Introduction to Linear algebra-Strang-Solutions-Manual_ver13

0 12 1 c 1 4 1 12 2 c 2 1 2 1 d 1 4 2 1 2 5 mean

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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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