Mathematics - 2020-02-16 (page 1 of 2)

12:00 AM

so what you're doing is this: you take a generator for each generator of $A$ (of course you can choose $S_A=A$ if you want) and a generator for each generator of $B$ and the only relations we're enforcing are the relations coming from $A$ and $B$ and we want the elements of $C$ to become trivial in the amalgamated product

so we're glueing together $A$ and $B$ along the common subgroup $C$

you have canonical homomorphisms $A \to A \ast_C B$ and $B \to A\ast_C B$

Rithaniel

Hmmm, alright, let me write something up and post it in here

Jacksoja

Is there a fast way to check if 2 is a generator of large prime ?

I think what am asking is , is there some criteria on the prime p such that 2 is a generator of F*_p

Ted Shifrin

It's unknown (according to experts who told me when I asked) how to predict who generates as a function of $p$.

Jacksoja

haha bad luck for me then

Rithaniel

If there were $a_3\in A_3$ such that $a_3\notin\text{Im}(f_2)$ then let $\text{Im}(f_2)=\langle F\mid R_1\rangle$ and $\langle a_3\rangle=\langle a_3\vert R_2\rangle$ where $F$ and $\{a_3\}$ are sets of generators and $R_1,R_2$ are sets of relations. Then let $D=\langle F, a_3\mid R_1\cup R_2\cup(\langle a_3\rangle\cap\text{Im}(f_2))\rangle$

Jacksoja

12:07 AM

@TedShifrin Why did you ask for that question ?

just curious

Rithaniel

Is that a correct way to describe the amalgamated free product?

Ted Shifrin

Because I wrote an exercise on the group theory of card shuffling and wanted to know a general formula for how many perfect shuffles it takes to return the deck to its original order.

Jacksoja

Neat :)

Ted Shifrin

That's precisely asking for the order of $2$ in the multiplicative group.

Leaky Nun

Artin's conjecture on primitive roots

In number theory, Artin's conjecture on primitive roots states that a given integer a that is neither a perfect square nor −1 is a primitive root modulo infinitely many primes p. The conjecture also ascribes an asymptotic density to these primes. This conjectural density equals Artin's constant or a rational multiple thereof. The conjecture was made by Emil Artin to Helmut Hasse on September 27, 1927, according to the latter's diary. The conjecture is still unresolved as of 2019. In fact, there is no single value of a for which Artin's conjecture is proved. == Formulation == Let a be an integer...

Under GRH, 2 is a generator of 37.4% of the primes

Jacksoja

12:16 AM

It seems like most theorems about primes are probalistic

Leaky Nun

because they're related to distributions

distribution <=> measure theory <=> probability

Jacksoja

the other theorems are unsolved

what I mean is , there is too much we dont know yet about primes

Semiclassical

12:40 AM

@Jacksoja for some philosophical remarks on that, see the first few paragraphs of this Terry Tao blog post: terrytao.wordpress.com/2015/01/04/…

Jacksoja

@Semiclassical thanks

Lukas Heger

12:54 AM

hey @Daminark

topologicalmagician

In general topological spaces

does the set of limit points of a set $A$

contain its limit points?

Leaky Nun

@topologicalmagician I think so

Lucas Henrique

1:17 AM

hey chat

I just saw a question about matrix multiplication and that lead me to another question: the ring $\mathcal M_n(F)$ of $n\times n$ matrices over a field $F$ can be seen as the ring of linear transformations from $F^n$ to itself. Is there any analogy with modules to talk about this when it's $\mathcal M_n(R)$, where $R$ is a commutative ring?

Lukas Heger

@LucasHenrique yes

just replace $F^n$ with the module $M^n$

Leaky Nun

$R^n$

Lukas Heger

yeah, that

Lucas Henrique

I suppose that most of the theorems from linear algebra won't work, though

Leaky Nun

you still have Cayley--Hamilton

Lukas Heger

1:28 AM

things get a lot more subtle, yes

for example, for vector spaces, there's one notion of finiteness: finite dimension. For modules you have a whole zoo of finiteness conditions: finitely generated, finitely presented, coherent, finitely cogenerated, Noetherian, Artinian, finite length etc.

for vector spaces, these all agree

Rithaniel

1:49 AM

Note to self: Chat doesn't have xymatrix enabled

I've been trying to prove exactness in this one sequence for, like, 3 hours now. I've gotten everything except the fact that $\text{ker}(f_2)\subseteq\text{Im}(f_1)$

(That's always the most difficult bit)

I am given that $0\to\text{Hom}_R(A_3,D)\to\text{Hom}_R(A_2,D)\to\text{Hom}_R(A_1,D)$ is exact for all $D$

Where the homomorphisms are the induced homomorphisms from $f_1$ and $f_2$. Ie $g_1(\psi)=\psi\circ f_1$ where $g_1:\text{Hom}_R(A_2,D)\to\text{Hom}_R(A_1,D)$ and $\psi\in\text{Hom}_R(A_2,D)$

topologicalmagician

@LeakyNun it only holds for $T_1$ spaces

Let $(X,d)$ be a metric space.

Show that if $q\in B(x,r)$ and $q\neq x$ then there exists $0<R<d(p,q)$ such that $B(q,R)\subseteq B(x,r)$.

Rithaniel

@LukasHeger I don't want to bother you, but do you have any insight into the above problem? I've reached the point of being tired of beating my head against this particular wall

Lukas Heger

@Rithaniel what do you want to prove?

Rithaniel

That $\text{ker}(f_2)\subseteq\text{Im}(f_1)$

Well, more generally, that $A_1\to A_2\to A_3\to 0$ is an exact sequence

where $f_i:A_i\to A_{i+1}$

Lukas Heger

the Yoneda functor $R-Mod \to [R-Mod^{op},Ab]$ is faithful and additive and hence it reflects exactness

Rithaniel

2:01 AM

I suspect this might be a bit beyond my knowledge

topologicalmagician

How do I prove this:

Let $(X,d)$ be a metric space.

Show that if $q\in B(x,r)$ and $q\neq x$ then there exists $0<R<d(q,x)$ such that $B(q,R)\subseteq B(x,r)$.

I mean if $B(q,r')\subseteq B(x,r)$ it might not be the case that $r'\leq r$

Rithaniel

Well, thank you for the help, Lukas

topologicalmagician

@Rithaniel @LukasHeger any ideas?

Rithaniel

My brain is a mess right now, so I might not be the best person to ask. I am not fully understanding what I read right now

Daminark

Whaddup

@lukas how are you doing?

New Student

2:19 AM

guys which book shows set theory $\cap$ and $\cup$ from logic operators like $\wedge$ , $\vee$ etc.
Its hard to find such a book.

Rithaniel

Heya Daminark

Daminark

How's it going Rithaniel?

Rithaniel

I have a headache trying to figure out a category theory problem using what I know from algebra

Also, I'm eating grapes. How about you?

Daminark

Trying to read through some Silverman

He might get better later but his makeshift intro to varieties and algebraic curves is... definitely makeshift

Rithaniel

Might be that he has his own understanding of these things and describes it as he understands them

(I've realized that my approach to this problem isn't going to get me anywhere, I think)

Daminark

2:33 AM

It's not quite that, more that he's trying to hit that "just enough" spot

Rithaniel

Ah, so he's trying not to spend too much time on the introduction

Daminark

You kinda need some algebraic geometry setup for elliptic curves but not that that much

In general you can define an elliptic curve as a smooth projective curve of genus 1 with a specified basepoint, and then prove that you can change variables and put such a curve in Weierstrass form

Also it ends up being noteworthy that a certain group attached to general varieties happens to be isomorphic to the elliptic curve as a group

But anyway my impression is that once you sorta get through that part, you just work with the Weierstrass equation and things are more explicit

Lukas Heger

@Rithaniel take $D=A_2/\mathrm{Im}(f_1)$

then the projection $\pi:A_2 \to D$ is in the kernel of $g_1$

as by construction $0=\pi \circ f_1=g_1(\pi)$

so by exactness we find $\psi \in \mathrm{Hom}_R(A_3,D)$ such that $\psi \circ f_2 = \pi$

Rithaniel

Huh, I think I see it. I was trying to come up with an appropriate $D$ but thought I'd keep ending up with dead ends. But moving the $f_1$ into the $D$ might change things

Oooooooh, got it!

Thank you again, Lukas

Lukas Heger

2:48 AM

@Daminark you already worked with elliptic curves, right?

for the $(a+bi)(c+id)$ thing

Daminark

Yeah, specifically I was gunning for Kronecker-Weber for $\mathbb{Q}(i)$ using $y^2 = x^3 - x$

Lukas Heger

nice

Daminark

I got through the proof that you always get an abelian extension when you adjoin torsion points of that guy to $\mathbb{Q}(i)$, started trying the other direction but didn't make it all too far

Lukas Heger

yeah I imagine it's not easy

Daminark

I might revisit it soon, I think I still have the document

Ted Shifrin

3:35 AM

Well, howdy, Demonark @Daminark

Daminark

Hey Ted, how's it going?

uhoh

5:35 AM

5

Q: Was this spherical device used in weather forecasting in the 1960's?

The New York Times obituary Robert M. White, Who Revolutionized Weather Forecasts, Dies at 92 shows the image below. It looks like it might be a model of the Earth but I can't tell for sure. Can the sphere be identified, and if so, is it related to weather forecasting? click for full size ...

The image in the question reminds me of the term "projective geometry". Any thoughts?

7einadh

6:40 AM

@anakhro I think I have fixed the mistake now - a sequence of points in the open neighborhood's points there isn't a convergent because it is supposed to be an infinite sequence.

Sha Vuklia

8:42 AM

@TedShifrin Hi Ted! Thanks for your message. I had also found Fulton's book:D (good to know you're recommending it too).

Alessandro Codenotti

@Lukas Are you around by any chance? I'm having an hard time understanding what (co)products look like in a subobject category

Rithaniel

10:17 AM

Is $\prod\limits_{i=1}^\infty\mathbb{Z}$ isomorphic to any more other less-difficult-to-type-out groups?

Mats Granvik

10:36 AM

The entries in the sequence of the partial sums of the Möbius inverse of the Harmonic numbers is a subset of the inverse error function when put in the context of tuples in combinatorics.

northerner

10:54 AM

Hello

Leaky Nun

@Rithaniel I guess not

@AlessandroCodenotti what is a subobject category?

northerner

Just to double check, division and multiplication have the same level of precedence, correct? It's not division is before multiplication?

Leaky Nun

right

northerner

BEDMAS confused me...I thought since D comes before M it meant division is first :\

Leaky Nun

The Order of Operations is Wrong

►

northerner

11:57 AM

That's a good video thanks

Alessandro Codenotti

@LeakyNun You have a category $C$ and an object $X\in C$. The category of subobjects of $X$ has as objects objects $A$ of $C$ that have a mono $A\to X$, up to equivalence

MadSpaceMemer

If the span of an empty set is equal to the zero vector, what is the span of the subset containing the zero vector.

Is it also, the zero vector?

Lukas Heger

@Alessandro I know what coproducts look like if $C$ is abelian

Alessandro Codenotti

Well that's not quite accurate, you really want monomorphisms into $X$ up to equivalence

Lukas Heger

not sure in general

Alessandro Codenotti

12:08 PM

@LukasHeger I'm trying to understand why this categorical presentation of a topology actually is a topology mathoverflow.net/questions/265522/…

But I don't quite see what product and coproducts in the subobject category are

@MadSpaceMemer yes

Lukas Heger

okay so if you have two subobjects $i:A \to X$ and $j:B \to X$, then we can take the pullback $A \times_X B \to X$, monos are closed under composition and pullbacks (exercise) so that's actually a subobject

Alessandro Codenotti

I guess you just get unions and intersections here?

Lukas Heger

if you do this in the category of sets (or groups or topological spaces) you do get intersections yeah

I think for coproducts it's more complicated

Alessandro Codenotti

Wait

Lukas Heger

I mean, you do get unions for coproducts in Set

Alessandro Codenotti

12:12 PM

The subobject category of $X$ is a subcategory of the slice category of $X$ right?

Lukas Heger

yes

Alessandro Codenotti

So products in this category are just fibered products in the original category

Lukas Heger

xwhich is what I said

but you need to make sure that you do get a monomorphism

Alessandro Codenotti

@LukasHeger Yes but I know from AG that this works out :P

Lukas Heger

but the general construction for coproducts requires some form of epi-mono factorization I think

take the coproduct $A \sqcup B \to X$, that won't be a monomorphism

Alessandro Codenotti

12:14 PM

Right

Lukas Heger

but if $\mathcal C$ admits canonical epi-mono factorization, then we can factor it as a composition of epi by mono

so just take the mono part

in Set, that gives you unions

in Ab, that gives you sum of subgroups

etc.

Alessandro Codenotti

Makes sense

Thanks

Lukas Heger

more generally, if $\mathcal C$ admits images in the sense of category theory, then we can take the image of $A \sqcup B \to X$

see

Image (category theory)

In category theory, a branch of mathematics, the image of a morphism is a generalization of the image of a function. == General Definition == Given a category C {\displaystyle C} and a morphism f : X → Y {\displaystyle f\colon X\to Y} in C {\displaystyle C} , the image of f {\displaystyle f} is a monomorphism m : I → Y ...

Leaky Nun

I mean products are just "intersections" right

Lukas Heger

yes

Leaky Nun

12:18 PM

@AlessandroCodenotti play?

Lukas Heger

@LeakyNun but intersections are just fiber products along monomorphisms, right?

Leaky Nun

@LukasHeger you're not wrong

Leaky Nun

12:39 PM

@LukasHeger should the plural of "anno domini" be "annis domini"?

Lukas Heger

yes

it's ablative

Alessandro Codenotti

1:02 PM

@LeakyNun I wasn't there, sorry

Leaky Nun

now?

Alessandro Codenotti

Just a game or two, I have to study

Henlo

Hi @Edward

What's up? :)

1:03 PM

Send me a challenge @Leaky

@EdwardEvans Not much, just reading some cool set theory, what about you?

Edward Evans

Just reading ANT notes lol

In the UK

Alessandro Codenotti

Nice

I'm also back to Italy

Nice :D

Hi @Edward

Hey @Lukas

1:14 PM

Hey, I have operation
1 / 1000 / 60 / 60 / 24

how to make it shorter?
i.e. making shorter the
1 / 2 / 2

would be:
1 / 4

I am performing multiple divisions and seek to perform just one

Edward Evans

@Lukas I emailed Studiumsverwaltung about the Semestergebühren lol, the bureaucrats at the university must thing I'm the most disorganised person alive

Lukas Heger

@EdwardEvans tbf you should have gotten an email one month prior to the deadline

the time span to pay your fees was 15.01.-15.02.

Edward Evans

Yep :P Unfortunately I didn't know I was obligated to check my university email address since everything else is registered via my personal email address (this is common in the UK too!)

and I'm a disorganised person anyway rofl

Adam Sandller

Hmm sorry I thought its math chat not uni chat cya

Lukas Heger

lol, we can chat about whatever we want

Edward Evans

1:20 PM

(Basically I'm a moron)

Leaky Nun

@AlessandroCodenotti ggs

Alessandro Codenotti

gg @Leaky

You had a chance to capture my passed pawn in the first game when I first offered a rook trade

Lukas Heger

@AlessandroCodenotti as a logician, are you interested in categorical logic at all?

I think it seems cool

Alessandro Codenotti

I don't even know what that is

Leaky Nun

@AlessandroCodenotti I'm sure we missed a lot of things in both games

Alessandro Codenotti

1:22 PM

I'm not really interested in logic to be fair (as in studying modal/infinitary/other weird logics), just in set theory

@LeakyNun Stockfish says I missed mate in 10 in the second game lmao

And that even ignoring that I had a big advantage that I wasted, the position was equal when I won on time

Leaky Nun

oh and there were back rank issues which means I couldn't capture your passed pawn

Lukas Heger

@Alessandro the idea of categorical logic, as I understand is, is to rephrase logical concepts in categorical terms and that allows you to apply logical reasoning to give really intuitive proofs for some things in certain categories. You can also use categorical algebra to prove logic things, but I know even less about that

Alessandro Codenotti

@LeakyNun Oh right, I didn't even notice. But I had the right bishop to attack the pinned knight on the back rank

Lukas Heger

the "internal logic of a category" is in general really weird though

Alessandro Codenotti

@LukasHeger Uh, weird, I know nothing about any of this

Is this related to the elementary topos things?

Lukas Heger

1:26 PM

yes, absolutely

Leaky Nun

@AlessandroCodenotti exactly

Alessandro Codenotti

I see

I want to learn some topos theory from a logician point of view eventually

Lukas Heger

@AlessandroCodenotti there's the book "Topoi - A categorical analysis of logic"

we could read it together, I'd be down

Alessandro Codenotti

Because I've seen that there's some super weird generalization of forcing over topoi

@LukasHeger Who is the author? I can't find it on the usual Russian website

Lukas Heger

Goldblatt

it's the categorical analysis not a

I misremembered

So if $X$ is a reduced scheme and $\mathcal F$ is a quasi-coherent sheaf of finite type over $X$, then there is a dense open subset $U$ of $X$ such that $\mathcal F|_{U}$ is locally free

that's a hard exercise in algebraic geometry

but using categorical logic, you can "trivialize" it

Jacksoja

1:32 PM

Hello, if we have rs dividing , x ( x^k - 1 ) , rs are primes

can we say that either one of x and x^k -1 divides rs?

Lukas Heger

in the internal logic of the Zariski topos associated to $X$, the statement corresponds to the fact that "every finitely generated vector space does not not possess a basis"

which is a theorem in intuitionistic logic

Jacksoja

is there categorical logic?

Lukas Heger

yes

Jacksoja

logic was not hard enough ??

haha

Logic is mathematic of mathematics

I do not want to know what categorical logic would be like

Lukas Heger

$\mathcal F$ being a quasi-coherent sheaf of finite type over a reduced scheme just translates to a vector space over a field in the internal logic of the Zariski topos

"not not" translates to "on a dense open subset"

not that for example "$\lnot \lnot \lnot \lnot P \Rightarrow \lnot \lnot P$ is a theorem of intuitionistic logic

Alessandro Codenotti

1:35 PM

@LukasHeger Looks interesting, but I already have a lot of stuff to work on

My thesis, an exam in March, I'm reading a set theory book...

Lukas Heger

which means that if something holds on a dense open subset of a dense open subset, it holds on a dense open susbet

if interpreted in the Zariski topos

Jacksoja

@LukasHeger do you have a particular book about categorial logic in mind?

You made me curious about the topic

Lukas Heger

@Jacksoja there's Goldblatt "Topoi - The Categorical Analysis of Logic"

Jacksoja

lemme check it!

adesh mishra

2:07 PM

@StupidQuestionsInc I’m enjoying that book very much, it’s a book I was searching for long time. Thank you for introducing me with that book.

Alessandro Codenotti

3:04 PM

@Lukas Is there a way to see an HNN extension as a pushout in Grp? I think there should be since you can view gluing a space to itself along subspaces as a pushout in Top

Lukas Heger

@Alessandro I'm not sure

I don't think you can

even semidirect products don't admit a nice description as a colimit in Grp

they're 2-colimits though

probably a HNN extension is some 2-colimit, too

Alessandro Codenotti

Hmm so is there some other categorical construction which becomes an HNN extension in Grp and a "gluing a space to itself along two subspaces" in Top? I'm just wondering because they seem like very close analogues so maybe they're two instances of a general construction

Lukas Heger

the problem is that the defining relation of an HNN extension uses elements

since you want the isomorphism to be given by conjugation

Alessandro Codenotti

Right

I see why category theory doesn't like that

Lukas Heger

well, 1-category theory doesn't

consider Grp as a subcategory of Cat, the category of small categories

the latter comes with a notion of natural transformations

that's why it's naturally a 2-category

Alessandro Codenotti

3:11 PM

@LukasHeger I know very little about that, but I'm following so far

Lukas Heger

if you restrict that structure to Grp, you can ask yourself what a natural transformation between two group homorphisms $f_1,f_2:G \to H$ is

as it turns out, it's just an element $h \in H$ such that $hf_1(g)=f_2(g)h$ for all $h$

i.e. $hf_1h^{-1}=f_2$

this looks more adequately equipped to handle HNN extensions

so any group presentation trivially implies a universal property for that group

Alessandro Codenotti

@LukasHeger Aha, this is looking promising

Lukas Heger

in the case of HNN extensions, if you have two subgroups $K,H$ of $G$ and an isomorphism $\alpha:H\to K$, the universal property of the HNN extension $W$ (what's the notation here) it should be this: let $i_1:H \to G$ and $i_2:K \to G$ be the inclusions, then homomorphisms $f:W \to T$ are in natural bijection to homomorphisms $f:G \to T$ and a natural transformation of $f \circ i_1 \to f \circ i_2$

I think this means that we have some form of 2-colimit

but I'm not sure

you should ask this on MO

(or I can)

if you do, be sure to reference this question:

27

Q: Are semi-direct products categorical (co)limits?

Products, are very elementary forms of categorical limits. My question is whether in the category of groups, semi-direct products are categorical limits. As was pointed in: http://unapologetic.wordpress.com/2007/03/08/split-exact-sequences-and-semidirect-products/ Bourbaki (General Topology, Pr...

ct.category-theory gr.group-theory

Alessandro Codenotti

3:36 PM

Very interesting, thanks

Alessandro Codenotti

4:26 PM

Hi @Balarka, how was the talk?

Balarka Sen

Good! Everyone enjoyed it

I printed pictures of the Morin-Francis eversion out and distributed it at the end of the talk

These pictures: 1, 2

Ajay Mishra

Given that a matrix has eigenvalue 1,2 and 3, I am required to find out the determinant of $A^2 + A^T$

Alessandro Codenotti

Nice

Cool pictures too

Balarka Sen

Yeah

Ted Shifrin

4:53 PM

Congrats, a @Balarka! :)

Hi, demonic @Alessandro.

Leaky Nun

@BalarkaSen did you shout "Gromov is God"?

Balarka Sen

5:06 PM

@LeakyNun Unfortunately no

Mike Miller

I need to update my triangulation talk

Ted Shifrin

Does that make it a rectangulation talk?

Balarka Sen

Lmao

What's the abstract, @MikeMiller

adesh mishra

@TedShifrin I have misconception about surface integral.

topologicalmagician

Consider subintervals of $[0,1]$ such that

$E_k$ which has $9^k$ intervals, each of length $10^{-k}$, with total gaps in between the intervals being greater than or equal to $10^{-k}$.

Let $F=\cap_{k=1}E_k$ be non-empty.

Suppose $0<\delta<1$ is such that $\frac{1}{10^{k+1}}\leq \delta <\frac{1}{10^k}$ and that any interval of diameter at most $\delta$ , may intersect 2 intervals. Further suppose each intersected interval contains point of the intersection, $F$, **it follows that the least number of sets in any $\delta$ cover of $F$ is greater than or equal to $\frac{9^k}{2}$.**

Ted Shifrin

5:17 PM

What misconception is that, @adesh?

topologicalmagician

May someone elaborate on the bold?

the last sentence

please

adesh mishra

@TedShifrin Let's consider a vector field $\mathbf A$, draw a closed surface $\sigma$ and let $\mathbf A$ is continuous in/on $\sigma$. Let $A_n$ represents the outward normal component at every point on the surface then the surface integral is defined as $$\oint_{\sigma} A_n~d\sigma$$

Ted Shifrin

You don't need a closed surface to talk about that, by the way. I would write $\mathbf A\cdot\mathbf n$, but yes.

adesh mishra

And now consider this figure

I thought that for every point inside and on the surface I got take the dot product with the outward normal but today someone told me that when we do surface integral we do dot product (or take the component along the outward normal) only for point on the surface not inside.

Have I made myself clear?

Ted Shifrin

We're talking about things on the surface. $\mathbf n$ doesn't even make sense inside.

You're measuring how much of $\mathbf A$ is going across $\sigma$.

And remember my comment. This makes sense for any oriented surface — it needn't be closed and it needn't even have an "inside."

adesh mishra

5:24 PM

MY GOD! You're awesome.

Please keep on explaining

I'm loving it

Alessandro Codenotti

Hi @Ted

Ted Shifrin

LOL, @adesh. You can watch my YouTube videos on surface integrals. There are lots of examples. (But they do use the language of differential forms, as well as the classical flux discussion.)

adesh mishra

@TedShifrin Can you please call me "Demonic Adesh" ROFL

Ted Shifrin

LOL, Alessandro earned that appellation.

topologicalmagician

Hi @TedShifrin

How are you?

Ted Shifrin

5:26 PM

Hi, @topologicalmagician.

topologicalmagician

@TedShifrin do you have any thoughts on the problem I wrote above?

I did draw pictures this time

Ted Shifrin

I don't have thoughts. It seems sloppily written. "may intersect 2 intervals"? What intervals?

adesh mishra

@TedShifrin Good Night

topologicalmagician

@TedShifrin in $E_k$

Ted Shifrin

Night, @adesh.

@topologicalmagician: The problem is incomplete. Do we know $E_{k+1}\subset E_k$?

topologicalmagician

5:31 PM

@TedShifrin yeah

Ted Shifrin

You realize that you should post a complete and clear question if you're going to bug us?

topologicalmagician

@TedShifrin yeah, sorry. I gathered the info from a particular proof.

Ted Shifrin

That doesn't help me.

topologicalmagician

Ok, the problem is to find the box dimension of the subset of $[0,1]$ which does not contain the digit 5 in their decimal expansion

Ted Shifrin

What is box dimension?

topologicalmagician

5:35 PM

$lim_{\delta\rightarrow 0} \frac{N_d(F)}{-log d}$ where $N_d(F)$ is the least amount of sets of diameter at most d

@TedShifrin

Ted Shifrin

Balls of diameter $\delta$?

This looks a lot like Hausdorff dimension.

topologicalmagician

@TedShifrin no, not balls. Just sets.

But I am able to compute part of it. The part I\m unable to understand is the following:

Ted Shifrin

Presumably you get the smallest $N_\delta$ when you take an interval (or ball) in your case.

topologicalmagician

Consider subintervals of $[0,1]$ such that

$E_k$ which has $9^k$ intervals, each of length $10^{-k}$, with total gaps in between the intervals being greater than or equal to $10^{-k}$.
Note that $E_{k+1}\subseteq E_k$. for each k.
Let $F=\cap_{k=1}E_k$ be non-empty.

Suppose $0<\delta<1$ is such that $\frac{1}{10^{k+1}}\leq \delta <\frac{1}{10^k}$ and that any interval of diameter at most $\delta$ , may intersect 2 intervals in $E_k$. Further suppose each intersected interval contains point of the intersection, $F$, **it follows that the least number of sets in any $\delta$ cover of $F$ is…

(I edited it)

Yeah, the only thing I'm unable to understand, is why $N_d(F)\geq \frac{9}{2}$

Ted Shifrin

That dimension definition makes no sense to me. You mean $N_\delta\ge 9^k/2$?

topologicalmagician

5:43 PM

@TedShifrin yes, if by $N_\delta$ you mean the least amount of sets of diameter at most delta required to cover the set, $F$.

Ted Shifrin

So what does that make the box dimension in this case?

topologicalmagician

log9\log10

but that's because $N_d(F)\leq 9^k$ for $0<\delta<1$ (This I'm able to prove)

Ted Shifrin

Where did the /2 go?

Mike Miller

@BalarkaSen What's a triangulation, what's the triangulation conjecture, high-D stuff reducible to homology cobordism in 3D, modern tools to attack hcob

topologicalmagician

@TedShifrin observe that for $0<\delta<1$, there exists a $k\in \mathbb{N}$ such that $\frac{1}{10^k}\leq \delta < \frac{1}{10}^{k-1}$ and so every interval of diameter at most delta is a delta cover for $F$, since there are $9^k$ such intervals, $N_{\delta}(F)\leq 9^k$

Ted Shifrin

5:50 PM

I do not follow.

topologicalmagician

@TedShifrin we were told to only consider $0<\delta<1$

Ted Shifrin

I'm not going to be able to help. My brain isn't working for this kind of stuff today.

topologicalmagician

@TedShifrin alright. No worries. I'll try to figure it out.

@TedShifrin oh! I got it

I mean I understood it

because he's removing intervals

the picture I just drew helped

@TedShifrin fractal geometry is interesting but annoying lol

Stupid Questions Inc

@adeshmishra glad you found it useful! and best wishes for what comes next ;)

Ted Shifrin

Cool @topologicalmagician

Poline Sandra

5:59 PM

hello, if $a\in [0,\pi]$ and $b\in [\pi/2,\pi]$ how to prove that cos(a)=cos(b) ?

Ted Shifrin

That is nonsense, @Poline.

Poline Sandra

I want to find $a\in[0,\pi]$ such that $cos(a)=cos(b)$ and $b\in[\pi/2,\pi]$

Ted Shifrin

Obviously, then, take $a=b$.

Poline Sandra

I don't see why a=b

that is my problem

Ted Shifrin

Look at the graph.

Stupid Questions Inc

6:02 PM

@PolineSandra if you want something that works then $a=b$ satisfies your conditions

also look at the graph as Ted suggeste

Ted Shifrin

The point is that it's the only thing that works.

Cosine is one-to-one on $[0,\pi]$.

Then there's a unique $a\in [\pi/2,\pi]$ with $\cos a = \cos b$.

And you can figure it out from the unit circle picture.

Poline Sandra

@TedShifrin yes it is true I forget this

and if $b\in [\pi,3\pi/2]$ ?

Ted Shifrin

Why did you erase this? I answered it already.

Poline Sandra

your answer is about $\pi/2,\pi]$

not about $[\pi,3\pi/2]$

Ted Shifrin

No, it was about this question. You need to review basic trigonometry.

Poline Sandra

6:07 PM

@TedShifrin is it -b ?

@TedShifrin I'm sorry I don't read carefully

Ted Shifrin

Where does $-b$ live if $b\in [\pi,3\pi/2]$?

Poline Sandra

$b\in [\pi,3\pi/2]$ but there is c symmetric to b in $[\pi/2,\pi]$

@TedShifrin

or $a=b-\pi$ @TedShifrin

Ted Shifrin

6:22 PM

No. Stop making guesses and look at either the unit circle or the graph of the cosine function

What does "c symmetric to b" mean?

Poline Sandra

$c=-b\in [\pi/2,\pi]$

Ted Shifrin

GROWL.

Come on. $-b\in [-3\pi/2,-\pi]$. We're talking numbers here, not points on the unit circle.

So you're close but you need to fix it.

Poline Sandra

if $b\in [\pi,3\pi/2]$ then $b-\pi\in[0,\pi]$

but $cos(a)\neq cos(b-\pi)$

Ted Shifrin

You're back to writing symbols instead of thinking.

Poline Sandra

I draw the circle and c but I can see what is equal c

Ted Shifrin

6:38 PM

How do you get c from b?

Poline Sandra

symmetry from the axe of abscise

Ted Shifrin

OK, reflecting across the $x$-axis. So the angle $b\in [\pi,3\pi/2]$ goes to what angle when you do that?

Poline Sandra

[\pi/2,\pi]

Ted Shifrin

Yes, but I want you to use the geometry to give me a correct formula.

Poline Sandra

I don't know how ?

Ted Shifrin

6:48 PM

What is the angle from the negative $x$-axis to the ray at angle $b$?

Poline Sandra

I don't know if I understand exactly (I speak French) but it is $\pi+b$

Ted Shifrin

Non, on commence au point $(-1,0)$ et on va jusqu'au point $(\cos b,\sin b)$.

C'est quel angle?

Poline Sandra

$3\pi/2-b$

Ted Shifrin

Vraiment? On a commencé à l'angle $\pi$, n'est-ce pas?

Poline Sandra

on pardon j'ai vu (0,-1), du point pi jusqu'a b c'est b-\pi

Ted Shifrin

6:56 PM

Oui, c'est ça. Et pour commencer à $(-1,0)$ et arriver à $c$, on fait quoi?

Poline Sandra

$\pi-b$

Ted Shifrin

Non. Il faut penser à ce qu'on vient de discuter.

Poline Sandra

je me suis trompé de sense

c'est $3\pi/2+b$

non pardon

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$Mathematics$ Mathematics