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Unformatted text preview: jxi C yi j jxi j C jyi j. Sum on i D 1 to n:
kx C y k1 kx k1 C ky k1 . 18 jx1 j C 2jx2 j is a norm but min.jx1 j; jx2 j/ is not a norm. kx k C kx k1 is a norm; kAx k is a norm provided A is invertible (otherwise a nonzero vector has norm zero;
for rectangular A we require independent columns to avoid kAx k D 0). 19 x T y D x1 y1 C x2 y2 C .max jyi j/.jx1 j C jx2 j C / D jjx jj1 jjy jj1 .
20 With j D 2 2 cos.j=nC1/, the largest eigenvalue is n 2 C 2 D 4. The smallest
2
is 1 D 2 2 cos.=nC1/ nC1 , using 2 cos 2 2 . So the condition number
is c D max =min .4= 2 / n2 , growing with n.
11
1
q
n
21 A D
has A D
with q D 1 C 1:1 C C .1:1/n 1 D
0 1:1
0 .1:1/n
n
n
n
n 0 10
.1:1
1/=.1:1 1/ 1:1 =:1. So the growing part of A is 1:1
with
01
p
jjAn jj 101 times 1:1n for larger n. Problem Set 9.3, page 489
1 The iteration x k C1 D .I A/x k C b has S D I and T D I A and S 1 T DI A. Solutions to Exercises 93 2 If Ax D x then .I A/x D .1 /x . Real eigenvalues of B D I A have j1 j < 1
provided is between 0 and 2.
1
1
3 This matrix A has I A D
which has jj D 2. The iteration diverges.
1
1
4 Always kAB k kAkkB k. Choose A D B to ﬁnd kB 2 k kB k2 . Then choose A D
5 6
7 8 9
10
11
12 B 2 to ﬁnd kB 3 k kB 2 kkB k kB k3 . Continue (or use induction) to ﬁnd kB k k
kB kk . Since kB k max j.B/j it is no surprise that kB k < 1 gives convergence.
Ax D 0 gives .S T /x D 0. Then S x D T x and S 1 T x D x . Then D 1 means
that the errors do not approach zero. We can’t expect convergence when A is singular
and Ax D b is unsolvable!
101
1
with jjmax D 1 . Small problem, fast convergence.
Jacobi has S T D 3
3
10
"
#
1
03
1
GaussSeidel has S 1 T D
with jjmax D 9 which is .jjmax for Jacobi)2 .
01
9
1
a
0
b
0
b =a
1
Jacobi has S T D
D
with jj D jbc=ad j1=2 .
d
c
0
c =d
0
1
a0
0
b
0
b =a
GaussSeidel has S 1 T D
D
with jj D jbc=ad j.
cd
0
0
0
b c=ad
So GaussSeidel is twice as fast to converge .or to explode if jbc j > jad j/.
p
3/ 1:07.
Set the trace 2 2! C 1 ! 2 equal to .! 1/ C .! 1/ to ﬁnd !opt D 4.2
4
The eigenvalues ! 1 are about .07, a big improvement.
GaussSeidel will converge for the 1; 2; 1 matrix. jjmax D cos2 .=n C 1/ is given
on page 485, with the improvement from successive over relaxation.
If the iteration gives all xinew D xiold then the quantity in parentheses is zero, which
means Ax D b. For Jacobi change x new on the right side to x old .
A lot of energy went into SOR in the 1950’s! Now incomplete LU is simpler and
preferred. 13 uk =k D c1 x 1 C c2 x 2 .2 =1 /k C C cn x n .n =1 /k ! c1 x 1 if all ratios j=1 j <
i
1
1. The largest ratio controls the rate of convergence (when k is large). A D 01
10 has j2 j D j1 j and no convergence.
14 The eigenvectors of A and also A 1 are x 1 D .:75; :25/ and x 2 D .1; 1/. The inverse
power method converges to a multiple of x 2 , since j1=2 j > j1=1 j.
j
= 2 sin nj1
nC1
C
j
cos nC1 . Then
nC1 1/
C1/
sin .jnC1
sin .jnC1 .
The last two terms combine into 2 sin...
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This note was uploaded on 09/25/2012 for the course PHY 103 taught by Professor Minki during the Spring '12 term at Korea Advanced Institute of Science and Technology.
 Spring '12
 Minki
 Mass

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