Mathematics - 2012-08-07 (page 3 of 6)

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10:03 AM

now what i am particularly interested in is the dependency or lack thereof between different heuristics, regardless of whether or not they take the same form as my examples. part of my question is: "What are some other examples of heuristics that are not independent (e.g. two statements that are both heuristically true but contradict each other)?"

Ok, here is one part of a diagram chase:

user image

user image

Bbl : )

glad i never took that nonsense seriously.

oh sorry i didn't intend to become part of the cycle of abuse!

what i should have said is that i just don't understand category theory so i don't take it seriously.

user19161

10:32 AM

@JonasTeuwen Just to add, many parents abuse their kids in various ways, physically, sexually and psychologically. Parents don't always love their kids you know.

...

wow my toof is hurting from all the snideness perpetuated in this forum (e.g., Jonas, Matt, myself).

(not referring to you at all Jasper...)

i think it is a problem that the etiquette guidelines meta.math.stackexchange.com/questions/3890/… say only what not to do and not what to do... and also not to not do the things that people routinely do here.

(speaking as both a victim and a perp...)

can't we all get along yo?

user19161

Often what others perceive as love, I perceive as something else. For a great book discussing what is love and what seems to be love but is not love, see The Road Less Traveled by David M Scott Peck.

user19161

@DanBrumleve Who is not getting along? I see no problem in here.

a better problem would be something like why can i easily prove that $e^2$ is irrational, but i have to resort to more abstract thinking (calculus) to prove that $e^3$ is irrational?

@Jasper if you scrollback you will see I took offense to a comment.

also more generally i feel under the gun here when i show up to say whassup.

user19161

@DanBrumleve Well, I think that was just a general remark not really saying it applies in your case.

10:40 AM

yeah and i'm saying it doesn't apply period.

user19161

OK, then with this understanding there is no problem, period.

Hey Dan, I just saw that you might've taken offense by some of my comments. I know nothing about you so I cannot say anything personal about you or your situation hence in particular, do not apply anything I say to yourself and keep in mind that I do not intend to offend you by anything I say.

K?

@Matt we're cool.

phew

i'm just on a chat culture trip now.

10:41 AM

Thanks.

user19161

Phew!

like wtf are we all doing?

i'm 34, am i the oldest guy in here?

where are the old school dudes like Gerry Myerson?

No, I think some in here are older than 50.

David Wheeler is 54, iirc.

He said so, at least.

user19161

Old John just turned 60 today!

user19161

@MattN. 51.

10:43 AM

Close enough : )

i'm tryna be mature here but i feel badly influenced by y'all.

user19161

@DanBrumleve Maturity is an often stupid concept. People change and grow every day in various ways.

Haha, scapegoat technique : ) I like that.

... but of course i love the mainstream math.SE culture and that's why i'm here, duh.

i'm afraid this chat culture pushes away the people who know stuff.

user19161

@DanBrumleve I don't think so. You should know that mathematicians can be very weird people...

10:47 AM

@DanBrumleve What do you mean by this exactly?

it is actually a really nice webchat system and it is a shame that more math.SE users don't come here is all i'm sayin'.

user19161

@DanBrumleve Not everyone has the time to come to chat you know.

extended-comment-nag worked to drag me here a couple times but i don't think that's really the best solution.

@Jasper eh few top users are ever in chat but many are seen within hours.

@DanBrumleve The thing is: if they sit in here they are prevented from being on main which means a reduction of helpful answers on main and a reduction in productivity of theirs.

fine we can't have people in chat saying pi=3 because they passed the law in indiana

10:51 AM

Come to think of it: I should go and compute some $\mathrm{Tor}$s and $\mathrm{Ext}$s!! I'll see you all later!

cool cya!

@MattN. sounds like fun (?)

user19161

@robjohn Hi! You're early as usual!

@JasperLoy early for what? I thought I was up late.

user19161

11:06 AM

Anyone here tried washing a keyboard in a dishwasher? Someone did and dried the keyboard and it was OK.

well i disagree with Matt because i think top users should be asking good questions (and editing top answers) rather than answering new questions (which should be answered by new users). so it doesn't make sense to talk about the lost "productivity" of top users from chatting.

user19161

@DanBrumleve The same applies. Less chatting, more asking and editing.

well i would argue that almost all creative activities are social.

of course it's hard to define what being a "top user" should mean and "being creative" doesn't cover it. some top users answer and never ask.

user19161

@DanBrumleve True, we need good questions.

i think it is cool that some top users answer but never ask, also it is fine that there are fly-by-night users who ask and never answer, but i think the SE way should be to encourage participating in both and for that we need a social context, i.e. chat.

11:23 AM

Hey all

I'm having an epic mental blank on something really basic, hopefully someone can see what I can't

With coupled recurrences we can usually combine them into a single recurrence (albiet of greater degree). I want to combine $d_{n+1} = 2d_n - a_n$ and $a_{n+1} = d_n - a_n$ but can't do it !!

Oh wow, I've been trying stupid things for 30minutes and as soon as I typed that out for everyone, I see it finally. THanks anyway everyone!

subtract them?

:)

yup $a_{n+1} = d_{n+1} - d_n$.

I actually meant so that it is a recurrence for a single one of a_n or d_n

This is what I did:

$d_{n+1} = 2d_n - a_n = d_n + (d_n - a_n) = d_n + a_{n+1}$ so $a_{n} = d_n - d_{n-1}$ and then putting it back: $ d_{n+1} = 2d_n - d_n + d_{n-1} = d_n + d_{n-1}.$

So in the end I find $d_n$ satisfies a fibonacci type recurrence! Which is what I wanted.

yeah i really didn't get it yet, doh.

was just working fruitlessly on $d_{n+2} = ...$

oh but i guess i was almost there, from the other side.

I did like 5 pages of complex calculations to find a formula for $d_n$ and at the end it all ended up nicely being a fibonacci type sequence, and that was definitely not a coincidence so I went looking for a trick to get there quicker. Wasted so much time :(

@JasperLoy Yes, I know. Irrelevant here.

@MattN. Sup?

Scary stuff.

12:00 PM

@robjohn At the moment the answer is of Schroedinger type. I will answer that on Thursday, after I looked into the box.

@JonasTeuwen Showing exactness stuff. What about you?

BBL

@MattN. Chillin'.

12:57 PM

Would OS X 10.8 be worth the upgrade?

Meh, only 15EUR. Let's see.

@JonasTeuwen That was only the first third of the lecture :,( And that's the third I sort of know what it's about. :,(

My chances are very bad :,(

1:14 PM

@MattN. Don't worry, just do as much as you can (and don't try more) and see what that gives. Will give you information about how to approach it next time. Good luck, bro :-).

OS X WAI U IGNORIN MA SHELL VAR?

: )

@JonasTeuwen You know, in theory, I think I should be able to pass by knowing exactly half of the lecture. But that does unfortunately not correspond to reality. Otherwise I'd be chilling it now.

@MattN. Yep. Just continue working on it. See how far you get.

Needing a break right now. Just realised that I forgot about one of the fundamental theorems of the lecture (right after the snake lemma).

I forgot that a sequence of short exact sequences of chain complexes gives you a long exact sequence of homologies.

user19161

@JonasTeuwen You upgrading your Mac? I did not know you are Apple boy too!

@JasperLoy I have it at work.

user19161

1:27 PM

@JonasTeuwen I never used one in my life. If I get a new comp, I think I will still stick to Windows.

Macs are so awesome.

Computers are like religion.

You just have to iBelieve and iLike.

Can anyone think of a nice way to show that the maximum of $f(x) = 2^{x-1} + 2^{n-x-1}$ over $[1,n-1]$ occurs at the endpoints of that interval. I'm having a blank :(

@JasperLoy Dude, did you have to rub it into my face that you're not one of the cool kids?

user19161

@MattN. I use Linux. I am cool!

Now you say. Before you said you'd stick to Windows!

Meow.

Time time to prove that SES give me LES in homology!

bbl

1:35 PM

@MattN. iBelieve. Meow to you too.

user19161

I am a Belieber.

user19161

Justin Bieber. Just believe.

Perfect.

@MattN. Hey

Did you see my question on applying van kampen?

@MattN. math.stackexchange.com/questions/179797/…

1:59 PM

Quickie: $\mathrm{Hom}(M,-)$ is exact iff $M$ is projective. Do we have $\mathrm{Hom}(-,N)$ iff $N$ is injective?

@MattN. Yes

2:16 PM

@ZhenLin I just got a solution!!!!!

good

@ZhenLin I was trying to prove that there is no retract from the mobius strip to its boundary circle

proof:

if there were such a retract

it would induce a homeomorphism between the mobius strip/boundary and the boundary circle

@ZhenLin applying $\pi_1$ to both of these guys and you would have that $\Bbb{Z} \cong \Bbb{Z}/2\Bbb{Z}$, contradiction.

I'm quite sure the Möbius strip has an infinite fundamental group...

@ZhenLin have you done this problem

@ZhenLin it's mobius strip mod out by the boundary

that is $RP^2$

with fundamental group $\Bbb{Z}/2\Bbb{Z}$

hmmm

maybe it works

I wouldn't have argued that way

2:21 PM

@ZhenLin hahahahahahahahahahahaha

I never thought about this

by the point is a retract is a quotient map

that is what gives you the homeomorphism!!!!!

I don't see why it would give a homeomorphism. One space is two-dimensional, the other is one-dimensional.

munkres corollary 22.3

You'll have to state that in full for me.

@ZhenLin Let $g : X \to Z$ be a surjective continuous map. Let $X^\ast$ be the following collection of subsets of $X$:

$X^\ast = \{g^{-1}(z) : z \in Z\}$

Give $X^\ast$ the quotient topology

then the map $g$ induces a homeomorphism $f : X^\ast \to Z$ which is a homeomorphism iff $g$ is a quotient map

in our case take $X$ to be the mobius strip, $Z$ the boundary circle and $g$ is our retract

we already know $g$ is a quotient map

that doesn't make any sense at all

are you sure you're reading it right?

2:27 PM

yes

sorry rather $g$ induces a bijective continuous map $f$ that is a homeomorphism iff $g$ is a quotient map

it's roughly saying that we have a quotient map if and only if we have a quotient map

which seems like a silly thing to say

@ZhenLin so yes

anyway

I don't see the connection to your claim.

well

$X^\ast$ is just the quotient space

the quotient of the mobius strip by its boundary circle

nope

2:30 PM

it is

that's not what $X^*$ is

the whole of the boundary has been sent to a point

nope

hmmm

@ZhenLin you see our retract $g$ is constant on $A$

what is $A$?

2:33 PM

@ZhenLin sorry $A$ is the boundary circle

@ZhenLin so then $g$ is constant on the fibers of the canonical projection onto $X/A$

It's the identity on $A$, by definition.

Being the identity map is not the same as being a constant map!

hmmm yes

I think I got mixed up there....

@ZhenLin Thank you!

@ZhenLin then what is the quotient space here?

Sorry rather what is $X^\ast$?

Probably homeomorphic to the circle.

Like I said, the corollary you cited is very uninteresting and probably not relevant.

2:38 PM

why do you say so

@ZhenLin but you see it will give us a homeomorphism between $X^\ast$ and the boundary circle

and if you apply $\pi_1$ and their fundamental groups are different

you get a contradiction there

I don't see why their $\pi_1$ should be different.

why not?

How did you calculate them in the first place?

by identifying what the spaces are?

Yeah, but I just said $X^*$ is just the circle, possibly with a finer topology.

2:41 PM

how do you know that?

Because that's what your corollary says.

no

I don't get how you deduced $X^\ast$ being just the circle

There is a continuous bijection. So it's just the circle, possibly with a finer topology.

ok....

@ZhenLin I'm off to bed

bye

bye

2:47 PM

@OldJohn: Happy $111100=\mathrm{LX}$! I tried for Cuneiform, but \unicode(x12400) to \unicode(x1247F) don't seem to render, at least not on my computer.

3:08 PM

Another quickie: when computing $\mathrm{Tor}(M,N)$, it doesn't matter whether you take a projective resolution of $M$ or of $N$, right?

@robjohn Thanks!! - I will forgive the lack of cuneiform :)

@MattN. Yes. I proved this in an answer a few days ago.

user19161

@robjohn That must be in binary I guess.

user19161

3:24 PM

@OldJohn Will there be a big party later?

@ZhenLin Very cool, heads out to find Zhen Lin's answer

@JasperLoy A quiet meal in a restaurant tonight - but we are having an "open house" weekend at the end of August to celebrate 2 birthdays and 2 retirements ... so if anyone happens to be in NW UK at the end of the month, you are invited!

@ZhenLin But only for f.g. modules. I meant for arbitrary $R$-modules.

That's a bit more involved than I can read now, I have to do some other stuff first.

3:49 PM

@JasperLoy Windows?

Halp.

@MattN. How?

How do I compute the image of $\overline{d_1} = \cdot n$ in
$$ 0 \to \mathrm{Hom}(\mathbb Z , \mathbb Z / m \mathbb Z) \xrightarrow{\cdot n} \mathrm{Hom}(\mathbb Z , \mathbb Z / m \mathbb Z) \to 0$$

Crazy stuff...

No, it's basic algebra.

But when there are numbers I get confused.

3:52 PM

@MattN. The same proof works for non-f.g. modules, but the question was about f.g. modules.

So I have $f: \mathbb Z \to \mathbb Z / m \mathbb Z $. Then I map it to $f(n \cdot x)$.

So my image looks like $\mathrm{Hom}(n \mathbb Z , \mathbb Z / m \mathbb Z)$?

Zhen, you know the answer to that, right?

@ZhenLin Cool, plus-one'd it already : )

It's better not to write it that way...

How should I write it instead?

I think I have to "see" what it looks like and then write it differently.

Or not?

Or is $\mathrm{Hom}(n \mathbb Z , \mathbb Z / m \mathbb Z)$ indeed the image of $\cdot n$?

Well, remember what I said a few weeks ago about $\mathbb{Z} \cong n \mathbb{Z}$...

You said it is an isomorphism.

So $\mathrm{Hom}(n \mathbb Z , \mathbb Z / m \mathbb Z) \cong \mathrm{Hom}( \mathbb Z , \mathbb Z / m \mathbb Z)$?

So $\cdot n$ is surjective?

4:00 PM

Well, that's the point. $\textrm{Hom}(n \mathbb{Z}, \mathbb{Z} / m \mathbb{Z})$ doesn't mean what you think it means!

So it does not mean $\mathrm{Hom}(\mathbb Z , \mathbb Z / m \mathbb Z)$?

Shouldn't because then $\mathrm{Ext}^1$ would be zero.

They are isomorphic, yes. But it isn't the image of $\cdot n$.

Hah!

@ZhenLin The image is $\mathrm{Hom}(\mathbb Z , n \mathbb Z / m \mathbb Z)$, right?

No : (

That can't be right. That homo needs to be applied to the domains, not the ranges.

And yet it works. :p

Formally you should write $n \mathbb{Z} / (m \mathbb{Z} \cap n \mathbb{Z})$.

But I can't apply $\cdot n$ to the ranges. It's the $\mathrm{Hom}(-, N)$ functor I applied. So the $\overline{d}$ maps get applied the other way around.

4:04 PM

Peoples of the math.

I have a question, but it is a very simple one.

About notation. And order theory.

I suspect it has to do with lattices.

Something weird happens here, yes, but basically you push the $n$ through the homomorphism to the codomain.

head asplode

After all, $f(n x) = n f(x)$.

Oh, right!

Ok, I'm going to see if I can finish my sums.

@ZhenLin Thank you!!

np

4:09 PM

@ZhenLin Why do we call the space $X=[0,\infty)$ with topology $\{X,\varnothing,(a,\infty):a\geq0\}$ the arrow?

I have no idea.

@ZhenLin The author uses this symbol for it:

(I have to draw it, hold on)

Here

I'm busy with something else right now. You should ask someone more familiar with general topology.

user image

@ZhenLin Ok.

@JasperLoy binary and Roman

4:27 PM

@Peter Tamaroff that picture goes to another space

@DavidWheeler Yes, I misread.

the only thing i can think of, is that the normal partial order of a topology (induced by inclusion) on that particular X induces a total order on X: a < b iff (b,∞) is contained in (a,∞).

@DavidWheeler Right.

all of the intervals are essentially "neighborhoods of ∞"

@PeterTamaroff That looks like the Constellation of Corvus viewed upside down (or from the Southern Hemisphere).

4:47 PM

it's just a picture of the lattice of a 4-point topology

@DavidWheeler Yes!

I'm already home at 6 PM, must be something wrong with me 8-).

@DavidWheeler how does that contradict my claim?

Another (and very important T_T) quickie: in general, do I have $\mathrm{Hom}(\mathbb Z / n \mathbb Z, \mathbb Z / m \mathbb Z) \cong \mathbb Z / \mathrm{gcd}(m,n) \mathbb Z$, just like I have $\mathbb Z / n \mathbb Z \otimes \mathbb Z / m \mathbb Z \cong \mathbb Z / \mathrm{gcd}(n,m) \mathbb Z$?

That looks wrong because then $M \otimes N $ would be isomorphic to $\mathrm{Hom}(M, N)$.

: (

So what is $\mathrm{Hom}(\mathbb Z / n \mathbb Z, \mathbb Z / m \mathbb Z) $ isomorphic to?

5:03 PM

@robjohn Context does =P

@PeterTamaroff No, it still looks like Corvus, no matter what the context. :-p :-p

@robjohn who said it did?

@DavidWheeler I did.

you said it contradicted your own claim? how odd.

@DavidWheeler No, I asked, "how does that contradict my claim?" which was meant to imply that I don't think it does contradict my claim.

5:10 PM

who said anything about contradicting a claim? did i use the word "no"?

@DavidWheeler You're discussing over something so irrelevant!

@PeterTamaroff lol, conversation doesn't have to be "relevant" to be fun :P

I feel asleep.

Oh? Only 20 minutes 8-).

5:46 PM

heya

Is there any "nice" sum/series for representing $\sin(\log(x+1))$ ?

What do you want to do with it?

So $$ 0 \to \mathrm{Hom}(\mathbb Z , \mathbb Z / m \mathbb Z ) \xrightarrow{\cdot n} \mathrm{Hom}(\mathbb Z , \mathbb Z / m \mathbb Z ) \to 0 $$
looks like $0 \to \mathbb Z / m \mathbb Z \xrightarrow{\cdot n} \mathbb Z / m \mathbb Z \to 0$. What is the image, $n \mathbb Z / m \mathbb Z $ isomorphic to?

@MattN. Looks like magic.

6:01 PM

@JonasTeuwen Approximate $$\int_1^2 \sin(\log x) \, \mathrm{d}x$$ yes, I know it is easy to evaluate it.

@N3buchadnezzar So you don't need the full series? Have you looked at Euler-MacLaurin?

@MattN um, that doesn't look right

@DavidWheeler Right. Just noticed.

I need to know what the image of $\mathbb Z / m \mathbb Z \xrightarrow{\cdot n} \mathbb Z / m \mathbb Z$ looks like.

How do I compute that?

I can't write what I wrote because $m Z$ is not necessarily a subgroup of $n Z$.

how do you know that even defines a homomorphism?

Perhaps it does not.

It's gotta map zero to zero.

6:14 PM

the 60 million dollar question is where it maps 1

To $n$.

But that doesn't mean it's a homo.

but what "n" is in Z/mZ depends on m.

Yes, if $m$ divides $n$ it's zero and if they are coprime then it maps to a generator.

But what about the cases in between?

well it always maps to a generator of "some subgroup", but again: if you don't have a homomorphism, why are you doing this?

But it is a homomorphism!

6:18 PM

suppose m = 3, and n = 2

Then 1 maps to 2, which is a generator of the whole group. So it's an isomorphism.

Right?

yes, but that's not nZ/mZ...that doesn't even make sense unless m divides n.

Yes, I know, I wrote that a few lines above^

: )

I want to do research on the south pole.

Or Greenland.

@JonasTeuwen Harmonic analysis on Penguins?

6:24 PM

No, just in a hut.

ok...if you look at the map $x \to nx$, you'll get an automorphism of $\Bbb Z/m\Bbb Z$

if gcd(m,n) = 1

Yes, I know, I wrote that a few lines above^

: )

but if they have some common divisor, you'll get a subgroup of $\Bbb Z/m\Bbb Z$ as the image

Say d divides both n and m.

n should have order m/gcd(m,n) i think

6:30 PM

Then the image is isomorphic to $Z / d Z$?

i think that's right. like if m = 6, n = 4 then gcd(4,6) = 2, and you get a subgroup of order 3: {0,2,4}

1 actually maps to 4, but <4> = <2>.

I think it's wrong: try m=12 and n=2. Then the subgroup is order $12/2=6$.

So your first idea with gcd is probably right.

gcd(2,12) = 2...still works out.

Exactly.

So I have to somehow memorise that if we have multiplication by n on Z mod m then the image looks like $Z / (m / (m,n)) Z$.

Ewww.

And is $Z / (m/(m,n)) Z \cong (m,n) Z / m Z$?

no, that doesn't look right...the top isn't "big enough"

6:36 PM

Hm... but I think it has to be that.

Let me post it nicely:

user image

$$ \mathbb Z / \left ( m / \mathrm{gcd}(m,n) \right ) \mathbb Z \cong \mathrm{gcd}(m,n) \mathbb Z / m \mathbb Z$$

i think the RHS is "upside-down"

now wait, it's ok

smaller numbers generate bigger groups/ideals...that always trips me up

Numbers trip me up, generally.

This is like being on a death row.

so $\Bbb Z/3\Bbb Z \cong 2\Bbb Z/6\Bbb Z$, for example.

6:42 PM

@MattN. Kickass?

Not really.

@DavidWheeler Yep, so it seem to be correct.

it's clear we have 3 cosets: integers of the form 6k, 6k + 2, and 6k + 4

Might be a good idea if I post this on main.

It simply means you can take a common factor out of numerator and denominator. So $AB/AC\cong B/C$

The whole Ext computation associated with it.

@anon With group quotients this doesn't seem so obvious to me.

6:44 PM

but "bigger" picture: what the image you're looking for looks like, depends on the relationship between m and n

if they have a larger common factor, there will be some "collapsing" going on

and that means you won't have an exact sequence

Thanks!

@skullpatrol You're welcome :-).

user image

Hi all

hi

6:57 PM

Hi.

Hey :,(

Bad timing. I'm having an "it's all futile" moment.

@MattN. That's only a moment for you? Cool.

especially resistance...

But nice to see you.

So, what's particularly futile?

6:59 PM

Well, if Ajay asks the right questions it will look as if I knew absolutely nothing and then of course he cannot pass me.

And I cannot hope to get asked lucky questions. Because in oral exams I always get the worst.

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