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Math Help - taylor series problem

  1. #1
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    taylor series problem

    Hi.
    Evaluate.

    \lim_{\ x\to 0}\frac{\log{cosx}-xsinx}{2e^{x^2}-cos2x-1}
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  2. #2
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    Quote Originally Posted by Shock View Post
    Hi.
    Evaluate.

    \lim_{\ x\to 0}\frac{\log{cosx}-xsinx}{2e^{x^2}-cos2x-1}
    use L'Hopital ... twice.
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  3. #3
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    Quote Originally Posted by skeeter View Post
    use L'Hopital ... twice.


    The prof said we have to solve it with taylor series.
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  4. #4
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    Quote Originally Posted by Shock View Post

    The prof said we have to solve it with taylor series.
    hmm ... I think you'll still have to use L'Hopital with the polynomials.

    maybe someone knows a shortcut. I apologize that I do not.
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  5. #5
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    Look, this is kinda' news to me but it seems to fall out right.. So if we can obtain the Taylor series:

    \log(\cos(x))-x\sin(x)=-3/2 x^2+x^4/10+\cdots

    2e^{x^2}-\cos(2x)-1=4x^2+x^4/3+\cdots

    and then use long division:

    <br />
\begin{array}{rc@{}c}<br />
& \multicolumn{2}{l}{\, \, \, } \vspace*{0.12cm} \\<br />
\cline{2-3}<br />
\multicolumn{1}{r}{4x^2+x^4/3+\cdots \hspace*{-4.8pt}} & \multicolumn{1}{l}{ \hspace*{-5.6pt} \Big) \hspace*{4.6pt} -3/2 x^2+x^4/12+\cdots} \\<br />
& \multicolumn{2}{l}{\phantom{xxxxxxxxxxxxxxxxxxxxxx  }}<br />
\end{array}<br />

    note the first term is -3/8 and I assume the remainder will have powers of x so that as x->0, the limit is -3/8 which is the limit given by Mathematica.

    I'm not entirely sure about the validity of this method with the other one since \sqrt{x} has a branch-point at the origin.
    Last edited by shawsend; March 27th 2010 at 09:42 AM. Reason: not sure about doing this with other one
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  6. #6
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    Quote Originally Posted by shawsend View Post
    Look, this is kinda' news to me but it seems to fall out right.. So if we can obtain the Taylor series:

    \log(\cos(x))-x\sin(x)=-3/2 x^2+x^4/10+\cdots

    2e^{x^2}-\cos(2x)-1=4x^2+x^4/3+\cdots
    not by "hand" I reckon ... ?
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  7. #7
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    Quote Originally Posted by skeeter View Post
    not by "hand" I reckon ... ?
    Why not skeeter? They're both analytic at zero so it's easy to compute the first few terms of the Taylor series for each just be taking a few derivatives at x=0. I guess all this is ok. Never solved a limit this way before. Maybe some issues of uniform convergence is required for it to be valid though.
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  8. #8
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    Quote Originally Posted by shawsend View Post
    Why not skeeter? They're both analytic at zero so it's easy to compute the first few terms of the Taylor series for each just be taking a few derivatives at x=0. I guess all this is ok. Never solved a limit this way before. Maybe some issues of uniform convergence is required for it to be valid though.
    ... still seems like a pain-in-the-@$$ way to do it, IMHO.
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