Iterative solution to achieve convergence

Hello,

I would like to ask you to help me to correct this interative solution. I would like to achieve solution with precision with 3 decimal places. Unfortunately this does not work.

Best regards

Michal

tol = 3;   % tolerance


for i = 1:1:10
f=i^2;
delta = tol-f;
j = i;
if abs(delta) >= 0.1;
    for j = j-1:0.1:j;
        f=j^2;
        delta = tol-f;
        k=j;
        if abs(delta) >= 0.01;
         
        end
    end
    end
end;

[Torsten]

Could you explain what you are trying to do in your code ?

I must admit that I don't understand it.

[M Teaxdriv]

Hello Torsten,

I am trying to calculate x in equation: f = x^2 in iterative way so that f was close to 3 with precision 3 decimal places from bottom.

For illustration I am writing here a code which does succesfully what I want. A problem with this code is that instead of f=n^2 I am starting external script and using this approach entire iteration becomes extremely slow and therefore can no be used. It requires too many cycles to converge to given precision. Therefore I am searching for faster way to achieve converence. I mean with "faster' to converge with smaller number of cycles. I think that a good solution could be the script which showed in the first message above, but it does not work as expected so I am asking you for help.

n = 1; % initial value
f=n^2; % funtion to calculate
tol = 3;   % tolerance
accuracy = 0.0000001; % accuracy
while f <= tol
    n = n+accuracy;
    f=n^2;
end

n = n-accuracy;
f=n^2;

disp(['n = ' num2str(n,10)])
disp(['f = ' num2str(f,10)])

[M Teaxdriv]

Hello,

I have corrected the code from the beginning. I would like to ask you to help me to correct it,make it densier and better looking. Apart from that this code has a problem because l , m and n are the same therefore it does not give solution which I want.

limit = 3; % maximum value for f in equation f =x^2

for i = 0:0.1:2;
f = i^2;

if f >= limit;
    break       
end
end
;

j=i-0.1;
f = j^2;

disp(['j = ' num2str(j,10)])
disp(['f = ' num2str(f,10)])

for j = j:1e-2:2;
f = j^2;

if f >= limit;
    break       
end
end
;


k=j-1e-2;
f = k^2;

disp(['k = ' num2str(k,10)])
disp(['f = ' num2str(f,10)])

for k = k:1e-3:2;
f = k^2;

if f >= limit;
    break       
end
end
;

l=k-1e-3;
f = l^2;

disp(['l = ' num2str(l,10)])
disp(['f = ' num2str(f,10)])

for l = l:1e-4:2;
f = l^2;

if f >= limit;
    break       
end
end
;

m=l-1e-4;
f = m^2;

disp(['m = ' num2str(m,10)])
disp(['f = ' num2str(f,10)])

for m = l:1e-5:2;
f = m^2;

if f >= limit;
    break       
end
end
;

n=m-1e-5;
f = n^2;

disp(['n = ' num2str(n,10)])
disp(['f = ' num2str(f,10)])

[Walter Roberson]

If your function is monotonic then consider using a binary search. Iterate with larger steps until you are able to bracket the solution, then reduce the step size and keep bracketing.

[M Teaxdriv]

Hello Walter,

could you show me binary search on this little example f = x^2 (precision 3 decimal places)?

[M Teaxdriv]

Hello, can anyone help me?

[Walter Roberson]

There are a number of posts showing binary search, several with complete code.

Finding a target value f(x) = t is often rewritten as g(x) = f(x) - t, and at that point you are looking for a zero crossing for g(x). The place where g(x) is 0 is the place where f(x) is the target value. Use any convenient root finding techniques.

The kind of situation where you might use iteration is a case where you are required to find the smallest x>x0 such that f(x) = t, and you are given a minimum distance between matching values but there may be a large number of matches. Figuring out whether there are an even number of matches in an interval can be awkward.