Solve Diff. Eq. using power series

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Homework Help Overview

The discussion revolves around solving the differential equation \((1-x)y'' + y = 0\) using power series methods. Participants are exploring how to combine summations that arise from this equation and are examining the implications of singular points in relation to the method of Frobenius.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the challenge of combining two summations in the context of the differential equation. There are suggestions to write out the first few terms of each sum to gain insight. Questions arise about the distribution of terms and the significance of singular points in the problem.

Discussion Status

There is active engagement with multiple participants providing insights and corrections. Some participants have offered guidance on how to approach the distribution of terms, while others are questioning the understanding of singular points and their relevance to the method of Frobenius.

Contextual Notes

Participants are navigating the complexities of the power series method and the implications of singular points, with references to Taylor and Laurent series. There is an acknowledgment of the original poster's uncertainty regarding the distribution of terms and the nature of the singular point in the context of the problem.

JamesonS
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Homework Statement


\begin{equation}
(1-x)y^{"}+y = 0
\end{equation}

I am here but do not understand how to combine the two summations:
Mod note: Fixed LaTeX in following equation.
$$(1-x)\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n+\sum_{n=0}^{\infty}a_nx^n = 0$$
 
Last edited by a moderator:
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JamesonS said:

Homework Statement


\begin{equation}
(1-x)y^{"}+y = 0
\end{equation}

I am here but do not understand how to combine the two summations:

$$(1-x)\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n+\sum_{n=0}^{\infty}a_nx^n = 0$$

Here is your equation using PF-compatible TeX:
$$
(1-x)\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2} x^n+\sum_{n=0}^{\infty}a_nx^n = 0
$$
Just replace the "\ begin {equation} ... \ end {equation} " by "$ $ ... $ $ " (no spaces between the initial and final $ signs). Also: write "\infty", not "\infinity".

As for your question: write out the first 3 or 4 terms, to see what you get. That will give you insight into what you should do next.
 
Last edited by a moderator:
Thanks for the response and Latex help. Writing out the first few terms of each sum:
$$
(1-x)[2a_2+6a_3x+12a_4x^2+...]+[a_0+a_1x+a_2x^2+...]
$$
I am not sure what to do with the (1-x) term outside the first sum...
 
Last edited:
JamesonS said:
Thanks for the response and Latex help. Writing out the first few terms of each sum:
$$
(1-x)[2a_2+6a_3x+12a_4x^2+...]+[a_0+a_1x+a_2x^2+...]
$$
I am not sure what to do with the (1-x) term outside the first sum...

What is preventing you from "distributing out" the product? That is, ##(1-x) P(x) = P(x) - x P(x).##
 
Last edited:
Been a while since I did DE, but doesn't the OP have to be aware of the singular point in this problem? Hence he has to use the method of Frobenius?
 
MidgetDwarf said:
Been a while since I did DE, but doesn't the OP have to be aware of the singular point in this problem? Hence he has to use the method of Frobenius?
But the singular point isn't at ##x=0##.
 
Thanks for the correction. It's been a while. I remembered that there is no singular point if we take the Taylor expansion about x=0? Correct?
 
JamesonS said:
I am not sure what to do with the (1-x) term outside the first sum...

Practically after your first equation, or at any later stage, just multiply it out. You have shown that you know how to write a sum of different powers of x in terms of xn by changing the subscript appropriately.
 
MidgetDwarf said:
Thanks for the correction. It's been a while. I remembered that there is no singular point if we take the Taylor expansion about x=0? Correct?
That's backwards. If you expand about a regular point, then the solution can be written as a Taylor series. If there's a singular point, then you can use the method of Frobenius and end up with a Laurent series.
 
  • #10
vela said:
That's backwards. If you expand about a regular point, then the solution can be written as a Taylor series. If there's a singular point, then you can use the method of Frobenius and end up with a Laurent series.

Thanks Vela!

It has been a while. I may pop open a differentials to bring the memory back.
 

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