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Exponential function
Function
| definition | $\exp: \mathbb C\to\mathbb C$ |
| definition | $\exp(z):=\sum_{k=0}^\infty \frac{1}{k!} z^k $ |
Discussion
Theorems
| $\mathrm{e}^z = \exp(z) $ |
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Because per definition $\mathrm{e}^z:=\exp(z\cdot \mathrm{ln}(\mathrm{e}))$.
| $\mathrm{e}^z \neq 0 $ |
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| $\frac{\mathrm d}{\mathrm d z}\mathrm{e}^{f(z)} = \frac{\mathrm d}{\mathrm dz}f(z)\cdot \mathrm{e}^{f(z)} $ |
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$a,b,r,\theta\in\mathbb R$
| $\exp(i\theta)=\cos(\theta)+i\sin(\theta)$ |
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| $\forall a,b.\ \exists r,\theta.\ a+ib=r\mathrm e^{i\theta} $ |
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Remarks
We have
$\left(x+y\right)^m=\sum_{k=0}^m \dfrac{n!}{k!\,(m-k)!} x^k y^{m-k}$
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: Exponential function, Matrix exponential, Exponential map
Refinement of
Context
