Mathematics - 2014-10-16 (page 1 of 4)

Mike Miller

00:01

so?

Balarka Sen

Not exactly sure :P

Mike Miller

You want to show it converges. So first figure out what the limit "should" be, and then prove that it converges to that limit.

Balarka Sen

But $a_i$ is arbitrary mod 5!

Mike Miller

yes

Balarka Sen

It's 5:30 here, so let me sleep on it. I can guess you're leading me somehow to the algebraic definition of $\mathbf{Z}_5$, but let's see.

Byes for now.

I'll be back.

MSEoris

00:25

hey can someone help me see where contradiction in the solution of 4a comes from? math.ucdavis.edu/~hunter/m201a/sola4.pdf

Pedro Tamaroff

01:30

@TedShifrin

I wonder why one cares about an holomorphic embedding of $B(0,1)$ in $\Bbb C^3$.

=D

Ted Shifrin

As a closed submanifold, @Pedro?

anon

ah, @Ted

I has question

Ted Shifrin

Heya mr @anon

anon

${\rm SL}_2(\Bbb Z)$ is gen'd by $[\begin{smallmatrix}1&1\\0&1\end{smallmatrix}]$ and $[\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}]$. The "continuous extensions" of these are matrices of the form $[\begin{smallmatrix}1&h\\0&1\end{smallmatrix}]$ & $[\begin{smallmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{smallmatrix}]$. I am wondering if these generate ${\rm SL}_2(\Bbb R)$, in particular if we can write $[\begin{smallmatrix}\lambda&0\\0&\lambda^{-1}\end{smallmatrix}]$ with em.

Ted Shifrin

01:46

Continuous extension = one-parameter subgp

anon

indeed

Pedro Tamaroff

@TedShifrin Remmert's second volume says there is a holomorphic embedding. No more.

@anon Hello! Did you check Serge Lang's book on that group?

@anon You can write that using elementary matrices. That's the Whitehead identity!

Ted Shifrin

@anon: I always defer to you on algebra :) i don't know offhand, but Imthink I have in my youth.

anon

naively, two one-parameter subgroups might seem to necessarily generate a 1+1=2 dimensional space, but this intuition is pretty shaky.

@PedroTamaroff don't have it handy, don't feel like dling it atm.

Pedro Tamaroff

@anon K.

anon

01:50

@PedroTamaroff ah, "whitehead's lemma" gets google results

Ted Shifrin

Not really, @anon, as Lie bracket will add dimensions because of non- integrability ...

Pedro Tamaroff

@anon Yes, that. Sorry. =)

It's actually pretty relevant!

Ted Shifrin

@Pedro: I think Remmert means that. Otherwise you have dopey $(z,0,0)$

Pedro Tamaroff

@TedShifrin Let me check

I can't even $\Bbb C^3$ in my head though.

anon

@PedroTamaroff indeed, it answers my question quite exactly.

you study lie algebras or something for that?

Pedro Tamaroff

01:52

@anon Oh my! I'm glad.

@anon Nah, I once was caught up with elementary matrices and the Steinberg relations.

@anon Funny as it is, I cannot see how it answers it. I can see it for the Jordan form, but not for the rotation.

Oh, sorry, I'm totally overpinging.

You're trying to write the diagonal thingy with rotations and the Jordan block yes?

anon

Yes, I was trying to write squeeze mappings as compositions of transvections and rotations

Pedro Tamaroff

@TedShifrin Ah! Here it is. Remmert calls a map $(f_1,\ldots,f_n)$ from $D$ to $\Bbb C^n$ an imbedding if it is smooth, meaning the at no point the derivatives vanish simultaneously, if it closed and finally if it is injective.

@anon So, how did you solve this?

anon

@PedroTamaroff the lemma is already there, and the lower triangular matrices I already figured out are conjugations of transvections by rotations

Pedro Tamaroff

@anon Oh, cool.

What are you working on, stranger?

Ted Shifrin

@anon, the Lie bracket of your generators is $\begin{bmatrix} 1\\ &-1\end{bmatrix}$.

anon

02:00

@PedroTamaroff first I was reading about modular forms, but got sidetracked by hyperbolic geometry. unfortunately I had an important history midterm on monday, and I had just bought the book on friday, so I really crammed over the weekend still feel like I'm deflating

Ted Shifrin

Right, @Pedro: Image is a closed submanifold. Does he give an explicit example?

noahnu

I'm having some trouble understanding what a question is asking. Let $\epsilon\,>0$. Find $N > 0$ such that, if $x > N$, then: $$\left|\frac{x^2 + 1}{x^2 - 1} - 1\right| < \epsilon$$ What am I actually supposed to solve for in this question? I'm assuming it's similar to a delta-epsilon proof (the topic we're currently covering). In that case, do I say $|x-N| < \delta$ and then attempt to re-arrange $|\frac{x^2 + 1}{x^2 - 1} - 1|$ into $|x-N|$?

anon

you're supposed to do exactly what it says to do: find an N (depending on epsilon) for which you can prove x>N implies |blah|<epsilon

Pedro Tamaroff

@TedShifrin He gives some crazy construction. =P

@anon Did you do well?

anon

I calculated that worst-case scenario I still got an A on it. plus up to 20 % points of our final grade can be comprised of extra credit work...

Ted Shifrin

02:04

Such things are called Stein manifolds, @Pedro. An important class of noncompact complex manifolds.

Pedro Tamaroff

Hehe, you go t this. =)

@TedShifrin I read that when I scanned the book.

I might be getting into deep water here. Hehehehe....

Ted Shifrin

@anon: I think my Lie bracket computation does it ...

anon

@Ted Do you know a good way to motivate the construction of any of the models of hyperbolic geometry? They all seem to be done out of thin air. I mean like: given "I want a geometry satisfying the first four postulates but not the parallel postulates," how to arrive at one of the constructions. the best hint I've found was a historical comment about a "sphere of negative radius"

@TedShifrin I am not sure I follow how the computation yields a formula for squeeze maps in terms of rotations and transvections.

Pedro Tamaroff

@anon Did you read Lobachevsky's or Bolyai's original writings?

anon

nope

Pedro Tamaroff

02:12

Or Gauss. Let's not make Him mad.

I think they are out there. I have a book that has them. Dover.

anon

the pseudosphere seems related

I will do more reading

Ice Boy

hi @AlexanderGruber

Alexander Gruber

evening @ice

Ice Boy

how's the neck feelin'?

Mike Miller

@anon I know a nice construction of $H^n$ as a Riemannian manifold that makes sense of your "sphere of negative curvature" comment, but I'm not sure if that's what you're looking for

nevermind, it's not the saddle, it's a hyperboloid

apparently it's called the "hyperboloid model" on wikipedia and whoever wrote that article has probably thought about this a lot more than me

Ice Boy

02:32

University wants scientists to make their research open access and resign from publications that keep articles behind paywalls Tuesday 24 April 2012

Ted Shifrin

02:51

That's in Lorentz space, @Mike.

@anon, in the last chapter of my algebra book, I did projective geometry and then tried to motivate hyperbolic after the spherical model of projective. If you email me, I'll send you a pdf of the section.

Hi @Alex

Mike Miller

What's the practical difference, @Ted? It looks like their metric is just the negative of mine.

Ted Shifrin

Huh? It's the Lorentz metric that induces a constant negative curvature metric on the (right) hyperboloid.

Euclidean hyperboloids have nonconstant negative curvature.

Mike Miller

Oh, @Ted, I thought I wrote down the details in here, but I wrote them in notepad. Yes, I know one pulls back the pseudometric of $\Bbb R^{1,n}$ - I just thought the name was Minkowski space, not Lorentz.

Ted Shifrin

Lorentz metric, I guess. No difference intended.

Committing to a challenge

03:47

Thank you for your response @RobertCardona. Yes I have been keeping a study log for about three years now(I was doing electrical engineering before switching to math), and I have tried the Pomodoro method before, but didn't find it working well for me. I am currently trying a method in which I time how long it takes to do each problem I come across in my worksheets, and retrial it at a later date. I have found trying to solve things as quick as possible has been improving my solution methods.

Ice Boy

04:36

welcome @Aram

Kaj Hansen

Hey hey

Ice Boy

hello

Kaj Hansen

@IceBoy, in regards to your above post, what is your opinion about publications behind paywalls?

Ice Boy

@KajHansen I would like freedom of access for all, but that is not realistic.

What is your opinion @KajHansen?

Kaj Hansen

@IceBoy, my biggest problem is that my taxes fund a decent fraction of the research that is then published and stuck behind paywalls. I feel that publicly funded research should be freely available.

Ice Boy

04:48

Fair enough.

Kaj Hansen

I feel like the world could be a better place in the absence of overreaching patent law, copyright law, and the like. In other words, information being more freely available.

That may or may not have negative unforeseen consequences though. I'm not sure.

I'm not too big on policy. There was a time in my life when I was very, very politically active, but nowadays I just prefer sticking to my math.

Ice Boy

Indeed, some of the big publishers have stopped publishing newer editions of their best sellers in book form, and instead have opted for these paywall versions :(

Kaj Hansen

That's truly a shame. I will always be partial towards physical books, and I cannot stand reading from a screen. Ironically, though, I do try to find the .pdf versions of all my textbooks for when I'm in a pinch.

Ice Boy

But now access has been lost to the public even through the library of the newer editions.

This leads to all sorts of pirating and hacking into the publisher's web sites.

Kaj Hansen

05:07

Indeed. You heard about the Reddit founder committing suicide after being prosecuted for downloading tons of docs from MIT's library?

I definitely disagree with the ridiculous sentences some of these people face. It's very disproportionate.

Ice Boy

Oh really? I never heard about that.

Cyber Law sounds so futurama :-)

Kaj Hansen

@IceBoy, he was facing $1 million in fines and 35 years in prison. See here: en.wikipedia.org/wiki/Aaron_Swartz

I guess it was actually papers from JSTOR accessed through MIT's network.

Hey there @RandomVariable

Ice Boy

@KajHansen thanks for the link

Kaj Hansen

Sure thing

Random Variable

@KajHansen Hello.

Ice Boy

05:17

National borders aren't even speed bumps on the information superhighway.

Shanika Samarasinghe

05:32

Hello people

Committing to a challenge

Hello

Kaj Hansen

Hey there @Shanika

Random Variable

@IceBoy I've been tempted to pay for articles I couldn't gain access to in any other way. But I never have.

Kaj Hansen

One of the most annoying things is when the abstract is just a little too vague and you're not sure if the article is what you want.

Ice Boy

yes^ intentionally vague >8(

Committing to a challenge

05:34

Don't most universities give access to all journals to their students?

Aram

Hi

Committing to a challenge

Hello @aram

Kaj Hansen

@Committingtoachallenge, indeed.

Ice Boy

@ShanikaSamarasinghe Welcome

Kaj Hansen

spikedmath.com/545.html

3

Committing to a challenge

05:47

@KajHansen Haha, two reasons pop into my head, the person gets called about having a 'telephone number (math)' every now and then, or it sounded like a prank on its own, being that it sounds silly

Kaj Hansen

:P

I'm a huge fan of SpikedMath comics. Very high-brow humor.

Ice Boy

Me too :-)

Ice Boy

06:03

@RandomVariable yea, it's easier to "Mooch." :D

Random Variable

@IceBoy Yesh!

Random Variable

06:46

@IceBoy I think that was the first time that anyone has made a reference to my avatar.

@r9m Hi.

r9m

@RandomVariable Hello :)

Ice Boy

@r9m hi pal

r9m

@IceBoy helloo :)

Random Variable

@r9m Did you take down your blog?

r9m

@RandomVariable yes :( .. one of my classmates criticized (made fun of) it so heavily that I took it down :'(

Random Variable

06:56

@r9m There was nothing about it to make fun of. It sounds like he was jealous.

Ice Boy

start another blog and don't tell them

use a different name

r9m

already did that :-) .. but I feel a lot less motivated to make new posts :(

Ice Boy

the motivation will return with your confidence my friend

Committing to a challenge

@r9m Did this class mate greatly dislike you? Was it listed under your actual name?

r9m

@Committingtoachallenge It was not listed as my actual name .. but he knew it was my blog

Committing to a challenge

07:07

What was there to be criticized? It was a math blog I assume

Random Variable

@Committingtoachallenge And a very good one.

r9m

@RandomVariable :) you are too kind :-)

Gustavo Montano

08:02

Anyone familiar with General Relativity?

Committing to a challenge

Why is that Gustavo?

Gustavo Montano

I have a few questions I wish to ask.

Unfortunately the Physics chat room is quite empty.

Balarka Sen

@Gustavo

Committing to a challenge

Ask away, I will see if there is anything I can answer

Gustavo Montano

Well this is what I asked:

Hi, I am a mathematician who has taken interest in Einstein's theory of General Relativity. I have a few questions in regards to definitions I've seen about. First, in this magnificent spacetime, I understand that gravity at a point is defined by the curvature at that same point. I also understand that geodesics are defined to be curves that are locally minimised. Now, the bending of these geodesics are a consequence of their interaction with a curved segment of spacetime - that is gravity.

I also have that "in general relativity, test bodies move along geodesics." What is a test body?
What is an example of a mass that is not a test body?

@BalarkaSen. Heya :D.

Balarka Sen

08:10

@GustavoMontano $(X, d)$ be a metric space. $(X, \bar{d})$ be another metric space, the metric defined as $\bar{d}(\bullet, \bullet) = d(\bullet, \bullet)/(1+d(\bullet, \bullet))$. I want to prove that $(X, d)$ and $(X, \bar{d})$ have the same open sets. Simmons hints me to look at the open spheres of both, but I am not sure how that helps. Can you give me walkthrough (not a solution?)?

Committing to a challenge

@GustavoMontano physics.stackexchange.com/questions/102910/…

Balarka Sen

The open spheres of $(X, d)$ around $x_0$ are $d(x, x_0) < r$

The open spheres of $(X, \bar{d})$ around $x_0$ are $d(x, x_0) < r/(1-r)$

Gustavo Montano

Interesting problem I don't think I have completed a question like this before. But given our knowledge of general topology - lets give it a go. So to say that both metric spaces have the same open sets would mean that every sphere in $M=(X,d)$ is also in $\bar{M} = (X,\bar{d})$. Would you agree?

Balarka Sen

The book says that $(X, d)$ and $(X, \bar{d})$ have the same open sets with one exception.

I don't get it.

@GustavoMontano Right.

Gustavo Montano

Ok. So we want to prove that every open sphere in $M$ is also contained in $\bar{M}$.

And vise-versa.

Balarka Sen

08:13

Simmons says that there is one exception.

??

Gustavo Montano

Has he explicitly mentioned it?

The exception, that is.

Balarka Sen

It was given in the hint

In any case, let's move forward.

The spheres are :

2 mins ago, by Balarka Sen

The open spheres of $(X, d)$ around $x_0$ are $d(x, x_0) < r$

2 mins ago, by Balarka Sen

The open spheres of $(X, \bar{d})$ around $x_0$ are $d(x, x_0) < r/(1-r)$

WATNOW

Gustavo Montano

Yes, so lets take a random point $x$ in $M$ such that the sphere around $x$ has radius $r$.

Can you prove to me, that this open sphere is also inside $\hat{M}$?

Balarka Sen

You mean a sphere around $x$ in $\overline{M}$ with radius $r$?

We want $\bar{d}(x, x_0) < r \Rightarrow d(x, x_0)/(1+d(x,x_0)) < r \Rightarrow d(x, x_0) < r + r \cdot d(x, x_0)$

Gustavo Montano

Ummmm. Let me think a little.

Just putting it on paper atm.

Balarka Sen

08:18

OK

Gustavo Montano

Just want to clarify "the same open sets"

Would you agree we want to show that an arbitrary sphere is located in both sets?

Balarka Sen

I don't get the "same" idea. Does he wants us to prove that every open sphere on radius $r$ in $M$ around $x_0$ is also in $\bar{M}$ with the corresponding metric?

Gustavo Montano

Yes, that is interesting the change in metric is a bit of a problem.

Let me think a little more

Ok, I've shed a bit of light on to this.

So, let $U$ be an open set in $M$. That means for all $x \in U$ $B(x) \subseteq U$.

Balarka Sen

Right.

Don't reveal the solution though

Gustavo Montano

And we want to show that this same $U$ is open in $\bar{M}$.

Balarka Sen

08:31

Ah.

That'd do. Let me see.

"the same pen sets" part was confuzzling

Gustavo Montano

So think about this new metric space, what changes is made to $U$, and furthermore, show that the requirements of the open definition - are still in tact.

It is a bit confusing. We're going from one structure to another, and want to show that the definitions required are still preserved.

Balarka Sen

Thanks for the hints, @Gustavo. I'll look into it.

@GustavoMontano Essentially this would prove that $(X, d)$ and $(X, \bar{d})$ are topologically equivalent, no?

Yet not isometric, i.e., length-preserving.

Weird.

Gustavo Montano

Is that the definition of topological equivalence?

Same Open sets? It's all about definitions at the end of the day :D.

Balarka Sen

@GustavoMontano Well the open sets are the same.

Gustavo Montano

If that is the definition - then I will agree with you.

Balarka Sen

08:34

A topology consists of a base set and it's open sets.

Base set is $X$ on which the metrics act, and the open sets are same too.

Gustavo Montano

Not sure about that one. Finish this question first :)

Balarka Sen

@Huy Huy.

Gustavo Montano

I must go @BalarkaSen. Indoor soccer :D. Good luck with your solution - be sure to share it when you're done!

Balarka Sen

Byes, @Gustavo. Will ping you when I'm done.

Chris's sis

Greetings

r9m

08:40

@Chris'ssis greetings ;)

Chris's sis

@robjohn I'm trying to prove that

$${\large\int}_0^1\frac{\ln^2(1-x)\,\operatorname{Li}_2 \left(\frac{1+x}2\right)}x dx = \frac{81\ ,\zeta(5)}{32}+ \frac{5\pi^2}{16} \zeta(3)- \frac{\zeta(3)}8\ln^22+\frac1{15} \ln^52\\-\frac{\pi^2}{ 18} \ln^32- \frac{\pi^4}{15}\ln2+2\operatorname{Li}_5 \left(\tfrac12\right)+ 2\operatorname{Li}_4\left(\tfrac12\right)\ln2$$

it's from the last answer of Cleo. I'm going to show all this stuff is very easy.

@r9m Hi

Ice Boy

Greetings

r9m

@Chris'ssis Incredible !!! that will solve a lot of Alternating Euler Summations too !! :D Awesome !!!!

Chris's sis

@r9m What do you think it's the first step to do above?

r9m

@Chris'ssis IDK :O .. but the RHS is a LC of results that you get from computing closed form of Alternating ES .. so an independent evaluation of the integral should nail the ESs too ..

Committing to a challenge

08:50

I don't like that Tunk Fey has changed the description on his Cleo account such that noone can say anything about the answers, without being morally unacceptable, despite the arrogance Cleo displayed in now deleted comments, which seems entirely contradictory

r9m

@Committingtoachallenge 'Tunk Fey has changed the description on his Cleo account' ?!! .. how did you know that ?!!! :O

Committing to a challenge

He answered numerous questions after Cleo by 3-18 hours with the same answer, he then answered a question incorrectly on both, with the same unlikely error, was questioned about it, gave no response, and then both of them started answering questions away from each-other.

I questioned him on knowing the gender of Cleo, despite no message ever being left related to 'her' gender, and he gave no response, now it is in the description. I also questioned his answering questions directed at her, and he gave no response. He then changed 'her' description to a bitcoin donation address, and immediately left a message to her saying he wished he had some bitcoins at the moment because it would make sense to donate to her

r9m

@Committingtoachallenge aha ! ;) but that does not prove the fact that they are in fact the same person in real lives .. they might be acquaintances in real life and might have shared or discuss problems with each other :)

robjohn

09:13

@Chris'ssis I don't imagine that Cleo has started providing any work with her integrals.

Committing to a challenge

@r9m If you look at Cleo's description, this isn't possible ;)

"I have a medical condition that makes it very difficult for me to engage in conversations"

Chris's sis

@robjohn No, but from all activity on main I can see there is a very strong connection between these accounts: Anastasya - Tunk-Fey - Cleo.

r9m

@Committingtoachallenge that might very well be the truth .. there is no evidence pointing otherwise

Committing to a challenge

@r9m There was bragging from Cleo that is now deleted, that said nothing of this, and was contradictory

r9m

@Committingtoachallenge yes !! that is fishy alright !!

@Chris'ssis is the race still on ? :-) (I'm being childish I know .. but I'm just curious to know your name =P =))

Chris's sis

09:22

@r9m That paper you showed me yesterday annoyied me terribly. I love my work very much ... :-)

Balarka Sen

@Chris'ssis Oh? I don't believe Anastasiya is related to Tunk-Fey or Cleo.

Look at the language accent.

Chris's sis

@BalarkaSen You can believe anything you want to, in my opinion there is a whole family of mathematicians involved in this story.

Balarka Sen

Ana can't speak english very well, while Tunk-Fey knows how to speak english.

r9m

@Chris'ssis I'm sorry if that troubled you ... but I bet you can surpass the contents of that paper by a factor if ten or more if you put your work in one single paper (people will go crazy just over the diversity of stuff)

Chris's sis

@BalarkaSen This means absolutely nothing.

Balarka Sen

09:25

shrugs

Chris's sis

@IceBoy Hi

r9m

r9m green !!

Chris's sis

@r9m My point is this: if you somehow suggested that I'm inspired from that paper, this is completely wrong. You don't now me in the real life, but I'm very proud of my work, and I'd never take things from elsewhere and then say that is my work. That proof of Au-Yeung is totally coming from my mind, it's my creativity spirit at work, and it's the best proof in the world in my opinion.

The mathematicians I use to talk to often send me papers and say: "well, you're not the first that found this, here is a paper for you to see". I often find results that were previously discovered, but this is not my fault (sometimes days in a row).

The proof of Au-Yeung series is totally original, it's the most brilliant thing of mine I've created in the last months.

Balarka Sen

@GustavoMontano We have the two metric space $(X, d)$ and $(X, \bar{d} = d/(1+d))$. We want to prove that the open sets of the two are the same. Consider the open sphere $S_r(x_0)$ on $(X, d)$. $r > d(x, x_0) > d(x, x_0)/(1+d(x, x_0)) = \bar{d}(x, x_0)$ so we get an open ball $S'_r(x_0)$ in $(X, \bar{d})$. Conversely, $S_r(x_0)$ be the ball in $(X, \bar{d})$. Then $\bar{d}(x, x_0) < r$ implies $d(x, x_0) < \frac{r}{1-r}$.Pick $r_1 = r/(1-r)$ which gets you a ball $S_{r_1}(x_0)$ in $(X, \bar{d})$

Done.

Ice Boy

50 mins ago, by Ice Boy

Greetings

Balarka Sen

09:32

@Chris'ssis I know that feeling. But I eventually get over it.

r9m

@Chris'ssis I had no idea that the paper even existed until 2 days .. and I know you always mention source if you are sharing problems form someone else's book/paper .. and there was never any doubt in my mind that your work was independent and original .. a lots of discoveries have been rediscovered that were lost over time (so this is nothing new)

Chris's sis

@r9m Maybe one day you'll discover something amazing and one will give you such a paper. Only this way you can understand that shitty feeling ... (especiall when you know you worked hard, you did research and so on)

Balarka Sen

Been there done that. It feels terrible.

Get over it @Chris'ssis. You are capable of discovering more stuff than any of the guys working with integral and series I know of.

Chris's sis

@r9m For instance I also discovered a nice series representation of $$\int_0^1 x^{n-1} \log(1-x) \log(1+x) \ dx$$

@r9m I've never ever seen such a stuff. I wonder myself if I'm the first that did it ...

@BalarkaSen :-)

Balarka Sen

=) We know that your works are original. No need to fret over it.

Ice Boy

09:37

@BalarkaSen it gets harder to just "get over it" as you become older

Chris's sis

@r9m I also discovered the version for $$\int_0^1 x^{n-1} \log^2(1+x) \ dx$$ and here you need to use the polylogarith, you cannot only make use of harmonic series, but only with some conditions.

Balarka Sen

@IceBoy "older" as in "mathematically"?

yes, I agree that I am not nearly as mathematically matured as half the people in here. I have mentioned that before.

Ice Boy

@BalarkaSen no as in age, young people bounce back quicker

Balarka Sen

oh

Ice Boy

Thus, you do not have a valid point in this case; but since @Chris'ssis has shown me, in particular, her true colors when you did have a valid point, by ignoring my "Greetings," her side of the story does not interest me in the slightest.

Chris's sis

09:48

@IceBoy Did I ever ignore you? When?

Ice Boy

1 hour ago, by Ice Boy

Greetings

This^ is the THIRD time

Chris's sis

@IceBoy First I said Greetings, and then you said Greetings. Should I say Greetings more times in a row?

Committing to a challenge

@IceBoy Hi - Chris's

@Chris'ssis Here @Iceboy

Ice Boy

Hi

Chris's sis

@Committingtoachallenge Ah, yeah, thanks.

@IceBoy you see ...

Ice Boy

09:51

@Chris'ssis see what?

Chris's sis

@IceBoy I responded to you.

Ice Boy

@Chris'ssis he acknowledged my "hello" NOT you

Chris's sis

@IceBoy I don't get your point

Ice Boy

@Chris'ssis never mind, it's no big deal :-)

Chris's sis

OK

Ice Boy

09:54

this internet interaction is so not like real life in some ways

@Chris'ssis I apologize

Chris's sis

@IceBoy No need for that. ;)

Ice Boy

OK

Balarka Sen

Who cares about a wretched hello anyway.

Ice Boy

in real life I do

most of the time

Committing to a challenge

Hey @Iceboy

Balarka Sen

10:04

Let's all greet @IceBoy

Hello @IceBoy

:P

LOL

Ice Boy

I surrender.

Balarka Sen

Party hard, but pick up the trash cans after the party.

3

Ice Boy

well said^

Committing to a challenge

What is the fastest way to solve 5 linear equations?

r9m

@Chris'ssis you are making it sound like I dug out the paper to hurt you intentionally .. if you really felt that way I really feel sorry ... but sometimes (like this one) the way you talk hurts me :( .. don't get the wrong impression .. I was only trying to share what I found with you and I was just as surprised and sudden struck (it was in the list of reference of a NT paper I was reading)

Chris's sis

10:12

@r9m my work wasn't there at all, it was just something that is similar with a tiny bit of my work. They have absolutely no connection. Sorry if I hurt you in the past, but you can let me know when this happens. You seem always happy and hard to be touched by words. :-)

This was my work (initial one)

and 4.2.44. from here is similar to what I have in my proof

arxiv.org/ftp/arxiv/papers/0710/0710.4023.pdf

I need to find an example to explain that well. I was studying logarithmic integrals, and when I managed to computed that one, that is $$\int_0^1 x^{n-1}\log^2(1-x) \ dx$$ I immediately saw the amazing thing there. I was instantaneously struck by a possible connection with Au-Yeung series.

As I said above I studied more integrals from this family, some I also posted here. For instance, I have a proof where I used $$\int_0^1 x^{n-1} \log(1+x) \ dx$$ that I also discovered that.

r9m

@Chris'ssis yes .. applying the $x \to 1-x$ in that integral and writing the $(1-x)^{n-1}$ with binomial expansion and evaluating the integrals gives the identity)

Chris's sis

@r9m I and a mathematician proved that identity in a different way. Actually he managed to proved it that way. I discovered the identity playing with beta function. For publishing the paper, I needed a very clever way of proving it.

r9m

@Chris'ssis if you remember .. I put that approach in my blog after I saw your proof :-) (I had no idea/knowledge that such a paper existed at that point in time)

Chris's sis

@r9m What you wrote seem a very clever way I didn't think of.

Ice Boy

@r9m the same blog you were talking about earlier?

r9m

10:23

@IceBoy yas :)

Chris's sis

@r9m I wonder if that guy, Donal F. Connon, really knows my integral, if he is aware of it. If you have an e-mail of him, you can ask him. I doubt he was aware of it!

@r9m The other mathematician asked his colleagues when I sent the paper, and he concluded my integral result is a new one.

r9m

@Chris'ssis Addendum 1 here :-)

Chris's sis

@r9m Nice that way, I saw it. I had no idea to go that way.

r9m

@Chris'ssis the idea I employed there is discussed at length in Connon's paper .. while I got the idea after looking at the integral identity that you proved with beta-functions in your proof :-)

Committing to a challenge

Is that your wordpress @r9m

r9m

10:30

@Committingtoachallenge yes (I changed the domain name :P)

Committing to a challenge

I thought you said you deleted it?

So now noone knows?

Chris's sis

@r9m hehe, I found his e-mail here, [email protected]. So, if you want, you can email him and personally ask him if he even was aware of my proof, of that logarithmic integral. If you want to you can attach the doscument above with my whole proof. I'd be really glad to see his answer.

r9m

@Committingtoachallenge now you know :) ..

Chris's sis

I bet on NO. It's simply too awesome not to include such a piece in your work with over 200 pages.

r9m

@Chris'ssis (+1) agreed !!

Committing to a challenge

10:34

@Chris You seem very defensive, and it seems unwarranted

Chris's sis

@Committingtoachallenge This is one of my best proofs I've ever created. Of course I care of it. And it's even more than this, it's the best proof in the world in my opinion. How to give up such a creation?

Committing to a challenge

The... Best... Proof... In... The... World?

ONE of your best, yet it is the best in the world?

Chris's sis

@Committingtoachallenge Exactly, it's the best proof in the world to Au-Yeung series (in my opinion).

Committing to a challenge

You are a student aren't you @Chris's What year?

r9m

@Chris'ssis opinions vary ! I still worship Sandham and Kneser's proof ! =)

Chris's sis

10:39

If I said other thing I wouldn't be sincere. This is what I strongly believe on it.

@r9m Yeah, this is also nice and agree the opinions vary.

@Committingtoachallenge No, I'm not a student, no math background here.

Committing to a challenge

No math background?

r9m

welcome to the Jungle ! (moments when raw talent beats cultivated inteligence ;) .. )

Chris's sis

@Committingtoachallenge True.

Committing to a challenge

So what do you do as a job?

Chris's sis

@Committingtoachallenge accounting and tutoring (this one not as a job)

Committing to a challenge

10:41

No math background? - Accounting?

26 years old or so?

Hey @JasperLoy

Chris's sis

@Committingtoachallenge I don't love to talk about me here :-), but about integrals, series and limits.

Committing to a challenge

Characteristic of a fake persona xD.

Chris's sis

Sorry?

Committing to a challenge

Nothing :)

r9m

@Committingtoachallenge or of one who is tired of fake personae and has eventually succumbed to it :L

Chris's sis

10:45

@Committingtoachallenge I didn't know that being anonymous and not sharing all your life on internet has such a meaning.

Committing to a challenge

:)

Chris's sis

Once (or more times?) I lost an interview because I didn't want to talk about me (in details).

r9m

^ ya .. that happened to me too ... some interviewers ask too much about what's going on in our tiny little brains ..

Chris's sis

I'm here because I appreciate very much people that come to this site, I feel as if I'm home if you know what I mean.

Jasper Loy

I am here to talk nonsense with people, lol.

Committing to a challenge

10:47

Don't be too defensive, it would make you seem dishonest :)

Ice Boy

54 mins ago, by Ice Boy

this internet interaction is so not like real life in some ways

Chris's sis

@r9m Someone even asked me why I use a certain description to my email, and that was crazy to me. I'm not an opened door to the world as regards my personal life.

Committing to a challenge

Oh, your age would have been far too much to share, that makes sense

With that I would know what you look like and where you are

Ice Boy

in life

Committing to a challenge

Normally one would think your career choice says more about you than your age, but I guess using tracking technology with age as the only input

But of course, telling people facts like, what your age is, makes keeping the lie harder :). That gives you a birthday, and means you have to know when events occurred or not\

r9m

10:53

@Chris'ssis I don't know why .. mine was a not a job interview .. (an interview for joining an institute) they wanted me to share intuitions behind the solutions I was writing down on the board (like how/why did you think like that) .. but I had no idea what was an acceptable response to such a question .. :P so they kept on bumping the difficulty of the problems to a point that I almost felt they were harassing me :P LOL

Chris's sis

@Committingtoachallenge I like to believe that the passions say more about people. I know many engineers that spend more time on TV than doing something really amazing, nice like mathematics, research ...

Committing to a challenge

I didn't say they did not

Ice Boy

hi @DanielFischer

Chris's sis

@Committingtoachallenge And I know engineers that are not able to solve a quadratic equations.

Committing to a challenge

I doubt that Chris

Chris's sis

10:56

I know doctors that are not able to keep into account all details when give you a certain medicine. I experienced that.

Committing to a challenge

Atleast any engineer in Australia must take Calc 1&2, aswell as PDEs

Chris's sis

@Committingtoachallenge You shouldn't doubt.

@r9m :-) Did you manage to join there?

Jasper Loy

@Committingtoachallenge I think ODEs too?

Committing to a challenge

ODEs are a subset of calc1-2 usually

Jasper Loy

@Committingtoachallenge Oh I see. Just very simple ones I think.

r9m

10:57

@Chris'ssis I rejected their offer and joined the insti I'm currently studying in :P

Jasper Loy

I applied to Cambridge, but I did not get a place for Mathematics, though they gave me a place in Education with Mathematics, which was not what I wanted.

@Chris'ssis About half the doctors I have seen I consider stupid, making stupid deductions. They did not really listen to what I was trying to tell them.

Committing to a challenge

Doctors in Australia must take 9 years in Medicine and are often the most dedicated people you will meet, might just be a difference in standards

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$Mathematics$ Mathematics