3 ms·
But remember that "we" chose the generalization which this all depends on.
by amelius 1y ago
But remember that "we" chose the generalization which this all depends on.
- mjburgess 1y agoWell, 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...