Mathematics - 2016-03-19 (page 1 of 2)

Ted Shifrin

00:00

shocked to have made any sort of algebraic comment to Karl :D

Karl Kronenfeld

Much more beautiful explanation than mine where I envisioned the ends of the interval overlapping each other at the end.

Ted Shifrin

I guess you've been philosophizing a bit too much, @Karl :)

Karl Kronenfeld

I doubt I would have seen your idea had I not been philosophizing too much hah

Ted Shifrin

I dunno: You're pretty clever.

Or you were :)

lopata

what are additive inverses?

Karl Kronenfeld

00:05

correlation versus causation

Ted Shifrin

Things that add up to 0, @lopata.

LOL, @Karl.

lopata

ok

thanks for the help

Ted Shifrin

When you work in the mod world, each element has another one so that their sum is 0.

lopata

oh I see

Is a modular system always perfect and each value can have a additive inverse?

Ted Shifrin

Sure. If you're working mod n, then the additive inverse of a is n-a .

lopata

00:07

I understand better now

Karl Kronenfeld

(Derives from the "perfection" of rings)

lopata

Is an abelian group a ring?

Karl Kronenfeld

Good point. groups also have that property

I was about to ramble some more, but why not stop here.

Ted Shifrin

LOL ... Good to have you back, @Karl. I need to leave and go meet friends for drinks and dinner.

lopata

ok

Karl Kronenfeld

00:11

@TedShifrin Cya, have fun.

lopata

Is there any way to keep this randomness in a mod otherwise than adding the values?

or substracting them from 0 ?

Karl Kronenfeld

Multiplication will work with some added conditions (length has to be prime, have to leave out zero)

lopata

length of what?

Karl Kronenfeld

of the given interval

lopata

you mean the mod value ?

Karl Kronenfeld

00:13

yes

lopata

ok I see

very helpful!

isn't there values that are not prime where it could still work?

I am learning so much

Karl Kronenfeld

You could leave out more values than just your zero.

Anything that is not relatively prime to your mod value has to be removed.

lopata

ok

but it won't be perfect like + or will it ?

Karl Kronenfeld

after that is done, it will behave "perfectly" like +

lopata

so, if The mod N is prime I just leave the 0

Karl Kronenfeld

00:17

(We're implicitly working with the rings Z_n and their units)

lopata

If it's not I leav the 0+ anything that's isn't prime with it?

Karl Kronenfeld

Yep.

lopata

Ok great! Is there any other way to get a random value in N according to different arbitrary values?

Karl Kronenfeld

You mean N as in the natural numbers, right?

lopata

N is the mod

Karl Kronenfeld

00:20

Ah, that makes more sense given what we've been doing hah.

lopata

for the moment the aribitrary values were <N

but is there any way to generate 1 with them <N or >N

Karl Kronenfeld

I guess I don't see what you're getting at.

lopata

Like if we use the + method there would be any values <N or <2xN

Yeah nevermind

thanks already for that

Karl Kronenfeld

Yeah, both addition and multiplication would still work as long as you are picking out random numbers from ranges that consist of a multiple of N numbers.

lopata

ok

what I thought, but it would be pointless since we already have the lowest N

I mean

could it be a method to get a random value within N by adding some squared values?

with some restrictions on the values

Karl Kronenfeld

00:25

N would have to be odd, if I remember my basic number theory correctly.

But, besides that you should be ok

lopata

N being odd any squared random values would generate a random one?

Karl Kronenfeld

Anyway, I am going to take off @lopata. Might be worth numerically investigating some of these claims too.

lopata

Ok dokie

thanks! bb

Timothy

02:25

Hello. When looking at the roots of $x^2-2x+1$, I find that there are repeated roots at $x=1$, this is because the parabola has it's minimum attained on $y=0$. Will this be the case in general? I was wondering what geometrically, repeated roots mean, in $\Bbb R^2$.

Eric Stucky

Yep!

If you have a polynomial (of any degree) with an even number of identical roots, the graph will look like it 'kisses' the $x$-axis and then turns around.

If there are an odd number of identical roots, then it will pass through the $x$-axis

However, you can still usually distinguish the 3-roots case (and higher) from the 1-roots case, because in the (3+)-roots case, the graph will level out before it crosses.

Timothy

Awesome, thanks, this makes sense. I was considering $x^2+r$ and as $r$ decreases(below $0$), our roots move away from the origin, and at $r=0$ we kiss the origin, at $r>0$ our roots become complex, and in $\Bbb R^2$ we never intersect $y=0$

Eric Stucky

ye

Timothy

Yep I see now with the $3$ degree case, just looking on wolframalpha at (x-1)(x-1)(x+1)

Very nice

how do 2015a and 2015b matlab differ?

Nvm, it's just another release

rorty

06:06

Name a field ($\neq \mathbb{R,C}$) which contains a root of $x^5 + 2x^3 + 4x^2 + 6$

my naive approach was to think of what the roots are of that polynomial but that is very hard

any advice?

Eric Stucky

What would you do if you had a root?

rorty

well i know that by eisenstein, the polynomial is irreducible over Q. Therefore Q(root) would be a valid field

Eric Stucky

Okay, so you can just do that, but without knowing what the root is

Existence comes from the fact that $\Bbb C$ is closed.

Actually, you don't even need to go that far.

rorty

don't i have to know what the root is to name the field? or can i just say $\mathbb{Q}[x]/<x^5+2x^3+4x^2+6>$?

Eric Stucky

Yeah, as long as you know that's a field, you're good

(A priori it's only a ring)

rorty

06:16

got it

thanks eric

Eric Stucky

npnp

rorty

how are you related to math if i may ask (occupation, hobby)?

Eric Stucky

occupation, hopefully :)

1st year grad student

so I guess technically I get paid to teach, at the moment

rorty

wow i wish i could be you

unfortunately i'm on a career path that has zero relevance to math, but i discovered it's beauty and devote my free time to it

Eric Stucky

And yet you're doing field theory :D

That's awesome

rorty

06:28

i'm trying to get through a solid undergraduate curriculum

i am so happy math exchange exists because without it i have nobody in my circles to talk to about math

Eric Stucky

Yeah, it's a truly fantastic resourc

and the chat is a perk :)

rorty

is the chat a recent development? i remmber using stack before and wanting desperately to be able to talk with the answerers of my questions

Eric Stucky

"recent" is a bit strong but I don't think it was here when MSE was in beta

The chat room record goes back ~6 years

rorty

wow

Eric Stucky

Aug 2010

rorty

06:35

also, i don't understand this bit in the description of the chat room: "Rarely if ever expressible as a ratio of integers. "

is that a joke?

is it an answer to an FAQ?

Eric Stucky

Yes :)

what types of numbers are expressible as a ratio of integers? :P

rorty

rarely if ever RATIONAL?!?!?

wow mind blown

rorty

07:02

"If $p(x)$ is irreducible and has degree 2, prove that $F[x]/<p(x)>$ contains both roots of $p(x)$."

Say $c,d$ are the roots of $p(x)$. Then I know $F(c)$ is isomorphic to $F(d)$.

(^because each is isomorphic to F[x]/<p(x)>)

I'm having trouble getting to the end of this proof...

Eric Stucky

Yeah, so you have to do some interpretation here.

I mean, it's rather unlikely that $F[x]/(p(x))$ literally contains both roots of $p(x)$; elements of the former are equivalence classes of polynomials, but the roots exist in the closure of $F$.

I guess then you want something like: if $\phi$ is the homomorphism $F[x]/(p(x))\to \overline F$ given by $x\mapsto c$, then $d$ is in the image of $\phi$?

If you know that $F(c)\cong F(d)\cong F[x]/(p(x))$, then you've just about got it.

iwriteonbananas

08:52

Quick question if you don't mind, @DanielFischer. The closed graph theorem says if $X,Y$ are Banach spaces, $T:X\to Y$ linear and closed, then $T$ is continuous. I'm looking for an example where $Y$ is only assumed to be complete, and where $T$ is discontinuous. Can't seem to cook one up. Do you have a suggestion?

Eric Stucky

You're looking for a counterexample when $Y$ is a complete TVS which is not normed?

iwriteonbananas

Sorry, what's a TVS?

Eric Stucky

topological vector space

iwriteonbananas

@EricStucky Yes, indeed

Lembik

hi..anyone got any ideas about math.stackexchange.com/questions/1704098/… ?

rorty

09:05

@EricStucky I'm still failing to connect the dots. if $x$ is already "assigned" to $c$ under $\phi$, how am I supposed to find which element in $F[x]/<p(x)>$ is assigned to $d$?

Eric Stucky

If you know the isomorphisms you suggested, you can just compose the isomorphisms appropriately.

If not, you have to figure something out.

Since you know $p$ is quadratic, it's probably not so bad to just bash it through

rorty

what method are you referring to by bash it through?

Eric Stucky

I mean, to literally find the element that goes to $d$.

it's got to be $ax+b$ for some $a$ and $b$

and you know that $p(ax+b)=0 \text{ mod } p(x)$, so

PVAL

@iwriteonbananas Have you looked at Conway? A lot of these results end up being true more generally. I think the closed graph theorem is one.

JC574

yo

Albas

09:14

Hello

JC574

got a question about the symmetric algebra

on a finite dim vector space $V$

iwriteonbananas

@PVAL Ok, I'll see if I can find something in Conway

JC574

hmm i think maybe this is obviously false the way i've written it - is $S(V)$ isomorphic to $T(V)/\left< v \otimes w \in V \otimes V : v \otimes w = -w \otimes v \right> $

PVAL

iirc correctly Conway proves a lot of the standard theorems ( I've really only ever looked at the Banach space proofs) for general convex TVS's

Eric Stucky

@rorty: I'm coming up with something like $2bx+(1/a)(b+bp_1+p_2-ap_2)=0$, where $p(x)=x^2+p_1x+p_2$; anywho it's a linear equation and you're assuming that $b\neq 0$.

iwriteonbananas

09:22

Ah, Conway provides two nice examples. One for $X$ non-complete and another for $Y$ non-complete

not quite what I was looking for but interesting regardless

JC574

anyone able to help me with my problem?

rorty

@EricStucky Thanks for your help. I'll think about what you suggested some more.

Eric Stucky

JC: that looks right, although the notation is a little strange

Normally you quotient by an ideal, so $T(V)/\langle v\otimes w + w\otimes v\rangle$.

JC574

yeah it's confusing me

i'm looking at this

ams.org/journals/tran/2008-360-07/S0002-9947-08-04373-0/…

I know the definition you've given

wait

are you sure normally we don't quotient by $\left< v \otimes w - w \otimes v \right> ?

Eric Stucky

yes

erm

no

:P

JC574

09:28

haven't you given the exterior algebra or something

anyway umm

Eric Stucky

indeed

But the thing I wrote is the same as the (first) thing you wrote, so yeah these should be different.

JC574

definition 2.7 in that link

seems to give what I gave if sigma just flips things

is it likely that they're round the wrong way?

oh WAIT

I see

no what I gave seems to be the symmetric algebra

well

not what i gave

but i see what they're doing

if $\sigma$ flips things then the ideal they have is the correct ideal

because eg $\sigma(v \otimes w - w \otimes v) = - (v \otimes w - w \otimes v) $

OK to test something

hmm no i can do this

thanks!

Eric Stucky

Anytime. I'm slightly more helpful than a rubber duck :)

lopata

10:28

hi

are N and N-1 always coprime?

lopata

10:39

I found the proof on here

because if d divides n and d divides n−1, then d divides n−(n−1)=1

I have another question

is (A x B (mod N)) x C (mod N) <=> A x B x C (mod N)

Silent

11:05

@robjohn Please let me know what is this metric $[\sum_{i=1}^n|x_i-y_i|^p]^{1/p}$ is called.

Eric Stucky

11:15

That's the $\ell^p$ metric, Silent.

lopata

13:08

Anyone here?

Timothy

13:56

If $x'(t)>y'(t), \forall t$, and $y(t)\to \infty$ as $t\to a$ ($a\in \Bbb R^+$), it is fair to say that $x(t)\to\infty$ as $t\to a$, correct? I mean even if $x(t)<<y(t)$, it must be finitely far below $y(t)$, and since $y(t)\to\infty$ as $t\to a$, that difference shouldn't matter

$a>0$

Mary Star

14:12

Hello!!

Does it hold that an element of prime order generates a cyclic subgroup?

Timothy

An element of prime order in where?

@MaryStar Like $3\in (\Bbb Z_7,\times)$?

$3=3,3^2\equiv 2, 3^3\equiv 6, 3^4 \equiv 4, 3^5\equiv 5, 3^6\equiv 1, 3^7\equiv 3$

$(3)_{\Bbb Z_7}=\Bbb Z_7$

Mary Star

We have that $g \in P$, for some $P \in \text{Syl}_p(G)$, then $g$ has order $p^k$ for some $k$.

We have that $f:G\rightarrow H$ is a homomorphism. That implies that the order of $f(g)$ divides $p^k$, so it is $p^m$ for some $0 \leq m \leq k$.

Do we cocnlude then that $f(g)$ generates a cyclic subgroup of $H=f(G)$ ? @Timothy

Timothy

What do you think? What does $f(g^2)$ look like

What does $g^{p^k}$ look like and what does $f(g^{p^k})$ look like?

Mary Star

We have that $f(g^{p^k})=f(e_G)=e_H$, right?
Since $m\leq k$ we have that $f(g^{p^m})$ and $f(g^2)$ are not equal to $e_H$, right? @Timothy

porton

I have 1575 reputation at math.SE. Should I trade 100 or 150 bounty for a question important for my research?

Timothy

14:25

@MaryStar Does our homomorphism give us $f(g)f(g)=f(g^2)$?

@MaryStar Seems right to me

Mary Star

@Timothy Yes.

Timothy

So $f(g^n)=f(g)^n$, and $f(g^{p^k})=f(e_G)=e_H$ and $f(g^{p^k})=f(g)^{p^k}=e_H$

lopata

Hi

Timothy

So we have identity, we have inverses

We have closure and we have associativity by default

lopata

porton no

not worth it

someone help me ? math.stackexchange.com/questions/1704316/…

Timothy

14:41

What's the point of your question @lopata

lopata

lets say we have 53 x 12 x 47 = 29892

how to reach 29892 with a self-increasing variable according to 53, 12 and 47

slowly

Timothy

I don't know what you mean, what is the self-increasing variable?

lopata

imagine A = 1

then I do A = A* (C)^3

C is a small constant >1

if A> 53,12 or 47 , then the power decrease by 1 ...etc until we find 29892

it could work if all C were the primes

but it's too fast

and if C = 1.23..etc, but then I couldn't fall on 53 exactly

or a multiple of it

Timothy

Well 53 is prime, so you can only get 53 by $1\times 53$, and it looks like your recursive $A$ works by multiplication

lopata

yes

if C=2, it wouldn't work either

Timothy

14:54

I can't help you how it is currently written I suspect, part of writing it in math, is that it is universally understood

lopata

I would like $A$ to increase slowly to 29892 (>1000 steps) by adding or multiplying it and comparing to {53,12 and 47}

ok

Timothy

Also you wrote $C^3$ above, but it seems you meant $3C$ from the question?

You meant $C^3$ in $(\Bbb Z_p, +)$

lopata

$3C$ is for when 29892 is made of values added like 28060+1830+2

I want $C^3$ because now is made of multiplied values

Timothy

I have no idea what that means sorry, are each of these C's even the same thing?

lopata

yes when doing $3C$

When doing $C^3$ I don't know

Timothy

14:59

What does "$3C$ is for when 29892 is made of values added like 28060+1830+2"

Mean??

lopata

it mean $3 \times 5% of 29892$ for instance, C is a small arbitrary constant that determines how fast the A will increase

Mary Star

What do we get from that? I got stuck right now... @Timothy

Mike Miller

morning

lopata

hi

DanielSank

Hello

I would like to solve for the statistics of a product of two independent Gaussian distributed random variables.

Posted on main here.

Anyone good at that sort of thing?

I worked it down to an integral I can't do :\

Whoever just down voted my question about Doob's Theorem... why?

Timothy

15:11

They downvoted it 5 hours ago it seems

DanielSank

@Timothy Ah, good call.

lopata

15:28

I was thinking of expending A by doing the log of some values, but the problem is I will never fall exactly on a multiple of my values so I can't divide by some number

But thank you anyway I'll try to find a solution myself

lopata

16:47

someone can help me? math.stackexchange.com/questions/1704316/…

Dun Peal

17:10

Is there an "about" page for math.stackexchange.com?

Some page providing basic details about the scope of allowed questions, maybe a FAQ, etc?

DanielSank

@DunPeal Every Stack Exchange site has a help center.

math.stackexchange.com/help

Dun Peal

@Dan

@DanielSank: thanks!

DanielSank

Sure.

Semiclassical

17:27

@mikemiller hmm. remember that $C(\phi)$ question we had talked about a while back? you had said it was related to the Calabi homomorphism

apparently the author removed it, which is a tad disappointing

Mike Miller

C(phi) is the Calabi homomorohism :)

Semiclassical

well, if the question was still there, maybe i'd have remembered that :P

anyways, kind've annouying. especially since i don't remember the question properly.

lopata

hello

let's say we have N values coprime to M, How to find what is equal $ \sum Ni $ mod M, in a maximum steps ?

ops

let's say we have N values coprime to M, How to find what is equal $ \prod Ni $ mod M, in a maximum steps ?

Albas

17:48

Hello

Balarka Sen

Given a projective variety $X \in \Bbb P^n$, one can define it's homogeneous coordinate ring $k[T_0, \cdots, T_n]/I$ where $I$ is the homogeneous ideal for $X$. This homogeneous coordinate is embedding-dependent: the Veronese variety in $\Bbb P^3$ and the standard 2-skeleton $\Bbb P^1$ in $\Bbb P^3$ are biregular, but their homogeneous coordinate rings are vastly different. So my questions are the following:

1) In affine category, given an affine k-algebra, one can recover the variety it's the coordinate ring of by taking $\text{mSpec}$. What is the analogue here? Given a homogeneous coordinate ring, how much data do we need to know about that ring to recover the variety it corresponds to? I suspect the natural grading of the ring is important here.
2) Here's a more specific question: $X$ and $Y$ be biregular varieties inside $\Bbb P^n$. What can one say about the relation between the homogeneous coordinate rings of $X$ and $Y$? More precisely, which (graded) rings appear as homogeneous coordin…

Mike Miller

Post it on main. I'm not gonna read that.

Balarka Sen

I figured you wouldn't! Hi, by the way.

Albas

18:03

Hello@Balarka

Balarka Sen

Hi.

Mary Star

18:25

We have that $|G|\geq |H|$.
At the prime factorization of $|G|$ and $|H|$ can it be that the power of a prime at $|H|$ is larger than that of $|G|$ ?

Mary Star

18:48

Does anyone have an idea?

Pedro Tamaroff

@BalarkaSen From the algebraic point of view, homogeneous coordinate rings are not too nice.

user189740

Hello, I am having a bit of a conceptual problem. I am having a hard time understanding why $\mathbb{R}^3$ minus the origin is simply connected. I would think that if you drew a circle laying on the x-y plane, and shrunk it enough, you would eventually enter into the origin which isn't there.

Balarka Sen

@PedroTamaroff Can you elaborate on "algebraic point of view" and "not too nice"?

Mary Star

@Timothy do you have an idea about my question above?

Balarka Sen

@byteofthat If you have a circle on the xy-plane, if you try to shrink it along the disk it bounds on the xy-plane, then you'll never be able to do it, yes. But what if you also let it move along the z-axis as you shrink?

Pedro Tamaroff

18:58

@byteofthat You're observing that $\Bbb R^2-\rm point$ is not simply connected, but here you have more room.

@BalarkaSen Never mind.

user189740

@BalarkaSen so I can "wiggle" my curve around, as long as I can maneuver past the place not part of my domain?

Pedro Tamaroff

@BalarkaSen I have a problem for you.

Construct a connected countably infinite Hausdorff space.

Mike Miller

fuck that

what's wrong with you

Balarka Sen

Huh. I don't know one off the top of my head.

Pedro Tamaroff

rubs hands

Balarka Sen

19:04

lol.

Pedro Tamaroff

ams.org/journals/proc/1953-004-03/S0002-9939-1953-0060806-9/…

Apparently Urysohn constructed one and the description was four pages long.

user189740

Hmm, @BalarkaSen thanks for the conceptual hint, I think it gave me enough to make sense of it

Pedro Tamaroff

@byteofthat In fact, @byteofthat, you should be able to show that 3-space minus a point is homotopic to $S^2$, the 2-sphere.

Can you do this?

Balarka Sen

@PedroTamaroff cool.

sorry, I meant, totally not cool.

Pedro Tamaroff

If $E$ is $3$-space minus a point, consider $r: E\to S^2$ that sends $x\to x/|x|$.

Balarka Sen

19:07

@PedroTamaroff 3-space minus pt is not homeomorhic to S^2.

Pedro Tamaroff

Show that this map is a strong deformation retraction.

Mike Miller

@PedroTamaroff: Maps are homotopic. Spaces are homotopy equivalent.

Balarka Sen

I am queasy about people using homotopic for homotopy equivalent.

Pedro Tamaroff

@MikeMiller Hm.

user189740

@PedroTamaroff that might be a bit of a step above what I am doing. I have no idea what a strong deformation retraction is.

user189740

19:10

For the map you provided though, am I to assume $x$ is a point and $|x|$ is the length of a vector pointing to that point?

Pedro Tamaroff

@byteofthat Right.

Draw it for 2-space minus a point and the circle.

Mike Miller

@byteofthat: He's trying to show you why your space being simply connected is the same as the 2-sphere being simply connected.

But then one has to figure that out.

Pedro Tamaroff

You're pushing points inside $S^1$ outwards and those outside $S^1$ inwards.

The same works in $n$-space, of course.

user189740

a 2-sphere is a circle, no?

Mike Miller

n-sphere means n-dimensional sphere.

A circle is 1-dimensional.

user189740

19:14

and I suppose then 2-space means I have an x, y, and z axis?

Pedro Tamaroff

2-space is 2 dimensional.

So two axes.

@Mike check your FB

user189740

doesn't it take 2 axes to represent a circle?

Pedro Tamaroff

Well, yes, but you can parametrize it as a $1$-dimensional curve, namely $t\mapsto (\cos t,\sin t)$.

Balarka Sen

Algebraic geometry is hard.

Pedro Tamaroff

Yes, Balarka.

I was meaning to learn some this quarter, but I decided other things are more important.

Balarka Sen

19:18

What are you learning then?

Axoren

Aww, man... A question I wanted to see the answer to has since been deleted.

Pedro Tamaroff

Differential geometry, functional analysis, and more homological algebra and combinatorics.

@Axoren Which one?

I could try and find it.

Balarka Sen

Cool. Differential geometry as in curves & surfaces, or as in Riemannian geometry?

Axoren

@PedroTamaroff It's too much of a bother now.

But it was an integral problem

user189740

@PedroTamaroff I get conceptually now why $\mathbb{R}^3$ minus a point is simply connected, but I don't think I am able to follow your mapping/example. I don't know what it means to be homotopic, for example.

Pedro Tamaroff

19:20

Differential geometry as in learning the dictionary to start learning differential geometry.

Axoren

@PedroTamaroff This one: math.stackexchange.com/questions/1703659/…

Balarka Sen

Fair enough. I don't know anything about either anyway.

Tobias Kildetoft

@byteofthat If you don't know what it means to be homotopic, how do you know what it means to be simply connected?

Hi @BalarkaSen @PedroTamaroff @MikeMiller

Axoren

I gave an answer in regards to bounds on the integral, but the author didn't clarify if they only wanted the asymptotic bounds or a closed-form expression for the integral. I was hoping someone could have given either tighter bounds or a closed-form.

user189740

@TobiasKildetoft I know as much as the definition in my calculus textbook. In that, given some curve, I can shrink the curve to any point on that curve without falling out of the domain.

Balarka Sen

19:24

Hi @TobiasKildetoft.

Tobias Kildetoft

@byteofthat Ahh, that "shrinking" is a homotopy of paths

Brody

In elementary calc we were just introduced to "simply connected regions" as a necessary criterion for determining whether a vector field is conservative.

Balarka Sen

I recently learnt that $\pi_1$ of smooth complex algebraic varieties is a birational invariant, i.e., birational varieties have isomorphic $\pi_1$. Surprising.

user189740

@Brody that is what is happening in my course right now

user189740

@TobiasKildetoft just looked up homotopy on wikipedia, the basic definition makes sense to me

Balarka Sen

19:26

I literally have no idea how the topology and algebraic geometry interacts on $\Bbb C$. Not that I know anything about algebraic geometry.

Brody

@byteofthat In my course we've yet been taught a decent definition for the term... or even an informal, intuitive one for that matter.

Axoren

Does anyone here understand fractional calculus?

user189740

@Brody my lectures are in Icelandic, so I can only make so much sense of them (as I'm not a fluent speaker), so when the prof tries to offer up an intuitive definition, sometimes it works, sometimes I look to my book and online

Brody

Is the open unit disk in $\mathbb{R}^2$ minus the origin simply connected?

Tobias Kildetoft

@byteofthat You mean you have lecture notes in Icelandic, but you don't speak the language?

user189740

19:30

@TobiasKildetoft lecture notes, and the lectures themselves are in icelandic. I can speak it, just not particularly well. Enough to make sense of it for the most part.

Brody

@byteofthat First time I've heard of math lectures in Icelandic. Are their conventions similar to European math?

Tobias Kildetoft

@byteofthat Unusual choice of place to study then (I assume this is actually on Iceland rather than the lecturer picking some weird language given the place)

user189740

@Brody I assume their conventions are similar to those used in the west. Iceland isn't a large place, so they tend to do as their neighbours in such fields.

user189740

@TobiasKildetoft Yes, I am studying in Iceland. I decided to enroll in the university after working here as a software developer for a few years.

Tobias Kildetoft

@byteofthat Ahh, neat

Brody

19:33

@byteofthat Sounds pretty neat tbh

user189740

I am taking a degree in Discrete Mathematics and Computer Science. I have the latter part down for the most part, so I am taking all the math classes I can as I have always wanted to study math.

user189740

Then I also get a degree that recognizes my comp sci skills, as I am just a self taught kinda guy

Tobias Kildetoft

cool

@byteofthat Is it free to enroll in University on Iceland?

user189740

@TobiasKildetoft Its about 650USD a year, regardless of your nationality. I study at Reykjavik University, its about 3300USD a year. It had the program I wanted, and its smaller so I figured I could more personal relationships with the staff there which has helped with the language barrier.

Tobias Kildetoft

@byteofthat I see. I know it was a long time since they were part of Denmark, but not that they had drifted so far :)

user189740

19:37

I understand it's entirely free in Denmark, no?

Tobias Kildetoft

@byteofthat You even get an automatic scholarship that can almost cover your cost of living

(by automatic, I mean that it is not merit based)

user189740

@TobiasKildetoft interesting. Generally the Icelandic government seems to want to have things as the other nordic countries. I think the issue is just a super small tax base.

Tobias Kildetoft

@byteofthat Plus of course their banks breaking apart a few years ago

user189740

@TobiasKildetoft True, but Iceland has recovered pretty well all things considered.

user189740

Anyway, thanks for the chat and help everyone. Have a wonderful evening!

Brody

19:42

Likewise @byteofthat :)

Having zero knowledge of topology, these Wiki articles on paths and connectedness are dulling me.

lopata

2=1+1
$? = \sqrt ? \times \sqrt ? $

?

Brody

@lopata What is the question?

lopata

How to make the $ \prod $ of values within a range $ \pmod$

Tobias Kildetoft

@lopata Try formulating a complete question

preferably using words rather than symbols

lopata

I made the question already, please do not downvote it thanks

-1

Q: What's the longest way to compute a product?

Good day, We have $N$ values $\in \mathbb{Z}^+$, all co-prime with $M$ and $ Ni < M$ How can we calculate $ \prod Ni \pmod M$ with the maximum steps possible (without using $ \times 1$ or operations that stops or make the result redundant)? For instance to calculate calculate $ \sum Ni \pmod M...

Mike Miller

19:52

Hi @TobiasKildetoft.

@BalarkaSen: $H^3(X;\Bbb Z)$ is also a birational invariant.

Balarka Sen

Interesting.

Mike Miller

Sorry, the torsion subgroup of that.

Balarka Sen

Ah, OK. I don't have many examples of higher dimensional non-smooth varieties in mind, but are you restricting to smooth varieties? 'Cause it is not true that $\pi_1$ is birational invariant for non-smooth (e.g. $y^2 = x^3$ and $\Bbb P^1$: the former is $S^1 \vee S^2$ or something of that sort).

Mike Miller

Sure.

Balarka Sen

OK, great. Thanks: very interesting.

Mike Miller

19:57

I don't remember if it's still true non-smoothly. Probably it is.

Brody

test: $\boldsymbol{A},\mathbf{A}$

Transcript for

$Mathematics$ Mathematics