MyWikiBiz, Author Your Legacy — Thursday September 26, 2024
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230 bytes added
, 17:34, 10 March 2009
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| 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. |
| | | |
− | <pre>
| |
| 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))
| + | {| align="center" cellpadding="8" width="90%" |
− | | + | | |
− | Eg = ((u + du, v + dv))
| + | <math>\begin{array}{lll} |
− | | + | \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 ))} |
| + | \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 |