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
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:
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