% Finding the square root of a number using an iterative formula


% EXERCISE#1
% Using fixed number of iterations
R=4900;
xold=-6.0;
nmax=120

for i=1:1:nmax
    xnew=0.5*(xold+R/xold);
    xold=xnew;
end

% EXERCISE#2
% Checking for tolerance using the for loop! 
R=4900;
xold=-6.0;
nmax=120
tol=0.00005;
for i=1:1:namx
    xnew=0.5*(xold+R/xold);
    epsa=abs((xnew-xold)/xnew)*100
    if epsa<=tol
        break;
    end
    xold=xnew;
end
i
xnew
epsa


% EXERCISE#3
% Checking for tolerance using the while loop! 
R=4900;
xold=-6.0;
tol=0.00005;
nmax=120;

% initializing the loop variable
i=0;
% initializing epsa to be more than the tolerance so that the loop works
% the first time
epsa=2*tol

%comparing for meeting the tolerance and the maximum number of iterations
while epsa>=tol & i<=nmax
    xnew=0.5*(xold+R/xold);
    epsa=abs((xnew-xold)/xnew)*100;
    xold=xnew;
    i=i+1;
end

i
xnew
epsa

% EXERCISE#4
% Do it now!