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MyWikiBiz, Author Your Legacy — Tuesday March 11, 2025
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In their application to this logical transformation the operators <math>\operatorname{E}</math> and <math>\operatorname{D}</math> respectively produce the ''enlarged map'' <math>\operatorname{E}F = (\operatorname{E}f, \operatorname{E}g)</math> and the ''difference map'' <math>\operatorname{D}F = (\operatorname{D}f, \operatorname{D}g),</math> whose components can be given as follows, if the reader, in the absence of a special format for logical parentheses, can forgive syntactically bilingual phrases:
 
In their application to this logical transformation the operators <math>\operatorname{E}</math> and <math>\operatorname{D}</math> respectively produce the ''enlarged map'' <math>\operatorname{E}F = (\operatorname{E}f, \operatorname{E}g)</math> and the ''difference map'' <math>\operatorname{D}F = (\operatorname{D}f, \operatorname{D}g),</math> whose components can be given as follows, if the reader, in the absence of a special format for logical parentheses, can forgive syntactically bilingual phrases:
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{| align="center" cellpadding="8" width="90%"
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|
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<math>\begin{array}{lll}
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\operatorname{E}f & = & \underline{((}~ u + du ~\underline{)(}~ v + dv ~\underline{))}
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\\ \\
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\operatorname{E}g & = & \underline{((}~ u + du ~,~ v + dv ~\underline{))}
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\\ \\
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\operatorname{D}f & = & \underline{((}~ u ~\underline{)(}~ v ~\underline{))}~ + ~\underline{((}~ u + du ~\underline{)(}~ v + dv ~\underline{))}
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\\ \\
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\operatorname{D}g & = & \underline{((}~ u ~,~ v ~\underline{))}~ + ~\underline{((}~ u + du ~,~ v + dv ~\underline{))}
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\end{array}</math>
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|}
    
<pre>
 
<pre>
Ef  =  ((u + du)(v + dv))
  −
  −
Eg  =  ((u + du, v + dv))
  −
  −
Df  =  ((u)(v))  +  ((u + du)(v + dv))
  −
  −
Dg  =  ((u, v))  +  ((u + du, v + dv))
  −
   
But these initial formulas are purely definitional,
 
But these initial formulas are purely definitional,
 
and help us little to understand either the purpose
 
and help us little to understand either the purpose
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