User:Dan Nessett/Sandboxes/Sandbox 2

In mathematics and physics, an associated Legendre function Pℓm is related to a Legendre polynomial Pℓ  by the following equation
 * $ P^{m}_\ell(x) = (1-x^2)^{m/2} \frac{d^m P_\ell(x)}{dx^m}, \qquad 0 \le m \le \ell $.

Although extensions are possible, in this article ℓ and m are restricted to integer numbers. For even m the associated Legendre function is a polynomial, for odd m the function contains the factor (1&minus;x &sup2; )&frac12; and hence is not a polynomial.

The associated Legendre functions are important in quantum mechanics and potential theory.

According to Ferrers the polynomials were named "Associated Legendre functions" by the British mathematician Isaac Todhunter in 1875, where "associated function" is Todhunter's translation of the German term zugeordnete Function, coined in 1861 by  Heine, and "Legendre"  is in honor of the French mathematician Adrien-Marie Legendre (1752–1833), who was the first to introduce and study the functions.

Differential equation
Define:
 * $ \displaystyle \Pi^{m}_\ell(x) \equiv \frac{d^m P_\ell(x)}{dx^m}$, where Pℓ(x) is a Legendre polynomial.

Differentiating the Legendre differential equation:
 * $ \displaystyle (1-x^2) \frac{d^2 \Pi^{0}_\ell(x)}{dx^2} - 2 x \frac{d\Pi^{0}_\ell(x)}{dx} + \ell(\ell+1) \Pi^{0}_\ell(x) = 0 $,

m times gives an equation for &Pi;ml
 * $ \displaystyle (1-x^2) \frac{d^2 \Pi^{m}_\ell(x)}{dx^2} - 2(m+1) x \frac{d\Pi^{m}_\ell(x)}{dx} + \left[\ell(\ell+1) -m(m+1) \right] \Pi^{m}_\ell(x) = 0 $.

After substitution of: $ \displaystyle \Pi^{m}_\ell(x) = (1-x^2)^{-m/2} P^{m}_\ell(x) $ and after multiplying through with $(1-x^2)^{m/2}$, we find the associated Legendre differential equation:
 * $ \displaystyle (1-x^2) \frac{d^2 P^{m}_\ell(x)}{dx^2} -2x\frac{d P^{m}_\ell(x)}{dx} + \left[ \ell(\ell+1) - \frac{m^2}{1-x^2}\right] P^{m}_\ell(x)= 0 $.

One often finds the equation written in the following equivalent way
 * $ \displaystyle \left( (1-x^{2})\; y\,' \right)' +\left( \ell(\ell+1) -\frac{m^{2} }{1-x^{2} } \right) y=0 $, where the primes indicate differentiation with respect to x.

In physical applications it is usually the case that x = cos&theta;, then the  associated Legendre differential equation takes the form
 * $ \displaystyle \frac{1}{\sin \theta}\frac{d}{d\theta} \sin\theta \frac{d}{d\theta}P^{m}_\ell +\left[ \ell(\ell+1) - \frac{m^2}{\sin^2\theta}\right] P^{m}_\ell = 0 $.

Extension to negative m
By the Rodrigues formula, one obtains


 * $ \displaystyle P_\ell^{m}(x) = \frac{1}{2^\ell \ell!} (1-x^2)^{m/2}\ \frac{d^{\ell+m}}{dx^{\ell+m}}(x^2-1)^\ell.$

This equation allows extension of the range of m to: &minus;m &le; ℓ &le; m.

Since the associated Legendre equation is invariant under the substitution m &rarr; &minus;m, the equations for Pℓ&plusmn;m, resulting from this expression, are proportional.

To obtain the proportionality constant we consider
 * $ \displaystyle (1-x^2)^{-m/2} \frac{d^{\ell-m}}{dx^{\ell-m}} (x^2-1)^{\ell} = c_{lm} (1-x^2)^{m/2} \frac{d^{\ell+m}}{dx^{\ell+m}}(x^2-1)^{\ell},\qquad   0 \le m \le \ell $,

and we bring the factor (1&minus;x&sup2;)&minus;m/2 to the other side. Equate the coefficient of the highest power of x on the left and right hand side of
 * $ \displaystyle \frac{d^{\ell-m}}{dx^{\ell-m}} (x^2-1)^{\ell} = c_{lm} (1-x^2)^m \frac{d^{\ell+m}}{dx^{\ell+m}}(x^2-1)^{\ell},\qquad   0 \le m \le \ell $,

and it follows that the proportionality constant is
 * $ \displaystyle c_{lm} = (-1)^m \frac{(\ell-m)!}{(\ell+m)!} ,\qquad  0 \le m \le \ell $,

so that the associated Legendre functions of same |m| are related to each other by
 * $ \displaystyle P^{-|m|}_\ell(x) = (-1)^m \frac{(\ell-|m|)!}{(\ell+|m|)!} P^{|m|}_\ell(x) $.

Note that the phase factor (&minus;1)m arising in this expression is not due to some arbitrary phase convention, but arises from expansion of (1&minus;x&sup2;)m.

Orthogonality relations
Important integral relations are:
 * $ \displaystyle \int\limits_{-1}^{1}P_{l}^{m} \left( x\right) P_{k}^{m} \left( x\right) \ dx =\frac{2}{2l+1} \frac{\left( l+m\right) !}{\left( l-m\right) !} \delta _{lk} $,

and:
 * $ \displaystyle \int_{-1}^{1} P^{m}_{\ell}(x) P^{n}_{\ell}(x) \frac{d x}{1-x^2} = \frac{\delta_{mn}(\ell+m)!}{m(\ell-m)!}, \qquad  m \ne 0  $.

The latter integral for n = m = 0
 * $ \displaystyle \int_{-1}^{1} P^{0}_{\ell}(x) P^{0}_{\ell}(x) \frac{d x}{1-x^2} $

is undetermined (infinite). (see the subpage Proofs for detailed proofs of these relations.)

Recurrence relations
The functions satisfy the following difference equations, which are taken from Edmonds.


 * $ \displaystyle (\ell-m+1)P_{\ell+1}^{m}(x) - (2\ell+1)xP_{\ell}^{m}(x) + (\ell+m)P_{\ell-1}^{m}(x)=0 $


 * $ \displaystyle xP_{\ell}^{m}(x) -(\ell-m+1)(1-x^2)^{1/2} P_{\ell}^{m-1}(x) - P_{\ell-1}^{m}(x)=0 $


 * $ \displaystyle P_{\ell+1}^{m}(x) - x P_{\ell}^{m}(x)-(\ell+m)(1-x^2)^{1/2}P_{\ell}^{m-1}(x)=0$


 * $ \displaystyle (\ell-m+1)P_{\ell+1}^{m}(x)+(1-x^2)^{1/2}P_{\ell}^{m+1}(x)- (\ell+m+1) xP_{\ell}^{m}(x)=0 $


 * $ \displaystyle (1-x^2)^{1/2}P_{\ell}^{m+1}(x)-2mxP_{\ell}^{m}(x)+ (\ell+m)(\ell-m+1)(1-x^2)^{1/2}P_{\ell}^{m-1}(x)=0 $


 * $ \displaystyle (1-x^2)\frac{dP_{\ell}^{m}}{dx}(x) =(\ell+1)xP_{\ell}^{m}(x) -(\ell-m+1)P_{\ell+1}^{m}(x) $
 * $ \displaystyle =(\ell+m)P_{\ell-1}^{m}(x)-\ell x P_{\ell}^{m}(x) $