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Math Help - HELP! solving limit with taylor and young

  1. #1
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    Exclamation HELP! solving limit with taylor and young

    i really need someone to solve this limit for me, cause im not sure if i did it right, it has to be done using taylor and young

    lim senhx^3 - senx^3 /x^9
    x-> 0
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  2. #2
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    I must plead ignorance. What is the Taylor and Young method of limits?. I even googled it and found nothing. Do you mean using Taylor series?.
    Also, what is sen and senh?. I assume perhaps sin and sinh?.

    Is this it:

    \lim_{x\to 0}\frac{sinh(x^{3})-sin(x^{3})}{x^{9}}
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by clar View Post
    i really need someone to solve this limit for me, cause im not sure if i did it right, it has to be done using taylor and young

    lim senhx^3 - senx^3 /x^9
    x-> 0
    \text{senh}(x)?

    Using the fact that i is above the middle finger as is e I assume your question is

    \lim_{x\to{0}}\frac{\sinh\left(x^3\right)-\sin\left(x^3\right)}{x^9}

    Note that \sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!} we can see two things

    \sin\left(x^3\right)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{6n+3}}{(2n+1)!}

    And

    \begin{aligned}\sinh\left(x^3\right)&=\sin\left(ix  ^3\right)\\<br />
&=\sum_{n=0}^{\infty}\frac{(-1)^n\left(ix\right)^{2n+1}}{(2n+1)!}\\<br />
&=\sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+1)!}\end{  aligned}

    So your limit is equivalent to

    \begin{aligned}\lim_{x\to{0}}\frac{\sinh\left(x^3\  right)-\sin\left(x^3\right)}{x^9}&=\lim_{x\to{0}}\frac{\l  eft(x^3+\frac{x^9}{6}+\cdots\right)-\left(x-\frac{x^9}{6}+\cdots\right)}{x^9}\\<br />
&=\lim_{x\to{0}}\frac{\frac{x^9}{3}+\cdots}{x^9  }\\<br />
&=\frac{1}{3}\end{aligned}
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  4. #4
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    yes! thats right, sorry its just that im from costa rica, we speak spanish so its diferent
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  5. #5
    Eater of Worlds
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    So, in Coats Rica Taylor series are called Taylor and Young series?. That's a new one I will have to remember.
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  6. #6
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    Not exactly, were supose to solve the limit using the usin taylor series and (resto de Young) which i dont know exactly how to translate into english, it would be like the rest of young which sounds dumb. But its basically the error o(x^n). Sorray i dont really know how to explain.

    But basically the problem i have with the limit is the answer i dont understand why its one third, what happened to the factorial?
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by clar View Post
    Not exactly, were supose to solve the limit using the usin taylor series and (resto de Young) which i dont know exactly how to translate into english, it would be like the rest of young which sounds dumb. But its basically the error o(x^n). Sorray i dont really know how to explain.

    But basically the problem i have with the limit is the answer i dont understand why its one third, what happened to the factorial?
    \frac{1}{3!}+\frac{1}{3!}=\frac{1}{6}+\frac{1}{6}=  \frac{1}{3}
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  8. #8
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    jaja oops , i feel really dumb right now, thank u so much
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  9. #9
    Eater of Worlds
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    Oh, I see. It translates as Big O. The terms which are more or less extraneous. i.e. Not needed because they become small.
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  10. #10
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    ah big O, ill remember that, thanks a lot
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