Chapter 3

Here are 5-year averages of the worldwide temperature anomaly as compared to the 1951–1980 average (source: NASA).

year = 1955:5:2000
temp = [ -0.0480, -0.0180, -0.0360, -0.0120, -0.0040,
       0.1180, 0.2100, 0.3320, 0.3340, 0.4560 ]
    
scatter(year, temp, label="data",
    xlabel="year", ylabel="anomaly (degrees C)", 
    legend=:bottomright)

A polynomial interpolant can be used to fit the data. Here we build one using a Vandermonde matrix. First, though, we express time as decades since 1950, as it improves the condition number of the matrix.

t = @. (year - 1950) / 10
n = length(t)
V = [ t[i]^j for i in 1:n, j in 0:n-1 ]
c = V \ temp

The coefficients in vector c are used to create a polynomial. Then we create a function that evaluates the polynomial after changing the time variable as we did for the Vandermonde matrix.

using Polynomials, Plots
p = Polynomial(c)
f = yr -> p((yr - 1950) / 10)
plot!(f, 1955, 2000, label="interpolant")

As you can see, the interpolant does represent the data, in a sense. However it’s a crazy-looking curve for the application. Trying too hard to reproduce all the data exactly is known as overfitting.

Here are the 5-year temperature averages again.

year = 1955:5:2000
t = @. (year - 1950) / 10
temp = [ -0.0480, -0.0180, -0.0360, -0.0120, -0.0040,
          0.1180, 0.2100, 0.3320, 0.3340, 0.4560 ]

The standard best-fit line results from using a linear polynomial that meets the least-squares criterion.

V = [ t.^0 t ]    # Vandermonde-ish matrix
@show size(V)
c = V \ temp
p = Polynomial(c)

f = yr -> p((yr - 1955) / 10)
scatter(year, temp, label="data",
    xlabel="year", ylabel="anomaly (degrees C)", leg=:bottomright)
plot!(f, 1955, 2000, label="linear fit")

If we use a global cubic polynomial, the points are fit more closely.

V = [ t[i]^j for i in 1:length(t), j in 0:3 ]   
@show size(V);

size(V) = (10, 4)

Now we solve the new least-squares problem to redefine the fitting polynomial.

p = Polynomial( V \ temp )
plot!(f, 1955, 2000, label="cubic fit")

If we were to continue increasing the degree of the polynomial, the residual at the data points would get smaller, but overfitting would increase.

a = [1/k^2 for k=1:100] 
s = cumsum(a)        # cumulative summation
p = @. sqrt(6*s)

scatter(1:100, p;
    title="Sequence convergence",
    xlabel=L"k",  ylabel=L"p_k")

This graph suggests that maybe $p_k\to \pi$, but it’s far from clear how close the sequence gets. It’s more informative to plot the sequence of errors, $\epsilon_k= |\pi-p_k|$. By plotting the error sequence on a log-log scale, we can see a nearly linear relationship.

ϵ = @. abs(π - p)    # error sequence
scatter(1:100, ϵ;
    title="Convergence of errors",
    xaxis=(:log10,L"k"),  yaxis=(:log10,"error"))

The straight line on the log-log scale suggests a power-law relationship where $\epsilon_k\approx a k^b$, or $\log \epsilon_k \approx b (\log k) + \log a$.

k = 1:100
V = [ k.^0 log.(k) ]     # fitting matrix
c = V \ log.(ϵ)          # coefficients of linear fit

In terms of the parameters $a$ and $b$ used above, we have

a, b = exp(c[1]), c[2];
@show b;

b = -0.9674103233127929

It’s tempting to conjecture that the slope $b\to -1$ asymptotically. Here is how the numerical fit compares to the original convergence curve.

plot!(k, a * k.^b, l=:dash, label="power-law fit")

Because the functions $\sin^2(t)$, $\cos^2(t)$, and 1 are linearly dependent, we should find that the following matrix is somewhat ill-conditioned.

t = range(0, 3, 400)
f = [ x -> sin(x)^2, x -> cos((1 + 1e-7) * x)^2, x -> 1. ]
A = [ f(t) for t in t, f in f ]
@show κ = cond(A);

κ = cond(A) = 1.8253225426741675e7

Now we set up an artificial linear least-squares problem with a known exact solution that actually makes the residual zero.

x = [1., 2, 1]
b = A * x;

Using backslash to find the least-squares solution, we get a relative error that is well below κ times machine epsilon.

x_BS = A \ b
@show observed_error = norm(x_BS - x) / norm(x);
@show error_bound = κ * eps();

observed_error = norm(x_BS - x) / norm(x) = 1.0163949045357309e-10
error_bound = κ * eps() = 4.053030228488391e-9

If we formulate and solve via the normal equations, we get a much larger relative error. With $\kappa^2\approx 10^{14}$, we may not be left with more than about 2 accurate digits.

N = A' * A
x_NE = N \ (A'*b)
@show observed_err = norm(x_NE - x) / norm(x);
@show digits = -log10(observed_err);

observed_err = norm(x_NE - x) / norm(x) = 0.021745909192780664
digits = -(log10(observed_err)) = 1.6626224298403076

Julia provides access to both the thin and full forms of the QR factorization.

A = rand(1.:9., 6, 4)
@show m,n = size(A);

(m, n) = size(A) = (6, 4)

Here is a standard call:

Q,R = qr(A);
Q

R

If you look carefully, you see that we seemingly got a full $\mathbf{Q}$ but a thin $\mathbf{R}$. However, the $\mathbf{Q}$ above is not a standard matrix type. If you convert it to a true matrix, then it reverts to the thin form.

Q̂ = Matrix(Q)

We can test that $\mathbf{Q}$ is an orthogonal matrix:

opnorm(Q' * Q - I)

The thin $\hat{\mathbf{Q}}$ cannot be an orthogonal matrix, because it is not square, but it is still ONC:

Q̂' * Q̂ - I

We’ll repeat the experiment of Demo 3.2.1, which exposed instability in the normal equations.

t = range(0, 3, 400)
f = [ x -> sin(x)^2, x -> cos((1 + 1e-7) * x)^2, x -> 1. ]
A = [ f(t) for t in t, f in f ]
x = [1., 2, 1]
b = A * x;

The error in the solution by Function 3.3.2 is similar to the bound predicted by the condition number.

observed_error = norm(FNC.lsqrfact(A, b) - x) / norm(x);
@show observed_error;
@show error_bound = cond(A) * eps();

observed_error = 4.665273501889628e-9

error_bound = cond(A) * eps() = 4.053030228488391e-9

We will use Householder reflections to produce a QR factorization of a random matrix.

A = rand(float(1:9), 6, 4)
m,n = size(A)

Our first step is to introduce zeros below the diagonal in column 1 by using (3.4.4) and (3.4.1).

z = A[:, 1];
v = normalize(z - norm(z) * [1; zeros(m-1)])
P₁ = I - 2v * v'   # reflector

We check that this reflector introduces zeros as it should:

P₁ * z

Now we replace $\mathbf{A}$ by $\mathbf{P}\mathbf{A}$.

A = P₁ * A

We are set to put zeros into column 2. We must not use row 1 in any way, lest it destroy the zeros we just introduced. So we leave it out of the next reflector.

z = A[2:m, 2]
v = normalize(z - norm(z) * [1; zeros(m-2)])
P₂ = I - 2v * v'

We now apply this reflector to rows 2 and below only.

A[2:m, :] = P₂ * A[2:m, :]
A

We need to iterate the process for the last two columns.

for j in 3:n
    z = A[j:m, j]
    v = normalize(z - norm(z) * [1; zeros(m-j)])
    P = I - 2v * v'
    A[j:m, :] = P * A[j:m, :]
end

We have now reduced the original to an upper triangular matrix using four orthogonal Householder reflections:

R = triu(A)