Changes

→‎Note 3: markup
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{| 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)
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|}
 
|}
    +
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}
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\operatorname{E}f(p, q, \operatorname{d}p, \operatorname{d}q)
 +
& = &
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(p + \operatorname{d}p)(q + \operatorname{d}q)
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& = &
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\texttt{(} p, \operatorname{d}p \texttt{)(} q, \operatorname{d}q \texttt{)}
 +
\end{matrix}</math>
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|-
 +
| align="center" |
 
<pre>
 
<pre>
In the example f<p, q> = pq, the enlargement Ef is given by:
  −
  −
  Ef<p, q, dp, dq>
  −
  −
  =  [p + dp][q + dq]
  −
  −
  =  (p, dp)(q, dq)
  −
   
o-------------------------------------------------o
 
o-------------------------------------------------o
 
|                                                |
 
|                                                |
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| Ef =            (p, dp) (q, dq)                |
 
| Ef =            (p, dp) (q, dq)                |
 
o-------------------------------------------------o
 
o-------------------------------------------------o
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</pre>
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|}
    +
<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
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