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Home, home again
I like to be here when I can
And when I come home cold and tired
Its good to warm my bones beside the fire
Far away across the field
The tolling of the iron bell
Calls the faithful to their knees
To hear the softly spoken magic spells.
who wrote that?
I have a somewhat insane question. How can one show that a finite set of numbers contains a maximum? (Let's say that the numbers are reals for concreteness.)
I guess induction on the size of the set works.
And also points up that my statement was not quite correct.
18:16
@MarkDominus Why do you think it is insane?
I'm not sure.
It seems a lot less insane now that I have a proof.
I was gonna prove it by producing an algorithm that finds the maximum, but then I couldn't prove that the algorithm worked unless the set contained a maximum element.
For finite sets with no maximum, the obvious algorithm might fail.
@MarkDominus How would you construct a finite set with no maximum element?
@Ilya These two Didiers are the same.
There isn't one, I just said that.
(modulo my minor error)
Carlin $=\text{Awesomeness}^{G}$ $G$ is the Graham number.
18:41
Is it correct that Davide Cervone is the inventor of MathJax?
19:20
@MarkDominus I guess so. He's the developer
Any ideas on how to integrate $$\int\limits_0^{\pi /2} {\frac{{\sin xdx}}{{1 - {a^2}{{\sin }^2}x}}} $$
for $|a|<1$?
I'm trying Weiertrass substitution now.
I think it is workign.
19:37
@skullpatrol A little DSotM... nice :-)
@PeterTamaroff $u=\cos(x)$
@robjohn ! =)
@PeterTamaroff I was just about to ask :-)
Good evening
@robjohn I think I overreacted. I was reviewin the GF of $$\int_0^{\pi /2} \sin^{2n+1}\theta d\theta =\frac{n!^2}{(2n)!}\frac{2^{2n}}{2n+1}$$
Good afternoon! :-)
19:40
And I'm fiddling around a bit.
@tb Good afternoon, fine sire.
@MarkDominus you could just say finite sets are compact and the identity function assumes a maximum on them ;)
@tb Excellent, thanks!
@tb I answered a question too quickly this morning, and leo sensed something was missing, and I had inadvertently assumed that a function was continuous. It took a while to fix the answer keeping the same idea. I think it is okay now.
@robjohn The one of the $500$ bounty?
I gave it a try, but I kept it simple because I had never heard of variation before.
@robjohn I saw the question and your deletion. I haven't looked closely, yet, but I will do that a bit later. Thanks for the link!
19:45
Hmm, I finally did something productive today :-).
I has another theorem! Or lemma.
Maybe I should call it sublemma.
@JonasTeuwen Yay!
@JonasTeuwen Wait. NO.
Don't make those nomenclatures more tortuous than they are already.
It is something that is already known in many other cases 8-). The duality of some cool spaces.
Lemma, corollary, scholium, theorem, note, blah blah
@JonasTeuwen cool! it is nice to prove something :-)
@JonasTeuwen is the dual of a cool space a hot space?
@robjohn It is some kind of $(H^1)^* = \text{BMO}$ for non-doubling measures :-).
Now I must interpolate!
19:48
@JonasTeuwen does that eliminate the use of maximal functions?
@robjohn No.
It is just a bit more tricky.
As what you do is split up the operators in "doubling" and "non-doubling" parts.
@JonasTeuwen how do you handle the non-doubling part?
The doubling parts are easy as the tricks there are well known, the non-doubling has some weird off-diagonal estimates.
okay. I guess that is dependent on the space.
And as always you estimate on spherical shells...
19:49
Arturo is jogging on my nerves.
Yes, I have it now for one particular non-doubling measure.
And, hola.
Where I have an explicit integral kernel, but I would like to do this without a change of variables to the Euclidean case.
@Gigili hopefully, not with spikes.
You see, now I take the Gaussian measure, and then I just use the $e^{-x^2}$ in my estimates, but I would prefer it if it does not rely on that change of variables that much.
But more on just the measure.
But maybe that is nitpicking... But I think it would be a nicer result as it is easier to generalize.
19:52
@Gigili Why?
@robjohn or snow shoes :D
@robjohn I heavily rely on the fact that my operator to define the maximal functions has an integral kernel for its semigroup... Anyway, you should check out that stuff, it is very interesting and gives you googolplex equivalent definitions for our Hardy space. You can take many cool elliptic operators.
Hi @Eugene
@Eugene Hey there-
20:00
He answered the same question I answered and he's got more upvotes.
That's how he has 130k by now.
</angry>
@Gigili Oh. He has a fan club.
But I guess he deserves it, right?
Has he ever dropped by this chat room?
@skullpatrol I haven't seen him
@PeterTamaroff Me neither.
@PeterTamaroff Left.
20:03
@Gigili BFG?
BFG.
@Gigili Yes or no?
I have to think about it.
@Gigili It's not worth getting upset about such things. He has 130k by now partly because he has been around a long time and partly because his answers are of extremely high quality.
But 130K reputation points and $1.69 will buy you a cup of coffee at Wawa.
Is that what I think it is @Peter?
20:07
So why stress out about it?
@Gigili Road Dhal's book I mean.
That's right, but I can't help it.
My mind can't make a priority list of things to be stressed out about.
@robjohn How was it you showed $$\log \left( {\frac{{\sqrt {{x^2} - 1} - x}}{{\sqrt {{x^2} - 1} + x}}} \right) = {\sin ^{ - 1}}x$$?
I'm stuck!
@robjohn Should one first go with $${\sin ^{ - 1}}x = {\tan ^{ - 1}}\frac{x}{{\sqrt {1 + {x^2}} }}$$?
@PeterTamaroff where did I use that?
@robjohn Let me get it for you.
20:12
thanks :-)
I notice that Arturo got his answer in before yours too.
@Teddy thanks for earlier today.
@MattN no problem at all. Hope you recovered from the shock :)
20:20
Privet everybody :)
One minute only, I was thinking if I should post my answer since he has the same thing.
Doesn't matter, I was just complaining.
He appears just everywhere.
@Gigili So were you talking about the Big Friendly Giant or not?
The right/left pun.
@tb Yes I did. I even read til Nakayama's lemma today. I think if you hadn't "saved" me I'd've gone home and not looked at any more of it for at least a day.
I was talking about it, left.
@MattN I see, very good. Nakayama is cool.
20:25
Are there some experts on dynamical systems? I have a problem with an attractor's basin
@tb There is too much noise in some people's output. I can't fill my already muddled brain with random blah when I'm trying to learn something.
@MattN Maybe these MO threads: one, two on Nakyama's lemma are helpful.
I'll be back in a moment.
Since $|x-i\sqrt{1-x^2}|=|x+i\sqrt{1-x^2}|$
$$
\begin{align}
\log\left(\frac{x-i\sqrt{1-x^2}}{x+i\sqrt{1-x^2}}\right)
&=i\arg\left(\frac{x-i\sqrt{1-x^2}}{x+i\sqrt{1-x^2}}\right)\\
&=2i\arg\left(x-i\sqrt{1-x^2}\right)\\
&=2i\arctan\left(-\frac{\sqrt{1-x^2}}{x}\right)\\
&=-2i\arccos(x)\\
&=-\pi i+2\arcsin(x)
\end{align}
$$
@tb Thank you. I'll read them tomorrow, I'm too tired now. Sorry for starring.
Oh, ok. I think I found an easier way. I'll show it to you when its done, I think you'll like it.
20:30
left out an $i$
$$
\log\left(\frac{\sqrt{x^2-1}-x}{\sqrt{x^2-1}+x}\right)=2i\arcsin(x)
$$
@PeterTamaroff Most ways I can think of are geometric.
WAIT
I wrote something wrongly.
@tb I don't know yet. But given that it has a name it seems like a likely exam question.
But enough CA for today.
@MattN Hey, the temperature is a nice 80°F in CA now :-)
@robjohn heads out to use google to convert to celsius
@robjohn The thought of it gives me a headache.
20:45
26°C or about, closer to 27°C
Actually, I already had one : )
@robjohn Yes, 26.66666667
@robjohn Today I got another upvote for my top answer. : )
@MattN What is you top?
@PeterTamaroff 24.
Thanks Peter, I'll assume it was you.
@MattN Yes.
This answer of mine seemed to rocket in hte last days, its a pity it is CW.
@PeterTamaroff Did you edit too many times or did you turn it into CW yourself?
20:54
Having a nicely-upvoted CW answer is obligatory if you want to be one of the kool kids.
@MattN I CWed it. I considered it appropriate.
@JonasTeuwen According to a survey? : )
@anon I never wanted to be cool.
@MattN Not cool, but *k*ool. Learn the difference.
Exactly. That's exactly what you're supposed to say in order to get cool points.
I guess I just am B-)
21:00
@robjohn nice and clean, according to me. What does the TeX-command at the beginning achieve? (you can avoid that ugly space before Break by moving the command at the end of the first line, I guess). Two tiny typos: partiion $P$, $(11)$ says (right before the first equation $(12)$) which leads me to the second typo: two equations $(12)$.
@robjohn Check th is
with a url like robjohn.jpg I thought it was going to be a really mean-looking rectangle
Maybe the first Betaish expression can be used to yield more results.
That looks as if written using a graphic tablet.
@anon Ha no, I just name the files like that to remember who I'm supposed to give it to. Not that I'm 10 seconds Tim of something, but it has become a "tradition".
@MattN It's plain birò, scanned.
21:04
@PeterTamaroff : )
It tweaked it with paint (sigh) so it got blurred.
Ah, that explains it.
I guess the only downside to plain white pages is my tendency to write "upstream".
I'll go study some epistemology stuff for tomorrow. Tag me if rob comes back, and we can all maybe talk about other GFs that can be retrieved with the beta integral.
Okay. I won't be here for much longer either though, just wanted to pop in for a quick hello before going to bed.
And then I ended up hanging out in here for an hour.
@tb This bug cropped up again, so that code was modified from the answer.
21:12
@PeterTamaroff He seems to be back.
@robjohn Did you see the links I gave you?
@PeterTamaroff I was just looking at them after I replied to tb.
Good night xxx
@MattN Night, sire.
Good night. (No ping necessary, I guess)
21:43
@tb Thanks. I have fixed the typos and altered some wording to be more precise.
@robjohn Yesterday I was thinking about some trivial, so to say, interpretations of the result of that question.
I'll show you in a sec.
@PaulSlevin You do some commutative algebra right?
well I have an exam on it on Wednesday, whether I know any algebra is a different question
why do you ask ?
21:50
3
Q: Exercise 2.17(d) of Eisenbud's Commutative Algebra text

Benjamin LimFirst some notation: Let $P$ be a prime ideal of a $\Bbb{Z}$ - graded ring $R$, $U$ the multiplicative subset of all homogeneous elements not in $P$. Suppose that there exists a homogeneous element $f$ of degree $1$ that is not in $P$. The problem in Eisenbud is to show that the image of $P$ (...

I can't prove some image of a prime ideal is a prime ideal....
is the quotient a domain
sorry, I dont know what a graded ring is
hmm
ok
thanks anyway
I have an exam tomorrow so I am not much use at solving problems just now
right paul you should concentrate on your exams and not worry on stuff like that
I just dont think I can study any more so I am jst relaxing for now
21:52
right ok :D
Hopefully that will help negate the profound terror in my heart haha
I went to see my mentor yesterday and even he can't solve it
@PeterTamaroff when you were working that out, were you working towards the series for $\sin^{-1}(x)^2$?
@robjohn Indirectly, I suppose.
I like this question.
good luck with your question
21:57
@PaulSlevin thanks
@PeterTamaroff It should be clear what obvious means
@robjohn You're saying the question is a little weak?
I can't get the word.
@PaulSlevin If I don't get an answer do you think I can cross post on MO?
Not sure I have never used MO
I knew there was a dupe
22:01
it is either obvious nor trivial that obvious means trivial
i have a question regarding the wiki entry on curve orientation - not sure it's worth an actual question on here
i know that to test whether three points in the plane are taking a left turn or a right turn, you look at the sign of the determinant of a matrix formed by the three points
however, some things seemed unclear to me from the entry @ wikipedia: en.wikipedia.org/wiki/…
specifically, in this paragraph: " In computations, the sign of the smaller angle formed by a pair of vectors is typically determined by the sign of the cross product of the vectors. The latter one may be calculated as the sign of the determinant of their orientation matrix. In the particular case when the two vectors are defined by two line segments with common endpoint, such as the sides BA and BC of the angle ABC in our example, the orientation matrix may be defined as follows:"
the sign of the smaller angle? i'm not exactly sure what they're saying?
hmm...
so negative -> clockwise, positive -> ccw
anticlockwise is positive yeah
ahh ccw = counterclockwise
Hope that helps. Im afraid i must go. ciao
 
1 hour later…
@tb Like using Carleson's theorem to get a sequence of $C^\infty$ functions that converge pointwise ae to a bounded measurable function?
Not quite :)
More like using Urysohn's lemma to prove that there are compactly supported continuous functions on $\mathbb{R}^n$.
@tb Hmm, could you also use it to prove that there is a $C^\infty$ function? 8-).
Also, I prefer $\mathbf R^d$.
@JonasTeuwen No, for that we have Whitney's extension theorem.
@tb Great!
Does that give compact support? Otherwise we might have to take convolutions and Urysohn...
23:38
No, no. You take the function which is constant equal to one on $|z| \leq 1$ and $0$ on $|z| \geq 2$, of course!
Then you hit that monkey with Whitney. To paraphrase you.
Great!
But maybe it is better if you hit a continuous compactly supported function with Whitney? So we can be sure it is integrable as we have this theorem that continuous functions are measurable and with compact support we get a finite $\sup$ norm on a compact set.
You forgot the step to pass to the space of distributions and then use decay to find out that it actually is a smooth function.
Oh right.
I should write that up.
Are you sure that is hard enough?
But something's fishy. We haven't used elliptic regularity, yet.
Exactly.
We could set up a PDE and then use some embedding to prove that the solution is smooth enough.
Then we should give it some compact support.
23:46
enough...
Haha 8-).
I'll talk with Ilya tomorrow about this, he might know something...
I'm using the Iverson bracket in my paper.
leo
leo
hi there!
Hi. Sup?
leo
leo
@robjohn, now you almost convinced me
@JonasTeuwen inf? min? or something like that? or soup?
Exactly.
leo
leo
23:54
so что???!!!
@JonasTeuwen wonderful!
Hmm, I use $[\complement D]$ for $[(y, t) \in \complement D]$.
For whom?
For our bro.

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