Mathematics - 2019-05-19

Balarka Sen

1:44 AM

Hi @loch, @Ted

Ted Shifrin

hi a @Balarka

Is it actually morning there?

Balarka Sen

Yeah 7:12 AM lol

Ted Shifrin

Must be 7:15 ... the half hour always throws me off.

Balarka Sen

About right

Regarding your comment here, $[Y]$ would in general be represented by a continuous image of a delta complex, and it seems believable that you can write any delta complex as quotient space of a single simplex.

loch

Hi @BalarkaSen @TedShifrin

Ted Shifrin

1:49 AM

Hmmm, how do I give the fundamental class of a many-holed torus with one?

hi @loch

Or maybe I need something higher-dimensional.

Balarka Sen

@TedShifrin The surface of $g$ is quotient space of some fundamental polygon, so no problem.

Map a simplex homeomorphically to the polygon.

Ted Shifrin

This doesn't sound believable in general.

how do you even do $\Bbb CP^n$?

Balarka Sen

That's quotient of a sphere, which is quotient of a disk - map a simplex homeomorphically to a disk!

You can even do that simplicially.

I'd like to see an example of a delta complex $D$ such that there is no simplicial quotient map from a triangulation of $\Delta^n$ to $D$. I think this is always true.

Ted Shifrin

OK, we need something with a lot more topology.

I've never heard of such a thing.

Balarka Sen

Me neither but sounds like a cool question

Ted Shifrin

1:54 AM

Meanwhile, the OP was confuzled by my comment.

Perhaps you can aid him.

Balarka Sen

Intuitively it feels that you can break $D$ apart into simplices and rearrange them appropriately to get a triangulated $\Delta^n$ with lesser identifications.

skullpetrol

@TedShifrin are you planning to teach any more AoPS courses in the near future?

Balarka Sen

@TedShifrin I can write an answer but then I have to sneak it under the rug that complex projective varieties admit triangulations ;)

I don't actually know a proof of that. If one can get a Whitney stratification, that's enough, by a theorem of Goresky. I don't know how to Whitney stratify a general complex projective variety: the singular stratification need not be one (eg the Whitney umbrella). I think one has to pass to some appropriate finer stratification than the singular one, but I don't know the details at all

Silent

2:34 AM

@Daminark Thank you very much.

Carrick

2:49 AM

Good morning everyone! I had a small doubt in the application of Leibniz integral rule

How do I differentiate $\int_0^x tf(x-t)dt$ where f(x) is a function.

With respect to x

Balarka Sen

3:53 AM

@TedShifrin Anyway, so topologically, any $n$-dimensional connected finite CW complex $X$ can easily be seen to be a ball quotient. Impose a CW structure with a single $0$-cell: this is always possible. Then $\varphi_i : D^n \to X$ be the characteristic maps of the various $n$-cells; $\bigvee_i \varphi_i : D^n \vee \cdots \vee D^n \to X$ is all of them at once, sending the wedge point to that single $0$-cell. This is a quotient map. Now a wedge of a bunch of $D^n$'s is a quotient of $D^n$.

I don't know how to do this simplicially though, because if I have a quotient map $D^n \to X$, I can $\epsilon$-homotope it to make it simplicial modulo appropriate simplicial structures on domain/codomain, but this need not be a quotient map any longer.

Silent

Let $G$ be a group with identity element $e$. And $N$ be a normal subgroup. Let $[G:N]=12$. Then $x^{24}\in N$ for all $x\in G$.

I think this is because, for any $x\in G$, $x^{12}N=(xN)^{12}=N$, hence $x^{12}\in N$ hence $x^{24}\in N$. Right?

Balarka Sen

(Well, there's a small problem above because the image of the attatching maps might not be all of the $(n-1)$-skeleton but that's easily fixable: you can homotope the attatching maps so that it goes "round and back again" around it)

Balarka Sen

5:23 AM

Ok, so this problem has a nice solution, actually. I have a simplicial complex $K$ and I want a triangulation on the $n$-simplex such that $f : \Delta^n \to K$ sends $i$-simplices to $i$-simplices. Basically this is going to be a sort of "unfolding" of $K$ (think of unfolding a cube into a cross of squares).

Just take the "dual graph" to $K$, which is given by joining the centers of all the $n$-simplices by edges. Take a maximal tree of this, and mark the simplices of $K$ which doesn't intersect this maximal tree. Cut $K$ along those simplices to get a simplicial complex $K'$ from which $K$ is obtained by gluing some pairs of faces togather. $K'$ deformation retracts to the maximal tree, so is topologically a ball.

Put up a homeomorphism $\Delta^n \to K'$ and pull the triangulation back. You have obtained $K$ as a simplicial quotient of $\Delta^n$.

Leaky Nun

duality yay

Balarka Sen

5:34 AM

By simplices above I meant the (n-1)-simplices, i.e., the faces. The idea is you cut along faces to get something which is a ball

user131753

6:24 AM

I am trying to show that if $K$ is a simplicial complex and the geometric carrier of $K$ has $k$ many path components, then the zeroth homology group of $K$, i.e., $H_0(K)$ is isomorphic to $\mathbb{Z}^k$.

user131753

My only observation so far is that the elements of $C_0(K)$ (the $0$-dimensional chain group of $K$) is isomorphic to $\mathbb{Z}^{|\text{vert}(K)|}$ where $\text{vert}(K)$ denote the set of all vertices of all simplexes in $K$.

user131753

However, I don't know how can I related this to $k$ (the number of path components of $K$). Any ideas?

Alessandro Codenotti

When are two (singular) 0-simplices the border of a (singular) 1-simplex?

user131753

@AlessandroCodenotti: Is this comment in response to my comment(s)?

user131753

(Sorry if it is not. But I just thought that I should confirm.)

Alessandro Codenotti

6:36 AM

To your question more than your comments but yes

user131753

@AlessandroCodenotti What are singular 0-simplices? I don't know about it.

user131753

However I think that $0$-simplexes are "endpoints" of $1$-simplex iff each of the simplex is a vertex of the one simplex.

user131753

I am not really fond of geometry, so I was trying to think of a purely algebraic solution of the question is possible.

noob

How to generate standard normal coefficients in excel?

Basically I am doing Functional Data Analysis. Where I need to calculate a polynomial which is a function of time and has coefficients which are from standard normal distribution.

should I just generate a random number between 0 and 1 in excel for it to be from standard normal distribution?

https://www.youtube.com/watch?v=jxuXh3KOvo0

I guess this answers the question

In the question I uploaded I have one query

How do I calculate the variance

in order to calculate the error term

I know what is beta of t what is the beta of s in the equation defining variance

Also what is E[X(t)X(s)] ?

Please help

is E the expected/average value of X(t) and X(s) ?

Alessandro Codenotti

7:05 AM

@user170039 how do you define $H_0$?

(or $H_n$ in general)

user131753

@AlessandroCodenotti Same as here.

Alessandro Codenotti

Which chain complex are you using?

user131753

Do you mean the definition of $C_n(K)$ for all $n\in \mathbb{N}$?

Alessandro Codenotti

Yes

(well to be precise the chain complex is the whole of the $C_n$ and maps between them)

user131753

Let $K$ be a oriented simplicial complex. Denote by $\bf{O}_n(K)$ the set of all oriented $n$-simplexes of $K$. Let $f:\bf{O}_n(K)\to \mathbb{Z}$ be a map such that $f(+\sigma)+f(-\sigma)=0$ for all $n$-simplex $\sigma\in K$.

user131753

7:16 AM

Call such a map a $n$-chain. For $n\in \mathbb{N}$ define $C_n(K):=\{f:f~\text{is a}~n\text{-chain map}\}$.

user131753

Let $+:C_n(K)\times C_n(K)\to C_n(K)$ be defined in the usual way of function addition.

user131753

Do I need to elaborate the boundary homomorphisms also?

Alessandro Codenotti

Well what is the boundary of a $1$-chain? To compute $H_0$ you need $\ker\partial_0$, but that is the whole of $C_0(K)$, so you only need to figure out what's the image of $\partial_1$

user131753

7:47 AM

As per the notes which I am reading $\partial_q(\widehat{+\sigma^q})=\displaystyle\sum[\sigma^q,\sigma^{q-1}]\widehat{+\sigma^{q-1}}$ where $\widehat{+\sigma^{q}}$ is the elementary $q$-chain (defined below) and $[\sigma^{q},\sigma^{q-1}]$ is the incidence number.

user131753

The elementary $p$-chain is defined as follows, $$\widehat{+\sigma}(\tau)=\begin{cases}1&\text{if}~\tau=+\sigma\\-1&\text{if}~\tau=-\sigma\\0&\text{else}\end{cases}$$

user131753

Here the sum runs over all the positively oriented $(q-1)$-faces of $\sigma^q$.

user131753

This definition is then extended by linearity to whole $C_q(K)$

noob

Can anyone please help me with that question

skullpetrol

9:29 AM

$Mathematics$ JEE Maths Zone

Related Rooms: [Physics](chat.stackexchange.com/rooms/54160/p...

Sebastian Alexander B Nielsen

12:34 PM

The following expression reduced is apparently equal to x, can anyone explain how that is the case?
$$\frac{x^2+2x}{x+2}$$

When I reduce the expression, I get the answer to be equal to $x^2$. This is how I would go about reducing it:

What did I do wrong?

skillpatrol

@SebastianAlexanderBNielsen you can not cancel the 2s like that.

jublikon

Hi, I have a small question:

Does $\neg (f (a,b) \neq x )\equiv (f(a,b) = x) \equiv(\neg f(a,b)\neq x)$ hold, where $f$ is a mapping and $a,b,x$ are elements of the underlying universe?

skillpatrol

Step 1 factor the x out of the numerator

Step 2 divide the numerator and denominator by (x + 2)

Sebastian Alexander B Nielsen

But how? I can "break apart" the denumerator $x+2$.

uh wait, thanks

skillpatrol

12:50 PM

np

Subhasis Biswas

1:01 PM

Hello, this seems to be a bit problematic for me

https://math.stackexchange.com/questions/3231729/if-sum-a-n-converges-then-sum-a-n-fracnn1-converges-as-well?noredirect=1#comment6648204_3231729

I don't know whether or not it will be valid to write $\lim_{n \to \infty}
\frac{a_n^{n/(n+1)}{a_n} $

$\lim_{n \to \infty} \frac{a_n^{n/(n+1)}{a_n}$

$\lim_{n \to \infty}$ $ \frac{a_n^{n/(n+1)}}{a_n}$

=1

hola

2:30 PM

Does someone know how to refer to a system that consists of several equations in IEEE? For example i have a system described by equation (1) and a controller for system (1) described by equation (2). I basicaly want to refer to the whole dynamics (1) and (2). I was thinking if i could say: Dynamics (1,2).

Secret

3:52 PM

May 11 at 16:52, by Tobias Kildetoft

If you lecture long into the void, the void starts lecturing back.

[Random]

Problem solving surface: The Path of sudden least resistance

noob

Secret

Sometimes given a mathematical problem. Suppose we can represent it as a n dimensional surface, and let some point p be the solved state of the problem

noob

In the question I uploaded I have one query
How do I calculate the variance
in order to calculate the error term
I know what is beta of t what is the beta of s in the equation defining variance
Also what is E[X(t)X(s)] ?
Please help
is E the expected/average value of X(t) and X(s) ?

If anyone can even provide the approach to solve that question it would be helpful

Secret

Let p be a set of curves that can be plotted on the surface. For some problem which is very difficult, there exists highly non smooth surfaces such that only certain curves will lead to p

This thus translate to a highly specific idea or understanding, such that the hard problem becomes straightforward to solve

What may be interesting is that: Given a problem surface, when will solutions to them be solution sets of finite cardinality or even of cardinality 1

@noob I do not know the other question, but E[X(t)X(s)] is going to be the expectation value of the product X(t)X(s)

user76284

6:37 PM

Is "modular arithmetic", i.e. Peano arithmetic where $\neg \exists x (Sx = 0)$ is replaced with $\exists (Sx = 0)$ $\omega$-inconsistent?

schn

7:19 PM

Why is the falling factorial $(m)_n=\frac{m!}{(m-n)!}$? I'd really appreciate a derivation. I define the falling factorial as given on Wikipedia.

Thorgott

$m!$ is the product of all integers from $1$ to $m$. $(m-n)!$ is the product of all integers from $1$ to $m-n$. So what's left after diving $m!$ by $(m-n)!$ is the product of all integers from $m-n+1$ to $m$.

schn

7:38 PM

@Thorgott Nice. Thanks!

schn

8:33 PM

What does it mean for a function to be defined on an open neighbourhood of a point? I'm familiar with the term without the word 'open' or 'open' being replaced by 'punctured'. Does it simply imply an open interval around that point?

Ryan Unger

@schn in a general topological space you have neighborhoods

Balarka Sen

GOD it feels nice when you draw a commutative diagram

3

Leaky Nun

@schn a subset $A$ of $\Bbb R$ is open iff every point $x \in A$ is contained in an open interval $(x-\varepsilon,x+\varepsilon)$ which in turn is contained in $A$

so you have a "full set of points" near each point

@BalarkaSen ikr

schn

@LeakyNun Alright, so saying a function is defined in an open neighbourhood at a point would mean an open interval $(a,b)$ which contains that point? Thanks!

Alessandro Codenotti

8:48 PM

@BalarkaSen D:

Balarka Sen

lmao

i mean it's more like the joy you feel when you're crushing a cockroach under your shoes

Alessandro Codenotti

I have to draw a commutative diagram and take a direct limit in my set theory seminar talk in a couple of days, I'll bring shame upon my family and my cows

Leaky Nun

@schn the former would imply the latter and indeed the latter would imply the former...

GFauxPas

9:23 PM

If a real valued function $C^1$ (or maybe just differentiable is enough) has a derivative bounded by $L$, does that mean that $L$ is a Lipschitz constant for the function? Or is it possible you need a larger number than $L$?

Leaky Nun

@GFauxPas $L$ is enough. Differentiable is enough.

GFauxPas

Thanks, useful to know

GFauxPas

9:50 PM

I can use help with this question

The system is $\begin{bmatrix} \dot x \\ \dot y \end{bmatrix} = \begin{bmatrix} t^{-1} & 0 \\ 1 & 3t^{-1} \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}$

I have the general form of the solution

$\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} c_1 t \\ c_2 t^3 - c_1 t^2 \end{bmatrix}$

I have a matrix $\Pi(t,1)$ with two columns spanning the solution space satisfying $\Pi(1,1) = I$

$\Pi(t,1) = \begin{bmatrix} t & 0 \\ t^3 - t^2 & 1 \end{bmatrix}$

now the problem is:

Solve $\begin{bmatrix} \dot x \\ \dot y \end{bmatrix} = \begin{bmatrix} t^{-1} & 0 \\ 1 & 3t^{-1} \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} + \begin{bmatrix} 1 \\ 0 \end{bmatrix}$

since that's part 2 of the problem, I assume that $\Pi(t,1)$ is supposed to play a role here

I don't think I'm supposed to set up a system of two equations, I think it's supposed to be simpler than that

Ryan Unger

@BalarkaSen I thought this was illegal in India

GFauxPas

oh I found the theorem

interesting, the inhomogeneous part of the solution is given by

$\displaystyle \int_{t_0}^t \Pi(t,s)g(s) \, \mathrm ds$ where $g(s)$ is the the inhomogeneous part, in this case $(1,0)$

that's interesting

Balarka Sen

@ryan cows, not roaches

Ryan Unger

10:11 PM

@BalarkaSen there was a guy in the hbar who denouncing killing of bugs

I believe he had authority to speak for the whole country

Balarka Sen

that does sound very indian to me

everyone here has authority to speak for the whole country

@RyanUnger Why are you lurking in "Jee Advanced physics solving strategies"

Ultradark

If you want to understand a complex function do you study it's roots and poles?

Isa

10:28 PM

In this answer math.stackexchange.com/questions/3230803/finding-borel-sets by Daniel Wainfleet, why $f^{-1}(U\cap \Bbb R)=g^{-1}(U\cap \Bbb R)$

Semiclassical

@Ultradark that's sufficient for something like $f(z)/g(z)$ where $f(z),g(z)$ are polynomials in $z$. that's a very common scenario

however, that will not help you understand something like $f(z)=\ln z$ very well. The behavior at zero is neither that of a root or a pole, but of a branch point

$f(z)=e^{1/z}$ is another example where things go weird at zero in a way other than a pole or a root

Ultradark

@Semiclassical Is it understood this weirdness?

Isa

I think it's not true because U might contain the point 0, thus $g(x)\not\in U$, am I correct?

GFauxPas

There's a theorem that, correct me if I get the hypotheses incorrect

that if $f(z)$ is holomorphic except at a point $z_0$ where it has an essential singularity, than $f(z)$ attains every complex number in a neighborhood of that point, with possibly one exception

Semiclassical

@Ultradark the classification is understood: en.wikipedia.org/wiki/…

(note that a root is a place where $f(z)$ is well-behaved, so it's not a singularity)

GFauxPas

10:36 PM

that's the case with $e^{1/z}$

Semiclassical

@GFauxPas yeah, picard's theorem (i forget if it's the little or big one)

GFauxPas

Great Picard's Theorem: If an analytic function f has an essential singularity at a point w, then on any punctured neighborhood of w, f(z) takes on all possible complex values, with at most a single exception, infinitely often.

which is pretty damn cool

Ultradark

that means all zeros of the analytic function are also mixed in

Semiclassical

another fun example: The series $f(z)=z+z^2+z^4+\cdots+z^{2^n}+\cdots$ converges everywhere inside the unit circle. But it has no analytic continuation outside the unit circle!

GFauxPas

That's a typo there Semi

$z^+$

Semiclassical

10:40 PM

thx

GFauxPas

that's cool. What about on the circle?

Semiclassical

doesn't converge anywhere on the unit circle iirc. the unit circle thereby serves as a "natural boundary of analytic continuation".

Leaky Nun

@GFauxPas convergent series tend to 0

Semiclassical

16

Q: What is the Riemann surface of $y=\sqrt{z+z^2+z^4+\cdots +z^{2^n}+\cdots}$?

The following appears as the second-to-last problem of Stewart's Complex Analysis: Describe the Riemann surface of the function $y=\sqrt{z+z^2+z^4+\cdots +z^{2^n}+\cdots}$. This problem intimidated me when I first saw it as an undergrad, as the series under the root isn't the expansion of a...

Leaky Nun

$z^{2^n}$ doesn't tend to 0

Ultradark

10:41 PM

does the essential singularity encode information about the analytic function or something?

GFauxPas

oh right

durr

if it has an essential singularity it's not analytic there

Semiclassical

@LeakyNun not on the unit circle, anyways

GFauxPas

but it will be expressible as a laurent series on an appropriate annulus

Semiclassical

see also mathoverflow.net/questions/178549/…

Leaky Nun

$$\sum_{n=1}^\infty \frac1{2^n} = \sum_{n=1}^\infty \frac1{2^n \ln(2^n)}$$

Ultradark

10:45 PM

@LeakyNun how is that an equivalence?

Leaky Nun

@Semiclassical can we find the pdf of Y=g(X) where X~Uniform(-1,1) and... g(x)=sin(1/x)?

Mathphile

0

Q: There cannot be more than three primes of the form $n^{n^2}-k$ for the same $k$?

I was searching for primes of the form $n^{n^2}-k$ on PARI/GP and noticed that primes of this form for same $k$ are quite rare. The probability of finding a prime of this form is $\frac {1}{n^2 \log (n)}$, which makes me believe that the number of primes for same $k$ of this form would be finite....

elementary-number-theory prime-numbers conjectures

any ideas?

guys

Leaky Nun

if X has a pdf, what are sufficient conditions on g such that g(X) also has a pdf?

Ultradark

what is a punctured neighborhood of $w$

Balarka Sen

@LeakyNun "$g$ is a diffeomorphism" :)

GFauxPas

10:59 PM

Neighborhood with a hole in the middle

One point removed in the interior

Balarka Sen

You can obtain a valid pdf of $g(X)$ whenever $g$ satisfies the necessary conditions on the range of $X$ for the change of variables formula to hold.

Eg, $g^{-1}$ exists and is injective on $\text{Range}(X)$, is $C^1$ and $dg^{-1}$ is full rank.

You can weaken these assumptions, of course, because a pdf is well-defined upto a measure zero set.

So for example if you start out with $X \sim N_2(\mathbf{0}, I)$ standard multivariable normal, and $g(x, y) = (\sqrt{x^2 + y^2}, \tan^{-1}(y/x))$ which is defined on the whole plane minus the non-negative real axis, say, then $g(X)$ is still absolutely continuous. Simply because even though $g$ wasn't defined on the full range of $X$, it was defined on a full measure subset, and $g^{-1}$ is a local diffeomorphism (it's a covering map we all know and adore)

We picked out a large local section of the covering map defined on a full measure open set, that's all. Incidentally, $g(X)$ turns out to be jointly distributed as $(R, \Theta)$ where $R$ and $\Theta$ are independent (!) and $\Theta \sim \text{Unif}(0, 2\pi)$ (!), indicating the rotational symmetry of the standard bivariate normal. $R$ is the "radial part" which is called the Rayleigh distribution; it's squareroot of $\text{Exp}(1)$.

Speaking of square roots, a way simpler example: $g : \Bbb R \to \Bbb R$, $g(x) = x^2$. $g^{-1}$ is piecewise defined on the positive and negative real axes, where it satisfies all the necessary hypothesis for change of variables, so this also causes no problem.

Abstractly, all this means is that absolutely continuous random variables with values in $\Bbb R^n$ are in 1-1 correspondence with $L^1$ $1$-densities on $\Bbb R^n$ with total integral = 1, the bijection being $X \leftrightarrow f_X(x) |dx|$. This changes correctly according to transformation rules, because if $X$ corresponds to $\omega = f_X(x) |dx|$ then $g(X)$ corresponds to $(g^{-1})^*\omega = f_X(g^{-1}(x)) |\det dg^{-1}(x)| |dx|$

You're pulling back by $g^{-1}$, so maybe you'd want to say it's a contravariant correspondence.

Ryan Unger

11:33 PM

@BalarkaSen I entered planning to dunk on some kids

turns out its abandoned

why did you go there

Balarka Sen

11:51 PM

@RyanUnger whim

$Mathematics$ JEE Maths Zone

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