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MyWikiBiz, Author Your Legacy — Monday November 25, 2024
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We may analyze <math>X^2 = \{ (u, v) : u, v \in X \}</math> into two parts, specifically, the ordered pairs <math>(u, v)\!</math> that lie on and off the diagonal:
 
We may analyze <math>X^2 = \{ (u, v) : u, v \in X \}</math> into two parts, specifically, the ordered pairs <math>(u, v)\!</math> that lie on and off the diagonal:
   −
: <math>X^2 = \{ (u, v) : u = v \}\ \cup\ \{ (u, v) : u \ne v \}</math>
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: <math>\begin{matrix}
 +
X^2 & = & \{ (u, v) : u = v \} & \cup & \{ (u, v) : u \ne v \}.
 +
\end{matrix}</math>
   −
In symbolic terms, this partition may be expressed as:
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This partition may also be expressed in the follwing symbolic form:
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: <math>\begin{matrix}
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X^2 & \cong & \operatorname{diag}(X) & + & 2 \tbinom{X}{2}.
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\end{matrix}</math>
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: ''X''<sup>2</sup> <math>\cong</math> Diag(''X'') + 2 * Comb(''X'', 2),
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The separate terms of this formula are defined as follows:
   −
where:
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: <math>\begin{matrix}
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\operatorname{diag}(X) & = & \{ (x, x) : x \in X \}.
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\end{matrix}</math>
   −
: Diag(''X'') = {‹''x'', ''x''› : ''x'' &isin; ''X''},
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: <math>\begin{matrix}
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\tbinom{X}{k} & = & X\ \operatorname{choose}\ k & = & \{ k\!\mbox{-sets from}\ X \}.
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\end{matrix}</math>
   −
and where:
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Thus we have:
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: Comb(''X'', ''k'') = "''X'' choose ''k''" = {''k''-sets from ''X''},
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: <math>\begin{matrix}
 
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\tbinom{X}{2} & = & \{ \{ u, v \} : u, v \in X \}.
so that:
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\end{matrix}</math>
 
  −
:  Comb(''X'', 2) = {{''u'', ''v''} : ''u'', ''v'' &isin; ''X''}.
      
We can now use the features in d<font face="lucida calligraphy">X</font> = {d''x''<sub>''i''</sub>} = {d''x''<sub>1</sub>,&nbsp;&hellip;,&nbsp;d''x''<sub>''n''</sub>} to classify the paths of ('''B''' &rarr; ''X'') by way of the pairs in ''X''<sup>2</sup>.  If ''X'' <math>\cong</math> '''B'''<sup>''n''</sup> then a path in ''X'' has the form ''q'' : ('''B''' &rarr; '''B'''<sup>''n''</sup>) <math>\cong</math> '''B'''<sup>''n''</sup> &times; '''B'''<sup>''n''</sup> <math>\cong</math> '''B'''<sup>2''n''</sup> <math>\cong</math> ('''B'''<sup>2</sup>)<sup>''n''</sup>.  Intuitively, we want to map this ('''B'''<sup>2</sup>)<sup>''n''</sup> onto ''D''<sup>''n''</sup> by mapping each component '''B'''<sup>2</sup> onto a copy of '''D'''.  But in our current situation "'''D'''" is just a name we give, or an accidental quality we attribute, to coefficient values in '''B''' when they are attached to features in d<font face="lucida calligraphy">X</font>.
 
We can now use the features in d<font face="lucida calligraphy">X</font> = {d''x''<sub>''i''</sub>} = {d''x''<sub>1</sub>,&nbsp;&hellip;,&nbsp;d''x''<sub>''n''</sub>} to classify the paths of ('''B''' &rarr; ''X'') by way of the pairs in ''X''<sup>2</sup>.  If ''X'' <math>\cong</math> '''B'''<sup>''n''</sup> then a path in ''X'' has the form ''q'' : ('''B''' &rarr; '''B'''<sup>''n''</sup>) <math>\cong</math> '''B'''<sup>''n''</sup> &times; '''B'''<sup>''n''</sup> <math>\cong</math> '''B'''<sup>2''n''</sup> <math>\cong</math> ('''B'''<sup>2</sup>)<sup>''n''</sup>.  Intuitively, we want to map this ('''B'''<sup>2</sup>)<sup>''n''</sup> onto ''D''<sup>''n''</sup> by mapping each component '''B'''<sup>2</sup> onto a copy of '''D'''.  But in our current situation "'''D'''" is just a name we give, or an accidental quality we attribute, to coefficient values in '''B''' when they are attached to features in d<font face="lucida calligraphy">X</font>.
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