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Revision as of 20:48, 12 June 2009
, 20:48, 12 June 2009→Note 3: markup
{| align="center" cellpadding="6" width="90%"
{| align="center" cellpadding="6" width="90%"
|
| align="center" |
<math>\begin{matrix}
<math>\begin{matrix}
\operatorname{E}f(p, q, \operatorname{d}p, \operatorname{d}q)
\operatorname{E}f(p, q, \operatorname{d}p, \operatorname{d}q)
|}
|}
In the example <math>f(p, q) = pq,\!</math> the enlargement <math>\operatorname{E}f</math> is computed as follows:
{| align="center" cellpadding="6" width="90%"
| align="center" |
<math>\begin{matrix}
\operatorname{E}f(p, q, \operatorname{d}p, \operatorname{d}q)
& = &
(p + \operatorname{d}p)(q + \operatorname{d}q)
& = &
\texttt{(} p, \operatorname{d}p \texttt{)(} q, \operatorname{d}q \texttt{)}
\end{matrix}</math>
|-
| align="center" |
<pre>
<pre>
o-------------------------------------------------o
o-------------------------------------------------o
| |
| |
| Ef = (p, dp) (q, dq) |
| Ef = (p, dp) (q, dq) |
o-------------------------------------------------o
o-------------------------------------------------o
</pre>
|}
<pre>
Given the proposition f<p, q> over X = !P! x !Q!, the
Given the proposition f<p, q> over X = !P! x !Q!, the
(first order) "difference" of f is the proposition Df
(first order) "difference" of f is the proposition Df
