- #1
gikiian
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Consider the ODE [itex] [/itex] [itex]y''+P(x)y'+Q(x)y=0[/itex].
If [itex]\stackrel{limit}{_{x→x_{o}}}P(x)[/itex] and [itex]\stackrel{limit}{_{x→x_{o}}}Q(x)[/itex] converge, can you call [itex]x_{o}[/itex] a 'regular singular point' besides calling it an 'ordinary point'?
I am saying this because if [itex]\stackrel{limit}{_{x→x_{o}}}P(x)[/itex] and [itex]\stackrel{limit}{_{x→x_{o}}}Q(x)[/itex] converge, then [itex]\stackrel{limit}{_{x→x_{o}}}(x-x_{o})P(x)[/itex] and [itex]\stackrel{limit}{_{x→x_{o}}}(x-x_{o})^{2}Q(x)[/itex] will also converge. And for a second-order linear ODE for which [itex]\stackrel{limit}{_{x→x_{o}}}(x-x_{o})P(x)[/itex] and [itex]\stackrel{limit}{_{x→x_{o}}}(x-x_{o})^{2}Q(x)[/itex] converge, [itex]x_{o}[/itex] is termed as a regular singular point.
If [itex]\stackrel{limit}{_{x→x_{o}}}P(x)[/itex] and [itex]\stackrel{limit}{_{x→x_{o}}}Q(x)[/itex] converge, can you call [itex]x_{o}[/itex] a 'regular singular point' besides calling it an 'ordinary point'?
I am saying this because if [itex]\stackrel{limit}{_{x→x_{o}}}P(x)[/itex] and [itex]\stackrel{limit}{_{x→x_{o}}}Q(x)[/itex] converge, then [itex]\stackrel{limit}{_{x→x_{o}}}(x-x_{o})P(x)[/itex] and [itex]\stackrel{limit}{_{x→x_{o}}}(x-x_{o})^{2}Q(x)[/itex] will also converge. And for a second-order linear ODE for which [itex]\stackrel{limit}{_{x→x_{o}}}(x-x_{o})P(x)[/itex] and [itex]\stackrel{limit}{_{x→x_{o}}}(x-x_{o})^{2}Q(x)[/itex] converge, [itex]x_{o}[/itex] is termed as a regular singular point.
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