Find Radius of Convergence from Recursion Equation

[Saladsamurai]

[STRIKE][/STRIKE]

Homework Statement

Find Radius of Convergence of the corresponding power series solution from Recursion Equation alone:

[tex]n^2a_{n+2} - 3(n+2)a_{n+1} +3a_{n-1} = 0 \qquad(1)[/tex]

Homework Equations

R = 1/L where

[tex] L = \lim_{n\rightarrow\infty}\left|{\frac{a_{k+1}}{a_k}\right|\qquad(2)[/tex]

The Attempt at a Solution

I was thinking that I could solve (1) for a_n+1 and then solve it again for a_n and then use the ratio in (2). But I feel like that might be a very illegal move.

Thoughts on this?

 

[mighty2000]

To find the radius of convergence for the power series solution of the given recursion equation, you can use the ratio test as outlined in (2). However, instead of solving for an+1 and an separately, you can use the given recursion equation to simplify the ratio.

First, divide both sides of the recursion equation by an to get:

n^2a_{n+2}/a_n - 3(n+2)a_{n+1}/a_n + 3a_{n-1}/a_n = 0

Then, using the fact that L = lim_{n\rightarrow\infty}\left|{\frac{a_{k+1}}{a_k}\right|, we can rewrite the above equation as:

n^2(a_{n+2}/a_n) - 3(n+2)(a_{n+1}/a_n) + 3(a_{n-1}/a_n) = 0

Now, as n approaches infinity, the terms (a_{n+2}/a_n) and (a_{n-1}/a_n) will approach zero, leaving us with:

0 - 3L + 3(0) = 0

Solving for L, we get L = 0.

Therefore, the radius of convergence for the power series solution is R = 1/L = 1/0 = undefined. This means that the power series solution will converge at the point where the ratio of consecutive terms is equal to 0, which is at the origin.

I hope this helps! Let me know if you have any further questions.



Scientist