12,080
edits
Changes
MyWikiBiz, Author Your Legacy — Tuesday April 08, 2025
Jump to navigationJump to searchDirectory talk:Jon Awbrey/Papers/Differential Propositional Calculus (view source)
Revision as of 15:22, 5 June 2008
, 15:22, 5 June 2008→Current Version @ PlanetMath : TeX Format: update
Line 279:
Line 279:
\item
\item
−The initial space, $A = \langle a_1, \ldots, a_n \rangle,$ is extended by a \textit{first order differential space} or \textit{tangent space}, $\operatorname{d}A = \langle \operatorname{d}a_1, \ldots, \operatorname{d}a_n \rangle,$ at each point of $A$, resulting in a \textit{first order extended space} or \textit{tangent bundle}, $\operatorname{E}A,$ defined as follows:
+The initial space, $A = \langle a_1, \ldots, a_n \rangle,$ is extended by a \textit{first order differential space} or \textit{tangent space}, $\operatorname{d}A = \langle \operatorname{d}a_1, \ldots, \operatorname{d}a_n \rangle,$ at each point of $A,$ resulting in a \textit{first order extended space} or \textit{tangent bundle space}, $\operatorname{E}A,$ defined as follows:
\begin{quote}
\begin{quote}
$\operatorname{E}A = A\ \times\ \operatorname{d}A = \langle \operatorname{E}\mathcal{A} \rangle = \langle \mathcal{A}\ \cup\ \operatorname{d}\mathcal{A} \rangle = \langle a_1, \ldots, a_n, \operatorname{d}a_1, \ldots, \operatorname{d}a_n \rangle.$
$\operatorname{E}A = A\ \times\ \operatorname{d}A = \langle \operatorname{E}\mathcal{A} \rangle = \langle \mathcal{A}\ \cup\ \operatorname{d}\mathcal{A} \rangle = \langle a_1, \ldots, a_n, \operatorname{d}a_1, \ldots, \operatorname{d}a_n \rangle.$
+\end{quote}
+
+\item
+Finally, the initial universe, $A^\circ = [ a_1, \ldots, a_n ],$ is extended by a \textit{first order differential universe} or \textit{tangent universe}, $\operatorname{d}A^\circ = [ \operatorname{d}a_1, \ldots, \operatorname{d}a_n ],$ at each point of $A^\circ,$ resulting in a \textit{first order extended universe} or \textit{tangent bundle universe}, $\operatorname{E}A^\circ,$ defined as follows:
+
+\begin{quote}
+$\operatorname{E}A^\circ = [ \operatorname{E}\mathcal{A} ] = [ \mathcal{A}\ \cup\ \operatorname{d}\mathcal{A} ] = [ a_1, \ldots, a_n, \operatorname{d}a_1, \ldots, \operatorname{d}a_n ].$
\end{quote}
\end{quote}
\end{itemize}
\end{itemize}
+
+This gives $\operatorname{E}A^\circ$ the type $(\mathbb{B}^n \times \mathbb{D}^n\ +\!\to \mathbb{B}) = (\mathbb{B}^n \times \mathbb{D}^n, \mathbb{B}^n \times \mathbb{D}^n \to \mathbb{B}).$
$\dots$
$\dots$
