NEWTON-RAPHSON METHOD EXAMPLE 1: Find the root of the function ?(?) = cos (?) using NRM having initial approximation of 1 Input the following codes to your MATLAB edit window and name as    “newton.m” % Newton-Raphson Algorithm                syms x    %Input section  y = input('Enter the given function: ');  yd = input('Enter the derivative of the given function:  ');  p0 = input('Enter initial approximation: ');  n = input('Enter no. of iterations, n: ');  tol = input('Enter tolerance, tol:  ');      i = 1;  while i <= n     d=(eval(subs(y,x,p0)))/eval(subs(yd,x,p0));     p0 = p0 - d;      if abs(d) < tol         if i == 3             fprintf('\nApproximate solution of %s is xn=  %11.9f on %drd iterations  \n\n',y, p0, i);         elseif i == 1             fprintf('\nApproximate solution of %s is xn=  %11.9f on %dst iterations  \n\n',y, p0, i);         elseif i == 2             fprintf('\nApproximate solution of %s is xn=  %11.9f on %dnd iterations  \n\n',y, p0, i);         else             fprintf('\nApproximate solution of %s is xn=  %11.9f on %dth iterations  \n\n',y, p0, i);         end        break;     else        i = i+1;     end  end   Run the program and enter the following. Enter the given function:  cos(x) Enter the derivative of the given function:  -sin(x) Enter initial approximation:  1 Enter no. of iterations, n:  30 Enter tolerance, tol:   0.0001 Write the output below:

NEWTON-RAPHSON METHOD

EXAMPLE 1: Find the root of the function ?(?) = cos (?) using NRM having initial approximation of 1

Input the following codes to your MATLAB edit window and name as    “newton.m”

% Newton-Raphson Algorithm            
 
 syms x
 
 %Input section
 y = input('Enter the given function: ');
 yd = input('Enter the derivative of the given function: 
');
 p0 = input('Enter initial approximation: ');
 n = input('Enter no. of iterations, n: ');
 tol = input('Enter tolerance, tol:  ');
 
 
 i = 1;
 while i <= n
    d=(eval(subs(y,x,p0)))/eval(subs(yd,x,p0));
    p0 = p0 - d; 
    if abs(d) < tol
        if i == 3
            fprintf('\nApproximate solution of %s is xn= 
%11.9f on %drd iterations  \n\n',y, p0, i);
        elseif i == 1
            fprintf('\nApproximate solution of %s is xn= 
%11.9f on %dst iterations  \n\n',y, p0, i);
        elseif i == 2
            fprintf('\nApproximate solution of %s is xn= 
%11.9f on %dnd iterations  \n\n',y, p0, i);
        else
            fprintf('\nApproximate solution of %s is xn= 
%11.9f on %dth iterations  \n\n',y, p0, i);
        end
       break;
    else
       i = i+1;
    end
 end

Run the program and enter the following.

Enter the given function:  cos(x)
Enter the derivative of the given function:  -sin(x)
Enter initial approximation:  1
Enter no. of iterations, n:  30
Enter tolerance, tol:   0.0001

Write the output below: