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

Revision as of 09:15, 12 April 2015

Theorem: If $p$ is continuous and $\mu$-regressive then $$e_p(t,s)=\hat{e}_{\frac{p^{\rho}}{1+p^{\rho}\nu}}(t,s)=\hat{e}_{\ominus_{\nu}(-p^{\rho})}(t,s),$$ where $e_p$ denotes the $\Delta$-exponential and $\hat{e}$ denotes the $\nabla$-exponential.

Proof: █

@@ Line 1: / Line 1: @@
 <div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
-<strong>[[Relationship between delta exponential and nabla exponential|Theorem]]:</strong> Let $p \in \mathcal{R}$ and $t,s \in \mathbb{T}$, then the following formula holds:
+<strong>[[Relationship between delta exponential and nabla exponential|Theorem]]:</strong> If $p$ is [[continuous]] and [[mu regressive | $\mu$-regressive]] then
-$$\left( \dfrac{1}{e_p(\cdot,s)} \right)^{\Delta} = -\dfrac{p(t)}{e_p^{\sigma}(\cdot,s)}.$$
+$$e_p(t,s)=\hat{e}_{\frac{p^{\rho}}{1+p^{\rho}\nu}}(t,s)=\hat{e}_{\ominus_{\nu}(-p^{\rho})}(t,s),$$
+where $e_p$ denotes the [[Delta exponential|$\Delta$-exponential]] and $\hat{e}$ denotes the [[nabla exponential|$\nabla$-exponential]].
 <div class="mw-collapsible-content">
-<strong>Proof:</strong>  █
+<strong>Proof:</strong> █
 </div>
 </div>