Mathematics - 2015-08-23 (page 2 of 3)

2:00 PM

nah, dunbother about it. i am , most of the time, the most unhelpful in this chatroom. [see, @anon asked something, and i was at the point of answering something very wrong]

d'you know the proof of Hilbert's finiteness of invariants theorem? asking because i want to hear the general idea of the proof

Boni

Don't know if it is the same proof as the one you are reading. The one I know of is basically generalising a construction of the elementary symmetric polynomials.

Balarka Sen

OK, I'm interested.

Boni

You can get the coefficients of the ESP (elementary symmetric polynomials) as the coefficients of $p(t) = \Pi_{g\in\Sigma_n} (t-x_1\cdot g)$.

Balarka Sen

right

anon

get ~~the coefficients of~~ the ESPs as the coefficients of [...]

Boni

2:05 PM

So, for the proof, you start of by replacing $\Sigma$ with $G$ and construct $n$ polynomials of that form, by replacing $x_1$ with each of the algebraic generators of $A$.

@anon Yea, woops.

Balarka Sen

quick non-serious side-question : is it natural for invariant theorists to write group elements acting on things in a wacky way? Eisenbud uses $(g, x) \mapsto g^{-1}(x)$, you are using $(g, x) \mapsto x \cdot g$ (i.e., right-action), etc.

anon

dual vectors might explain why one may be using g^-1s or right actions

Boni

My supervisor uses right action in the dual space. So I am following his convention.

anon

(consider the contragredient action of a group on a function space, induced from the action of the group on the domain)

Balarka Sen

@Boni ah. and how do you prove that the coefficients you get really generates $A^G$?

Boni

2:06 PM

It doesn't always do that. It generates a subalgebra.

Balarka Sen

@anon oh, ok.

Boni

But you can show that $A^G$ is finitely generated as a module over this subalgebra.

So if you take those ESP coefficients together with the module generators, you can get finite set of generators.

They might not be algebraically independent though.

So it might not be as neat as the ESP for $\Sigma_n$.

For the finite generation of $A^G$ as a module, you consider $x_1$ as a root of the polynomials $p(t)$.

And similarly for the rest: $x_j$.

It lets you write $x_1^n$ as a linear combination of lower powers of $x_1$ over the subalgebra, because the coefficients are the generators of the subalgebra.

Oh, it should be $x_1^{|G|}$, not $x_1^n$.

ADG

hello anyone know sup here?

hello

Boni

@BalarkaSen Right um... as @anon has said, the group action was originally defined as linear maps on $V=<e_1,\cdots,e_n>$. The group action on $A$ is defined by extending the definition on the dual space: $x_j(g\cdot e_k) = [x_j \cdot g](e_k)$.

anon

2:22 PM

hmm, $(f\cdot g)(x)=f(gx)$, I like that notation

Balarka Sen

whoops, I was away.

let me read that

anon

if you interpret a function $X\to Y$ as a subset of $X\times Y$ (its graph) and have $G$ act on the first coordinate of $X\times Y$ hence act on functions, you get the contragredient action. Since $\{(gx,f(x)):x\in X\}$ equals $\{(x,f(g^{-1}x)):x\in X\}$ we have $(g\cdot f)(x)=f(g^{-1}x)$. (This is when people want a left action.)

Balarka Sen

@Boni ok. how do you prove that?

Boni

@BalarkaSen Consider substituting $t=x_1$ into $p(t) = \Pi_{g\in G} (t - x_1 \cdot g)$.

If you expand that product, you get a polynomial in $t$ over the subalgebra.

The substitution gives you a sum of the form $x_1^{|G|} = \sum_{k=0}^{|G|-1} \sigma_{1,k} x_1^k$.

Balarka Sen

mhm

Boni

2:28 PM

So, all elements in $A$, not just $A^G$, can be written as a linear combination of some finite powers of $x_j$ over the subalgebra.

Hm... there is some extra steps here I am forgetting.

Oh. Hilbert's basis theorem. A finitely generated algebra, namely $A$, is a Noetherian ring.

So, $A^G$ being a subalgebra of $A$, is also finitely generated over the um... subalgebra from before. I should've named it.

ADG

if sup(A) =a, sup(B) =b so a+b is an upper bound of (A+B) but how do i prove it is s.t. sup(A+B)=a+b

Boni

@ADG What's the underlying set? I think you usually use the $\epsilon-\delta$ definition. Split the $\epsilon$ in two and then use minimum of the two $\delta$-s.

ADG

I have $\sup A+\sup B<a+b+2\epsilon$

and let us have another upper bound n then

$a+b\le n$

so $\sup A+\sup B< n+2\epsilon$ but that doesn't mean $\sup A+\sup B\le n$ @boni

anyone plz help/

Ted Shifrin

Hello @Balarka, @anon.

Mike Miller

Morning.

Ted Shifrin

2:41 PM

@ADG: Can't you make $\epsilon$ as small as you want?

Goodnight, @MikeM.

ADG

$\epsilon>0$

Ted Shifrin

Right, but as close to $0$ as you want.

ADG

we have not reached limits formally again in college

Ted Shifrin

Prove the following lemma: If $x<y+\epsilon$ for every $\epsilon>0$, then $x\le y$.

Hint: Draw a picture on a number line.

ADG

let a\me 0

since a\ge0 and a\ne 0 implies a>0

choose e=a/2

0<e/2<a

there exists e in R s.t. 0<e<a

Ted Shifrin

2:44 PM

OK, there you go. Here $a=y-x$.

ADG

therefore a=0

Ted Shifrin

Right, so a simpler version of the lemma would be: Suppose $0\le a<\epsilon$ for every $\epsilon>0$. Then $a=0$.

Balarka Sen

@Boni OK, right. That's an interesting proof you're sketching. Do you know of a reference for this proof?

Thanks for the brief sketch, btw!

hi @TedShifrin

ADG

Balarka Sen

what you said earlier was right, choosing a minimal set of basis for a module is a tough job. The correct way to go about it is to pick a basis for $M/\mathfrak{m}M$ over $A/\mathfrak{m}$, and then pull this back to a set of generators for $M$.

Ted Shifrin

2:46 PM

@ADG: I have absolutely no earthly desire to look at that!

ADG

@TedShifrin thankyou

OK

it'd tough if i were there in the foreign

Balarka Sen

(recall : $M$ is a f.g. (projective) $A$-module where $(A, \mathfrak{m})$ is a local ring)

ADG

@TedShifrin even if you look is it correct?

Ted Shifrin

I'm not looking!

ADG

:( OK

Ted Shifrin

2:48 PM

That doesn't seem obvious to me, either, @Balarka. But I haven't thought about this stuff in centuries.

Ask @anon.

Balarka Sen

nah, I have figured this one out. It's an exercises in the book I am working through. But you're right, I should ask anon all this.

ADG

so you mean to say if sup A+sup B<a+b+2e and a+b<=k that would mean supA+sup B<k=2e for all e>0 so supA+supB-k=0??

Kevin Driscoll

Anyone know of a good reference for learning tensor networks?

Mike Miller

@Ted: It's a standard technique. The fact that you can do this, IIRC, uses Nakayama.

Kevin Driscoll

The literature is incomprehensible

Ted Shifrin

2:50 PM

He used Nakayama later in his argument, @MikeM. But, yes, it looks very Nakayama-ish.

Mike Miller

Well, you are doing physics...

Ted Shifrin

I don't know what those are, @Kevin.

ADG

@Ted so you mean to say if sup A+sup B<a+b+2e and a+b<=k that would mean supA+sup B<k=2e for all e>0 so supA+supB-k=0?? Sorry for any unintentional rudeness

Kevin Driscoll

OK, I wasnt sure if its something math folks were working on or not

ADG

I respect you as a person

Ted Shifrin

2:51 PM

@ADG: You're writing too much stuff. It follows from our lemma above if you just substitute. But sometimes it's better to have simpler things to look at.

You have a typo with $k=2e$ where it should be $k+2e$.

ADG

yes that is true

would that mean every upper bound is equal to supremum?

Ted Shifrin

That's why I stated it with fewer letters. $x=\sup A+\sup B$, $y=a+b$.

Kevin Driscoll

@ADG Maybe I'm just cofnused but I don't see why $\sup(A+B)$ should be exactly $a+b$. Shouldn't it be $\sup(A+B) \le a+b$?

Ted Shifrin

@ADG Very much NO.

Balarka Sen

@KevinDriscoll I don't even know what are tensor networks.

Ted Shifrin

2:53 PM

Yes, we said $\le$ earlier.

Kevin Driscoll

@TedShifrin ADG has said $=$ a few times before you got here, I think

Ted Shifrin

Oh. I don't even know the original question. I was just addressing what I stated as the Lemma.

Balarka Sen

@TedShifrin I was in a ergodic number theory seminar a few days ago. I didn't understand much, plus the lecturer said there is a natural interpretation of all this using "horocycle flows on modular curves". So much to learn :'(

Ted Shifrin

Well, I know what horocycles are and I know what modular curves are, but that's it.

Kevin Driscoll

@BalarkaSen Not knowing things is just something you will get used to as you spend longer and longer in academics. No one can learn everything

Boni

2:57 PM

@BalarkaSen Uh... I got this from some notes my supervisor gave me and I also looked at a book by Benson (LMS). Let me see if there are any online proofs. The theorem might also be called Hilbert-Noether.

Balarka Sen

The weird thing is that so much extraterrestrial machinary is used to prove a quite number theoretic statement, @Ted (it's the Oppenheim-Davenport conjecture, which says for every nondegenerate quadratic form $Q$ of $n \geq 3$ variables which isn't a rational multiple of a rational quadratic form, $Q(\Bbb Z^n)$ is dense in $\Bbb R^n$)

ADG

i'm studying real analysis and we prove 1+1=2 and x.0=0 :(

Ted Shifrin

Look at how much hard math goes into Fermat's Last Theorem!!! Nothing could be a simpler statement.

Well, @ADG, I used to do $x\cdot 0 = 0$ the first day of both abstract algebra and Spivak's Calculus with Theory course.

It's a good thing to make students understand that this is not an axiom.

Balarka Sen

well, at least the proof of FLT uses things surrounding algebraic number theory (iirc?). the proof of Oppenheim-Davenport uses ergodic theory -- heaps of analysis out of nowhere..!

Mike Miller

"It's just a statement about the integers - and heaps of algebraic geometry out of nowhere!"

Balarka Sen

3:01 PM

I am regretting not learning analysis before, in fact.

Ted Shifrin

There's all sorts of algebraic geometry stuff in FLT. But there are LOTS of connections between number theory and hyperbolic geometry, so I'm not the least surprised.

Mike Miller

Twin primes stuff also used analysis. It should not be surprising that some numver theoretic statements take analysis to prove, some take algebraic geometry, ...

Boni

Invariant Theory of Finite Groups, page 5

Ted Shifrin

Combinatorial analysis has become one of the hot areas. Terry Tao plus several of my (former) colleagues plus many others.

ADG

all my mates did supA>=a and supB>=b so supA+supB>=a+b so sup(A+B)=supA+supB because it is a ">=" are they correct or should we use e-d

Balarka Sen

3:03 PM

I didn't know FLT used a lot of algebraic geometry. I guess I was just too narrow-minded, and I am realizing it.

@MikeMiller er, yeah, you're right.

Ted Shifrin

How do you get $\sup(A+B)=\sup A+\sup B$, @ADG?

heya mr @Kaj !

Balarka Sen

hi @Kaj

ADG

A+B={a+b,a in A and b in B}

didn't we talked over this before?

Ted Shifrin

You really need to show $\sup(A+B)\le \sup A+\sup B$ and $\sup(A+B)\ge \sup A+\sup B$, @ADG.

I haven't been here for earlier discussion.

ADG

OK anyways I did that something similarly. now i'm finding sup and inf of 2<x^2<3 for x in Q but i don't think there is one. sqrt2 and sqrt3 are irr so we can find another and contracdiction

Kaj Hansen

3:06 PM

Hey @BalarkaSen, @TedShifrin

Ted Shifrin

Correct, @ADG, in $\Bbb Q$ there are no sup and inf. Yes, you were talking about this a week ago.

So, how did your first week go, @Kaj?

Balarka Sen

What kind of math have you been thinking about, then, @Kaj?

Kaj Hansen

A lot of avatars aren't rendering for me. Is that just me?

ADG

What do you people do for timepass ?

Ted Shifrin

Was a Ted-less math building fun? :D

ADG

3:08 PM

hello kaj

we met a long itme ago

Ted Shifrin

They're all rendered for me, @Kaj.

ADG

I hate real analysis I know it's obvious and I can think the outline of proof but can't write it

Kaj Hansen

Oh I know what it is. I fixed it.

Remember me

Hey@Kaj

Ted Shifrin

Have you taken abstract algebra already, @ADG?

Remember me

3:09 PM

Howdy @Ted

ADG

no I'm just joined college some weeks back for bachelor degree

Remember me

@ADG you have abstract algebra in IIT?

Kaj Hansen

First week went pretty well @TedShifrin. I'm about halfway through the first algebra set, and almost done with my first cryptography set

ADG

no not yet

Remember me

What all do they offer?

ADG

3:10 PM

Computer science students study real analysis and matlab and linear algebra

Balarka Sen

What will the algebra course cover, @KajHansen?

ADG

and mechanical and electrical ones study advanced calculus and same matlab and lin alg

the calc is really adv they have multivariable thingos

Remember me

And mathematics students?@ADG

ADG

there are none in bachelor's srry

Kaj Hansen

Apparently I'm grading Math 3000 as well @TedShifrin. This'll be an experience...

ADG

3:11 PM

masters there are some

Ted Shifrin

I would recommend doing algebra before real analysis, @ADG.

ADG

i know algebra from +2

Balarka Sen

@TedShifrin Really?

Remember me

C'mon ... That's a pain.. I won't do any good going to iit then @ADG

ADG

why a pain

Ted Shifrin

3:12 PM

If you wrote proofs in linear algebra and some in algebra, then proofs in analysis are not so bad. But the sentences in analysis are more complicated than the sentences in algebra.

ADG

are you nri or indian resident

Kaj Hansen

@BalarkaSen, the official description says this: "Some topics covered include the Sylow theorems, solvable and simple groups, Galois theory, finite fields, Noetherian rings and modules."

Idk how closely we'll follow that though

Ted Shifrin

That'll be fine, @Kaj. I know Dr. Graham was looking for lots of graders. I gave him names from 3510 last spring, too.

ADG

actually i spent the last two years solving multiple choice questions

Remember me

@ADG Indian

ADG

3:13 PM

OK

now you look good Kaj

Balarka Sen

@KajHansen Sylow theorems are good.

Kaj Hansen

haha @ADG

Balarka Sen

So are Noetherian rings. You'll enjoy both.

Ted Shifrin

@Kaj: Of the two teachers, which style do you find more interesting?

Balarka Sen

I was just telling Kaj a few days ago that he looked like Bill Thurston

Ted Shifrin

3:14 PM

@BalarkaSen I basically disagree with that one.

Remember me

I would like a pure mathematics course more @ADG ... I don't care about the applications of math

@Ted by algebra you mean abstract also?

Ted Shifrin

Yes, that's what I said earlier, @Remember.

Balarka Sen

@TedShifrin Well, the hairstyles are pretty similar.

ADG

these are just the basics for the next three years of computer science

Kaj Hansen

My hair's actually short now lol

Ted Shifrin

3:14 PM

Thurston, albeit one of the most brilliant mathematicians of the century, was not a very good looking guy.

ADG

if you want maths then do msc not btech

Balarka Sen

fair enough, @Ted

Remember me

Yes I guess I will be doing that

I agree with @Balarka @Kaj you are looking a lot like Bill Thurston

ADG

hello

i changed my photo as well

Kaj Hansen

I think it's becoming a tradition on this chat for people to liken me to other people :P

ADG

3:18 PM

im going to dinner and then robotics

ill do my homework of real analysis later and study for inorganic quiz and maybe do physics practical filework of diffraction and interference of light and maybe then sleep i have a class tom at 9 AM @Rememberme life here is tough

Ted Shifrin

Well, @Kaj, two more people told me I look like Richard Dreyfuss :P

Balarka Sen

@KajHansen Yes, we're just waiting for your avatar to change.

Ted Shifrin

So, do you prefer Lorenzini or Nakano? :)

Remember me

@Balarka after understanding isomorphism theorems will I be able to understand what shot exact sequences are ?

Mike Miller

At least it's not inspector Dreyfus.

Ted Shifrin

3:20 PM

LOL, one of my favorite movie characters, @MikeM. :)

Mike Miller

"Ah, Ted, your telltale twitch reminds me of someone..."

Kaj Hansen

@TedShifrin, that's hard to say. I can tell I'm really going to enjoy them both as lecturers.

Ted Shifrin

OK @Kaj.

All right, I'm outta here.

Balarka Sen

@Rememberme A short exact sequence is a chain of homomorphisms $A \stackrel{f}{\to} B \stackrel{g}{\to} C$ such that $f$ is injective, $g$ is surjective, and $\ker g = \text{im} f$

byebyes, @Ted

Kaj Hansen

Have a good day @TedShifrin

Remember me

3:23 PM

Why do we care about them ? @Balarka

Bye @Ted

Balarka Sen

Some people cares, some people don't. You may not care if you like.

It's just a convenient way to write down certain things.

Remember me

Ok

anon

why we care about exact sequences is something you learn after exposure to many situations in nature where they are important and meaningful

Remember me

Situations like?

Balarka Sen

@anon a good answer to the question I recently heard was that it's a geometric way to see algebraic things

@Remember situations in homological algebra, say

Remember me

3:26 PM

Okay

anon

group theory and (co)homology are prime examples off the cuff

Balarka Sen

the point is that certain things get very easy once you write down a short exact sequence. for example, to write down all the information give in the short exact sequence mentioned above in words, you have to say : $f : B \to C$ is a map, which has kernel $A$, so that it induces the isomorphism $B/A \cong C$ by the 1st isom theorem.

that's too much to write down.

also, one of the beautiful facts about short exact sequences is the splitting lemma. i mean, you can write down the statement of the splitting lemma in words if you want but that'd be too much writing.

to illustrate, to write "$f : A \to B$ is surjective" in paper you need 17 characters, but to write "$0 \to A \stackrel{f}{\to} B$" you just need 6 characters.

Remember me

3:42 PM

Splitting lemma ... @Balarka what does it say about ?

Balarka Sen

@Remember You should get used to short exact sequences before knowing what splitting lemma says.

Give me an example of a short exact sequence first.

ADG

hello

Remember me

Okay... I have to go now.. I will return with an example

evinda

4:08 PM

Hallo @Alessandro

Alessandro

4:25 PM

Hallo @evinda wie geht's es dir?

evinda

@Alessandro Mir geht es ganz ok. Dir? :)

Alessandro

@evinda gut, danke, obwohl ich zurück nach Italien umziehen musste (wegen burokratischer Probleme konnte ich nicht diesen Jahr in Deutschland studieren :( )

evinda

@Alessandro Schade :( Und was hast du jetzt vor?

Alessandro

Übermorgen schreibe ich die Aufnahmeprüfung der Universität Trento, Ich werde meinen Bachelor in Italien absolvieren

evinda

@Alessandro Hast du dich dafür vorbereitet? Braucht man für den Bachelor drei Jahre?

Alessandro

4:34 PM

Ich habe eine Probeprüfung geschrieben und sie mit 33/35 bestanden, deshalb glaube ich dass ich die Prüfung ohne Probleme bestehen soll

Random Variable

@DanielFischer Are you in the mood for a question? You might find it interesting (or perhaps not).

Alessandro

Ja, drei Jahre für den Bachelor und zwei für den Master, aber ich will den Master in Deutschland machen (wenn möglich) @evinda

evinda

@Alessandro Gut!!! Viel Glück für die Aufnahmeprüfung!!! :)

Alessandro

Danke :D jetzt muss ich weg, bis später @evinda

evinda

Bis später!!! @Alessandro

Chris's sis the artist

5:27 PM

Maybe some of you are interested in sharing ideas here

0

Q: Double integral with a product of dilog $\int _0^1\int _0^1\text{Li}_2(x y) \text{Li}_2((1-y) x)\ dx \ dy$

One of the integrals I came across these days (during my studies) is $$\int _0^1\int _0^1\text{Li}_2(x y) \text{Li}_2((1-y) x) \ dx \ dy$$ that can be turned into a series, or can be approached by using the integration by parts, but these ways do not look like as a promising way to go, or I migh...

Stan Shunpike

5:38 PM

@TedShifrin got any good problems to teach me about adjoints? Was reviewing my quantum mechanics today and think I need to understand adjoint and self-adjoint operators better

Owatch

Hello.

Stan Shunpike

Hello

Alessandro

5:54 PM

Does someone have a PDF with a proof of the normality of Champernowne's constant? The original article on the journal of london mathematical society's website needs to be payed

Mike Miller

E-mail me the link to the article. My email is in my profile.

Alessandro

I sent you an email @MikeMiller

Mike Miller

Got it, replied.

Alessandro

I received your reply, thanks a lot!

Daniel Fischer

6:26 PM

@RandomVariable Depends on the question. What is it about?

ccorn

Celebrating my 4444 rep, here is an answer on iterating squareroots in $\mathbb{F}_p$

Random Variable

@DanielFischer Taylor series

Daniel Fischer

@RandomVariable Broad topic.

Mike Miller

@DanielF: Modulo details the suggestion you gave worked. Thanks!

Daniel Fischer

@MikeMiller What sort of details?

Mike Miller

6:36 PM

@DanielF: I just meant checking the details. I guess "Modulo" was the wrong word. Pretend I said "After checking the".

Daniel Fischer

'kay

@RandomVariable In case you didn't understand, that meant "tell me more".

Random Variable

@DanielFischer I know. I was typing it up.

@DanielFischer This is something I came across about 2 years ago, but I've never been able to understand: For what sort of functions can $f(a+z)$ be expanded in the form $f(a+z) = \sum_{k=0}^{\infty} c_{n} e^{-kz}$? In other words, for what sort of functions could you find another function with a Maclaurin series expansion such that $f(a+z) = g(e^{-z})$? One author claimed this was possible for $(a+z)^{-n}$.

Kevin Driscoll

6:51 PM

@RandomVariable Do you mean for the $k$ to be integers or could they bet some arbitrary sequence of increasing real numbers?

Random Variable

@KevinDriscoll integers

Kevin Driscoll

Ok, pretty specific then

Daniel Fischer

@RandomVariable Since $z\mapsto e^{-z}$ has period $2\pi i$, it is necessary that $f$ has period $2\pi i$. And if $f$ is holomorphic and $2\pi i$-periodic on a strip $s < \operatorname{Re} z\rvert < t$, it induces a holomorphic function $g$ on the annulus $e^{-t} < \lvert w\rvert < e^{-s}$ via $w = e^{-z}$.

Now when we look at the desired expansion, $g$ has zero principal part, so $g$ extends to a holomorphic function on the disk $\lvert w\rvert < e^{-s}$, and that means that $f$ has an analytic continuation to the half-plane $s < \operatorname{Re} z$, and such that $\lim\limits_{\operatorname{Re} z \to +\infty} f(z)$ exists.

So, since $(a+z)^{-n}$ is not $2\pi i$-periodic, that doesn't work.

Random Variable

7:09 PM

@DanielFischer Are there any restrictions you could place on $z$ that would make it possible?

Daniel Fischer

@RandomVariable The right hand side is $2\pi i$-periodic. No power function $z \mapsto (a+z)^b$ for $b \neq 0$ is $2\pi i$-periodic.

Random Variable

@DanielFischer That what I thought, but I had to ask.

Clarinetist

Oops

0

Q: Understanding degrees of freedom in relation to rank for $\sum_{i=1}^{n}(y_i-\bar{y})^2$

So I'm looking at this website which states: One of the questions an instrutor [sic] dreads most from a mathematically unsophisticated audience is, "What exactly is degrees of freedom?" It's not that there's no answer. The mathematical answer is a single phrase, "The rank of a quadratic form....

linear-algebra statistics

user159870

Hi. Do you know which is the origin of the name Ted Zmagin?

Random Variable

@DanielFischer Basically what he ends up using is $ (a+it)^{-n} = \sum_{n=0}^{\infty} c_{n} e^{-ikt}$.

Daniel Fischer

7:23 PM

@RandomVariable Fourier series perhaps? For $\lvert t\rvert \leqslant \pi$ or $0 \leqslant t < 2\pi$?

Random Variable

@DanielFischer $t \in \mathbb{R}$

Daniel Fischer

Thought so. Probably a Fourier expansion.

Mary Star

Does someone of you know what an algebraic differential equation is?

sonicboom

7:41 PM

I need to "distort" a sigmoid function, here is its wolfram alpha code: y(x) = (1/(1 + e^(-8(x - 0)))) from -1 to 1 .....I need the same "kind" of function except want it to start "ramping up" further along the x-axis. That is, it should be close to zero until almost 0.6 on the axis..and then ramp up rapidly. Anyone know how to manipulate the above function to do that? Or can give me code for a function that does what I need?

Random Variable

@DanielFischer He never uses the word "Fourier". He simply replaces $z$ with $it$.

Clarinetist

**REALLY** basic linear algebra problem here. It has been years since I've done Gauss elimination. I am trying to calculate the rank of $$M = \begin{pmatrix}
1-\frac{1}{n} & -1/n & \cdot & -1/n \\
-1/n & 1-\frac{1}{n} & \cdot & -1/n \\
\cdot & \cdot & \cdot & \cdot \\
-1/n & -1/n & -1/n & 1-\frac{1}{n}
\end{pmatrix}$$

Someone told me on SE that a standard way to do this is to do Gauss elimination

and start by multiplying it by $n$

Daniel Fischer

@RandomVariable That looks wrong. But the context may justify it. It's however impossible as an identity of analytic functions.

Clarinetist

So let's do that: $n M = \begin{pmatrix}
n-1 & -1 & \cdot & -1 \\
-1 & n-1 & \cdot & -1 \\
\cdot & \cdot & \cdot & \cdot \\
-1 & -1 & -1 & n-1
\end{pmatrix}$

Now if memory serves me right, I'm supposed to do some sort of elementary row operations

Daniel Fischer

@Clarinetist An $n\times n$ matrix?

Clarinetist

7:45 PM

@DanielFischer Yes

Now if I recall, I have to do some row operations

Daniel Fischer

@Clarinetist Then you can do better than Gauß elimination. What eigenvalues (with multiplicities) has an $n\times n$ matrix with all entries $1$?

Clarinetist

@DanielFischer I honestly don't remember, but I know if I were to multiply $nM$ by $-1$, I would get a diagonal matrix with eigenvalue $(1-n)$, multiplicity $n$... I think. grabs book

@DanielFischer Well, actually, wouldn't that just be $1$ with multiplicity $n$? (complete guess)

Daniel Fischer

@Clarinetist That would be for the identity matrix, all diagonal elements $1$, all off-diagonal elements $0$.

Clarinetist

Ah, that's right

Yes, diagonal matrix means stuff in the diagonal, $0$ elsewhere. Completely forgot

Daniel Fischer

More abstract, a matrix of the form $u\cdot v^T$ with vectors $u,v$?

Clarinetist

7:55 PM

@DanielFischer I worked with the $2 \times 2$ case. So for a matrix of all $1$s in the $2\times 2$ case, I get $\lambda = 0, 2$ each with multiplicity $1$. But the $\lambda = 0$ is trivial, if I recall, and is not included as an eigenvalue

Daniel Fischer

@Clarinetist No, $\lambda = 0$ also is an eigenvalue. The $3\times 3$ case would be more informative here than $2\times 2$.

Clarinetist

@DanielFischer Or is it the set of eigenvectors that I'm thinking of, where you don't include the vector $\mathbf{0}$?

Daniel Fischer

That.

Clarinetist

K. Doing the $3 \times 3$

@DanielFischer If my algebra is right, $\lambda = 1, 1-\sqrt{3},1+\sqrt{3}$. So I would guess that for a $n \times n$ matrix entirely of $1$s, there are three eigenvalues: $1$, $1+\sqrt{n}$, and $1 - \sqrt{n}$

Daniel Fischer

@Clarinetist You miscalculated. For $3\times 3$, you should have got one eigenvalue $3$, and two eigenvalues $0$.

Clarinetist

8:03 PM

@DanielFischer Ah, you

're right

I forgot to add $2$

to the characteristic equation (I think that's what it's called?)

Daniel Fischer

A revised guess for the general case?

Clarinetist

@DanielFischer One eigenvalue $n$, $n-1$ eigenvalues $0$?

Daniel Fischer

Yeppa.

Clarinetist

@DanielFischer So now what? Heh

Daniel Fischer

Now, multiplying that with $-\frac{1}{n}$, we have ...?

Kevin Driscoll

8:05 PM

@Clarinetist I guess you could prove the general case by considering the rank of the matrix, its trace, and its determinant

Clarinetist

@DanielFischer An $n \times n$ matrix of $\dfrac{-1}{n}$s as the entries?

Daniel Fischer

@Clarinetist Yes, and the eigenvalues of that are?

Clarinetist

@DanielFischer Gut reaction is to think $-1$ and $\dfrac{1-n}{n}$, but I'm not 100% positive on that

Oh wait

Daniel Fischer

@Clarinetist If you multiply a matrix with a scalar, say $c$, what happens to the eigenvalues?

Clarinetist

@DanielFischer They get divided by $c$?

Daniel Fischer

8:09 PM

@Clarinetist Multiplied, not divided.

$Ax = \lambda x \implies (cA)x = c(Ax) =c\lambda x$

Clarinetist

@DanielFischer Ah, I was making it too complicated

So my guess was right

Daniel Fischer

So the eigenvalues of our matrix with entries $-\frac{1}{n}$ are?

Clarinetist

$-1$ (multiplicity $1$), $0$ (multiplicity $n-1$; my bad, I thought $n-1$ was the eigenvalue last time)

@DanielFischer

Daniel Fischer

Right. And now we add the identity matrix, then we obtain $M$. What eigenvalues do we get?

Clarinetist

@DanielFischer AH, clever. Let me think about this one

$Ax = \lambda_{A}x$, $Bx = \lambda_{B}x$ so $(A+B)x = Ax + Bx = (\lambda_{A}+\lambda_{B})x$

@DanielFischer So then we get $0$ with multiplicity $1$, $1$ with multiplicity $n-1$.

Kevin Driscoll

8:16 PM

only if the eigenvectors are the same!

Clarinetist

@KevinDriscoll That's true

Daniel Fischer

But for a multiple of the identity, all vectors are eigenvectors, so there it always works.

@Clarinetist Right. So we have rank $n-1$.

Clarinetist

@DanielFischer So you're telling me that the rank can also be seen as the total multiplicity of non-zero eigenvalues?

Daniel Fischer

@Clarinetist For diagonalisable matrices.

Clarinetist

@DanielFischer I haven't seen diagonalization in at least 3-4 years. Heh.

Balarka Sen

8:20 PM

This is a silly question, which would possibly have a silly answer : Consider the affine space $\Bbb A^n_3$. Take two linked circle in your space. This is a variety, and the corresponding coordinate ring is $k[x, y, z]/((y^2 + z^2 - 1)(x^2 + (y-1)^2 - 1))$. Can we say something about linking of the two subvarities -- the two circles -- from the coordinate ring?

Daniel Fischer

In general, it's more complicated, but the rank is always the dimension minus the multiplicity of $0$ as an eigenvalue.

Random Variable

@DanielFischer I might have badly misinterpreted what he was talking about. He might be simply assuming that $f(a+it)$ can be expanded in a Fourier series of the form $f(a+it) = \sum_{k=0}^{\infty} c_{k} e^{-ikt}$ that is valid for all $t \in \mathbb{R}$.

Clarinetist

@DanielFischer Rank-nullity theorem?

Daniel Fischer

@Clarinetist Yes.

Balarka Sen

Actually, I think not -- the sheaf of rings shouldn't detect anything about linking. The coordinate ring (global section of the ring) might just be coordinate ring of two disjoint things.

Kevin Driscoll

8:21 PM

@Clarinetist Diagonalization is often paramount when considering eigenvalues because if $A$ is diagonalizable there is a similarity transform such that $A = U^{-1} D U$ where $D$ is diagonal and has the eigenvalues of $A$ as its diagonal elements

Daniel Fischer

@RandomVariable That can well be.

Balarka Sen

OK, so a bit more nontrivial question : what should be the correct algebraic way to detect linking of the two subvarities? Is there even a way to do it?

Clarinetist

@KevinDriscoll That looks so familiar.

Daniel Fischer

@KevinDriscoll $U^\dagger$? Thinking of unitary diagonalisation? Then you need a normal matrix.

Clarinetist

I wish stats people would use more linear algebra. It seems like it'd be to my benefit to relearn it since it's all matrix work

Kevin Driscoll

8:23 PM

Sorry forgot that the genreal case is inverse, not dagger. I'm used to all the matricies being unitary or possibly even hermitian

Clarinetist

Some day, if I ever find the time and money, I'll go do some M.S. classes in math

Kevin Driscoll

You could recall all this stuff by taking a linear algebra class on Coursera for free!

Balarka Sen

gah, I just realized that I typoed $\Bbb A_k^3$ above.

and global section of the *sheaf. i guess i really am sleepy

Mike Miller

From the coordinate ring of the subvariety? No. I don't see any reason in general that you should have a good notion of linking.

Nor really why you would want one.

I think it's folly to jump into structure sheaves without first being comfortable with the coordinate rings themselves. That doesn't take much time. But that's just my opinion.

Balarka Sen

8:41 PM

I dunno. You can extract a lot of information about the variety itself from looking at it sheaf of rings (e.g., as I got to know a few days ago, you can tell about "smoothness" of a variety from regularity of the stalk of the standard sheaf at that point). So a natural question would be if you can tell something about the subvarities from it too.

Kasper

@Huy the new release of euclid the game is ready for download

btw, anyone else here interested in testing out a mathematical iOS app ?

> Euclid: The Game gamifies the 2300 years old book "The Elements" written by the ancient Greek mathematician Euclid in Alexandria.

> Euclid's Elements has been referred to as the most successful and influential textbook ever written. The first level of the game is exactly the first theorem of this ancient book. Throughout the levels you unlock constructions, once you prove you are able to make them.

> The web game (http://euclidthegame.com) is played by 500.000 users in in 213 different countries, we hope that as much people will enjoy the much improved iOS game (to be released september…

Little info about the game above. We are now actively searching for beta testers. You can see screenshots of the app in the facebook page: facebook.com/euclidthegame

Mike Miller

Maybe after I get bored of Fallout Shelter.

Kasper

The source code is open source (MIT licensed), and will be available from github.com/euclidthegame in a couple of days.

@MikeMiller Ah, well, if you (or anyone else) is interested, you can send your email adress (associated with your appstore account) to beta@euclidthegame.com. And I will make sure you can download it.

Mike Miller

So, I'll probably email you in about a week.

Kasper

Awesome :)

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$Mathematics$ Mathematics