0

I'm trying to derive Rodrigues' formula for Chebyshev Polynomials, by using Chebyshev DE $$(1-x^2)y''-xy'+n^2y=0. \quad \quad (1)$$ Rodrigues' formula for Chebyshev Polynomials is $$T_n(x)=(-1)^n2^n\frac{n!}{(2n)!}\sqrt{1-x^2}\frac{d^n}{dx^n}(1-x^2)^{n-1/2}$$ The problem has been addressed also in Derive the Rodrigues' formula for Chebyshev Polynomials with a different solution. Unfortunately, I can't explain why, replacing in (1) $y$ by $\sqrt{1-x^2}\frac{d^n}{dx^n}(1-x^2)^{n-1/2}$ don't obtain an identity. In fact, if I set, in (1), $$y(x)=\sqrt{1-x^2}\frac{d^n}{dx^n}(1-x^2)^{n-1/2}$$ I obtain $$\sqrt{1 - x^2} \left[(-1 + n^2) \frac{d^n}{dx^n}(1-x^2)^{n-1/2} - 3 x \frac{d^{n+1}}{dx^{n+1}}(1-x^2)^{n-1/2} + (1 - x^2) \frac{d^{n+2}}{dx^{n+2}}(1-x^2)^{n-1/2} \right]=0$$ which is not an identity.

Mark
  • 7,841
  • 6
  • 38
  • 72
  • This might be helpful: https://math.stackexchange.com/questions/1960306/deriving-rodrigues-formula – Chee Han Dec 27 '21 at 19:01
  • @Chee Han: thank you! I read the post but in my case I have also a function that mulplies the multiple derivatives. – Mark Dec 27 '21 at 20:19

0 Answers0