3 ms·
All of these assume your background metric is Euclidean. If your background 2D metric is a projection of a warped 3D space, you can make π as big as you want b
by TheOtherHobbes 3y ago
All of these assume your background metric is Euclidean.
If your background 2D metric is a projection of a warped 3D space, you can make π as big as you want by tugging on the centre of the circle.
- azeemba 3y agoThere is no concept of the "background metric" here. Both the radius and the circumference are measured in the defined metric itself. Any metric that "pulls on the origin" compared to Euclidean distance will have to do the mapping in a continuous way. This will basically result in both the radius and circumference being expanded in that metric. Matter of fact, I linked an article that proves that for _all_ metrics, the value of π is always between 3 and 4 (inclusive). Unfortunately the article might have gotten the hug of death so here is an alternative link: https://www.researchgate.net/publication/353330827_Extremal_Values_of_Pi https://www.researchgate.net/publication/353330827_Extremal_...
- charlieyu1 3y agoHow is circumference defined? And I can think of a counterexample on a sphere, just using Euclidean distance on the surface. Consider a circle with centre at North Pole and radius being the distance from the North Pole to a point on the equator. For this circle it is easy to find out that pi=2
- Filligree 3y agoHmm. And if you keep increasing the radius, pi will shrink all the way to 0.
- azeemba 3y agoThanks for your example! I have been thinking about it. Your observation is correct and the surface of the sphere is a metric. The ratio of radius to circumference is not constant with that metric though so I feel like something should disqualify it. But I am not sure how. So I think your observation shows that we need a stronger constraint than just being a metric. Other commenters have hinted that you need a normed vector space but I am not sure if that's sufficient.
- jimmySixDOF 3y agorelatively completely off topic but everything I understand about pi has come from 3D gif models I never saw in school they should be a core part of the learning curve much further to the start of it than 3B1B
- seanw444 3y agoThose GIFs really do make it super simple. I learned it the same. The unit circle made absolutely no sense to me, and appeared as yet another dogmatic arbitrary "rule" shoved down my throat in school. Had they made an attempt to make it intuitive by showing one single GIF, it'd have all come together for me much quicker. Math is far more elegant than public school allows it to appear. https://raypatrick.xyz/blog/2023/10/27/were-you-mathematically-abused-as-a-child/ https://raypatrick.xyz/blog/2023/10/27/were-you-mathematical...
- Miiko 3y agoIt's not the background metric but the space geometry is assumed Euclidean - in non-Euclidean geometry the ratio of of the circumference of any circle to the diameter of that is not a constant, it depends on such diameter (so you simply cannot define 'pi' in that case)
- cjfd 3y agoWell.... One still obtains pi for the ratio of circumference vs diameter in the limit that the diameter goes to zero.