@pedro: that would just be $\frac{1}{(1-x^3)(1-x^4)(1-x^5)....}$ then

@MathyPerson Yessir.

Hmm, @Mathy. I thought the original question said that $P(x)/R(x)$ should be a polynomial. Now I'm confuzled.

@pedro: and that is .... R(x)?

@Balarka Go do every problem in chapter 1 of Hatcher.

Was it $R(x)/P(x)$ instead?

@TedShifrin it was r(x) / p(x)

@Mike: There are some hard problems in there.

You lied to me, @Mathy. Tut tut :D

@MikeMiller By the way, how do you think Hatcher compares to Bredon?

@TedShifrin perhaps a typo

@MikeMiller I have to read the damn thing from head to foot then.

BBL

There are some nontrivial problems in there which can only be approached from stuff written across the book.

@PedroTamaroff i believe r(x) / p(x) is (1-x)(1-x^2), then?

Aha.

Or $(1-x)^2(1+x)$.

sigh I still have only two upvotes on my veeeery looong answer on Gal(\bar Q/Q)

Oh, hush @Pedro.

A proof for what, @PedroTamaroff?

Evening, everybody.

@PedroTamaroff Thanks for the help!

hi @Huy

@Huy it's only 2 over here

It's about 4 am here.

@MathyPerson: 9pm here.

You healthy now, @Huy?

@TedShifrin: A lot more than I used to be. I think it's the vodka.

LOL

I did a lot of my grading at school, so I drank less than usual during final-exam grading :D

@TedShifrin: Don't worry. Just a nice pre-christmas dinner with my parents and I had a "Le Colonel" for dessert. That's lemon sorbet with vodka, in case you've never heard of it.

No, I hadn't heard of it. Sounds like a fun summer dessert ... I'd prefer a nice tarte for winter :)

@TedShifrin: It is very good for digestion, and I ate a lot, so I preferred that. :D

Point taken.

We have very nice apple pie with vanilla sauce, in winter, in restaurants.

Or apple tart, not sure.

I'm not much of a dessert expert.

pie is more American; tart is more European :P

@MikeMiller Congratulations on Red Baron. BTW Solstice is already available, any upvoted post or comment.

Oh, ok.

@Behaviour: Are you goading @Mike into yet more hats?

Hats are hats.

@Behaviour DANG.

Did you all know that it should be 'he got his just deserts' and not 'he got his just desserts'?

That's a suave hat.

I don't believe so, @Jasper. You don't get stranded in the Sahara.

@TedShifrin Which is why it's time to look up a dictionary and see for yourself!

Oh, and Bill Lumbergh! 5 answers until the end of UTC Saturday (again each needs an upvote; implicit quality requirement).

Interesting, @Jasper. Totally archaic remnant: Deserts here is the plural of desert, meaning "that which one deserves". "Desert" is now archaic and rarely used outside this phrase.

So it's coming from the French ...

<-- owes Jasper an apology.

How could I format a code?

I tried it like that code but with no success...

@evinda I think there is a code feature on the main site. Oh for math, maybe you could try using the quote feature for code.

@JasperLoy I meant here.. Isn't it possible?

@evinda In chat? I dunno.

indent
    four
    spaces
to get
    readable
    code

@evinda ^^

@DanielFischer So do we have to do it like that?

if (A[j]!=y*A[i]){
   p=position-1;
   l=position-2;
   while (p>=0 and l>=0 and abs(A[l])!=abs(A[p])*y){
             if (p!=l){
                 if (abs(A[l])<y*abs(A[p])) l--;
                 else p--;
             }
             else p--;
   }

What is a rational function?

@JasperLoy Hello.

@MathyPerson One that can be expressed as a ratio of two polynomials.

@evinda you do not want nested loops for $O(m \log m)$

@JasperLoy: Do you know what proof Chris'ssis was referring to earlier? I read something about Euler but I didn't find out more.

@evinda You are now at the stage where you should compile, fix any errors and warnings until you get a clean compilation, and then run the code to check if it works as it should.

@Huy I do not know.

@ccorn This is for the subarray where we have negative integers. We just look at it only if we haven't found two elements of which the quotient is equal to y from the array at which we have positive integers, right @DanielFischer ?

@evinda Right. Of course you could also do the negatives first.

@Behaviour I've got two up voted answers today. Busy Christmas shopping. I'm going to try to get my other three when I get home.

@DanielFischer And now we have to check the case when we have y<0, right?

@Ted I know there are some hard problems in there; I did them all when qual prepping.

Is the (number of partitions of 0 into an even number of parts) - (number of partitions of 0 into an odd number of parts) = 0?

@evinda Right. That case is different, but uses the same basic idea.

@MathyPerson No, it's $1$.

There's one partition of $0$, and that has an even number ($0$) of parts.

@DanielFischer So now do we have to have two indices, one at the subarray with the negative intergers and one at the subarray with positive integers, right?

@DanielFischer but isn't there only one part (0), so it would equal -1?

@evinda Right. And you need two such traversals (if you don't find one in the first). Why?

What does it mean for a $n \times n$-matrix to preserve a quadratic form? Is it simply that $A(q(x)) = g(x)$, where $g(x)$ is of the same form?

@MathyPerson No, the parts are by definition positive integers, $0$ can never be a part.

@AndrewThompson More likely $q(Ax) = q(x)$.

@Andrew, if the quadratic form is given by $x^\top Ax$, then it should be that your matrix $B$ satisfies $B^\top AB = A$.

@MathyPerson If we allowed $0$, every number would have infinitely many partitions, since we could always add more $0$-parts.

Which is equivalent to what @DanielF said.

Ah, yes, that does indeed make more sense. Thanks, Daniel and Ted.

@Behaviour Thanks, by the way. I'm very happy about my Baron answer; I think the original question was a good question, just hard to read.

@DanielFischer Because it can be that a number from the left subarray is at the numerator but it can also be that it is at the denominator, right?

@evinda Bingo!

Oooh, I just got the necromancy badge :P

I think I owe it all to @Mike.

What'd I do to you now?

You made me answer a question, for which I received the necromancy badge :P (more than 5 upvotes on a 60+-day old question, or something like that).

What was the question...?

the geometric approach to linear algebra question

Ah, right.

@DanielFischer If we then have A[p]<y*A[r] what index do we have to change? :/

@Mike, I see you never answered that $\int x^n f(x)\,dx$ question.

Not home yet.

But I gave an upvote to the comment that $L^1$ does not imply $L^2$.

hmm, how do we negate "implies"?

$\not\implies$

see what I mean, @Andrew? :P

The closest I can do on top of my head, although that looks ugly.

Yes.

well, @Mike, it's too late — someone gave what I believe is a correct answer.

I answered a cute linear algebra quesrion earlier: if $S$ is a finite subset of a (finite dimensional) vector space over an infinite field, there's a hyper plane that doesn't intersect $S$.

($0\not\in S$, of course)

That shouldn't be too tough.

That's precisely the proof I was going to write, @Ted.

It wasn't tough, but it was cute.

For someone who learned what a hyperplane was a couple of days ago, that looks fun.

No spoilers.

Well, don't look up my answer :)

@TedShifrin what was the top mark in your probability course?

there were a number of high grades, @skull.

Any one get 100%

288/320, 270/320, 270/300, 276/320, 265/300

No ... That would be expecting a lot.

I ended up with 11 A's and A-'s out of 29.

How many F's?

7 D's and F's

and two C-'s (which are equivalent to D's for the major)

Not bad.

I found the 9 non-passing grades very upsetting. Two were people who gave up on the course ages ago.

Don't see any obviously interesting problems, which is messy, given the time constraint.

Huh, @Mike?

Need to get three more answers within three hours, @Ted :)

well, stop goofing off and answer 'em.

Gotta find something worth answering.

While driving?

When else?

:O

@evinda That depends on which is the negative and which the positive element.

Where are y'all driving now, @Mike?

I'm home.

"Home"?

We didn't find anything whilst shopping.

ah, your family's home

@MikeMiller Too late anyway, already answered by PhoemueX.

It's where the heart is, @Ted

uh huh ... ok, I'm outta here for now.

Aye, @DanielF, so I heard. It's almost identical to what I was going to write.

Too late to shop now.

@MikeMiller, if $V$ is an $n$-dimensional vector space and $S \subset V$ as in the problem, I may still view a hyperplane as a coset of some $(n-1)$-dimensional subspace $K$, correct?

@MikeMiller Thought so, otherwise it wouldn't be too late.

@AndrewThompson Ah, I was lying. I should have just said $(n-1)$-dimensional subspace.

If you can do a hyperplane, you could just pick one so far out from the identity that it doesn't contain any of $S$.

Yes, or simply get a contradiction with finiteness.

(Which is equivalent, I suppose.)

(The next line on my paper now says "Mike was lying, the problem is a tad different.")

@DanielFischer What do we have to do if A[p] is the negative element? :/

Good, @AndrewThompson, since that's the truth.

@evinda Then $A[p-1] \leqslant A[p] < y\cdot A[r]$, so if we decrement $p$, we don't get nearer to equality. Hence?

@PedroTamaroff what is the first thing you are going to do if you become a mod?

Ban everyone, I assume.

Made it, Ma! Top of the world.

@DanielFischer So we have to increment r, right?

@skullpatrol I will ban Mike.

For life.

I knew this day was coming.

@evinda And what happens then?

Hey, wait your turn @MikeMiller I'm first.

@DanielFischer Then we will have A[p]<=y*A[r+1], right?

@evinda Maybe, maybe not. But from $y < 0$ and $A[r+1] \geqslant A[r]$, we can deduce something.

Btw don't rain on Chris'ssis's parade. It does look nice until the votes are counted @PedroTamaroff

@DanielFischer What can we deduce from that?

@evinda The relation between $y\cdot A[r]$ and $y\cdot A[r+1]$.

Seen any good questions, @Daniel? Decent? Legible?

@MikeMiller Not many.

Damned holidays.

@r9m you might like to know that I have a very nice proof to the Basel problem. I'd like to know if you find it in some papers. :-)

@DanielFischer Could you explain it further to me?

@evinda Which is larger?

@DanielFischer $y\cdot A[r]$ is larger, right?

hello everyone. would anyone like to help with a basic complex analysis problem?

Maybe if you post it on main :P

lol

@daOnlyBG people just askaway here

@daOnlyBG OK.

awesome, thanks

@evinda Yes. So decrementing $p$ decreases the left hand side, and incrementing $r$ decreases the right hand side. Then if we have $A[p] < y\cdot A[r]$, incrementing $r$ gives us a smaller right hand side and we have $A[p] ? y\cdot A[r+1]$. There are three possibilities then. We could have $A[p] = y\cdot A[r+1]$, and we're done. Or we could have $A[p] < y\cdot A[r+1]$, then we're closer to an equality, but not yet there. Increment $r$ again. Or we could have $y\cdot A[r+1] < A[p]$, then the

step was too large and we need to decrement $p$ next.

so I need to find a function $g(x,y)$ that is harmonic on {$1<x^2+y^2<9$} such that $g(x,y)=1$ when $x^2+y^2=1$ and $g(x,y)=8$ when $x^2+y^2=9$

@Mike Your selfish greed is beyond belief!

Believe it, @Ted

I know what the definition of a harmonic function is, of course

@daOnlyBG: Do you know some basic harmonic functions?

sure

any function of the form $a+bxy$, where $x,y$ are variables, is harmonic

also, any function of the form $ax^2+by^2$

Any that are radially symmetric?

that second one?

No, your latter example only works for certain $a$ and $b$

ah, okay

hmm

An idea please : math.stackexchange.com/questions/1075970

I'm not sure, then

Hey @Chris'ssis so are you going to email me?

There's a very important one, @daOnlyBG: $\ln(x^2+y^2)$

Is it mathematically sound to say that: $ F_{2n}=aF_{2n-2}+bF_{2n-4} = F_{2(n)} = aF_{2(n-1)}+bF_{2(n-2)}$ to try to relate it to $F_{k} = F_{k-1} + F_{k-2}$ (by the way, F is the Fibonacci numbers)

@TedShifrin thanks

Sure thing 😀

@DanielFischer Would it be the same way for n=1, or is that just special for n= 0?

what can ya do without that fella

Who're you mumbling at, @Mike?

You, re: $\ln |z|$

Ah, 2-D Gauss's law :)

How did you do that, @Ted?

Yes @Khallil

@KhallilBenyattou: Yes.

He's on his phone, @KhallilBenyattou

@KhallilBenyattou Yes, it is as simple as combining.

No, he's definitely not, @Mike

So I'd just say that $f$ and $g$ being injective and surjective $\implies g \circ f$ being bijective and that'd be the end of my proof, @DanielFischer?

iPad then

@MathyPerson Sorry, I don't understand what you're aiming at.

Waiting for the physics, @Khallil 😛

anyway if $ln(x^2+y^2)$ is harmonic,

@KhallilBenyattou Yes.

I probably need to find some a,b to make that work, no?

would I look for $ln(ax^2+by^2)$?

Yes, @daOnlyBG. Shouldn't be so hard.

That's not radially symmetric!

No, that'll mess it up ... Try the chain rule

Probably not harmonic either, but I deleted to be safe.

hm

@TedShifrin Try the chain rule to find the right harmonic function for the problem?? Too much work

You want to preserve radial symmetry, @daOnlyBG

No, @Mike, to see it messes it up.

ok, let me think a little more

You might be thinking too hard

nods

I decided to be blue again.

do I literally let $g(x,y)=ln(x^2+y^2)$, perhaps multiplied by some scalar $c$?

Partly, @daOnlyBG

Well, $c\ln |z|$ spits out $0$ for $|z| = 1$...

right

@Chris'ssis Basel problem ?! I have seen a couple of proofs :) !

if $g(x,y)=ln(x^2+y^2)=1$ when $x^2+y^2=1$, then we need $g(x,y)=ln(x^2+y^2)+1$, no?

Get ready for a good one @r9m :-)

@skullpatrol :-)

@Mike: You need to learn to be patient and give only slight hints.

(removed)

I didn't see his hint- I'm trying to figure it out. I know this shouldn't be too hard or anything, I just haven't done complex analysis in a while

Or you can be a popular teacher and give everything away and give all As 😁

but am I on the right track? am I essentially just looking for some $a \cdot ln(x^2+y^2)+b$?

@TedShifrin: You can also be a popular teacher not by giving everything away and all As, right?

Yes!!

Today's questions are crap.

@BalarkaSen: Where?

I wouldn't know, @Huy

I need something to answer, @BalarkaSen, and there's nothing there

@TedShifrin was that "Yes!!" directed toward me?

Yes!!

Yes @daOnlyB

No, @Khallil, there's nothing at all

@BalarkaSen Nah man ... there were a couple of good ones around noon ... rest of the day I didn't find anything interesting :(

@r9m Which noon? Local or UTC?

I only answered two of 'em @Mike

I've got two more UTC hours and two questions to answer. Didn't get any the first hour; someone just upvoted one I did earlier todya.

@DanielFischer Local ..

I need 5 for a hat.

Nothing at all? So all the questions aren't interesting, @Mike?

@BalarkaSen hmm .. my thoughts throughout the day :P

Hats are making this place worser and worser.

They're either not interesting or they're in fields I can't work in, @KhallilBenyattou

Just unignore me @Ted. It'd be a better place.

@TedShifrin No, it's the holiday season that's doing it for once :P

hi @blondblau

@TedShifrin Plusungood and plusungooder?

thank you very much @TedShifrin and @MikeMiller. I got it from here

Like what, @Mike?yes @DanielF

How many UTC hours are there before Sunday?

@Ted No questions because everyone is having too much fun

sure thing, @daOnly

Two, @Balarka

Holy crap.

:P lol

And I've yet got three questions to answer to.

@DanielFischer I meant: Does the ( number of partitions of 1 into an even number of parts ) - (number of partitions of 1 into an odd number of parts) = 1?

@MathyPerson Trivially yes, because there's only one partition of 1

You came to the right place @blondblau ;-)