I Legendre polynomials and Rodrigues' formula

Tags:
1. Sep 3, 2016

TimeRip496

Source: http://www.nbi.dk/~polesen/borel/node4.html#1
Differentiating this equation we get the second order differential eq. for fn,
$(1-x^2)f''_n-2(n-1)xf'_n+2nf_n=0$ .............(22)
But when I differentiate to 2nd order, I get this instead,
$(1-x^2)f''_n+2(n-1)xf'_n+2nf_n=0$

Applying General Leibniz formula to (22) we easily get,
$(1-x^2)f^{(n+2)}_n-2xf^{(n+1)}_n+n(n+1)f^{(n)}_n=0$ ,................(24)
How did the author get to (24)? I thought he get that by subbing the Leibniz formula into (22) by replacing the f with fn but it doesn't get to (24) and it doesn't make sense at all.
What I get,
$(1-x^2)f^{(n+2)}_n+2(n-1)xf^{(n+1)}_n+2nf^{(n)}_n=0$

which is exactly Legendre's differential equation (1-49). This equation is therefore satisfied by the polynomials
$$y=\frac{d^n}{dx^n}(x^2-1)^n$$......................(25)
The Legendre polynomials Pn(x) are normalized by the requirement Pn(1)=1. Using y=2nn! for x=1,
Is equation (25) supposed to be the Legendre polynomial? And why do we normalized with x=1?Besides, attempt to normalize the Legendre polynomial with x=1 doesn't get me anywhere.
$$P_l(1)=\sum^n_{k=0} (-1)^k \frac{(2n-2k)!}{2^nk!(n-k)!(n-2k)!}*1$$

Last edited: Sep 3, 2016
2. Sep 3, 2016

mathman

Your initial statement, that the text is wrong, is correct. I haven't gone further.