Difference between revisions of "Exponential functions"
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=$\nabla$-exponential Functions= | =$\nabla$-exponential Functions= | ||
| + | Define the function $\hat{\xi}_{h} \colon \mathbb{C}_h \rightarrow \mathbb{Z}_h$ by | ||
| + | $$\hat{\xi}_h(z) = \dfrac{1}{h} \log(1-zh).$$ | ||
| + | Define the $\nabla$ exponential function for $s,t \in \mathbb{T}$ by | ||
| + | $$\hat{e}_p(t,s) = \exp \left( \displaystyle\int_s^t \hat{\xi}_{\nu(\tau)}(p(\tau)) \nabla \tau \right).$$ | ||
Revision as of 10:37, 21 September 2014
The classical exponential function $e^x$ is the unique solution to the initial value problem $$y'=y; y(0)=1.$$
Contents
$\Delta$-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
- The Gaussian bell
| $\mathbb{T}=$ | $e_p(t,s)=$ |
| $\mathbb{R}$ | $e_p(t,s)= \left\{ \begin{array}{ll} \exp \left( \displaystyle\int_s^t p(\tau) d \tau \right) &; t>s \\ 1 &; t=s \\ \exp \left( -\displaystyle\int_t^s p(\tau) d\tau \right) &; t<s \end{array} \right.$ |
| $\mathbb{Z}$ | $e_p(t,s) = \left\{ \begin{array}{ll} \displaystyle\prod_{k=s}^{t-1} 1+p(k) &; t > s \\ 1 &; t=s \\ \displaystyle\prod_{k=t}^{s-1} \dfrac{1}{1+p(k)}&; t < s \end{array} \right.$ |
| $h\mathbb{Z}$ | $ 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.$ |
| $\overline{q^{\mathbb{Z}}}, q > 1$ | $e_p(t,s) = \left\{ \begin{array}{ll} \displaystyle\prod_{k=\log_q(s)}^{\log_q(t)-1} 1 + p(q^k)q^k(q-1) &; t > s \\ 1 &; t=s \\ \displaystyle\prod_{k=\log_q(t)}^{\log_q(s)-1} \dfrac{1}{1+p(q^k)q^k(q-1)} &; t < s \end{array} \right.$ |
| $\overline{q^{\mathbb{Z}}}, q < 1$ | $e_p(t,s) = \left\{ \begin{array}{ll} \displaystyle\prod_{k=\log_q(s)}^{\log_q(t)-1} 1 + p(q^k)q^{k-1}(1-q) &; t > s \\ 1 &; t=s \\ \displaystyle\prod_{k=\log_q(t)}^{\log_q(s)-1} \dfrac{1}{1+p(q^k)q^{k-1}(1-q)} &; t < s \end{array} \right.$ |
| $\mathbb{H}$ | $ 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} \dfrac{1}{1 + \dfrac{1}{k+1} p \left( \displaystyle\sum_{j=1}^k \dfrac{1}{j} \right)} &; t < s \end{array} \right.$ |
$\nabla$-exponential Functions
Define the function $\hat{\xi}_{h} \colon \mathbb{C}_h \rightarrow \mathbb{Z}_h$ by $$\hat{\xi}_h(z) = \dfrac{1}{h} \log(1-zh).$$ Define the $\nabla$ exponential function for $s,t \in \mathbb{T}$ by $$\hat{e}_p(t,s) = \exp \left( \displaystyle\int_s^t \hat{\xi}_{\nu(\tau)}(p(\tau)) \nabla \tau \right).$$
