# Thread: First three terms of hyperbolic function

1. ## First three terms of hyperbolic function

Find the first three terms of the series
$\ln (\cosh x)$

$\cosh x=\displaystyle \sum^{\infty}_{r=0} \frac{x^{2r}}{(2r)!}=\sum^{\infty}_{r=0}\frac{x^{2 r}}{(2r)!}+1$

$\ln (\cosh x)=\ln\left(\displaystyle \sum^{\infty}_{r=0}\frac{x^{2r}}{(2r)!}+1\right)=\ displaystyle \sum^{\infty}_{n=1}\left(\frac{(-1)^{n+1}(\displaystyle \sum^{\infty}_{r=0}\frac{x^{2r}}{(2r)!})^n}{n}\rig ht)$
This is as far as I got, I don't think I'm on the right track though.
Thanks!

2. The easiest way is to find the coefficients as...

$\displaystyle a_{n} = \frac{f^{(n)} (0)}{n!}$ (1)

... so that is...

$\displaystyle f(x) = \ln \cosh x = 0$, $x=0$

$\displaystyle f^{'} (x) = \tanh x = 0$, $x=0$

$\displaystyle f^{''} (x) = 1 - \tanh x = 1$, $x=0$

$\displaystyle f^{'''} (x) = \tanh x -1 = -1$, $x=0$

... and the series expansion required is...

$\displaystyle \ln \cosh x = \frac{x^{2}}{2} - \frac{x^{3}}{6} ...$ (2)

Kind regards

$\chi$ $\sigma$