Difference between revisions of "Quantum q greater than 1"
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(Created page with "Let $q>1$. The set $\overline{q^{\mathbb{Z}}}=\{\ldots, q^{-2}, q^{-1}, 1, q, q^2, \ldots \}$ of quantum numbers is a time scale. {| class="wikitable" |+$\mathbb{T}=\over...") |
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− | Let $q>1$. The set $\overline{q^{\mathbb{Z}}}=\{\ldots, q^{-2}, q^{-1}, 1, q, q^2, \ldots \}$ of quantum numbers is a [[time scale]]. | + | Let $q>1$. The set $\overline{q^{\mathbb{Z}}}=\{0, \ldots, q^{-2}, q^{-1}, 1, q, q^2, \ldots \}$ of quantum numbers is a [[time scale]]. |
{| class="wikitable" | {| class="wikitable" | ||
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|- | |- | ||
|[[Delta_derivative | $\Delta$-derivative:]] | |[[Delta_derivative | $\Delta$-derivative:]] | ||
− | |$f^{\Delta}(t) = \ | + | | $f^{\Delta}(t)= \left\{ \begin{array}{ll} |
+ | \dfrac{f(qt)-f(t)}{t(q-1)} &; t\neq 0 \\ | ||
+ | \displaystyle\lim_{h \rightarrow 0} \dfrac{f(h)-f(0)}{h} &; t=0 | ||
+ | \end{array} \right.$ | ||
|- | |- | ||
|[[Delta_integral | $\Delta$-integral:]] | |[[Delta_integral | $\Delta$-integral:]] |
Revision as of 04:46, 18 May 2014
Let $q>1$. The set $\overline{q^{\mathbb{Z}}}=\{0, \ldots, q^{-2}, q^{-1}, 1, q, q^2, \ldots \}$ of quantum numbers is a time scale.
Generic element $t\in \mathbb{T}$: | For some $n \in \mathbb{Z}, t =q^n$ |
Jump operator: | $\sigma(t)=qt$ |
Graininess operator: | $\begin{array}{ll} \mu(t)&=qt-t\\ &=t(q-1) \end{array}$ |
$\Delta$-derivative: | $f^{\Delta}(t)= \left\{ \begin{array}{ll} \dfrac{f(qt)-f(t)}{t(q-1)} &; t\neq 0 \\ \displaystyle\lim_{h \rightarrow 0} \dfrac{f(h)-f(0)}{h} &; t=0 \end{array} \right.$ |
$\Delta$-integral: | $\begin{array}{ll} \displaystyle\int_s^t f(\tau) \Delta \tau &= \displaystyle\sum_{k=\log_q(s)}^{\log_q(t)-1} q^{k-1} (1-q) f(q^k) \\ \end{array}$ |
Exponential function: | $\begin{array}{ll} e_p(t,s) &= \exp \left( \displaystyle\int_{s}^{t} \dfrac{1}{\mu(\tau)} \log( 1 + p(\tau) \mu(\tau) ) \Delta \tau \right) \\ &= \exp \left( \displaystyle\sum_{k=\log_q(s)}^{\log_q(t)-1} \mu(q^k) \dfrac{1}{\mu(q^k)} \log(1 + p(q^k)\mu(q^k)) \right) \\ &= \exp \left( \displaystyle\sum_{k=\log_q(s)}^{\log_q(t)-1} \log(1 + p(q^k)\mu(q^k)) \right) \\ &= \displaystyle\prod_{k=\log_q(s)}^{\log_q(t)-1} 1 + p(q^k)q^{k-1}(1-q) \end{array}$ |