# Thread: maclaurian's expansion of 4 sinh x

1. ## maclaurian's expansion of 4 sinh x

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..

2. Originally Posted by aero-me
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..
The answer is $4\sinh(x)=4\sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+ 1)!}$

To derive it either use the fact that $\sinh(x)=\frac{e^x-e^{-x}}{2}$ and use both those Maclaurin series and add them together...or even more slickly note that $\frac{\sin(ix)}{i}=\sinh(x)$

3. Originally Posted by Mathstud28
The answer is $4\sinh(x)=4\sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+ 1)!}$

To derive it either use the fact that $\sinh(x)=\frac{e^x+e^{-x}}{2}$ and use both those Maclaurin series and add them together...or even more slickly note that $\frac{\sin(ix)}{i}=\sinh(x)$
No,

$\sinh(x)=\frac{e^x-e^{-x}}{2}$

But the series still looks OK

CB

4. hey thankx fer yur post......its relli worked out now..