This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 leastsquares 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+ 062 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 + 35263 +§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 CauchySchwarz 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~2f3g 2f
“II—projuhll ‘ llh2f—39ll 5 ‘ 5 one: g+§h Orthonorrnal basis: éf,g, —§f — £9 + é}; ...
View
Full
Document
 Fall '05
 HUI
 Math, Linear Algebra, Algebra

Click to edit the document details