Changes

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:

Ef  = ((u + du)(v + dv))

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Eg  = ((u + du, v + dv))

 

\operatorname{E}f & = & \texttt{(( u + du )( v + dv ))}

Df  = ((u)(v)) + ((u + du)(v + dv))

 

\operatorname{E}g & = & \texttt{(( u + du ,~ v + dv ))}

Dg  = ((u, v)) + ((u + du, v + dv))

\operatorname{D}f & = & \texttt{((u)(v)) ~+~ (( u + du )( v + dv ))}

\operatorname{D}g & = & \texttt{((u,~v)) ~+~ (( u + du ,~ v + dv ))}

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