# Derivative of cosh x

1. Jan 10, 2013

### Michael_Light

Hi... I read this from my lecture note, but i couldn't understand:

i) the red part, shouldn't it be k =0 instead of k=1?

ii) the third line to the fourth line... i.e.

I have no idea how to change from the third line to the fourth line..

Can anyone enlighten me? Thanks a lot..

2. Jan 10, 2013

### HallsofIvy

Staff Emeritus
It doesn't matter. Since the terms being summed have a factor of "k", the k= 0 term has value 0.

In general, for power series, $f(x)= \sum_{k=0}^\infty a_kx^k$, the derivative can be written as either $f'(x)= \sum_{k=0}^\infty ka_kx^{k- 1}$ or as $f'(x)= \sum_{k=1}^\infty ka_kx^{k- 1}$ because the k= 0 term is 0.

By making the change of index, n= k- 1, so that k= n+1, the second can be written $f'(x)= \sum_{n=0} (n+1)a_{n+1}x^n$.

3. Jan 16, 2013

### THSMathWhiz

The first sum can be rewritten as
$\displaystyle \sum_{k=1}^\infty (2k)\frac{x^{2k-1}}{(2k)!}=\sum_{k=1}^\infty \frac{x^{2k-1}}{(2k-1)!}$​
since $(2k)!=(2k)(2k-1)!$. To change the base index, replace $k$ by $k+1$. You then get the sum
$\displaystyle \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}.$​

4. Jan 17, 2013

### mathman

This seems to be a very roundabout way of getting this result.

cosh(x) = (ex + e-x)/2
The derivative = (ex - e-x)/2 =sinh(x).

5. Jan 17, 2013

### HallsofIvy

Staff Emeritus
IF cosh(x) has been defined as $(e^x+ e^{-x})/2$, yes, that is more direct. However, if cosh(x) has been defined as $\sum_{i=0}^\infty x^{2i}/(2i)!$, as is quite possible and apparently as done in the first post, the argument given is more direct.