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The study of mathematics cannot be replaced by any other activity that will train and develop man's purely logical faculties to the same level of rationality. The American Mathematical Monthly, 56, 1949, p. 19. |
Loci: ConvergencePeano on Wronskians: A TranslationSurvey of Infinitesimal AnalysisThe original French passage of Résumé du Cours d’Analyse Infinitésimale de l’Université de Gand (Survey of Infinitesimal Analysis, Ghent University) can be found at [PM], while our English translation can be found in Appendix 2. Mansion began his passage by stating:
This statement is only partially valid, since Peano’s counterexample disproves this statement as well (Passage 2). We have already noted that the Wronskian of a set of functions will be zero if the functions are linearly dependent and Mansion gives the same argument we provided earlier.
The problem, then, must be in the proof of the “vice versa”. Mansion proceeded by induction, and while the induction step is (essentially) valid, there is a problem with the base case.
The first mistake in this passage is simply assuming that the theorem is true for a \( 3\times 3\) Wronskian. While we know the theorem is false for two functions, technically Peano did not give a counterexample for three. However, the theorem is indeed false for three functions as well, as Bocher showed in [B3]. Despite the problem with the base case, we will reproduce the rest of Mansion’s argument, filling in details for clarity.
These relationships can be derived by taking the determinants of the matrices: \[ \left[ {\begin{array}{*{20}c} r & s & t & u \\ r & s & t & u \\ {r'} & {s'} & {t'} & {u'} \\ {r''} & {s''} & {t''} & {u''} \\ \end{array}} \right]\quad \left[ {\begin{array}{*{20}c} {r'} & {s'} & {t'} & {u'} \\ r & s & t & u \\ {r'} & {s'} & {t'} & {u'} \\ {r''} & {s''} & {t''} & {u''} \\ \end{array}} \right]\quad \left[ {\begin{array}{*{20}c} {r''} & {s''} & {t''} & {u''} \\ r & s & t & u \\ {r'} & {s'} & {t'} & {u'} \\ {r''} & {s''} & {t''} & {u''} \\ \end{array}} \right]\quad \left[ {\begin{array}{*{20}c} {r'''} & {s'''} & {t'''} & {u'''} \\ r & s & t & u \\ {r'} & {s'} & {t'} & {u'} \\ {r''} & {s''} & {t''} & {u''} \\ \end{array}} \right] \] Each of these matrices has a zero determinant, and expanding along the top rows gives the equations \( \left( {{\mathcal{Z}}_1}\right), \left( {{\mathcal{Z}}_2}\right), \left( {{\mathcal{Z}}_3}\right), \) and \( \left( {{\mathcal{Z}}_4}\right), \) except for having alternating signs. It appears this was an error in the original printing, because the argument will work if the signs are treated carefully. The next step is to differentiate each equation \( \left( {{\mathcal{Z}}_1}\right), \left( {{\mathcal{Z}}_2}\right), \left( {{\mathcal{Z}}_3}\right), \) and \( \left( {{\mathcal{Z}}_4}\right), \) making use of the product rule and simplifying at each step. Again, the equations should have alternating signs.
Notice that this is simply a system of linear equations in the variables \( {k^{\prime}},{m^{\prime}},{n^{\prime}}, \) and \( {p^{\prime}}. \) Mansion solved for these variables. One way of doing so would be to rewrite the corrected system as \[ {k^{\prime}}r- {m^{\prime}}s+ {n^{\prime}}t= {p^{\prime}}u, \] \[ {k^{\prime}}{r^{\prime}}-{m^{\prime}}{s^{\prime}}+{n^{\prime}}{t^{\prime}}={p^{\prime}}{u^{\prime}}, \] \[ {k^{\prime}}{r^{\prime\prime}}-{m^{\prime}}{s^{\prime\prime}}+{n^{\prime}}{t^{\prime\prime}}={p^{\prime}}{u^{\prime\prime}}, \] \[ {k^{\prime}}{r^{\prime\prime\prime}}-{m^{\prime}}{s^{\prime\prime\prime}}+{n^{\prime}}{t^{\prime\prime\prime}}={p^{\prime}}{u^{\prime\prime\prime}}. \] and then apply Cramer’s rule to see that: \[ k' = \frac{{\left| {\begin{array}{*{20}c} {p'u} & { - s} & t \\ {p'u'} & { - s'} & {t'} \\ {p'u''} & { - s''} & {t''} \\ \end{array}} \right|}}{{\left| {\begin{array}{*{20}c} r & { - s} & t \\ {r'} & { - s'} & {t'} \\ {r''} & { - s''} & {t''} \\ \end{array}} \right|}} = \frac{{p'\left| {\begin{array}{*{20}c} s & t & u \\ {s'} & {t'} & {u'} \\ {s''} & {t''} & {u''} \\ \end{array}} \right|}}{{\left| {\begin{array}{*{20}c} r & s & t \\ {r'} & {s'} & {t'} \\ {r''} & {s''} & {t''} \\ \end{array}} \right|}} = p'\frac{k}{p} \] This yields the formulas given by Mansion in the following passage. These relationships are all derivatives of the logarithms of the functions, and so their integrals will all differ by a constant. This step will require that not only are \( k \) and \( p \) not zero, but neither are their derivatives. Integrating and keeping track of the constants will give a relationship between \( r,s,t,\) and \( u. \)
Next page >> On the Wronskians Pages: | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | Engdahl, Susannah M. and Adam E. Parker, "Peano on Wronskians: A Translation," Loci (March 2011), DOI: 10.4169/loci003642 |