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Legendre Polinomlarının Negatif Mertebe İlişkisi

Negatif mertebeli Legendre polinomunu açıkça yazalım:

PL−m(x)=12LL!(1−x2)−m/2dL−mdxL−m(x2−1)LP_L^{-m}(x) = \frac{1}{2^L L!} (1 - x^2)^{-m/2} \frac{d^{L-m}}{dx^{L-m}} (x^2 - 1)^L

Ayrıca biliyoruz ki:

dL−mdxL−m(x2−1)L=(L−m)!(L+m)!(x2−1)mdL+mdxL+m(x2−1)L\frac{d^{L-m}}{dx^{L-m}} (x^2 - 1)^L = \frac{(L - m)!}{(L + m)!} (x^2 - 1)^m \frac{d^{L + m}}{dx^{L + m}} (x^2 - 1)^L

Bu ifadeyi yerine koyarsak:

PL−m(x)=12LL!(1−x2)−m/2(L−m)!(L+m)!(x2−1)mdL+mdxL+m(x2−1)LP_L^{-m}(x) = \frac{1}{2^L L!} (1 - x^2)^{-m/2} \frac{(L - m)!}{(L + m)!} (x^2 - 1)^m \frac{d^{L + m}}{dx^{L + m}} (x^2 - 1)^L

(x2−1)m=(−1)m(1−x2)m(x^2 - 1)^m = (-1)^m (1 - x^2)^m olduğundan:

PL−m(x)=(−1)m(L−m)!(L+m)!12LL!(1−x2)m−m/2dL+mdxL+m(x2−1)LP_L^{-m}(x) = (-1)^m \frac{(L - m)!}{(L + m)!} \frac{1}{2^L L!} (1 - x^2)^{m - m/2} \frac{d^{L + m}}{dx^{L + m}} (x^2 - 1)^L

Parantez içindeki ifade (P_L^m(x)) olduğundan:

PL−m(x)=(−1)m(L−m)!(L+m)!PLm(x)\boxed{P_L^{-m}(x) = (-1)^m \frac{(L - m)!}{(L + m)!} P_L^m(x)}

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