math294_hw7fa06 - New 201 ‘1 3 —2 and :E" is...

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Unformatted text preview: New 201 ‘1 3 —2 and :E" is the exact solution; the error a 1155*” is 0. 22. Using Fact 5.4.6, we find CE" = [ ] and I; —- A?“ = (1. This system is in fact consistent 30. We attempt to solve the system (:0+0c‘]=1,or 1 0 co+0c1=O 1 l) 0 C0 1 1= 1 e0+1c1=1 1 1 \ - . . C“ This system cannot be solved exactly; the least-squares solution 18 [ O] 2: [ MIN (01“ . The line that fits the data. points best is f*(t) = + Figure 5.18: for Problem 5.4.30. The line goes through the point (1, 1) and “splits the difference” between (0, O) and (O, 1). See Figure 5.18. C0 32. we want cl of f(t) : CD + 01t+ C2152 such that Figure 5.19: for Problem 5.4.31. 27 Z en +061+002 1 0 0 CO 27 O=Co+ 161 +1620r 1 1 1 cl : 0 0=co+261+462 1 2 4 c 0 0=Co+361+962 1 3 9 2 0 If we call the coefficient matrix A, we notice that ker(A) = {6} so * 27 CO 0 25.65 ‘ c1 = (ATA)‘1AT O = ~28.35 so f*(t) = 25.65 _ 28.35t + 6.75252. 02 O 6.75 Co 38. We want (:1 such that C12 110=C0+261+C2 .l 2 1 110 180 2 CO + 1261+ 06-2 1 12 0 C0 180 120 = C0 + 561 + C2 01‘ l 5 1 C] 2 120 160 2 co + 11C1+ C2 1 11 1 (:2 160 160 2 co + 6C1+ 062 1 6 0 160 (:0 * 125 The least—squares solution is c1 = 5 , so that w" 2 125 + 5/2. ~ 259. C2 ~25 For a general population, we expect c0 and c; to be positive, since cu gives the weight of a 5’ male, and increased height should contribute positively to the weight. We expect C2 to be negative, since females tend to be lighter than males of equal height. 10. A function g(t) 2 o, + bt + ct2 is orthogonal to f(t) = t if 1 ' i 1 2 .fi . m (fig) 2/ (01+ bt‘z + ct3)dt z [gtz + 35-263 +§t4L1= 3b = 0, that it, If I) _ 0. —1 Thus. the functions 1 and 1‘2 form a basis of the space of all functions in P2 orthogonal to f (t) = .. To find an orthonormal basis 91(t),gg(t), we apply Gram—Schmidt. Now 1 22—03% _ 9—; - 1111] = %/ W = 1, so that we can let 91(t) = 1. The“ 92“) 3 TV —<1~t§>‘fu ” lie-$11 “’ —1 €561? — 1) Answer: g1 (t) = 1,gg(t) : ——‘é—§(3f2 — 1) . at We start with the standard basis 1, t and use the Gramechmidt process to construct an orthonormal basis 91(t),gg(t). I 1 :1 = _ _ t—(l,t)1 _ t—é /0 dt 1, so that we can let gl(t) —— 1. Then 92(t) — W .. m : fiat — 1). Summary: 91(23) 2 1 and 92(t) 2 flat -— 1) is an orthonormal basis. b. We are looking for projP1(t2) : (g1(t),t2)gl(t) + (92(t),t2)gg(t), by Fact 5.5.3. 1 . 1 . 1 A 3 We find that (91(t),t2) = f t2 dt 2 and (92(t),t2) = (22:3 —- t2)dt 2 lg; so 0 0 that projp1 t2 2 % + 15(225 — 1) : t ~ See Figure 5.21. Figure 5.21: for Problem 5.5.16b. 1 1 2 1 20. a. , = [$1362] [2 8] [0] = 3:1 +2222 = 0 when 331: —2$2. This is the line spanned by vector _i . b. Since vectors [1] and ["2] are orthogonal, we merely have to multiply each of them 0 1 1 2 1 2 1 1 with the reciprocal of its norm. Now i [O] = [1 0] [2 8] [O] z 1, so that [0] 2 -—2 _ _ 1 2 —2 - ~2 [ I] _[ 2 ll,[2 8][ 1]f4,sothat [ I] . Thus [ , [ _i] is an orthonormal basis. ' 2 is a unit vector, and 1:2. 22. Apply the Cauchy-Schwarz inequality to f (t) and g(t) = 1; note that = 1: 2 . |<f,g>lslllelgH=HfllOHM/>2:[1sz ( melt) st<f(t>)2dt. 24. a. (f,g+h) :(f,g)+(f,h) =O+8=8 b. ug+hn = flg+h,g+h) : x/(g,g)+2(g,h)+(h,h)=\/1+6+h50:\/57 G. Since (fig) 2 0, = 1, and 2 2, we know that gig is an orthonormal basis of spam (f, g). Now proth = (gm) 53 + (9,1299 = w w + (9, h>g : 2f + 39. d. From part c we know that f, g are orthonormal, so we apply Fact 5.2.1 to obtain the third polynomial in an orthonormal basis of spa11(f, g, h): h—pl‘Oj,..h h—2f—3g __ h~2f-3g 2f “II—projuhll ‘ llh-2f—39ll 5 ‘ 5 one: g+§h Orthonorrnal basis: éf,g, —§f — £9 + é}; ...
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math294_hw7fa06 - New 201 ‘1 3 —2 and :E&amp;quot; is...

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