A Generating Function for Legendre Polynomials

The technically correct generating function for Legendre polynomials is obtained using the equation

$$\frac{1}{\sqrt{1-2xu + u^2}} = \sum_0^\infty P_n(x)u^n$$ (1)

We expand the denominator using the binomial theorem,

$$\frac{1}{(1+y)^m} = 1 - my + \frac{m(m+1)}{2!}y^2 - \frac{m(m+1)(m+2)}{3!}y^3 + \cdots$$

where $m=\frac{1}{2}$ and the series converges when y<1. Notice that it is an alternating series. Identifying y = u²-2xu we have

$$\begin{eqnarray*}\frac{1}{(1-2xu+u^2)^{\frac{1}{2}}} = 1 - (\frac{1}{2})(u^2-2xu... ...}{2})((\frac{1}{2})+1)((\frac{1}{2})+2)}{3!}(u^2-2xu)^3 + \cdots \end{eqnarray*}$$

which we now re-arrange in powers of u (in the mode required by Equation 1), obtaining

$$\begin{eqnarray*}\frac{1}{(1-2xu+u^2)^{\frac{1}{2}}} = 1 - \frac{u^2}{2} + xu + ... ...(\frac{1}{2})(\frac{3}{2})(\frac{5}{2})}{3!}(u^2-2xu)^3 + \cdots \end{eqnarray*}$$

which is, rearranging

$$\begin{eqnarray*}\frac{1}{(1-2xu+u^2)^{\frac{1}{2}}} = \\ 1 + xu -\frac{u^2}{2... ...})}{3!}\left ( u^6 - 6xu^5 + 12x^2u^4-16x^3u^3 \right ) + \cdots \end{eqnarray*}$$

or, collecting terms in powers of u, we have

$$\frac{1}{(1-2xu+u^2)^{\frac{1}{2}}} =$$ (2)

$$1 + xu +\frac{1}{2} \left (3 x^2 -1 \right )u^2 +\left (- \frac{(... ...)}{2!} 4x + \frac{(\frac{1}{2})(\frac{3}{2})(\frac{5}{2})}{3!}16x^3 \right )u^3$$

$$+ \left (\frac{(\frac{3}{4})}{2!} - \frac{(\frac{1}{2})(\frac{3}{2})(\frac{5}{2})}{3!} 12x^2 \right ) u^4$$

$$- \frac{(\frac{1}{2})(\frac{3}{2})(\frac{5}{2})}{3!} (-6xu^5) - \frac{(\frac{1}{2})(\frac{3}{2})(\frac{5}{2})}{3!} u^6 + \cdots$$ (3)

where the quadratic terms (in u, i.e. 3x²-1) is the familiar spherical harmonic associated with the d_z² orbital, among other things.