Difference between revisions of "Derivation of delta exponential T=isolated points"
From timescalewiki
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$$\begin{array}{ll} | $$\begin{array}{ll} | ||
e_p(t,s) &= \exp \left( \displaystyle\int_s^t \dfrac{1}{\mu(\tau)} \log(1+\mu(\tau)p(\tau)) \Delta \tau \right) \\ | e_p(t,s) &= \exp \left( \displaystyle\int_s^t \dfrac{1}{\mu(\tau)} \log(1+\mu(\tau)p(\tau)) \Delta \tau \right) \\ | ||
| − | &= \exp \left( \displaystyle\sum_{k=\pi(s)}^{\pi(t)-1} \log(1+\mu(t_k)p(t_k) \right) \\ | + | &= \exp \left( \displaystyle\sum_{k=\pi(s)}^{\pi(t)-1} \log(1+\mu(t_k)p(t_k)) \right) \\ |
&= \displaystyle\prod_{k=\pi(s)}^{\pi(t)-1} 1+\mu(t_k)p(t_k). | &= \displaystyle\prod_{k=\pi(s)}^{\pi(t)-1} 1+\mu(t_k)p(t_k). | ||
\end{array}$$ | \end{array}$$ | ||
Latest revision as of 23:35, 9 June 2015
Let $\mathbb{T}$ be a time scale of isolated points. For $t>s$, find $e_p$ by computing $$\begin{array}{ll} e_p(t,s) &= \exp \left( \displaystyle\int_s^t \dfrac{1}{\mu(\tau)} \log(1+\mu(\tau)p(\tau)) \Delta \tau \right) \\ &= \exp \left( \displaystyle\sum_{k=\pi(s)}^{\pi(t)-1} \log(1+\mu(t_k)p(t_k)) \right) \\ &= \displaystyle\prod_{k=\pi(s)}^{\pi(t)-1} 1+\mu(t_k)p(t_k). \end{array}$$
