===== Exponential function =====
==== Function ====
| @#FF9944: definition  | @#FF9944: $\exp: \mathbb C\to\mathbb C$ |
| @#FF9944: definition  | @#FF9944: $\exp(z):=\sum_{k=0}^\infty \frac{1}{k!} z^k $ |

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=== Discussion ===
=== 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

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

=== References ===
Wikipedia: 
[[http://en.wikipedia.org/wiki/Exponential_function|Exponential function]], 
[[http://en.wikipedia.org/wiki/Matrix_exponential|Matrix exponential]], 
[[http://en.wikipedia.org/wiki/Exponential_map|Exponential map]]

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=== Refinement of ===
[[Matrix exponential]]
=== Context ===
[[Infinite sum of complex numbers]], 
[[Factorial function]]