Mathematics - 2016-11-20 (page 6 of 9)

Secret

17:00

The interesting thing is that while there are additive inverses, thus normally one will expect collapse will occur. However because 0 is left absorbing, it mop up those problematic zero terms thus allowing the results to be consistent

meow-mix

@Secret did someone use mspaint to make this...

Secret

Nope I ues powerpoint

SteamyRoot

@Adeek It may be helpful to also share the theorem/proposition this is a corollary of.

meow-mix

please use LaTeX :(

Balarka Sen

@Adeek They're proving that there's no proper normal subgroup of $A_5$. Is there anything particular you're having trouble with?

SteamyRoot

17:01

And, well, don't they just prove that a proper normal subgroup does not exist in that proof?

That's literally the definition of simple group, no?

Adeek

yeah I don't understand how this proves it @BalarkaSen

SteamyRoot

They give all the options for $|N|-1$ where $N$ is a proper normal subgroup

heather

how do I solve $\int \frac{x}{x}$?

or some function over a function

Balarka Sen

@Adeek He's getting obstruction from the class equation. Please be particular about what you're having trouble with.

Adeek

@BalarkaSen so we are proving that there are no normal subgroups of $A_5$ then we got the divisors of 60. But what is the other things ? what are 1,2,3,4,5,9,11,14,19,29 ?

Null

17:04

how does one read sums that have not a clear end?

user379685

Hello could someone help me with 1/(x sin(x))-1/x^2 when x goes to 0 without using l'hospital's rule

Secret

-> "This message is too long." Fine I will shut up, bleh

Adeek

I agree that a normal subgroup is union of conjugacy classes.

Balarka Sen

@Adeek The possible orders of the normal subgroup, except the identity.

Adeek

ok I see but then why this doesn't happen ?

how do we know that ?

Balarka Sen

17:06

What are the size of the conjugacy classes of A5?

Adeek

ohh I see

I see the class formula for $A_5 = 1 + 15 + 20 + 12 + 12$

we can't get any of the normal subgroups from the conjugacy classes.

arctic tern

@heather x/x=1 has antiderivative x+C. there is no general rule for integrating f(x)/g(x).

Balarka Sen

Ya

arctic tern

@Null what do you mean?

Adeek

oh ok I understand that is pretty cool

heather

17:08

oh, okay @arctictern

Balarka Sen

@Adeek Yes. You can't sum 1, 15, 20, 12, 12 to anything that divides 60 except 1 and 60 itself.

That's the whole point

Adeek

yes]

Balarka Sen

Hi @Ted

Ted Shifrin

Hi @Balarka, Karim, @heather, Tern ...

Adeek

hi @TedShifrin

heather

17:12

@TedShifrin, hello

Astyx

@Ted hi

Ted Shifrin

Salut, @Astyx

Astyx

Tu parles Français couramment ? @Ted

Ted Shifrin

Oui, @Astyx ... au moins, je parlais couramment il y a longtemps :P

Astyx

Cool ! :)

Null

17:14

@arctictern how would you read $\sum ab+cd+(xy)^2$ ?

Ted Shifrin

@Null: What does that even mean?

arctic tern

Is $\sum ab+cd+(xy)^2$ something you actually saw somewhere?

Notation should either be unambiguous, or its meaning should be clear from context.

Ted Shifrin

Sum over what?

meow-mix

@TedShifrin currently working on your exercises

Ted Shifrin

LOL @meow ... And hating me, I'm sure :P

meow-mix

17:16

a tiny bit, maybe ;)

Ted Shifrin

Section 2 is very long with lots of interesting exercises (including some introductory algebraic geometry stuff)

meow-mix

I'm probably overthinking some of these, but AFAIK the proofs are right.

Null

@TedShifrin sum over what is actually uninteresting. i meant where does the sum end. at ab or at $(xy)^2$. @arctictern ok thanks

heather

@TedShifrin, which would you say comes first: multivariable or vector calculus?

Ted Shifrin

well, it's ok, @meow — you join generations of students who've hated me :)

meow-mix

17:16

@heather multivariable calculus

heather

@meow-mix, okay, thanks

Ted Shifrin

@heather: to me it's the same thing ... sometimes colleges teach a vector calculus class for engineers.

arctic tern

@Null your question still makes no sense.

ask something that makes sense instead

Null

$\sum_{i=1}^{n}x_iy_i+x_{n+1}^2y_{n+1}^2$

in the above case its clear

since n+1 doesnt belong to the sum

arctic tern

yes

Ted Shifrin

17:17

Which means that my question is totally a propos.

Null

i just wondered what if it is not clear, that's all ;)

@TedShifrin $\sum_{i=1}^{n}a_i+b_i=(\sum_{i=1}^{n}a_i)+b_i$ ?

Balarka Sen

That doesn't make sense, so no

Mary Star

An other question is the following:
Let $(a_n)_{n=1}^{\infty}$ be a real sequence and let $(a_n)_{n=1}^{\infty}$ a sequence in the set of limit points of $(a_n)_{n=1}^{\infty}$, $L(a_n)$.
There is also a $a_0\in \mathbb{R}$ with $a_n\rightarrow a_0$ for $n\rightarrow \infty$.

I want to show that then $a_0\in L(a_n)$.

But isn't this obvious, since $a_0$ is the limit of $a_n$ ?

Balarka Sen

If you are indexing over both $a$ and $b$, both are part of the sum

Ted Shifrin

$i$ the index of summation, so obviously $b_i$ is included in the sum

If you wrote $\sum\limits_{i=1}^n a_i+3$, then I would agree it's ambiguous.

And I would insist on parentheses if I meant the $3$ to be inside the sum.

Astyx

17:21

@MaryStar Not necessarily

Null

ah ok, so in such a case a small note would be helpful?

(or paranthesis)

Ted Shifrin

@Mary: Don't you mean for $(b_i)$ to be a sequence in the set of limit points? You can't reuse the same letter.

@Null: 95% of people would say $\sum a_i + 3$ is unambiguously $(\sum a_i)+3$.

If you intend $\sum (a_i+3)$, then don't write any notes; use parentheses, as mathematics intends you to.

Astyx

@MaryStar If you have not studied topology yet, I guess you would need to do it with $\epsilon$

Ted Shifrin

@Mary: I really do not understand the question because you have used $(a_n)$ with two different meanings.

Astyx

@Ted She wants to show the set of limit points of a sequence is closed I think

Ted Shifrin

17:25

Well, most likely. :) But it's important to write stuff that makes sense, first of all :)

Astyx

Very true :)

Ted Shifrin

That's one of the hardest things for people learning mathematics. I speak from 40+ years of experience on this one.

Balarka Sen

@TedShifrin "I have a topology and a Morse function on it. What is the second homotopy group of the limit set?"

Ted Shifrin

LOL

Whom are you quoting?

Balarka Sen

None particularly. Perhaps me from 3 years ago.

Ted Shifrin

17:27

So I can say it's ununderstandable garbage?

Astyx

@MaryStar A first step would be to show that a is a limit point iff $\forall N \in \Bbb N, \forall \epsilon \gt 0, \exists n\ge N, |a_n - a| \lt\epsilon$

Ted Shifrin

Hi @Alessandro !

That's false, @Astyx.

Alessandro

Hi! @Ted

Balarka Sen

@TedShifrin Yeah, feel free to :P

Astyx

Yes, but I didn't give the definition of the limit of the sequence, did I ?

Semiclassical

17:30

hi chat

Ted Shifrin

Oh, I apologize. I hate quantifier symbols and didn't read them correctly.

heather

@Semiclassical, hello

Astyx

@Ted No problem, I perfectly understand you :p

Adeek

@TedShifrin There is a seminar next week about hodge theory

it seems interesting

Ted Shifrin

It'll probably be pretty sophisticated, Karim.

hi @Semiclassic

Semiclassical

17:32

Main thing I know about Hodge stuff (beyond the hodge star) is the notion of a Hodge decomposition

Ted Shifrin

@Astyx: Isn't that unnecessarily complicated, though?

Semiclassical

But I wouldn't say I know that particularly well

Balarka Sen

I went to one which proved the Hodge theorem. Didn't understand much except the theorem statement.

Adeek

yeah I will attend but probably not understand anything but just go get an idea of high level math

Ted Shifrin

Hard to do the Hodge Theorem without lots of analysis, @Balarka.

Astyx

17:33

@Ted If you haven't studied topology I don't see how else you'd proceed

Adeek

@BalarkaSen yeah that is me in most of grad seminar talks

Balarka Sen

@TedShifrin Yeah... Sobolev spaces and elliptic operators everywhere....

Ted Shifrin

Can't you leave out the $N$? For every $\epsilon>0$, there is some $n$ with $|a_n-a|<\epsilon$?

Cool stuff, though, @Balarka.

Astyx

That's not sufficient to construct a subsequence converging to $a$

Balarka Sen

Seems like it. Hopefully I'll get to know some one day. Not today though.

Ted Shifrin

17:34

It seems not to be, but it is :)

Astyx

You don't have equivalence anymore

Ted Shifrin

But a limit point is not defined in terms of convergent subsequences, is it?

Astyx

I think it is ? We don't call them limit points in french so I'm not sure

Balarka Sen

Ted is right, I think. Limit points are things around which stuff accumulates.

So having at least one $a_i$ in an arbitrarily small nbhd suffices.

Astyx

a is a limit point iff there exists an extraction $\phi : \Bbb N \to \Bbb N$ such that $a_{\phi(n)} \to a$

Ted Shifrin

17:36

But we can, in the case of a sequence, construct a subsequence converging thereto.

(Working in a metric space here.)

That's not the definition I learned/taught, @Astyx. But we can prove it's equivalent in a metric space.

Null

what does $a<<b$ mean?

Balarka Sen

Agreed.

Astyx

a is much smaller than b

Ted Shifrin

There must, in fact, be infinitely many points in the $\epsilon$ neighborhood, and so we can choose one with as large an index as we need.

Null

@Astyx thanks

Balarka Sen

17:38

Yep. That fails in nonmetric worlds though.

Astyx

@Ted since you know french : fr.wikipedia.org/wiki/Valeur_d'adh%C3%A9rence :)

So yeah you'd be right

Mary Star

@Astyx @TedShifrin There were a typo... The statement should be as follows:
Let $(a_n)_{n=1}^{\infty}$ be a real sequence and let $(b_n)_{n=1}^{\infty}$ a sequence in the set of limit points of $(a_n)_{n=1}^{\infty}$, $L(a_n)$.

There is also a $b_0\in \mathbb{R}$ with $b_n\rightarrow b_0$ for $n\rightarrow \infty$.

I want to show that then $b_0\in L(a_n)$.

Ted Shifrin

Thanks, @MaryStar :)

This is now a good thing to prove :)

Astyx

@Ted so what I said would be necessary if we are in a metric space

Balarka Sen

@Astyx What you said would be equivalent to what Ted said if we are in a metric space. Not "necessary".

Astyx

17:43

@BalarkaSen Yes, I meant that the quantified line I gave would be necessary for $a$ to be a limit point in a metric space

Ted Shifrin

You just need to know that the sequence is not effectively constant, as, at least for me, when you define a limit point $a$ you must have points $x$ from the set, $x\ne a$, in arbitrarily neighborhoods of $a$.

No, @Astyx. Modulo what I just said.

But your definition of point d'adhérence allows the sequence to be sitting at the limit point.

Null

@Astyx an infimum is either the existing minimum or...? how do you find the infimum if no minimum exists?

Ted Shifrin

Oh wait ... those two reformulations are not equivalent.

Wiki is wrong.

1) allows the subsequence to be constantly at the limit point. 2) will allow one to rule that out.

Astyx

@Ted Okay I see

Adeek

just a quick question about split extensions.

Astyx

17:46

@Null It is the biggest minorant

Adeek

Null

@Astyx do all those terms only make sense in ordered sets?

Astyx

@Ted so your definition of limit point would be what we call "un point d'accumulation" fr.wikipedia.org/wiki/…

Ted Shifrin

Karim: I don't like using the same notation for that subgroup of $G$. I would introduce $H'\cong H$ with $H'\subset G$.

Adeek

ok so I see that we can think of N as a subgroup of G by exact sequence. Then we say something is a split extension if we can think of H as a subgroup of G such that they intersect trivially ?

Ted Shifrin

17:47

@Astyx: d'accord. But I'm still saying that 1) and 2) are not equivalent.

Astyx

@Null I think so

Ted Shifrin

Yes, @Null.

SteamyRoot

@Adeek it means there's a morphism $\psi: H \to G$ such that $p \circ \psi = id_H$

Mary Star

Could you maybe give me a hint how we could show it? @TedShifrin

Adeek

just a sec I am trying to activate latex something wrong happened with my broweser

Astyx

17:48

@Ted I fail to see why. (you do mean 1) and 2) of the first paragraph, right ?)

Here they are working with real sequences

Balarka Sen

I don't like your definition, @Adeek. But yes.

Ted Shifrin

@Astyx: It's quite likely I'm being stooopid again. But in 1) you could take all $u_n=y$, whereas in 2) we're supposed to have infinitely many elements near $y$.

Adeek

Yeah I agree notation is a bit weird

Astyx

@Ted we have $|y-y|\lt \epsilon$ ? Or did I misunderstand you

Ted Shifrin

@Mary: What is the definition of limit point you're working with? Start there.

@Astyx: Well, isn't $0<\epsilon$ for any $\epsilon>0$? :D

Astyx

17:51

Exactly

Adeek

@SteamyRoot yeah nice way to express it.

Astyx

So in 2) the set would indeed de infinite

Ted Shifrin

So in 1) a constant sequence $(u_n)$ would work. It violates 2), however.

Balarka Sen

That's the standard way to say it, @Adeek

Astyx

@Ted Why does it violate 2) ?

Ted Shifrin

17:51

DogAteMy!!

Akiva Weinberger

Hi! What's happening?

Adeek

I see @BalarkaSen

Ted Shifrin

Oh, it's a set of $n$ that's infinite, not a set of $u_n$. AGH. OK. We're ok.

Astyx

Ok right, you scared me :p

SteamyRoot

The identification in your definition, @Adeek, would be the image of $\psi$.

Ted Shifrin

17:52

What's happening is that Ted is being stooopid, DogAteMy.

Adeek

yeah

hi @AkivaWeinberger

Astyx

Okay so the issue is that "limit point" is not the english word for "valeur d'adhérence", as wikipedia suggests it

It should link "limit point" with "point d'accumulation"

Ted Shifrin

In English we have both limit point and accumulation point.

Akiva Weinberger

So we're talking about limit points and split SESses?

Ted Shifrin

But definitions are not universally consistent across all textbooks.

Akiva Weinberger

17:54

(What's the plural of SES)

Ted Shifrin

DogAteMy, @MaryStar is trying to prove that the set of limit points of a real sequence is a closed set.

SESs

Astyx

SES ?

Ted Shifrin

short exact sequence

Astyx

Ah thanks

And so what is an "accumulation point" in english ?

Ted Shifrin

same as your point d'adhérence

I personally have never taught that term.

Astyx

17:58

Argh

Very inconvenint terminology

Ted Shifrin

d'accord

Astyx

Anyway thanks for the info @TedShifrin @BalarkaSen

Akiva Weinberger

18:14

Something you might find humerous: satirev.org/harvard/…

(Also, my neighborhood is in national news! For the worst reasons.)

s.harp

@TedShifrin I hope it doesn't bother you, but I was wondering if you knew something about the universal coverings of "surfaces of genus infinity with $n$ ends", for example with $n=2$ its an an open cylinder with a countable amount of "Torus holes".

they should be some sort of tiling of the poincare disk, but I can't find any literature on the internet or something

I imagine that there should be super pretty pictures, stuff that looks like this for example: en.wikipedia.org/wiki/Apeirogon#Apeirogons_in_hyperbolic_plane

Balarka Sen

We were discussing about these a few days ago.

s.harp

yeah I know

(weeks)

I had searched and thought a bit about it since then, but not gotten anything new out of it

Mary Star

18:34

@TedShifrin We have that $b_n\in H(a_n)$ is a limit point of $(a_n)$ if every neighbourhood of $b_n$ contains at least one point of $(a_n)$ different from $b_n$ itself, right?

Vrouvrou

Someone help me on this please

0

Q: Orlicz-Funtion and $\Delta_2$ condition

I have that $\Phi$ is a $N-$function (or Orlicz function) if If i have that $$\lim_{n\rightarrow\infty}\int\Phi(|u_n|)dx =\int\Phi(|u|)dx$$ and $\Phi$ satisfy the $\Delta_2$ condition i.e., $$\Phi(2t)\leq C \Phi(t),~~\forall t\geq0$$ How to prove that $$\int \Phi(|u_n-u|)dx\rightarrow0$$ Than...

user379685

18:56

Could someone help me solve ln(1+1/x)=1/x

Monad

Is a set of points, a set of numbers?

Alessandro

Has anyone seen the terminology "fixed point of a filter" before? Google doesn't bring up anything and the author never defined it, my guess it's that it is an element of $\bigcap\limits_{F\in\mathcal{F}}F$ since the filter is fixed if this intersection is nonempty

Maks

Hi !

Mats Granvik

@user379685 Reduce::nsmet: This system cannot be solved with the methods available to Reduce. >>

Semiclassical

1/x > ln(1+1/x) for all real x

(to see that, let u=1/x. then that's just u > ln(1+u) which follows from ln(1+u) being concave down and u being the tangent line of ln(1+u) at u=0)

Akiva Weinberger

19:08

@s.harp You should know that @MikeMiller was thinking about these also. I don't know if he's thought about it since then, though.

@Semiclassical Alternatively, that follows from $e^x\ge x+1$ for all $x$, my favorite inequality. (Sub in $x\mapsto1/x$, as you did.)

Semiclassical

yeah, i was trying to remember the name of that

PVAL-inactive

@MikeMiller I'm pretty certain I'd seen the talk quite a few times so I sort of ignored it.

s.harp

Wait, @AkivaWeinberger was this the conversation I was also a part of?

Akiva Weinberger

Oh. Whoops. I forgot. Sorry

But yeah

I think so

s.harp

ok, searching the chat for genus infinity does not return many resutls :=) (actually only the message I had just written and a joke that is funny)

(and now that one probably)

Akiva Weinberger

19:12

@s.harp We actually kept on talking for a while after you left, I think

Null

@NaCl did you advance with your problem?

Akiva Weinberger

(Searching "apeirogon" should bring the conversation back up)

NaCl

@Null not at all

There was one guy who answered me and his answer was essentially useless

Akiva Weinberger

So I was reading about the Jones polynomial in knot theory, and the introduction on how it was discovered is quite interesting (page 46 of this lovely set of notes)

(*knotes)

Null

@NaCl mmh, tough luck. Do you have only this as an axcercise?

NaCl

19:16

basically, yes

I solved all others

Alessandro

What was the problem? @NaCl

NaCl

@Alessandro It's not a math problem

Null

@NaCl i would implement it the way you would do without restrictions. or is it the task to do it without "datatype"/"exception"?

NaCl

@Null We did not had datatype/exception in our lecture, so we are not allowed to use it

Null

@NaCl I see..

Balarka Sen

19:19

I wish I knew knot theory. But there's so many things I wish I knew.

Akiva Weinberger

Essentially, the Jones polynomial is this miraculous and amazing object that was discovered by accident by someone who was working in a completely different area of math.

@BalarkaSen The set of notes I linked to above seem to be good

Balarka Sen

I think it has a physics background, IIRC?

Akiva Weinberger

(if not entirely typo-free)

Null

@BalarkaSen ywhy exactly knot theory?

Balarka Sen

Yeah, I mean, I can go ahead and read something if I wanted to. But life and time are both cruel enough.

Akiva Weinberger

19:21

@Null 'Cause I mentioned it above

Alessandro

There's only time for differential geometry in life

2

Null

@NaCl well, not to sound dumb, but did you compile your code? Does it what you want it does?

@NaCl do you even know what you want?

Balarka Sen

@Alessandro Not doing much of that either of late

NaCl

@Null ...what are you trying to tell me?

It's like python, you write the code an then run the interpreter

Akiva Weinberger

@Null Chocolate

NaCl

19:22

Yes, I always save the file I edited

Alessandro

@Balarka because you're doing other maths or because you're too busy to do any kind of maths?

Astyx

@NaCl what is it you're trying to do ?

Balarka Sen

the latter. studying for exams is tiring enough that when i actually get some free time i spend time either being lazy, or looking at less mentally straining things

NaCl

@Astyx do you know any functional programming language?

Astyx

Yeah

Null

19:26

I am not sure if this makes sense: $f~n~x~y :=f\;(n-1)\;(f\;n\;x\;(y-1))\;((f\;n\;x\;(y-1)) + y)$

Astyx

@NaCl Oops I have to go and eat, I'll look at your problem afterwards ;)

Mike Miller

@PVAL Fair enough. I hadn't.

Akiva Weinberger

Fun fact: ~ functions as a space in LaTeX

Null

@Astyx my problem solved itself eventually, (the cauchy stuff)

NaCl

it's ok

Null

19:28

@AkivaWeinberger you learn something new everyday :)

Mike Miller

@AkivaWeinberger It is still an open question whether the Jones polynomial detects the unknot.

Astyx

@Null Cool

Akiva Weinberger

@MikeMiller I know (it's mentioned a few pages later)

But there are other things that we do know detect the unknot

Mike Miller

yes

Akiva Weinberger

which I think is pretty amazing

Mike Miller

19:29

usually those are absurdly expensive and time-consuming; there are much better algorithms than 'calculate these invariants'

Akiva Weinberger

I'm not sure how an algorithm is different from an invariant in this case

Balarka Sen

In general I don't understand how people comes up with these combinatorial invariants, not that I have studied them in any level of depth

Null

My question is: is induction overkill for the Cauchy-Schwarz-Inequality? I have read proofs without induction, and they seem more meaningful to me

Akiva Weinberger

@BalarkaSen The Jones polynomial, apparently, was discovered in connection to a completely unrelated field. So the answer to your question, I guess, is that people get really lucky

Mike Miller

@AkivaWeinberger an algorithm doesn't spit out a number, or a group, or whatever

it just tells you the answer

Akiva Weinberger

19:31

Also the tricolorability may or may not have to do with the fundamental group? I haven't gotten to that bit yet

@MikeMiller If it spits out something in $\{\sf Yes,No\}$, it's still an invariant, no?

Mike Miller

meh

technically correct but everybody working in the field will roll their eyes at you

Balarka Sen

@AkivaWeinberger huh, fair enough

Akiva Weinberger

It's a function whose domain is the set of diagrams and which is invariant under knot isotopy

Null

I'd say an algorithm gives you one answer to one question. If you don't yet know the question things might get messy.

Akiva Weinberger

The question is, "Is this the unknot?".

Mike Miller

19:36

in any case, computing the jones polynomial or khovanov homology or heegaard floer homology are almost always unnecessarily expensive operations if you want to check if it's the unknot

Mary Star

@Astyx I have edited my question... http://math.stackexchange.com/questions/2022192/limit-of-sequence
Is my idea correct so far?

Akiva Weinberger

OK

And we know non-expensive ways to do it?

Balarka Sen

@AkivaWeinberger maybe you can check if it's tricolorable in a non-expensive way.

Mike Miller

tricolorability doesn't tell you if it's the unknot...

Balarka Sen

oh, the question is if it's the unknot, not if it's not the unknot. ok.

Mike Miller

19:39

there are algorithms using what's called normal surface theory which are exponential time in principle but "usually" low degree polynomial time

Akiva Weinberger

@BalarkaSen You can - there's a linear algebra way

@MikeMiller Interesante

Mike Miller

27

Q: What part is left unsolved in the Unknotting problem? (after results of Bar-Natan, Khovanov, Kronheimer and Mrowka)

Kronheimer and Mrowka showed that the Khovanov homology detects the unknot. Bar-Natan showed a program to compute the Khovanov homology fast: there was no rigorous complexity analysis of the algorithm, but it is estimated by Bar-Natan that the algorithm runs in time proportional to the square ro...

knot-theory computational-topology

Balarka Sen

I don't know if one can parameterize the knot and compute the curvature efficiently in a computational sense.

Then one can invoke the Fary-Milnor theorem.

Akiva Weinberger

What's that?

Balarka Sen

If total curvature of a knot is less than or equal to 4pi it's an unknot.

Null

19:41

@AkivaWeinberger what is a good example for "non polynomial time"? something like: "it either stops eventually or not"?

Mike Miller

again, that's not good enough

you can certainly make the unknot very curvy

Krijn

Do you know the joke about knot theory?

I guess you do

Null

me neither

Mike Miller

it's knot theory! a ha ha

@BalarkaSen No, you don't understand me. It wouldn't solve the problem.

There are unknots with curvature greater than 4pi.

Balarka Sen

Ah, yes.

I'm dumb.

Akiva Weinberger

19:44

@Null Analysing generalized chess (on an $n\times n$ board) is exponential time, because games can get exponentially larger as the board's size increases

Mike Miller

All the algorithms for unknot recognition are nontrivially complicated. So I wouldn't suggest coming up with one right now.

Akiva Weinberger

The best known algorithms for the Traveling Salesman Problem (and every NP-hard problem) are nonpolynomial @Null

Null

@AkivaWeinberger but that fact alone doesn't prove that there isn't a polynomial one right?

Akiva Weinberger

I think the chess one is provably exponential

though I don't know the proof

Null

I see

Mike Miller

19:47

It takes exponential time to write out all the numbers from $1$ to $n!$.

Akiva Weinberger

…this is true

Null

and polynomial time?

Akiva Weinberger

It was recently proven that determining whether a number is prime is polynomial time in the number of digits

heather

@Null multiplying numbers

Akiva Weinberger

Right?

LeGrandDODOM

19:49

@DanielFischer I've seen in your answers that you know Lusin's theorem (measure theory). Is Lusin's theorem true for functions that may have infinite values ? Is it also true for functions with unbounded domain ?

Mike Miller

@Null That can't take polynomial time (in $n$), because there are more than polynomially many numbers from $1$ to $n!$.

Akiva Weinberger

This was proven in a paper with the borderline ungrammatical title "PRIMES is in P"

I mean, it's grammatical, but it doesn't look like it

Krijn

@AkivaWeinberger Wasn't this done by three students or something

Akiva Weinberger

Was it? I don't know

Null

and polynomial time algorithms are expected to get a quicker solution than nonpolynomial algs?

Krijn

19:52

Must have raised a few eyebrows

Akiva Weinberger

Would be amazing if true though

"For our thesis, we prove one of the hardest major unsolved problems in math"

Mike Miller

It's true. A couple undergrads and a professor.

Krijn

Also, recently was in 2002

Akiva Weinberger

@Null Generally, especially as $n$ gets large. Because, for any polynomial, once $n$ gets large enough, $e^n>{\rm polynomial}(n)$

Exponential time algorithms quickly take longer than the age of the universe to compute, is my understanding

Unless the exponential base is really small and we don't care about huge $n$

Mike Miller

Let $\mathcal K$ be the space of smooth embeddings of $S^1$ into $S^3$ that are isotopic to the unknot. Then it's a theorem that there is a smooth deformation retraction of $\mathcal K$ onto the space of great circles in $S^3$ (aka, $SO(4)$). In principle, following this deformation retraction would give one a way of showing that something is the unknot, if it is (and if the "deformation retraction process" makes sense for all knots, just that it'll end up singular for the rest of them).

Null

19:55

@AkivaWeinberger so for some m<n, it might be true that exp-algorithms do it faster or at least in the same time?

Mike Miller

Unfortunately, nobody knows how to explicitly do this yet.

Akiva Weinberger

@Null Yeah, sure.

Definitely.

heather

anyone here have texmaker on linux and is willing to provide some help?

Balarka Sen

@MikeMiller That's a neat fact.

Krijn

Why are some people so crazy genius

Null

19:56

@Krijn law of big numbers?

Krijn

Reading up on Herbrand, who made contributions to mathematical logic and class field theory, and died at 23

Null

i mean, give me a big enough population and there will to 100% be someone who is smart.

Krijn

Yeah but it's unfair

Mike Miller

@Null That's not quite how the law of large numbers works...

Null

@Krijn i like Galois.

Krijn

19:58

@Null Yeah, another one like that!

Mike Miller

Lily Allen - Not Fair

►

Null

@MikeMiller well, it is, "unrigorous", i agree :P

Maks

Hey ! How can I prove that the set of polynomials of grade less than n is a subspace of R[x] ??

Mike Miller

with gusto

Akiva Weinberger

@Krijn Misread that, thought you wrote "Why are some crazy people so genius"

Krijn

19:59

@AkivaWeinberger Ted Kaczynski comes to mind

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$Mathematics$ Mathematics