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12:00 PM
my thought is for example $C^n $ and a A= line in $C^n$ so $C^n - A$ but the zero polynomial is zero everywhere.
 
What's the difference of two algebraic sets?
 
Hello everyone! I need a little help (and further links) with a certain question: Why are there 24 hours in a day? I have a basic idea that the number of factors of a number expressed as a fraction of the number is large, but I'm looking for some further reading here... Can someone help out?
Everywhere I look this up, I'm getting historical answers... I'm looking for a mathematical one... Can anyone please enlighten me? :)
 
12 is just a very nice number I guess
The only numbers coprime to 12 are 1,5,7,11
(The only numbers between 1 and 12 inclusive, I mean)
 
@AkivaWeinberger Yeah, that's true in a way... It has a lot of factors (1,2,3,4,6) which is more than what 10 has (1,2,5)
 
I wonder if Native American civilizations also divided the day into the same number
 
12:08 PM
@AkivaWeinberger Exactly!
@AkivaWeinberger Not quite... This system, AFAIK, was used by the Egyptians and Indians
Check this out for History -> nist.gov/pml/walk-through-time-early-clocks
But I'm here for the math XD
 
I'm trying to look up Mesoamerican timekeeping, and I'm only getting calendars (and units larger than a day)
 
0
Q: Formal group law and koenigs function conjecture !?

mickLet $f(x,y)$ be a symmetric real function and a formal group law. $G(x + y) = f(G(x),G(y)) $ Consider the equation $$ h(2x) = f(h(x),h(x)) = A(h(x)) $$ This equation has many solutions. Compute a solution to that equation with the fixpoint at $0$ and its koenigs function, and call the soluti...

Any ideas ?
 
hmm... what is:
💥(conjecture)
 
Wiki koenigs if necc
 
Actually, I need a new operator, 💥 is too powerful to be useful
meanwhile, my complex analysis sucks too much to answer that question
I have not even wrap my head around branch cuts yet
 
12:19 PM
A highly composite number, also known as an anti-prime, is a positive integer with more divisors than any smaller positive integer has. The term was coined by Ramanujan (1915). However, Jean-Pierre Kahane has suggested that the concept might have been known to Plato, who set 5040 as the ideal number of citizens in a city as 5040 has more divisors than any numbers less than it.The related concept of largely composite number refers to a positive integer which has at least as many divisors as any smaller positive integer. The name can be somewhat misleading, as two highly composite numbers (1 and...
@AbhigyanC ^This may be of interest
 
Wow!!! Thanks @AkivaWeinberger
 
Define 🌀 to be the cyclone operator such that given any two objects M, N under the ordering induced by 💥, the follow is true
 
Exactly what I was looking for!
 
$$M < 🌀(N)$$
 
See also this concept:
In mathematics, a superior highly composite number is a natural number which has more divisors than any other number scaled relative to some positive power of the number itself. It is a stronger restriction than that of a highly composite number, which is defined as having more divisors than any smaller positive integer. The first 10 superior highly composite numbers and their factorization are listed. For a superior highly composite number n there exists a positive real number ε such that for all natural numbers k smaller than n we have ...
@AbhigyanC
which is the same thing but with "superior" added onto it
 
12:23 PM
@AkivaWeinberger 'Superior' thank you XD
 
Thus while 💥(composite number) cannot fit within this chat, the following is true:
🌀🌀(composite number) = 🌀(highly composite number) = superior highly composite number
Due to the nature of 💥, the image of 🌀 is unstable, i.e. it changes depends on context
Now with the cyclone operator, I should be able to do explosive generalisation in a more controlled fashion
Some identities:
🌀💥=💥
💥🌀=family of higher cyclone operators
 
@Secret What are you doing??
 
Trying to upgrade my notions that I previously used in this chat. Akiva had already covered your query
 
Okay... Go ahead
 
 
4 hours later…
4:25 PM
Whats the real Lie algebra of $\text{SL}_2(\mathbb{C})$ considered as a real Lie group?
 
4:50 PM
Take 3 on infinity:
Let $X$ be an object, $M$ be an object of type $\text{Fin}$, $S$ be an abstract operator that capture a notion of size, $f$ be an operator that scales the notion of size as defined by $S$. Then $X$ is potentially infinite if:
$$S.(X,X)\neq S.(X,f.M)$$
$f.M$ is interpreted to be all possible strings consists of $M$, as sufficiently many scaling operators that changes the notion of size given by $S$. Thus if $S$ is denoted by doubling the number of elements in the set and $M$ is a set, then $f.M$ includes all $2^nM$ for all $n \in \Bbb{N}$, as well things like $2^n MM, 2^nM2^mM^2$ and so on
The basic idea is that potential infinity, when explosively generalised, is an abstract mathematical object that is boundness in the most general sense, that is given any fixed non infinite object and its scaled version such that it stays non infinite, it never contain said object
Now, $X$ is actually infinite if in addition it satisfies:
$$S.(X,f.X)$$ is not of type $\text{Fin}$
Thus 💥(Infinity) is charaterised by unboundedness by finite objects and scale invariance
For comparison, the class $S$ such that the elements satisfy the relation $x_{i+1}=x_i+1$ does not biject with any finite natural number, and e.g. $n\aleph_0=\aleph_0$ for natural number $n$
9 hours ago, by Akiva Weinberger
In other words, "what are near-counterexamples"
Thus, an object that can model a notion of the "edge of infinity" will look something like this:
The object Y can be bounded by at least one object K of type Fin, however, of the all possible f.Y, in a partial ordering, as we transverse up through any chain, the K that can successfully bound Y will started to become increasingly larger, eventually after some finite object, said representative of f.Y can no longer be bounded
For a real value function example, it is any function in the 1st quadrant such that its x asymptote is positive, in other words, finite time blowup functions
 
5:20 PM
i want to prove that the set $C^n -V$ where V is an algebraic proper subset of $C^n$ cant be algebraic
this means there doesnt exist Ideal $I$ of $C[x_1,...x_n]$such that generates $C^n -V$
 
5:40 PM
What is the word for 'percentile' when there are $n$ bins?
 
 
1 hour later…
7:05 PM
@ManolisLyviakis Is there way to write C^n in terms of the ideal which defines V anc C^n\V
 
@PaulPlummer Thanks for the information! It doesn't surprise me: the Smith theory argument is inaccessible.
I don't know an obstruction in the smooth case either, though, but it seems implicit in your statement that one exists.
In the future it will be easier to get in touch with me via my email, smmiller@ucla.
 
7:23 PM
@MikeMiller are math blogs citable?
 
I guess? I wouldn't try to get it in a published paper.
 
I got a mail from a student who said that my blog helped him a lot for his thesis, so he's asking me how to cite it
 
Just find the relevant fact elsewhere; I doubt the full details first appeared in a blog.
 
yeah, that's true
 
Oh, for a student? Seems fine.
They're probably writing a paper for a class.
 
7:28 PM
tbh, I wouldn't exactly recommend them to cite me, since I worked the stuff out on my own (though it's probably all well-known), so I don't know if my notation and terminology is standard
 
Ah, okay, I can email instead. I had been reading some stuff/slides by Kathryn Mann, which I can't find right now and there where some results which imply Diff(M) can't contain high rank lattice subgroups and things like that. Are you saying that p-group paper is inaccessible(Some parts definitely looked complicated, but assuming fixed points didn't look to bad, assuiming some smith theory) @MikeMiller
 
thanks @MikeMiller
 
7:48 PM
Sorry, @Paul, I didn't look at the paper. I mean that Smith theory is about fixed points of finite group actions, so the torsion-free assumption renders it useless to us.
 
8:02 PM
@MatheinBoulomenos Can one prove that the free commutative ring over a set is an integral domain (if one doesn't know that multivariable polynomials exist)?
i.e. just using the universal properties?
 
8:14 PM
1
Q: Formal group law and koenigs function conjecture !?

mickLet $f(x,y)$ be a symmetric real function and a formal group law. $G(x + y) = f(G(x),G(y)) $ Consider the equation $$ h(2x) = f(h(x),h(x)) = A(h(x)) $$ This equation has many solutions. Compute a solution to that equation with the fixpoint at $0$ and its koenigs function, and call the soluti...

Any ideas ?
 
8:50 PM
Hey guys, if $\Omega$ is a compact set, then $C_0^{\infty}(\Omega) \subset H^2(\Omega)$?
 
What's that 0 ?
And H ?
 
compact support and sobolev space
 
why would you need $\Omega$ compact ?
 
It is true if $\Omega$ is not compact?
 
I think it is
Since $C^\infty_0(\Omega)\subset L^2(\Omega)$ and the derivative of an element of $C^\infty_0(\Omega)$ is in $C^\infty_0(\Omega)$
 
9:00 PM
Astyx is right. I can't think of any definition of Sobolev space where this isn't immediate :)
What is yours?
 
I got confused with something, the thing is, if we don't have compact support do we need $\Omega $ to be compact?
 
@LeakyNun I can't think of a proof like that
 
At least $\Omega$ bounded would do the work i believe
 
And isn't $C^{\infty}_0(\Omega)$ just $C^{\infty}(\Omega)$ when $\Omega$ is compact ?
 
Ya it is, i feel silly rn :(, and $\Omega$ bounded, what u think
 
9:08 PM
@MatheinBoulomenos sad
 
I don't think so
 
I don't think $\Omega$ bounded works. Consider somthing like $x^{1/2}$ on $(0,1)$
 
Take the open ball of radius 1, there are continuous functions on it that are not $L^2$ (something like $1\over (1-\Vert x\Vert)^k$)
Or that yeah
 
I meant $x^{-1/2}$
 
What are some good heuristics to follow for problem-solving?
with respect to math problems specifically
 
9:29 PM
Say I want to find the points on $R = sin(3
Say I want to find the points on $R = sin(3\theta)$ closest to and furthest from the origin. I set $dR/d\theta$ equal to $0$, and end up with $3\theta = arccos(0)$, (the multifunction arccos). So $3\theta = \pi/2 + n\pi$, thus $\theta = (2n + 1)*\pi/6$, in other words, the odd integer multiples of $\pi/6$. This expression seems to find only the maximum values of R, not the minimum values at the even multiples of $\pi/6$, which should also be critical points. Why is this?
 
Do you know what $R = \sin(3\theta)$ looks like ?
hint: $\sin 3\theta$ can be negative
 
Yeah, I graphed it on Desmos
 
When is is closest to 0 then ?
The thing is, you're not trying to know when $\sin(3\theta)$ is minimal, but when $|\sin3\theta-0|$ is minimal
 
9:45 PM
$\theta = n\pi/3$, I think
 
That's right
 
I don't see how negative values differ, they just trace out the rose in the opposite direction, so to speak.
 
Yeah
But if you allow $R$ to be negative, you're not trying to find it's minimal points any more
But the ones closest to 0
 
Hmm, I was thinking of $R$ as being non-negative and $\theta$ as going full circle.
 
Yeah, but $\sin$ takes negative values
 
9:58 PM
Okay, this sorta makes sense.
 
sorta
 
So I'm thinking of the $R$-coordinate of a point as the length of the vector that points there, but I guess that's slightly off.
 
yeah, but length are positive, and your $R$ takes negative values
 
Does the polar coordinate $R$ always take signed values by default, unless you convert $(R, \theta)$ to make $R$ positive?
 
No, in polar coordinates, $R$ is positive, otherwise $(R, \theta)$ is the same point as $(-R, \theta+\pi)$
 
10:04 PM
Right, that was the conversion I had in my mind.
 
The number of tabs on my phone has steadily been going down
They're at 85 now
 
You didn't even reach 100
You won't get the achievement
 
@Astyx No, they're going down
They were at 100 for a good while
 
pics or it didn't happen
Last time I checked they were at 97
 
Above 100 it stops counting
It just does a smiley face
 
10:20 PM
fair enough
What happens at 1k though ?
 
That, I have yet to see, I admit
 
;)
or maybe :D
 
>:(
 
10:21 PM
UwU
 
I still don't know the answer to your puzzle by the way
 
Which one
 
Circles and triangles
 
Oh the points, every triple of which is contained in a unit circle?
Right OK so there's some circle of smallest radius containing the points
(Maybe multiple tied for smallest, whatever)
If you can show that that circle is also the smallest circle containing some triple, you're good
Finish from there
(Also: think about what the smallest circle containing three points looks like. It's different if they form an acute or obtuse triangle @Astyx)
 
10:28 PM
I kinda got there on my own, but how do you do that ?
 
How many points are on that minimum circle?
Clearly zero and one lead to a contradiction
 
Agreed
 
Two are possible, but only if they form a diameter. Otherwise, they lead to a contradiction
$n>2$ is possible, but only if they don't all lie on one side of some diameter
So basically your only options are, there are two points on the circle that form a diameter, or there's at least three points on the circle that aren't all on one side of some diameter (in other words, they form an acute triangle)
 
Right
 
In the former case, choose some other point from the set at random, forming an obtuse triangle
The smallest circle containing that triangle is our minimum circle, and thus our minimum circle has size l/e 1 by hypothesis
In the latter case, choose three points on the circle that form an acute triangle
Oh by the way
I'm using the fact that the only way to get a right angle is like that^ (i.e. with a diameter)
Right so in the latter case, choose three points on the minimum circle that form an acute triangle. Then the smallest circle containing those three points has size less than or equal to 1 by hypothesis
and so we're done
 
10:36 PM
You mean more than ?
Oh right, no
I get it
 
The hypothesis is that every triple is in a unit circle, so the smallest circle containing a given triple is unit or smaller
 
Yeah I wa thinking it the other way around
Like, assume the smallest circle containing them is of radius >1, then there has to be three points with minimal circle of radius >1
 
I guess you could argue that way as well
You still get that you either get two points on that circle forming a diameter, or three points on that circle forming an acute triangle
 
let's say we totalize division for a field with the axiom x/0=0
 
I kinda had that reasoning but I thought you meant something must more elementary in terms of geometry
 
10:40 PM
then identities like a/b+c/d=(ad+bc)/(bd) suddenly become false
and I have no idea how best to fix this
 
Yeah I guess "there exists a minimum radius" is more of a compactness argument
though you could probably do it by induction
 
I don't know too much about minimal circles of triangles too
 
Me neither, but intuitively if it's very obtuse the circle gets stuck on the longest edge
By the way
To prove that the angle subtended by a diameter is always right
reflect the triangle across O
BOOM rectangle
 
You could also use Pythagoras
 
Something like that
 
10:45 PM
however we still have -a/b = -(a/b)
 
@LeakyNun I don't see an issue here: a/0+b/0=0+0=0 by zero reciprocal axiom you defined, meanwhile (a0+b0)/(00)=(0+0)/0=0/0=0
 
Comes from the fact that if you take a circle and draw two diameters at random, the points always form a rectangle
 
@Secret 1/1+1/0 = 1+0 = 1; (1x0+1x1)/(1x0) = 1/0=0
 
which, by the way, is the way to draw a right angle if all you have is a marked ruler
Take a random point, draw two lines through it, mark off 1 inch away from the point on the lines
 
I'm wondering if there's a nice solution to the problem using circle inversion
 
10:47 PM
@LeakyNun do you want x/0=0 for a specific x, or all x?
Because that axiom is basically saying, take the zero reciprocal of any number has the same effect as multiplying by zero
 
for all x
 
How to draw a right angle given only a marked ruler (a way to draw straight lines and to mark off 1 unit from points on those lines)
Or use the corner of the ruler itself but that's less fun
 
The I don't see any issue. Given any x, x/0=0 implies (x/0)(1/x)=(x1)/(0x)=1/0=x/0=0
once you have 1/0=0 derived, all x/0 has to go to zero
 
Plagiarism: the worship of Plagiar and all He stands for
10
(and would you believe that I stole this joke from somewhere else)
 
x/0=0 is a relatively tame axiom, it basically saying: project all zero reciprocals to zero. I think it might had something to do with projective geometry
-a/b = -(a/b) thus has to result from that since you are basically do the same as multiplying both sides by zero
It's only when you have nonzero y and then axiomise x/0=y for some pairs of (x,y) that will get you in trouble if you do not throw away associativity
O wait I misread
y/1+1/0 = y/1+0 = y; (yx0+1x1)/(1x0) = (0+1)/0 =1/0=0
hmm...
y/1+0=y/1; y/1+(0z)/z=(yxz+((0z)1+1))/(1x1)=(yz+1)/1=y+1
typo
y/1+0=y/1; y/1+(0z)/z=(yxz+((0z)1+1))/(1xz)=(yz+1)/z=y+1/z
y/1+a=y/1; y/1+(az)/z=(yxz+((az)1+1)/(1xz)=(yz+az1+1)/z=y+a1+1/z
::::
y/1+1/0 = y/1+0 = y; (yx0+1x1)/(1x0) = (0+1)/0 =1/0=0
y/1+0 = y/1+(0z)/z = y; (yxz+(0z)x1)/(1x0) = (yz+0z)/0 =yz/0+0z/0=0+0=0
y/1+0 = y/1+(0z)/z = y; (yxz+(0z)x1)/(1xz) = (yz+0z)/z =yz/z+0z/z=y+0=y
::::
(y/1+1/0)(1x0)=yx0+1x1; y0/1+0/0=y0+0/0=y0
ah...
@LeakyNun Your issue is because 0/0=/=1 ,while for all nonzero x/x=1, thus you have a discontinuity there
Thus (x/0)0=00; x0/0=0; 0/0=0
This is why the product of reciprocal axiom fails for the zero reciprocals, leading to the formation of a new theorem y=y+1
since 0/0 should be 1, but it cannot, thus the only way for the system to adapt, is to have y=y+1
This is common phenomenon in many zero term algebra, that you end up with the theorem y=y+1
Protip: Whenever one meddles with some forms of division by zero, always write out the zero terms in full to see what's going on
So what you have here is not only that a/b+c/d=(ad+bc)/(bd) fails, but because of y=y+1 (and similarly y=1+y) for all y
and that means you have 1=0, thus you collapse into the trivial ring
In order to save this structure, you might want to figure out how to nuke 0/0 from the expansion by modifying the sum of fraction axiom
Alternately, you can discard one of the distributive laws and thus end up with two one sided identities 0 and 1
or, you can modify the distributive laws in the same way as wheels to deal with the occurrence of zero terms
 
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