Knowing $P_1(x) = x$ and
$Q_1(x) = \frac{1}{2}xlog\frac{1+x}{1-x} - 1 $
I need to verify they satisfy the Legendre equation:
$(1-x^2)y'' - 2xy' + n(n+1)y = 0$
so far I've gotten the derivatives of $P_1(x) = x$, $P'_1(x) = 1$, $P''_1(x) = 0$
and $Q_1(x) = \frac{1}{2}xlog\frac{1+x}{1-x} - 1 $, $Q'_1(x) = \frac{x}{1-x^2}\frac{1}{2}xlog\frac{1+x}{1-x}$, $Q''_1(x) = \frac{2}{(x^2 - 1)^2} $
so I sub in $y'' = P''_1(x)$ and $y' = P'_1(x)$ and $y = P_1(x)$ but I don't get 0 because of the $n(n+1)y$ term...what am I doing wrong?
Need some LaTeX practice so here goes :-)
$(1-x^2)P_1^{''}-2xP_1^{'} + 2P_1 = 0-2x+2x=0$
\begin{cases} Q_1^{'} = \frac{1}{2}log\frac{1+x}{1-x} + \frac{x}{2}(\frac{1}{1+x} + \frac{1}{1-x}) = \frac{1}{2}log\frac{1+x}{1-x} + \frac{x}{1-x^2} \\ Q_1^{''} = \frac{1}{1-x^2} + \frac{1-x^2 -x\cdot-2x}{(1-x^2)^2} = \frac{1}{1-x^2} + \frac{1+x^2}{(1-x^2)^2} = \frac{2}{(1-x^2)^2} \end{cases} $$(1-x^2)Q_1^{''}-2xQ_1^{'} + 2Q_1 = \frac{2}{1-x^2} - \left(xlog\frac{1+x}{1-x} + \frac{2x^2}{1-x^2}\right) + \left(xlog\frac{1+x}{1-x} - 2\right) = 0$$