Question

\[\sum_{n=0}^{\infty} (coshx)' =\sum_{n=0}^{\infty} sinhx\] how ????

Answer 1

knowing that coshx=\[\sum_{n=0}^{\infty} x^2 x^n /( 2n)!\]

Answer 2

dose it make any sense to u m8 ?

Answer 3

did the derivation and didnt have answer that make any sense

Answer 4

Jeez ... \( (\cos hx)' = \sin hx \)

Answer 5

i had 2n x^(2n-1)/2n!

Answer 6

i am talking about the sum of an infinite series here

Answer 7

\[ \cos hx = \frac{e^x + e^{-x}}{2}\]
\[ \sin hx = \frac{e^x - e^{-x}}{2}\]

Answer 8

Oh .. sorry ... i thought it was hyperbolic function ... I'm out of my mind today

Answer 9

is that 'h' or 'n' ??

Answer 10

the derivation of the sum of the coshx is sinhx its true ,

Answer 11

it seems that u didnt take the calculus II course yet :P

Answer 12

ok what is the derivative for x! factorial of x ? ?

Answer 13

you should be taking the derivative with respect to x, not n. anything with just an n is a constant.

Answer 14

The only thing you should be applying to the derivative of that power series is the power rule.

Answer 15

\[\cosh (x) = \sum_{n=0}^\infty \frac{x^{2n}}{(2n)!}\]The derivative of this is:\[\frac{d}{dx}\left(\frac{x^{2n}}{(2n)!}\right)=2n\cdot \frac{x^{2n-1}}{(2n)!}=\frac{x^{2n-1}}{(2n-1)!}\]The infinite summation of what you have left will be sinh(x)

Answer 16

what's the point of summing if you don't have 'n' ...??