Mathematics - 2014-11-27 (page 2 of 3)

@TedShifrin Consider the p-adic solenoid $X$ defined by the inverse limit of inverse system of circles with the morphisms being multiplication by some prime $p$. I have proved that $\mathbf Z_p$ acts properly discontinuously and fixed point freely on $X$. i almost believed that there is a ses $1 \to \pi_1(X) \to \pi_1(X/\mathbf{Z}_p) \to \mathbf{Z}_p \to 1$ right when mike pointed me out that $X$ is not path connected :(

such a waste

Ted Shifrin

@Balarka: I have never in my life thought about p-adic stuff.

Balarka Sen

i am thinking of patching this up by looking at the monodromy assosiated to the natural map $X \to S^1$

it should be Z_p, and there must be some generalization of some such by replacing fundamental groups by monodromies in this setting dunno

@TedShifrin well not many geometers are interested in p-adics.

user2179021

4:38 PM

hi

anyone got any ideas about math.stackexchange.com/questions/1039829/… ?

Ted Shifrin

Arithmetic geometers are ...

user2179021

I thought about turning it into an integral maybe

Balarka Sen

@TedShifrin my objective is to have geometric interpretation for profinite groups, that's why i am looking at Z_p as an example. the actual goal is to do this for gal(\bar q/q) but that's probably too much to hope for

Alizter

If I say a finite set $X$ has carnality $n$ do I really mean that there is a bijection between $X$ and $n$ (with the von neumann ordinal construction)?

Mike Miller

That's the definition, yes. Not my preferred one, but my preferred one isn't equivalent, so...

anon

4:43 PM

a finite set $X$ has carnality $n$ if it has had sex with $n$ partners

2

Balarka Sen

@MikeMiller!

anon

@BalarkaSen hmm, interesting

Balarka Sen

@anon phew glad to know someone thinks that

Jasper Loy

@anon I hope you do not get flagged.

Alizter

@MikeMiller What is your preferred definition and why.

anon

4:44 PM

interview tiem

Jasper Loy

@anon With who?

Alizter

@JasperLoy himself.

Balarka Sen

@anon i am free to suggestions, just to let you know

my initial idea to patch this up was to look at some embedding of $X$ in $\Bbb R^n$ (which should naturally be a manifold) and look for some action of Z_p on the embedding. but my prof pointed out that it's currently a conjecture that Z_p cannot act on manifolds by homeomorphisms

in fact, he said that it has been settled that Z_p cannot act on manifold by diffeomorphisms

Mike Miller

@Alizter A set is finite if any injection to itself is a bijection. I find this more natural than "is in bijection with some certain set".

@Balarka I would be very surprised if the last result was known. If it is, it should be very recent.

Balarka Sen

it is recent, afaik

Mike Miller

4:49 PM

Also, wait, you're trying to make $X$ into a manifold?

Balarka Sen

@MikeMiller no the complement.

but i have refuted the idea

Mike Miller

Oh, sure. I think that's how you construct the Whitehead manifold.

Balarka Sen

i am now looking at the monodromy associated to the morphism $X \to S^1$ by contracting the fiber onto a point

the monodromy group should be $\mathbf Z_p$, giving some hope to think about this analogy of fundamental groups

Alizter

@MikeMiller How do you say that $\{1, 2, 3\}$ has cardinality $3$ then. Because I am trying to understand how $|.|$ is defined.

Mike Miller

What do you mean by mondromy here?

Alizter

4:52 PM

Because if $3$ is a set then how can I find it's cardinality without being paradoxical.

Mike Miller

@Alizter Oh, that's a different question. Remember - cardinality means "class of sets that are in bijection". To say that we have cardinality (something)... we must have some set in mind when writing it down.

Balarka Sen

@MikeMiller i dunno, it's an intuitive idea. locally the map X \to S^1 is not even a covering space (is it?) but for example if you loop around the base S^1 then the endpoint of the lift of the loop moves from one leaf to another in X

you have to realize X as the fiber product of R and S^1 to do this

Mike Miller

@Balarka But it's not a fiber product of R and S^1.

Balarka Sen

sorry i mean R and Z_p

Mike Miller

@Alizter So when we say cardinality n, we do mean that von Neumann ordinal, yes.

Alizter

4:55 PM

@MikeMiller OK :)

And so does that make the VN construction superior over a construction like $\{S\}$?

Mike Miller

The "problem" with my definition of finite... is that not every finite set needs cardinality n.

Alizter

Zermelo's construction I think it is called?

Mike Miller

I don't know what you mean by the latter construction. If they're both equivalent, it doesn't matter.

@Balarka Anyway, if you write down the standard definition of monodromy of a covering wrt a basepoint you'll see you're in trouble.

Alizter

@MikeMiller How can a finite set not be bijective to an ordinal?

Balarka Sen

i know i would be @Mike. X \to S^1 is not even a covering

Mike Miller

4:59 PM

I'm pretty sure it is, @Balarka. But if you say so.

Balarka Sen

hmm

ok, let me think about it

Mike Miller

@Alizter If you use my definition, "finite if any injection is a bijection", it's not necessarily true.

Alizter

@MikeMiller The wording is a bit unclear for me. Are you saying if all injections are bijections or if there exists a bijective injection?

Mike Miller

Huh? The identity map is a bijective injection. I'm saying if all injections are bijections.

Balarka Sen

@MikeMiller it has also been settled for 3-manifolds : here. the referenced paper [4] settles the diffeomorphism case. the author of [20] is my professor.

Mike Miller

5:07 PM

Hm, the diffeomorphism case is old. I must have a narrower view of the problem than I thought.

@Balarka So I was wrong; of course thats not a covering space, since by definition the fibers need to be discrete. But totally disconnected fibers is not an entirely different scenario. People have probably written about this before.

Alizter

@MikeMiller Is that not problematic for sets with cardinalities larger than $1$?

wait

I still don't see a problem

Mike Miller

@Alizter No, every set you would call finite, I would still call finite. But there can be more finite sets.

@Balarka Here's some food for thought with lots of references. Jeremy Brazas spends a lot of time thinking about things that are topologically fucked up, which your situation lies in.

Alizter

@MikeMiller Explain :)

Balarka Sen

@MikeMiller right so it's a covering space only if we are talking about the cantor set with discrete topology, which is rather boring. can you refer me to some articles for the last statement you made?

language, man

but thanks anyway

Mike Miller

@Alizter There's no reason to believe a set for which all injections are bijections has a bijection with [n] for some n. That's silly! If you have some form of choice, it's true.

You'll survive, @Balarka. You should get very comfortable with the standard case first.

Alizter

5:14 PM

@MikeMiller I am scared to let go of C. We have choice.

Also @MikeMiller why are equivalence classes sets? or rather, why call them classes?

Or is it unfortunate naming?

Mike Miller

I dunno. I dun care.

Alizter

Ah

bi den

Mike Miller

@Alizter So we have choice. We can inductive define an injection from [n] to our set X by mapping the new element to something that hasn't been mapped to yet. If this terminates, we're your kind of finite.

Alizter

ok

Mike Miller

Suppose it doesn't. Then this defines an injection $\Bbb N \to X$. Now consider the map from $X$ to itself given by fixing everything not in the image of $\Bbb N$, and sending $f(n)\mapsto f(n+1)$. This is an injection that's not a bijection.

QED. I don't think we need the full power of choice to do this, but don't quote me on that.

Balarka Sen

5:20 PM

@MikeMiller I understand that I need to get used to the "civil" cases first before diving onto pathological spaces and their topological properties. "you'll survive" is that supposed to be some american expression that is completely out of my understanding?

Jasper Loy

@alizter I feel so sick of life. I wanna scream AAAAAAAAAHHHHHHHHHHHHH.

Alizter

@JasperLoy Go walk.

Mike Miller

@Balarka I mean my language won't kill you.

You will survive having heard it.

Balarka Sen

i am not so sure of that. i will die of chest congestion anyway so no problem.

Mike Miller

Ah, okay. That's fine.

Balarka Sen

5:24 PM

@MikeMiller you're fine with me dying of chest congestion?

Mike Miller

Sure, as long as it's not from my language.

Alizter

Ooo harsh

Balarka Sen

ah that MO post is very helpful, thanks very much @Mike

@MikeMiller another way might be to look at $X$ as a subspace of the infinite torus $\Bbb T^\infty = \bigoplus S^1$

And then taking the compliment of $X$ in $\mathbb{T}^\infty$.

Is Hilbert-Smith also true for infinite manifolds?

Vrouvrou

5:42 PM

@robjohn in the end $\delta $ depend only on $\varepsilon$, $|x_0|$ vanish right ?

Chris's sis

I was seriously thinking of creating a society of integration ...

Jasper Loy

@Chris'ssis How is the progress of your book? Will it be ready for Christmas?

Chris's sis

@JasperLoy No. Well, I'm a bit depressed these days, my outcome is poor ...

Jasper Loy

@Chris'ssis OK. I am depressed every day. I think you running every day helps.

Chris's sis

@JasperLoy Yeah, I know ...

Mike Miller

5:47 PM

@Balarka Things are bad in infinite dimensional manifolds.

Balarka Sen

ok a little googling says that people have considered infinite dimensional Hilbert-Smith

not sure if it's positive or negative though

googles further

Mike Miller

It's not even known for actual manifolds...

Balarka Sen

@MikeMiller true, but don't you think it's natural for it to fail for infinite manifolds?

Jasper Loy

@BalarkaSen Your math is too advanced for most people in this chat. You should join MO instead, lol.

Balarka Sen

that's what i am trying to find right now

@JasperLoy i am 0 compared to some guys in here. MO is too advanced for me

for example, Mike knows a lot more than I do and he has no problem understanding any of my statements, except the vague ones ;)

Mike Miller

5:50 PM

@Balarka Not entirely. I suspect you won't find an action of $\Bbb Z_p$ on an infinite dimensional manifold. If you do find something, it should be by something you might call an "infinite dimensional Lie group".

Jasper Loy

@BalarkaSen If you are 0, then I am -1, lol.

Mike Miller

So perhaps it fails in an obvious way. But you should be able to generalize the conjecture in some sense like the above to make it reasonable, and unknown.

Jasper Loy

@mike Have you decided what to write for your thesis?

Balarka Sen

@Mike dis

John Jack

Hi all, does anyone know how to produce the symbol # in latex?

Balarka Sen

5:52 PM

@JohnJack \#

Jasper Loy

@JohnJack Use \#

John Jack

Kewl thanks

Jasper Loy

That would be 100 dollars.

Mike Miller

@Balarka Ok, here's an obvious faithful action on an infinite dimensional manifold by something that's not a Lie group: $\Bbb R^\infty$ is a topological group in the obvious way, so let it act on itself by the group action.

evinda

@anon @MikeMiller @MartinSleziak @MaryStar @TedShifrin @DanielFischer @BalarkaSen
Hey!!!
If $(I_a)_{a \in A}$ a family of ideal of $K[x_1,x_2, \dots, x_n]$,
I have the following definition in my notes:

$$\sum_{a \in A} I_a=\{ a_{i1}+a_{i2}+ \dots+ a_{ij} | a_{ij} \in I_{a_j} \}$$

Is it right, or is there a typo here: $ a_{ij} \in I_{a_j}$ ?

Jasper Loy

5:55 PM

@evinda You should not ping people randomly to answer your question in chat.

evinda

@JasperLoy Sorry...

Jasper Loy

@evinda No problem, lol. I am glad you did not ask me, because I know nothing, lol.

Balarka Sen

but you know that you know nothing so that's a contradiction @JasperLoy

evinda

@BalarkaSen :D

Jasper Loy

I am gonna try to answer some lhf.

Balarka Sen

5:59 PM

@MikeMiller why should that not generalize to arbitrary lie groups?

Mike Miller

Hm? I don't understand.

Balarka Sen

G be a lie group. it is a manifold. let it act on itself.

Mike Miller

Oh, that claimed counter example doesn't work. It's not locally compact. There are a lot of problematic things here.

Yeah, what's the problem with that, @Balarka?

Chris's sis

Give me ideas for a name to an integration society that attends limits, series and integrals ... Just Integration Society? I think I wanna have something more complex.

Balarka Sen

oh right local compactness is the condition.

Mike Miller

6:02 PM

I don't want to think about the generalization so much. But I still don't get your problem with Lie groups.

Balarka Sen

forget about it

Studentmath

Complex Integration Society is more complex

I guess he doesn't trust them, @Mike @Balarka

Chris's sis

@Studentmath Yeah, it sounds nicer ... (or? Let me think some more):-)

Balarka Sen

integrated society of integrators @Chris'ssis

Studentmath

I am drowning in Topology definitions :/ It's just definitions upon definitions, no actual proofs..

And then I get a question and I have no idea what to do with it

Balarka Sen

6:04 PM

@Studentmath draw the pictures

Chris's sis

@BalarkaSen That sounds even nicer ... :D

anon

@evinda they shouldn't overuse the letter $j$ as they have, you're right

Jasper Loy

@anon What interview was that?

Chris's sis

brb

Jasper Loy

@Studentmath What book are you using?

evinda

6:06 PM

@anon So, should it be like that?

$$\sum_{a \in A} I_a=\{ a_{1}+a_{2}+ \dots+ a_{i} | a_{i} \in I_{a} \}$$

Or should it be something else?

Studentmath

@Balarka nah, for example I get the idea of 'meager' subset - I already knew about dense from Set Theory.. but I don't get the way they define it

@Jasper It's the uni's book

They wrote it

Balarka Sen

@Studentmath use Simmons

Studentmath

I need to learn by their defintions if I want to pass any test..

Balarka Sen

then digest them, don't complain.

Studentmath

I like whining

I feel good at it

Jasper Loy

6:09 PM

Yay! I like whining too.

anon

@JasperLoy read the message right above the one where I said "interview," yeash

@evinda heh, now you're overusing $i$

evinda

@anon How should it be? Should I use also $j$ ?

anon

figure it out

do you understand why $i$ is overused in the thing you gave?

Jasper Loy

J is the most beautiful letter of the alphabet.

Balarka Sen

it's because J for Jasper

evinda

6:13 PM

@anon Why is $i$ overused at $$\sum_{a \in A} I_a=\{ a_{1}+a_{2}+ \dots+ a_{i} | a_{i} \in I_{a} \}$$ ?

anon

well, the first instance of $i$ is to say how many terms are in that specific sum, and the second instance of $i$ is supposed to range over the values $1,2,\cdots,i$. also your set doesn't say which $a$ is being used, or if it's supposed to be varying as well. are all the $a_i$ values from one $I_a$, or do the $a$s vary? these things are not specified in your notation.

Mike Miller

@Studentmath That's what I found frustrating about graph theory :) Every week a new set of definitions I never understood why I cared about.

Studentmath

@Mike Amusing :P I always found connections with chemistry, and other ideas, games, probability, proving things in linear algebra and etc. via it - so I always found it interesting

But it was a lot of definitions, too, yeah

Mike Miller

Anyway, there shouldn't be too many definitions in introductory topology, because there's not really that much worth saying about arbitrary topological spaces, despite what some people say.

If you have questions, feel free to ask. I like thinking about topplogy sometimes.

Balarka Sen

me too

Studentmath

6:16 PM

I don't think there really are that many definitions - it's just I've went through about 20~25 definitions, and they were used to prove about two-three nontirivial general things

evinda

@anon Should it be like that: $$\sum_{a \in A} I_a=\{ a_{1}+a_{2}+ \dots+ a_{i} | a_{j} \in I_{a} \}$$ ?

Studentmath

all the other exercises where just asking whether some definition holds, or doesn't hold for a specific space

Mike Miller

Jeez, what defijitions are those? I can't even think of 20 definitions I care about.

Where're you learning?

Studentmath

It still feels shaky, that's all, but thanks, I will probably bother this chat with some questions as soon as I hit real exercises

@Mike Still in Metric spaces.

Balarka Sen

@Studentmath metric spaces are good

Mike Miller

6:17 PM

I mean what source

Studentmath

My own Uni's book, they wrote it

It's from 2007, usually their books are much older (199ish) and unreadable

Mike Miller

I'd like to hear the long list of definitions... I can tell you which ones don't mstter.

Balarka Sen

@MikeMiller I am interpreting theorems on covering spaces from munkres in terms of galois theory, which makes them as obvious as cake :P

it's the most fun part of alg topo

Mike Miller

And does it make the proofs obvious, too? :)

Balarka Sen

nope

proofs are different beasts

but yeah sometimes if you think about galois groups and think about the proof in that context, you can use some of the ideas to use in the context of covering spaces

Studentmath

6:26 PM

@Mike well, it's long, but:
Metric, Pseudometrics, discrete spaces, $l_1^n, l_2^n, l_\infty^n$ and $l_1, l_2, l_\infty$, $C[a,b]$, isometry, embedding, bounded metric space, bounded functions, bounds, diameter, distance, open and closed balls, interior, exterior, closue, boundary, open and closed subspaces, neighbourhood, intervals, Cantor set, $F_\sigma$ and $G_\lambda$, dense, nowhere-dense (meager), separeable - and that's where I stopped

There were more minor definitions I skipped as they were mostly used for a single example

Hippalectryon

@Chris'ssis Hello :-)

Studentmath

out of these they proved really nothing besides other ways to define these definitions

Balarka Sen

@Studentmath those definitions are pretty intuitive

Studentmath

@Balarka they all are

Balarka Sen

so why should you have any problem remembering those?

Studentmath

6:28 PM

It's not a problem of remembering them

I feel like they are just thrown at me with no purpose

Balarka Sen

@Studentmath if you want to do exercise, pick up Simmons and do the exercises from head to foot.

Studentmath

I'm really complaining for no reason :P I will go find some interesting exercises instead

I think I will try to do that, yeah

Mike Miller

Most of those don't matter, @Studentmath

Well, most do. But a chunk don't.

It sucks that they're all thrown at you at once. I bet most books Probablt don't do that.

bobby

I have a problem I'd like to put in a problem solving competition motivated by one of Balarka Sen's comments on this chat, but I'm afraid it's too hard or ill-formed, would you guys judge the question for me?

"Construct an explicit one-parameter family of homotopic curves between the top half and bottom half of the unit circle, and a continuous one-parameter family of Riemannian metrics such that at any fixed value of the parameter the homotopic curve is a geodesic for that metric."

Balarka Sen

oh you're still thinking about that stuff

bobby

6:39 PM

I said it to my friend that day and he liked it so much he wanted to put it in his problem solving competition two weeks later, he asked me earlier what it was hehe so he liked it too :)

Balarka Sen

cool, glad to know that

bobby

Do u think the question makes sense?

Chris's sis

Back.

@Hippalectryon Hi. How are you doing? :-)

Hippalectryon

@Chris'ssis Ok I guess :-) what about you ?

bobby

My intuition is that we simply want to deform the unit circle into a family of ellipses using x = arcos(0) & y = brsin(0) with a = 1 & b as the parameter, then the metric and the family are functions of b, solved hehe

Balarka Sen

6:41 PM

@bobby looks alright, but i don't happen to be an expert in riemannian manifolds.

Chris's sis

@Hippalectryon Just back from jogging.

Balarka Sen

it looks good when interpreted in terms of geodesic metric spaces though

so yeah, it's good.

Mike Miller

@bobby Looks nice. I suspect if someone can write down the metric for the top curve they can probably do it for all of them.

bobby

Well I mean a geodesic in the plane is a straight line, okay - but what is a geodesic on the circle? It's a straight line along the arc of the circcle right? So if u have a polar coordinate metric a geodesic is the arc of the circle right?

Balarka Sen

right

bobby

6:43 PM

So if u give the plane polar coordinates, a geodesic is not a straight line it's an arc of a circle

wow what a cool question

anon

wat

bobby

hehe nice

Mike Miller

@anon I'm staying out of this one

Balarka Sen

@anon the original question (of me) was if a notion of "geodesic homotopy" can be defined, in analogy with straightline homotopy.

i.e., homotoping two maps through the geodesics on your space.

Hippalectryon

@Chris'ssis Got any new awesome result ?

anon

6:46 PM

@BalarkaSen ah, sure

Chris's sis

@Hippalectryon Not really. I mainly focus on some points that arose in the answers to my bounty question here math.stackexchange.com/questions/1021647/…

Hippalectryon

Ok. I see.

anon

just assume $I\times\{x\}\to Y$ is a geodesic for each $x\in X$ I guess

Mike Miller

doubt it's possible in general

even if you assume really nice things about $Y$

anon

@MikeMiller what if we assume it's path-connected?

that should mean there is a geodesic between any two points

(I think)

Balarka Sen

6:53 PM

it does

anon

presumably if we fix $y$ and let $y'$ range freely, "the" geodesic between $y$ and $y'$ should "vary smoothly" as $y'$ does, outside a surmountable set of exceptions. so for each $x\in X$ and $f,g:X\to Y$ we should just create "the" geodesic from $f(x)$ to $g(x)$, and then put these geodesics together, and hopefully we get a geodesic homotopy

bobby

Well thanks for looking into it guys, I feel nobody in the session will solve it hehe

The big words will confuse

I've never made a problem up before (and admittedly only half-made this one up) but it gets you thinking about how problems are made and how their solutions are found

This is the problem with problem-solving though, I've always thought humans have a simple answer then they cloud it in mystery so you must wipe off the human smog to find the solution, but real problems in nature do not have human smog covering them up hehe

Balarka Sen

And I've never solved a self-made problem before haha

Mike Miller

7:11 PM

@anon If the two paths are close enough fhis should be possible but in general there's no canonical way to choose a geodesic between two points

actually it's not even true that every Riemannian manifold has a geodesic connecting two points

Jasper Loy

I hope one day I can make a big discovery in math...

Balarka Sen

just use geodesic metric spaces guys

Mike Miller

$SL_2(\Bbb R)$ with a bi-invariant metric is an example

if you demand your manifold be complete you do always get a geodesic between two points but I don't see a reason to believe you can pick this canonically or always vary it smoothly.

Vincenzo Oliva

@Mike An answer to a question of mine uses $$\underset{p\leq x}{\prod}\frac{p}{p-1}=\frac{1}{2}e^{\gamma}\log\left(x\right)+e^{-c\sqrt{\log\left(x\right)}}$$, is it true?
I think not since dividing by $e^\gamma \log \log x $ both sides one gets $$\frac{1}{e^{\gamma}\log\left(x\right)}\underset{p\leq x}{\prod}\frac{p}{p-1}=\frac{1}{e^{\gamma+c\sqrt{\log\left(x\right)}}\log\left(x\right)}+\frac{1}{2}$$, whose LHS goes to 1, whereas the RHS goes to $\frac{1}{2}$.
Am I mistaken?

(I hope I didn't miswrote anything, I'm using my mobile)

*miswrite

Hippalectryon

@VincenzoOliva 'goes to 1' when x->oo ?

Mike Miller

7:21 PM

hmm, what I just said was incorrect about SL. it doesn't carry a bi-invariant metric.

but the thing I said about non-complete things is true.

Vincenzo Oliva

@Hyppalectryon Yep

@Hippalectryon

Mike Miller

I'm not sure anybody cares about anything I'm saying, so I'll stop.

Hippalectryon

@VincenzoOliva 'I think not since dividing by eγloglogx both sides one gets'

log log ? did you mean log ?

@MikeMiller I don't know who you're talking to :/

Vincenzo Oliva

@Hippalectryon Urgh, sure. Typo

Mike Miller

anon, mostly

Hippalectryon

7:24 PM

@VincenzoOliva Well if the LHS does go to $1$ (that's the only thing I can't check) then Indeed there is an error

Vincenzo Oliva

@Hippalectryon To see it does you just need to know Mertens' third theorem

Hippalectryon

@VincenzoOliva Indeed then

Vincenzo Oliva

@Hippalectryon Thanks

Hippalectryon

8:01 PM

@Chris'ssis Do you think one day you'll look at other fields of mathematics ?

Chris's sis

@Hippalectryon Sure, I might also look at other fields of mathematics ... :-)

Hippalectryon

That would be great :D You could find other kinds of awesome results

Chris's sis

Probabilities & applied mathematics?

Number theory is also very important too.

@Hippalectryon Before going there I wanna learn complex analysis very well and bring my contribution in this area (even if a bit).

Hippalectryon

I don't know much about probabilities, I have only done a little

@Chris'ssis Maybe some more broad calculus concepts ?

@Chris'ssis Vector spaces, Sheaves, ...

Chris's sis

@Hippalectryon Sure, yeah! I only need good books, I can learn alone.

Hippalectryon

8:06 PM

:D

robjohn

@Chris'ssis: I amended my answer too late to catch Venus. I don't know if it is any clearer.

Vincenzo Oliva

Does PNT yield $$\underset{p\leq x}{\sum}\log\left(p^{n+1}\right)=\left(n+1\right)\left(c_{1}x+c_{2}\frac{x}{\log‌\left(x\right)}+o\left(\frac{x}{\log\left(x\right)}\right)\right)$$ ?

Chris's sis

@robjohn How do you pass from the first blue expression to the second blue expression? This is what Venus wanted to know.

robjohn

@Chris'ssis Standard orthogonality of trig functions. The only ones that don't cancel are when $j=k$

Chris's sis

@robjohn Yeah. This is the point Venus wanted to know I think.

@robjohn Maybe you should mention it somewhere in your answer.

Vincenzo Oliva

8:14 PM

@Hippalectryon That's another statement of the answerer. Do you happen to know the implications of PNT well enough? (I don't yet)

Hippalectryon

@VincenzoOliva I haven't done much NT at all

Vincenzo Oliva

Uhm... @robjohn May I ask your help?

robjohn

@Chris'ssis I was doing just that.

@VincenzoOliva with?

Chris's sis

@robjohn Great.

Vincenzo Oliva

@robjohn Is it true thay PNT yields $$\underset{p\leq x}{\sum}\log\left(p^{n+1}\right)=\left(n+1\right)\left(c_{1}x+c_{2}\frac{x}{\log‌\left(x\right)}+o\left(\frac{x}{\log\left(x\right)}\right)\right)$$ ?
Apologies for any typo, I'm writing on my mobile.

Chris's sis

8:21 PM

@robjohn It looks like your way is the simplest way.

robjohn

@Chris'ssis I've had a lot of experience with those kinds of sums recently. :-)

Chris's sis

@robjohn Yeah, I know. :-)

@robjohn How does it seem to you this integral as difficulty? It produced some difficulties on main in some answers.

robjohn

@VincenzoOliva I believe $c_1=0$ and $c_2=1$, but that looks right.

@Chris'ssis I guess that without the right ideas, one can employ approaches that make it pretty difficult. The best thing I think to do is to get rid of the $x$ in the original integral and deal only with the trig functions.

Chris's sis

@robjohn Definitely.

robjohn

If you get that taken care of, it is not too difficult, but I don't know how intuitive that is.

@Chris'ssis But judging from some of the answers, It would give it a high difficulty.

Chris's sis

8:35 PM

@robjohn It looked like until Felix Marin posted the closed form no one believed it had a nice closed form since known people good at integration didn't work on it (or they worked some, but failed)

This is an integral from which one can learn a lot of lessons.

robjohn

@Chris'ssis Once I had some time to work on it, it didn't take too long, but as I said, I have had just the right experience for that question.

Vincenzo Oliva

@robjohn I see, thanks.

Chris's sis

@robjohn Yeah, without experience it is pretty hard to attend.

user2179021

would anyone be so kind as to translate some math for me please? I am reading the final comment of fedja at mathoverflow.net/questions/168474/… which starts

"Our matrix is degenerate iff the corresponding polynomial P(z) has a root that is also the n'th root of unity. Let q|n be the order of that root of unitOur matrix is degenerate iff the corresponding polynomial P(z) has a root that is also the n'th root of unity. Let q|n be the order of that root of unit"

what does degenerate mean?

does it just mean singular?

evinda

@anon Is it $\sum_{a \in A} I_a=\{ a_{1}+a_{2}+ \dots+ a_{i} | a_{j} \in I_{a} \}$
or have I understood it wrong? :/

evinda

9:18 PM

Does it stand that $I\cup J \subset I \Rightarrow V(I \cup J) \supseteq V(I)$, where $V$ is the set of roots and $I,J$ ideals?

Vrouvrou

9:58 PM

@robjohn please why $|x+x_0|\le \frac32|x_0|$ is $|x|\leq |x_0|$ ? please

robjohn

@Vrouvrou I don't understand. That was never part of the derivation

Chris's sis

@Hippalectryon ...

Hippalectryon

@Chris'ssis ?

Chris's sis

@Hippalectryon use for that oen of series you gave me but using $2n+1$ instead od $2n$ ...

$$\sum_{n=1}^{\infty} (-1)^n \frac{(z)_{2n+1}}{(2n+1)!} t^{2n+1}=\frac{\sin(z\arctan(t))}{(1+t^2)^{z/2}}$$

@Hippalectryon ^^^

Hippalectryon

:-)

Chris's sis

10:03 PM

:-)

Hippalectryon

That would look better with some $\Gamma(n/2+1)$ :P

Chris's sis

Today I go to sleep earlier ... much work to do tomorrow morning ...

Hippalectryon

How did the interviews go ?

Chris's sis

@Hippalectryon Everything is perfect until we talk about salary (with all my interviews)

Hippalectryon

Ugh :/ Salary

Let me Guess

Interviewer -- What salary are you looking for ?
@Chris'ssis -- Uhm... Shall I write it as polygamma series ?

:DD

Chris's sis

10:07 PM

@Hippalectryon lol, no :-)

Hippalectryon

/month ?

Chris's sis

@Hippalectryon Yeah.

Hippalectryon

Wow that's pretty low

Well, I'm living in France after all

Chris's sis

@Hippalectryon It's hard to ask more at the beginning, since you need to prove in the company that you're really valuable. That's fair after all.

Hippalectryon

@Chris'ssis About how many hours/ months ?

Chris's sis

10:10 PM

@Hippalectryon 8-10 hours every day, it depends. Besides that you also need to work at home often.

Hippalectryon

In France, for 150 hours/month, you get minimum 1445€ (raw, you need to deduce the taxes)

Chris's sis

@Hippalectryon I'm talking about hardcore jobs.

Hippalectryon

What do you mean 'hardcore' ?

Like, mining ? :c

Chris's sis

@Hippalectryon You attend difficult stuff, like the integrals, series and limits I like. :-)

Hippalectryon

:-)

I don't work though

Chris's sis

10:12 PM

@Hippalectryon I'm a very hard worker (no matter the situation).

Hippalectryon

That's a good thing

Chris's sis

Anyway.

I leave.

Hippalectryon

:-) sleep well

Chris's sis

@Hippalectryon Thanks. ;)

evinda

Could you tell me when the equality $V=V(I(V))$ stand?

$V$ is the set of roots, $I$ is an ideal.

evinda

10:33 PM

I have that $x \in V \Rightarrow x \in V_1 \cup V_2 \Rightarrow x \in V_1 \text{ or } x \in V_2$

Does this imply that $\forall f \in I(V_1):f(x)=0 \text{ or } \forall g \in I(V_2): g(x)=0$, because $V_1=V(I(V_1))$ and $V_2=V(I(V_2))$ or is there an other reason?

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$Mathematics$ Mathematics