Numerical Estimation of Orthogonal Polynomials

Numerical Estimation of Orthogonal Polynomials#

Walter Gautchi (a CS professor at Purdue) had created a Fortran package (ORTHPOL) that constructs orthogonal polynomials given an arbitrary weight function. The package is available at here. I have recoded the package in Python and added some additional features.

The package is based on two important theorems. In what follows, there is a random variable \(\Xi\) and we are working within the Hilbert space \(L^2(\Xi)\) with the usual inner product. The polynomials we construct are named \(\pi_k(\xi)\).

Theorem 1: There is a unique set of (monic) orthogonal polynomials:

\[ \pi_k(\xi) = \xi^k + \text{lower order terms}, \]

satisfying:

\[ \langle \pi_k, \pi_l \rangle = 0 \quad \text{if} \quad k \neq l. \]

The other important theorem is the three-term recurrence relation. This relation enables us to construct and evaluate the polynomials efficiently.

Theorem 2: The polynomials satisfy the three-term recurrence relation:

\[ \pi_{k+1}(\xi) = (\xi - \alpha_k) \pi_k(\xi) - \beta_k \pi_{k-1}(\xi). \]

where:

\[ \alpha_k = \frac{\langle \xi \pi_k, \pi_k \rangle}{\langle \pi_k, \pi_k \rangle} \]

and:

\[ \beta_k = \frac{\langle \pi_k, \pi_k \rangle}{\langle \pi_{k-1}, \pi_{k-1} \rangle} \]

and:

\[ \pi_{-1}(\xi) = 0 \quad \text{and} \quad \pi_0(\xi) = 1. \]

Orthonormal polynomials constructed in this way are called polynomial chaos. The name has nothing to do with chaos theory of dynamical systems.

Orthojax#

You will have to install the package orthojax:

!pip install orthojax --upgrade

Example: Uniform random variable on \([0, 2]\)#

Let \(\Xi \sim U([0,2])\). The PDF is:

\[ p(\xi) = \frac{1}{2} \quad \text{for} \quad \xi \in [0,2]. \]

Here is how we can construct the polynomials:

import orthojax as ojax

degree = 10
pdf = lambda xi: 0.5
poly = ojax.make_orthogonal_polynomial(degree, left=0.0, right=2.0, wf=pdf)

Here is how the object looks like:

poly

OrthogonalPolynomial(
  alpha=f32[11],
  beta=f32[11],
  gamma=f32[11],
  quad=QuadratureRule(x=f32[100], w=f32[100])
)

Here are the recursion coefficients:

poly.alpha, poly.beta, poly.gamma

(Array([1.        , 1.0000025 , 0.99999017, 1.0000262 , 0.999944  ,
        1.0001017 , 0.9998344 , 1.0002512 , 0.9996395 , 1.0004972 ,
        0.9993389 ], dtype=float32),
 Array([1.        , 0.5773435 , 0.51641864, 0.5070478 , 0.50403017,
        0.50240046, 0.5019109 , 0.5010617 , 0.5012555 , 0.50043714,
        0.50102293], dtype=float32),
 Array([1.        , 0.5773435 , 0.51641864, 0.5070478 , 0.50403017,
        0.50240046, 0.5019109 , 0.5010617 , 0.5012555 , 0.50043714,
        0.50102293], dtype=float32))

Let’s plot the polynomials:

import numpy as np

xis = np.linspace(0.0, 2.0, 200)
phi = poly(xis)
phi.shape

(200, 11)

fig, ax = plt.subplots()
ax.plot(xis, phi)
ax.set(xlabel=r"$\xi$", ylabel=r"$\phi_i(\xi)$")
sns.despine(trim=True)

../../_images/298331414c0c5e71333cc5b48e25dd3e710069658fdf4947e2085ef8cf71bfed.svg

Are they really orthonormal? Let’s check using a quadrature rule to calculate the required inner products. Recall that a quadrature rule is a tuple of points and weights:

\[ \left\{ \xi_i, w_i \right\}_{i=1}^n \]

such that:

\[ \int_{\Xi} f(\xi) p(\xi) d\xi \approx \sum_{i=1}^n w_i f(\xi_i). \]

You learn about them in ME 581. Typical quadrature rules are the trapezoidal rule, Simpson’s rule, and Gauss quadrature. Here we are using the Fejer quadrature rule. The rule is exact for polynomials of a certain degree.

In particular, we can approximate the norm of a basis function by:

\[ \|\phi_i\|^2 = \langle \phi_i, \phi_i \rangle \approx \sum_{j=1}^n w_j \phi_i(\xi_j)^2. \]

x, w = poly.quad
phis = poly(x)

norms = np.einsum("i,ij->j", w, phis ** 2)

for i in range(degree + 1):
    print(f"‖φ{i}‖ \t= {norms[i]:.2f}")

‖φ0‖ 	= 1.00
‖φ1‖ 	= 1.00
‖φ2‖ 	= 1.00
‖φ3‖ 	= 1.00
‖φ4‖ 	= 1.00
‖φ5‖ 	= 1.00
‖φ6‖ 	= 1.00
‖φ7‖ 	= 1.00
‖φ8‖ 	= 1.00
‖φ9‖ 	= 1.00
‖φ10‖ 	= 1.00

Let’s check orthogonality. We approximate the inner product by:

\[ \langle \phi_i, \phi_j \rangle \approx \sum_{k=1}^n w_k \phi_i(\xi_k) \phi_j(\xi_k). \]

for i in range(poly.num_terms):
    for j in range(i + 1, poly.num_terms):
        inner_phi_i_phi_j = np.sum(w * phis[:, i] * phis[:, j])
        print(f"<φ{i},φ{j}> \t= {inner_phi_i_phi_j:.2f}")

This is pretty close to zero.

Example: Hermite Polynomials#

Let \(\Xi \sim N(0,1)\). We can construct the polynomials like this:

import math

degree = 10
pdf = lambda xi: np.exp(-xi ** 2 / 2) / np.sqrt(2.0 * np.pi)
poly = ojax.make_orthogonal_polynomial(degree, left=-math.inf, right=math.inf, wf=pdf)

Let’s just plot them:

xis = np.linspace(-3.0, 3.0, 100)
phi = poly(xis)

fig, ax = plt.subplots()
ax.plot(xis, phi)
ax.set(xlabel=r"$\xi$", ylabel=r"$\phi_i(\xi)$")
sns.despine(trim=True)

../../_images/78fd36e7eb54392667da60545d78e7e466cef674d03455895170285c02a6162b.svg

Example: Mixture of Gaussians#

Let’s do a non-trivial example where we construct orthogonal polynomials for a Gaussian mixture. The random variable is:

\[ \Xi \sim 0.5 N(-1, 0.5^2) + 0.5 N(1, 0.5^2). \]

The PDF is:

\[ p(\xi) = 0.5 \frac{1}{\sqrt{2 \pi} 0.5} \exp \left( - \frac{(\xi + 1)^2}{2 (0.5)^2} \right) + 0.5 \frac{1}{\sqrt{2 \pi} 0.5} \exp \left( - \frac{(\xi - 1)^2}{2 (0.5)^2} \right). \]

Here we go:

degree = 10
pdf = lambda xi: 0.5 * np.exp(-(xi + 1.0) ** 2 / 2 / 0.5 ** 2) / np.sqrt(2.0 * np.pi * 0.5 ** 2) \
    + 0.5 * np.exp(-(xi - 1.0) ** 2 / 2 / 0.5 ** 2) / np.sqrt(2.0 * np.pi * 0.5 ** 2)
poly = ojax.make_orthogonal_polynomial(degree, left=-math.inf, right=math.inf, wf=pdf)

And here is how the look like:

xis = np.linspace(-2.0, 2.0, 100)
phi = poly(xis)

fig, ax = plt.subplots()
ax.plot(xis, phi)
ax.set(xlabel=r"$\xi$", ylabel=r"$\phi_i(\xi)$")
sns.despine(trim=True)

../../_images/b4c9930e3bbb380e804d42e21bea2de8bc70ac62a7478603888d2e1d34fa53be.svg

Example: Projecting a function onto the orthogonal basis#

Let’s say we have a function \(f(\xi)\) and we want to project it onto the orthogonal basis. The projection is:

\[ f(\xi) = \sum_{k=0}^n \langle f, \pi_k \rangle \pi_k(\xi). \]

So, just need to be able to do the inner products. Let’s demonstrate with the Legendre polynomials.

poly = ojax.make_legendre_polynomial(degree).normalize()
x, w = poly.quad

# Just a function to project
f = lambda x: np.sin(10 * x) / (10 * x)

# The projection
proj = np.einsum("i,ij,i->j", f(x), poly(x), w)
proj

array([ 1.6583475e-01, -4.9767550e-09, -1.5908138e-01, -3.1719683e-06,
        1.8076573e-01,  2.2900640e-05, -2.1764600e-01, -8.9398352e-05,
        1.2994759e-01,  2.0104554e-04, -4.2646304e-02], dtype=float32)

Let’s visualize the projection:

proj_f = lambda xi: np.einsum("k,ik->i", proj, poly(xi))

xis = np.linspace(-1.0, 1.0, 100)
fig, ax = plt.subplots()
ax.plot(xis, f(xis), label=r"$f(\xi)$")
ax.plot(xis, proj_f(xis), label=r"$\tilde{f}(\xi)$")
ax.legend(loc="best", frameon=False)
sns.despine(trim=True)

../../_images/e4c7e3501a1a855aa982548f1bcfc5eedccb505d4e1ba1189d3043c203450675.svg

Okay, pretty good.

Example: Calculating statistics of a random variable#

Let’s use the polynomial chaos expansions to calculate statistics of a random variable.

Take:

\[ \Xi \sim N(0, 1), \]

and then:

\[ Z = f(\Xi) = \sin(2 \Xi) / (2\Xi). \]

We want the mean and variance of \(Z\).

First, we will project \(f\) on the polynomial basis:

\[ f(\xi) = \sum_{k=0}^n \langle f, \pi_k \rangle \pi_k(\xi) = \sum_{k=0}^n f_k \pi_k(\xi). \]

The mean is just the first coefficient:

\[ \mathbb{E}[Z] = \mathbb{E}[f(\Xi)] = \langle f, \pi_0 \rangle = f_0. \]

Recall that \(\pi_0 = 1\) (the constant polynomial).

The variance is:

\[ \text{Var}[Z] = \mathbb{E}[Z^2] - \mathbb{E}[Z]^2 = \sum_{k=1}^n f_k^2 \]

degree = 11
poly = ojax.make_hermite_polynomial(degree)
x, w = poly.quad
f = lambda x: np.sin(2 * x) / (2 * x)
proj = np.einsum("i,ij,i->j", f(x), poly(x), w)

mean = proj[0]
print(f"mean: {mean}")

var = np.sum(proj[1:] ** 2)
print(f"variance: {var}")

mean: 0.5981440544128418
variance: 0.14386773109436035

Let’s compare with the Monte Carlo estimates:

xis = np.random.randn(10_000)
samples = f(xis)
mc_mean = np.mean(samples)
mc_var = np.var(samples)
print(f"MC mean: {mc_mean}")
print(f"MC variance: {mc_var}")

MC mean: 0.5966954762780894
MC variance: 0.1448722874750317