4 ms·
Well, it's pi parameterised by the distance metric, Pi(d) You can parameterise it by other concerns if you wish, and other things follow. But as a matter of fa
by mjburgess 1y ago
Well, it's pi parameterised by the distance metric, Pi(d)
You can parameterise it by other concerns if you wish, and other things follow. But as a matter of fact, this is how pi depends on the distance metric.
- amelius 1y agoYes, but we're looking at a specific set of distance metrics. It'd be interested in the set where Pi(d) is constant and equal to Pi. (disclaimer: IANAM and I haven't given it much thought)
- frotaur 1y agoThis is certainly an interesting question; however, it's probably very hard. This is because the 'set of all possible metrics of the 2d plane' is extremely large, and I am not sure if we have a good characterization for it. There are a bunch of very strange metrics, e.g. a metric for which d(x,x) = 0, d(x,y)=1, that is, all points are at a distance of 1 to each other (this satisfies all axioms).
- srean 1y agoYou might like the discussion here https://math.stackexchange.com/questions/254620/pi-in-arbitrary-metric-spaces https://math.stackexchange.com/questions/254620/pi-in-arbitr...