Graphics Programs Reference
In-Depth Information
30 2
|
z
| =
+
10 2
+
20 2
=
37
.
417
30
.
0
0
.
80177
z
1
v
=
=
417 =
10
.
0
0
.
267 26
|
z
|
37
.
20
.
0
0
.
534 52
Second iteration:
=
30
10
20
0
.
80177
37
.
416
z
=
Sv
=
10 0
50
0
.
267 26
24
.
053
20
50
10
0
.
534 52
34
.
744
37
|
z
| =
.
416 2
+
24
.
053 2
+
34
.
744 2
=
56
.
442
37
.
416
0
.
662 91
z
1
v
=
=
442 =
24
.
053
0
.
426 15
|
z
|
56
.
34
.
744
0
.
615 57
Third iteration:
=
30
10
20
0
.
66291
36
.
460
z
=
Sv
=
10 0
50
0
.
42615
20
.
362
20
50
10
0
.
61557
40
.
721
36
|
z
| =
.
460 2
+
20
.
362 2
+
40
.
721 2
=
58
.
328
36
.
460
0
.
62509
z
|
1
v
=
=
328 =
20
.
362
0
.
34909
z
|
58
.
40
.
721
0
.
69814
Atthis point the approximation of the eigenvalue we seek is
328 MPa (the
negativesign is determinedbythe sign reversalof z betweeniterations). This is
actually close to the second-largest eigenvalue
λ =−
58
.
39 MPa! By continuing the
iterative process we would eventually end up with the largest eigenvalue
λ 2 =−
58
.
λ 3 =
70
.
94
MPa. Butsince
are rather close, the convergence is too slow from this
pointon for manuallabor. Here is aprogram that does the calculationsforus:
| λ 2 |
and
| λ 3 |
% Example 9.4 (Power method)
S = [-30 10 20; 10 40 -50; 20 -50 -10];
v=[1;0;0];
fori=1:100
vOld=v;z=S*v;zMag=sqrt(dot(z,z));
v = z/zMag; vSign = sign(dot(vOld,v));
Search WWH ::

Custom Search