Mathematics - 2014-12-20 (page 1 of 8)

N3buchadnezzar

12:03 AM

Anyone here know python?

Mike Miller

hi @DanielRust

Mathy Person

Math.SE should consider making two separate rooms--one for general off topic discussion and one for math questions

Behaviour

@MichaelAlbanese When you come across an old unclear unanswered question that nobody ever cared about, closing as unclear makes sense -- but downvoting from 0 to -1, when applicable, is a lot easier and leads to auto-deletion.

Alexander Gruber

We hear you, @MathyPerson. But nobody wanted to answer it.

robjohn

Darn... I was trying to cap before the day change, so that I could erase the -2 from a random downvote. I didn't make it, so now I am back at 3 mod 5.

Mathy Person

12:08 AM

@AlexanderGruber I know, I figured :p

Daniel Rust

@robjohn just downvote 3 bad answers :)

robjohn

@DanielRust ack... and ruin my clean downvote record ;-)

Daniel Rust

Ah, you're one of those voters :P

N3buchadnezzar

@robjohn Downvote record?

robjohn

@N3buchadnezzar look at the bottom of my profile

N3buchadnezzar

12:10 AM

@robjohn You are such a downer.

Or not.. ;)

Daniel Rust

I think most of my downvotes are for questions tbh, or CW answers

Michael Albanese

@Behaviour When is it applicable?

Behaviour

@MichaelAlbanese When the question has score of 0, no answers, and you are sure the site is better off without it.

robjohn

@DanielRust I'd rather comment and try to get the answer/question improved. I know that downvotes motivate people, but there are enough downvotes out there.

Michael Albanese

@Behaviour Is number of answers relevant? I thought a similar thing happens if there are no answers with positive score.

N3buchadnezzar

12:12 AM

@robjohn It is always interesting seeing questions with 6 or 7 downvotes.

Daniel Rust

@robjohn True. In theory I'm happy to convert a downvote when an edit is made, and it improves the post... But I rarely check back on posts I've previously downvoted.

Alizter

I am getting a headache from this stuff

Behaviour

@MichaelAlbanese That's on closed questions only. My point was, some poor abandoned questions don't even have to be closed; if they have no answer, a downvote is enough.

Michael Albanese

@Behaviour Oakie dokie, thanks for the tip.

robjohn

@DanielRust to me, not checking on the posts one has downvoted removes some of the meaning of the downvotes. I know when I get downvotes, they are rarely commented on and rarely reversed. The ones that are commented on, do tend to get reversed when I handle their concern.

Daniel Rust

12:22 AM

@robjohn you've given me the motivation to check through my downvote history for any that may have been improved/ or for which I can comment

Behaviour

@DanielRust My opinion is that if the first version of the post was bad, the author deserves the downvotes they got for it. If they improve later, fine -- that will prevent more downvotes and may bring more upvotes. But they should still pay the price for putting others through reading their first version. I rarely reverse downvotes.

Posting a bad first version followed by an okay 2nd version is still worse than getting it right from the beginning. There should be an incentive to get it right at the beginning, not after others spend time asking for clarification or editing the post.

Daniel Rust

@Behaviour I'm more of the opinion that a softer approach encourages people to stay on the site and hopefully improve their contributions, but I can respect your approach too

robjohn

@DanielRust then I consider it a good day :-)

Daniel Rust

@robjohn one vote already reversed

this answer was fixed

robjohn

@DanielRust Great. However, one downvote still remains.

Alexander Gruber

12:35 AM

Okay so

Daniel Rust

@robjohn Yeah, that's a shame.

Alexander Gruber

I've got $n$ points on a unit circle

and I'm taking $n\to\infty$

They aren't random but I guess lets say they are cause I can't figure out how they work.

i believe what happens is that the maximum distance between any two points goes to $0$ almost surely, as $n\to\infty$

Daniel Rust

@AlexanderGruber How are the sequence of points defined?

Alexander Gruber

@DanielRust it's this

there may be some way to analyze what's happening to the roots algebraically but i've given up on that avenue for now

I just thoroughly digested Will Sawin's answer for the first time since he posted it, and he's shown that the set of roots converges to the unit circle

Daniel Rust

@AlexanderGruber Ah I remember this question. The setup was a bit too complicated for me.

Alexander Gruber

12:41 AM

(Well, what I mean is, there is an $N$ for which all the roots of $p_n$ for $n\geq N$ have magnitude $1\pm \epsilon$)

@DanielRust Right. the polynomial itself is a big mess

so i'm trying to think about how to formalize what I want to know

We know everything happens on the unit circle in the limit. What I wonder is whether there are accumulation points for roots on the circle

or just in general how to describe "which points are roots in the limit"

epimorphic

@robjohn Do you happen to have a gallery of your astronomy photos online? I'd love to see them.

Alexander Gruber

but this needs to be formalized

Daniel Rust

@AlexanderGruber Sounds like the kind of thing an ergodic theorist might be able to help with.

Alexander Gruber

@DanielRust yeah that is probably true. i don't really know anything about ergodic theory.

This is all pretty much magic to me as a combinatorialist.

robjohn

@epimorphic I have a few on Flickr...

Daniel Rust

12:49 AM

@AlexanderGruber I wonder if there's a nice way of posing the question in terms of algebraic geometry.

What with the main object of study being roots of polynomials (rational functions)

robjohn

@epimorphic did you see the comment?

Alexander Gruber

@DanielRust Yeah that's what I was wondering to begin with

Let me think so

How would we do that

epimorphic

Yep.

robjohn

@epimorphic enjoy... I have to go out for a bit BBL

epimorphic

later

Kaj Hansen

1:16 AM

Hey guys

Mathy Person

1:30 AM

@Studentmath Original statement: "Five vertices are labeled 1,2,3,4,5. In how many ways can edges be drawn between some pairs of these vertices so that the result is a connected graph?" . I got 193. Is that correct?

Can anyone else here see if 193 is the right answer?

Alexander Gruber

@MathyPerson how bout up to isomorphism

Mathy Person

@AlexanderGruber isomorphism?

what is that?

Alexander Gruber

Well, if you have a function $f$ between two graphs $G$ and $H$

Mathy Person

@AlexanderGruber i'm not familiar with that term

Alexander Gruber

what term

(oh, isomorphism. Sorry, I was continuing)

if you have a function $f$ between two graphs $G$ and $H$, and $f$ has the property that $\{f(g),f(h)\}$ is an edge of $H$ if and only if $\{g,h\$ is an edge of $G$

then that function is an isomorphism

Mathy Person

1:40 AM

hmm

how would i apply it then?

Alexander Gruber

@MathyPerson if an isomorphism exists between two graphs, they are "the same."

in other words, the only difference is what the labels of the vertices are

Mathy Person

how would using isomorphism help, though?

is there another method?

Alexander Gruber

@MathyPerson I was asking you kind of an extended question. (There is actually an extended method you could use but it is probably a little too complicated to be worth the effort on a graph on 5 vertices.)

Mathy Person

@AlexanderGruber I'm not really that familiar with graph theory (only the basics), so the simpler the better!

I found the cases for edges from 4 to 10 and got 193

Alexander Gruber

@MathyPerson have you had group theory yet by any chance?

Mathy Person

1:44 AM

@AlexanderGruber nope

Alexander Gruber

ok cool

So just one more question, let me clarify, by "ways," you do mean without isomorphism? For example, is the graph with edges {1,2}, {2,3}, {3,4}, {4,5}, {5,1} the same or different from the graph with edges {1,3}, {3,5}, {5,2}, {2,4}, {4,1}?

We would count those as two graphs or one graph towards your total?

Mathy Person

@AlexanderGruber hold on, i was given a picture as an accompaniment

those two are considered distinct

@AlexanderGruber

Alexander Gruber

Ah I see, okay. So, then that means that you are considering graphs as "not being unique up to isomorphism."

Kaj Hansen

@AlexanderGruber, if I understand the question correctly, is the answer essentially "Number of spanning trees of $K_5$ + .... of $K_4$ + ..... of $K_3$ ...." ?

Mathy Person

@KajHansen Are you referring to the question @AlexanderGruber and I are discussing?

Kaj Hansen

1:51 AM

Yes @MathyPerson

Mathy Person

@KajHansen This was the question: "Five vertices are labeled 1,2,3,4,5. In how many ways can edges be drawn between some pairs of these vertices so that the result is a connected graph?"

Alexander Gruber

@KajHansen Well let's see you can definitely start with the spanning trees of $K_5$.

Kaj Hansen

Oh wait, the stipulation is just "connected". For some reason I read cycle-less somewhere up above.

Alexander Gruber

Let me work on this on paper for just one moment

Mathy Person

@AlexanderGruber Ok, sure!

I was given a hint: Try complementary counting and recursion. How can a non-connected graph be built out of smaller connected graphs? @KajHansen @AlexanderGruber

Kaj Hansen

1:58 AM

A non-connected graph can be built from connected graphs either by deleting edges from the original or adding new vertices but no new edges.

Alexander Gruber

@MathyPerson actually i think this is much more difficult if the graphs are not considered up to isomorphism. I'm having a pretty hard time not using algebra.

Kaj Hansen

@AlexanderGruber, so given this no-isomorphism condition, does this mean there are $4!$ distinct $C_4$ graphs?

Alexander Gruber

@KajHansen Yeah.

Kaj Hansen

Oh man...

Mathy Person

Do you both know of any other members on here who have experience with graph theory that I can ask?

Alexander Gruber

2:01 AM

@MathyPerson i am experienced with graph theory, this is just tedious.

Mathy Person

I can show what I did to get 193 , if that is helpful

Alexander Gruber

You should show me

However, I can already tell you that that is wrong

Kaj Hansen

I don't think the isomorphism stuff is well-defined just yet. That picture may imply that graphs are the same up to rotations, but not reflections. Anyways...

Mathy Person

@AlexanderGruber I used C(n-1,2) + 1 to get a minimum of 7 edges up to a max of 10 edges, so i ended up getting $C(10,7) + C(10,8) + C(10,9) + C(10,10)$, but since every connected graph has at least n-1 edges, I needed to have at least 4, 5, 6 edges as well.

if i made a mistake it was most likely in the 4, 5, 6 edges part

Alexander Gruber

Well, the thing is, take a step back for a moment and look at this wholistically

Mathy Person

2:05 AM

alright

Alexander Gruber

there are $2^{{5 \choose 2}}=1024$ graphs on $5$ vertices

It couldn't be true that only $193$ are connected

Mathy Person

true...that does seem like a stretch..

Alexander Gruber

OK I am doing alright

Here's my approach so far

We fix 1 and make it a special vertex. It can be connected to either one, two, three, or four of the other vertices. Call that number $k$.

Mathy Person

alright

Alexander Gruber

There are two kinds of connected graphs we can make this way: graphs where $2,3,4,5$ would be connected if $1$ weren't there, and graphs where $2,3,4,5$ would not be connected if $1$ weren't there

We'll start with the connected ones.

If $K$ is a connected graph on four vertices, we can add $1$ to it in any of the above ways, and it will be a connected graph. (By the above ways, I mean that it can be connected to any number of them.)

Mathy Person

2:21 AM

any number of vertices, you mean?

Alexander Gruber

right.

Mathy Person

hmm. ok

Alexander Gruber

There are exactly four ways we could attach $1$ to a connected graph on $4$ vertices using one edge.

(that sentence should read less obviously than it is. if we only have one edge, which do we want it to go to? there are four options.)

There are 12 ways we could attach $1$ to a connected graph on $4$ vertices using two edges.

First we select one vertex, then we select a different vertex. Thus 4*3=12 ways

Similarly, there are $4*3*2$ ways to connect $1$ to the connected graph on $4$ vertices using $3$ edges, and there are also $4*3*2$ ways to do it using four edges.

(In general, the formula would be ${4\choose k}k!=\frac{4!}{(4-k)!}$, because it's the number of ways to pick which vertices you connect $1$ to, multiplied by the number of ways they can be permuted.)

Wait, crap I'm double counting. We don't need to permute them. There's just ${4 \choose k}$ ways for each connection, so, $4, 6, 4, 1$ ways.

The reason that's important is that, because the connection to $1$ has no bearing on whether or not the graph is connected, we can just let $C(4)$ be the number of connected graphs on four vertices, and we know there are at least $(4+6+4+1)C(4)$ connected graphs on five vertices.

So, that takes care of the connected case.

For the disconnected case, let's start with $k=4$.

Any disconnected graph will become connected if it is connected up to $1$ by $k=4$ edges.

Now, for $k=3$, the three points that $1$ connects to must be disconnected without $1$. There are only $4$ disconnected graphs on three vertices. There is one vertex $v$ which is not connected to $1$, so it must connect to the three vertices adjacent to $1$ also.

There are six ways for it to connect to the empty graph on three vertices, then for each of the three disconnected graphs with one edge there is one option, so that is nine ways.

So for $k=3$ we'll have 9*4=36 disconnected graphs.

For $k=1$ we're just counting connected graphs on three vertices (there are four) because we need to connect an isolated point to a connected set of three vertices.

Mike Miller

2:45 AM

holy moley

Alexander Gruber

IM COUNTING MIKE

Mathy Person

@AlexanderGruber Hi, sorry I haven't been responding. Someone called and I had to talk to them. I'm reading your counting now

Why do we have to count both connected and disconnected graphs, again?

Nameless

Hi

Mathy Person

Ok im done reading

Alexander Gruber

@MathyPerson Because we want the whole thing to be connected, so it always works if the 2,3,4,5 graph is already connected, but it could also work if the 2,3,4,5 graph is not connected, but the edges added going to 1 makes it connected

Mathy Person

2:50 AM

hmm ok

What were you referring to via $4, 6, 4, 1$ ways?

@AlexanderGruber

Alexander Gruber

Finally, for $k=2$ there should be $12$. So there should be $4\times 6+4\times 6 + 4\times 9 + 26=158$ ways from the disconnected case. For the connected case, first we have to count the connected graphs on four vertices... using the same drawings i'm getting $C(4)=38$ there. So that would be $(4+6+4+1)\times 38=570$ ways. Adding that up I get $728$ connected graphs on $5$ vertices. Whew.

@MathyPerson Those are ${4 \choose k}$ for $k=1$ to $4$. I'm selecting which of the other four vertices I want to connect $1$ to.

Mathy Person

I see

Wow, 728 is a lot

Thanks for all your help...I know that it must have taken a long time..!!

@AlexanderGruber

Alexander Gruber

@MathyPerson No problem. Try to draw some pictures when you go through it

Mathy Person

Ok, will do. I have a similar problem, but this time it is with 4 vertices, so I'll try doing it using the similar method that you used

Mike Miller

3:35 AM

@Ted Where can I see a proof that (M orientable supports a spin structure) $\iff$ ($w_2(M)=0$)?

Ted Shifrin

3:48 AM

Hmm, Milnor? Steenrod? @Mike

It might be in Lawson/Michelson.

Mike Miller

It's not in Milnor/Stasheff, if that's what you mean. I'm taking a look at steenrod soon for other reasons so I'll check there.

Thanks for the references!

Ted Shifrin

Surely must be in Lawson/Michelson.

Mike Miller

That's the book titled "spin geometry", yes?

Looks like yet another thing I should read...

I don't really understand yet why spin manifolds are a natural object of study, but I suppose I will.

N3buchadnezzar

4:14 AM

Hey captain jack

The Artist

4:56 AM

@robjohn Yep lol

@robjohn and received 22 downvotes....never seen a question before get so much, maximum I have seen before is -12

N3buchadnezzar

5:12 AM

Anybody here?

How close is this math.stackexchange.com/questions/701557/… to a duplicate of this math.stackexchange.com/questions/100495/…?

It really follows directly since

$$ \int_0^1 \frac{x^a - b^b}{\log x}\,\mathrm{d}x = \int_0^1 \frac{x^a-1}{\log x} - \int_0^1 \frac{x^b-1}{\log x}\,\mathrm{d}x-$$

HAH

I FOUND THE DUPLICATE

4

Q: Integrate $\int_0^1 {\frac {x^a-x^b} {\ln x} dx}$

We are given parameters $a > 0, b > 0$. Task is to integrate that: $\displaystyle \int_0^1 {\frac {x^a-x^b} {\ln x} dx}$. I have tried approaching problem from different angles with no luck. I tried integration by parts(tried all combinations of possible $v$ and $u$), u-substitution with no luck...

Kaj Hansen

6:37 AM

@N3buchadnezzar, so vigilant!

N3buchadnezzar

@KajHansen There are like 4 similar questions too

skullpatrol

6:54 AM

He's our chatroom resident vigilante ;-)

en.m.wikipedia.org/wiki/Vigilante

Kaj Hansen

"Vigilantes may assault targets verbally, physically, vandalize property, or even $\textbf{kill}$ individuals."
We need to keep an eye on him.

N3buchadnezzar

Psycological terror is not a punishable crime

skullpatrol

Good point^

All's fair in love and psychological war fare.

en.m.wikipedia.org/wiki/Psychological_warfare

skullpatrol

7:22 AM

WTF "In Hampshire, England, during 2006, a vigilante slashed the tires of more than twenty cars, leaving a note made from cut-out newsprint stating "Warning: you have been seen while using your mobile phone".[21] Driving whilst using a mobile is a criminal offense in the UK, but critics feel the law is little observed or enforced."

Kaj Hansen

You're still reading those wiki articles @skullpatrol? :P

skullpatrol

:P

I have always been a slow reader.

robjohn

8:02 AM

@TheArtist I have added links to the comments to which you are referring. It is best to use these when referring to thing said long ago or when talking in a busy chat.

Mike Miller

This question was originally pretty poorly written; I rewrote it to fit with what I think is the author's intent. I can't figure out what good tags are for this, though. Any thoughts?

(its only tag originally was constants, which really had nothing to do with the question...)

Kaj Hansen

Oh man....I had a good chucke at $\displaystyle \frac{d\text{BugsOnGround}}{dt}$.

Mike Miller

(I also hope you upvote it if you think it's a good question... I think it got a lot of downvotes just because it was poorly written, but there was a great question hidden there.)

I want to add a diffeq tag just because any answer should talk about diffeq.

skullpatrol

8:21 AM

+1

Balarka Sen

8:36 AM

Heh. I got Naruto.

Mike Miller

push it further up on your avatar, I think

it's too far down as is

Balarka Sen

looks good now?

Mike Miller

I dig it

Balarka Sen

it takes time for it to change in the chat

@MikeMiller consider this covering space of $S^1 \vee S^1$ : you have the bunch of circles at each integer on the x-axis copy of the real line and you have the bunch of circles at each integer on the y-axis copy of the real line.

this guy is not galois, am i right?

Mike Miller

It's too late for me to think anymore

sk me tomorrow if you're still not sure :)

Balarka Sen

8:42 AM

k. go to sleep.

hmm. i guess $\Sigma_4$ also forms a galois cover of $\Sigma_2$

oh sure it does.

Kevin Driscoll

8:55 AM

QUick question for you chat regarding sets with infinite elements. Suppose I generated a string of letters in the patter A, B, B, A, A, B, B, B, B etc etc

where the number of Bs is always double that of the As the precede it

am I correct in say that that the probability of drawing an A at random from this infinite sequence is the same as the probability of drawing a B at random from this sequence, since both the subset of As and the subset of Bs have the same cardinality?

math110

9:59 AM

hello,everyone ,Today just finished the Chinese mathematical olympiad first day

Samurai

@math110 In which grade do you study?

math110

10:16 AM

Hello?

I don't understand you mean

Now I 'm china teacher,and Responsible for the China's math competition and other questions

Venus

@math110 Are you a teacher or a student?

math110

teacher

Just graduated from the graduate student

Venus

@math110 Do you major in math?

math110

Hello,I'm graduate my boss is reseaher PDE

Venus

@math110 I mean do you graduate from mathematics faculty?

math110

10:21 AM

Yes

Venus

@math110 In which city do you live in China?

math110

Now I'm in shanxi

are china too?

you?

sorry,the english is not good

Venus

@math110 No, I'm from Malaysia but I'm Chinese by descent

@math110 It's okay, my English is not good either

math110

oh,Hello

do you know chinese ?

Venus

@math110 Ya, hello there

math110

10:25 AM

language

Venus

@math110 Not much, only a bit

math110

哈哈

很高兴认识你

Venus

@math110 Should I call you math110 laoshi?

@math110 Sorry, I can't read Chinese characters

math110

Haha year ,I call you venus

?

Venus

@math110 Ya, you may call me Venus

Ni hao

math110

10:28 AM

ni hao

哈哈

are you interesting I post CMO problem inequality?

Venus

@math110 What language do you speak? Is it Mandarin or Cantonese?

@math110 What is CMO? China Math Olympiad?

math110

Yes

today is first day

totall two days exam

today It is said is not hard ,and I post ineuqlaity on ME

Venus

@math110 So, do you teach your students to prepare for CMO?

math110

Yes. a few students

Venus

@math110 How many students?

math110

10:32 AM

three student

Venus

@math110 Are they junior or senior high school students?

math110

rom all over the country more than 300 students attend

senir high school

Venus

@math110 Only 300 students from all over China? Are you serious?

math110

Each province about seven to eight students

I don't know how to explain
because I can't use English say

The 300 students, selected about fifty people training team in China

this fifty student attend china(hard exam :TST)

From 50 people choose 6 people as a national team again

Venus

@math110 Don't worry about the language, I can still understand you.

math110

10:39 AM

Oh,Thank you

I know you interesting integral?

Venus

@math110 These 6 students will be trained again in Beijing?

@math110 Yes, I'm interested in integral problems just as a hobby

math110

yes,this 6 students attend IMO

Venus

@math110 Who will train them in Beijing?

math110

Hello,Bejing is one of province

Huy

Morning, @Venus.

math110

10:42 AM

china have 34 province ,each province have 7-8 students attend the CMO

Venus

@math110 Ya, I know that but where these 6 students will be trained as IMO contestants if not in capital city of your country?

math110

yes,That's my mean

Haha

each year about (100000)student attend senior school mathmatical competition around the china eaxm

Venus

@Huy Guten morgen

math110

then Result row in front of 300 people are expected to attend CMO

Venus

@math110 How many students do you teach in each class?

math110

10:46 AM

50-60

Venus

50-60 students in each class? Wow!

math110

yes,china class all it

and you ?

Venus

How old are you if I may know?

math110

How many it?

oh,27

Venus

@math110 I'm not a teacher but here we have about 25-30 students in each class

Is this your hometown?

math110

10:49 AM

oh,in our countries 50-60 is called little class, at most 100-120 a class

Venus

@math110 You're kidding right?

math110

yes,Now I job here

Huy

@Venus: Our freshmen lectures usually are attended by 350-500 students.

Venus

One teacher for 100-120 students?

math110

yes

I don't kidding right

Venus

10:50 AM

@Huy This is high school! Not college

Huy

oic

20-25 then

my class only consists of 14 ^_^

Venus

@math110 Have you ever taught more than 100 students?

@Huy You should be grateful then

Huy

I am

math110

No,my school don't have 100 students a class

at most 76 students a class in my school

Venus

@math110 Can you control all of your students in class room?

math110

10:53 AM

He nan province have many people ,it is said there are more 100 students a class

Venus

I mean to handle 60-70 high school students are not easy job

math110

Make trouble I don't know, they don't talk in class

Huy

@Venus: I always imagine Chinese kids to be much more disciplined because they know from when they're a child that their future depends heavily on their education.

Venus

@math110 So I suppose you can handle them. Salute!

@Huy I don't think so, but I don't know

@math110 How strict your government control their people?

math110

Hello,

chinese language:jihuashengyu

One-child policy

The family planning policy, known as the one-child policy in the West is a population control policy of the People's Republic of China. The term "one-child" is a misnomer, as the policy allows many exceptions and ethnic minorities are exempt. In 2007, 35.9% of China's population was subject to the one-child restriction. The policy is enforced at the provincial level through fines that are imposed based on the income of the family and other factors. "Population and Family Planning Commissions" (计划生育委员会) exist at every level of government to raise awareness and carry out registration and inspection...

Transcript for

$Mathematics$ Mathematics