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Math Help - Radius of convergence..Taylor series..

  1. #1
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    Radius of convergence..Taylor series..

    Hey guys..I was revising and stumbled onto this question which the lecturer talked about..but I cant seem to get to how he got the Radius of convergence...heres the question..

    Find the Taylor series for f(x)=e^x about x=1..from this use the ratio test to find the radius of convergency...

    could someone please show me how this is done..Thanks..
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  2. #2
    MHF Contributor Calculus26's Avatar
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    The kth term is x^k/(k!)


    lim(ck+1)/ck =

    lim|x^(k+1)*k!/[(x^k*(k+1)!]|= |x| lim1/(k+1) < 1

    |x|< lim(k+1) = infinity

    So the radius of convergence is inf and the IC (-inf,inf)
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  3. #3
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    Quote Originally Posted by Calculus26 View Post
    The kth term is x^k/(k!)


    lim(ck+1)/ck =

    lim|x^(k+1)*k!/[(x^k*(k+1)!]|= |x| lim1/(k+1) < 1

    |x|< lim(k+1) = infinity

    So the radius of convergence is inf and the IC (-inf,inf)
    But the expansion is about x= 1 so the kth term would be

     <br />
\frac{e(x-1)^k}{k!}<br />

    The interval of convergence is still all real x.
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  4. #4
    MHF Contributor Calculus26's Avatar
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    good catch---didn't even notice the expansion was about x = 1
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