from numpy import *
from scipy import linalg
from scipy.linalg import norm
from matplotlib.pyplot import *
from prettytable import PrettyTable
from timeit import default_timer as timer
import sys
sys.path.append('fncbook/')
import fncbook as FNC

# This (optional) block is for improving the display of plots.
# from IPython.display import set_matplotlib_formats
# set_matplotlib_formats("svg","pdf")
# %config InlineBackend.figure_format = 'svg'
rcParams["figure.figsize"] = [7, 4]
rcParams["lines.linewidth"] = 2
rcParams["lines.markersize"] = 4
rcParams['animation.html'] = "jshtml"  # or try "html5"

2.1.Polynomial interpolation¶

The United States carries out a census of its population every 10 years. Suppose we want to know the population at times in between the census years, or to estimate future populations. One technique is to find a polynomial that passes through all the data points.^[1]

As posed in Definition 2.1.1, the polynomial interpolation problem has a unique solution. Once the interpolating polynomial is found, it can be evaluated anywhere to estimate or predict values.

2.1.1Interpolation as a linear system¶

Given data $(t_i,y_i)$ for $i=1,\ldots,n$ , we seek a polynomial

p(t) = c_1 + c_{2} t + c_3t^2 + \cdots + c_{n} t^{n-1},

(2.1.1)

such that $y_i=p(t_i)$ for all $i$ . These conditions are used to determine the coefficients $c_1\ldots,c_n$ :

\begin{split} c_1 + c_2 t_1 + \cdots + c_{n-1}t_1^{n-2} + c_nt_1^{n-1} &= y_1, \\ c_1 + c_2 t_2 + \cdots + c_{n-1}t_2^{n-2} + c_nt_2^{n-1} &= y_2, \\ c_1 + c_2 t_3 + \cdots + c_{n-1}t_3^{n-2} + c_nt_3^{n-1} &= y_3, \\ \vdots \qquad & \\ c_1 + c_2 t_n + \cdots + c_{n-1}t_n^{n-2} + c_nt_n^{n-1} &= y_n. \end{split}

(2.1.2)

These equations form a linear system for the coefficients $c_i$ :

\begin{bmatrix} 1 & t_1 & \cdots & t_1^{n-2} & t_1^{n-1} \\ 1 & t_2 & \cdots & t_2^{n-2} & t_2^{n-1} \\ 1 & t_3 & \cdots & t_3^{n-2} & t_3^{n-1} \\ \vdots & \vdots & & \vdots & \vdots \\ 1 & t_n & \cdots & t_n^{n-2} & t_n^{n-1} \\ \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \\ \vdots \\ c_n \end{bmatrix} = \begin{bmatrix} y_1 \\ y_2 \\ y_3 \\ \vdots \\ y_n \end{bmatrix},

(2.1.3)

or more simply, $\mathbf{V} \mathbf{c} = \mathbf{y}$ . The matrix $\mathbf{V}$ is of a special type.

Polynomial interpolation can therefore be posed as a linear system of equations with a Vandermonde matrix.

Example 2.1.1 (Linear system for polynomial interpolation)

We create two vectors for data about the population of China. The first has the years of census data and the other has the population, in millions of people.

year = arange(1980, 2020, 10)   # from 1980 to 2020 by 10
pop = array([984.736, 1148.364, 1263.638, 1330.141])

It’s convenient to measure time in years since 1980.

t = year - 1980
y = pop

Now we have four data points $(t_1,y_1),\dots,(t_4,y_4)$ , so $n=4$ and we seek an interpolating cubic polynomial. We construct the associated Vandermonde matrix:

V = vander(t)
print(V)

[[    0     0     0     1]
 [ 1000   100    10     1]
 [ 8000   400    20     1]
 [27000   900    30     1]]

To solve a linear system $\mathbf{V} \mathbf{c} = \mathbf{y}$ for the vector of polynomial coefficients, we use solve, located in scipy.linalg:

from scipy import linalg
c = linalg.solve(V, y)
print(c)

[-6.95000e-05 -2.39685e-01  1.87666e+01  9.84736e+02]

The algorithms used by solve are the main topic of this chapter. As a check on the solution, we can compute the residual $\mathbf{y} - \mathbf{V} \mathbf{c}$ , which should be small (near machine precision).

Tip

Matrix multiplication in NumPy is done with @ or matmul.

print(y - V @ c)

[0. 0. 0. 0.]

By our definitions, the coefficients in c are given in descending order of power in $t$ . We can use the resulting polynomial to estimate the population of China in 2005:

p = poly1d(c)          # construct a polynomial
print(p(2005 - 1980))     # apply the 1980 time shift

1303.0119375

The official figure was 1303.72, so our result is rather good.

We can visualize the interpolation process. First, we plot the data as points. Then we add a plot of the interpolant, taking care to shift the $t$ variable back to actual years.

scatter(year, y, color="k", label="data");
tt = linspace(0, 30, 300)   # 300 times from 1980 to 2010
plot(1980 + tt, p(tt), label="interpolant");
xlabel("year");
ylabel("population (millions)");
title("Population of China");
legend();

2.1.2Exercises¶

Exercise 2.1.3

⌨ Here are population figures (in millions) for three countries over a 30-year period (from United Nations World Population Prospects, 2019).

	1990	2000	2010	2020
United States	252.120	281.711	309.011	331.003
India	873.278	1,056.576	1,234.281	1,380.004
Poland	37.960	38.557	38.330	37.847

(a) Use cubic polynomial interpolation to estimate the population of the USA in 2005.

(b) Use cubic polynomial interpolation to estimate when the population of Poland peaked during this time period.

(c) Use cubic polynomial interpolation to make a plot of the Indian population over this period. Your plot should be well labeled and show a smooth curve as well as the original data points.

Exercise 2.1.4

⌨ Here are the official population figures for the state of Delaware, USA, every ten years from 1790 to 1900: 59096, 64273, 72674, 72749, 76748, 78085, 91532, 112216, 125015, 146608, 168493, 184735. For this problem, use

t = \frac{\text{year} - 1860}{10}

as the independent (time) variable.

(a) Using only the data from years 1860 to 1900, plot the interpolating polynomial over the same range of years. Add the original data points to your plot as well.

(b) You might assume that adding more data will make the interpolation better. But this is not always the case. Use all the data above to create an interpolating polynomial of degree 11, and then plot that polynomial over the range 1860 to 1900. In what way is this fit clearly inferior to the previous one? (This phenomenon is studied in Chapter 9.)

Footnotes¶

We’re quite certain that the U.S. Census Bureau uses more sophisticated modeling techniques than the one we present here!
↩