prove two identities regarding chebyshev polynomials and derivatives 
The source of this problem is putnam and beyond #198, so I tagged it under contest-math.


Let $T_n(x)$ be the sequence of polynomials defined by $T_0(x) = 1, T_1(x)=x, T_n(x)=2xT_{n-1}(x)-T_{n-2}(x)$ for $n\ge 2.$ Let $U_n(x)$ be the sequence of polynomials defined by $U_0(x)=1, U_1(x)=2x, U_n(x)=2xU_{n-1}(x)-U_{n-2}(x)$ for $n\ge 2$. For $n\ge 1,$ prove the following identities:


*

*$\dfrac{T_n(x)}{\sqrt{1-x^2}} = \dfrac{(-1)^n}{1\cdot 3\cdots (2n-1)}\dfrac{d^n}{dx^n}(1-x^2)^{n-1/2}$

*$U_n(x) \sqrt{1-x^2} = \dfrac{(-1)^n(n+1)}{1\cdot 3\cdots (2n+1)} \dfrac{d^n}{dx^n} (1-x^2)^{n+1/2}$
It is known that $\cos (n\theta) = T_n(\cos \theta)$ for all $n\ge 0$ (e.g. by induction using the fact that $\cos(n\theta)$ is a polynomial of $\cos \theta$ and that the polynomial expressing $\cos (n\theta)$ as a polynomial of $\cos \theta$ and $T_n$ agree at infinitely many points) and $\dfrac{\sin ((n+1)\theta)}{\sin (\theta)} = U_n(x)$ for $n\ge 0$.
I think one possible approach is to show that the two expressions on each side of the first identity have the same recurrence relation and base case. When $n=1,$ equality holds in both cases. For the left-hand side (LHS), we have the recurrence $T_n(x)/\sqrt{1-x^2} = 2x\dfrac{T_{n-1}(x)} {\sqrt{1-x^2}} - \dfrac{T_{n-2}}{\sqrt{1-x^2}}.$ In other words, if $a_n$ denotes the LHS, then $a_n(x) = 2xa_{n-1}(x)-a_{n-2}(x).$
Then we have
$\begin{align*}\dfrac{d^{n+1}}{dx^{n+1}} (1-x^2)^{n+1-1/2} &= \dfrac{d^n}{dx^n} (n+1-1/2)(1-x^2)^{n-1/2}(-2x)\\
&= \dfrac{d^n}{dx^n} -(2n+1) x (1-x^2)^{n-1/2}\\
&= \dfrac{d^{n-1}}{dx^{n-1}} (-(2n+1) (1-x^2)^{n-1/2} -(2n+1)x\dfrac{d}{dx}(1-x^2)^{n-1/2})\\
&= -(2n+1)\dfrac{d^{n-1}}{dx^{n-1}} (1-x^2)^{n-1/2}
\end{align*}$
We need to show that if $t_n(x)$ denotes the right-hand side, then $t_n(x)= 2xt_{n-1}(x) - a_{n-2}(x).$
For the second identity, it could be useful to differentiate both sides and use the fact that if two functions have the same initial conditions and the same derivative, then they must be equal. We have for $|x|<1$ that $(1-x^2)^{n+1/2} = \sum_{i=0}^\infty {n+1/2\choose i} (-x^2)^i.$ Here, for a real number x and integer $j\ge 0$, ${x\choose j} := \dfrac{x(x-1)\cdots (x-j+1)}{j!}.$ Let $u_n(x)$ denote the RHS of the second identity. Again it may be useful to prove that $u_n(x)$ satisfies the same recurrence relation as the LHS, but I'm not sure about the details for this.
Note that $-xU_{n-1}(x) + 2(1-x^2) U'_{n-1}(x)=-nT_n(x),$ which can be shown by showing that both sides equal each other whenever $x=\cos \theta$ for some $\theta\in [0,2\pi)$.
 A: I will sketch a proof for $T_n$ and show how to modify it for $U_n$.
Assume we have a positive inner product (positive bilinear form) $<>$ on the space of polynomials with real coefficients say, and two sets of polynomials $A_n, B_n, \deg A_n=\deg B_n=n, A_0=B_0=1$ (in particular none of these are the zero polynomial since that has degree $-\infty$) st $<A_n,A_m>=<B_n,B_m>=0, m \ne n$ then $A_n=k_nB_n$
Proof by induction noting that $A_0=B_0$ so $A_1=k_1B_1+k_{10}B_0$ so taking the inner product with $A_0=B_0$ we get $0=<A_1,A_0>=k_{10}<B_0,B_0>$ and $<B_0,B_0> \ne 0$ since the inner product is positive so $k_{10}=0$ hence $A_1=k_1B_1, k_1 \ne 0$
Continuing it is clear that writing $A_n=k_nB_n+k_{n,n-1}B_{n-1}+...k_{n0}B_0$ and using $A_m=k_mB_m, 0 \le m \le n-1, k_m \ne 0$, taking the inner products with $B_m$ cancels out all $k_{nm}, m <n$ so we remain with $A_n=k_nB_n$ and $k_n \ne 0$ since $A_n \ne 0$ as it has degree precisely $n$ as noted above
Let $P_0(x)=1, P_n(x)=\sqrt{1-x^2}\dfrac{d^n}{dx^n}(1-x^2)^{n-1/2}, n \ge 1$
Applying Leibniz rule to $\sqrt{1-x}\sqrt{1+x}\dfrac{d^n}{dx^n}((1-x)(1+x))^{n-1/2}$ it is obvious that $P_n$ has degree precisely $n$
Now we take the $(-1/2, -1/2)$ positive inner product on the space of real polynomials given by $<P,Q>=\int_{-1}^1(1-x^2)^{-1/2}P(x)Q(x)dx$ and notice that $T_n$ are orthogonal wr to it since with $x=\cos \theta, dx=-\sin \theta d\theta, \theta \in [0, \pi]$ we have $<T_n, T_m>=\int_0^{\pi}\cos n\theta \cos m\theta d\theta=0, m \ne n$
Hence it is enough to prove that $P_n$ above are orthogonal wr to the same inner product as then $P_n=k_nT_n$ since $P_0=T_0=1$
But if $m<n$ it is obvious that $$<P_n,x^m>=\int_{-1}^1x^m\dfrac{d^n}{dx^n}(1-x^2)^{n-1/2}dx=0$$ since integrating by parts $m$ times we get that $\dfrac{d^{n-k}}{dx^{n-k}}(1-x^2)^{n-1/2}, k=1,..m$ evaluated at end points $\pm 1$ is zero and the final integral being $m!\int_{-1}^1\dfrac{d^{n-m}}{dx^{n-m}}(1-x^2)^{n-1/2}dx=0$ since $n>m, n \ge 1$
Since $P_m$ has degree $m$ it follows $<P_n, P_m>=0$ for $n>m$ hence $P_n=k_nT_n$
But when we apply the Leibniz rule to $\sqrt{1-x}\sqrt{1+x}\dfrac{d^n}{dx^n}((1-x)(1+x))^{n-1/2}$ we notice that the only term that doesn't cancel at $-1$ is the one where we differentiate only $(1+x)^{n-1/2}$ so we get that $P_n(-1)=2^n(n-1/2)...(1-1/2)=1\cdot 3\cdots (2n-1)$ while $T_n(-1)=(-1)^n$ hence we get that $T_n=\dfrac{(-1)^n}{1\cdot 3\cdots (2n-1)}P_n$ and we are done!
For $U_n$ the same proof works using the $(1/2,1/2)$ inner product given by $<P,Q>=\int_{-1}^1(1-x^2)^{1/2}P(x)Q(x)dx$ for which $U_n$ are orthogonal by a simple change of variables and showing that $Q_0(x)=1, Q_n(x)=(1-x^2)^{-1/2}\dfrac{d^n}{dx^n} (1-x^2)^{n+1/2}$ (polynomials of degree $n$ by Leibniz rule) are orthogonal wr to it which follows similarly as above since the $(1/2,1/2)$ weight precisely cancels the $(1-x^2)^{-1/2}$ factor of $Q_n$ allowing integration by parts to do its job. Then $Q_n=c_nU_n$ and looking at the values at $-1$ again gives the required result immediately
A: There is a general  Rodrigues formula which covers many classical orthogonal polynomials. In case of Jacobi polynomials the formula is here. We get $T_n$ for $\alpha=\beta={-1/2}$ and $U_n$ for $\alpha=\beta={1/2}.$ The proof of the orthogonality  in general Rodrigues formula is straightforward, and can be performed by induction and by applying one integration by parts.
