Prof. W. Gander Masaryk University Fall 2014
Least Squares
1. Determine the parameters a and b such that the function f(x) = aebx
fits the
following data
x 30.0 64.5 74.5 86.7 94.5 98.9
y 4 18 29 51 73 90
Hint: If you fit log f(x) the problem becomes very easy!
Solution: Taking the logarithm of the function we get
ln y = ln a + bx.
With the unknown c = ln a the least squares problem becomes








1 30.0
1 64.5
1 74.5
1 86.7
1 94.5
1 98.9








c
b
≈








ln 4
ln 18
ln 29
ln 51
ln 73
ln 90








.
x=[30.0 , 64.5 , 74.5 , 86.7 , 94.5 , 98.9]’;
y=[ 4 , 18 , 29 , 51 , 73 , 90]’;
A = [ones(size(x)), x]
b=log(y);
p = A\b
a=exp(p(1))
b=p(2)
The solution is
b = 0.04524310648 and c = 0.00789262406 ⇒ a = 1.00792752.
2. Consider the plane in R3
given by the equation
x1 + x2 + x3 = 0.
Construct a matrix P which projects a given point on this plane. Hint: consider
first the orthogonal complement of the plane.
Solution: The normal of the plane is n =


1
1
1

 and the orthogonal complement
is therefore Ax =


1
1
1

 x. Because R3
= R(A) ⊕ N(A ) we have to compute
the projector PN(A ) = I − AA+
.
A+
= (A A)−1
A =
1
3
(1, 1, 1)
⇒ P = I − AA+
= I −
1
3


1 1 1
1 1 1
1 1 1

 =
1
3


2 −1 −1
−1 2 −1
−1 −1 2

 .
3. Given the matrix
A =




1 2 6
1 3 7
1 4 8
1 5 9




Compute a Householder matrix P such that
PA =




σ x x
0 x x
0 x x
0 x x




It is sufficient to determine σ and the Householder-vector u.
Solution: P = I − uu with u 2 =
√
2.
u =
x − σe1
x 2(|x1| + x 2)
, x =




1
1
1
1



 , σ = − x 2 = −2 (avoid cancellation)
Thus
u =
1
√
6




3
1
1
1



 , Px = x − u(u x) =




1
1
1
1



 −
1
√
6




3
1
1
1




6
√
6
=




−2
0
0
0




4. Consider the plane in R3
given by the equation
2x1 − 2x3 = 0.
Construct a matrix P ∈ R3×3
which reflects a given point at this plane (computes
the mirror image).
Solution: The normal vector of the plane is
u =


1
0
−1


Since u 2 =
√
2 the solution is given by the Householder matrix
P = I − uu = I −


1 0 −1
0 0 0
−1 0 1

 =


0 0 1
0 1 0
1 0 0

 .
5. Let the measured points (tk, yk) for i = k, . . . , m be given. We want to fit the
function f(a, b) = aebt
such that
m
k=1
aebtk
− yk
2
−→ min
using the Gauss-Newton method. Write up the system of equations for the
first iteration step.
Solution: We want to solve the nonlinear least squares problem f(x) ≈ 0 where
fi(x) = x1ex2tk
− yk, k = 1, . . . , m
For Gauss-Newton we expand f(x + h) ≈ f(x) + J h with the Jacobian
J =



...
...
∂fi/∂x1 ∂fi/∂x2
...
...


 =



...
...
ex2tk x1tkex2tk
...
...


 .
The system of equations for the correction h is a linear least squares problem



ex2t1
x1t1ex2t1
...
...
ex2tm
x1tmex2tm



h1
h2
≈ −



x1ex2t1
− y1
...
x1ex2tm
− ym


