Mathematics - 2017-07-20 (page 2 of 4)

Daminark

06:02

Hey @Astyx!

Astyx

06:36

Ohi

What's up ?

Daminark

Not much, how about you?

Astyx

Same

Daminark

Mygr8 to dc real quick

Astyx

@LeakyNun You probably can compute the order of all elements in $(\Bbb Z/n\Bbb Z)*$ in $\mathcal O(n\ln n)$. I'm not convinced you can a priori compute the order of a specific element out of the blue in $\mathcal O(\ln n)$ (at best $\mathcal O(n)$)

I might be totally wrong though

Daminark

Oh also hai @Alessandro

The Raiders of Las Vegas

07:11

@arctictern bring back @anon

arctic tern

le sigh

The Raiders of Las Vegas

:(

anon

back

2

The Raiders of Las Vegas

Welcome @anon :-D

thnx

A room needs its owner.

Alessandro Codenotti

08:17

I wasn't really there, hi @Dami

Astyx

How are exams going ?

Alessandro Codenotti

It's over, it's done (cit. Frodo after destroying the one ring)

Astyx

Precious.

When will you have the results ?

Alessandro Codenotti

I'll have an oral exam in analysis in September, but the written part went very well and I have no other exams

Astyx

Oh cool !

Alessandro Codenotti

08:20

I already did, just 30 minutes ago

What about yours? Did you get the results?

Astyx

I got my marks (but not my rank) for Polytechnique

And I think I have good chances of being accepted

Alessandro Codenotti

that's great

when will you know for sure?

Astyx

In a week or so

Not too sure

The derivative of a uniformly continuous function is assymptotically bounded right ?

Alessandro Codenotti

what does asymptotically bounded mean? eventually smaller than a constant?

Astyx

You can't have $(u_n)_{n\in \Bbb N}$ going to $\infty$ and $f'(u_n) \to_{n\to\infty}+\infty$

There is a real $m$ and a bound $M$ such that for all $x\ge m$ $|f'(x)|\le M$

(I am making this terminology up)

Alessandro Codenotti

08:24

hmmm I'm not sure

Astyx

No, it seems I'm wrong

Only the converse is true

Alessandro Codenotti

I mean $\sqrt{x}$ is uniformly continuous and has unbounded derivative in a nbhd of $0$, can one costruct a uniformly continuous function that "looks like" $\sqrt{x}$ in $0$ at all integers? I think so

Astyx

But its derivative is bounded around $\infty$

Vincenzo Oliva

@Astyx Boundedness of the derivative implies Lipschitz continuity which implies uniform comtinuity, but not the other way around. Consider $f(x)=\sin (x^n)/(1+x^2)$ for $n\ge4$, as $x\to\pm\infty $

Alessandro Codenotti

yeah, but I want to construct a function that has at all integers the same behaviour that $\sqrt{x}$ has in $0$ @Astyx

Astyx

08:27

Oh right I see why I am wrong

Vincenzo Oliva

I omce asked a pretty comprehensive question on the matter, you may want to look at it: math.stackexchange.com/questions/1837681/…

Astyx

@VincenzoOliva Wait, the derivative of that function isn't bounded is it ?

Vincenzo Oliva

It should be read "in order", that is first number (0), then (1) etc.

@Astyx Indeed. Didn't you ask if u.c. implies boundedness of the derivative?

The answer is no

Secret

[Old news]

Astyx

Oh yeah I'm being silly

Thanks a lot !

Daminark

08:31

Hey @AlessandroCodenotti

Vincenzo Oliva

@Astyx Nah, you're welcome!

Alessandro Codenotti

@Vincenzo I see you're Italian as well! Where do you study?

Secret

There are at least two types of maths people:
1. Amateur mathematicians which explores maths by intuition and inventing their own exploration tools
2. Professional and/or amateur mathematicians which explores maths by following a systematic procedure of proving and constructions

quantamagazine.org/marjorie-rices-secret-pentagons-20170711

NB It is an open question whether there are professional mathematicians that explores maths mostly intuitively

Astyx

I would never be able only to follow procedures @Secret, to do math I need to feel what's going on

(this being said, I'm not a professional, yet)

Secret

I suspect this guy might fit the token for that open question:

Inter-universal Teichmüller theory

In mathematics, inter-universal Teichmüller theory (IUT) is a theory created by Shinichi Mochizuki (2012a, 2012b, 2012c, 2012d). It is an arithmetic version of Teichmüller theory for number fields with an elliptic curve, introduced by Shinichi Mochizuki (2012a, 2012b, 2012c, 2012d).This theory is considered to be the most fundamental development in pure mathematics in several decades. Other names for the theory are arithmetic deformation theory and Mochizuki theory. Several previously developed and published theories in the previous 20 years by Shinichi Mochizuki are related and used in many ways...

Vincenzo Oliva

08:35

@Alessandro Yep, in Milan. What about you? I'm investigating on the procedure of making students handwrite what the say in an oral exam. This is very often done in my university, does it occur in yours, at least for undergraduates? I feel like it would be better to use a keyboard and a monitor, provided one has been taught LaTeX

Secret

For me, I am an amateur who kinda straddles somewhere in the middle of the two extremes

Vincenzo Oliva

Then the printed document would be manually signed. I even asked a question in the math educators Stack Exchange, I'm interested in what happens in other countries, too.

Daminark

I think research mathematicians operate by intuition more than you think @Secret

They just get really good at translating that intuition into formality

But it's p rare that people operate immediately via formality, heuristics typically precedes that

Alessandro Codenotti

@VincenzoOliva I'm in Trento. I've had a few oral exams where I had to write on the blackboard and a few more where I had to write on a sheet of paper, some professors want you to write everything you say, others are satisfied with much less as long as what you say is complete and correct, I don't think there's a precise standard

Secret

For me:
structures and questions I constructed and raised are guided by prior experience, generalisations, dreams and memory correlations, but when exploring these structures, I tried to be rigorous by following well establish principles. In the event I need to invent things, they are phrased in a language that mathematicians can understand (though to be honest, my mastery of the terminology is still poor. Combined with my weird personality, people can sometimes have trouble understand my attempt in translating my ideas into the maths language)

Vincenzo Oliva

08:41

@Alessandro The blackboard does make sense, but how about writing in a sheet of paper? What do you think of typesetting instead? Personally, that extra care I have to put for clear but fast writing is kinda distracting, so I'm looking for other experiences

Secret

@Daminark Hmm, guess I still need work to better understand mathematicians and maths people culture and thought process

Daminark

shrugs

BORN TO LEARN

this could be.

Alessandro Codenotti

@VincenzoOliva I don't know, that's not a problem for me

Astyx

Hi Steamy

abenthy

08:58

Is every connected semisimple subgroup $G$ of $GL(n,\mathbb{R})$ a subgroup of $SL(n,\mathbb{R})$? I read that $G$ is perfect, i.e. $[G,G] = G$, and hence each $g \in G$ has to have determinant one in Mostow's book on Rigidity. But I don't quite understand why $G$ must be perfect when it is semisimple (My Idea: Maybe $G/[G,G]$ is semisimple too and since its abelian, it must be trivial, i.e. $[G,G] = G$?)

Ah and $G/[G,G]$ must be connected since $G$ is connected. This could actually work out...

Daminark

Hey @s.harp

s.harp

Hello @Daminark

and everybody else

Daminark

How've you been?

s.harp

confused

Daminark

About?

s.harp

09:03

Usually about my thesis, sometimes about people and social interactions! With non-negligible mass also about myself

Daminark

People are definitely pretty confusing. What's your thesis on?

Also hey @Tobias

s.harp

The general idea is to find a procedure that allows one to optimise a "dirty" system in a non-generic way.

Abcd

How would you represnt $\sqrt{-121}*i$ in terms of $i$ and why?

Daminark

I see

Tobias Kildetoft

@Abcd Using the sqrt sign is a poor idea when dealing with complex numbers

Abcd

09:07

@TobiasKildetoft Why?

I am new to complex numbers (Started this chapter today)

I think the solutions are: $-11$ and $+11$

Never mind...

Bye

Abcd

09:30

I don't understand where the $i$ disappeared in this proof.

Can someone please tell why there's no i in the proof?

(Please ping me)

Tobias Kildetoft

@Abcd I am not sure where you mean the $i$ disappeared

Abcd

09:46

@TobiasKildetoft in the proof

anon

which part of the proof don't you understand?

Tobias Kildetoft

@Abcd But there is not supposed to be any $i$ there

Abcd

@TobiasKildetoft after "therefore"

@TobiasKildetoft why?

Tobias Kildetoft

The definition of the norm does not involve an $i$

Abcd

@anon finding the modulus of the complex number -a-bi

anon

09:48

well, -a-bi is (-a)+(-b)i. do you know what the definition of |x+yi| is?

Abcd

Yes, why does the definition not involve $i$.

anon

it's sqrt(x^2+y^2) isn't it? well, in that case, |(-a)+(-b)i| must be sqrt((-a)^2+(-b)^2)

Tobias Kildetoft

@Abcd Because that is the way the definition is

anon

|z| is defined to be the magnitude of z. length is a real number.

The Raiders of Las Vegas

Proof by definition.

Abcd

09:50

@TobiasKildetoft Does eliminating $i$ help us in finding the complex number's distance from zero. If yes, how?

The Raiders of Las Vegas

The reason is the definition.

anon

length is a real number. it can be found with the pythagorean theorem. draw a triangle.

Tobias Kildetoft

@Abcd Now you are asking something completely different

You were asking about this specific proof, which uses the definition of the norm. Why the norm is defined as it is is not related to this proof

Abcd

@TobiasKildetoft So please answer that so that I can inter-relate stuff

Daminark

Basically, multiplication of complex numbers should add angles and multiply lengths

Tobias Kildetoft

09:52

@Abcd anon already answered that part

Abcd

@anon I don't know how to draw triangles for complex numbers @TobiasKildetoft

anon

the ray there is the hypotenuse. x is the base, y is the height.

Daminark

@Abcd If you have a complex number $a+bi$, you can think of it as an ordered pair of real numbers, so on $\mathbb{R}^2$ this becomes $(a,b)$

Lol sniped

Abcd

@anon Interesting!

@Daminark "ordered pair of real numbers"?

Daminark

@anon no you had sniped me with the picture :P

Tim The Enchanter

09:55

@Abcd We define complex numbers in terms of ordered pairs of real numbers

Daminark

So, you know a complex number by its real part and imaginary part, yeah?

Abcd

@anon what next?? How to prove?

@Daminark Yes

anon

yes, as we learn in intermediate algebra, an ordered pair of real numbers (x,y) represents a point in the plane

Tim The Enchanter

@Abcd You may know the diagram as the representation of a complex number in the Argand plane.

The Raiders of Las Vegas

P(x,y).

Abcd

09:56

@TimTheEnchanter Oh. I haven't reached Argand plane yet

anon

@Abcd the proof that |-z|=|z| is already there in the picture...

@Daminark I deleted and edited because you cut off my response

Tim The Enchanter

@Abcd Well the picture anon posted is it.

Daminark

So, in a sense, if we christen the first coordinate as being the real one, and the second as the imaginary one, you can represent a point in the plane using two numbers as in the picture anon drew

@anon sorry :/

The Raiders of Las Vegas

Work with the point P(x,y). @Abcd

Abcd

@anon why is modulus of z = x^2 + y^2

anon

09:57

oh, you mean the proof the length is sqrt(x^2+y^2)? that follows from the pythagorean theorem. this is something you should have learned well before complex numbers...

Abcd

@anon But there's an i too

@TheRaidersofLasVegas why not iy instead of y

Tim The Enchanter

@Abcd What do you think the modulus is?

The Raiders of Las Vegas

i^2 = -1

Abcd

@TimTheEnchanter distance from origin

@TheRaidersofLasVegas so yes it should become x^2 - y^2

Tim The Enchanter

@Abcd Must distance be a real number?

Abcd

09:59

@TimTheEnchanter Umm....yes(unsure)

anon

a^2+(bi)^2 would indeed be a^2-b^2, but we're not talking about a^2+(bi)^2

Tim The Enchanter

@Abcd Exactly, that's why we define it as the length of the hypotenuse of the right triangle with $real$ length sides.

So the 'i' doesn't enter the diagram at all here.

anon

draw a+bi as a point in the plane, draw the triangle, the sides have length a and b (no i, lengths don't involve i), so by the pythagorean theorem the magnitude is sqrt(a^2+b^2)

The Raiders of Las Vegas

^@Abcd

Abcd

Okay, thanks @anon and @TimTheEnchanter and others too.

Tim The Enchanter

10:03

Also I just realized that Ted's name was Shifrin. For some reason I thought that his name was Shrifin all this time.

SteamyRoot

o.O

Daminark

ohi @Steamy

@Tim :O

SteamyRoot

ohi

Actually, his name is Theodore :^)

Daminark

Also @Tim: youtube.com/watch?v=vJQhDkpT5aY

Tim The Enchanter

@SteamyRoot -,-

SteamyRoot

10:05

Nonabelian cohomology hurts my brain >.<

The Raiders of Las Vegas

That's sooo Balarka like @Daminark :P

Daminark

Oh yeah he showed me that

The Raiders of Las Vegas

Me too.

Tobias Kildetoft

@SteamyRoot Well, you only get a few ones anyway

Tim The Enchanter

@Daminark This is the dankest thing I've seen in a long time.

Daminark

10:07

@Steamy So, I'm guessing this isn't $K(G,n)$ stuff

Tobias Kildetoft

@SteamyRoot And most of them are not even groups

SteamyRoot

@Daminark What's $K(G,n)$ again? :P

Daminark

Wait hmm, are there cohomology theories for which there does not exist classifying spaces?

$K(G,n)$ is the Eilenberg-Maclane space

Which is defined to have all of its homotopy groups be trivial, except for the nth one, which is $G$

Tobias Kildetoft

@Daminark You can do cohomology with no spaces at all. No sure if one can somehow still get classifying spaces

Daminark

Ordinary cohomology can be given by $H^n(X;G) = [X,K(G,n)]$

SteamyRoot

10:10

Oh, meh.

I just work with cycles and cocycles and such

Daminark

What does "cohomology with no spaces" entail?

@Tobias

SteamyRoot

And, yeah, as soon as you get to $H^n$ with $n \geq 1$ they're not groups in general anymore

Tobias Kildetoft

@Daminark cohomology is something you associate to a chain complex

Daminark

@Steamy Eh, Peter's influence

SteamyRoot

But they are pointed sets, which means there's a notion of exact sequences and neutral elements

Tobias Kildetoft

10:12

@SteamyRoot Don't you also have to stop at like $n=2$ or $n=3$ for the nonabelian stuff?

SteamyRoot

Not necessarily, but it gets really whacky there.

Tobias Kildetoft

@SteamyRoot Hmm, I thought I recalled them no longer being defined for larger $n$

SteamyRoot

Serre just goes "One would also like to define $H^n$ for $n \geq 2$ but I'm not going to bother, here's a reference"

Daminark

@Tim truly, I've used it a good bit. That and the MLG violin, especially for sniping (which happens a good bit in this chat :P)

SteamyRoot

Apparently Dedecker and Giraud did some work on that.

Luckily, in my case, I don't need to deal with them.

Tim The Enchanter

10:16

@Daminark You are doing the Lord's work.

3

Daminark

@Tobias Peter mentioned once that you could sorta define homotopy on simplicial sets without thinking about them yet as topological spaces, is it that kinda thing?

SteamyRoot

I'm only dealing with $$1 \to H^0(\mathbb{Z},N) \to H^0(\mathbb{Z},G) \to H^0(\mathbb{Z},G/N) \to H^1(\mathbb{Z},N) \to H^1(\mathbb{Z},G) \to H^1(\mathbb{Z},G/N) \to 1$$

for $N \triangleleft G$

The Raiders of Las Vegas

@TimTheEnchanter vengeance?

Tobias Kildetoft

@Daminark homotopy is not (co)homology. (co)homology can be seen as a purely algebraic thing (or in fact an abelian category thing)

@SteamyRoot Hmm, so in some sense the LHS spectral sequence with the entries switched

Tim The Enchanter

@TheRaidersofLasVegas Dankness must be shared.

Daminark

10:20

Ah. Lol I guess I'm too used to it as overlaying on homotopy theory or something. But huh, that's nifty

SteamyRoot

@TobiasKildetoft Yup.

The nice thing is, while $H^1$ is (in general) not a group but an orbit space, you can define the "neutral element of $H^1(G)$" as the orbit of the neutral element in $G$.

And if you define $\operatorname{ker}(f) = f^{-1}([e])$, the preimage of this orbit, then the sequence is exact

Tobias Kildetoft

10:35

@SteamyRoot Neat. I never really did any nonabelian cohomology, though apparently it has some uses for algebraic groups (there is a paper of Stewart on it that I keep meaning to read)

BAYMAX

11:11

any1 got a ping?

I got a ping but unable to find out by whom and where?

user84215

when?

BAYMAX

just a min ago!

Tobias Kildetoft

@BAYMAX Was not in this room. Might be in another room

BAYMAX

ok

I never tried MSE on Android !

any1 can share their experience?

Lucas Henrique

12:12

Yo

does anyone know where can I find @Ted 's book on abstract algebra?

Leaky Nun

@LucasHenrique amazon.com/Abstract-Algebra-Geometric-Theodore-Shifrin/dp/…

2

Lucas Henrique

Maybe in portuguese? :p

Akiva Weinberger

We've decided the sequel should be called Lovecraftian Mythos: The Things That Should Not Be Known Approach

Erik M

12:36

Good morning

Akiva Weinberger

Good morning to you / Good morning to you / You look kind of drowsy / In fact, you look lousy / Is this the right way / To start a new day?

Secret

12:51

@AkivaWeinberger My idea of trying to derive a "generalised comoving frame" might be similar to what you interpret form that message. Picture the usual euclidean metric as a regular grid of squares. Now as time progress, the grid become distorted by some continous deformation parametrised by time.

For the simple case of train stations which I investigated with semi, it seemed to simulate the impression of objects moving about because there distance literally changes due to the deforming metric tensor (and hence the metric)

The open question is whether it can always be done for arbitrary motions of n objects, by picking a suitable continous deformation that is parametrised by one parameter such as time

If the answer is positive, then we can always find a frame in a dynamical system such that all objects of interest become frozen in place, and the movement is implicated by the time varying metric tensor

PS: I don't think I know what a homeomophism class is

user84215

Why do the bold command and shift + Enter not work with each other properly ?

Leaky Nun

@aminliverpool only single-line messages can be formatted

user84215

I mean for example

user84215

**hjkkjjkj** njknjknjkbnjk knknknk
jnjknjnj jnjnj kn **knknkn**

user84215

13:07

I used one bold command for each line.

Leaky Nun

I know

2 mins ago, by Leaky Nun

@aminliverpool only single-line messages can be formatted

The message is multi-line, so it cannot be formatted

user84215

Thanks.

Federico

13:47

I need to use Fourier's integrals in C++. Is it possible to compute them with any of the new built-in functions of C++17 (http://en.cppreference.com/w/cpp/experimental/special_math) ? or do I have to implement my own solver?

Said otherwise: can Fourier's integrals be expressed as a function of some elliptic integrals (and others, see link above)?

[is this a question I should ask on the main site?]

pardon, brainfart on my side, FRESNEL integrals, not Fourier

skullpatrol

14:02

::sprays chat room with air freshener::

Federico

sorry ::flogs himself::

skullpatrol

::brings out the whip:: How many lashes should we give him?

Daminark

7

skullpatrol

::1, 2, 3, ... 6, 7::

Let that^ be a lesson to you.

Fourier indeed.

Federico

14:21

ouch :(

but I guess that the answer to the question is a "no", right?

Akiva Weinberger

14:36

@Secret Two things are in the same homeomorphism class if they're homeomorphic

(And here "class" is appropriate since I guess it's a proper class and not a set)

Did you see the conversation I had with Balarka?

Speaking of, I have a simpler solution which I'll send him later when he rejoins the chat

Last I saw him he said he was about to travel 2000 kilometers

skullpatrol

Nobody's in the room right now to help pal. They'll usually ping you @Federico

Akiva Weinberger

(2 megameters?)

Federico

@skullpatrol ok, thanks :)

skullpatrol

np

You could ask on the main site @Federico

and then post a link to your question in here

A lot of people do that.

Sorry for the whipping :-)

Federico

Done

0

Q: Can Fresnel integrals be expressed as a function of other functions/integrals?

I need to use Fresnel integrals in C++. Is it possible to compute them with any of the new built-in functions of C++17? or do I have to implement my own solver? Said otherwise: can Fresnel integrals be expressed as a function of some elliptic integrals (and others, see link above)?

prays not to have messed up the tags

Secret

14:56

@AkivaWeinberger The connection between our questions with ricci flow seems quite interesting, though I must admit my topology is still poor thus part of the conversation I don't quite get it, but I will get there eventually...

Erik M

Can someone help me understand an answer? I don't understand the steps after $f(x,y)=y^TC_{n-1}x^TC_n-x^TC_{n-1}y^TC_n\leq 0$. Here is the post: math.stackexchange.com/questions/1771525/…

Secret

The part of the discussion about knots seemed to make sense to me intuitively, but how the manhattan metric fail to embed in higher dimensional space is not quite without reading a proof

and suppose I understood the part about ricci flow correctly, our questions both concern about a 1 parameter deformation but without the need to satisfy some kind of diffusion equation like in ricci flow

Akiva Weinberger

In any case, manifolds are homogenous, which means (I believe) there's always a homeomorphism that sends one set of $n$ points to any other set of $n$ points

So we can take the metrics inherited from those, and then average them as discussed

Secret

So you mean the averaged metric will have the required property that its homeomophism class will be preserved by any continuous 1 parameter deformations?

Akiva Weinberger

*connected manifolds

By averaged metric I mean $\lambda d_1+(1-\lambda)d_2$, so $\lambda$ is the parameter

But yeah, if $d_1$ and $d_2$ are homeomorphic, so is the averaged thing, even as $\lambda$ changes

Secret

15:26

Interesting. Hmm... so feeding this into my idea, if I have n mutually incompatible reference frames (due to accelerations, I cannot find a single inertial frame for all moving objects) and in each frame I try to write a time varying metric tensor to capture all the motions seen in the frame, then I am basically introducing a one parameter deformation to my given metric.
Now, if the metric tensor in these n reference frames are homeomophic then I can potentially average all n of them to get the averaged metric tensor which by the above argument should preserve the homeomophism class

Simply Beautiful Art

16:11

This doesn't appear to be on the starboard yet, so...

5 problems, $1000 each, they look fun, made by Conway.
https://oeis.org/A248380/a248380.pdf

2

Oh yes, and the last problem has already been solved, so actually 4 problems remain

@Federico Oh, hi

Akiva Weinberger

16:35

Whoa, cool

$13532385396179=13\times53^2\times3853\times96179$

Simply Beautiful Art

Yeah lol

Federico

@SimplyBeautifulArt hi! and thanks for the "real only" tip! :)

Akiva Weinberger

so it solves the last problem as claimed

Simply Beautiful Art

:P np @Federico

Akiva Weinberger

More details on the first problem: https://en.wikipedia.org/wiki/Sylver_coinage

Abcd

16:53

Hey!

I just needed little help. When I am in an elevator going upwards with acceleration with a. What is the apparent gravitational force g+a or g-a. Just wish to understand this intuitively so that I can later derive Atwood machine formulas.

Research Effort: Had discussion with a person on physics.SE, read 3 books - Mechanics Part 1, Concepts of Physics and Resnick Halliday

Akiva Weinberger

Have you ever been in an elevator

Do you feel heavier or lighter

Abcd

@AkivaWeinberger It's been a while since I have been in an elevator :(

Akiva Weinberger

(when it's just starting to ascend or ending a descent)

@Abcd …Where do you live?

Abcd

@AkivaWeinberger India. There is no dearth of elevators here but I rarely use them. Moreover, I am so busy at home and school that I have no time to visit Elevator places.

Akiva Weinberger

In any case, it's $g+a$ if it's accelerating upwards

or, actually, is $g$ a vector here?

Abcd

16:58

@AkivaWeinberger Yes

Akiva Weinberger

I don't know if I should treat it as $\approx9$ or $\approx-9$

Abcd

@AkivaWeinberger How?

@AkivaWeinberger Downwards negative 10 and upwards +10

Akiva Weinberger

Well, if it's accelerating upwards at 9.8 m/s^2, then you don't feel weightless

That only happens if it's going down 9.8 m/s^2, aka freefall

Abcd

@AkivaWeinberger Yes.

Simply Beautiful Art

@AkivaWeinberger Depends which way I'm going

Akiva Weinberger

16:59

But now I'm not sure if $g$ and $a$ even have the same sign when you're accelerating upwards

Abcd

So how do I know the apparent gravitational force?

Simply Beautiful Art

@AkivaWeinberger Very unsafe elevator

Transcript for

$Mathematics$ Mathematics