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niemand_sequence [2016/07/13 13:27] nikolaj |
niemand_sequence [2016/07/22 15:13] (current) nikolaj |
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| == 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!} $ | + | >todo: compare the following construction with the two q-exponentials |
| - | $\sum_{n=0}^N a_{n,N} = \left(1+c(N)\dfrac{z}{N}\right)^N$ | + | $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 $ |
| - | For $c(N)=k$ | + | ${\mathrm e}_N(z):=\sum_{n=0}^N a_{n,N} = \left(1+\dfrac{1}{1-\frac{k\zeta}{N}}c(N)\dfrac{z}{N}\right)^N$ |
| + | |||
| + | 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] k z)/(1 - (k \[Zeta])/N))^n/n!, {n, 0, N}] // simple |
| </code> | </code> | ||
