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Revision as of 14:18, 11 March 2009
, 14:18, 11 March 2009→Computation Summary : <math>f(u, v) = \texttt{((u)(v))}</math>: markup
<math>\operatorname{D}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 bring about a change in the value of <math>f,\!</math> that is, if you want to get to a place where the value of <math>f\!</math> is different from what it is where you are. In the present case, where the reigning proposition <math>f\!</math> is <math>\texttt{((u)(v))},</math> the term <math>\texttt{uv} \cdot \texttt{du~dv}</math> of <math>\operatorname{D}f</math> tells you this: If <math>u\!</math> and <math>v\!</math> are both true where you are, then you would have to change both <math>u\!</math> and <math>v\!</math> in order to reach a place where the value of <math>f\!</math> is different from what it is where you are.
<math>\operatorname{D}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 bring about a change in the value of <math>f,\!</math> that is, if you want to get to a place where the value of <math>f\!</math> is different from what it is where you are. In the present case, where the reigning proposition <math>f\!</math> is <math>\texttt{((u)(v))},</math> the term <math>\texttt{uv} \cdot \texttt{du~dv}</math> of <math>\operatorname{D}f</math> tells you this: If <math>u\!</math> and <math>v\!</math> are both true where you are, then you would have to change both <math>u\!</math> and <math>v\!</math> in order to reach a place where the value of <math>f\!</math> is different from what it is where you are.
Figure 1.4 approximates <math>\operatorname{D}f</math> by the linear form <math>\operatorname{d}f</math> that expands over <math>[u, v]\!</math> as follows:
{| align="center" cellpadding="8" width="90%"
| <math>\operatorname{d}f ~=~ \texttt{uv} \cdot \texttt{0} ~+~ \texttt{u(v)} \cdot \texttt{du} ~+~ \texttt{(u)v} \cdot \texttt{dv} ~+~ \texttt{(u)(v)} \cdot \texttt{(du, dv)}</math>
|}
<pre>
<pre>
Figure 1.5 shows what remains of the difference map Df
Figure 1.5 shows what remains of the difference map Df
when the first order linear contribution df is removed:
when the first order linear contribution df is removed:
