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niemand_sequence [2016/07/13 13:27]
nikolaj
niemand_sequence [2016/07/14 01:27]
nikolaj
Line 35: Line 35:
  
 == Exponential function == == Exponential function ==
-$a_{n,N} = \left(\prod_{k=1}^n\left(1-\dfrac{k-1}{N}\right)\right) ​c(N)^n\dfrac{z^n}{n!} $+$a_{n,N} = \left(\prod_{k=1}^n\left(1-\dfrac{k-1}{N}\right)\right) \dfrac{1}{n!}\left(\dfrac{C(N)\,​k\,​z}{1-\frac{k\zeta}{N}}\right)^n ​$
  
-$\sum_{n=0}^N a_{n,N} = \left(1+c(N)\dfrac{z}{N}\right)^N$+${\mathrm e}_N(z):=\sum_{n=0}^N a_{n,N} = \left(1+\dfrac{1}{1-\frac{k\zeta}{N}}c(N)\dfrac{z}{N}\right)^N$
  
-For $c(N)=k$+Fulfills 
 + 
 +${\mathrm e}_N'​(\zeta)=k\,​{\mathrm e}_N(\zeta)$ 
 + 
 +(only for $z=\zeta$, not for all of $z$ like the exponential function) 
 + 
 +and with $c(N)=1$, we have
  
 $\lim_{N\to\infty}\sum_{n=0}^N a_{n,​N}={\mathrm e}^{k\,z}$ $\lim_{N\to\infty}\sum_{n=0}^N a_{n,​N}={\mathrm e}^{k\,z}$
  
 <​code>​ <​code>​
-Sum[Product[1 - (k - 1)/N, {k, 1, n}] c[N]^n z^n/n!, {n, 0, N}] // Simplify+Sum[Product[1 - (k - 1)/N, {k, 1, n}] ((C[N] z)/(1 - (k \[Zeta])/​N))^n/n!, {n, 0, N}] // simple
 </​code>​ </​code>​
  
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