Legendre Polynomials

Legendre polynomials $P_l$ are a sequence of orthogonal polynomials that are solutions to the Legendre differential equation:

$(1-x^2)\frac{d^2P_l}{dx^2}-2x\biggr(\frac{dP_l}{dx}\biggr)+l(l+1)P_l=0\;\;\;\;\;\;\;\;332$

which can also be expressed as

$\frac{d}{dx}\biggr\[(1-x^2)\frac{dP_l}{dx}\biggr\]+l(l+1)P_l=0\;\;\;\;\;\;\;\;332a$

When $x=0$ , eq332 simplifies to a form that resembles the simple harmonic oscillator equation, which has a power series solution. This implies that we can use ${P_l(x)=\sum_{n=0}^{\infty}a_nx^n}$ to solve eq332 around $x=0$ . Substituting ${\frac{dP_l}{dx}=\sum_{n=0}^{\infty}na_nx^{n-1}}$ , ${\frac{d^2P_l}{dx^2}=\sum_{n=0}^{\infty}n(n-1)a_nx^{n-2}}$ and ${x^2\sum_{n=0}^{\infty}n(n-1)a_nx^{n-2}=\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^{n+2}}$ in eq332 yields

$\sum_{n=0}^{\infty}n(n-1)a_nx^{n-2}-\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^{n+2}-2\sum_{n=0}^{\infty}na_nx^{n}+l(l+1)\sum_{n=0}^{\infty}a_nx^{n}=0$

To simplify this equation, substitute $n=n'+2$ and $n=n'-2$ for the first sum and second sum, respectively, and then change the indices for both sums back to $n$ to give

$\begin{align}&\[l(l+1)a_0+2a_2\]x^0+\[6a_3-2a_1+l(l+1)a_1\]x\\&+\sum_{n=2}^{\infty}\[(n+2)(n+1)a_{n+2}-n(n-1)a_n-2na_n+l(l+1)a_n\]x^n=0\end{align}$

To satisfy the above equation for all $x$ , all coefficients must be zero. Therefore, $(n+2)(n+1)a_{n+2}-n(n-1)a_n-2na_n+l(l+1)a_n=0$ , which rearranges to

$a_{n+2}=a_n\frac{n(n+1)-l(l+1)}{(n+2)(n+1)}=0\;\;\;\;\;\;\;\;333$

Eq333 is a recurrence relation. If we know the value of $a_0$ , we can use the relation to find $a_2,a_4,a_6,\cdots$ . Similarly, if we know $a_1$ , we can find $a_3,a_5,a_7,\cdots$ .

Comparing the recurrence relations for even $n$ ,

$a_{2k}=(-1)^k\frac{a_0}{(2k)!}\prod_{i=0}^{k-1}\[l(l+1)-(2i+1)(2i)\]\;\;\;k=1,2,\cdots\;\;\;\;\;334$

Question

Prove that eq334 is consistent with eq333 by induction.

Answer

For $k=1$ , eq334 becomes $a_2=-\frac{a_0}{2!}l(l+1)$ , which is consistent with eq333 when $n=0$ . Let’s assume eq334 is true for $k=m$ , i.e.,

$a_{2m}=(-1)^m\frac{a_0}{(2m)!}\prod_{i=0}^{m-1}\[l(l+1)-(2i+1)(2i)\]\;\;\;\;\;\;\;\;334a$

is consistent with eq333 when $n=2m-2$ .

We need to prove that eq334 holds for $k=m+1$ , i.e.,

$a_{2(m+1)}=(-1)^{m+1}\frac{a_0}{\[2(m+1)\]!}\prod_{i=0}^{m}\[l(l+1)-(2i+1)(2i)\]\;\;\;\;\;\;\;\;334b$

is consistent with eq333 when $n=2m$ .

Substituting eq334a in eq334b gives eq333 when $n=2m$ .

Similarly, mathematical induction proves that the recurrence relations for odd $n$ can be expressed as

$a_{2k+1}=(-1)^k\frac{a_1}{(2k+1)!}\prod_{i=1}^{k}\[l(l+1)-(2i+1)(2i)\]\;\;\;k=1,2,\cdots\;\;\;\;\;335$

Hence the general form of the power series is:

$\begin{align}P_{k}(x)=&\;a_0+\sum_{k=1}^{\infty}(-1)^k\frac{a_0}{(2k)!}\prod_{i=0}^{k-1}\[l(l+1)-(2i+1)(2i)\]x^{2k}\\&+a_1x+\sum_{k=1}^{\infty}(-1)^k\frac{a_1}{(2k+1)!}\prod_{i=1}^{k}\[l(l+1)-(2i+1)(2i)\]x^{2k+1}\;\;\;\;\;336\end{align}$

To see how eq336 behave for large $k$ , we carry out the ratio test for the coefficients. With respect to eq333, the numerator $n(n+1)-l(l+1)\approx n^2$ for large $n$ . Hence $\lim\limits_{n \to \infty}\biggr\vert\frac{a_{n+2}}{a_n}\biggr\vert=1$ , which implies that $\lim\limits_{n \to \infty}\biggr\vert\frac{a_{n+1}}{a_n}\biggr\vert=1$ . This means that the coefficients will not vanish as $n$ increases. In fact, for large $n$ , the behaviour of $P_k(x)$ resembles the Taylor series expansion $\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$ , which diverges at $x=1$ . Therefore, to ensure that $P_k(x)$ is square-integrable, we need to truncate either one of the series after some finite terms and let all the coefficients of the other series be zero.

Question

Can we instead truncate both series of eq336 after some $k$ ?

Answer

Since the value of $k$ is arbitrary, the sum of two truncated series, each with finite terms, may still have an infinite number of terms. The only way to guarantee that $P_k(x)$ has finite terms is to truncate one of the series and let $a_0$ , if the odd series is truncated, or $a_1$ , if the even series is truncated, be zero.

To truncate either series, we let $n=l$ for the numerator of eq333 so that every successive term in the selected series is zero. The solution to eq332 then becomes two separate equations of eq366, each associated with one truncated series. To further show that these two equations can be combined into one, substitute $n=l-2k$ in eq333 and rearrange it to give:

$a_{l-2k}=-\frac{(l-2k+2)(l-2k+1)}{2k(2l-2k+1)}a_{l-2k+2}\;\;\;\;\;\;k=1,2,\cdots,m\;\;\;\;\;\;337$

where $m=\frac{l}{2}$ for the even series and $m=\frac{l-1}{2}$ for the odd series.

Eq337 for some values of $k$ are presented in the table below. In contrast with the table above, the coefficients of eq337 are expressed in the reverse order.

Comparing the recurrence relations in the above table, we have

$a_{l-2k}=\frac{(-1)^k}{2^kk!}\frac{l(l-1)(l-2)\cdots(l-2k+1)}{(2l-1)(2l-3)\cdots(2l-2k+1)}a_{l}\;\;\;\;\;\;\;\;338$

Since $l!=l(l-1)\cdots(l-2k+1)\cdots 1$ , we have $l(l-1)\cdots(l-2k+1)=\frac{l!}{(l-2k)!}$ . So, eq338 becomes

$a_{l-2k}=\frac{(-1)^k}{2^kk!}\frac{l!}{(l-2k)!(2l-1)(2l-3)\cdots(2l-2k+1)}\frac{(2l-2)(2l-4)\cdots(2l-2k)}{(2l-2)(2l-4)\cdots(2l-2k)}a_{l}$

$a_{l-2k}=\frac{(-1)^k}{2^kk!}\frac{l!}{(l-2k)!(2l-1)(2l-3)\cdots(2l-2k+1)}\frac{2^k(l-1)(l-2)\cdots(l-k)}{(2l-2)(2l-4)\cdots(2l-2k)}a_{l}\;\;\;\;\;339$

Substituting $(l-1)(l-2)\cdots(l-k)=\frac{(l-1)!}{(l-k-1)!}$ in eq339 yields

$a_{l-2k}=\frac{(-1)^k}{k!}\frac{l!}{(l-2k)!(2l-1)(2l-2)\cdots(2l-2k+1)(2l-2k)}\frac{(l-1)!}{(l-k-1)!}a_{l}\;\;\;\;\;340$

Since $(2l-1)!=(2l-1)(2l-2)\cdots(2l-2k)\cdots 1$ , we have $\frac{1}{(2l-1)(2l-2)\cdots (2l-2k)}=\frac{(2l-2k-1)!}{(2l-1)!}$ . So eq340 becomes

$a_{l-2k}=\frac{(-1)^k}{k!}\frac{l!(2l-2k-1)!}{(l-2k)!(2l-1)!}\frac{(l-1)!}{(l-k-1)!}a_{l}\;\;\;\;\;341$

By convention, the leading coefficient is selected as $a_l=\frac{(2l)!}{2^l(l!)^2}$ (so that $P_l(1)=1$ ), which when substituted in eq341 gives

$a_{l-2k}=\frac{(-1)^k}{k!}\frac{(2l-2k)!}{2^lk!(l-2k)!(l-k)!}$

Therefore, eq336 can be rewritten as

$P_l(x)=\sum_{k=0}^m\frac{(-1)^k(2l-2k)!}{2^lk!(l-2k)!(l-k)!}x^{l-2k}\;\;\;\;\;\;\;\;342$

where $m=\frac{l}{2}$ for even $l$ , $m=\frac{l-1}{2}$ for odd $l$ and $P_l(x)$ are known as the Legendre polynomials.

The first few Legendre polynomials are:

$\begin{multline}P_0(x)=1\\P_1(x)=x\\P_2(x)=\frac{1}{2}(3x^2-1)\\P_3(x)=\frac{1}{2}(5x^3-3x)\\P_4(x)=\frac{1}{8}(35x^4-30x^2+3)\\P_5(x)=\frac{1}{8}(63x^5-70x^3+15x)\end{multline}$

Legendre Polynomials

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