Differences
This shows you the differences between two versions of the page.
Both sides previous revision Previous revision Next revision | Previous revision Last revision Both sides next revision | ||
infinite_geometric_series [2016/07/10 22:33] nikolaj |
infinite_geometric_series [2016/07/22 15:20] nikolaj |
||
---|---|---|---|
Line 24: | Line 24: | ||
$\sum_{k=0}^\infty\left(1-\dfrac{1}{z}\right)^kX^k = z+(X-1)\left(1+\dfrac{1}{z}\right)\dfrac{1}{z+(1-X)}$ | $\sum_{k=0}^\infty\left(1-\dfrac{1}{z}\right)^kX^k = z+(X-1)\left(1+\dfrac{1}{z}\right)\dfrac{1}{z+(1-X)}$ | ||
- | == 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} $ | ||
Line 32: | Line 37: | ||
$ \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 === |