Mathematics - 2024-07-26

Zac U.

2:26 AM

Where is an appropriate forum to ask how to make money as an independent researcher of mathematics?

Xander Henderson

@ZacU. No idea. Certainly not SE, unless you are a lot more specific. And probably not Math SE, even then.

But, like, "independent researcher" is not really a viable career path for all but a tiny, tiny, tiny number of people, and I would imagine that none of those people are in mathematics.

(Depending, I suppose, on what, precisely, you mean by "independent researcher".)

Zac U.

It's peculiar to me as for why it's so difficult to be thought as even potentially prolific without usual credentials given how drastically the landscape must have changed in last few decades

Xander Henderson

@ZacU. I'm sorry. I'm afraid that I don't follow.

Zac U.

I mean that companies don't really trust that you could be some Ramanujan type despite the flowing ideas and new tools

Xander Henderson

@ZacU. Okay, but Ramanujan wasn't really an independent researcher, and he certainly wasn't making money as an independent researcher.

Insofar as he made money as a mathematical researcher, it was under the umbrella of Hardy.

(Hardy... not Hilbert. All names which begin with the same letter are isomorphic.)

Zac U.

2:34 AM

yeah I get it, it's not like everyone has need for some grandiose ideas, but it just seems far too many doors are shut entirely closed for folks who grew up with the Internet. To me it's entirely different

Jakobian

@XanderHenderson see $H^k$ discussion with Leslie

Zac U.

I believe it's partially because the types who would have been the financiers of off the beaten path researchers have assumed that with these new tools they will be the next great pioneer

Xander Henderson

@ZacU. I don't follow.

Zac U.

Wolfram as an example. Idk outliers gonna outlay

leslie townes

under the umbrella of Nicolski. got it

Xander Henderson

2:43 AM

@ZacU. But even Steven Wolfram was not an independent researcher in mathematics. He got started at CalTech...

He left academia only after he was able to make money selling Mathematica.

And his "pure research" output since then seems to be greatly diminished. He's primarily a businessman, not a mathematical researcher.

@leslietownes Gah! Now you're going to confuse me with Cyrillic names?!

OH NO!

(or, perhaps, Боже мой!)

Zac U.

You make good points. It shouldn't be assumed being an independent researcher of math will pay handsomely.

Xander Henderson

@ZacU. No, you shouldn't. Indeed, you should assume that being an "independent researcher in mathematics" (whatever that means) isn't going to pay a dime. Because it almost certainly won't.

leslie townes

you'd have to do what people did hundreds of years ago, which is, find some rich sponsor

Xander Henderson

@leslietownes "I'm trying to resolve RH. Please sign up for my Patreon to support my efforts."

leslie townes

well not the patreon approach as much as finding one big whale

Xander Henderson

2:52 AM

"For \$10 per month, I'll you get a sticker with Gauss's portrait!"

leslie townes

then you only have to please one person instead of a fickle internet audience

Xander Henderson

@leslietownes I dunno... the Patreon approach might be fun!

leslie townes

sigh i guess it's time to put on the crop top and do math in a hot tub on twitch again

Zac U.

@leslietownes that's what I'm trying to do. It's proving absolutely difficult. Even with quite solid findings and ingenious approaches to demonstrate

leslie townes

even isaac newton eventually had a day job as master of the mint

Xander Henderson

2:58 AM

@leslietownes You need an OnlyFans. :D

leslie townes

that's where all of this is headed

Xander Henderson

For only \$25, I'll give you the sexiest proof of the Pythagoean Theorem.

leslie townes

for $50 i'll prove it with my feet

Xander Henderson

@leslietownes How much to see you prove it with your *****?

leslie townes

i think we're basically proving that the funding of independent mathematics research is an unresolved problem

Xander Henderson

3:13 AM

Heh.

leslie townes

if i were being mercenary about this i'd say suck up to some rich guy who thinks your math has something to do with cryptocurrency

there just isn't a ton of money out there chasing mathematical results, it is the other way around

Zac U.

True.

Joako

6:02 AM

Hi everyone. Recently I found a YT channel I think is interesting, but in my ignorance I am not 100% sure if it is mathematically rigurous, or instead if cherry-picking concepts in order to plot something controversial. Hope you can take a look and comment if it looks fine. youtu.be/7fRfxiyTKS0?si=MD3q_1MFk2MedGnN

Soumik Mukherjee

6:54 AM

In Eckmann-Hilton argument, we require (a×b)•(c×d)=(a•c)×(b•d) for all a,b,c,d. So I was wondering if we try to state the argument with less variables. Say (a×b)•(b×c)=(a•b)×(b•c) holds for all a,b,c. Will the conclusion still hold? Now the problem is that the proof of the original argument is one liner and I don't see any way of copying the proof. I am unable to construct any counterexamples as well. Any ideas?

🧐

CowperKettle

8:40 AM

> AI achieves silver-medal standard solving International Mathematical Olympiad problems

x.com/PoShenLoh/status/1816500643940511803

Jakobian

8:56 AM

@SoumikMukherjee from this you can prove that both operations are the same. But then the equality doesn't tell you anything

So this is equivalent to those operations being the same

Soumik Mukherjee

Okay so we don't get commutativity?

Jakobian

That's what I said

Soumik Mukherjee

Okay

Thanks

Zac U.

9:28 AM

@leslietownes I reflected on the quote you seem to have created about math theorems and money. In practice it should seem to be a bit misguided functionally, I now think. It comes down to whether you're more likely to use the math... Or is the math more likely to use you. Who was it that mentioned his equation as being smarter than he upon hearing of its utility in predicting physical phenomena

Soumik Mukherjee

Matrix multiplication is an example where the operation holds but it is not commutative

psie

10:09 AM

@psie I think we should be able to write $$S_N = \sum_{j = 1}^{\infty}X_j \mathbf1_{j \leq N},$$ so $S_N$ is $\sigma((Y_n)_{n\in\mathbb N})$-measurable, where $Y_1=N, Y_2=X_1, Y_3=X_2, \ldots$.

Jakobian

10:45 AM

@SoumikMukherjee any non-commutative operation with identity element

Once you know the two operations are equal, $(ab)(bc) = (ab)(bc)$ is what the equation reads, this is always satisfied

Soumik Mukherjee

11:08 AM

Yes

psie

11:42 AM

I am looking at this library of modified Bessel functions of the 2nd kind. For instance, in equation 10.32.9 they write on the right that $|\operatorname{ph}z|<\frac12\pi$, where $\operatorname{ph}z$ is the phase of $z$. Is this when the integral will converge?

Jakobian

12:01 PM

@psie yes. This seems to be equivalent to $\text{Re}(z) > 0$ which is the factor that makes integral of $e^{-z\cosh(t)+...}$ converge. The $\cosh(t)$ part is dominative and the only thing that matters for convergence is the real part

There might be other values of $z$ with $\text{Re}(z) = 0$ for which it converges, I guess

If $\text{Re}(z) = 0$ then this seems to depend on the value of $\nu$ as well

i.e. the general area in which this converges is $\text{Re}(z) > 0$, but the formula might hold for some points with $\text{Re}(z) = 0$ depending on value of $\nu$

psie

12:28 PM

Nice, thanks!

Jakobian

2:03 PM

@Thorgott math.stackexchange.com/q/2927185/476484

Zac U.

3:14 PM

Anyone here have an idea about what it means to supposedly be unable to know something in math? Is this the correct and usual framing?

Xander Henderson

@ZacU. No? I don't understand your question...

Zac U.

I've heard but can't recall how the general terminology goes. But even beyond that hearing that something like Chaitin's Omega constant cannot be computed or whatever beyond the first few digits just seems off

Though I suppose if an algorithm did compute it but couldn't be verified it wouldn't count?

Xander Henderson

I still really don't understand what you are asking.

Zac U.

@XanderHenderson are you familiar at all with experimental mathematics? Sometimes beauty and potentially/ hopefully probability can be the guide

Xander Henderson

@ZacU. "At all"? In the sense that I know it exists... sure.

Zac U.

3:30 PM

That's about all I know too xD

psie

4:12 PM

If $E|Y|^n<\infty$ for some $n\in\mathbb N$, then $\varphi_Y^{(k)}(0)=i^k\cdot EY^k$ for $k=1,2,\ldots, n$, where $\varphi$ is the cf of $Y$. Suppose $E|S_N|<\infty$, where $S_N$ is the random sum of random variables ($N$ independent of the terms, which are i.i.d.). We may thus state $$ES_N=\frac{\varphi_{S_N}'(0)}{i}.$$

I'm trying to show the RHS indeed gives the expectation. Now, the cf of $S_N$ is given by $\varphi_{S_N}(t)=g_N(\varphi_X(t))$, where $g_N$ is the pgf of $N$ and $X$ is a term in the sum. We know $g_N'(1)=EN$ where the derivative is understood as a left-hand derivative. From chain rule, $$\varphi_{S_N}'(t)=g_N'(\varphi_X(t))\cdot\varphi_X'(t).\tag1$$But how do I get $\varphi_X(t)$ approaching $1$ from below when it may be complex-valued?

Thorgott

@Jakobian the answer to that question is no

Jakobian

@ZacU. I have an idea what it can mean

@Thorgott oh really? Why

Zac U.

@Jakobian awesome ill take note if you'd like to share

Thorgott

the inclusion of a point into a manifold (neither compact nor with/-out boundary matters) is always a closed cofibration and a closed cofibration is a homotopy equivalence iff it is a deformation retraction

the technical result this boils down to is that manifolds are ANRs

Jakobian

@ZacU. For example, Godel have proven that in certain formal systems there exist statements that can be neither proven nor disproven. This is Godel's first incompleteness theorem

@Thorgott by "it is a deformation retraction" are you referring to deformation retracts as in Hatcher or wikipedia?

also what is "it". I understand that closed cofibration is some kind of map $f:A\to X$. Do you mean to say that we are only taking inclusions here, and it refers to $A$?

Thorgott

4:29 PM

deformation retract in the strongest possible sense

yes, I'm referring to the map, as cofibrations are automatically embeddings

so, if I am to be very precise, a closed cofibration $i\colonA\rightarrow X$ is automatically a closed embedding and it is homotopy equivalence iff $X$ deformation retracts (in the strongest possible sense) to its subspace $i(A)$

Jakobian

I see, thanks

Thorgott

or, alternatively $i$ has a homotopy inverse $X\rightarrow A$ in the slice category under $A$ :P

Jakobian

I found that question because I was searching for a contractible space $X$ such that $\{x\}$ for $x\in X$ is never a strong deformation retract. So far, I found some discussion online about this but without much details, all referring to Hatcher

Thorgott

hmm, I could swear I've seen that question before

ah wait, the standard example from Hatcher works, I must've been thinking of something works

it's the zig-zag graph with hairs growing from it

probably not an adequate description lol

Jakobian

that works as an example such that there is $x\in X$ with $\{x\}$ not a strong deformation retract, but there are $x\in X$ such that $\{x\}$ is a strong deformation retract

there are some posts online that give examples of compact $X$ in $\mathbb{R}^3$ with the above property, but they are kinda complicated

冥王 Hades

4:47 PM

@CowperKettle This isn't too surprising, and contrary to what many think, most IMO problems are not "non-standard/obscure/never before seen problems". They're very much based on well-known and studied patterns, in fact that's how IMO-contestants are trained. That's how I solve many Olympiad problems. With this in mind, training an AI on a bunch of well-known patterns isn't impossible to achieve.

On the other hand, when you do get a truly non-standard problems not based on any previously known standard patterns, you get the IMO 2011 Windmill situation. Not the hardest* problem on the test by mile, but the one with the lowest correct solve rates, even by would-be gold medalists.

Hardness here can be defined as what the question number is. Q3's and Q6's are typically the hardest. This problem was Q2.

Jakobian

@冥王Hades that's interesting because in the post they're saying that they've been a coach for American IMO team for 10 years or so, and hold a contrary opinion

冥王 Hades

@Jakobian It's not exactly contradictory. Look at his very last sentence, where he talks about how the coaches would look at old, obscure competitions with similar problems.

Jakobian

sure, there is some agreement but that's not exactly what I've been pointing at

冥王 Hades

It's also worth noting that a lot of these problems do involve some non-standard tricks that you've either seen before (by training extensively on problems of similar difficulty), or are just lucky enough to have that Eureka moment in the allotted time.

I was able to solve the 2011 Windmill problem without hints or help, but it took me 3 hours, the duration of the entire exam so I most likely wouldn't have solved it in the actual contest.

Jakobian

5:08 PM

For me it doesn't matter how fast you solve it

It's still admirable, good job

Thorgott

@Jakobian which $x$?

Pizza

5:24 PM

Hi!

Sine of the Time

@Pizza hey

long time no see

Jakobian

@Thorgott $(0, 1)$

ah no this is the comb space

this is what I get for speaking from memory

冥王 Hades

6:17 PM

One day I'll get ChatJax working on my PC, I swear I will.

CowperKettle

@冥王Hades Ah! So it's a bit of hype then, about AI being super-smart to solve it.

The unexpectedly hard windmill question (2011 IMO, Q2)

►

robjohn

6:40 PM

@冥王Hades do you have the bookmarklet installed, or are you having trouble installing it?

Jakobian

@XanderHenderson math.stackexchange.com/questions/4950907/…

something you'd like perhaps?

Thomas Finley

6:56 PM

0

Q: Help needed in understanding the definition of a continuous random variable.

I was studying Mathematical Statistics when I stumbled upon the definition of a continuous random variable. It confused me badly. I will precisely state the issue. But before that, I think I should make the definitions that I am using, for various terms, clear, in order to avoid ambiguity from m...

Need some help with this.

Jakobian

@ThomasFinley and the question is?

I don't believe that you can't type it in here in just one sentence

Thomas Finley

@Jakobian Sure, why not! But as they say, giving the most references and context helps the asker as well as the answerer. I tried to be very clear in the post, to the best of my ability.

The question is: what is a continuous random variable?

Jakobian

If you present me with walls of text, even if its clear, I'm not going to read it

Unless perhaps I'm interested in it, but in this case I'm not

@ThomasFinley a continuous random variable has two definitions depending on who you ask

Thomas Finley

@Jakobian may I ask what are they?

Jakobian

some people distinguish between absolutely continuous random variables and singular continuous random variables

Thomas Finley

7:05 PM

Never heard it that way, tho

Jakobian

others say that continuous random variables refer to the absolutely continuous random variables

Thomas Finley

Ok, then what's that?

Jakobian

I am going with the convention that a continuous random variable can be either absolute or singular

Thomas Finley

What does absolute and singular mean in this context?

Jakobian

then a random variable $X$ is continuous when $P(X = x) = 0$ for all $x\in\mathbb{R}$ or equivalently $F(x) = P(X \leq x)$ defines a continuous function $F:\mathbb{R}\to [0, 1]$

for $X$ to be an absolutely continuous random variable it means that there is a measurable function $f:\mathbb{R}\to\mathbb{R}_+$ such that $P(X\in A) = \int_A f$ for any measurable set $A$

and finally, for $X$ to be singular continuous random variable it means that $X$ has no "absolutely continuous part" i.e. there exists a set $A_0\subseteq\mathbb{R}$ such that $\mathbb{R}\setminus A_0$ is a null set and $P(X\in A_0) = 0$, yet $X$ is continuous

Thomas Finley

7:10 PM

@Jakobian yes, I have seen some definitions like that in a few books. But umm...is there something simpler than that? For eg, I am studying from the book, by John E Freund and there there is no mention of continuity (of real analysis) while speaking about continuous ra dom variables till the portion I have studied so far... But thanks! Once I advance a bit more in the course, I'll surely make it a point to grasp the things you wrote.

Jakobian

Simple is subjective. This is simple to me

Sine of the Time

Can't think about a simple example in $\mathbb R$ that disproves the claim

2

Q: Does $f \in L^1 \implies f \log f \in L^1$?

Let $\Omega$ be a bounded domain in $R^3$. Suppose $f \in L^1(\Omega)$ with $f>0$ a.e. Prove or disprove: $f \log f \in L^1(\Omega)$. Attempt: I am stuck at this for a while and unable to make any progress. This is the only inequality that I could find but it does not help to conclude anything:...

real-analysis integration lebesgue-integral

Jakobian

If you don't like that a continuous random variable has a reference to continuity, then go with the definition that it means $P(X = x) = 0$ for all $x\in \mathbb{R}$

I am simply providing you the information, you decide how you want to think about these things, its something you can handle

Thomas Finley

@Jakobian this sounds simpler.

Jakobian

Its also included in above

6 mins ago, by Jakobian

then a random variable $X$ is continuous when $P(X = x) = 0$ for all $x\in\mathbb{R}$ or equivalently $F(x) = P(X \leq x)$ defines a continuous function $F:\mathbb{R}\to [0, 1]$

Thomas Finley

7:14 PM

@Jakobian maybe I should call this an encapsulation of the elaboration you presented there :P

But nevertheless, I'll try thinking about this and let you know.

leslie townes

8:54 PM

as a background thing, i wouldn't expect a lot of sources to give a coherent definition of 'continuous random variable' or even 'random variable'

and i would be inclined to push back against the idea that this concept has some kind of meaning independent of a specific source you are getting it from

qiaochu yuan's recently submitted answer gives a lot of useful background but maybe obscures the point that i would make, which is, a whole ton of people are going to use these words without any specific reference to anything in mind

such that it is, in general, error to assume that these words are a reference (if only an imprecise reference) to some specific background concept that we all know

if a specific textbook uses this term without defining it, that is telling you something, and something very different from "i mean to refer to a background precise definition of this term that you can infer from considering other sources"

Jakobian

11:33 PM

@leslietownes didn't I provide the meaning?

or do you mean this difference between continuous and absolutely continuous random variable

leslie townes

well, i certainly meant that, but also, have you ever opened a shitty statistics book? you will not find real definitions in there

and not because the authors all know what the real definition is and are just forgetting to include it

but yeah the difference between continuous and absolutely continuous is something that a lot of books don't even raise, let alone meaningfully address

and more specifically, a lot of that asker's questions above, begin from a place of 'we should be able to figure this out' where i don't think it is warranted

'what is a quasilinear differential equation' being another one

leslie townes

11:53 PM

"is a constant random variable normally distributed" another example of this kind of question

:)

Jakobian

@leslietownes yeah, they probably just don't know it. Which is funny if you think about it

its like a hidden shitpost of life

leslie townes

a lot of it is just in the realm of "i can give you what an answer ought to be"

Derso

Is statistics allowed in the chat?

leslie townes

@Jakobian funny or frightening

Jakobian

My type of humour is that anything absurd in life is worth enough to get a laugh out of it

leslie townes

11:59 PM

derso it depends, i don't think anything would be expressly forbidden but you might find that the audience for a question is greatly reduced by how much stat background knowledge you presume in a question

and whether an answer to your question would involve a number with a decimal point in it

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$Mathematics$ Mathematics