derivative of cosh x

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..

HallsofIvy

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, [itex]f(x)= \sum_{k=0}^\infty a_kx^k[/itex], the derivative can be written as either [itex]f'(x)= \sum_{k=0}^\infty ka_kx^{k- 1}[/itex] or as [itex]f'(x)= \sum_{k=1}^\infty ka_kx^{k- 1}[/itex] 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 [itex]f'(x)= \sum_{n=0} (n+1)a_{n+1}x^n[/itex].

THSMathWhiz

The first sum can be rewritten as
since [itex](2k)!=(2k)(2k-1)![/itex]. To change the base index, replace [itex]k[/itex] by [itex]k+1[/itex]. You then get the sum

mathman

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

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

HallsofIvy

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