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exponential_function [2016/07/10 15:04]
nikolaj
exponential_function [2016/07/10 15:08] (current)
nikolaj
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 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
  
 $\left(1 + \dfrac {x} {n} \right)^n = \sum_{k=0}^n \left( \dfrac {n!} {(n-k)!\,​n^k} \right) \dfrac {x^k} {k!} = \sum_{k=0}^n a_k(n)\dfrac {x^k} {k!}$ $\left(1 + \dfrac {x} {n} \right)^n = \sum_{k=0}^n \left( \dfrac {n!} {(n-k)!\,​n^k} \right) \dfrac {x^k} {k!} = \sum_{k=0}^n a_k(n)\dfrac {x^k} {k!}$
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