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infinite_geometric_series [2016/07/10 22:33] nikolaj |
infinite_geometric_series [2016/07/22 15:19] 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 $1$ is defined as | ||
| + | |||
| + | $\sum_{k=0}^\infty f(z^k)\,z^k=\dfrac{1}{1-z}\int_0^1 f(s){\mathrm d}_zs$ | ||
| + | |||
| + | == 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]]. |
| === References === | === References === | ||
