Legendre equation is

$$ (1-x^2) y'' - 2xy' + \lambda y=0$$

We are interested in finding solutions in the range $[-1,1]$. We seek solutions around the ordinary point $x=0$ $$ \sum_{n=0}^\infty c_n x^n $$ Since legendre equation has regular singular points at $-1,1$ , this series is guaranteed to converge in $|x|<1$. Substituting this power series into the differential equation one arrives at the following recursive relation

$$c_{n+2} = \frac{n(n+1)-\lambda}{(n+2)(n+1)}$$

Putting $c_0 = 1$ and $c_1 = 0$ gives the solution $y_1(x)$

Putting $c_0 = 0$ and $c_1 = 1$ gives the solution $y_2(x)$

Now if $\lambda = l(l+1)$ with $l$ a non-negative integer one of these series terminates and reduces to a polynomial and it can be shown that a solution to legendre equation is the legendre polynomial $$P_l(x) = \frac{1}{2^l l!} \frac{d^l}{dx^l}(x^2-1)^l$$ which is analytic everywhere.

However, if $\lambda$ is not of this form i'm not sure if the solutions

$$y_1(1) = \sum_{n=0}^{\infty} c_{2n}$$ $$y_2(1) = \sum_{n=0}^{\infty} c_{2n+1}$$ $$y_1(-1) = \sum_{n=0}^{\infty} (-1)^n c_{2n}$$ $$y_2(-1) = \sum_{n=0}^{\infty} (-1)^n c_{2n+1}$$

converges at the end points.



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