# Difference between revisions of "Nabla derivative"

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 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:

## Theorem

Let $\mathbb{T}$ be a time scale and let $f \colon \mathbb{T} \rightarrow \mathbb{R}$. If $f$ is $\Delta$-differentiable and $f^{\Delta}$ is rd continous on $\mathbb{T}^{\kappa}$, then $f$ is $\nabla$-differentiable on $\mathbb{T}_{\kappa}$ and $$f^{\nabla}(t) = f^{\Delta}(\rho(t)).$$

## Theorem

Let $\mathbb{T}$ be a time scale and let $f \colon \mathbb{T} \rightarrow \mathbb{R}$. If $f$ is $\nabla$-differentiable on $\mathbb{T}_{\kappa}$ and $g^{\nabla}$ is ld continuous on $\mathbb{T}_{\kappa}$, then $f$ is $\Delta$-differentiable on $\mathbb{T}^{\kappa}$ and $$g^{\Delta}(t) = g^{\nabla}(\sigma(t)).$$