Mathematics - 2019-12-26 (page 3 of 3)

Semiclassical

16:00

from the book I've got on my table right now:

nbro

In my opinion, many mathematicians are lost in abstraction and then do not really understand the intuition behind many concepts.

tigre

Why don't you open a textbook nbro?

Alessandro Codenotti

I'd argue that intuition was not the main driving force behind the formalization of probability in a measure theoretic setting

nbro

@tigre What are you talking about?

tigre

It sounds like you've been confused for days, whereas a mathematician would just get it all down from a textbook

Balarka Sen

16:01

Any mathematician worth their salt is pretty clear on what intuitions behind the terms in basic probability is

The pdf is simply not a measure, if you call a false statement an "intuition", that's pretty dangerous

the pdf is what is known as a Radon-Nikodym derivative

Semiclassical

"Let $X$ be a random variable from a probability space $(\Omega,\mathcal{F},P)$ to a measurable space $(\Psi,\mathcal{G})$. For $B\in\mathcal{G}$ let $Q(B)=P(X^{-1}(B))$. Then $(\Psi,\mathcal{G},Q)$ is a probability space. The probability measure $Q$ is called the distribution of the random variable $X$ and is said to be induced by $X$ (or by $X$ from $P$)."

With a comment at the bottom of the page: "Note that every distribution is a probability measure, and that every probability measure is a distribution (induced by the identity function), so 'distribution' and 'probability measure' are essentially synonymous terms that tend to be used in somewhat different contexts."

nbro

You just talk about definitions because you look them up and you follow the rules. My impression is that you don't know many times what you're talking about. In fact, you didn't even mention this simple thing $p_X(x) = \mathbb{P}(X=x)$, which clearly shows that you don't know what you were talking about and then you were claiming that a probability distribution is a probability measure, as if this was the absolute definition, but it isn't.

Balarka Sen

I didn't mention it because the pmf is a useless notion for continuous distributions

Semiclassical

Oh hey star board

Balarka Sen

I also imagined anyone conversing about measure theory would know the definition of a pmf

which is a fair thing to assume in general

nbro

16:05

Whatever!

Secret

ok let me try again:
$\bar{a}b = a \bar{b} = \longline{a \bar{b}}$ so $\bar{a}b$ is real
$ab = cdr$
Multiply both sides by $\bar{a}\bar{b}$
$a\bar{a}b = \bar{a}cdr$
$(a\bar{b})(\bar{a}b) = \bar{a}\bar{b}cdr$
so the LHS is real. Since $r$ is real, $\bar{a}\bar{b}cd$ must be real

tigre

lol

@Secret This I agree with, in the case that $(a,b)\sim (c,d)$

nbro

Conclusion: the expression "probability distribution" can also refer to a p.m.f. or even, indirectly, to a p.d.f., but it almost always refers to a c.d.f. and a probability measure, depending on the nature of the r.v. (discrete vs continuous).

Semiclassical

i mean, yes, knowing the p.m.f. is basically equivalent to knowing the probability distribution for a discrete random variable. doesn't mean that it's sensible to equate the concepts in general

Alessandro Codenotti

If people here have no idea what they're talking about (I have to agree with you that there is at least one user fitting that description) why do you keep coming back to the chat instead of consulting a more reputable source?

4

nbro

16:09

Because I think it's important to discuss (for the purposes of learning)

tigre

@nbro You're not going to like the abstract notion of a correspondence between two sets lol

Balarka Sen

well theres no point in discussing if you are assuming a priori that the person you are discussing with is incompetent

Semiclassical

Just because there's contexts where knowing A is equivalent to knowing B, doesn't mean it's true in all contexts

nbro

I am not saying you're incompetent. I am just saying that often you're lost in abstraction (i.e. definitions) and you fail to understand the intuition behind the concepts and connect them.

Semiclassical

if they really are equivalent, great. But if they're not, then it's not helpful to insist that they're one and the same when they're really not

Secret

16:11

@tigre I am not sure how to argue from this that there are only two equivalence class, but it seems the two equations kind of fix a,b,c,d to only two degrees of freedom

Semiclassical

and for the purposes of measure-theoretic probability, there are good reasons to insist that the pmf of a discrete random variable is not the same concept as its distribution.

namely, because not every random variable has a pmf or a pdf.

tigre

@Secret One thing you can do is show that $(a,b)\sim (l_1a,l_2b)$ for $l_1\in \Bbb R-\{0\}$ and $l_2\in\Bbb R_{>0}$ (possibly)

nbro

Btw, I really appreciate your feedback. This is just my opinion, so it is just an opinion!

@Semiclassical Yes, but you can safely say "for a discrete r.v., the expression probability distribution almost always refers to a p.m.f."

Secret

Hmm...

tigre

@Secret I.e. that you can scale them individually

Semiclassical

16:14

insofar as that represents a common abuse of notation / convention, yes

Secret

$(a\bar{b})(\bar{a}b) = \bar{a}\bar{b}cdr$
Let $s = \frac{r}{(a\bar{b})(\bar{a}b)}$
Then we have
$1 = \bar{a}\bar{b}cd s$
$\frac{1}{\bar{a}\bar{b}} = cd s$

Balarka Sen

@nbro that's correct, and I'd scratch the "almost always". This is essentially because the measure induced by the discrete distribution is indeed the p.m.f.

So all the various notions we have discussed so far agrees

Semiclassical

I think the tricky bit there is "refers to"

If I draw a graph of a function, is that the function?

Secret

so $\bar{ab}$ is proportional to $cd$ by some real number

$a,b, d \neq 0$

tigre

brb

Semiclassical

16:16

certainly, if I have a good enough graph of a simple function, I can figure out what that function is doing and write out its definition

Secret

There are for variables a,b,c,d and 2 equations, thus only 2 degrees of freedom, thus we can split up the real number $s$ and get the scaling individually

Semiclassical

but it seems a bad idea to identify a function with its graph, if only because some functions are not going to have any nice graph

Secret

and hence only 2 equivalence class

Balarka Sen

@Semiclassical It's a weirdly basic pedagogical point but maybe you're right that something like that is indeed the source of the confusion for nbro

tigre

That's what I was referring to when I joked that he won't like the 'abstract' notion of a correspondence between sets, since he seems to be suggesting that avoiding conflating things is 'getting lost in abstraction' (where abstraction for some reason means caring about definitions, which sounds fairly concrete to me)

@Secret That's a really nonrigorous argument :P

Secret

16:22

simple hard question now in MSE:

0

Q: How to prove the roots of cosine is dense or sparse when a tend to infinity?

So I am a bit surprised that there isn't a question on this. Recall the first time when you came across the cosine function $\cos (ax)$ as $a \to \infty$ graphically, it intersect the x axis more and more frequently, result in the limit to diverge Now, more familiar with the different kinds of i...

algebra-precalculus

Putting the high school memory of calculus into focus

Semiclassical

for a simpler version of that, you can write your cosine function as $\cos(\pi n x)$ for integer $n$

Secret

true

then we have a sequence of nowhere dense sets

thus the question boils down to whether the limit of this sequence of sets is also nowhere dense

which many are now warning that defining that limit is not trivial

Semiclassical

at which points the roots are of the form (m+1/2)/n

for integer m

Alessandro Codenotti

I see no reason why that limit should exist

Secret

how so?

using semiclassical's case, as $n \to \infty$ the spacing between roots tends to zero

while each step stays nowhere dense

Alessandro Codenotti

16:27

I see no reason why that limit should be nonempty rather then

Semiclassical

what is true is the following: if I pick any point on the real line, then I can choose n large enough that one of the roots is arbitrarily close to the chosen points.

but that's not really about density

Secret

true

Semiclassical

to pick a simpler example, suppose I consider numbers with n digits in their decimal expansion

for any number $r$ and any $\epsilon>0$, I can always pick $n$ large enough that there's a decimal expansion within $\epsilon$ of $r$

but for any particular finite $n$, I just have a finite list of rational numbers which is certainly not dense

Secret

right, so some irrationals will be missed regardless of how big n is

Balarka Sen

16:56

The core of the Mazur manifolds deformation retracts to a dunce cap, which is essentially a D^2 attached to S^1 by a degree 1 map.

Ted Shifrin

Hi, a @Balarka, @Semiclassic, demonic @Alessandro

Semiclassical

hi

Balarka Sen

This can be seen by observing that $S^3$ minus an unknot deformation retracts to a torus with a disk glued along the meridian. Since the torus gets glued, longitude-to-meridian to the boundary of the deleted knot in $S^2 \times S^1$ which represents the generator of $H_1$, after collapsing a little more we get that the core is a disk attached to a circle by an attaching map that "follows the knot"

Alessandro Codenotti

Hi @Ted

Balarka Sen

Which is degree 1

Hi @TedShifrin!

Ted Shifrin

16:59

@Semiclassic The actual pedantic definition of a function is actually its graph.

Thorgott

I know I'm late to the party, but I think the important is the following: If you have a pmf, a pdf or a cdf (the former two only in case of their existence), then they uniquely determine the distribution (and, essentially, the reverse holds too). The definitions are very clear (and this has been communicated time and time again). However, once you've learned the subject a bit, you should be able to freely translate between these concepts and then all those worries don't matter.

Balarka Sen

Uh oh

you will awaken the demon inside Alessandro now

Semiclassical

...huh

Ted Shifrin

What else is new, a @Balarka?

Semiclassical

i mean, there's a graph as in "a picture I make" and the formal definition as a subset of $X\times Y$

but that definitely sweeps my rhetoric out from under me

Balarka Sen

17:01

~~~~~~~

sin(x)

Ted Shifrin

Well, I mean, graphing the ruler function (I forget its fancy name) is not something any of us can do literally ...

Or the characteristic function of the rationals.

But still ...

Semiclassical

fair

Thorgott

constant function

Balarka Sen

___-----_____

Semiclassical

constant function over [a,b]

Balarka Sen

17:02

Unif(1, 2)

Ted Shifrin

I wonder who opened this Pandora's box.

Balarka Sen

That my friend is a distribution

:3

Ted Shifrin

One of the three different sorts?

Thorgott

--^--

Unif(1,2)+Unif(1,2)

Balarka Sen

Oh let's prove central limit theorem

Ted Shifrin

17:04

Wow, a @Balarka sure has changed.

Alessandro Codenotti

@BalarkaSen lol

Balarka Sen

@TedShifrin To be perfectly clear on this my favorite proof of the strong law of large numbers is by the Birkhoff ergodic theorem

:3

Thorgott

the Irwin-Hall distribution is a cool illustration of the CLT

Ted Shifrin

I probably once knew that theorem, a @Balarka, but no longer.

Balarka Sen

It's how I initially guessed that something like this must be the statement of CLT before my course mentioned them

Given an ergodic transformation, the average of a measurable function over an orbit of the transformation is the average of the function on the full space

Popularly known as time-average = space-average

Really just a pumped up law of large numbers

Probability is just an absolutely beautiful branch of math. I wish I knew more

Ted Shifrin

17:11

Ah, that's intuitively the meaning of ergodicity anyhow. So in the CLT setting, what's the ergodic transformation?

oops. SLLN, I mean.

Balarka Sen

It's the shift operator on R^N, with the Kolmogorov measure (compatible with the Lebesgue measures of R^n)

I am being sloppy

Here's what it is

Suppose $X_1, X_2, \cdots$ are an iid sequence of random variables with finite mean

Then jointly they define a function $(X_1, X_2, \cdots) : \Omega \to \Bbb R^{\Bbb N}$

Ted Shifrin

Yes, I imagined it thusly.

Balarka Sen

Push the probability measure $\Bbb P$ on $\Omega$ forward to $\Bbb R^{\Bbb N}$ by this map. By i.i.d-ness, the induced measure on $\Bbb R^N$ is $\Bbb P^{\Bbb N}$, where this means countable product of the measures (this can be done and what requires Kolmogorov's result. Essentially the measure of the open sets in the product topology are the product of the measures of the intervals appearing in the product)

Then define $f : \Bbb R^{\Bbb N} \to \Bbb R$ to be first projection. This is your measurable function

If $T : \Bbb R^{\Bbb N} \to \Bbb R^{\Bbb N}$ is left-shift, this is ergodic with respect to $\Bbb P^{\Bbb N}$

Space average of $f$ is of course $\Bbb E X_1$, and time average is $\lim_n (X_1 + \cdots + X_n)/n$

Because $f(T^n(x))$ is the $n$-th projection, which has the distribution of $X_n$

(In light of previous pedagogy: This is another reason "distribution" is such an important word and should not mean the random variable in general. Two random variables from two distinct measure spaces can have the "same distribution". Their induced measures are the same on the real line!)

@TedShifrin I have long wondered if there's a second order analogue of ergodic theorems, because CLT feels like one

There are all kinds of CLTs for stuff which doesn't have finite variance as well, but have polynomial tails like the Pareto distribution

Instead of $n^{-1/2}$ you have to scale by $n^{-\alpha}$

The limiting distributions are called "symmetric $\alpha$-stable distributions", which feels like these basins in the space of distributions which attract other distributions toward it under scaled limits

I dunno if this is a thing/can be made precise

Ted Shifrin

17:29

Balarka, I really never studied this stuff. I did do dynamical systems in grad school, but never took a course in ergodic theory. Sounds like you're really interested :)

Thorgott

I'm not sure what the result by Kolmogorov says exactly, but doesn't Carathéodory suffice?

Balarka Sen

@Thorgott Yeah just knowing the measures on the open sets of the product topology suffices here, which is $\mu([a_1, b_1] \times \cdots \times [a_n, b_n] \times \Bbb R \times \Bbb R \times \cdots) = \prod_{i = 1}^n (b_i - a_i)$

In general though you can still define countable product of $\sigma$-finite measure spaces, in some level of generality

Thorgott

those sets look closed to me, but yes

I think you can even do uncountable products

at the very least for probability spaces

Balarka Sen

Ya I meant open intervals there

@Thorgott Yeah probability spaces is fine, but you can do it beyond those

Secret

I think nowhere continous functions are ungraphable

like the dirichlet function

Ted Shifrin

17:40

@Secret: Just we can't draw a picture doesn't mean that their graph makes no sense. They are graphable; we just can't draw the picture.

Balarka Sen

For example $\Bbb R^{\Bbb N}$ has a measure which is compatible with the Lebesgue measures on $\Bbb R^n$ for all $n \geq 1$

This is known as Kolmogorov extension theorem

Thorgott

Is that different from extending the above it via Carathéodory?

Balarka Sen

How would you extend it?

Secret

I see, still I am wonder if there are really functions that its graph is not a subset of $X \times Y$ and hence making them ungraphable.

something uncomputable seemed to be a good candidate

Ted Shifrin

If you have a function from $X$ to $Y$, then by definition its graph is a subset of $X\times Y$ (with additional properties, of course).

Secret

17:43

ah ok, so it is already ok at the definition level

Ted Shifrin

Whether the function is computable or not is irrelevant. If you know it is a function, then it is a function.

Balarka Sen

@Thorgott Sorry, I'm getting distracted by other things. (1) The Lebesgue measure on $[0, 1]^{\Bbb N}$ can be defined by setting $\mu(\prod (a_i, b_i) \times \prod [0, 1]) = \prod (b_i - a_i)$. This is strictly beasure $\mu([0, 1]) = 1$. (2) It's not clear how to do this for $\Bbb R^{\Bbb N}$, because the Lebesgue measure is not even a finite measure let alone a probability measure on $\Bbb R$.

So I don't know what the "Caratheodory trick" for $\Bbb R^{\Bbb N}$ is. Nonetheless, there is such a measure, which is papa K's theorem

I hope that clears it up

Semiclassical

which K is that?

Balarka Sen

Gotta go now, see ya

@Semiclassical Kolmogorov :P

Semiclassical

of course it is

Ted Shifrin

17:46

Bye, a @Balarka.

Semiclassical

as compared with uncle K, who would be Khinchin

Balarka Sen

Hah well said

Alright now I'm gone

Alessandro Codenotti

17:59

@BalarkaSen There's also some more general versions of the extension theorem. I forgot how they work though, some stuff about projective systems of measure spaces

user736948

In this post why does it suffice to show that the polynomial has exactly one root in the first quadrant? math.stackexchange.com/questions/3027908/…

Ted Shifrin

What part of their justification do you not understand, @user736948?

user736948

"It suffices to show that there is exactly one root in the first quadrant because it is a polynomial with real coefficients, and zeros of polynomials come in conjugate pairs."

Ted Shifrin

So non-real roots are complex numbers and their conjugates.

user736948

Oh I get it... being stupid.

Ted Shifrin

18:06

OK :)

user736948

The polynomial has exactly four roots, if it has exactly one root in the 1st quadrant then it has exactly one root in the 4th quadrant. So 2 roots must be in quadrants 2 and 43 , but as roots come in conjugate pairs, we must have exactly one root in each...

Ted Shifrin

There you go.

user736948

cheers

Ted Shifrin

:)

Hi, DogAteMy.

Akiva Weinberger

Hi

Ted Shifrin

18:14

Happy Chanukah.

Akiva Weinberger

Happy Boxing(?) Day

Eran

18:36

Solve the following initial boundary value problem using the parallelogram identity
utt − ux x = 0 0 < x < ∞, 0 < t < 2x,
u(x, 0) = f (x) 0 ≤ x < ∞,
ut(x, 0) = g(x) 0 ≤ x < ∞,
u(x, 2x) = h(x) x ≥ 0,

Can someone please help me with this question? I tried to draw it out and it's not working for me.

I thought it can be proven with D'Alambert only

Simply Beautiful Art

18:56

0

Q: Verifying tetration properties

In my previous question I asked about the numerical instability and convergence of my tetration. It would seem to be the case that it converges, but suffers from catastrophic cancellation. The definition of my tetration is provided as: $${}^xa=\lim_{n\to\infty}\log_a^{\circ n}({}^\infty a-({}^\i...

sequences-and-series limits tetration

o.o don't mind me advertising, but it seems my question hasn't had much attention, though it could just be the holidays

Semiclassical

19:08

@Eran what’s the parallelogram identity?

Thorgott

$2\lVert x\rVert^2+2\lVert y\rVert^2=\lVert x+y\rVert^2+\lVert x-y\rVert^2$?

Semiclassical

that’s the usual one, yes, but I don’t know what it means in the context of PDEs

Might be equivalent of course

Context:

40 mins ago, by Eran

Solve the following initial boundary value problem using the parallelogram identity
utt − ux x = 0 0 < x < ∞, 0 < t < 2x,
u(x, 0) = f (x) 0 ≤ x < ∞,
ut(x, 0) = g(x) 0 ≤ x < ∞,
u(x, 2x) = h(x) x ≥ 0,

Ted Shifrin

@Semiclassic: The connection didn't leap to my mind, either.

Eran

@Semiclassical : users.math.msu.edu/users/yan/847ch4.pdf, Lemma 5.1 on page 3

Ted Shifrin

This has nothing to do with the vector identity you wrote down, @Eran.

It's just about the geometry of the PDE (note reference to slope $\pm 1$).

Eran

19:24

What do you mean? It sure does. We use the parallelogram identity and D'Alembert to find the solution u for the wave equation.

It's just an example of slope +-1 but it can be any constant that is equal on both sides.

Ted Shifrin

I don't agree with you. Where is the vector identity used in their presentation?

Eran

I didn't write the question "Solve using the parallelogram identity". It is in intro to pde book

Ted Shifrin

They're referring to that lemma as the parallelogram identity. It has nothing to do with the vector identity.

Eran

x-ct=const, x+ct=const

Ted Shifrin

You just don't get it. Never mind.

Eran

19:26

What vector identity ?

Ted Shifrin

Oh, Thorgott pasted that in.

The book is using unusual terminology, because that term is used for a totally different thing that we all thought of.

Eran

Yes, that's why I added an explanation when @Semiclassical asked...

Ted Shifrin

OK, sorry. Thorgott muddled everything.

Semiclassical

Parallelogram property, not parallelogram identity, to match the book’s terminology

Eran

Don't know why he did that, probably tried to help but it confused us all.

Ted Shifrin

19:29

Ah, yes, the book says PROPERTY, and you typed IDENTITY.

That also confused us.

But OK, enough.

Semiclassical

Is it a problem from the prof or from the book?

Eran

The question I read from (different book) wrote identity.

Semiclassical

ahhh

But, it’s the same theorem?

Eran

From the book An Introduction to Partial Differential Equations - Y. Pinchover, J. Rubenstein

yes yes, the same.

Semiclassical

Kk.

I don’t immediately see how to use it, in any case

Eran

19:32

Thanks. When I learned about this property it was for the purpose of finding the solution u where we can't use D'Alembert. But I think here I can use it everywhere.

use it= using D'Alembert

Ted Shifrin

Isn't it a rectangle with sides parallel to the axes?

Ah, so no slopes $\pm 1$. Bingo.

Eran

Why sides parallel to the axes?

Thorgott

lol, I'm sorry

Ted Shifrin

Oh, I see. Their example shows how they're applying it. This is just the method of characteristics in disguise, I guess.

Eran

I tried to create triangles for using D'Alembert and thought that we can reach all (x,t) s.t 0<t<2x in this way.

Ted Shifrin

19:37

No, you have to use rectangles with sides with slopes $\pm 1$ to apply that. But you can get anywhere you need to that way.

Semiclassical

The characteristics have slope pm 1 here, while the t<2x boundary has slope 2

Eran

So this is the reason I can't use D'Alembert?

Ted Shifrin

You can still put one of the vertices of the parallelogram on that line.

You just can't put a side of the parallelogram on that line. :)

Semiclassical

Try drawing it in accordance with what we’re saying

Eran

I'm actually not really understanding what you're saying, sorry. Can you please explain more?

Semiclassical

19:39

We’re saying your picture is wrong, to start with

Eran

Ok. Why? h(x) is right, right?

Semiclassical

The black lines should have slope +-1 , but the red line should have slope 2

Eran

Ok. So the red line is okay, but why should they have slope +- 1?

Semiclassical

because your pde is u_xx=u_tt

Not u_xx=4u_tt or something else

Eran

Right!

So something like this (sorry for my bad drawing lol)

Ted Shifrin

19:44

Get rid of that dark parallelogram.

All your lines must have slope $\pm 1$. I.e., you need tilted rectangles.

Eran

Ok just a sec

Anonymous

Hi. Any idea if a self-adjoint operator, in general, can be fully-reconstructed from its spectrum, in the infinite-dimensional case? I suppose the answer is no, unless the operator can be diagonalized in some orthogonal basis. But I'm not sure.

Ted Shifrin

@SanchayanDutta: Look up the spectral theorem. It's a thing in Hilbert space, too.

Eran

Anonymous

@TedShifrin Thanks. I'm reading the page on Spectral theorem from Wikipedia. It appears that for compact self-adjoint operators, an orthogonal basis exists, but it's not immediately clear to me whether the same holds for bounded or unbounded self-adjoint operators in general...

Eran

19:56

Okay I think it goes like this: I can use D'Alembert for all the triangles (all area under x=x). And I need to use the property of parallelogram for all area above that and under x=2x. Am I right?

Anonymous

I see this though: $$A = \int_{\sigma(A)} \lambda dE_{\lambda}$$

Semiclassical

@Eran yeah, that seems right

Putting that together seems to give a reasonably simple formula in the region x<t<2x

Eran

@Semiclassical @TedShifrin Thanks very much :)

@Semiclassical I'll try it now. Thank you!!

Anonymous

20:33

@TedShifrin Nevermind, Hall's textbook apparently covers this topic in detail. Now my confusion is along the lines of - the unique projection valued measures required to reconstruct the self-adjoint operators from requires knowledge of the nature of the self-adjoint operator and not just its spectrum. So it seems it's not correct to say that self-adjoint operators can be reconstructed from only their spectrum

Anonymous

Or perhaps not, because maybe we could somehow run through all possible projection valued measures and find one set that reconstructs a particular self-adjoint operator

Anonymous

But then perhaps multiple self-adjoint operators could be constructed from a given spectrum. The mapping doesn't seem one-one

Anonymous

Then again it could be that those multiple self-adjoint operators are necessarily unitarily equivalent

Anonymous

I'll have to read this properly. The book seems huge :)

Ted Shifrin

I'm afraid I'm not a good help on this stuff. Come back to me with differential geometry :)

Anonymous

20:43

Heh. Thanks, nonetheless! :D

Shine On You Crazy Diamond

21:12

0

Q: Formula for $M_{p^k} = \{ x \in \Bbb{Z} : x^2 = 1 \pmod {p^k}\}$?

Let $M_n = \{ x \in \Bbb{Z}: x^2 = 1 \pmod n\}$. It is a multiplicative submonoid of $\Bbb{Z}$. When $\gcd(a,b) = 1$ then we have: $$ x^2 = 1\pmod {a,b} \iff \\ x^2 = 1\pmod {ab} $$ by the Chinese remainder theorem, which means $M_{ab} = M_a \cap M_b$. My question is, do we have a formula fo...

abstract-algebra elementary-number-theory prime-numbers integers monoid

Eran

21:40

@Semiclassical what would be your formula for this?

Maximilian Janisch

23:49

@LeakyNun I know this is 4 days late but the answer to [this](chat.stackexchange.com/transcript/message/53009045#53009045 ) is Rc6+ followed by ... Rxh7

Leaky Nun

@MaximilianJanisch nice

Maximilian Janisch

Really cool puzzle

@LeakyNun Are you Chessnoobz on lichess?

Leaky Nun

yeah

Maximilian Janisch

wait i'll follow you there :D

Transcript for

$Mathematics$ Mathematics