Difference between revisions of "Nabla derivative"

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Let $f \colon \mathbb{T} \rightarrow \mathbb{R}$ and let $t \in \mathbb{T}_{\kappa}$. The $\nabla$-derivative of $f$ at $t$ is denoted by $f^{\nabla}(t)$ to be the number such that given any $\epsilon > 0$ there is a neighborhood $U$ of $t$ and $s \in U$,  
 
Let $f \colon \mathbb{T} \rightarrow \mathbb{R}$ and let $t \in \mathbb{T}_{\kappa}$. The $\nabla$-derivative of $f$ at $t$ is denoted by $f^{\nabla}(t)$ to be the number such that given any $\epsilon > 0$ there is a neighborhood $U$ of $t$ and $s \in U$,  
 
$$|f(\rho(t))-f(s)-f^{\nabla}(t)[\rho(t)-s]|\leq \epsilon|\rho(t)-s|.$$
 
$$|f(\rho(t))-f(s)-f^{\nabla}(t)[\rho(t)-s]|\leq \epsilon|\rho(t)-s|.$$
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==Properties of the $\nabla$-derivative==
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<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
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<strong>Theorem:</strong> If $f$ is $\nabla$-differentiable at $t$, then $f$ is [[continuity | continuous]] at $t$.
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<div class="mw-collapsible-content">
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<strong>Proof:</strong>  █
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</div>
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</div>
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<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
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<strong>Theorem:</strong> If $f$ is continuous at $t$ and $t$ is left-scattered, then
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$$f^{\nabla}(t) = \dfrac{f(t)-f(\rho(t))}{\nu(t)}.$$
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<div class="mw-collapsible-content">
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<strong>Proof:</strong>  █
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</div>
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</div>
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<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
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<strong>Theorem:</strong> If $t$ is left-dense, then (if it exists),
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$$f^{\nabla}(t) = \lim_{s \rightarrow t}\dfrac{f(t)-f(s)}{t-s}.$$
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<div class="mw-collapsible-content">
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<strong>Proof:</strong>  █
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</div>
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</div>
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<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
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<strong>Theorem:</strong> If $f$ is differentiable at $t$, then
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$$f(\rho(t))=f(t)+\nu(t)f^{\nabla}(t).$$
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<div class="mw-collapsible-content">
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<strong>Proof:</strong>  █
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</div>
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</div>
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<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
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<strong>Theorem (Sum rule):</strong>
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$$(f+g)^{\nabla}(t)=f^{\nabla}(t)+g^{\nabla}(t).$$
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<div class="mw-collapsible-content">
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<strong>Proof:</strong>  █
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</div>
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</div>
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<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
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<strong>Theorem (Constant rule):</strong> If $\alpha$ is constant with respect to $t$, then
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$$(\alpha f)^{\nabla}(t) = \alpha f^{\nabla}(t).$$
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<div class="mw-collapsible-content">
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<strong>Proof:</strong>  █
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</div>
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</div>
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<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
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<strong>Theorem (Product rule,I):</strong> The following formula holds:
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$$(fg)^{\nabla}(t)=f^{\nabla}(t)g(t)+f(\rho(t))g^{\nabla}(t)).$$
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<div class="mw-collapsible-content">
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<strong>Proof:</strong>  █
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</div>
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</div>
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<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
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<strong>Theorem (Product rule,II):</strong> The following formula holds:
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$$(fg)^{\nabla}(t) = f(t)g^{\nabla}(t)+ f^{\nabla}(t)g(\rho(t)).$$
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<div class="mw-collapsible-content">
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<strong>Proof:</strong>  █
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</div>
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</div>
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<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
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<strong>Theorem (Quotient rule):</strong> The following formula holds:
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$$\left( \dfrac{f}{g} \right)^{\nabla}(t) = \dfrac{f^{\nabla}(t)g(t)-f(t)g^{\nabla}(t)}{g(t)g(\rho(t))}.$$
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<div class="mw-collapsible-content">
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<strong>Proof:</strong>  █
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</div>
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</div>
  
 
=References=
 
=References=
 
[http://wwwp.cord.edu/faculty/andersod/p20.pdf Nabla dynamic equations on time scales - D. Anderson, J. Bullock, L. Erbe, A. Peterson, H. Tran]
 
[http://wwwp.cord.edu/faculty/andersod/p20.pdf Nabla dynamic equations on time scales - D. Anderson, J. Bullock, L. Erbe, A. Peterson, H. Tran]

Revision as of 21:07, 20 October 2014

Let $\mathbb{T}$ be a time scale. If $\mathbb{T}$ has a right-scattered minimum $m$, then define $\mathbb{T}_{\kappa}=\mathbb{T} \setminus \{m\}$, otherwise let $\mathbb{T}_{\kappa}=\mathbb{T}$. Define the backward graininess function $\nu \colon \mathbb{T}_{\kappa} \rightarrow \mathbb{R}$ by $$\nu(t) = t - \rho(t).$$

Let $f \colon \mathbb{T} \rightarrow \mathbb{R}$ and let $t \in \mathbb{T}_{\kappa}$. The $\nabla$-derivative of $f$ at $t$ is denoted by $f^{\nabla}(t)$ to be the number such that given any $\epsilon > 0$ there is a neighborhood $U$ of $t$ and $s \in U$, $$|f(\rho(t))-f(s)-f^{\nabla}(t)[\rho(t)-s]|\leq \epsilon|\rho(t)-s|.$$

Properties of the $\nabla$-derivative

Theorem: If $f$ is $\nabla$-differentiable at $t$, then $f$ is continuous at $t$.

Proof: █

Theorem: If $f$ is continuous at $t$ and $t$ is left-scattered, then $$f^{\nabla}(t) = \dfrac{f(t)-f(\rho(t))}{\nu(t)}.$$

Proof: █

Theorem: If $t$ is left-dense, then (if it exists), $$f^{\nabla}(t) = \lim_{s \rightarrow t}\dfrac{f(t)-f(s)}{t-s}.$$

Proof: █

Theorem: If $f$ is differentiable at $t$, then $$f(\rho(t))=f(t)+\nu(t)f^{\nabla}(t).$$

Proof: █

Theorem (Sum rule): $$(f+g)^{\nabla}(t)=f^{\nabla}(t)+g^{\nabla}(t).$$

Proof: █

Theorem (Constant rule): If $\alpha$ is constant with respect to $t$, then $$(\alpha f)^{\nabla}(t) = \alpha f^{\nabla}(t).$$

Proof: █

Theorem (Product rule,I): The following formula holds: $$(fg)^{\nabla}(t)=f^{\nabla}(t)g(t)+f(\rho(t))g^{\nabla}(t)).$$

Proof: █

Theorem (Product rule,II): The following formula holds: $$(fg)^{\nabla}(t) = f(t)g^{\nabla}(t)+ f^{\nabla}(t)g(\rho(t)).$$

Proof: █

Theorem (Quotient rule): The following formula holds: $$\left( \dfrac{f}{g} \right)^{\nabla}(t) = \dfrac{f^{\nabla}(t)g(t)-f(t)g^{\nabla}(t)}{g(t)g(\rho(t))}.$$

Proof: █

References

Nabla dynamic equations on time scales - D. Anderson, J. Bullock, L. Erbe, A. Peterson, H. Tran

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