Difference between revisions of "Square integers"
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Revision as of 05:31, 18 May 2014
The set $\mathbb{Z}^2 = \{0,1,4,9,16,\ldots\}$ of square integers is a time scale.
| Generic element $t\in \mathbb{T}$: | For some $n \in \mathbb{Z}, t=n^2$ |
| Jump operator: | $\begin{array}{ll} \sigma(t) &= (\sqrt{t}+1)^2 \\ &= t + 2\sqrt{t} + 1 \end{array}$ |
| Graininess operator: | $\begin{array}{ll} \mu(t) &= (t+2\sqrt{t}+1)-t \\ &= 2\sqrt{t}+1 \end{array}$ |
| $\Delta$-derivative: | $f^{\Delta}(t)=\dfrac{f(t+2\sqrt{t}+1)-f(t)}{2\sqrt{t}-1}$ |
| $\Delta$-integral: | $\displaystyle\int_s^t f(t) \Delta t = \displaystyle\sum_{k=\sqrt{s}}^{\sqrt{t}-1} (2k+1)f(k^2)$ |
| Exponential function: | $\begin{array}{ll} e_p(t,s) &= \exp \left( \displaystyle\int_s^t \dfrac{1}{\mu(\tau)} \log(1 + p(\tau) \mu(\tau)) \Delta \tau \right) \\ &= \exp \left( \displaystyle\sum_{k=\sqrt{s}}^{\sqrt{t}-1} \mu(k^2) \dfrac{1}{\mu(k^2)} \log ( 1 + p(k^2)\mu(k^2)) \right) \\ &= \exp \left( \displaystyle\sum_{k=\sqrt{s}}^{\sqrt{t}-1} \log ( 1 + p(k^2)\mu(k^2)) \right) \\ &= \displaystyle\prod_{k=\sqrt{s}}^{\sqrt{t}-1} 1 + p(k^2)(2k+1) \end{array}$ |
