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Directory:Jon Awbrey/Papers/Differential Analytic Turing Automata (view source)

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→‎Computation Summary : <math>g(u, v) = \texttt{((u,~v))}</math>: markup

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==Note 19==

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===Computation Summary : <math>g(u, v) = \texttt{((u,~v))}</math>===

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===Computation Summary : <math>g(u, v) = \texttt{((u, v))}</math>===

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Figure 2.1 shows the expansion of <math>g = \texttt{((u,~v))}</math> over <math>[u, v]\!</math> to produce the expression:

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Figure 2.1 shows the expansion of <math>g = \texttt{((u, v))}</math> over <math>[u, v]\!</math> to produce the expression:

{| align="center" cellpadding="8" width="90%"

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Figure 2.2 shows the expansion of <math>\operatorname{E}g = \texttt{((u + du, ~v + dv))}</math> over <math>[u, v]\!</math> to produce the expression:

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Figure 2.2 shows the expansion of <math>\operatorname{E}g = \texttt{((u + du, v + dv))}</math> over <math>[u, v]\!</math> to produce the expression:

{| align="center" cellpadding="8" width="90%"

| <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{E}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 end up in a place where <math>g\!</math> is true. In this case, where the prevailing proposition <math>g\!</math> is <math>\texttt{((u, v))},</math> the component <math>\texttt{uv} \cdot \texttt{((du, dv))}</math> of <math>\operatorname{E}g</math> tells you this: If <math>u\!</math> and <math>v\!</math> are both true where you are, then change either both or neither of <math>u\!</math> and <math>v\!</math> at the same time, and you will attain a place where <math>\texttt{((du, dv))}</math> is true.

<pre>

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~~Eg tells you what you would have to do, from where you are in the~~

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~~universe [u, v], if you want to end up in a place where g is true.~~

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~~In this case, where the prevailing proposition g is ((u, v)), the~~

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~~component uv((du, dv)) of Eg tells you this: If u and v are both~~

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~~true where you are, then change either both or neither u and v at~~

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~~the same time, and you will attain a place where ((u, v)) is true.~~

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Figure 2.3 expands Dg over [u, v] to obtain the following formula:

Dg = uv (du, dv) + u(v)(du, dv) + (u)v (du, dv) + (u)(v) (du, dv).

Jon Awbrey

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