Mathematics - 2025-01-25

leslie townes

00:00

i write it "\zeta"

:)

CroCo

@pie These are Arabic letters. What you are looking for is $\zeta$.

Thorgott

grab your pen (or piece of chalk) so hard that your hand starts shaking uncontrollably and then perform a downwards motion

this is how professional mathematicians do it

Ben Steffan

do not ask about $\xi$

Thorgott

same, but shake harder

pie

@CroCo I don't know how to write zeta so I write an arabic letter and hope no one notice it :)

@BenSteffan WOW...

CroCo

00:12

@pie There is an online tool that recognizes your drawing and tries to find the symbols.

detexify.kirelabs.org/classify.html

pie

@CroCo Well, it didn't recognise my "zeta"

@CroCo This look very good tbh

CroCo

@pie Well, I guess that is not zeta.

pie

I guess that I should stop using ع as zeta then..

CroCo

Well, it is not zeta, and you shouldn't type it that way. Don't overlook the skills of Tex enthusiasts; they often pick it up much quicker than I did!

pie

does this look good enough?

CroCo

00:22

What did the tool say?

pie

@CroCo it gives zeta as a fourth option.. idk if this is good enough or not

CroCo

Are you attempting to convince me, or are you trying to accommodate the tool?

Thorgott

upxi is crazy

CroCo

My point is when you write letters resembling "zeta", people will notice them.

Ben Steffan

I would read that as a xi

@Thorgott sigh what's upxi?

CroCo

00:29

You're fortunate that the tool ranked it as the fourth option, even though it's terribly inaccurate.

pie

@BenSteffan well I think that I am the first one who learned how to write xi before zeta lol

Thorgott

@BenSteffan I don't know lol

Ben Steffan

@Thorgott wrong answer. correct answer would have been "not much, how about you?" :)

pie

@CroCo well the arabic ح look like zeta enough for the tool :)

Sine of the Time

not muchxi

pie

00:31

Ben Steffan

see sine got it

CroCo

@Thorgott I'm considering buying the reMarkable 2 tablet. Is that a bad idea?

Thorgott

@BenSteffan touché

CroCo

@pie I'm not entirely convinced either.

Thorgott

@CroCo never heard of it

Ben Steffan

00:33

detexify is, well

it's good that it exists but there is room for improvement

CroCo

@Thorgott it is planning to replace the papers as they claim.

@BenSteffan The tool will cry if it is used by doctors.

Sine of the Time

paper + pencil can't be replaced :D

Ben Steffan

@CroCo I know quite a few people here who use one or something comparable and they seem pretty happy with it I think

CroCo

there has been some time without any papers at all.

Ben Steffan

I personally use an ipad, at the cost of, well more

CroCo

00:37

@BenSteffan I don't personally use it. I check the lists and find the symbol.

Ben Steffan

@CroCo yes, that's usually my approach, too

CroCo

@BenSteffan I have an iPad, but I find it inconvenient to recharge the pen frequently. Additionally, the surface feels frictionless.

Ben Steffan

@CroCo ah, so you have the older generation of pen, then? mine charges wirelessly and is perfectly convenient

there's these paperlike.com for the friction problem, but they wear out rather quickly

Homesick Iguana

How do you read math? Do you read everything out in words or do you internalize the symbols while reading the words?

Ben Steffan

not sure I understand the question

CroCo

00:41

I have iPad 12.9 pro (first version).

works fine.

@BenSteffan nice. I will definitely try it.

Homesick Iguana

@BenSteffan for example: "$\chi_T$". Would you read it as "chi sub t"? or just glance over it while comprehending?

Ben Steffan

the latter I guess

I don't really have an inner narrator when reading

Homesick Iguana

oh okay thanks

Thorgott

@Ben Achim Krause confirmed that everything seems to be in order with my conclusion

Ben Steffan

splendid

sad that you got so few upvotes on that :/

on the other hand getting a "Very cool!" from Achim Krause is nice

Thorgott

00:49

yeah lol

upvotes are secondary, plus it's only been a few hours

pie

BTW do you guys read math magazines?

Ben Steffan

math magazines?

I have a look at the Notices every now and again

Homesick Iguana

What do you make of Mellin pairs that satisfy the same functional equation?

pie

@BenSteffan I think it is something that MAA publish between now and then I find solution of certain exercise in problem book lead to it in reference

Homesick Iguana

say $P=(f,g)$ denotes the pair and $f\ne g$

pie

00:56

It is annoying since I can't access it and now I don't have a solution. some of my mse questions are posted because of this

Homesick Iguana

This seems different from functions whose fourier transform are themselves - but there is a parallel invariance

Ben Steffan

the notices are an inadvertently funny publication, at least to a european

they are quintessentially american

Homesick Iguana

like the functional equation is invariant (not the functions themselves).

Ben Steffan

also they have incredible ads

Thorgott

???

Ben Steffan

01:12

?

Thorgott

im just perplexed by the ad

Ben Steffan

it's great isn't it

Thorgott

yes, great and surreal

Ben Steffan

yes, that's why you should read the notices :)

that and the "we couldn't find a dissenting opinion so here's a take by Zeilberger"

Maxime Jaccon

01:59

log(1+2+3) = log(1)+log(2)+log(3) hehe

Ben Steffan

(source: Mochizuki)

Thorgott

02:26

certified Mochizuki moment

Ben Steffan

I think it's actually fairly reasonable,

whether it hits the core of the high schooler's struggle with logarithms is debatable but surely that's at least part of the problem

I guess I chose "live long enough to see yourself become the villain" :)

Thorgott

are you gonna start talking about mutually alien copies

Ben Steffan

yes

Thorgott

oh no...

Ben Steffan

I've found a flaw with your answer: You need to consider the full poly-homotopy equivalence of $K(G, n)$'s...

Thorgott

02:38

a shocking revelation

Homesick Iguana

03:05

Let $$R(s)=\frac{2}{\sqrt{s}}\sum_{n=1}^\infty n K_1\big( 2n\sqrt{s} \big).$$

Then how do you show that $$\frac{1+2R(s)}{1+2R(1/s)}=1/s^t$$? I.e. show that the ratio is a rational function for some $t>0$?

$K$ is the K-bessel function

I guess it's something with Poisson summation..

weeab00

07:34

Any high-rep user can kindly share screenshots of this deleted question? math.stackexchange.com/questions/1821046/…

hbghlyj

08:32

From the answer math.stackexchange.com/a/3743490 it is possible to embed Penrose triangle into curved three dimensional space, something called "nil geometry". I wonder if the impossible cube also can be embedded into curved three-dimensional space.

2

A three-dimensional cube, with adjacent squares linking number 1

one potato two potato

10:21

huh, that is an interesting feature of nil geometry.

Thorgott

13:06

@weeab00 it's asking that a collection of smooth functions in the neighborhood of a point $p$ on a smooth manifold can be extended to a chart around $p$ if and only if their differentials are linearly independent at $p$

आर्यभट्ट

13:19

Help me in this I do not know where I goes wrong :

---

Question:

Consider the following sequence of numbers: 85, 87, 89, 91, 95, 96. We will explore the behavior of the last two digits when these numbers are multiplied.

1. Step 1: Multiply 95 and 96
Start by multiplying 95 and 96. Observe the last two digits of the result. The last two digits of the product are 20 (since 95 × 96 = 9120).

2. Step 2: Multiply the Result by 87
Now, multiply the product from Step 1 (which ends in 20) by 87. Notice how the units and tens digits of the product evolve. The units digit of the product will be…

Number are multiplied

Sorry do not reply I got my answer a silly mistake

It ends 40 then then 60 then 00

Sine of the Time

14:43

:|

Ryder Rude

15:28

Trace (linear algebra)

In linear algebra, the trace of a square matrix A, denoted tr(A), is the sum of the elements on its main diagonal, a 11 + a 22 + ⋯ + a n n {\displaystyle a_{11}+a_{22}+\dots +a_{nn}} . It is only defined for a square matrix (n × n). The trace of a matrix is the sum of its eigenvalues (counted with multiplicities). Also, tr(AB) = tr(BA...

> Let v be in V and let g be in V*. Then the trace of the indecomposable element $v\otimes g$ is defined to be g(v); the trace of a general element is defined by linearity.

does this definition make it obvious that trace is basis independent?

it is supposed to be a basis independent definition

i think it is a basis independent definition for undecomposable elements

Ben Steffan

well do you see any bases around?

Ryder Rude

but is it fishy for decomposable elements? because there would be more than one way to decompose it

and we have to show that the definition by linearity is unique

like there are no inconsistencies

so we are back to the usual basis dependent proof?

@BenSteffan no..

Ben Steffan

good, so it's basis independent

there you go

the real question you should ask if the definition works, i.e. if this gives you something well-defined

Ryder Rude

yes. The definition is basis independent. i guess what we really have to show is if it is consistent

@BenSteffan yes!

Ben Steffan

right, and you should try your hand at this :)

Ryder Rude

15:35

thanks. i will :)

Thorgott

something something universal property

PM 2Ring

16:05

Given any irrational real number $r$ it's possible to find a linear Diophantine approximation for any other incommensurable irrational $s$ in the form $z=ar+b$ with integers $a,b$ such that $|s-z|<\epsilon$ for any given $\epsilon>0$. I have a fairly simple algorithm to find $a,b$ pairs, based on the continued fraction convergents of $r$ and the extended Euclidean algorithm. But I'm not sure how optimal it is, and I'm curious about alternative algorithms.

Eg, $-489030\sqrt{2} + 691596 \approx 3.141592685 \approx \pi$

Ryder Rude

Scott Aaronson: The Greatest Unsolved Problem in Math

►

@Thorgott I'm not familiar with it yet

i am linking something called the Newcomb's paradox

it is a timestamp

it is about decisions and predictions and free will

PM 2Ring

I do have an alternative algorithm when $r$ is a quadratic irrational, based on Pell's equation. Given $p^2 - nq^2=1$ then $p-\sqrt{n}q=1/(p+\sqrt{n}q)$, so it's easy to find $p-\sqrt{n}q<\epsilon$.

@RyderRude It has a Wiki page. en.wikipedia.org/wiki/Newcomb%27s_paradox I remember seeing it in Martin Gardner's column in Scientific American when I was a kid. I find it a bit annoying, because of how it depends on an omniscient being.

Thorgott

16:21

@Ben this should be it math.stackexchange.com/questions/2015592/…

@RyderRude surely you are familiar with the fact that specifying a linear map $V\otimes W\rightarrow U$ is the same as specifying a bilinear map $V\times W\rightarrow U$, that is the entire purpose of introducing tensor products

Ryder Rude

@PM2Ring oh. Aaronson's version is about a machine which can predict what you will do

i found it interesting. i would choose only one box

@Thorgott oh

@Thorgott lemme think

psie

In Rudin's PMA, Theorem 1.37, when he establishes that $|x+y|\leq|x|+|y|$ for $x,y\in \mathbb R^k$, he starts with $$|x+y|^2=(x+y)\cdot(x+y)=x\cdot x+2x\cdot y+y\cdot y=\ldots$$but nowhere has he clearly written that the inner product is bilinear, i.e. the second equality above. This makes me sad. :(

Ryder Rude

@Thorgott the billinear map is (0,2) tensor, yes (assuming U is the field)

@Thorgott but I'm not familiar with the first idea. it seems to be a linear map defined from the space of (2,0) tensors to U. This should also be a (0,2) tensor then

so both ideas define (0,2) tensors

@Thorgott so you are saying that a map $V* \times V -->R$ uniquely defines a map $V* \otimes V -->R$. this is also something like what wikipedia mentions in the "universal property" section

Thorgott

a bilinear map of the first kind uniquely determines a linear map of the second kind and vice versa

Ryder Rude

oh

@Thorgott i haven't studied this motivation of tensor products yet

but thanks for bringing it up

Thorgott

16:35

in your case, these are (1,1)-tensors on V and the trace is the same thing as contraction

@RyderRude you just know it under a different name

Ryder Rude

@Thorgott i will have to think

the billinear map $V* \times V ---> R$ given by $f(v)$ $f\in V* , v\in V$. i think is the same map as the kronecker delta tensor $\delta ^a _b$

and then the Kronecker delta also defines a map $V*\otimes V--->R$

taking the trace of a tensor $T^a_b$ is literally the same thing as doing $T^a_b \delta ^b _a$ !!!

omg why did no else teach it this way

@Thorgott have i arrived at the correct conclusions

and the Kronecker delta remains Kronecker delta in any basis. so ofc this is basis independent

Ben Steffan

17:05

Please, for the sake of everyone's eyes here, use \to

@Thorgott it is done, then

good job

Ryder Rude

@BenSteffan sorry..

user193319

17:20

For path connected spaces $X$ and $Y$, we have that $\pi_1(X \times Y) \cong \pi_1(X) \times \pi_1(Y)$ (omitting base points). Is there a counterexample where this breaks down when at least one of the spaces is not path connected?

Ben Steffan

Well, ask yourself whether your notation makes sense for path-disconnected spaces

Then determine the path components of $X \times Y$ and write down the analogous version of that isomorphism

user193319

I don't know if the notation still makes sense...I thought it did.

Ben Steffan

Ok, question: What's $\pi_1(S^1 \sqcup (S^1 \vee S^1))$?

user193319

Depends on where the base point is. Either $\mathbb{Z}$ or $\mathbb{Z} \ast \mathbb{Z}$.

Ben Steffan

Good, good

But there's no base-point in your notation, so the answer is: the question is not well-defined

you're only allowed to write $\pi_1(X)$ if $X$ is p.c. in the first place

and even then it's a slight abuse

What you should write is $\pi_1(X, x)$

user193319

17:25

Okay, is there a counterexample for $\pi_1(X \times Y, (x_0,y_0)) \cong \pi_1(X,x_0) \times \pi_1(Y,y_0)$?

Ben Steffan

3 mins ago, by Ben Steffan

Then determine the path components of $X \times Y$ and write down the analogous version of that isomorphism

Do this exercise. This is completely elementary, and it's very important that you understand the dependence on the base point yourself.

user193319

Okay.

Thorgott

17:38

@Ben hopefully the last one mathoverflow.net/questions/486605/…

I'll need a couple months without seeing any Eilenberg-MacLane spaces after this odyssey

Ben Steffan

you could just about write a paper about this lol

Thorgott

17:50

perhaps there'll be an excuse a couple years down the line to put this in the appendix of some paper

psie

18:28

I'm reading about the construction of the reals from Dedekind cuts in Rudin's PMA. We have $R$, which is an ordered set by proper inclusion, and at one point, in proving the least-upper-bound property, the author considers a nonempty subset $A\subset R$. Then he says that since $A$ is nonempty, there is a cut $\alpha_0\in A$ and $\alpha_0$ is nonempty. Why is $\alpha_0$ nonempty? I guess I'm asking if it's true that $\{\varnothing\}=\varnothing$?

Because if say $A=\{\varnothing\}$, then this would be considered empty I guess.

Xander Henderson

Is that a cut? What is the definition of a cut?

psie

ah yes, a cut shoud be not empty

Soumik Mukherjee

@XanderHenderson ctrl + x

Sine of the Time

hi chicken

Soumik Mukherjee

Hi sine

Sine of the Time

18:35

how's life going?

Soumik Mukherjee

e4 e5 ke2, like that

What about you?

Sine of the Time

e4 e5 Bd3

Soumik Mukherjee

That's not that bad, if you follow up by c3 bc2 and d4

Sine of the Time

A bit slow tho

but yeah the structure can be ok

Soumik Mukherjee

Yeah

Have you seen this game?

White resigned in a winning position, not just any winning position, it was mate in 1

Sine of the Time

19:12

@SoumikMukherjee what...

it reminds me of a game of Naka vs Hou Yifan I think

Soumik Mukherjee

which game?

Sine of the Time

it was something online

Soumik Mukherjee

ooh

Sine of the Time

Naka threatens mate in one but Hou can promote and protect with the queen (only way to protect mate)

but she resigned

If I find it I'll send it

Soumik Mukherjee

okay okay I think i can remember it

It was a b/c pawn that was gonna promote

Sine of the Time

19:21

yeah something like that

maybe Qc8 protecting h3

Soumik Mukherjee

Yes you're right

found it

pie

Hey guys I have a question: It is known that If $a_n$ is a sequence of real numbers and if $\lim\limits_{n \to \infty} \frac{a_{n+1}}{a_n} =l$ the $\lim\limits_{n \to \infty} \sqrt[n]{a_n}=l$, I wonder if this can be used is limits where $l =\infty $ i.e assume we need to find $\lim\limits_{n\to\infty}\frac{\sqrt[n]{a_n}}{b_n}$ and $\sqrt[n]{a_n},b_n\to \infty$ and $\lim\limits_{n\to \infty}\frac{a_{n+1}}{a_nb_n}=C$ does this imply that $\lim\limits_{n\to\infty}\frac{\sqrt[n]{a_n}}{b_n} =C$?

Sine of the Time

@SoumikMukherjee yes!

Sine of the Time

19:35

@pie are you looking for something like this?

pie

19:55

@SineoftheTime $ \le \liminf a_n^{1/n} \le \limsup a_n^{1/n} \le \limsup(a_{n+1}/a_n)$ says nothing when $\liminf(a_{n+1}/a_n)=\limsup(a_{n+1}/a_n)=\infty$ my question is when we want to find $\lim\frac{\sqrt[n]{a_n}}{b_n}$ $\sqrt[n]{a_n},b_n\to \infty$ and we able to find $\lim\limits_{n\to \infty}\frac{a_{n+1}}{a_nb_n}=C$ does this imply that $\lim\limits_{n\to\infty}\frac{\sqrt[n]{a_n}}{b_n} =C$?

I think this is false but I can't find a counter example

psie

20:54

I'm reading about the construction of the reals from Dedekind cuts in Rudin's PMA. We have $R$, the set of cuts which is an ordered set by proper inclusion. Fix an $\alpha\in R$ and let $\beta$ be the set of all $p$ with the following property: $$\text{there exists }r>0\text{ such that }-p-r\notin\alpha.$$

Rudin shows $\beta\in R$ and that $\alpha+\beta=0^\ast=\{\text{the negative rationals}\}$. When proving $\supset$ in $\alpha+\beta=0^\ast$, the author picks $v\in 0^\ast$ and puts $w=-v/2$. Then he claims that $w>0$ and there is an integer $n$ such that $nw\in\alpha$ but $(n+1)w\notin\alpha$. He says this depends on the archimedean property. How?

From the definition of a cut, $\alpha\neq Q$. At the same time, $Q$ does not have the least-upper-bound property, so how do we know such an integer exists with the claimed property?

I don't understand how the archimedean property is used here.

Thorgott

you're looking at the set $\{n\colon nw\in\alpha\}$ of integers

you want to show it has a maximal element

to do so, it suffices to show that it is non-empty and bounded above

the latter uses the Archimedean property

psie

ok 👍 let me think through what you said

@Thorgott so is $\{n\colon nw\in\alpha\}$ bounded above because $\alpha\neq Q$?

I think you wanted to say "the latter uses the Archimedean property [and $\alpha\neq Q$]"

Sine of the Time

21:24

@pie why not? If $\liminf(a_{n+1}/a_n)=\limsup(a_{n+1}/a_n)=\infty$ then $\infty\le \liminf a_n^{1/n} \le \infty$ and $\infty\le \limsup a_n^{1/n} \le \infty$. Am I missing something?

Thorgott

22:39

@psie yes, and that uses the Archimedean property

psie

ok, thanks!

psie

23:07

actually, showing it is nonempty uses the archimedean property too; since $\alpha\neq\varnothing$, there exists $p\in\alpha$. Now, if $w\leq p$, then $1\in\{n:nw\in\alpha\}$ (since $\alpha$ contains all $q<p$). And if $w>p$, then either $p\geq0$, in which case $-1\in\{n:nw\in\alpha\}$, and if $p<0$, then $-p>0$ and we can find an integer $m$ such that $mw>-p\iff -mw<p\implies -m\in\{n:nw\in\alpha\}$.

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$Mathematics$ Mathematics