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Math Help - Taylor Series on a limit

  1. #1
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    Taylor Series on a limit

    I'm generally ok with these sorts of problems, but these two have me stumped. Our professor hasn't gone over these particular cases (taylor series on a limit/integral) so any help is appreciated.


    Here I have to use Taylor series to evaluate:



    L'hopital doesnt get me anywhere on this limit. My Ti-89 says the limit is equal to 3 but that doesnt make sense. All the derivatives f'(3), f''(0) etc.. would just be 0 after making Taylors series pointless.




    and,

    Find the first four terms for the MacLaurin (Taylor series evaluated around x = 0) series for



    thanks again in advance,
    Sumner
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  2. #2
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    In the first one, you're supposed to use the Taylor series for e^{x^2} and \sin{x}; namely, as e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots, then e^{x^2}=1+x^2+\frac{x^4}{2}+\frac{x^6}{6}+\cdots; and as \sin{x}=x-\frac{x^3}{6}+\frac{x^5}{120}-\cdots, then c=x^2-\frac{x^4}{6}+\frac{x^6}{120}+\cdots.

    Thus \lim_{x\to0}\frac{e^{x^2}-1-x^2}{x\sin{x}-x^2}=\lim_{x\to0}\frac{\frac{x^4}{2}+\frac{x^6}{6}  +\cdots}{-\frac{x^4}{6}+\frac{x^6}{120}+\cdots}=\lim_{x\to0}  \frac{\frac{1}{2}+\frac{x^2}{6}+\cdots}{-\frac{1}{6}+\frac{x^2}{120}+\cdots}=-\frac{6}{2}=-3.

    --Kevin C.
    Last edited by TwistedOne151; April 8th 2009 at 11:27 PM. Reason: typos
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by white-tt View Post
    I'm generally ok with these sorts of problems, but these two have me stumped. Our professor hasn't gone over these particular cases (taylor series on a limit/integral) so any help is appreciated.


    Here I have to use Taylor series to evaluate:



    L'hopital doesnt get me anywhere on this limit. My Ti-89 says the limit is equal to 3 but that doesnt make sense. All the derivatives f'(3), f''(0) etc.. would just be 0 after making Taylors series pointless.
    follow the instructions. after writing out the Taylor expansions for e^{x^2} and \sin x, there will be no need of L'Hopital's rule. it will simply be a ratio of polynomials, which you should be able to deal with

    and,

    Find the first four terms for the MacLaurin (Taylor series evaluated around x = 0) series for



    thanks again in advance,
    Sumner
    do you remember how to find the coefficients of the Taylor series? Hint: for the first derivative, use the second fundamental theorem of calculus
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  4. #4
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    In the second, remember the definition of the Maclaurin series for f(x):
    f(0)+f'(0)x+\frac{f''(0)}{2}x^2+\frac{f'''(0)}{6}x  ^2+\cdots. Now, if f(x)=\int_{0}^{x}\sqrt{1+t^3}\,dx, then we see f(0)=\int_{0}^{0}\sqrt{1+t^3}\,dx=0. The Fundamental theorem of calculus tells us that f'(x)=\sqrt{1+x^3}, and from there you can find the higher derivatives, and thus the Maclaurin series.

    --Kevin C.
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  5. #5
    MHF Contributor chisigma's Avatar
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    A good alternative for arriving to the Mc Laurin expansion of...

    f(x)= \int_{0}^{x} \sqrt{1+t^{3}}\cdot dt (1)

    ... is the following. You can start with the binomial series...

    \sqrt{1+a}= 1 + \frac{1}{2}\cdot a - \frac{1}{2\cdot 4}\cdot a^{2} + \frac{1\cdot 3}{2\cdot 4\cdot 6}\cdot a^{3} - \dots (2)

    ... then you set in (2) a=t^{3} and finally integrate 'term by term' ...

    Kind regards

    \chi \sigma
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  6. #6
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    Wow! Thanks for your help everybody, I didnt think of using expansions on the first one, that makes sense now.

    You guys are a lifesaver!

    Thanks again!!

    Sumner
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  7. #7
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    Hi

    These 2 problems have already been posted recently

    http://www.mathhelpforum.com/math-he...or-series.html

    http://www.mathhelpforum.com/math-he...-integral.html

    I guess generic1 and you are in the same class !
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  8. #8
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    Quote Originally Posted by running-gag View Post
    Hi

    These 2 problems have already been posted recently

    http://www.mathhelpforum.com/math-he...or-series.html

    http://www.mathhelpforum.com/math-he...-integral.html

    I guess generic1 and you are in the same class !

    haha doesnt surprise me! Everybody's been pretty lost on these. Sorry for the double post then, guys! I tried searching but didnt come across those.
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