Question

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sthat this n is hardly different from the scalar version in Section 4.3 . The root estimates are stored as columns in an array . The Newton step is calculated using a backslash. The function norm is used for the magnitude of a vector, instead magnitude of a scalar. Function 4.5,1 (netonsys) Newron's method for a system of equations function x=newtonsys (f,x1) 2 % NETONSYS Newton's method for a system of equations 3 % Input : function that computes residual and Jacobian matrix initial root approximation (n-vector) 5%x1 6 % Output array of approximations (one per column, last is best) %x 9 %Operating parameters. 10 funto1=1000.eps ; xto1= 1000 * eps; maziter.49; 13 [n]=f(x1); 16 7 while (norm(dx)xtol) && (norm (y) >funtol)&&Ck < maxiter) 18 dx -(J\y); % Newton step 20 21 22 23 24 k=k+1 ; [y,J]=f(x(: ,k)); end 25 if kemaxiter, warning('Maximum number of iterations reached. '), end

Answer 1

Matlab code for system of equations clear all close all All initial guess for x1 %solution using newton systems x = newtonsys ( @( x1 ) f(x1 ) , x1) ; fprint f ( 'For initial guess x-82.2f ' y 82.2fn',... 2.2f , z- 82.2f and lambda- fprint f ( \nroot of the equation is x-%2.2f , y-%2.2f ,z- %2.2f and Lambda-%2 .2 f \n' , x (1,end), x (2,end),x (3,end), (4,end)) xl 50:40:30;1001; solution using newton systems x = newt on sys ( @ ( x1 ) f (x1 ) , x1) ; forint f("\n\nFor initial guess %2.2f\n', x=%2.2f y=%2.2f %2.2f and lambda- , z= , x1 (1), xl (2), xl (3), x1 (4)) fprint f (.inroot of the equation is x-%2.2f , y-%2.2f ,z» %2.2f and lambda-%2 .2 f \n' , . * . x (1,end), x (2,end), x (3,end),x (4,end)) 응음움各各各各各各各各各各음음各各음음各各음음各各음음各各음음各各음음各各음음各各음음各各음음各各음음各各음음各各 Matlab function for Newton system function x - newtonsys(f,x1) funtol 1000*eps; xtol-1000 *eps maxitr-40; dx-Inf; k-1 while (norm( dx) > xtol) && (norm(y)> funtol) && (k< maxitr) ly, j]- f (x,k)) end if k--maxitr warning 'Maximum number of iterations reached. ) end end 888888888888888388888888888888888888888888888

%function for jacobian and residual function ty,j1- f(xl) Y (2,1) - x1(2)-4-(x1(4)*x1 (2))/16, y(3,1x1(3)-3-(x1 (4)x1 (3))/9 Y(4,1(((xl(1)).2)/25)+(((x1 (2)) 2)/16)+0x1 (3)).2)/9)-1i j= [ 1-((x1(4))/25) 0 0 (-x1(1))/25; 0 1-((x1(4))/16) 0 (-x1(2))/16;. 0 0 1-( (x1 (4))/9) (-x1 (3))/9;... (2*x1 (1))/25 (2*x1(2))/16 (2*x1(3))/9 01 end 용음음응용음음응용음음응용음음응용음음용용음음용용음음용용음응용용음응용용융응용용음응용욤융응용욤응응용욤응응용욤 For initial guess x-5. 00, y-4.00,z- 3.00 and lambda-1.00 root of the equation is x-3 .42 , y-2.33 ,ZF 1.32 and lambda:-11.51 Warning: Maximum number of iterations reached. For initial guess x-50.00, y 40.00z30.00 and lambda-100.00 root of the equation is x#6.04 , y-16.58 ,ZF-3.21 and lambda-15.20 Published with MATLABE R2018a

Let 丁

%Matlab code for system of equations
clear all
close all

%All initial guess for x1

x1=[5;4;3;1];
%solution using newton systems
x = newtonsys(@(x1) f(x1),x1);
fprintf('For initial guess x=%2.2f , y=%2.2f ,z= %2.2f and lambda=%2.2f\n',...
    x1(1),x1(2),x1(3),x1(4))

fprintf('\nroot of the equation is x=%2.2f , y=%2.2f ,z= %2.2f and lambda=%2.2f\n',...
    x(1,end),x(2,end),x(3,end),x(4,end))

x1=[50;40;30;100];
%solution using newton systems
x = newtonsys(@(x1) f(x1),x1);
fprintf('\n\nFor initial guess x=%2.2f , y=%2.2f ,z= %2.2f and lambda=%2.2f\n',...
    x1(1),x1(2),x1(3),x1(4))

fprintf('\nroot of the equation is x=%2.2f , y=%2.2f ,z= %2.2f and lambda=%2.2f\n',...
    x(1,end),x(2,end),x(3,end),x(4,end))

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%Matlab function for Newton system
function x = newtonsys(f,x1)

    funtol = 1000*eps; xtol = 1000*eps; maxitr=40;
    x =x1(:);
    [y,j]=f(x1);
    dx=Inf;
    k=1;
    while (norm(dx) > xtol) && (norm(y)> funtol) && (k< maxitr)
      
        dx= -(j\y);
        x(:,k+1) = x(:,k) + dx;
        k=k+1;
      
        [y, j]= f(x(:,k));
    end
    if k==maxitr
        warning('Maximum number of iterations reached.')
    end
  
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%function for jacobian and residual
function [y,j]= f(x1)

    y(1,1) = x1(1)-5-(x1(4)*x1(1))/25;
    y(2,1) = x1(2)-4-(x1(4)*x1(2))/16;
    y(3,1) = x1(3)-3-(x1(4)*x1(3))/9;
    y(4,1) = (((x1(1)).^2)/25)+(((x1(2)).^2)/16)+(((x1(3)).^2)/9)-1;
  
    j=[1-((x1(4))/25) 0 0 (-x1(1))/25;...
        0 1-((x1(4))/16) 0 (-x1(2))/16;...
        0 0 1-((x1(4))/9) (-x1(3))/9;...
        (2*x1(1))/25 (2*x1(2))/16 (2*x1(3))/9 0];
end

  
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%