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Revision as of 03:04, 11 March 2009
, 03:04, 11 March 2009→Computation Summary : <math>f(u, v) = \texttt{((u)(v))}</math>: markup
| <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>
| <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>
|}
|}
<math>\operatorname{E}f</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>f\!</math> is true. In this case, where the prevailing proposition <math>f\!</math> is <math>\texttt{((u)(v))},</math> the indication <math>\texttt{uv} \cdot \texttt{(du~dv)}</math> of <math>\operatorname{E}f</math> tells you this: If <math>u\!</math> and <math>v\!</math> are both true where you are, then just don't change both <math>u\!</math> and <math>v\!</math> at once, and you will end up in a place where <math>\texttt{((u)(v))}</math> is true.
<pre>
<pre>
Figure 1.3 expands Df over [u, v] to end up with the formula:
Figure 1.3 expands Df over [u, v] to end up with the formula:
Df = uv du dv + u(v) du(dv) + (u)v (du)dv + (u)(v)((du)(dv)).
Df = uv du dv + u(v) du(dv) + (u)v (du)dv + (u)(v)((du)(dv)).
