QUESTION ONEUse newtons interpolation divided difference formulae algorithm 3.2 to construct interpolation polynomial of degree one, two and three for the following data. Approximate the specific value using each of the polynomials of degree one, two and three for the following data. Approximate the specified value using each of polynomials.

Let

1st divided difference

2nd Divided difference

2rd Divided difference

QUESTION TWOThe composite Simpsons rule to approximate the integral use

SOLUTION

00.237425780.548311360-2.1932484-5.9356445-9.8696044

QUESTION THREEUse the Runge-Kutta for Systems Algorithm to approximate the solutions of the following higher-order differential equations, and compare the results to the actual solutions.

If

Question4Use S0R method with =1.2 to solve the linear system with a tolerance TOL= in the l 4x1-x2-x4`=0-x14x2-x3-x5=0-x24x3-x6=0-x14x4-x5=-6-x2-x44x5-x6=-2-x3-x5+4x6=6 AnswerUsing w= 1.2

Xi=(1-w)+For x

(1)

(2)

.(3)

(4)

(5)

(6)

kX1X2X3X4X5X6

0111111

10.42.020.7062.020.7122.0254

21.1321.8611.024721.94921.008282.00482

30.916662.0126981.00031141.9876420.9998921.9990970

41.016772.00255241.00043252.00747021.00275752.0011376

50.99965282.00034241.00035751.99922910.99966121.9997781

60.99994091.99991940.99983782.000034810.99998751.9999920

Checking for Error

Ans==4.0284 x

Hence TOL =is satisfied

5. Find the first two iterations of the SOR method with for the following linear systems:

Answer+ cos(x1x2)=0=G ()=X(K-1) J(x(k-1))-1 F(x(k-1))

Jacobi matrix is:

Using The iterative method produces the table belowk

022

11.9686825641.4789055410.521094459

21.7227713371.6422622570.245911227

31.7999277181.7519248330.109662576

41.7723831111.7711625500.027544607

51.7724556931.7724520381.89488 x

61.7724572051.7724504921.5460 x

71.7724538511.772453851

81.7724538511.7724538510

-And satisfies the error bound

Question6function [alpha,beta,gamma,delta]=finite_function(x,h)alpha=-1-(h/2)*p(x);beta=2+(h^2)*q(x);gamma=-1+(h/2)*p(x);delta=-(h^2)*r(x); function p=p(x)%p=-2/x; %Example 1/P663%p=-3; %11.3.3.a/P665%p=-4/x; %11.3.3.b/P665%p=-(x+1); %11.3.3.c/P665%p=1/x; %11.3.3.d/P665p=-(x+1); function q=q(x)%q=2/(x^2); %Example 1/P663 q=2; %11.3.3.a/P665 %q=2/(x^2); %11.3.3.b/P665 %q=2; %11.3.3.c/P665 %q=3/(x^2); %11.3.3.d/P665 function r=r(x)%r=sin(log(x))/(x^2); %Example 1/P663 r=2*x+3; %11.3.3.a/P665 %r=(-2/(x^2))*log(x); %11.3.3.b/P665 r=(1-x^2)*exp(-x); %11.3.3.c/P665 %r=(log(x)/x)-1; %11.3.3.d/P665 disp('Finite difference method for linear problems') clearformat long a=input('Input a=');b=input('Input b=');ya=input('Input y(a)=');yb=input('Input y(b)=');h=input('Input h='); NumberInterval_First=(b-a)/h;NumberInterval=fix(NumberInterval_First);if NumberInterval_First-NumberInterval==0 NumberUnknow=NumberInterval-1; Result_matrix=zeros(NumberInterval+1,2); Result_matrix(1,1)=a; Result_matrix(1,2)=ya; Result_matrix(NumberInterval+1,1)=b; Result_matrix(NumberInterval+1,2)=yb; %Result_matrix(1,3)=g(a); %Result_matrix(1,4)=abs( Result_matrix(1,3)-Result_matrix(1,2) ); %have not a,b in array x(i) x=zeros(1,NumberUnknow); for i=1:NumberUnknow x(i)=a+i*h; Result_matrix(i+1,1)=x(i); end A=zeros(NumberUnknow); b=zeros(NumberUnknow,1); for i=2:NumberUnknow-1 [A(i,i-1),A(i,i),A(i,i+1),b(i)]=finite_function(x(i),h); end [temp1,temp2,temp3,temp4]=finite_function(x(1),h); A(1,1)=temp2; A(1,2)=temp3; b(1)=b(1)+temp4+(-1)*temp1*ya; [temp1,temp2,temp3,temp4]=finite_function(x(NumberUnknow),h); A(NumberUnknow,NumberUnknow-1)=temp1; A(NumberUnknow,NumberUnknow)=temp2; b(NumberUnknow)=b(NumberUnknow)+temp4+(-1)*temp3*yb; solution_w=A\b; for i=1:NumberUnknow Result_matrix(i+1,2)=solution_w(i);% Result_matrix(i+1,3)=exact_func(x(i));% Result_matrix(i+1,4)=abs( Result_matrix(i+1,3)-Result_matrix(i+1,2) ); end else disp('h is not appropriate because [(b-a)/h] is not integer') end disp(' x(i) Finite_difference_w(i) ') Result_matrix % filename = 'Numerical_ass.xlsx';% xlswrite('Numerical_ass.xlsx',A)

>>finitediffFinite difference method for linear problemsInput a=0Input b=1Input y(a)=-1Input y(b)=0Input h=0.1 x(i) Finite_difference_w(i)

0 -1.000000000000000 0.100000000000000 -0.814229721703067 0.200000000000000 -0.654773599432834 0.300000000000000 -0.518308412475371 0.400000000000000 -0.401904443119574 0.500000000000000 -0.302980806647425 0.600000000000000 -0.219265711250556 0.700000000000000 -0.148761111240269 0.800000000000000 -0.089711275448616 0.900000000000000 -0.040574844868567 1.000000000000000 0QUESTION SEVENApproximate the solution to the following partial differential equations using the backward difference algorithm.

SOLUTION

becomes

becomes

Therefore from 1 substituting all these valueEquation1

Equation2

Equation3

Equation4

Equation5

Equation5

Equation6

Hence equation

So the backward substitution ijxiwwijU(xi,w)

110.50.050.632950.65204

211.00.050.895130.88394

311.50. 050.632950.62504

120.50.10.566570.55249

221.00.10.801260.78134

321.50.10.566570.55249