Mathematics - 2018-01-03 (page 1 of 2)

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12:00 AM

$you may not stay out past 11$

what kind of restrictions? I often just use the vertical bar in my keyboard, like so: $S = \{n \in \mathbb{N}| n < 5 \}$

Akiva Weinberger

@Adeek Innit just $f|_{A}$

Oh wait no

okay sure I thought there is a fancy way to do this

$\upharpoonright$?

Akiva Weinberger

Hm that does seem fancy ^

12:01 AM

yeah

\upharpoonright

$f\upharpoonright_A$

detexify.kirelabs.org/classify.html

You can draw stuff and it will guess what you want

Akiva Weinberger

> In mathematics, the restriction of a function $f$ is a new function ${\displaystyle f\vert _{A}}$ obtained by choosing a smaller domain $A$ for the original function ${\displaystyle f}$. The notation ${\displaystyle f{\upharpoonright _{A}}}$ is also used.

- da wiki

sure

okay

The naming upharpoonright is rather annoying lol.

Akiva Weinberger

$\downharpoonleft\downharpoonright\upharpoonleft\upharpoonright$

12:06 AM

I hope you have macros for typing those

12:24 AM

$\upharpoonright\upharpoonleft$y goodness.

I want to start writing my proofs such that I refer to elementary results in obtuse language.

Like, instead of de Morgan's laws, call them "a theorem of de Morgan".

Karl Kronenfeld

12:44 AM

"Let C be a point equidistant from A and B, which exists by the first proposition proved by the most illustrious Euclid in his most exceptional work, the Elements, and refined by the great David Hilbert"

I aspire to such terseness.

"We now take a local isometry of this surface, which preserves Gaussian curvature by a useful and egregious theorem of differential geometry."

isn't the modern use of that word the opposite of the meaning of the latin word

Connotation-wise, yes.

Egregius meant "exceptionally good" in Latin, but means "exceptionally bad" in English; still, the "exceptional" sense of the word did survive.

i wonder how that happened

probably from people using it ironically

Karl Kronenfeld

lol

12:56 AM

egregious (adj.)
1530s, "distinguished, eminent, excellent," from Latin egregius "distinguished, excellent, extraordinary," from the phrase ex grege "rising above the flock," from ex "out of" (see ex-) + grege, ablative of grex "a herd, flock" (see gregarious).

Disapproving sense, now predominant, arose late 16c., originally ironic. It is not in the Latin word, which etymologically means simply "exceptional." Related: Egregiously; egregiousness.

Karl Kronenfeld

that's like ultra interesting

1 hour later…

2:01 AM

God damn

I woke up at 7:30

What happened to me?

I finally fixed my sleep schedule

@Balarka Relatable.

It's like all the time in the world has suddenly increased

I feel fantastic loool

@BalarkaSen :O

Hey @Daminark, and @Fargle too

2:17 AM

How's it going?

Going to finish writing the hyperbolic dynamics notes. The reading course is probably going to jump to circle diffeomorphisms today

I see

3:04 AM

@Daminark @BalarkaSen

I really love algebraic geometry I mean it is like a universal mathematics capturing complex analysis, algebra, geometry,etc.

Such cool subject

such a cool subject *

It's not universal mathematics. Just happens to lie in the intersection of those.

yeah sure

2 hours later…

4:51 AM

because answer key reads so. — cmi 3 mins ago

end my life

lmfao

are they homeomorphic though?

please check the question number 3.6. This is the most prestigious phd screening test in India. I don't think they would give wrong answer. — cmi 2 mins ago

can't be arsed to check honestly

I'm stuck between this and my head

my head tells me they are not

one looks like a line with endpoint and one looks like a line without

you mean a line with a corner?

methinks X_1 has both positive x- and y-axes, with the origin connecting them

4:54 AM

like $[0,\infty)$

Wait, lol. So isn't they $y$-axis also part of the first one?

oh!

yeah they are homeomorphic

I was tricked by desmos

desmos didn't show me the vertical part

I correct pjs on that but then he deleted his comment lol.

4:55 AM

@Dair $X_1$ is a bent line and $X_2$ is half of a hyperbola, yeah

just straighten it up

bloomshilkcp

that's the sound the homeomorphism makes

5:15 AM

What do we answer if your answer is correct? — Kenny Lau 2 mins ago

Seriously. "Is my answer correct?"

If I answer "Yes it is", I get downvoted and flagged

proof-verification is a tag that is specifically meant for this purpose. I usually just answer by "That is correct" and giving extra insights if possible

I have no extra insights to provide

Then don't answer :P

Don't bully people if you have no answer to provide

If you actually care about helping people and not shooting for that extra rep, comment below the answer whether it's correct or not

@BalarkaSen I usually do that only when I think the question is too stupid

then I leave the answer as a comment there

You're just too pretentious pal.

3

5:20 AM

user image

@LeakyNun Regarding the question, it was not with any other motive, perhaps users can show some other approach to solve it, well if you believe there exist different ways to solve the same problem. Well its good regarding expressing your opinion.

@BAYMAX I do not see other ways

perhaps others will

You may not some one will may be in this ite

6:15 AM

@LeakyNun hi

@Liad hi

remember my question from yesterday ?

about formulas describing unique equivalent classes of order n

yes?

how would you do the unique part?

for every other n-tuples of elements satisfying the given condition, they are a permutation of the original n-tuple

6:21 AM

so the formula would be a long one? @LeakyNun :P

yes

(Has nothing to do with maths) I think this post of Balarka is strange enough to my subconscious that last night dream, there's an anime game generated completely from scratch

23 hours ago, by Balarka Sen

@Secret What is this fantastic anime? I must watch it

Basically, the backstory of the anime goes as follows:

The world that Lane and the other characters lives in (the city inside that cubic structure) has some unspecified spiritual properties. Memories from Lane's visit to the zoo (and the zoo is said to be somewhere in reality, suggesting the whole world is either Lane's dream, imagination, or that the world interacts with itself in some self referential manner) in the past have contaminated the world and this is bad in some unspecified way. Sana, which is seen at the intro, is actually not human but some kind of spirit guardian of this wor…

We learnt that Lane get an extra bedroom as some kind of base or hub by murdering the previous inhabitant. It is never specified the motivation nor the details of this crime

Kane is a friend of Lane and she had been helping Lane on her life for some time

The people met by the tree was planning to infiltrate Memory, which is linked to the background of the anime or Lane, in order to destroy it or something. In order to do so, they have to ensure both Chasity and (forgot) followed their programmed routes in the game (which centre around that big tree with the 4 quadrants) to gain access to …

user image

(O and my drawing sucks, thus it does not look anime)

6:38 AM

@Secret @LeakyNun my next version

drive.google.com/file/d/1uH23x_ivfqAYYvobULqPWWtX52_Bt9Gz/view

i proved that one tricky proposition

it is now on page 12

hi @AlessandroCodenotti

that paper above is about a way of dealing with the floor function in differential equations

Narcissus jewel

@Secret I thought this would have something to do with the MSE eras :P.

hopefully it ends up being intriguing.

@Narcissusjewel Well, Balarka's response is indeed about my lore of the MSE eras, but somehow, my subconscious (via the dreams) seemed to held onto the strange emotions produced when I read that response of his and then generate an anime game unbeknowest to me

@Narcissusjewel whats up?

imma switch to phone

brb

Narcissus jewel

@Typhon Wait, what's up in the sense of 'what does this comment mean', or in the sense of 'What're you up to'?

6:48 AM

@Secret This is unironically an awesome anime plot.

Love it

$1 \longrightarrow Z(G) \longrightarrow G \longrightarrow \operatorname{Aut}(G) \longrightarrow \operatorname{Out}(G) \longrightarrow 1$

this is the most beautiful exact sequence I see

after $1 \longrightarrow 1+J(R) \longrightarrow R^* \longrightarrow (R/J(R))^* \longrightarrow 1$

cc @TobiasKildetoft @Daminark @MatheinBoulomenos

@Narcissusjewel what're you up to?

Narcissus jewel

7:13 AM

@Typhon Reading Raymond's text on differential analysis

@Narcissusjewel that sounds interesting

i assume differential refers to differential equations

Narcissus jewel

user image

that looks complicated

Narcissus jewel

@Typhon Nah, I think it just needs a little background.

more like a lot

@AkivaWeinberger hi

7:27 AM

differential analysis... sooo this is the umbrella term of all that incomprehensible stuff I found between Mike and Balarka and Danu

Now that the umbrella term is known, I can start to get forward to make it more comprehensible to myself in the future

Narcissus jewel

Actually, I'd never heard the term before this textbook title. So I don't know how reasonable it is to use it.

Well, the three subheadings there are indeed the topics that dominate most of the maths chat history

Narcissus jewel

Wait, are the users 'waiting' and 'simple art' distinct people?

Just reading back what you wrote here:

yeah

Narcissus jewel

yesterday, by Secret

....With the recent spread of the topology epidemic, more and more users of this chat were converted to topology. It would not be long before everyone be converted to algebraic topology and differential geometry, and becoming slaves of Balarka's rein...

7:33 AM

Waiting aka IDoMathArt aka Chris sis are one of the users who done the series and integral stuff with some series methods that are not found in the literature

Robjohn is another example (but he covers most calculus questions)

Narcissus jewel

Oh, I just thought that person was also Simple Art.

Simpleart aka SimplyBeautifulArt does the googlogy and ordinal notations. He developed many ordinal collapsing functions to write ordinals with

i do implied differential equations stuff

and also polygonal surface geometry

as well quadratic rings, periodic differential algebra and other number theory stuff

and algebraic number theory

7:39 AM

As for me, I sometimes construct some very weak abstract algebraic structures, analysing general trends on integral and series closed forms (and closed form of any given proposition that interested me in general), global relationships between various mathematical objects, philosphy of mathematics, some set and type theory

But my major speciality in this chat is comprehending weird ideas and translating some of the weird maths I saw in dream into reality compatible formats

Narcissus jewel

@Secret Wouldn't you have a much easier time doing that if you first learned some more advanced mathematics?

That's indeed what is happening right now. I was part of user21820's class as he said I must get my first order logic foundation firm before touching other maths

Narcissus jewel

Class? User21820?

I and lastironstar and batman(forgot) were taking logic classes in the logic chat

drive.google.com/file/d/1uH23x_ivfqAYYvobULqPWWtX52_Bt9Gz/view

Narcissus jewel

7:41 AM

Oh. Why not learn ring/field/galois theory and then category theory from the handbook of categorical algebra?

my current compilation of implied derivative stuff

User21820 is a computer science background (professor(?)) but he is well recognised in MSE to be proficient at mathematical logic

@Narcissusjewel I had Maclane as my category theory text, but I had not read it yet

yes

professor indeed

Narcissus jewel

@Secret I think learning some proper abstract algebra, prior to learning category theory, would be wise.

That I have Pintar suggested by 0celo

I plan to read it as soon I got that data analysis batch running

7:45 AM

i should probably study abstract algebra as well at some point

but id probably only need a day or two

i tend to pick up on things quick

The only book I have read more than 2 chapters so far is Munkres, and that (combined with Alessandro and Leaky's teaching) is where my topology background is from

Narcissus jewel

@Typhon This is weak bait.

> Definition. The nth implied derivative of a function f is defined to be f
→n
= (f
→n−1
)
→

when n > 1 and it is defined to be f
→1
= f
→ when n = 1.

It might help to extend that to negative indices for implied antiderivatives, analogous to the ordinary counterparts

@Narcissusjewel bait?

Narcissus jewel

@Typhon Bait.

7:53 AM

...

Narcissus jewel

'''

perhaps im underestimating AA

Narcissus jewel

Uh huh, that's what that comment was!

its within the error value of 100% so we're all good.

Narcissus jewel

Next you'll have a 50 page write-up on galois theory ;).

7:55 AM

im not in grad school

Narcissus jewel

I was referencing something from ages back, but never mind. I shouldn't get distracted atm.

thiugh i did write a 50 page geometry paper last year cause i went overboard on the class paper

we had... 5 weeks

or rather i did. it was for an extra project

> and that the right derivative at

x is the right limit of ((g +C1)−(h+C2))→((g +C1)−(h+C2))→ = (g −h)

→.

duplicated expression ((g +C1)−(h+C2))→

i will find that

thanks

if you dont mind it is late so i am going to hop off and watch some videos to wind down

i hope you enjoy the paper

or enjoyed if you already finished

> Proposition. Let C(x) be a piecewise constant function. Then if for all real numbers z
we have that y(x, z) is a solution to the implied differential equation f(x, z, y, y→, y→→, · · · , y→n
) =
0 then y(x, C(x)) is also a solution to the implied differential equation f(x, C(x), y, y→, y→→, · · · , y→n
) =

0.

Sure, but you will also need another proposition to relate the solutions to the implied diff eqt to the ordinary diff eqt in order to use it, and for that one, is where $\mathcal{C}^n$ is important

otherwise, looks fine so far except you might want to clarify to the reader that y=C(x) for some of the proofs of the later propositions when you do the chain rule step

and I am guessing the part where you said you need measure theory is the fact that there are only at most countably many discontinuities?

3 hours later…

10:49 AM

@Typhon are you the user formerly known as "the great duck" ?

Manolis Lyviakis

Let A be a subspace of $R^{n}$ let $h:(A,a_0)→(Y,y_0) $. Show that if h is extendable to a continuous map of $R^{n}$ into $Y$ , then $h∗$ is the trivial homomorphism (the homomorphism that maps everything to the identity element)

for every loop base in $a_0$ $h∗ =h \circ f =[e_{y0}] $

Alessandro Codenotti

is $A$ a linear subspace?

Manolis Lyviakis

and that is where im stuck

$A$ is just a subspace of $X$

no of $R^n$

i know $h∗ = [h\circ f]$ where $f$ is a loop based on $a_0$ and that must be $e$ in order to be trivial

on thing i can do is say that every loop on $A$ is homotopic to the expantion of $h$ for there is only one element in the first fundamental group and since$ h∗ $ is a map between fundamental groups and a homomorphism so its image must be a subgroup of the other fundamental group and since the only 1-element subgroup is the trivial ...

Alessandro Codenotti

if $g:\Bbb R^n\to Y$ is a continuous extension of $h$ then $g^*$ is clearly trivial, what's the relation between $g^*$ and $h^*$?

Manolis Lyviakis

11:07 AM

i dont know

Alessandro Codenotti

Neither do I, I was just thinking loudly :P

Manolis Lyviakis

i know that $h∗ = [h \circ f ]$ where $f$ is a loop

i can find a homotopy between the $g_*$ and the constant

but i dont know if $h∗ =g_*$

ohhh what i know is that they are homotopic

soo $h∗ \equiv g_* \equiv y_0 $

hm... but in definition of homotopic functions they must be continouus and $h$ is not

unless we know $h$ is at least continuous in $A$

Manolis Lyviakis

12:08 PM

0

Q: Prove the fundamental group of A is trivial $A= \{(x,y,z)\mid z \geq 0 \} \backslash \{(x,y,z)\mid y=0, 0 \leq z \leq 1 \}$

As stated i need to prove that the fundamental of the upper half of $R^3$ minus the line $y=0$ and the line segment $ 0 \leq z \leq 1 $ has a trivial fundamental group.Im just started in these kind of things. So i dont know how to start. I know i have to prove that given a random loop on $A$ wi...

general-topology algebraic-topology fundamental-groups

i sketched a proof i just need some to check if it is right

and i didnt justify the fact that my problem lies only to certain loops...

12:25 PM

0

Q: Primes of the form $ 2 x^2 + 2 xy + 3 y^2 $

Why is every prime $3,7 (mod \space 20)$ of the form $$ 2 x^2 + 2 xy + 3 y^2 $$ ?? I do not think that form is the norm of an abelian ring ? How to prove this ?

elementary-number-theory polynomials prime-numbers

Daniel Guldberg Aaes

12:38 PM

Quick question, I know this graph is transitive but is it also reflexive? I'm not sure if it is enough that x -> y and y -> x when there also is an z.
https://imgur.com/Gx7WbxF

1:21 PM

Turns out I have to learn some measure theory

Narcissus jewel

Why?

Haar measure?

Alessandro Codenotti

Measure theory is cool, don't worry

None of that algebraic bullshit :P

Narcissus jewel

@BalarkaSen >:(!

>>>>:((((((((((((

There is a dynamical invariant for a given circle homeomorphism called the rotation number

The proof that rotation number makes sense is a measure theoretic one...

Alessandro Codenotti

1:24 PM

What kind of measure theory is needed?

@Alessandro Ah yes you're a measure theory dude, I forgot

Alessandro Codenotti

Not really, we just saw the measure theory needed to talk about integral wrt to a measure but then concentrated mostly on integrals wrt the Lebesgue or Hausdorff measures. I really liked the measure theory we did though

Well I only followed the proof superficially but here's how it comes to the story

Suppose $f : S^1 \to S^1$ is an orientation preserving circle homeomorphism

Then it lifts to a monotonically increasing homeomorphism $F : \Bbb R \to \Bbb R$ which commutes with the deck transformations of the cover $p : \Bbb R \to S^1$, i.e., $F(x + 1) = F(x) + 1$.

Alessandro Codenotti

You're talking Arabic to me

Ah, well, what I mean is simply that there exists a monotonically increasing homeomorphism $F$ of $\Bbb R$ such that $p \circ F = f \circ p$.

It's a true fact, not hard to prove. You can take it as an exercise.

(It fits within the story of lifting maps in covering space theory.)

And somehow $F$ commutes with translations of $\Bbb R$, because $p(F(x + 1)) = f(p(x + 1))$, but $p(x + 1) = p(x)$ by the definition of the covering map $p$. So $p(F(x + 1) = f(p(x)) = p(F(x))$. So $F(x + 1)$ and $F(x)$ must differ by an integer; using the fact that $f$ is a homeomorphism you can infer that difference is $1$.

Otherwise $f$ would end up wrapping around the circle, which breaks injectivity, so on

In any case, notice that this means - purely formally - $\phi(x) = F(x) - x$ is $1$-periodic. So if everything was $C^1$, we could take the Fourier expansion of $\phi$.

The constant coefficient term of that Fourier expansion should have been the "rotation number" of $F$.

If you write it down, that's just $\alpha_0 = \int_0^1 \phi(x) dx$

Alessandro Codenotti

1:41 PM

Ok, this makes sense (it feels similar to how Hatcher proves that $\pi_1(S^1)=\Bbb Z$)

Oh that translation thingy? Exactly right.

That's just standard algebraic topology

In any case, the right notion of rotation number, however, is as the limit $\displaystyle \lim_{n \to \infty} \frac1n \sum_{k = 1}^{n-1} \phi(F^k(x))$ (which simplifies to $\lim_{n \to \infty} (F^n(x) - x)/n$)

That's like that integral definition, just averaging over an orbit of $F$ instead of over the whole interval $[0, 1]$.

So you'd have to prove this limit 1) exists 2) is independent of the orbit you average over, i.e., independent of $x$

2) is a small exercise; the dynamics of $F$ is somehow "bounded" so two nearby orbits don't diverge apart

1) is kinda complicated. The point is that the sum in that limit is a measure (average over the dirac measures at each point in the orbit).

I think the proof revolved around taking the limit of those measures in the weak topology

Alessandro Codenotti

I don't know anything about convergence of measures, sorry, but maybe Daminark does

2:00 PM

Ah ok. Well I think I lack more fundamental things anyway

2:14 PM

5

Q: Irrationality of $\pi^2$ and $\pi^3$

I wonder if there is any book and/or article you can recommend on the topic "Irrationality of $\pi^2$ and $\pi^3$" for me to study on. In case you are curious about why I ask these particular exponents, it's because this is a project that my lecturer gave me to study on and then present to the cl...

reference-request book-recommendation pi rationality-testing

Alessandro Codenotti

@BalarkaSen Actually I do know how to turn the set of signed measures into a Banach space, I suppose the same construction also works for just measures

I don't think you get a Banach space anymore

Just a topological vector space

Alessandro Codenotti

Hmmm, so for signed measures on a measurable space $(M,\mathcal A)$ the norm is given by $||\mu||=\sup\limits_{A\in\mathcal A}(\mu(A)-\mu(M\setminus A))$

Ah, no, I think in this case I think our topology was that $\mu$ and $\mu'$ are close if $\int_X \phi d\mu$ and $\int_X \phi d\mu'$ are close for all continuous functions $\phi$ on $X$

Alessandro Codenotti

(for signed measures $\mu(M)<\infty$ is assumed)

2:23 PM

Oh my measures are also all probability measures

$\mu(X) = 1$

Alessandro Codenotti

@BalarkaSen I see, that sounds like a very strong notion of closeness but my intuition doesn't work with analysis so I don't know :P

I don't know anything about this either so

But I have been recommended some stuff to read

looking at measure theory from either Folland or Royden

and a spicy book called "The Geometry of Fractal Sets" by Falconer

Alessandro Codenotti

My professor followed his own notes so I can't suggest any book in particular

Narcissus jewel

I've been recommended Folland myself

No worries. I might bug you about stuff if I get into it, though, @Alessandro :P

Alessandro Codenotti

2:34 PM

Considering how many times I've asked you stuff I'm looking forward to help if I can

@Narcissus I'm probably just going to read measure theory and not all of the book

Narcissus jewel

Chapters 1,7 and 11?

Hey guys. I've been asked about this integral by a friend, but it's about 20 years away for me. Any suggestions? Find the integral \int_R \sin((pi-z)^3) dV, where V = {(x,y,z) | z>=0, z<=x, x<=y, y<=pi}.

@Narcissus I haven't downloaded it yet, but let me copyleft it

Dude most of the book looks like measure theory

Narcissus jewel

Oh, for sure

It's so damn long though.

2:41 PM

1,2,3 are probably the basics

Narcissus jewel

11 has that beautiful haar measure ;).

Lets see what Royden has

hi everyone

what exactly does a 6-element exact sequence tell us?

nothing special

Alessandro Codenotti

That is has more than 5 elements but less than 7

2:43 PM

you can break it up into two 3-element short exact sequences

I mean, 1->A->B->C->1 basically means that B/A=C

but what do I get if I add one more element

break it up like i said

how?

1 --> A --> B -f-> C --> D --> 1 breaks into 1 --> A --> B --> im f --> 1 and 1 --> coker f --> C --> D --> 1

Hm, maybe that's not correct

So B/A=im f and C/(coker f)=D ?

2:49 PM

The first is true

I think I just meant 1 --> coker f --> D --> 1

D is isomorphic to coker f = C/im f = C/ker (next morphism), yup

Right, so you break into a 3-element short exact sequence and a 2-element short exact sequence. Sorry for misspeaking earlier

Hot damn Falconer's book looks awesome

Hey @Eric

Yo

Tfw you wake up late for class and forget to dress properly for the cold

Rip me I had a good run

rip in dab

also known as logan paul

but 2-element just means isomorphism

which doesn't tell us much

what?

ah, so it's basically 1->A->B->(B/A)->(B/A)->1

is this right?

3:04 PM

you literally said 1->A->b->C->1 means B/A = C

@BalarkaSen I mean, 1->coker f ->D->1 just means coker f is isomorphic to D

that's an isomorphism

that doesn't tell us much?

what the fuck?

alright, sorry

a 2-term exact sequence has the most information literally

yes it is an isomorphism, that's why it has the most information!!

anyway, the fourth isomorphism theorem should be $1 \longrightarrow Z(G) \longrightarrow G \longrightarrow \operatorname{Aut}(G) \longrightarrow \operatorname{Out}(G) \longrightarrow 1$

@BalarkaSen alright I get it

3:10 PM

What's with all the exact bois

fourth isomorphism theorem

Alessandro Codenotti

3:33 PM

Hi @Dami

Yo

hi

How's it going?

Alessandro Codenotti

I started studying logic rather than functional analysis, much better :P

@AlessandroCodenotti welcome to our camp

Alessandro Codenotti

3:43 PM

Have you read @Balarka wants to learn measure theory? Guess he got bored of chemistry :P @Dami

Measure theory is good shit

1 hour later…

5:08 PM

https://math.stackexchange.com/questions/2590487/pi-is-not-algebraic-of-degree-one-or-two
Can you please have a look at what I have just posted if you are familiar with the concept of the irrationality of pi.

2 hours later…

7:06 PM

@GabrielRomon yes, why?

@Secret indeed

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