Derive the Rodrigues' formula for Chebyshev Polynomials

Question

Rodrigues' formula for Chebyshev Polynomials is stated as $$T_n(x)=(-1)^n2^n\frac{n!}{(2n)!}\sqrt{1-x^2}\frac{d^n}{dx^n}(1-x^2)^{n-1/2}$$

I understand how the Rodrigues formula for all other special functions can be derived. One that for Laguerre polynomials is asked at Derive Rodrigues’ formula for Laguerre polynomials , but that for Chebyshev Polynomials is nowhere to be found.

The generating function for the Chebyshev polynomials is

$$ g(x,z)=\frac{1-zx}{1-2zx+z^2}=\sum_{n=0}^\infty T_n(x)z^n $$

Can I derive it starting from the generating function or is there an easier approach ?

Answer 1

I believe that the easiest approach is to check that both $\{T_n(x)\}_{n\geq 0}$ and $\{P_n(x)\}_{n\geq 0}$ (where $T_n$ is defined in the usual trigonometric way and $P_n$ is given by your formula involving $\frac{d^n}{dx^n}$) are sequences of orthogonal polynomials with respect to the inner product $$ \langle a(x),b(x) \rangle = \int_{-1}^{1}\frac{a(x)b(x)}{\sqrt{1-x^2}}\,dx = \int_{0}^{2\pi}a(\cos\theta)b(\cos\theta)\,d\theta$$ then check that $\deg T_n = \deg P_n = n$ and $T_n(1)=P_n(1)=1$ for any $n\geq 0$.

Answer 2

An alternative method is to check that the $T_n$, as you've defined them via the Rodrigues formula, have $\deg T_n=n$, $T_n(1)=1$ and $y=T_n$ solves the Chebyshev DE $$(1-x^2)y''-xy'+n^2y=0.$$

Answer 3

The weight function of the Chebyshev polynomials is $W(x)=\frac{1}{\sqrt{1-x^2}}$. According to the Rodrigues formula for classical orthogonal polynomials, $T_n(x)$ is proportional to $\frac{1}{W(x)}\frac{d^n}{dx^n}(Q(x)^nW(x))$,(https://en.wikipedia.org/wiki/Classical_orthogonal_polynomials).

Here, $Q(x)=(1-x^2)$ from @jlammy's answer: $Q(x)$ is the coefficient of $y''$ in the Chebyshev differential equation $(1-x^2)y''-xy'+n^2y=0.$

As you also discussed in the comments, it is enough to prove that the proportinality constant is $(-1)^n\frac{2^n n!}{(2n)!}$.

After taking derivative $n$-times and dividing by $W(x)$, the expression $\frac{1}{W(x)}\frac{d^n}{dx^n}(Q(x)^nW(x))$ becomes $$\sum_{k=0}^{n}\binom{n}{k}(-1)^k(n-\frac{1}{2})_{(k)}(n-\frac{1}{2})_{(n-k)}(1-x)^{n-k}(1+x)^k.$$ The above expression has one non-zero term at $x=1$ and it is $$\binom{n}{n}(-1)^n(n-\frac{1}{2})_{(n)}2^n=(-1)^n\frac{(2n)!}{2^nn!}.$$ Since, $T_n(1)=1$, we are done.