prove that √2+√3 is an irrational
Answers
Step-by-step explanation:
4x + y − 6
x + 2y + z − 7
y + 2z − 8
; Hessian =
4 1 0
1 2 1
0 1 2
.
(a) r3 = [1.5 2.75 2.625]T
, kg3k = 3.8017, error in position: [0.3 1.55 − 0.775]T
.
(b) r3 = [1.15 1.70 3.03]T
, kg3k = 0.7061, error in position: [−0.046 0.5 − 0.372]T
.
(c) r3 = [1.2 1.2 3.4]T
, kg3k = 0, error in position: [0 0 0]T
.
2. Comment: Using restart option, Fletcher-Reeves algorithm converges to a tolerance of 10−4
in 19 iterations.
3. p = 6.7, q = 6.9, r = 6.91. [Denoting the points as A, B, C etc, notice that AB and CD
are parallel. Then, DE, as extension of BD is the next (conjugate) direction. Next pair of
searches are then EF parallel to CD and FG parallel to DEDenoting a point on the line segment as s, a point on the triangle as t and the vector connecting
them as d, respectively; s = a + x1(b − a), t = p + x2(q − p) + x3(r − p); and
d = s − t = a − p + [b − a p − q p − r] x = a − p + Cx.
Minimize d
2 = x
T CT Cx + 2(a − p)
T Cx + ka − pk
2
subject to x1 ≤ 1, x2 + x3 ≤ 1, x1, x2, x3 ≥ 0.
KKT conditions: 2CT Cx+2CT
(a−p)+µ1
-
1 0 0
T
+µ2
-
0 1 1
T
−ν = 0; along
with feasibility conditions and µ1(x1 − 1) = 0, µ2(x2 + x3 − 1) = 0, νixi = 0, µ, ν ≥ 0.
A KKT point is a local minimum as well a global minimum. [The problem is convex.]
2. KKT conditions: x1 + x2 − π
2
x1x2µ1 − µ2 + µ3 = 0, x1 − π
4
x
2
1µ1 − µ4 + µ5 = 0 and gi(x) ≤ 0,
µigi = 0, µi ≥ 0 for i = 1, 2, 3, 4, 5. The area PQR in Fig. A.11 is the feasible domain. Point
P(12, 13.3) is a KKT point, while points Q(10.3, 18) and R(12, 18) are not.
3. Solution: x
∗
(15.81, 1.58), f(x
∗
) = 5, µ1 = 0.2, µ2 = 0. [Through enumeration and
verification with the KKT conditions, it is possible to conclude that g1 is the only active
constraint, i.e. g1(x
∗
) = 0 and µ2 = 0. Hessian turns out to be positive semi-definite.
1 1 1 1
1 1 −1 −1
1 −1 1 −1
1 −1 −1 1
.
Comment: For this problem, alternative solutions are possible, because the problem does not
ask for a Householder matrix.
2. (a) One eigenvalue is −1 with eigenvector w with the notation of page 73. All other eigenvalues
are 1 with corresponding eigenvectors forming an eigenspace orthogonal to w.
Determinant = −1.
(b) All eigenvalues are of absolute value of unity. First, there can be one or more unit (+1)
eigenvalue with real eigenvectors. Besides, on 2-d planes of rotation, there may be pairs of
eigenvalues in the form e
±iθ with associated eigenvectors in the form u ± iw. In a special case
of θ = π, u and w separate out as individual (real) eigenvectors. Finally, there may remain a
single negative eigenvalue −1 signifying reflection.
3. We cannot. With such an attempt, zeros set in left multiplication will not be preserved through
the right multiplication by the Householder matrix.
4. Eigenvalues of B : 0.59, 2, 3.41. Eigenvalues of C : −4.33, 1.29, 4.98, 7.05.
5. Five. [Examine the Sturmian sequence at zero.]qk+1, in that order, as above.
Finally, bn,n = q
T
n Aqn.
6. (a)
0.897 −0.259 0.359
0.341 0.921 −0.189
−0.281 0.291 0.914
. [Find unit vector c1 along the line. Assume c2 and c3
to complete orthonormal basis. Rotation matrix = C<
T
yzCT
. Analytical multiplication and
subsequent reduction can be used to eliminate c2 and c3. Alternatively, a consistent pair can
be chosen, e.g. through the Gram-Schmidt process (see Chap. 3), and the solution follows
from straightforward computation.]
(b) First verify that the given matrix is orthogonal and its determinant is +1. The analytical
expression of the matrix in part (a) shows that its trace gives the angle. Then the difference
of the given matrix and its transpose gives the axis. Alternatively, one can simply find its
eigenvalues and eigenvectors. Eigenvalues are 1 and e
±iθ
. Eigenvector corresponding to the
real unit eigenvalue is the axis of rotation and θ is the angle. The plane of rotation can
be developed either as the subspace orthogonal to the axis or through real combinations of
eigenvectors corresponding to the complex eigenvalues e
±iθ
.
(c) Direct problem: form orthogonal matrix C with pair(s) of columns spanning the plane(s)
of rotation, then < = CPT
12CT
. Inverse problem: Same as part (b) except that there is no
relevance of an axis, and there may be several plane rotations involved in mutually orthogonal
planes.
Comment: In higher dimensions, the second approach looks cleaner in each of the direct and1. Eigenvalues and eigenvectors: 0.620 + 0.785i, [−0.115 − 0.500i − 0.115 + 0.500i 0.688]T
;
0.620 − 0.785i, [−0.115 + 0.500i − 0.115 − 0.500i 0.688]T
; 1, [0.688 0.688 0.229]T
.
det(Q) = 1.
Yes, Q represents a rotation. [Check orthogonality.]
Plane orthogonal to the third eigenvector, or the plane of the first two eigenvectors, of which
the real plane can be obtained with the basis {[−1 − 1 6]T
, [1 − 1 0]T }.
2. 3.4142, 2, 0.5858 (after two sweeps, diagonal entries: 3.4134, 2.0008, 0.5858).
3.
5.00 3.00 0 0
3.00 −2.11 −3.28 0
0 −3.28 4.53 1.22
0 0 1.22 1.58
.
4.