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13:00
@robjohn Ah, OK. Great! :-)
@robjohn hi
@robjohn I saw your comment on my question just now.
@robjohn does the limit exist?
@Chris'ssis yes
@robjohn do you think your approach for the linked question can really help with my question?
@user2179021 which question?
13:11
@robjohn oh sorry.. my nick is screwed up.. math.stackexchange.com/questions/1059379/…
You are using a different name in chat?
@robjohn The stackexchange system can't cope it seems.. my nick randomly get screwed up in different places
@user2179021 That is how I would start thinking about it. I don't know if it will help
@user2179021 do you want different names on different sites, or just one?
13:13
@robjohn ok that is interesting.. my problem is this. It seems that the number of zeros you expect in w changes when you condition
@robjohn same name.. I think the real problem is that more than one person uses this browser
@robjohn and stackexchange tries really hard to hold on to your identity.. too hard in my view
@user2179021 do you log out of SE each time you leave the browser?
@robjohn back to the math :) So it seems that conditioning on the first inner product being zero, you expect there to be more zeros in w
@robjohn umm. I am not honestly sure if I do log out each time. But I know that even if you do, and then someone else wants to log in as someone else, you often have the situation where different parts of the SE network have different IDs for you
@user2179021 I haven't thought about it. I don't immediately see why, but it could be.
@robjohn I don't really mind :)
@user2179021 your chat account is linked to your SO account, and this is your name on SO
13:16
@robjohn on the math.. my intuition is that you are more like to get 0 inner product if you have more 0s in the vector. So if you know you got 0 inner product before then this skews the number of 0s you are likely to have
@robjohn ah ok thanks
@user2179021 I'm not saying you would get the same answer, I am just saying that I would look at it using the same methods.
@robjohn can you tell me if this question was deleted today? I'm not 10k yet :(
@robjohn ok thanks. I have tried to solve it myself with no luck so far. I even implemented a simulation!
@user2179021 I can change you chat account to be linked to you math account, if you like, but it's up to you.
@robjohn thanks!
13:19
@Integrator hang on...
@MatsGranvik Why don't you use $\LaTeX$ in your profile's about me section??
@robjohn $-6$? However, Mathematica suggests it doesn't exist.
@Integrator It was deleted by vote. You could go and ask it be reopened on meta
@Chris'ssis It is not $6$... but that is close, in a sense.
@robjohn No, It's fine thanks!
@user2179021 does that mean you want me to so link your chat account?
13:23
@robjohn yes please
@Integrator I never thought of that.
@robjohn And what kind of flag should be raised for this kind of answers, I've raised low quality flag!
@Lembik There... as soon as people refresh the page, they will see your name
thanks
@robjohn That question derives from a question I saw on main these days. I think the best way is to make use of Cosine Integral and the Sine Integral.
13:25
I feel so important now :)
if only I could also solve my prob problem
@Chris'ssis yes, directly from my answer, in fact :-)
@robjohn do you think people would be interested to see exact numerical results?
@Lembik I don't know. Only to check their theoretical answers.
@robjohn ok.. just wanted to find something to contribute
@robjohn Well, or by using complex analysis ...
13:27
@Lembik when you hover over someone else's comment, do you see a little bent arrow at the right of the comment field?
@robjohn ah yes
@robjohn I would have done it as Felix Marin since I'm not used to complex analysis.
@Chris'ssis I used complex analysis
@Chris'ssis yeah, but I was trying to simplify the integral, using Si and Ci does not seem to simplify.
It aids in entering it into Mathematica :-)
@robjohn Isn't it helpful to use their series representations?
@Chris'ssis I haven't looked into the various representations... if there is one that would make things simpler?
@Lembik thanks :-) that makes multiple conversations easier to follow (and I am often in multiple conversations).
13:31
@robjohn I didn't check things in details.
@robjohn my pleasure and thank you!
@robjohn Assuming[a>1,Integrate[Cos[a x]/(1+x^3),{x,0,Infinity}]]
MeijerG[{{1/3},{}},{{0,1/3,1/3,2/3,5/6},{1/6,1/2}},a^6/46656]/(2 Sqrt[3 [Pi]])
@robjohn the last form does not lead us to any finite limit and it suggests the limit doesn't exist.
@robjohn I'll check the details later on.
@robjohn Despite the bounty, my question is not getting many views, is it for the title? How could I improve it?
13:46
@VincenzoOliva Maybe improve the start? That is introduce the problem and some motivation before going through your working
@VincenzoOliva Also the problem statement is a little unwieldy. Can you split it up into at least two parts?
@VincenzoOliva Do you even need to ask both questions in the same question?
Hrm, I need some guidance with some sub-part of a big exercise question. I want to show the Sylow-2 subgroup of $S_7$ is isomorphic to $D_4 \times C_2$, and then count the number of elements of order $4$ in such a sylow-2 subgroup and conclude it can't have $Q_8$ as a subgroup. Anyhow, I am stuck in the first phase.
I understand that all the elements of the sylow-2 subgroup of $S_7$ must be permutations whose order is either 2 or 4, and necesserily not all of order 2.
I have no clue how to proceed, I just need a little kick in the right direction though.
@Lembik Thank you for helping me!

Uhm... what could I say besides that it is something I encountered in a new study of mine? (I'd like to avoid telling others what I'm up to; is it bad?)

You mean one part for each inequality? I think they really belong to the same problem, so I wouldn't think of asking another question
Also, what do you think of the title?
(going to lunch now)
14:10
Hi
Is there any body
@VincenzoOliva If you can think of a better title, otherwise, give some time for people to see it.
I made myself for a test , so I have to solve math and chemistry problem how can I increase my problem solving skill
I try to solve a lot of problem can help me
Can it help me to improve it
Is there a simpler way than using $\sin(a+b) = \cos a \sin b + \sin a \cos b$ for finding $\sin 3x$ given that $\sin x = 1/2$?
I got the answer 1, so I was thinking there were a simpler way.
fuuuuu, I just found it
silly me
14:26
Hello @robjohn !! Can I ask you further about the exercise:

Let $1 \leq p_1 \leq p_2 \leq +\infty$. Show that in a Lebesgue measurable $E\subset R^d$ with $0<m(E)<+\infty$ we have that $L^{p_2} \subsetneq L^{p_1}$.

??
15:04
This is, by far, the best question I've seen in a while.
Did you just find that $\sin x = \frac{1}{2}$ is satisfied by $x=\frac{\pi}{6} \implies 3x = \frac{\pi}{2}$ so $\sin (3x) = 1$, @N3?
@Khallil BTW, I solved the integral Venus posted yesterday! The answer is $\displaystyle \pi\log(4)$
No way! I got something to do with $\pi \log(2)$ and an integral I couldn't evaluate, @teadawg1337. How'd you do it? ^_^
15:20
@Khallil I dunno, I didn't write down the steps. I got to $\displaystyle \int_0^1\frac{\operatorname{Li}_1(w) \sqrt{\operatorname{Li}_0(w)}}{w}\mathbb{d}w$, but I did the rest in my head and wrote down the answer I got.
Ah. That's to do with those logarithmic integrals I've not heard of yet! Well done! That was a really tough one, @teadawg1337. ^_^
Are you an undergrad, @teadawg1337?
@Khallil Nope, I'm a high school grad. I'm starting college next semester
Woah! That's awesome! College is the equivalent of university in the UK, right @teadawg1337?
@Khallil Yes
Awesome! I didn't know that you could start in the second semester. Is it a common thing to do, @teadawg1337?
15:23
@Khallil Not to my knowledge, I put it off so I could study some mathematics independently
Suppose I know that $|a-b|>t$. Can this lead to an inequality of the form $|a-b|>f(t,\epsilon)?$
@Khallil For me, it'll be my first semester.
What's the significance of $f(t, \epsilon)$, @Eric?
Ah, I see. So you'll be a semester behind those that started before you then, @teadawg1337?
I mean, if $t \geq f(t, \epsilon)$, then you can arrive at that inequality. Otherwise, it doesn't seem possible at first sight. I might be wrong, @Eric.
@Khallil I suppose... I had another good reason to put off college for a semester: severe depression.
Ah, I see. I was speaking to my tutor and apparently it's not as uncommon as you'd think for mathematicians, @teadawg1337.
15:38
@khalil, f is just some function of epsilon and t
16:05
Anyone have a quick way to show $\displaystyle \int_0^\infty \bigg \vert \frac{\sin x}{x} \bigg \vert \, dx = \infty$?
In that case, as I said, you'd need to exploit the transitivity of inequalities, @Eric.
@RobertCardona This is a classical problem, and I was thinking of a different approach now: Is it useful to consider $$\int_0^{\infty} \frac{\sin^2(x)}{x} \ dx$$?
Well, we can show $\displaystyle \int_0^\infty \frac{\sin x}{x} \, dx = \frac\pi2$ using Dominated Convergence Theorem I think.
yeah, It would be useful to consider
I think I need to brush up on my Math, it's been a while....lol
$\sin^2(x) \leq \vert \sin x \vert$
so if we show your integral diverges, mine must as well
16:13
Apostol's Mathematical Analysis vs. Zorich's Analysis (I and II): which one is better? I have both, I'm just curious
I've been hearing a lot of people talking about Zorich's books lately
committingtoachallenge is using them
and I think will hunting as well
@RobertCardona Does it help to integrate by parts then?
Q.E.D.
@RobertCardona Welcome :-)
@Chris'ssis Do you have an opinion on those two texts?
16:16
@teadawg1337 I didn't read any of them. Sorry.
If you're interested. The problem came up as follows: I was trying to come up with $f \in L^1$ such that $\widehat f \notin L^1$. I chose $f = \chi_{[-1, 1]}$, and $\widehat f$ looks something like what I started with, but to show that it isn't in $L^1$, I need to show the integral of it's absolute value isn't in there.
@Chris'ssis It's fine
@robjohn did you manage to get some results on that integral?
@RobertCardona You can also simply observe that $$\int_{k\pi}^{(k+1)\pi} \left\lvert \frac{\sin x}{x}\right\rvert\,dx \geqslant \frac{1}{(k+1)\pi} \int_{k\pi}^{(k+1)\pi} \lvert \sin x\rvert\,dx = \frac{2}{(k+1)\pi}$$ for $k\geqslant 0$, and since the harmonic series diverges, $\frac{\sin x}{x} \notin L^1$.
16:27
Ugh... I hope to eventually get a copy of Rudin, but it's so dang expensive...
That's what libraries are for!
@teadawg1337 abebooks.com/servlet/… not too bad
@teadawg1337 which country are you in?
@Lembik I live in the US
@teadawg1337 do any of those prices look good?
@Lembik Those prices look a little sketchy imo...
16:33
@teadawg1337 in what sense?
I looked up the description of the Mass Market Paperback (the one that's cheapest in the link you've provided) on Amazon: "Paperback edition published in China. This book is completely in English with totally the same contents."
"... completely in English with totally the same contents."
1,
1,1,
1,1,1
1,1,1,1
1,1,1,1,1
...
7683417568?
16:49
@Chris'ssis sorry... I've been walking a dog, answering a question, and dealing with some moderator stuff. I still have it in a window to work on it.
@teadawg1337 That's because you clicked on the chinese language one! Look at abebooks.com/servlet/… instead
@Chris'ssis I know that it is $$\int_0^1\int_0^1(1-s-t)_+\frac{\mathrm{d}s}{\log(s)}\frac{\mathrm{d}t}{\log(t)‌​}$$
@robjohn I know that it is $$\int_1^{\infty} \int_1^{\infty} \frac{\Gamma(x+1)\Gamma(y+1)}{x y \Gamma(x+y+2)} \ dx \ dy$$
@Chris'ssis you don't know a closed form yet?
@robjohn Not yet.
16:59
@Chris'ssis yes. $-6$. I thought Mathematica might have some trouble with that one. I am sure it won't be able to handle $$\lim_{a\to\infty}a^6\left(6+a^4\int_0^\infty \frac{\cos(ax)}{1+x^3}\mathrm{d}x\right)$$
@robjohn Mathematica gives very different results.
0
Q: Evaluating the integral $\int_1^{\infty} \int_1^{\infty} \frac{\Gamma(x+1)\Gamma(y+1)}{x y \Gamma(x+y+2)} \ dx \ dy$

Chris's sisDuring the study of some integrals I came across a very interesting integral, that is $$\int_1^{\infty} \int_1^{\infty} \frac{\Gamma(x+1)\Gamma(y+1)}{x y \Gamma(x+y+2)} \ dx \ dy$$ Apparently nothing seems to work, but I bet you can do much more than me. So, what tools would you recommend me t...

I'm curious to see if anyone knows how to approach such integrals ... It seems a tough nut at first sight, but I might be wrong.
@Chris'ssis looks like a Beta Integral modified. The integrand is $\frac{\mathrm{B}(x+1,y+1)}{xy}$
@robjohn Yeah, that's true.
@robjohn I need to check some paper on gamma function integrals.
@DanielFischer Speaking of those users, your account and two of your comments were imported to Physics Overflow. I don't know how you feel about this, but I thought you should be aware.
@MikeMiller Hmmm. Perhaps I should send a "cease and desist" message?
17:15
@Chris'ssis Mathematica does handle it, actually.
What's the difference between "Baby" Rudin and Rudin's Real and Complex Analysis? Would diving straight into the latter be mental suicide?
@Chris'ssis FullSimplify[Limit[a^6 (6 + a^4 Integrate[Cos[a x]/(1 + x^3), {x, 0, Infinity}]), a -> Infinity]]
@robjohn Maybe with the proper code.
Their import practices are OK by the CC license we have. I'm more displeased by the account import than the content import, but both are properly attributed. You might see the recent discussion on Meta.MO that made me aware of this practice.
@Chris'ssis That answer simplifies to a two character answer.
17:18
@robjohn I test the code now.
@robjohn 362880, this is the answer I got
@Chris'ssis $9!$
@robjohn Yeap.
@Chris'ssis This follows easily from the asymptotic expansion in my answer.
Hello. Anyone here good with commutative algebra?
(and algebraic geometry?)
I'm not, but I know a little bit about it.
17:25
mmk
my problem is this:
"For an arbitrary field, show that $\sqrt{IJ} = \sqrt{I\intersect J}$
woops, one sec
there we go
I,J are ideals
Ideals in what?
some arbitrary field, $k[x_1, ... , x_n]$
You're looking for \cap, I think.
Ah okay, that's what it is
Are you able to get either of the inclusions?
(This is actually true for arbitrary ideals in a commutative ring, not just polynomial rings.)
17:30
"For ideals $I,J$ in an arbitrary field $k[x_1,...,x_n]$, show that $\sqrt{IJ}=\sqrt{I \cap J}$."
there we go- that looks better
well, here's what I have done so far
a theorem in my text allows me to make the claim that $\sqrt{IJ}=I(V(IJ))$
The real question I have regards what IJ actually means
is it the set of products of each polynomial in ideal I with each polynomial in ideal J?
Ah, this is an annoying terminology at first. What you just described is not always an ideal.
$IJ$ is the ideal generated by things of the form $ij$, where $i\in I, j \in J$.
mmkay, and just to be sure, $i,j \in k[x_1,...,x_n]$, correct?
Sure, because $I$ and $J$ are ideals of it.
This is the definition of a product of ideals in any ring, though.
17:38
What is Ideal?
Nevermind, I will google.
so $V(IJ)$ is the set of all points where the polynomials vanish, correct?
I think you might be thinking too hard about this. Can you show $\sqrt{IJ}\suvset \sqrt{I\cap J}$?
It's the set of points that are killed either by all the polynomials in $I$ or all the polynomials in $J$.
@robjohn Are you here?
(To prove what you want to prove, you don't need to use any properties of polynomials!)
17:42
@Studentmath generalization of normal subgroups in rings.
perhaps I missing something obvious
Write down the definitions. What does $f \in sqrt{IJ}$ mean?
@Balarka looks interesting. I have two more chapters to go, and then I am into Ring Theory.
what chapters are you doing atm?
$f\in \sqrt{IJ}$ is some polynomial in a set that has polynomials generated by the polynomials in I and J
17:44
Screw it, I'm buying both Baby Rudin and Real and Complex Analysis.
@teadawg1337 To answer your question on those books: they're different in that the first is a prerequisite of the latter.
perhaps I'm unsure of what $\sqrt{IJ}$ is supposed to mean
is it just the polynomials in IJ, with their exponents divided by 2?
i am reading a course on sequence and I don't really understand how can I found $lim \inf (a_n)$
(sorry if this is such a trivial question)
@daOnlyBG I don't quite get what you mean. The definition I know (and the usual one) is that for an ideal $I$, $f\in \sqrt{I}$ if $f^n \in I$ for some $n$.
17:46
Currently went back to Topology, but next two up are 'finitely generated ablean groups' and Normal sequences
hmm. it looks like Cech nerve of any open cover of the solenoid is the uncountable graph of $|\mathbf{Z}_p|$-many vertices obtained from taking the inverse limit of circles with $p^n$-th roots of unities marked (the nodes) and the arcs joining them as edges.
For instance, $x \in \sqrt{(x^2)}$, but $x \not\in (x^2)$.
@Studentmath ah.
@Mike Sounds like a blast!
Wish me luck, guys. I may or may not need it for reading Rudin
@MikeMiller thanks- let me try that again, perhaps that's the definition I was missing
what would an element of $IJ$ look like, then?
17:50
@MarcGato Sorry... I am back. Lots going on this morning
A sum of things of the form $ij$.
OK, good
one minute- let me draft something
I believe I have everything I need
Mathematics may not be for lazy people. But it is created by lazy people.
If you have everything you need, I won't say anything more.
lol I saw what you removed, and I was trying to do just that
17:52
@MatsGranvik Mathematics is totally not "created".
It's discovered.
@BalarkaSen ok
@Mike how about Rudin's Functional Analysis? Are Baby and R&C both pre-reqs?
Essentially, yes. Baby is a prerequisite to pretty much any form of analysis you want to do.
Some topics in R&C are used in FA.
analysis
it's fortunate that mike is ignoring me :P
Hey @MikeMiller Could I ask you a question about flexes of a curve?
18:04
My main draw to Rudin's texts is their alleged difficulty :)
Dunno what that is; can't help.
@MikeMiller I've been through Baby Rudin and Real and Complex Analysis, but I haven't read Functional Analysis.
OK, so here's what (I think) works:
Let $f^n \in IJ. Thus, f\in \sqrt{IJ}.$ This implies that $f^n$ is of the form $h_1\cdot i_1\cdot j_1, h_2\cdot i_2\cdot j_2,..., h_k\cdot i_k\cdot j_k$. If this is correct, I'll let you know what else I have-if not, a correction would be appreciated
@robjohn What should I expect for Baby and R&C? I'm almost done with Apostol's Analysis, and I'm skimming through Zorich as well
@robjohn I've only read bits and pieces of both... I stopped reading his complex half when I decided I liked other folks better.
18:14
@teadawg1337 I don't know what's in Apostol or Zorich, so I can't say
@MikeMiller I had an excellent teacher for complex analysis. We could have used almost any book, I think. I think we used Ahlfors, in that class. Perhaps it was Rudin for that class and then Ahlfors was for the graduate Complex Analysiis course that I took soon after
It's been a while, so it is hard to remember
That's always a nice experience; I had that experience for my first topology class.
@robjohn Apostol covers Riemann-Stieltjes integrals, infinite series/products, sequences of functions, Lebesgue integration Fourier series, and residue calculus (among other things). Not sure about Zorich, I just started skimming through the PDF an hour ago
My complex analysis didn't do enough for my taste, so I just kept reading... I like Ahlfors a lot
Hey all!
hi
@MikeMiller, can I ask you a couple follow up questions?
You can try :)
just posted a question; anyone see anything i've obviously left out / goofed on?
0
Q: Closed-form of sums in Fourier series of $\sqrt{1-k^2 \sin^2 x}$

SemiclassicalConsider the even, $\pi$-periodic function $f(x,k)=\sqrt{1-k^2 \sin^2 x}$ with Fourier cosine series $$f(x,k)=\frac{1}{2}a_0+\sum_{n=1}^\infty a_n \cos2nx,\quad a_n=\frac{1}{\pi}\int_0^{\pi} \sqrt{1-k^2 \sin^2 x}\,dx$$ as considered in this earlier question. In comments for that problem, the foll...

if $f\in \sqrt{IJ}$, then $f^n$ must have the form $h_1i_1j_1+h_2i_2j_2+...+h_ki_kj_k $, correct?
18:45
Where the $h_n$ are just arbitrary polynomials?
yes- sorry
$h_m \in k[x_1,...,x_n]$
also, is there a better name for "a periodic function whose Fourier coefficients are polynomials in a parameter"? i suspect not, but it'd be cute if there was
Two comments. First - yes, for some $n$. Second - the $h_n$ are redundant, because $I$ is an ideal; so by definition $h_ki_k\in I$, so you could just not write them at all.
@Semiclassical I saw the bounty you put on my question; appreciate it.
no problem. figured that since i'm the one who initially got the ball rolling on that topic, me putting a bounty on it was appropriate
though i'm not sure what kind of response it'll get anyways. it feels like a question which is almost-but-not-quite more suited for MathOverflow than MSE
figuring out the line between questions for MSE v. MO is something i have a hard time judging, though, so take that with a grain of salt
@MikeMiller gotcha. By the same reasoning, then for some polynomial $g$, $g^n \in \sqrt{I \cap J}$ implies $g^n$ has the form of whichever polynomials are in both I and J.
18:51
0
Q: Some combinatorial identity.

PrzemoLet $a_1 \ge 0$ and $a_2 \ge 0$ be real numbers and let $n_1 \ge 0$ and $n_2 \ge 0$ be integers. Finally let $m\ge 1$ be another integer. By using the method of generating functions I have shown that the following identity holds: \begin{equation} \sum\limits_{i=0}^{m-1} \prod\limits_{\xi=1}^2 \bi...

eww
well, mabye I'm just not used to those. But they look horrible
it's a typical Przemo question (which is not a bad thing IMHO, to be clear)
trying to show that the former is in the latter, though, is a bit weird
haha, I just noticed almost none of his questions get answered
yeah, heh. i like his questions, but they require a lot of work
anyone interested in a probability question? math.stackexchange.com/questions/1059379/…
19:07
@MikeMiller? I'm still struggling to see the inclusion
@daOnlyBG Sorry. So this one is a little bit trickier. If $f^n \in I \cap J$, then what can you tell me about $f^n \cdot f^n$?
$f^n \cdot f^n$ vanishes on the same points
and it has terms in both I and J
***has polynomials in both I and J
oh, wait
$f^m = i_1j_1+...+i_kj_k$ implies that $f^m$ is in both I and J
no?
ahhh man that's eays
***easy
But that's not what you're looking for
19:16
@MikeMiller Harro.
@MikeMiller explain
Because $f^n \in I \cap J$, $f^{2n} \in IJ$ :)
OK, here's what I meant
$f^m = i_1j_1+...+i_kj_k$. This directly implies that $f^m\in I$ and $f^m \in J$, hence $f^m \in \sqrt{I \cap J}$
thus, $\sqrt{IJ} \in \sqrt{I \cap J}$
er, subset of
Hi @PedroTamaroff!!! :)

Could I ask you a question about inflection points of a curve?
there's nothing wrong with that, right?
19:25
"Just ask; don't ask to ask"
@Semiclassical Ok.. I will... :)
@PedroTamaroff @Semiclassical How can I find which are the flexes of the cubic curve of Fermat

$$x^3+y^3+z^3=0$$

at $\mathbb{P}^2(\mathbb{C})$ ?
@MikeMiller not sure what the issue was
alright sorry- i'll just ask elsewhere
19:49
@daOnlyBG Sorry, I'm doing other stuff. The point I was making was that $f^n \in I, f^n \in J$, so $f^{2n}\in IJ$, so $f \in \sqrt{IJ}$.
Oh, oops! Sorry!
You were doing the other inclusion. Yes. Your argument is correct.
Thanks
Hi @Pedro
r9m
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20:44
@JorgeFernández actually it looks sweet ! ;) 'terribly sweet' =P
not really though
@PedroTamaroff Hello.
r9m
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@Chris'ssis I might know you ?! come on its not a clue at all =P I know lots of people ! what I do not know is how I could have known you :P
@Venus lol ! maybe :P
@Hippalectryon I have no idea what she is talking about ;)
Can anyone look at example 2 of this?
@r9m Let me show you something that you might like A LOT :-)
15
Q: Evaluating the integral $\int_1^{\infty} \int_1^{\infty} \frac{\Gamma(x+1)\Gamma(y+1)}{x y \Gamma(x+y+2)} \ dx \ dy$

Chris's sisDuring the study of some integrals I came across a very interesting integral, that is $$\int_1^{\infty} \int_1^{\infty} \frac{\Gamma(x+1)\Gamma(y+1)}{x y \Gamma(x+y+2)} \ dx \ dy$$ that is obtained by manipulating $$\int_0^1 \operatorname{li}(x) \operatorname{li}(1-x) \ dx$$ Apparently nothi...

20:52
What the hell is $C_\infty$?
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@Chris'ssis saw it ! I don't understand how legendre polys help in finding the closed form ! -_-
@r9m Neither do I. Maybe he will come with further explanations.
r9m
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@Chris'ssis for that I eagerly wait :D
@Chris'ssis do you know the closed form btw ?
@r9m No, but the intuition tells me there should be a nice closed form. :-)
r9m
r9m
@Chris'ssis I see :) okay !!
20:59
i'm not sure what's terribly confusing: he shows that $li(x)$ can be written as an infinite series in (shifted) Legendre polynomials (whose coefficients are evidently relatively 'nice'), uses this to do the same for $li(1-x)$, then multiplies them together and uses orthogonality to compute the integral
@BalarkaSen You neved solved my topology problem.
I think I gave you two.
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@teadawg1337 rep-magnets ! -_- sigh
@r9m I'm sorry?
r9m
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@teadawg1337 reputation magnet ! Q framed to gather reputations :P
21:11
Are you calling me a rep magnet, or the OP? I was just sharing a topic that I found interesting
r9m
r9m
@teadawg1337 the OP ofcourse ..
21:27
@PedroTamaroff Which one? The Lebesgue number one I already did.
And the other I don't recall.
Lebesgue number lemma is not hard. It says that for any given open cover of a compact space, any point has an open ball that fits into some elt of the open cover.
When the new moderators have been picked I will have been on this site for 100 consecutive days. (hopefully)
So assume not and collect a bunch of points. the whole space is compact, so the sequence converges to something. and as that something is contained in some elt of the open cover, there is a disk around it that fits inside that set.
however we can choose (as the sequence converges to that particular point) a disk around each point of the sequence that fits in the disk that in turn fits inside some open set of the cover.
bleh i just used words. woo.
r9m
r9m
@Alizter nice ! :-) and I'll be in this site for 300 consecutive days in a couple of hours ;)
21:46
@Alizter I'm once again fiddling with Gal(\bar Q/Q) :P
Hey guys! Do you know any good lecture notes on Binary Integer Programming? I'm not really looking for a complete reference, just a quick introduction to the main methods.
Uh. "Hey guys and girls" would have been better. Sorry for that!
Have there been any attempts to prove the Riemann hypothesis by contradiction? Easy points for whoever wants them: the answer is yes.
@JacopoNotarstefano I don't know what that is; sorry.
@r9m don't you wanna become a moderator? You have my vote.
@MikeMiller I'd be more interested in a fuzzier version of that question: What implications of RH and $\not$RH do people consider as (heuristic) evidence for or against RH?
r9m
r9m
@Chris'ssis :D Thank you for your confidence ! :D .. but ninjas are supposed to stay hidden and fight from shadows .. ;)
21:59
@r9m Cake ninjas? :-)
@Semiclassical But I claim this is either too vague or already answered. I would be astonished if there weren't questions on MSE and MO about "Why should I believe RH?"
r9m
r9m
@Chris'ssis I like cakes :P and the way they chop the cakes in air and lets them fall into pieces makes my go crazy (such a waste of nice cakes :P)
oh, i'm sure it's out there. i'm just saying that i find that more interesting than the present question @MikeMiller
@Semiclassical Oh, sure. I do not find it to be a very good question.
r9m
r9m
22:04
@Chris'ssis besides I have yet to understand a lot about the gears that run this site behind the face :-) .. so No rush !!
@r9m Yeah, I got your point perfectly. I also asked myself these days for what reasons I'd like to be a moderator for free? I don't find any reason to ever want such a position for free. I'm also very curious about the real face of the things with this site, but time will reveal it definitely.
@r9m I don't bother too much with these things, I'm here only for having some math conversations with a few people.
2
r9m
r9m
@Chris'ssis I might like to be a moderator some day considering the fact that this site has aided my growth since my participation tremendously !! currently I lack knowledge or the level of maturity required for such a responsibility ! but some day sure I'd like to run for a diamond star ;)
@r9m That back to the aid you're talking about I personally understand it by the questions and answers I posted. As far as I know, to be a moderator is a totally different story, it requires a lot of resources, time, but well, you know better how to spend your resources. To be honest, presently I don't know who will be able to beat you for such a position. Really! :-)
r9m
r9m
22:22
@Chris'ssis 'A leader is best when people barely know he exists, when his work is done, his aim fulfilled, they will say: we did it ourselves.'-Lao Tzu ;)
yo man @G.T.R ... got any mind boggling integral inequalities for me ? :-)
@r9m hi, sure
r9m
r9m
@G.T.R nice !! :-) share please then ... :)
@r9m First I wanna be a great leader of my thoughts, feelings, and all things related to my life, and then to be the leader of the others. To be a moderator is not really to be a leader in my opinion, you're restricted to a set of actions, and possibly only apply what you're told to apply. Never that kind of leader Lao Tzu talked about. World has changed much in the meantime ...
r9m
r9m
@Chris'ssis well when you say 'you're restricted to a set of actions, and possibly only apply what you're told to apply' .. I think of chess ;)
@r9m Let $E$ be the set of $C^{1}$ real functions over $[a,b]$. Prove that $\forall \epsilon>0, \exists D>0,\forall f\in E$, $\sup_{x\in [a,b]}f^2(x)\leq D \int_a^bf^2+\epsilon \int_a^bf'^2$
22:30
@r9m I didn't play chess for years though ... :-)
r9m
r9m
@Chris'ssis ;) I suck at chess though !
@r9m when you're done, prove that if $f:[0,1]\to \mathbb R$ is a continuous concave function with $f(0)=1$ then $\int_0^1xf(x) dx\leq \frac{2}{3} \left(\int_0^1 f(x)dx\right)^2$
@r9m I only played chees with my firends when I was pretty young, but I wasn't poor at it at all.
r9m
r9m
@G.T.R Nice ! .. I think its Block's Integral Inequality :-)
@G.T.R hmm .. is there a square on the rhs ?! $\int_0^1xf(x) dx\leq \frac{2}{3} \left(\int_0^1 f(x)dx\right)$ is true atleast !
@r9m there's a square for sure
r9m
r9m
22:38
@G.T.R I see .. okay :)
Guys... $\displaystyle \frac{\pi\operatorname{Li}_1(\frac12)(\Gamma(5)+1)}{\left[\Gamma(\frac32)\right]‌​^2\left(\frac{\Gamma'(1)}{\Gamma(1)}-\frac{\Gamma'({\frac12})}{\Gamma({\frac12})}‌​\right)}=50$
r9m
r9m
@G.T.R for these inequalities (related to quadrature and stuff) I think it suffices to verify it for piece wise linear functions (concave arrangement in this case) and claim its true for the limiting case .. but I'll try and see if there is any smarter approach :-)
@teadawg1337 that $\Gamma(5)$ on the top is a complete give-away ! ;) ;P
@r9m It's still amazing, though
$\Gamma(5) = 4!$, right @teadawg1337?
@Khallil Yep
r9m
r9m
22:47
@teadawg1337 :-) .. nice ! but not a shell-shocker ! ;)
That's one pretty impressive fraction, @teadawg1337!
How'd you work out the $\Gamma'$s?
Is it the derivative of the definite integral that $\Gamma(...)$ represents, @teadawg1337?
300 consecutive days, @r9m? :0 ^_^
r9m
r9m
@Khallil ya ! crazy right ?! ;) =P .. If I could stay with one girl for 300 days I'd marry her :P LOL
@Khallil $\displaystyle \Gamma'(n)=-\Gamma(n)\left(\frac1{n}+\gamma-\sum_{k=1}^n\frac{1}{k}\right)$
Since no one answered here's a question on the main site: math.stackexchange.com/questions/1061574/…
What does that even mean, @teadawg1337? Is that Euler's constant? I think I've investigated it before ...
That's not even a year, @r9m. Besides, time isn't really an important factor.
Happiness is the most important thing!
22:56
(How does one get a nice preview of a question in chat? Something like chat.stackexchange.com/transcript/message/19001681#19001681)
@Khallil $\displaystyle \gamma=\lim_{n\to\infty}\left(\sum_{k=1}^n\frac{1}{k}-\ln(n)\right)$
It's the Euler-Mascheroni constant
r9m
r9m
@Khallil you are right ! :)
@r9m are you still on Modafinil?

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