Mathematics - 2014-12-23 (page 2 of 10)

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2:00 AM

@teadawg1337 A tool?

@beginner Could be better, I'm still fighting this stupid sinusitis

tool 

One who lacks the mental capacity to know he is being used. A fool. A cretin. Characterized by low intelligence and/or self-steem.

My school made me do iq tests because I am weird

I scored borderline mentally-challenged on the WAIS-3

IQ tests don't mean much to me.

@user130018 What is that supposed to test?

2:01 AM

They didnttell me what I got but icant move up since I need ti work on social skills

@JasperLoy Just general cognitive ability

if you judge a fish by its ability to climb a tree, well, you're just retarded

@user130018 OK. Does that mean you are a slow learner?

@JasperLoy Yea

stupid question: is a map to the zero power the identity map or the zero map?

2:02 AM

I scored a 127 on the WAIS-IV, but i took the Stanford-Binet 5 and got a 138

Wow, that is really high @teadawg1337

If I remember, I got in the high 70s

I never took an IQ test, lol.

@JasperLoy From personal experience, they just reaffirm what you already know :P

My parents didn't say my score to me because they said the score is inflated for my age, but I really want to know

@teadawg1337 Maybe different tests test different things.

@beginner Well, they should at least let you see the report. They are really weird.

2:06 AM

They said it will worsen my social abilities

Oh dear. That is nonsense.

yep, looking at numbers definitely didn't worsen mine. it just didn't improve them.

It was after my state math exam which I didn't see results for either

For me, looking at numbers just made it even more challenging to fit in...

Social skills has got nothing to do with knowing your IQ, lol.

2:08 AM

That's what they said to me teadawg, I will have more trouble fitting in

@beginner My self-confidence has always been abysmal, don't use my experiences as an affirmation for your parents' (rather false) reasoning

@user130018 I am going to bed. I will see you in my dreams.

@JasperLoy Good night

@user130018 Remember not to study too hard.

I don think I have bad self esteem because the other students don't know how to work things out that I do all the time

2:10 AM

One thing I did learn while looking through old test scores is that I have an IQ of 149 specifically in mathematics, so I guess I am a mathematical genius...

That's awesome!!

I hope I scored that high

I'm a grade 7'er next year, that puts me as the biggest kids of the school :)

@RobertCardona: The identity map.

@teadawg and @beginner: Don't think about such scores. Just concentrate on being interested in good stuff and doing the best you can ... and, if possible, in having some friends :)

@TedShifrin Friends? But that means I'd actually have to socialize with people.... Who does that??? .... whistles

@teadawg: Math actually can be lots of fun when done with friends. It doesn't haven't to be a loner occupation. :) My students do a lot of collaboration and have gotten to be good friends with one another, many of 'em.

Hope you'll feel better soon, @teadawg.

I like making friends on this website

2:17 AM

but we're fake friends, mr eyeglasses :P

:(

Most of us don't even know one another's names ...

Are names important in a friendship?

@Ted coughs Theodore

My sense of humor is practically nonexistent today... :(

awww, it's the sinuses, @teadawg.

2:20 AM

@ted that is what I am aiming for :). I can see that z/abc is isomorphic to z/a X z/b X z/c using Chinese remainder theorem on some examples, but how can I prove it in general? Is there a small hint you can give me?

I guess I think of friendship as more than an anonymous typing thing, mr eyeglasses, although I try to be friendly to most of you :P

@beginner: Start small. Just do two. Assume $\gcd(a,b)=1$. The Chinese remainder theorem is the key. Do you know the fundamental homomorphism theorem?

Is that the f(ab)=f(a)f(b) for a multiplication relation and f(a+b)=f(a)+f(b) for the additive binary relation?

I think you need both to be defined on the group?

@Ted You want to hear the proof that $O(2)$ isn't linked?

sure @Mike

@beginner: Are we proving ring isomorphism or group isomorphism?

Oh whoops group is just armed with one, ring is two

2:26 AM

thanks @TedShifrin !

@beginner: What I'm referring to is this: If we find a homomorphism $\phi: R\to S$ mapping onto, then it induces an isomorphism $R/\ker\phi \cong S$.

Let me think about this, thank you ted!

So all I need to do is find a homomorphism $\Bbb Z\to \Bbb Z/\langle a\rangle \times \Bbb Z/\langle b\rangle$, check it's onto and find its kernel. Bingo.

Is that phi rulers totient function?

No, my $\phi$ is whatever ring (group) homomorphism I'm thinking of.

2:28 AM

I can't render latex on the tablet so give me a min to work it all out

You're entitled to all the minutes/hours you want, @beginner.

Hehe I just didn't want to seem like I wasn't doing it because you ignored someone for asking for help and then they didn't do it

I mean they asked for help, didn't do it and then were ignored in that order

It's not like I go around ignoring just everyone, @beginner. Geez.

Just mike and balarka

Maybe lol

LOL ... I've never ignored @Mike (although I should), and with Balarka it was a very different issue.

2:31 AM

@Ted Consider the map $T^2 \to S^3$ given by $$(v_1,v_2)\mapsto [v_1,v_2]/\|[v_1,v_2]\|$$. This is a smooth emvedding whose image contains $O(2)$. It is a theorem that every smooth torus embedding into $S^3$ extends to an embedding of the closed torus (aka, a torus can't be knotted on both sides)

It always is with balarka

Whoa, slow down, @Mike.

This embedding of the closed torus provides an extension of the embedding of $O(2)$ to an embedding of two discs, showing that it's not linked.

I guess I don't see why the closed torus stops two particular circles from being linked.

Oh, wait.

OK, how do you prove this theorem?

Hell if I know.

2:35 AM

It's pretty clear for the usual Clifford tori.

Where did you find such a theorem stated?

Can we prove it just in $\Bbb R^3$? Should be the same thing.

I'm no longer convinced by what I said about the solid torus. It would be fine if they were (0,1)-homology elements but they're (1,1). How do I extend to a disc?

Right, that's what was bothering me earlier.

I believe it's true - I think if two circles are contained in an embedded torus they shouldn't be linked.

The extension statement I thinj I read somewhere in Thurston's book. Let me try and find a reference online.

What about $(p,q)$ and $(p',q')$ torus knots? Hmm ...

Those will intersect

2:39 AM

Well, if they're disjoint, that limits my options.

The cup product of the corresponding elements of $T^2$ is nonzero.

I could certainly have parallel such for $p=p'$ and $q=q'$.

Unless you mean p=p' etc but I believe those to be unlinked.

Right.

Guys! I haven't given up on project euler 494. I will be top 50!

2:41 AM

A bit frustrated that proof idea fell apart.

We can cohomologically compute linking numbers, too, @Mike (although I usually do this with differential forms instead).

But I actually think this torus question is far more interesting than the O(2$ one.

Oh... this was an old qual question

We've been known to have false qual questions :P

I seem to recall the method presented being pretty damn uncomputable

No, I mean the linking number thing.

Ohhh ... sure, that part's right :)

Well, it's as "computable" as filling in a disk and intersecting or computing the degree of the chordal map ... in general :)

It comes up in defining the Hopf invariant in general, too.

2:44 AM

Yes, I recall.

@anon !! You survived your torrid teaching term.

yep

Enough to make you quit math right now? :P

nah

I wanna see my darn evals.

2:45 AM

Some of mine are making me decide retirement is right, @Mike :P

I guarantee you'll get some that make you upset.

So what do you get to take next term, @anon?

intro psych, moral philo, music hist, crim just survey

(gen ed reqs)

wow, well, a few of those sound interesting ... so no math at all?

nope

@anon are you a freshman?

Jorge Fernández

Could someone read this answer please math.stackexchange.com/questions/1078179/…?

I'm pretty happy about it.

2:48 AM

you done applying to grad school, @anon?

@DonLarynx junior

@TedShifrin haven't

@anon sounds like my senior year.

oh, wait, if you're only a junior, why not spread all the gen ed over the remaining semesters?

@Ted a friend of mine got a few "clearly doesnt know the material" (it was calc 1...). Curious if I get the same. I expect my Thursday evals to be quite good.

@TedShifrin you mean take them at the tail end of my student career? that's what I'm doing.

2:49 AM

I got several snide ones this term, @Mike. They seem to think that because I'm retiring I no longer care about teaching or students. They couldn't be wronger.

@anon: You mean you have three semesters of nothing but gen ed?

@Ted why are you retiring?

My students usually do one or two each semester.

Jorge Fernández

@anon I wouldn't like that.

@TedShifrin are you still going to be on the site often?

I don't know, @Jorge.

I don't follow your argument, @Jorge. How does adding points near the vertices of your equilateral triangle make all the points more than $\sqrt2$ apart?

The OP wanted points at least $\sqrt2$ apart ...

Jorge Fernández

I had a picture

user image

the points are like the ones on the left

so there are $12$ pairs of points.

The equilateral triangle has length 1.5

2:53 AM

I think the OP wanted a set of points where each pair is at least $\sqrt2$ apart. I'm confused.

Jorge Fernández

the OP wanted to maximize the number of pairs of points that where that far apart

Alexander Gruber

@JulianRachman I saw your deleted meta question; you may be interested in this.

Jorge Fernández

with the triangle arrangement every point is at such a distance from 4 points, so there are 12 pairs of points at that distance

@Jorge. I'm still confused. And do you really mean 12 pairs? or 12 points?

Jorge Fernández

12 pairs, there are 6 points.

well 24 pairs, if they are ordered.

2:55 AM

Whoa. $\binom 62 \ne 12$.

Jorge Fernández

yeah, but what she wants it to know what the max number of pairs at distance larger than $\sqrt2$ she can get if all points are ate distance less than $2$ from each other.

I say it is $12$.

so you can't achieve the hypothetical max which is $15$

I am not sufficiently energized to understand this problem tonight ...

Jorge Fernández

You mean the solution or the question?

Either one :)

All this confusion between points and pairs of points makes it worse.

I'll try tomorrow :)

Jorge Fernández

thanks

3:08 AM

@Ted I think I give up on the torus problem, but I find it fascinating.

Maybe I'll ask a knot theorist when I get back.

Of course - for a general embedding of a torus - there's no reason that the two circles shouldn't each be unknots; I only claim they're not linked.

Jorge Fernández

does anyone here like combinatorial geometry, I would like for someone to check a solution I gave.

Hey guys, I am looking for a solution to a very SIMPLE problem. Basically I have a 5 numbers. $271, 262, 98, 532,$ and 712. The total adds up to 1875. I'd like to split up the 712 and redistribute equally amongst $271, 262, 98 ,532$ . This seems easy enough. I can divide by 4 and take the floor function.

You may interpret "not linked" either as "null-homotopic in the complement of each other" or "equivalent to the disjoint Union of two knots". I guess these are the same but my ignorance is showing.

@masfenix I have edited your question

Ahh okay, I know Its hard to explain. Consider the 2 sets of data:
$$ 271, 262, 98, 532 | 712 \\ 342, 673, 83, 468 | 309 $$Note that the total adds up to 1875. How do I split up the number $712$ and $309$ and redistribute it to the first four numbers so that the total still adds up to 1875

I can't simply divide by 4 and redistribute since it will introduce fractions and I need to stick with integers.

I can't take floor or ceilings because that loses information. Any other method?

@robjohn do I make sense now?

3:29 AM

Ahh, my solution is kinda very inefficient but the numbers are small so it dosnt matter.

basically I loop through 1:712. and increment each of the first four numbers by 1.

i know this dosnt 100% redistribute evenly, but close enough.

Oh wow, there's a high IQ society called the ISI-Society... Don't mean to bring up politics, but boy is that name tarnished now....

Wait, I called that politics? I'm way too tired to talk right now...

Alexander Gruber

3:47 AM

@JorgeFernández sounds fun i'll bite

Jorge Fernández

0

A: Combinatorial Geometry Problem

There is a way to get $12$ pairs, take an equilateral triangle of side length $1.5$, for each vertex add an extra point right next to it.(there is an image at the bottom) We now prove there can't be more than $12$ pairs, if there where more than $12$ edges we would have two points $a,b$ such tha...

Alexander Gruber

@JorgeFernández alright so when you start talking about $c$ and $d$

Jorge Fernández

yes?

Alexander Gruber

Are you assuming there that $A,B$, etc are as in your diagram?

Jorge Fernández

no, the diagram is only for the existence of the arrangement with $12$ pairs

It is only of relevance to the first paragraph.

Alexander Gruber

3:56 AM

Okay, I see. You may want to edit it so the diagram comes directly afterwards, or move the 12 pairs paragraph to the end, just to reduce potential confusion

OK second, what is the proof of "if there where more than 12 edges we would have two points a,b such that the distance between each of these points and another is in the interval (2√,2]"?

Jorge Fernández

The sum of degrees is at least $26$ and if there are not $2$ with degree $5$ the max sum of degrees is $5+5*4=25$

@AlexanderGruber is it clear what I mean?

Alexander Gruber

@JorgeFernández Not quite. What precisely do you mean by edges in this context?

Jorge Fernández

4:16 AM

@AlexanderGruber well, there is an edge between two points if they are at distance greater than $\sqrt2$

Well, what I mean is that if there where at least $13$ pairs there would have to be two points such that all points are at distance greater than $\sqrt{2}$ from each of those points.

Let $d(p)$ be the number of points at distance larger than $\sqrt(2)$ from $d$. Then the number of pairs is half of $\sum_{p\in P}d(v)$. Where $P$ is the set of $6$ points. therefore that sum is at least $26$, if we didn't have two points with $d(v)=5$ the sum would be at most $25$. @AlexanderGruber

Feels like everybody went home for the holidays with a PDF of a problem set but without their textbooks. Lot's of "what does this mean?" questions, where their class definitely covered it. :)

2

LOL'd @ThomasAndrews

Jorge Fernández

@AlexanderGruber looking forward to your feedback

Alexander Gruber

@JorgeFernández Ahh I see, you should explain that in the post

Jorge Fernández

ok

Alexander Gruber

4:26 AM

@JorgeFernández Your math is correct, you should just explain the definition of the graph you're making.

Jorge Fernández

@AlexanderGruber thanks, I think it is ok now.

Alexander Gruber

@JorgeFernández I agree.

Jorge Fernández

I normally don't do questions that hard on MSE

I only answered that one because the OP seemed cute.

Alexander Gruber

@JorgeFernández Well, maybe if you get enough upvotes, she'll give you her number, eh?

Jorge Fernández

yes, that's what I was hoping for.

5:12 AM

@JorgeFernández It is somewhat disappointing that your hard work was left sitting under the totally nondescriptive title "Combinatorial Geometry Problem". (I edited the title now.)

@robjohn do we loose our hats after winter? I don't want to loose my hat :(

Yes, on January 5 all hats go back into SE vault. Take a lot of screenshots.

@Behaviour even using screenshots, I can't wear it like this again :/ coz u see my hat is over the profile picture (the top part)

Maybe I should ask the SE team for .psds of the hats.

That's true.

5:14 AM

Shouldn't be too hard to modify this (or the Lumbergh) to work in the limited space given.

We'd also need their JavaScript to place the hats on the avatars locally.

But why take away earned hats :( these look so cool

Oh, I was just going to actually photoshop them on.

Hats aren't special if they're all year, @TheArtist

Jan Dvorak still wears a hat from Winterbash 2013.

@MikeMiller okay but it's special such that we can earn them. I think we should be able to keep the earned hats, and next year earn more

5:16 AM

I don't.

And I will fight for your right to not have hats.

this will still be a special period where we earn hats.

@Behaviour ohhhhhh , his hat goes over the profile picture too. Do u kno How does he do that?

You mean the Sumo hat? All hats can go a bit outside of the avatar bounds. Yours does it too.

His last year's hat is the owl. That one is photoshopped, so it must be within the square.

@Behaviour oh within the square. I thought that the existing hat is last year's

Jorge Fernández

@Behaviour thank you.

5:58 AM

Are we not able to keep hats that we're wearing right now?

No.

Aw :(

Hats are finite, just like life. Cherish them!

3

@MikeMiller

HALP.

maybe

6:08 AM

You could make it into an avatar with a screen shot. @KajHansen

Suppose that $k[X_1,\ldots,X_m]$ is a ring of polynomials.

I'll try

That's true @skullpatrol

Suppose we take $x_1,\ldots,x_n$ elements of that ring, such that $k[X_1,\ldots,X_m]$ is integral over $k[x_1,\ldots,x_n]$.

yes

6:09 AM

Then it follows by localization that $k(X_1,\ldots,X_m)$ is algebraic over $k(x_1,\ldots,x_n)$.

I have a smooth argument that shows in the case $m=n$ that the $x_i$ are algebraically independent over $k$.

But I cannot seem to generalize it when $m\neq n$.

meh

@PedroTamaroff, your minimally sneaky groups should be called simply sneaky, imo

The argument is easy: suppose $f(X_1,\ldots,X_n)$ kills the $x_i$. Then there is an equation of the form $f^N=\sum_{i=0}^{N-1} p_i(x_1,\ldots,x_n)f^i$, and we may take it of smallest degree. Evaluating at the $x_i$ gives $p_0(x_1,\ldots,x_n)=0$, so we can lower the degree if we assume $f\neq 0$.

So it must be that $f=0$.

@KajHansen They are not mine.

They are Alex's.

I'm sorry bud but I'm not reading anymore

@MikeMiller =(

6:13 AM

Wait for someone smart to show

@AlexanderGruber, your minimally sneaky groups should be called simply sneaky, imo :D

@MikeMiller Feeling small tonight, Mike?

Did someone do something to you?

I'll fight them.

Yes, you showed me that problem.

hahaha

And I cried many tears.

6:15 AM

@TheArtist Yes... there will be a period of separation anxiety for all.

@masfenix what is the goal of the redistribution? Do you want to distribute the number evenly, proportionally to the number added to, what?

@MikeMiller Alright I just punched myself in the face.

:)

@MikeMiller Feeling better?

After I heard that you punched yourself in the face? Absolutely.

@MikeMiller You have forsaken me, Mike. All is lost.

6:37 AM

Vote For Pedro!

3

Now that he's wearing a hat, he deserves a vote.

:))

@MikeMiller I guess my question is pretty stupid and can be answer with basic stuff about field extensions.

Which I know nothing about.

It should be relevant that the $Y_i$ are algebraically independent over $k$.

I don't know anything about stuff.

Maybe I want to show that if $k$ is a fixed field, and $L= k(a_1,\ldots,a_n)$ is an extension with the $a_i$ algebraically indep. then any intermediate extension $L'=k(b_1,\ldots,b_m)$ with $L/L'$ algebraic has the $b_i$ algebraically independent.

Chantry Cargill

6:45 AM

I am reading a blog post located [here](https://algean2015.wordpress.com/2013/11/07/a-beautiful-convergent-series-ii/). I was wondering if someone here could explain how the author made this step. I'm not seeing it.
$$\frac{1}{4^k}\sum_{n=-2^{k-1}}^{2^{k-1} -1}\frac{1}{sin^2\frac{x + n\pi}{2^k}} = \lim_{k \to \infty}\sum_{-k}^k\frac{1}{(x + n\pi)^2}$$

What an unpleasant banner at the top of the pge.

@MikeMiller Indeed.

Cringeworthy.

Chantry Cargill

It's certainly colourful.

@ChantryCargill That equation is not correct.

He's probably missing a limit to the left.

I am assuming he's taking the limit on both sides, and using $$\frac{\sin t}t\to 1$$

At any rate, it is pretty fishy and he might as well just say that thing is true.

You can prove it rigorously without such juggling.

@Huy Fun! Anyway, Guete Morge ^^

Chantry Cargill

6:51 AM

@PedroTamaroff I guess I'd be better off trying to prove it myself?

@ChantryCargill Depends on what mathematics you know.

Using complex analysis, it is not hard.

But one needs some theorems.

Chantry Cargill

@PedroTamaroff Yeah, that's definitely not a strong area. I really do need to make more progress on rudin >.<. Thank you though.

7:12 AM

@PedroTamaroff What is le question?

What the hell is this guy talking about?

Fibre bundle? Serre fibrations? Huh? They prove path lifting on covering space using fibrations these days?

No.

@BalarkaSen #namedropping

heh, just like i used to do

@MikeMiller is it true that all knot complements in R^3 are connected?

Think about it.

if so, then we have another proof of nonhomeomorphism of R^2 and R^3

i am actually not sure @Mike. take an embedded S^1 in R^3 that space-fills the surface of the unit sphere.

7:31 AM

When you talk about things, please try to recall the definitions.

eh. i don't mean connected. path connected.

That is not the relevant definition you seem to have forgotten.

what definition do you want me to recall?

of connectedness?

Anyway, connected implies path-connected here.

the other way

there are connected spaces inherited from subspace topology of R^3 that are not path-connected

7:34 AM

The other way is always true. The one I wrote down is true here.

oh?

what about the warsaw circle

oh the complement

Open connected subsets of $\Bbb R^n$ are path-connected.

Anyway, I suggest you come back to knot theory after studying the definition of an embedding.

i dunno any knot theory, @Mike

The definition of an embedding is a pre-requisite to the definition of a knot.

so what you're saying is that, say, a warsaw circle is never an embedding of S^1 in R^3?

7:38 AM

8 mins ago, by Balarka Sen

i am actually not sure @Mike. take an embedded S^1 in R^3 that space-fills the surface of the unit sphere.

i think of embedding as a continuous map S^1 --> R^3

@robjohn hmmmmmmmmm

@BalarkaSen Do you know what an embedding is?

Okay, then you think of an embedding wrong.

Words have meanings.

@MikeMiller nope

googled. ok, so an embedding f : X --> Y is a continuous map s.t. X is homeomorphic to f(X).

I can't fall asleep. Yaaaay

7:44 AM

Actually, I no longer see an elementary way of showing that the complement of a knot (which is an embedded $S^1$) has connected complement in $\Bbb R^3$. I was thinking of smooth knots (a special case of tame knots, which are also good enough), for which it's easy to check that the complement is connected.

oh

guess i won't think about what i can't do.

Still true, but as far as I can tell seven hundred and three orders of magnitude harder than just calculating the fundamental group.

silently creeps away

:-)

Admiral Yu*

8:28 AM

Greetings

Hello

Greetings

Doess anyone here know what is English word for a person who (always) gives false hopes?

Is it false hope or fake hope? Which one is correct?

False hope, I think. I don't quite understand your question, though. I must be really tired

well you can ask in the Eng.SE chatroom, whatever you want to ask @Venus

8:39 AM

I love staying here & I don't wanna go anywhere :-(

@Venus So stay here :3

@Venus False hope is the term I've always used/heard

False hope is correct.

What you call that person depends on the context in which that person is giving false hope @Venus

here

8:55 AM

I'm so bored... You guys are boring me

@teadawg1337 I never get bored because I always work (well, maybe I should take a break sometime in this life...).

It's 3 AM here, I can't really do much work right now...

@teadawg1337 What kind of math are you interested in?

@Balarka Analysis, mostly

Ugh.

8:59 AM

I do a lot of number theory on the side, too

@Venus: Hey.

Aha!

Number theory is cool stuff @teadawg1337

@Venus: Sorry, I wasn't quite up yet. Guete morge to you too!

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