hie ...
I would like to know how do you solve maclaurian's expansion of 4 sinh x. Is it the same as maclaurian's expansion of 4 sin x or not ?
thankx..
please reply as soon as possible
The answer is $\displaystyle 4\sinh(x)=4\sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+ 1)!}$
To derive it either use the fact that $\displaystyle \sinh(x)=\frac{e^x-e^{-x}}{2}$ and use both those Maclaurin series and add them together...or even more slickly note that $\displaystyle \frac{\sin(ix)}{i}=\sinh(x)$