Equation 1: $$P_n(x) = \begin{cases} 1, & \text{ if } n = 0; \\ & \\ x, & \text{ if } n = 1; \\ & \\ \dfrac{1}{n}[(2n-1)xP_{n-1}(x)+(n-1)P_{n-2}(x)], & \text{ if } n \geq 2. \end{cases}$$
Equation 2: $$P'_{n+1} - 2xP'_n + P'_{n-1} = P_n$$
Using equation 1 and 2, show that for $n \geq1$, $$P'_{n+1}-P'_{n-1}=(2n+1)P_n$$
My attempt:
From equation 1: $$P_n(x) = 1 + x + \dfrac{1}{n} \left((2n-1)xP_{n-1}(x)+(n-1)P_{n-2}(x)\right)$$
Differentiate with respect to $x$: $$P'_n(x) = 1 + \dfrac{1}{n}\left((2n-1)xP'_{n-1}(x) + (2n-1)P_{n-1}(x) + (n-1)P'_{n-2}(x)\right) + P'_{n-1} \cdots (3)$$
Substitute (3) into equation 2:
$$P_n = P'_{n+1} - 2x\left(1 + \dfrac{1}{n}\left((2n-1)xP'_{n-1}(x)+(2n-1)P_{n-1}(x) + (n-1)P'_{n-2}(x)\right)\right) + P'_{n-1}$$
$$P_n = P'_{n+1} - 2x \left(1 + \dfrac{2n-1}{n}xP'_{n-1} + \dfrac{2n-1}{n}P_{n-1} + \dfrac{n-1}{n}P'_{n-2}\right)+ P'_{n-1}$$
Need advise if I have done correct so far & how should I proceed? Kind of stuck. Thanks
From your advice:
$$P'_{n+1} = \frac{1}{n}\left[(2n+1)(P_{n} + xP'_n ) + nP'_{n-1}\right]$$
$$P'_{n+1} - P'_{n-1} = \frac{1}{n}(2n+1)(P_{n} + xP'_n )$$
Replace $P_n$ with equation (2),
$$P'_{n+1} -P'_{n-1} = \frac{1}{n}\left( (2n+1)(P'_{n+1}-2xP'_n+P'_{n-1}+xP'_n) + nP'_{n-1}\right)$$
$$n(P'_{n+1}-P'_{n-1}) = (2n+1)(P'_{n+1}-xP'n+P'_{n-1})+ nP'_{n-1}$$
When I expand the above, it will be complicated and can't get $P'_{n+1}-P'_{n-1}=(2n+1)P_n$ still.
Technically you are doing it right, but expanding might not help. It'd be clearer to write as follows. Dropping the $(x)$ suffix and differentiating for Eq 1 in $(n+1)$
$$P'_{n+1} = \frac{1}{n}\left[(2n+1)(P_{n} + xP'_n ) + nP'_{n-1}\right]$$
$$P'_{n+1} - P'_{n-1} = \frac{1}{n}(2n+1)(P_{n} + xP'_n )$$
Can you take it from here?