Difference between revisions of "Relationship between nabla derivative and delta derivative"
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| − | <strong>Theorem:</strong> Let $\mathbb{T}$ be a [[time scale]] and let $f \colon \mathbb{T} \rightarrow \mathbb{R}$. If $f$ is [[Delta derivative|$\Delta$-differentiable]] and $f^{\Delta}$ is [[rd continous]] on $\mathbb{T}^{\kappa}$, then $f$ is [[nabla derivative|$\nabla$-differentiable]] on $\mathbb{T}_{\kappa}$ and | + | <strong>[[Relationship between nabla derivative and delta derivative|Theorem]]:</strong> Let $\mathbb{T}$ be a [[time scale]] and let $f \colon \mathbb{T} \rightarrow \mathbb{R}$. If $f$ is [[Delta derivative|$\Delta$-differentiable]] and $f^{\Delta}$ is [[rd continous]] on $\mathbb{T}^{\kappa}$, then $f$ is [[nabla derivative|$\nabla$-differentiable]] on $\mathbb{T}_{\kappa}$ and |
$$f^{\nabla}(t) = f^{\Delta}(\rho(t)).$$ | $$f^{\nabla}(t) = f^{\Delta}(\rho(t)).$$ | ||
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Revision as of 09:05, 12 April 2015
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: █
