That would be highly unlikely @Complexanalysis. You're basically saying $\lim\limits_{n\to\infty} f_n(x) = y$ and $\lim\limits_{n\to\infty} f_n(y) = x$, so, saying $\lim\limits_{n\to\infty} f_n = f$, we have $f(x)=y$ and $f(y)=x$. It can happen, of course, but it's unexpected.

@TedShifrin Does uniform convergence of $f_n $ ensure ?

No, no, @Complexanalysis.

a continuity condition won't ensure f(f(x))=x at x

@TedShifrin so there is no basic characterization to it ?

Nope, just what @seaturtles said is a way of rephrasing. If you know something about the limit function and $x$ and $y$ ... then there's hope :P

sometimes the simplest characterization of when something happens is "it happens when it happens"

i.e. one more or less already has the simplest characterization

anybody ever translated any books?

mr @Alex

@Daniel: Updated, for what it's (not) worth. :P

Darn, whole proof collapsed..

@TedShifrin yessir

The hair's grayer, @Ted, but it's definitely the same man. A couple of years older, I admit.

Probably not wiser, either, @Daniel. I think the radiation I had 2+ years ago zapped some of the hair :(

There will be some minor updates to the notes after this semester, @Daniel. I've gotten frustrated with students' "misreading" of several problems, so I'm rewriting for clarity. :)

I worked as a technical translator one summer during college, @Alex ... It was interesting, since I knew so little about computers, etc. That was the dark ages, too.

@TedShifrin Don't underestimate the ingenuity of people misunderstanding things.

@TedShifrin sounds neat :) was it for a company?

You speak from MSE experience, of course, @Daniel :) I had an elaborate footnote on the actual homework assignment about how they should do stuff with Mathematica, so they intentionally misunderstood the problem so that none of it would have to be used :P

Yeah, @Alex ... I found them down the road in Cambridge, MA, many a century ago.

@TedShifrin that's... MIT, right?

or harvard? i forget

of course, @Daniel, it's scary that I have some students who, at midpoint, don't yet know that the cross product needs to be a vector, the product rule, or how to turn something into a unit vector. Math major seniors? Guess what. They're not passing my course :P

both, @Alex, but I went to MIT. The company had nothing to do with either.

@TedShifrin That may be uncharitable, but I'm not sad to read the last sentence.

One of them was overjoyed to find out we're offering Complex Variables in the summer. I am embarrassed by some of the degrees we're giving out. Seriously.

@TedShifrin did you know you wanted to go into math before going to college?

Pretty much, @Alex ... although I wavered on languages/linguistics a bit.

My sister went into art history :P

There are similarities in the two

Yes, indeed, @Studentmath ... and music, too.

In a way more than physics and math hold.

that's interesting, I can't imagine. I didn't even think about applying to anywhere good in high school.

True, music too

Sorry, @Alex, I was an overachiever :D

When it came to a research career, I perhaps underachieved, and put way more energy into teaching. I don't regret it.

@TedShifrin i don't regret it ;) i just mean I can't imagine studying math at 18.

@Studentmath you're right, i was actually torn between linguistics at first too. if they'd had it as a major at my university i may never have gone into math.

These are supposed to be the best years to study math. Which makes me rather depressed, as that fact never helps me with the courses I take.

People work at different paces and get motivated by different things at different times. One of my hardest challenges is to back off pushing young students who are soooo talented but don't want to bury themselves (yet or ever?) in math.

in fact, the course that tied me down into being a math major was abstract algebra, and it was probably its similarities to linguistics which got me.

Ironic, @Alex, since I've never liked at all the symbol-pushing parts of math.

Interesting, my aunt works with linguistics, really good at it too, yet she claims to not be able to even look at math.

And refuses to listen whenever I mention that her presentations and books are full of mathematical logic and formulas..

Cultural linguistics is more anthropology, but a lot of the science side of linguistics is very mathy.

Chomsky's stuff ...

@TedShifrin when i first saw the abstract definition of a group, i basically heard it like "instead of doing math with numbers, let's just do it with whatever we want, and whatever operation we're doing with it let's just call it *"

that moment completely changed my view of mathematics. it was a linguistic revelation

Yeah, the world overestimates how much math people like or do numbers :P

I get it, @Alex. I was more hooked by the deltas and epsilons, ironically. I loved the notion of estimates. But I absolutely was fascinated by group actions in algebra.

@TedShifrin i'm still terrified by deltas and epsilons, man.

see, now you sound like Studentmath's aunt.

i really don't know how you guys do it. inequalities of any kind are the death of me. i am profoundly bad at analysis.

It's a matter of attitude, @Alex.

Just like her.

I wrote an algebra book partly to effect a different viewpoint to the teaching/learning of algebra, but I'm not at all a natural algebra thinker ... despite what my colleagues tease me for.

The guy I wrote my first paper with (a Frenchman) accused me of being an algebraist because I liked to do differential geometry with differential forms. Sigh.

@TedShifrin it is something i'm working on.

@TedShifrin guy's got a weird definition of "algebraist"

I wonder, have you ever truly struggled with some question in the B.Sc/college Math courses? Or with figuring out a mathametical concept?

He was scared of differential forms, as are, indeed, MOST mathematicians.

Of course, @Studentmath. Plenty. I'm teaching probability next fall ... have never taken it or taught it ... and I'm quite petrified :P

Perhaps you can find joy in the fact that it comforts me :P

That's why I told you :P

Not that you are petrified, but that it happened to you too

@TedShifrin algebraic topology has a lot of them things right?

I'm not really petrified, but I'm excited by the prospect of getting confused and messing it up :P

i have been reading a book about that, casually. it's above my level, i don't understand it yet.

Excited is always good

Slightly, @Alex. One can interpret cohomology using them, though, but they show up more in geometry and algebraic geometry (in a very abstract way).

Look at rational homotopy theory, @Alex. I'm actually on a Ph.D. oral committee with that on Monday.

@TedShifrin what is that?

Some of the most fun I've had, @Studentmath, has been teaching stuff I didn't really know beforehand. Learned a lot.

It's the best way to learn, sometimes - to teach.

It's homotopy theory done over $\Bbb Q$ in terms, to a great extent, of differential forms. I'm not sure what the guy is doing for his lecture ... I'll find out Monday :P

Yes, @Studentmath, as long as one does it responsibly. Some teachers stay a day ahead of the students. Generally doesn't go well.

True. Also, the demotivation a teacher can cause with a single statement..

Oh, there's an actual Alex here now. I guess I need to get more of Mr. Gruber's full name in there.

It's a real responsibility

Students can tell pretty easily when teachers care (a) about the subject (b) about the students, eh, @Studentmath? :D

@Chris'ssis I don't think that would be much of a challenge for ethan :)

Yes..

If the students themselves care about learning, at least

I only recall how I did my best to skip a certain teacher classes, he really cared about the subject, but every single solution he started with "that was a very easy problem, you shouldn't have had any problems with it". Every single solution.

@Studentmath "Proof: Obvious."

Sometimes I do tell students in office hours: "Here's a hint. You're making the problem too hard." Most students actually get that it's an actual hint. :P

I only use the word "trivial" in its technical sense (i.e., the trivial solution).

Precisely, @alexander and that is usually a great hint

@seaturtles it's just for fun (no challenge).

And is usually true, too, for these courses at least

Well, @Studentmath, I do get pissed off when I announce well before the test what particular sorts of problems students need to know how to do, I put them on there, and some students bomb. I have no sympathy.

I wouldn't have either

@TedShifrin alright, so, i have a pretty broad, possibly dumb, question.

I sound like an old fogie ... But one of my strong points as a teacher used to be that I motivated students to work twice or thrice as hard for me as for other teachers. With a good number of students now I'm failing ... but it still is true of plenty.

Some would say there are no dumb questions, @Alexander :P

so i'm taking this algebra class. we're doing commutative algebra, with (I guess) some basic algebraic geometry (because the guy is an algebraic geometer). i'm keeping up, do understand some things

who, out of curiosity?

@TedShifrin Richard Crew.

commutative algebra needs alg geo or number theory to motivate things :P

ok, don't know him.

@TedShifrin he's mostly in $p$-adic stuff from what I understand

oh, so arithmetic geometry or alg number theory

I know nothing of that

so we have things like, local rings, $\mathfrak{m}$-adic and Zariski topology, polynomials in infinite variables

and from what i gather from seeing other questions about algebraic geometry and reading other sources this tends to be what people study in algebraic geometry

polynomials in infinitely many variables are more for algebraic counterexamples (non-Noetherian)

@BalarkaSen @Danny: I am back. Did you still have questions?

@TedShifrin right- lots of discussion of Noetherian-ness

and its various consequences

local rings, Zariski topology, for sure ... localization is like choosing a coordinate neighborhood on a manifold

heya @robjohn

so all of this is obviously very algebra-ish

@TedShifrin howdy

indeedy

Hey all

hi @Brittany

but when i step back and think about all this business i'm left with a weird questions which is, how exactly is any of this geometry

good question, and you should ask him

with the Nullstellensatz, etc., you're relating algebra to geometric objects, namely varieties

@TedShifrin we... can't really have productive conversations.

localization is about giving "local" coordinates on a variety

@AlexanderGruber Is this algebraic geometry? Algebra is all about abstracting things past recognizability ;-)

smacks robjohn

Speaking of which, if anyone could hint/direct me towards a path to proof: I've been trying to prove that for a given digraph G of order n (with n vertices), and A being its adjacency matrix, $A^n$=0 <--> G has no cycles. I went around proving that digraph G of order n is a cycle itself if it has a walk of length n, only to realize that entire proof works only if G is a simple digraph (should've sensed it's getting it over-complicated).

@TedShifrin you can tell I am not an algebraist?

Nor am I, robjohn, nor am I ... but I am on the analytic/geometric side of it :P

cycles correspond to the nullspace of $A$ or $A^\top$, @Studentmath.

I am thinking about using the degree-sum of edges equations, and the fact that if it has no cycles there is at least one vertex with inner degree of zero (and another/same with outer degree of zero), yet I am not sure how to move along from there...

@robjohn i thought it was all about making cooler symbols, like commutative diagrams.

What do you mean, Ted?

The Kernel?

yes, nullspace is "elementary linear algebra" speak for kernel.

@AlexanderGruber Yes, and those look more like organizational charts than anything to do with exact sequences :-)

Homological algebra -- cool as it can be -- is universal in math ...

@Studentmath you mean no $n$-cycles or no cycles?

No cycles at all. Doesn't contain any cycle.

Hmm ... the kernel should be $1$-dimensional, so I don't buy it.

I managed to show that in $A^n$ aij=number of walks of length n from vi to vj

absolutely right, @Studentmath

directed walks in the case of digraph

but you need to think about what the kernel of (probably) $A^\top$ means.

So now I only have to show that if it has a walk of length n, it must have a cycle. That's where I get stuck.

Hm

It means the inner degree (i.e. row) is zero.

Well, luckily for me, I need to go cook dinner. So y'all will have fun without me.

See you @TedShifrin, thanks for the input!

@Studentmath well, if it doesn't have any walks of length $n$ from vertex $i$ to vertex $i$ it doesn't have any cycles of length $d$ for $d\mid n$

so, we've got that right?

Correct, yes

so if $A^n=0$ then $A^{r}=0$ for each $r\geq n$

you see what i'm thinking?

Yes. The diag of A..

Many thanks Alex!

mhm

@Chris'ssis $$\sum_{n=1}^\infty \frac{x^n\phi(n)}{1-x^n}=\sum_{n=1}^\infty (1*\phi)(n) x^n=\sum_{n=1}^\infty nx^n=\frac{x}{(x-1)^2}$$

@Chris'ssis So it simplifies to $$\frac{x}{(x-1)^2}=1$$

(r(e(m(o)v)e)d)

@Chris'ssis Which has the two solutions $\frac{1}{2}(3\pm \sqrt{2})$

[r[e[m[o]v]e]d]

@robjohn hold on

@Danny holding

{r{e{m{o}v}e}d}

It just means write it factored out

write it as $C\prod_{\rho}(x-\rho)$

@Brittany $f(a)=0$ if and only if $x-a$ is a factor of $f$

where each $\rho$ is a zero of your polynomial

$x-a$ has degree one so it's called "linear", this is what it means by linear factors.

This chat is enabled with TeX

@Brittany there's a thing called LaTeX which displays math differently

You see the $ signs because you havn't enabled it

If it were enabled you would see it formatted correctly

Instead of $ signs

look over at the sidebar at "LaTeX in chat"

How can I enable that? Im googling it to learn about it

I see it @AlexanderGruber thank you

@Brittany once you have it activated and you can see it, visit this link to learn how to write in TeX.

and welcome to MSE. :)

enjoy your stay

TY =))

they live without it in the irc channels #math etc.

@Brittany Here is a great place to start.

Got it

@Ethan i can't imagine

What is the difference ebtween rendering MathJax and starting ChatJax?

@robjohn math.stackexchange.com/questions/695041/…

@Brittany MathJax should work automatically in your browser on regular Math StackExchange, like, the main site. ChatJax is something you have to use to see it in chat here, too.

you won't have to install anything, it's just a bookmark you add and click while you're in here.

Got it, I have it enabled, if thats the right wording for that. So i can now see the stuff you guys wrote earlier, thank you!

$(r(e(m(o)v)e)d)$

@Brittany $$\mathfrak{Welcome}\,\,\mathfrak{to}\,\,\mathfrak{the}\,\,\mathfrak{Mathematics‌​}\,\,\mathfrak{Illuminati,}\,\,\mathfrak{Brittany.}\,\,$$

show off :)

lol

:D

LAS?

nvm

Why is this called Math.SE?

stack exchange

Just because its linked with Stackexchange?

ok

yea

'=)

@Brittany did you look at the link I gave you?

Yee

Its a lot

:)

A probably stupid question, but if tr($A^k$)=0, can I state it certainly means tr(A)=0? It seems true to me, yet..

I am unsure.

Nevermind, not always true it seems.

do you mean for all k?

even for all k isn't sufficient if the dimension is a multiple of the characteristic (take A=id)

however if char doesn't divide dim then tr(A^k)=0 for all k implies A=0

how @seaturtles

all its eigenvalues would be zero (one uses the fundamental theorem of elementary symmetry polynomials to show this)

for $k\to \infty$ we can construct a matrix with the given property . bur for finite $k$ i guess it holds .

Not only k is finite, but also every eigenvalue is nonnegetive

huh?

in.. the given matrixes.

so you have some given A that you are talking about

I am again going over-complicated, nevermind :) Just need to figure out the simple solution for the question

Yes of course, but this is not the way to solve it :) over-complicating it

ok so you are talking about a specific matrix

I am still trying to show that $A^n$=0, where A is the adjacency of a digraph G, iff G have no cycles..

digraph G having n vertices. So A is of nxn, and every aij of A is a natural number