Mathematics - 2018-12-04 (page 1 of 3)

Mike Miller

00:00

Oh man

I misread the problem pretty badly

Ted Shifrin

@maths: Did you mistype? And what's your definition of the real numbers?

Mike Miller

No that was right

maths researcher

I just realized that thats the definition of a limit

@TedShifrin

Balarka Sen

I want to translate B by rationals to have disjoint copies getting inside a set of finite measure

Like the Vitali trick

Mike Miller

You are trying to prove that either A or R / A is null; in either case the point is that a null-measure set is null-measure in every subinterval

So a full-measure set is full-measure in every subinterval

Ted Shifrin

00:02

@maths: You typed rationals the first time and then reals the next time.

Mike Miller

So your function $f(\epsilon, x)$ is actually constant in both $\epsilon$ and $x$, you don't even need to take the limit

Drew Brady

yeah I see what you're saying

maths researcher

It wasn't a typo, but I also noticed that the reals are dense over the reals, so that holds also for the reals

@TedShifrin

Ted Shifrin

It's a better question with the rationals, @maths, or else you could take a constant sequence of reals. The reals case is way easier, shrinking distance by 1/2 each time.

maths researcher

@TedShifrin yeah, I realized that after I asked the question. It seems though as though that holds for any dense subset of R, am I correct?

Ted Shifrin

00:07

That's what dense means, actually, @maths.

user76284

Which probability distribution minimizes the sum of KL divergences to a set of distributions?

Balarka Sen

@Mike @Drew We're taking the function f(x, y) = m(A \cap [x, x + y])? f is constant for rational x, y - is that the line you're thinking about?

user76284

I recall reading somewhere that the Jensen-Shannon divergence between P and Q is the minimum of KL(R,P) + KL(R,Q) over all possible distributions R.

maths researcher

@TedShifrin I meant that for any x in the reals, and any subset that's dense, there will always exist a sequence from the dense set that will converge to x

Mike Miller

@BalarkaSen We were talking about the limit of your $f(x - t, x+t)/2t$.

Ted Shifrin

00:10

Yes, @maths, I know that's what you meant. Check why dense implies that's always true.

Mike Miller

I called that f(x,t).

Balarka Sen

Ah, ok

Drew Brady

You mean f(t, x)

maths researcher

@TedShifrin because of the squeeze theorem?

Mike Miller

Sure, whatever. :)

maths researcher

00:13

so if i have an element, say a in R then for all n in N there exists an x_n in the dense set such that, a-1/n<x_n<a+1/n

@TedShifrin

because the left side and the right side of the inequality converge to a, x_n must also

Ted Shifrin

Right, @maths. That'll do nicely.

maths researcher

Thanks @TedShifrin.

@TedShifrin I was wondering about a theorem, I think it's about coordinates gaussian normal coordinates. Where on any coordinate system, you'll always find a patch where the covariant derivative is just 0. There seems to be a bilinear transformation involved to prove the statement, but i'm not sure why that particular transformation has been chosen

@TedShifrin Why was a bilinear transformation, not a linear one required?

Ted Shifrin

No, you need the exponential mapping, which is based on geodesics.

This is pretty fancy stuff.

Drew Brady

Well, define f(x) = m(A \cap [0, x)). Then if 0 < x < y < 1, we have for large n, f(y + 1/n) - f(y + 1/n) = m(A \cap [x, y)) .

*oops, f(y + 1/n) - f(x + 1/n) = m(A \cap [x, y))

Similarly, f(y - 1/n) - f(x - 1/n) = m(A \cap [x, y))

CaptainAmerica16

@TedShifrin It's official, I can't think of one.

Drew Brady

00:23

So f(y + 1/n) - f(y - 1/n) = f(x + 1/n) - f(x - 1/n) for all x, y in (0, 1).

maths researcher

@TedShifrin Why is the exponential mapping required?

Ted Shifrin

Try to draw pictures, not use formulas, @CaptainAmerica. But, yeah, it's surprising, isn't it?

Drew Brady

Passing to limit as n -> 0, this means that f' is constant when it exists.

Ted Shifrin

@maths: That's the most natural way to do it. I can give a different argument using differential forms and differential equations, but it's geometrically less natural.

Drew Brady

Oh wait, I think that gives the result.

@MikeMiller, here's what I'm thinking. Let's look at B := A \cap [0, 1), and let us take g(x) = m(B \cap [0, x)). Note that if 0 < x < y < 1, for large n, we have f(y + 1/n) - f(x + 1/n) = m(A \cap [x, y)), by the property A + Q = A.

erm. m(B \cap [x, y))

Similarly, f(y - 1/n) - f(x - 1/n) = m(A \cap [x, y)).

Thus, we have that f(y +1/n) - f(y - 1/n) = f(x + 1/n) - f(x - 1/n) for all x, y in (0, 1) (assuming large enough n).

So when f'(y) := lim f(y +eps) - f(y - eps)/2eps exists (on (0, 1)), it must be constant.

maths researcher

00:28

@TedShifrin I have another question, if I have a surface embedded in the ambient space , why is it that I can't use the definition of the christoffel symbol where its the covariant derivative dotted with the bases? Why do I have to consider the christoffel symbol thats dependent on the metric tensor?

Drew Brady

So supposing that m(B) > 0, we know that this constant must be 1 by Lebesgue density theorem.

Ted Shifrin

You don't have to. The beautiful result is that you have two ways of computing them — one extrinsic, one intrinsic. Are you reading my notes? This sounds like fancier language than you need.

Drew Brady

And now the result follows from the fundamental theorem of calculus: m(B) = f(1) - f(0) = \int_0^1 f' = 1.

And then we conclude that m(R \ A) = 0.

Ted Shifrin

Providing the FTC applies to $f$, @Drew.

Not that I'm following carefully ...

Drew Brady

oh, but it does.

CaptainAmerica16

00:30

@TedShifrin That is surprising. It feels like I'm missing something...

Drew Brady

because f is Lipschitz hence absolutely continuous

so the FTC applies by Lebesgue theorem.

Ted Shifrin

OK, @Drew. I just wanted to be difficult. :)

Drew Brady

No worries @Ted, I think this works

Ted Shifrin

@CaptainAmerica: Why are you missing something? Cuz it's not obvious? it's not.

But you should be able to convince yourself with pictures.

Mike Miller

It seems good to me but I did not read carefully.

CaptainAmerica16

00:32

@TedShifrin I always second guess myself with stuff like this. Now I'll have something to draw when I'm bored.

Ted Shifrin

I don't want to ruin your fun by giving hints yet.

This is an exercise in Spivak, too, of course.

maths researcher

@TedShifrin I haven't read your notes yet. But if I consider the one with the dot product wouldn't the curvature of the surface and the euclidean position vector, give a vector that is perpendicular to the surface? So I wouldn't be able to parallel transport it because it doesn't lie in the same tangent space

?

Ted Shifrin

That didn't make any sense to me. Try again?

What are you reading?

maths researcher

@TedShifrin Ive read intro to tensor analysis and the calculus of moving surfaces

Ted Shifrin

That doesn't help me much. I think you should start with my notes. They're less abstract and there are lots of examples and intuition in there.

maths researcher

00:36

A first course in curves and surfaces?

I'll start reading it

Ted Shifrin

Yes.

maths researcher

What prequisites are required it to fully grasp the material?

CaptainAmerica16

I'm on chapter 2 of Spivak now.

maths researcher

@TedShifrin your notes I mean

Ted Shifrin

@maths: Multivariable calc and 2-D linear algebra. Of course, stuff at the level of my blue book gives a deeper understanding at places. And occasionally you need some basic analysis (like the maximum value theorem for continuous functions on compact spaces).

LOL, @CaptainAmerica. Yippee.

CaptainAmerica16

00:50

@TedShifrin Slow and steady wins the race.

Ted Shifrin

Regarding that two function thing, remember not to try to draw another twice-differentiable function when you draw pictures.

CaptainAmerica16

I taught my little sister pig latin and now she keeps calling people oehay.

CaptainAmerica16

01:09

I'm gonna peace out, I have homework. Thanks for the problem Ted.

Ted Shifrin

Night, @CaptainAmerica. Stay calm.

CaptainAmerica16

Lol, I'll try.

Balarka Sen

I love how in most blogpost of Terence Tao there will be at least one person who writes a comment starting with "Dear Prof. Tau"

Ted Shifrin

It sounds right, @Balarka.

Jeff

01:41

Hi folks. Two questions please. How does one solve the system $2x=Ly, 2y=Lx, xy=1$? (Answers are (1,1) and (-1,-1)). And then how does one find the "endpoints" of $xy=1$?

(oh yeah, L is $\lambda$)

Ted Shifrin

Well, note that $\lambda = \dfrac{2x}y = \dfrac{2y}x$, so, cross multiplying, $x^2=y^2$, so $y=x$ or $y=-x$. (Note that we can't have $x$ or $y$ equal to $0$, since $xy=1$. So I didn't divide by $0$.)

Jeff

OK. Good, got that. Now the hard part.

(and TY, @ted

Ted Shifrin

There are no endpoints.

The curve goes on forever ...

Jeff

That's what I thought...

Ted Shifrin

Usually with Lagrange multipliers one doesn't try to give a proof that one's found the global max/min. But one can make arguments based on the problem at hand.

Jeff

01:48

This is a Lagrange multiplier question: Minimize $x^2+y^2$ subject to $xy=1$. The instructions say to solve that system, then find the endpoints .

@TedShifrin OK, I think I understand what you mean.

Ted Shifrin

Find the endpoints makes absolutely no sense.

You can make an argument that you move off to infinity on $xy=1$, the function $x^2+y^2$ gets bigger and bigger, so you're certainly never going to minimize it by going far away.

Jeff

OK. Got it.

So now what does it mean (from a different problem) to find the endpoints of $x+y=20$. They seem to assume the endpoints are $(20,0), (0,20)$. But I beg to differ. Note that the problem does NOT say "in the first quadrant" or any other such restriction.

Ted Shifrin

Then it's garbage.

What's the function they're trying to extremize? Is there some reason its domain should be just the first quadrant?

Jeff

Instructions: Find the required constraint extrema (<-- maybe this is why)
Question: Maximize $x^2+y^2$ subject to $x+y=20$.

Ted Shifrin

For that function, it's natural to restrict to nonnegative numbers ... because of symmetry.

Jeff

01:54

OK, that makes sense

Ted Shifrin

But Lagrange multipliers doesn't work on things with endpoints ... so I don't know why they're playing this game.

Jeff

Maybe cuz it's introductory?

Ted Shifrin

introductory?!

I've taught for 40+ years and I've never done stuff like this.

Jeff

It's a Calc 2 book. Not a graduate level book.

Ted Shifrin

This has nothing to do with grad level. Maybe there's some discussion somewhere in the text of what they're doing. But I certainly have never seen it anywhere.

If this is Calc II, I'm guessing it's a calculus for business/economics type book.

Jeff

01:56

@TedShifrin Yes, that's right.

Ted Shifrin

I have no idea what sort of stuff the authors wrote.

It seems like they're asking for more of a rigorous justification than most Calc III courses do.

Jeff

It's kinda foreign to me, too. I'm racing through it fast. The subsection heading is "Constrained Optimization Problems". Authors Smith, Strauss, Toda.

Ted Shifrin

Well, you'd better race slower and see if they're explaining what they're doing with this endpoints stuff.

There are zillions of calculus books. I certainly don't know this one.

Anything else? I'm about to disappear.

Jeff

No. Thanks!!

Ted Shifrin

See you next time!

Prashin Jeevaganth

02:12

Does anyone know what's the general technique to draw a diagram to understand the question

it seems the solution for this is very dependent on the accuracy of your diagram

The Great Duck

0

Q: Does $\int_{0}^{x} f(g_1(t),g_2(t)) dt - \int_{0}^{x} f(g_1(t),g_2(x)) dt$ reduce in any meaningful way?

I'm trying to figure out what $\int_{0}^{x} f(g_1(t),g_2(t)) dt - \int_{0}^{x} f(g_1(t),g_2(x)) dt$ reduces to, if anything. I know it does simplify to $\int_{0}^{x} f(g_1(t),g_2(t)) - f(g_1(t),g_2(x)) dt$ but I curious as to whether any useful integration identities stem from this. I know that i...

integration

Adam

03:16

what campus is stack exchange math community mostly based? like ive noticed a lot of people discuss having to go here and there etc

Dair

There are a couple of U of Chicago people on here.

Not me tho... (lol)

Let's see, Daminark is U of Chicago I believe, the name escapes me, but there is someone else there. Mike Miller is at UCLA according to his profile. Kenny is at Imperial College in London...

Balarka Sen

03:36

Eric is from there as well

Adam

And Balarka?

Balarka Sen

Adam

also what is up with being allowed to edit Wikipedia being a reputation point based privilege on stack exchange? how does that work

like I am studying a theorem that I originally found on Wikipedia, and I checked Wolfram and it calls something completely different by the same name, and the Wikipedia page, which states the one I was already aware of is missing a heap of crucial information that's needed to make the theorem true

but in saying that I'm not entirely sure I should be allowed to edit Wikipedia at will. That seems to be a dark road to travel for humanity in general

well id find it hilarious but there would be a lot of angry people at some point because of something I put when drunk

ok gj

Dair

04:01

@Adam You could get IP banned if you did malicious edits. Depending on your seniority you are supposed to go to the talk pages before you edit anything. "That seems to be a dark road to travel for humanity in general" It can be, especially depending on the subject. There are some popular and controversial pages on Wikipedia that go back and forth on edits.

Also, I'm not exactly sure how you would get a good reputation system going there...

Adam

06:11

@Dair ok so you get what I am saying I don't necessarily want the freedom to edit any given article at any given time, what I am saying is there is one in particular that is very short and vague, and to make matters worse the other public encyclopedia, being wolfram, has an article under the same name for what appears to be a different theorem at first glance, but of course I could be mistaken and it's just been very different phrased with different terminology,

anyway I'm not saying either resource is better than the other I've used both a huge amount and for anyone without constant access to other journals the are pretty invaluable, I just know I would be able to clarify things for the reader on that particular subject, but it tells me I should be able to after x amount of points here, then it switches and says I have access to review ques,

which as far as I can tell is just an empty list with a message saying something along the lines of @#$#@ off we already did everything

ques? cues? they both seem right

sorry, to clarify, I just want to know the process for me to submit an addition to a Wikipedia article, but I also feel as if those that review it should have to argue with me directly if they are going to decline it's approval

Adam

07:18

also finding it difficult today to see how $0$ is in $\mathbb Q$, $\mathbb R$ or $\mathbb C$ for that matter, I mean sure ok I get it is obviously in all three but for starters we expect all the others in $\mathbb Q$ to come up with two precise integers for which they are the ratio of. and then with $0$ it can just have itself on the numerator and any random stranger can get beneath it really 0 isn't fussed

that was not what I wanted to write in wiki fyi

ÍgjøgnumMeg

08:17

@Adam elements of $\Bbb Q$ are equivalence classes of pairs of integers. The equivalence class of $0$ is then just $(0, k)$ with $k \in \Bbb Z\setminus \lbrace 0 \rbrace$

(note the relation here is that $(a, b) \sim (c, d) \iff ad - bc = 0$, so you can see that $(0, k) \sim (0, k^\prime)$ for any $k, k^\prime \in \Bbb Z\setminus \{ 0\}$)

Theantomc

Hi guys, anyone can help me with Chernoff bound?

Secret

09:51

What about it

Theantomc

10:07

@Secret cs.stackexchange.com/questions/100870/…

Secret

> Use a Chernoff bound.
> Use the fact that variance is additive for independent random variables.
> You might as well use the stronger Chernoff inequality.
> (User have no idea what a Chernoff inequality is)
> (Yuval drops out of discussion, just like many of the past people who are in the middle of answering a question, but refuses to acknowledge that they refuse to refuse to help the user)
> Traps @Yuval into a Klein bottle

Theantomc

i did a dimostration using Chernoff bound but isnt so easy argument for me

but i dint use a variance @Secret

i followed this paper people.eecs.berkeley.edu/~satishr/cs270/sp17/rough-notes/…

(last part)

Secret

I don't understand why a Chernoff bound is used here since this has nothing to do with time? Also what is $\Theta$?

Theantomc

$\Theda(d) in this case is largest distance (approximation)

a sort of bound @Secret

i m using this in vectorial space reduction, in computer science... So i m with Chernoff bound i m so far

Secret

10:27

hmm...

$E[(x-y)^2]=\text{Var}(x-y)+E[x-y]^2$

Theantomc

i tried to calculate the variance but in my case is negative ...so i wrong some in calculation

Yuval hell me in expecation here cs.stackexchange.com/questions/100770/…

and after i calculate with 1/6 the variance

integrate (x-y)^4 - (1/6)^2

i m not looking for solution, i just want understand how work, the conceptual aspect

Secret

I am trying to understand how the exp is not involved in the expectation value

thinking

Theantomc

thanks for help @Secret

Leaky Nun

@AlessandroCodenotti cof(L) = cof(A[L]) = cof(K) = K = A[L], so cof(L)=L, so K=A[L]=A[cof(L)]=A[K]

Secret

10:43

hmm... the uniform distribution of the d cube is $\frac{1}{2^d}$. Thus:

$$E [||x-y||^2] = \frac{1}{2^d}\sum_{n=1}^d\int_{-1}^{1}(x-y)^2dxdy$$

(sorry missed out another $\int_{-1}^1$)

Theantomc

For the variance?

I cant use integrate E(x)^2 - u^2?

I need to see good you formulate for expectation

Secret

$\text{Pr}((x-y)^2 \geq (1-\epsilon)cd) \leq \min_{t > 0} e^{-t(1-\epsilon)cd} \prod_i e^{t(x_i-y_i)^2}$

hmm...

each $(x_i-y_i)^2$ is part of the uniform distribution of the d cube, so:

$$\text{Pr}((x-y)^2 \geq (1-\epsilon)cd) \leq \min_{t > 0} e^{-t(1-\epsilon)cd} \prod_i e^{t(x_i-y_i)^2} = \min_{t > 0} e^{-t(1-\epsilon)cd}e^{dt(x-y)^2}$$

now what is epsilon...

Prashin Jeevaganth

11:03

Does anyone have good books for calculus mathematical modelling? I can't seem to get any rate of change or growth model question correct because I don't understand what the english is talking about

Akiva Weinberger

@Semiclassical Hey, remember Truchet tiles, which we were talking about a while ago?

Dec 5 '16 at 21:46, by Semiclassical

Dec 6 '16 at 4:24, by Semiclassical

It occurs to me that if you add two "crossing tiles", you can get knots, like this:

I don't know what to do with this, but it was just a random thought

(I wonder if the picture is the smallest number of tiles required to make a trefoil)

Theantomc

why @Secret?

Akiva Weinberger

Hm, there are papers on these: mypages.iit.edu/~krawczyk/rjkisama11.pdf

Theantomc

11:19

you use a sort of bound , but i dont understand how @Secret

because with product you use a sort of momentum

Akiva Weinberger

Oh, this is neat

10 PRINT CHR$(205.5+RND(1)); : GOTO 10

►

Basically, approximate this sort of tiling by maxing a computer randomly print forward- and backslashes in a monospace font

One line of BASIC

Oh, what, there's a one hour talk on it?

10 PRINT CHR$(205.5+RND(1)); : GOTO 10

►

I'm gonna have to look at that 'cause I have no idea how you'd talk about it for an hour

> This book takes a single line of code—the extremely concise BASIC program for the Commodore 64 inscribed in the title—and uses it as a lens through which to consider the phenomenon of creative computing and the way computer programs exist in culture.

> The authors of this collaboratively written book treat code not as merely functional but as a text—in the case of 10 PRINT, a text that appeared in many different printed sources—that yields a story about its making, its purpose, its assumptions, and more.

> They consider randomness and regularity in computing and art, the maze in culture, the popular BASIC programming language, and the highly influential Commodore 64 computer.

From here

Theantomc

11:34

@Secret this is standard form, the result will be 1/6 for the 3 dimensional (cube). I agree on this

Akiva Weinberger

11:47

Whoa: 3D Trechet

Why does this 3D regular tiling remind me of Escher:

(And also, what's the connection to Trechet?)

From here

@Secret I remember you showed interest in this last time it was mentioned here

You tried to draw a tiling by hand

Secret

12:05

@Theantomc Ok I thought I got even that wrong, so at least I have the correct expectation value. But what is $\epsilon$. To apply the Chernoff bound, we need the tail of the probability. What I am not sure is how to derive the upper bound of the expectation value $(1-\epsilon)cd$

smci

Can I get an opinion from you regular users before I go ahead make a post on Math.Meta? Arising from the (ahem) differences in opinion on what position (if any) Math.SE takes on 'cultural meta-tags' of math-related questions, such as sangaku, 'gematria', 'vedic'. I don't think Math.SE has any clear position, so I'm going to start a Meta discussion (Note:

(Note: I'm not making any proposal. Just starting a discussion placeholder so that Math.SE can figure out if it has a coherent position. Before we start to get meta-tags like the >120 different genres of Japanese logic puzzles, just for constraint-satisfaction logic puzzles on nxn grid

Secret

@AkivaWeinberger Yes, I recall I have talk something about them, but I need to catch up again by locating that post before I can comment furthr

Akiva Weinberger

Hm, if we do the hexagonal version on this, we could see the whole pattern at once:

Secret

that looks nothing like a Truchet pattern

it's too tidy looking

rather than maze like

Akiva Weinberger

Or, alternatively, if we put the original (square version) here:

@Secret Here

Mar 26 at 0:08, by Secret

And what I meant by "do the hexagonal version" is

Dec 6 '16 at 4:31, by Semiclassical

Replace each hexagon with three parallel lines, like that

which, on the plane, gives this:

Dec 6 '16 at 4:24, by Semiclassical

Secret

12:14

Ah I see

Mar 26 at 0:09, by Secret

One reason I exposed myself to fine art is to be able to cultivate an intuition so I can drew these labyrithine patterns easily

hmm...

51 mins ago, by Akiva Weinberger

10 PRINT CHR$(205.5+RND(1)); : GOTO 10

►

I wonder, if what most of us considered as totally random noise, actually has some kind of indefinable structure in an artistic intuitive perspective. That might explain why they look so structurally complex despite they can be generated with simple rules

alan2here

whats this family of rules called? I want to apply it to things that are not reals

Secret

29 mins ago, by Akiva Weinberger

3D is the smallest number of dimensions needed to introduce knots, thus it is reasonable that 3D truchets can be a lot more knotty

Akiva Weinberger

@alan2here What rules?

alan2here

@Secret yes, for example the normal distribution

Akiva Weinberger

@Secret I started with 2D knots

1 hour ago, by Akiva Weinberger

Trefoil, made using four types of tiles

It's a projection of a knot (a knot diagram), at least

Secret

12:18

That's technically 3D else you will have a self intersection, but yeah

"2.5D"

Akiva Weinberger

I wonder if you can get a knot with fewer than 12 such tiles

alan2here

2 points moving smoothly (meaning without teleporting) along a line starting at A and B and then moving to each others starting locations have to pass each other.

A very obvious statement, but you can build on it to make the following statements as well.

---

A convex shape is a shape where any two points on the shape must be able to be joined with a straight line that stays inside the shape. Convex shapes include rectangles and star shapes and the shape of a comma. The shape of the letter 'c' is not convex, because it's got an alcove.

Adam

sorry this does get messy when people are in an object unlabelled kind of mood .cannabis kills people there more you know

Secret

@AkivaWeinberger That page shows all things tiling, not just Truchet patterns. Btw, the last few diagrams have physical relevances. It is one reason meshes are pretty resilient and strong materials despite being porous. Those interlocking give strength

Akiva Weinberger

> Convex shapes include rectangles and star shapes and the shape of a comma.

Stars aren't convex

although there is a weaker notion of a "star-convex shape", in which there is at least one special point which can be connected to anything else without leaving the set

Balarka Sen

12:21

They are, well, star-convex :D

Akiva Weinberger

Snipe

Balarka Sen

Rip

alan2here

thanks, I'll fix that mistake

Akiva Weinberger

> Any three reasonably overlapping convex shapes can be cut with the same cut such that all the shapes are cut in half, it's two parts with the same area.

I don't think I've heard of this, actually

Adam

@ÍgjøgnumMeg ok so in the same manner we can separate the integers into a partition consisting of two elements, being the set of even integers and the other the set of odd, what is the generalized name for these, Binary relation?

Akiva Weinberger

12:23

The wobbly table theorem (if that is it's real name) and the Borsuk–Ulam theorem are famous

alan2here

The convex shape one is called the ham sandwich problem :) A great name. All the shapes can even be different from each other.

Akiva Weinberger

Ah

I've heard the name but forgot the statement

alan2here

@Adam "equivalence classes" is a more general idea like this, where a set is divided in it's entirety into smaller non-overlapping sets

@AkivaWeinberger There must be a name for this property, it's bassicly the same property in all my examples.

Akiva Weinberger

Intermediate Value Theorem

alan2here

thank you :)

Secret

12:30

@smci If there is no similar question on this, feel free to make a meta question on it to get the community to discuss

alan2here

I assume that more intricate formulations based on the "intermediate value theorem" such as the I'm hungry and want a ham sandwich problem, table and planet examples must work in R and R<sup>2</sup>. And presumably therefore in vectors. And even more presumably in tensors and in the complex plane.

smci

@Secret Sure. Would appreciate any advice on how to format said post to make it 120% clear I'm not advocating for/against meta-tags on Math.SE, just creating a place for that discussion to happen. Because posts on meta-sites invariably get cranky/anonymous downvotes from people who misunderstood the post, and react based on assumptions.

alan2here

But maybe it all works in wilder examples, such as for (nodes and edges) graphs too, for example.

Secret

smci: That I am not sure, for I have very little experience on meta since I am not active there. You might be better off checking with other regulars

Akiva: https://chat.stackexchange.com/transcript/message/47887452#47887452
I am wondering, does a hyperbolic Truchet tiling exist

semanticscholar.org/paper/Hyperbolic-Truchet-Tilings-Dunham/…

OOOoooooo

alan2here

I might make a question now :)

Akiva Weinberger

12:36

It surely does, just draw a // or \\ on each square in the order-5 square tiling, just like you'd do for the Euclidean version

@Secret Oh, cool

alan2here

I like the / \ basic program posted above :) It's very elegant.

Astyx

12:55

Hi chat

ÍgjøgnumMeg

@Adam this is just $\Bbb Z / 2\Bbb Z$. An integer is $0$ if it is even and $1$ if it is odd.

or rather, the image of an integer in the quotient

Theantomc

@Secret you see this paper? people.eecs.berkeley.edu/~satishr/cs270/sp17/rough-notes/… at last page speak how to apply Chernoff at hypercube

@Secret for me \theta(d) is the distance x (the larger one, from to point choose random), and i think that 1-\epsilon is an approximation for (i dont want wrong) lower bound

Theantomc

13:21

hi @Astyx :)

alan2here

13:56

What do you think of my theorem?

0

A: Extenstion of Intermediate Value Theorem.

I expect such a theorem to already exist, for if it doesn't, I posit the Intermediate Slice Theorem. It makes statements analogous to the intermediate value theorem. $d$ is a member of N. P = $\mathbb{R}^{d}$ Q is an $\mathbb{R}^{d - 1}$ dimensional line in P. A 0 dimensional line is a point,...

I tried to relate this to functions like in the original 1D case, but I wasn't able.

Somone tell me if this is just too ridiculous before I leave the answer there :-P

Secret

14:39

The windmill saddle $\frac{x^2-y^2}{x^+y^2}$ will be a good counterexample to your intermediate slice theorem

In fact, in 2 dimensions and above, you need to fix lines, curves etc. in all directions to ensure the resulting surface can cut through exactly one point. Even that is not enough as if you allow the function to curve a bit, it can easily oscillate and ruin the intended result of the intermediate slice theorem

This is one reason why boundary conditions of PDES are so hard to solve

rschwieb

wave

Secret

@Theantomc Ok I understood where to look for the information. I am not sure whether I can get back to you in time as it will took me some time to figure out where the epsilons and etc. are to be plugged into the inequality correctly to produce the required bound. At least not tonight I can reply. Check with other probability users in the meantime.

At least I knew I am roughly on the right track as I get the correct standard form of $E[(x-y)^2]$

Theantomc

14:55

@Secret so i try to good deep in a Chernoff bound (because i need for my proof)

@Secret i see that for Chernoff need a Variance .. I just have an expectation (1/6)

So i need to compute this part too

@Secret i m not sure if Yuval here cs.stackexchange.com/questions/100770/… calculate the variance too

because use this formula ( x-E(x))^2

Semiclassical

@AkivaWeinberger cute. i can see some interesting statistical questions, e.g. if I make a random n-by-n tiling with crossing tiles then what kinds of knots can I expect to see

Theantomc

i can apply this E(X1−X2)^2=2Var(X1). @Secret

Semiclassical

dunno how one would implement that on a computer tho. creating a random tiling of that kind isn't particularly hard, but getting the computer to recognize the knot structure seems far hairier

Akiva Weinberger

15:20

@Semiclassical The Jones polynomial seems pretty compatible with this sort of thing

Basically, say your four tiles are //, \\, -|-, and — (I can't really draw the last one well with characters)

so right-leaning (smooth), left-leaning (smooth), vertical overcrossing and horizontal overcrossing

Semiclassical

yeah

Akiva Weinberger

There are two ways to "smooth" a crossing: A, where you rotate the overcrossing a tiny bit counterclockwise and cut along there, and B, where you rotate it the other way

so -|- has A-smoothing \\ and B-smoothing //

and — has A-smoothing // and B-smootning \\

$\div$

In the Jones polynomial, what you do is, you replace each crossing with the formal sum of $A(\text{A-smoothing})+A^{-1}(\text{B-smoothing})$

so if there are $n$ crossings in your knot or link, you end up with $2^n$ terms

each term being a disjoint union of loops, with a coefficient being $A$ raised to some power

and then, if your term has $u$ disjoint loops, it becomes the expression $(-A^2-A^{-2})^u$

so the reason we do this is because

Semiclassical

the jones polynomial falls squarely in the realm "stuff I've heard of but don't know about"

i know it's a knot invariant, which is handy

Akiva Weinberger

Naively we might just do $A(\text{A-smoothing})+B(\text{B-smoothing})$ and then each disjoint loop in a totally smoothed term contributes a factor of $C$

Semiclassical

(bah, why does this matlab program I grabbed keep telling me that it can't solve my problem when I know it can)

Akiva Weinberger

15:28

but it turns out, if we want it to be invariant under the Reidemeister II move, we need it to have $B=A^{-1}$ and $C=-A^2-A^{-2}$

Reidemeister II is the second one^

And this makes it invariant under the Reidemeister III move as well

Semiclassical

i have too impure a mind to look at II without snickering a bit

Akiva Weinberger

It's not invariant under the Reidemeister I move, but a Reidemeister I move just multiplies it by $A^3$ or $A^{-3}$ (depending on which way you do it) so it's not too bad

There's something called the "writhe" which counts the Reidemeister I moves, in a sense, so you can multiply your thing by $A^{-3w}$ and it cancels out that effect

I realize this probably isn't the best way to describe the Jones polynomial

and if I had a better ability to draw lots of pictures it would make more sense

The Wikipedia page doesn't do too bad a job at describing it but it doesn't explain why $A^{-1}$ and $-A^2-A^{-2}$ are used

Semiclassical

the main trouble I can see computationally is the number of terms involved

i.e. 2^(# of crossings) diagrams

Akiva Weinberger

I think computing the Jones polynomial is NP-hard? Or something similar

Semiclassical

sounds right

no one ever said knots were easy

Akiva Weinberger

15:35

#P-hard, apparently

not that I know what that means

I like the idea of, instead of having a tile in a certain position, you have a formal linear combination of tiles

Adam

@ÍgjøgnumMeg ok here is the situation in full I must formally describe: suppose for each index $p \gt 1$ we have a set $S_p={\{(n,m):f(n,m) \in {\{1,2,3,...,p}\}}\}$. So as can be expected one can partition this into disjoint subsets, ie $S_p={\{(n,m):f(n,m)=1}\}\cup{\{(n,m):f(n,m)=2}\}…\cup{\{(n,m):f(n,m)=p}\}$. now the most important aspect here most of all, is that $S_p$ is the union of exactly $2$ such subsets if and only if $p$ is prime.

Akiva Weinberger

It doesn't quite turn this into a "ring", but it gives some sort of structure

The distributive property says that, if a certain square is \\ plus // (say), it equals the entire tiling with \\ plus the entire tiling with //

but there isn't really a "multiplication" here, only various sorts of concatenation

Semiclassical

@AkivaWeinberger i'll admit, it appeals to the quantum person in me insofar as it's like you're considering a superposition of all possible smoothings of your diagram

Akiva Weinberger

Heh

Adam

@ÍgjøgnumMeg ie $S_p={\{(n,m):f(n,m) =1}\}\cup{\{(n,m):f(n,m) =p}\}$ if and only if $p$ is prime

@ÍgjøgnumMeg the quantity of partitions for which $S_p$ has proportion to the number of distinct prime factors of $p$

user131753

15:47

Please have a look into the following answer,

user131753

12

A: Maximum number of edges in a non-Hamiltonian graph

Found a lot easier proof for the harder part, using Ore's theorem: If $G$ is complete, there's a Hamiltonian cycle. So suppose it's not. We'll take some non-neighbor vertices $v,w$, and delete them from the graph. For the resulting graph $G'$ - $|E(G')| \geq \binom {n-1}{2} + 2 - (d(v)+d(w))$. In...

user131753

Can anyone explain to me how is it true that "For the resulting graph $G'$ - $|E(G')| \geq \binom {n-1}{2} + 2 - (d(v)+d(w))$."?

Adam

@ÍgjøgnumMeg the cardinality of the partition of $S_p$, or the total number of disjoint subsets it can be split into with this predicate,has proportion to the number of distinct prime factors of $p$

Akiva Weinberger

@Semiclassical

Here's explanation of where the constants come from

In the box is our "hope" for what our invariant will look like

We hope we can replace each crossing with a linear combination of its smoothings

The second line is the same as the first, just rotated 90 degrees

and the third line is saying that each disjoint loop contributes a factor of C

So now we try out our hopeful invariant on the "poked" Reidemeister II move

Collecting terms, we get that our thing will be invariant under that move if $A^2+ABC+B^2=0$ and $AB=1$

This forces $B=A^{-1}$

which means the first equation becomes $A^2+C+A^{-2}=0$

and so $C=-A^2-A^{-2}$

There's a pretty neat way to see that, if we have Reidemeister II, we have Reidemeister III

Let me draw it

user131753

Ok. Nevermind.

Adam

15:57

where constants come from? lol

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