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Unformatted text preview: c D .2; 3; 8/ the Ph.D. takes 4 hours .x1 C x2 C 2x3 D
4/ and charges $8. The teacher in the dual problem now has y 2; y 3; 2y 8 as constraints
AT y c on the charge of y per problem. So the dual has maximum at y D 2. The
dual cost is also $8 for 4 problems and maximum D minimum.
7 x D .2; 2; 0/ is a corner of the feasible set with x1 C x2 C 2x3 D 4 and the new
constraint 2x1 C x2 C x3 D 6. The cost of this corner is c T x D .5; 3; 8/ .2; 2; 0/ D
16. Is this the minimum cost?
Compute the reduced cost r if x3 D 1 enters (x3 was previously zero). The two
constraint equations now require x1 D 3 and x2 D 1. With x D .3; 1; 1/ the new Solutions to Exercises 86
cost is 3:5
optimal. 1:3 C 1:8 D 20. This is higher than 16, so the original x D .2; 2; 0/ was Note that x3 D 1 led to x2 D 1 and a negative x2 is not allowed. If x3 reduced
the cost (it didn’t) we would not have used as much as x3 D 1. 8 y T b y T Ax D .AT y /T x c T x . The ﬁrst inequality needed y 0 and Ax b 0. Problem Set 8.5, page 451
1 R 2
0 h i2
sin..j Ck/x/
j Ck
0 R 2
D 0 and similarly 0 cos..j k /x/ dx D 0
R 2
k ¤ 0 in the denominator. If j D k then 0 cos2 jx dx D . cos..j C k/x/ dx D Notice j 2 Three integral tests show that 1; x; x 2 1
3 are orthogonal on the interval Œ 1; 1 :
R1
1
.1/.x/ dx D 0; 1 .1/.x
dx D 0; 1 .x/.x 2
/ dx D 0: Then
1
3
1
2
2
2
2
2x D 2.x
/ C 0.x/ C 3 .1/. Those coefﬁcients 2; 0; 3 can come from integrating
3
f .x/ D 2x 2 times the 3 basis functions and dividing by their lengths squared—in other
words using aT b=aT a for functions (where b is f .x/ and a is 1 or x or x 2 1 ) exactly
3
as for vectors.
p
1
3 One example orthogonal to v D .1; 2 ; : : :/ is w D .2; 1; 0; 0; : : :/ with kwk D 5.
R1
R1
1
3
4
c x/ dx D 0 and 1 .x 2 3 /.x 3 c x/ dx D 0 for all c (odd functions).
1 .1/.x
R1
Choose c so that 1 x.x 3 c x/ dx D Œ 1 x 5 c x 3 1 1 D 2 c 2 D 0. Then c D 3 .
5
3
5
3
5
R1 R1 2 1
/
3 5 The integrals lead to the Fourier coefﬁcients a1 D 0, b1 D 4= , b2 D 0. 6 From eqn. (3) ak D 0 and bk D 4=k (odd k ). The square wave has kf k2 D 2 . 1
Then eqn. (6) is 2 D .16= 2 /. 12 C 31 C 51 C /. That inﬁnite series equals 2 =8.
2
2 1; 1 odd square wave is f .x/ D x=jx j for 0 < jx j < . Its Fourier series in
equation (8) is 4= times Œsin x C .sin 3x/=3 C .sin 5x=5/ C . The sum of the ﬁrst N
terms has an interesting shape, close to the square wave except where the wave jumps
between 1 and 1. At those jumps, the Fourier sum spikes the wrong way to ˙1:09
(the Gibbs phenomenon) before it takes the jump with the true f .x/: 7 The This happens for the Fourier sums of all functions with jumps. It makes shock
waves hard to compute. You can see it clearly in a graph of the sum of 10 terms.
p
8 kvk2 D 1 C 1 C 1 C 1 C D 2 so kvk D 2; kvk2 D 1 C a2 C a4 C D 1=.1 a2 /
2
4
8R
p
p
2
so kvk D 1= 1 a2 ; 0 .1 C 2 sin x C sin2 x/ dx D 2 C 0 C so kf k D 3 .
9 (a) f .x/ D .1 C square wave/=2 so the a’s are 2 =3 , 0, 2=5 , . . . (...
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 Spring '12
 Minki
 Mass

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