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    taylor series

    Find the taylor series centered at c.

    f(x)= 1/(1-x) c=5
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    Quote Originally Posted by Link88 View Post
    Find the taylor series centered at c.

    f(x)= 1/(1-x) c=5
    Dear Link88,

    First write the Taylor series for f(x)=\frac{1}{1-x}.

    f(x)=f(c)+f'(c)(x-c)+\frac{f''(c)}{2!}(x-c)^2+..........

    f'(c)=\frac{1}{(1-c)^2}~,~f''(c)=\frac{2}{(1-c)^3}........

    Therefore, f(x)=\frac{1}{1-c}+\frac{1}{(1-c)^2}(x-c)+\frac{2}{2!(1-c)^3}(x-c)^2+...........

    Since c=5,

    f(x)=\frac{1}{1-5}+\frac{1}{(1-5)^2}(x-5)+\frac{2}{2!(1-5)^3}(x-5)^2+\frac{2\times{3}}{3!(1-5)^4}(x-5)^3+\frac{2\times{3}\times{4}}{4!(1-5)^5}(x-5)^4...........

    f(x)=\frac{1}{1-5}+\frac{1}{(1-5)^2}(x-5)+\frac{1}{(1-5)^3}(x-5)^2+\frac{1}{(1-5)^4}(x-5)^3+\frac{1}{(1-5)^5}(x-5)^4...........

    f(x)=-\frac{1}{4}+\frac{1}{4^2}(x-5)-\frac{1}{4^3}(x-5)^2+\frac{1}{4^4}(x-5)^3-\frac{1}{4^5}(x-5)^4...........

    Hope this helps you.
    Last edited by Sudharaka; July 2nd 2010 at 11:59 PM. Reason: Mistake shown by Faliure
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  3. #3
    Super Member Failure's Avatar
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    Quote Originally Posted by Link88 View Post
    Find the taylor series centered at c.

    f(x)= 1/(1-x) c=5
    An alternative method is to use a geometric series expansion like this:

    \frac{1}{1-x}=\frac{1}{1-5-(x-5)}=-\frac{1}{4}\cdot \frac{1}{1+\frac{x-5}{4}}=-\frac{1}{4}\cdot\sum\limits_{n=0}^\infty \left(-\frac{1}{4}\right)^n(x-5)^n=\sum\limits_{n=0}^\infty\left(-\frac{1}{4}\right)^{n+1}(x-5)^n
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    Super Member Failure's Avatar
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    Quote Originally Posted by Sudharaka View Post
    Dear Link88,

    First write the Taylor series for f(x)=\frac{1}{1-x}.

    f(x)=f(c)+f'(c)(x-c)+\frac{f''(c)}{2!}(x-c)^2+..........

    f'(c)=\frac{1}{(1-c)^2}~,~f''(c)=\frac{2}{(1-c)^3}........
    The signs of the terms must alternate in the taylor series, which might seem a little surprising since
    \frac{1}{1-x}=1+x+x^2+\cdots
    for |x|<1, but such is life...
    Last edited by Failure; July 3rd 2010 at 05:34 AM.
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    Quote Originally Posted by Failure View Post
    Almost, but not quite: f''(c)=-\frac{2}{(1-c)^3}
    The signs of the terms must alternate in the taylor series, which might seem a little surprising since
    \frac{1}{1-x}=1+x+x^2+\cdots
    for |x|<1, but such is life...
    Dear Failure,

    Thanks. Corrected the earlier post.
    Last edited by Sudharaka; July 2nd 2010 at 11:57 PM.
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