Flattening a curve -- Stereographic Projection

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[tomasgarau]

Michael is so brave, he didn't even flinch with the sniper aiming at him the whole video

[rublade]

Yeah 6:21 sqrt( 1 - x squared) also at 8:35

[ReallyAmateurPianist]

Fun note (about the "north pole" case): You can map the circle to represent an open interval (0, 2*Pi), where any point in the circle corresponds to the distance from "north pole" (when traveling the circle clock-wise).

With the bijection given in this video, it shows that the open real interval (0, 2*Pi) has the same cardinality as the real numbers.

If you do these same things for a smaller circle (circumference r>0), you can show that any open real interval (0, r) has the same cardinality as the real numbers.

It's also quite trivial to extend this to hold for any open real interval (a, b), where a<b.

[goodplacetostop]

0:03 Animation
0:21 Michael with the « Good Place To Stop » shirt
25:28 Homework
25:55 This looks like a very cool concept, not gonna lie

[terryendicott]

You could introduce compact sets and then you are one step away from showing that the reals can be made compact by adding a "point at infinity". Next go the R^2 and add a "point at infinity".

[lenskihe]

We just did that a couple days ago in my projective geometry course.
The only difference is, that the point (0, 1) get's mapped to the point at infinity. Also, we did it for general conic cections.

[AntonBourbon]

2-dimensional stereographic projection is well known to every photographer making spherical panoramas. The "f()" projection (with the centre in the North Pole, i. e. "as seen from above") gives a photo a "little planet" look, while the "g()" projection ("as seen from below") usually looks horrible, but in rare cases it's a surreal "well". All this is impossible to explain without looking at the actual photos, which I do recommend to the curious (e. g. by googling images for "little planet panorama" for f() and "reverse little planet panorama" for g()
P. S. An excellent video, thank you!

[s]

5:34 shouldn't that be x² in the radical instead of x?

[fordtimelord]

In complex analysis, we used stereographic projection from a sphere onto the complex plane, adding a point on the complex plane called the point at infinity which maps to the North Pole. A one to one an onto map from a 3 dimensional object to a 2d object. This complex stereographic projection deeply connected to mobius transformations. I seem to remember some old physics professor of mine mentioning mobius transformations were related to General Relativity Theory, but it has been a very long while since I looked into this.

Maybe at some point you can do a video or two on mobius transformations and their effect on the stereographic projection of the sphere?

[xyzzy]

i'd love to see a series on algebraic geometry.

[pequalsnpsquared]

6:21 sqrt( 1 - x squared) or am I being dumb?

[manucitomx]

I very much enjoyed this topic.
Thank you, professor!

[omograbi]

Thank u for the vid, but think you missed some squares if I'm right.

[ingiford]

Always liked this. This projection is used a lot. It is the basis for Weierstrass Substitution (though it uses (-1, 0) and the y axis) just like the rationalization of a circle stuff used in various Pythagorean Triple m & n proofs.

[winky]

I wish I'd had this video when I took a class in Celestial Mechanics. Not difficult calculations, but a great study of elementary mapping.

[isaacgonzalez]

Talking about the Frenet-Serret would be beautiful.

[freddupont]

Wonderful - everything always leads to circle inversion!

[draklorlab]

Great video! I wonder if this will lead to a follow-up with the Mercator Projection. Would be very interesting, in my opinion.

[pwmiles]

Interesting topic but... I find it's neater to handle the northern and southern parts of the circle together. In fact you can completely avoid non-rational functions, thus:.
The forward projection is X=x/(1-y). To do the inverse projection, use the condition that (x, y) falls on the line from (0, 1) to (X, 0). So put
   (x, y) = (1-s)*(0, 1) + s*(X, 0)  for some s. I.e.
   x = s*X
   y = 1 - s
Use the circle condition x^2+y^2 = 1 =>
   s^2*X^2 + (1-s)^2 = 1
Rearrange and simplify
   s^2*(X^2 + 1) - 2*s = 0
The crucial point is that the quadratic in s divides through by s, since s=0 also gives a point of the circle. So
   s*(X^2 + 1) - 2 = 0
   s = 2/(X^2 + 1)
   x = 2*X/(X^2 + 1))
   y = (X^2 - 1)/(X^2 + 1)  
This approach works for all quadratic curves and also for quadratic surfaces in higher dimensions e.g. spheres, ellipsoids etc.

[Iomhar]

So sometimes it's x squared and sometimes it's just x?