Difference between revisions of "Relationship between nabla exponential and delta exponential"
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| − | + | ==Theorem== | |
| − | + | If $p$ is [[continuous]] and $\nu$-regressive then | |
| − | $$\hat{e}_p(t,s)=e_{\frac{ | + | $$\hat{e}_p(t,s)=e_{\frac{p^{\sigma}}{1-p^{\sigma}\nu}}(t,s)=e_{\ominus(-p^{\sigma})}(t,s),$$ |
| − | + | where $\hat{e}_p$ denotes the [[nabla exponential|$\nabla$-exponential]] and $e_p$ denotes the [[Delta exponential|$\Delta$-exponential]]. | |
| − | + | ||
| − | + | ==Proof== | |
| − | + | ||
| + | ==References== | ||
| + | |||
| + | [[Category:Theorem]] | ||
| + | [[Category:Unproven]] | ||
Latest revision as of 22:22, 9 June 2016
Theorem
If $p$ is continuous and $\nu$-regressive then $$\hat{e}_p(t,s)=e_{\frac{p^{\sigma}}{1-p^{\sigma}\nu}}(t,s)=e_{\ominus(-p^{\sigma})}(t,s),$$ where $\hat{e}_p$ denotes the $\nabla$-exponential and $e_p$ denotes the $\Delta$-exponential.
