Legendre polynomials recurrence relation

Question

How can i get? $$P_{n+1}=xP_n(x)-\frac{1-x^2}{n+1} P'_n(x)$$ $n>=0$ Also know as the leadder equation of the legendre polinomials i tried to use de recurrence relations as: $$P_n(x)=P_{n+1}'(x)-2xP'_n(x)$$ and $$nP_n(x)+P_{n+1}'(x)-xP'_n(x)$$

Answer 1

Differentiating the generating function $$ g(x,t)=(1-2xt+t^2)^{-1/2}=\sum_{n=0}^\infty P_n(x)t^n,\ |t|<1 $$ with respect to $x$, one has $$ \frac{\partial g(x,t)}{\partial x}=\frac{t}{(1-2xt+t^2)^{3/2}}=\sum_{n=0}^\infty P_n’(x)t^n. $$ From this we obtain $$ (1-2xt+t^2)\sum_{n=0}^\infty P_n’(x)t^n-t\sum_{n=0}^\infty P_n(x)t^n=0 $$ which leads to $$P_{n+1}’(x)+P_{n-1}’(x)=2xP_n’(x)+P_n(x)\tag 1$$ Differentiating the following recurrence relation (Bonnet’s recursion formula) $$ (2n+1)xP_n(x)=(n+1)P_{n+1}(x)+nP_{n-1}(x)\tag 2 $$ with respect to $x$, and adding 2 times $\frac{\operatorname{d}}{\operatorname{d}x}$(2) to $-(2n+1)$ times (1), we get $$ (2n+1)P_n=P_{n+1}’(x)-P_{n-1}’(x).\tag 3 $$ $\frac{1}{2}[(1)+(3)]$ gives $$ P_{n+1}’(x)=(n+1)P_n(x)+xP_n’(x).\tag 4 $$ $\frac{1}{2}[(1)-(3)]$ gives $$ P_{n-1}’(x)=-nP_n(x)+xP_n’(x).\tag 5 $$ Replace $n$ by $n−1$ in (4) and add the result to $x$ times (5): $$ (1-x^2)P_n’(x)=nP_{n-1}(x)-nxP_n(x).\tag 6 $$ Finally we have $$ P_{n}(x)=xP_{n+1}(x)+\frac{1-x^2}{n+1}P_{n+1}'(x). $$