Mathematics

Akiva Weinberger

1:00 PM

Y suena como algo que yo haría a los francehablantes

:P

Astyx

I'll be back

DHMO

@AkivaWeinberger :/

@Astyx bis bald

Akiva Weinberger

@DHMO I mean, were they reporting your comments?

DHMO

@AkivaWeinberger I don't know whose comment was reported

Akiva Weinberger

@DHMO I wouldn't actually do it, that was just a joke

DHMO

1:01 PM

@AkivaWeinberger yo lo se

Akiva Weinberger

Ah, I see (clicked on the link)

DHMO

@AkivaWeinberger a-vlá-mo-sa-sí-pa-ra-ke-la-gén-te-no-pwé-dau-sar-Google Translate :p

user84215

I read somewhere that singularities in geometric theories may give rise to examples of non-factoriality. Can you explain it ?

user84215

any suggestion ?

user84215

1:18 PM

I wonder that nobody is interested to my questions.

DHMO

@aminliverpool nobody here knows the field involved in your question

if you asked about ordinals (like in here) then more people here can answer you

ccorn

Projective geometry: Does anyone know a reference for the following: Given a reference triangle with Steiner point $S$ and a point $E$ at infinity, then $E$'s isogonal conjugate and $E$'s isotomic conjugate are collinear with $S$. This fell out from this thread, and a reference might be a useful addition.

Astyx

@DHMO The tense rule for "sembler" is really complicated

DHMO

@Astyx it isn't a tense; it's a mood

Astyx

Right

DHMO

1:25 PM

tense (English) < tens (Old French) > temps (French)

Astyx

Oui je sais

DHMO

let's continue

36 mins ago, by DHMO

Can you give me a sequence $(b_n)_0^\omega$ of open intervals whose lengths sum to a value strictly less than 1

36 mins ago, by DHMO

but which fully covers $A$?

user84215

lets do research altogether

Astyx

Not right now @DHMO

Anonymous

@DHMO what is difference between +E and -E effect?(in short) E for electromeric effect

DHMO

1:34 PM

@Anonymous context?

@DHMO chemistry

no idea

@DHMO now??

> Electromeric effect can be classified into +E and -E effects based on the direction of transfer of the electron pair.

When the electron pair moves towards the attacking reagent, it is termed as the +E effect. The addition of acids to alkenes is an example of the +E effect. After the transfer takes place, the reagent gets attached to the atom where the electrons have been transferred to.

The -E effect can be found in reactions such as addition of cyanide ion to carbonyl compounds. In these reactions, the electron pair moves away from the attacking reagent.

Electromeric effect

Electromeric effect refers to a molecular polarizability effect occurring by an intramolecular electron displacement (sometimes called the ‘conjugative mechanism’ and, previously, the ‘tautomeric mechanism’) characterized by the substitution of one electron pair for another within the same atomic octet of electrons. However, this term is now considered and this effect is considered along with the inductive effect. This effect is shown by those compounds containing multiple bonds. When a double bond or triple bond is exposed to an attack by a reagent, a pair of bonding electrons involved in the...

kayak

Hey @DHMO Long time no see

DHMO

1:39 PM

annyeong @kayak

Anonymous

I need to upload picture :/ I can't

BAYMAX

Guys quick question: is $\frac{\sin(x)+\cos(x)}{exp(x)}$ always less than 1 ?

DHMO

@BAYMAX desmos.com/calculator/pnhxclnezl

Anonymous

exp (x) means?

DHMO

@Anonymous $e^x$

Astyx

1:41 PM

@Baymax I doubt so

As $x\to-\infty$

anonymous

I need to change my name I guess..I get pinged everytime the other Anon is pinged :P

Suggest me a new name someone :'D

not_anonymous

lol :P

@anonymous anaam

@DHMO means ? :D

1:42 PM

@anonymous I thought you speak Hindi

anonymous

@DHMO Oh, that word seems weird when typed in English

In Hindi the pronunciation is quite different :)

DHMO

sure

[ənaːm]

anonymous

I will think of some more and change it tonight :)

Haha :D

DHMO

ohne Name

(without Name)

user84215

lets do research altogether

BAYMAX

1:46 PM

anonymous how about your name to be bnonymous !! he he

anonymous

changed it to 2anonymous :P

it begins with a digit

DHMO

@2anonymous hi

BAYMAX

@DHMO yeah it blows up aswell as down..

@Astyx yeah it blows to infinity for this case

Anonymous

anonymous

@Anonymous What's the question?

math.stackexchange.com/users/404484/2017?tab=profile

Anonymous

1:48 PM

@DHMO so isn't in ex: of -E attacking reagent is getting attached?

DHMO

@Anonymous I don't know

Anonymous

@anonymous sorry I took your name,once I will get banned you will be alone with your name

anonymous

@Anonymous It is just like nucleophilic addition rkn

I changed my name, why isn't it appearing !

In chat

Maybe cache is not cleared

DHMO

@anonymous you need a moderator to refresh

Anonymous

I don't know how it is nucleophilic addition reaction @anonymous

anonymous

1:52 PM

@Anonymous CN- acts as a nucleophile here....

It is negatively charged

It seeks positive centres..

Any Mods...please refresh my chat profile ! :)

Anonymous

@anonymous @DHMO I got it! , CN is attaching to left side but electrons are moving to right side!

Semiclassical

hi chat

Anonymous

Hi

anonymous

@Anonymous Exactly. When the negative charge shifts to the right then a positive center is created...

user84215

I read somewhere that singularities in geometric theories may give rise to examples of non-factoriality. Can you explain it ?

Anonymous

1:59 PM

@anonymous math.stackexchange.com/questions/2120668/… in this question is it differentiation of LHS = RHS?

anonymous

Oh that was @DHMO's question

Secret had solved it if iirc

Anonymous

Wait,I will try first,tell me is it differentiation of LHS = RHS?

anonymous

If forgot the solution

@Anonymous What? It is a differential equation...

$y'=dy/dx$

Anonymous

Problem needs integration concept?

anonymous

@Anonymous You need to solve for y...

Do you know how to solve DE's ?

Anonymous

2:02 PM

No

anonymous

Then I'm afraid you won't be able to solve it unless you know the basics

Anonymous

I don't know integration… i can do differentiation

anonymous

For solving DE you need to know both...

Why not learn integration and the basics of DE and then come back to the problem?

Anonymous

OK,that's not my business then

anonymous

Hmm

BAYMAX

2:16 PM

Guys i am unable to find resources about the order of convergence of fixed point method , any help ?

Mats Granvik

2:35 PM

The alternating sequence +1,-1,+1,-1,+1,-1,... shares essentially the same recurrence over the divisors as the characteristic function of twin primes, but with different starting values.

Alaa Mohamed

for the topic of y=fx anyone know how to find the image of (3,1) under the transformation 3f(3x)

DHMO

let g(x)=3f(3x)

we know that f(3)=1

so g(1) = 3f(3) = 3

so the image is (1,3)

Balarka Sen

I don't actually know how to write down the Freedman's E8 manifold

But I understand that's not what you are asking.

Danu

2:50 PM

What do you mean by write down?

Balarka Sen

How's it constructed?

Danu

derp, I accidentally deleted instead of edited

Repost: @MikeMiller Do you know/is there a not-terribly-hard-to-understand (simply connected) topological 4-manifold that realizes the $E_8$ Cartan matrix as its intersection form?

@BalarkaSen I think that's what I'm asking too

I know how to do all of them except $E_8$... I suppose it's hard.

Balarka Sen

Well then you'd probably want simply connected

Danu

Yeah, of course

Balarka Sen

Simply connected ones are uniquely determined by intersection form, not in general.

OK, got it

Danu

2:53 PM

Yeah

Balarka Sen

I'd like to hear an answer to that too

Danu

I have no idea about the non-simply connected ones (not that I have any idea besides quoting Freedman about simply connected ones; best I can do is mostly understanding Whitehead-Milnor modulo a lot of important details)

eurocoder

Say I have f(a, b, n) = |a - b|^n where a and b are complex numbers, and n is an integer greater than 0 - Is it possible to expand f(a, b, n) as a series/summation somehow?

Balarka Sen

I wish I knew anything about manifolds :(

Danu

Shaddap :P

You shouldn't be complaining---with all that you already know now, you have so much extra time compared to the rest of us!

Balarka Sen

3:02 PM

we'll die anyway

there's no time. so let's spend whatever's there procrastinating

Mike Miller

3:23 PM

No, there's no easy example.

Danu

@MikeMiller :'(

Zach Hauk

4:05 PM

hi and bye

Random Variable

4:53 PM

@DanielFischer When you have time, could you please explain why the asymptotic expansion of the Bessel function of the first kind does not provide enough information to determine if (and when) the integrals on the vertical sides of the contour vanish in the limit? Doesn't the expansion tell us that $J_{1}(az) = J_{1}(a(x+iy))$ remains bounded as $x \to \pm \infty$ for fixed values of $y$? Is this apparent fact not useful? I've been thinking about this question nonstop for hours.

Semiclassical

@Danu would that be the same as an E_8 singularity?

I don't have an answer regardless but I wanted to check if I had the terminology right

BAYMAX

5:28 PM

Guys .. $(1-x)^-1 = 1 + x + x^{2} + x^{3} + ......$

Semiclassical

Yes. (If $|x|<1$.)

BAYMAX

But how to do this $(1+x)^{-\frac{1}{2}}$

Intuitively

Semiclassical

Binomial theorem.

BAYMAX

yes @Semiclassical

But i think about the first one as a Geometric series sum ... like this how can i write this one in easier way instead of calculating the coefficients by Binomial theorem

Semiclassical

Who says you can?

BAYMAX

5:31 PM

?

Semiclassical

The geometric series one is an especially nice case of the binomial theorem.

Why should $(1+x)^{-1/2}$ have so simple an explanation?

Mike Miller

@Danu Actually I guess this probably doesn't involve the full force of Friedman's work. You start by taking the solutions to $x^2 + y^3 + z^5 = 0$ inside $\Bbb C^5$. Intersecting this with a small sphere around zero you get the Poincare homology sphere $\Sigma(2,3,5)$. The variety is singular at zero, but you can resolve the singularity by taking what's called the "Milnor fiber".

Semiclassical

oh hey, milnor fibration

I sorta (read: mostly don't) know what those are.

Mike Miller

Doing so gets you a compact 4-manifold with boundary $\Sigma(2,3,5)$ and intersection form $-E_8$.

Now all we need is to get a contractible manifold whose boundary is $\Sigma(2,3,5)$ and we're done.

BAYMAX

yes , but i thought it me might be there and i don't know that one but it's ok

Semiclassical

5:34 PM

I think a counting interpretation of the series for $1/\sqrt{1+x}$ can be found, but it requires some work and isn't obvious.

BAYMAX

ok..

nice

Semiclassical

By contrast, a counting interpretation for $1/(1-x)$ is obvious.

Mike Miller

it is a theorem of Freedman that for every homology sphere $Y$ there is a proper smooth embedding of $Y \setminus \{pt\}$ into some exotic $\Bbb R^4$. Taking one point compactifications you get an embedding of $Y$ into some space homotopy equivalent to $S^4$. It is a theorem of Kirby that this embedding is automatically locally flat.

And it is Freedman's theorem that this space homotopy equivalent to $S^4$ is indeed $S^4$.

BAYMAX

Yes @Semiclassical

Mike Miller

I guess you do need Freedman's big machine to do this. Oh well.

Anyway, the complement of your locally flat embedding of $Y$ in $S^4$ is a homology ball but I'm pretty sure it's also just contractible.

So you can glue that thing onto your original thing and be done with it.

@BalarkaSen See above.

Semiclassical

5:39 PM

About the only possibility I see off the top of my head is the following. Writing $y(x)=1/\sqrt{1-x}$, we have $$y(-2x)=1/\sqrt{1+2x}=1+x+\frac32 x^2+\frac52 x^3+\cdots = 1+x+3\frac{x^2}{2!}+15\frac{x^3}{3!}+\cdots$$

(making use of Mathematica to facilitate the computation)

If I plug the coefficients 1,1,3,15... in at OEIS.org, it indicates that these are the double factorials of odd numbers.

So evidently $\displaystyle y(x)=\sum_{k=0}^\infty (2k+1)!!\frac{(-2x)^k}{k!}$

Danu

@Semiclassical No, I think :P

Semiclassical

As to why that should be the case...oh look, a convenient distraction! exit stage left

Danu

Well, maybe it is cf. Mike's stuff

Thanks for your extended reply @Mike

Way over my head though

MickLH

5:57 PM

Guys. Who studies modular subset product problems?

I feel like this path should already be well beaten, where da asymptotics at yo?

Balarka Sen

@MikeMiller Fun, I didn't know one could make that singular variety into a nullcobordism.

Mike Miller

Milnor map

In mathematics, Milnor maps are named in honor of John Milnor, who introduced them to topology and algebraic geometry in his book Singular Points of Complex Hypersurfaces (Princeton University Press, 1968) and earlier lectures. The most studied Milnor maps are actually fibrations, and the phrase Milnor fibration is more commonly encountered in the mathematical literature. The general definition is as follows. Let f ( z 0 , … , z n ...

Balarka Sen

6:13 PM

Hmm, I am not sure if I get the picture. Eg, for $zw = 0$ in $\Bbb C^2$ the link is a Hopf link in $S^3$. What's the fibering of the complement there?

Mike Miller

It's the decomposition of $S^3 \setminus L$ into pieces corresponding to $\text{arg}(zw)$.

Topologicalife

Hi.

Balarka Sen

Ah, ok. So I guess the fibers are Siefert surfaces of the Hopf link?

Mike Miller

Yes, that's what we mean for any fibered knot.

Balarka Sen

Ok, understood. So it indeed makes sense that the closure of the fibers are the links. So one just remove a small nbhd of the singularity and caps off with that.

Nice.

Topologicalife

6:17 PM

If $f$ is a decreasing function defined on $(a,b)$ and we divide $(a,b)$ in $n$ subintervals, then $x_i = \inf \left(\dfrac{i-1}{n}, \dfrac{i}{n}\right)$ is the right hand end-point of the interval $ \left(\dfrac{i-1}{n}, \dfrac{i}{n}\right)$, i.e: $x_n = \dfrac{i}{n}$, right?

Mike Miller

@BalarkaSen Did you see my ping?

Balarka Sen

Is there a non-technical reason why that nullcobordism of $\Sigma(2, 3, 5)$ has intersection form $-E_8$? I'm prepared to believe what you said at the end.

No, I didn't. I just looked and I have it now.

Mike Miller

nah.

LeGrandDODOM

@DanielFischer Given $H$ a Hilbert space and $C$ a closed convex subset of $H$, I want to find the Frechet derivative of $x\mapsto (d(x,C))^2$. I found that $$d(x+h,C)^2=d(x,C)^2+2\langle x-p_C(x),h\rangle + 2\langle x-p_C(x),p_C(x)-p_C(x+h)\rangle + \|p_C(x)-p_C(x+h)+h\|^2$$ I'm thinking the derivative is $h\mapsto 2\langle x-p_C(x),h\rangle$ but I can't prove that $2\langle x-p_C(x),p_C(x)-p_C(x+h)\rangle + \|p_C(x)-p_C(x+h)+h\|^2$ is $o(\|h\|)$. What do you think ?

Balarka Sen

Also an exotic R^4 is homeomorphic to R^4; taking one point compactification should give something homeomorphic to one point compactification of R^4, aka S^4, not? Not sure why you say it's just homotopy equivalent to S^4.

MickLH

6:30 PM

Lets say I found a paper claiming exactly a result I'm looking for, but it's 17 pages of solid Pi Sigma Integrate Sigma Pi Integrate Integrate Sigma Pi Pi Sigma Pi... Am I a terrible person if I don't rigorously verify their argument, but just build on it instead? (The result is intuitively obvious, imo)

Daniel Fischer

@RandomVariable I haven't yet considered the integrals, I've so far only thought about the series. For fixed $y$, the integral over the vertical sides tends to $0$ as $\lvert x\rvert \to \infty$, provided we stay away from the poles of $\cot$. Then the $\cot$ factor remains bounded, the Bessel factor remains bounded (it even tends to $0$), and the $1/z$ makes it vanish in the limit.

Mike Miller

@BalarkaSen Freedman actually proved that it was proper homotopy equivalent to $\Bbb R^4$. He didn't know it was homeomorphic before the big paper.

Balarka Sen

Ahhh. Got it.

LeGrandDODOM

Daniel Fischer

@LeGrandDODOM Once you know that $p_C$ is Lipschitz, you have $\langle p_C(x) - p_C(x+h), x-p_C(x)\rangle + O(\lVert h\rVert^2)$. Then you need that $p_C(x) - p_C(x+h)$ is "approximately orthogonal" to $x - p_C(x)$ for small $h$. I'm not sure how to show that off the top of my head.

Random Variable

6:44 PM

@DanielFischer That was exactly the argument I was using to justify that the integrals vanish on the vertical sides of the contour as $N \to \infty$. Where then in the evaluation does the value of $a$ come into play?

LeGrandDODOM

@DanielFischer I concur. I don't know how to prove that either though

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$Mathematics$ Mathematics