BACHELOR'S DEGREE PROGRAMME (BDP) Term-End Examination June, 2017

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1 No. of Printed Pages : 8 MTE-10 BACHELOR'S DEGREE PROGRAMME (BDP) Term-End Examination June, E- 27D ELECTIVE COURSE : MATHEMATICS MTE-10 : NUMERICAL ANALYSIS Time : 2 hours Maximum Marks : 50 (Weightage : 70%) Note : Answer any five questions. All computations may be done up to 3 decimal places. Use of calculators is not allowed. Symbols have their usual meanings. 1. (a) Perform two iterations of the Birge-Vietta method to find a root of the polynomial P(x) = 2x3 - x2 + 2x - 2 = 0. Take the initial approximation p c, = (b) Solve the system of equations x x x3 6 using LU factorisation method. Us eu11= u22 =u33 = 1. 5 MTE-10 1 P.T.O.

2 2. (a) Use Lagrange interpolation method to find the value of y, where x = 6, from the following table : 3 x y (b) Prove that : ,8 = (A + V) 2 (c) Solve the initial value problem y,_ y + 2x, y(1) = 2 y + 3x using third order classical Runge-Kutta method. Find y(1.2) taking h = (a) Determine the spacing h in a table of equally spaced values of the function f(x) = between 1 and 2, so that interpolation with a second degree polynomial in this table yields accuracy 5 x (b) Evaluate 2 X dx 1 x 2 + x + 1 using trapezoidal rule with 2 and 3 nodal points. Obtain the improved value using Romberg integration. 5 MTE-10 2

3 4. (a) The population of a town in the decennial census is given below. Estimate the population for the year 1915 : Year : x Population : y (in thousands) (b) Using three iterations of the inverse power method, find the eigenvalue nearest to 5.5, [1 5 of the matrix. Also, find the 2 4 corresponding eigenvectors. Assume the initial approximation to the eigenvector as v( ) = [ T. 5. (a) Perform two iterations of the Newton-Raphson method to obtain the approximate value ofṭ8. Assume the initial approximation to the root as 4. 4 (b) From the data x f(x) interpolate the value for f(3.2) using the Newton's backward difference formula. 6 MTE-10 3 P.T.O. F

4 6. (a) Find the solution of the system of equations 6x1 + 2x2 - x3 = 3 x1 + 3x2 + 2x3 = 3 x1-3x2 + 5x3 = 0 using three iterations of Gauss-Seidel method. Assume the initial approximation as x (o) 0.3, x ) = 0-6, x ( ) = 0-3. Find 2 3 the iteration matrix and hence, find the rate of convergence of the method. (b) Construct a fixed point iteration form x = g(x) for the equation x 3 + x2-1 = 0, so that the method converges in the interval 10, (a) Determine the constants a, 13, y in the differentiation formula y"(x0) = a y(x0 - h) + R y(x0) + y y(xo + h) so that the method is of the highest possible order. Find the order and the error term of the method (b) The iteration method xn+1 1 [5x 5N N2 + - n = 0, 1, 2,... 9 xn xn where N is a positive constant, converges to N1/3. Find the rate of convergence of the method. 4 MTE-10 4

5 chitisto4 -T4, 2017 trntig Li cr 9 04 : 1 1 ICI 71:r.tt -10 : 9t4 qui HAN : 2.1t 12 3TrEIWTA. 3 : 50 (Fff : 70%) 41? 47e- ev 517 -j drflef-47 / Fift arrtiw-e7 3 cwsirict F-2TRY rich uw-d- f cl,(-35c.?ej ht q 375T6- Tif 37E74 Hisifrq 3121ff / 1. () --N- r-d-r f4rt -4-Trq (IN) P(x) = 2x3 x2 + 2x 2 = 700 4rFq I 31TR t411c4 4.1 pc, = 0.5 #11-A7 (Pii),TH) faf x x2 4 qi0 W-fichtut -WM x3_ 6 EM Arr-4 I ull = u22 = u33 = 1 Mq 11 A117. I 5 MTE-10 5 P.T.O.

6 2. () ( R fart g RI i-11-iicirgi fftf-0-*1 y TIR Vc-I W-47, x = 6 t: 3 x y (IN) FITZ : 2 1 (A + V) = 2 (IT) ORE FIU rclfdd *4-111 fak A Y = y + 3x, y(1) = 2 3.T1t ITR 1=P11:41 -Q* A t* h = 0.2 c. ct-) y(1.2) Vci AtN7 I 5 3. () 1 2* 1:FO 1 f(x) = fr-dwf.) 3i-dT h 711c1 A177 -N1:114 TP- fez' 41c1T alca 3rq 5 x ff- ftwe tl \IR I 5 (1: f) 23 3 *TR Pie414 A 2 x dx itedch-1 W-A7 I 1 x 2 + x + 1 W-11chd-E Itl RIJII4-I TTIT I 5 MTE-10 6

7 4. (Ti) t zi-wseu *r -f1-415if4 Tf I ai 1915 : 4-44 : x \TH(4{Sel I : y (311t. 11) (t f) felt T1tI1 31T T 5.5* Pichdoil tiilh alr&4ffrqr ift ;Ho W4 3Tr04Titz* 37T tiro te.-1 = [ T cM -r d-ri 6 5. (T)k--e4-11:1TE4 fea TRifieit,is TfrqT 3Trif *rr--a-r al-r 4 HI-Ichk 4 ( -u) &go 31TT0 x f(x) - FO tv TT -5RTiTT TTT f(3.2) ITT 1TF *a -47R *t-r-4t I 6 MTE-10 7 P.T.O.

8 :11fai 1 tq. - ~ti~r~~~ fir 6x1 + 2x2 - x3 = 3 x 1 + 3x 2 + 2x 3 = 3 x1 3x2 + 5x3 = 0 fir.1wm. * I x (i. ) = 0.3, x( ) = x ( ) = r r ici a did atfirr-or Tim*'i -r--a-r I (t i) iv-fit:mu! x3 + X2 1 = 0 X = g(x) vcr ff % [0, 1] i 3Triwit-ff 4,1 7. () y"(x0) = a y(xso h) + y(x0) + y y(x,0 + h) R.K atw y X11 d*1-rqr aftr*--d-r Trnra I re. trq 311 ;11(1 *IF-4R I TTF1 14.N xn+1 1 5N N 2 9 5xn ± 2 5 x n x n n = 0, 1, 2, N 1/3 * rilTatd *th t, N.) tratt-it- t I MT -*t atnr:ruf 7 old I 4 MTE ,000