Difference between revisions of "Delta derivative"

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We sometimes use the notation $\dfrac{\Delta}{\Delta t} f(t)$ or $\dfrac{\Delta f}{\Delta t}$ for $f^{\Delta}(t)$.
 
We sometimes use the notation $\dfrac{\Delta}{\Delta t} f(t)$ or $\dfrac{\Delta f}{\Delta t}$ for $f^{\Delta}(t)$.
  
==Properties of the $\Delta$-derivative<ref>Bohner, Martin ; Peterson, Allan. Dynamic equations on time scales. An introduction with applications. Birkhäuser Boston, Inc., Boston, MA, 2001,p.8.</ref>==
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==Properties of the $\Delta$-derivative==
 
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[[Delta derivative implies continuous]]<br />
{{:Delta derivative implies continuous}}
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[[Relationship between nabla derivative and delta derivative]]<br />
 
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[[Relationship between delta derivative and nabla derivative]]<br />
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<strong>Theorem:</strong> If $f$ is continuous at $t$ and $t$ is right-scattered, then
 
$$f^{\Delta}(t) = \dfrac{f(\sigma(t))-f(t)}{\mu(t)}.$$
 
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<strong>Proof:</strong>  █
 
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<strong>Theorem:</strong> If $t$ is right-dense, then (if it exists),
 
$$f^{\Delta}(t) = \displaystyle\lim_{s \rightarrow t}\dfrac{f(t)-f(s)}{t-s}.$$
 
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<strong>Proof:</strong>
 
</div>
 
</div>
 
 
 
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<strong>Theorem:</strong> If $f$ is differentiable at $t$, then
 
$$f(\sigma(t))=f(t)+\mu(t)f^{\Delta}(t).$$
 
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<strong>Proof:</strong>  █
 
</div>
 
</div>
 
 
 
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<strong>Theorem (Sum rule):</strong>
 
$$(f+g)^{\Delta}(t)=f^{\Delta}(t)+g^{\Delta}(t).$$
 
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<strong>Proof:</strong>  █
 
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</div>
 
 
 
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<strong>Theorem (Constant rule):</strong> If $\alpha$ is constant with respect to $t$, then
 
$$(\alpha f)^{\Delta}(t) = \alpha f^{\Delta}(t).$$
 
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<strong>Proof:</strong>  █
 
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<strong>Theorem (Product rule,I):</strong> The following formula holds:
 
$$(fg)^{\Delta}(t)=f^{\Delta}(t)g(t)+f(\sigma(t))g^{\Delta}(t)).$$
 
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<strong>Proof:</strong>  █
 
</div>
 
</div>
 
 
 
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<strong>Theorem (Product rule,II):</strong> The following formula holds:
 
$$(fg)^{\Delta}(t) = f(t)g^{\Delta}(t)+ f^{\Delta}(t)g(\sigma(t)).$$
 
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<strong>Proof:</strong>  █
 
</div>
 
</div>
 
 
 
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<strong>Theorem (Quotient rule):</strong> The following formula holds:
 
$$\left( \dfrac{f}{g} \right)^{\Delta}(t) = \dfrac{f^{\Delta}(t)g(t)-f(t)g^{\Delta}(t)}{g(t)g(\sigma(t))}.$$
 
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<strong>Proof:</strong>  █
 
</div>
 
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{{:Relationship between nabla derivative and delta derivative}}
 
{{:Relationship between delta derivative and nabla derivative}}
 
  
 
== References ==
 
== References ==
<references/>
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* {{BookReference|Dynamic Equations on Time Scales|2001|Martin Bohner|author2=Allan Peterson|prev=Induction on time scales|next=Delta differentiable implies continuous}}: Definition 1.10

Revision as of 05:12, 10 June 2016

Let $\mathbb{T}$ be a time scale and let $f \colon \mathbb{T} \rightarrow \mathbb{R}$ and let $t \in \mathbb{T}^{\kappa}$. We define<ref>Bohner, Martin ; Peterson, Allan. Dynamic equations on time scales. An introduction with applications. Birkhäuser Boston, Inc., Boston, MA, 2001,p.5.</ref> the $\Delta$-derivative of $f$ at $t$ to be the number $f^{\Delta}(t)$ (if it exists) so that there exists a $\delta >0$ so that for all $s \in (t-\delta,t+\delta) \bigcap \mathbb{T}$, $$|[f(\sigma(t))-f(s)]-f^{\Delta}(t)[\sigma(t)-s]| \leq \epsilon |\sigma(t)-s|.$$ We sometimes use the notation $\dfrac{\Delta}{\Delta t} f(t)$ or $\dfrac{\Delta f}{\Delta t}$ for $f^{\Delta}(t)$.

Properties of the $\Delta$-derivative

Delta derivative implies continuous
Relationship between nabla derivative and delta derivative
Relationship between delta derivative and nabla derivative

References