Mathematics - 2014-11-09 (page 1 of 4)

12:07 AM

If an increasing function is the addition of a positive function and its square, does that mean that the positive function is increasing too?

Mike Miller

It's true of the increasing function is differentiable; probably true in general

UserX

Can I prove it?

Mike Miller

In general, or for differentiable functions? It's pretty easy for differentiable ones.

Daniel Fischer

@UserX Suppose $x < y$ and $f(x) > f(y) \geqslant 0$. What can you then say about $f(x) + f(x)^2$ in relation to $f(y) + f(y)^2$?

Mike Miller

how silly of me

Alizter

12:21 AM

Hello all!

UserX

Got a new prob

Jasper Loy

@Alizter Still not sleeping?

UserX

But it seems kinda trivial, although it's placed last(on an increasing difficulty manner). Let $f(x)>0\forall x\in\Bbb R$ be a polymomial with real coefficients, show that there exist polynomials A(x),B(x) such that $f(x)=A^2(x)+B^2(x)$.

Oh wait squares. That just deleted my "trivial" part. How should I approach this?

Alizter

@JasperLoy I could but eh. I have to finish my physics but that is likely not to be done.

Mike Miller

12:38 AM

No idea, @UserX. Maybe solve some special cases and see if you can find an algorithm in general from these.

Like do it for $x^4+2x^3+2x^2+1$.

UserX

@MikeMiller I can't. How will I square a polynomial and get that cubed term?

Mike Miller

Dunno. But it satisfies the conditions of your problem.

UserX

Yea...

Mike Miller

So solve that one first and see if it gives you any ideas :)

UserX

It doesn't satisfy the conditions

Mike Miller

12:48 AM

Yes it does.

UserX

Mike Miller

You must have typed it in wrong.

UserX

Ah yea

Then I got no idea how this could ever work

Rafflesia arnoldii

@G.T.R The users who favorited that were 133569, 166405, 189024, 187230, 139265, 189040. Two of them are named Thierno M. SOW... (Posting this just for the record; the post itself is deleted, and the voting data will go out of SEDE shortly)

I used the simple query Who favorited this post? <-- may be useful for future occasions like this.

Mike Miller

1:04 AM

@UserX I've found a way to write it as a sum of two squares, but it's quite ugly. Looks to be a hard problem.

UserX

@MikeMiller I'm on my way. I got A(x)=x^2+x+c

Mike Miller

Right, that's one of the terms. You'll have to solve for $c$ and it's not so pretty.

UserX

Trying to find a fitting c to cancel the linear part from squaring B(x)

What did you find?

Alizter

I feel better now that I have seen bad questions on PhSE. One of them was titles "If there is no gravity on the moon how is the american flag waving".

8

Mike Miller

It's $(x^2+x+a)^2 + (bx+c)^2$ where $b = -\sqrt{1-2a}$, $c=\sqrt{1-a^2}$, and $a$ is the root of $2x^3-2x^2-2x+1$ that's approximately $0.4$.

Pedro Tamaroff

1:11 AM

@MikeMiller Dude.

UserX

How do I prove this for all polynomials lol

Pedro Tamaroff

Is there any deep reason as to why the residues of a function in all the Riemann sphere sum up to cero?

@Alizter CHECKMATE ATHEISTS.

UserX

The difficulty after understanding the problem highrocketed

Mike Miller

@UserX No idea.

UserX

Should I post it on MSE?

Mike Miller

1:12 AM

Up to you

Pedro Tamaroff

@UserX What are you having problems with?

Mike Miller

It doesn't make sense to talk about the residue of a function on the sphere, @Pedro.

Pedro Tamaroff

${\rm res}\,(f,\infty)=-{\rm res}\,(f(z^{-1})z^{-2},0)$.

I am saying that for a meromorphic function $\sum {\rm res}\,(f,c)=0$ where $c\in \Bbb C^*$.

Mike Miller

How are you defining residue?

Pedro Tamaroff

The $-1$ term of the Laurent expansion.

Mike Miller

1:16 AM

That's going to depend on the chart you choose around $\infty$. Your definition above is coherent but that's because you've chosen a specific chart rather than define residue in a way that's invariant under picking different charts.

Alizter

@PedroTamaroff Such elegance.

Pedro Tamaroff

@MikeMiller Well, the residue should be invariant under charts since it is defined as an integral.

Mike Miller

You don't integrate functions on a Riemann surface, you integrate 1-forms :)

So if you want to do this you should be talking about why the sums of residues of 1-forms are 0.

Pedro Tamaroff

@MikeMiller It's the same fucking thing Mike, don't play smart on me.

=D

Mike Miller

No, it's not.

Pedro Tamaroff

1:18 AM

Yes, it is.

Alizter

Fight fight fight!

UserX

@Pedro math.stackexchange.com/questions/1012733/…

Mike Miller

Good luck using Stokes' theorem with functions.

Rafflesia arnoldii

don't say homework question.. or something like that, just drop an answer please. — avDec25 6 mins ago

Alizter

@Rafflesiaarnoldii Answer as "No."

Pedro Tamaroff

1:19 AM

@MikeMiller A $1$ form $\omega =f dz$ corresponds to the function $f$. We can go back and forth. You're being pedantic.

Alizter

@Rafflesiaarnoldii I commented accordingly.

0

Q: Square root in $\mathbb{Z}_p{}^*$

Given a prime $q$, and another prime $p$ = 20q + 1, I am able to find generators in $\mathbb{Z}_p$. Does $-1$ have a square root in $\mathbb{Z}_p{}^*$? Thanks!

UserX

I love it when two graduate students fight and I don't understand anything apart from some words

Rafflesia arnoldii

OP deleted the comment, sadly.

Mike Miller

@PedroTamaroff You wanted a deep reason, and in fact if you define the residue on a general Riemann surface, where 1-forms and functions are not in bijection, it's still true. Because of Stokes' theorem.

Alizter

@Rafflesiaarnoldii I guess a mod might have fun.

Pedro Tamaroff

1:21 AM

@MikeMiller OK, hit me.

Mike Miller

Google it, it's been written down before.

Pedro Tamaroff

@MikeMiller You're particularly grumpy today.

Alizter

this is a stupid place, if you can give an answer for such a question then do it.. if i will ask this question in some other way, then one of you will provide answer.. example: as a computer program.. there is no meaning like this — avDec25 25 secs ago

UserX

Link me alizter

I gotta drop my "MSE is a privilege, not a right" line

Alizter

@UserX click on the 25 secs ago

Mike Miller

1:25 AM

You were right when you said your definition was chart-independent, @Pedro. Sorry about that.

Alizter

why should one waste time in writing all of that, this will make the answer more specific to just one person. asking like this way, may attract answers with different approaches. — avDec25 27 secs ago

This user is fun and constructive to talk to.

UserX

Isn't that self-contradictory?

@MikeMiller I guess this problem is quite interesting but hard too. It reached 6 upvotes and not a single comment.

Pedro Tamaroff

@MikeMiller How are your classes going so far?

Mike Miller

They're good, @Pedro, mostly.

Pedro Tamaroff

@MikeMiller That's nice.

Mike Miller

1:31 AM

Not a fan of one of them but I'm just reading other books on the subject on the side instead.

Pedro Tamaroff

@MikeMiller Cool. I do that often. One of my goals is to be able to read Stanley's "Commutative Algebra and Combinatorics." It's Chapter 0 is basically all my university's undergrad courses.

Mike Miller

lol

Pedro Tamaroff

Yeah, I'm totally serious. If you can, take a look.

Mike Miller

I saw a paper with the title. Glanced at it.

Pedro Tamaroff

@MikeMiller Saw this?. Super cute result.

UserX

1:33 AM

Can someone explain me how the close-reopen system works?

Mike Miller

A book I like that you might is "Classical topology and combinatorial group theory"

Yeah I've seen it, @Pedro. Nice question but I'll never upvote anything that adjoins elements to a module.

Pedro Tamaroff

@MikeMiller Let me look.

Alizter

@UserX Ask @Rafflesiaarnoldii

Pedro Tamaroff

@MikeMiller LOL.

UserX

@Rafflesiaarnoldii how does the close-reopen system work? If I choose "close" for example, will it close or I'll have voted it to close and if all the voters reach a certain vote count and end up in a consensus it closes?

What's the algorithm behind it?

Mike Miller

1:35 AM

If there are 5 close votes, a question is put on hold.

Alizter

That is equivalent of the question walking the plank.

Then after some time it jumps.

Reopening is sending a rope down and pull up the drowning question.

Mike Miller

Well, sometimes the questioneers edit their question and the pirates let them walk back.

Alizter

Yar. They be keepin' extra booty.

Kevin Driscoll

Only when that question requests 'parlay'

Alizter

Booty = Context. In most cases.

Kevin Driscoll

1:37 AM

'parley' I mean

Mike Miller

Sometimes some stupid jerks think the guy who got thrown off the plank didn't deserve it and throw him a rope without him editing anything, but that's mutiny, arrr, and they'll be thrown off next!

Kaj Hansen

@Alizter, I did give it to marko

Pedro Tamaroff

@KajHansen lolled at your a) b) proof.

Mike Miller

Hey, Kevin's back!

Alizter

@KajHansen Yar. Matey cough yes I saw. I think it was a good choice.

Pedro Tamaroff

1:37 AM

@KajHansen I have a problem for you, while you're with analysis.

Kaj Hansen

Glad I could provide some laughs @PedroTamaroff :P

Alizter

@PedroTamaroff I read that originally as "I have a problem with you".

Kaj Hansen

Maybe I can solve it? hahaha

Mike Miller

Did you find the proof via Stokes, @Pedro?

Pedro Tamaroff

@MikeMiller I did not google it.

You broke my heart, Mike.

There's a cab waiting out front.

Mike Miller

1:38 AM

Yeah but you broke mine first.

Pedro Tamaroff

Don't make a big fuzz out of it.

Mike Miller

So you deserved it.

Pedro Tamaroff

</3

@KajHansen You should.

Kevin Driscoll

@Mike Howdy. I do come and go a bit.

Alizter

Hi @robjohn

Pedro Tamaroff

1:40 AM

Let $M$ be a metric space with the property every sequence in $M$ has a convergent subsequence, equivalently (prove it?) every infinite subset of $M$ has an accumulation point in $M$. Then given any open cover $\mathcal O$ of $M$ there is a $\varepsilon >0$ such that for any $x\in M$ there is $O\in \mathcal O$ for which $B(x,\varepsilon)\subseteq O$.

Kevin Driscoll

@Alizter About bad questions over on Physics.... there are some real crackpots from time to time as well. They want to come onto Physics.SE and tell us all how their new theory changes all of established physics.....

Pedro Tamaroff

@MikeMiller Take me back, Mike.

►

Mike Miller

@Pedro You'll have to become a topologist first.

Pedro Tamaroff

@MikeMiller Does algebra/topology/combinatorics do it?

Kaj Hansen

Interesting problem @PedroTamaroff. I'll think about it while getting dinner :D

Pedro Tamaroff

1:42 AM

@KajHansen Hint Attack it by contradiction.

I'll say no moar.

Kaj Hansen

Sure thing

Alizter

@KevinDriscoll We had another Riemann Hypothesis proof the other day. Cross posted on MO as well :P

@PedroTamaroff does contrapositive

Kevin Driscoll

@Alizter Ridiculous! At least people don't come around trying to tell you all of single-variable calculus is incorrect, though.

Alizter

You lectured me about that well last year.

Mike Miller

@PedroTamaroff Lern some algebraic and differential topology son.

Alizter

1:44 AM

@KevinDriscoll Occasionally Goldbach, RH and P=NP is proved at once.

UserX

How do I link in comments such that the link is hided behind something I type?

Alizter

[behind me](link)

?

Kevin Driscoll

@Alizter Doesn't impress me! You need to solve at least 4 seminal unsolved problems, so you can win all 4 Field's medals at the same time [/troll]

Alizter

@KevinDriscoll Why not go ahead and solve all the millennium prize problems while you are at it.

Mike Miller

@PedroTamaroff In my dream world you become an operator algebraist.

Kevin Driscoll

1:46 AM

@Alizter Good point. I could use 6 million dollars.

Alizter

@KevinDriscoll That way you can buy more hair gel for crackpot hair?

Kevin Driscoll

@Alizter It's kind of funny because the prizes are certainly not independent! If one could show how to obtain a mass gap from a Yang-Mills theory, they'd certianly win the Nobel prize in physics as well!

Mike Miller

@KevinDriscoll If you both prove and disprove P = NP you get 2 million.

Pedro Tamaroff

@MikeMiller What is that?

Mike Miller

@PedroTamaroff People who study $C^\ast$ algebras.

datalava

1:48 AM

@MikeMiller lol

Pedro Tamaroff

@MikeMiller I will, in time.

Alizter

@KevinDriscoll And a Nobel peace prize for ending crackpot claims.

Mike Miller

Which are like the lovechild of algebra and analysis.

People often study them using tools of algebraic topology.

It's like so you.

Alizter

Pedro is flattered.

Pedro Tamaroff

@MikeMiller Cool.

I am still bedazzled by combinatorics mixed with algebra and topology.

Mike Miller

1:49 AM

Do you know functional analysis yet?

Pedro Tamaroff

@MikeMiller No, I will take that in a year or so.

Kevin Driscoll

@Mike If i'm allowed to prove P=NP and P=/=NP I get a lot more than 2 million! Using reductio ad absurdum I also solve all 5 other Millenium problems, some of them in the affirmative and the negative. I could conceivably collect 12million!!

Pedro Tamaroff

Next semester I'll take real analysis, which is basically measure theory.

Alizter

@MikeMiller I know it is mistagged. A lot.

Mike Miller

Don't you already know measure theory?

Kevin Driscoll

1:51 AM

@Alizter I would look ridiculous with crackpot hair! I have shoulder-length hair but Im also starting to go bald. I think I'd look worse than the Ancient Aliens guy.

Rafflesia arnoldii

@UserX Also, see math.stackexchange.com/help/closed-questions

datalava

Is anyone who frequents here a linear algebraist

Mike Miller

I use linear algebra sometimes, does that count?

Kevin Driscoll

I let Mathematica do all my linear algebra. Perhaps ask Steven Wolfram?

datalava

@MikeMiller if you want it to.

Mike Miller

1:52 AM

@PedroTamaroff He wrote it oddly, it's not an odd degree polynomial.

Alizter

@KevinDriscoll If you have shoulder length hair could you not just wrap it over?

datalava

@MikeMiller I'm just asking because it's the field I'm most interested in

Pedro Tamaroff

@MikeMiller What does he mean?

Mike Miller

I don't know people that still do research in linear algebra...

@PedroTamaroff Read it more carefully.

$x^{90}+x^3+50$

Pedro Tamaroff

Oh, LOL

I misread.

Mike Miller

1:53 AM

You should just upvote my comment. :P

Kevin Driscoll

@Alizter I could try, but I feel like somehow the 'hair ball theorem' works against me.

Mike Miller

@KevinDriscoll Your head is not a sphere, so you do not need to worry about the hairy ball theorem.

Alizter

@MikeMiller How do you know? I mean homology and all... but that does not really exist right?

Mike Miller

Huh?

Alizter

Gets slap from MO

Kevin Driscoll

1:55 AM

@Datalava The only kind of research that goes on in Linear Algebra that I know of is what one might call applied linear algebra. Things like making ever more efficient solvers for large systems of equations. Some stuff with like integer programming. It's mostly computer science and engineering. not mathematics.

Mike Miller

I don't understand your joke.

Alizter

Too much reading math that I do not understantd

asdfijaf

Good night all

Pedro Tamaroff

@MikeMiller Google "The Jacobson Lemma."

That's lovely.

Kevin Driscoll

@Pedro You're in your 2nd year now? or 3rd? What math classes are you being forced to take?

Mike Miller

What do you want me to read, @Pedro?

Pedro Tamaroff

1:57 AM

@MikeMiller jankobracic.files.wordpress.com/2011/02/…

That's interesting linear algebra.

@KevinDriscoll Second year. Finally taking complex analysis, =)

Mike Miller

Looks like a pile of equations I do not want to read, so won't.

datalava

@KevinDriscoll people do research in matrices, fields, and other things related to linear algebra

Mike Miller

Pedro's in his second year and knows as much as I do.

Pedro Tamaroff

@MikeMiller OK. Prove the following. Suppose that $A,B$ are matrices, that $A$ is diagonalizable and $[A,[A,B]]=0$. Then $B$ is nilpotent.

I had that in one of my linear algebra midterms.

Kevin Driscoll

@Pedro Looks like Quantum Mechanics to me! Particularly related to the so-call Baker-Campbell-Hausdorff lemma

Pedro Tamaroff

1:58 AM

@MikeMiller Well, I don't know. Mike knows a lot of fancy pansy stuff.

Kevin Driscoll

@Pedro FINALLY! Now you can help me computer contour integrals. Thank the math lords.

Mike Miller

@KevinDriscoll Related to the BCH how, exactly? That it has brackets involved? :p

UserX

@MikeMiller you would be interested in seeing the answer

Pedro Tamaroff

Here, feed my ego.

datalava

@PedroTamaroff brackets meaning dual?

Kevin Driscoll

1:59 AM

@Mike If [A,[A,B]]=0 then Exp[A+B] = Exp[A]Exp[B]

and maybe there's a commutator in there I dont remember

Pedro Tamaroff

@datalava $[A,B]=AB-BA$.

@KevinDriscoll ORLY.

datalava

@PedroTamaroff oh

Pedro Tamaroff

Cool..

Mike Miller

@KevinDriscoll Not usually how it's stated. Or used. :P

Pedro Tamaroff

Whenever someone speaks of the "nullity" of a matrix, an elf kills a baby reindeer.

Kevin Driscoll

2:00 AM

@Mike it is in physics!

I don't know the actual pure math version of BCH

But I'd imagine B being nilpotent is somehow important

or I should say, B being nilpotent is perhaps sufficient to derive the usual manipulations that Ive used BCH for

Ah yes, I did drop a commutator! Exp(A+B)= Exp(A)Exp(B)Exp([A,B])

@Pedro You asked a question about an open cover earlier. An open cover is just a set of open sets which divide up a space into some (possibly-overlapping) pieces, no?

Pedro Tamaroff

@KevinDriscoll $\mathcal O$ is a colection of open sets that cover $M$: $M=\bigcup \mathcal O$.

Mike Miller

In math it's a statement about Lie algebras, that $e^xe^y = e^z$ for some infinite formal power series $z$ in $x$ and $y$ (where $xy := [x,y]$).

We rarely need the actual explicit formula, as far as I can tell.

We used it once (the explicit formula, I mean) so far in this Lie algebras class, to show that it converges, and we'll probably never use it again.

Kevin Driscoll

@Mike Ya, we use the explicit form somewhat frequently. Mostly for the easy case where [x,[x,y]]=0=[y,[x,y]]

Pedro Tamaroff

@MikeMiller The problem in my midterm actually was: $AB-BA=A$. If $B$ is diagonalizable, $A$ is nilpotent.

Mike Miller

meh/10

Kevin Driscoll

2:12 AM

@Pedro If B is diagonalizable it has an inverse, no?

WOW linear algebra was a long time ago....

Pedro Tamaroff

@KevinDriscoll $0$ is diagonalizable.

You want nonzero eigenvalues.

Kevin Driscoll

Ok, I wasnt sure if that counted as diagonalizable

Mike Miller

@datalava My impression is that if they exist there are not many of them.

Kevin Driscoll

In retrospect, it has to. Excluding those doesn;t makes sense.

Mike Miller

There aren't any here, at least.

On the other hand I doubt there's a single researcher whose work doesn't frequently use linear algebra.

datalava

2:23 AM

@MikeMiller Yes, I'm sure that's true.

@MikeMiller My current matrix professor has done a lot of research in matrices. I have found it very interesting

Pedro Tamaroff

@datalava You have a "matrix professor"? What is that?

datalava

@PedroTamaroff lol, Matrix Analysis professor

Pedro Tamaroff

@MikeMiller Prove that $\Bbb C\otimes_{\Bbb R}\Bbb C$ is not a domain.

Mike Miller

2:47 AM

@Pedro This is a corollary of a problem I posed to you a while back.

Pedro Tamaroff

@MikeMiller Oh?

What did it say?

Mike Miller

Indeed, if $L/K$ is a finite extension, $L \otimes_K L$ is not a domain.

It's a problem I like a lot.

"Special case", not corollary

Pedro Tamaroff

@MikeMiller Oh. I am not sure I proved that.

Wait, maybe we did?

Mike Miller

Indeed, call the latter ring $R$. It's a finite $K$-algebra hence Artinian. So if it's a domain its a field.

Pedro Tamaroff

Cool.

Mike Miller

2:50 AM

Now consider the map $R \to L$ given by multiplocation. If $R$ was a field it would be injective as it's clearly not 0. But $K$-algebra morphisms are in particular vector space morphisms and it's impossible for such a morphism to be injective for reasons of dimension.

Of course your specific case can be seen by showing that $(i \otimes 1-1\otimes i)(i \otimes 1+1\otimes i)=0$.

Pedro Tamaroff

@MikeMiller Yay Mike.

Mike Miller

@Pedro Do you know Haar measures?

@Kevin I forget, what area of physics are you in?

Kevin Driscoll

3:07 AM

@Mike Cold atom theory

@pedro @mike do either of you know what a Grassman algebra is?

Pedro Tamaroff

@KevinDriscoll Rings a bell, I could google it.

Do you mean exterior algebra? Google gives that.

Kevin Driscoll

Apparently they;re the same thing. I had no idea.

Mike Miller

@Kevin So none of the stuff that calls itself physics but is really math, then.

Kevin Driscoll

@Mike Like string theory? No not that.

Pedro Tamaroff

@KevinDriscoll What does "cold" mean there?

Kevin Driscoll

3:10 AM

@Pedro micro to nano Kelvins

Mike Miller

I'm slowly beginning to appreciate that stuff from a mathematical POV. Because it's really useful in math. (not so much in physics)

@Pedro Atoms that live in Canada.

Kevin Driscoll

@Pedro Basically the scale at which Quantum Mechanics becomes the dominant description

@Mike Have a specific example?

Pedro Tamaroff

@KevinDriscoll Cool.

Mike Miller

@Kevin More gauge theory than string theory in my case. I'm taking a gauge theory course next quarter.

That stuff is huge in low-dimensional topology nowadays

Kevin Driscoll

@Mike Ah yes. This stuff is pretty cool. I always wonder exactly how much math one needs to know to say that they understand it.

Mike Miller

3:18 AM

It seems like a physicist's understanding is different than a mathematician's. Probably you don't need to view it the same way we do.

Kevin Driscoll

@Mike I have to agree! I think the difference in perspective is at least part of the reason it takes 100 years for current mathematics to become current physics. Maybe we've knowcked a few decades off that estimate though, recently.

Mike Miller

Certainly the other way is hard too!

Definitely the communication is faster now, though.

Kevin Driscoll

UGH. It seems like $\int_a^b x f(x) dx$ and $\int_a^b f(x) dx$ should have some very simple relationship. But I just can't find one.

Mike Miller

3:34 AM

I don't see why they should. Take $f(x)=e^{-x^2}$. The former has a trivially obtainable expression and the latter is not expressible in terms of elementary functions.

Kevin Driscoll

Ya I know. It just seems weird to me. Because you're weighting the function everywhere by such a simple factor. It seems like the change in the area should be obvious. But of course it isn't.

Mike Miller

I was just trying to convince you to give up on finding one :)

Kevin Driscoll

@Mike the reason I care is that I'm working with a particular integral transform where the parseval/plancherel identity for this transform involves $\int_0^{\infty} x \lvert f(x) \rvert^2 dx$

but physically, the relevant quantity is $\int_0^{\infty} \lvert f(x) \rvert^2 dx$

Mike Miller

Gross.

Mike Miller

5:12 AM

Hi @Ted

Integrator

5:37 AM

Oops Nobody here! :(

r9m

@Integrator I'm here :P .. I see you are 14 and from India ! where are you from (which state ?)

Integrator

@r9m That's great!

@r9m MH!

r9m

great !! :)

The Artist

2

Q: Solve This Crazy Puzzle

Something found in this picture that made him believe that his wife was cheating on him? Solve ?

logic puzzle

It has two UPVOTES :o

@r9m ^^^ check that out :o we should all get together and start hitting downvote

r9m

@TheArtist LOL :P Rofl !! I'm upvoting that shit if you don't mind !! :P

The Artist

5:51 AM

@r9m yesterday I saw a question has 12 downvotes :o this is worst :o and it has two upvotes :o wow

r9m

@TheArtist what was the question with 12 downvotes ?

The Artist

@r9m can't remember now......seriously 4 upvotes? -.-

r9m

@TheArtist haha !! :P ... no matter how irrelevant and gross it is ... its damn funny :P

The Artist

@r9m :p did u actually find the object? :p

r9m

@TheArtist nope :P ... I followed chicharito's answer ... that was creepy funny :P lol

ah .. Q deleted !! lol sad ..

The Artist

6:03 AM

@r9m deleted? :p

YAY :D

@r9m have U read "interesting Integrals" ?

r9m

@TheArtist interesting integrals ? is that a book ? or is it some math.se link you are talking about ? :)

Nick

6:17 AM

Greetings math-amigos,

Does anyone have a clever way to factorize $10001$ ?

Mike Miller

@Nick I think the comment to my long division comment is the best you'll get

And I don't think it's very convincing :P

Nick

@MikeMiller $$10^4+1=100^2+1=100^2+1000+25-1024=100^2+2 \cdot 100 \cdot 5+5^2-2^{10}=(100+5)^2-(2^5)^2$$ — N. S. 1 hour ago

Mike Miller

Right.

Nick

Well that's clever.... for someone clever, I guess.

Jasper Loy

@mike I forgot, so did you study French or German in the end?

Mike Miller

6:21 AM

French.

@Nick I doubt someone would come up with that without already knowing the factorization.

Jasper Loy

@MikeMiller Did you use any books?

Mike Miller

A french to english dictionary.

The Artist

@r9m book

Jasper Loy

@MikeMiller I see. So you are basically doing translation only, lol.

Mike Miller

I can read French fairly well, not speak it.

r9m

6:23 AM

@TheArtist are you talking about Nahin's book .. Inside interesting integrals ? :D

Nick

@MikeMiller This form $10^{n \in \Bbb N} + 1$ is interesting to me because there are so many primes of the form. Right?

Nah, I'm just joshing, after $11$ and $101$, there aren't any more found and it's been checked upto $10^{16777216} + 1$ by some guy.

Mike Miller

It's conjectured that there are infinitely many repunit primes

(repunit meaning numbers consisting only of 1s)

Nick

You mean numbers of the form: $$\frac{10^{n \in\Bbb N} - 1}{9}$$ Cool, Repunit, that's a reputable name :D

The Artist

@r9m yes :D im reading it these days , fallen in love wit it

r9m

@TheArtist yes ,,, its a beautiful book :-) lots of fun integrals discussed !! :D

The Artist

6:30 AM

@r9m Yep :D

Mike Miller

@Nick "Repunit". Stands for "repeated unit".

Ah, you corrected it.

Nick

@MikeMiller Ah, a portmanteau. How quaint :D

Sawarnik

6:42 AM

Hi @Nick @r9m

You are spending too much time on chat these days? :P @Nick

Nick

@Sawarnik No, I'm multitasking to a greater extent. I feel like I'm getting more done if I talk to people while I do stuff.

Sawarnik

Great.me does it too.

so bye.

Nick

Hi - bye ; the modern conversation.

r9m

@MikeMiller cool factorization !! :-)

@Sawarnik you are on ignore until you are ready to something about those downvotes ! :(

Andreas Bonini

6:58 AM

hey I want to ask a really simple question that is almost impossible to google.. What does the | mean in P(A | B), where P is probability?

Integrator

@r9m math.stackexchange.com/questions/1011837/…

r9m

@Integrator Cool find !! thanks !! :D

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$Mathematics$ Mathematics