Taylor series for $\sinh1$ I am doing taylor series and I want to do it on $\sinh1$.
is there a way to make this problem really simple before I begin?
note: $\sinh x= \cfrac{e^x - e^{-x}}2$  
Any ideas are really helpful thanks. I know how to do taylors btw, just don't want to waste lots of time if there is an easier way to do it then that above $e^x$ stuff.
 A: *

*Write the Taylor series for $e^x$.  

*Write the Taylor series for $e^{-x}$, which you can get by replacing $x$ with $-x$ in (1)

*Subtract (2) from (1), term by term

*Divide all the coefficients by 2.

*Put $x=1$.


This should not take more than a few minutes.
A: If you happen to already know the Taylor series for the sine function
$$\sin z = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}z^{2n+1},$$
then you can use the fact that $\sin iz = i\sinh z$ to get $\sinh 1 = -i\sin i$ to get
$$\sinh 1
= -i\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}i^{2n+1}
= -i\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}(i)^{2n}\cdot i$$
$$= \sum_{n=0}^{\infty}\frac{1}{(2n+1)!}$$
since $-i\cdot i = 1$, and $i^{2n}=(i^2)^n=(-1)^n$, and $(-1)^n(-1)^n = 1$.
The upshot is that the series for $\sinh z$ looks exactly like the series for $\sin z$ except that it isn't alternating. Ditto for $\cosh z$ and $\cos z$, by the way.
A: Here you should simply use the taylor series for the exponential: $$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$$ which also tells us that $$e^{-x} = \sum_{n=0}^\infty \frac{(-1)^n x^n}{n!}.$$
Then the rest is pretty straightforward.
A: You can use that


*

*$\frac{d}{dx} \sinh(x) = \cosh(x)$  

*$\frac{d}{dx} \cosh(x) = \sinh(x)$


This is easy to show using the definition:
$\frac{d}{dx} \sinh(x) = \frac{d}{dx}(\frac{e^x - e^{-x}}{2}) = \frac{e^{x} + e^{-x}}{2} = \cosh(x)$
Then do the taylor series and sort for $\sinh$ and $\cosh$.
A: Since $\sinh 1$ is a constant function, its Taylor series is simply $\sinh 1$, or ${e^1+e^{-1}}\over 2$.
