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

Revision as of 09:29, 12 April 2015

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: █

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-<strong>[[Relationship between delta exponential and nabla exponential|Theorem]]:</strong> If $p$ is [[continuous]] and [[mu regressive | $\mu$-regressive]] then
+<strong>[[Relationship between delta exponential and nabla exponential|Theorem]]:</strong> If $q$ is [[continuous]] and [[mu regressive | $\mu$-regressive]] then
-$$e_p(t,s)=\hat{e}_{\frac{p^{\rho}}{1+p^{\rho}\nu}}(t,s)=\hat{e}_{\ominus_{\nu}(-p^{\rho})}(t,s),$$
+$$e_q(t,s)=\hat{e}_{\frac{q^{\rho}}{1+q^{\rho}\nu}}(t,s)=\hat{e}_{\ominus_{\nu}(-q^{\rho})}(t,s),$$
-where $e_p$ denotes the [[Delta exponential|$\Delta$-exponential]] and $\hat{e}$ denotes the [[nabla exponential|$\nabla$-exponential]].
+where $e_q$ denotes the [[Delta exponential|$\Delta$-exponential]] and $\hat{e}_q$ denotes the [[nabla exponential|$\nabla$-exponential]].
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 <strong>Proof:</strong> █
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 </div>