Mathematics - 2015-11-08 (page 3 of 3)

user143442

18:00

he said he's openly gay

0celo7

Good for him?

user143442

bad for you

0celo7

Not really.

MickLH

Only when ${{\left(\sum_{i=0}^{{\it Users}}{{{{\it Intelligence}_{i}}\over{
{\it MemoryEfficacy}_{i}}}}\right)\,\int {{\it IdeaFragmentRate}
\left(t , {\it Topic}\right)}{\;dt}}\over{{\it Users}}}\geq
{\it IdeaNotabilityThreshold}$, provided that $Topic = Mathematics$

user143442

the worst thing for a homophobic must be being helped by a gay

0celo7

18:01

@user There are worse things.

user143442

like what

Agawa001

wth ^^

0celo7

@user Talking to you.

Being in a room with you.

3

Etc.

anon

@0celo7 stop feeding

user143442

:( I feel discriminated

anon

18:02

@user consider this message a warning to stop trolling

10

user143442

@anon Ok I'm sorry, it won't happen again

user143442

but I wish there was a plataform to chat with the users here about other things besides maths

Remember me

@user cannot be @Jasper. Jasper is much more formal and not a troll

user143442

formal in what sense

A Googler

Can someone help me finding integral of lnx/(x^2+2x+2) from 0 to infinity?

Agawa001

18:07

@AGoogler latex !

A Googler

oh yes

user143442

$\int_0^\infty \frac{\ln x}{x^2+2x+2}dx$

user143442

this one is alike math.stackexchange.com/questions/808678/…

A Googler

Nice but there the denominator has a quadratic without a constant term. If I delete the constant from my problem , WA says the integral does not converge.

MickLH

@AGoogler Is this a homework problem?

Agawa001

18:22

@AGoogler a question alike

A Googler

@MickLH It was on my test today.

Agawa001

see clearly that any polynomial can be written as $(x+t)²-y²$

A Googler

wait i can complete the square

yeah dumb me

MickLH

Ok then I'll share my result :P

$$\int_{0}^{\infty }{{{\log x}\over{x^2+2\,x+2}}\;dx}={{\pi\,\log 2
}\over{8}}$$

A Googler

that's right

@Agawa001 but then the formula doesn't seem to work? $a=1$ right?

@MickLH What method did you use?

MickLH

18:25

horribly unorthodox methods involving Li and limiting values of ArcTan

A Googler

@Agawa001 Besides what about the $x$ numerator? it isn't $x+1$

@MickLH Can it be done by series?

MickLH

I wouldn't know, I am a horribly unorthodox person

Krijn

How would I TeX that?

\mathcal{Cl}(K) $\mathcal{Cl}(K)$ does not do the trick

In LATeX

0celo7

$\mathcal{C}\ell(K)$

simple

MickLH

lol I thought that was an alpha in the picture

Krijn

18:31

Thanks, did not know $\ell$

MickLH

The contents of any one panel are dependent on the contents of every panel including itself. The graph of panel dependencies is complete and bidirectional, and each node has a loop. The mouseover text has two hundred and forty-two characters.

0celo7

Any Riemannian geometry nerds here?

Américo Tavares

18:50

15

Q: What is the purpose of this site?

What is (and what should be) the purpose of math.SE? As far as I can say, various users have different views on this question. Some users view it as a repository of knowledge. Some users approach it as teaching opportunity. Similar to previous, but slightly different: It could be understoo...

discussion

felipa

anyone know anything about matrix norms and can help with math.stackexchange.com/questions/1517103/… ?

Agawa001

@AGoogler did you find a solution or not yet ?

A Googler

@Agawa001 nope

Agawa001

@AGoogler try to post in main, a solution would be longer to fit here

make sure to use latex functionlities

user174558

I like my colour.

Joe Stavitsky

19:07

So our instructor says continuity does not apply to discrete stuff, but one of the books I have contains ctrlv.in/665297

Agawa001

@Jasper i like more the elephant appearing in your avatar !

Huy

@JoeStavitsky: where's the discrete stuff in that excerpt?

Joe Stavitsky

@Huy: my bad, it is from material on series and sequences

Huy

@JoeStavitsky: it just says that continuous functions respect limits, nothing else

that's a statement about functions which are continuous already in the first place

Joe Stavitsky

In other words, only referring to underlying function of the series?

Agawa001

19:17

@AGoogler where r u stuck

Mike Miller

@JoeStavitsky: Let $x_n$ be a convergent sequence, in particular suppose $x_n \to 0$. Now if $f$ is a continuous function - for instance $\cos$ - we know that $\lim_{n \to \infty} f(x_n) = f(\lim_{n \to \infty} x_n)$. In our particular case, this means $\lim \cos(x_n) = \cos(\lim x_n) = \cos(0) = 1$.

And $x_n = \sin(n)/n^2$.

MickLH

@AGoogler Sorry if it's teasing you, but I found a really fun generalization: $$\int_{0}^{\infty }{{{\log x}\over{a\,x^2+b\,x+c}}\;dx}={{2\,
\arccos \left({{b}\over{2\,\sqrt{a}\,\sqrt{c}}}\right)\,\log \left(
{{\sqrt{c}}\over{\sqrt{a}}}\right)}\over{\sqrt{a}\,\sqrt{{{4\,a\,c-b
^2}\over{a}}}}}$$

Agawa001

@MickLH you calculate so fast !!

MickLH

assuming $b > 0$ and $c > 0$ anyways

Joe Stavitsky

@MikeMiller, how do we know $(limn→∞f(xn)=f(limn→∞xn)$?

on related note, how can you input and copy/paste mathjax?

Mike Miller

19:21

@JoeStavitsky: This is true for every continuous function $f$. We're using continuity.

See LaTeX in chat on the right.

Agawa001

i proposed him to post in main and wait a stream of nice diverse solutions

MickLH

@Agawa001 I work with computer algebra tools lol

Mike Miller

Oh, copy paste? That's trouble.

Joe Stavitsky

ah ok

Agawa001

@MickLH which ?

Joe Stavitsky

19:22

that sounds like a fun project

MickLH

@Agawa001 maxima (MACSYMA spiritual successor)

Agawa001

seems powerful !!

MickLH

it is, I learned math from it after I dropped out of school lol, that's why my methods are so ridiculously unorthodox

it could probably crack the whole thing by itself really...

ouch... it can, but....

see for yourself: $$\int {{{\log x}\over{a\,x^2+b\,x+c}}}{\;dx}={{2\,\left(\log x\,
\arctan \left({{2\,a\,x+b}\over{\sqrt{4\,a\,c-b^2}}}\right)-{{\left(
i\,\log \left({{a\,x^2}\over{c}}\right)-2\,{\rm atan2}\left({{\sqrt{
4\,a\,c-b^2}\,x}\over{2\,c}} , -{{b\,x}\over{2\,c}}\right)-2\,i\,
\log \left({{x}\over{a}}\right)\right)\,\log \left(4\,a^2\,x^2+4\,a
\,b\,x+4\,a\,c\right)+2\,\arctan \left({{2\,a\,x+b}\over{\sqrt{4\,a
\,c-b^2}}}\right)\,\log \left({{a\,x^2}\over{c}}\right)+4\,i\,
{\it li}_{2}({{-2\,i\,a\,x+\sqrt{4\,a\,c-b^2}-i\,b}\over{\sqrt{4\,a

Agawa001

holy capuccino !!!

i must assume nothing apart a machine can come with that out

MickLH

Yeah it's horrible, but all I had to do was tell it that $4\,a\,c-b^2>0$ and 2 seconds later it had the result :P

Balarka Sen

19:28

@DanielFischer You're there?

A Googler

@Agawa001 Sorry , gotta sleep now. I'll post tommorow.

@MickLH Cool!

Agawa001

ok

Daniel Fischer

19:45

@BalarkaSen Depends. What for?

Ted Shifrin

Hi @DanielF

Daniel Fischer

Hi @Ted

Balarka Sen

There is a user (heh) intentionally spamming the algebraic topology chatroom, even after a warning (I'm a room owner there). I used kick-mute on him, but it turns out that just mutes him for 1 minute.

Is there a way to permanently mute that guy? Or should I flag it for mods when he does that?

hi @TedShifrin.

Daniel Fischer

@BalarkaSen I don't think you can permanently (or even for a longish time) mute him. Flag when somebody is behaving badly, and can't be reasoned into stopping.

Ted Shifrin

Hi Balarka

Balarka Sen

19:52

@DanielFischer Oh well. Thanks!

Loong

@BalarkaSen If you have to kick again, the ban duration will increase: meta.stackexchange.com/a/239226/271002

user143442

@TedShifrin I just wanted to apologize about yesterday

Balarka Sen

ah, thanks @Loong.

Ted Shifrin

Ok, user, now spend your time studying instead of accusing everyone of being homophobic.

user143442

Ok sorry

user143442

19:55

:(

0celo7

@TedShifrin hey

Balarka Sen

@TedShifrin It's useless to talk to him, just ignore. He's a serial troller.

I have him on ignore in this chat right now.

0celo7

Still stuck on that geodesic ball thing

user143442

I'm going to study, bye

0celo7

@TedShifrin I understand that I can get a finite cover of geodesic balls

Ted Shifrin

19:57

I have to go play duplicate bridge after I make lunch, so I'm not here for more than a few seconds.

0celo7

@TedShifrin But that doesn't prove that the minimum radius of the finite cover is actually a radius that works for all points

evinda

Hello @AméricoTavares
Could I ask you something about an inequality?

Balarka Sen

I never understood how to play bridge. It's confusing.

Have fun though!

Ted Shifrin

it's intellectually challenging, Balarka :)

Balarka Sen

inteklectually, indeed.

Ted Shifrin

19:59

Look at other books, @0celo.

I can't type on the iPad.

0celo7

I checked Lee and do Carmo.

Balarka Sen

@TedShifrin I have to learn how to play it someday.

0celo7

Neither have this result.

People in the physics chat haven't heard of it.

@TedShifrin do you have a specific book in mind?

Ted Shifrin

It's in the books. Look for injectivity radius.

0celo7

"the books"

looking

@TedShifrin do Carmo mentions it, but it's embedded in a chapter on the sphere theorem

@TedShifrin Is the result I'm looking for "the injectivty radius of a compact manifold has a nonzero lower bound"?

Agawa001

20:16

@BalarkaSen that same user was been ranting about ted, his comment was deleted but kickin him permanently can prevent more damages

Balarka Sen

what kind of damage?

Agawa001

personal

Balarka Sen

I don't care, I don't want my chatroom spammed.

Agawa001

i was banned because of him, and i thought he was sent specially to scew my mood

Alessandro

good evening everybody

Agawa001

20:18

g. evening

evinda

Could you take a look at my question?

0

Q: How do we get the right constant?

We consider the initial - boundary value problem $$u_t(t,x)=a(t,x) u_{xx}(t,x)-c(t,x) u(t,x) \forall t \in [0,T_f], x \in [a,b] \\ u(0,x)=u_0(x) \forall x \in [a,b] \\ u(t,a)=0=u(t,b) \forall t \in [0,T_f]\\ a,c \in C^1, a(t,x)>0, c(t,x) \geq 0$$ $$\tau=\frac{T_f}{N_t}, t_n=n \tau, n=0,1, \dots...

calculus inequality numerical-methods taylor-expansion

Mike Miller

21:40

@BalarkaSen: It is probably a bit much to suggest you carry out the construction I outlined. The hard part is extending the triangulation over the handlebodies. Let's look at the easiest case: 1-handles. You're gluing on $S^0 \times D^{n-1}$ to the boundary of $M$. To make this match up with the triangulation, you're asking that every triangulation of $S^0 \times D^{n-1}$ extends to a triangulation of $D^1 \times D^{n-1}$.

This is the same thing as saying that triangulations of $D^{n-1}$ are unique up to concordance. This is not trivial, but is possible to prove. But it's the simplest case. If $X_k$ is the "space of triangulations of $D^k$", then somehow this is saying that $\pi_0 X_{n-1} = 0$. Then being able to extend the triangulations of a $k$-handles is saying that $\pi_{k-1} X_{n-k} = 0$. Because I claim you can do this in any dimension for any handle, I'm asking you to prove that $X_k$ is contractible.

Balarka Sen

I'm unsure what space of triangulations of $D^k$ mean.

Mike Miller

Again, possible, not easy. No doubt this was done in or before the 60s. Wouldn't recommend spending a day on it. At best take my outline as a good reason to believe that intersection is dual to cup product

I put it in scare quotes for a reason.

But do you see the analogy between what I said and why that should be "$\pi_0 X_{n-1}$"?

Balarka Sen

Yes, because if you have two triangulations on $D^{n-1}$, it says you can extend the triangulation on $S^0 \times D^{n-1}$ (with two ends having those two triangulations) to a triangulation on $D^1 \times D^{n-1}$. So in a sense, triangulations on $D^{n-1}$ can be connected by paths in the space $X_{n-1}$.

Mike Miller

Yes. That's why what I said follows morally. To be rigorous about it, $X_k$ is some sort of simplicial set.

But you should take this as a motivating statement as opposed to a challenge. The point is that carrying out my outline would be technically challenging.

Balarka Sen

Right, it makes sense, but I have no idea how to make it rigorous. Thanks!

user174558

21:48

I wish user would stop coming to chat.

Mike Miller

(I think one can avoid all this by choosing the handles you're gluing along very carefully, and being willing to subdivide the original triangulation as you go. But, again, don't bother.)

Balarka Sen

maybe piece $D^1 \times D^{n-1}$ up by triangulating $D^{n-1}$ and $D^1$ both, then subdividing the prisms to get a triangulation, and then as you close on to the two ends, namely $S^0 \times D^{n-1}$, barycentrically subdivide like mad. I don't know, certainly I cannot make this rigorous.

Mike Miller

That one is easier than the other ones. It's just a matter of proving that any two triangulations of $D^{n-1}$ have a common subdivision.

Balarka Sen

Ah, ok.

Mike Miller

Which, again, not trivial, since it's not true for a general triangulation of a manifold.

$D^k$ is special.

Balarka Sen

21:53

@MikeMiller oh, right, Hauptvermutung.

Ashley Davies

Hi everyone; just hoping I can get a textbook recommendation. I'm really struggling with the basics of sequences & series, and now I've had questions incl. the convergence of the infinite sum of 1 / (i * ln(i) * ln(ln(i))) and have literally no idea what to do about it (I'm not seeking an answer to this question though). digital resources are fine and if you have no recommendation but know what field of maths would cover this kind of question that information would be excellent too! :)

Mike Miller

Here's how I would actually do this: I can always move around the places I'm gluing the handles. So pick the discs I'm gluing along to be a single $(n-1)$-simplex. It's pretty obvious how to extend the trivial triangulation of the disc to $D^{n-1} \times I$.

Do something similar, but still harder, for higher handles.

Balarka Sen

ok, that makes more sense.

why can't you do the same thing for higher dimensions? $D^k$ can be thought as a $k$-simplex.

Mike Miller

You don't know, or have forgotten, what a handle is. I'm gluing $D^k \times D^{n-k}$ to the boundary along $S^{k-1} \times D^{n-k}$.

So there is an obvious "model triangulation" of this, but the attaching map can be weird. I'm allowed to isotope it, but can I isotope it so that it coincides with an appearance of the model triangulation on the boundary? Think about $n=3$, $k=2$. The answer there appears to be "yes". But it should highlight the difficulties.

Balarka Sen

Although I have looked up the definition a couple of times, admittedly I have forgotten it everytime because I don't understand handlebodies. ok, I see.

Mike Miller

21:59

They take practice.

I feel like you're tricking me into making this argument rigorous.

Balarka Sen

No, haha, not really. I probably wouldn't understand the whole rigorous argument anyway. I am trying to understand the argument a bit better.

@MikeMiller oh, yes, that makes sense.

Mike Miller

The technical bit is "You can extend this triangulation to the whole manifold". Take that part for granted.

Balarka Sen

Yeah, I kind of assumed it in my "proof".

Mike Miller

The only proof I know fully worked out uses tools you don't have yet, and also takes more work to establish an intuitive understanding. Triangulations are good for that last part.

Balarka Sen

OK, thanks for telling me all this. It's very helpful.

Mike Miller

22:04

I suggest understanding this picture, while taking for granted the above assumption, to then understand the claim. Then come back another day some time from now to see the more standard proof.

Balarka Sen

Yeah, the person I talked to told me all kind of things about tubular neighborhood theorem and geodesics and Riemannian manifold. Yeesh.

Mike Miller

Don't know the proof they're thinking about. Tubular neighborhoods is half of what I was thinking of.

There's probably a different proof for most ways of thinking about manifolds.

Balarka Sen

@MikeMiller Which part do you think I should work in understanding? I can prove the duality between cup and intersection for $k + l = n$ (taking what you said as granted), but not sure when the manifolds are not of dual dimension.

Mike Miller

Oh, I mean, that's probably good enough. You really mostly need to know that it's true.

And you appreciate pictures.

Balarka Sen

Alright. Thanks. And the proof the person talked about was a souped-up version of the proof that every smooth manifold admits a triangulation, he said.

He put a Riemannian metric on the manifold first and then he covered the manifold up with balls (something geodesic), used the covering dimension and took the nerve of the cover. That's all I remember, although I understand nothing.

Mike Miller

22:09

Sounds reasonable. Most people prove the existence of triangulations with Morse theory, but that argument sounds like something one could do.

Balarka Sen

That person was prof, by the way. Admittedly he's more of a geometer than a topologist :P

0celo7

@MikeMiller You mind helping me with my geodesic ball problem?

Balarka Sen

Thanks again @MikeMiller.

Mike Miller

22:27

@BalarkaSen: I just checked the details, you definitely can extend those to the whole handle. It would be obvious if I could draw a picture. The key word is "Take the cone on the triangulation, then take the cone twice more." Try to see if you see what I mean in the $k=2$ $n=3$ case when you're trying to extend a triangulation of the cylinder to $D^2 \times D^1$.

@BalarkaSen I like interesting questions. No problem.

Balarka Sen

Hrm.

Taking cone thrice is getting a bit too crazy for me to visualize. I can see what you get when you take cone once. Unsure what you're trying to accomplish by taking cone two more times.

Mike Miller

@BalarkaSen: Look at the top half and the bottom half. I have a triangulation on the disc, which I want to extend over something homeomorphic to the cone of the disc...

This only works precisely for $k=2, n=3$. You have to work harder for other cases (as I just noticed with $k=1, n=3$). But the moral is correct. And I am going to stop checking the details on the rest.

Eric Tressler

Hi

Gridley Quayle

Any hints with the following problem if R(n) is the 'repeated form of n' (such that if n=12, r(n) =1212) then is R(n) ever a perfect square?

Eric Tressler

22:43

well, depending on the number n of digits x has, R(x) is x + 10^n*x

Gridley Quayle

I know that 10^k +1 divides r(n) if n has k digits so I was thinking of an argument involving digits

yeah, would it be prudent to consider the digits in the square of some x?

And check if its possible for the digits to match?

Balarka Sen

Ohh, I see it @MikeMiller

That's a pretty neat way to do it.

Eric Tressler

I was thinking more like find the factors of 10^n + 1. I'm not sure; I suspect it's either impossible for trivial reasons or else there's a small example

I guess 0 is the trivial example

If 10^n + 1 has no repeated factors, you're done. The first one with a repeated factor is

100000000001

you would need a 10^n + 1 with a repeated factor of 2 or 3

and it can't have either

since it isn't even and the decimal digits sum to 2

Jaken Herman

Can anybody help me with a problem I asked? It didn't get much love or attention after I posted it and I really can't wrap my head around it. It's calculus. math.stackexchange.com/questions/1519622/…

Gridley Quayle

We need 10^n+1 to have a prime factorisation with even powers. What does it not having factors of 2 and 3 matter?

Gridley Quayle

23:08

For odd n 10^n+1 is divisible by 11. I suspect its only once but I don't have a proof.

Tien-Cheng Huang

Could someone give me some exercises involving both analysis and abstract algebra on undergraduate level? Like this one mathb.in/46103

Simeon

Are there norms (other than the taxicab norm) which are linear in the sense that $||x+y|| = ||x||+||y||$?

Tien-Cheng Huang

@Simeon Thank you! Because I need to prepare some worked exercise to discuss with TAs in recitation :-)

Mike Miller

@Simeon I mean, that's not true of any norm. You need to be more precise. (Take $y=-x$.)

Simeon

@Tien-ChengHuang LOL! I was asking this as a question, but if you find it interesting, you're welcome.

skull petrol

23:21

Are you @MikeMiller going to participate in the hat activities this year?

user174558

23:33

@Tien-ChengHuang There are so many exercises in books.

Tien-Cheng Huang

@Jasper Could you suggest one book to me? I am on undergrad level, and the exercises in my analysis books are all about concepts in analysis only, and the exercises in my abstract algebra books are all about concepts in abstract algebra only too...

user174558

@Tien-ChengHuang I see. Sorry, I misunderstood your question, LOL.

Tien-Cheng Huang

And this one mathb.in/46103 is the only exercise involving both analysis and abstract algebra I have ever seen.

user174558

Sometimes, I think this site should not exist.

user147690

@Jasper Why?

user174558

23:44

@AlexClark Does it really help anyone?

skull petrol

Another site would exist in its place.

user147690

@Jasper It's helped me many times

user174558

@AlexClark OK. I know it helps you because you met me xD

user147690

@Jasper Exactly

user174558

It's a holiday tomorrow, Deepavali.

skull petrol

23:47

The Festival of Lights?

user174558

Yes. Krishna slays the demon.

skull petrol

We all have our own demons to slay.

user174558

@skullpetrol Have you found peace?

skull petrol

no

I was saddened about Nash and his wife's accident @Jasper

user174558

I was sad, but not too sad. When one is dead, one is dead.

user174558

23:53

It is not death that is frightening, but life.

user174558

I think when one dies, sadness mostly applies to his loved ones, not he himself, because he has died.

user174558

Heh, I think ELU should be renamed to EL @skullpetrol.

skull petrol

Death does put an end to one's pain. That thought can be comforting to the loved ones @Jasper

user143442

hello, it's me

skull petrol

Hi pal

user143442

23:56

hi pal

skull petrol

How are you?

user143442

sick of taking fluoxetine

skull petrol

Then slow down.

user143442

I'm gonna be my psychiatrist on Wednesday

skull petrol

Good plan pal :-)

user143442

23:59

but he gave me fluoxetine

user143442

last month

Transcript for

$Mathematics$ Mathematics