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

From timescalewiki
Jump to: navigation, search
 
Line 1: Line 1:
<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
+
==Theorem==
<strong>[[Relationship between nabla exponential and delta exponential|Theorem]]:</strong> If $p$ is [[continuous]] and $\nu$-regressive then
+
If $p$ is [[continuous]] and $\nu$-regressive then
 
$$\hat{e}_p(t,s)=e_{\frac{p^{\sigma}}{1-p^{\sigma}\nu}}(t,s)=e_{\ominus(-p^{\sigma})}(t,s),$$
 
$$\hat{e}_p(t,s)=e_{\frac{p^{\sigma}}{1-p^{\sigma}\nu}}(t,s)=e_{\ominus(-p^{\sigma})}(t,s),$$
 
where $\hat{e}_p$ denotes the [[nabla exponential|$\nabla$-exponential]] and $e_p$ denotes the [[Delta exponential|$\Delta$-exponential]].
 
where $\hat{e}_p$ denotes the [[nabla exponential|$\nabla$-exponential]] and $e_p$ denotes the [[Delta exponential|$\Delta$-exponential]].
<div class="mw-collapsible-content">
+
 
<strong>Proof:</strong> █
+
==Proof==
</div>
+
 
</div>
+
==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.

Proof

References