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

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==Theorem==
<strong>[[Relationship between delta exponential and nabla exponential|Theorem]]:</strong> Let $p \in \mathcal{R}$ and $t,s \in \mathbb{T}$, then the following formula holds:
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If $q$ is [[continuous]] and [[mu regressive | $\mu$-regressive]] then
$$\left( \dfrac{1}{e_p(\cdot,s)} \right)^{\Delta} = -\dfrac{p(t)}{e_p^{\sigma}(\cdot,s)}.$$
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$$e_q(t,s)=\hat{e}_{\frac{q^{\rho}}{1+q^{\rho}\nu}}(t,s)=\hat{e}_{\ominus_{\nu}(-q^{\rho})}(t,s),$$
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where $e_q$ denotes the [[Delta exponential|$\Delta$-exponential]] and $\hat{e}_q$ denotes the [[nabla exponential|$\nabla$-exponential]].
<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 22:22, 9 June 2016

Theorem

If $q$ is continuous and $\mu$-regressive then $$e_q(t,s)=\hat{e}_{\frac{q^{\rho}}{1+q^{\rho}\nu}}(t,s)=\hat{e}_{\ominus_{\nu}(-q^{\rho})}(t,s),$$ where $e_q$ denotes the $\Delta$-exponential and $\hat{e}_q$ denotes the $\nabla$-exponential.

Proof

References