Math 151B Homework #7 – Due Friday 3/6
1. least squares
[Note: feel free to use octave/matlab to compute matrix inverses – you don’t have to do them by
hand.]
Consider the data y(−2) = 3, y(−1) = −2, y(0) = 0, y(1) = 4, and y(2) = 5. We want to
find the least-squares quadratic polynomial for these points.
a. Write down the curvature matrix.
Answer:
The curvature matrix is
Pm
i=1 x0
iPm
i=1 x1
iPm
i=1 x2
i
Pm
i=1 x1
iPm
i=1 x2
iPm
i=1 x3
i
Pm
i=1 x2
iPm
i=1 x3
iPm
i=1 x4
i
=
5 0 10
0 10 0
10 0 34
b. What is the condition number of this matrix.
Answer:
||A||∞= 44 and ||A||2= 37.114.
||A−1||∞= 0.62857 and ||A−1||2= 0.5302.
K∞(A) = 27.657 and K2(A) = 19.678.
c. Use the Jacobi preconditioner, and compute the new condition number.
Answer:
˜
A=D−1A
=
1/5 0 0
0 1/10 0
0 0 1/34
5 0 10
0 10 0
10 0 34
=
1 0 2
0 1 0
10/34 0 1
||A||∞= 3 and ||A||2= 2.46.
||A−1||∞= 7.3 and ||A−1||2= 5.98.
K∞(A) = 21.9 and K2(A) = 14.71.
d. Repeat with the Gauss-Seidel preconditioner.
Answer:
˜
A= (D+L)−1A
=
1 0 2
0 1 0
000.412
