Difference between revisions of "Delta derivative"

Revision as of 21:13, 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 $t$.
• If $f$ is continuous at $t$ and $t$ is right-scattered, then

$$f^{\Delta}(t) = \dfrac{f(\sigma(t))-f(t)}{\mu(t)}$$

• If $t$ is right-dense, then (if it exists),

$$f^{\Delta}(t) = \displaystyle\lim_{s \rightarrow t}\dfrac{f(t)-f(s)}{t-s}.$$

• If $f$ is differentiable at $t$, then

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

• 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 $\mathbb{T}=[0,1] \bigcup \{2,3,4,\ldots\}$, then we see $$\sigma(t) = \left\{ \begin{array}{ll} 0 &; t \in [0,1) \\ 1 &; t \in \{1,2,3,\ldots\}. \end{array}\right.$$ This function is clearly not continuous at $t=1$ and hence it is not $\Delta$-differentiable at $t=1$.