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12:28 AM
Anyone know a cute algorithm for ideal membership in a boolean ring?
Without enumerating the ring (it's finite though)
I have so far, $x \in (i) \iff x = ri$ lel
for some $r \in A$ the boolean ring
There is a poset ordering on elements $\subset$
and a maximal element $X \supset$ each other element
It has up to $O(|s|^2)$ elements where $|s| \in \Bbb{N}$ is the size of my SGP algorithm input
I.e. the element $X$ is itself a set, and all other elements in $A$ are subsets of $X$.
1 hour later…
so, the previously mentioned long-ass paper with my name on it is online now
so that's neat
2:05 AM
@Semiclassical congrats, you're all called michael
hence why our name for it is the 3M paper
Does the diagonal representation of a quadratic form preserve scale?
1 hour later…
3:22 AM
Why isn't 1 a solution to the following?
$\sqrt{4x} = x - 3$
Isn't -2 a possible solution to $\sqrt{4}$?
-2 is a square root of 4
but $\sqrt4 = 2$
$\sqrt x$ means the positive square root of $x$
thats really annoying
that's life
so if it said "solve for all values of x" it would only be 9?
not 1 and 9?
3:25 AM
alright, thanks. Will know for next time I guess.
2 hours later…
4:57 AM
@AkivaWeinberger hi
@Semiclassical Oh that's the company what invented post-its
this usage of "what" feels weird
"what" is followed by a predicate to make it a noun
I would say "that" / "which" here
5:55 AM
Hello. I'm finding good references for abstract polytope. I choose Peter McMullen, Egon Schulte, <Abstract Regular Polytopes> and I'll read it. But, well, one review in Amazon said "there have been huge advances in the field of abstract polytopes." I suddenly want to know what is huge advances.
6:11 AM
Q: linearity of transformation

maths studentProve that a function $f: V \rightarrow \mathbf{R}^{m}$ is linear if and only if each component of $f$ is linear. Determine the form of a linear map $T: \mathbf{R}^{n} \rightarrow \mathbf{R}^{m} .$ I get the the form of a linear map $T: \mathbf{R}^{n} \rightarrow \mathbf{R}^{m} .$ which is conse...

7:01 AM
Having trouble figuring this old problem out
Q: Asking a previous question about a Round Robin Tournament with $2^n$ teams

Harold SmithRound Robin Tournament I understood the hint involving the Pigeon Hole Principle but I'm confused on how to apply it to the inductive step. Any help is appreciated. What I have so far: We will prove by induction on $n$. Base Case: $n=0$, so there is only 1 team. Base case holds as we can easil...

3 hours later…
9:51 AM
I tried to find books for studying some speicifc topics, is it okay t ask it in M.SE or M.O?
Sure, why not?
(be sure to check the tag first)
Thank you
Can anyone help me?
10:23 AM
@LeakyNun I've been influenced by the Yorkshiremen I see on the YouTubes
I'm completely lost on this round-robin proof
2 hours later…
12:00 PM
If someone is in 11th and if I wanted to study geometry which is higher than 12th, then what and how should I study?
I have done some Projective geo, from Rey Casse, which is relatively easier to understand than most other Projective geo book.
So, given a set of convex polyhedra such that these polyhedra will tessellate the plane and one of the polyhedra is a regular n-gon, can you determine the smallest cardinality that set can have in terms of n?
@Rithaniel nope.
(Also, random-15, you'll probably want to ask Ted when he's in, as he is the resident geometer, but I'll predict that it depends very heavily on who you are as a person)
I guess this means I should study more projective and then ask. Thanks!
@Rithaniel oh okay.
Also, my question was an open question, not directed at you, apologies
12:05 PM
@Rithaniel oh okay. I thought you were asking from PG, and I didn't understand your question, so I thought this means I should study more before asking.
Nah, you're good. What I asked is a potentially difficult question, from what I know. Someone in here might have an easy answer, though, so I figured I would drop it in here, just in case.
1 hour later…
1:20 PM
Given a continuous function $f\colon\mathbb{R}^m\rightarrow\mathbb{R},\,m\ge1$ and $x,y\in\mathbb{R}^m$, s.t. $f(x)\neq f(y)$ and a $c$ between $f(x)$ and $f(y)$, can one always find a $m-1$-dimensional topological submanifold $M\subseteq\mathbb{R}^m$, s.t. $f(M)=\{c\}$ and $\mathbb{R}^m\setminus M$ consists of exactly two connected components, one of which contains $x$ and the other contains $y$?
2:20 PM
Are there any good examples of stochastic games with a discrete state space of cardinality on the other of 100 or less?
2:36 PM
@Thorgott Originally misread that as $f^{-1}(c)=M$
@Thorgott I'm a bit worried you could have something be 0 on the topologist's sine curve (the graph of $y=\sin(1/x)$ together with the interval $\{x=0~\&~-1\le y\le1\}$), positive above it, and negative below it
@AkivaWeinberger but the components are connected
it's just the graph that isn't path-connected
It's not a manifold, is the point
It's not an $m-1$-dimensional topological submanifold $M\subseteq\Bbb R^m$
that's the counter-example then :P
2:46 PM
(John Preskill being one of the pioneering researchers in the field, particularly the one who coined the phrase "quantum supremacy")
Thanks for the counter-example. Do you haven an idea about the case in which the manifold condition is dropped?
3:03 PM
I think it's true without that but I don't have a proof
Part of me wonders if you could generate a counterexample using a Newton fractal
but I'm not seeing one
3:20 PM
As noted here, the mean minimizes the mean squared error and the median minimizes the mean absolute error. Is there a bounded loss function whose minimizer is also the mean?
Take any bounded function and compose it with the squared error
Let such a function be $f$. Then $f \circ \mathrm{sqrt}$ is also bounded, but minimizing it would be minimizing a bounded function of the absolute error, no? And by the same logic wouldn't its minimizer be the median rather than the mean?
3:36 PM
$f\circ\sum\rm sqrt$
What does that mean
Sum of square roots?
Oh wait why sqrt
Mean squared error is the mean of squares, yeah?
So like $\sum x^2/n$, ish
and then we can do $f(\sum x^2/n)$
I'm looking for a bounded loss function $L$ such that $\operatorname*{argmin}_{\hat{x}} \mathrm{E}_x[L(\hat{x},x)]$ is the mean.
$L(\widehat{x},x) := (\widehat{x}-x)^2$
If $L(\hat{x},x) = |\hat{x}-x|^2$ we get the mean. If $L(\hat{x},x) = |\hat{x}-x|$ we get the median.
3:45 PM
so you just found it
I'm looking for a bounded $L$.
bounded in which variable?
Both. I guess I should say bounded as a function of $|\hat{x} - x|$.
Since that'll satisfy the properties I expect.
Another example is, in the limit as $p \rightarrow \infty$, $\operatorname{argmin}_{\hat{x}} \mathrm{E}_x [|\hat{x} - x|^p]$ yields the mid-range, and in the limit as $p \rightarrow 0$ it yields the mode.
4:19 PM
Posted the question here if anyone's interested.
Sounds like the minimizer is a real monster
Was that a pun? lol
4:37 PM
Guess what
I'mma gloat here for a moment if you're OK with that
Just got back my math midterm
Only 100 in the course
100/100? Nice
It's not nice to gloat, DogAteMy! @Akiva
But congratulations.
Send me the exam, out of curiosity.
5:03 PM
Suppose we want to predict the probability of some future event. We can set up a prediction market where people trade binary options that pay $1 if the event occurs and $0 if it doesn't. The market price is then a good estimate for the probability of the event.

But what if we're interested in predicting a real-valued random variable X rather than a binary-valued one? I propose setting up a prediction market where people trade binary options of the form "X will be less than x" for arbitary x. It seems this would yield a good estimate for the *full* probability distribution (more precisely,
The motivation for my earlier question was the case where you're just interested in the mean of the distribution and want to set up a prediction market to trade options with payoffs between $0 and $1, in such a way that the price will be a good estimate for the mean.
Idea: instead of a box for a QED symbol, end proofs with your name
Thus $x$ equals the infimum. -Akiva Weinberger
(Don't actually do this)
Hm - maybe at the end of theorem statements is better
@user76284 Neat vídeo that talks about some related ideas:
Yeah, the logical induction paper used the market analogy IIRC.
I don't know enough about the stock market but I think while there are mechanisms for predicting the expected value of something, I'd be surprised if there were mechanisms for predicting the probability distribution
But your idea seems to make sense
@AkivaWeinberger that's Bayesian inference
What is, the video?
5:13 PM
no, predicting the probability distribution
Q: Which geometries are valid?

Ultradark$$X:=\Bbb H^2 \times \Bbb R$$ is one of the valid Thurston geometries. But can you do something like: $$\zeta:=\Bbb R^4\times\Bbb C^2?$$ What geometries can be combined like this and which ones cannot?

Quantum Electronic Dynamics (QED)
I never quite understood exactly what the Thurston geometries were
3 dimensional
@Ultradark What do you think of my comments on your lens question?
ZenoRogue posted some videos of the weirder ones on Twitter and YouTube and they are incredibly weird
and perhaps haunted
5:15 PM
by the Great Spectrum of the Ring
I have good reason to believe that a certain object can only be embedded up to $\Bbb R^4 \times \Bbb C^2$
@user76284 I will check to see your comment
I accepted your answer
In other words there is a maximal embedding space
Hello @TedShifrin can you help me?
I now want to know what a Thurston geometry is defined as
Oh yeah, I had a question earlier that you might be interested in: What is the smallest set of polyhedrons containing a regular n-gon, such that the set will tessellate the plane? Can it be expressed in terms of n? @AkivaWeinberger
5:38 PM
why do animations like that video above always move
These are some neat visualizations zenorogue.blogspot.com/2019/09/…
@Ultradark if it didn't move it wouldn't be an animation
@Ultra: I have no idea what you're talking about, but there is never a maximal embedding space. You can always set $\Bbb R^n$ into $\Bbb R^N$ with $N>n$ by fixing the last coordinates.
@Random-15 You shouldn't just randomly pin people. What is this about?
@Ultradark Just pause the video :P
hi @Mathein
5:47 PM
hi @Ted
Q: Minimizing an element in a finite Boolean ring.

Shine On You Crazy DiamondLet $A$ be a finite Boolean ring. Suppose we had a size function $\phi: A \to \Bbb{N}$, which means $\phi(x) = 0$ is not possible. The size function is such that $\phi(x +y) \leq \phi(x) + \phi(y)$. I'm looking for other properties required for $\phi$ to efficiently find a minimimum $x \in A$ ...

Can you crack the puzzle? As you can see there are many open problems in the area
@Mathein how did ya find Rösner's lecture?
@ÍgjøgnumMeg I liked it even if it was calculation-heavy
@Mathein yeah I thought it was cool but near to the end there were too many errors and I got lost hahaha
but it's in Neukirch so I'll take a look again and try to follow through my notes
6:03 PM
Good evening @Ted @Mathei @ÍgjøgnumMeg
Hey @Alessandro :)
good evening @Alessandro
So how's living and studying in Germany? @ÍgjøgnumMeg
@BalarkaSen Every oriented 3-fold is surgery on a link. You need to be able to do non+oriented 0-handle attachments too to get non-oriented guys
@AlessandroCodenotti RP^{2^n} fails to embed in R^{2^{n+1} - 1}
6:09 PM
@Alessandro it's been alright so far, I get the feeling I'm gonna need to work super hard to stay on top of things because of my somewhat weak background/rusted skills in some areas lol, otherwise it's going fine :) I spoke to one of my Klassenkameraden earlier and it turns out he's not finding the exercises that easy either and I was actually able to help him with some stuff
which is nice because I thought I was just weaker than everyone lol
so it's "nice" but.. obviously in a weird way
But Whitney's theorem is not sharp in all dimensions
I see, thanks @Mike
Iirc for instance it's only sharp in dims m = 2^n, and otherwise you can embed into dim 2m-1
Don't you do the exercises in small groups in Heidelberg?
@MikeMiller but for some m it could be even lower?
@Alessandro we have to hand them in in pairs, but my ANT practice session isn't until tomorrow so I don't have a partner yet, altho I know one person in the group so I'll ask her tomorrow
since approaching unknown people is of course forbidden
6:13 PM
I see
nah I saw some guy working on the ANT problem sheet in some open area today so I approached and asked how he was finding the exercises, but it turns out he's attending a different practice session to me
That shouldn't be an issue?
well it's not really but idk
I've handed in exercises with people going to other sessions before
maybe he was just finding an excuse to decline my offer
6:17 PM
lol could be
nah it's fine anyway
@Alessandro how's your semester going? :)
Quite well, I'm learning a lot of set theory and started working on my thesis
hi demonic @Alessandro
Consider the problem of finding the minimum of the quadratic form $r(\textbf{x})=x_1^2+x_2^2+x_3^3$ subject to the constraint $q(\textbf{x})=x_1^2+3x_2^2+x_3^3+2x_1x_2-2x_1x_3-2x_2x_3=1$. The diagonal representation of $q$ is $y_2^2+4y_3^2$ and that of $r$ stays the same. Solving the problem, one can simply insert $q$ in $r$ and minimise.
Though, why is this possible? Aren't the diagonal representations based on two different basis, i.e. the basis of eigenvectors of $q$ and $r$ respectively, and so represent different coordinate systems?
6:22 PM
@Alessandro that's great
I'm also finally learning diffgeo properly lol
lol nice, I wanted to take diff top but it clashed with modular forms
Careful when you say that around me, @Alessandro.
and I probably would#ve died in it
Hi @Ted
Hi @ÍgjøgnumMeg
6:23 PM
It's a real, honest course, no abstract nonsense @Ted we're following Lee
@Alessandro: Differentiable manifolds? I still don't call that differential geometry, but I've complained about this for years.
Both Lee's smooth Manifolds and Riemannian Geometry are listed as references, but we're working on topics from the former at the moment
6:37 PM
if a division ring satisfies Dedekind's Linear Independence of Characters, muss es denn ein Körper sein?
@LeakyNun I don't even understand how to interpret this
so our division ring $H$ satisfies the property that for every monoid $M$ the set of monoid homs $M \to H$ is linearly independent over $H$
6:56 PM
What does it mean to say that a $*$-homomorphism between $C^*$-algebras is norm decreasing?
Q: Elementary proof that $*$homomorphisms between C*-Algebras are norm-decreasing

Luiz CordeiroA lecturer once gave a very elementary proof that $*$-homomorphisms between C*-algebras are always norm-decreasing. It is well-known that this holds for a $*$-homomorphism between a Banach algebra and a C*-algebra, but all the proofs I find involve the spectral radius and so. If I remember it w...

@AlessandroCodenotti Yes
If $f(m)$ is the least integer so that any closed $m$-manifold embeds into $\Bbb R^{f(m)}$ then what's known is $f(2^n) = 2^{n+1}$ and $f(m) \leq 2m-1$ for $m \neq 2^n$
But the function itself is certainly not known in any explicit way
7:21 PM
@user193319 that $\|\phi(x)\|\leq \|x\|$ for all $x$, where $\phi$ is the homomorphism
I see, thanks @Mike
@AlessandroCodenotti sniped?
The proof is a straightforward computation from the *-property
double sniped?
8:16 PM
Does anybody know how I can find a moment generating function for $Y$ if I have the moment generating function for $X$
how is $Y$ related to $X$?
given that $Y=2(1-p) X$
lets say that $M_X(t) = \frac{a}{1-e^tp}$
$M_Y(t) = E[\exp(tY)] = E[\exp(2(1-p)tX)] = M_X(2(1-p)t)$
Hmm I don't understand the last part above
why does $E[exp(2(1−p)tX)]=MX(2(1−p)t)$
$M_X(t) = E[\exp(tX)]$ for all $t$
in particular for $2(1-p)t$ it also holds
8:19 PM
ah so our new $t$ is just $(2(1-p)t$?
But $2(1-p)t=t$ implies $(1-2p)t=0$
Makes sense, I overcomplicated the hell out of this
@MatheinBoulomenos you messed up
8:20 PM
Our moment generating function should be greater than 0 right
well at least 0
because $\exp > 0$
right doesn't that give us strictly greater than 0? ]
I guess
I forgot that our domain must have measure $1$
@LeakyNun is there any use for ordinal arithmetic?
it seems pretty esoteric
goodstein theorem
8:29 PM
okay maybe outside logic
no its completely useless
I dont know
I am looking for an equivalent property to this for a set $A$ on the real line, could anyone help please:
Another set $B$ on the real line intersects $A$ if and only if $\sup B\in A$.
Note that all open sets have this property.
$A$ must be convex since $a\leq b$ are two points in $A$ and $a\leq x\leq b$ then $\{a,x\}$ intersects $A$ and $x\in A$ by the property above.
Ok so this characterize connectedness on the real line. This property is motivated from this question:
Q: Inverse Image of $\sup$

ToposIn Folland's Real Analysis, he says that $$ g_1^{-1}((a,\infty]) = \bigcup_{j = 1}^{\infty}f_j^{-1}((a,\infty]) $$ where $g_1(x) = \sup_j f_j(x)$, but he doesn't explain why. Why is this true?

Now let me check it in general for order topologies.
8:52 PM
it must be (-infty,a]
Oh I haven’t checked in detail for other convex sets. Let me see.
Actually $(a,+\infty)$ also works right?
Same for $[a,+\infty)$?
$[a,+\infty)$ doesn’t work. My bad.
So we require that the left endpoint is open.
We can’t say any convex subset in partially ordered set is an interval in general right?
9:11 PM
@LeakyNun What would you recommend for an introduction to topology through normed vector spaces?
9:27 PM
@WilliamSun What's your definition of convex if you're not a vector space?
9:51 PM
@AlessandroCodenotti In any partially ordered set $A$ we define a subset $C\subset A$ to be convex if for any two points $x,y\in C$ we have $x\leq a\leq y$ implies that $a\in C$.
Then it doesn't need to be an interval
So if we look at the linear continuum that should be an interval.
But yes in general it is not.
It is an interval for linear orders
I just did a little more in this topic. The reason why we have to restrict ourself on a good ordered set(like a set with linear order) is because the definition of supremum of a set of functions from $X\to Y$ where $X,Y$ are partially ordered, does not behave naturally with the initial lattice structure(that is, the complete lattice $P(X)$)...
So it should not happen in general that the supremum and inverse image functor commutes.
Where the supremum refers to either of the set Hom(X,Y) or of P(X), the union.
I think P=NP
Here is the intuitive reason:
The CFG's generating a single string $s$ correspond in a way to elements in a ring.
Any one care to comment?
Donald Knuth also believes P = NP
10:30 PM
guys, I'm confused by the notation (x,y)=(r cos theta, r sin theta)
what exactly is the polar coordinate? is it (r,theta)->(r cos theta, r sin theta), or the other way around?
judging from "a point $p$ whose polar coordinate representation is $(r,\theta)=(2,\pi/2)$", I'm guessing it's the other way around as I wrote
but the notation is still confusing
yea ok, I had it differently in my mind, but it makes sense this way
10:58 PM
It's not confusing, @Sha. $(r,\theta)$ are the polar coordinates. The mapping from polar to cartesian coordinates is $f(r,\theta) = (r\cos\theta,r\sin\theta) = (x,y)$.
@TedShifrin What would you recommend for an introduction to topology through normed vector spaces?
Maybe Simmons' book that does both topology and analysis. It's a nice book.
Hello, quick question: Can a model interpret a formula both true and false or not interpret it in first order logic? Looking at the definition, it looks like the model is inheriting the internal logic of the meta-object/model. Just one or two sentence of guidance would be nice.
Any other options?
What are you actually looking for? It's hard to tell what to suggest if one suggestion is objectionable.
What do you want to know about topology? Why? Is there a specific approach you're looking for?
11:07 PM
If the density of a solid ball depends on the formula $k(2a-R)$ (where $k$ is a constant), is it obvious that the radius of this sphere then is $a$? The problem is related to evaluating a triple integral, and the integral w.r.t. to $R$ has limits from $0$ to $a$? (Shouldn’t $R=2a$?)
Well I know virtually nothing about topology. I'm taking a course. Yes, one that focuses on normed vector spaces. @MikeMiller
Edit: shouldn’t $R$ range from $0$ to $2a$?
It sounds more like you're looking for a book on functional analysis. Lax's is appropriate. Ted's suggestion is a fine book and does cover Banach spaces.
Thank you all
@schn if a is the radius and you're working with spherical coordinates then you know it should range from 0 to a
11:27 PM
@FuzzyPixelz Okay. It is not given in the problem that $a$ is the radius (in the solution it is), only that the density in the ball depends on the formula given. Looking at the formula, it would make more sense if the radius was 2a, since then anything beyond 2a of radius would have 0 density.
i.e. outside the ball.
11:48 PM
You could sensibly have a radius R=a: then the density would be highest in the middle, but still finite at the ball’s surface @schn
And if they meant the density to vanish at the surface, I’d have guessed they’d write it as k(a-R)
So my guess would indeed be that the radius is a not 2a
@Semiclassical Okay. Wouldn't the density be highest at the origin?
By middle I meant center ie origin

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