Difference between revisions of "Delta derivative"

Revision as of 20:58, 19 May 2014

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 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 $s \in (t-\delta,t+\delta) \bigcap \mathbb{T}$, $$|[f(\sigma(t))-f(s)]-f^{\Delta}(t)[\sigma(t)-s]| \leq \epsilon |\sigma(t)-s|.$$

Properties of the $\Delta$-derivative

• If $f$ is $\Delta$-differentiable at $t$, then $f$is continuous at $t$.
• $f(\sigma(t))=f(t)+\mu(t)f^{\Delta}(t)$
• Sum rule:

$$(f+g)^{\Delta}(t)=f^{\Delta}(t)+g^{\Delta}(t)$$

• Constant rule:if $\alpha$ is constant with respect to $t$, then

$$(\alpha f)^{\Delta}(t) = \alpha f^{\Delta}(t)$$

• Product Rule I

$$(fg)^{\Delta}(t)=f^{\Delta}(t)g(t)+f(\sigma(t))g^{\Delta}(t))$$

• Product Rule II

$$(fg)^{\Delta}(t) = f(t)g^{\Delta}(t)+ f^{\Delta}(t)g(\sigma(t))$$

• Quotient Rule:

$$\left( \dfrac{f}{g} \right)^{\Delta}(t) = \dfrac{f^{\Delta}(t)g(t)-f(t)g^{\Delta}(t)}{g(t)g(\sigma(t))}$$

Interesting Examples

The jump operator $\sigma$ is not $\Delta$-differentiable on all time scales. Consider $\mathbb{T}=