Polynomial interpolation - Fundamentals of Numerical Computation

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 of 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

Julia

MATLAB

Python

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

It’s convenient to measure time in years since 1980. We use .- to subtract a scalar from every element of a vector. We will also use a floating-point value in the subtraction, so the result is also in double precision.

A dotted operator such as .- or .* acts elementwise, broadcasting scalar values to match up with elements of an array.

t = year .- 1980.0
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:

An expression inside square brackets and ending with a for statement is called a comprehension. It’s often an easy and readable way to construct vectors and matrices.

V = [ t[i]^j for i=1:4, j=0:3 ]

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

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 ascending-degree order for the interpolating polynomial. We can use the polynomial to estimate the population of China in 2005:

The Polynomials package has functions to make working with polynomials easy and efficient.

p = Polynomial(c)    # construct a polynomial
p(2005-1980)         # include the 1980 time shift

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.

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

scatter(t,y, label="actual", legend=:topleft,
    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. The dot after the name of the polynomial is a universal way to apply a function to every element of an array, a technique known as broadcasting.

The range function constructs evenly spaced values given the endpoints and either the number of values, or the step size between them.

Adding a dot to the end of a function name causes it to be broadcast, i.e., applied to every element of a vector or matrix.

# Choose 500 times in the interval [0,35].
tt = range(0,35,length=500)   
# Evaluate the polynomial at all the vector components.
yy = p.(tt)
foreach(println,yy[1:4])

Now we use plot! to add to the current 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.

By convention, functions whose names end with the bang ! change the value or state of something, in addition to possibly returning output.

plot!(tt,yy,label="interpolant")

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:

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.

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.

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.

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")

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)

To solve a linear system $\mathbf{V} \mathbf{c} = \mathbf{y}$ for the vector of polynomial coefficients, we use solve (imported from numpy.linalg):

c = solve(V, y)
print(c)

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).

Matrix multiplication in NumPy is done with @ or matmul.

print(y - V @ c)

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

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¶

Suppose you want to interpolate the points $(-1,0)$ , $(0,1)$ , $(2,0)$ , $(3,1)$ , and $(4,2)$ by a polynomial of as low a degree as possible.
(a) ✍ What is the maximum necessary degree of this polynomial?
(b) ✍ Write out a linear system of equations for the coefficients of the interpolating polynomial.
(c) ⌨ Solve the system in (b) numerically.
(a) ✍ Say you want to find a cubic polynomial $p$ such that $p(-1) =-2$ , $p'(-1) =1$ , $p(1) = 0$ , and $p'(1) =-1$ . (This is known as a Hermite interpolant.) Write out a linear system of equations for the coefficients of $p$ .
(b) ⌨ Numerically solve the linear system in part (a) and make a plot of $p$ over $-1 \le x \le 1$ .
⌨ 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.
⌨ 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
$t = \frac{\text{year} - 1860}{10}$
(2.1.4)
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.)

	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

Footnotes¶

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

Preface

Linear systems of equations

Preface

Computing with matrices