Mathematics - 2014-06-26 (page 2 of 4)

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9:00 AM

@DanielFischer Cool. Is it also true that, given a continuously differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$, we have $V(f,[a,b]) \le C \sup_{[a,b]}|f'(x)|$ for some constant $C$ which depends only on the length of $[a,b]$?

My reasoning is: fix an $[a,b]$ and some $\alpha$, $\beta$, and consider all $f$ with $f(a) = \alpha$, $f(b) = \beta$ and with the same $V(f,[a,b])$. Then there are lots of ways different $f$s can "wiggle" in $[a,b]$ -- some might have a graph which is 'flat' for most of the interval and then cram all its variation into some tiny subinterval, but this will skyrocket $\sup|f'(x)|$.

The one which minimizes $\sup|f'(x)|$ seems like it should oscillate "evenly" over $[a,b]$, but even then, the variation dictates that a certain number of turns get squeezed into the interval, which should drive $\sup|f'(x)|$ up, the smaller the interval the higher it should go. That's pretty fast and loose, not sure how to turn that into a rigorous statement.

First time I've played with these ideas of bounded variation so I may be confused a bit.

if $f$ is $\mathbb{C^1[a,b]}$ we could write $\displaystyle V(f,[a,b]) = \int_a^b |f'(x)|\,dx$

@AndrewG $$\sum_{k=1}^n \lvert f(a_k) - f(a_{k-1})\rvert = \sum_{k=1}^n \left\lvert \int_{a_{k-1}}^{a_k} f'(t)\,dt\right\rvert \leqslant \sum_{k=1}^n \int_{a_{k-1}}^{a_k}\lvert f'(t)\rvert\,dt = \int_{a_0}^{a_n} \lvert f'(t)\rvert\,dt$$

Ah

Cool thnx

3 hours later…

11:38 AM

Greetings

Greetings

@Chris'ssis Greetings :D

@r9m :D

@Chris'ssis okay .. where do you live ? :-)

@r9m sure. Now I live in a small village from Romania. :-)

11:54 AM

@Chris'ssis Nice :) .. I live in a big-polluted-densely populated city in India =)

@r9m Where? (Which city?)

@r9m Nice! :-) -> transitionsabroad.com/listings/travel/women/articles/…

The first thing that I found on google.

destination360.com/asia/india/kolkata

@r9m Well, the bad thing is that no one here likes to attend things I like. In the future I plan to move to Timisoara city.

(in the near future I mean)

@r9m see here: google.ro/…

@Chris'ssis WOW :D

@r9m I'm so amazed by the things I discover every day (maybe just rediscovered by me, but this is less important).

Starting from simple things one can reach some really amazing results. Well, it's such a profound pleasure doing that ...

I really hate Sylow.

The worst part of Sylow theory is probably determining homomorphisms.

12:09 PM

@Chris'ssis :D great ... things that I discover everyday turns me more evil and twisted .. =P and I find great pleasure in doing that :P

@r9m :-)))

@r9m @robjohn did you try it? $$\int_0^1 \frac{\left(\operatorname{Li_2}(x)\right)^2}{x} \ dx$$

It's absolutely awesome!!!

@Chris'ssis I haven't. Didn't you just say it was not that difficult?

@robjohn Sure, I said that. I initially thought it's difficult, but then I realized I was completely wrong. It's a very beautiful question, very beautiful.

@Chris'ssis I have not tried it. I had been working on another problem for the last day or two. I finished that and am working on a new one.

Hardy -- "There is no ugly mathematics"

12:17 PM

@robjohn @Chris'ssis have you seen this ? $$\lim_{n\to+\infty}\left(\frac{2}{3}\right)^n\sum_{k=0}^{[n/3]}\binom{n}{k}2^{-‌k}=\frac{1}{2}$$

And that sure is ugly, @r9m

@r9m hehe, it looks nice. :-)

I'm planning to ask it on main :| .. I'm stumped :(

well, well, i'm off.

@BalarkaSen Give it a try when you have some time. You'd like it very much.

12:19 PM

@Chris'ssis What? That Li integral?

@BalarkaSen Yes.

@r9m I think I've done something like it...

@Chris'ssis I think we both agree that our likings on mathematics are totally off from opposite poles.

But I'll look at it nevertheless.

@BalarkaSen You wont' regret. It's very nice! :-)

@r9m I think I attended something like that in the past (I mean a more general case).

@Chris'ssis Hmm. What if I just remove the square term and think of $$\int_0^1 \frac{\text{Li}_2(x)}{x} dx$$

I am not sure if that converges though.

No idea about polylogarithmic growth.

12:23 PM

@BalarkaSen You just changed the problem. :-)

I know I did.

I am wondering what happens to the final result if I remove the squared term.

@Chris'ssis gr8) .. will you answer it if I put it on the main ? :D

@BalarkaSen What you don't know is that what you did contains a key to the solution to the problem.

@Chris'ssis Oh?

@BalarkaSen that is $\text{Li}_3(x)$

12:25 PM

Interesting.

Yeah, it is!

hey math gurus. can someone tell me why d Log(V(x)) = d V(x)/V(x) ?

You're right.

:-)

@Sosi Not a guru, but $(\log(x))' = 1/x$

@r9m I'll check my old papers to see if I find it.

12:26 PM

@Chris'ssis Thanx :D

@Chris'ssis I'll definitely have a look at that problem, but I am struggling a lot with a little group theory.

let me go through that first.

Speaking of the devil, here he comes @PedroTamaroff

@BalarkaSen OK, no hurry.

@PedroTamaroff I am doing some semidirect product stuffs. Can you help?

@BalarkaSen OK.

@BalarkaSen by the chain rule you then have that dLog(V(x)) = dV(x)*1/x

is this correct?

12:32 PM

I am classifying all groups of order $30$, here's what I did :. The direct product of $3$-sylow and $5$-sylow is a cyclic group of order $15$ and has index $2$ in $G$, i.e., normal. You have a split exact sequence $$1 \to \Bbb Z_{15} \to G \to \Bbb Z_2 \to 1$$ The group $\Bbb Z_2$ act on $\Bbb Z_{15}$ by automorphisms

So we look for homos $\Bbb Z_2 \to \rm{Aut}(\Bbb Z_{15})$, the automorphism group is isomorphic to $\Bbb Z_4 \times \Bbb Z_2$

Now I can't find the homomorphisms, except the trivial one which gives abelian groups. This is my first excercise in semidirect products so it'd be great if you can help me.

Hold on, hold on.

OK.

Your group is of order $pqr$; $p<q<r$ primes.

Groups of order $pq$, $p<q$ primes are not hard to classify.

Right. In this case, it's cyclic.

Well, there's also the $6\times 5$ option. Or the $3\times 10$ option.

Let me think about it for a second.

12:37 PM

Why care about them?

You just need a normal subgroup to get going, no?

And I have it right in front of my nose.

Nah, man. You're not guaranteed the product of two subgroups is a subgroup.

?

Product of two subgroups with less than groups order is a subgroup, I believe

No.

You're not guaranteed the product of two subgroups is a subgroup.

Example time.

Well, you need none of them to be normal. If either of them is normal, you do have a subgroup.

12:41 PM

Yeah, you're right.

But that I guess can be taken care of.

OK, at any rate look at $3$-Sylows.

There is either $1$ or $10$.

Yup.

But counting shows there is no 2-Sylow, if there are $10$.

Contradiction.

If there are $10$, there are $10\times 2=20$ elements of order $3$.

@PedroTamaroff No, what's the order of the 3-Sylow?

@BalarkaSen $3$.

So $10\times (3-1)=20$.

12:43 PM

$3$, right? And the groups are of prime order, so intersection is trivial.

Yes, but don't overcount the identity.

Oh, OK.

So now what?

Well, if there are $10$ $3$-Sylow subgroups, there must be $1$ normal $5$-Sylow.

Oh, I didn't count the $2$-Sylow.

Yeah, $2$-Sylows ain't helping here.

12:46 PM

Right. So I guess one can also prove vice-versa.

Yes.

But tell me about the homomorphisms

What homomorphisms?

$\Bbb Z_2 \to \Bbb Z_4 \times \Bbb Z_2$

At any rate, @BalarkaSen, what we have proven is that there is always a subgroup of index $2$.

Hence normal.

And has order $15$.

12:47 PM

That's all very fine, I can finish with a number theoretic argument here

But what I need is semidirect products

@BalarkaSen A morphism $C_2\to C_4\times C_2$ is determined by where you map the generator of $C_2$.

And if it is nontrivial, you must map it to an element of order $2$ in $C_4\times C_2$.

Ah, right right.

I can do this now.

You have $(y^2,x)$, $(1,x)$, $(y^2,1)$ and none more.

Thanks again for help, @Pedro.

So $4$ morphisms total, counting the trivial morphism.

@BalarkaSen No problem. Don't overcomplicate things and don't do false claims!

12:50 PM

OK. Will keep in mind.

@Chris'ssis I assassinated an interesting integral ineq :D .. but the question its so old that the OP possibly doesn't give a rat's a** about it anymore =P

@r9m Which one? :-)

@Chris'ssis \m/

@r9m That looks nice ...

@Pedro I am scared, it's so close. You have yours on the 3rd, right?

12:56 PM

@Chris'ssis :D .. its sweeeet :) I like it :} .. the OP solved a related problem and conjectured this one :)

@Studentmath $3,8,10,11$.

>not being ganged up on by midterms.

Jeez. 30, 3, 9, 14, 17 - slightly better..

@PedroTamaroff

@BalarkaSen What is that?

Your exam dates.

12:59 PM

A finite sequence associated with the Lie algebra C_3.

3,8,10,11

Oh, LOL.

=P

Mine isn't associated with anything..

@Studentmath No, and that's very bad.

1:00 PM

Thanks @Balarka, that's comforting

It indicates that you'll fail all your examination

If I do fail all, I will blame you

The truths revealed by my Inner Eye (OE Eye S) is always harsh, dear.

For a science student, I'm very focused on 'luck'

Hi, I have two quadratic forms $q$ and $q'$ on a complex vector space $E$, if I assume that $q^{-1}\{0\}=q'^{-1}\{0\}$. Does it implies that $q=q'$ ?

1:02 PM

Don't think so..

@Nico Equal kernels implies equal forms? Hmm...

equal inverse kernels

@Nico Unlikely.

Though you shouldn't trust me, as my knowledge on these stuffs is infinitesimal.

I guess I should go and study more group theory. Byes.

@PedroTamaroff Yep I concluded too quickly. Do you think we have some interesting consequences ?

To a false claim?

1:08 PM

no, if I assume that $q^{-1}\{0\}=q'^{-1}\{0\}$. What can I say about $q$ and $q'$ ?..

Oh, well I don't really know. When $q,q'$ are linear forms, you get they are scalar multiples of each other.

right. It's something I have to think about it so. Thanks for your time. :)

@Chris'ssis haha .. the constant $24$ could be replaced by $40$ :) and still its not equality condition .. seems it can be improved even more :D

@r9m Nice and interesting :D

@Chris'ssis yas .. as I'd quote in Japanese - Omoshiroi =P

means highly interesting :P

1:23 PM

@r9m hmmm, I didn't know that. :-)

Wait ...

translate.google.com/#ja/en/%E9%9D%A2%E7%99%BD%E3%81%84

@r9m see above :-))))))

@Chris'ssis wow ... how did you get the Japanese kanji ?! :o

@r9m I only entered things and then all turned automatically into Japanese kanji . :-)

@Chris'ssis ah .. ic :P I thought you knew Jap

@r9m lol, NO :-)))

1:50 PM

yo

2:13 PM

@Chris'ssis There I posted it ^_^

Now I'm off to assaulting some more G.T.R's old problems .. ;)

@r9m Nice!!! (+1)

:-)

@r9m By the way, I have somewhere a nice integral inequality for you.

@Chris'ssis Gimmie gimme :-)

@r9m :D

@r9m I'm sure you're very used to this type of question.

@r9m Prove without computing the integral that $$\int_0^{\infty} e^{-x^2} \ dx \le\frac{4+\pi e}{4e} $$

@Chris'ssis ^^' .. I've seen it already and have it scribbled somewhere in my notebook :-)

@r9m Well, this version is newly created, especially forged for you. :D

2:29 PM

@Chris'ssis aha like .. forged in the deeps of Mount Doom .. the one ring ?! =D

@r9m hehehe :D It's a very nice question!

@Chris'ssis and I'm already feeling a chill down my spine :o .. may the $4^{th}$ be with me :-)

:-))))

@r9m One ring to rule them all and in the darkness bind them.

@BalarkaSen yes ... muahahaha !! :D

I'm listening to the ABBA albums after a long time :-)

2:32 PM

LOTR was probably one of the best books I ever read.

But evidently perhaps not the best.

@r9m ABBA?

ABBA were a Swedish pop group formed in Stockholm in 1972, comprising Agnetha Fältskog, Björn Ulvaeus, Benny Andersson, and Anni-Frid Lyngstad. ABBA is an acronym of the first letters of the band members' first names (Agnetha, Benny, Björn and Anni-Frid) and is sometimes stylized as the registered trademark ᗅᗺᗷᗅ. They became one of the most commercially successful acts in the history of popular music, topping the charts worldwide from 1975 to 1982. They are also known for winning in the Eurovision Song Contest 1974, giving Sweden its first victory in the history of the contest and bei...

OK, I don't wanna know.

lol XD .. they are heavenly :-)

Gil galad was an elven kind of whom the harpers sadly sing the last whose realm was fair and free beneath the mountains and the Sea

Hey! I still recall it!

@r9m Did you read the book or just seen the movies?

@BalarkaSen Just the movies :( .. I wanna read the books too .. but I'm a very slow reader

2:37 PM

@r9m The movies actually sucked.

At least the first part. I've never seen any other movies after then.

The book is good. You can't appreciate LOTR unless you've read the book.

omg ... now you are making me wanna read it more than ever .. :(

'cause one should. Tolkien was a great writer.

But... if you haven't read the book, how could you have understood the movie, @r9m?

The Hobbit was the first part of the book no?

yes

And 'til recently, they didn't make it a film.

So you have read Hobbit, @r9m?

no .. I haven't read anything .. just the movies

2:41 PM

But you can't understand LOTR without seeing/reading Hobbit.

And they haven't finished making Hobbit yet.

K, gotta go.

I got a Revival Badge for that ?! .. -_- sigh

@Chris'ssis $$\int_0^\infty e^{-x^2}\,\mathrm{d}x\le\int_0^1\frac1{1+x^2}\,\mathrm{d}x+\int_1^\infty e^{-x}\,\mathrm{d}x$$

3

@robjohn GREAT!!! :-)

2:57 PM

@Chris'ssis but I think you get a better bound using $$\int_0^\infty e^{-x^2}\,\mathrm{d}x\le\int_0^1\frac1{1+x^2}\,\mathrm{d}x+\int_1^\infty xe^{-x^2}\,\mathrm{d}x$$

let me see

$\frac\pi4+\frac1{2e}$ instead of $\frac\pi4+\frac1e$

@robjohn Nice.

@robjohn :D !! Nice !!

Boy I like those green [+10] bubbes =D =P I know I'm sounding silly .. but its the truth :P

@TedShifrin I managed to write the thing properly.

3:17 PM

Ascents? @PedroTamaroff

@BalarkaSen Did you read the question...?

An ascent happens when $a_i<a_{i+1}$.

Ah.

Oh, by the way, @Pedro, I figured the groups out.

Cool.

We already determined it was a semidirect product of a group of order $15$ and $C_2$, right?

Eh?

Did we?

@robjohn How do you bound it on 0 to 1?

3:23 PM

I get $\Bbb Z/3\Bbb Z \times (\Bbb Z/2\Bbb Z \ltimes \Bbb Z/5\Bbb Z)$, for example, @PedroTamaroff

@Chris'ssis have you seen this one ? $\displaystyle \int_{0}^{2\pi}e^{\sin{x}}dx<2\pi e^{\frac{1}{4}}$

@N3buchadnezzar $e^x\ge1+x$ for all $x$, so $e^{-x^2}\le\frac1{1+x^2}$

@BalarkaSen I have to go, sorry.

Bu-byes.

@robjohn right thanks

3:27 PM

@r9m No.

@Sawarnik Did you try that problem?

@Chris'ssis Its nice too .. its among one of Math110's questions :-)

@BalarkaSen Not yet, I have been very busy, I will disappear in a few minute now. Tommorow is the deadline!

@Sawarnik Deadline for what?

@r9m $$ \displaystyle 2\pi < \int_{0}^{2\pi}e^{\sin{x}}\,\mathrm{d}x<2\pi e^{\frac{1}{4}}$$

3

3:29 PM

@BalarkaSen Sorry: I already told you, for any group $G$ of order $30$ there is a normal subgroup $H$ of order $15$. There is always a cyclic subgroup $C$ of order $2$, so $G$ is a semidirect product $G=H\rtimes C$.

Meaning there exists a real number $\lambda \in (0,1/4)$ such that $\displaystyle \int_{0}^{2\pi}e^{\sin{x}}\,\mathrm{d}x = 2\pi e^{ \lambda } $

This follows from $H\lhd G$, $HC=G$, $H\cap C=1$.

Bye.

@PedroTamaroff Yes, I know.

@N3buchadnezzar yas .. the lower bound cos $e^x \ge 1$ for $x \ge 0$

@BalarkaSen Well, use that, it is easier.

3:30 PM

@N3buchadnezzar ah ... Nice :-)

@N3buchadnezzar Fun. I conjecture : $\lambda$ is transcendental.

@BalarkaSen Come to the other room

@r9m He posted some nice questions. One of them is this one math.stackexchange.com/questions/508525/… that I'm going to solve elementarily. (I hope I can really do it)

W|A reports $$\displaystyle \int_{0}^{2\pi}e^{\sin{x}}\,\mathrm{d}x = 2\pi I_0(1)$$

@Alizter OK.

@Chris'ssis :D ... slay dragons :D !!

3:47 PM

@BalarkaSen Does it say what $I_0$ is?

@robjohn Modified bessel function of the first kind.

@BalarkaSen Okay, that is pretty much by definition.

@robjohn Hey

Any idea how to prove that $\exists \, ! \, \lambda \in (1/5,1/4)$ such that $\int_0^{2\pi} e^{\sin x}\,\mathrm{d}x = 2\pi e^{\lambda}$ ?

Which is the same as wanting to prove that $1/5 <\log I_0(1) < 1/4$, not sure if that is easier.

Matlab gives $\lambda \approx 0.235914358882689$.

@BalarkaSen I get $2\pi\sum\limits_{n=0}^\infty\frac1{n!^24^n}$

Any graph theorists around?

3:59 PM

@BalarkaSen That converges pretty fast

@N3buchadnezzar You might be able to show it from this series

So $$ \sum_{n=0}^\infty \frac{1}{5^k k!} <\sum_{n=0}^\infty \frac{1}{n!^2 4^n} < \sum_{n=0}^\infty \frac{1}{4^k k!} $$

4:14 PM

Hi people! I believe this question could be of some interest for you! Ok, ti probably seems ridiculous to you, not interesting, but it would be very helpful if you gave me some insight... Thanks a lot!

4:34 PM

What's the name of the module of all the integer lattice points on the place?

4:51 PM

$\mathbb Z^2$?

I thought that implied a group, but OK.

a group is a $\mathbb Z$-module

too late to edit that: abelian group

can anyone tell me of a book which gives a intutive understanding of the concept of "e"

5:10 PM

@EvilWarrior As in the exponential constant?

5:41 PM

@Chris'ssis chat.stackexchange.com/transcript/36?m=16277596#16277596

@r9m Yeah, I know, but I didn't work on that anymore. I need to put things on paper and then I let you know. :D

@EvilWarrior For what purpose?

@Chris'ssis okay :D

@EvilWarrior Have you read this?

@r9m I've just proved a beautiful result :D

5:47 PM

@Chris'ssis great :D !! .. and I have sharpened the integral inequality to the best constant =D

$$\sum_{i=1}^{\infty} \sum_{j=1}^{\infty}(-1)^{i+j} \frac{H_{i+j}}{i+j}=\frac{\pi^2}{12}-\frac{\log(2)}{2}-\frac{\log^2(2)}{2}$$

9

@Chris'ssis Nice !!! :-)

@r9m :D

@Chris'ssis What's the use of that result?

@r9m 3.103. by Ovi. Once I saw it I instantly knew what I needed to do.

5:51 PM

@Shisui Wrecking chaos and unleashing pandora box in heaven =P

@Chris'ssis :) (y) !

@Shisui That is an art (of beauty).

@Chris'ssis That doesn't make sense.

It seems to me that one of the most offensive things to say to a scientist, mathematician, or artist is "oh, that's nice... but what's its use?"

4

;)

"What's the use of that result?"
"That is an art (of beauty)."

@anorton That's d*mn true!

5:54 PM

So is it just for curiosity's sake @Chris'ssis?

@Shisui What's the use of living your life? Just asking.

@Shisui Define "use"?

To find out more about the world around us.

@Chris'ssis

@Shisui It is interest in itself.

@Shisui You're going to study maths you should know that it's beautiful! ;)

@Shisui Summation like what Chris's sis can do is an art form. It receives justification simply by its beauty. Not for curiosity’s sake nor for interpreting the world around us--it is an end to itself, rather than a means to something else.

2

5:56 PM

@Shisui What is useful today may not be tomorrow, and vice versa.

Fair enough.

Why do people climb moutains?

How do you view LaTeX on this chat on a Mac?

@Shisui math.ucla.edu/~robjohn/math/mathjax.html

@Shisui The art is a part of our life, we need it to feed our souls.

3

5:58 PM

Jun 20 at 0:53, by anon

Chat guidelines | $\LaTeX$ in chat

@Alizter Um?

(curious as to why you posted that)

hello

hi

@anorton Oh. I see now. :P

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