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Dan Sommers <dan at tombstonezero.net> wrote: > > Given: > > > > x = INF > > y = INF > > assert x == y > > > > there is a reason to pick atan2(y, x) = pi/4: > > > > Since x == y, the answer should be the same as for any other pair of x == y. > > When x == y == 0, then atan2(y, x) is 0. This is the only solution (0) where atan2 return sometime else than pi/4. And INF is not 0, it can be lot of values, but not zero. Yes x=INF and y=INF do not mean x=y (in math world) but INF is a frontier that can be reach, so it tend to the maximum value. So the tendancy is always atan2(y,x) tend to pi/4 if you looks at lot od y and x that will be grater and greater each time : the final frontier would always be pi/4, even if t take a long time to reach it. -- Pierre-Alain Dorange <http://microwar.sourceforge.net/> Ce message est sous licence Creative Commons "by-nc-sa-2.0" <http://creativecommons.org/licenses/by-nc-sa/2.0/fr/>

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