2025pr8 - EEQRmLHm. , MW "$mgvawgygmgvémma a)...

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Unformatted text preview: EEQRmLHm. , MW "$mgvawgygmgvémma a) WWW, &m%mmg , ;.b7) WLZWLL __;;g;._c;:_fl§C..m). . Wec&im £69 36% CE'Vf‘M’mCéGfifiL .{D-mgacdi-iflfnp? 3 Raw% . i‘o) 33. (L235) }. c) ?..L2t._w2\%... ' Lem w Cw” I IR?“ 1: X,“ 23. :+..% , arthafjém—Q . _ ~qu 1. 3.3.0., i, 50, 3 , W CL} “#1., . , . Q) 73:1th 0W0. w :33; 3, u) $014603 ML). .1312, t, n h w .. 2b.)..WMM ia :LWQ, xii“)? 27;). .anatonmmm Angwmygw “7;. :2 9,5. ax. .1 (5.5; .. 119).... ml) 113(5) ; ‘P __ L22? W is: : . . (3%?)003 _ _ w in?! k £Wn$£§iwh§ . _. r951 ficflwi‘éfi .. flail ...-€,m:(~£~c§~ 19%, km ' WQO’be/c/Rm wg 7 _.¢L_ a»; ._c..cm:;4;g§;g ' :2 .CSEWQWMWCK CQMW. \ QM . ' Qty CL \ - I i a My {Miami.v\.c\__..._t£mghom$ “054% W, N 6 Viwfi; kw: as .. (:3 +1???) M. CQRLLWQM , r 3 Jam, .. . . ‘ iii) é . .. . _ H Math 2025, Quiz #6 (Fall 2003) Name: 1) Denote by W the piano {(m,y,z) E R3 : r}; + y + z 2 0}. a) Find the formula for the orthogonal projection P : R3 _—2. W‘ P($=ZJ=~Z) : “i1” GDP?” 3) “Xeay mE)-‘><-7 +25?) m :: (xv-“(3‘) Qorm om Mikosomq (V3 LAD UL + {V} M2.) Main 3 1 Wu” 2 L42 . m van-Kai‘s Mk: (K, \““2\ Max “2. km “FCW‘W\ M Oftho‘fia’mk-Q @poth-{cw JIM/n 13(er . ~ 3. in) \mluz‘r 1+ HM.) : 6) man"; ‘Z+C__nm____2' 566) (CX,%’}\)CUJI*‘2—fl=¢ Rwy-22 (Cumm‘Caj-iiofl = W‘fi- ><+5 "'92:. iv) Mafia") : é (44‘?) 7(— a; (0”; ‘0 M ' X1. ~22 X“ W kw: “Qatar-3x43, gap—.2 _22 2x -~)’"‘$ “Jr "*‘ "a": - Q, = _--Z-—-Q : Wo‘ "" WW AM S‘aw CMWXMM form W COWAW. b) WThat is P(l,0., #1) = (/1 0E “(J) W61, (“AMI 9M2} 7%: We we. 3M (4'. (42);?wmo/i—1QIMM (gain/flaw 2) Let W’ be the space of functions on [0} 1) generated by the stop functions Mail/2) and xii/2,1]. Notice that those two fonctione form an orthogonal baeis for W and “Marl/2mg m “pm/.23”? : 1/2. Find the orthogonal projection of the function x 33:2 onto W. - \ M. “L 4...? PW“ :i‘fia m 0“ “‘5' "K H») 0‘) é” 3:- m “:3 1 ' (x l/P@X1)O(}: [Amagevmcafigx wen [0;4)3r>(£01,/¢) ) 2 i 4» LAWAEQ vaaW-ec'ggx 5M" 331103 Viagra“) V I 2 f = [9x 8:3 *ZCLXEXKQJX) + [2 92.” Emmi-3.1) H SSOIXCXX:2‘X\?O 3 “7 Recall: Let V be a vector space with inner product < 3- >. Let W C: V be a finite dimensional subspace. Let 101,.”ij be an orthogonal basis for W. Then the orthogonal projection Pm : V ——-> W is given by “will? H. llw'n-il2 If v E V then the point in W closest to v is exactly PW Let us recall how we get this result: 1) if the piont H) 2 51101 + 521.92 + . . . +- snwn E 'W is such that <vwwfiwj >20 foralljz hum, Then the coefficients sj are given by ‘ < ’U.’in > “will Whe? Let us see What our assumption < 7; ~ wywj >m O for all j m 1, . . . ,n implies: We take the inner product with each of the vectors leg and get using that the inner product is linear in the first factor: n < 1') _ 211,2ri >m< inn/'5: > —E Si < 101710;; > 1'21 But < when >= 0 iii # j and < inking? >33 ijl|2. Hence the above reduces to 0 m< v w 10,103; >=< v,in > wsj Hun-H2 or < 11,2113; > 53- 2 W ilell which is exactly What the above formula says. 2) If ’U i w l w, j = 1,. . . ,n then 1; w w is perpenticnlar to all points in the plane W. Why? How does an arbitary point a in the plain W looks like? By our assumption the vectors wl, . . . , ten are a basis. Hence u m @1101 + . . . + (1an for some numbers {23- E R (or aj E Hence TL <vww,u>:<niw,g ejwj> 2'21 because the inner product is conjugate linear in the second variable. Why is the orthogonal projection Pw(v) the point in the plane W closest to e? Let- u be another point in the plain. We want to show that llv * Plelll S [iv * UH 26 or «i which is the same (Why?) 7 Hr) — PVT/(3mg E H’U — UH- Let us just caicuiete this: H'U -" EH? 3 Hv "" PM?” + (PWW — 11»ng :< v # Pw(’U) + (Pr/12(1)) w u),v m PW(U)+ < Pw(?)) — u, PW(U) — u > :< v —-— Im/(21),“u — Pw(v) > + < ‘Pm/(o) — the — PW(v) > + < PWCU) ~41th * PW(U) > min < PW(o) m u,PW(o) —— u > 3 Hv "" PEA/(UNI? + lleW ~ UH?“ Where we have used that PIA/(o) e u is orthogonal to the plain W and that the lenght of a, vector is always positive. 1) Let V m R2 and let W be the line RH, 1). Find the orthogonal projection of the following vectors to W: Notice first that mi, 1W =1+ 1 x 2 a) PW(1,2) : Was) 2 girl) = (1571.5) MUN? b) row/(3, m5) 2 1) = 32 (13.1) = w(1,1) c) we 1) = 1) = 341, 1) : (45,45) 2) Find the point on the hue RQ, m1 closest to the point (5, 2). Sotution: T16 point on the line closest to (5,2) is the orthogonal projection of (5, 2) onto the line. Hence this point is given by <(5,2),(2,—1)> mii _ W(2"“1)““ 22+ (4)2(2’ 1) mime—1) The final answers is therefore: The point on the tine, closest to (5, 2) is the point (16/5, MS/S). 3) Find the point on the line 3y m 23: closest to the point (—47 Solution: The first thing we need to do is to find one vector 7 cell it to ~ on the line. If we take it : 1 then y : Hence (1, 2/3) is on the line. But then also (3? 2) is on the line. Let u; = (3, 2). 27 Then “10112 m 9 —1— 4 m 13. Hence the point on the fine, closest to {—4, 3) is given by < (~4,3),(3,2) > _ ~12+ 6 13 (3’2) ‘ 13 m6 m 1—3812) 4) Let V = R3 and let W be the piene generated by the vectors (1,0,1) and (0,1,0). Find the orthogonal projection of the following vectors to W: a) PW(1,3, #1) x b) PW(1,1,1) : c) PW(4, ~21) 2: (3:?) Solution: Notice first that the vectors (1,0,1) and (0, 1, G) are orthogonal. Furthermore 13010, 1);)? m 2 and 1|(011101112 *1 Hence for <(1,3,m1),(1,0,1)> <(1,3,w1),(e,1,0)> Puree, 1) = 2 (1,0,1) 1" 1 (0,1,0) 1 ——1 z 2 (1,0,1)+3(0,1,e) = (0,3,0) 13) We have [DH/(1,1,1) m L31(1,0,1)+1(0,1,0) (1,0,1) + (0,1,0) Ii x (1,1,1) Hence we get the point back again. That simply meens, that (1., 1, 1) 2 (1,0,1) + (G, 1,0) is in the planin spenneé by the vectors (1,0,1) and (0, 1, O). c) We get in the same way: 4 1 Pwe, —2, 1) 2 £411}, 1) — 2(e,1,0) 5 2(1,0,1) — (0,2,0) 5 5 * (NE)- 5) Find the point on the line R11, ~1, 2) ciosest to the points: 28 a) (4,2,w1) m(&—Lm) Solution: Recall that the point on the line generated by (1,—1,2) is given by the orthogonal projection, which we will denote by P. Next we calculate ML—meml+1+4me a)Weheve 42ml 1w}? 3 1 gfl(5 5 fi~2m2 2 1—12 6 (J 3) m0. b) In this case we get: PBrLunmffiijfigg£:Eai 33+231WLML3 6 7 =§e¢43) (1, M12) 6) Let W be the subspace in R3 generated by the vectors (1, 1, 0) and (l, “13 1). Find the point in W closest to the points: a) (1,2,3) b) {w]‘,0, 1) b) (1,0,1) Solution: First we notice that <(1,1,8),(1,AE,1)>21M~1:U. Hence the vectors (151,0) and (1,w1,1) are orthogonal. Then we calmllate the lenght of the generating vectors. Those are IKLLOW m 1+1 $2 and i1: Z 29 a) The point is given by < (1,2,3), (1,2,0) > 1,. 2 «to 31 i 1:: 3;; + 1)) Same calculation gives (32w, 338). (I) Here we get: 1 2 7 1 2 — 1 1 m — m m m u- . 2(5 7) Let V x PCQO, 1]) be the space of (real valued) piecewise continuous function on the closed interval [0,1]. Let W be the wavelet space generated by 9051) and (pill. Thus W is the space of all functions of the form 509(223) + 51<p('2t w 1). Find the orthogonal projection of the following functions fit) onto W: a) = 1 for all t. Mf@:% c) f 3 415 + 2 ermmwwwaae—u e) f0?) : 2%415) + 499(415 — 1) + 390(4t — 2) + 9<p(4t —— Solution: Let us first recall that the inner product in PC([0,1]) is given by <fe>=flfmflflfi In particular the norm of a function in POGO, 1 n WWm£f@%t ( is given by Let us start by finding the norm of toéljfi) x cp 2t) and @905) 2 99(21‘, — 1.). A simple calculation show that 1 0§t<06 0 else 1 ’ l aremvee:{ and l . i 0.5gt<1 ekwxeflm: 0 else Hence 1/2 2 1 lhmlmf “Newm— a 2 38 and 1 ll W2: M1531“ my}: 3 901 1/2 ‘ll/2 2 (1) 3) Here we simply notice (or evaluate) that f :2": 1 2:: gag + (pg) 6 V1. 2te<t<$ 2twé”(t)== { " 2 1)) First we notice that 0 else and 2t %‘<t < 1 fleflfi): 2W 0 else Hence 1/2 < f tag) > m/ 2tdt o _ 2 1/2 W 1 W 7: l0 1 and f2 2 1 = t l1/2 1 :1-— 4 _.§ “'4 Hence the grojection is given by 1/4 3/4 P t = " 2f; Wm 23 i 1 or) U¢r>+u2e ) 1 3 3 m’ 215 w 21:“1 . 2w£ )t-2¢( ) c) There are at least two different ways to solve this part. One way is to do the calculation directly or netice that the progection operator is linear. Hence by using (a) and (b) we get me + 2) : 2pm) + 213(1) 2 :3e@e+4ememn. 22<§Mmy+iM%—U)+2wwfl+eQe—n) (1) Let us first figure out what the products 90(413900215)E @(41‘. — 1)cp(2t), 99(4thth — 1), and 90(47? — 1)<p(2t m- 1). By drawing the graphs are evaluate directly we get 1 0 g t<:§ 0 else Memeexweem{ 31 1 S t < 0 else .MH Mir—d “‘ x {70(415 _ 1) m{ and cp(4t)<,0(2t w 1) : 90(4t W1)gp[2t m1) 3 U. The inner products are therefore "1/4 1/2 1 1 1 1 <f1eé”>:/ 1dt+/ Idtmg+§mggm 0 1/4 2 < f, to?) >m 0. The projection is therefore given by Pm = W. e) The same aneiysis gives in this case PU) m 392(215) + 699(21: — 1). (Do you recognize those numbers from the wavelet transform theory?) 8) Let V a: V2 be the space of functions of the form 3099(4t) + 3199(415 w 1) wt 529(th w 2) + 3399(4t — 3); 85, 31, 52,33 6 Let V1 be the subspace generated by @223) and 90(215 e 1). Finally iet 1% be the subspace of V; of the form toiMZt) + tit/4215 w 1). 21) Find the orthogonal projection of fit) :2 3099(421') + 3190(4t — 1) + 3290(41‘ — 2) + 3390(4t w 3) onto 1/1. 13) Find the orthogonal projection of 3099(41‘) + 5190(4t7 1) + 329.9(4t 7 2) + 539441: e 3) onto W1. Solution: Here is just the final answere: a) The projection gives the Haar wavelet transform Pv1(f)(t) : Wt) "i" 33) This is the detail—transform 80+Si 82+53 2 {pat —- 1)‘ SD M“ 81 PWitf)(f)= 2 #42233; 82. W ' 2 1/2(2t— 1). ...
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This note was uploaded on 11/23/2011 for the course MATH 2025 taught by Professor Staff during the Spring '08 term at LSU.

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2025pr8 - EEQRmLHm. , MW &amp;quot;$mgvawgygmgvémma a)...

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