Why do we use different methods for finding power series solutions?

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Discussion Overview

The discussion revolves around the different methods used for finding power series solutions to differential equations, specifically focusing on the forms of the series: \(y(x)=\sum^{\infty}_{n=0}a_nx^n\) versus \(y(x)=\sum^{\infty}_{n=0}a_nx^{n+1}\). The scope includes theoretical considerations and the implications of these choices in the context of differential equations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants suggest that the choice between the two forms of power series may not have a particular reason and that one form may yield neater equations than the other.
  • It is proposed that both forms will work as long as certain conditions, such as \(y(0) = 0\), are satisfied.
  • One participant raises a condition related to the differential equation \(\alpha(x)y''(x)+\beta(x)y'(x)+\gamma(x)y(x)=0\), stating that if \(\alpha(0)=0\), one must work with the series of the form \(\sum^{\infty}_{n=0}a_nx^{n+k}\), though they express uncertainty about this claim.
  • A later reply clarifies that this form is equivalent to another series \(\sum^{\infty}_{n=0}b_nx^n\) with specific definitions for \(b_n\), but the participant remains unsure about the implications.

Areas of Agreement / Disagreement

Participants express differing views on the necessity and implications of using different forms of power series. There is no consensus on the best approach or the conditions under which each form should be used.

Contextual Notes

Participants note that the choice of series form may depend on specific conditions of the differential equations being solved, such as the value of \(\alpha(0)\). However, the discussion does not resolve the implications of these conditions or the equivalence of the series forms.

matematikuvol
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Why sometimes we search solution of power series in the way:
[tex]y(x)=\sum^{\infty}_{n=0}a_nx^n[/tex]
and sometimes
[tex]y(x)=\sum^{\infty}_{n=0}a_nx^{n+1}[/tex]?
 
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hi matematikuvol! :smile:
matematikuvol said:
Why sometimes we search solution of power series in the way:
[tex]y(x)=\sum^{\infty}_{n=0}a_nx^n[/tex]
and sometimes
[tex]y(x)=\sum^{\infty}_{n=0}a_nx^{n+1}[/tex]?

no particular reason …

sometimes one gives neater equations than the other …

they'll both work (provided, of course, that y(0) = 0) :wink:
 
I think that in the case when
[tex]\alpha(x)y''(x)+\beta(x)y'(x)+\gamma(x)y(x)=0[/tex]
if ##\alpha(0)=0## you must work with ##\sum^{\infty}_{n=0}a_nx^{n+k}##, but I'm not sure.
 
matematikuvol said:
I think that in the case when
[tex]\alpha(x)y''(x)+\beta(x)y'(x)+\gamma(x)y(x)=0[/tex]
if ##\alpha(0)=0## you must work with ##\sum^{\infty}_{n=0}a_nx^{n+k}##, but I'm not sure.

but that's the same as ##\sum^{\infty}_{n=0}b_nx^n## with ##b_n = a_{n-k}## for n ≥ k, and ##b_n = 0## otherwise :wink:
 

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