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Thread: Maclaurin series with hyperbolic functions

  1. #1
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    Maclaurin series with hyperbolic functions

    Hi,

    I am having trouble understanding a question:

    I know the maclaurin series up to $\displaystyle x^4$ for cosh(x) goes

    $\displaystyle \cosh(x) = 1 + \frac{x^2}{2} + \frac{x^4}{24}$

    But what happens when it is $\displaystyle \cosh(x^2)$ my understanding of it makes me think that the outcome would be this:

    $\displaystyle \cosh(x^2) = 1 + \frac{x^4}{24}$

    Does anyone know if that is correct

    Thanks for any help
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  2. #2
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    Quote Originally Posted by Beard View Post
    Hi,

    I am having trouble understanding a question:

    I know the maclaurin series up to $\displaystyle x^4$ for cosh(x) goes

    $\displaystyle \cosh(x) = 1 + \frac{x^2}{2} + \frac{x^4}{24}$

    But what happens when it is $\displaystyle \cosh(x^2)$ my understanding of it makes me think that the outcome would be this:

    $\displaystyle \cosh(x^2) = 1 + \frac{x^4}{24}$

    Does anyone know if that is correct

    Thanks for any help
    No that is not correct. How did you get it?

    If $\displaystyle cosh(x)= 1+ \frac{x^2}{2}+ \frac{x^4}{24}$
    (to 4th degree)
    then it should be obvious that
    $\displaystyle cos(x^2)= 1+ \frac{(x^2)^2}{2}+ \frac{(x^2)^4}{24}$
    $\displaystyle = 1+ \frac{x^4}{2}+ \frac{x^8}{24}$.

    If you are still asking about "up to fourth degree" it would be just
    $\displaystyle cos(x^2)= 1+ \frac{x^4}{2}$
    NOT over 24.
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