118 reputation
6
bio website mailto:micaroni3@gmail.com
location Campinas, Brazil
age 31
visits member for 2 years, 1 month
seen Nov 29 at 6:46

Java, Android, Scheme, Javascript etc.


Oct
26
revised How to find the function $f$ that satisfies $f(x, y) = f(x^{-1}, y^{-1})^{-1}$ and $f(x, y)$ is $\approx$ $average(x, y)$?
added 80 characters in body
Oct
26
revised How to find the function $f$ that satisfies $f(x, y) = f(x^{-1}, y^{-1})^{-1}$ and $f(x, y)$ is $\approx$ $average(x, y)$?
adding more conditions
Oct
24
awarded  Citizen Patrol
Oct
23
comment How to find the function $f$ that satisfies $f(x, y) = f(x^{-1}, y^{-1})^{-1}$ and $f(x, y)$ is $\approx$ $average(x, y)$?
It does not fit $min(x, y) < f(x, y) < max(x, y)$ e.g.: $f(5, 8) = \frac{64}{5} > max(x, y)$. I will use $\sqrt{xy}$ with a slight modification (see the link on pastebin above). It works for me. --- I could use $f(x, y) = x$ as you suggest (actually it was my first thought) but despite being close, it would favor one side, and my algorithm is about disputes.
Oct
23
awarded  Autobiographer
Oct
23
comment How to find the function $f$ that satisfies $f(x, y) = f(x^{-1}, y^{-1})^{-1}$ and $f(x, y)$ is $\approx$ $average(x, y)$?
$f(x, y) = x^2y^{-1}$ does not work for me because $f(x, y) \neq f(y, x)$. I just forget to put this condition because (not sure why) I unconsciously assumed it was a natural consequence to first condition. Accidentally the solution $f(x, y)=\sqrt{xy}$ also satisfies $f(x, y) = f(y, x)$. DanielV, could you please send me an email, I just want to thank you for your effort and useful help.
Oct
22
comment How to find the function $f$ that satisfies $f(x, y) = f(x^{-1}, y^{-1})^{-1}$ and $f(x, y)$ is $\approx$ $average(x, y)$?
Finally I solved my problem. Thank you for everyone that helped me. If you guys are interested how looks like the "final function" that fits to my problem, I put it here: pastebin.com/KcqaKNSk
Oct
22
revised How to find the function $f$ that satisfies $f(x, y) = f(x^{-1}, y^{-1})^{-1}$ and $f(x, y)$ is $\approx$ $average(x, y)$?
deleted 30 characters in body
Oct
22
comment How to find the function $f$ that satisfies $f(x, y) = f(x^{-1}, y^{-1})^{-1}$ and $f(x, y)$ is $\approx$ $average(x, y)$?
Great! I didn't realize that before, but in my case $x \approx y$. So, the first aprox. that you and Macavity have proposed to me $f(x, y) = \sqrt{xy}$ works very well to solve my problem. --- But now, I have a collateral effect: the system works only with rational numbers, and using square root I introduce irrational, what brings me another problem that I didn't solve yet. I am thinking in "cut" the number and transform: e.g. 999.999499999875 in 9999994/10000, but not sure yet if it will work to all cases. Is it possible to find a solution that also satisfies $f(x, y) \in\mathbb{Q}$ ?
Oct
22
revised How to find the function $f$ that satisfies $f(x, y) = f(x^{-1}, y^{-1})^{-1}$ and $f(x, y)$ is $\approx$ $average(x, y)$?
better notation
Oct
22
revised How to find the function $f$ that satisfies $f(x, y) = f(x^{-1}, y^{-1})^{-1}$ and $f(x, y)$ is $\approx$ $average(x, y)$?
using min/max
Oct
22
awarded  Scholar
Oct
22
accepted How to find the function $f$ that satisfies $f(x, y) = f(x^{-1}, y^{-1})^{-1}$ and $f(x, y)$ is $\approx$ $average(x, y)$?
Oct
22
comment How to find the function $f$ that satisfies $f(x, y) = f(x^{-1}, y^{-1})^{-1}$ and $f(x, y)$ is $\approx$ $average(x, y)$?
@Macavity thanks and congratulations! Please put it as answer and I will accept that. It works for me. If you want, explain in the answer how did you get that.
Oct
22
comment How to find the function $f$ that satisfies $f(x, y) = f(x^{-1}, y^{-1})^{-1}$ and $f(x, y)$ is $\approx$ $average(x, y)$?
@Macavity wow! (sqrt (* x y)) is pretty amazing! How did you get that? In my tests does not work, but the difference is very very close to zero. Maybe something is wrong with the sqrt fn I am using. Maybe I need a more precise sqrt function.
Oct
22
awarded  Editor
Oct
22
comment How to find the function $f$ that satisfies $f(x, y) = f(x^{-1}, y^{-1})^{-1}$ and $f(x, y)$ is $\approx$ $average(x, y)$?
Thank you! The average is not strictly necessary. It is fair if x < f(x, y) < y
Oct
22
revised How to find the function $f$ that satisfies $f(x, y) = f(x^{-1}, y^{-1})^{-1}$ and $f(x, y)$ is $\approx$ $average(x, y)$?
added 47 characters in body
Oct
22
awarded  Student
Oct
22
asked How to find the function $f$ that satisfies $f(x, y) = f(x^{-1}, y^{-1})^{-1}$ and $f(x, y)$ is $\approx$ $average(x, y)$?