MAE 502Engineering Analysis IIProblem Set No. 311/04/2014

Ong, Anthony 498911.95%Define the entries of the coefficient matrix and the right-hand side%vecotr as inline functionA11 = inline('2*x','x','y'); A12 = inline('2*y','x','y');A21 = inline('4*x','x','y'); A22 = inline('6*y','x','y');f1 = inline('x^2+y^2-1.2','x','y');f2 = inline('2*x^2+3*y^2 - 3','x','y');tol = 1e-4; kmax = 10;v(:,1) = [0.25;1];for k = 1:kmax, A = [A11(v(1,k),v(2,k)) A12(v(1,k),v(2,k)); A21(v(1,k),v(2,k)) A22(v(1,k),v(2,k))]; b = [-f1(v(1,k),v(2,k));-f2(v(1,k),v(2,k))];% Check to see if a root has been found, while two successive vectors do% not satisfy terminating condition due to the nature of function(s) near a root if abs(b(1)) < tol && abs(b(2)) < tol, root = v(:,k); return end delv = A\b; v(:,k+1) = v(:,k) + delv; %update solution if norm(v(:,k+1)-v(:,k))> v v = 0.2500 1.3250 0.8889 0.7819 0.7746 0.7746

1.0000 0.8000 0.7750 0.7746 0.7746 0.7746

12.16

function [a2, a1, a0] = QuadraticRegression(x,y)% The user-defined function calls [a2 a1 a0] = QuadraticRegression(x,y)% The method uses the quadratic least-squares regression approach to find% the second-degree polynomial that best fits a set of data.Sumx = sum(x); Sumxx = sum(x.*x); Sumx3 = sum(x.^3); Sumxxxx = sum(x.^4);Sumxy = sum(x.*y); Sumx2y = sum(x.^2.*y);Sumy=sum(y);n=length(x);A = [n Sumx Sumxx;Sumx Sumxx Sumx3;Sumxx Sumx3 Sumxxxx];b = [Sumy;Sumxy;Sumx2y];c=A\b;a0 = c(1,1); a1 = c(2,1); a2 = c(3,1);% for i = 1:n,% l(1) =a2*x(i)^2+ a1*x(i) + a0;% endxnew=linspace(0,1.6); m=length(xnew);l = zeros(100,1);for i = 1:m, l(i) =a2*xnew(i)^2+ a1*xnew(i) + a0;endplot (x,y,'o')hold onplot (xnew,l)end>> [a2 a1 a0] = QuadraticRegression(x,y)

a2 =

1.9554

a1 =

-0.8536

a0 =

2.9777

12.19function [a1 a0] = LinearRegression(x,y)n = length(x);Sumx = sum(x); Sumy = sum(y); Sumxx = sum(x.*x); Sumxy = sum(x.*y);den = n*Sumxx - Sumx^2;a1 = (n*Sumxy - Sumx*Sumy)/den;a0 = (Sumxx*Sumy - Sumxy*Sumx)/den;l = zeros(n,1);for i = 1:n, l(i) = a1*x(i) + a0;endplot (x,y,'o')hold onplot (x,l)endfunction [a2, a1, a0] = QuadraticRegression(x,y)% The user-defined function calls [a2 a1 a0] = QuadraticRegression(x,y)% The method uses the quadratic least-squares regression approach to find% the second-degree polynomial that best fits a set of data.Sumx = sum(x); Sumxx = sum(x.*x); Sumx3 = sum(x.^3); Sumxxxx = sum(x.^4);Sumxy = sum(x.*y); Sumx2y = sum(x.^2.*y);Sumy=sum(y);n=length(x);A = [n Sumx Sumxx;Sumx Sumxx Sumx3;Sumxx Sumx3 Sumxxxx];b = [Sumy;Sumxy;Sumx2y];c=A\b;a0 = c(1,1); a1 = c(2,1); a2 = c(3,1);% for i = 1:n,% l(1) =a2*x(i)^2+ a1*x(i) + a0;% endxnew=linspace(min(x),max(x)); m=length(xnew);l = zeros(m,1);for i = 1:m, l(i) =a2*xnew(i)^2+ a1*xnew(i) + a0;endplot (x,y,'o')hold onplot (xnew,l)end>> x = [0 1 2 3 4 5]; y = [2 4.1 5.5 6.7 8.8 9.1];

>> [a1 a0] = LinearRegression(x,y)

a1 =

1.4514

a0 =

2.4048

>> [a2 a1 a0] = QuadraticRegression(x,y)

a2 =

-0.1107

a1 =

2.0050

a0 =

2.0357

12.36

function yi = NewtonInterp(x,y,xi)%n = length(x); a = zeros(1,n); a(1) = y(1);DivDiff = zeros (1,n-1);for i = 1:n-1, DivDiff(i,1) = (y(i+1) - y(i))/(x(i+1) - x(i));endfor j = 2:n-1, for i = 1:n-j, DivDiff(i,j) = (DivDiff(i+1,j-1) - DivDiff(i,j-1))/(x(j+i) - x(i)); endendfor k = 2:n, a(k) = DivDiff(1,k-1);endyi = a(1);xprod = 1;for m= 2:n, xprod = xprod*(xi-x(m-1)); yi = yi + a(m)*xprod;end>> x = [1 1.5 2.5 3 4 4.5 5 6]; y = [1.3956 1.6487 2.3010 2.7183 3.7937 4.4817 5.2945 7.3891];

>> format long

>> yi = NewtonInterp(x,y,4.25)

yi =

4.123376224998191