason: Fixed the sigma
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    yes?

    ?
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  4. #4
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    Quote Originally Posted by khuezy View Post
    ?
    Do you know what a Taylor series looks like? If you do, how can you possibly think that what you posted looks like one.
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    zing!

    just kidding!
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by khuezy View Post
    Can i make 1/(x^2) into -1/(((1- ((x-1)^2)+2x)))
    and use f(x) = 1/(1-u) where u = ((x-1)^2)+2x))
    and get -sigma : ((x-1)^2 + 2x))^n
    ??
    Or is that illegal
    How about 1/(x^2) = d/dx [-1/x]
    but what is the taylor series for 1/x
    You can just apply the definition of Taylor series.

    f(x)=\frac{1}{x^2}\implies f(1)=1

    f'(x)=-\frac{2}{x^3}\implies f'(1)=-2

    f''(x)=\frac{6}{x^4}\implies f''(1)=6

    f'''(x)=-\frac{24}{x^5}\implies f'''(1)=-24

    f^{(4)}(x)=\frac{120}{x^6}\implies f^{(4)}(1)=120

    \therefore \frac{1}{x^2}=1-2(x-1)+\frac{6(x-1)^2}{2!}-\frac{24(x-1)^3}{3!}+\frac{120(x-1)^4}{4!}+\dots =1-2(x-1)+3(x-1)^2-4(x-1)^3+5(x-1)^4+\dots+(-1)^n(n+1)(x-1)^n

    \therefore \sum_{k=0}^{\infty}(-1)^k(k+1)(x-1)^{k}=\frac{1}{x^2} at the point x=1.

    Does this make sense?
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