Mathematics

1:15 AM

Does the equation $\prod_{i=1}^{\infty}(1+c/n^2) = \frac{sinh(\sqrt{c}\pi)}{\sqrt{c}\pi}$ hold for general c? Sorry in advance for bad latex notation if that doesn't render properly

CroCo

1:27 AM

@user193319 thank you. that makes sense.

user616128

2:10 AM

I wonder why this room is becoming less active over time. I guess there are many other places to chat now about mathematics?

Semiclassical

2:55 AM

meh, natural ebb and flow

Secret

3:58 AM

This room is like super active compared to most others

5

Akiva Weinberger

5:38 AM

11 hours ago, by manooooh

@AkivaWeinberger concerning to our proof of $un=n$, which kind of proof method is it: a direct proof, counterreciprocal or argumentum ad absurdum? I think it is a direct proof, right?

@manooooh I think it's a direct proof, too

@Semiclassical So IIRC the amount by which two random variables fail to be independent is measured by E(XY)-E(X)E(Y)

The covariance

Is there a way to measure how three variables fail to be independent?

(Like the example from before where (X,Y,Z) is chosen from the coordinates of the vertices of a tetrahedron, i.e. X and Y are chosen from {0,1} independently at random and Z is their sum mod 2)

You can tell that I've never actually taken a class on probability

Secret

researchgate.net/post/…

Probably one might need a 3rd order Covariance tensor to capture all those triple correlations

Akiva Weinberger

6:23 AM

I think E((X-E(X))(Y-E(Y))(Z-E(Z))) measures how "tetrahedral" X, Y, and Z are

Say they're taken from {-1,1} for simplicity (so E(X)=0 etc)

So we have E(XYZ)

This means we want XYZ to be positive, so an even number of them should be -1

Akiva Weinberger

6:58 AM

And that expands to E(XYZ)-ΣE(X)E(YZ)+2E(X)E(Y)E(Z)

So that equals E(XYZ)-E(X)E(Y)E(Z) iff ΣE(X)E(XY)=3E(X)E(Y)E(Z)

which is true if (sufficient but not necessary) X, Y, and Z are pairwise independent

(otherwise E(XYZ)-E(X)E(Y)E(Z) isn't translation independent)

Silent

7:48 AM

My book says:

Laurent’s series of an analytic function in an annular region can be
differentiated term-by-term. As a consequence, since Log z is not analytic
in any annulus around 0, it cannot be represented by a Laurent series
around 0

I don't get why 'not analytic' implies no Laurent series

Tobias Kildetoft

8:00 AM

@Silent because of the "around $0$"

Akiva Weinberger

8:48 AM

Log z isn't even continuous in an annulus around zero, right?

if you're taking the principal logarithm

Leaky Nun

9:03 AM

if you're taking any logarithm at all

also this is the basis of like the entirety of complex analysis

$\displaystyle \prod_{n=1}^\infty \frac{2^n (\exp(2^{-n} t) - 1)}t$ converges for all $t \in \Bbb R$ right

Dattier

9:28 AM

Hi

if $a_{1} < a_{2} < ... < a_{n}$ integers $(X - a_{1})(X - a_{2})...(X - a_{n}) - 1$ is it irreductible on $\mathbb Z[X]$?

Akiva Weinberger

10:23 AM

I have such a bad cough

I can't get two words into a sentence without coughing

If I could infect people through a chat room you'd all be sick now

@LeakyNun In other words $\displaystyle\prod_{n=1}^\infty\dfrac{\exp(2^{-n}t)-1}{2^{-n}t}$?

I know that $\exp(x/2)\le\dfrac{\exp(x)-1}x\le\exp(x)$ (at least for $x\ge0$)

Like so

Manolis Lyviakis

10:55 AM

suppose i have an "arc-ed" cross meaning a bended cross how do i show exactly that it is contractible

that its identity map is contractible

which homotopy to use

Silent

@AkivaWeinberger I would advice that, you peel out pomegranate, and boil that peel, let it cool naturally, and drink.

Akiva Weinberger

That's an idea

Sounds like so much effort though

Silent

its just two step process :)

Akiva Weinberger

Does it taste like the seeds when you do it like that?

Silent

it tastes a little bad, but that's effective

Akiva Weinberger

10:58 AM

Arright

Secret

11:50 AM

3

Q: Infinite subset contains finite subset of any size

Is it true in ZF that given any infinite set $X$ and any natural number $n$, there is a subset of $X$ with cardinality $n$ (i.e. equinumerous with $n$)? Here, $X$ is defined to be an infinite set if $X$ is not equinumerous to any natural number $n$. This may turn out to be quite trivial, but I ...

set-theory

It may be cool to have an "Strictly Dedekind infinite set" where all its subsets are infinite, but to do that will require circumventing "finite choice" by somehow nuking induction, which is way too costly to be worthwhile

Akiva Weinberger

12:22 PM

@ManolisLyviakis Like ナ?

user616128

Is the topology on the manifold that the regular value gives you the subspace topology?

user193319

Problem: Let $A \subseteq \Bbb{R}$ be a measurable set with finite measure. Then $F(x) = m(A \cap (-\infty, x))$ is continuous...Proof: Let $y < x$. Then $$|m(A \cap (- \infty,x)) - m(A \cap (-\infty,y))| = m(A \cap [y,x)) \le m([y,x)) = |x-y|$$
So $F$ is actually Lipschitz and hence continuous.
Does this sound right?

Silent

12:39 PM

I don't get why sinz and cosz are the only entire functions that can assume the values sinx and cosx, respectively, along the real axis or any segment of it.

Manolis Lyviakis

1:06 PM

@AkivaWeinberger yes

how can i see a graph as the quotient space of the disjoint union of vertice and intervals

so consider X to be a graph

Graph (topology)

In topology, a subject in mathematics, a graph is a topological space which arises from a usual graph G = ( E , V ) {\displaystyle G=(E,V)} by replacing vertices by points and each edge e = x y ∈ E {\displaystyle e=xy\in E} by a copy of the unit interval [ 0 , 1 ] {\displaystyle [0,1]} where 0 {\displaystyle 0...

what does it mean it is the quotient space of the disjoint union

Manolis Lyviakis

1:21 PM

it just mean that is the union of these set but gluing the vertices to the egdes?

Leaky Nun

1:33 PM

@AkivaWeinberger 10/10

Akiva Weinberger

@Silent They're uniquely determined by their values over any arc in $\Bbb R$

Leaky Nun

@AkivaWeinberger let $X \sim U(0,1)$

$X=1-X$, so $2X=1$, so $X=\frac12$

Semiclassical

1:51 PM

Trying to figure out good terminology for something

Akiva Weinberger

@ManolisLyviakis Yeah. Imagine taking $A=[0,1]\cup[2,3]\cup[4,5]$ and taking the equivalence relation $x\sim y$ if either $x,y\in\{0,2,4\}$ or $x=y$

Then $A/{\sim}$ is $A$ with $0$, $2$, and $4$ glued together

and it's a graph

Semiclassical

Suppose I’ve got a poset. I think, in general, you can always assign numbers to the elements of the poset in a way that agrees with the partial ordering

Akiva Weinberger

specifically, the star with three edges

Semiclassical

Not in a unique way, of course

But is that instinct sound? I’m coming at this from an example, so my knowledge of posted is pretty bad

Akiva Weinberger

If it's a finite poset then it's always a subset of the order relation on a finite subset of $\Bbb N$, I'm pretty sure

(I guess I should've said it's always a subset of a total order on the same amount of elements)

('cause it doesn't matter if they're in $\Bbb N$ or $\Bbb Q$ or $\Bbb R$ or whatever)

Semiclassical

1:57 PM

Yeah, thought so

Akiva Weinberger

'Cause you can do this inductively, can't you? Like, inductively assign to each thing a rational number

Choose a random thing and call it 0

Semiclassical

The other intuition is that, if you draw the Hasse diagram

Akiva Weinberger

Move things less than it to one pile, things greater than it to another pile, and put things incomparable to it in whichever pile you want

and then choose a random thing in the "greater" pile and call it 1

and something in the "lesser" pile and call it -1

and in the next step you assign -2, -½, ½, and 2

etc etc

Semiclassical

In my particular case the poset is finite with unique greatest and smallest elements

Akiva Weinberger

and then, if it really bothers you at the end that you have labels in $\Bbb Q$ and not $\Bbb N$, you can map the labels order-isomorphically to some set of the form $\{1,2,\dots,n\}$

Semiclassical

2:00 PM

@AkivaWeinberger nah, that’s not an issue

For me anyways

I’m more interested in what such orderings are possible, subject to some conditions on what values are assigned on the boundary of the poset

Eg take your poset to be 3-letter words from the alphabet 0,1,2

Akiva Weinberger

I guess you could look at the minimum "range" (distance from the largest label to the smallest) but that'd probably just be the length of the largest chain

(…minus one?)

Semiclassical

With one word covering another if all it’s letters are greater or equal than the other’s

So 210 covers 110 covers 100 etc

Akiva Weinberger

Mhm

The largest chain would be 000-001-011-111-112-122-222 I think?

Or that's tied for largest anyway

Semiclassical

Looks right

Akiva Weinberger

meaning the "range" of the labels has to be at least 6

Well I mean

Just label each word with the sum of the entries

Semiclassical

2:07 PM

When you draw the Hasse diagram you get 7 levels, containing 1,3,6,7,6,3,1 elements respectively

Akiva Weinberger

so 122 gets a label of 1+2+2=5 for example

Semiclassical

Right

Akiva Weinberger

and then 000 gets a label of 0 and 222 gets a label of 6, so that gives you a labeling with the smallest possible range

Semiclassical

What I’m particularly interested in is the following

Let $f$ be a mapping from each word to [0,1]

Akiva Weinberger

such that f respects the ordering?

Semiclassical

2:09 PM

Yeah

Akiva Weinberger

And we probably want f(000)=0 and f(222)=1 or something like that

Semiclassical

Well, I only require the latter

But I also require f(221)=2/3, f(211)=1/3, and permutations thereof

Trying to remember if that’s all.

I think it is?

Obviously that allows a lot of freedom. The boundary conditions prescribe the values for 7 of the 27 words

Leaky Nun

sanity check: F'(x/2)+F'((1-x)/2)=1 means 2F(x/2)-2F((1-x)/2) = x+F(0)-F(1) right

oh I spotted the mistake the very moment I posted it

sanity check: F'(x/2)+F'((1-x)/2)=1 means 2F(x/2)-2F((1-x)/2) = x+2F(0)-2F(1/2) right

Secret

[Very random]

A dedekind sunset is a sunset that has indefinite number of stages which can arbitrarily reach the horizon, but never able to reach it

Akiva Weinberger

2:30 PM

Autocorrect replaces "subset" with "sunset" a lot for me

Secret

It does especially on my mac. Because it is so common, I decided to capitalise on the autocorrect and fabricate this concept of dedekind sunset

as a joke

Akiva Weinberger

4:14 PM

ÍgjøgnumMeg

i feel attacked

Semiclassical

4:28 PM

@AkivaWeinberger funny thing is, while I do say the letter z as "zee" (as is typical of Americans)

user76284

I remember being so confused when I moved to Canada as a kid and people used the word "zed". Felt like I had entered a different world.

Semiclassical

I sorta wish "zed" was the standard here as well

if only to avoid confusing "z" and "c" and "see"

user76284

Ha, you should try Mandarin.

Semiclassical

oof

More Anonymous

Hey all!

So glad one of my formulas gives the right result!!:
https://math.stackexchange.com/questions/3173284/what-is-the-limit-of-this-dirichlet-series

Rithaniel

5:23 PM

"five, pip pip, cheerio" is my new favorite number.

Akiva Weinberger

I have to admit, though, I never really gave much thought to the pronunciations of $k[x]$ and $k(x)$

Alessandro Codenotti

Just read the two letters and add "with round brackets" in the second case

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$Mathematics$ Mathematics