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Math Help - 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 x^4 for cosh(x) goes

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

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

    \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 x^4 for cosh(x) goes

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

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

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

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