Mathematics - 2024-07-20

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Obliv

2:13 AM

In computer science, a data structure is a data organization, and storage format that is usually chosen for efficient access to data. More precisely, a data structure is a collection of data values, the relationships among them, and the functions or operations that can be applied to the data, i.e., it is an algebraic structure about data. Data structures serve as the basis for abstract data types (ADT). The ADT defines the logical form of the data type. The data structure implements the physical form of the data type. Different types of data structures are suited to different kinds of applications...

apparently data structures are algebraic structures. huh the more you know

2 hours later…

copper.hat

4:04 AM

@Obliv They pretend to be. Even other structures, Haskell has monads.

Obliv

4:59 AM

@copper.hat I've seen that word before in very different contexts (Leibniz' monad philosophy and somewhere in nlab) I'm guessing monads are a computer science thing?

Q_p

@XanderHenderson was referring to your statement that the RH is basically unsolvable in 5 pages.

If you have a solid background in basic analytic number theory, you can verify the proof for yourself. It won't take much of your time.

2 hours later…

Soumik Mukherjee

6:38 AM

@Obliv "A monad is just a monoid in the category of endofunctors"

4 hours later…

Ryder Rude

10:20 AM

Jakobian

11:10 AM

@Obliv and category theory. In philosophy its something else

psie

11:23 AM

I asked a question on main concerning the generalized gamma distribution and its mgf, for those interested.

One of the questions is, could $\beta$ and $\alpha$ both be negative?

Another question is, how does one show the integral $\int_{-\infty}^\infty e^{tx}x^{\beta-1}e^{-x^\alpha}\, dx$ diverges for all $t>0$, and for appropriate values of $\alpha$ and $\beta$?

Jakobian

It diverges for $0<\alpha<1$

Xander Henderson

@Q_p I don't find it fun or interesting or worth my time to read other people's broken resolutions of RH. If I were asked by a journal editor to read such a proof as part of a peer review process, is take the time, but that isn't the case, here.

Besides: while I have a stronger background in analytic number theory than the average bear (thanks to my advisor taking a fractal-based run at RH some 20 years ago), my expertise is in fractal geometry.

psie

11:54 AM

@Jakobian how do you see this? :)

Jakobian

12:50 PM

@psie you still need $\beta > 0$

3 hours later…

psie

3:22 PM

@Jakobian for the pdf to converge I believe too, but for the $n$th moment to exist, we don't. $\beta>-1$ will do in that case.

I know that's what you said yesterday kind of. Since if you proceed by a substitution, $y=x^\alpha$ and rewrite the integral in terms of a gamma integral, then according to the convergence of the gamma integral, it'll converge if $\beta/\alpha>0$. But if you have the extra factor $x^n$ from the moment calculation, you'll get convergence if $(\beta+n)/\alpha>0$, so $\beta>-1$ since it should hold for $n=1$.

psie

3:39 PM

what about $\alpha,\beta$ both being negative? would that work?

I guess that would cause problems with the moments to exist...

Jakobian

3:52 PM

@psie ?

How is the nth moment of something that's not even a random variable and doesn't exist is supposed to converge

No. $\beta > 0$

psie

what about $\beta<0$ and $\alpha<0$?

Jakobian

@psie well of course we end up with some kind of distribution then as well

psie

Jakobian

I guess you could call it "generalized inverse gamma distribution"

if it already doesn't exist under some other name

will moments be different? All the calculations you did for $\alpha, \beta > 0$ hold for $\alpha, \beta < 0$

psie

4:09 PM

well, we'll have the condition $(\beta+n)/\alpha>0$, and moments will certainly be different then, or?

Jakobian

$\int_0^\infty x^r\cdot x^{\beta-1}e^{-x^\alpha}dx = \int_0^\infty y^{(\beta+r)/\alpha-1} e^{-y}dy$

the condition for $r$th moment $E[|X|^r]$ to exist is that $\beta+r < 0$ in the case that $\alpha, \beta < 0$

so only some moments will exist, but for large enough $r$ they won't

psie

interesting

Jakobian

this was kind of the same for your case, you just end up with $\beta+r > 0$ which is true for $r > 0$ anyway

(unless we want to look at the negative moments i.e. we allow $r < 0$)

$E|X|^r$ exists for $r > -\beta$, when $\alpha, \beta > 0$

psie

yeah

3 hours later…

user 85795

7:16 PM

@copper.hat will you be watching any of the Olympic Games?

:^)

copper.hat

@user85795 probably not, i don't have cable.

user 85795

Catch it on YouTube.

Wikipedia has the entire schedule.

Q_p

7:58 PM

@XanderHenderson you could have simply put tour point across without such condescending language. To be honest, you don't sound like an expert of any sort in analytic NT.

*your

2 hours later…

Xander Henderson

9:49 PM

@Q_p (1) You don't like my tone. Oh well. But you need to understand what you have done: you came into this room, and you asked a large group of people---none of whom can be expected to be experts in analytic number theory---to give opinions on a non-peer reviewed, self-published paper which purports to resolve RH. You are asking other people to spend their precious time on something that (a) they are almost certainly not expert in, and (b) is almost certainly nonsense.

(2) I have never claimed to be an expert in analytic number theory. What I said is that I likely know more numerical analysis than the average bear, but my expertise is in fractal geometry. My phd thesis spends more pages than average on $p$-adic analysis, largely based on Tate's thesis, and I took a couple of courses on analytic number while in graduate school. Given that most mathematicians don't do this, I am confident in my claim that I know more analytic number theory than average.

But I don't claim that this makes me an expert.

So your condescending remark---"To be honest, you don't sound like an expert of any sort in analytic NT"---kind of misses the point.

Soumik Mukherjee

10:05 PM

This is a random 3am question: My right hand is my dominant hand. But when I ride bicycle with one hand, it feels that it is much easier to do with the left hand. I get a better control at balancing. So I was wondering how is my non dominant hand better at it than my dominant hand?

Xander Henderson

10:32 PM

Your dominant hand is dominant in the sense that it is better at fine motor tasks. Steering a bicycle is kind of more of a gross motor task, perhaps?