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12:17 AM
@Srivatsan go ahead...
I should be here for a bit now. We are watching the Rose Bowl, and so I go afk once in a while.
And it doesn't matter since Srivatsan is gone :-(
 
@robjohn Back =)
 
@Srivatsan Hurray! :-)
 
I have one specific question. What is the role of dimension in Euclidean and non-Euclidean geometry?
When I do synthetic geometry, it's clear that I am talking about R^2.
 
@Srivatsan There are several ways that it impacts geometry...
 
@robjohn One second, let me complete what I want to say. =)
 
12:24 AM
One definition of dimension (Hausdorff) has to do with volumes
@Srivatsan Okay
 
Yes, I'm coming to that.
Let's take some random sampling of topics. (1) Synthetic geometry. (2) Analysis. (3) Algebra (though I don't see how it's related here). (4) Measure theory (for fun).
In this, I somehow find that synthetic geometry alone somehow presupposing that we are going to study R^2, may be R^3.
I'm not sure if I am making sense :-). Anyway...
 
Again, what do you mean by synthetic geometry?
 
@robjohn High-school geometry. Doing it the way Euclid did.
 
@Srivatsan okay
So yes, dim = 2
 
For Euclid it might have been ok because the world is after all 3-dimensional. But surely we would want a theory that can generalise to any $n$, no?
I forgot topology in the list. The list doesn't matter.
I'm fine with assuming that synthetic geometry studies R^2 or R^3. May be you can write axioms defining R^4, but that's alright. In fact, I don't even care that much.
Here's what I do care about.
So what about non-Euclidean geometry? I know of the traditional account that you get it by taking Euclid's axioms minus the parallel postulate, adding some other replacement (I'm not sure about this).
But does it mean that non-Euclidean geometry is somehow "two-dimensional"?
Ok, I'm done with my talking. If you need some clarifications about what I said, then I will elaborate. Otherwise, you may take the stage now. =)
 
12:33 AM
@Srivatsan Not necessarily. Consider Einstein's space-time that is at least 4 dimensional.
I think most of the non-Euclidean generalizations are differentiated by the curvature tensor
 
@robjohn Well, I don't know this stuff well. (despite being a Physics minor in my undergrad :).)
 
This is because the Theorema Egregium opens the door for curvature that is measured from within the manifold, not depending on the embedding.
@Srivatsan I don't know this well either, so I am not sure how well the TE generalizes to higher dimensions.
 
TE works in three dimensions?
 
I think that General Relativity deals with curvature, and I don't believe it depends on how R^3 is embedded in R^4
I could be sorely mistaken.
It probably depends on finding a suitable definition for curvature that obeys the TE
by "obey the TE" I mean it is a curvature like quantity that is invariant under the embedding
Asking me is probably like asking a blind man to describe a sunset. :-)
 
But does my question make any sense at all? I am not sure I have a precise enough question, but probably you saw what I was getting at.
 
12:43 AM
about non-euclidean geometry being 2 dimensional?
 
@robjohn Yes. Actually, let me try this question:
Take a particular generalisation of non-Euclidean geometry, say Riemannian. Is there a notion of dimension in RG?
For Euclidean, there is. One can place points, draw lines, spheres, and whatnot in any dimension, i.e., R^n.
 
@Srivatsan Perhaps you mean is there an idea of what higher dimensions of Riemannian geometry might be?
 
@robjohn Yes. That is a better phrasing.
 
I think that Riemannian geometry knows it is 2 dimensional when it is.
 
(That sentence does not parse properly...)
 
12:48 AM
I think that higher dimensions of Riemannian geometry have a connection (g) that gives it a positive or negative curvature
@Srivatsan which sentence? that Riemannian geometry knows it is 2 dimensional when it is (2 dimensional)?
 
@robjohn OK, you just clarified. Thanks.
@robjohn In two dimensions, the curvature is zero?
 
@Srivatsan no, that is what the TE is about, do you mean 1 dimension?
I don't think there is enough structure in g (which is a scalar or 1x1 matrix) for curvature to be defined.
In 1 dimension, the curvature is defined by the embedding. afaik
 
@robjohn "higher dimensions ... positive or negative curvature" -- what is higher here?
 
manifolds define dimension by which R^n a small patch is mapped to.
bbl have to go afk
 
 
3 hours later…
3:48 AM
hmm. leaving to go home. See whoever is still here in an hour or two.
 
QED
4:43 AM
Hi
 
@robjohn there?
hi QED
 
QED
Hello
 
@robjohn, If you are are free enough to continue the Riemannian geometry conversation, ping me.
 
QED
4:59 AM
I think he's between terminals
 
Yes, that should be the case. Just the hope he will see it on coming back home.
hi Dylan
In this post, do you see what does the OP mean by transform of a matrix? Could it be transpose?
 
QED
5:25 AM
I don't know about matrices
all those stsructure theorems and stuff are pretty hard work
I've only needed that stuff once for the abelian group theory
 
5:47 AM
@Srivatsan I saw that. Transpose seems likely.
You could leave a comment. He seems to check the site still.
Lots of clever answers there.
I really enjoy Bill Dubuque's answers but sometimes he seems to rag on other responses. Maybe we should all just have thicker skin and be able to take criticism, for our betterment. I don't know.
4
 
@DylanMoreland True that.
But I am not at all a fan of one-correct-way-of-seeing-this-thing.
 
I'm sort of surprised that the popular commutative algebra books don't really deal with graded rings much.
 
@DylanMoreland They don't? That would indeed be surprising.
 
I forgot why you do what you do when localizing a graded ring at a homogeneous prime and it isn't really in Atiyah-Macdonald, Eisenbud, Matsumura, what have you.
Maybe this isn't very deep but I always find it confusing.
A-M talk about it for a bit when they do completions.
 
Has Bill Dubuque written any textbooks (at least lecture notes)? It will be interesting experience learning from it.
 
6:02 AM
I would read that.
 
Heh, is this for real? =)
Gosh, I am a stalker now. I even got an IMDB account. All I wanted was some webpage where he has put up notes and stuff; doesn't look like he has one.
 
QED
if you're just googling his name you probably found other people with the same name?
 
@QED Yes, I'm just googling his name. Dubuque isn't that common a surname I thought, but I could be wrong, of course.
 
QED
> The development team, at this time, included Jeff Golden (language, compiler, etc), Bill Gosper (special functions, summation), Howard Cannon (user interface, optimization), and consultant Bill Dubuque (integration, equation solving, database, optimization). Other developers made major contributions in numerical analysis, graphics, and help systems.
Cool! He worked on Macsyma, Inc
 
Ah, I shouldn't slander Eisenbud. He has a few exercises that are good.
And everything is in EGA. Maybe that goes without saying.
 
6:12 AM
There would be a lot to say about this question but its original form is somewhat off-putting...
 
@tb Original form was gone within a few minutes; why are you still "off-put" by that? =)
Question: Doesn't clearing the browser cache clear incomplete posts?
 
QED
I think that stuff is saved on the server
 
@Srivatsan what do you mean?
 
@tb Say you write an answer but don't post it; then it's saved, right? Won't it disappear if I clear the browser cache? I thought everything is stored locally, but QED says otherwise.
 
@Srivatsan I think the last question and the last answer you drafted is stored on the server.
(Let me look for the relevant thread)
Point 3. here
 
6:20 AM
@tb Oh, thanks!
Are Scribd links ok in this site?
 
@Srivatsan I would be wary. Usually, the most obvious Google search will lead you there anyway, so I'm avoiding them.
 
OK, I will remove it then. Thanks.
 
QED
yeah when you load the page it sends you the HTML for the box with the content already in it
Scribd is pretty horrible
IIRC they set up their site as a free way to host PDFs.. then changed it so people had to pay later
pay to access the PDFs that people uploaded thinking it was free
or am I confusing it with a different site...
 
Good morning.
 
I thought Scribd stuff was still free?
They used to have some terrible Flash interface but that's gone now.
 
6:36 AM
Um, after solving a system, it says $x_1=3.8750 (5S)$ and $y_1=3.1667 (5S)$ .. Is it enough @tb?
 
@Gigili No, unfortunately I still can't parse this.
 
Somewhere else, it says $m \approx 2.9985 (4D)$ as final answer.
I guess that 'D' is related to the digits after decimal point, but no clue about 'S'
Is it on-topic on the main site? I don't know how to google it to find out what it is.
 
@Gigili I'm a bit dubious about its suitability. What is the source exactly?
 
A book on numerical analysis.
 
What book?
 
6:48 AM
Um, not a famous one and not in English. It's in my native language, "numerical analysis".
 
The problem is that such a question can only be reasonably answered when having the full context available. Why don't you just say what book it is?
 
You mean the author? the title is "numerical analysis" as I said
 
I am tempted to "fix" the TeX in Bill's old posts; e.g., this one.
 
@Gigili Yes, I mean a unique identification of the book. I'm pretty sure that in every written language there exist at least a dozen books with that title. I honestly don't understand the point of being secretive about it.
 
6:57 AM
Is this circular?
 
@Srivatsan You know that Bill is very particular about it... Do you mean to add \rm everywhere?
 
@tb Yes, make them upright. Just one post; just for the kicks. :P
 
Maybe think about the risk-reward of that one.
 
But I'll mostly end up shooting myself in the toe. I guess I won't do it. :)
 
QED
@DylanMoreland, good question
I don't think that is circular
 
6:59 AM
@DylanMoreland I'm quite certain it is.
 
@DylanMoreland It is not circular but it doesn't answer anything. It just hides the difficulty in step 2 without mentioning how it is done. A blatant non-answer, nothing smart-ass about it.
 
QED
@tb, but can't you just use the fact that matrices represent linear transformations i.e. invertible functions?
 
The real question is this: If a square matrix A has a left inverse, then it has a right inverse and they coincide. There's an irreducible core of difficulty here; and I don't find it addressed in that post.
 
QED
and get the fact that it's a group instantly
I'll write out the details of this proof and see where I get stuck, because I don't see any difficulty from this distance
 
@QED You might get a correct proof; I'm not disagreeing with that. What I am thinking is that the all the group-theory stuff mentioned there will turn out to be extra stuff that is not relevant to this problem.
 
QED
7:04 AM
In a group, all elements are invertible
It is not known that A and B are invertible
 
For example, in a solved example, finding the roots of $x- \sin x=0$ by NR method .. it says $\frac{x_{n+1}-x_n}{x_n - x_{n-1}}$ .. putting $n=6$then $ x_7=0.02909 x_6=0.04364 x_5=0.06547$ , from $\frac{x_7-x_6}{x_6 - x_5}$ which is equal to $\frac{m-1}{m}$ m would be $m \approx 2.9985 $ (4D)
 
QED
or maybe AB = I proves that they are invertible
 
@QED No. A matrix is invertible if there exists a B such that AB and BA are both identities.
 
Maybe circular is the wrong word.
 
QED
ok so the missing step of his proof is that A and B are not proved invertible
 
7:07 AM
But somehow none of the actual work seems to come from group theory.
 
QED
so you don't know that they are elements of the matrix group yet
 
@tb Nothing secretive, I just think the context wouldn't help in this case ... I know there are a dozen books with that title but it's not translation, it's originally in Persian .. "Numerical analysis , Ismail babelyan " I just don't know how to type the author's name in English .
 
@DylanMoreland Exactly; it's linear-algebraic somehow.
 
QED
I added this comment "It's not known that A and B are invertible - so you have to prove that before using group theory. Proving A and B invertible solves the problem without any group theory"
I'm impressed you can spot something like that...
 
Maybe I should read the answers again: did anyone say that $AB = I$ implies $A$ surjective and $B$ injective, hence $A$ and $B$ are bijective, inverses are unique, blah blah?
 
QED
7:10 AM
no
 
Maybe the last part is important to see. I think some people used this.
 
@DylanMoreland I thought one or two answers said something like that. Is there a catch to this?
Oh, that is clearly wrong. You at least need to invoke rank-nullity theorem.
 
What is?
 
@Gigili I see. Since I don't read Persian I can't help, of course... Doesn't the author say anywhere (maybe at the beginning of the solutions) what those suffixed parentheses are supposed to mean? Maybe (4S) could mean four significant digits, or something like that.
 
@DylanMoreland If A is surjective and B injective, it is not at all clear that A and B are bijective. In fact, again that is what the question asks basically.
 
QED
7:12 AM
I'd like to see a nice simple proof otf this theorem
 
Of course. That would be the meat of it.
I guess I just mean that in a group, if $xy = e$ then $y = x^{-1}$. Some statement like that.
 
QED
I just find matrices pretty ugly
 
That's helpful. I could imagine a beginner not realizing that.
 
QED
since I don't understand them yet
 
@DylanMoreland On the contrary, that will only confuse a beginner in this thread. Note that there's the subtlety of left-inverse, right-inverse and honest to goodness inverse.
 
7:15 AM
You just can't avoid using finite-dimensionality and the rank-nullity theorem in some guise.
 
QED
the definition of a group is important
to reject the group theory proof
 
Yes, but out of place here.
 
QED
> For each a in G, there exists an element b in G such that a • b = b • a = 1G.
 
I don't necessarily mind answers that the OP may not fully understand. As long as there's an appropriate answer already there...
 
@DylanMoreland Sure, that's a valid point. :-)
But what do you want to say?
 
7:18 AM
Here's another thread with the same mistaken answer.
 
QED
I'm very pleased to see this subtle mistake brought up
it's really the sort of thing that is so easy to fall for
 
@QED That's why I didn't say the identification of the book cause I thought it wouldn't be any of help. No, there's nothing about it but I'm wondering if it's relevant to the first chapter about errors and ways they can be arisen , something like that .. Or maybe round to n significant digits? So you think the question on the main site would be vague or off-topic?
 
@Gigili I haven't read that many numerical analysis texts, but as I said, I've never seen such a thing. The problem is that this may well be a notational peculiarity of the author and as such I think it's not a very good question for the main site. But that's just my opinion, I might be completely mistaken.
 
I feel bad for Andrew L. I think he's kept his head down lately, so that's good. I hope he's getting math done.
 
QED
What's Andrew L.?
 
7:26 AM
@tb Thank you for your patience, I'd ignore it then.
 
@Gigili You might try to ask J.M. when he's around. He's well-versed in numerical analysis.
@QED Mathemagician1234's MO handle
 
QED
oh
 
@tb I will, thank you.
 
Now I'm around...
 
@JM can you make sense of this and this?
Hello, by the way!
 
7:37 AM
Hello to you too, t.b.!
@Gigili I'm looking at those. The missing information in there is that the seed was not given. On the other hand, the slow convergence is expected since $x=0$ is a triple root for the function $x-\sin\,x$ (note the Maclaurin series).
 
@JM Aha, do you know what do 'S' and 'D' mean?
 
I haven't gotten to that part of the transcript; wait...
 
@Gigili: I'm guessing it's respectively "significant figures" and "decimal places". I'm not sure why the notation is needed; I'd think one'd already have the habit of displaying only the figures you can justify...
 
QED
I had some of my comments secretly removed
well just one
 
7:48 AM
@QED I see. I'm sure this has happened to me once in the past.
That makes it twice now.
 
e.g. if your starting data is only good to six or so figures, then there are rules on how many figures should you be retaining in the final result. That being said, one should always do the rounding at the end, and not in intermediate stages.
 
It means the number has 4 decimal places (I mean it's written up to four decimal places) or it's rounded to 4 decimal places? If it makes sense ..
 
@Gigili If it's the final result of a computation, the latter.
 
Hm, are the resident functional analysts around?
 
@ZhenLin maybe I can help?
 
7:58 AM
Just the person I was looking for!
 
@JM Thank you.
 
@ZhenLin Very good, then (I'm flattered). What's up?
 
Someone (elsewhere) just asked about an exercise I couldn't do back when I did Linear Analysis. The question was, given a unitary operator $U$, how do we deduce that its spectrum is non-empty by considering the operator below? $$i (U + \textrm{id})(U - \textrm{id})^{-1}$$
It seems to be hinting at a reduction to the spectral theorem for self-adjoint operators, but we only learned the spectral theorem for compact self-adjoint operators in that course.
 
@Gigili No worries. :)
 
@ZhenLin It's the inverse Cayley transform. However, this is an unbounded self-adjoint operator in general (note that $1$ might be in the spectrum of $U$, so the inverse isn't defined everywhere.
 
8:03 AM
If $1 \in \sigma(U)$ then we're done, so I guess we're free to assume $1 \notin \sigma(U)$...
 
Oh, right...
Silly me.
Let me think for a moment.
 
(On the other hand, that does look like something that somebody too lazy to write many Pochhammer symbols would do... it would probably pop up if you're dealing with those generalizations of hypergeometric functions.)
 
@tb Don't worry about it. If I can't remember the solution then that probably means it was non-trivial / involved mysterious tricks / couldn't be solved.
 
@ZhenLin I almost have it, just a second.
So, if $1 \notin \sigma(U)$ then $(1-U)$ is invertible and it's not hard to check that $S$ is actually self-adjoint. However I don't think it's compact in general (the spectrum of $U$ is mapped bijectively to the spectrum of $S$ via $\lambda \mapsto i(\lambda+1)(\lambda-1)^{-1}$ and there's no reason for the image to be a sequence converging to $0$).
 
8:20 AM
It's easy enough to produce a counterexample: let $U = i \, \textrm{id}$, then $S = \textrm{id}$...
 
@ZhenLin In fact, the Cayley transform $\kappa(S) = (S-i)(S+i)^{-1}$ is a bijection from the densely defined symmetric operators and the isometries $U$ on $H$ with the property that $(U-I)$ has dense range. Moreover, $U$ is unitary if and only if $S$ is self-adjoint.
 
Hm, maybe it's one of those proof-by-cases thing. If $S$ is compact, then we're done, so what if $S$ is not compact?
 
Well, it's not hard to prove that the spectrum of $S$ is non-empty: this is a standard application of the resolvent map and Liouville's theorem.
 
Ah, I remember looking that up last time. But I don't remember using it...
 
 
2 hours later…
10:05 AM
@Srivatsan: this might be close to what you were saying earlier:
$
\begin{align}
\sum_{m^a\le n^b}\frac{1}{m^a+n^b}
&\ge\sum_{m^a\le n^b}\frac{1}{2n^b}\\
&\ge(n^{b/a}-1)\frac{1}{2n^b}\\
&=\frac{1}{2n}-\frac{1}{2n^b}\text{ when }\frac{1}{a}+\frac{1}{b}=1
\end{align}
$
 
Right, that's correct. Upto the last floor. =)
 
that's what the $n^{b/a}-1$ is for
 
Thanks. I didn't expect you to do it though...
 
$n^{b/a}-1\le \lfloor n^{b/a}\rfloor$
 
@robjohn Yes, down with those pesky floors!
 
10:08 AM
@Srivatsan It's just easier to handle -1 than floor in this case
 
Sure, sure.
 
So, whose homework did we do :-)
 
Nobody's?
 
It just had the feel of a homework problem, that's all.
 
@robjohn I was walking when the series $\sum \frac{1}{m^2 + n^2}$ struck me. I was asking myself if this could end up being rational and so on. :)
@robjohn I did say puzzle, no?
 
10:12 AM
and then it was generalized
 
@robjohn Well, if you notice, that series doesn't even converge. Oh well.
 
when I say generalized, I meant asking when $\sum\frac{1}{m^a+n^b}$ converges
 
QED
finding expressions for sums seems really impossibly difficult
unless you're Euler
 
Woo! Windy! 9 Bft :D.
 
@robjohn and that is precisely when $\frac{1}{a}+\frac{1}{b}<1$
@JonasTeuwen Bft?
 
10:16 AM
Beaufort. Windforce.
 
@JonasTeuwen Hmm... learn something new... never heard of it
 
Maybe it is something European?
 
@robjohn: hi. I think I'm happy now with your Stirling answer. I still need to do some scribbling (I still would like to figure out how your approximation relates to the error function I'm used to: $$\mu(z) = \log \Gamma(z) - (z - \frac{1}{2})\log{z} + z - \frac{1}{2} \log{(2\pi)} = -\int_{0}^\infty \frac{w - [w]-\frac{1}{2}}{z+w}dw,$$ but I'll manage to do this myself.)
 
hi everybody
@Jonas, what would you advise to take a look at in Antwerpen? I wanna show it to my parents, but I've never been there myself
 
QED
Hi ilya
 
10:17 AM
@robjohn en.wikipedia.org/wiki/Wind_force here is a scale.
@Ilya Hmm, the center at least.
Plus there is a very good coffee bar at Hopland 42 (?), Caffènation. Not sure if you like that. :P.
 
@JonasTeuwen Yes, I was just looking at en.wikipedia.org/wiki/Beaufort_scale
 
@JonasTeuwen what's there?
 
@tb That hopefully gives you something close to the asymptotic expression I gave.
 
Is there a smooth/higher-dimensional version of Morera's theorem?
 
@robjohn yes, pretty close, but not exactly, that's what I meant with figuring out how the two are related.
 
10:21 AM
@tb But I am still working on the easiest way to finish off the answer. I had tried to give the general idea without cluttering with what I think are usual details, but I guess they are not so usual to everyone.
 
@Ilya The center? It is nice.
@Ilya In 2-3 hours we will get wind speeds ~100km/h.
 
@tb: the proof become less concise when you have to worry about details :-)
 
@JonasTeuwen nope, I linked my message to Caffentaion. Is there just coffee?
@JonasTeuwen we're today visiting Delft, they like it a lot. Tomorrow we'll go to Brussels and then on Thursday to Antwerp
 
@Ilya Not "just" coffee. It is a great coffee. The best in Antwerp!
Oh, nice.
Who is "they"?
 
@JonasTeuwen well... parents?
 
10:24 AM
Oh, right. Cool!
 
I was to Rotterdam for the 3rd time yesterday, one of the most amazing cities in this country
maybe, the most
 
@Ilya parents visiting?
 
@robjohn aha :) how was your New Year?
@JonasTeuwen do you know how much is a train ticket from Brussels to Antwerp? They fail to tell it me here, at NS
 
@robjohn I mainly managed to confuse myself a lot by staring at the graph of $u^2/2$ without actually thinking... It's certainly not your fault. I really like the recursion for the approximating coefficients that you can give.
 
@Ilya One-way?
 
10:26 AM
@JonasTeuwen yes
 
@Ilya pretty good. we watched a parade and some football games. How was yours?
 
What is the age of the people and is there any discount?
@Ilya You can check out the railroads page of Belgium: b-rail.be/main/E You can also go to the ticket boots and ask what would be the cheapest since for groups it can be cheaper if you buy a "10x ticket".
One moment, be right back.
 
@tb I did like that by adding the sum to that with reversed indices, you get a simpler recursion.
@Srivatsan That series is like integrating $\frac{1}{1+r^2}$ in $\mathbb{R}^2$
 
@JonasTeuwen thanks a lot. could you tell me, where is the city center? From central station I can go either in the Zoo direction or Stadspark direction
 
This question reminded me of an exercise I forgot how to do: Given a function which is analytic and non-injective in some ball $\{|z| \lt \rho\}$ then there are two points $z_1 \neq z_2$ with $|z_1| = |z_2|$ such that $f(z_1) = f(z_2)$.
 
10:33 AM
That sounds like it should have something to do with the local picture of analytic functions as maps $z \mapsto z^n$...
 
As I said, I don't remember how to do it at all. However, the assumptions do not prevent the function to be injective on some small ball, so I don't see how the local picture helps.
 
Morning!
 
@Matt heyhey
 
Hm, true. But then the claim is suspect, since 0 isn't very special...
 
Morning, @Matt, @Ilya.
 
@robjohn we followed the tradition and celebrated it at home first at 9 pm (Moscow time) and then went to the street at 12pm
 
@Ilya Ah, you are talking about the evening before. We spent that with some sparkling cider with cheese and crackers before midnight. We packed it in soon after midnight.
 
@Matt what's wrong? Looks fine to me.
 
@tb Well $s$ is a simple function and already positive, so what's $s^-$? I don't see why the triangle inequality needs to be mentioned.
 
QED
0
 
10:42 AM
But maybe I just haven't had enough coffee yet.
 
@Matt I don't understand what you say: you try to reduce to $f \geq 0$. So assume it's already solved and let $f$ be arbitrary. Write $f = f^+ - f^-$ with $f^{\pm} \geq 0$ and approximate them with simple functions $s^{\pm} \geq 0$. Then approximate $f$ by $s = s^+ - s^-$ and use the triangle inequality.
 
@Ilya Well, in front of the station you have "Astridplein" and there to the right is the Zoo.
 
aha
 
Now you also have the exit "Keyserlei", that's to your left if you come from the tracks.
If you walk that one you will get to the Meir which is part of the center.
 
@JonasTeuwen thanks, then we will just walk around
 
10:46 AM
(Sorry, Ilya and Matt, not for your eyes.)
 
@Srivatsan ah
 
@Ilya How have you been?
 
@Srivatsan I'd like to get such an approved stamp, too :)
 
@Ilya And I'm not sure if you were around during new year time. Happy new year any way.
 
@Srivatsan how have I been? what do you mean? :)
 
10:49 AM
@Matt I think the point that is being made is just that you weren't very explicit about the reduction and commenter spells it out and tells you that you should do so, too.
 
@Srivatsan oh, I don't remember as well - Happy New Year to you as well
seems that I've spent all my 'as well' for today
 
@Srivatsan The question is, is he at $1$ with the universe?
 
@robjohn No, at 0.99999....998
 
@tb Indeed :-)
 
@tb how many nines?
 
10:51 AM
@Ilya a lot.
 
How many dots? =)
 
How much do you hate the Romans?
 
@Srivatsan a lot.
 
@Srivatsan I've just realized in which meaning Asaf uses 'your mother'
 
@Ilya There is a sense?
 
10:53 AM
that's a very basic obscene phrase in Russian too
 
@Ilya Is it nice in any language?
 
@Ilya I don't get it: if it's so obscene, why did it take so long to realise? (honest question)
 
@robjohn well, at least in Russian it is a cut version of 3-words phrase with a verb before 'your'
@Srivatsan because there are no grammatical cases in English which differ the word itself
 
Oh, I see.
 
I thought his 'your' is another 'your'
anyway, let's skip it
 
10:57 AM
Right. So Asaf has been throwing obscene words all the while?
Not like it's any surprising =)
 
QED
you're mother
 
@Srivatsan indeed ) but maybe I'm wrong
@QED that may be surprising ;)
 
@tb Got it. Thank you. You don't want $s$ here to be a simple function because simple functions are positive so writing them as positive and negative parts is unnecessary. If you assume that it's solved then $s \in L^1$ with finite support.
 
@Srivatsan If it wasn't already obvious, he's nuts. ;)
 
@QED I am not!
 
10:58 AM
@robjohn yet? who knows, maybe in 10 years all of us will be mothers
 
@Ilya Are you indicating that at some point I will become QED's mother?
 
@robjohn Oh, I am sorry to hear that. Truly.
 
@robjohn just some new terminology may appear
 

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