@FrankScience Is Science your real last name? :-).

@JonasTeuwen Actually No. I have no English name, but Frank is my nickname.

Oh right.

@MartinSleziak Could you show me how to write it in the three cases? I never do a thing like this before

I'll just write $g'(x)$.

@Martin: Before I forget, thanks for adding the link and finding Steve Watson’s paper; I’d completely forgotten about it.

@MartinSleziak It seems that writing such a proof is not easy, especially in English.

@unNaturhal We have $$g'(x)=\begin{cases} -2x & x\in(0,1)\\ 2-2x &\text{otherwise}\end{cases}.$$

@MartinSleziak Wait, the exponent is $x-|x^2-x|$ how could be this the two derivatives?

I hope I did not mess up signs somewhere.

@MartinSleziak : In product rule, is it necessary that the two functions should be differentiable or only one?

@RajeshD I'd say both.

surprising

@unNaturhal It's not two derivatives. It's only one function written by cases.

Similarly you could define absolute value as: $$|x|=\begin{cases} x & x\ge 0\\ -x & x<0 \end{cases}$$

@MartinSleziak $f(x)$ seems continuous, so we can use try to use $\lim_{x\to x_0}f'(x)$ instead of $f'(x_0)$ if $\lim_{x\to x_0}f'(x)$ exists, but if not, we should check them by definition of derivative.

You've seen such definition of $|x|$ haven't you?

@MartinSleziak Yeah, I seen on my book

@MartinSleziak What if one function is not differentiable, and the other differentiable function is zero at that point?

this is the case i want to know

@RajeshD If it is zero identically, the answer is obvious.

@RajeshD What about $g\circ f$ with $g(x)=x$ and $f(x)=|x|$? Both of them are differentiable at 0, both have value 0, but the composite is not differentiable at 0.

not zero identically but only at that point

@MartinSleziak : I am talking about product rule not chain rule

@JM I see you are toasting non-Möbius bagels :-)

Sorry. I was confused, because right before this we were talking about chain rule with unNaturhal.

$g(x)=g'(0)x+o(x)=o(x)$

yes thats how i got into this question

@MartinSleziak

@FrankScience only if $g'(0)=0$

@FrankScience: About this part of your answer: Now, let's prove that $g$ is continuous. For all $\epsilon>0$, we have $|f(x)-g(x)|<\epsilon$ on the open interval $I_\epsilon(x)$. Let $y\in I_\epsilon(x)$, and $x\to y$, we have $|g(y)-g(x)|\le\epsilon$, hence $g$ is continuous.

@robjohn Ah, you rang? :) Still having trouble coloring that up properly...

@JM Good to see you :-)

(Hi, y'all.)

Notation seems to be confusing - you use $x$ for two different things. And what exactly $x\to y$ means?

@JM I was just spying on Mma chat.

Hi @JM

You wrote that you have $|f(x)-g(x)|<\epsilon$ for each point in $I_\epsilon(x)$.

So you denote by $x$ the middle of the interval and arbitrary point of that interval.

@JM How are things? I haven't seen you in a while.

@FrankScience But even if I read this as $|f(y)-g(x)|<\epsilon$ for each $y\in I_\epsilon(x)$, you should explain why this is true. (At least I don't see it immediately.)

@robjohn Well, somewhat busy on- and offline. I still peek at questions here. but too many (new!) people answer quickly. :)

I checked the transcript just now. @Jonas, you needed help with Hermite?

@JM I just used the proof of Gautschi's Theorem for an answer today.

@MartinSleziak edited

@FrankScience Did you want to write $y\to x_0$ there? Or perhaps $y_n\to x_0$?

@robjohn Very useful, that.

@JM I point it out since you asked the Gautschi question

@MartinSleziak exactly $x\to y$

@JM It comes in handy :-)

I still don't think it's correct.

I would guess this would be easier to discuss in a separate chatroom, what do you say @FrankScience?

We would not distract other people and they would not be distracted by our discussion.

@MartinSleziak Unless we were watching the discussion :-)

@robjohn Dang, reminds me, I still need to do that write-up on Gautschi's paper. I have a draft, but it still looks like more of a quote than a proper summary.

@MartinSleziak i just applied some medicine on my hand, so it's awful for me to write a lot

@MartinSleziak what's your question

@FrankScience Ok.

@JM Glad to remind you of the things you need to do that will make your day more burdened :-)

@MartinSleziak try saying briefly

Well, hopefully this week. I haven't written anything on main for quite a while now...

@JM I'm trying to do enough to get to 30K before my 1 year anniversary :-) July 28

about 7 and a half hours until the leap second!

I read Gautschi's Inequalty as Goat Cheese Inequality :-)

@PeterTamaroff good day!

@robjohn Hello! 1:30 PM here

@PeterTamaroff Does that mean I can't wish you a good day?

@robjohn No, no, just letting you know what my timezone is. Thank you, good day to you too!

@robjohn That inequaility reminded me of some Gamma asymptotics

Let me check my notes.

@PeterTamaroff That means you're in very eastern Canada or South America :-)

@robjohn With you, not too hard... :)

@robjohn Here it is

For $x >0$

@JM I was having some pretty lean time for a while.

$$\mathop {\lim }\limits_{n \to \infty } \frac{{\Gamma \left( {x + n} \right)}}{{{n^{x - 1}}n!}} = 1$$

@robjohn I'm in Argentina.

@JM I am not like Brian Scott, who gets 300+ a day with some regularity.

@PeterTamaroff That looks to be a reformulation of Gautschi's inequality, yes.

@PeterTamaroff that is a weaker form of Gautschi, Gautschi gives better bounds.

@robjohn A machine, I tell you... :D (Hi @Brian!)

@robjohn I see.

@BrianMScott: hey there, I didn't see you :-) howdy!

@robjohn I guess that is because the proof of the above doesn't use the log convexity, but some easy integral manipulations. I understand log convexity is an important property of the Gamma function.

@robjohn Not surprising: I’m barely here! How goes it?

@JM I learned a lot about adding sine waves while writing an answer yesterday. I never knew that not only does the amplitude vary, but so does the frequency (but usually very slightly).

@PeterTamaroff Indeed it is. It is not too hard to show from some "easy integral manipulations' :-)

@BrianMScott Pretty good. I get a dose of humility looking at your daily rep. If I ever feel too good about myself... :-)

@robjohn I’ve been taking it easy for the last week or so. Though I will admit to being chuffed to get the gold general topology badge.

@BrianMScott There's a gold for that now? 'grats!

@BrianMScott Gold badges are really hard to get. Congrats on that!

@robjohn For what is worth, the "trick" is splitting $\Gamma$ into $\int_0^n $ and $\int_n^{\infty}$ and using $0<t \leq n \Rightarrow t^{x-1}\geq n^{x-1}$ and $t \geq n \Rightarrow t^{x-1}\leq n^{x-1} $ , where $0<x<1$

"taking it easy" - IIRC Arturo has said those exact same words when asked about his rep... :D

@JM Thanks! (I suspect that in principle there’s a gold for any tag.)

@BrianMScott Oh, there is one indeed. Having a gold badge means your topology answers are wildly popular...

The general result then follows by an "inductive" argument. (Basically using $\Gamma(x+1)=x \Gamma(x)$

@JM Hi!!!

@JM I did.

@JonasTeuwen Hey. You needed Hermite help, eh? What about?

@JM Do you have some time? Say 10 minutes? Then I can explain.

(In total 8-)).

@robjohn Well, pretty much a simplification of how AM radio works... ;)

@JonasTeuwen Sure, fire away.

@JM I said the same 8-).

Anyone knows where the notation $GL(n,K)$ where $GL(n,K)=\{ A \in K^{n\times n} : A \text{ is invertible } \}$ comes from?

Since it is a Spanish text $GL$ might mean "Grupo Linear" which translates to "Linear Group".

@JM Right, I want to compute some "Mehler like kernel". $\sum_n f(n) H_n(x) H_n(y)$, okay? Good. So what I do obtain in the end is some finite sum of $H_{m_1} H_{m_2} H_{m_3}$ in some variables with $|m|$ constant. So, I was wondering... These addition formulas, are they not a special case of the thing I actually want to compute? As in, if I can do this, I would be able to compute the direct sum directly as well.

@PeterTamaroff More likely German than Spanish, if memory serves.

@JonasTeuwen If they are indeed a special case, it's not readily obvious to me. I've seen those addition formulae, but I've always found them unwieldy to use.

Hmm, Galois introduced the word "group".

But he talked about permutation groups.

@JM What you do is put stuff in the generating function and compute like a madman right?

So perhaps... If you can compute it you can go back and see how to do it directly. But then I might feel like an idiot because I didn't realise this in the first place 8-).

@JM What does $GL$ stand for?

General Linear.

@JonasTeuwen Yes, the straightforward but tedious route... :D

@JM :D.

@JM It sucks.

@JonasTeuwen Yeah, I know. I'll check my notes on this when I see them again. I'll ping you if I figure out a shortcut.

@JM Great. I found many papers on the Mehler kernel, some do it combinatorically... It is very very cool. Would you use something like that in an analysis paper? 8-). Zeilberger does this by arguing about homosexual relationships between Hermite polynomials.

I'm impressed by this downvoting!

What a consensus!

@JM Thanks for listening. Made me feel better 8-).

@JonasTeuwen I dunno, the combinatoric route sometimes makes for a nice guide on how to proceed from an analysis viewpoint...

@JM The analysis viewpoint is often so ugly.

The argument is perfectly fine too. It just sounds weird if you call polynomials homos, but then you are missing the point.

@JM Yes, but the frequency is modulated as well. That is what surprised me.

@JonasTeuwen ugly?

@JonasTeuwen Well, you know what they say about beauty and the corresponding beholders... :)

@PeterTamaroff If that impressed you, you should see the most downvoted post ever on math.SE ...

@robjohn I mean the proof is totally insightless, just manipulate symbols.

The combinatorics gives a concrete interpretation.

@JonasTeuwen Yeah, that's a point for combinatoric routes; they're easier to "see".

@JM Oh, link? (I'm confused by what Doug Spoonwood wants to obtain through his question)

@PeterTamaroff The most downvoted post ever.

@JM Well, much harder to actually find the combinatorical interpretation. From there on it is usually quite easy, just counting stuff you know 8-).

@JonasTeuwen "Obvious in hindsight", yes. :)

@JM Freaking hilarious!

"...it seems I have invented new field of mathematics."

@JM That answer and the question it answers :-) although there are 4 upvotes to the question.

"(Notice the 8 in the end)"

@PeterTamaroff That's the part that got the downvoters' goats, if memory serves.

The bulk of the downvotes came after that edit.

@PeterTamaroff the 8 in whose end? :-)

Haha, naming numbers after yourself, that is just excellent.

@robjohn In the question.

@PeterTamaroff I know, I was just being funny (in the head)

@robjohn I don't get the pun... "The ate in whose end?"

"Magma" is such a cool word. Mathematicians should come up with those more often.

@JM But you might have the feeling: how the bloody monkey did they guy find out this...

Magma is not cool! :-)

And then you are like hehe, he doesn't seem my 27 pages of odd unsightless analytical manipulations and then I only understood the problem 8-). That will mess with their mind for sure!

@BrianMScott Why not!?

@PeterTamaroff Magma.

@BrianMScott Hahahhahahahhaaahahah I'm slow today!!!!

Does anyone have any clue why this answer got two downvotes? Is there some subtle error I overlooked?

@BillDubuque Bill, you didn't answer my question on the binomial coefficients. Do you or don't you think that it is trivial, from a combinatorial view, that they are integers?

@BillDubuque I don't see a reason why to downvote that!

I upvoted it a while ago.

@BillDubuque Nothing obviously wrong. I thought it was a nice touch to mention the Wronskian...

@JM Side with you.

@Peter I don't know how to respond since I don't know what "natural" means to you (nor did I want to further clutter WimC's answer with tangential discussions).

@PeterTamaroff "trivial" - depends on how you're defining them, of course. ;)

@BillDubuque My guess would be that the downvoters thought it rather unhelpful to the OP.

@JM That's my point =)

@Brian I wondered about that. But even if so, do people downvote simply because they think answers might be too general?

@BrianMScott Probably because Bill gives rigorous and detailed answers that might take a while to understand, but I think they are great.

In any case, I asked here because if there is any error then I would certainly like to fix it. Sometimes it is hard to spot one's own errrors!

@BillDubuque I don’t think that it’s the generality per se. I think that it comes back to the difference between those who are interested first in building a reference library and those who are interested first in answering the specific question at the level at which it was asked.

(i.e. they downvoted for philosophical reasons, not mathematical ones, assuming Brian's right.)

I suppose. Perhaps I should have emphasized further just how fundamental a role that Wronskian's play in such matters. This becomes clearer when one studies differential Galois theory. But, alas, if I write more, I fear my post might end up with a negative vote total!

@BillDubuque That sounds like an answer for a different question. :)

@BillDubuque Interesting field!

@JM In some sense, it's a differential analog to why determinants are useful in theoretical (vs. algorithmic) linear algebra, a question which was recently asked.

In any case, as we all know well, voting can be very random on MSE. I just wanted a few more sets of eyes there to make sure I wasn't overlooking something obvious. Thanks for the feedback.

i implies ii implies iii are clear, and iii implies i because iii states the n column vectors making up A are linearly independent, so det A is not 0 and A is invertible. I know it makes more sense to skip the determinant step, but thats just how I remember it.

I’d rather argue that (iii) says that the kernel of the linear transformation is trivial, so the transformation is an isomorphism.

I guess that's an even higher level view of the "most natural" reason why it is true.

I had forgotten the utility of this simple theorem since I hadn't used it much since I had first learned it, but it came up nicely when I was trying the exercise: If A and B are square matrices such that AB is invertible, show A and B are each invertible.

That result falls even quicker to the determinant!

Ahh yes!

oh gosh.

hey brian

Yeah, if you switch to determinants, it's exceedingly straightforward... :D

In my defence, that exercise appeared in the chapter right after the invertible matrix theorem and before determinants!

@Eugene Hullo.

@BrianMScott rare to find you here in the afternoon

@RagibZaman OTOH, you know the bit about determinants coming first before matrices, yes? ;)

@Eugene My schedule is shot to hell lately, so I’ve become very unpredictable.

@JM No? I'm not sure what you mean.

@BrianMScott we once had a discussion on your "about me" section. isn't emeritus synonymous with retired?

I guess one could define the determinant of a linear transformation by the signed volume of the image of the unit cube.

Another thing that many folks don't know is that Wronskians prove fundamental in Diophantine approximations to solutions of ODEs and transcendence theory. For example, Mason's abc theorem and the consequent high-school-level proof of FLT for polynomials has a natural Wronskian view.

@Eugene No: it implies retired, but I could be retired without being emeritus. That status is granted by the trustees.

@RagibZaman I meant historically speaking; determinants were being studied before matrices, and then it was decided that the matrices were more fundamental than determinants. Nevertheless, it would have been transparent to people who just knew determinants.

@BrianMScott ah i see.

How could determinants have been studied before matrices?

@BrianMScott So your "shot to hell" schedule is due to an increased demand for your services, I presume? :)

What definition were people using before?

@RagibZaman I don't have my nice reference on this, but wiki seems to have a good summary.

@JM No, it’s a combination of worrying about a friend and having to adjust my schedule a couple of times to deal with the rest of the world.

@JM Oh wow, "a determinant was defined as a property of a system of linear equations. The determinant "determines" whether the system has a unique solution (which occurs precisely if the determinant is non-zero)". That just made a whole lot of sense lol. I should think more before I ask things I guess.

I really like the OP’s response to Jonas Meyer in the comments here.

LOL!

Another nice exercise I saw was to determine if there were any solutions to AB - BA = I, where A and B are square matrices.

Turns out there's an entire book on pre-matrix history. :D

@BrianMScott Oof. I hope your pal's going to be fine.

"I have tried thinking hard and long." - don't we all? On the other hand, after seeing some recent questions, well...

@JM She will be eventually; it’ll just take time.

Anyway, I should be off. It was nice seeing you guys. Later.

cya J.M

@JM Sorry to have gone off for a while. See you later. Thanks for dropping by. :-)

any idea on showing how two functions have the same fixed point?

@Clash What are the functions?

$1/2(x+e^{-x})$ and $1/6(5y+e^{-y})$. I don't know why but I noticed that a part of the formula speed of convergence is the same for them. It may be related to the fact that they have the same fixed point. The number $\frac{|x^1-x^0|}{1-q}$ is the same for both. en.wikipedia.org/wiki/Banach_fixed-point_theorem

I have already proven it's a contraction, $q_x=0.31$ and $q_y=0.77$

Why do you have $x$ in one and $y$ in the other?

crap, can't edit anymore... $x^0$ was supposed to be the starting value, which is $0$. $x_1=1/2$ (second iteration)

@PeterTamaroff hm I dont know, ask my teacher :D

$\frac{|x^1-x^0|}{1-q}$ gives me $0.724$ in both functions

@Clash Hm, I can't help you, don'tknow much about matric spaces and stuff, sorry.

ok, thanks!

@unNaturhal It seems to be okay. What does W|A give?

@BrianMScott Are you around?

@BrianMScott See my answe here. What do you think? The OP is asking for another solution but I guess (as Kcd states) the use of monotone convergence suffices.

@Brian! Could you point me to a (short) book explaining the main ideas in descriptive set theory?

@JonasTeuwen Let me think about that for a few minutes.

@BrianMScott Oh, no problem. Even a couple of weeks is okay 8-). Or not at all! Thanks.

@PeterTamaroff Your proof by induction of the boundedness above has me perplexed.

@robjohn Am I messing up?

@JonasTeuwen this is a pretty set: it has nicely shaped elements and a pleasing arrangement.

@robjohn The easy joke, haahhaahha.

@robjohn 8-).

@robjohn Is it wrong?

@PeterTamaroff I don't follow why $\sqrt{r+k}=r$

@robjohn It has an equal sign.

$\sqrt{r+k}=r$

@robjohn Because it is the root of the equation $x^2-x-k=0$.

@PeterTamaroff Ah, I missed what $r$ was, okay

I state it in the beginning.

@robjohn Maybe I should clarify it again, just in case.

@PeterTamaroff wouldn't hurt, but I was probably reading it too quickly.

@robjohn The proof is pretty ansatzish, but it works out nicely.

@PeterTamaroff The Monotone Convergence Theorem usually refers to a theorem about integration.

"If $\{ a_n \}$ is a monotone sequence of real numbers, then this sequence has a finite limit if and only if the sequence is bounded."

It is right there in the page, I don't follow.

@PeterTamaroff However, there is this Monotone Convergence Theorem

@robjohn Rob, you're linking to the same page.

By the way, where are you from?

@PeterTamaroff I don't think I have usually called it that, but I guess it is.

@robjohn Maybe I should clarift that, too! =)

Hi all

@tb Long time no see!

@tb I just just asking Brian to see this answer of mine. The user is asking to use the Banach Fixed point theorem which I know nothing about, so I suggest another way to solve it.

Hi Peter, yes, I'm quite successful with cutting down my procrastinating on this site :)

@PeterTamaroff That's odd, I thought I had copied this address, but I copied from the top of the page and not the section.

@tb Hey there!

@tb we haven't been active at the same time for a while :-)

@tb Hahah, for the time being, it helps me a lot, though I have a hard time stopping. I had a physics midterm today and I almost had to turn off the screen to sit down and study yesterday night!

@JonasTeuwen A very nice and excellently written intro is given in chapter III of Arveson's book on C*-algebras (ca 20 pages). That's what I always recommend to my students. Srivastava's Course on Borel set is quite short and nice, too. But I would definitely look at Kechris. It's not short, but it's simply brilliant.

@JonasTeuwen It’s probably my weakest area of set theory, so take what I say with a grain of salt. You could do a lot worse, I think, than read the introduction and first chapter of Yiannis N. Moschovakis, Descriptive Set Theory, the second edition of which is available here as a 3MB PDF.

@tb Hi! Thanks man!

@BrianMScott Excellent. Lovely.

@BrianMScott I second that, but it definitely doesn't pass the "shortness test" (and it's farthest from Jonas's interests, I guess)

I’ve never read Kechris, but it has a very good reputation. There are also a couple of sets of notes online, one by Marker (with a link from the WP article on descriptive set theory) and one by Zapletal.

@tb That’s why I specified the intro and Ch. 1.

@tb Well, I actually want to figure out how bad the null sets in Carleson's theorem can be.

That is the motivation for me to look into this.

@JonasTeuwen Can't they obviously be arbitrarily bad?

@tb Why is it obvious? I doubt it.

I mean, it is very easy to have a null set $E$ and find a Fourier series that diverges on $E$.

But there is no way you guarantee that it only diverges on $E$.

How do you select a Borel representative on $L^2$?

A short observation made me believe it is not worse than $\Sigma_3^0$

Teddy bear!!

Hi, Matt!

Missed you.

@tb Hmm, what do you mean with that?

@MattN Hi Bro.

Hi Bro!

@JonasTeuwen I think your question is ill-posed in the sense that it depends on the representative you choose of your $L^2$-function.

Take $0$ and perturb it by an arbitrary null set.

@tb Oh, right. Well then rephrase it for continuous functions?

Problem solved by... change of assumptions? 8-).

I might miss something, I'm new to crazy stuff like this.

(Silently I was already thinking about continuous functions)

You have too much cool knowledge to stay away too long from this chat! 8-). I like all mathematics which I can relate to my original stuff, so for the moment I added set theory, Lie groups and combinatorics (and special functions etc).

@BrianMScott Yiannis is at UCLA, and was chair when I taught there. I actually found a mistake in one of his preprints and corrected it. I don't know if it was any that are on that page; the formatting is quite different now.

@robjohn I’ve never met him, but I’ve known his name since grad school.

@BrianMScott KP told me: Be careful or you end up with doing a PhD in set theory. 8-).

@JonasTeuwen :-) He just might be right!

Oh, man, that would suck so much. That would mean Asaf was right.

@BrianMScott He taught there when I was an undergrad, too.

But! It would be cool if he ends up finishing a PhD in analysis.

@robjohn So he is like... really old?

@JonasTeuwen He is professor emeritus from some university in Greece, I think. He is probably near retirement at UCLA.

He was born in 1938.

@BrianMScott So 74, not as old as I thought.