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Associated Legendre Polynomials for Negative Index m 📂Functions

Associated Legendre Polynomials for Negative Index m

Formulas

The Associated Legendre Polynomials obey the proportionality relation below, depending on the sign of mm.

Pl−m(x)=(−1)m(l−m)!(l+m)!Plm(x) P_{l}^{-m}(x)=(-1)^{m}\frac{(l-m)!}{(l+m)!}P_{l}^{m}(x)

(1−x2)d2ydx2−2xdydx+(−m21−x2+l(l+1))y=0 (1-x^{2})\frac{ d^{2}y }{ dx^{2} }-2x \frac{dy}{dx}+\left( \frac{-m^{2}}{1-x^{2}}+l(l+1) \right)y=0

Explanation

Looking at the associated Legendre differential equation, the section regarding mm is shown as m2m^2, so whether mm is positive or negative doesn’t affect the solution. Thus, the associated Legendre polynomials can be derived as follows.

Plm(x)=(1−x2)∣m∣2d∣m∣dx∣m∣Pl(x)=(1−x2)∣m∣2d∣m∣dx∣m∣[12ll!dldxl(x2−1)l] \begin{align*} P_{l}^{m}(x)&= (1-x ^{2})^{\frac{|m|}{2}} \frac{ d^{|m|} }{ dx^{|m|} } P_{l}(x) \\ &=(1-x ^{2})^{\frac{|m|}{2}} \frac{ d^{|m|} }{ dx^{|m|} }\left[ \dfrac{1}{2^l l!} \dfrac{d^l}{dx^l}(x^2-1)^l\right] \end{align*}

Therefore, according to the formula above, Plm(x)=Pl−m(x)P_{l}^{m}(x)=P_{l}^{-m}(x). However, rather than seeing them as equal, they are considered to be proportional. Since merely multiplying the solution of a differential equation by a constant still results in a solution, assuming AA to be an arbitrary constant, then Pl−m(x)P_{l}^{-m}(x) is expressed like APlm(x)AP_{l}^{m}(x). This article will demonstrate that A=(−1)m(l−m)!(l+m)!A=(-1)^{m}\frac{(l-m)!}{(l+m)!} holds.

Derivation

First, applying the Leibniz rule to the differentiation part of the associated Legendre polynomials gives

dl+∣m∣dxl+∣m∣(x2−1)l=dl+∣m∣dxl+∣m∣[(x−1)l(x+1)l]=∑k=0l+∣m∣(l+∣m∣)!(l+∣m∣−k)!k!dl+∣m∣−kdxl+∣m∣−k(x+1)ldkdxk(x−1)l \begin{align*} \frac{ d ^{l+|m|}}{ dx^{l+|m|} }(x^{2}-1)^{l} &= \frac{ d ^{l+|m|}}{ dx^{l+|m|} }[(x-1)^{l}(x+1)^{l}] \\ &=\sum \limits _{k=0}^{l+|m|} \frac{(l+|m|)!}{(l+|m|-k)!k!}\frac{ d ^{l+|m|-k}}{ dx^{l+|m|-k} }(x+1)^{l}\frac{ d ^{k} }{ dx^{k} }(x-1)^{l} \tag{1} \end{align*}

Case 1. m>0m>0

Consider the case of m>0m>0. Then, l+∣m∣−k=l+m−k=l+(m−k)l+|m|-k=l+m-k=l+(m-k). The index kk ranges from 00 to l+ml+m. In the case when k<mk < m, since l+(m−k)>ll+(m-k)>l, the number of differentiations exceeds the highest order term’s degree

dl+∣m∣−kdxl+∣m∣−k(x+1)l=0 \frac{ d ^{l+|m|-k}}{ dx^{l+|m|-k} }(x+1)^{l}=0

Therefore, the RHS of (1)(1) for 0≤k<m0\le k <m is entirely 00. Similarly, if k>lk>l, then dkdxk(x−1)l=0\frac{ d ^{k}}{ dx^{k} }(x-1)^{l}=0, so only in the case of k≤lk\le l does a term other than 00 exist on the RHS. That is, for kk that is not m≤k≤lm\le k \le l, the RHS is 00, hence ∑k=0l+m=∑k=ml\sum\limits_{k=0}^{l+m}=\sum\limits_{k=m}^{l} holds. Therefore,

dldxl+m(x2−1)l=∑k=ml(l+m)!(l+m−k)!k!dl+m−kdxl+m−k(x+1)ldkdxk(x−1)l=∑k=ml(l+m)!(l+m−k)!k!l(l−1)⋯(l−l−m+k+1)(x+1)l−l−m+kl(l−1)⋯(l−k+1)(x−1)l−k=∑k=ml(l+m)!(l+m−k)!k!l!(k−m)!(x+1)k−ml!(l−k)!(x−1)l−k=(l!)2∑k=ml(l+m)!(l+m−k)!k!1(k−m)!1(l−k)!(x+1)k−m(x−1)l−k \begin{align*} \frac{ d ^{l}}{ dx^{l+m} }(x^{2}-1)^{l} &= \sum \limits_{k=m}^{l}\frac{(l+m)!}{(l+m-k)!k!}\frac{ d ^{l+m-k}}{ dx^{l+m-k}}(x+1)^{l}\frac{ d ^{k}}{ dx^{k} }(x-1)^{l} \\ &= \sum \limits_{k=m}^{l}\frac{(l+m)!}{(l+m-k)!k!}l(l-1)\cdots(l-l-m+k+1)(x+1)^{l-l-m+k} \\ &\qquad l(l-1)\cdots(l-k+1)(x-1)^{l-k} \\ &= \sum \limits_{k=m}^{l}\frac{(l+m)!}{(l+m-k)!k!}\frac{l!}{(k-m)!}(x+1)^{k-m}\frac{l!}{(l-k)!}(x-1)^{l-k} \\ &=(l!)^{2}\sum \limits_{k=m}^{l}\frac{(l+m)!}{(l+m-k)!k!}\frac{1}{(k-m)!}\frac{1}{(l-k)!}(x+1)^{k-m}(x-1)^{l-k} \tag{2} \end{align*}

Case 2. m<0m<0

In this scenario, equation (1)(1) becomes

dl+∣m∣dxl+∣m∣(x2−1)l=dl−mdxl−m(x2−1)l=∑k=0l−m(l−m)!(l−m−k)!k!dl−m−kdxl−m−k(x+1)ldkdxk(x−1)l=∑k=0l−m(l−m)!(l−m−k)!k!l!(k+m)!(x+1)k+ml!(l−k)!(x−1)l−k=(l!)2∑k=0l−m(l−m)!(l−m−k)!k!1(k+m)!1(l−k)!(x+1)k+m(x−1)l−k \begin{align*} \frac{ d ^{l+|m|}}{ dx^{l+|m|} }(x^{2}-1)^{l}&=\frac{ d ^{l-m}}{ dx^{l-m} }(x^{2}-1)^{l} \\ &= \sum \limits _{k=0}^{l-m} \frac{(l-m)!}{(l-m-k)!k!}\frac{ d ^{l-m-k}}{ dx^{l-m-k} }(x+1)^{l}\frac{ d ^{k} }{ dx^{k} }(x-1)^{l} \\ &= \sum \limits _{k=0}^{l-m} \frac{(l-m)!}{(l-m-k)!k!}\frac{l!}{(k+m)!}(x+1)^{k+m}\frac{ l! }{ (l-k)! }(x-1)^{l-k} \\ &= (l!)^{2}\sum \limits _{k=0}^{l-m} \frac{(l-m)!}{(l-m-k)!k!}\frac{1}{(k+m)!}\frac{1 }{ (l-k)! }(x+1)^{k+m}(x-1)^{l-k} \end{align*}

Changing the index to k=k−mk=k-m gives

dl−mdxl−m(x2−1)l=(l!)2∑k=ml(l−m)!(l−k)!(k−m)!1k!1(l+m−k)!(x+1)k(x−1)l+m−k(3) \frac{ d ^{l-m}}{ dx^{l-m} }(x^{2}-1)^{l}= (l!)^{2}\sum \limits _{k=m}^{l} \frac{(l-m)!}{(l-k)!(k-m)!}\frac{1}{k!}\frac{1 }{ (l+m-k)! }(x+1)^{k}(x-1)^{l+m-k} \tag{3}

Combining (2)(2), (3)(3) results in

dl−mdxl−m(x2−1)l=(l−m)!(l+m)!(x+1)m(x−1)mdl+mdxl+m(x2−1)l(4) \frac{d^{l-m}}{dx^{l-m}}(x^{2}-1)^{l}=\frac{(l-m)!}{(l+m)!}(x+1)^{m}(x-1)^{m}\frac{ d ^{l+m}}{ dx^{l+m} }(x^{2}-1)^{l}\tag{4}

Also, when m>0m>0,

Plm(x)=(1−x2)m212ll!dl+mdxl+m(x2−1)l P_{l}^{m}(x)=(1-x ^{2})^{\frac{m}{2}} \dfrac{1}{2^l l!} \dfrac{d^{l+m}}{dx^{l+m}}(x^2-1)^l

but we are currently considering the case of m→−mm\rightarrow-m, so by substituting −m-m for mm,

Pl−m(x)=(1−x2)−m212ll!dl−mdxl−m(x2−1)l P_{l}^{-m}(x)=(1-x ^{2})^{-\frac{m}{2}} \dfrac{1}{2^l l!} \dfrac{d^{l-m}}{dx^{l-m}}(x^2-1)^l

Substituting (4)(4) then yields

Pl−m(x)=(1−x2)−m212ll!(l−m)!(l+m)!(x+1)m(x−1)mdl+mdxl+m(x2−1)l=(1−x2)−m(l−m)!(l+m)!(x2−1)m[(1−x2)m212ll!dl+mdxl+m(x2−1)l]=(1−x2)−m(l−m)!(l+m)!(x2−1)mPlm(x)=(−1)m(x2−1)−m(l−m)!(l+m)!(x2−1)mPlm(x)=(−1)m(l−m)!(l+m)!Plm(x) \begin{align*} P_{l}^{-m}(x) &= (1-x ^{2})^{-\frac{m}{2}} \dfrac{1}{2^l l!} \frac{(l-m)!}{(l+m)!}(x+1)^{m}(x-1)^{m}\frac{ d ^{l+m}}{ dx^{l+m} }(x^{2}-1)^{l} \\ &= (1-x ^{2})^{-m } \frac{(l-m)!}{(l+m)!}(x^{2}-1)^{m}\left[(1-x ^{2})^{\frac{m}{2}}\dfrac{1}{2^l l!}\frac{ d ^{l+m}}{ dx^{l+m} }(x^{2}-1)^{l} \right] \\ &= (1-x ^{2})^{-m } \frac{(l-m)!}{(l+m)!}(x^{2}-1)^{m}P_{l}^{m}(x) \\ &= (-1)^{m}(x ^{2}-1)^{-m } \frac{(l-m)!}{(l+m)!}(x^{2}-1)^{m}P_{l}^{m}(x) \\ &= (-1)^{m} \frac{(l-m)!}{(l+m)!}P_{l}^{m}(x) \end{align*}

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