Mathematics - 2020-12-01 (page 1 of 2)

Annamalai Sriram

01:08

21 hours ago, by Annamalai Sriram

in JEE Maths Zone, 2 days ago, by Annamalai Sriram

This is the solution provided to the problem

If anyone explain the first part of solution that's enough.

copper.hat

01:21

solve y' = y/x

not sure why that didn't occur to me yesterday when you asked.

Ted Shifrin

The first two lines come from working out the algebra they give you with the tangent and normal lines.

monoidaltransform

01:59

hi

If $X_n\in \mathcal{L}^1$ are random variables, how do I show that for each $n$, there exists $c_n$ such that $\mathbb{P}(|\frac{X}{c_n}|\geq \frac{1}{n})\leq \frac{1}{2^n}$

?

epic_math

02:35

I, the god of mathematics, came here to see how modern people do mathematics.

Which book on analytic number theory should I read after Apostol?

copper.hat

03:28

@monoidaltransform it is equivalent to showing that $\lim_{c \to \infty} P[|X| \ge c \epsilon ] = 0$ for any $\epsilon>0$. this follows from continuity of measure.

Mike

Hello, can anyone take a look at my post regarding mutually orthogonal probability measures? Thank you.

0

Q: Show that two orthogonal probability measures have the following representations for disjoint intervals on $[0,1]$

Problem: $\alpha$ and $\beta$ are two mutually orthogonal probability measures in the sense that for some Borel subset $E\subset[0,1]$, $\alpha(E)=0$ and $\beta(E)=1$. If $F(x)=\alpha\{[0,x]\}$ and $G(x)=\beta\{[0,x]\}$ for $0<x\leq 1$, show that for any $\varepsilon>0$, there is finite collectio...

user 6232128

03:44

@AlessandroCodenotti shouldn't all historical events be starred , in this chatroom, for future reference?

Ted Shifrin

@user6232128 Certainly not the jokes.

user 6232128

:p

Rithaniel

I'd actually disagree. Jokes deserve to be immortalized the same as any other thought. (Within reason, of course)

user 6232128

when is your next math club zoom presentation?

Rithaniel

We have them on Fridays. Different speaker each week

user 6232128

03:51

thanks

Rithaniel

So, my next presentation would probably be next semester

If you want, I can actually talk to the organizer and see if we can invite people from off campus

user 6232128

nah, I just like the idea in general

Rithaniel

Yeah, I still haven't watched the video of my presentation back

There's just something nerve wracking about it

user 6232128

you did a fine job

Rithaniel

Well, I'm glad about that

In other news, it's the last week before finals where I am, and I made a game-assignment for my kids to compete against each other while practicing math. However, it looks like it might not be played because I over-estimated the enthusiasm of Freshmen in a math course

I'm thinking about spreading it around to others so that it doesn't go to waste

user 6232128

04:10

yeah, providing games so close to the final exams is being a bit overly optimistic :P

Rithaniel

Well, that's a thing for me to remember for next time

user 6232128

Right now, most of them, are scrambling to find practice final exams.

Ted Shifrin

04:41

Heya @Rithaniel

Leonhard Euler

Hello everyone

Leonhard Euler

05:35

0

Q: Evaluating the sum $\sum_{n=1}^{\infty}\frac{1}{F_n}$

Let $F_n$ be the $n^\text{th}$ Fibonacci number. I wanted to calculate $$\sum_{n\ge1}\frac{1}{F_n}$$ I simplified it to $$\sqrt{5}\sum_{n\ge1}\frac{1}{\varphi^n-\phi^n}$$ But this didn't seem to help. The value is approximately $3.35988566624317755317201130291892...$. The OEIS entry of this cons...

sequences-and-series fibonacci-numbers

Does someone know the answer?

Annamalai Sriram

05:51

@TedShifrin Thanks I came up with equations but I have a small doubt that is if we consider the equation of the tangent y = mx + c , the equation comes up only when y = f(x) but don't the y in tangents equation represent the trajectory of tangent

Ted Shifrin

I don't understand. Write the equation of the tangent line to $y=f(x)$ at $x=a$. Same for $g$. The equation for them to intersect on the $y$-axis is an equation with $f$, $g$, $f’$, $g’$ all evaluated at $a$. They then rewrote it with $x$ instead of $a$.

Annamalai Sriram

06:08

but I think only the slope "m" comes from the y = f(x)

at x=a

copper.hat

06:23

the equation of the tangent line at a is given by f(a)+f'(a)(x-a), so the value of the y-axis intercept is when x=0, so that gives f(a)-af'(a)

hence f(x)-xf'(x)=g(x)-xg'(x)

Ted Shifrin

@copper That first one is not an equation.

feynhat

@TedShifrin How much longer do you plan to stay up?

Annamalai Sriram

Ah! Yes, I got it. Generally in the equation of tangent y is a variable but at point x=a the "y" of the tangent is equal to that of "y = f(a)" of curve so it is equal to f(a)

Ted Shifrin

No, that's not right. But use the point-slope formula for lines.

Why? @feynhat

feynhat

I had a couple of questions.

Sections of line bundles and stuff.

Annamalai Sriram

06:33

yes it comes with slope formula y = mx +c as you said I've misunderstood that point

Ted Shifrin

Around for a half hour or so.

feynhat

Okay. Give me a minute.

Annamalai Sriram

@TedShifrin @copper.hat Thank you very much.

copper.hat

@TedShifrin sry, i meant the function t(x)=f(x)+f'(x)(x-a)

@AnnamalaiSriram it might help to draw a little diagram. my research advisor used to call such things as 'proof by picture'.

Ted Shifrin

Better to write the equation since we want to set the $y$-intercepts equal.

feynhat

06:42

Let $F(x, y) = x^2 - y^2$. I want to show that the section of $\gamma_{\mathbb{CP}^1}^{\ast\otimes 2} \to \mathbb{CP}^1$ (ie. the second tensor power of the dual of the tautological line bundle. I believe its also sometimes written as $\mathcal{O}(-2)$) induced by $F$ is transverse to the zero section.

Lets denote the induced section by $s_F$. I need to check that at the points where $F=0$, $s_F$ is a submersion. The point in $\mathbb{CP}^1$ where this happens is $[1, 1]$.

What does the induced section look like in charts? Let $U_i$ denote the standar atlas of $\mathbb{CP}^1$. In $U_0$, the polynomial is just the dehomogenization $f(z) = F(1, z)$. Now, the target of the section is $\gamma_{\mathbb{CP}^1}^{\ast\otimes 2}$ which has a standard trivialization over $U_0$.

So, the section can be thought of as a map $U_0 \to U_0 \times \mathbb{C}$. I will ignore the the first output because its just identity. So, basically, this is just $U_0 \to \mathbb{C}$. What is this map? Is it just $F(1, z) = 1 - z^2$?

Ted Shifrin

No, it’s written $\mathscr O(2)$.

There are two zeroes, of course.

feynhat

right the other being [1,-1].

Ted Shifrin

Right.

Yes, that's the dehomogenization.

And the derivative is non-zero at $z=\pm 1$.

feynhat

Thanks. It seems a bit hacky to me. Like I missing some detail.

Ted Shifrin

Hacky?

Amin Idelhaj

06:53

Hey Ted!

Ted Shifrin

Hi Demonark

Amin Idelhaj

How are you doing?

feynhat

So, if I want to check if a section induced by a homogeneous polynomial is transverse to the zero section, I just need to see if look at the roots of its dehomogenization and see that all of those roots are simple (derivative doesn't vanish at those roots.)

Ted Shifrin

Still alive.

Amin Idelhaj

That's good, same here

Ted Shifrin

06:55

In general, what does it mean for a section to be transverse to the zero section?

For any line (vector) bundle?

feynhat

At the points where they intersect, the images of the derivatives should span the tangent space at those points.

Ted Shifrin

So if the section is given locally by a function, the graph if that function should be transverse to the graph of the zero function. Just go back to first principles.

feynhat

Hmm... so locally, the sections are given by dehomogenization of the polynomial, so my assertion about roots being simple is true.

Ted Shifrin

Yes

Rithaniel

Heya Ted. Sorry about the delay in response
(I find myself on a nocturnal schedule once again)

Ted Shifrin

07:05

No problem. Balarka breaks everyone’s sleep.

Rithaniel

Hehe, fair enough. I break my own sleep, though. Of course, I'm thirty now and it's becoming more difficult to switch between different sleep schedules

Ted Shifrin

Old man!

Amin Idelhaj

Now that I think about it my sleep schedule was better before I met Balarka

10

Correlation \implies causation and all that

Ted Shifrin

You've been around here since late your first year in college, I think.

Amin Idelhaj

Maybe mid second year, I think I already finished a quarter of analysis by then?

Ted Shifrin

07:09

Approximately.

feynhat

07:55

@SayanChattopadhyay Hi.

Have you returned to Mohali?

Leonhard Euler

08:47

2

liked it?

Alessandro Codenotti

10:23

@Balarka I noticed only now but King Gizzard put out a new album a couple of weeks ago

Abhigyan Chattopadhyay

12:57

Hi all! I'm currently a 2nd year EE student. I plan on taking an introductory Graph Theory course in my next semester and wanted some insights from people who possibly have more experience in that field.

Would anyone say that Graph Theory is, in general, a heavy course for someone who has only done Multivariable Calculus and Partial Differential Equations, in terms of mathematical maturity?

Any and all insights are appreciated

Edward Evans

learn how to do inductive proofs

Abhigyan Chattopadhyay

So that's what constitutes the bulk of Graph theory, is it?

Edward Evans

13:21

Well

it'll be proof heavy

and a lot of proofs in a first course on graph theory are inductive

idk how it is in more advanced courses

but it's definitely an important technique in graph theory

Leonhard Euler

Hello

I noticed something

I was just calculating some sequences

Let $\omega(n)$ be the number of distinct prime divisors of n.

And $\Omega(n)$ the number of prime divisors of n, counted with multiplicity

Let $$l(x)=\frac{\pi(x)}{\Omega(n)-\omega(n)+1}$$

Can every positive integer be represented in the form $l(n),n\in\Bbb{N}$?

It seems to be true from computational evidence

Edward Evans

13:38

3

Astyx

yikes

Mike Miller

@Leonhard if p_n is the n'th prime then I(p_n) = n

Leonhard Euler

okay thanks

It is l, btw

not I

Rithaniel

@EdwardEvans Amazing

Edward Evans

lel

Leonhard Euler

13:45

@EdwardEvans Please tell the name of the book, it seems very rigorous and intuitive.

Ugly duckling theorem

The Ugly duckling theorem is an argument showing that classification is not really possible without some sort of bias. More particularly, it assumes finitely many properties combinable by logical connectives, and finitely many objects; it asserts that any two different objects share the same number of (extensional) properties. The theorem is named after Hans Christian Andersen's 1843 story "The Ugly Duckling", because it shows that a duckling is just as similar to a swan as two duckling are to each other. It was proposed by Satosi Watanabe in 1969. == Mathematical formula == Suppose there are...

Edward Evans

14:01

what

Leonhard Euler

btw if someone wants more fractals images ask me

I have tons of those

See this mandelbrot:

I zoomed out further and got this image:

Mike Miller

that seems more like a glitch

2

Leonhard Euler

oh okay

I wondered what was this

user2103480

@EdwardEvans seems like a normal lemma in analysis

Leonhard Euler

And a remark of a normal analytic number theory book

Edward Evans

14:16

lol

Leonhard Euler

okay just one more fractal image I want to share:

Someone tell me the creator of this one

Really hilarious

this is called 'heart'

user2103480

looks more like an upside-down pear to me

Leonhard Euler

when someone thinks like you, his mind is called 'the mind of a scientist'

congrats :P

Edward Evans

congrats @user2103480

user2103480

thanks for the gold kind stranger

ah wait wrong cringy forum

Edward Evans

14:28

heh

Leonhard Euler

okay so give a name of this:

other than axe

Edward Evans

axe

Leonhard Euler

*other than axe

Astyx

who're you calling cringy nerd?

anakhro

worm escaping my apple

user2103480

14:29

empty straw next to puddle of ket

aka a british lunch

4

Edward Evans

looool

Leonhard Euler

@anakhro worm is making a hole

WoRmHoLe

Edward Evans

ket coke and piiiills

anakhro

Okay, now let's discuss actual math.

Anyone learn anything neat recently?

Edward Evans

why should I care about Hilbert 90

Like if i have a field $K$ with group $G$, why is $H^1(G, GL_n(K)) = 0$ an interesting result

Rithaniel

14:31

I learned what a Banach space is, after months of working with Hilbert spaces

anakhro

@Rithaniel I don't believe you.

Edward Evans

Taking the Banach dem du dich in einem vollbesoffenen Zustand blamiert hast

Leonhard Euler

not learned recently, but it is $$e^{\pi i}+1=0$$

And I learned recently that 'formal definition' of a fractal

anakhro

@Rithaniel you might be very happy with learning about Banach algebras.

Leonhard Euler

a fractal is...

user2103480

14:36

@anakhro I need to do semigroup theory for finding mild solutions of SDEs and SPDEs now

Leonhard Euler

a set for which the Hausdorff dimension strictly exceeds the topological dimension

user2103480

and other PDE stuff that I'm not used to yet

anakhro

@user2103480 is this for research towards a thesis?

user2103480

it's for a master's course on SPDE, and tangentially for courses on stochastic processes in neuroscience/fluids

Leonhard Euler

just wanted to know, what is your (you all's) favorite branch of mathematics?

Mine is Analytic NT

anakhro

14:39

@user2103480 Very fancy sounding. How does it relate to neuroscience/fluids, roughly?

user2103480

@anakhro for stochastic processes in neuroscience, one begins with basically a circuit model like in electronic circuits, and add's a noise with drift that is a first stop towards formalising the randomly incoming inputs to the neuron

so we arrive at a stochastic differential equation, and we're interested in spiking times

The time depenedent density of such a process then satisfies the fokker-planck equation, a form of continuity equation

Leonhard Euler

bye

Rithaniel

@anakhro Nah, really. I'd seen the term bandied about before, but I had never actually gotten around to looking at the definition. My linear analysis course has been working almost exclusively with Hilbert spaces (I think this is in part due to the fact that the presence of an inner product makes things easier)

Leonhard Euler

have to go

user2103480

and we're interested in finding densities for the times when spikes occur, which can be imagined as kinds of poisson processes

Leonhard Euler

14:44

Euler would come after some time

user2103480

So we already have some SDE and PDE stuff at hand, and try to find differential equations that densities of stopping times, depending on starting potential, satisfy, and reasonable boundary conditions. Then we do laplace transforms etc. It's hairy

Alessandro Codenotti

@user2103480 noice

user2103480

When one goes much further than me, one arrives at SPDE models for neuron firing and how the spikes are propagated between neurons

And for those, the semigroup stuff - also related to markov processes and their generators - is helpful to define forms of solutions to certain types of SPDEs

anakhro

@user2103480 Ah, I see. Do you enjoy seeing this side with the applications? Or are you more interested in the pure SPDE side.

@Rithaniel Well it's a great type of space, and now it opens up all the Gelfand-Naimark theory stuff.

user2103480

I did enjoy the preliminaries, i.e. constructing stochastic integrals on function spaces (formally on hilbert spaces, but in the context of SPDEs, function spaces are mostly relevant)

I couldn't imagine doing "pure" PDE

Rithaniel

14:53

Oh, new term to look up

user2103480

all the intuition comes from the scenarios where the PDEs come from and I lack intuition for differential equations in any case

so I think I need applications to understand what that is all about

anakhro

@user2103480 I suppose most differential equation stuff comes up as physical problems. Then you want to be able to solve these specific problems. Then they come up with some mathematical tools. But these tools have limits, and so forth.

user2103480

the stochastic processes in fluids stuff is more focused on correlation functions of complex-valued stochastic processes and the dynamics that these are related to, so lots fourier transforms

apparently this is all in the hamiltonian dynamics/thermodynamics setting but I don't get that yet

anakhro

With the fluids stuff, does it get at all symplectic?

user2103480

no, thankfully not

anakhro

15:01

Because in like hydrodynamics where the Hamiltonian setting is a given, most stuff seems to revolve around symplectic geometry.

user2103480

Hansen and McDonald: Theory of simple liquids (Academic Press, 2006).

thats our main book

Sayan Chattopadhyay

Hmm let me see if I remember this correctly. If you take a manfiold M as your configuration space, then the setting for fluid dynamics is you take G to be Diff(M), which a lie group, look at it's lie algebra. It's dual has a Poisson structure and you restrict to the symplectic leaves.

Edward Evans

is a finitely generated module over a discrete valuation ring flat?

I want it to be flat

anakhro

@user2103480 Thanks, I will have to check it out.

@SayanChattopadhyay that basically just summarizes integrable systems in general, not just fluid dynamics.

user2103480

@SayanChattopadhyay thankfully, we're not doing that

Sayan Chattopadhyay

15:14

Yeah @anakhro I was thinking about integrable systems and changing the lie groups was giving you various physical systems. I think one can derive the Euler equations from here by looking at the coadjoint orbits and stuff, I may be wrong

anakhro

Yes, I think that sounds about right.

Sayan Chattopadhyay

@user2103480 ah I see. There's a book my Arnold on fluid dynamics isn't there?

anakhro

if it has the word dynamics in it, Arnold probably wrote something on it

Alessandro Codenotti

When people say "the topology generated by a family of pseudometrics" what is meant exactly?

anakhro

@AlessandroCodenotti link to the context?

Alessandro Codenotti

15:20

I was thinking about finite intersections of balls but I'm not sure

Astyx

That's what I meant to say

It's the roughest topology such that the pseudonorms are continuous

IIRC

Alessandro Codenotti

Right but they are functions on the product so it's annoying

Maybe it's the same as something like the coarsest topology such that for all $x\in X$ and all pseudometrics $d$, the function $y\mapsto d(x,y)$ is continuous

Leaky Nun

@EdwardEvans can't you use STFGMPID

to reduce it to checking whether R/π^n is flat

and I assume they aren't?

love_sodam

16:04

How can I show one sheeted cone without the origin is non-extendable?

nonextenable: If S is a connected regular surface then $S$ is called non-extendable if it is not a proper subset of a larger connected regular surface

Edward Evans

@Leaky well the quotients are finite rings in this case .. idk if that helps, my Alg2 weakness is showing

oh wait

I think my ring is free over the base ring lmaoo

Leaky Nun

ergo ut finitus potest?

Edward Evans

it's f.g. free over

the dude

Leaky Nun

you said free dude

so it's flat

Edward Evans

yeah exactly

that's why "lmaoo"

Leaky Nun

16:11

ok

anakhro

@love_sodam This cone extends infinitely in the one direction?

Edward Evans

it's a general ANT fact that if you have an extension L/K then O_L is a free O_K-module

of rank [L:K]

anakhro

Or is it just an open edge.

Leaky Nun

@EdwardEvans it is not

Edward Evans

wat

Leaky Nun

16:12

A Non-Free Relative Integral Extension by Keith Conrad

love_sodam

@anakhro $\{(x,y,z)\in\Bbb R^3:z = \sqrt{x^2+y^2}\}$

And remove the origin

Edward Evans

damn

anakhro

Alright, so it extends infinitely upwards.

Edward Evans

ruin the basis of my existence why don't you

love_sodam

Yea

anakhro

16:13

So where could you possibly extend it, @love_sodam?

Edward Evans

so what if I have the additional assumption that my fields are local?

Leaky Nun

then I think you're fine

@EdwardEvans c'est mon plaisir

Edward Evans

rofl

I mean

yeah it works in the local case

love_sodam

@anakhro maybe downside..? If I add a handle on the surface of the cone then what is the problem?

anakhro

@love_sodam how do you add a handle? Walk me through the steps.

love_sodam

16:18

@anakhro maybe I will draw it

anakhro

I know what it would look like, but to see the problem, you have to explicitly write what you are doing.

Because adding a handle starts with a very problematic operation on manifolds.

love_sodam

I thought If I add a handle then it's connected regular surface that properly contains the original cone

anakhro

Well how do you add a handle?

Walk me through the steps.

What is the first thing you have to do when adding a handle to a surface?

Or put another way, a handle is an gluing of a tube or cylinder to the surface---where am I gluing this tube to on the cone?

love_sodam

On the side of the cone..?

Edward Evans

@Lukas don't look at the erroneous claims I made about the integers of extension fields

love_sodam

16:26

I didn't think specifically about that. Just wonder why that's not a connected regular surface containing the cone.

anakhro

@love_sodam how do I glue it on the side of the cone?

Edward Evans

@Lukas it works in the local case but Leaky gave an example where it doesn't work in the global case

Lukas Heger

it's true if $O_K$ is a PID and $K$ is a local or global field or if you assume that $L/K$ is separable

Edward Evans

but I only care about the local case atm so screw that paper

Lukas Heger

ofc in the local case $O_K$ is always a PID

Edward Evans

16:29

The other day you helped me prove my result in the p-torsion case and now I've extended it to the p^n torsion case, but the result rested on this flat extension

anakhro

@love_sodam after you figure out how you glue it to the side of the cone, you will most definitely realize why it's not a connected regular surface containing the cone.

Lukas Heger

@EdwardEvans which flat extension?

love_sodam

@anakhro just draw two disjoint circle on the side of the cone and connect those two by a tube

Edward Evans

$\mathcal{O}_K \to \mathcal{O}_{\widehat{K^{ur}}}$

Lukas Heger

yes that's flat

note that over PID, flat is equivalent to torsion-free

Edward Evans

16:30

the right guy is free over the left guy and I wanted left tensoring over $\mathcal{O}_K$ to be exact

anakhro

@love_sodam okay, so what appears to be the problem with that?

Lukas Heger

and clearly $\mathcal{O}_{\widehat{K^{ur}}}$ is a torsion-free $\mathcal O_K$-module

Edward Evans

yeah lol

but freeness gives you the flatness as well

Lukas Heger

I don't see why it's free

love_sodam

@anakhro Maybe the circle part..?

Edward Evans

16:32

ah shiet

Lukas Heger

there's a completion here, and completions are in general not free, e.g. $k[[x]]$ is not free over $k[x]_{(x)}$

Edward Evans

damn

Why does torsion free over PID imply flat?

anakhro

@love_sodam suppose you are an ant living on your tubed cone. If you lived on the edge of a circle that you just attached a handle to, what incredibly un-surface-like feature do you notice about your home?

Lukas Heger

there's a criterion for flatness which say that a left module over a ring $R$ is flat iff for all f.g. left ideals $I$, the map $I \otimes_R M \to IM$ is injective

over a PID $I=(r)$ is isomorphic to $R$ via $s \mapsto rs$ and using that isomorphism and $R \otimes_R M \cong M$, the map $I \otimes_R M \to IM$ becomes multiplcation by $r$: $M \to rM$

Edward Evans

ah nice

Lukas Heger

16:35

and clearly torsion-free means that multiplication by any scalar is injective

Edward Evans

stick that in the bank

yeah

thanks again lol

just do my pset for me

Lukas Heger

if you use that $M$ is flat iff $M_{\mathfrak{p}}$ is flat for all primes, you can also make torsion-free iff flat work over a Dedekind domain

Edward Evans

@Lukas just as a sanity check.. if I have an exact sequence and Hilbert 90 holds, that "means" that taking Galois invariants is exact right?

because I get my long exact sequence from Galois cohomology and if the first cohomology vanishes then zeroth cohomology is exact

Lukas Heger

yes

Edward Evans

yay then my solution works

lmao

love_sodam

16:39

@anakhro No nbd diffeomorphic to R^2 at that point?

anakhro

@love_sodam why the question mark?

love_sodam

@anakhro No nbd diffeomorphic to R^2 at that point.

anakhro

@love_sodam okay, so that is problematic, right?

love_sodam

@anakhro Yes

@anakhro Then the only extendable way is to put the origin?

anakhro

@love_sodam alternatively, you might try to add a handle by cutting out holes, then attaching the tube to the boundary of the holes you punched (this is the usual way of adding a handle, usually to existing holes), but this is problematic to.

And yes, I'd say the only convenient "edge" on which to extend your cone is the edge, but this is obviously problematic.

love_sodam

16:49

@anakhro Thanks. I got it

Astyx

17:22

o/ Daminark

Edward Evans

17:34

@user2103480 Meinst du "Linkstensorifikazifizierung" ist zulässige Sprache?

love_sodam

If S is a connected complete regular surface and define $diam(S) = \sup\{d(p,q)|p,q\in S\}$ where $d$ is a intrinsic distance. If $diam(S)$ is finite then $S$ is compact.
I need to show $S$ is closed. So if $\{x_n\}$ be a sequence in $S$ that converges to $x$. Then I want to show this sequence is d-cauchy so that $x\in S$. So suppose not. Then, there is $\epsilon>0$ such that $d(x_i,x_j)>\epsilon$ for any $i,j$. I want to show $d(x_1,x)$ is infinite. How can I?

Edward Evans

hahahaha

Alessandro Codenotti

@user2103480 @BalarkaSen probability nerds do you know about the Lovasz local lemma by any chance?

Anush

17:56

in this puzzle, what is S_n? "n red points r_i in the plane and n blue points b_j. Show there is a matching \sigma \in S_n such that none of the straight line segments joining a_i to b_\sigma(i) cross. "

Astyx

Symmetric group

the group of permutations of n elements

Anush

@Astyx hmm... is this hard to solve?

Astyx

I'm not sure, but it doesn't look impossible at first sight

if (r,b) and (r', b') cross, then look at (r, b') and (r', b)

And you can hope that applying this algorithm over and over again gives you such a matching

There might be pathogenic cases however, if you have four points in a line, blue blue red red

I'm not sure whether that counts as crossing

Alessandro Codenotti

There's a mindblowingly neat solution to this puzzle, I discussed it here before

Astyx

I'm all ears (or eyes, rather)

Alessandro Codenotti

18:08

But yeah you do need to exclude sets with 3 or more collinear points

Ok here's the trick

There are finitely many ways to place $n$ segments between those two sets of points. I claim that one minimizing the sum of the lengths of the segments works.

To see that is enough to, given a matching with a crossing, produce one with lower total length. Suppose that the segments between $r,b$ and $r',b'$ cross. Replace them with those between $r,b'$ and $r',b$, by the triangle inequality total length went down

Astyx

Oh yeah that's neat

Alessandro Codenotti

It's my favourite tricky puzzle :P

user2103480

18:51

so the basics of thermodynamics are still a bit opaque to me. This post (https://terrytao.wordpress.com/2007/08/20/math-doesnt-suck-and-the-chayes-mckellar-winn-theorem/) has a pretty good summary, and the math isn't what's confusing me, but the "objects" we're working in.
When he says " the number of possible microstates that the system S could be in is finite", what is meant here? I imagine a microstate to be something like an extremely small region in space

@EdwardEvans linkstensorifikation eher

Edward Evans

:/ Das wäre ein neuer Fachbegriff gewesen

user2103480

@AlessandroCodenotti heard of it, not more

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$Mathematics$ Mathematics