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infinite_geometric_series [2016/07/10 21:58]
nikolaj
infinite_geometric_series [2016/07/10 22:33]
nikolaj
Line 1: Line 1:
-===== Niemand ​series ===== +===== Infinite geometric ​series ===== 
-==== Note ==== +==== Function ​==== 
-We consider ​$a_{n,N}$, viewed as a sequence (in $N$) of sequences ​(in $n$).+| @#FF9944: definition ​ | @#​FF9944: ​$Q_\infty: \{z\in{\mathbb C}\mid \vert{z}\vert<​1\}\to\mathbb C
 +| @#FF9944: definition ​ | @#​FF9944: ​$Q_\infty(z):​=\sum_{k=0}^\infty z^k $ |
  
-The map+-----
  
-$a_{n,​N} ​\mapsto ​\sum_{n=0}^N a_{n,N}$+$Q_\infty(z)=\dfrac{1}{1-z}$
  
-removes the $n$-index.+This can also be written as
  
-Then+$\sum_{k=0}^\infty\left(\dfrac{1}{1+z}\right)^k = 1+\dfrac{1}{z}$
  
-$\sum_{n=0}^N \lim_{N\to\infty} a_{n,N}$+and
  
-removes the second. ​+$\sum_{k=0}^\infty\left(1-\dfrac{1}{z}\right)^k = z$
  
-=== Examples === +orfor $z>0and $X<1+zresp. $X<​z/​(z-1)$
-== a{n,N} constant in N == +
-For $a_{n,N}constant in $N$, the series ​$\sum_{n=0}^N a_{n,N}is just the sequence of partial sums.+
  
-== Riemann integral == +$\sum_{k=0}^\infty\left(\dfrac{1}{1+z}\right)^kX^k ​1+\dfrac{1}{z}+(X-1)(z-1) \,z\dfrac{1}{z+X(1-z)}$
-$f$ a function and $x_0,x_1numbers.+
  
-With+and
  
-$a_{n,N}:=(x_1-x_0)\dfrac{1}{N}\,f\left(x_0+(x_1-x_0)\dfrac{n}{N}\right)$+$\sum_{k=0}^\infty\left(1-\dfrac{1}{z}\right)^kX^k = z+(X-1)\left(1+\dfrac{1}{z}\right)\dfrac{1}{z+(1-X)}$
  
-we have that $\sum_{n=0}^N a_{n,N}$ is the Riemann integral.+== Notes ==
  
-== Exponential function == +$z \sum_{k=0}^\infty\left(z^{-1}(z-1)\right)^k ​= \sum_{k=0}^\infty\left(1-z^{-1}\right)^k = \sum_{k=0}^\infty \sum_{m=0}^k {k \choose m}(-z)^{-m} $
-$a_{n,N} = \prod_{j=1}^k\left(1-\frac{k-j}{n}\right) \cdot \dfrac{z^k}{k!} $+
  
-$\sum_{n=0}^N a_{n,N} = \left(1+\frac{z}{n}\right)^n$+In fact
  
-$\lim_{N\to\infty}\sum_{n=0}^N a_{n,N}={\mathrm e}^z$+$ \sum_{k=0}^{n-1\sum_{m=0}^k {\choose m}(-z)^{-m} = z\left(1-\left(\dfrac{z-1}{z}\right)^n\right)$
  
-<​code>​ +See also [[Niemand series]].
-Sum[Product[1 - (k - j)/n, {j, 1, k}] z^k/k!, {k, 0, n}] // Simplify +
-</​code>​ +
- +
-What's interesting here is that also +
- +
-$\sum_{n=0}^\infty \lim_{N\to\infty} a_{n,N} = {\mathrm e}^z$ +
- +
-== Inverting z == +
-$a_{n,N} = \dfrac{n}{k+1} \prod_{j=1}^k\left(j-n\right) \cdot \dfrac{z^k}{k!}$ +
- +
-$\sum_{n=0}^N a_{n,N} = \dfrac{1}{z} - \dfrac{1}{z}(1-z)^n$ +
- +
-<​code>​ +
-Sum[Product[1 - (k - j)/n, {j, 1, k}] z^k/k!, {k, 0, n}] // Simplify +
-Sum[Product[-(n - j), {j, 1, k}(1/(k + 1)) z^k/k!, {k, 0, n}// Simplify +
-</​code>​+
  
 === References === === References ===
Link to graph
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