@N3buchadnezzar What language is that? What Tallteori means?

Number theory? no.wiki

@MartinSleziak Tallteori means Number theory. Tall=number, and teori=theory

@MartinSleziak Good guess, its in Norwegian =)

@N3buchadnezzar I am just ignoring jordan

it's this typical attitude of oh i'm lousy the rest of the world is having it easy

Well, not a guess, that was the first Google Search result for Tallteori.

@MartinSleziak I'm sick of all the whining from Jordan

@MartinSleziak can he be suspended or something?

@BenjaminLim So I must have a lot of luck, because I heard almost nothing like that from him. But I'm not very often in chat.

@MartinSleziak Hahahahahahahahahahaha

IIRC he mentioned himself he was banned from chat for some shorter period a few times.

just go slightly above and look at the transcript :D :D :D :D

@MartinSleziak Because of all the whining

it's just this sick attitude of

"oh the rest of the world is having it so easy why am I so unlucky bla bla bla"

people have their own problems

I don't think that's good reason to ban somebody from chat.

I have problems with trivial questions like whether or not a space is sequentially compact

@MartinSleziak You have no idea of the level it gets to

topological spaces? metric spaces?

What does $P(R \cap S)$ mean, does it mean S first then R or opposite? Regarding probability

peter tamaroff is sick

@MartinSleziak The question that day on $C[0,1]$

so yes in a metric space

Oh, I see.

BTW thanks for accepting my answer...

do I whine and whine?

Much better would be whine and dine.

@N3buchadnezzar I am told in norwegian "sylow" is pronounced "see-low" and not "sigh-low"

@BenjaminLim Silo ;)

We actually pronounce the letters like they should be pronounced, not like the mad british.

Phosphorous, I mean for real?

@N3buchadnezzar the "sy" part of "sylow"

how do you say it?

like "see" as in to see something?

its like see yes. You form your mouth to a smile. The I in I

Wait until you hear how Lie groups are pronounced.

Galois theory ?

How in the world am I supposed to know its pronounced galoa

English is like that anyway. It's full of words borrowed from other languages. Sometimes you just have to know.

@MattЭллен My son sent me this: English doesn't borrow from other languages. English follows other languages down dark alleys, knocks them over and goes through their pockets for loose grammar.

lol

ops, wrong chatrooom.

@N3buchadnezzar :-) pretty though

Yeah, I was wondering if there was any improvements I could make.

@BenjaminLim: an Australian in our department was surprised that Lie groups are not lying groups but rather pronounced as in the American "Lee".

(That was an anecdotal point to the discussion on how to pronounce Sylow...)

Yeah, people sometimes pronounce "Joule" with a soft J, as if it were French.

Godement, Leray, ...

Erdös, ... List goes on.

I had a girlfriend who refused to even try to pronounce Erdős, and just called him "Erds".

Erdøs :D

"That asshole with the beard."

"You know the one I mean."

@MarkDominus Captain Haddock ?

@MarkDominus Where is that from ?

@MB Who's this?

That's Chebyshev.

@JasperLoy it's french and I speak french

lurh - beg

@BenjaminLim Who's who?

the person the australian in your dpt

Post.doc. PhD from Queensland University of Tech.

@MB Name?

Hmm, I should have said Smullyan.

@MarkDominus it's Galwah

in french "oi" is pronounced "lwah"

Like

La loi - the law

à la fois - ah la fwah

@BenjaminLim He's in statistics. PhD from 2009, you think you know him?

no

I am not from QUT

@BenjaminLim It was N3buchadnezzar who was unsure how to pronounce Galois.

ah sorry!

@N3buchadnezzar see above

How do you pronounce Høst in french then ?

huh?

what's with the $\emptyset$?

Just think of it as ö.

dunno

never seen that

Scandinavian languages carry three extra letters.

But the ø is very closely related to ö in German

the ø is pronounced as the i in bird.

Actually the vowel in Erdős is not an ö. Hungarian has an ö, but ő is different.

ő has the same relation to ó that ö has to o.

@N3buchadnezzar how to pronounce angelveit?

It seems that people have very different opinions about what tags should be used.

If you have a look at revision history here, this question changed a few tags already, including logic, cardinals, elementary-set-theory, real-analysis, functions.

@MartinSleziak I was really unsure how to tag the question, but I was sure that logic and cardinals were wrong.

I tried to be content that I was at least getting the ball closer to the hole.

(Angeltveit)

I've added elementary-set-theory, but someone changed my edit back.

elementary-set-theory makes sense to me. I would not have removed it.

@MB how???

Ang-el-tveit

I would have guessed an-gelt-veit.

With a hard g.

@MarkDominus I've added it back.

Is there any way to makie

$$\binom{50}{20}\left(\frac{2}{5}\right)^{20}\left(\frac{3}{5}\right)^{30}$$

Look prittier ?

What do you dislike about it?

I think I would prefer either to only have powers or only have binomial terms, but I guess that is not possible here.

I might add a \cdot between the binomial expression and the fractions since otherwise they tend to look too similar.

Do you like $\binom{50}{20} (2/5)^{20} (3/5)^{30}$ better?

You meant prettier in the sense of typesetting or in the sense of neater expression?

@MartinSleziak Neater expression. I know prettier is purely subjective, but I have been looking at math for quite some time now. and I guess lost sense of what is neat and not.

ok, I thought you're talking about typesetting (and based on what he wrote, Mark probably thought the same)

I don't think that expression can get much simpler than this.

B(20, 50, 2/5)

^^

So the normal distribution and the poisson distribution can be used to approximate the binomial distribution. That's neat.

Is there a way to tell the bounty system that I want my bounty to go unawarded? The only choice I see is to let the grace period expire and then let the automatic system award it the way it wants to.

Is there any preference to whether using poisson or normal distribution to approximate the binomial distribution ?

yes.

@MB For appropriate B :-)

@MarkDominus I think that is the way it is. Once it is out there, the bounty will be given except if there are no accepted answers and no answers with at least two votes.

Thanks.

@MarkDominus It will go to TonyK unless you award it to Eric Wong

I thought it would go to nobody, as neither of those answers was posted after the bounty was etablished.

So far I have not had any success with bounties.

@MarkDominus Oh, I didn't notice that. Then it won't be awarded. It would have gone to Eric actually since the OP is not the bounty offerer

and only half would have gone to Eric

Hmm, how would I make sense of $\frac1{\Delta^n}$? 8-).

As an operator?

Yes.

I try the binomial expansion on my favorite operator.

Bounty FAQ

@JonasTeuwen $\binom{O}{U}$

Ornstein-Uhlenbeck? 8-).

Right. What I wanted to know is if there was a way to tell the system that I am not going to award the bounty, and let it skip the grace period, or even to explicitly not award the bounty.

Maybe I just do $\Delta^n f = g$, so then $f = \frac{1}{\Delta^n} g$.

@MarkDominus Nope. I think as soon as you give the points to the system, you can only award the points. I don't think there is any more control. You'd have to ask a mod or dev for more info.

Now I need to think about what that means... or how it makes sense.

@JonasTeuwen It looks ugly on the FT side

Yes.

It needs some functional calculus.

Okay, that is quite Horrible®.

The heat semigroup behaves very nicely, but its generator sucks.

Well, it is also quite nice.

But too little diffusion and too little potentials.

At least the OU has some diffusion! When I'm done with this I'll give it some more potential!

Applied analysis?

Then what is pure analysis?

Hmm...

So anything that has nothing to do with differential operators? Okay.

The heat semigroup is just called the heat semigroup because it is the vector valued ODE version of the heat equation which has the Laplacian.

@JasperLoy what about "wave"?

@robjohn Hmm, that can contain heat too!

My advisor just said that I'm good at calculating things. I'm not sure if that is a good thing or not, so are calculators 8-).

I have been computing kernels for about two months, so at least it can be said that I have some stamina 8-).

Hmm, because I can compute... kernels?

I saw a very cool big name (in a few years 8-)). Hytönen!

That guy has a very big box full of tricks, and he knows very well how to use them.

They are not so huge, but then I need to look up the Unicode for the smaller 8.

Then what are no glasses?

Just -)?

That's more like: "I'm blind and smiling!".

@robjohn Hmm, given any null set $A$ can we construct an $L^2$ function such that its Fourier series converges on $\mathbf R \setminus A$?

Related to this, I feel like asking a strange question.

Is not the term "for almost all x" rather vague @JonasTeuwen ?

@MarkDominus There was already a case when a used did not want get the bounty which was awarded to him. But it seems that mod's can do much about something like that: Is it possible to remove the bounty on this question?

@JonasTeuwen You mean so that $A$ is its set of divergence?

Because any nice function converges on $\mathbb{R}\setminus A$.

and do you mean $\mathbb{T}\setminus A$ or do you mean Fourier Transform?

@JasperLoy Thanks for the clarification =)

@JonasTeuwen Take a look at the last line in this section

@JonasTeuwen I used Katnelson's book when I learned Fourier Series with Krantz.

@JasperLoy Indeed.

@robjohn You mean the remark about Katznelson? (The rest I know of course...)

Huh, really?

@JasperLoy Yes.

@robjohn Well, yes, that is the result I think. But does it only diverge on those points or can it also diverge elsewhere?

I should look it up!

Because this would answer my question above :-). If you can make the thing even continuous...

Those books are lovely. And be careful, he was robjohn's advisor at UCLA 8-).

@JasperLoy :-p

@JonasTeuwen mentor, really. Undergrads didn't officially have advisors.

@robjohn Oh right, my mistake.

@JasperLoy Exactly when did you sell your soul? :-)

I found a mistake in my calculation. Bloody monkey, I keep making mistakes! 8-).

Mein gut!

He writes backwards

@N3buchadnezzar Bloody monkey! That is not so hard by the way.

@JonasTeuwen Not hard? It`s bloody impossible I tell you!

8-).

:/

@N3buchadnezzar By the way "a.e." or "a.s." (almost surely) is not "vague". A function is for example "almost everywhere" $1$ if the set where the function is not $1$ is a null set, here that would mean that given that set $A$ and any $\varepsilon > 0$ we could find a countable sequence of intervals $[a_i, b_i]$ such that their union covers $A$ and the sum of their lengths is smaller than $\varepsilon$.

So it is pretty damn small. But it still can be pretty damn big in cardinality! It can happen that you can biject it with $\mathbf R$!

But for us, integrators, a null set is pretty damn small.

But here you specifies where the function is not $1$.

If someone says to me: This walkway has almost no holes, and the holes it has is pretty small." I still would say that it is a vague statement, even more so if the walkway was over a pit of deadly snakes. Knowing where the pitfalls is essential.

No it is not.

If the holes are very small they will not harm you, using your analogy.

A snake would have positive measure, I would say! 8-).

@robjohn I supplied another solution to our dilemma.

See here

@JonasTeuwen I invite you to see it, too.

I'll do it when I'm home. Need to do some groceries now, it is 8:12.

Bye guys!

Who designs Wikipedia sites like this?

@N3buchadnezzar Awesome people.

@robjohn nuts. i should have seen it's bounded by the tangent line. thanks robjohn!

@Eugene That was stated in the question, wasn't it ?

@robjohn it was! all the more reason i should've seen it... I was thinking of the slope but didn't think of the line equation for some reason...

@robjohn thanks again!

@Eugene np :-)

@Eugene I have this thing to prove: $$(ah,bk)=\left(\frac{a}{(a,b)},\frac{k}{(h,k)} \right) \left(\frac{b}{(a,b)},\frac{h}{(h,k)} \right)$$

Should it be hard?

@JasperLoy I bet he parts his hair on the left :-)

@PeterTamaroff looks more tedious than difficult

@robjohn this is really funny

@Eugene I know! Maybe I can use $(a,bc)=1 \Rightarrow (a,b)=1 ,(a,c)=1$ right? :P

Or something of the sort.

@robjohn The only hard thing would be mirror writing, but then it is cool

@PeterTamaroff well you don't have that here.

@PeterTamaroff as I said, he probably parts his hair on the left.

@robjohn What does "parts his hair on the left" mean?

@PeterTamaroff that is, the video is flipped.

@robjohn Oh! LOL

@robjohn Did you see my new solution?

@Jordan Jordan, I study in a public university, and I'm in first year of college. I'd like you to stop with that nonesensical ranting, it is tiring me. We're not here to be your counselors and eat up your tears. I enjoy helping anyone in math as long as they compromise. I have helped you many times, but, seeing you talk like that about me (and Benjamin) I'll have to tell you: 1. I won't help you any more, unless you change your attitude. 2. I'll flag any subsequent comments of that nature.

@PeterTamaroff Algebra is first year of college math :P

@Eugene I forgot something there.

It is $$\left( {ah,bk} \right) = \left( {a,b} \right)\left( {h,k} \right)\left( {\frac{a}{{\left( {a,b} \right)}},\frac{k}{{\left( {h,k} \right)}}} \right)\left( {\frac{b}{{\left( {a,b} \right)}},\frac{h}{{\left( {h,k} \right)}}} \right)$$

Hmm

@PeterTamaroff Doesn't taking the Laplace Transform of certain functions use complex integration?

I am doing some simple statistics and got $Z = -0.144$ what should I look up in the table?

@robjohn In the one I used, no. $\varphi$ is to be real.

@PeterTamaroff also you use the integral of the sinc function, which is usually gotten by complex methods.

@robjohn Sinc? Where?

@PeterTamaroff I am not saying that $\varphi$ is complex, but that to compute that Laplace Transform, complex methods are used

@robjohn Not always, rob. In this case, the improper integral is found by usual integraton.

@PeterTamaroff You cited $\int_0^\infty\frac{\sin(x)}{x}\mathrm{d}x=\frac{\pi}{2}$

@robjohn That's not true. What are you looking at?

I'm using only $$\int\limits_0^\infty {\frac{{dx}}{{{s^2} + {x^2}}}} = \frac{\pi }{{2s}}$$

@PeterTamaroff why all the unnecessary parenthesis?

@robjohn Oh. Well, complex integration is not the only way to obtain that. Anyways, I'm not claiming I'm using only real results.

@N3buchadnezzar MathType.

@PeterTamaroff Sigh..

@robjohn Multiply the integral by $\exp(-sx)$ and the denote the integral by $I(s)$, now evaluate $I'(s)...$

@PeterTamaroff Okay, I thought you said you were using a non-complex method to get the result.

You surely know the drill.

@N3buchadnezzar Yeah.

@robjohn But bear in mind that integral is not exclusively evaluated with complex tools.

Am I helping here?

You would have to justify the change of integral and differentiation though.

@PeterTamaroff you still don't have coprimality

@Eugene Right, right. Maybe I should then use this:

$$\left( {\frac{a}{{\left( {a,h} \right)}},\frac{h}{{\left( {a,h} \right)}}} \right) = 1$$

Or something of the sort. I don't really know.

Mathtype. It burns!

@N3buchadnezzar I renders OK. Don't be a Michael Hardy!

@robjohn I am unable to obtain the DeLeeuw, Katznelson and the other guy their paper about those null sets :-(. Are you?

@PeterTamaroff I am not saying there is anything wrong with your solution, I am just looking at how much machinery is being used to get it.

@robjohn Yeah, I like to avoid big ammo when it is not necessary, too.

Are residues fairly simple? I know they relato to improper integrals over $\mathbb R$

And the contour integral is equal to the sum of the residues enclosed by the path, something like that right?

@PeterTamaroff Once you learn contour integrals, you never go back :-)

@JonasTeuwen I see the result is also proven in Katznelson "An Introduction to Harmonic Analysis" (Chapter II, 3.4). I don't see anything about being restricted to the set.

@robjohn Ah, thanks!

Should find that book.

@JonasTeuwen I think it is available online, but I guess not for free.

@robjohn 8-). Got it!

@JonasTeuwen where?

Our library.

@JonasTeuwen Ah! :-) okay

I need to figure out precisely which semigroups have integral kernels.

And understand why it can be possible that they don't!

@robjohn Hmm... II.3.4 introduces my favorite spaces!

3.. 2.. 1.. Go! The first one to the answering location wins!

@N3buchadnezzar there are children watching, and they may be hungry

@robjohn The eyes, they see.

@robjohn A homogeneous Banach space is a Banach space isomorphic to each of its infinite dimensional subspaces right?

@robjohn know any good movies? ;)

So... for example $C(X)$.

@Eugene Carnage was good.

@PeterTamaroff enough with that movie!

@Eugene Mystic River

The Tree of Life

Catch me if You Can

Tarzan (Disney's)

Thanks Robjohn, I edited and updated =)

tarzan? seriously?

huh. i guess question was more stupid than i thought

@Eugene Yeah, it kicks ass. And WALL-E

@PeterTamaroff Wall E was baller

@N3buchadnezzar What does baller mean?

@PeterTamaroff Kick-ass

@robjohn But the result does not hold for $L^2$ does it? If I recall correctly we have that for a Hilbert space $H$ that there is no complemented subspace of the bounded operators on $H$ such that this is isomorphic to $c_0$. So for $B(H) = L^2$ we have that $C_0$ the functions vanishing at infinity are not isomorphic to a complemented subspace of $L^2$. So...?

@N3buchadnezzar Oh, sure.

@PeterTamaroff i watched tarzan....

@Eugene Then I don't see what you don't think it is absolutely awesome.

WHY DOES TARZAN HAVE LONG HAIR BUT NO BEARD?

@PeterTamaroff lol

IS HE INDIAN???

there's one

@N3buchadnezzar He shaves, silly.

@PeterTamaroff With what? monkey doo doos?

@N3buchadnezzar With fire.

@PeterTamaroff ....

@N3buchadnezzar it is spooky how close Rahul and my answers are, and our comments about the reflection of light.

@PeterTamaroff I am just saying, Tarzan would be more baller with a kickass beard.

@N3buchadnezzar maybe hahah

@robjohn This question was actually taken from a recent exam, I am sort of sketching out solutions for the youngsters.

@robjohn Can't Snells Law be used?

@PeterTamaroff Clever!

@PeterTamaroff snell's law is for refraction

@robjohn Well, isn't there an analogous for reflection? :P (I thought it could be adapted)

@PeterTamaroff the property that I commented about equal angles of incidence and reflection

You can derive it from Maxwell's equations! 8-).

The difference between the reflection and refraction laws is that when you reflect you're generally reflecting into the same material...

Snell's law and the reflection rule of equal angles both derive from the same principle of minimization, though. (My favorite way of seeing this for Snell's law is to imagine yourself at the beach, trying to get from a point some distance up the shoreline to a point a ways out in the water; since you run faster than you swim, it's in your best interests to put more of your distance on land and 'bend' your straight path at the shoreline)

@StevenStadnicki That's true, they are both minimization of distance, but they can also be derived independently using Maxwell, which just proves that math and the universe work well together :-)

i'm thinking of watching the big lebowski so i was wondering if anyone here had heard about it

@Eugene I have the disc, but I have not watched it. We started watching it, but my son, who was too young at the time, was in the room.

@robjohn Ah! I've got the result. The theorem above theorem 3.5?

@JonasTeuwen That sounds about right.

@robjohn ah. is it a movie that sounds appealing to you however?

@robjohn "Every set of measure zero is a set of divergence for $C(\mathbf T)$"?

@robjohn There must be a typo in the proof... What are the $l_n$? (not $I_n$). Or you don't have the book near?

@Eugene It is supposed to be good.

@robjohn cool. i'll take your word for it and go for it then. i wasn't sure about it but i heard things similar to what your are saying as well.