# Thread: Power series solution - are these Recurrence relationships correct?

1. ## Power series solution - are these Recurrence relationships correct?

I'm working with solving an ODE via power series, and the recurrence relations seem like they're more complicated than they should be. Can someone check what I've done so far and let me know if I've screwed up royally somewhere?

$\displaystyle (3 - x^2)y'' - 3xy' - y = 0 \qquad x_0 = 0$

$\displaystyle y = \sum_{n=0}^{\infty} a_n x^n$

$\displaystyle y' = \sum_{n=1}^{\infty} na_n x^{n - 1}$

$\displaystyle y'' = \sum_{n=2}^{\infty} n(n - 1) a_n x^{n - 2}$

$\displaystyle (3 - x^2) \sum_{n=2}^{\infty} n(n - 1) a_n x^{n - 2} - 3x \sum_{n=1}^{\infty} na_n x^{n - 1} - \sum_{n=0}^{\infty} a_n x^n = 0$

$\displaystyle 3 \sum_{n=2}^{\infty} n(n - 1) a_n x^{n - 2} - \sum_{n=2}^{\infty} n(n - 1) a_n x^n - 3 \sum_{n=1}^{\infty} na_n x^n - \sum_{n=0}^{\infty} a_n x^n = 0$

$\displaystyle 3 \sum_{k=0}^{\infty} (k + 2)(k + 1)a_{k + 2} x^k - \sum_{k=2}^{\infty} k(k - 1) a_k x^k - 3 \sum_{k=1}^{\infty} ka_k x^k - \sum_{k=0}^{\infty} a_k x^k = 0$

$\displaystyle 3 \sum_{k=2}^{\infty} (k + 2)(k + 1)a_{k + 2} x^k - \sum_{k=2}^{\infty} k(k - 1) a_k x^k - 3 \sum_{k=2}^{\infty} ka_k x^k - \sum_{k=2}^{\infty} a_k x^k + 6a_2 + 18a_3x - 3a_1x - a_0 - a_1x = 0$

$\displaystyle \sum_{k=2}^{\infty} \left[ x^k \left[ 3(k + 2)(k + 1)a_{k + 2} - k(k - 1) a_k - 3ka_k - a_k \right] \right] + 18a_3x + 6a_2 - 4a_1x - a_0 = 0$

$\displaystyle 18a_3x + 6a_2 - 4a_1x - a_0 = 0$

$\displaystyle a_3 = \frac{-6a_2 + 4a_1x + a_0}{18x}$

$\displaystyle 3(k + 2)(k + 1)a_{k + 2} - k(k - 1) a_k - 3ka_k - a_k = 0$

$\displaystyle a_{k + 2} = \frac{k(k - 1)a_k + 3ka_k + a_k}{3(k + 2)(k + 1)}$

It's that first recurrence relationship for $\displaystyle a_3$ that seems odd. The RR for the orphan terms has always evaluated to something simple, and certainly never with three levels of $\displaystyle a$ terms, nor with $\displaystyle x$ making an appearance.

It makes me think that I made a mistake somewhere. Did I?

2. ## Re: Power series solution - are these Recurrence relationships correct?

Originally Posted by Lancet
$\displaystyle 18a_3x + 6a_2 - 4a_1x - a_0 = 0$
$\displaystyle a_3 = \frac{-6a_2 + 4a_1x + a_0}{18x}$

It should be: $\displaystyle 18a_3x + 6a_2 - 4a_1x - a_0 = 0 \Leftrightarrow \begin{Bmatrix} 18a_3-4a_1=0\\6a_2-a_0=0\end{matrix}\Leftrightarrow \begin{Bmatrix} a_3=2a_1/9\\a_2=a_0/6\end{matrix}$

3. ## Re: Power series solution - are these Recurrence relationships correct?

Originally Posted by FernandoRevilla
It should be: $\displaystyle 18a_3x + 6a_2 - 4a_1x - a_0 = 0 \Leftrightarrow \begin{Bmatrix} 18a_3-4a_1=0\\6a_2-a_0=0\end{matrix}\Leftrightarrow \begin{Bmatrix} a_3=2a_1/9\\a_2=a_0/6\end{matrix}$
Fascinating!

Why? How do you know which terms to form a sub-split with?

4. ## Re: Power series solution - are these Recurrence relationships correct?

What about the uniqueness of the series expansion?

5. ## Re: Power series solution - are these Recurrence relationships correct?

Originally Posted by FernandoRevilla
What about the uniqueness of the series expansion?
I'm afraid I don't understand...

6. ## Re: Power series solution - are these Recurrence relationships correct?

Originally Posted by Lancet
I'm afraid I don't understand...
$\displaystyle f(x)=\sum_{n=0}^{+\infty}\alpha_nx^n=\sum_{n=0}^{+ \infty}\beta_nx^n$ for all $\displaystyle x\in (-R,R)$ implies $\displaystyle \alpha_n=\beta_n=\dfrac{f^{(n)}(0)}{n!}$ for all $\displaystyle n$ .