Introduction to Linear algebra-Strang-Solutions-Manual_ver13

# X y z 0 ending in 0 not 1 in homogeneous coordinates

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Unformatted text preview: b) a0 D R 2 0 1 , 2 0, 0; : : : and the b ’s are 2= , 0, x dx=2 D  , all other ak D 0, bk D 2 =k .  to  or from 0 to 2 (or from any a to a C 2 ) is over one R 2 complete period of the function. If f .x/ is periodic this changes 0 f .x/ dx to R R0 R 0 f .x/ dx C  f .x/ dx . If f .x/ is odd, those integrals cancel to give f .x/ dx D 0 over one period. 10 The integral from 11 cos2 x D 1 2 C 1 cos 2x ; cos.x C  / D cos x cos  2 3 3 sin x sin  D 3 1 2 cos x p 3 2 sin x . Solutions to Exercises 87 2 32 32 32 3 1 0 00 00 0 1 6 cos x 7 6 sin x 7 6 0 0 1 0 0 76 cos x 7 76 76 76 7 d6 6 sin x 7D6 cos x 7D6 0 1 0 0 0 76 sin x 7. This shows the 12 6 76 76 76 7 differentiation matrix. dx 4 cos 2x 5 4 2 sin 2x 5 4 0 0 0 0 2 54 cos 2x 5 sin 2x 2 cos 2x 00 02 0 sin 2x 13 The square pulse with F .x/ D 1= h for x  h=2  x is an even function, so all sine coefﬁcients bk are zero. The average a0 and the cosine coefﬁcients ak are Z h=2 1 1 a0 D .1= h/dx D 2 h=2 2    Z 2 kh 1 kh 1 h=2 .1= h/ cos kxdx D sin which is sinc ak D  h=2 kh 2  2 (introducing the sinc function .sin x/=x ). As h approaches zero, the number x D kh=2 approaches zero, and .sin x/=x approaches 1. So all those ak approach 1= . The limiting “delta function” contains an equal amount of all cosines: a very irregular function. Problem Set 8.6, page 458 T 1 The diagonal matrix C D W W is † 1 D &quot; 1 1 # with no covariances (inde1=2 pendent trials). Then solve AT CAb D AT C b for this weighted least squares problem x (notice C t C D instead of C C Dt ): &quot; # &quot;# 0C C D D 1 01 1 C Ax D b is 1C C D D 2 or 1 1 b D2: D 2C C D D 4 21 4      32 6 C 10=7 T T A CA D A Cb D bD x D : 2 2:5 5 D 6=7 2 2 If the measurement b3 is totally unreliable and 3 D 1, then the best line will not use b3 . In this example, the system Ax D b becomes square (ﬁrst two equations from Problem 1):  01 11       C 1 C 1 D gives D : The line b D t C 1 ﬁts exactly: D 2 D 1 3 If 3 D 0 the third equation is exact. Then the best line has C t C D D b3 which is 2C C D D 4. The errors C t C D b in the measurements at t D 0 and 1 are D 1 and C C D 2. Since D D 4 2C from the exact b3 D 4, those two errors are D 1 D 3 2C and C C D 2 D 2 C . The sum of squares .3 2C /2 C .2 C /2 is a minimum at 8 D 5C (calculus or linear algebra in 1D). Then C D 8=5 and D D 4 2C D 4=5. Solutions to Exercises 88 4 0; 1; 2 have probabilities 1 ; 1 ; 42 1 4 and  2 D .0 1/2 1 C .1 4 1 1/2 2 C .2 1 /2 1 D 1 . 4 2 1 2 5 Mean . 1 ; 1 /. Independent ﬂips lead to † D diag. 1 ; 1 /. Trace D total D 2 . 22 44 6 Mean m D p0 and variance  2 D .1 p0 /2 p0 C .0 p0 /2 .1 p0 / D p0 .1 p0 /. 2 2 2 2 2 7 Minimize P D a/ at P D 2a1 2.1 a/2 D 0I a D 2 =.1 C 2 / recovers equation (2) for the statistically correct choice with minimum variance. 2 a2 1 C .1 2 2 2 0 A/ 1 AT † 1 †† 1 A.AT † 1 A/ 1 D P D .AT † 1 A/ 1 . 9 The new grade matrix A has row 3 D row 1 and row 4 D row 2, so the rank is 7. The nullspace of A now includes .1; 1; 1; 1/ as well as .1; 1; 1; 1/. Compare to the grad...
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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.

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