Difference between revisions of "Isolated points"
From timescalewiki
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|[[Delta_integral | $\Delta$-integral:]] | |[[Delta_integral | $\Delta$-integral:]] | ||
| − | | | + | | $$\displaystyle\int_{t_s}^{t_n} f(\tau) \Delta \tau = \left\{ \begin{array}{ll} |
| + | \displaystyle\sum_{k=s}^{n-1} \mu(t_k)f(t_k) &; n > s \\ | ||
| + | 0 &; n=s \\ | ||
| + | -\displaystyle\sum_{k=n}^{s-1} \mu(t_k) f(t_k) &; n < s | ||
| + | \end{array} \right. $$ | ||
|- | |- | ||
|[[Exponential_functions | Exponential function]]: | |[[Exponential_functions | Exponential function]]: | ||
| − | | | + | | If $t_n > t_s$, $$\begin{array}{ll} |
| + | e_p(t_n,t_s) &= \exp \left( \displaystyle\int_{t_s}^{t_n} \dfrac{1}{\mu(\tau)} \log(1 + p(\tau)) \Delta \tau \right) \\ | ||
| + | &= \exp \left( \displaystyle\sum_{k=s}^{n-1} \log(1+\mu(t_k)p(t_k)) \right) \\ | ||
| + | &= \displaystyle\prod_{k=s}^{n-1} \left( 1+\mu(t_k)p(t_k) \right) \\ | ||
| + | \end{array}$$ | ||
|} | |} | ||
Revision as of 20:04, 4 September 2014
Let $\mathbb{T}$ be a time scale. We say that $\mathbb{T}$ is a time scale of isolated points if there exists $\epsilon > 0$ such that for all $t \in \mathbb{T}$, $\mu(t) \geq \epsilon$. Let $\mathbb{T}=\{\ldots,t_{-1},t_0,t_1,\ldots\}$ be a time scale of isolated points with $t_k > t_n$ iff $k>n$. Define the bijection $\pi \colon \mathbb{T} \rightarrow \mathbb{Z}$, $\pi(t_k)=k$.
| Generic element $t\in \mathbb{T}$: | For some $n \in \mathbb{Z}, t=t_n$ |
| Jump operator: | $\sigma(t)=\sigma(t_n)=t_{n+1}$ |
| Graininess operator: | $\mu(t)=\mu(t_n)=t_{n+1}-t_n$ |
| $\Delta$-derivative: | $f^{\Delta}(t)=f^{\Delta}(t_n) = \dfrac{f(t_{n+1})-f(t_n)}{t_{n+1}-t_n}$ |
| $\Delta$-integral: | $$\displaystyle\int_{t_s}^{t_n} f(\tau) \Delta \tau = \left\{ \begin{array}{ll} \displaystyle\sum_{k=s}^{n-1} \mu(t_k)f(t_k) &; n > s \\ 0 &; n=s \\ -\displaystyle\sum_{k=n}^{s-1} \mu(t_k) f(t_k) &; n < s \end{array} \right. $$ |
| Exponential function: | If $t_n > t_s$, $$\begin{array}{ll} e_p(t_n,t_s) &= \exp \left( \displaystyle\int_{t_s}^{t_n} \dfrac{1}{\mu(\tau)} \log(1 + p(\tau)) \Delta \tau \right) \\ &= \exp \left( \displaystyle\sum_{k=s}^{n-1} \log(1+\mu(t_k)p(t_k)) \right) \\ &= \displaystyle\prod_{k=s}^{n-1} \left( 1+\mu(t_k)p(t_k) \right) \\ \end{array}$$ |
