Could someone explain the reason behind using Legendre Polynomials?

Where $$x = \cos(\phi)$$ and therefore (and this is very important) $$x=[-1,1]$$, we have the (special such that m = 0) associated Legendre Equation $$(1-x^2)\frac{d^2 y}{dx^2} - 2x \frac{dy}{dx}+ky = 0$$, and the power series solution obtained via the recurrence relation $$c_{n+2} = \frac{[n(n+1)-k]}{(n+2)(n+1)}$$c_n is:

$$y(x) = c_0\bigg[1-\frac{k}{2!}x^2-\frac{k(6-k)}{4!}x^4+...\bigg] + c_1\bigg[x+\frac{2-k}{3!}+\frac{(2-k)(12-k)}{5!}x^5+...\bigg]$$

The problem with the series is that the radius of convergence is just outside of the domain of $$x$$ i.e. $$|x|<1$$ (which translates to being outside of $$\phi = [0,2\pi]$$). To get around this we utilize the following values of k in order to truncate the series:

$$k = l(l+1)$$

and setting the constant of the corresponding series to 0. To illustrate, the first couple values of $$l$$ would yield the following polynomials:

• $$l = 0 \rightarrow k = 0 \rightarrow y_0 = 1$$
• $$l = 1 \rightarrow k = 2 \rightarrow y_1 = x$$
• $$l = 2 \rightarrow k = 6 \rightarrow y_0 = (1-3x^2)$$
• $$l = 3 \rightarrow k = 12 \rightarrow y_1 = (x-\frac{5}{3}x^3)$$

Where $$y_0$$ and $$y_1$$ are our basis functions which make up our solution $$y(x) = c_0 y_0+c_1y_1$$

Why can't I just pick a relatively large value for $$l$$ and have that resulting polynomial approximately be our basis function?

As some additional context for why I'm confused, the text which I am reading this from does not offer any underlying reason for moving to Legendre Polynomials, it instead says that it is "customary to normalize them (unusually) by setting $$P_l(1) = 1$$", without going into detail as to why which has left me lost. I don't know much about "normalization" either, so if that is at all important, it would be appreciated if some attention could be given there as well in the explanation.

• They form an orthonormal basis and they're complete in that any piecewise continuous function can be well approximated by them. It's much easier to project onto an orthonormal basis which is essentially what the approximation is doing. Commented Jun 1 at 23:52
• Regarding the choice of normalization, an obvious reason is that referenced by GEdgar in an older answer. Namely, it ensures that the Rodrigues' formula and the generating function both come out nicely. (For the first, the point is basically that $\frac{d^n}{dx^n}(1-x^2)^n$ evaluates to $2^n n!$ at $x=0$.) Commented Jun 2 at 0:24
• @CyclotomicField "project onto an orthonormal basis", is that what it means to "normalize" the polynomials left after truncating the series? I haven't taken a course on Linear Algebra presuming that's where this is all coming from Commented Jun 2 at 0:29
• @ResearcherR yes it's from linear algebra. Normalizing a vector means to scale it by some number so that it has length $1$. So say I have the vector $(2,0)$ then the normalized vector would be $(1,0)$ since it has length $1$ and is in the same direction. Orthonormal means that the vectors are all of length $1$ and orthogonal to each other. Many of the ideas related to vector spaces can be argued geometrically. Like say you have some three dimensional shape you might ask what it looks like if flattened onto a plane and that's what orthogonal projection does. Commented Jun 2 at 0:38
• To clarify one further point: The sense of the word "normalization" is (superficially) more expansive for orthogonal polynomials than in linear algebra. In the latter, normalization means "rescale to have length 1". But in orthogonal polynomials it means to rescale each orthogonal polynomial to satisfy some condition. (I call this "superficial" because you can think of this as defining your inner product in such a way that the two senses of normalization coincide. See the Wikipedia article on classical OPs for more.) Commented Jun 2 at 2:46

The Legendre operator

$$-(1-x^2)\partial_{x,x}+2x \partial_x =-\partial_x (1-x^2) \partial_x$$

obviuosly is a symmetric operator in the Hilbert space on the interval $$(-1,1)$$ with measure $$d\mu = (1-x^2) dx$$. Interpreting $$x=\cos\theta$$ we have $$dx = \sin \theta \ d\theta,\quad (1-x^2) = \sin^2 \theta , \quad \partial_\theta ´=(\partial_\theta x)\ \partial_x = \sin \theta \ \partial_x$$

such that $$\int_{-1}^1 dx \ (-f^* \partial_x \left( (1-x^2 ) \ \partial_x g \right)= \int_{-1}^1 dx \ \partial_x f^* \ (1-x^2)\ \partial_x g =\int_0^\pi d\theta \ \sin \theta \ \partial_\theta f^* \ \ \partial_\theta g$$ for all bounded smooth functions along the meridian of the unit sphere, because the measure supresses the boundary values at the poles for integration by parts.

In order to get smooth rotational invariant function on the sphere. the eigenfunction of the Legendre operators have to have derivative zero at the poles. Its naturally therefore, to normalize to $$P_l(\cos (0,\pi)) =\pm 1$$.

Since these types of eigenfunction series of linear selfadjoint second order equations obey a 2-term recursion, starting with $$P_0=1,P_1=x=\cos \theta$$ fixes all coefficients.

The norm conventions were practical when Legendre polynomials were used for numerical approximations and Gauss-Legendre integration.

After the reinvention of the associated Legendre polynomials as quantum eigenstates of the Laplacian in the Hilbert space over the unit sphere, a normalization to $$\int_{-1}^1 dz \ P_l(z) \ P_m(z) = \delta_{lm}$$ would have saved forests of trees for the paper of QM books and exercises.