Orthogonal functions

Legendre polynomials (important in spherical harmonics)
There are three general ways of forming these polynomials.
1. Schmidt orthogonalisation of $\{x^n\}$ on $[-1,1]$
Let $p_1(x)=\dfrac{1}{\lVert 1\rVert}$. Then $\lVert 1\rVert^2=\int_{-1}^1 1dx=2$ and thus $\boxed{p_1(x)=\dfrac{1}{\sqrt{2}}}$. Notice that $\int_{-1}^1 f(x)dx=0$ for any odd function $f(x)$. So $x$ is orthogonal to $1$. We then find $p_2(x)=\dfrac{x}{\lVert x \rVert}$. Since $\lVert x \rVert^2=\int_{-1}^1 x^2dx=\dfrac{2}{3}$, we have $\boxed{p_2(x)=\sqrt{\dfrac{3}{2}}x}$. Although $x^2$ is orthogonal to $x$ (and hence $p_2$), it is not orthogonal to $1$. Let $q_3(x)=x^2-\langle x^2,p_1 \rangle p_1$, and then normalise to get $p_3=\dfrac{q_3}{\lVert q_3 \rVert}$. Now $\langle x^2,p_1 \rangle=\dfrac{1}{\sqrt{2}}\int_{-1}^1 x^2 dx=\dfrac{\sqrt{2}}{3}$, so $q_3(x)=x^2-\dfrac{\sqrt{2}}{3}\dfrac{1}{\sqrt{2}}=x^2-\dfrac{1}{3}$. $\lVert q_3 \rVert^2=\int_{-1}^1 (x^2-\dfrac{1}{3})^2dx=\dfrac{8}{45}$. Thus $\boxed{p_3(x)=\sqrt{\dfrac{45}{8}}(x^2-\dfrac{1}{3})=\sqrt{\dfrac{5}{2}}(\dfrac{3}{2}x^2-\dfrac{1}{2})}$.
The whole orthonormal set of functions is $\{\sqrt{\dfrac{2n+1}{2}}P_n(x)\}$, where $P_n(x)$ signifies the Legendre polynomial of rank $n$.

2. Solution of Legendre's differential equation
$$\dfrac{d}{dx}[(1-x^2)\dfrac{df}{dx}]+l(l+1)f=0,$$ where $l$ is a positive integer or zero.
One can verify that the polynomials that are generated by orthogonalisation of $\{x^n\}$ on $[-1,1]$ are solutions of this differential equation. Now we attempt a power series solution $\sum_n a_nx^n$ for $f$. Then $$\dfrac{d}{dx}[(1-x^2)\sum_n na_nx^{n-1}]+[l(l+1)\sum_n a_nx^n]=0\\ \dfrac{d}{dx}[\sum_n na_nx^{n-1}-\sum_n na_nx^{n+1}]+\sum_n l(l+1)a_nx^n=0\\ \sum_n n(n-1)a_nx^{n-2}-\sum_n n(n+1)a_nx^n+\sum_n l(l+1)a_nx^n=0.$$ Collecting all similar powers of $x$, we obtain $$\sum_n x^n\underbrace{[(n+2)(n+1)a_{n+2}-n(n+1)a_n+l(l+1)a_n]}_{(*)}=0.$$ Since the functions $\{x_n\}$ are linearly independent, $(*)$ must be zero. This gives the recurrence relation for the expansion coefficients: $$a_{n+2}=\dfrac{[n(n+1)-l(l+1)]a_n}{(n+2)(n+1)}.$$ This equation allows us to find every other coeffcient when we are given the first. For example, if $a_0=1$, then $a_2=\dfrac{-l(l+1)}{2},a_4=\dfrac{6-l(l+1)}{12}\cdot \dfrac{-l(l+1)}{2}$, and so on.

If $l$ is even, eventually $n$ will equal $l$ for some term, then $n(n+1)$ will equal $l(l+1)$ and the following coefficients will also equal zero. So the even terms of the power series cut off at $n=l$ whereas the infinite series of odd powers diverges at $x=\pm 1$. The finite series of even powers gives $$\begin{align}l=0\quad &a_0=1\quad &f_0=1\\ l=2\quad &a_0=1\\ &a_2=-3\quad &f_2=-3x^2+1,\end{align}$$ which are proportional to the Legendre polynomials derived by orthogonalisation. We can choose $a_0$ such that the polynomial $f_n=1$ at $x=+1$. Then $$\begin{align}l=0\quad &a_0=1\quad &P_0=1\\l=2\quad &a_0=\dfrac{-1}{2}\\&a_2=\dfrac{3}{2}\quad &P_2=\dfrac{3}{2}x^2-\dfrac{1}{2}.\end{align}$$
If $l$ is odd, the series in odd powers terminates at $n=l$, and the series in even powers is an infinite series, divergent at $x=\pm 1$. Again, we choose $a_0$ such that $f_n(+1)=1$. Then $$\begin{align}l=1\quad &a_1=1\quad &P_1=x\\ l=3\quad &a_1=\dfrac{-3}{2}\\ &a_3=\dfrac{5}{2}\quad &P_3=\dfrac{5}{2}x^3-\dfrac{3}{2}x.\end{align}$$
3. Generating function
A generating function for a set of functions is a function of two variables such that when this function is expanded as power series in one of the variables, the coefficients are the set of functions in other variable. Often many useful relations may be derived from a generating function; nonetheless, there are no general methods for constructing a generating function.

The generating function for Legendre polynomials is $$F(x,t)=(1-2xt+t^2)^{-1/2}=\sum_n P_n(x)t^n.$$ It can be expanded as a binomial series $$F(x,t)=1+\dfrac{1\cdot 1}{2\cdot 1!}(2xt-t^2)+\dfrac{1\cdot 3\cdot 1}{2\cdot 2\cdot 2!}(2xt-t^2)^2+\frac{1\cdot 3\cdot 5\cdot 1}{2\cdot 2\cdot 2\cdot 3!}(2xt-t^2)^3+\cdots$$ and then rearranged in powers of $t$: $$\begin{align}F(x,t)&=1+(x)(t)+(\dfrac{-1}{2}+\dfrac{3}{2}x^2)(t^2)+(\dfrac{-3}{2}x+\dfrac{5}{2}x^3)(t^3)+\cdots\\&=\sum_n P_n(x)t^n.\end{align}$$
Hermite polynomials (important in Quantum Mechanics)
They are solutions of the differential equation
$$y''-2xy'+2py=0\quad y(0)=1,\quad y'(0)=1.$$
Chebyshev polynomials (important in numerical analysis)
They are solutions of the differential equation
$$(1-x^2)y''-xy'+n^2y=0\quad y(0)=1,\quad y'(0)=0.$$
They can also be represented as $$T_n(x)=\cos (n\cos^{-1} x).$$
Applications

$\min \int_{-1}^1 |x^3-a-bx-cx^2|dx$
Geometrically, the minimum, call it $Q$, is the square of the distance from $x^3$ to the subspace $S$ spanned by the functions $1,x,x^2$. We can compute the orthogonal projection $\phi(x)$ of $x^3$ into $S$ and use the Pythagoras' theorem: $Q^2=\lVert x^3 \rVert^2-\lVert \phi(x)\rVert^2$. We first need to transform the basis $\{1,x,x^2\}$ into an orthonormal one, namely $\{p_1,p_2,p_3\}$ as above. $\phi(x)=\langle x^3,p_1\rangle p_1+\langle x^3,p_2 \rangle p_2+\langle x^3,p_3 \rangle p_3$. $x^3$ is orthogonal to $p_1$ and $p_2$ by oddness. Since $\langle x^3,p_2 \rangle=\sqrt{\dfrac{3}{2}}\int_{-1}^1 x^3x dx=\sqrt{\dfrac{3}{2}}\dfrac{2}{5}=\dfrac{6}{5}$, we have $\lVert\phi(x)\rVert^2=\dfrac{6}{25}\lVert p_2 \Vert^2=\dfrac{6}{25}$. With $\lVert x^3 \Vert^2=\int_{-1}^1 x^6 dx=\dfrac{2}{7}$, we have $Q=\dfrac{2}{7}-\dfrac{6}{25}=\dfrac{8}{175}$.

physical applications [later]

The Legendre polynomials are important in some quantum-chemical problems. They are the basis for the wave functions for angular momentum, and thus occur in problems involving spherical motion, such as that of the electron in a hydrogen atom or the rotations of a molecule. They are also important in describing the angular dependence of one-electron atomic orbitals; this dependence, in turn, forms the basis of the geometry of chemical compounds.

Reference:
Mathematics for Quantum Chemistry by Jay Martin Anderson