Difference between revisions of "Relationship between nabla exponential and delta exponential"
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<strong>[[Relationship between nabla exponential and delta exponential|Theorem]]:</strong> If $p$ is [[continuous]] and $\nu$-regressive then | <strong>[[Relationship between nabla exponential and delta exponential|Theorem]]:</strong> If $p$ is [[continuous]] and $\nu$-regressive then | ||
| − | $$\hat{e}_p(t,s)=e_{\frac{q^{\sigma}}{1-q^{\sigma}\nu}}(t,s)=e_{\ominus(-q^{\sigma})}(t,s) | + | $$\hat{e}_p(t,s)=e_{\frac{q^{\sigma}}{1-q^{\sigma}\nu}}(t,s)=e_{\ominus(-q^{\sigma})}(t,s),$$ |
| + | where $\hat{e}$ denotes the [[nabla exponential|$\nabla$-exponential]] and $e_p$ denotes the [[Delta exponential|$\Delta$-exponential]]. | ||
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<strong>Proof:</strong> █ | <strong>Proof:</strong> █ | ||
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Revision as of 09:15, 12 April 2015
Theorem: If $p$ is continuous and $\nu$-regressive then $$\hat{e}_p(t,s)=e_{\frac{q^{\sigma}}{1-q^{\sigma}\nu}}(t,s)=e_{\ominus(-q^{\sigma})}(t,s),$$ where $\hat{e}$ denotes the $\nabla$-exponential and $e_p$ denotes the $\Delta$-exponential.
Proof: █
