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infinite_geometric_series [2016/07/10 22:33]
nikolaj
infinite_geometric_series [2016/07/22 15:20]
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]].
  
 === References === === References ===
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