Good day! Tell me please, what does mean notation $\| \cdot \|_{X}$ in locally compact Hausdorff space X?

@Nimza: which is not necessary linear?

btw, I saw you on dxdy.ru

@Ilya I saw you too I think (Gortaur?)

sure

I didn't visit it for a long time

:) Yes, X is not necessary linear

no idea then :) what is $\cdot$? element of $X$?

yes

$f(\cdot)$ means that $f$ is a function

I though maybe it refers to the norm on the space of function with X as a domain

Yes, it it the norm on $C_{0}(X)$

so \cdot is not an element of X

oops :) yes, it's an element of $C_0(X)$

:)

sup-norm maybe then - since you don't have any other kind of structure

C_0 are bounded, right?

Here is the sentence: "$C_{0}(X)$ denotes the set of real functions on X which vanish at infinity with $||\cdot||_{X}$".

maybe this norm was defined above, or in the notation list. Since you don't have any other kind of structure, seems to be a sup-norm

Thanks. But I don't know what does mean vanish at infinity in this case. What is "the infinity" for an arbitrary LC Hausdorff space? Yes, it is defined above, but I have to pay money to see this :)))

I found, it is sup-norm, you're right

ahaha. don't pay it. Make them bankrupt!

:)))

but what does mean "vanish at infinity"?

if X is $R^n$, it is clear. But in general case? We have no norm and no distance on X

I am not so good in this topic, maybe $\lim\limits_{n\to\infty}\sup\limits_{x\in K_n^c}f(x) = 0$ where $K_n$ is any increasing sequence of compacts covering $X$

checkout wikipedia, I hear $C_0(X)$ used for LC Hausdorff spaces very often here, so it should be famous

Thanks, I've found in the "Locally compact space" article

@Nimza: where are you studying?

@Ilya MSU, and you?

@Nimza VolSU, then Halmstad University (Sweden) and now in TU Delft

cool

I wish I've studied on mechmat :)

why?

I feel some lack of background in math now, but I have a lot of projects in my PhD and small time to educate myself

I'm not studying in mechmat, I'm in cmc (ВМК)

also cool

I'm not sure.. If I had a choice now I would choose mipt

Probability theory on 2 cource here is awful, for example. Because lections on functional analysis start on 3 cource

emmm

why do you need funcan for probability?

For classical probability I don't need funcan

But in general case it uses Lebesgue integration theory etc

Hi all.

Good evening

Morning here :-)

:)))

@Nimza Lebesgue theory should be given in measure and integral course

which is then used both by funcan and PT

@Ilya I'm agree. But some basics of Lebesgue integration theory may be given in funcan course. PT?

probability theory

but you had measure and theory course, right?

I had measure theory course at the second half of 3-rd year and at first part of 4-th year, that says, after half year of funcan. And I had funcan for a whole 3 course (I think it's very unsufficient)

@Ilya I got it in the topology and measure part of real analysis.

@robjohn good for you, what can I say :) I wish our real analysis to be the same

*analysis

@Nimza that's weird, we had ТМИ in the 3rd semester

sorry, I was just adding a data point

:)

I'm gonna sleep now

@Ilya It isn't weird for CMC, beacuse the distributions on cathedras (how do we call this in english) take place on 3rd course. And a half of cathedras don't deal with mathematics, it is programming. So first 2 coursec we had a lot of programming disciplines and we didn't have a time for a lot of math

@Ilya good night)

Holy monkeys, I'm grading calculus 2. A battlefield!

@JM Back. Thinking about your problem, I napped. :(

Can I tell more about that problem?

So, I want to know the number of linearly independent Eigen vectors of Eig. val. $1$.

@JonasTeuwen Red F`s everywhere, everywhere!

@N3buchadnezzar Yup.

I am taking that course now.

I find Calculus 3 much more challenging

Solving for $AX=X$ : $$\sim\begin{bmatrix}2&-3&1 \\ 0&0&0 \\ 0&0&0 \end{bmatrix}$$

eg multidimensional calculus :(

What the fu.....!^

Some Skype users, how to rectify that please!?!!! (@Jonas @robjohn)

@KannappanSampath Heh, I would kill explorer.exe, should work.

If it doesn't restart it, restart it from the task manager. That is what should work in Windows.

@JonasTeuwen Ah, finally!

Thank you, it is OK now!

Back square one now! I need someone to discuss JNF with . :'(

Stupid nitpick, no?

@robjohn Working through your example now!

what is JNF?

Jordan Normal form^ @DavidWheeler

oh. similarity classes over the algebraic closure of the underlying field :).

are you trying to find the JNF of that matrix you wrote above?

Yes.

The char. poly is $t^3-20t^2+50t-140$

Hmm that has complex roots. 8-)

that's why sometimes you have to pass to the algebraic closure

a real matrix may not have a real JNF

Yes, I am over $\Bbb C$ but just amazed I have three distinct eigen values over $\Bbb C$ and so there is nothing much.

Hi @Henning

@KannappanSampath Hi.

What's with the sudden influx of questions that ask a reasonable geometry question, but then want the answer in the form of the sum of the numerator and denominator of the fraction in lowest term that answers the reasonable question?

E.g. here and here from two different users.

@KannappanSampath check your calculation of the char. ply.

wolframalpha.com/input/…

@HenningMakholm The name sounds like they come from the same place.

@DavidWheeler I got signs wrong. sigh

So, there in fact is something.

it can be hard to keep the signs straight in calculating determinants

Hi Henning!

@JM Hi.

@DavidWheeler Yeah, that's why Cramer is a mess for linear equations in more than three variables.

@KannappanSampath yes 2 is an eigenvalue of (algebraic) multiplicity 3...so now you need to check the dimension of the eigenspace

@DavidWheeler The Eigen space is of dimension $1$.

ok, so what does that tell you about the JNF?

@KannappanSampath Yeah, probably within the same quarter or so of the world's population.

@DavidWheeler So, I have one block of size $1$.

Now, wait. I would like to get something clear:

and the other block must be 2x2, right?

Yes. ^

But wait, please.

Now, I know $2$ is the only E-value.

so you can write down the 2 possible JNF you could have, without further calculation.

So, to find Eigen vectors, I solve $AX=2X$ and find the dimension of the solution space.

i like solving (A-2I)X = 0 better...same thing, though

What would $x-3y+8z=0$ mean now?

if you pick x and y (for example), z is then determined

OK. True dat.

But, then in terms of the eigen vectors?

well, make it easy on yourself, pick y = 0 to start with.

That is the system $A-2I$ reduces to a single equation: $x-3y+8z=0$.

1 equation, 3 unknowns, 2 degrees of freedom, one constraint.

So, what is the dimension of Eigen space then?

$2$ right?

the same as the degrees of freedom.

OK.

in this case, we can pick x and y independently of each other.

well, (x,0,z) and (0,y,z) are going to be LI, no matter what

Yeah. So,

I use the relation I get from $8z=3y$

and $x=-8z$ right.

Or Equivalently, $(-8,0,1)$ and $(0, 1, 3/8)$ are E. vectors.

you can multiply the 2nd vector by 8 to get integers values, but sure.

and those are eigenvectors, not eigenvalues.

Corrected, Thanks.

so they form a basis for the eigenspace $E_2$, right?

Yes.

So, $(T-2)$ is a nilpotent operator on $E_2$.

Now, how do I go about blocks?

Hi @Aarthi Some problem around here, flags or some such? =)

Something like that. I have a story for you guys

You all know this question?

Hah! Great. Please share that with us!

right...so if you're trying to find a generalized eigenvector, you need something in $\mathrm{ker}(T - 2I)^2$ but not in $\mathrm{ker}(T-2I)$

@KannappanSampath I just received an email from the subject of the question, Mr. Nahar, explaining his work to the asker.

I see.

Hi Aarthi, fancy seeing you here... :)

Hello @JM :D

@KannappanSampath I am unsure how to proceed, heh.

@Aarthi He sent an e-mail to the team@SE account?

@JM Yes.

That is weird, no?

@KannappanSampath It's because the question is protected. I think I may need to un-protect the question

I'll do it. I did the protection after all...

Oh, I see.

Done.

brilliant.

If he posts yet another jumble, then we can redo protection...

May be we should upvote Aarthi's comment so it stays above the fold.

ok, if i understand it correctly, what has been "discovered" is a way to compute approximations of cube roots?

hmm allow me to provide magic

one moment please muzak

here's what he sent me

@DavidWheeler Yes, but it doesn't seem to be new. Or earth-shattering for that matter.

it sounds as if the hype is media-generated, not necessarily Mr. Nahar's fault. if it converges faster than Newton's method, that's a plus, sometimes Newton's method converges fairly slowly.

I cannot make sense out of certain things there!?! Is it just me?

@DavidWheeler Unfortunately, for iterative methods that converge faster than Newton, he's been beaten to it by at least a decade...

For instance how is $\sqrt[3]{2000}$ evaluated without any error? It is not rational I guess!

@KannappanSampath I also edited my comment to include links to the pages.

In step $(3)$ I mean

@JM that does not surprise me, but in all fairness, that does not mean his method is devoid of interest

@Aarthi Oh, sure, another reason why we should upvote that comment! :-)

@KannappanSampath :D I DO MY BEST! hahaha

I am sure from there alone that the $\%$ error is calculated from unsophisticated tools!

@KannappanSampath it could be that the error is below the mesh size of his error measurement system. it happens. any "calculator" (computer) can only store numbers so big (memory is finite), and thus only reciprocals so small.

@DavidWheeler Right; that's why arbitrary precision is needed for full investigations of things like these...

@DavidWheeler But, I'd argue it should not be difficult to evaluate the error upto three significant digits.

And, what JM said.

It is plainly not zero. So, evaluate that thingy more precisely.

@KannappanSampath but if one is "calculating it" one needs to be able to handle arbitrarily large exponents...i don't think we can do that....

is thingy the technical term, then? :P

@Aarthi Oh, we sometimes use stuff if we really want to be technical... :D

for example, if your error is on the order of $\frac{1}{10^{f(n!)}}$ for some function f...

Well, I am not into Numerical Analysis as yet. So, \me not sure.

@JM Heh! Right! =)

@DavidWheeler I would love to see a method with that kind of error behavior... :)

Usually we're already quite happy if the error decreases geometrically.

take f(x) = -10^12 for every x

But, sure this is not. Firstly, a method developed on Well-established Math does not use the O/o notation to substantiate its claim.

It is not written in $\TeX$.

Then, whatnot. :P

I wonder who would pick the phone if I called the phone number there. 8-)

(I am in India, so it won't hurt.)

@KannappanSampath An answering machine? :)

looking at his "published" values, it appears as if he does have error decreasing "geometrically" (or thereabouts). without knowing his exact algorithm, i can't speak for "ease of use", which would be another factor

@JM Sounds quite likely, yeah!

So many cranks around me. I better keep away!

Bye for now, guys. @JM I could not do your example to completion, nor did I do @robjohn's example too. feels bad.

@KannappanSampath but you were so close!

@DavidWheeler I was? I'll think over it then . Thank you.

you can already write down the JNF, the only thing you lack is if you want the transforming invertible matrix to get a (generalized) eigenbasis

@KannappanSampath Dang, I built it to make you sweat a bit, but the answer is actually simple... :)

@JM what "answer" was being sought?

@DavidWheeler I was asking him for the Jordan decomposition of the Frobenius companion matrix of $(x-1)^3$... :)

If he did everything correctly, he should be seeing a Jordan block and a certain familiar triangle of numbers...

So why does everyone claim to have a formula, but like 99% of the time they just are able to do approximations?

I mean, approximating roots is fine. But when you claim to have a formula, you better dish out that formula. Or me gets angry, and people do not like me when I am angry.

@N3buchadnezzar It's really a problem of "truth in advertising". If you say that your formula is a (good) approximation at the outset, we have no problems.

This is mathematics, not sign advertisement along the 66.

If one claims to give "exact" results from simple arithmetic, even if the result is expected to be irrational, then we have a problem.

@N3buchadnezzar Oh, it sounded like "my formula is better than yours, bleh!" to me... ;)

Well better carries litt meaning here yes? Do you mean a beautiful formula, a fast converging formula, an easy to use formula, a formula that requires few iterations etc

@N3buchadnezzar ...and often you can only choose one or two of those options.

Indeed, but choosing zero is often far easier :D

this all seems like making a mountain out of a molehill. is his formula useful? dunno. is it a ground-breaking discovery...doesn't seem like it. does the news media not understand what is, and isn't important math? apparently.

How long does it take the team normally to reply to emails @Aarthi?

today i convinced myself that any set of p-2 distinct p-th roots of unity (p a prime) is LI over Q. it was a bit more involved than i hoped it would be.

I sent an email two days ago and still no response

@Gigili hi (insert affectionate yet socially unobjectionable term of endearment here)!

Hi David (is there a term like that?)

oh, i certainly hope so! if not, i am SO up the creek without a suitable means of locomotion.

@DavidWheeler friend?

see? i knew there was something! whew! saved from social awkwardness yet again. life can continue, now.

can x be a zero-divisor in R[x]? it seems to me like it cannot, but there are some strange rings?

@tb Thank you Teddy. But that also scrolls for me. Maybe there is really no way to avoid it.

I think I've not seen the teddy answer a linear algebra question.

even those mathematicians who claim they never do linear algebra occasionally do.

Oh, he didn't claim anything.

I was just observing. : )

well, then, there you are!

@DavidWheeler No: multiplying any nonzero polynomial by x preserves the leading coefficient.

Hey Henning!

Did you see this? If you'd like to answer it I'll delete my answer. I'm sure you can write a much better answer than me.

Yay : ) The teddy. Ello.

'ello 'll

@MattN right. Was in the mood for some lhf :)

@tb : ) If you do another $2$ you'll cap.

(just need to pick the right ones)

yeah, it would be good to refresh my knowledge on lcm and gcd...

(refresh is a big word...)

It's not as much of a jump as Asaf answering an analysis question, I think... :)

@Henning well i figured if x(f(x)) = 0, then in particular cx = 0 (where c is the constant term of f), but i still don't see how that shows c is non-zero.

@JM @tb Hi!

hey, JT!

@JonasTeuwen Hi!

Poor me. He doesn't say hi to me. :,(

:-).

@MattN I didn't see you 8-). Hi!

@DavidWheeler I assumed you were talking about formal polynomials rather than polynomial functions. The R[X] notation unambiguously means the former in my experience.

@JonasTeuwen : D

@Teddy How are your feet? It's been cold today. Not quite as cold as yesterday but still.

@DavidWheeler Perhaps I'm misreading your comment, though.

@MattN Oh, woolen socks and hot water bottle did the trick both today and yesterday...

: )

My argument is that f=ax^n+(terms of lower degree) with a nonzero, then xf = ax^{n+1}+(terms of lower degree), which is nonzero because a is nonzero and no terms of lower degree can change that.

no, it's fine...and i appreciate your reply. i think what you and i are saying are the same thing, in R[x], cx = 0 (that is cx = 0x + 0) means c = 0 by the definition of equality of formal polynomials

(two formal polynomials are equal iff all their coefficients in R are equal)

your answer is "from the front" mine is "from the back"

Yes, but then you need some kind of induction to show that there cannot be other nonzero coefficients in f.

i agree, your argument is cleaner.

You could do it "from anywhere" by saying: assume that f is nonzero. Then by definition there is an n such that the nth coefficient of f is nonzero. But that is the same as the (n+1)th coefficient of xf, which means that xf is also nonzero.

Can you take the log even though the exponent is to the $a$?

right...but we may as well take that to be the "leading coefficient"

@MattN What would be the problem? The log just becomes (logn)^a - 2logn.

@HenningMakholm How? Sorry I don't understand.

@HenningMakholm Thanks for your answer to my question. May I ask how you came up with it? I can't see how one can find such a structure.

Oh!

@HenningMakholm (Not necessarily now. I don't want to interrupt your discussion.)

@HenningMakholm Ok. I see.

you still have "e to the something"....

@ymar I just tried fiddling. It didn't work anyway...

...done for today. Hopefully everything is figured out by Friday. See you guys (hopefully soon)!

@JM Good luck! And bye.

@HenningMakholm OK, I just found it impressive that you could see distributivity would hold.

Hi, Matt and David.

Hi ymar! : )

@ymar hullo there

@tb I was cold too today. So I thought I'd make myself a soup (some packet of powder soup that came as a free sample with something). Turned out to be inedible and stank the entire flat out with ready-soup stench. (I have a sensitive nose) So I got to choose between feeling sick or opening a window to let in the cold...

@ymar Well, that was by design. The underlying concept between the multiplication was (a,b)*(c,d) = (ac,ad) for bits a,b,c,d -- so I just had to consider the various ways each side of the multiplication could arise as a sum.

the homomorphic image of an ideal is only necessarily an ideal if the homomorphism is surjective, yes?

the counter-example i am thinking of is the inclusion of R[x] in R[x,y]....

@MattN Oy! That's awful. One of my flat mates in my first years of Zurich would come home with the last bus and make those instant pasta thingies... So I imagine I know the ghastly smell... I tend to open the windows, stick a hot water bottle between the sheets and while letting in the cold air I take a hot shower.

as we followers of FSM say: ramen, brother!

@tb Sounds like a good idea. But how do you survive from the shower to the bed? : D

If I come out of the shower I'm even colder than before if the windows are open.

@HenningMakholm Interesting idea! I have to think about this multiplication, thanks!

Hi all of you.

@MattN The trick is to jump out of the shower, run and close the windows and then run back in the bathroom (of course you need to take care to have the bathroom window closed while you're closing the windows). It takes years and years of training to reach perfection, but I'm sure you'll do just fine.

@DavidWheeler Yes, when the homomorphism isn't surjective it may fail.

Hi, Kannappan.

@ymar How are you? I hope those devilish ideas are at rest now.

@ymar...that's a good example, yes?

@KannappanSampath Umm, not really.

@DavidWheeler Yes!

@ymar Then, shall I do more talking? (I know I'll make a fool of me, but never mind.)

So, you see, @ymar, .....

@DavidWheeler You might like this question

@tb Shall we talk about that Jordan Canon form now, please say yes. :)

@KannappanSampath Well, I have to make up my mind. I know my limitations and I know my hopes. Now I need a decision. But I have to make it myself.

@tb Running out of the shower doesn't sound like a good idea. : ) Images of unconscious teddy bears who slipped and hit the back of their head against the tub appear in my mind.

@ymar I sincerely believe you would make a positive one.

Hey Kannappan. Are you pissed off at me because of the quaternion balls-up?

@KannappanSampath What does Erdos have to do with it? :)

@ymar Changed to an isomorphic statement.

@MattN What?

@KannappanSampath You seemed a bit grumpy the other day so I was wondering : )

No not at all. I was upset such an error crept in and felt bad. But, why should I be pissed of at you?

@MattN yeah, I wasn't clear enough... but you got the idea: as long as you're in the steamy bathroom there's no hurry, then take care that the bathroom remains steamy while you're closing the windows. :)

@KannappanSampath Because I snipped the example out of some random pdf on the web so I thought it's my fault and you hate me for the embarrassment caused by it.

@KannappanSampath I'm a bit tired, but we can try.

@tb Just being obtuse : )

@MattN I would share the fault. Come on. I should have looked too. But, still, my mmind is not clear about those issues.

@tb Right, that example of yours cleared up a lot.

So, we begin there:

Now, the those $x$ and $y$ are to be determined systemmatically, right?

Right. Gimme a minute, I'll be right back.

Sure.

@KannappanSampath what are $x$ and $y$?

@DavidWheeler They were vectors lin. ind. from $v, Tv, T^2 v$ where $v \in V \setminus \ker T^2$.

is this the same T as earlier?

No, $T$ lemme look up:

okay, I'm back.

This thread here is very interesting.

@tb Hmm. I read the question and I have a feeling like "duh!", am I missing something?

@tb Y u no write answer and cap? : )

ai u no wrait cap?

I have an excellent new coffee grinder for my home coffee bar. The Mahlkönig Vario. I could obtain it quite cheap :-) (415€).

It is only suitable for filter coffee though.

@MattN well four answers are enough for this triviality, don't you think?

ah, ok so $T^3 = 0$

@tb No?

Because it is trivial many people can answer it.

Apparently he's got a shorter solution. So I see no reason why he shouldn't post it.

Will be back in a few minutes.

But you're all probably right: there are already 4 answers.

@JonasTeuwen I thought you didn't do filter coffee!

Huh? I do all kinds of coffee.

I thought for coffee otakus the real deal comes from a (proper) machine only : ) Where proper means the ones with a little cup thingie that you fill with coffee and then push/screw it on.

hi everyone.

Hi Jeff.

How do you define the supremum of complex numbers?

that is, if $M=$supremum of $|f(z)|$ on $C$ (where $C$ is circle $|z|=R$). Is the $M$ just $R$?

The what? That's a supremum of a real number.

The complex numbers are no ordered field.

@JonasTeuwen Right. that's what's confusing me. You can't really have a supremum of a complex number.

But $|f(z)|$ is a real number.

And it doesn't have to be $R$. Why should it be?

@JonasTeuwen OK. I realized that as I was typing it. :D But thanks for confirming

okay, I'll call it a day for now. See you guys!

@tb Good night!

I'm going for a completely text-based computer experience.

@tb Good night!

@tb later dude!

I keep missing people. :-(

Recently you've not been around as much as you used to. Maybe that's why?

@jonas The supremum of $|z^2|$ over $|z|=R$ is also $R$, right? Because $Z^2$ just moves the point $z_0$ "twice as far" around the circle.

@Jeff $|z^2| = |z|^2$.

Ahh. And, in fact, the supremum of $|z^n|$ for all positive integer $n$ is still $R$ (over $|z|=R$)

@Jeff That's only true when $R=1.$

or n=1

OK, right. I've been experimenting only with R=1. Otherwise it's going to be $R^n$.

Night folks.

@MattN Babai!

if you subtract 1 from every n-th root of unity except 1, take the modulus, and then multiply all of those moduli together, you get n. is it feasible to use induction to prove this?