Exponential functions

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

Time Scale Exponential Functions

$\mathbb{T}=$	$e_p(t,s)=$
$\mathbb{R}$	$e_p(t,s)= \left\{ \begin{array}{ll} \exp \left( \displaystyle\int_s^t p(\tau) d \tau \right) &; t>s \\ 1 &; t=s \\ \exp \left( -\displaystyle\int_t^s p(\tau) d\tau \right) &; t<s \end{array} \right.$
$\mathbb{Z}$	$e_p(t,s) = \left\{ \begin{array}{ll} \displaystyle\prod_{k=s}^{t-1} 1+p(k) &; t > s \\ 1 &; t=s \\ \displaystyle\prod_{k=t}^{s-1} \dfrac{1}{1+p(k)}&; t < s \end{array} \right.$
$h\mathbb{Z}$	$ e_p(t,s) = \left\{ \begin{array}{ll} \displaystyle\prod_{k=\frac{s}{h}}^{\frac{t}{h}-1} (1+hp(hk)) &; t > s \\ 1 &; t=s \\ \displaystyle\prod_{k=\frac{t}{h}}^{\frac{s}{h}-1} \dfrac{1}{1+hp(hk)} &; t < s \end{array} \right.$
$\mathbb{Z}^2$	$ e_p(t,s) = \left\{\begin{array}{ll} \displaystyle\prod_{k=\sqrt{s}}^{\sqrt{t}-1} 1 + p(k^2)(2k+1) &; t > s \\ 1 &; t=s\\ \displaystyle\prod_{k=\sqrt{t}}^{\sqrt{s}-1} \dfrac{1}{1+p(k^2)(2k+1)} &; t < s \end{array} \right.$
$\overline{q^{\mathbb{Z}}}, q > 1$	$e_p(t,s) = \left\{ \begin{array}{ll} \displaystyle\prod_{k=\log_q(s)}^{\log_q(t)-1} 1 + p(q^k)q^k(q-1) &; t > s \\ 1 &; t=s \\ \displaystyle\prod_{k=\log_q(t)}^{\log_q(s)-1} \dfrac{1}{1+p(q^k)q^k(q-1)} &; t < s \end{array} \right.$
$\overline{q^{\mathbb{Z}}}, q < 1$	$e_p(t,s) = \left\{ \begin{array}{ll} \displaystyle\prod_{k=\log_q(s)}^{\log_q(t)-1} 1 + p(q^k)q^{k-1}(1-q) &; t > s \\ 1 &; t=s \\ \displaystyle\prod_{k=\log_q(t)}^{\log_q(s)-1} 1+p(q^k)q^{k-1}(1-q) &; t < s \end{array} \right.$
$\mathbb{H}$	$ e_p(t,s) = e_p\left( \displaystyle\sum_{k=1}^n \dfrac{1}{k}, \displaystyle\sum_{k=1}^m \dfrac{1}{k} \right) = \left\{\begin{array}{ll} \displaystyle\prod_{k=m}^{n-1} 1 + \dfrac{1}{k+1} p \left( \displaystyle\sum_{j=1}^k \dfrac{1}{j} \right) &; t > s \\ 1 &; t=s \\ \displaystyle\prod_{k=n}^{m-1} 1 + \dfrac{1}{k+1} p \left( \displaystyle\sum_{j=1}^k \dfrac{1}{j} \right) &; t < s \end{array} \right.$