Mathematics - 2017-11-13 (page 3 of 5)

DaenerysDracarys

05:00

Shoveling snow. Boo.

Anonymous

@Semiclassical I mean is $z=\infty$ an essential singularity? I don't think it is

Anonymous

For $\frac{\cot(\pi z)}{(z-a)^2}$

DaenerysDracarys

In any case, have fun and stay schwifty.

Semiclassical

well, first off

Kasmir Khaan

all right handsome folks

kasmir gonna go Watch videos of mit on sylow

Semiclassical

05:01

note that $1/(z-a)^2$ definitely doesn't have an essential singularity at infinity

Kasmir Khaan

wish him good luck

anon

read a couple really cool proofs of $\zeta(2)=\pi^2/6$. one was inspired by light intensity in physics, which approximated $\Bbb Z$ on the number line by equally-spaced points on a circle using euclidean geometry to obtain the inverse squares series representation for $\csc$, and another which used random unimodular lattice statistics and hyperbolic geometry.

Anonymous

@Semiclassical Agreed

Semiclassical

so if that's to have an essential singularity, it'll have to be from cot(pi*z)

Anonymous

Yup

Kasmir Khaan

05:02

@anon anon can I ask you personal question ? :D

anon

I may not answer, but okay

Kasmir Khaan

thats fine =p

how old are you? =p

anon

mid-twenties

Kasmir Khaan

nice :D

Semiclassical

i'm trying to remember the definition of essential singularity tbh

Kasmir Khaan

05:03

you seem to know alot of stuff, like proffesor level or more

such skill is aquired at late age

like 40 +

Secret

well if you think anon knows a lot from such age, then barlaka is even more so, cause he is high school

anon

indeed

Secret

But he has the knowledge base of a professional topologist

Kasmir Khaan

i know balarka and I know anon , not same level tbh =p

anon is awesome _

but that being said -.- ill work hard to get to that level :D

Secret

same

Daminark

05:05

I don't think comparisons of the sort rare particularly helpful

Kasmir Khaan

thats the right attitude :D

Daminark

Like, both are brilliant for their own reasons

Kasmir Khaan

its not comparison its just my opinion

to be fair to balarka he is good too

Secret

What I need is to brush up my logic so I will stop making funny mistakes in my reasoning

Anonymous

@Semiclassical Well, points $z=n$ are poles, right?

Semiclassical

05:06

yeah. easiest way to see that is to invoke the periodicity of cot(pi*z)

if it's got a pole at z=a, it's got a pole at z=a+1 as well. and so forth

Anonymous

So, then my textbook says, as $n\to \infty$, poles accumulate at $\infty$ and thus $z=\infty$ is an isolated essential singularity

Anonymous

I don't know if that makes sense

Semiclassical

I suspect it's valid, but I don't remember well enough

Semiclassical

05:19

@Blue this looks relevant? math.stackexchange.com/q/135458/137524

Anonymous

05:29

@Semiclassical Ah, thanks

Anonymous

That explains it

io_cantor

05:43

What field of math are you guys most interested in?

Personally, I like $\mathbb{F}_2$.

anon

$\Bbb F_1>\Bbb F_2$

io_cantor

$\mathbb{F}_1$ is really hard to understand

Daminark

06:24

$\mathhbb{F}_{57}$

57 is the best prime

Kasmir Khaan

><

io_cantor

$\mathbb{F}_57$

$\mathbb{F}_{57}$

studrayght5

What's the use of max/min in epsilon/delta convergence proofs? I can't understand why we take something like $N=\max(k, a/(b\epsilon))$. Shouldn't $N = a/(b\epsilon)$ work? What's the role of $k$?

Kasmir Khaan

06:41

@anon anon how does one prove the first sylow theorem, using the action of G on all subsets of G ?

tried to follow the lecturer , was not that clear

Semiclassical

presumably it's because there's a scenario where $N=a/(b\epsilon)$ would fail to establish the correct conclusion, where $N=k$ would.

Daminark

06:58

@Kasmir so I'm gonna go to bed soon because I have an algebra midterm tomorrow (not Sylow business tho)

math.uchicago.edu/~may/REU2016/REUPapers/Idelhaj.pdf

studrayght5

@Semiclassical For $|1/10^n| \le \epsilon$, they take $N = \max\left\{\lceil{\log_{10}(1/\epsilon)\rceil}, 1\right\}.$ Do they take this value because if, say $\epsilon = 10$, then the ceiling bit would be $-1$, which doesn't even belong to $\mathbb{N}$? So we wouldn't have $n \ge N$.

Kasmir Khaan

@Daminark thanks dami ill read that

Daminark

I wrote a paper on this some time back, try reading it to see if it's at all clear. The first bit is about group actions, you can skip it if you already know

Semiclassical

sounds right. the 'obvious' scenario is with $\epsilon$ small enough exceed 1. but if you pick $\epsilon$ to be large instead of small, then the obvious scenario doesn't apply

Kasmir Khaan

okay the proof the lectures used , was using combinatorial idea

m p^n chose p^n

Daminark

07:01

Nice, so it's probably the one you did

studrayght5

Thanks @Semiclassical I

Kasmir Khaan

from that idea gives us subsets not subgroups

><

Anyway good luck on your exam dami :D

studrayght5

Is the $1$ arbitrary though? What if I took $2$ instead?

Daminark

Thanks! Have a good night everybody! :)

Semiclassical

i suspect it's not arbitrary, but I don't have a sense for it

studrayght5

07:18

I suppose it's from the fact that $\dfrac{1}{10^n} < 1$ for $n > 0$. So it's also less than $2,3, \cdots$ so perhaps any natural bigger would also work. I'm gonna think on it.

Secret

08:08

[Random]

010101010101010101010101...

001100110011001100110011...

000111000111000111000111...

000011110000111100001111...

000001111100000111110000...

NV-US

consider the set $X = [1,2] \cup [3,4]$ with the usual metric, then $[1,2]$ is open in $X$. how? help.

nevermind, got it

Silent

08:44

Please someone look at this! I dont know if there exists another way to prove $\det A^t=\det A$ using row reduction.

Secret

[Random]

What kind of new mathematical phenomenon do you guys will think will open the door to a completely new discipline of mathematics?

user104729

@TedShifrin Was that paper thing possible?

Secret

09:11

Shinichi Mochizuki

Shinichi Mochizuki (望月新一, Mochizuki Shin'ichi, born March 29, 1969) is a Japanese mathematician working in number theory and geometry. He is the leader of and the main contributor to one of major parts of modern number theory: anabelian geometry. His contributions include his famous solution of the Grothendieck conjecture in anabelian geometry about hyperbolic curves over number fields. He initiated and developed several other fundamental developments: absolute anabelian geometry, mono-anabelian geometry, and combinatorial anabelian geometry. Among other theories, Mochizuki introduced and developed...

Hmm, it seems they are getting some progress

I really like this notation, because it looks like a glyph rather than a word

However, I don't understood this because I don't even have background on algebraic varieties

but anyway, let's see what the google search returns for my previous query...

forums.xkcd.com/viewtopic.php?t=122649

Timeline of mathematics

This is a timeline of pure and applied mathematics history. == Rhetorical stage == === Before 1000 BC === ca. 70,000 BC – South Africa, ochre rocks adorned with scratched geometric patterns (see Blombos Cave). ca. 35,000 BC to 20,000 BC – Africa and France, earliest known prehistoric attempts to quantify time. c. 20,000 BC – Nile Valley, Ishango Bone: possibly the earliest reference to prime numbers and Egyptian multiplication. c. 3400 BC – Mesopotamia, the Sumerians invent the first numeral system, and a system of weights and measures. c. 3100 BC – Egypt, earliest known decimal system ...

abenthy

10:02

Can the cup product be defined on cohomology with compact support? $H_c^p(M) \times H_c^q(M) \to H_c^{p+q}(M)$?

Kirill

11:17

hey guys!, need help

how would you read $B(\mathbb{R}^n)$ in the context of Borel sets?

user8469759

hi guys

if I define $\lVert A \rVert_2 = \sqrt{\sigma_{max}(A)}$ (namely max abs value of eigen values)

is it true that $\lVert A \rVert_2 \to \infty \Rightarrow \lVert A^{-1} \rVert_2 \to 0$?

my guess would be yes

and I was trying to exploit the singular value decomposition of $A$ to prove it

because the determinant of $A$ is the product of all the eigenvalues

and the determinant of the inverse is the product of the inverse of the eigenvalues, but I've the feeling I'm being too naive

Balarka Sen

11:44

@abenthy I think so. Compactly supported cohomology is the homology of the cochain complex $C^i_c(X) \subset C^i(X)$ consisting of cochains $\phi : C_i(X) \to \Bbb Z$ which takes zero on the simplices outside of some compact $K \subseteq X$.

The cup product map $C^i(X) \times C^j(X) \to C^{i+j}(X)$, $(\phi_1 \smile \phi_2)([v_0, \cdots, v_{i+j}]) = \phi_1([v_0, \cdots, v_j])\phi_2([v_{i+1}, \cdots, v_{i+j}])$ restricts to $C^i_c(X) \times C^j_c(X) \to C^{i+j}_c(X)$ because if $\phi_1, \phi_2$ are compactly supported on $K_1$ and $K_2$, then the product vanishes if the simplex fed to it is outside $K_1$ or $K_2$

So it should be compactly supported on $K_1 \cap K_2$

The "dual" of cup product in compactly supported things should be intersection of open submanifolds and such. That's well defined only if you don't push intersections to infinity, for example. That kinda explains why the compact support gets smaller as you take cup product.

I guess the dual to $H^i_c$ is the Borel-Moore theory

Also I think the cup product is reasonably obvious for de Rham cohomology with compact support. The cochain groups are just differential $k$-forms with compact support there. Wedge of differential forms with compact support is still a differential form with compact support, so the DGA structure is still retained.

ÍgjøgnumMeg

12:55

are the notations $\Bbb Z[\sqrt{-5}]$ and $\Bbb Z \oplus \Bbb Z \sqrt{-5}$ equivalent?

Balarka Sen

As rings? Surely not, as the latter has zero divisors; multiplication is defined as (a, b)*(c, d) = (ac, bd).

ÍgjøgnumMeg

Oh yeah

Balarka Sen

multiplication in Z[sqrt(-5)] is different.

ÍgjøgnumMeg

indeed, cheers

Balarka Sen

:thumbs up:

ÍgjøgnumMeg

12:59

The book i'm using uses the notation $\Bbb Z + \Bbb Z\sqrt{-5}$ instead of $\Bbb Z[\sqrt{-5}]$

Tung Nguyen

13:25

How do i prove that 2^n>n^2+ C without using limit or log ( calculus stuffs in general )

Preferably with induction

Was able to prove that 2^n>n^2 for sufficiently large n

But i stuck when including the constant

anon

so, you want "for all C, there exists an N such that 2^n>n^2+C for all n>N" I'm guessing?

Leaky Nun

Do the objects in a category have to be sets?

anon

no

Leaky Nun

But most categories that I see have the objects as sets

except the "preorder" category

in the category $(\Bbb N, \le)$, can I have two arrows pointing from $1$ to $2$?

anon

you say "the" category, which presumably means the one where there is exactly one arrow between objects

but of course there is another category with the same objects and arrows and one extra arrow from 1 to 2

Leaky Nun

13:40

eh, can I define an alternative category with two arrows pointing from $1$ to $2$?

sniped :c

Alessandro Codenotti

Concrete categories is probably the keyword to google here

Balarka Sen

right

hTop is not concretizable

Tung Nguyen

@anon: yes

MatheinBoulomenos

A similar example to the "preorder" category is turning a monoid (e.g. a group) into a one-object category

In that case, the single object might be set, but it doesn't matter what it is

anon

or a monoid action

Leaky Nun

13:53

@AlessandroCodenotti how would you translate "maledettissimo"?

@MatheinBoulomenos (cc ^) apparently "verdammtest" isn't a word

MatheinBoulomenos

Indeed, it isn't

Leaky Nun

then how would you translate it to German?

MatheinBoulomenos

"vermaledeitest" maybe, but that sounds a bit old-fashioned

Leaky Nun

@MatheinBoulomenos :o is the "deit" preserved Vulgar Latin pronunciation that wasn't even attested

I know it as the intermediate from the Latin "dictum" to Italian "detto"

(and also French "dit", Spanish "decho", etc.)

MatheinBoulomenos

"vermaledeit" comes from maledictus, I have no idea about Vulgar Latin

Leaky Nun

14:00

the "dictus" part turned into "deito" before going to other Romance languages

but that part was reconstructed by linguists and was never written down

it would make sense if the Germans borrowed that word from that period

MatheinBoulomenos

yeah, makes sense

Leaky Nun

(but if it turns out that it is a recent borrowing then you can disregard the entirety of what I said)

(recent = less than a millennium)

MatheinBoulomenos

I don't think it's recent

Leaky Nun

do you know any other -deit words?

MatheinBoulomenos

gebenedeit

Leaky Nun

14:01

@_@ what's with those prefixes

nvm that's the regular prefix

MatheinBoulomenos

I think these two are the only ones

Leaky Nun

hmm, I think I'm wrong

benedeien is from benedīgen

so the "ei" might just be regular development from ī (which was preserved from the Latin long vowel)

I don't think it inherited the irregular past participle from Latin

Mockingbird

Hi everyone. Can you give me a hint for integrating this function!

Leaky Nun

@Mockingbird x = pi/2 - u

Semiclassical

14:17

To put the point in more generic terms: when a complicated integral comes out simple, look for a symmetry

abenthy

15:03

@BalarkaSen Thank you for writing this up, this makes a lot of sense.

Leaky Nun

Can one define "gcd" in terms of nonsense?

I think morphisms would be "x divides y"

it's the product of x and y isn't it

how ironic

(cc @BalarkaSen @MatheinBoulomenos)

Huy

@BalarkaSen: any time for the geometry exercise?

Leaky Nun

@Huy would you have any idea? sorry for bothering you

Huy

what about the gcd

Leaky Nun

if the ring is a category and the morphism between x and y is "x divides y", then is gcd the product of x and y?

Huy

15:13

sorry I don't know categories

Leaky Nun

ok

Alessandro Codenotti

If the gcd exists

Leaky Nun

@AlessandroCodenotti let's say it does

then it is the product of x and y?

Lagranian

Can someone recommend a site or forum?

Alessandro Codenotti

More generally any poset can be interpreted as a category where meet and joins are product and coproducts

Lagranian

15:21

to ask my physics questions

Alessandro Codenotti

(So that a poset as a category has products and coproducts if it is a complete lattice)

Leaky Nun

@AlessandroCodenotti now we're getting abstract

Alessandro Codenotti

The divisibility relation is a partial order

Leaky Nun

so the coproduct is LCM?

Alessandro Codenotti

Yes

Leaky Nun

15:22

@AlessandroCodenotti how ironic that the product is a divisor

Alessandro Codenotti

What do you mean?

Lagranian

@LeakyNun I'm tryna find anohter.

not physics.se

Leaky Nun

@AlessandroCodenotti well the gcd is the product

@AlessandroCodenotti is there any categorical implication of that?

Alessandro Codenotti

Dunno, I just know the basic definitions of category theory

Leaky Nun

ok

Secret

15:29

[Chemistry] Another part of the workflow automated. Meanwhile, the AI guys have not replied yet

Balarka Sen

@LeakyNun So you are not just into abstract nonsense, but pure nonsense these days?

Secret

All that left is to write up the data analysis interface and then all that data will be analysed in a few clicks

Balarka Sen

@Huy I was really busy studying various things as my exams are coming up. Let me try it a bit now.

Huy

sure,np

Balarka Sen

I have pen and paper available at least

Secret

15:32

[Random]

Balarka Sen

So I have a triangle ABC, I have to construct a line PQ with P $\in$ AB and Q $\in$ AC such that BP = PQ = QC?

booya

Anyone here is familiar with the CoxPH model and Tensorflow?

Leaky Nun

@BalarkaSen yes I am

Secret

link.springer.com/chapter/10.1007/978-3-642-55682-1_4

Leaky Nun

so is there any categorical implications of gcd being product in Euclidean domain?

Balarka Sen

15:34

i dont care bro

Leaky Nun

ok

Huy

@BalarkaSen: yes. the triangle is given by $a = 6, b = 8.5, c = 7.5$ but I don't think it's relevant

Balarka Sen

OK. Imma try find out what I can know about such a configuration

I guess angles and similarity are relevant

Huy

don't know. no pythagoras or trigonometry

chapter is central dilation

(so similarity should be highly relevant)

Balarka Sen

Hmm, alright

If the angles CP and BQ make with AC and AB respectively are $\theta_1, \theta_2$ respectively then the angles of APQ at the base PQ are $2\theta_1$ and $2\theta_2$

That's a pretty immediate observation

Mockingbird360

15:55

Hello, everyone. is it necessary to master general topology before learning algebraic topology?

Balarka Sen

Master? No. But you need to have a working knowledge of general topology, yes.

Semiclassical

it does depend on what level you're wanting to understand it

if you're wanting to do any kind of proofs, then I don't think you can avoid it

Huy

@BalarkaSen does that help though?

Semiclassical

(if you're just wanting to see what a fundamental group is etc you can get away with less, but it definitely limits you)

Balarka Sen

@Huy I'm pondering it. Can't put it to any use as of now

I kinda want to draw the angle bisectors of angle P and angle Q (which are again theta_1 and theta_2) and iterate the same construction on APQ

It's a crazy idea though

Secret

16:06

1

Q: probability on countable infinite sets

My question relates to probabilities on countable infinite sets. For example, what is the probability of choosing an even number from the positive integers. Believe it or not I am interested in this question from a practical standpoint. I am writing a paper on Boltzmann's brains (Brains that o...

ok, so that means a probability distribution need to be assigned

[O,OO,OOO,OOOO,OOOOO,OOOOOO,...]

Suppose the 1st place of every clump is occupied with 1s. i.e.

[1,1*,1**,1***,1****,1*****,]

then this is the largest possible number without decimal places

abenthy

What is $GL_n(\mathbb{R}^m)$ ? Is this something like $GL_{n \cdot m}(\mathbb{R})$?

Secret

Now we want to find whether there are infinitely decreasing chains in this list

abenthy

Or better question, what is $GL_n(A \otimes B)$ for rings $A$ and $B$?

Secret

Alessandro Codenotti

@abenthy $A\otimes B$ is a ring, so I guess this is the group of invertible matrices over $A\otimes B$?

Secret

16:19

Since each clump has a string length of a natural number, it follows that all of them will always have a first letter corresponding to the smallest element in ordering 2

I need to figure out how to write the dots connected by the black line as a tree.

The idea is that, if we fill in the dots along the black line in order, we will end up with something that looks like a countable binary sequence

abenthy

@AlessandroCodenotti Why is $SL_2(\mathbb{R} \otimes_\mathbb{Q} \mathbb{Q}(\sqrt{3}))$ isomorphic to $SL_2(\mathbb{R}) \times SL_2(\mathbb{R})$?

Secret

that is, let no dots be the sequence 000000000...., then if the lowest dot is filled in, we get 10000000...., next we add 1 to the sequence and then we will get 01000000...., then 11000000..., then 001000000....

This sequence is getting bigger as the 1 move further to the right

While each level (e.g. the 1st, digit, the 2nd-3rd, digits, the 4th-6th digits and so on) has only a natural number length and thus always finite, it is not clear when the whole hierarchy is populated, then an infinitely decreasing chain will result

Leaky Nun

@abenthy because $\Bbb R \otimes \Bbb Q(\sqrt3)$ is isomorphic to $\Bbb R \times \Bbb R$?

abenthy

16:35

@LeakyNun I guess you mean via $\lambda \otimes (a+\sqrt{3}b) \mapsto \lambda (a+\sqrt{3}b, a-\sqrt{3}b)$, right?

As what algebraic objects are $\Bbb R \otimes \Bbb Q(\sqrt3)$ and $\Bbb R \times \Bbb R$ isomorphic?

And then one can use that $SL_2(A \times B) = SL_2(A) \times SL_2(B)$?

Leaky Nun

@abenthy as rings, as is the proper type of the thing inside $SL$

$SL_n(R)$ where $R$ is a ring is the group of all special endomorphisms of $R^n$

abenthy

Ah right, it makes sense that $SL_2(A \times B) = SL_2(A) \times SL_2(B)$ where $A \times B$ is the product of the algebras $A,B$.

But I still feel strange about the isomorphism $\Bbb R \otimes \Bbb Q(\sqrt3) \to \mathbb{R} \times \mathbb{R}$...

Does this work in general, like $A \otimes_\mathbb{Q} \mathbb{Q}(\sqrt{3}) \cong A \times A$ for a $\mathbb{Q}$-algebra $A$?

Secret

So the ordering is roughly something like this

where the base is nonconstant in the sense that, getting to a higher position becomes exponentially harder

The issue is to find whether there's an infinitely decreasing chain somewhere...

In usual number base $b$, given a string, each positions will increment with a carryover whenever the digit $b-1$ is reached

Here, the number base is nonuniform in that for the nth position to increment that will result in a carryover is given by $n$

More clearly, the numbers increment as follows:

$0,1,10,11,20,21,2^20,2^21,100,101,110,111,120,121,12^20,12^21,200,201,210,211, 220,221,22^20,22^21,300,301,310,311,320,321,32^20,32^21,1000,...$

Ok, I got the tree representation

It's a tree where the levels increase in the form $n!$

5

Q: Factorial of Infinite Cardinal

I have been thinking about the following problem: Let $A$ be a set of cardinality $k$ and denote $\sum_A$ the set of all bijection from $A$ to $A$. Also denote $k! = \mathrm{card}\left(\sum_A \right)$. Prove that $k!=2^k$. My proof consists of finding a bijection $F:\sum_A\to P(A)$ whi...

elementary-set-theory cardinals

and therefore, the $\omega$ th level has $\beth_1$ many nodes

However we are back to square one because this tree cannot be enumerated just like the binary tree

user193319

17:06

Is the set of all sequences that converge to zero in the box topology connected? How about path-connected?

Vrouvrou

can someone help me: math.stackexchange.com/questions/2518540/…

abenthy

@LeakyNun What is the inverse to $\Bbb R \otimes \Bbb Q(\sqrt3) \to \Bbb R \times \Bbb R$?

Leaky Nun

@abenthy not sure

MatheinBoulomenos

17:22

@abenthy No, that doesn't work in general. For example $\Bbb Q(\sqrt{2}) \otimes_{\Bbb Q} \Bbb Q(\sqrt{3}) \cong \Bbb Q(\sqrt{2},\sqrt{3})$

Leaky Nun

@MatheinBoulomenos so gcd can be interpreted as categorical product... is there any categorical implication of that?

MatheinBoulomenos

yeah, gcd is associative, commutative, $\gcd(a,0) = a$ (everything up to a unit)

that all follows from general properties of products

Secret

Ok, I need a ternary tree, binary tree looks really messy

will try this tomorrow, theoretically, we can deal with the $\omega$ th level by using the natural bijection of $\Bbb{Z}\to\Bbb{N}$

MatheinBoulomenos

@LeakyNun now for something more obscure: if you have funtions $f,g: R \to R$ such that $\forall a,b \in R$ $a \mid g(b) \Leftrightarrow f(a) \mid b$, then for any $x,y \in R$ such that $\gcd(x,y)$ exists, we have that $\gcd(g(x),g(y))$ exists and $g(\gcd(x,y))=\gcd(g(x),g(y))$. The analogous statement if you replace "g" by "f" and $\gcd$ by $\operatorname{lcm}$ also holds

Leaky Nun

@MatheinBoulomenos is 0 the terminal object?

MatheinBoulomenos

17:33

yes, every element divides $0$

Leaky Nun

@MatheinBoulomenos what is the concrete version of this?

MatheinBoulomenos

I don't have any example of this phenomenon :D

Vrouvrou

@AlessandroCodenotti hello

MatheinBoulomenos

But I guess you can modify the statement slightly and use it to prove that $\gcd(ax,ay)=a\gcd(x,y)$

I forgot that $f$ and $g$ also need to satisfy $a \mid b \Rightarrow f(a) \mid f(b)$

Leaky Nun

what is $\mid$ in the nonsense?

MatheinBoulomenos

17:42

The general concept behind this is a natural isomorphism between homsets

this whole thing is called adjunction

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$Mathematics$ Mathematics