Mathematics - 2022-03-20

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leslie townes

00:07

when it was over livvy jumped down, meowed, and washed her face on the couch.

1 hour later…

robjohn

01:12

@leslietownes I'm impressed.

user4539917

I was impressed today also by finding out that the Mayor of Kyiv is Vitali Klitschko who is the first professional boxer ever to carry a doctor title - honorary titles not included.

copper.hat

I find that hard to believe.

Ted Shifrin

What? A boxer with a doctorate?

Rithaniel

01:39

Hi math chat

user4539917

Hi

Rithaniel

It might be that copper.hat finds it difficult to believe that Vitali is the first

user4539917

Calling Dr Klitschko

The first ever!

😳

Ted Shifrin

Hi, Rith.

Rithaniel

Hola, Ted

I've been getting into Hyperbolic geometry, and my lack of "being-a-geometer-ness" has been shooting me in the foot

Ted Shifrin

01:53

Leslie will comfort you.

user4539917

@copper.hat his brother also has a PhD.

Rithaniel

Like, I wanna find the curvature of the plane you can tile with five quadrilaterals around each point (and the space analog), but everything so far has turned into a case of "well, I gotta find out what that means, now."

Ted Shifrin

02:52

Geodesic quadrilaterals? You know angles? Sounds like little Gauss-Bonnet.

Rithaniel

03:31

Yeah, geodesic quadrilaterals. I will read up on the Gauss-Bonnet theorem (and little Gauss-Bonnet?)

Ted Shifrin

Get my text for free.

There’s a section on hyperbolic geometry, too.

Rithaniel

Diff Geo: A first course in curves and surfaces?

Yep, has a section on hyperbolic stuff (I assume planes and not spaces, though. It seems the generalization to spaces is non-trivial, too)

Ted Shifrin

Well, going up in dimension you have to learn what the curvature tensor is. But hyperbolic stuff generalizes pretty easily.

Rithaniel

Yeah, I suppose the thing that's been tripping me up is understanding the curvature tensor. I imagine, once you know it, it's simple, but it looks like Greek to me, right now

Ted Shifrin

04:00

Start with surfaces. Then curvature is geometric. The curvature tensor is understood by sectional curvatures, which are curvatures of surfaces.

copper.hat

04:50

@user4539917 Harold Reitman

robjohn

05:20

@copper.hat I doubted the "only" claim

copper.hat

06:17

Not to detract from either accomplishment...

1 hour later…

user4539917

07:42

I concur.

barista

Second order linear PDE $au_{tt}+2bu_{xt}+cu_{xx} = 0$ can be transformed into wave equation $v_{rr} - e^2v_{ss} = 0$ if $a$ or $c$ is nonzero by coordinate transformation. If PDE is of the form $bu_{xt} = 0$ where $b\neq 0$ then still can we transform this equation to wave equation?

Ignore the above chat.

1 hour later…

Alessandro Codenotti

09:16

On behalf of the DMV (German mathematical society) the university of Münster has set up a web page where Ukrainian mathematicians can apply for temporary positions in Germany and German institution can offer such positions. Please share this link with anyone who might be interested.

PM 2Ring

10:04

@Rithaniel Latvian mathematician Daina Taimiņa developed a way of illustrating hyperbolic geonetry using crochet. She has made the 4th edition of Experiencing Geometry: Euclidean and Non-Euclidean with History open source.

barista

user image

I wonder if the argument make sense. The graph is the domain of influences. I divided regions to apply d'Alembert's formula.

In R_1,R_10,R_7, u(x,t) = 0 also by d'Alemberts' formula

1 hour later…

barista

11:23

Oh R_10 is not 0

barista

11:55

it's 1/4

1 hour later…

Jakobian

13:11

@leslietownes This is called survivorship bias

Survivorship bias

Survivorship bias, survival bias or immortal time bias is the logical error of concentrating on the people or things that made it past some selection process and overlooking those that did not, typically because of their lack of visibility. This can lead to some false conclusions in several different ways. It is a form of selection bias. Survivorship bias can lead to overly optimistic beliefs because failures are ignored, such as when companies that no longer exist are excluded from analyses of financial performance. It can also lead to the false belief that the successes in a group have some special...

@AlessandroCodenotti where can I learn more about pseudo-circles, pseudo-arc and circle of pseudo-arcs?

iamacomputer

13:29

Hey there,

I'm looking for some vocabulary to look up a problem. I'm sure this is an established problem, but I don't know what it would be called:

I have an ID. It is made up of N numbers: { 1, 6, 3, 19 }
I then have another ID. It is made up of the same quantity N numbers { 7, 3, 2, 9 }
I then have many more of these IDs. ID#1, ID#2, ID#3, etc.

I want to make a bound of a set of IDs, such that the bound will be the minimum "hyper-sphere?" which will contain the IDs.

I believe I should be able to calculate the center of each of N numbers, and calculate the radius of each of N numbe…

leslie townes

14:11

iamacomputer: it may be that this is known, but some aspects of your setting are unclear. is it important that N = 4 in the example? from what set are the numbers taken? by a 'bound,' do you just mean you want to count the set of allowable IDs?

Alessandro Codenotti

@Jakobian For the pseudoarc there is a nice exposition from a REU by Wandsnider, googling will take you to the pdf

For the others I'm not really aware of any book where they are described well, I'm afraid one has to look at the original papers

3 hours later…

Koro

16:49

Suppose that F is a field and E is a field extension of F. If a in E is transcendental then F[x] is always a field?

But this is false clearly if we take F=Q and E=R and a= pi. Q[x] is not a field.

But then what is wrong with the following: suppose that $\phi$ is defined from F[x] to F(a) as: $\phi(p(x)=p(a)$ then ker $\phi=\{0\}$. Noting that: $\phi(x)=a$, it follows that range $\phi=F(a)$. So by first isomorphism theorem, $F[x]/\{0\}\simeq F(a)$

Hence F[x] is isomorphic to F(a) and therefore F[x] is also a field.

Joe Shmo

this dude is really pissing me off

leslie townes

i'm seeing both a and x here, i'm not fully sure of what's going on with that.

why is the range of phi F(a)?

Joe Shmo

asks a stupid question, changes it halfway through asking it, and is complaining that I am not answering his idiotic non-questions

leslie townes

the usual proof that F[a] = F(a) involves a being algebraic over F.

Koro

@leslietownes I think this is because: $\phi$ is identity on F and takes x to a.

leslie townes

16:56

koro, OK, but the range is clearly the set {f(a): f is a polynomial} which many of us would recognize as F[a]. but why is this F(a)? i.e. why is it a field?

Koro

and so range phi contains a and F and since range phi is in F(a), and F(a) is the smallest field that does that, I said range phi=F(a).

leslie townes

ok, but you seem to be assuming that the range of phi must be a field. it will be if a is algebraic over F, but there's no general reason to expect that it would be.

Koro

@leslietownes ahh

range phi need not be a field.

Thank you so much :-).

leslie townes

:)

my daughter carried the cat up the stairs this morning. then before putting the cat down, she went down the stairs again.

it's still a mystery to me why the cat allows this.

Koro

@JoeShmo what did I do ? :(

Joe Shmo

16:58

what did you do?

leslie townes

joe if you wanna yell at me in person i'm right here. no reason to talk about 'this dude.'

Koro

yeah :-)

does this ever happen that someone posts answer to your question and you are not notified about it?

say you asked a question and then accepted an answer. After the acceptance, someone posts an another answer to your question, are you not notified about it?

Joe Shmo

I'm passive aggressive like that

and I'm not gonna name names here, Leslie.

leslie townes

this is what makes the joke so good. it could very well be about me. you don't deny it!

Joe Shmo

Koro, it might have happened once or twice

Koro

17:03

that I annoyed you once or twice? :P

Joe Shmo

user image

no there's this dude who asked a stupid question in broken english, then went back through all the comments and I think edited the whole thing to make it look like what he was asking is less stupid than what it actually is

Koro

That looks like Havier Peña from Narcos show.

Joe Shmo

but it's still stupid however way you slice it, and I don't know why it got to me the way that it did, instead of just having ignored it

Koro

*Javier

Joe Shmo

also, @leslietownes the day I get to answer YOUR questions, is a good day.

so it can't be about you, unfortunately.

leslie townes

17:05

it's true that i haven't asked a lot of questions on MSE.

Joe Shmo

I respect his mustache, Koro

Koro

@JoeShmo but one can always see history of the post to see what changes were made in the post so I think you don’t need to delete your comments if you left any.

Joe Shmo

I have 0 intention of deleting anything

iamacomputer

17:19

@leslietownes

N can be any number.
Each number is a positive integer, most likely restricted 2^64, but not necessarily so.

I want to group the IDs so that all IDs within a group a contained within a bound.
I also want to split this group into two groups, most likely by finding some centroid, and sorting.

1 hour later…

robjohn

18:28

@leslietownes only 4. That's even less than me.

leslie townes

i have one drafted up and ready to go. i will drop it in 2023.

robjohn

@leslietownes must not be anxious for an answer.

leslie townes

yes, it's more of a musing than a question.

robjohn

is it an amusing musing?

or just a common query?

AMDG

18:54

Hey, quick question. What's a good way to map $f: \Bbb{R}^{+}\mapsto [0, \infty]$ to $g :[0, \infty]\mapsto [0, 1]$ where $\frac{d}{dx} Cf' = \frac{d}{dx} Cg' = C$ for $C$ an arbitrary constant?

i.e. You begin with a linear function $f$ over all positive reals and I want a linear function $g$ which maps all positive reals to the range [0, 1]. More generally, that $g$ preserves the curvature of $f$ such that their domains and codomains are congruent.

Ted Shifrin

@leslie I have posted zero questions.

General announcement: \mapsto should NOT be used in place of \to.

@AMDG That general announcement was in particular to you. There is no linear function $g$ as you describe. You know better.

leslie townes

ted: that makes sense.

Ted Shifrin

19:10

Which makes sense?

leslie townes

that you've posted zero questions.

mapsto not being used for to also makes sense. as does the nonexistence of a linear function doing such-and-such above. but i was commenting mainly on your not having posted any questions.

Ted Shifrin

Oh, yeah. I did post one on SO. It turns out it was a stooopid question, and Robert Bryant pointed out how it was obvious, in hindsight.

1 hour later…

AMDG

20:25

@TedShifrin Fair enough

At least I still have modular reduction :smiling_face_with_tear:

On the contrary, knowing how to map from $[0, \infty]$ to $[-\infty, \infty]$ would be nice for generalizing Gauss's Diagamma Theorem.

robjohn

21:30

@TedShifrin How about $\newcommand{\mapsfrom}{\mathrel{\unicode{x21a4}}}\mapsfrom$?

PM 2Ring

@AMDG One option is to take the logarithm.

AMDG

Even for complex-valued Digamma?

PM 2Ring

I didn't see any complex numbers in those intervals. ;)

AMDG

Ah, right, I forgot to make the infinities spanish.

PM 2Ring

Logarithm is a bit annoying with complex values, since it's multi-valued. It conventionally has a branch cut on the negative real axis.

AMDG

21:43

To what extent is it true that algebraically manipulating a complex-valued function as though it were real-valued is correct?

robjohn

Lots of case-by-case is in that question. It depends on what you are doing.

afk bbl

AMDG

aight

In that case, it is subjective. My limited knowledge has me understanding that the issue would be division as one concern. $\frac{\alpha}{\beta}$ is not like $\frac{x}{y}$.

If I were to then find some closed form identity of $\frac{1}{\Gamma(x)}$, then $\frac{1}{\Gamma(z)}$ may not be a valid complex-valued identity. That is my concern.

(Just as an example)

AMDG

22:01

My search for a "nice" closed form of complex Gamma has led me to consider the Barnes G function--particularly the formulas 11 and 12 in mathworld.wolfram.com/BarnesG-Function.html

as well as formula 16 in mathworld.wolfram.com/HurwitzZetaFunction.html

As I mentioned before, the intent in finding a closed form instead of looking specifically for any existing identity to approximate is to find something for which there is uniform and rapid convergence over the entire domain of $\Bbb{C}$ (which is made more difficult by the fact that sometimes the formulas listed in MathWorld are either for real part greater than zero, or the domain is not specified).

robjohn

22:25

@AMDG If there is a fast algorithm for $\operatorname{Re}(z)\gt0$, it can be fairly quickly extended to $\operatorname{Re}(z)\lt0$ by the recurrence relation.

AMDG

Thanks, I wasn't aware.

robjohn

$\Gamma(z+1)=z\Gamma(z)$

AMDG

Honestly, it's somewhat difficult for me to develop an intuition for the substance of Gamma itself.

It's easy to tell what the analytic continuation of $a^b$ to all reals is after you look at it for a few minutes, but Gamma is the analytic continuation of factorial.

@robjohn I'm not quite sure about how this recurrence relation works. Could you please explain how it can be used to extend to the rest of the complex plane for me?

Unfortunately, the best fast-converging identity is for an incredibly limited range greater than zero and I don't have the knowledge necessary to extend the domain.

robjohn

Using that recurrence you can compute $\Gamma(z)$ for all values of $z$ if you know it for $0\lt\operatorname{Re}(z)\le1$. For instance, if you know that $\Gamma\!\left(\frac12\right)=\sqrt\pi$, you can get $\Gamma\!\left(-\frac12\right)=-2\sqrt\pi$, since $-\frac12\Gamma\!\left(-\frac12\right)=\Gamma\!\left(\frac12\right)$

AMDG

22:42

Hm, alrighty, well what about formula 28 on page 14 of this? csie.ntu.edu.tw/~b89089/link/gammaFunction.pdf

It was the best I could find for the time being.

Practically every function and operation there can also be subject to argument reduction, although I wouldn't know how to apply argument reduction for Zeta there.

From the same paper, there's a function $\Lambda(s)$ which I tried messing with to find a closed form, and indeed I found an identity, but not one that I could really use with what little I know. You can see that stuff beginning at cell 129 if that sounds remotely curious, here: desmos.com/calculator/5rtfetofyt

The function is found in the paper on page 17.

Anyways, if I substitute $x$ for a complex argument in formula 28, will it still be correct is my only remaining question.

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