Difference between revisions of "Nabla derivative"
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Revision as of 00:53, 23 August 2016
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 $s \in U$, $$|f(\rho(t))-f(s)-f^{\nabla}(t)[\rho(t)-s]|\leq \epsilon|\rho(t)-s|.$$
Contents
Properties of the $\nabla$-derivative
Nabla differentiable implies continuous
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 $$f^{\nabla}(t) = f^{\Delta}(\rho(t)).$$
Proof
References
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 $$g^{\Delta}(t) = g^{\nabla}(\sigma(t)).$$
Proof
References
References
Nabla dynamic equations on time scales - D. Anderson, J. Bullock, L. Erbe, A. Peterson, H. Tran
