BACHELOR'S DEGREE PROGRAMME (BDP) Term-End Examination December, 2012 ELECTIVE COURSE : MATHEMATICS MTE-10 : NUMERICAL ANALYSIS

Transcription

1 No. of Printed Pages : 11 MTE-10 L. BACHELOR'S DEGREE PROGRAMME (BDP) Term-End Examination December, 2012 ELECTIVE COURSE : MATHEMATICS MTE-10 : NUMERICAL ANALYSIS Time : 2 hours Maximum Marks : 50 (Weightage 70%) Note : Answer any five questions. All computation may be done upto 3 decimal places. Use of calculator is not allowed. 1. (a) Find an interval of unit length which 3 contains the smallest positive root of the equation f (x)= x3 5x +1= 0. Taking the end points of this interval as initial approximations, perform 2 iterations of the secant method. (b) Evaluate the integral 5 d x I= J O 3 + 2x using trapezoidal rule with 2 and 4 subintervals. Determine the minimum number of subintervals required if the error in magnitude is less than MTE-10 1 P.T.O.

2 (c) A particle is moving along a straight line. 2 The displacement x at sometime instant t is given below t x Find the velocity of the particle at t =3. 2. (a) Perform 3 iterations of the Gauss-seidel 4 iteration method to solve the system of equations. 4x + 2z = 6 5y + 2z = 3 5x +4y +10z =11 Take the initial approximation as x(()) = 1.5, y(0) =-- 0.6, z( ) = 1.1. (b) Using divided differences, show that the 3 data. x f (x ) represents a second degree interpolating polynomial. Hence, obtain the polynomial. (c) Consider the Taylor series expansion for the 3 5/ function f (x) = X/2 in [ 1, 1] about x =0. Check if the bounds for the error (i) R2(x) (ii) R3(x). exists. Give justification for your answer. MTE-10 2

3 3. (a) Find an approximate value of y (1.2) for the 6 initial value problem y1= x2 + 2y2, y (1) =1 using the classical Runge-kutta fourth order method with h = 0.1. (b) Prove that 11 = \ I where 6 is the 2 central difference operator andµ is the mean operator. 1 (c) For the function f (x) =- x, show that 2 second order divided difference based on the points x0, x1, x2 is 1 f I x o, xi, x21 = xo xl x2 4. (a) Using Taylor series third order method solve 5 the initial value problem y1 = x + y2, y (0) =1 upto x = 0.4 with h = 0.2. (b) The solution of the system of equations 5 (1 4 (x) 6 2) (y) 5 8 is attempted by the, Gauss - Jacobi and Gauss - Seidal iteration schemes. Set up the two schemes in the matrix form. Will the iteration scheme converge? Justify your answer. 'E-10 3 P.T.O.

4 5. (a) Solve the system of equations 5 4x + y + z =4 x +4y 2z = 4 3x + 2y 4z = 6 using LU factorization method. Take the diagonal elements of U as 1. (b) The equation f = 18 x3-33x2 + 2x +5=0 3 has 3 real roots. Find the intervals which contain each of these roots. Perform 2 iterations of the bisection method to obtain the negative root. (c) Show that A + V A V where A and A V are the forward and the backward difference operators, respectively (a) Find the largest eigen value in magnitude of the matrix A = using the power method. Take the initial approximation to the eigenvector as z0)= ]T and perform 4 iterations. MTE-10 4

5 (b) Using Lagrange interpolation and the 3 following data : x f (x ) find the approximate value of f (1). (c) Using synthetic division, find f '(3) where 2 f (x) = X5 3X4 2X (a) Solve the system of equations 3 x + 2y + 3z = 5 2x + 8y + 22z = 6 3x + 22y + 82z = 10 Using Gauss - elimination method with 5 pivoting. (b) The method 1 N N2 X n+1 = 9 5xn + n Xn 2 xn 5 n =0, 1, 2,... where N is a positive constant, converges to N/ 3. Find the rate of convergence of the method. MTE-10 5 P.T.O.

6 (c) Find the interval which contains all the 2 eigen values of the matrix A = using Gerschgorin bounds MTE-10 6

7 717.&f-io rn~ 31Tai chl4ch1-1 Qt-th- t-1,11.1 TRI-arr -kfizrt, 2012 :Tifurd 714.t : 443e4ircich cw;qui 77-4-q: 2 Firr - 37ṈWT=I 3W : 50 *2- : f.77 17NTRqd!-H Or,fq I Tr2fi37firwN7/ TP-1T-q1 047 Timm apt rch 3T: t I (TO. chl 70% ) `I --@fa?-i chi Tr*If WT 1. (a) RM-T c T q TIM \TR 3.H 4-11ch (TIT f (x)= x3 5x +1=0 t Tr-4A TilTīTW T chi 3# RE tic T t 31:1 3T 1-) 3Trfq cht tkwr fafg c ' 1-f-A7 (b) td7T chc-1-1 t--1 4-KiGi 5 figg "g1t1 dx I= x fir wte. HI.1 71Icf ch) c clfqfiT4rch d chl y f t 4-11u chl-i t I MTE-10 7 P.T.O.

8 (c) to-r t'k-ur tl TOTP=Ftt -ci{, 2 chur cni fa-t-21-frffx f-zrf nui t : t x t -cr -11T chl ( c1 I 2. (a).(9 41 *01 fi-*-17 4x+2z=6 5y+ 2z= 3 5x + 4y 1-10z = 4 ml Sri t 1-c,N fq.fn X11-1 rricirtigitlf-a7 I tli:?iche.i 11 x(0)=1.5, y((l) = 0.6, 2(1)=1.1 I (b) fdl-ttf-a7 3 chl 7t7T.T-{t f T 14, 3-1T-Wt 3 x f (x ) f-g-dtzi f3f-d-41-f c ritc\ryi t 31m gild s I (c) x=0 tic 3-f-d7T 1, Th- TS f (x) = x 3 chi cl 411a rch chi fd. (i) R2(x) R3(x) t -crft-qm t 3174 MTE-10 8

9 3. (a) h=0.1 rd d tc;' -17 farq 6 gill3-trfq TPWIT yl = x2 + 2y2, y (1) =1 t IC1 i. y (1.2) 11?1 oh d TIN 79. I (b) ft =1Z TIF-r-A7fc : 2 11, = \ atd7tii Itch, 3 jj grrzi 4enRen (c) -cfcl 1 f (x) l itqt5is fe f 4-34-I 2 x x x atrinf-cd fir fqitrf-a-u 3f-d-T o, 1, 2 f xo, xi, x2 = 1,2 t1 4. (a) 7171 ('-R 44 PARTART h= 0.2 c cht 5 x=0.4 cich3-itfq :11W-IT yl = x + y2, y (0) =1 TT 6(.1 511H.Ttr--4R I (b) TT1Z11.-IT 771Ifff fdfqz4 5 glt1 f.1h chtul fiwp4 C6 4 ( x (5) 2) y) 8) -5rrff faftrz4 en-1 3 fi R.Trfcru -4-F---4R fqfirzer 34-filTrrf-cd 1d1? 3T 3-it (*CI MTE-10 9 P.T.O.

10 5. (a) L/ t. 741# fqtcri 31 d-tq ch ,(, LurcV-1)\A-1 fqfq- ch fiwp4 4x+y+z=4 x+4y-2z=4 3x + 2y 4z = 6 chf -1;17-ci ; 5 (b) chtui f (x) = 18 x3 33x2 ± 2x +5=0 t 3 cik-ci ; TE-0-3Tri-d7 affft c1 -WT4 TP1-frr9 fqfq t (c) : 2 A + V V V A 4,4-;w4-; t' 31R v 3 1=1Vr: 3-111c1( 34rT 1 Icl 3ta7 6. (a) 3-1T-- A t -cr-ft4t-cri P 41-{ 5 TITM -11c-1 t Fri )-1 I ci fdf -f TIVIFT-NTtrR7 A = C 0 2_, zo) = [I 1 ; MTE-10 10

11 (b) riukri3-tdi-79' 3 F-11-1 Fri RA d 31T-* 3 x f (x) chit f (1) 4, (c) ch f T T =hf 70T f ' (3)71i 2 6 f (x) = X5 3X4 2X (a) m1rich IIIi FCIO)4 Ṉfq A F-P-irrirod 3 f9-*-71 ail Sri : x+2y+3z=5 2x+8y+22z=6 3x + 22y + 82z = -10 (b) fqfg 5 1 X n +1 = 9 5x, + 5N - x 2 N2 x11 5,n =0, 1, 2,... 6 N q q7 t N A 31)7 3-1-Ṉ fit ~lcl1 ti fqfq eb1 3-TfiTATur (c) Tr-F-Iftft 13417T 34TV A= tici Zip 317ffiTrt 4-> 3tdf-47.a-ffr MTE-10 11