Difference between revisions of "Relationship between nabla derivative and delta derivative"

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==Theorem==
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<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  
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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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<strong>Proof:</strong> █
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==Proof==
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==References==
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[[Category:Theorem]]
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[[Category:Unproven]]

Latest revision as of 06:02, 10 June 2016

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

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