Problem Statement: Define the inner product of two real functions $f(t)$ and $g(t)$ in the following way:
$$(f, g) = \int\limits_{-1}^1 f(t)g(t)\;dt$$
Let $P_0(t)=\frac{1}{2}$ and define the other elements $\{P_n(t)\}$ by the condition that $P_n(t)$ is of degree $n$ and $(P_n,P_m)=0$, $m\neq n$, and $(P_n,P_n)=1$. Prove that
$$\left(\left(\frac{d}{dt}\right)^n (1-t^2)^n, t^m\right)=0,$$
$0\leq m\leq n-1$, and thus express $P_n$ in terms of the expression $\left(\frac{d}{dt}\right)^n (1-t^2)^n$.
Attempt: By the binomial theorem, we know
$$(1-t^2)^n=\sum\limits_{k=0}^n (-1)^{n-k} \binom{n}{k} t^{2(n-k)}$$
and so
$$\left(\frac{d}{dt}\right)^n (1-t^2)^n=\left(\frac{d}{dt}\right)^n \sum\limits_{k=0}^n (-1)^{n-k} \binom{n}{k} t^{2n-2k}$$
$$=\sum\limits_{k=0}^{\lfloor \frac{n}{2}\rfloor} (-1)^{n-k}\binom{n}{k}(2n-2k)\cdots (2n-2k-n+1) t^{n-2k}$$
$$=\sum\limits_{k=0}^{\lfloor \frac{n}{2}\rfloor} (-1)^{n-k}\binom{n}{k}\binom{2n-2k}{n}(n-2k)!t^{n-2k}.$$
From here, I know I want to show
$$P_n(t)=\alpha\left(\frac{d}{dt}\right)^n (1-t^2)^n$$
for some constant $\alpha$, so then we have
$$\left(\left(\frac{d}{dt}\right)^n (1-t^2)^n, t^m\right)=\alpha(P_n,t^m)=0$$
as $P_n$ is orthogonal to $t^m$ by construction. However, I am not sure how to construct the sequence $P_n$ - I tried explicitly, but I don't see the pattern. Any help is appreciated with establishing the correspondence!
Source: "Introduction to Matrix Analysis" by Richard Bellman. FYI, I am working through all the problems in the book for enjoyment on break, not coursework.