Question

a) i. A numcrical scheme is known to perform best if the height of the uniform subinterval is small enough in the interval (0.2]. The equation that gives the height h of the uniform subinterval is given by f(h) = 3 +h-1=0 Use the fixed point method of finding roots of equations to find the height h for which the scheme performs best. Conduct only four (4) iterations. 6 Marks ii. find algebraically all the fixed points of the function f(x) = x In(x + 3) - I and describe their nature. 4 Marks b) Solve using the Gauss-Seidel method, the system of linear equations below, + 1032 2:13 1021 7 9 6 2.12 + 1024 Take initial vector x = 0 and do only four iterations. Compute also the residual, Ax - b 2 for cach iteration and explain whether or not your solution is converging 12 Marks! c) Write a MATLAB code for the problem in b above. Let your code output all the results you had in b including the residual computation 8 Marks

Solve b and c only

Answer 1

b)

Ans- (b) -2 lon 1002- 2037 9 -272 t 1013 6 m2 diagonally The coefficient mateix is not dominant los Reawranging - 9 - + 10xa-ang-7 -272 7 10 *3= 6 long na (1) кер 97 x 12 (2) K. (2) (3) - 0) ta к NKA 7 + 2 XR 10 (3) NK! 16 +20 421 10 (x), z (1) (3)) = OD a Initial grauss Ist approximation (1) 2 (970) 0.9 10 x 7.9 -0.79 10 10 (37 ni (4+ 0.9+ 2(0)) = 46 +240.79), 2 O 0.758

and afperoximation 2 o-14 ) : ng 0 -979 Tь (2) С, 11 | ++ (ре+ 1); 2 [o •158)) (о с • Чеѕ (+ = (р. 1995). - о - 78 99 10 3ud approx (0) - — 1.3- 9 + p. 495) - о: 97s (р (2) х, р + p. 9 9 5 г (о: 28 94 ) О • 951S (2) (в г (о • 9 575)) - О: 2915 10 Ч ордно. tol (1) 그나 а+ Ф• 9515) - О. 1957 () (+ + О• 99 55+ (о•74)ѕ)) о Ф• 9 599 1) 75 [ 6+2 (р. 9574) • о • 8 9 10

For ut Let iteration Ax-b 0-79 - 105.16 0 1.7095 HAxb) for 2nd iteration Ar-b = -0.1595 -0.0638 o Ian-b/l = 0.1718 3ad iteration An-b = -0.008 -0.0032 0 Han-bll 0.0086 4th iteration An-b= -000003987 - 001595 O 11 An-bll= -0.0004 CS Scanned with CamScanner
c)

x=0;
y=0;
z=0;
A=[10 -1 0;-1 10 -2;0 -2 10];
b=[9;7;6];
e=zeros(4,1);
for i=1:4
    x=(9+y)/10;
    y=(7+x+2*z)/10;
    z=(6+2*y)/10;
    X=[x;y;z];
    fprintf('%i th iteration gives:',i);
    fprintf("\n")
    disp(x);
    disp(y);
    disp(z);
    e(i)=norm(A*X-b)
end