Mathematics - 2017-07-16 (page 3 of 3)

Felix.C

18:04

can someone pls give me a quick hint how to show, that $(n!)^{1/n}$ diverges?

Lucyfer Zedd

Uhm, this might be very easy, but can anyone give me hints on how to prove that if a series of positive harmonic functions converge uniformly on an open set of a Riemann surface,then the series converges uniformly on all compact subsets of the surface?

I tried applying harnack's lemma

Astyx

@Felix.C Take the logarithms

Then apply some kind of Cesaro probably

Balarka Sen

$n! > n^n$, so $(n!)^{1/n} > n$

Ted Shifrin

Salut @Astyx

oh, Balarka' still exists ... ?

Astyx

Salut

Balarka Sen

18:11

Indeed

Astyx

$n! \gt n^n$ ?

Ted Shifrin

@EricSilva: Just wanted to let you know I finally heard from Mike Spivak. He's fine. Apparently Amazon "misplaced" a box of 60 of the physics books, so it is temporarily out of print. He will get more copies printed when he has the money to do so.

@Astyx That would be interesting.

Astyx

Corollary : Balarka is interesting

Balarka Sen

@Astyx Yeah

Ted Shifrin

Well, let's not get carried away, Astyx.

Felix.C

18:12

@Astyx Never heard of that guys...I'm also not sure how to use logarithm..do you mean something with series expansion?

Balarka Sen

Any number smaller than $n$ is greater than $n$

Ted Shifrin

Over there >>>>^^^^ it says he's dumb.

Balarka Sen

That was Soham starring my message :(

And I helped him!!

Ted Shifrin

I'm really at my wit's end dealing with this OP. I wonder if mathematics is an appropriate thing for him to be studying.

So the answer clearly is Yes, @Balarka'.

Balarka Sen

sad reacts only

Astyx

18:14

@Felix.C A way to go around this is to fix $N\in\Bbb N$ and notice $n! \ge (N-1)!N^{n-N}$

(for $n$ big enough)

Ted Shifrin

The moral of the story, @Balarka', is to stop helping people. Like Mike and I have. :)

Balarka Sen

Nah Soham is a good friend

Ted Shifrin

"friend"

Balarka Sen

So I don't really know how to prove $(n!)^{1/n}$ diverges other than my troll-kek-proof.

Ted Shifrin

I don't get to use Stirling?

Astyx

18:16

I've given two possible ways (which are equivalent really)

Balarka Sen

Yeah without that

Astyx

@Ted You can't put equivalents to the power $1\over n$ can you ?

Ted Shifrin

How 'bout weak Stirling just from essentially the integral test (with $\log$)?

What does "equivalents" mean, @Astyx?

Astyx

I mean that it's the same computations, but not expressed in the same way

Ted Shifrin

Was that answering the question I just asked?

Astyx

18:18

No, Balarka's

Ted Shifrin

So the inequality you get from integrating $\log$ from $1$ to $n$ should suffice, I think.

Lucyfer Zedd

if a series of positive harmonic function converges uniformly on an open set, does it converge normally?

Astyx

$\ln (n!^{1/n}) = {1\over n}\sum_{k=1}^n \ln k$

Balarka Sen

Oh, I see.

Astyx

And the latter diverges to $+\infty$

(Cause of Cesaro)

Balarka Sen

18:19

Hmm. GM > HM says (n!)^(1/n) > n/H_n

I guess I can use the O(log n) bound on H_n

Astyx

HM ?

Balarka Sen

Harmonic Mean

Astyx

Don't know of this

Balarka Sen

It's just inverse of the AM of the reciprocals

GM > HM is a consequence of AM > GM applied to reciprocals

It's "easy"

Astyx

Right, fair enough

Balarka Sen

18:22

So, yeah, H_n < log(n + 1) + gamma

That means (n!)^(1/n) > n/(log(n+1) + gamma)

Astyx

However GM/HM seems a bit nonelementary given it comes from convexity

Balarka Sen

That does diverge

Astyx

But sure it would work

Ted Shifrin

SO none of you has advice on what further I should(n't) say to the person who can't see a linear relation among $t$, $1/t-t$, and $1+1/t$?

Balarka Sen

@Astyx AM-GM can be proved by elementary means?

Induction 101. You prove it for powers of 2 and fill in the rest.

Astyx

18:23

Huh, the induction is hardcore IIRC

Like you prove it for powers of two

SNIPED

Balarka Sen

It's tedious but fine

Astyx

REVENGE

Alessandro Codenotti

hi @Ted

Ted Shifrin

Heya @Alessandro

Balarka Sen

@Astyx SNIPED

GET REKT

Astyx

18:25

Pfff, so low ..

Balarka Sen

It's a technique devised by Daminark

Astyx

@Ted Don't lose your nerves on this question, you can say that you don't have time right now and move on

Balarka Sen

Hi @PaulPlummer

Paul Plummer

Hi @BalarkaSen

Ted Shifrin

I'm not losing nerve(s). Other than just writing down the answer, I don't see what to do. But it makes me sad.

Astyx

18:26

Yeah, probably advise him to take some time off and think about it with a clearer mind

Balarka Sen

@PaulPlummer What's new?

Also hi @Araske

Araske

hello

Astyx

Hi

And bye, I need to go and eat

Ted Shifrin

Me too soon. Bon appétit!

Astyx

Merci à toi aussi !

Balarka Sen

18:28

See ya

Alessandro Codenotti

bye

Paul Plummer

Nothing much @BalarkaSen There is a question I have been thinking about, but haven't gotten anywhere with it. The question is asking if the non seperating curve graph for surfaces $S_{g,n}$ is $\delta_g$-hyperbolic (independent of $n$, and say $g>2$).. How about yourself?

Ted Shifrin

hi @PaulP.

Semiclassical

hi chat

Ted Shifrin

Hi @Semiclassic.

PVAL-inactive

18:38

Hi @Ted

Hi @Semi

Paul Plummer

Hi @TedShifrin

Ted Shifrin

Hi @PVAL

Paul Plummer

@TedShifrin I was thinking of starting to play bridge, just for the fun, would you recommend not thinking to much about elaborate betting stuff when starting out, or do you think a lot of the fun is in that.

PVAL-inactive

19:10

the entire game is bidding...

Paul Plummer

"elaborate betting", I guess all the different conventions is what I am talking about, or just develop strategies through more "organic" means

Tobias Kildetoft

19:26

@PaulPlummer You need to know the conventions, or your partner will get way too confused about what you mean

Balarka Sen

@PaulPlummer I see. Well, I have been studying Riemannian geometry

Paul Plummer

cool how has that been going? @BalarkaSen

@TobiasKildetoft Well I was thinking about playing with people who have about as much experience as I have, which is pretty much 0 and if we had anything we would agree on it before hand. I guess I was thinking a more "logical" play would feel more satisfying than have some arbitrary coded messages, but I haven't really played except for a test game or two to try to get some people interested in playing more often.

Tobias Kildetoft

@PaulPlummer If you play with a sufficiently basic system, it is far from arbitrary

(not that the more complicated systems are arbitrary, they just might seem so at first glance)

Paul Plummer

Alright, that makes sense, I guess I just need to really start playing, and looking into different setups to get a feel for it.

Tobias Kildetoft

19:43

@PaulPlummer There is a pretty good set of programs called Learn to Play Bridge which walks you through a basic bidding system and also teaches some of the basic techniques of the game itself

Paul Plummer

@TobiasKildetoft Cool, I will have to check them out

Daminark

Hey PVAL, Tobias, Paul, and Balarka!

Paul Plummer

HI @Daminark

Tobias Kildetoft

Hi @Daminark

Balarka Sen

@PaulPlummer It's going, but slowly

I am sitting down and doing a bunch of exercises today

Hi @Daminark, @TobiasK

Daminark

19:46

How's everything going?

Paul Plummer

Were you going to do some things in GGT? How is that going (if I remember correctly)? @Daminark

Daminark

I may yet do some stuff, but I am currently still in the, building up background stage

Paul Plummer

Makes sense

Daminark

This summer I'm gonna do a bit of diffgeo on curves/surfaces

And then I'm gonna take an actual group theory class in the fall, so maybe sometime in the winter or spring would be when I'd do such a reading a course

Balarka Sen

I think your background on dynamics might end up helping you more on GGT than diffgeo of surfaces as background

But of course @PaulP would be able to tell you about that better than me

Daminark

20:01

Hmm

Alessandro Codenotti

What's GGT?

Daminark

Geometric Group Theory

Paul Plummer

I might agree with @BalarkaSen, but no reason not to learn some diffgeo anyways, and it does come up. Although there is probably a lot of of avenues to go having both in your pocket. Plus I am sure whoever Daminark is talking to (if anybody) knows better than I do!

Daminark

At one point I messaged Danny and he recommended me this one book on Hyperbolic Geometry to start with

Which I miraculously found the same day in this pile of free books

But that book wants you to know Riemannian geometry and the like

(Perhaps it was his specific brand of GGT?)

Paul Plummer

Yah, having some knowledge of hyperbolic geometry is definitely important. And if Danny recommended it then you should probably just ignore whatever I or Balarka say.

And you should have knowledge of diff geo, you just don't need enough knowledge to be a Riemannian geometer or whatever

Balarka Sen

20:07

100% agree with everything Paul said.

Paul Plummer

What book is it, out of curiosity? @Daminark

Daminark

Benedetti-Petronio

Lectures on Hyperbolic Geometry

Balarka Sen

@EricSilva Quick, remind me the formula that defines Riemannian connection uniquely out of a given metric

[This is a social test to prove Riemannian geometers have good memory]

(Well, symmetric Riemannian connection, but yeah)

{I don't have any other comments to add but I just thought using different types of parenthesis would be cool}

Daminark

/kek\

Astyx

<-Taking it too far->

Balarka Sen

20:15

@Things get really swirly now@

Astyx

=I'm running out of characters=

Paul Plummer

\\begin{chattext} That looks like a really nice book @Daminark, and having enough diff geo to understand that should be very rewarding \\end{chattext}

Astyx

123By the way, maybe they have good memory, but they're quite slow to answer :p321

SteamyRoot

Daminark

$\langle$ So it seems @Astyx, and yeah I'd like to try the stuff out anyway @Paul $\rangle$

Paul Plummer

20:17

I have just looked at the table of context, have no idea about how it is presented though

Balarka Sen

%&They're busy, unlike us, who are also busy, but busy procrastinating&%

Astyx

$\int$Fair enough$\int$

Balarka Sen

M8TW0T I should be doing work now M8TW0T

Paul's parenthesis are clearly the most original

SteamyRoot

If by Riemannian connection you mean Levi-Civita, I guess you want
$$\langle \nabla_XY,Z\rangle = \frac12 \left( X(\langle Y,Z\rangle + Y(\langle Z,X \rangle) - Z(\langle X,Y\rangle) + \langle [X,Y],Z \rangle - \langle [Y,Z],X \rangle - \langle [X,Z],Y\rangle) \right)$$

Alessandro Codenotti

$\bigg<\text{what's going on?}\bigg>$

SteamyRoot

20:21

(correctness of signs not guaranteed)

Balarka Sen

@SteamyRoot Yeah, that's it. I could actually easily find it on doCarmo, I was just kidding with Eric. Thanks for writing it out, that helps!

Daminark

Your memory is good @Steamy!

You've even retained that memory after you joined the light! :P

Balarka Sen

Steamy was a Riemannian geometer before he became a group theorist

SteamyRoot

Haha :P

Daminark

"SteamyRoot"
Occupation: A Group Theory

SteamyRoot

20:23

Well, once you know that it's like a cyclic sum of vector function of metric and metric of Lie brackets

Balarka Sen

I was trying to imagine how a person can be a group theory

SteamyRoot

All you have to know is the signs :P

Daminark

Maybe a cohomology theory works better

SteamyRoot

Does becoming a theory mean I lose my physical form? D:

Daminark

Like, just take someone and make him satisfy the Eilenberg-Steenrod axioms

Yeah @Steamy, I think the process of making you fit into a long exact sequence will probably mess up your physical form

Balarka Sen

20:29

He's the Hochschild homology theory

SteamyRoot

Well, I'd rather be in a long exact sequence than in a short one.

Because weing in the middle of a short exact sequence means you get things injected into you.

Daminark

I mean May actually gave us the short exact sequence version, and derived a long exact sequence

Using the suspension axiom

Mary Star

Suppose we have three lines of the form $a_ix+b_iy+c_i=0, i=1,2,3$. We consider the vectors (a,b,c). When the three vectors are linearly dependent, what information do we get about the place of the three lines?

Eric Silva

oh geez @Balarka it's long and i am too caught up in something else to type it out rn

i do remember it though

Balarka Sen

how do you even do that man

Paul Plummer

20:42

They just love nasty notation

Eric Silva

it has enough symmetry that i can figure out what it is in my head

Balarka Sen

what a mutant

i need a page from my binded notebook to derive it out

Eric Silva

as riemannian calculations go it's an easy one tbf

by contrast i cannot do the taylor expansion of the metric in my head

Daminark

Key phrase: "As Riemannian calculations go"

Eric Silva

it's a pretty calculational subject maaan

Balarka Sen

20:49

i am failing at a basic calculation rn

i'm terribad

Eric Silva

a lot of calculations in riemannian geometry stuff are simple conceptually but really really long

Akiva Weinberger

21:05

@MaryStar Probably that they have a common intersection or something

Oh, yeah, definitely.

Well, it could also mean two of the lines are equal, I guess

Mary Star

@AkivaWeinberger How do we conclude that?

Akiva Weinberger

or that all three are parallel

Suppose $(c_1,c_2,c_3)$ is a linear combination of $(a_1,a_2,a_3)$ and $(b_1,b_2,b_3)$

For the case where $a_1x+a_2y+a_3=0$ and $b_1x+b_2y+b_3=0$ intersect at a point (that is, they're not parallel)

If $(x,y)$ is in the intersection of those, it also satisfies $c_1x+c_2y+c_3=0$ 'cause you can do the same linear combination

I guess you probably need to check the converse of that

and also the parallel case

The example I'm thinking of is the three lines $x+1=0$, $~y+1=0$, and $x+y+2=0$

all three of which are satisfied at $(-1,-1)$

and note that $(1,0,1)$, $~(0,1,1)$, and $(1,1,2)$ are linearly dependant (the latter is the sum of the first two)

Mary Star

21:25

Since $(c_1,c_2,c_3)$ is a linear combination of $(a_1,a_2,a_3)$ and $(b_1,b_2,b_3)$, we have that $(c_1,c_2,c_3)=m(a_1,a_2,a_3)+n(b_1,b_2,b_3)$.

At the case where $a_1x+a_2y+a_3=0$ and $b_1x+b_2y+b_3=0$ intersect at a point, say $(x_0, y_0)$ we have the following:
$$c_1x_0+c_2y_0+c_3=(ma_1+nb_1)x_0+(ma_2+nb_2)y_0+(ma_3+nb_3)=m(a_1x_0+a_2y_0+a_3)+n(b_1x_0+b_2y_0+b_3)=0+0=0$$
Therefore the thirs line intersects the other lines also at $(x_0,y_0)$.

At the case where $a_1x+a_2y+a_3=0$ and $b_1x+b_2y+b_3=0$ are parallel, we have that $(a_1, a_2)$ is parallel to $(b_1,b_2)$. That means that $(…

Daminark

21:53

Okay so let's say $A$ is a non-negative matrix

Also not the 0 matrix

And let $v\ne 0$ be a non-negative vector

We take its max-mod eigenvalue $\lambda$

Why is it true that for all such $v$, there are positive constants $C_1$ and $C_2$ such that $C_2(v)\lambda^k \le \|A^kv\| \le C_1(v)\lambda^k$?

One direction, I'd think is true because given some non-negative vector $v$ like so, we should have $\|A^kv\| ≤ \lambda^k \|v\|$

Semiclassical

22:41

@Daminark It probably isn't sufficiently rigorous, but if I can write the given vectors in the eigenbasis $\{v_n\}$ as $v=\sum_n a_n v_n$ , then $A^k v=\sum_n \lambda_n^k a_k v_n$

blah, should have been $a_nv_n$ in the last expression

Daminark

That assumes the matrix is diagonalizable which I don't think is necessarily the case

Semiclassical

yeah.

Daminark

Okay so we figured it out, and apparently you need that the matrix is primitive as well

Hushus46

23:00

Hello

Daminark

23:14

Hey @Hushus46!

Hushus46

Don't mind me, its 1am I'm just pondering whats going on in chat

Daminark

23:26

@Balarka when you see this, I think I may have an idea how to prove without invoking the classification of compact connected curves that continuous retractions of a manifold to its boundary are impossible

Since the closed disk is contractible, all homotopy groups are trivial

Now, let $i:S^n\to D^{n+1}$ be the inclusion, and so a retraction would be a function $r:D^{n+1}\to S^n$ such that $r\circ i$ is the identity

But then the induced maps on their nth homotopy groups would give us the identity homomorphism on $\mathbb{Z}$ as factoring through the trivial group, which shouldn't be possible

So this should give you continuous Brouwer immediately

ALannister

0

Q: If $R$ is a subring of a field $F$, then the ring of fractions is embedded in $F$

Let $R$ be a subring of a field $F$ and $D$ a multiplicative subgroup of $R$. I need to prove that the ring of fractions $D^{-1}R$ is embedded in $F$. To do so, I began by supposing that $R$ is such a subring of $F$ and $D$ a multiplicative semigroup of $R$. Then, I noted that since the subring ...

In case anyone is able and willing to help <3

Hushus46

Goodnight every1

Ted Shifrin

23:41

sure, Demonark.

Daminark

Also found this cool proof of FTA

Let $f(x) = x^n + c_1x^{n-1} + \ldots + c_n$. Assume $f$ has no zeroes for $\|x\| \le 1$. We have an induced map $\hat{f} = \frac{f}{\|f\|}: S^1\to S^1$, so we want to compute its degree

Using $h(x,t) = \frac{f(tx)}{\|f(tx)\|}$, we can homotope $\hat{f}$ to the constant map $\frac{c_n}{\|c_n\|}$, so $\hat{f}$ has degree 0

Ted Shifrin

This is the standard proof, same with (smooth) degree ...

I guess I should say a standard proof, as there are so many.

Semiclassical

The one I always think of is the complex analysis proof.

Daminark

But we can also use $H(x,t) = \frac{k(x,t)}{\|k(x,t)\|}$, where $k(x,t) = x^n + t(c_1x^{n-1} + tc_2x^{n-2} + \ldots + t^{n-1}c_n)$, so that we're homotoping $\hat{f}$ to $x^n$, so $\hat{f}$ has degree $n$. But that implies $n=0$, so we're done

@Semi using Liouville?

Semiclassical

Right.

Ted Shifrin

23:54

There are about 4 using complex analysis, offhand.

Semiclassical

Yeah, true.

But the Liouville proof is the one I meant.

Ted Shifrin

Rouché is essentially the homotopy one Demonark is citing.

Semiclassical

Neat.

Daminark

Well, I know this one and one of a basically identical nature using winding number. I know how to proving it using Liouville but we're a bit behind so haven't reached it yet

Ted Shifrin

You're applying Liouville to $1/f$?

Semiclassical

23:56

I should probably give a nod to the argument-principle proof, if only because how essential the argument principle idea is to numerical analysis.

Ted Shifrin

Rouché is basically a homotopy idea, but says specifically that if you change $f$ by something $g$ of smaller magnitude on a curve, then $f$ and $f+g$ have the same number of roots inside. It's basically saying that the dog you're walking goes around the block the same number of times you do. :P

Argument principle is another one.

Maximum principle is another one.

Daminark

And we had the proof which basically says that outside of some large ball around the origin, you can be sure that $|f|$ is much larger than the infimum, but inside the ball you use compactness so you achieve a minimum, so that if you hop to the image, it's basically tangent but intersecting the disk of radius $\min |f|$, which isn't possible

Semiclassical

Following up on what I said, the argument principle idea is nice because it suggests nice methods of numerical root-counting and root-finding.

Daminark

In class we proved it was impossible by directly dipping into the ball via term playing, and on the pset we proved using inverse function theorem that polynomials are open mappings, which is also no good

Ted Shifrin

Yeah, I usually showed both diff top and complex analysis a Mathematica animation doing that.

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$Mathematics$ Mathematics