@r9m I asked my former teacher by mail for the book, i'll give you the answer

@IceBoy Ask @Chris'ssis, she fight titans. Da real math warrior

@TheGame Why not just look in [blah]?

I looked i some similar website, but didn't find it

@BalarkaSen en.bookfi.org/s/…

en.bookfi.org/s/?q=Tenenbaum+number+theory&t=0

blah

@TheGame There is no book on earth that's not in bookfi

you just need the right keywords.

:P

@TheGame lol. In a way you're right, I mean I have the easiest solution to the Au-Yeung series in the world. :-)

@Chris'ssis >:c Y u make we want your book so much

mean

Hope to be published soon in one of the papers it was sent to ...

Monthly rejected it ...

AMM requires rather general articles.

@Chris'ssis :O

@Chris'ssis Why ?

@TheGame For the reason mentioned by Balaraka.

The wretched name

Ah :/

@Chris'ssis Send me da paper >:c

@TheGame I can't do it now since it's going to be published in other magazine. I did it together with another mathematician.

Ah I see

Then we will see your name ? :P

@TheGame I don't think so.

why not?

:D

@TheGame :D

Tell me when it's out anyway @Chris'ssis

you have to publish your real name in a paper, @Chris'ssis

@TheGame OK

@BalarkaSen Yeah, I know.

we'll get you at last evilgrin

@BalarkaSen Well, there's an exception for Dark Va... oops I said it

flees

huh, @TheGame?

@BalarkaSen hahahahaha :D

Or should it be voldemo...arghhh

he published a paper?

On resurrection

where is la paper?

Harry burnt it

usenix.org/legacy/event/fast12/tech/full_papers/Sumbaly.pdf

^ Not that xD

There is a sad things though, I don't manage to find someone like me here (where I live), with my ambitions, the willing to work in the research area of integrals, series and limits, I mean someone 100% dedicated to such things.

@Chris'ssis That's the same nearly everywhere

Actually, no one of my friends understands anything of my work. Where I live no one knows anything of what I'm doing.

she has no friends

aah don't bash smack me

@Chris'ssis That's the same in France

@TheGame Ted smacks, not bash

@TheGame I mean they don't understand the stuff I like.

That's a horrible thing ...

Chocolate ?

@TheGame Puleeeze

French are like super-genius mathematicians

@BalarkaSen That doesn't mean that every single guy is doing maths

That would be soooo far from the realty

It's true that we have some geniuses, but that's as far as it goes.

There aren't a lot of people who do maths

Then kick the heck out of them from the country.

As anywhere else in the world

@BalarkaSen -__-

We, the mathematicians, are the 99% .00000001%

Harsh reality

You are gonna make me cry, @Hippa

I've got feelings, you know

snif

HAHAHAHAHAHAHAHAHA

Such maths

I believe this: I think every person that knows nothing about integrals, series and limits and that by some magic thing would be able to suddenly see the beauty of this stuff I wanna say all of them would be SHOCKED by the beauty of mathematics.

They would need medical care, really! Imagine you enter the paradise (suddenly) ...

@TheGame HAHAHAHHHhAHAH

@Chris'ssis The paradise of $1+2+3+...=-1/12$? :P

@TheGame LOL

Back in 20 min.

@TheGame oh ! thank you very much !!

away

@Balarka! @Thegame!

@TheGame Such math. Much physics. Very Casimir.

@Studentmath \o

How goes it?

Not bad. Wataboutyou?

Okay, doing some elementary number theory before my modern algebra course

Coolio.

Is there any nice way to prove 5,11,17,23... has infinie number of primes and infinite number of non-primes?

Let me know if you get stuck on anything.

@Studentmath What's the sequence?

Thanks @Balarka

+6 everytime

starting at 5

ah. i am not good at tracking patterns.

so they're like $5 + 6k$.

Yeah

@Studentmath evoke dirichlet's theorem. end of the game :P

non primes are obvious

Oh gee. Let me try to see if I have that theorem

take $k$ to be a multiple of $5$.

actually you don't need dirichlet @Studentmath :P

Yeah I figured a nice way just now

@Studentmath Euclid-like, I presume?

Yep

OK, what is your way?

I know all numbers are 5 modulo 6

And it rather immediately follows from there

@Studentmath huh?

Wait, let me formulate it well and I will explain

proof by intimidation is not a proof, man

ITS TRUE OK?

assume there are finitely many of them. if there are even number of them, multiply and add 4, otherwise multiply and subtract 2. you have a number 5 mod 6 in both cases.

now note that every 5 mod 6 number must be divisible by a 5 mod 6 prime.

but this number you have constructed is not divisible by any of the primes you have assumed to be there, thus it must be divisible by a prime not any of them, a contradiction.

@Studentmath That is le proof, not "immediately follows from there".

oops no that is flawed. consider the primes $p_1, \cdots, p_n$ of the form 5 mod 6, and consider $6p_1 p_2 p_3 \cdots p_n + 5$. that should do.

@Studentmath you may say thank you to me now.

@Balarka that's awesome.

elementary and patently obvious

That's the kind of way I was looking for

Would you expect me to explain why a 5 mod 6 number is only divisible by a 5 mod 6 prime?

Heck why not, I'll be thorough

I never said that

A 5 mod 6 number is divisible by at least one 5 mod 6 prime

think about it a bit. i have to go and eat.

oh and also, mull over the difficulty of in general primes in arithmetic progressions by thinking about 4 mod 7 numbers. are there infinitely many such primes?

I'd say no

Spoilers; there are

First of all, I wanna say I appreciate very much any book that is specialized in treating integrals, series and limits, and that means I also like very much the book you may see below.

amazon.com/…

However, I wanna show you a very interesting challenge from the author's side you might like to take.

I finished that integral in less than 5 minutes. Does it seem hard to you?

@Chris'ssis I thinbk I can do it in 5

I treated a whole class of such integrals

@N3buchadnezzar Great. Let me know when you're done.

I'm working on a second solution.

@Chris'ssis Splitt in half

use $x \mapsto 4 - x$ on the second half

Done. (in this proof I proceeded as the author)

Let me try a 3rd proof ...

@Chris'ssis complex

@Chris'ssis hmm .. does not easy :|

10 mins and I'm still thinking about it.

@r9m It is very easy (honestly).

I've staring at it for 10+ mins now .. I got nothing :(

back

Don't tell the way ;) (trying!)

OK :-)

One important hint: the ways are obvious.

I guess then the Prob is that everyone is thinking too hard...

@Tharindu I'm done in a 3rd way.

@Chris'ssis Do you like geometry ?

@TheGame Yeah, but after I do a bit practice since I didn't do for a while.

I'll post that for anyone who wanna help me :)

I'm awful at geometry

@TheGame I'll take it when I return back to this area (+1)

:)

Im getting ln (0) as a term :/ , other than that it' looks simple . How can this be easy :/

Any math student should meet this integral at least once in life ... MY ALL AWE TO THE INTEGRAL

@TheGame can you explain the two phrases 'symmetric of a triangle's vertexes by the opposite side' and 'aligned' .. I'm not familiar with them

@r9m It may be a bad translation from the French, correct me if the grammar/vocab is wrong

@r9m Basically take one vertex

Construct its symmetric by the opposite side

You get a point

Do that for all 3 vertexes

And see when they are aligned

Ok wow @Chris, Now I feel like begging you for the answer

@TheGame Construct its symmetric by the opposite side .. thats where I'm confused .. what does that mean

It's too nice to tell you the answer. Just try some more. :-)

Ya well ;) don't tell yet :)

not here please .. not while I'm staring at the chat screen (begs like a rabbit)

Yep

@r9m The reflection of a point on a line ?

@TheGame aha .. now it makes perfect sense .. thank you :)

@r9m Look at A,A' in the second picture

@r9m Edit it so that it looks more english if you want

@TheGame: Formulated/posted a nice way to think about the approach I gave to your problem. Hope you like it!

@Semiclassical i'll look at it when I have some time :)

@Chris's sis , Nope still gets ln (0).

@Tharindu The answer is $0$.

Ok now how do we do this :/

@Tharindu Don't give up so easily, here you begin to acquire the skills for other integrals.

Ok lemme give another try ;) since u emphasized its 0 ;)

Yeap.

@Tharindu She's lying, the answer is $\sum_{k=1}^\infty k+\frac{1}{12}$ :P

@TheGame lol, there would a nice thing to have a movie posted on youtube with the students that see that stuff you posted above for the first time. I mean a video with their faces when looking at that ... :-)))))))))))))))))))))))))

@Chris's sis , tell me if I'm close ;) Can I remove the integral :p coz its zero , like differentate both sides of the formula that we have to show? (Sorry if this sounds stupid)

@Chris'ssis Your face -> :-)))))))))))))))))))))))))) , their face -> >:cccccccccccccc

@TheGame hahahaahahahahaha :-)

thats a stack of chins .. chins.stackexchange :Pj

@TheGame by he way, did you know that a Romanian took the Nobel prize for chemistry this year?

@Chris'ssis $$\sum_{k=0}^\infty \frac1{x^3+1}$$

@Chris'ssis >:c How did he manage to steal it ? I thought those places were well guarded ...

Uh no

Stefan W. Hell, German citizen. Born 1962 in Arad, Romania. Ph.D. 1990 from the University of Heidelberg, Germany. Director at the Max Planck Institute for Biophysical Chemistry, Göttingen, and Division head at the German Cancer Research Center, Heidelberg, Germany.

@Chris'ssis What did he find/work on ?

nobelprize.org/nobel_prizes/chemistry/laureates/2014/press.html

Chemicals

@Chris'ssis Have you seen that sum before?

One might say $$1/486 (-27 \zeta(3)+153 \zeta(5)+90 \pi^2-2 \pi^4)$$

@Alizter Noo

@Alizter use digamma

@TheGame His work is about optical microscopy with resolution at the nanometer scale. I cannot tell you the details in English.

Which is barely an approximation

That was found with mathematica

@Alizter $\dfrac{1}3\left(\gamma+\sqrt[3]{-1}\psi{-\sqrt[3]{-1}}-(-1)^{2/3}\psi{(-1)^{2/3‌​}}\right)$

@TheGame eh how?

I GIVE UP :(

@Alizter Yes, I saw it. There is some kind of generalization on MSE.

@Chris'ssis oh really?

let me see

@BalarkaSen looks like rationality is on steak here =P

SOMEONE TELL ME THE SOLUTION PLEASEEEEEEE

THE PROOOOFFFF PLEASEEEEEEE

nooooooooooooooo.com

PLEASEEEEEE I NEED TO SLEEP :p

@Tharindu what is the problem?

@Alizter, Ok wait lemme see if I can attach it here

@Chris'ssis I cannot find the post on MSE

@Alizter WAIT

you don't grow until you learn to take the days problems to the bed ;)

$$\int_0^1 \frac{\log(x)}{\sqrt{4x-x^2}}\ dx$$

show that $$\int_{0}^{1} \frac {ln x}{\sqrt{4x- x^2 }} dx = 0$$

@Alizter HE HAS THE "BELIVIOUS PEOPELIUS FARAWAIUS LOUDIUS SPEAKIIUS" AS WELL (speaking loudly, because you believe people are far away)

@Tharindu What have you tried

@Tharindu I know how to solve

Show :) Ok wait ummmm one part of my mind is saying keep struggling at this, while the other side jus wants to know it.....well I tried many things, and the hardest thing for my Brain is that it seems to be the easiest thing (according to Chris's Sis)

I don't what to do :/ Ok clues ? :D

Don't kno*

Wot to do in the sense, read the proof by asking you :) or struggle harder /longer

@r9m , hahah :D ;)

;) :P

Ok wait a minute did you @r9m figure it out? :) Coz I remember you were staring at the screen with me

not yet ..

Btw where are you people from :) (I mean country & univ)

@Tharindu Romania (financial accounting, no background in math)

lies lies lies .. I don't believe her :|

You gotta be kidding me, you did some 3 methods in 15 mins

@r9m lol, no lie :-)

And a reputation of 13.6k , with no math background ? O.o

swallows the bitter truth pill with a glass of water

@Tharindu lol, this is a very small reputation. Look at @robjohn here.

$\LARGE 125,856$

phew .. done !! ^^ (peacefully goes back to texing assignment)

Not small enough to believe what you said :p

@r9m Did you work on my inequality?

I will guess. your a Professor in Mathematics

@Chris'ssis I will .. give me some more time (there's going to be an end semester exam within 5 days .. X_x) .. and I need to read (my mid-sem performance was poor :( .. )

hmm. anyone recognize this binomial coefficient identity? $$\prod_{k=1}^{n-1} k^{2k-n}=\prod_{k=1}^{n-1}\binom{n-1}{k}$$

@Tharindu me? lol, no way. I'm not good enough to be a professor.

@r9m , whats your field?

@Tharindu BSc math :) (undergrad)

Ok then a phD student.....

@Tharindu No

@r9m , Coool :) guessed it ;)

@Chris's sis , hmmm

Trying my last and final method I could think of :)

@thegame: BTW, if that binomial identity i asked about above is true, then my answer agrees entirely with Yulia's

Almost finished with that problem

I get $\displaystyle 2\int_{-\pi/6}^{3\pi/2}\log(1+\sin \theta)d\theta$

kinda bored with it now

Ok I am Done...Please tell me the answer :)

@Chris's sis, please tell how to prove :)

@Tharindu I can't take people's pleasure of finishing the problem. :-)

That is $$ \int_0^1 \frac{\log(x)}{\sqrt{4x-x^2}}\ dx$$

Best part of it is , it took only 5 mins for you to solve and it's very simple -.- and so does the image describe it simple

@Tharindu Chris'sis has had much much practice in this

@Alizter , do you know Chris'sis? :)

WAIT

@Tharindu over this chat yes. IRL no

Hello @MikeMiller

It is $$\int_0^4 \frac{\log(x)}{\sqrt{4x-x^2}}\ dx$$

Wot? :o

The question is changed? :o

@Tharindu No, I only put the wrong uper integration limit. Now it's like in the picture you saw above.

log?

But you changed ln x to log x...the problem for me was the ln becoming ln (0)

Ok doesn't make much of a difference :p sorry :)

@Tharindu mathworld.wolfram.com/NaturalLogarithm.html

@Chris'ssis did you do it woporpap ? :D

@r9m What the heck does it mean? :-) "woporpap"

@Chris'ssis ah !! you forgot ? .. without pen or paper .. ofcourse :P

@Chris's sis, what's up wit the link :p

@r9m It can be done without pen and paper. :-)

@Tharindu For natural logarithm we can use bot $\ln(x)$ and $\log(x)$ notations. Mathematicians usually use $\log(x)$.

@Chris's sis, yes I understood that :)

Now can someone solve :/

Your a mathematician ;)

@Chris'ssis certainly seemed so after I wrote it down on my cupboard =) but I couldn't do it until I used the chalk :)

@Tharindu What's your first thought on it?

@r9m lolll :-)

@Chris's sis, my brain is empty :o

@Tharindu How about letting the variable change $x=4y$? See what happens.

whaaa ? .. just a simple one line trig sub solves it in a pair of lines

@r9m :-)

@Chris'ssis do you know if the book (the one you linked to earlier) is available on the internet ? :-)

@r9m Initially I saw the possibility of using beta function maybe combined with $\lim_{s\to0} \frac{x^s-1}{s}=\log(x)$

@r9m Check gen.lib.rus.ec first.

@Chris'ssis oo !! sweet !!!!

@r9m Then when you try to solve $\int_0^1 \frac{1}{\sqrt{x-x^2}} \ dx$ you see what else you can do.

(geometric interpretation, beta function, Euler substitutions and so on)

@Chris'ssis yes !! .. indeed :)

@r9m I mean the problem pushes you in the right direction.

@r9m I don't know if it's available on some site. I have received a couple of pictures with those.

I get $\frac{log 4+log y}{\sqrt{y-y^2}}$ , how do I continue? :)

@JasperLoy SWEET !!! thanks !! got it !! :D

awesome site man !!

@r9m Did you find it?

Haha got it

@Chris'ssis yes :) jasper's linked site had it .. the download speed is awfully slow .. its downloading as we speak !! :-)

Wait but this is longer than 5 mins. I timed myself

@Chris's sis :))))

@Tharindu Maybe I moved myself a bit faster. ;)

@r9m I can't load the page ...

@Chris'ssis libgen.org/get?md5=dd3891c740af26fb79ab93e5eb7ec95f

Not working either,...wait a min is this that book @Chriss mentioned at start?

Ok wait I guess it's my net...it's loading :)

Thanks for the link

@Chris's sis, thank you for the great problem and helping us solve

It was really nice talking to you guys....gtg peace :) wil meet again over here ;)

@Tharindu Welcome ;)

ak

@Chris'ssis still cant find it

@Alizter to find what?

$$\sum_x \frac1{x^n+1}$$

@Alizter Ah ...

@Chris'ssis :(

I discovered a new representation of the harmonic number

@r9m ^^

hmmm, I wonder how to make it useful, like applying it in some problems ...

@robjohn are you there?

@Chris'ssis I am now...

@Chris'ssis What is that?

@Chris'ssis Are you trying to kill us ??

@TheGame Why? :-)

@Chris'ssis Too beautiful >:c

Mindblown

@TheGame :D

Hi, some easy question in a notes that I've read I find the following: show that if we have X and Y two topological spaces s.t. X it has the trivial topology. Then a function is continuous iff is constant. One inclusion is completely trivial but I think that without more structure in Y the other inclusion is not true. For example let X=Y=\{0,1\} both with the trivial topology, and f the identity map. Thus f is continuous but is not constant.

Am I right?

Thanks

Someone?

I think the notes there is a mistake in the notes...

@JoseAntonio If $Y$ also has the trivial topology, every map $f\colon X \to Y$ is continuous. If $Y$ is $T_1$, only the constant maps $X\to Y$ are continuous.

:18063099 I don't think so

@DanielFischer: Since there is not more structure in $Y$, therefore is a mistake, right?

@JoseAntonio Yes, there can be non-constant continuous $f\colon X\to Y$ if we impose no restrictions on $Y$.

@DanielFischer So it suffices to assume the extra hypothesis that $Y$ has to be $T_1$. Thanks

Now everything makes sense

@JoseAntonio Actually, $T_0$ suffices.

So @DanielFischer asaf has quite a large group of followers.

he should become a mod, and use his powers for good

@IceBoy Umm, followers? He's not the head of a sect, is he?

@DanielFischer A MSE sect :)

@IceBoy And how do you know how many followers he has?

@DanielFischer I don't know how many he has, but what I have seen on the threads shows that people do quickly join his side.

He has skill in that sense.

@IceBoy He could also just choose the reasonable side most of the time.

@DanielFischer True.

@DanielFischer In the same manner, in other part in the notes this says if $(X,T)$ and $Y$ both has card $\ge 2$ and $Y$ is Hausdorff and every function is continuous then $T=P(x)$. I think that is suffices fo assume that $Y$ is $T_0$ using the same idea that before

@JoseAntonio If every function $f\colon X\to Y$ is continuous, it suffices that $Y$ has one open set that is neither $\varnothing$ nor $Y$. Say $\varnothing \neq U \neq Y$. Let $u\in U$ and $v\in Y\setminus U$. Pick a subset $A\subset X$, and let $f_A(x) = u$ if $x\in A$, and $f_A(x) = v$ if $x\notin A$. Then $A = f_A^{-1}(U)$ is open.

Indeed I prove showing that the points are open but the argument is exactly the same just. Thanks for the generalization. Thanks.