is there a name for a matrix that can be divided into 9 equal parts

Anyone know a good undergrad PDEs book, that focuses more on theory than specific applications?

@SirCumference maybe Fritz John

i used Evans but it's big , people like Taylor but it is also big

@ÉricoMeloSilva I see, thanks

Why does Mathematica's FindGeneratingFunction give up so easily? Anyone know how to make it search harder?

big books suck. (Ok more realistically, not my preference)

@user76284 Idk, have you considered asking the mathematica chat?

I have the correct generating function and it's not complex, but Mathematica refuses to find it even when I specify the FunctionSpace and increase the TimeConstraint.

@Dair Yeah I'll post it there if no one knows here.

It's visited less often.

@user76284 I would suspect the answer would be hard given that it's proprietary software.

hence, I don't know what method they're using to come up with the solution.

@user76284 There is a param for TimeConstraint

@user76284 Click on the details and options picture thing: reference.wolfram.com/language/ref/FindGeneratingFunction.html

Also, consider narrowing the function space.

Yeah I narrowed it down to RationalFunction and increased the TimeConstraint. Strangely, increasing the TimeConstraint doesn't increase the amount of time Mathematica takes until it gives up, which leads me to believe it's ignoring it.

:/

I'm going to assume you tried changing the method as well?

Not sure how I should change Method, just left it at Automatic.

FindGeneratingFunction[
 Table[SeriesCoefficient[-((
    x (1 + 2 x + x^2 + 2 x^3 + x^4))/((-1 + x)^3 (1 + x))), {x, 0,
    n}], {n, 0, 4}], x, FunctionSpace -> "Rational",
 TimeConstraint -> 99999]

There we go haha.

lol

Maybe I'll write my own searcher.

or determine what Method is being used.

I've used mathematica like once ever so, basically I'm just giving general debugging advice at this point.

I would almost certainly be down voted to oblivion on SO.

@user76284 You could almost certainly ask this as a question on mathematica website, explain that you can't find a solution to a seemingly simple generating function. Explain you tried changing time and limited it to just rational functions and that you have no idea what the method should be.

Done: mathematica.stackexchange.com/questions/190806/…

i am not liable for any down votes you may receive. :P

We'll see :)

Have you tried setting the time constraint to like 100

lol i just thought of that now.

the one you set might be too large for mathematica to accept and then it just defaults back to like 10.

Yeah. It always gives up immediately.

Almost like it's saying "Nope, this 5-element list is too long, I'm not gonna try". Works fine for a 4-element list though.

Actually, hold on. It works for the first 4 elements but only if I remove the FunctionSpace and TimeConstraint parameters altogether.

oh lol what does Options[Method, FindGeneratingFunction] give?

{Method -> Automatic}

lol

Ok. So lol u'r f'ed

Mathematica's weird. I'm gonna call it a day and check for answers tomorrow.

Thanks for the help anyway. Good night!

G'night. Sorry I couldn't figure anything else lol.

gosh darn hecking proprietary software.

hi @Daminark

Yo @Dair, what's up?

@Daminark Nothing much. Just bugging people on chat as always. Lol. You?

Doing a topology pset

sounds fun.

i did a fair share of abstract algebra review today, brain is fried.

@Daminark any good probs

i forgot so much of that stuff.

Hey all! Need help with this puzzle - The digit 4 is removed from a number system. So it's like 1,2,3,5,6,7... 38,39,50,51,... So given a number (n) in this number system, what would it be in decimal system. Eg. 5 -> 4, 50->39. If I wanted to write a program that solves this, how do I go about it? Thanks in advance!

The two I've done so far were standard fare. One was to show that degree of a composition is product of degrees, the other (which I'm typing up now) is to show that compact convex subsets of $\mathbb{R}^n$ with non-empty interior are homeomorphic to $\overline{B}^n$, so Brouwer applies to self-maps of those guys as well

If you're interested, here is the whole pset. I've only done problems 1 and 3 so don't spoil any others

@sP_ how efficient do you need the program?

fun

@Dair Not sure, I got asked in an interview today. I solved it by brute force by ignoring all 4s encountered. I had only 20 mins to solve this so couldn't come up with a pattern in time.

Yup. Seems like the point of 4 is to demonstrate why you want compactness for Schauder fixed point, 6 seems real slick

i hate the dVol

Or I mean idk how the proof is gonna go, probably will just be "Force it to be Brouwer"

But the result seems nice

@ant Well ill assume you got bored of that

@sP_ That's probably what I would do for an interview.

@Dair Hmm.. That's like O(n). Seems like there would be a pattern to solve it efficiently.

@sP_ Things aren't always as they seem.

but lol idk.

Hmm.. Okay thanks! Hope what I did is correct haha.

By correct I mean, what they expect.

@sP_ There really might be a better way, but then again, even if there is, it doesn't mean they expect it under interview conditions.

the best thing i can think of is if there is some modularity trick, but I'm to tired right now to work out the details.

@Dair Haha np, thanks for looking at it :)

so every number can be written as $\sum_{i=1}^n 10^i \cdot a_i$ for $a_i \in \{ 0, 1, \ldots 9 \}$. Then suppose that for some $a_i$ there is a $4$. Can we mod the thing by some number such that it shows us a $4$ appeared somewhere.

Why is that for a continuous function $f$, between any two successive (local) maxima, there is exactly one (local) minimum, and vice versa? I see it via a picture but how to make it rigorous?

@user330477 So say the function isn't constant, what would it even mean to have no local minima between two local maxima? What would happen to f'(x) throughout this?

@user616128 Call the two local maximas $a$ and $b$. Then, there exists no point between a and b such that $f''>0$ i.e. $f'' \leq 0 $ over $[a,b]$ So, $f'$ is decreasing over this interval, and if $f'(a)=0$, then we can't have $f'(b)=0$ as $a \neq b$. But how do I show that the local minima is unique?

One can usually furnish intuition into a proof. The intuition here is that after the first maxima, the function is decreasing. Eventually the function must start increasing again to reach the second maxima. Is there some minima necessarily then, yes. If there are two minima, wouldn't there be a maxima between these two? But then that maxima would be the one immediately following the previous maxima, so we are done?

@user616128 Thank you for your answer. Is my reasoning above correct?

@user330477 did you get f''>0 with the sign the wrong way, and this carried through everything else?

Also, the function is decreasing for some time, and then it is increasing for some time, does your description of f' agree with this intuition?

Also, there may be (and are) points that aren't critical between a and b, such that f''>0 and f''<0 (i.e. you have to be careful with using your criterion when f'\ne 0)

@user616128 i understand what you mean. But I was asking if my proof of existance using contradiction was right or not?

It's not right for the three reasons I list above

$

and look at that graphic

Is there a mathematical structure for things like matchup combinations?

Like forming multiple pairs of teams of 5 people from a pool of $N$ people?

Such that you end up with each person appearing exactly $K$ times?

I feel like this is some sort of Stars and Bars construct

So if it exists, does someone else know what it is called?

Hey

whose here

eyes here

btw who is -> who'is -> who's

My here.

^

Grading for a linear algebra course, and I've discovered that I hate problems which just require a theorem to be invoked in order to be answered.

"By the law of blissful ignorance, the problem is no longer a problem. QED."

Hi there! can someone please tell me that how can i prove distributive laws of set theory without using 'Venn-diagram' and $x\in A$ inplies that $x\in B$ properties.

You don't want to prove it by studying where elements live? @nurunnesha

Like you are forbidding saying stuff like $a\in A\cap (B\cup C)$ so $a\in A$ and $a\in (B\cup C)$

?

Does the metric space defined by the distance function $d(x,y)=\frac{1}{\sqrt {x^2+y^2}+1}$ if $x\neq y$ and $d(x,y)=0$ if $x=y$ have a name? (Or is it perhaps not even a metric space because of something I missed?)

(On mobile, so I don't know if my formatting screwed up.)

I have added a number of people to the room owner list. If anybody disagrees with any one of these choices, please voice a complaint and I will gladly listen. The intent was to choose a number of people in different timezones, of various interests.

@Semiclassical @AkivaWeinberger @AlessandroCodenotti @Daminark @TobiasKildetoft You are room owners now. There is no real obligation on you to do anything, but if you are uncomfortable with this you can either remove yourself or ask me to remove you and I will be glad to oblige.

At some point in the future I am liable to try to develop community rules for the chatroom, which will be enforced given 2/3 consensus of chat users over the span of a week. There should be plenty of discussion about such here ahead of time.

I'm I bit rusty on maths, and I was studying reading this slide set on linear programming. The author describes generating rows/variables to a linear programming model via mapping. I'm a little bit confused that what does $2^{[1,\ldots,n]}$ mean as a domain of the function? See the picture attached below:

@MikeMiller Why is the star on this comment different from the other stars?

what is this sorcery.

@MikeMiller Now I will need to decide whether to be a benevolent ruler, or if it is time to bring out the iron fist.

@TobiasKildetoft Better do it now before 2/3 consensus is enforced.

@TobiasKildetoft Well, there are currently no rules, so until those exist we should probably not go full Machiavelli.

@Dair It is "pinned", which means it won't go away over time until I remove it.

@MikeMiller I suppose. But I was just in the mood for that, since I am signing exam grades, forever dooming the students (or at least one of them)

Marc seems to be a StackExchange employee so I am sure he will have no difficulty banning us all without being a room owner.

@Blue I suspect that would be useful in the future (but not right now). It might be worth creating now, though.

Well, I just mean, may as well open it up and invite everyone, and then we'll talk if/when necessary.

I'm just off to bed soon, so I am not about to start a productive discussion. :)

For sure, I know the law of the land. :) thanks!

@Akiva Consider the two involutions on R^3, given by reflection along the planes z=0 and z=1. There are no relations among these beyond that they square to the identity, so this generates an action of Z/2 * Z/2 on R^3.

Thurston calls the quotient "the barbershop". It has fundamental domain $z \in [0,1]$ and the quotient in essence makes each wall function as a mirror - you are in a room with two mirrors facing one another.

You can think through this to see that it gives geometric meaning to the fact that we see infinitely many copies of ourselves through the mirror

Did you mean to ping me

Yes, you like cute pictures

Oh OK

Hi Mike

are you a teacher?

I was confused because it sounded like the continuation of something

I just like that one can encode what happens in a barbershop this way

oups wrong guy haha

I'm a finishing graduate student. Sometimes I teach. I don't this quarter.

Neat @MikeMiller

How old are you if I may ask ?

Off to bed.

So that's the symmetries of $\Bbb Z$, yeah?

@AkivaWeinberger Yes, which I guess one imagines as modelling and infinite regular polygon - this is also the infinite dihedral group

Good observation I didn't notice

There. Grades sent, students doomed.

So many owners :o

Didn't they doom themselves long before you started grading?

@AlessandroCodenotti Sure, that is also a way to look at it

@MikeMiller Except it doesn't act on $\Bbb Z$ the way you expect - reflection across the plane $z=n$ corresponds to reflection across the point $\{n/2\}$

@AkivaWeinberger what group are we talking about?

looks like $D_\infty$ to me

Ya

@AkivaWeinberger This fits with the infinite polygon picture

He was thinking of the group of symmetries of 3-space generated by reflection across two parallel planes

@MikeMiller Yeah - vertex or edge

surely reflection across $z=n$ corresponds to reflection across $\{n\}$

it sends $0$ to $2n$ though, so the formula looks like $z \mapsto 2n-z$

He had the group of symmetries generated by reflection across the planes $z=0$ and $z=1$

The group of symmetries of $\Bbb Z$, on the other hand, is generated by reflection across the points $0$ and $\frac12$

oh no

Otherwise, if you just take reflections across integers, parity is always conserved

maybe just have the generators be $z \mapsto -z$ and $z \mapsto z+1$, problem solved

I love how algebraic this is

Leaky :D

My man

It's also isomorphic to the subgroup of $S_\infty$ generated by $\prod(2n,2n+1)$ and $\prod(2n-1,2n)$

every group has a copy in S_n Akiva

Like, $\dotsb(12)(34)(56)\dotsb$ and $\dotsb(01)(23)(45)\dotsb$

I see

@KasmirKhaan thanks for your useful remark

Arright I'm gonna go do something else

Bye all

bye

@LeakyNun you are very bad man

what, I thanked you and i'm bad?

you are being sarcastic

very obvious

you are a very dark man leaky

I dont know what made you like this and I do not want to know

it takes one man to be sarcastic

@KasmirKhaan but it also takes one man to understand sarcasm

@LeakyNun it takes at least two persons to make a conversation -.-

that is what you are saying or maybe you wanted to say, it takes one to know one ?

aha, yes that's the phrase, lmfao

either way , kasmir has to go study now ! Ill see you very soon my brotha <3

ok

@LeakyNun If I have $\text{Spec}(\Bbb Q)\to \text{Spec}(\Bbb{Z}_{(q)})$, must that be the generic point? Otherwise it would have to be a morphism from $\text{Spec}(\Bbb F_p)$ right (mapping to $p\Bbb Z$)?

@ant is that a morphism of schemes or just continuous map?

schemes

then just pass it to the rings lol

$\Bbb Z \to \Bbb Z_{(p)}$ is an epimorphism btw

but $\Bbb Z$ is initial

so there's only one map $\Bbb Z_{(p)} \to \Bbb Q$

and that's the inclusion

Very smart

thank you

no problem

You can't tile space with pentagonal symmetry but you can damn well try

(See also: Penrose tiling)

looks pretty rhombic as well

@LeakyNun Do you know what the inertial degree of a ring map $R\to S$ is?

I know the inertial index of a given prime $p\Bbb Z$ if this is just $\mathcal{O}_K$ over $\Bbb Z$ ($\text{Spec}(\mathcal{O}_K)\to\text{Spec}(\Bbb Z)$), but not for an arbitrary ring map

@ant no idea

no worries

@AkivaWeinberger can you F-transform that?

Let $G$ be a group and $G\ge H$ is it correct that $C_G(H) \ge H$ ?

@Eran that's the same as claiming that everything in $H$ centralizes $H$

i.e. everything in $H$ commutes with everything in $H$

i.e. $H$ is abelian

now can you find a counter-example?

found it already, thanks

What is the intuition about the centralizer? Why is it called centralizer, I thought it would be like the normalizer of H where it is the largest subgroup of G where H is normal.

because invariant and fixed are different concepts

if I swap your left arms and right arms then you are invariant, but you pretty much can't use neither arms

if I keep your left arm on the left and your right arm on the right then you're fixed

So how would you describe the centralizer of a subgroups in words?

$N_G(H)$ is the elements of $G$ that leave $H$ invariant

$C_G(H)$ is the elements of $G$ that fix $H$

both under the action of conjugation

The normalisers elements may 'permute' the elements, but it atleast keeps them within H, the centralisers elements fix things pointwise, which is much stronger

What is the centralizer of $V_4$ in $S_4$?

what is V_4?

@ant klein four group, {e,(12)(34),(13)(24),(14)(23)}

$V_4 = \{e,(12)(34),(14)(23),(13)(24)\}$

@Eran you know how to compute conjugation right

using disjoint cycle notation

@Rudi_Birnbaum I don't know what that means

but I found this picture

@AkivaWeinberger E8?

@Eran i.e. what is $\sigma \tau \sigma^{-1}$ where $\tau=(135)(46)$ and $\sigma=(1234567)$?

@AkivaWeinberger Fourier-Transformation

@AkivaWeinberger if you want a fun question: find $\left|\operatorname{Aut} \left( \underbrace{C_2 \times C_2 \times \cdots \times C_2}_{\text{$n$ times}} \right) \right|$

@LeakyNun $\sigma \tau \sigma^{-1} = (\sigma(1)\sigma(3) \sigma(5))(\sigma(4) \sigma(6))$?

yes

and can you now deduce the answer to your previous question?

I'm actually trying to find an example where the centralizer is not all G, not H but contains H

hmm

@Eran Try in $S_5$

let me check

lol

@TobiasKildetoft you win :P

can't you do like centraliser of upper triangulars in GL_2 or something?

and its the diagonals maybe?

or just scalars

@LeakyNun All of $S_4$

@TobiasKildetoft What in $S_5$?

@Eran $V_4$ in $S_5$

yay, I was close.

I hate when my brain keeps providing questions instead of letting me sleep. It's 5am and I'm stuck on trying to disprove the existence of a bijective function from the integers to the integers such that it composed with itself equals x+1

Actually, could such a function could exist? I suddenly think that maybe it could.

lol

So, g(x) can't equal x or x+1, so just pick any other integer n and say that g(0)=n while g(n)=1. From there the entire function just sort of falls into place, doesn't it?

g(0)=n, g(n)=1, g(1)=n+1, g(n+1)=2, ... eventually they collide and you can't escape

Okay, then a "half-function" isn't always assured to exist? That's what I think my brain was after.

@Rithaniel Better to think of it as a square root, since that makes it more clear why we should not expect it to exist.

It's kind of like generalizing exponents, you mean? Or considering the square root function as a "half-function?"

@I mean the function you would like to exist is like a square root of the one given by $x\mapsto x+1$

@TobiasKildetoft any idea about above question?

no

do you seen the question?

yes

no problem ; do you know anyone who works in hermitian category theory?

no

What is your profession by the way?

unemployed

are you doing post doc or phd?

My postdoc contract ran out last week

ok, have you apply for any teaching position or planning to do something else?

I am moving out of academia, since there are no jobs within commuting distance and I don't want to move any more

what are your thoughts about doing phd after master in matsh?

*maths?

Is it good or bad from carrer prospective?

depends on what you want to do afterwards

whats in your case/

I have loved working as a postdoc, and I could not have done that without a PhD, so it was definitely the right choice for me.

Gears and magnets are kinda the same, aren't they

nice @TobiasKildetoft best luck for your future ;

Clockwise meshes with counterclockwise

North attracts south

You can't make a triangle out of bar magnets and have them all attract, the same way you can't get three gears put together to turn

Or maybe I want electric charges instead, because they're points?

You could also think of jigsaw puzzle pieces, with indents inwards and outwards, they work the same way

I would like to announce that today Introduction to Riemannian Manifolds (second edition) by the great John Lee arrived in my mailbox after I waited for it the past 7 years.

@TobiasKildetoft Too bad, I was waiting for you to win the Fields medal. =)

How can I prove that for every prime $p>2$ there exists $x\in \mathbb{Z}[\sqrt{-2}]$ such that $N(x)=p$, where $x = a + b\sqrt{-2}$ and $N(x) = a^2 +2b^2$?

Hey, you know the function $N:\Bbb Z+\Bbb Z\sqrt2\to\Bbb Z$ given by $a+b\sqrt2\mapsto a^2-2b^2$?

What does the graph of that look like?

Can we say anything about $N(x+y)$ given $N(x)$ and $N(y)$?

@user1729 I realized/decided after the fact I should try not to engage in discussions in that room: it's intended for users to engage with moderators, and my presence mainly clutters it. So I won't respond there.

But you made a good argument as to why a moderator shouldn't delete that, and I appreciate the point.

@AkivaWeinberger The norm here is defined differently? why defining the norm this way?

@Eran your norm is for a different ring

@AkivaWeinberger well it certainly will be very discontinuous

oh wait nvm I misread

in that case your graph should be really nice

and I think it's a norm, as in it satisfies the triangle inequality

wait I completely misread

you have $\sqrt2$ and the person above you has $\sqrt{-2}$ and I keep mixing the two number

Hello guys!!

Very basic question

Why if we know that for two numbers $x,y$, its greatest common divisor is $6$, then $x=6a$ and $y=6b$ for some $a,b\in\Bbb Z$? Where does "$x=6a$ and $y=6b$" come from?

What is a divisor?

Yes

That is not a yes/no question...

A number $x$ is divible by $n$ if there exists a $k\in\Bbb Z$ such that $x=kn$

Right?

Sounds right.

Good

Is the greatest common divisor a divisor?

@MikeMiller ohhh yes! I could not relate a $\mathrm{gcd}$ with the definition of a divisor! Thank you!!

When does space $X$ admit a measure? If I have a sigma algebra $B$ on the space $X$, would it mean that I can always define a measure on $B$? Can I always form a sigma algebra from $X$? (Not assuming X is topological or countable)

@quallenjäger Yes, given $B$ you can always find a measure such that all sets in $B$ are measurable if you have no requirements that it should interact with other structures on $X$ or anything like that

Simply because there always are (boring) measures such that every set in $2^X$ is measurable and you can always restrict them to $B$

@AlessandroCodenotti "2^X" is only measurable if the space is countable isnt it?

Not necessarily

@Shaun Note that one answers the question "order 2n" and one "order at least 2n".

@AlessandroCodenotti So I can actually always build a measure space on $X$?

Sure, we can find a measure $\mu$ such that the Vitali set will be measurable but one of translation invariance and $\mu([a,b])=|b-a|$ will fail

Fix $x_0\in X$ and define $\mu(A)=1$ if $x_0\in A$, $\mu(A)=0$ otherwise, check that this defines an outer measure and that every subset of $X$ is measurable wrt to it

A slightly more interesting example would be the counting measure

I see, thanks

Usually you're interested in measures with extra properties though, especially if $X$ is a topological (or maybe even metric) space

I am currently working on an totally ordered set $(X,<)$, I want to work out a condition on the space $X$, such that I can have a measure preserving the order, i.e. if $a<b,a,b \in X \rightarrow \mu(a) < \mu(b)$.

So you want all singletons to have positive measures?

Yes

$X$ has to be countable in this case then

Why?

Ah, no, wait, that is not true in general

But if $\mu(X)$ is finite then $X$ needs to be countable

So your question really has little do with measures, you're just asking which totally ordered sets can be embedded into $\Bbb R\cup\{-\infty,\infty\}$, right?

Right

But back to the question above, why if $\mu(X)$ is finite then $X$ needs to be countable?

That's a general fact: if $(X,\Sigma,\mu)$ is a finite measure space and $(X_i)_{i\in I}$ is a family of pairwise disjoint measurable sets then only countably many $X_i$ can have positive measure

I see

So back to my original question, What I have thought is, I can only have countable many non singleton intervals in $\Bbb R$

Anyway I have to go now, but regarding embeddings of total orders into $\Bbb R$ the only thing I can say off the top of my head is that every countable total order embeds into $\Bbb R$ (into $\Bbb Q$ actually) but this is clearly false if we look at total orders of cardinality $\leq |\Bbb R|$

I see, Thank you for your help

Is the definition of the integral generated by $(x,y)$ in $F[x,y]$ (F is a field) = $$(x,y) = \lbrace rx+sy | r,s \in F[x,y] \rbrace$$ ?

ideal: yes

@Eran You can generalize away any reference to fields/polynomial rings. Given elements $x_1,\dots,x_n$ of a ring $R$, the ideal generated by $x_1,\dots,x_n$, denoted $(x_1,\dots,x_n)$ is the set $\{r_1x_1+\dots+r_nx_n \mid r_1,\dots,r_n\in R\}$.

(Compare the subspace of a vector space spanned by a set of vectors)

Maybe not the right chat for this but does anyone see the pattern in this sequence and know what number would come next: 26, -17, 6, 5, -2, 3, 2, 1,

@MikeMiller So it does, thank you. I only gave it a cursory glance.