Main Content

This example shows that extrapolating data using polynomials of even modest degree is risky and unreliable.

This example is older than MATLAB®. It started as an exercise in *Computer Methods for Mathematical Computations*, by Forsythe, Malcolm and Moler, published by Prentice-Hall in 1977.

Now, MATLAB makes it much easier to vary the parameters and see the results, but the underlying mathematical principles are unchanged.

Create and plot two vectors with US Census data from 1910 to 2000.

% Time interval t = (1910:10:2000)'; % Population p = [91.972 105.711 123.203 131.669 150.697... 179.323 203.212 226.505 249.633 281.422]'; % Plot plot(t,p,'bo'); axis([1910 2020 0 400]); title('Population of the U.S. 1910-2000'); ylabel('Millions');

What is your guess for the population in the year 2010?

p

`p = `*10×1*
91.9720
105.7110
123.2030
131.6690
150.6970
179.3230
203.2120
226.5050
249.6330
281.4220

Fit the data with a polynomial in `t`

and use it to extrapolate the population at `t = 2010`

. Obtain the coefficients in the polynomial by solving a linear system of equations involving an 11-by-11 Vandermonde matrix, with elements as powers of scaled time, `A(i,j) = s(i)^(n-j)`

.

n = length(t); s = (t-1950)/50; A = zeros(n); A(:,end) = 1; for j = n-1:-1:1 A(:,j) = s .* A(:,j+1); end

Obtain the coefficients `c`

for a polynomial of degree `d`

that fits the data `p`

by solving a linear system of equations involving the last `d+1`

columns of the Vandermonde matrix:

` A(:,n-d:n)*c ~= p`

If

`d < 10`

, then more equations than unknowns exist, and a least-squares solution is appropriate.If

`d == 10`

, then you can solve the equations exactly and the polynomial actually interpolates the data.

In either case, use the backslash operator to solve the system. The coefficients for the cubic fit are:

c = A(:,n-3:n)\p

`c = `*4×1*
-5.7042
27.9064
103.1528
155.1017

Now evaluate the polynomial at every year from 1910 to 2010 and plot the results.

v = (1910:2020)'; x = (v-1950)/50; w = (2010-1950)/50; y = polyval(c,x); z = polyval(c,w); hold on plot(v,y,'k-'); plot(2010,z,'ks'); text(2010,z+15,num2str(z)); hold off

Compare the cubic fit with the quartic. Notice that the extrapolated point is very different.

c = A(:,n-4:n)\p; y = polyval(c,x); z = polyval(c,w); hold on plot(v,y,'k-'); plot(2010,z,'ks'); text(2010,z-15,num2str(z)); hold off

As the degree increases, the extrapolation becomes even more erratic.

cla plot(t,p,'bo') hold on axis([1910 2020 0 400]) colors = hsv(8); labels = {'data'}; for d = 1:8 [Q,R] = qr(A(:,n-d:n)); R = R(1:d+1,:); Q = Q(:,1:d+1); c = R\(Q'*p); % Same as c = A(:,n-d:n)\p; y = polyval(c,x); z = polyval(c,11); plot(v,y,'color',colors(d,:)); labels{end+1} = ['degree = ' int2str(d)]; end legend(labels, 'Location', 'NorthWest') hold off