Legendre polynomials recurrence relation 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
$$nP_n(x)+P_{n+1}'(x)-xP'_n(x)$$
 A: 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
$$
P_{n}(x)=xP_{n+1}(x)+\frac{1-x^2}{n+1}P_{n+1}'(x).
$$
A: Here is a corrected version of alexjo's answer (which I found very useful):
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 (5) and subtract $x$ times (4):
$$
P_n’(x)-x[(n+1)P_n+xP_n']=-(n+1)P_{n+1}.\tag 6
$$
Finally we have
$$
P_{n+1}(x)=xP_n(x)+\frac{1-x^2}{n+1}P_n'(x).
$$
