## Exponential function

### Function

 definition $\exp: \mathbb C\to\mathbb C$ definition $\exp(z):=\sum_{k=0}^\infty \frac{1}{k!} z^k$

#### Theorems

$\mathrm{e}^z = \exp(z)$

Because per definition $\mathrm{e}^z:=\exp(z\cdot \mathrm{ln}(\mathrm{e}))$.

$\mathrm{e}^z \neq 0$
$\frac{\mathrm d}{\mathrm d z}\mathrm{e}^{f(z)} = \frac{\mathrm d}{\mathrm dz}f(z)\cdot \mathrm{e}^{f(z)}$

$a,b,r,\theta\in\mathbb R$

$\exp(i\theta)=\cos(\theta)+i\sin(\theta)$
$\forall a,b.\ \exists r,\theta.\ a+ib=r\mathrm e^{i\theta}$
##### Remarks

We have

$\left(x+y\right)^m=\sum_{k=0}^m \dfrac{m!}{k!\,(m-k)!} x^k y^{m-k}$

so

$\left(1 + b(n)\,x \right)^n = \sum_{k=0}^n \left( b(n)^{-k}\dfrac {n!} {(n-k)!} \right) \dfrac {x^k} {k!}$

(Note that here the summands depend on the upper sum bound $n$, this sum doesn't make for an infinite sum of partial sums - the to be partial sums are all different)

So

$\left(1 + \dfrac {x} {n} \right)^n = \sum_{k=0}^n \left( \dfrac {n!} {(n-k)!\,n^k} \right) \dfrac {x^k} {k!} = \sum_{k=0}^n a_k(n)\dfrac {x^k} {k!}$

with $a_k(n)=\prod_{j=1}^{k}\left(1-\dfrac{k-j}{n}\right)$

also

$= \sum_{k=0}^n \prod_{j=1}^{k}\left(\dfrac{1}{j}-\dfrac{1}{n}\left(\frac{k}{j}-1\right)\right)x$