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Revision as of 17:34, 10 March 2009
, 17:34, 10 March 2009→Note 15: texttt
For <math>F = (f, g)\!</math> we have <math>\operatorname{d}F = (\operatorname{d}f, \operatorname{d}g),</math> and so we can proceed componentwise, patching the pieces back together at the end.
For <math>F = (f, g)\!</math> we have <math>\operatorname{d}F = (\operatorname{d}f, \operatorname{d}g),</math> and so we can proceed componentwise, patching the pieces back together at the end.
We have prepared the ground already by computing these terms:
We have prepared the ground already by computing these terms:
{| align="center" cellpadding="8" width="90%"
|
<math>\begin{array}{lll}
\operatorname{E}f & = & \texttt{(( u + du )( v + dv ))}
\\ \\
\operatorname{E}g & = & \texttt{(( u + du ,~ v + dv ))}
\\ \\
\operatorname{D}f & = & \texttt{((u)(v)) ~+~ (( u + du )( v + dv ))}
\\ \\
\operatorname{D}g & = & \texttt{((u,~v)) ~+~ (( u + du ,~ v + dv ))}
\end{array}</math>
|}
<pre>
As a matter of fact, computing the symmetric differences
As a matter of fact, computing the symmetric differences
Df = f + Ef and Dg = g + Eg has already taken care of the
Df = f + Ef and Dg = g + Eg has already taken care of the
