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exponential_function [2016/07/10 15:04] nikolaj |
exponential_function [2019/10/06 17:26] (current) nikolaj typo |
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We have | We have | ||
- | $\left(x+y\right)^m=\sum_{k=0}^m \dfrac{n!}{k!\,(m-k)!} x^k y^{m-k}$ | + | $\left(x+y\right)^m=\sum_{k=0}^m \dfrac{m!}{k!\,(m-k)!} x^k y^{m-k}$ |
so | so | ||
- | $\left(1 + b_n\,x \right)^n = \sum_{k=0}^n \left( \dfrac {1} {b_n^k}\dfrac {n!} {(n-k)!} \right) \dfrac {x^k} {k!}$ | + | $\left(1 + b(n)\,x \right)^n = \sum_{k=0}^n \left( b(n)^{-k}\dfrac {n!} {(n-k)!} \right) \dfrac {x^k} {k!}$ |
- | so | + | (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 | ||
$\left(1 + \dfrac {x} {n} \right)^n = \sum_{k=0}^n \left( \dfrac {n!} {(n-k)!\,n^k} \right) \dfrac {x^k} {k!} = \sum_{k=0}^n a_k(n)\dfrac {x^k} {k!}$ | $\left(1 + \dfrac {x} {n} \right)^n = \sum_{k=0}^n \left( \dfrac {n!} {(n-k)!\,n^k} \right) \dfrac {x^k} {k!} = \sum_{k=0}^n a_k(n)\dfrac {x^k} {k!}$ |