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Directory:Jon Awbrey/Papers/Differential Analytic Turing Automata (view source)
Revision as of 03:12, 12 March 2009
, 03:12, 12 March 2009→Computation Summary : <math>g(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>
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|}
<math>\operatorname{D}g</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>g,\!</math> that is, if you want to get to a place where the value of <math>g\!</math> is different from what it is where you are. In the present case, where the ruling proposition <math>g\!</math> is <math>\texttt{((u, v))},</math> the term <math>\texttt{uv} \cdot \texttt{(du, dv)}</math> of <math>\operatorname{D}g</math> tells you this: If <math>u\!</math> and <math>v\!</math> are both true where you are, then you would have to change one or the other but not both of <math>u\!</math> and <math>v\!</math> in order to reach a place where the value of <math>g\!</math> is different from what it is where you are.
<pre>
<pre>
Figure 2.4 approximates Dg in the proxy of the linear proposition
Figure 2.4 approximates Dg in the proxy of the linear proposition
dg = uv (du, dv) + u(v)(du, dv) + (u)v (du, dv) + (u)(v) (du, dv).
dg = uv (du, dv) + u(v)(du, dv) + (u)v (du, dv) + (u)(v) (du, dv).