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infinite_geometric_series [2016/07/10 22:33] nikolaj |
infinite_geometric_series [2019/09/23 21:19] nikolaj |
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| $\sum_{k=0}^\infty\left(1-\dfrac{1}{z}\right)^kX^k = z+(X-1)\left(1+\dfrac{1}{z}\right)\dfrac{1}{z+(1-X)}$ | $\sum_{k=0}^\infty\left(1-\dfrac{1}{z}\right)^kX^k = z+(X-1)\left(1+\dfrac{1}{z}\right)\dfrac{1}{z+(1-X)}$ | ||
| - | == Notes == | + | == q-Integral == |
| + | For a function $f$, the q-integral from $0$ to $1$ ("$z$-integral" if we stick to our notation above) is defined as | ||
| + | |||
| + | $\sum_{k=0}^\infty f(z^k)\,z^k=\dfrac{1}{1-z}\cdot\int_0^1 f(s)\,{\mathrm d}_zs$ | ||
| + | |||
| + | == Related notes == | ||
| $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} $ | $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} $ | ||
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| $ \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)$ | $ \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)$ | ||
| - | See also [[Niemand series]]. | + | See also [[Niemand sequences]]. |
| + | |||
| + | == Powers == | ||
| + | <code> | ||
| + | Sum[Binomial[-s, k] x^k, {k, 0, \[Infinity]}] | ||
| + | Series[(1 - x)^-s, {x, 0, 4}] | ||
| + | Binomial[-s, 3]; | ||
| + | % - (-s)!/(3! ((-s) - 3)!) // FullSimplify | ||
| + | %% + 1/6 (2 s + 3 s^2 + s^3) // FullSimplify | ||
| + | </code> | ||
| === References === | === References === | ||
