User:Ben Spinozoan/Wronskian&Independence

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[edit] Proof

  • In the language of first-order logic, the set of functions \{f_1,f_2,...,f_n\}\, is linearly independent, over the interval \Omega\, in \R, iff:
\forall \boldsymbol{\alpha}.\,\Big[\boldsymbol{\alpha}\in\R^n\and\, \boldsymbol{\alpha}\ne0\,\to\,\exists x\in\Omega\,.^\neg P(\boldsymbol{\alpha},x)\Big] , where P(\boldsymbol{\alpha},x) \equiv \Bigg[\sum_{i=1}^n\alpha_i\,f_i(x)=0\Bigg] .
Expressed in disjunctive normal form (DNF) the above definition reads:
LI(\Omega)\,\equiv\,\forall \boldsymbol{\alpha}.\,\Big[\,\boldsymbol{\alpha}\notin\R^n\,\or\,\boldsymbol{\alpha}=0\,\or\,\exists x\in\Omega\,.^\neg P(\boldsymbol{\alpha},x)\,\Big] ,
where LI(Ω) is shorthand for the statement occurring immediately before the iff (note that negation of LI(Ω) gives the correct statement for linear dependence).
  • The text of our theorem "If the Wronskian is non-zero at some point in an interval, then the functions are linearly independent on the interval", now translates as
\exists y\in\Omega.\;W(y)\ne0\,\to\,LI(\Omega),
or, in DNF,
(1)\quad \forall \boldsymbol{\alpha}.\,\bigg[\,\forall y.\,\Big[\,y\notin\Omega\,\or\,W(y)=0\,\Big]\or\,\boldsymbol{\alpha}\notin\R^n\,\or\,\boldsymbol{\alpha}=0\,\or\,\exists x\in\Omega\,. ^\neg P(\boldsymbol{\alpha},x)\,\bigg],
where W(y) is the value of the Wronskian at the point y.
  • The following statement summarizes the situation when Cramer's rule is applied to the linear system associated with the Wronskian:
\forall \boldsymbol{\alpha}.\,\Bigg[\,\boldsymbol{\alpha}\in\R^n\to\bigg[\,\exists x \in \Omega.\, \Big[\,P(\boldsymbol{\alpha},x)\,\and\, W(x)\ne0\,\Big] \to\, \boldsymbol{\alpha}=0\; \bigg]\,\Bigg],
or,
(2)\quad \forall \boldsymbol{\alpha}.\,\bigg[\,\forall x. \, \Big[\, x\notin\Omega \,\or\;W(x)=0 \;\or\, ^\neg\!P(\boldsymbol{\alpha},x)\,\Big] \or\,\boldsymbol{\alpha}\notin\R^n\,\or\, \boldsymbol{\alpha}=0\;\bigg] .
  • In first-order logic, the statement \forall x. \,\big[ A(x) \or B(x)\big] , entails the statement \forall x.A(x) \or \exists y.B(y) . Consequently, statement (2) entails statement (1), and the theorem is proved.