Question

state the prove rodrigue's formula​

Answer 1

The most known applications of Rodrigues' type formulas are the formulas for Legendre, Laguerre and Hermite polynomials:

Rodrigues stated his formula for Legendre polynomials {\displaystyle P_{n}} P_{n}:

{\displaystyle P_{n}(x)={1 \over 2^{n}n!}{d^{n} \over dx^{n}}\left[(x^{2}-1)^{n}\right].} P_{n}(x)={1 \over 2^{n}n!}{d^{n} \over dx^{n}}\left[(x^{2}-1)^{n}\right].

Laguerre polynomials are usually denoted L0, L1, ..., and the Rodrigues formula can be written as

{\displaystyle L_{n}(x)={\frac {e^{x}}{n!}}{\frac {d^{n}}{dx^{n}}}\left(e^{-x}x^{n}\right)={\frac {1}{n!}}\left({\frac {d}{dx}}-1\right)^{n}x^{n},} L_{n}(x)={\frac {e^{x}}{n!}}{\frac {d^{n}}{dx^{n}}}\left(e^{-x}x^{n}\right)={\frac {1}{n!}}\left({\frac {d}{dx}}-1\right)^{n}x^{n},

The Rodrigues formula for the Hermite polynomial can be written as

{\displaystyle H_{n}(x)=(-1)^{n}e^{x^{2}}{\frac {d^{n}}{dx^{n}}}e^{-x^{2}}=\left(2x-{\frac {d}{dx}}\right)^{n}\cdot 1} H_{n}(x)=(-1)^{n}e^{x^{2}}{\frac {d^{n}}{dx^{n}}}e^{-x^{2}}=\left(2x-{\frac {d}{dx}}\right)^{n}\cdot 1

Answer 2

Answer:

Step-by-step explanation: