MATH 251 Exam 2 with Solution, Spring 2009

AMERICAN UNIVERSITY OF BEIRUT Mathematics Department-FAS MATH 251 TEST 2 SPRING 2008-2009 Closed Book, 75 mn STUDENT NAME SotlvutTioN ID NUMBER Problem | Grade 1 /15 2 710 3 714 4 /ii TOTAL 50 (b) -(4 points) Approximate f(1.2) using Newton’s Interpolating poly- nomial based on the given set of data D, and the coefficients found in a Aso 13 Porzsa() = “e +Aye (mw ms? + Myo (a —aedln a h Wat " 4 uy aN 2 (we nw Vu 02 ) = ae (m — moh amyl uz \ = (w- igen t cA.z ¥ f (1.2) = ween 10 1. Let A be the following invertible 3 x 3 matrix. 2-1 7 4 4 0 -7|* 23 343 (a) (5 points) Apply on this matrix, Gauss Elimination with the scaled partial pivoting strategy, showing the status of the matrix after each elimination, i.e. each of the pivot rows and the corresponding multipliers should be identified and circled. Update the index vector at each reduction. Iv= C4A,2,% =\ Rew ction A, neigh ‘A ees $=C+,4,% ma idiot inienenee C3,2,4 > =e Rds Lwot equals —» AN= (b) (5 points) Specify the elements of the 3 by 3 matrices P, L and U which verify: PA=LU )2\ Lecce IVs C3,2,4\= é: (e) Ra (c) (5 points) Using the LU decomposition of A, find the last row of A~!.(Do not compute the entries of A~ that are not required). AAT! si = SrA” =f Loa = 2 /s LY Le, ot, c3d= Tes,e2, er) - + cocmvel % © LU oe. =&3 2 Te ber Ars Snve “ 4 tia eee ee 3) LV c3 sea To @o ule ss. \orr Row 9 An’, eo \e -e3 2 by FS Ua = aa fete TacesTe 33 =&3 4 79 @® LUases : ler Ur =% Ua ayaa y &S ao \(>\ [2 \ au” a (4 | (2 , 2 }ide]r la 4370) 3 \ > 1 2 5 , a a =p 2s € a 3 ae (:) = i 39 o 4 -\3 wh \ o 9 3/3 ud 3 w) Fe) 3 was i a. wy (< 6 fa \(=)* (i\— 38 age-= = 3° 38 Aq ° o 38/% m3 ood kd 3 ia 7 7 = Br Kae SS i @ eset! Ler Ucz sy =o Solve Uvega oe OR is ‘ o\f Ys A 7 _ og ay 2) a \2(3) mR = 4" (*) rn) ce 4e 43 = l3 2. (10 points)The following algorithm takes as input an invertible matrix A, performs scaled partial pivoting and returns an index vector IV up- dated at each reduction, a scale row-vector s, and determines the pivot row and pivot element at each reduction. Fill in for each of the following the least number of Matlab commands. function{[IV, s , pivbt | = ScaledpartialPivoting(A) (i) - Create the Index vector IV [m a = Bize CA); LV = \:n, (ii) - Create the Scale vetcor s va te1i9 gta) = ACL); (iii) - Determine the Pivot Row at each reduction You lee 1: aH) pivds AC TMi) i ke) A. maxi = ACV) | s (Iv); (iv) - Update the Index Vector at each reduction f= Iv(k)} Ey («) = Ivey, ANC) 2 vr; end end (v) - Determine the Pivot Element at each reduction Pon Ce et A(zVG) ,u) = A (TV, &) [er vot for ya ktlin 4(sv@),;) = AGVO),a) — A (ZV(aj, k)¥ AlZv00),;) i end 3. (14 points) Derive the equations of the Natural Cubic Spline interpolant S(a), for the following set of data Do, then draw an approximate graph representing S(x). i 0 j1]2 \ — wie a 4 x; |-1/2|0] 1 wie t a ee f(z) | 1 [24-1 Mezed Wise M274 z Lo ahs SU imi ews (es ea —& ee Waa aewi => ore = SS 7 ie 2 Cytnc SFLINE ered on Nz. 3 Qo C-Yo ods Qala N= A +5 (nm +3, ) ~— A ah = Q as aS ee And Qo £0,433 Salwv= +3 3 Approximate Graph of S(x) y | sy Wo A 2</a o 2 |ul4 | 8/3 A 1%/qa] o wad 1 4. (a) - (7 points) The following set of data D, is associated with the 0 order Bessel’s functions of the 1st kind : a 0 1 2 3 4 Li 1.0 1.3 1.6 1.9 2.2 Yi = f(x) | 0.76519 | 0.62008 | 0.45540 | 0.28181 | 0.11036 Define the difference operators A*y; recursively, then construct the Dif- ferences -Table (the entries being the Difference functions {A*}) based on D4. Ay; = ose e Pe — a Mcosssreerersuenrnumnermnrnnnaiad i= 9,A,---% -4 k-) AK-A Aty, = cl MA weet! 4 ee a By ea te for k= phe EO ae Differences -Table : i Xy Mig || geese... M tereacces | oneneseeeas, A? ssiteamenes: | nowwencane Ae desseuten | axegessy) At seoseeene 0] 1.0 | 0.76519 x x x * t 1| 1.3 | 0.62008 Ay: a ONS - x % x 2| 1.6 | 0.45540 Als - 0.16468 Byo2-0.01987 M \ 3 ze 3]1.9]o.28181) Ayas-ONSND YD o0ongs [Pre= 0.01066 4 \ 3 20, 3 4} 2.2} 0.11036] Sys z-0,.migiO'y~, o.o0214 DM ponos Ay. peed] _Nyo z= 0.91454 j A¥e= Oy - Myo Alyo= wey? => OWS Bro = Ay = 0.01066 A’ yp = ye-4 7 7 O.MNOE Abus Aha -My 0 008911 N 4 4 4s -\e =. 617 354 i Atya = Nys- qo? O,002\y Fi bAy = a Ay, I 43 = anys = 0-18 14S ) eos Alyse Sy) - Sys ur nein =