Mathematics - 2015-05-21 (page 3 of 6)

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10:00 AM

@TobiasKildetoft but for that you already have to know what you are looking for .. I was talking about the problem columns of journals

Tobias Kildetoft

@r9m Ahh, I guess I don't really read those sort of journals, so I didn't consider that those were a special case

@TobiasKildetoft okay

@r9m I am a great supporter of copyleftists (!), but I guess it's not sensible to complain about why journals and stuff aren't free, etc. Where d'you think will the funding come from if they make them free? Copyleftists don't complain, they just apply the act and upload whatever they think would be worth uploading into the internet.

3

@BalarkaSen It's also true, but one can sell more copies at lower prices such that the funds preserves at a certain level. I just say it would be great if the prices were lower.

Saying is easier, doing isn't. AMS is a big association.

user147690

10:10 AM

So in saying $(x-1)$ is an ideal, you mean:

$$(x-1)(a_0+a_1x+a_2x^2+\cdots+a_nx^n)$$
I have any choice of real coefficients here as the ideal?
$$=-a_0+x(a_0-a_1)+x^2(a_1-a_2)+\cdots+x^n(a_{n-1}-a_n)+x^{n+1}a_n$$

Tobias Kildetoft

@AlexClark I can't quite see what that is supposed to reply to

user147690

@TobiasKildetoft It was just to Balarka in general. I was asking for ideals of $\Bbb R[x]$

$(x - 1)$ is, by definition, the set $\langle f(x)(x - 1) : f \in \Bbb R[x] \rangle$.

I don't know what you mean when you write that horrendous expression.

user147690

@BalarkaSen Just an arbitrary $n$ degree polynomial(multiplied by x-1)

You can check that the set I mention is in fact an ideal.

@AlexClark What do you want to show?

10:13 AM

@BalarkaSen I see your point ..

user147690

Hmmm What does an arbitrary element of the ideal $(x-1)$ look like?

iwriteonbananas

hey balarka and alex

user147690

Hey bananas

$f(x) (x - 1)$, where $f$ is any polynomial in $\Bbb R[x]$

i.e., a multiple of $x - 1$

hi @iwriteonbananas

Tobias Kildetoft

@AlexClark An alternative description is that it consists of all those polynomials in which $1$ is a root

user147690

10:14 AM

Hmmm yes, true, okay

This is actually a maximal ideal of $\Bbb R[x]$. Something which is not maximal, as you had asked for, would be $(x^2 + 1)$.

Tobias Kildetoft

(being the kernel of the map to $\mathbb{R}$ given by evaluation at $1$)

@Chris'ssis one way of generalizing the O. Furdui integral $\int_0^1 \log^2 (\sqrt{1+x} - \sqrt{1-x})\,dx$ is trying to compute $\int_0^1 \log^2 (\sqrt{1+x^k} - \sqrt{1-x^k})\,dx$ :)

user147690

@BalarkaSen How does one prove something is a maximal ideal?

@r9m It looks appealing! :-)

user147690

10:19 AM

Show all other ideals are within it?(Is this even true?)

Tobias Kildetoft

@AlexClark No, other way around (i.e. no larger ideals)

@Chris'ssis it shouldn't be :)

Tobias Kildetoft

But often it is easier to show that the quotient is a field

Equivalently, a better way is to show that the quotient ring is a field.

user147690

Hmmm I guess I have to learn quotient rings then

Tobias Kildetoft

10:20 AM

@AlexClark There is a name for rings with a proper ideal containing all other proper ideals. They are called local

@r9m lol, did you try it for some specific values? Let me check for $k=2$ ...

@iwriteonbananas doing anything interesting?

iwriteonbananas

is this the power series development of the complex logarithm about a point $z_0\in \Bbb{C}\setminus (-\infty,0]$? $$\sum_v \frac{(-1)^{v-1}}{v} z_0^{1-v}(z-z_0)^v$$

@BalarkaSen meh, did some diagram chasing today, and now i need to do some complex analysis

diagram chased what?

user147690

@TobiasKildetoft Thanks, I'll read up on these

Tobias Kildetoft

10:23 AM

@AlexClark You should probably wait till you have a better general background in rings

@Chris'ssis all is in my head now (just an idea) .. don't trust my instincts .. they are wrong most of the time :P

iwriteonbananas

five-lemma and another theorem whose name i dont know

trying to think of it (the name contained the word "sum")

user147690

@TobiasKildetoft Yep, I just add these things to my (ever growing) list of things I should learn

@r9m Note that $$\int_0^1 \log (\sqrt{1+x^2} - \sqrt{1-x^2})\,dx=\frac{\log (2)}{2}-\frac{\Gamma \left(\frac{1}{4}\right)^2}{4 \sqrt{2 \pi }}-1$$

3

:D

@iwriteonbananas five-lemma is a cool thing.

iwriteonbananas

10:24 AM

yeah, it is

@Chris'ssis hmm .. interesting :D very interesting indeed!!! :D

@r9m :D

iwriteonbananas

lol i love these 3-dimensional diagrams

user image

yeah, i saw that Hatcher edited his .pdf again to include those

iwriteonbananas

when was that edit?

10:27 AM

very recently (about 2-3 months ago), i think.

i don't know, since i used my hard copy for a long time

iwriteonbananas

i see

yeah, i got mine 3-4 months ago

Soham Chowdhury

@Chris'ssis @r9m what is $\int^0_1 \log^2(\sqrt{1+x}+\sqrt{1-x})$ equal to?

a very cool application of five lemma is equivalence of simplicial and singular homology

Soham Chowdhury

Something pretty?

iwriteonbananas

yeah, i skimmed that part, it looked cool

hatcher embeedded the five lemma into the proof of equivalence of simplicial and singular homology

10:29 AM

speaking of which, i have to read simplicial approximation theorem at some point : i heard that using that you can directly prove that simplicial homology is homotopy-invariant without proving the equivalence with singular homology

18

Q: An integral by O. Furdui $\int_0^1 \log^2(\sqrt{1+x}-\sqrt{1-x}) \ dx$

The following integral was proposed in a paper by O. Furdui, namely $$\int_0^1 \log^2(\sqrt{1+x}-\sqrt{1-x}) \ dx$$ and then the generalization $$\int_0^1 \log^2(\sqrt[k]{1+x}-\sqrt[k]{1-x}) \ dx$$ As regards the first integral, my approach was to combine the integration by parts and the va...

calculus real-analysis integration definite-integrals closed-form

i didn't read 2.C.

iwriteonbananas

ahh

that's pretty cool

so it sounded.

iwriteonbananas

what's the gist of simplicial approximation?

im not familiar w/ it

10:31 AM

you can even prove Mayer-Vietoris in simplicial homology (it's an exercise in 2.2) -- the hardest part of proving that it's really a homology theory is the homotopy invariance

iwriteonbananas

every homology theory has mayer vietoris

and i dont think it's particularly hard to prove that

@iwriteonbananas well, we want to prove that simplicial homology is really a homology theory :P

iwriteonbananas

yeah, that's hard :P

well, prove it if you don't think it's hard ;)

@iwriteonbananas me neither.

i haven't read that stuff at all.

iwriteonbananas

+1

10:36 AM

i need to know about lefschetz fixed point stuff though -- i've heard that riemann hyptohesis over finite fields is all about proving lefschetz fixed pt theorem for etale cohomology (i am not planning to read those : but knowing what it is in standard homology theory would be at least an infinitesimal progress :P)

iwriteonbananas

yikes

so you wanna prove riemann hypothesis

no, i absolute don't.

iwriteonbananas

haha

user147690

He isn't a crank

I just want the $1M :3

iwriteonbananas

10:38 AM

whew

i just want to read what's all this hubbub about "etale" stuff, but i guess that's far future

iwriteonbananas

i dont know the word "etale"

me neither.

iwriteonbananas

lol

user147690

I do

user147690

10:40 AM

It's a joyous story that is transmitted electronically

"eat alex"?

user147690

:'(

user147690

Pls no

Don't be sad, it won't hurt.. too much...

user147690

I have another complex assignment when I thought the last one was the last one...

10:41 AM

assignments are complex, yes.

Tobias Kildetoft

@BalarkaSen as far as I recall, etale means something like "spread out"

@TobiasKildetoft In French it means to spread

user147690

@BalarkaSen - complex analysis if you weren't just joking(can't tell haha)

If there exists an axis of symmetry, then the figure is axially symmetric. What is the English name for something which has center of symmetric. (Which is symmetric w.r.t point. I.e., rotational symmetric for angle $\pi$.)

Tobias Kildetoft

@Hippalectryon Ahh, close enough

Soham Chowdhury

10:44 AM

@Chris'ssis What's G?

iwriteonbananas

@SohamChowdhury it's a letter

@AlexClark i was.

Soham Chowdhury

@iwriteonbananas . . .

@SohamChowdhury $\sum\limits_{k=1}^{\infty} \frac{(-1)^{k-1}}{(2k-1)^2}$ called Catalan's constant

user147690

@SohamChowdhury The 7th letter specifically in English

Soham Chowdhury

10:46 AM

Woo.

@iwriteonbananas you should really read up the proof of excision theorem at some point, though. that's the only nontrivial thing among the axioms of homology.

@SohamChowdhury You can also meet it as $K$, $C$ with the same meaning. So, $3$ notations for Catalan constant.

iwriteonbananas

homotopy invariance wasnt trivial

that too, fair enough. but it's not as hard and cool.

Soham Chowdhury

Bananas, are you and Balarka talking about topology?

iwriteonbananas

10:51 AM

yeah, i will definitely read up on excision proof though

@SohamChowdhury yeah

Soham Chowdhury

oh

i am still not satisfied with my understanding of chain homotopies, though

there is more to it than what meets the eye.

@r9m I'm not sure if you got my point about Knuth's problem. In the left side you can also use another generating function with the same effect, the equality remains there. :-)

iwriteonbananas

do you mean chain maps?

they are natural transformations

no, chain homotopies between chain maps.

ok, now you're losing me.

10:53 AM

@Chris'ssis I don't understand .. what's the other generating function?

can you tell me about the tensor product of chain complexes, bananas?

iwriteonbananas

hmm im just realizing something

@r9m hehe, no worry about that, it's about the effect of log^2() that makes thing still working. :-)

let's move on to the algebraic topology room, why don't we?

iwriteonbananas

sure

10:56 AM

@Chris'ssis stop playing mischiefs on my weak and sensitive heart .. are trying to give the poor kid a heart attack or what?

@Chris'ssis aha! :) ofcourse

@r9m :D

@Chris'ssis :P there's nothing to keep secrecy about that ,. it's obvious :P

@r9m It's not about any secrecy. :-)

@Chris'ssis you just deleted the last comment -_-

@r9m Instincts ... :-)

10:58 AM

@Chris'ssis :P LOL

:D

What are you plotting >:o

@Chris'ssis btw what do you think about my solution to 11821?

@r9m I didn't manage to look at it entirely, I was trying to finish something. It is very nice as far as I can saw! :-)

@Chris'ssis :) the first solution is robjohn's :) The second is mine :)

11:04 AM

@r9m I think Faà di Bruno's formula might be avoided. :-)

@Chris'ssis maybe .. but it made life easier for me :P .. the other way involves stirling numbers of second kind and a nice combinatorial identity :-) I'll add that too but later.

@r9m Yeap. Anyway, it's a very nice way as I said. I'll look at details a bit later.

@r9m hehe, it's nice to see these problems smashed in many ways. :-)

I've gotta go now ... BBL :)

@r9m one thing - you can add your solution also on I&S if you're active there.

@r9m OK :-)

@Chris'ssis By the way, tell me when IMM publishes one of your solutions :-)

11:12 AM

@Hippalectryon IMM?

Uh not IMM. I never remember the names :c. What was it called ?

@Hippalectryon :D

@Hippalectryon AMM? ;)

Oh exactly

@Hippalectryon I think the solution I have now to Knuth's problem is hard to beat. :-)

Which is why I wanna see it when it comes out :P

11:14 AM

@Hippalectryon See here some approaches

math.stackexchange.com/questions/1148203/…

@Chris'ssis The last answer shows perseverance, to say the least

(Marko's)

Soham Chowdhury

^ Haha.

@Hippalectryon :-)

Soham Chowdhury

11:30 AM

Anyone here familiar with a little bit of category theory?

If we define a category where the set of morphisms $\text{Hom}(a,b) = \{(a,b)\}$ iff $a\sim b$, the identity is obviously just $1_a = (a,a)$.

@Chris'ssis already added a link to my blog :-)

@r9m Btw, what's your profile pic ? Alucard ?

@Hippalectryon yes

@r9m :D

Soham Chowdhury

What does it mean to say that "$1_a$ is an id with respect to composition"? I don't know of anything that wouldn't be painfully obvious.

@TobiasKildetoft @BalarkaSen, if you're online, please halp

11:35 AM

@r9m What do you mean? :-)

user147690

@SohamChowdhury It means the composition of functions acts like the identity: E.g. g(f(a))=a

@Chris'ssis I already put a link of my solution in the I&S page :)

@r9m nice blog

@r9m Nice :-)

Soham Chowdhury

@AlexClark I didn't get you. Composing any two functions gives me the id of $A$?

11:36 AM

@Hippalectryon yea I know (after all it's mine :P)

Soham Chowdhury

What's $A$?

user147690

@SohamChowdhury $f:A\to B$, $g:B\to A$ $g(f(a))=a$, $g\odot f: A\to A, g\odot f: a\mapsto a,g\odot f = Id_a$

@r9m by the way, some time ago here was I guy from India, very talented at integrals, series and limits, his username was Marvis.

@Chris'ssis Shivaram? :-)

@r9m Yeap.

user147690

11:38 AM

He is still on MSE

Soham Chowdhury

@AlexClark I think you didn't get the example properly. Wait.

user147690

Well he is back

@Chris'ssis I don't know him personally :)

user147690

9.5k 6 32

calculus integration limits real-analysis sequences-and-series elementary-number-theory number-theory

user147690

That's Marvis, he is back

11:39 AM

@AlexClark Yeah, I know, he answered a question of mine and gave him the bounty.

user147690

Oh okay sorry

user147690

Hey @Saw

@AlexClark I don't know if these days he is as active as he was in the past.

Soham Chowdhury

user image

user147690

@Chris'ssis He is the top rep as of recent

Soham Chowdhury

11:40 AM

user image

@AlexClark Indeed. I just noticed that.

Soham Chowdhury

@Alex, there you are.

@r9m a while I thought it was about you. :-)

user147690

@SohamChowdhury I haven't done essentially any category theory

user147690

What category are we working in?

Soham Chowdhury

11:42 AM

And @BalarkaSen is lost in his world of chain homotopies, and @TobiasKildetoft is away

Hi @Alex

@Chris'ssis ??

Soham Chowdhury

Look at the first picture, it defines the category.

user147690

@SohamChowdhury No it doesn't?

Hi @SohamChowdhury

11:43 AM

@r9m I mean I thought you were Marvis (in the past I mean).

Soham Chowdhury

"We can encode the data into a category"
He then proceeds to define the objects and morphisms of the category.

@Chris'ssis :P lol .. not even close :P That guy is a genius!!!

@Chris'ssis You also thought I was an expert :3

user147690

@SohamChowdhury Yes but we aren't in a category, like the category of groups(Grp,AbGrp,FinGrp) or the category of sets

@hipp Hey.

Soham Chowdhury

11:44 AM

$\text{Set}$

@SohamChowdhury Ellu

Soham Chowdhury

Hey, @Sawarnik

@Hippalectryon ;) I still believe that!

user147690

It says unlike in Set

@SohamChowdhury Did you give INMO, Soham?

11:45 AM

@SohamChowdhury If you define a category, you define a composition. This composition has to satisfy associativity and identity law.

Soham Chowdhury

I know. We're not working in Set. He defines a new category in the first pic.

@r9m never underestimate you :D

It means that for each object $A$ you must have some morphism $1_A\in \operatorname{Hom}(A,A)$ which behaves as an identity.

Soham Chowdhury

@Sawarnik Technical problems, my online form wasn't submitted, no RMO for me last year :'(

user147690

@SohamChowdhury It looks like he defines category in general

Soham Chowdhury

11:46 AM

@MartinSleziak So that $1_af = f1_a = f$ right?

Yes.

@SohamChowdhury :( .. didn't know you could fill online as well :/

@hipp

In your case composition is simply defined as $(b,c)\circ(a,b)=(a,c)$.

@r9m In my opinion, you're very talented too! I don't see such people every day.

Soham Chowdhury

You had to do both. Online *and* offline.
65-year-old prof insisted he was a Google Forms expert . . .

11:47 AM

Or $(a,b)\circ(b,c)=(a,c)$.

Depending on which convention (order of composition) is used in your text.

@Sawarnik ?

@Chris'ssis yea! I know! I have talent of messing things up :P

Soham Chowdhury

@MartinSleziak Yes. But what does the id "mean"? The equivalence relation is reflexive, so $(a,a) \bigcirc(a,b)=(a,b)$. Just that much?

@SohamChowdhury In any case, if for any two objects the set $\operatorname{Hom}(A,B)$ has at most one element, the identity law is in this case trivial.

@Hippalectryon he's just having fun typing 'hip' and it's extension 'hipp' :P

11:49 AM

hi, bit of a random question but has the reimmann hypothesis been proved yet?

@r9m hehe, it's not that bad making mistakes after all. From many mistakes I learned a lot of precious things, and I also did new discoveries. :-)

Yes. You need reflexivity to have at least one object in $\operatorname{Hom}(A,A)$.

@SohamChowdhury That's so odd :/

@r9m I've cut hands for less than that !

@r9m :D .. you could also say it is a de-extension of hippo :D

Soham Chowdhury

11:49 AM

@Sawarnik 65-year-old profs aren't always nice. "YOU SAY I NO KNOW HOW TO USE COMPUTER???" etc.

@Chris'ssis but I seldom learn from my mistakes :P .. I just go ahead and do them again and again :P

@r9m :D

heh.

Soham Chowdhury

@MartinSleziak thanks.

@SohamChowdhury By "id" you mean $1_a$?

Soham Chowdhury

11:51 AM

@Sawarnik RMO was soooo easy this time. :(

@MartinSleziak Yeah.

Ok, so I suppose your doubts are cleared Soham.

@SohamChowdhury Didn't you give before?

Soham Chowdhury

@MartinSleziak Yeah, thanks.

@Sawarnik Nope. You're in . . ?

Ah.

10th .. you?

Soham Chowdhury

11th.

11:53 AM

Then why didn't you gave the RMO before?

Soham Chowdhury

What's the composition symbol in LaTeX? Detexity says $\bigcirc$, which seems wrong.

@SohamChowdhury You mean $\circ$ \circ?

Soham Chowdhury

@Sawarnik Reasons. Mostly didn't really know about olympiads and all.

@MartinSleziak Yes, thanks. I need some fresh air, I'm getting stupid.

Okay.

@r9m Btw, is your hideous villanious murderous serial killing over?

Soham Chowdhury

brb

11:54 AM

BTW what category theory book was that excerpt from @Soham?

Just being curious.

Soham Chowdhury

^ Aluffi's algebra book. He introduces categories in the first chapter.

Haskell made me get interested in category theory.

And I had to learn algebra.

Two birds with one stone.

:)

@Sawarnik yes! resulting in the public display of their mutilated dead remains (my latest two blog posts :P)

Soham Chowdhury

^ You're freaking me out

Metaphors, I hope.

@r9m Yayay!

Soham Chowdhury

@Martin, are you a category theorist?

11:57 AM

@SohamChowdhury Well, the area I have closest to is general topology, but I used category theory quite often.

Soham Chowdhury

Oh.

But I don't think I can call myself general topologist (or category theorist). Maybe one day....

Soham Chowdhury

^ Prof? Just curious. Or still a student?

C'mon, don't tell me western civilization was destroyed again, @MikeMiller.

@SohamChowdhury Oh no, I'll have to buy a new bunker again

12:20 PM

@Hippalectryon A short integral to be done in the spirit of the art

$$\int_0^1 \frac{1-\sqrt{4 x^2-4 x+1}}{1-x} \, dx$$

@Chris'ssis thanks

@Hippalectryon it's cool! :-)

Something something log 4

@Hippalectryon :-)

Here is a small challenge I just created $$\sum _{n=1}^{\infty } \frac{H_{2 n-1}-H_n}{n (2 n+1)}$$

Soham Chowdhury

I always wanted to ask you something. Who's Chris? (Apart from just "my brother!")

12:30 PM

@SohamChowdhury A nice person. :-)

Soham Chowdhury

Well, that's nice.

I'm finally realising how the whole "every time you reread a chapter, you get the ideas better" thing works, with my algebra book. :)

Hi

@Chris'ssis How do you even keep track of those you have already solved ? Many look alike :P

@Hippalectryon I write things down on papers, files, books (when no paper free). All is kept. :-)

Soham Chowdhury

@MartinSleziak are you online?

I need a little help if possible.

Never mind, I got it.

12:38 PM

@Hippalectryon you know what? Let me make the question more beautiful.

$$\sum _{n=1}^{\infty } \frac{H_{2 n+1}-H_n}{(2 n+1)n}$$

:D

:D

That nearly makes 240 problems I got from you :-)

@Hippalectryon Are you serious? :-) lolllllll

Yep

12:40 PM

@Hippalectryon Awesome!!! :-)

Roughly one per day (more since I'm not always here)

@Hippalectryon You're amazing! :-)

A problem a day keeps the doctor away

3

@Hippalectryon lolllllllllll (+star) :-)))))

:-)

12:42 PM

@Hippalectryon Also consider $$\sum _{n=1}^{\infty } (-1)^{n+1}\frac{H_{2 n+1}-H_n}{(2 n+1)n}$$ but this is way easier.

Someone should really do integrals with fractional parts of $H_n$

@Hippalectryon Well, out there is something like that in terms of digamma function (if I recollect well).

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