Difference between revisions of "Exponential functions"
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| − | Define $\xi_h(z) := \dfrac{1}{h} \log(1+zh)$. Let $p \colon \mathbb{T} \rightarrow \mathbb{R}$ be a function | + | Define $\xi_h(z) := \dfrac{1}{h} \log(1+zh)$. Let $p \colon \mathbb{T} \rightarrow \mathbb{R}$ be a [[regressive_function | regressive function]]. The exponential function $e_p \colon \mathbb{T} \times \mathbb{T} \rightarrow \mathbb{R}$ is defined as |
$$e_p(t,s) := \exp \left( \displaystyle\int_s^t \xi_{\mu(\tau)}(p(\tau))\Delta \tau \right)$$ | $$e_p(t,s) := \exp \left( \displaystyle\int_s^t \xi_{\mu(\tau)}(p(\tau))\Delta \tau \right)$$ | ||
Revision as of 03:13, 18 May 2014
Define $\xi_h(z) := \dfrac{1}{h} \log(1+zh)$. Let $p \colon \mathbb{T} \rightarrow \mathbb{R}$ 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 $$y^{\Delta} = py; y(s)=1.$$
| $\mathbb{T}=$ | $e_p(t,s)=$ |
| $\mathbb{R}$ | $\begin{array}{ll} e_p(t,s) &= \exp \left( \displaystyle\int_s^t \displaystyle\lim_{h \rightarrow 0} \dfrac{1}{h} \log(1 + hp(\tau)) d\tau \right) \\ &\hspace{-10pt} \stackrel{\mathrm{L'Hôpital}}{=} \exp \left( \displaystyle\int_s^t \displaystyle\lim_{h \rightarrow 0} \dfrac{1}{1+hp(\tau)} p(\tau) d\tau \right) \\ &= \exp \left( \displaystyle\int_s^t p(\tau) d \tau \right) \end{array}$ |
| $h\mathbb{Z}$ | $\begin{array}{ll} e_p(t,s) &= \exp \left( \displaystyle\int_{s}^{t} \dfrac{1}{\mu(\tau)} \log(1 + hp(\tau)) \Delta \tau \right) \\ &= \exp \left( \displaystyle\sum_{k=\frac{s}{h}}^{\frac{t}{h}-1} \log(1+hp(hk)) \right) \\ &= \displaystyle\prod_{k=\frac{s}{h}}^{\frac{t}{h}-1} \left( 1+hp(hk) \right) \\ \end{array}$ |
| $\mathbb{Z}^2$ | $\begin{array}{ll} e_p(t,s) &= \exp \left( \displaystyle\int_s^t \dfrac{1}{\mu(\tau)} \log(1 + p(\tau) \mu(\tau)) \Delta \tau \right) \\ &= \exp \left( \displaystyle\sum_{k=\sqrt{s}}^{\sqrt{t}-1} \mu(k^2) \dfrac{1}{\mu(k^2)} \log ( 1 + p(k^2)\mu(k^2)) \right) \\ &= \exp \left( \displaystyle\sum_{k=\sqrt{s}}^{\sqrt{t}-1} \log ( 1 + p(k^2)\mu(k^2)) \right) \\ &= \displaystyle\prod_{k=\sqrt{s}}^{\sqrt{t}-1} 1 + p(k^2)(2k+1) \end{array}$ |
| $\mathbb{H}$ | $\begin{array}{ll} e_p(t,s) &= e_p \left( \displaystyle\sum_{k=1}^n \dfrac{1}{k}, \displaystyle\sum_{k=1}^m \dfrac{1}{k} \right) \\ &= \exp \left( \displaystyle\int_{ \sum^m \frac{1}{k}}^{\sum^n \frac{1}{k}} \dfrac{1}{\mu(\tau)} \log(1 + \mu(\tau) p(\tau)) \Delta \tau \right) \\ &= \exp \left( \displaystyle\sum_{k=m}^{n-1} \log \left[1 + \mu \left( \displaystyle\sum_{j=1}^k \dfrac{1}{j} \right) p \left( \displaystyle\sum_{j=1}^k \dfrac{1}{j} \right) \right] \right) \\ &= \exp \left( \displaystyle\sum_{k=m}^{n-1} \log \left[1 + \dfrac{1}{k+1} p \left( \displaystyle\sum_{j=1}^k \dfrac{1}{j} \right) \right] \right) \\ &= \displaystyle\prod_{k=m}^{n-1} 1 + \dfrac{1}{k+1} p \left( \displaystyle\sum_{j=1}^k \dfrac{1}{j} \right) \\ \end{array}$ |
