Difference between revisions of "Exponential functions"
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|[[Integers |$\mathbb{Z}$]] | |[[Integers |$\mathbb{Z}$]] | ||
|$e_p(t,s) = \left\{ \begin{array}{ll} | |$e_p(t,s) = \left\{ \begin{array}{ll} | ||
| − | \displaystyle\prod_{k=s}^t 1+p(t) &; t>s \\ | + | \displaystyle\prod_{k=s}^{t-1} 1+p(t) &; t>s \\ |
1 &; t=s \\ | 1 &; t=s \\ | ||
| − | \displaystyle\prod_{k=t}^s \dfrac{1}{1+p(t)}&; t<s | + | \displaystyle\prod_{k=t}^{s-1} \dfrac{1}{1+p(t)}&; t<s |
\end{array} \right.$ | \end{array} \right.$ | ||
|- | |- | ||
Revision as of 19:42, 19 May 2014
Let $\mathbb{T}$ be a time scale. Define $\xi_h(z) := \dfrac{1}{h} \log(1+zh)$. Let $p \in \mathcal{R}(\mathbb{T},\mathbb{R})$ be a regressive function. The exponential function $e_p \colon \mathbb{T} \times \mathbb{T} \rightarrow \mathbb{R}$ is defined as
$$e_p(t,s) := \exp \left( \displaystyle\int_s^t \xi_{\mu(\tau)}(p(\tau))\Delta \tau \right)$$
for $s,t \in \mathbb{T}$. It turns out that $e_p$ is the unique solution to the dynamic initial value problem $$y^{\Delta} = py; y(s)=1.$$
Properties of Exponential Functions
For all $p,q \in \mathcal{R}$ and $t,s \in \mathbb{T}$,
- $e_p(t,r)e_p(r,s)=e_p(t,s)$ (semigroup property)
- $e_0(t,s)=1, e_p(t,t)=1$
- $e_p(\sigma(t),s)=(1+\mu(t)p(t))e_p(t,s)$
- $\dfrac{1}{e_p(t,s)}=e_{\ominus p}(t,s)$
- $e_p(t,s)e_q(t,s)=e_{p \oplus q}(t,s)$
- $\dfrac{e_p(t,s)}{e_q(t,s)} = e_{p \ominus q}(t,s)$
- $\left( \dfrac{1}{e_p(\cdot,s)} \right)^{\Delta} = -\dfrac{p(t)}{e_p^{\sigma}(\cdot,s)}$
Examples of Exponential Functions
| $\mathbb{T}=$ | $e_p(t,s)=$ |
| $\mathbb{R}$ | $e_p(t,s)= \left\{ \begin{array}{ll} \exp \left( \displaystyle\int_s^t p(\tau) d \tau \right) &; t>s \\ 1 &; t=s \\ \exp \left( -\displaystyle\int_t^s p(\tau) d\tau \right) &; t<s>s \\ 1 &; t=s \\ \displaystyle\prod_{k=t}^{s-1} \dfrac{1}{1+p(t)}&; t<s \end{array} \right.$ |
| $h\mathbb{Z}$ | $\begin{array}{ll} e_p(t,s) &= \exp \left( \displaystyle\int_{s}^{t} \dfrac{1}{\mu(\tau)} \log(1 + hp(\tau)) \Delta \tau \right) \\ &= \exp \left( \displaystyle\sum_{k=\frac{s}{h}}^{\frac{t}{h}-1} \log(1+hp(hk)) \right) \\ &= \displaystyle\prod_{k=\frac{s}{h}}^{\frac{t}{h}-1} \left( 1+hp(hk) \right) \\ \end{array}$ |
| $\mathbb{Z}^2$ | $\begin{array}{ll} e_p(t,s) &= \exp \left( \displaystyle\int_s^t \dfrac{1}{\mu(\tau)} \log(1 + p(\tau) \mu(\tau)) \Delta \tau \right) \\ &= \exp \left( \displaystyle\sum_{k=\sqrt{s}}^{\sqrt{t}-1} \mu(k^2) \dfrac{1}{\mu(k^2)} \log ( 1 + p(k^2)\mu(k^2)) \right) \\ &= \exp \left( \displaystyle\sum_{k=\sqrt{s}}^{\sqrt{t}-1} \log ( 1 + p(k^2)\mu(k^2)) \right) \\ &= \displaystyle\prod_{k=\sqrt{s}}^{\sqrt{t}-1} 1 + p(k^2)(2k+1) \end{array}$ |
| $\mathbb{H}$ | $\begin{array}{ll} e_p(t,s) &= e_p \left( \displaystyle\sum_{k=1}^n \dfrac{1}{k}, \displaystyle\sum_{k=1}^m \dfrac{1}{k} \right) \\ &= \exp \left( \displaystyle\int_{ \sum^m \frac{1}{k}}^{\sum^n \frac{1}{k}} \dfrac{1}{\mu(\tau)} \log(1 + \mu(\tau) p(\tau)) \Delta \tau \right) \\ &= \exp \left( \displaystyle\sum_{k=m}^{n-1} \log \left[1 + \mu \left( \displaystyle\sum_{j=1}^k \dfrac{1}{j} \right) p \left( \displaystyle\sum_{j=1}^k \dfrac{1}{j} \right) \right] \right) \\ &= \exp \left( \displaystyle\sum_{k=m}^{n-1} \log \left[1 + \dfrac{1}{k+1} p \left( \displaystyle\sum_{j=1}^k \dfrac{1}{j} \right) \right] \right) \\ &= \displaystyle\prod_{k=m}^{n-1} 1 + \dfrac{1}{k+1} p \left( \displaystyle\sum_{j=1}^k \dfrac{1}{j} \right) \\ \end{array}$ |
