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| | The classical exponential function $e^{x-s}$ is the unique solution to the initial value problem | | The classical exponential function $e^{x-s}$ is the unique solution to the initial value problem |
| | $$y'=y; y(s)=1.$$ | | $$y'=y; y(s)=1.$$ |
| − | The standard way to generalize this to time scales is called the $\Delta$-exponential function, which is the solution of | + | The standard way to generalize this to time scales is called the [[Delta exponential | $\Delta$-exponential]] function, which is the solution of |
| | $$y^{\Delta}=y;y(s)=1.$$ | | $$y^{\Delta}=y;y(s)=1.$$ |
| | It generalizes the above equation in the sense that the classical derivative is replaced by the [[Delta derivative | $\Delta$-derivative]] on some time scale. If instead of using the $\Delta$-derivative one uses the [[nabla derivative | $\nabla$-derivative]] then the resulting exponential equation is | | It generalizes the above equation in the sense that the classical derivative is replaced by the [[Delta derivative | $\Delta$-derivative]] on some time scale. If instead of using the $\Delta$-derivative one uses the [[nabla derivative | $\nabla$-derivative]] then the resulting exponential equation is |
| | $$y^{\nabla}=y;y(s)=1,$$ | | $$y^{\nabla}=y;y(s)=1,$$ |
| − | and we call its unique solution the $\nabla$-exponential function.
| + | defining the [[nabla exponential | $\nabla$-exponential]] functions. |
| | | | |
| − | =$\Delta$-exponential functions=
| + | Generally speaking, given some kind of time scale derivative operator $D$, we can define exponential functions by the $D$-dynamic equation |
| − | 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 | regressive function]]. The exponential function $e_p \colon \mathbb{T} \times \mathbb{T} \rightarrow \mathbb{R}$ is defined as
| + | $$Dy=y; y(s)=1.$$ |
| − | | |
| − | $$e_p(t,s) := \exp \left( \displaystyle\int_s^t \xi_{\mu(\tau)}(p(\tau))\Delta \tau \right)$$
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| − | | |
| − | for $s,t \in \mathbb{T}$. It turns out that $e_p$ is the unique solution to the dynamic initial value problem
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| − | $$y^{\Delta} = py; y(s)=1.$$
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| − | | |
| − | | |
| − | == Properties of Exponential Functions ==
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| − | For all $p,q \in \mathcal{R}$ and $t,s \in \mathbb{T}$,
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| − | *$e_p(t,r)e_p(r,s)=e_p(t,s)$ (semigroup property)
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| − | *$e_0(t,s)=1, e_p(t,t)=1$
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| − | *$e_p(\sigma(t),s)=(1+\mu(t)p(t))e_p(t,s)$
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| − | *$\dfrac{1}{e_p(t,s)}=e_{\ominus p}(s,t)$
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| − | *$e_p(t,s)e_q(t,s)=e_{p \oplus q}(t,s)$
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| − | *$\dfrac{e_p(t,s)}{e_q(t,s)} = e_{p \ominus q}(t,s)$
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| − | *$\left( \dfrac{1}{e_p(\cdot,s)} \right)^{\Delta} = -\dfrac{p(t)}{e_p^{\sigma}(\cdot,s)}$
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| − | | |
| − | == Examples of Exponential Functions ==
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| − | *The [[Gaussian_bell | Gaussian bell]]
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| − | {| class="wikitable"
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| − | |+Time Scale Exponential Functions
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| − | |-
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| − | |$\mathbb{T}=$
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| − | |$e_p(t,s)=$
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| − | |-
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| − | |[[Real_numbers | $\mathbb{R}$]]
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| − | |$e_p(t,s)= \left\{ \begin{array}{ll}
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| − | \exp \left( \displaystyle\int_s^t p(\tau) d \tau \right) &; t>s \\
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| − | 1 &; t=s \\
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| − | \exp \left( -\displaystyle\int_t^s p(\tau) d\tau \right) &; t<s
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| − | \end{array} \right.$
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| − | |-
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| − | |[[Integers | $\mathbb{Z}$]]
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| − | |$e_p(t,s) = \left\{ \begin{array}{ll}
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| − | \displaystyle\prod_{k=s}^{t-1} 1+p(k) &; t > s \\
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| − | 1 &; t=s \\
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| − | \displaystyle\prod_{k=t}^{s-1} \dfrac{1}{1+p(k)}&; t < s
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| − | \end{array} \right.$
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| − | |-
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| − | |[[Multiples_of_integers | $h\mathbb{Z}$]]
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| − | | $ e_p(t,s) = \left\{ \begin{array}{ll}
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| − | \displaystyle\prod_{k=\frac{s}{h}}^{\frac{t}{h}-1} (1+hp(hk)) &; t > s \\
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| − | 1 &; t=s \\
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| − | \displaystyle\prod_{k=\frac{t}{h}}^{\frac{s}{h}-1} \dfrac{1}{1+hp(hk)} &; t < s
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| − | \end{array} \right.$
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| − | |-
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| − | | [[Square_integers | $\mathbb{Z}^2$]]
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| − | | $ e_p(t,s) = \left\{\begin{array}{ll}
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| − | \displaystyle\prod_{k=\sqrt{s}}^{\sqrt{t}-1} 1 + p(k^2)(2k+1) &; t > s \\
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| − | 1 &; t=s\\
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| − | \displaystyle\prod_{k=\sqrt{t}}^{\sqrt{s}-1} \dfrac{1}{1+p(k^2)(2k+1)} &; t < s
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| − | \end{array} \right.$
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| − | |-
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| − | |[[Quantum_q_greater_than_1 | $\overline{q^{\mathbb{Z}}}, q > 1$]]
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| − | | $e_p(t,s) = \left\{ \begin{array}{ll}
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| − | \displaystyle\prod_{k=\log_q(s)}^{\log_q(t)-1} 1 + p(q^k)q^k(q-1) &; t > s \\
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| − | 1 &; t=s \\
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| − | \displaystyle\prod_{k=\log_q(t)}^{\log_q(s)-1} \dfrac{1}{1+p(q^k)q^k(q-1)} &; t < s
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| − | \end{array} \right.$
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| − | |-
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| − | |[[Quantum_q_less_than_1 | $\overline{q^{\mathbb{Z}}}, q < 1$]]
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| − | | $e_p(t,s) = \left\{ \begin{array}{ll}
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| − | \displaystyle\prod_{k=\log_q(s)}^{\log_q(t)-1} 1 + p(q^k)q^{k-1}(1-q) &; t > s \\
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| − | 1 &; t=s \\
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| − | \displaystyle\prod_{k=\log_q(t)}^{\log_q(s)-1} \dfrac{1}{1+p(q^k)q^{k-1}(1-q)} &; t < s
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| − | \end{array} \right.$
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| − | |-
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| − | |[[Harmonic_numbers | $\mathbb{H}$]]
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| − | |$ 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}
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| − | \displaystyle\prod_{k=m}^{n-1} {1 + \dfrac{1}{k+1} p \left( \displaystyle\sum_{j=1}^k \dfrac{1}{j} \right)} &; t > s \\
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| − | 1 &; t=s \\
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| − | \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
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| − | \end{array} \right.$
| |
| − | |}
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| − | We say that a function $p \colon \mathbb{T} \rightarrow \mathbb{R}$ is $\nu$-regressive if
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| − | $$1-\nu(t)p(t) \neq 0$$
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| − | for all $t \in \mathbb{T}_{\kappa}$. We use notation $\mathcal{R}_{\nu}$ for the set of $\nu$-regressive functions.
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| − | =$\nabla$-exponential Functions=
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| − | Define the function $\hat{\xi}_{h} \colon \mathbb{C}_h \rightarrow \mathbb{Z}_h$ by
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| − | $$\hat{\xi}_h(z) = \dfrac{1}{h} \log(1-zh).$$ | |
| − | Define the $\nabla$ exponential function for $s,t \in \mathbb{T}$ by
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| − | $$\hat{e}_p(t,s) = \exp \left( \displaystyle\int_s^t \hat{\xi}_{\nu(\tau)}(p(\tau)) \nabla \tau \right).$$
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| − | | |
| − | ==Properties of $\nabla$-exponential functions==
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| − | The function $\hat{e}_p(\cdot,s)$ is the unique solution of the initial value problem
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| − | $$y^{\nabla} = py; y(s)=1.$$
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| − | For all $p,q \in \mathcal{R}_{\nu}$ and $t,s \in \mathbb{T}$,
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| − | *$\hat{e}_p(t,r)\hat{e}_p(r,s)=e_p(t,s)$ (semigroup property)
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| − | *$\hat{e}_0(t,s)=1, \hat{e}_p(t,t)=1$
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| − | *$\hat{e}_p(\rho(t),s)=(1-\nu(t)p(t))\hat{e}_p(t,s)$
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| − | *$\dfrac{1}{\hat{e}_p(t,s)}=\hat{e}_{\ominus_{\nu} p}(s,t)$
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| − | *$\hat{e}_p(t,s)\hat{e}_q(t,s)=\hat{e}_{p \oplus_{\nu} q}(t,s)$
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| − | *$\dfrac{\hat{e}_p(t,s)}{\hat{e}_q(t,s)} = \hat{e}_{p \ominus_{\nu} q}(t,s)$
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| − | *$\left( \dfrac{1}{\hat{e}_p(\cdot,s)} \right)^{\nabla} = -\dfrac{p(t)}{\hat{e}_p^{\rho}(\cdot,s)}$
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| − | | |
| − | <div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
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| − | <strong>Theorem:</strong> (Sign of the Nabla Exponential Function) Let $p \in \mathcal{R}_{\nu}$ and $s \in \mathbb{T}$. <br />
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| − | i.) If $p \in \mathcal{R}_{\nu}^+$, then $\hat{e}_{p}(t,s) > 0$ for all $t \in \mathbb{T}$. <br />
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| − | ii.) If $1-\nu(t)p(t) < 0$ for some $t \in \mathbb{T}_{\kappa}$, then
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| − | $$\hat{e}(\rho(t),s)\hat{e}_{p}(t,s)<0.$$
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| − | iii.) If $1-\nu(t)p(t) < 0$ for all $t \in \mathbb{T}$, then $\hat{e}_p(t,s)$ changes sign at every point of $\mathbb{T}$.<br />
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| − | iv.) The exponential function $\hat{e}_p(\cdot,s)$ is a real-valued function that is never equal to zero.
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| − | <div class="mw-collapsible-content">
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| − | <strong>Proof:</strong> proof goes here █
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| − | </div>
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| − | </div>
| |
The classical exponential function $e^{x-s}$ is the unique solution to the initial value problem
$$y'=y; y(s)=1.$$
The standard way to generalize this to time scales is called the $\Delta$-exponential function, which is the solution of
$$y^{\Delta}=y;y(s)=1.$$
It generalizes the above equation in the sense that the classical derivative is replaced by the $\Delta$-derivative on some time scale. If instead of using the $\Delta$-derivative one uses the $\nabla$-derivative then the resulting exponential equation is
$$y^{\nabla}=y;y(s)=1,$$
defining the $\nabla$-exponential functions.
Generally speaking, given some kind of time scale derivative operator $D$, we can define exponential functions by the $D$-dynamic equation
$$Dy=y; y(s)=1.$$
