Consider an electric charge $q$ placed on the $z$-axis at $z=a$.

Electric Potential

The electrostatic potential of charge $q$ is $$\varphi=\frac{1}{4\pi\epsilon_0}\frac{q}{r_1}.\ \ \ \ \ \mbox{(1)}$$ Using the Laws of Cosine, one can write $r_1$ in terms of $r$ and $\theta$:
$$r_1=\sqrt{r^2+a^2-2ar\cos\theta}$$
and thereby the electrostatic potential (1) can be written as
$$\varphi=\frac{q}{4\pi\epsilon_0}(r^2+a^2-2ar\cos\theta)^{-1/2}.\ \ \ \ \ \mbox{(2)}$$

Recall the Binomial Expansion Formula: Suppose that $x,y\in\mathbb{R}$ and $|x|>|y|$. Then
$$(x+y)^r=\sum_{k=0}^\infty\begin{pmatrix}r\\k\end{pmatrix}x^{r-k}y^k,\ \ \ \ \ \mbox{(3)}$$ where $\begin{pmatrix}r\\k\end{pmatrix}=\frac{r!}{k!(r-k)!}$.

Legendre Polynomials: If $r>a$ (or more specifically $r^2>|a^2-2ar\cos\theta|$), we can expand the radical to obtain:
$$\varphi=\frac{q}{4\pi\epsilon_0 r}\sum_{n=0}^\infty P_n(\cos\theta)\left(\frac{a}{r}\right)^n.$$ The coefficients $P_n$ are called the Legendre polynomials. The Legendre polynomials can be defined by the generating function
$$g(t,x)=(1-2xt+t^2)^{-1/2}=\sum_{n=0}^\infty P_n(x)t^n,\ \ \ \ \ \mbox{(4)}$$ where $|t|<1$. Using the binomial expansion formula (3), we obtain
\begin{align*}
(1-2xt+t^2)^{-1/2}&=\sum_{n=0}^\infty\frac{(2n)!}{2^{2n}(n!)^2}(2xt-t^2)^n\ \ \ \ \ \mbox{(5)}\\
&=\sum_{n=0}^\infty\frac{(2n-1)!!}{(2n)!!}(2xt-t^2)^n.
\end{align*}
Let us write out the first three terms:
\begin{align*}
\frac{0!}{2^0(0!)^2}&(2xt-t^2)^0+\frac{2!}{2^2(1!)^2}(2xt-t^2)^1+\frac{4!}{2^4(2!)^2}(2xt-t^2)^2\\
&=1t^0+xt^1+\left(\frac{3}{2}x^2-\frac{1}{2}\right)t^2+\mathcal{O}t^3.
\end{align*}
Thus we see that $P_0(x)=1$, $P_1(x)=x$, and $P_2(x)=\frac{3}{2}x^2-\frac{1}{2}$. In practice, we don’t calculate Legendre polynomials using the power series (5). Instead, we use the recurrence relation of Legendre polynomials that will be discussed later.

The Maxima name for Legendre polynomial $P_n(x)$ is legendre_p(n,x). The following graphs of $P_2(x)$, $P_3(x)$, $P_4(x)$, $P_5(x)$, $-1\leq x\leq 1$ is made by Maxima using the command:

plot2d([legendre_p(2,x),legendre_p(3,x),legendre_p(4,x),legendre_p(5,x)],[x,-1,1]);

Legendre Polynomials

Now expand the polynomial $(2xt-t^2)^n$ in the power series (5):
\begin{align*}
(1-2xt+t^2)^{-1/2}&=\sum_{n=0}^\infty\frac{(2n)!}{2^{2n}(n!)^2}t^n\sum_{k=0}^n(-1)^k\frac{n!}{k!(n-k)!}(2x)^{n-k}t^k\\
&=\sum_{n=0}^\infty\sum_{k=0}^n(-1)^k\frac{(2n)!}{2^{2n}n!k!(n-k)!}(2x)^{n-k}t^{k+n}.\ \ \ \ \ \mbox{(6)}
\end{align*}
By rearranging the order of summation, (6) can be written as
$$(1-2xt+t^2)^{-1/2}=\sum_{n=0}^\infty\sum_{k=0}^{[n/2]}(-1)^k\frac{(2n-k)!}{2^{2n-2k}k!(n-k)!(n-2k)!}(2x)^{n-2k}t^n,$$ where
$$\left[\frac{n}{2}\right]=\left\{\begin{array}{ccc}
\frac{n}{2} & \mbox{for} & n=\mbox{even}\\
\frac{n-1}{2} & \mbox{for} & n=\mbox{odd}.
\end{array}\right.$$

Hence,
$$P_n(x)=\sum_{k=0}^{[n/2]}(-1)^k\frac{(2n-k)!}{2^{2n-2k}k!(n-k)!(n-2k)!}(2x)^{n-2k}.\ \ \ \ \ \mbox{(7)}$$
In practice, we hardly use the formula (7). Again, we use the recurrence relation of Legendre polynomials instead.

Electric Dipole: The generating function (3) can be used for the electric multipole potential. Here we consider an electric dipole. Let us place electric charges $q$ and $-q$ at $z=a$ and $z=-a$, respectively.

Electric Dipole Potential

The electric dipole potential is given by
$$\varphi=\frac{q}{4\pi\epsilon_0}\left(\frac{1}{r_1}-\frac{1}{r_2}\right).\ \ \ \ \ \mbox{(8)}$$
$r_2$ is written in terms of $r$ and $\theta$ using the Laws of Cosine as
\begin{align*}
r_2^2&=r^2+a^2-2ar\cos(\pi-\theta)\\
&=r^2+a^2+2ar\cos\theta.
\end{align*}
So by the generating function (3), the electric dipole potential (8) can be written as
\begin{align*}
\varphi&=\frac{q}{4\pi\epsilon_0 r}\left\{\left[1-2\left(\frac{a}{r}\right)\cos\theta+\left(\frac{a}{r}\right)^2\right]^{-\frac{1}{2}}-\left[1+2\left(\frac{a}{r}\right)\cos\theta+\left(\frac{a}{r}\right)^2\right]^{-\frac{1}{2}}\right\}\\
&=\frac{q}{4\pi\epsilon_0 r}\left[\sum_{n=0}^\infty P_n(\cos\theta)\left(\frac{a}{r}\right)^n-\sum_{n=0}^\infty P_n(\cos\theta)(-1)^n\left(\frac{a}{r}\right)^n\right]\\
&=\frac{2q}{4\pi\epsilon_0 r}\left[P_1(\cos\theta)\left(\frac{a}{r}\right)+P_3(\cos\theta)\left(\frac{a}{r}\right)^3+\cdots\right]
\end{align*}
for $r>a$.

For $r\gg a$,
$$\varphi\approx\frac{2aq}{4\pi\epsilon_0 r}\frac{P_1(\cos\theta)}{r^2}=\frac{2aq}{4\pi\epsilon_0 r}\frac{\cos\theta}{r^2}.$$
This is usual electric dipole potential. The quantity $2aq$ is called the dipole moment in electromagnetism.