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infinite_geometric_series [2016/07/10 22:01]
nikolaj old revision restored (2016/07/10 17:44)
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)$
  
-We can get rid of the double sum as follows: +See also [[Niemand sequences]].
- +
-$ = \sum_{k=0}^{n-1}\left(\prod_{j=0}^k\dfrac{n-j}{1+j}\right)(-z)^{-k} $ +
- +
-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. +
- +
-There are some related sums, e.g. +
- +
-<​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>​ +
-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 ===
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