Difference between revisions of "Quantum q greater than 1"
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&= \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} \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) \\ | &= \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 | + | &= \displaystyle\prod_{k=\log_q(s)}^{\log_q(t)-1} 1 + p(q^k)q^{k}(q-1) |
\end{array}$ | \end{array}$ | ||
|} | |} | ||
Revision as of 06:39, 6 February 2015
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: | $\begin{array}{ll} e_p(t,s) &= \exp \left( \displaystyle\int_{s}^{t} \dfrac{1}{\mu(\tau)} \log( 1 + p(\tau) \mu(\tau) ) \Delta \tau \right) \\ &= \exp \left( \displaystyle\sum_{k=\log_q(s)}^{\log_q(t)-1} \mu(q^k) \dfrac{1}{\mu(q^k)} \log(1 + p(q^k)\mu(q^k)) \right) \\ &= \exp \left( \displaystyle\sum_{k=\log_q(s)}^{\log_q(t)-1} \log(1 + p(q^k)\mu(q^k)) \right) \\ &= \displaystyle\prod_{k=\log_q(s)}^{\log_q(t)-1} 1 + p(q^k)q^{k}(q-1) \end{array}$ |
