Differences
This shows you the differences between two versions of the page.
| Both sides previous revision Previous revision | Next revision Both sides next revision | ||
|
exponential_function [2015/12/09 15:10] nikolaj |
exponential_function [2015/12/09 16:06] nikolaj |
||
|---|---|---|---|
| Line 29: | Line 29: | ||
| so | so | ||
| - | $\left(1 + \dfrac {x} {n} \right)^n = \sum_{k=0}^n \left( \dfrac {n!} {(n-k)!\,n^k} \right) \dfrac {x^k} {k!} $ | + | $\left(1 + \dfrac {x} {n} \right)^n = \sum_{k=0}^n \left( \dfrac {n!} {(n-k)!\,n^k} \right) \dfrac {x^k} {k!} = \sum_{k=0}^n a_k(n)\dfrac {x^k} {k!}$ |
| - | $= \sum_{k=0}^n \dfrac {x^k} {k!}\cdot\prod_{j=1}^{k}\left(1-\dfrac{k-j}{n}\right)$ | + | with |
| + | $a_k(n)=\prod_{j=1}^{k}\left(1-\dfrac{k-j}{n}\right)$ | ||
| + | |||
| + | also | ||
| $= \sum_{k=0}^n \prod_{j=1}^{k}\left(\dfrac{1}{j}-\dfrac{1}{n}\left(\frac{k}{j}-1\right)\right)x$ | $= \sum_{k=0}^n \prod_{j=1}^{k}\left(\dfrac{1}{j}-\dfrac{1}{n}\left(\frac{k}{j}-1\right)\right)x$ | ||
