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infinite_geometric_series [2016/07/10 21:58] nikolaj |
infinite_geometric_series [2016/07/10 22:01] nikolaj old revision restored (2016/07/10 17:44) |
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| - | ===== 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 === | + | or, for $z>0$ and $X<1+z$ resp. $X<z/(z-1)$ |
| - | == 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 $x_0,x_1$ numbers. | + | |
| - | 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 {k \choose m}(-z)^{-m} = z\left(1-\left(\dfrac{z-1}{z}\right)^n\right)$ |
| - | <code> | + | We can get rid of the double sum as follows: |
| - | 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_{k=0}^{n-1}\left(\prod_{j=0}^k\dfrac{n-j}{1+j}\right)(-z)^{-k} $ |
| - | $\sum_{n=0}^\infty \lim_{N\to\infty} a_{n,N} = {\mathrm e}^z$ | + | But note that here the summands depends on the upper bound of the sum and for this we can't take the limit $n\to \infty$ of the summand term anymore. |
| - | == Inverting z == | + | There are some related sums, e.g. |
| - | $a_{n,N} = \dfrac{n}{k+1} \prod_{j=1}^k\left(j-n\right) \cdot \dfrac{z^k}{k!}$ | + | |
| - | $\sum_{n=0}^N a_{n,N} = \dfrac{1}{z} - \dfrac{1}{z}(1-z)^n$ | + | <code> |
| + | expFactor[n_] = (n - (k - j))/(j*f[n]); | ||
| + | invFactor[n_] = -(n - j)/(j + 1); | ||
| + | Sum[Product[expFactor[n]*z, {j, 1, k}], {k, 0, n}] // Simplify | ||
| + | Sum[Product[invFactor[n]*z, {j, 1, k}], {k, 0, n}] // Simplify | ||
| + | </code> | ||
| + | |||
| + | or written differently | ||
| <code> | <code> | ||
| Sum[Product[1 - (k - j)/n, {j, 1, k}] z^k/k!, {k, 0, n}] // Simplify | 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 | Sum[Product[-(n - j), {j, 1, k}] (1/(k + 1)) z^k/k!, {k, 0, n}] // Simplify | ||
| </code> | </code> | ||
