How to prove this generating function of Legendre polynomials?

Question

How to prove this generating function of Legendre polynomials? $$\frac{1}{\sqrt{1-2xt+t^2}}=\sum_{n=0}^{\infty}P_n(x)t^n$$ I found 2 proofs and they are different from each other and I don't understand them.

Answer 1

I do not know any quick proofs from scratch to show this formula but a general formula to find generating functions is the following: $$G(x,y) = \frac{\omega(z)}{\omega(x)}\frac1{1-y\cdot p'(z)}$$ Where $z = x + y\cdot p(z)$ and $p(z)$ is from the Legendre ODE $\frac{d}{dx}[p(x)y_n'(x)] + q(x)y_n'(x) + \lambda_ny_n(x) = 0$ in its standard form.

So here you have $p(z) = 1-z^2$ for the Legendre equation so you find that $p'(z) = -2z$ and solve $z = x+y + 1 - z^2$ to get $z$ as a function of $x$ and $y$. (Nasty but that's where you get the square root from.)

Sub back everything in the formula and you get your answer.

PS : Generating functions are not unique.

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ This is a very popular proof in Electromagnetism ( see, for example, Jackson book ): $\verts{\vec{r} - \vec{r}'}^{-1}$ satisfies the Laplace Equation when $\vec{r} \not= \vec{r}'$. It means $\verts{\vec{r} - \vec{r}'}^{-1}$ can be expanded in a serie of Legendre Polynomials: $$ \left.\nabla^{2}\pars{{1 \over \verts{\vec{r} - \vec{r}'}}} \right\vert_{\vec{r}\ \not=\ \vec{r}'} = 0 \quad\imp\quad{1 \over \verts{\vec{r} - \vec{r}'}} = \sum_{\ell = 0}^{\infty}A_{\ell}{\rm P}_{\ell}\pars{\cos\pars{\theta}} $$ where $\theta \equiv \angle\pars{\vec{r},\vec{r}'}$: $$ {1 \over \verts{\vec{r} - \vec{r}'}} = {1 \over r_{>}}\,{1 \over \root{1 - 2\pars{r_{<}/r_{>}}\cos\pars{\theta} + \pars{r_{<}/r_{>}}^{2}}}\,,\qquad r_{< \atop >} = {\min \atop \max}\braces{r,r'} $$ Set $r_{>} = 1$, $h \equiv r_{<}/r_{>}$ and $x = \cos\pars{\theta}$. We get $$ {1 \over \root{1 -2xh + h^{2}}} = \sum_{\ell = 0}^{\infty}A_{\ell}{\rm P}_{\ell}\pars{x} $$ With $x = 1$, $\pars{~{\rm P}_{\ell}\pars{1} = 1\,,\ \forall\ \ell = 0,1,2,\ldots~}$: $$ \sum_{\ell = 0}^{\infty}A_{\ell} = {1 \over 1 - h} = \sum_{\ell = 0}^{\infty}h^{\ell} $$

Answer 3

In general, a generating function for a sequence of functions $P_n(x)$, is a function $G(x,t)$, such that

\begin{eqnarray*} G(x,t)= \sum_{n=0}^{\infty} P_n(x) t^n, \end{eqnarray*} where, by matching equal powers of $t$, the Taylor series expansion of $G(x,t)$ provides the functions $P_n(x)$. In particular we find $G(x,t)$ when the $P_n(x)$ are Legendre polynomials. Historically, Legendre defined the polyomials $P_n(x)$ knowing that $G(x,t)$ was the Newtonian potential (gravitational potential) but here we assumed that $P_n$ were found from solving the Laplace equation with azimuthal symmetry and some data given on the surface of a sphere. That is, we set up the problem of solving

\begin{eqnarray*} \nabla^2 u(\mathbf{r}) =0 \end{eqnarray*} subjected to some boundary conditions. Let us assume the Dirichlet boundary condition at a sphere $S$, $|\mathbf{r}|=r_1$,

\begin{eqnarray*} u(\mathbf{r} \in S)=\frac{1}{|\mathbf{r}_1-\mathbf{r}_0|} \label{bcdiri} \end{eqnarray*} with $\mathbf{r}_0$ some fixed vector in $\mathbb{R}^3$. Then the solution of the Laplace's equation

\begin{eqnarray} u(\mathbf{r}) = \frac{1}{|\mathbf{r} - \mathbf{r}_0|} \label{sollap} \end{eqnarray} satisfies the Dirichlet boundary condition above. Since the solution of the Laplace's equation is unique with Dirichlet boundary conditions we claim that this equation is THE solution of the Laplace's equation with the given Dirichlet boundary conditions. Let us further assume that $\mathbf{r}_0$ is aligned with the positive $z$ axis.

The solution of the Laplace's equation using separation of variables provides

\begin{eqnarray*} u_{\mathbf{r}_0}(r, \theta, \phi) = \sum_{n=0}^{\infty} A_n r^n P_n( \cos \theta). \end{eqnarray*} Here $r$ is $|\mathbf{r}|$, $\theta$ is the polar angle that $\mathbf{r}$ makes with the positive axis $z$, and $\phi$ is the azimutal angle. The angle of the projection of $\mathbf{r}$ on the plane $z=0$ and the $x$ axis. It is then implicit here that we are assuming azimuthal symmetry. If there is no azimuthal symmetry we are forced to use spherical harmonic functions $Y_{nm}(\theta,\phi)$ instead of Legendre polynomials $P_n(\cos \theta)$ for this representation. Still, we show at the end how to get around this assumption. When solving the Laplace's equation using separation of variables the radial functions $r^n$ are solutions of the resulting Euler ODE and the polar functions $P_n(\cos \theta)$ are solutions of the Legendre differential equation. We have \begin{eqnarray*} |\mathbf{r}-\mathbf{r}_0|^2 = \mathbf{r} \cdot \mathbf{r} - 2 \mathbf{r} \cdot \mathbf{r}_0 + \mathbf{r}_0 \mathbf{r}_0 = r^2 - 2 r \cos \alpha + 1, \end{eqnarray*} where $\alpha$ is the angle between $\mathbf{r}$ and $\mathbf{r}_0$. However, since we assume that $\mathbf{r}_0$ is aligned with the positive $z$ axis $\alpha=\theta$ and the function

\begin{eqnarray*} \frac{1}{\sqrt{r^2 - 2 r \cos \theta + 1}}. \end{eqnarray*} satisfies the Laplace's equation with Dirichlet boundary conditions, so we can write

\begin{eqnarray*} \frac{1}{\sqrt{r^2 - 2 r \cos \theta + 1}} = \sum_{n=0}^{\infty} A_n r^n P_n(\cos \theta). \end{eqnarray*} We should now find $A_n$ to finish the problem. We show $A_n=1$. Since the identity should be valid for all angles, let us assume $\theta=0$, so $\cos \theta=1$ and we know already that $P_n(1)=1$, so

\begin{eqnarray*} \frac{1}{|r-1|} = \frac{1}{1-r} = \sum_{n=0}^{\infty} A_n r^n \quad, \quad \mathrm{please \; observe \; that \; we \; are \; assuming} \quad r<1 \end{eqnarray*} but

\begin{eqnarray*} \frac{1}{1-r}=1 + r + \cdots +r^n + \cdots, \end{eqnarray*} so $A_n=1$ and

\begin{eqnarray} \frac{1}{\sqrt{r^2 - 2 r \cos \theta + 1}} = \sum_{n=0}^{\infty} r^n P_n(\cos \theta). \label{generatingLeg} \end{eqnarray} Hence

\begin{eqnarray} G(x,r) = \frac{1}{\sqrt{r^2 - 2 r \cos \theta + 1}} = \frac{1}{\sqrt{r^2 - 2 r x + 1}} \label{torecur} \end{eqnarray} with $x=\cos \theta$ is the generating function for the Legendre polynomials.

At this point $x$ could represent any value between -1 and 1. In particular if $\mathbf{r}$ and $\mathbf{r}_0$ are any two vectors with $|\mathbf{r}_0|=1$, then $\mathbf{r} \cdot \mathbf{r}_0 = r \cos \alpha$, and we can write

\begin{eqnarray*} \frac{1}{\sqrt{r^2 - 2 r \cos \alpha + 1}} = \sum_{n=0}^{\infty} r^n P_n(\cos \alpha ), \end{eqnarray*} where now $\alpha$ is the angle between $\mathbf{r}$ and $\mathbf{r}_0$.

It is common in the literature to find the expansion of the generating function $G(x,r)$ in Taylor series in $r$ and from their coefficients verify that they correspond to $P_n(x)$. However this method, even though valid, implies more cumbersome algebra (infinity binomial expansions with fractional coefficients, Gamma functions with fractional coefficients...) and less physical insight.

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