# Maclaurin series with hyperbolic functions

• Nov 5th 2009, 04:13 AM
Beard
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
• Nov 5th 2009, 04:52 AM
HallsofIvy
Quote:

Originally Posted by Beard
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.