I need just a little help with this question. I almost solved it except for a minor detail at the end.

Here is the question.

The equation

$\displaystyle xy''(x) + (1-x)y'(x) +my(x) = 0$,

where m is a nonnegative interger, is called Laguerre's differential equation. Show that for each m, this equation has a polynomial solution of degree m. These polynomials are denoted by $\displaystyle L_{n}(x)$ and are called Laguerre polynomials. The first few Laguerre polynomials are

$\displaystyle L_{0}(x) = 1$

$\displaystyle L_{1}(x) = -x +1$

$\displaystyle L_{2} = x^2-4x+2$

Ok so I did a power series solution for the problem and arrived at

$\displaystyle a_{0} = constant$

$\displaystyle a_{1} = -ma_{0}$

$\displaystyle k \geq 1$ $\displaystyle a_{k+1} = \frac{a_{k}(k-m)}{(k+1)(k) +(k+1)}$

so

$\displaystyle a_{2} = a_{0}\frac{(-m)(1-m)}{4}$

$\displaystyle a_{3}=a_{0}\frac{(-m)(1-m)(2-m)}{(4)(9}$

$\displaystyle a_{4} = a_{0} \frac{(-m)(1-m)(2-m)(3-m)}{(4)(9)(16)}$

and this is where I get confused.

I see that there is a factorial pattern so I'm trying to use factorial notation to condense.

so would it be

$\displaystyle \sum_{n=2}^{\infty} \frac{(-m)_{n}}{(1)^2_{n}}x^n$

or

$\displaystyle \sum_{n=2}^{\infty} \frac{(-m)_{n}}{(1^2)_{n}}x^n$

So would the final solution be

$\displaystyle m \geq 0$

$\displaystyle y(x) = a_{0} -ma_{1}x +$ whatever series is correct?

This solution would show depending what m is chosen that the answer would cancel out certain terms.