Iterative solution to achieve convergence
2 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
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;
7 Kommentare
Walter Roberson
am 18 Mai 2022
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.
Antworten (0)
Siehe auch
Kategorien
Mehr zu Creating and Concatenating Matrices finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!