12,080
edits
Changes
MyWikiBiz, Author Your Legacy — Tuesday April 30, 2024
Jump to navigationJump to searchDirectory:Jon Awbrey/Papers/Cactus Language (view source)
Revision as of 04:50, 28 May 2009
, 04:50, 28 May 2009→Differential Logic: markup
Applying the enlargement operator <math>\operatorname{E}</math> to the present example, <math>f(x, y) = xy,\!</math> we may compute the result as follows:
Applying the enlargement operator <math>\operatorname{E}</math> to the present example, <math>f(x, y) = xy,\!</math> we may compute the result as follows:
{| align="center" cellpadding="6" style="text-align:center" width="90%"
{| align="center" cellpadding="6" width="90%"
|
| align="center" |
<math>\operatorname{E}f(x, y, \operatorname{d}x, \operatorname{d}y) \quad = \quad (x + \operatorname{d}x)(y + \operatorname{d}y).</math>
<math>\operatorname{E}f(x, y, \operatorname{d}x, \operatorname{d}y) \quad = \quad (x + \operatorname{d}x)(y + \operatorname{d}y).</math>
|-
|-
|
| align="center" |
<pre>
<pre>
o---------------------------------------o
o---------------------------------------o
|}
|}
Given the proposition <math>f(x, y)\!</math> in <math>U = X \times Y,</math> the (first order) ''difference'' of <math>f\!</math> is the proposition <math>\operatorname{D}f</math> in <math>\operatorname{E}U</math> that is defined by the formula <math>\operatorname{D}f = \operatorname{E}f - f,</math> that is, <math>\operatorname{D}f(x, y, \operatorname{d}x, \operatorname{d}y) = f(x + \operatorname{d}x, y + \operatorname{d}y) - f(x, y).</math>
Given the proposition f(x, y) in U = X x Y,
the (first order) 'difference' of f is the
proposition Df in EU that is defined by the
formula Df = Ef - f, or, written out in full,
In the example f(x, y) = xy, the result is:
In the example <math>f(x, y) = xy,\!</math> the result is:
{| align="center" cellpadding="6" width="90%"
| align="center" |
<math>\operatorname{D}f(x, y, \operatorname{d}x, \operatorname{d}y) \quad = \quad (x + \operatorname{d}x)(y + \operatorname{d}y) - xy.</math>
|-
| align="center" |
<pre>
o---------------------------------------o
o---------------------------------------o
| |
| |
| Df = ((x, dx)(y, dy), xy) |
| Df = ((x, dx)(y, dy), xy) |
o---------------------------------------o
o---------------------------------------o
</pre>
|}
<pre>
We did not yet go through the trouble to interpret this (first order)
We did not yet go through the trouble to interpret this (first order)
"difference of conjunction" fully, but were happy simply to evaluate
"difference of conjunction" fully, but were happy simply to evaluate
