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Revision as of 02:58, 12 March 2009
, 02:58, 12 March 2009→Computation Summary : <math>g(u, v) = \texttt{((u, v))}</math>: markup
<math>\operatorname{E}g</math> tells you what you would have to do, from where you are in the universe <math>[u, v],\!</math> if you want to end up in a place where <math>g\!</math> is true. In this case, where the prevailing proposition <math>g\!</math> is <math>\texttt{((u, v))},</math> the component <math>\texttt{uv} \cdot \texttt{((du, dv))}</math> of <math>\operatorname{E}g</math> tells you this: If <math>u\!</math> and <math>v\!</math> are both true where you are, then change either both or neither of <math>u\!</math> and <math>v\!</math> at the same time, and you will attain a place where <math>\texttt{((du, dv))}</math> is true.
<math>\operatorname{E}g</math> tells you what you would have to do, from where you are in the universe <math>[u, v],\!</math> if you want to end up in a place where <math>g\!</math> is true. In this case, where the prevailing proposition <math>g\!</math> is <math>\texttt{((u, v))},</math> the component <math>\texttt{uv} \cdot \texttt{((du, dv))}</math> of <math>\operatorname{E}g</math> tells you this: If <math>u\!</math> and <math>v\!</math> are both true where you are, then change either both or neither of <math>u\!</math> and <math>v\!</math> at the same time, and you will attain a place where <math>\texttt{((du, dv))}</math> is true.
Figure 2.3 shows the expansion of <math>\operatorname{D}g</math> over <math>[u, v]\!</math> to produce the expression:
{| align="center" cellpadding="8" width="90%"
| <math>\texttt{uv} \cdot \texttt{(du, dv)} ~+~ \texttt{u(v)} \cdot \texttt{(du, dv)} ~+~ \texttt{(u)v} \cdot \texttt{(du, dv)} ~+~ \texttt{(u)(v)} \cdot \texttt{(du, dv)}</math>
|}
<pre>
<pre>
Dg tells you what you would have to do, from where you are in the
Dg tells you what you would have to do, from where you are in the
universe [u, v], if you want to bring about a change in the value
universe [u, v], if you want to bring about a change in the value
