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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 column 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 = [1982; 2000; 2010; 2015]; 
pop = [1008.18; 1262.64; 1337.82; 1374.62];

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)

To solve for the vector of polynomial coefficients, we use a backslash to solve the linear system:

Tip

A backslash \ is used to solve a linear system of equations.

c = V \ y

The algorithms used by the backslash operator are the main topic of this chapter. As a check on the solution, we can compute the residual.

y - V * c

Using floating-point arithmetic, it is not realistic to expect exact equality of quantities; a relative difference comparable to $\macheps$ is all we can look for.

By our definitions, the elements of c are coefficients in descending-degree order for the interpolating polynomial. We can use the polynomial to estimate the population of China in 2005:

p = @(t) polyval(c, t - 1980);  % include the 1980 time shift
p(2005)

The official population value for 2005 was 1303.72, so our result is rather good.

We can visualize the interpolation process. First, we plot the data as points.

Tip

The scatter function creates a scatter plot of points; you can specify a line connecting the points as well.

scatter(year, y)
xlabel("years since 1980")
ylabel("population (millions)")
title("Population of China")

We want to superimpose a plot of the polynomial. We do that by evaluating it at a vector of points in the interval.

Tip

The linspace function constructs evenly spaced values given the endpoints and the number of values.

tt = linspace(1980, 2015, 500);    % 500 times in the interval [1980, 2015]
yy = p(tt);                        % evaluate p at all the vector elements
yy(1:4)

Now we use plot! to add to the current plot, rather than replacing it.

Tip

Use hold on to add to an existing plot rather than replacing it. The plot function plots lines connecting the given $x$ and $y$ values; you can also specify markers at the points.

hold on 
plot(tt, yy)
legend("data", "interpolant", "location", "northwest");

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!
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