the strangest circle

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[@sven3490]

🙏🙏🙏 Thank you, thank you, thank you! 🙏🙏🙏 I am an aerospace engineer veteran (CFD and so on). With this video you have provided me with an understanding of adiabatic, inviscid, compressible flow (a.k.a. the EULER equations) which I never had before! The change of the nature of the set of the underlying physical set of equations associated with the transition from a subsonic flow (elliptic), through the instable point of transsonic flow (parabolic) to supersonic (hyperbolic) has always worried/occupied me. And as I never had a thorough understanding, I felt a little uneasy when studying the eigenvalues/eigenspaces. But no more, because you have listened to our request for another video of this series! Thank you very much again, sir. And with a view on the future, is there any reasonable extension of your three-part set of lectures on hyperbolae, parabolae and ellipseis to PDE, i.e. the metod of characteristics according to Courant, Friedrichs and Levy? This is of exceptional significance in programming codes for solving the Euler or Navier-Stokes equations numerically. On a side note: the viscidity of of the Navier-Stokes equations is very often regarded as a "parabolizing" feature in supersonic compressible flows. Of late, I had a look into the set of the so-called Osipov/Lanchester equations which are hyperbolic. If any of the aforementioned applications were worth your consideration, I should be delighted.

[@Alan-zf2tt]

Request gets thumbs up from me. Sticking to traditional undergrad stuff is fine - great in fact/ Doing stuff for people no longer undergrads but with adult experience sounds like a great extension to Michael's content. Math for something associated with pedagogy is great. Math for something associated with adult requests and for love of math sounds equally good and far better in my opinion

[@Jooolse]

Random errors: 4:57 we end up with X^2+Y^2 = -1, not 0 11:40 non horizontal" 12:13 if a = 0, then it's gonna be the Y axis

[@schweinmachtbree1013]

error in your errata: it should be y^2 + v^2 = -1, not X^2 + Y^2 = -1

[@edmundwoolliams1240]

This is the natural habitat of Michael Chalk on TubeYou

[@krelly90277]

Michael Chalk is on KZbin and you are saying the natural habitat of Michal Chalk is, in fact, KZbin.

[@roberttelarket4934]

The natural habitat of Mike Penn is on KZbin.

[@orionspur]

And that's a good place to stop. 🛑

[@mathemagicalpi]

Mike Penn is cursed

[@roberttelarket4934]

@@orionspur: . . a good place to be except for the censorship.

[@stevefoster5476]

Thanks!

[@byronwatkins2565]

At 30:15, he actually factored out z^2 NOT 1/z^2.

[@eveeeon341]

I was a little bit confused as to why he chose to set y=1 for the equivalence relation, then proceeded to define the point at infinity where x=0, did he get mixed up and do this the wrong way around, should it have been set to x=1? (Which I think is more in line with convention)

[@andrewporter1868]

Michael _Penn_ yet no pen in sight. Something's up here...

[@zh84]

It reminds me of studying the behaviour of the trig functions in the complex plane. The circular functions become hyperbolic functions if you multiply the argument by i, and vice versa.

[@goodplacetostop2973]

30:59

[@minamagdy4126]

An interesting point is that the CP^2 transform between ellipses and hyperbolas works just as well in RP^2, since that affine transform only uses real numbers.

[@BennettAustin7]

Reminds me of habitat skateboard logo

[@stephenhamer8192]

Why did we have to go into complex projective space - my understanding is that conics in the real projective plane are all (projectively) equivalent. Conics in the complex projective plane are _affinely_ equivalent: is that it?

[@MDNQ-ud1ty]

They are all the same. Originally conics come from taking the general equation and seeing it as the interaction of a cone in 3 space with a plane: sum(a_ij x^iy^j) = 0 The cone comes from looking at the ^2 terms(circles) with z^2 (the radius) and what is left over is the plane that intersects the cone. Hence the name of the term. (z is introduced as the intersection between the cone and plane) Dividing everything by z gets one into the projective case. Expanding to complex number simply encodes two equations in one. It's not that they are "projectively" equivalent. It's that in the projective plane there is clearly a reduction in possibilities. Basically the projection occurs along the cone. If you just work out the equations and use the substitutions x->x/z and y->y/z and use complex versions you'll see how it all works out. After all, you are just introducing new variables and such so end up with the same thing + extra stuff. Technically you don't have to leave the xy plane. The point in lifting to higher dimensions is to give new insights and possibly see generalizations. The point of going to a cone in 3d is that one can see these various classes as simply the way the plane intersects with the cone. Hence they all are "equivalent" in that sense. Rather than seemingly very different things they can be seen as simply an intersection of two things with the angle of the plane defining the conic. Again, remember when you are lifting something you end up with the same thing + extra stuff. Sometimes it is somewhat useless because that extra stuff just gets in the way but people think it is cool because it wraps something boring in new cloths. Other times the lifting unifies things that seem quite distinct. All this means is that it becomes easier to think about because.

[@echoes6092]

does the same thing work with H2 (quarternion × quaternion) space and quadric surfaces (like paraboloids and ellipsoids)?

[@Alan-zf2tt]

All I can say is 5 star excellence! 🌟🌟🌟🌟🌟 That does not imply I understand it all but I enthuse about challenging my lack of understanding. Plus some other considerations.... @ 21:08 (orthogonally?) (ortho?) normalized on all values of z? (x, y, z ) maps to (x/x, y/z, 1) since, of course z/z = 1 even when z is zero @ 24:25 x² + y² = 1 represents a locus of points at distance one unit from origin in ℝ² BUT! x² + y² = 1 represents a locus of points at distance one unit from origin in ℝ² but extends to meeting at infinity thus forming a unit tube extending thru z and symmetric about all axes in ℝ³ so what ℂP² does looks almost frightening. Attractive and yet frightening at the same time. @ 28:40 handwaving is fine - getting down to x₁e₁ + x₂e₂ + x₃e₃ equations or (x₁e₁, x₂e₂, x₃e₃) or (x₁ê₁, x₂ê₂, x₃ê₃) in ℝ³ or ℝˣ is nitemarish but in ℂˣ - that reminds me - where is my fishing rod 🙂 EDIT and in ℂˣP I think I'll take my fishing rod and overnight kit with additional food and water 🙂

[@bjornfeuerbacher5514]

"of course z/z = 1 even when z is zero" No. When z is zero, you have 0/0, and that does _not_ give 1, but is an undefined expression. "x² + y² = 1 represents a locus of points at distance one unit from origin in ℝ² but extends to meeting at infinity thus forming a unit tube extending thru z and symmetric about all axes in ℝ³" That's right, but rather irrelevant for CP².

[@Alan-zf2tt]

@@bjornfeuerbacher5514 whatever you say @bjornfeuerbacher5514 whatever you say 🙂

[@Alan-zf2tt]

Potential topic for the future? It is all one but have many names circular inversion inversive geometry Geometric inversion

[@__christopher__]

I actually would have expected RP^2. What would go wrong there?

[@bjornfeuerbacher5514]

You need i for the transformation at 27:25, I don't see how that should work in RP².

[@__christopher__]

@@bjornfeuerbacher5514 That this specific transformation uses i does not imply that you cannot find a transformation that doesn't use i.

[@bjornfeuerbacher5514]

@@__christopher__ Well, try it - I think it does not work without using i.

[@__christopher__]

​@@bjornfeuerbacher5514 Well, if I had tried and failed, that would prove nothing because I might have failed to find a transformation that exists. But I did indeed try and succeed: Start with the projective unit circle, x^2+y^2=z^2. Remember, we are in RP^2 now, so x, y, z are real variables. Now let's apply the transformation (a:b:c)→(a:(b+c):(c-b)). That's quite obviously an allowed transformation in RP^2. By doing so, we get the set {(x:(y+z):(z-y)) | x^2+y^2=z^2}. Now, the equation of the standard parabola is y=x^2, or in RP^2, y z=x^2. That is, the standard parabola is give by the set {(u:v:w) | u^2 = v w} (where I renamed the variables). So let's see if the transformed circle fulfills that equation: v w = (y+z)(z-y) =(z^2-y^2). But we know that x^2+y^2=z^2, and therefore z^2-y^2=x^2, and thus v w = x^2 = u^2. To get an equation similar to the vide, we have to invert the transformation: x=u, y=(w-v)/2, z=(w+v)/2, and then we can insert and rename u,v,w to x,y,z to get: (x:(z-y)/2:(z+y)/2) If you now apply the original condition, you get x^2 + ((z-y)/2)^2 = ((z+y)/2)^2, or x^2 + (x^2 - 2 z y + y^2)/4 = (z^2 + 2 z y + y^2)/4. Add (2 y z - y^2 - z^2)/4 to both sides and simplify, and you get x^2 = y z. This is exactly the parabola, (y/z)^2 = 2(x/z), with one extra point at infinity to close the circle. As you can see, no imaginary unit to be seen anywhere.

[@bjornfeuerbacher5514]

@@__christopher__ I'm not sure if I can follow you... "If you now apply the original condition, you get x^2 + ((z-y)/2)^2 = ((z+y)/2)^2, or x^2 + (x^2 - 2 z y + y^2)/4 = (z^2 + 2 z y + y^2)/4." Here you use that the new variables u = x, v = (z-y)/2 and w = (z+y)/2 are a solution to the equation u² + v² = w². "simplify, and you get x^2 = y z" Here you use that x² + y² = z². Why should _both_ equations, u² + v² = w² and x² + y² = z², hold at the same time? Perhaps I'm missing something...

[@Alan-zf2tt]

I have read about this in past as a topic called Conic Sections. Basically a couple of cones with shared pointy bit at zero and stacked perfectly one on top of the other. My inquisitiveness asks why do they have to be arranged perpendicular with a central axis? Why not with wobbles? I accept the basic concept of ideal spaces but realistically how can we perturb an ideal balanced pair of cones? Say common zero at infinity ie open end of cone going to pointy bits at each extreme from common zero. Follow symmetrically along a line on a torus, follow symmetrically along along a 2π thing forming a chain of connected cones? Or even a perturbed concatenation of cones with ideals and wobbles going on finitely or infinitely with or without modular behaviors Any way I am glad Michael posted this coz thinking about yesterday's video and pondering ℂ² and ℝˣ Does ℂ² twiddle about as ℝ² or ℝ⁴? Naively I go for ℝ⁴ as it has 4 mutually orthogonal variables and 5 is we count common zeroes? Although these may be paired as two plus two too to one plus one plus one plus one too if you know what I mean. I think I posted a comment somewhere in the past about commonality of dimensions automatically implying a solution in one space is a solution in all spaces even though that might include "no solutions in this combination of spaces and dimensions" Just like conic sections appear magically in ℝ³ and above even if it is outside domain of a particular combination of axes it will intersect at Zeroes Anyway someone theorized and someone proved something about Completeness in math and someones else hinted at higher dimensions or lower dimensions or each dimension separately might provide paths to solutions. I am not implying it will be easier but ...

[@jakobr_]

It’s not two cones, “both” are part of the same object. It’s what you get if you take a line and revolve it about an axis that intersects it. It’s what you get if you take a circle and let its radius vary linearly with time, allowing for negative distances. If you perturb the two halves of the cone, you break the continuity of all of those straight lines that make it up. And I don’t think that makes much sense to do.

[@Alan-zf2tt]

@@jakobr_ Thank you - that makes things a little bit clearer now

[@roberttelarket4934]

The natural habitat of Mike Penn is on KZbin.