Taylor series for $\sinh1$

Question

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.

Answer 1

1. Write the Taylor series for $e^x$.
2. Write the Taylor series for $e^{-x}$, which you can get by replacing $x$ with $-x$ in (1)
3. Subtract (2) from (1), term by term
4. Divide all the coefficients by 2.
5. Put $x=1$.

This should not take more than a few minutes.

Answer 2

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.

Answer 3

You can use that

1. $\frac{d}{dx} \sinh(x) = \cosh(x)$
2. $\frac{d}{dx} \cosh(x) = \sinh(x)$

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 $\cosh$.

Answer 4

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.