No. of Printed Pages : 10 ArrE-10 BACHELOR'S DEGREE PROGRAMME (BDP)

r-1 4 Term-End Examination

1 3, June, 2014

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 upto 3 decimal places. Use of calculators is not allowed.

1. (a) Find an interval of unit length which contains the root of the equation x3

3x 5 = 0. Using Newton Raphsoes method, perform two iterations to find the approximate value of root.

(b) The area A of a circle of diameter d is given for the following values :

d 80 85 90 95 100 A 5026 5674 6362 7088 7854

Find approximate value for the area of circle of diameter 82 using Newton's forward difference formula.

3

4 MTE-10 1 P.T.O.

(c) Construct a fixed point iteration form x = g(x) for the equation x3 + x2 1 = 0 so that the method converges in the interval [0, 1]. 3

2. (a) Find the inverse of the matrix 2 2 4

2 3 2

1 1 1

using LU decomposition. 5

(b) The following table of values is given :

x 0.2 0.3 0.4 0.5 0.6

y(x) 1.8054 1.5769 1.2834 0.9483 0.5981

Using all possible values of h, find approximate value of y"(0.4) using the difference formula

y"(x) = 1 [y (x h) 2y (x) + y (x + h)]. 5

3. (a) Use Stirling's formula to find y35, given

Y20 = 512, y30

- = 439, v40 - v50 = 346, = 243. 5

MTE-10 2

(b) The Gauss Seidel method is used to solve the system of equations :

2 1 5 xl 15

4 8 1 X2 -21

4 1 2_ _

x3 _ _

1

Check whether the method converges for the given system of equations or not. 5

4. (a) Set up divided difference table for the following function and hence find the interpolating polynomial :

x 3 1 0 2 3

f(x) 9 5 3 11 33

(b) Use fourth order classical Runge Kutta method to approximate y2

when h = 0.1,

given that y = 1 when x = 0 and --Yd = x + y. 6 dx

5. (a) Use Lagrange's interpolation to find f(3) from the following table : 4

x 0 1 2 4 5 6 lx) 1 14 15 5 6 19

MTE-10 3 P.T.O.

(b) For the method

f'(x' = 2 1h I

, 3 flx0) + 4 f(xo + h) f(xo + 2h)) determine the optimal value of h based on the criteria : Max I Truncation error I =

Max I Round-off error

when f(x) = 1 , 1 x 2 and the 1 + x maximum round-off error in evaluating f(x) is 0.005.

6. (a) Use the Composite Trapezoidal rule to 0.5

fevaluate the integral f(x) dx where f(x)

0.1 is given by

x 0.1 0.2 0.3 0.4 0.5

f(x) 0.425 0.475 0.400 0.450 0.575 Integrate with all possible values of the step lengths. Improve the results obtained using Romberg integration.

(b) Express V1 + 11282 in terms of 62 and a constant. 3

6

5

MTE-10 4

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: iTfirrff

: titStiltlict) idW4t1u1

6414 : 2 eru2

3?0*-6-17 3IW : 50 (Tel : 70%)

Tie N'R FR. 4,07 / T 3rf4R7. Wiq. vvicia F277-4 uw P-026' *7 Hoc? f cici,25c,?e

n - 1

(c) If E A2 fk a A fn + b A fo then find the k = 0

values of a and b. 2

7. (a) Suppose that the roots of the equation x2 + px + q = 0 are a and b. A fixed point

iteration method is written as - xi.+1 = q

xi + p

Find the condition on a, b such that this iteration method converges to the root a. 4

(b) Using three iterations of the inverse power method, find the eigenvalue nearest to

1 5 5.5, of the matrix

. Assume the

2 4

initial approximation to the eigenvector as

A (o) = [0.6 0.51T.

MTE-10 5

(1:4) -PHRi (sio fli-11.t)kut -ftwrzr To. crA

TFITFf r WTI 4 dti '14)ti I TRU 2 1 5 x1 15

4 8 1 X2 21

4 1 2 x3 1

4N. *'rr--4R -17 .fro -ftc-zr (it; aTkErrtff .1c11 3421-or 5

4. (i) tsid 141-717ff atff Mr-ff'W Pa-rfcm ttN-R 311-( f ATiq Tff Aztr-47 4

x 3 1 0 2 3

f(x) 9 5 3 11 33

(w) e. RiurciFact I fqk T WiT cht y2 ti4zhiad *I.NR 7-0ft h = 0.1 t,

atIT Reu TrArr t 17-d

=x+y Ax= 0 TR dx y = 1 t I 6

5. (i) m11 )1I cbt 1-1rriR I Mr-K*1 A R3) W117 TM tri* : 4

x 0 1 2 4 5 6

f(x) 1 14 15 5 6 19

MTE-10 8

(TO chul x3 + x2 1 = 0 R-R x = g(x)* '411 rem a4Trir7 af-du

[0, 1] 3f d o 3

2. LUfa*AR. faRT chk4) 34T&kg 2 2 4

2 3 2

1 1 1

irr-AR 5

(IN) PHICI (go 1791 .tt Trzft t : x 0.2 0.3 0.4 0.5 0.6

y(x) 1.8054 1.5769 1.2834 0.9483 0.5981

aTK

Y"(x) = 1 [y (x h) 2y (x) + y (x + h)]

N ei 41 chk h A. 43-1-4 1 y"(0-4) 1:1171:Ifq 7114 *1NR I 5

3. () cti t y20 = 512, y30 = 439, y40 = 346

y50 = 243 ', c k.e.ioAt mq 41 ,:t-K y35 vd Jr4R I 5

MTE-1 0 7 P.T.O.

(w) f4RT 1

, 3 f(x0) + 4 f(x13 + h) f(xes

+ 2h)} f'(x' = 2h

Max I t;s1-1Q I= Max I i4 It I

011, m(4) h Pe.(11-1 1:17 Tff f(x) =

1 + 1

x , 1 e4 2 at f(x) 3.7-*--

3TRWIT i474-3Q 0.005 I

6. TizjT tv-m

n - 1 (TI) 44E A2 fk =aAfn +b4fn cia*b*

k = 0 Trfq 711c1 R I 2

7. TER -dINR chtul x2 + px + q = 0 * Irt

a 3 b Prvicl refrS Ful% i fT 1:1.

_ q

x j+1 = xi + p

a atR b -sed-4tT Tff 4=rr-4R 17E* fa% a *1 31-1I aftiftff I 4

"gra-d)14 v.ird WiT ct) 1 5

3.TT&O T 5.5 * Plebdc11-1 3-1-1-44qMq

2 4

Id -*F4R 3g0-4Erfqz vw) = 0.51T w 4'07 I 6

MTE-10 10 3,000

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