When you make it above 9000 let me know

Cool video! I am very excited about the recent einstein tiling discovery, I hope you'll manage to make a video about that!

Can you please do SSCG(3) next?

I miss the old days of the Eisenbud 17-gon

Woah, phrasing, pal

Indeed, his way of speaking creates a truly unique cogni-feast ambience, I could listen to him teaching for hours without losing focus.

Some of you guys are strange.

Math ASMR

Numberphile Cinematic Universe lore

can you give a timestamp as well?

This 1930 book inspired several other history books with similar titles and in the same sarcastic tone.

Oh, I was definitely not familiar with this fact! Is there something special about the year 1066 in that title?

​@@DukeBG That was the year that England was invaded by the Normans. It was supposed to be one of the only two dates taught in history which people actually remembered. I think that the other one was nineteen fourteen eighteen.

@@DukeBG it's important for english history - french,english and norman leaders battled over who would rule brittain

l put my hamster in a sock and slammed it against the furniture.

You mean "a line in the space of circles"? I thought that was a weird choice of phrase too but I guess that just means a one-parameter continuous family of circles

@@viliml2763 Yep, that's what I meant. And yes, that's exactly what it means. But to someone not used to abstracting things via geometry it's not obvious.

@@QuantumHistorian And I think this non-obviousness is well placed in that context. It makes people stop and think, "wait. What _is_ a line really?"

@@lonestarr1490Yes, exactly, except that because the speaker *doesn't* stop and instead continues the stream of new ideas, the viewer doesn't have the time to stop and think unless they manually pause (or, if they do, they'll fall behind the rest of the video). That's precisely why the speaker should spend a little bit of time clarifying what he means by that in order to give the viewer time to digest that alternative way of thinking about things.

+

Here, here! And the trig functions that prove the Pythagorian theorem too.

I can't wait to see Prof. Kaplan on Numberphile!

@@nosuchthing8 *hear, hear

"Collatz Conjecture guy" 🙄 This is David fucking Eisenbud

@@user-jc2lz6jb2e donald knuth goldbach?

@@user-jc2lz6jb2e it's supposed to be tongue and cheek; it's my ignorance not his. I bet he would get a kick out of the comment, not feel anger or anything. If I thought he would be upset by it I wouldn't have made the comment.

​@@user-jc2lz6jb2e nobody cares

Dang his mom really named him specifically

intriguing, could you say more?

Yeah, conic curves and anything with a graphical representation feel more appealing to me than random integers

Also at 11:40, there should be an xy term as well. That *would* mean there are six parameters instead of five, but really we should think of all of these equations up to a multiplicative scalar (e.g. x^2 - y = 0 is the same conic as 2x^2 - 2y = 0), which drops us back down to five independent parameters.

@@ravi12346 x^2-y=0 is the parabola y=x^2. But without the rectangular term xy u can't make a hyperbola

Hi! The point is that each of them is a double point (xy=0 and xy=z^2 gives xy=0 and z^2 =0, so the second equation gives twice the line at infinity). After all, as Eisenbud explained, you should expect 4 solutions in total, for the intersection of two conics. This is like a circle and an ellipse meeting at 4 points, or being tangent at 2 double points.

@@issoroloap You’re right! I neglected the fact that there are 4 points of intersection of two conics counted with multiplicity. Furthermore I realized that I can see the tangency by the fact that the slope of the line joining the origin and the point on the curve (this is y/x) doesn’t change sign as you wrap around. However, I can still make the point that merely having slopes converge is not a sufficient geometric interpretation of tangency. That’s just saying they intersect at infinity. Maybe physical distance approaching 0 on the other hand is sufficient for tangency (for algebraic curves), but I haven’t worked that out yet

Yes, it was used for ironing.

@@asheep7797 Lol, standard slapstick humor. Kudos :)

I couldn't stop staring at it!

I think it is a mistake, or perhaps intentionally sweeping it under the rug, although the conclusion that there are 5 degrees of freedom in the parameters is still correct. Generally, ax²+bxy+cy²+dx+ey+f = 0 seems to have six, but if we multiply the entire thing by a (nonzero) number we get a *different* equation for the *same* conic, so that is overcounting one degree of freedom. Your own argument that you could rotate away the xy seems correct, but I would object that you then would get a different (rotated!) conic. Maybe somebody else has a better justification for the choice in the video...

xy can be rewritten as 1/4 * [(x+y)^2- (x-y)^2]

@@diniaadil6154 Yeah, so that is a transformation to variables v=x+y and w=x-y, but if you do that, the x² and y² terms are going to reintroduce v*w terms, so you haven't (generally) lost the product term... Unfortunately, I think it is a bit more tricky than that.

I also spotted it and I have no clue other than it is a mistake.

It can be eliminated by rotation, yes. Not just v=x+y and w=x-y, but a more generic-looking matrix

Yes, the xy term should be included, but you can multiply all six constants by the same number, and you get the same conic. Thus there are really five constants to determine a conic.

xy isn't a constant, so it can't be altered in the same way

@@michaeltajfel But what if the xy term has a coefficient of 0?

The right answer is that there are 6 parameters but only up to scaling. The equation is axx+byy+cxy+dx+ey+f=0, but replacing (a,b,c,d,e,f) by (ga,gb,gc,gd,ge,gf) has the same solution set, so we get 6-1=5 parameters. They simplified for presentation and you caught it.

@@ipudisciple That's still only true for nonzero g.

I think "step-by-step" is the key word here. Algebraic geometry requires fairly broad background to really get into, however, this background material is often best understood knowing how it used in algebraic geometry! Commutative algebra is certainly the biggest culprit in this regard. From my experience of learning the subject (which is admittedly not so much, but this is consistent with what more experienced people have told me) what really helps is the willingness to revisit things with the new perspectives you gained. I don't think there is any need to be intimidated, just read what you find interesting, and fill in the background as you need it. Eventually you'll learn a lot!

This would be easier to explain with a picture, but I'll try. He mentioned two degenerate conics: a single point x^2 = 0 and two lines (cut the cone in half). There is a third which is one line. If you take any point but the tip and draw a line to the tip you will have a line that runs along the side of the cone. You can imagine the cutting plane as just touching that line, like you were preparing to wrap the cone with it. Any plane parallel to that plane by pushing inwards will intersect with a parabola. Any deviation from this angle would either tip to ellipse or hyperbola.

@@embryonicsuperfemme i think i got it now, even without the picture :D thank you for your reply/explanation!

Turn the paper 90 degrees so you can't see the front or the back and are looking at it edge on... the circle and the line are overlapping in that dimension?

You'd think, right? That's actually a common question. Turns out, because the cut approaches the wider part of the cone at a shallower angle, the section is in fact a perfectly symmetric ellipse. It is counterintuitive, I'll grant you.

Intersection theory on moduli spaces. There's not a simple explanation

​@@theflaggeddragon9472 What is the expression of the number given by that theory?

​@@soyokou.2810 I am no expert in enumerative geometry (I barely know basic algebraic geometry), but after skimming the relevant section, here's what I can say. The space of plane conics (as Eisenbud indicated in the video) is 5-dimensional; in fact it is P^5 (projective 5-space). This means the parameters are unrestricted and have no nontrivial relations (no two equations ax^2 + by^2 + cx+dy + e with different coefficients are isomorphic. Given a plane conic C, it's _dual_ C^* is the set of tangent lines, a smooth conic in the dual projective plane (space of lines in projective plane). The difficulty in narrowing 6^5 = 7776 to 3264 comes down to casting out "degenerations" of conics (double lines and such, as Eisenbud mentioned). This is technical and requires working on a _compactified_ moduli space of conics. This is the closure of the space of usual conics C in P^5. In fact, we take pairs (C,C^*) in P^5 x (P^5)^* and take the closure in there. We call this space X. Fix five general place conics C_,i, i = 1,...,5. The space of tangent conics in X is a hypersurface of degree 6. As mentioned in the video, taking a naive intersection gives a count of 6^5. The issue is the degenerate intersections occurring on the boundary of X. Now there is an object called the "Chow ring" A(X) = direct sum A^i(X), essentially formal integral sums of subvarieties of X modulo an equivalence relation. The A^i(X) encodes varieties of codimension i in X. Algebra in this ring allows us to compute intersections and many other things in algebraic geometry. On the open subset of smooth conics U in X, the hypersurface Z of conics tangent to a given conic has degree 6 (mentioned in the video). Let a,b in A^1(X) be pullbacks to X in P^5 x (P^5)^* of hyperplane classes on P^5 x (P^5)^*, and c,d in A^4(X) be classes of curves that are pulled back from general lines in (P^5)x(P^5)^*. One cna show that A^1(X) is generated by a,b over the integers. Take an equivalence class [Z] containing the hypersurfaces we want to intersect. The degree of its 5th power tells us the number of intersections (this is the point of working with Chow rings). Hypersurfaces lie in A^1(X), and it's free part has rank 2, so you can write [Z] = pa + qb for some p,q in Q and x,y forming a basis for A^1(X) (tensor Q). From basic properties of degree, you can show that [Z] = 2a + 2b in fact. Hence deg[Z]^5 = 32 deg(a+b)^5. So its enough to calculate the degree of a^ib^(5-i) for each i = 0 ,..,5. By symmetry, enough to do for i = 0,1,2. The calculations with explanation are on page 307-308 of 3264 and all that. All in all, you get deg([Z]^5) = 2^5deg(a+b)^5 = 2^5(5C0 + 2(5C1) + 4(5C2) + 4(5C3) + 2(5C4) + 5C5) = 2^5 * 102 = 3264. Reply

​@@theflaggeddragon9472 Blimey. That's what we get for asking questions! 😂

​@@theflaggeddragon9472 "no two equations ax^2 + by^2 + cx+dy + e with different coefficients are isomorphic" ax^2 + by^2 + cx+dy + e and k(ax^2 + by^2 + cx+dy + e) have different coefficients but are isomorphic the issue is forgetting the sixth, xy term

What do you define as algebraic geometry and what is just regular function graphings or like high school analytic geometry in the coordinate plane stuff? Eisenbud’s first numberphile video years ago was on the graph characteristics of odd functions.

@@stephenbeck7222 Algebraic geometry is the geometric study of multivariable polynomials like the curves in this video. Classically, it uses lots of projective geometry like in this video, but modern algebraic geometry uses a lot of commutative algebra. Eisenbud is himself a famous algebraic geometer as the author of the book Commutative Algebra With a View Towards Algebraic Geometry.

also, a proper 'cone' has six lobes. most of the ones you depict have 1, and only for the hyperbola do you finally show the classic 2. but if you use the correct number, 6, then your slices correctly illustrate all of the relationships which occur between conic sections. it's quite nice. and it also demonstrates very elegantly how thoroughly fubar modern philosophy of mathematics really is.

@3:54 the 2 imaginary solutions here are on the hyperbola. they're imaginary here because the lobe of the cone that the hyperbola slices is on an axis perpendicular to the lobe sliced by the circle. why wouldn't you just mention that? or did you just not know about this?

this isn't controversial, either. it's literally how Special Relativity works. the curve of the relationship between t and t' is circular for vc. this is why tachyons would require energy inputs not to accelerate.

the other closure of the parabola occurs at the same exact spot as the vertex you have. it just looks like a mirrored copy of the parabola you drew. there is nothing happening at infinity. that's nonsense. for the hyperbola you showed, there is no contact between the y-axis and the hyperola at y = +inf. the hyperbola reaches a height of +inf when x is the successor to 0, which contrary to Peano, is not 1. but this successor is a value that we use all the time without understanding it, since it's absolutely required for evaluating limits. when you do something like: lim x->0+ 1/x = +inf you obviously can't evaluate at x=0, because division by 0 is undefinable. further, we know this function is discontinuous, since for x0 we get a positive branch that grows in magnitude as we approach 0. so, when we take this limit and say that it gives us positive infinity, what we did is we evaluated it at the successor to 0. which I will notate as L(0). now, you can trivially see that L(0), 0 and -L(0) are completely distinct values, because 1/L(0) = +inf, 1/0 is undefinable, and 1/-L(0) = -inf. your hyperbola reaches L(0), and when it does its height is the largest possible infinity that exists, but it does not reach 0. and thus it is simply nonsense to claim that it touches the vertical asymptote, x=0, at y = infinity.

it's hilarious to hear someone speak of rigor in mathematics when it's been known for 92 years that modern mathematics cannot possibly be rigorous. Incompleteness demonstrates that your assumptions about how mathematics works are wrong, and yet you just carry on acting as if that never happened... ok, but when that's the choice you make, you don't get to speak of rigor.

It's not that it goes through two points (it only intersects the circle at a single point). It's that if you solve the underlying equations you get a repeated root. It's the same as, say, the equation x^2 = 0. This has one solution (x=0), but through the fundamental theorem of algebra we know every degree n polynomial has n roots: we can write it as a*(x-c_1)(x-c_2)...(x-c_n) = 0, where the c_i are complex numbers. So for x^2 = 0 this gives (x - 0)(x - 0) = 0. The c_i are, collectively, the roots, so in this example we have the roots 0 and 0 i.e. a repeated root at x = 0. The solutions are just the roots listed without repeats (without multiplicity, to use the maths term for it). Hope that helps.

Yes, but it should be thought of as being over the complex numbers, so it is more than just the origin.

I kept watching and realize that it’s the graph that was weird, not the equation

Actually, the proof by Argand seems to use an incomplete infinite descent argument, which I believe would need some sort of Weierstrass theorem or ODE method to work. So I don't even know who first proved this theorem.