Mathematics - 2017-02-26 (page 4 of 5)

Akiva Weinberger

13:00

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

13:01

@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

13:18

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

13:25

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

13:34

@Anonymous context?

Anonymous

@DHMO chemistry

DHMO

no idea

Anonymous

@DHMO now??

DHMO

> 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

13:39

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

13:41

@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

Astyx

not_anonymous

anonymous

lol :P

DHMO

@anonymous anaam

anonymous

@DHMO means ? :D

DHMO

13:42

@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

13:46

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

13:48

@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

13:52

@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

13:59

@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

14:02

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

14:16

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

Mats Granvik

14:35

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

14:50

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

14:53

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

15:02

we'll die anyway

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

Mike Miller

15:23

No, there's no easy example.

Danu

@MikeMiller :'(

Zach Hauk

16:05

hi and bye

Random Variable

16:53

@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

17:28

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

17:31

?

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

17:34

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

17:39

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

17:57

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

18:13

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

18:17

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

18:30

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

18:44

@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

Mahmoud

Hello there !

Daniel Fischer

@RandomVariable No idea yet. But if the numerical evaluations of the series are to be trusted, it plays a role. And looking at the differentiated series supports the point where things happen being $2\pi$.

Mahmoud

Can I please have some intuition behind this ? I know it didn't spawn randomly out of nowhere, so it must have some geometric intuition : $$\cos(a)+\cos(b)=2\cos(\frac{a+b}{2})\cos(\frac{a-b}{2})$$

Daniel Fischer

@LeGrandDODOM The "approximately orthogonal" part, the Lipschitz part, or both?

Topologicalife

18:58

Hi.

Is there a way to compute $\displaystyle \lim_{n\to \infty}\sum_{i=1}^{n} \dfrac{2}{n}\dfrac{1}{\left (-3+\dfrac{2i}{n}\right)^2}$?

Daniel Fischer

@Mahmoud Did you mistype? Shouldn't it be $\cos a + \cos b$ on the left?

Mahmoud

And it's more complicated friends of strange structure, please.

@DanielFischer Yes, thank you.

I don't want to memorize it $:($

Daniel Fischer

@Topologicalife It's a Riemann sum. For what integral?

Sha Vuklia

Consider the vector space $(\mathbb R,\mathbb R^{\mathbb N},+)$ of all sequences in $\mathbb R$. Consider the following subset:
$$
\{(x_n)_n\in\mathbb R^{\mathbb N}|\text{ only finitely many components $x_i\neq 0$}\}.
$$
I can verify that this is a subspace of $\mathbb R^{\mathbb N}$. However, my book says that this subspace is isomorphic with a "well known vector space". Which one would that be?

I can imagine that the dimensions of this vectorspace is countably infinite. I know the following vectorspaces: $\mathbb R^n,\mathbb R^{n\times m},\mathbb R[X]_{\leq n},C[X]$, and so on. But I don't know which one they're referring to.

Topologicalife

@Daniel yes.

For the integral $\displaystyle \int_{-3}^{-1} \dfrac{1}{x^2} dx $

Sha Vuklia

19:03

Actually, never mind. I'll post it on the forum (unless someone really wants to explain it here!)

Daniel Fischer

@Topologicalife More directly, for $$\int_0^1 \frac{1}{(-3+x)^2}\,dx.$$ But that's the same in green of course. So you evaluate that integral.

Semiclassical

@Mahmoud Ah, the product-to-sum identity. Yeah, that's a weird one.

Mahmoud

@Semiclassical Any ideas ?

Random Variable

@DanielFischer The formulas themselves don't make sense for large values of $a$. So we don't actually have to rely on numerical approximations to know that a restriction of some sort is necessary. We have a true mystery on our hands.

Topologicalife

@Daniel I will have a similar difficult evaluating the sum

LeGrandDODOM

19:08

@DanielFischer The "approximately orthogonal" part

Daniel Fischer

@Topologicalife Evaluating the sum is hard. Evaluating the integral is easy.

Semiclassical

Sure.

Start by considering what happens if you replace $a\to a+2\pi$ on the right-hand side of the identity.

Topologicalife

I need to evaluate the sum @Daniel :P

Nvm, I got an idea.

Daniel Fischer

@LeGrandDODOM I'd make a sketch and see whether that helps finding an argument.

@Topologicalife You asked for the limit of the sums, that's the integral.

Topologicalife

Yeah, sorry @Daniel, I was looking to evaluate the sum.

for*

i.e, I'm asking to evaluate the integral finding the riemann sum

Mahmoud

19:16

@Semiclassical It becomes mysteriously, $2\cos(\frac{a+b}{2})\cos(\frac{a-b}{2})$ $?!!$

Semiclassical

It should become $2\cos(\pi+\frac{a+b}{2})\cos(\pi+\frac{a-b}{2})$.

Yep. It returns to itself.

So the right-hand side is $2\pi$-periodic in $a$.

Mahmoud

But what's the essence ?

Semiclassical

And since that function is symmetric in $a,b$ you also have that it's $2\pi$-periodic in $b$.

Mahmoud

$a,b$ ?

Semiclassical

Sure. If you swap $a$ and $b$ then the function stays the same since $\cos(\frac{a-b}{2})=\cos(\frac{b-a}{2})$.

So the right-hand side is symmetric and $2\pi$-periodic in both variables.

Mahmoud

19:20

That's very nice.

Semiclassical

That suggests, though it doesn't prove, that the right-hand side can be written as $f(a)+f(b)$ where $f(\cdot)$ is $2\pi$-periodic.

If you plug in $a=b$, that becomes $2f(a)=2\cos(0)\cos(a)=2\cos a\implies f(a)=\cos a$.

Well, I mean that one way to get a function that's symmetric in a,b and $2\pi$-periodic in both is to guess that it's just additive.

If you assume that's the case, then the next line shows that it'd have to be $\cos a+\cos b$.

The only question then is to prove that that's actually true.

For an actual derivation I'd use the angle addition formula.

Another argument I like relies on differential equations, but if you're not versed in those then it won't help. @Mahmoud

Mahmoud

@Semiclassical Sadly.

Semiclassical

Mmkay.

There's probably a way to infer the sum-to-product identity from the main image here, if you relabel the angles: cut-the-knot.org/triangle/SinCosFormula.shtml

But honestly I don't remember the sum-to-product or product-to-sum identities so well as the angle addition formulas.

Mahmoud

@Semiclassical And you don't feel like you have to ?

Semiclassical

Yeah. Knowing them is definitely useful in some applications, but if I need them I can just derive them from the angle addition formulas.

Or using Euler's formula.

Mahmoud

19:31

Hmm

Danu

Derive them every time using Euler's formula haha

2

Mahmoud

Let me try,

Semiclassical

Euler's formula is probably the best way once you've got it under your belt, yeah.

Danu

There's a very small range of time (penultimate year of high school through middle of 2nd year of uni) when knowing them by heart is useful

Mahmoud

That's ironically my case, lol.

Balarka Sen

19:33

@Semiclassical Me neither!

Mahmoud

But I'm still not going to surrender to ''Learn Math by heart people'', even if that's going to hurt my grades.

iled

Hi all! I am studying numerical methods and have a basic question on some steps I am not understanding. Consider an q-stage IRK method with associated collocation scheme $\phi \in [\mathbb{P}_q] ^m$. We have the following identity:

$$ (\phi_i (t_{n+1}))^2 = (\phi(t_n))^2 + 2 \int_{t_n}^{t_{n+1}} \phi_i(t) \phi_i'(t) dt $$

Mahmoud

So $\cos x=\frac{e^{ix}+e^{-ix}}{2}$

But heh, I got stuck pretty fast, $\frac 12 (e^{ix}+e^{iy}+e^{-ix}+e^{-iy})$

iled

Sorry, hit enter by mistake. Here is the rest.

and we want to prove that $\mid u_n \mid = 1 ~\forall n \in \mathbb{N}$.

It follows like this

$\sum u_{{k+1}_i}^2 = \sum u_{k_i}^2 + \sum~2 \int_{t_n}^{t_{n+1}} u_{k_i} u_{k_i}' dt =\\
= 2 \int_{t_n}^{t_{n+1}} \sum u_{k_i} u_{k_i}' dt =\\
= 2 \int_{t_n}^{t_{n+1}} u_k^T u_k'i dt =\\
= 2 \int_{t_n}^{t_{n+1}} u_k^T B u_k dt$

In the last two steps, I do not understand how we got rid of the sum and the transpose comes up, and then B.

Mahmoud

Now what ?

MickLH

19:40

@Mahmoud Now we party. (Sorry lol couldn't resist. I just made good progress towards the result I want! :D)

Mahmoud

lol

But how can I simplify the rest ? I'm stuck.

Topologicalife

@DanielFischer I solved it (if you are interested).

Mahmoud

It's okay :D

Daniel Fischer

@Topologicalife How?

Semiclassical

@Mahmoud Start from the product side.

It's not immediately obvious what to do with the sum side, but the product side is transparent.

LeGrandDODOM

20:10

@DanielFischer I did some drawings but I can't come up with any intuition towards a bound in $o(\|h\|)$

Simple

20:50

Let $X(\geq 0)$ have probability generating function $G$ and write $t(n)=P(X>n)$ for the "tail" probabilities of $X$. Find the generating function of the sequence $\{t(n):n\geq 0\}$ is $T(s)=(1-G(s))/(1-s)$

Daniel Fischer

@LeGrandDODOM Step 0: If $x\in C$, the derivative at $x$ is clearly $0$ since $d(y,C)^2 \leqslant \lVert y-x\rVert^2$. Step 1: For $x\notin C$, consider the hyperplane $H$ perpendicular to $x - p_C(x)$ through $p_C(x)$. $C$ lies in the closed half-space determined by $H$ opposite to $x$. Now consider an $h$ such that $\lVert h\rVert \ll \lVert x - p_C(x)\rVert$, and draw a circle with centre $x+h$ and radius $\lVert (x+h) - p_C(x)\rVert$. $p_C(x+h)$ must lie in the corresponding ball.

The angle at which the circle (the sphere) intersects $H$ can't be large if $\lVert h\rVert$ is small.

Semiclassical

21:14

$X$ is a discrete variable? @Simple

To show it, start by writing out $T(s)$ and consider what $(1-s)T(s)$ would be.

Simple

yes

Semiclassical

(start by writing out the first few terms, I mean)

Simple

what is "tail" probability? I don't understand what does that mean

Semiclassical

It's just a definition. It's the probability that $X$ is bigger than some specified threshold $n$.

If you draw out the probability distribution (as a bar graph, say) it amounts to asking how much of the distribution is to the right of some specific. The distribution to the right of the cutoff is the tail.

Simple

If I just apply the definition of generating function, I will have $\sum_{n\geq 0}s^nP(X>n)=\sum_{n\geq 0}s^n(1-P(X\leq n)=\frac{1}{1-s}-\sum_{n\geq 0}s^nP(X\leq n)$

Semiclassical

21:28

Last should be $X\leq n$, but otherwise that's true.

But I don't think it's especially helpful. Try doing $(1-s)T(s)$ directly (i.e. write out the first few terms of $T(s)$ and multiply it out).

Simple

I see, it just like proving the sum of geometric series

Semiclassical

It's similar, yes.

For a geometric series, though, you'd have $$(1-x)(1-x)^{-1}=(1-x)(1+x+x^2+\cdots)=1+(1-1)x+(1-1)x^2+\cdots =1.$$

Here the difference between terms isn't quite so simple.

Simple

I get this

Vrouvrou

@DHMO hello, please i don't understand the lst step

Ali Caglayan

Hey @BalarkaSen are you around?

Balarka Sen

21:41

I am

Ali Caglayan

Do you know of a way to imagine S^3 as a smash product of a circle and a sphere?

I am trying to visualise it as S^1 x S^2

then I have a copy of S^2 sitting in that

Balarka Sen

Do you know how to see S^2 as smash product of S^1 and S^1?

Ali Caglayan

wedged with a copy of S^1

Yes

Mahmoud

How is smashing Math ?

Balarka Sen

Then you just generalize that, @Ali

Semiclassical

21:42

smash product is a math operation.

Mahmoud

Oh, my bad.

Ali Caglayan

I am trying but I am not really sure how to shrink the s^1 to a point

Its easy for S^1 and S^1 because we are shrinking lines

Semiclassical

As a point of reference, a Cartesian product of sets A,B is all possible ordered pairs (a,b) with a from A and b from B

Balarka Sen

The point is if you give S^n and S^m the usual cell structures with one vertex and an n (or m resp) cells then S^n x S^m has got a point, an n-cell, an m-cell and a n+m cell in the product CW complex.

arctic tern

@AliCaglayan visualize $S^3$ as $\Bbb R^3\cup\{\infty\}$, then consider all spheres in $\Bbb R^3$ tangent to the $xy$-plane at the origin (including the plane itself as a "generalized" sphere and the origin itself).

Balarka Sen

21:43

Then pinching S^n v S^m gives you D^(n+m) with boundary pinched to a point

That S^(n+m). Done

Semiclassical

A smash product is a specific way of taking two manifolds and getting a third one as output from them.

Balarka Sen

Not manifolds

CW complexes

Semiclassical

Ah.

I don't know smash products, so I didn't know better beyond the obvious "two come in, one comes outo"

Balarka Sen

also, yes, pointed CW complexes

Mahmoud

Can't wait to learn all the fancy terminology, or am I ?

Ali Caglayan

21:46

@arctictern your visualisation makes a bit more sense actuallly

Mahmoud

Anyway I was going to ask what would you do if you had this question at first glance, $$\lim_{x\to -\pi /_4}\frac{1+\tan(x)}{\sin(x)+\cos(x)}$$

arctic tern

yes, you can do the same thing with $S^2$ and all the other spheres. stereographic projection.

Semiclassical

Simplify.

Balarka Sen

this is the pencil of circles thing arctic likes a lot

Mahmoud

I was shocked that simply replacing $\tan$ with $\sin \over \cos$ solves it.

Semiclassical

21:49

Yeah, it's easy.

Ali Caglayan

Smash is just the tensor product in the category of pointed spaces right

Balarka Sen

@AliCaglayan Also if you want to do it that way, if you shrink S^2 to a point in S^2 x S^1 you get S^3 with two points identified in it

Ali Caglayan

so its left adjoint to hom and stuff

Balarka Sen

who cares

Ali Caglayan

but thats about all the nice properties you can get out of it I think

if you restrict to CW complexes then it acts nicer

Balarka Sen

21:53

nobody really cares about smash except smashing with circles

LeGrandDODOM

@DanielFischer I get it, thanks again! Your writing made me think of something geometrically obvious: if $H$ is some hyperplane, $x\in H^+$ and $y\in H^-$ then $\|x-p_H(x)\|\leq \|x-y\|$. Can you give me a hint on how to prove that ?

Ali Caglayan

I guess its also useful for constructing higher dimensional projective spaces

but I will have to think about that more

Balarka Sen

I don't see why you'd think that

Ali Caglayan

Not a useful construction then

Balarka Sen

Smashing with circles is useful, also known as the reduced suspension.

Also, yes, you're right that it makes a certain category of topological spaces into a tensor category

Daniel Fischer

21:57

@LeGrandDODOM What can you say about $\langle x - p_H(x), p_H(x) - y\rangle$? (Its real part if we're looking at a complex Hilbert space.)

Balarka Sen

It seems you want the category of compactly generated spaces

Ali Caglayan

Where do reduced suspensions come up

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$Mathematics$ Mathematics