now i got $0$

$(\beta^2-\alpha^2)x^2/6$ absorbs $\frac{1}{x}$

@Frank I didn't know about this kind of stuff, and also the theorem you have mentioned

Is there any book or references on the internet?

@Frank : Thanks for the help and patience

It's important in calculus.

And O-manipulation doesn't seem so welcome in calculus. You can find another book, say Concrete Mathematics, but I think it's better for you to learn calculus fisrt, where that book is based.

@RajeshD The limit and derivation is one-sided. I don't know the conventional notation for the one-sided one.

ok

I'll make an attempt

Maybe $\lim_{x\to x_0^+}F(x)$ is for right-sided limit.

But I don't know for the right-sided derivative.

How come I didn't find this theorem in many books I've browsed

I don't know. I'm reading a book which has not been translated into English.

The mean-value theorem might be useful to prove such stuff. I personally consider that it's more easy to understand by limit of a function in terms of sequences.

any hints? I am proceeding via limits

@Frank

Eh?

Given that $y=x\phi(z)+\psi(z)$

and $x\phi'(z)+\psi'(z)\neq0$.

Supposing that $\phi(z)$, $\psi(z)$ is well-behaved.

and $z$ is the function of $x$,$y$. Prove that $$\frac{\partial^2z}{\partial x^2}\cdot\left(\frac{\partial z}{\partial y}\right)^2-2\cdot\frac{\partial z}{\partial x}\cdot\frac{\partial z}{\partial y}\cdot\frac{\partial^2 z}{\partial x\partial y}+\frac{\partial^2 z}{\partial y^2}\cdot\left(\frac{\partial z}{\partial x}\right)^2=0$$

Is there any good idea? The pure calculation seems horrible.

Well, I'll post the question.

I need a sequence of continuous functions $f_n$ so that $f_n\to \chi_{\Bbb Q\cap[0,1]}$

@FrankScience don't see any other way

@robjohn are you?

@leo am I what?

@robjohn Can you help me?

That is a sequence that converges to the Dirichlet function. All the functions are continuous. The problem is that this sequence depends on two index

Take a diagonal.

@ZhenLin And it still converge to ?

Provided your first limit converges in the sense $(m, n) \longrightarrow (\infty, \infty)$ rather than as an iterated limit.

it does

Thanks @ZhenLin

Have you seen this limit before?

There is no reference in all the places I found it

It's probably considered obvious to the people in the know...

Hi

@Ilya oh, I've done it!

In Wikipedia there are some things with "Baire" about that

@leo Maybe this answer at MO might be useful.

@MartinSleziak thanks for the anbswer :-)

According to that link: Baire proved that a function defined on $\mathbb R$ is of Baire class 1 iff it is continuous everywhere except, possibly, for a meagre set.

So we cannot get $\chi_{\mathbb Q}$ which is discontinuous everywhere.

this doesn't work in wolfram alpha limit (cos(n! pi x))^(2n) as n->infinity

If you want reference for that double limit, see Dunham: The Calculus Gallery, p.197

@MartinSleziak thanks. You helped me a lot!

William Dunham did.

You're welcome!

Now wolfram give me This: Removed["$$Failure"] /. $Aborted[]

BTW Dunham's book is given as a reference in the Wikipedia article on Dirichlet function.

ha ha

something is wrong today!

Hello :) How's this for moving to maths?

@MartinSleziak I think that the convergence is uniform

I'll take some breakfast

@leo Uniform convergence preserves continuity. Uniform limit theorem

Hello, may I ask? I don't have enough reputation to comment (50) so I'll try to ask here. I found this question quite interesting, but I have just one newbie question: What exactly do they mean there by addition and subtraction of sets? I can't figure out what $A - A$ means.. :-/

In the Minkowski sense?

Uhh, can you be more specific? That math question looked pretty elementary, well suited for someone who wants to contribute but has not much advanced knowledge of math. :-) But I may be wrong. Thanks

@Jeyekomon It must be that $A - A = \{a - b : a, b \in A\}$. I admit that this is confusing.

But if it meant set subtraction then surely we'd always have $A - A = \varnothing$.

So for A = {1, 7, 8} we have A + A = {2, 8, 9, 14, 15, 16} ?? OK, now I understand. Thank you.

there should be 3 more values there I think

oh no, of course not. my mistake

@MartinSleziak I just realized that

@Jeyekomon Looks good.

Our government lies so much - they said they'd build 400.000 houses in 4 years. Either they are morons, or they think we are.

Ahoy

Soup ?

@N3buchadnezzar yohA!

@leo Mind having a look at something simple? The simplicity has blinded me..

@N3buchadnezzar :-)

why do you say that?

I have a glider i that goes from 0 to 8 + g*y(E)

So I can let i be any number between 0 and 8 + g*y(E)

Now I need to move a point H from (a,y(E)) to (a,0) in the time from 8 to g*y(D)

Glider? I was thinking about wingsuit jumping today.

But I have problems how to define H

I was thinking

@N3buchadnezzar I hope it didn't hurt too much.

H(a , y(D) - ( i - 8 )/g )

But this failed =(

@JonasTeuwen :p

9r 023t79å1e08y9u+¨vn304cmr9p8xo!§!!

GEOGEBRA IS UNABLE TO SAVE MY FILE, I SPENT TWO HOURS FIXING IT

@N3buchadnezzar What the...? Wai no TikZ?

@N3buchadnezzar now that stinks!

i feel for ya

Hi folks

@leo The Dirichlet function? really?

Yes. this is the reference

It's pretty interesting

@robjohn I think that in terms of descriptive set theory the null sets in Carleson can be at most $\Sigma_3^0$.

@OldJohn hi!

Yes, Carleson's theorem.

I mean, the Hunt part is almost for free.

@leo The limit as $n\to\infty$ is $1$ at all integer multiples of $\frac{2}{m!}$ and the limit doesn't exist for odd multiples of $\frac{1}{m!}$. So the limit is $1$ for every rational number and $0$ for every irrational number

Hytönen and Lacey already have like the most general version.

@JonasTeuwen what is $\Sigma_3^0$?

@robjohn Hmm, the complexity level of your Borel set :-).

@JonasTeuwen And is it proved that it can't be generalized?

Hello gentlemen. I have a question about gaussian fit that you may be able to help me out with.

@leo Uh,... not that I am aware of. That is my initial observation after reading some definitions and checking things out.

@JasperLoy Deep stuff bro.

Great. Shall we write about the names with the most J's in it?

Like some combinatorical problem with some analytic touch.

Ah, you're ambitious. I like that. Shall we also raft down the Niagara falls?

So the graph that I want to fit a gaussian to is shifted up (as in it doesn't start at zero). Now, I have a python function that does the gaussian fit, however it needs the data to be around zero. So if I try to add a constant to account for the upward shift, the python function is all messed up. (Sorry if this is a programming question, but I want to make sure initially that I'm not doing some math mistake).

How do you go about mathematically to account for the upward shift in a gaussian function?

I'm stuck with this one

The thing is that it is not allowed to use dominated convergence theorem or any Lebesgue stuff

:-|

By the above conversation, I was trying to use $\cos^{2n}(n! \pi x)$

@robjohn $10^4$

@JonasTeuwen any idea?

@leo Yes. There is a LDC for Riemann integrals as well. Stupid restriction btw.

It is like: Please dig a hole, but use your hands and not this shovel I have here.

Hi again :-)

@JasperLoy Similar to what I did - got a good chunk of my PhD by proving results that he missed in one of his own papers :-)

Sorry to disappoint you :-)

@JasperLoy Like which result?

@JonasTeuwen This question can't be right.

@PeterTamaroff It is so wrong it is right!

@JonasTeuwen It is trolling us!

That integral is ill defined for any $x\neq 0$

@PeterTamaroff I don't see why.

@JonasTeuwen I was confused. Don't pay attention to the above! =D

I find it very amusing that I have no votes on the answer to this question and Marvis has 11, yet my answer was accepted. I am glad that the OP read all the answers :-)

@JonasTeuwen this one?

@leo Hmm, seems like it.

@JonasTeuwen but then we need to proof 1. here

@PeterTamaroff with a cryptic answer, it would be nice to have a reference pointer :-) I have no idea to what $10^4$ refers.

@JasperLoy I'd like to think that mine was the best answer (and I do :-)

and that's where I'm having trouble

@robjohn My rep. Now $10^4+\epsilon$

@JasperLoy Okay, I am confused. Why does the J have anything to do with the best answer?

@robjohn What if Jesus was a mathematician?

Alright I guess, just started paratmetric equations and polar coordinates and series in class so I need to study

@JasperLoy Ah... but if you're talking nicknames, robjohn starts with an 'r'

@JasperLoy Sort of like when I answer a question? ;-)

@robjohn I was referencing my rep!

THe other day you wrote some power which was close to yours.

Thanks

@JasperLoy It stands for the $30$ in $10030$

@JasperLoy I participate daily...

@JasperLoy Oh, I learned English already. I sat for 5 cambridge IGCSE's

@PeterTamaroff I noticed yesterday that you had broked 10K.

@PeterTamaroff Mine was $13^4$ exactly for a while.

@robjohn broken =) But it happened today.

@PeterTamaroff about 2 and a half hours ago, when one of your answers was accepted.

@robjohn In what timezone are you?

@JasperLoy I had actually been waiting for it for about two weeks :-)

@PeterTamaroff It was today, but someone else went to 10K yesterday.

@robjohn Oh!

@JasperLoy What are interesting questions from ELU?

@JasperLoy No, I had been at $166^2=27556$ recently and noted that $169^2=13^4$ was also $1\pmod{5}$ so I could get that.

@JasperLoy I can quit any time I want. I will quit when my rep gets higher than Arturo's :-)

@JasperLoy I do? Perhaps a numerologist.

@robjohn Read the short story "S as in Sebatinsky" from Isaac Asimov

!

@JasperLoy so now I'm a nutcase?

@JasperLoy Über Science Fiction.

@JasperLoy It is a good time to start!

You'll enjoy it!

<< Certified >>

I am now 29K+1!

@JasperLoy but $29001=3\cdot7\cdot1381$ is composite

@JasperLoy It's clearly a multiple of 1381! :-)

@JasperLoy Yes, that is why I commented it was composite in the first place.

@robjohn can you help me?

@JasperLoy I'll do that then...

@leo I don't know, but I can try.

the problem is that it is a kind of homework

@leo tag it as such.

@leo What do you need help with?

@PeterTamaroff this

1. in there

since the other follows

$\chi_{\mathbb{Q}\cap[0,1]}$ is not Riemann integrable.

@robjohn but that do not implies that can't be a sequence so that $|f_n-\chi_{\Bbb Q\cap [0,1]}|^2$ is

I mean, there is functions $f$ so that $f$ isn't Riemann integrable but $f^2$ is

@leo I don't know if you considered that ${\Bbb Q \cap [0,1]}$ is countable.

Basically $\chi_{\Bbb Q \cap [0,1]}$ is zero almost everywhere, isn't it?

Do you know any theory to make use of that?

@PeterTamaroff Yes I did. The problem is that the corresponding indicatriz function is discontinuous everywhere. In the context of Riemann integral there is no much to do

I think I get it, but it is a lot of work...

@leo Roll up your sleeves then!!!

@PeterTamaroff :-)

lol

"Sin sacrificio no hay victoria."

eso es muy cierto

El problema es que a veces es frustrante no poder dar argumentos bellos o directos. Es decir, uno sabe que puede resolverlo pero se pregunta si es la mejor forma de hacerlo.

@leo Siempre se pueden pulir las impurezas... pero con algo hay que empezar. Creo que los argumentos directos salen después con la experiencia.

@PeterTamaroff Yep. :-)

@leo What do you have in mind, for the moment?