In this lecture, we derive some important recurrence relations of Legendre functions and use them to show that Legendre functions are indeed solutions of a differential equation, called Legendre’s differential equation.

Differentiating the generating function

$$g(x,t)=(1-2xt+t^2)^{-1/2}=\sum_{n=0}^\infty P_n(x)t^n,\ |t|<1\ \ \ \ \ \mbox{(1)}$$

with respect to $t$, we get

\begin{align*}

\frac{\partial g(x,t)}{\partial t}&=\frac{x-t}{(1-2xt+t^2)^{3/2}}\ \ \ \ \ \mbox{(2)}\\&=\sum_{n=0}^\infty nP_n(x)t^{n-1}.\ \ \ \ \ \mbox{(3)}\end{align*}

(2) can be written as

$$\frac{x-t}{(1-2xt+t^2)(1-2xt+t^2)^{1/2}}=\frac{(x-t)(1-2xt+t^2)^{-1/2}}{1-2xt+t^2}.$$

By (1) and (3), we obtain

$$(x-t)\sum_{n=0}^\infty P_n(x)t^n=(1-2xt+t^2)\sum_{n=0}^\infty nP_n(x) t^{n-1}$$ or

$$(1-2xt+t^2)\sum_{n=0}^\infty nP_n(x) t^{n-1}+(t-x)\sum_{n=0}^\infty P_n(x)t^n=0$$

which can be written out as

\begin{align*}

\sum_{n=0}^\infty nP_n(x)t^{n-1}-\sum_{n=0}^\infty &2xnP_n(x)t^n+\sum_{n=0}^\infty nP_n(x)t^{n+1}\\&+\sum_{n=0}^\infty P_n(x)t^{n+1}-\sum_{n=0}^\infty xP_n(x)t^n=0.\ \ \ \ \ \mbox{(4)}\end{align*}

In (4) replace $n$ by $n+1$ in the first term, and replace $n$ by $n-1$ in the third and fourth term. Then (4) becomes

\begin{align*}

\sum_{n=0}^\infty (n+1)P_{n+1}(x)t^n-\sum_{n=0}^\infty &2xnP_n(x)t^n+\sum_{n=0}^\infty (n-1)P_{n-1}(x)t^n\\&+\sum_{n=0}^\infty P_{n-1}(x)t^n-\sum_{n=0}^\infty xP_n(x)t^n=0.

\end{align*}

This can be simplified to

$$\sum_{n=0}^\infty[(n+1)P_{n+1}(x)-(2n+1)xP_n(x)+nP_{n-1}(x)]t^n=0$$

which implies that

$$(2n+1)xP_n(x)=(n+1)P_{n+1}(x)+nP_{n-1}(x).\ \ \ \ \ \mbox{(5)}$$

The recurrence relation (5) can be used to calculate Legendre polynomials. For example, we found $P_0(x)=1$ and $P_1(x)=x$ here. For $n=1$, (5) is

$$3xP_1(x)=2P_2(x)+P_0(x)$$

i.e.

$$P_2(x)=\frac{1}{2}(3x^2-1).$$

Continuing this using the recurrence relation (5), we obtain

\begin{align*}

P_3(x)&=\frac{1}{2}(5x^3-3x),\\

P_4(x)&=\frac{1}{8}(35x^4-30x^2+3),\\

P_5(x)&=\frac{1}{8}(63x^5-70x^3+15x),\\

\cdots.

\end{align*}

A great advantage of having the recurrence relation (5) is that one can easily calculate Legendre polynomials using a computer with a simple programming. This can be easily done for instance in Maxima.

Let us load the following simple program to run the recurrence relation (5).

(%i1) Legendre(n,x):=block ([],

if n = 0 then 1

else

if n = 1 then x

elseĀ ((2*n – 1)*x*Legendre(n – 1, x)-(n – 1)*Legendre(n – 2,x))/n);

(%o1) Legendre(n, x) := block([], if n = 0 then 1

else (if n = 1 then x else ((2 n – 1) x Legendre(n – 1, x)

- (n – 1) Legendre(n – 2, x))/n))

Now we are ready to calculate Legendre polynomials. For example, let us calculate $P_3(x)$.

(%i2) Legendre(3,x);

The output is not exactly what we may like because it is not simplified.

In Maxima, simplification can be done by the command *ratsimp*.

(%i3) ratsimp(Legendre(3,x));

The output is

That looks better. Let us calculate one more, say $P_4(x)$.

Now we differentiate $g(x,t)$ with respect to $x$.

$$\frac{\partial g(x,t)}{\partial x}=\frac{t}{(1-2xt+t^2)^{3/2}}=\sum_{n=0}^\infty P_n’(x)t^n.$$

From this we obtain

$$(1-2xt+t^2)\sum_{n=0}^\infty P_n’(x)t^n-t\sum_{n=0}^\infty P_n(x)t^n=0$$

which leads to

$$P_{n+1}’(x)+P_{n-1}’(x)=2xP_n’(x)+P_n(x).\ \ \ \ \ \mbox{(6)}$$

Add 2 times $\frac{d}{dx}(5)$ to $2n+1$ times (6). Then we get

$$(2n+1)P_n=P_{n+1}’(x)-P_{n-1}’(x).\ \ \ \ \ \mbox{(7)}$$

$\frac{1}{2}[(6)+(7)]$ results

$$P_{n+1}’(x)=(n+1)P_n(x)+xP_n’(x).\ \ \ \ \ \mbox{(8)}$$

$\frac{1}{2}[(6)-(7)]$ results

$$P_{n-1}’(x)=-nP_n(x)+xP_n’(x).\ \ \ \ \ \mbox{(9)}$$

Replace $n$ by $n-1$ in (7) and add the result to $x$ times (9):

$$(1-x^2)P_n’(x)=nP_{n-1}(x)-nxP_n(x).\ \ \ \ \ \mbox{(10)}$$

Differentiate (10) with respect to $x$ and add the result to $n$ times (9):

$$(1-x^2)P_n^{\prime\prime}(x)-2xP_n’(x)+n(n+1)P_n(x)=0.\ \ \ \ \ \mbox{(11)}$$

The linear second-order differential equation (11) is called Legendre’s differential equation and as seen $P_n(x)$ satisfies (11). This is why $P_n(x)$ is called a Legendre polynomial.

In physics (11) is often expressed in terms of differentiation with respect to $\theta$. Let $x=\cos\theta$. Then by the chain rule,

\begin{align*}

\frac{dP_n(\cos\theta)}{d\theta}&=-\sin\theta\frac{dP_n(x)}{dx},\ \ \ \ \ \mbox{(12)}\\ \frac{d^2P_n(\cos\theta)}{d\theta^2}&=-x\frac{dP_n(x)}{dx}+(1-x^2)\frac{d^2P_n(x)}{dx^2}.\ \ \ \ \ \mbox{(13)}

\end{align*}

Using (12) and (13), Legendre’s differential equation (11) can be written as

$$\frac{1}{\sin\theta}\frac{d}{d\theta}\left[\sin\theta\frac{dP_n(\cos\theta)}{d\theta}\right]+n(n+1)P_n(\cos\theta)=0.$$