Associated Legendre polynomials for negative order

[Rulonegger]

Homework Statement

I just need to deduce the expression for the associated Legendre polynomial $P_{n}^{-m}(x)$ using the Rodrigues' formula

Homework Equations

Rodrigues formula reads $$P_{n}(x)=\frac{1}{2^{n}n!}\frac{d^n}{dx^n}(x^2-1)^n$$ and knowing that $$P_{n}^{m}(x)=(-1)^{m}(1-x^2)^{\frac{m}{2}}\frac{d^m}{dx^m}\left[P_n(x)\right]$$

The Attempt at a Solution

Using the expressions above mentioned i get $$P_{n}^{m}(x)=\frac{(-1)^m}{2^{n}n!}(1-x^2)^{\frac{m}{2}}\frac{d^{n+m}}{dx^{n+m}}(x^2-1)^n$$
Then i see that -n≤m≤n, so the last expression must be useful to determine the $P_{n}^{-m}(x)$ polynomial, substituting m by -m, but i cannot find the relationship between the derivatives $\frac{d^{n+m}}{dx^{n+m}}$ and $\frac{d^{n-m}}{dx^{n-m}}$, that i know it must be $$\frac{d^{n-m}}{dx^{n-m}}(x^2-1)^n=(-1)^m\frac{(n-m)!}{(n+m)!}(1-x^2)^{m}\frac{d^{n+m}}{dx^{n+m}}(x^2-1)^n$$
Any help would be greatly appreciated.

 

[mighty2000]

Thank you for your inquiry. I would be happy to assist you in deducing the expression for the associated Legendre polynomial using the Rodrigues' formula. The first step would be to understand the meaning and derivation of the Rodrigues' formula. This formula is a general expression for the Legendre polynomial P_{n}(x) of degree n, which can be obtained through the application of the n-th derivative of the function (x^2-1)^n. This function is known as the generating function for the Legendre polynomials.

Now, let's focus on the specific case of P_{n}^{-m}(x). As you have correctly mentioned, the associated Legendre polynomials have the form P_{n}^{m}(x)=(-1)^{m}(1-x^2)^{\frac{m}{2}}\frac{d^m}{dx^m}\left[P_n(x)\right]. However, in order to obtain P_{n}^{-m}(x), we need to substitute m by -m in this expression. This will result in P_{n}^{-m}(x)=(-1)^{-m}(1-x^2)^{\frac{-m}{2}}\frac{d^{-m}}{dx^{-m}}\left[P_n(x)\right].

Now, we can use the property of the derivative operator, which states that \frac{d^{-m}}{dx^{-m}}=\frac{1}{\left(\frac{d}{dx}\right)^m}. Substituting this in the previous expression, we get P_{n}^{-m}(x)=(-1)^{-m}(1-x^2)^{\frac{-m}{2}}\frac{1}{\left(\frac{d}{dx}\right)^m}\left[P_n(x)\right].

Next, we can use the Rodrigues' formula for P_{n}(x) to obtain P_{n}^{-m}(x)=(-1)^{-m}\frac{1}{2^{n}n!}\frac{1}{\left(\frac{d}{dx}\right)^m}\left[(x^2-1)^n\right].

Finally, we can apply the chain rule to the derivative operator, which states that \frac{d}{dx}\left[f(g(x))\right]=f'(g(x))g'(x). Using this, we