Difference between revisions of "Relationship between delta exponential and nabla exponential"
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| − | + | ==Theorem== | |
| − | + | If $q$ is [[continuous]] and [[mu regressive | $\mu$-regressive]] then | |
$$e_q(t,s)=\hat{e}_{\frac{q^{\rho}}{1+q^{\rho}\nu}}(t,s)=\hat{e}_{\ominus_{\nu}(-q^{\rho})}(t,s),$$ | $$e_q(t,s)=\hat{e}_{\frac{q^{\rho}}{1+q^{\rho}\nu}}(t,s)=\hat{e}_{\ominus_{\nu}(-q^{\rho})}(t,s),$$ | ||
where $e_q$ denotes the [[Delta exponential|$\Delta$-exponential]] and $\hat{e}_q$ 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]]. | ||
| − | + | ||
| − | + | ==Proof== | |
| − | + | ||
| − | + | ==References== | |
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| + | [[Category:Theorem]] | ||
| + | [[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.
