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r9m
r9m
11:01
@Sawarnik haha :P ... bring it on kid :P
and also even though i don't like football
@SanathDevalapurkar I can't be sure since I know absolutely no topos theory.
Perhaps you can explain what they are to me in simpler language?
@r9m Argentina will loose.
Muller will blow them to bits. like they did to Brazil.
I mean, if they play like yesterday.
Huy
Huy
@G.T.R I'm working 12 hours per week and make enough for a living. Just need to organise your finances.
Awwwwwwwwwww....Balarka knows some sports stuff!
And he knows it right!
@Sawarnik I like football.
11:04
"Muller will blow them to bits. like they did to Brazil." Exactly.
@Huy do you have a wife and kids ?
@BalarkaSen I think its lose.
Ah, Bengal Board students these days.
@Sawarnik yeah, but potato pohtato.
Huy
Huy
@G.T.R I do not, but if I did have a wife and a kid, both working 12 hours would easily suffice for the three of us, imo.
@G.T.R He's 21.
11:05
typos.
@BalarkaSen :(
@Sawarnik If you saw the Brazil-Netherlands match, you'll see that Thomas Muller contributed to each of the goals they made. All 7.
@Huy how much do you make now with 12 hours a week ?
@BalarkaSen Mistakes again? :|
@BalarkaSen Do you play?
Huy
Huy
Roughly 2700€ per month before taxes.
11:07
@Sawarnik Nah. I am a silent observer.
Let $\mathbf{x}\in\mathbb{R}^n$. Then $\langle \mathbf{x}, \mathbf{x} \rangle = \lVert \mathbf{x} \rVert^2$. If $A\in\mathbb{R}^{n\times n}$, what would be the $A\langle \mathbf{x}, A\mathbf{x} \rangle$? cc @DanielFischer, @robjohn [Thanks!]
@BalarkaSen :P Cricket is better though.
I don't watch cricket, sorry.
That's insanely high ! What field are you in ? @Huy
@Sawarnik I don't like insects which make noise.
11:09
@BalarkaSen Yeah, that's what I say, cricket is better.
Huy
Huy
@G.T.R I'm just teaching maths at a high school.
@BalarkaSen Nor do I.
@Sawarnik Crickets belong to that class.
Darn typos.
@BalarkaSen Cricket does not.
Well, you're better off in Swiss then. You wouln't get past 1500€ here :(
11:11
@Sawarnik Do you have any good problem? Functional equation? Transcendence theory?
@BalarkaSen I didn't see it.
Huy
Huy
@G.T.R The minimum wage for a teacher - even without any diploma such as BSc - is about 70'000€ divided into 13 months before taxes - that is working full-time.
Tune into Star Sports 1 ... India vs England .. :D
Huy
Huy
If I had my MSc plus teaching diploma, the wage would start at 100'000€.
11:13
Dhoni set to make a century, after Vijay .. yay!
Huy
Huy
@G.T.R I'm a bit shocked to find out to be honest. After all, when I was in Paris, stuff didn't seem to be cheaper at all, it tended to be the other way around to be honest.
(and I would expect living expenses to be way lower if wages differ by so much)
r9m
r9m
@BalarkaSen ? kya ?
If anybody's good with mathematica, I could use some help. I want to count the number of ordered pairs of 3x3 matrices over F_2 which are both nilpotent and commute with each other. my formula gives 344 pairs, a different formula gives 400.
@Huy take a look at the gap between Swiss and France :O
@r9m The first thing to do is to inspect the behavior at $n = \exp(k)$.
Huy
Huy
11:18
@G.T.R I see. :P
Also, prime number theorem tells that $\text{deg} Q(x) > \text{deg} P(x)$
@G.T.R Not going to happen. It's kind of reasonable, and it works (in the Netherlands). So absolutely zero chance that the Germans will even seriously ponder how they could introduce such a model.
@BalarkaSen Quickie, find a function discontinuous only on ${1,1/2,1/3...}$.
Huy
Huy
@G.T.R That's just average though. As I live in Zurich, wages in all kinds of fields are a lot higher than on average. :)
r9m
r9m
@BalarkaSen certainly so :-)
11:20
@nullgeppetto $\langle \mathbf{x}, A\mathbf{x}\rangle$ is a scalar. So $A\langle \mathbf{x}, A\mathbf{x}\rangle$ would be a matrix. Might however be a typo and $\langle A\mathbf{x}, A\mathbf{x}\rangle$ was intended.
@DanielFischer Are the Germans not reasonable as a whole ? Also it seems according to the figures above that teaching is well-paid in Germany, lucky you
@DanielFischer If you have a chance could you please look at my MSE post
Dunno .. I had another example in mind.
Huy
Huy
Yeah, I forgot the fraction and now it doesn't converge.
r9m
r9m
@BalarkaSen yup ... :) but didn't want to use PNT if possible .. all I wanted to do is just use the definition of $\pi(n)$ as the functions that counts primes .. the rest is just real-analysis :P
11:23
$$f(x)=\frac1{\lfloor\frac1{x}\rfloor}$$
@r9m Equality is highly unlikely. Any indefinite integral with rational integrand is in the differential extension $\Bbb C(x, \log)$ while the logarithmic integral is not in there (and you can't approximate it arbitrarily well either, $\log(x)$ always grows like $o(s^\varepsilon)$), yet with an error of $\ll \sqrt{x}\log(x)$ (assuming RH). I do believe there is a differential approximation theorem somewhere.
@r9m I stay out of real analysis whenever possible.
@Sawarnik Good find.
Huy
Huy
@Sawarnik: What about $x=0$?
@G.T.R The Germans, so-so. The German politicians - unreasonable beyond any hope of redemption. It is holy dogma for the CSU that family means the wife stays home and takes care of the children (you know, they had child-care in the GDR, so child-care is TEH EVIL), and a large part of the CDU shares that dogma. The reasonable part of the CDU never dares challenge these dogmas.
@Huy Undefined.
So not continuous there, obviously.
@Huy Do the little modifications.
Undefined on x>1 as well, tweak that again.
r9m
r9m
11:26
@BalarkaSen Boy I didn't understand a single word you said there :P .. I fell dumb .. can you simplify that for me ?
Huy
Huy
@Sawarnik: And it is continuous at $x=1$, no?
@Huy Yes. Add $f(x)=1$ for $x>1$.
@r9m An integral with rational integrand is elementary and expressible using $\log$s.
Huy
Huy
No, wait, the upper limit doesn't exist.
@r9m Search up [Li integral]
11:27
Coming in some time.
It's about the best approximation possible for prime counting function.
Huy
Huy
@Sawarnik Right.
r9m
r9m
@BalarkaSen okay
@cc looking good
What I am saying is that Li is not an elementary integral (in it's indefinite form) while a rational integrand integral is. And you can't approximate the integrand in Li integral too well with rational function as $\log(n)$ grows like $o(n^\varepsilon)$ (read up [asymptotic notations + little o])
@r9m This might just be an overkill but it actually would answer a more general question about the order of $|\pi(x) - \int_1^{\log(x)} P(x)/Q(x) dx|$
11:34
@DanielFischer You don't have feminists voicing their concerns ? We have a lot (chiennes de garde, ni putes ni soumises, femen,...)
@DanielFischer that was a typos... I need $\langle A\mathbf{x}, A\mathbf{x} \rangle$, or $\langle \mathbf{x}', \mathbf{x}' \rangle$, where $\mathbf{x}'=A\mathbf{x}\in\mathbb{R}^n$. Thanks!
r9m
r9m
@BalarkaSen IC .. can you give me some reference where I can read up similar ideas ? :)
@r9m I am not sure, I just sprouted something off the top of my head.
I think I have read up a question in MSE concerning a link between differential galois theory and number theoretic approximations. I can give you the link if you want.
@G.T.R We have various brands voicing their concern, but that doesn't mean they have influence. (Although, the situation improves a bit since the EU forces some degree of equal treatment. But tradition is a strong force.)
r9m
r9m
@BalarkaSen ya sure :)
11:40
@r9m this and this
r9m
r9m
@BalarkaSen okay :) .. thanks .. although I don't unerstand a word there .. sorry :|
You can try reading up some differential galois theory if you are interested.
r9m
r9m
@BalarkaSen boy that's beyond my knowledge and current level of understanding .. :|
I have only scratched it's surface ones but I have a copy of a book on Picard-Vessiot so I can read up whenever I want.
r9m
r9m
I hope I will learn enough in a few years to begin understanding these advanced topics :)
11:45
@r9m Just get yourself Alekseev to learn galois theory/diff galois theory/algebraic geometry all at once from basic groundworks. It's a compilation of Arnold's notes. He taught some highschool kids in paris about galois theory.
So it depends not on your knowledge but your interest.
@DanielFischer Will you look at my MSE post if you have a chance?
Heya @TedShifrin
r9m
r9m
@BalarkaSen okay :) ..
@r9m It's a beautiful book. Starts with basic group theory and topology and ends with abel-ruffini theorem. All through exercises.
@JohnDoe I am looking. The first equality (you have a typo in your transcription, by the way) is just the parallelogram identity. The second, I don't see yet.
11:47
Actually, it doesn't take the classical route. What you'll learn there is a modern version of galois theory but surprisingly easier to understand.
r9m
r9m
okay :)
@DanielFischer Yes the confusion just comes with regard to the second equality. Will check for the typo now.
@huy back.
Huy
Huy
Welcome back.
@Sawarnik le probleme?
11:48
@JohnDoe $u_j + u_l$ instead of $u_j - u_l$ on the right hand side.
Its raining torrentially here!
Huy
Huy
It is raining over here as well. :(
@BalarkaSen Dosen't know French.
Huy
Huy
And I'm getting a headache studying functional analysis.
@DanielFischer Yes I see, will correct.
11:49
@Sawarnik What about your problem?
The functional equation/transcendece theory?
Uuh, I see Vijay has departed after such a bad decision! :(
Stop cricketing.
And flurry of sixes by Jadeja in test matches.
@BalarkaSen Why?
@Sawarnik You saw my solution to the strict tangent problem?
Triviality.
Huy
Huy
@Sawarnik: What are you refering to ?
11:55
@G.T.R
@BalarkaSen yes ?
0
Q: Quora vs. StackExchange when asking about mathematics

Martin SleziakIn SE network we have math.SE and there is also MathOverflow for research-level questions. Plenty of questions about mathematics are also asked and answered on Quora. I would like to know what are the differences between Quora and StackExchange, especially from the viewpoint of someone asking an...

Ok, another quickie:
$\displaystyle \int_{a}^{b} f^2(x) \, \mathrm{d}x$ = $\displaystyle \int_{a}^{b} f^4(x) \, \mathrm{d}x$ = $\displaystyle \int_{a}^{b} f^3(x) \, \mathrm{d}x$. Find all possible $f$ if its continuous on $[a,b]$.
@G.T.R Have you looked at the IMO 2014 problems?
@Sawarnik I pass.
No wish to get integrals.
r9m
r9m
@Sawarnik IMP .. you know Cauchy Schwarz for integrals ?? :O .. how fast do you learn stuff ? :o !!
Huy
Huy
@G.T.R People use Quora to ask questions about mathematics? O_o
11:56
@BalarkaSen Its much easier, just algebra.
@r9m Actually no ... its much easier :P
@Sawarnik I am not interested.
@BalarkaSen I took a cursory glance at the first one, and plan to spend some time on it Sunday. The rest was geometry/combinatorics that I despise
@BalarkaSen Have you looked?
@G.T.R What was le problem?
@Sawarnik No.
r9m
r9m
@Sawarnik Its one liner with CS inequality .. :P
11:58
$
\begin{align}
\int_a^b\left[f(x)^2-f(x)\right]^2\,\mathrm{d}x
&=\int_a^b\left[f(x)^4-2f(x)^3+f(x)^2\right]\,\mathrm{d}x\\
&=0
\end{align}
$
4
@r9m Proof of CS is not one-liner, AFAIK
@Sawarnik Holy crab
Let $\{a_0<a_1<...\}$ be an infinite sequence of positive integers. Prove that there exists a unique integer $n \geq 1$ such that $\displaystyle a_n<\frac{a_0+a_1+...+a_n}{n} \leq a_{n+1}$
Hi all :)

I'm still having trouble with GAP. It's such a simple thing, there's bound to be someone here who knows what to do.

I can save work spaces alright with "SaveWorkspace(file-name);" but I still can't load them. Using the command line "gap -L file-name" doesn't work.

Please help :)
r9m
r9m
@Sawarnik yup baby .. thats the other one :P .. I was typing that right now :D .. how did you figure it out ? :o
@G.T.R This is easy.
11:59
Ok bye.
It's not just my laptop now either :/
@MatsGranvik @robjohn @AlexanderGruber You guys I think have CAS knowledge that might help me out. I need to know the number of ordered pairs of matrices mod 2 which are both nilpotent and commute. my formula gives 344, a different formula gives 400, I want to know which is correct. any help?
@G.T.R Hmm. Or maybe not.
@r9m hee .. dark magic .. actually stole it from robjohn :|
@BalarkaSen yes?
12:02
@Huy yeah, their layout looks sloppy and most questions on the main page are opinion-based
r9m
r9m
@Sawarnik I meant its just CS all the same :)
@VibhavPant having fun with modular arithmetic?
@BalarkaSen yep
r9m
r9m
@BalarkaSen ya .. its 1/2 liner :P
12:03
@r9m oh well. i am not meant for inequalities.
@BalarkaSen you have a question?
@VibhavPant I have to think of one.
r9m
r9m
@DanielFischer are you busy ? :-)
heh, sure
quora.com/Higher-Education/… Harvard Math 55 course has some reputation
12:08
Sorry, @VibhavPant, I think most of the problems I have are non-modular arithmetic ones.
hehe
no problem
@VibhavPant For any prime $p$ show that $a^p = b^p \pmod{p}$ implies $a^p = b^p \pmod{p^2}$
I have a batter book in front of me from which I am posing these problems.
@BalarkaSen I was going to ask this problem here
12:10
I did these problems years ago.
@VibhavPant STAHP.
Do that yourself.
attempts again
I will try it too. I forgot how I did it.
Actually, this is obvious by fermat's little theorem but if I am correct it can be done without fermat's little theorem.
@r9m Just posted an answer, so at the moment, I'm not busy.
@JohnDoe I think I understand what the author intended. Posted an answer.
yes, yes, I remember this one, @Vibhav! This is a very beautiful problem.
r9m
r9m
@DanielFischer can you help me here :-) ..
12:18
which one?
this problem.
@DanielFischer, I need your help too!!!
the one I just posed.
@r9m No idea whether I can. Needs thinking.
12:19
NTisboring.
@Vibhav You need some extra machinaries to do this.
@nullgeppetto What with?
@DanielFischer, with many things :) but for the time being with my last question here, Previously I did make a typo,
Never mind, everyone; I've figured a way around it!

There was no point saving the work space. Instead, I wrote a notepad document with the commands in to just use "Read("file-name");".

Thanks anyway :)
@DanielFischer thanks will have a look.
12:21
@BalarkaSen Ill try anyway :)
@Vibhav ask me if you want a hint.
r9m
r9m
@DanielFischer okay .. thanks for taking a look ... *ps : I'll add an additional !100 bounty if I get an answer to that Q :D
@BalarkaSen sure
@nullgeppetto Which was your last question?
@VibhavPant Prove that a^p - b^p = (a - b)^p modulo p.
12:22
@BalarkaSen Are you having the same book I have?
@DanielFischer, you're right! Sorry!
Let $\mathbf{x}\in\mathbb{R}^n$. Then $\langle \mathbf{x}, \mathbf{x} \rangle = \lVert \mathbf{x} \rVert^2$. If $A\in\mathbb{R}^{n\times n}$, what would be the $\langle A\mathbf{x}, A\mathbf{x} \rangle$?
@VibhavPant What do you have?
"Challenge and Thrill of Pre-College Mathematics"
@VibhavPant Nah man, I have a hard-core number theory book.
12:25
@nullgeppetto Well, $\langle A\mathbf{x}, A\mathbf{x}\rangle = \lVert A\mathbf{x}\rVert^2$. Beyond that, I don't know much to say. What's the context?
@VibhavPant So are you trying to get along with my hint?
trying
@BalarkaSen two cases: $p$ divides $a$ and $p$ and $a$ are coprime. I can deal without trouble with the first case, but for the second case, I can only prove that $a=b \pmod p$
@G.T.R $a = b \pmod p$ regardless of these two cases.
@DanielFischer, it's something that I need to figure out for a few days now... It would be nice if you could help: math.stackexchange.com/questions/839956/…
12:27
@r9m He copied it from my answer :-)
@G.T.R See my hint.
@DanielFischer, it concerns my second post there...
r9m
r9m
@robjohn LOL :D .. XD .. can't blame him for that :D .. its really a cute one line killer :D
Sawarnik copy-pastes stuffs.
A lot.
So if you google, you'll find the answer right away.
@BalarkaSen He should copy-paste the link so I can get points ;-)
12:29
Heh
I'd think 100k+ users don't care about rep points anymore.
@BalarkaSen Ugh! the other answer, which is not quite right, got accepted. Oh, well.
@BalarkaSen Except for deriving $a=b \pmod p$, I don't get how your hint helps
@G.T.R Well, it helps to derive a = b mod p
@G.T.R $b^p = (a+kp)^p =$ binomial expansion.
r9m
r9m
@robjohn why isn't the other answer right ? :o
12:32
And that's all you need to cut through the problem.
@DanielFischer That's new. I proved it in some other way.
@BalarkaSen Interesting. How?
@DanielFischer a - b = 0 mod p implies (a - b)^(p-1) = o mod p.
@r9m It only cites being integrable, and seems to indicate that is all that is needed. the continuity is necessary. The question does not ask which constant functions are possible, so I won't fault him the fact that he doesn't say it has to be $0$ or $1$.
Now a^p - b^p = (a - b)(a^(p-1) + a^(p-2)b + ... + ab^(p-2) + b^(p-1))
@blue cool, lookin' forward to it
12:35
@r9m I said not quite right... :-)
:16515173 Right. I meant to say a = b mod p implies a^k = b^k mod p
r9m
r9m
@robjohn ya .. he implicitly used continuity when he said $f^2 = cf$, without that we have the same result but a.e
And the sum a^(p-1) + a^(p-2)b + ... + ab^(p-2) + b^(p-1) is then $\sum_{k =1}^p a^{p-k} a^{p-1} = \sum_{k = 1}^p a^{p-1} = 0 \mod p$
r9m
r9m
@robjohn okay ;-)
@r9m but he says that implies $f=c$...
12:38
Hence $a^p - b^p = 0 \pmod {p^2}$
@r9m I mention this in my comment.
Heya @AlexanderGruber
r9m
r9m
@robjohn right :D .. the OP wasn't looking carefully enough :P
@DanielFischer Actually you can do it in many different ways. I think one in my notebook uses the fact $\binom{p-1}{k} = (-1)^k \pmod{p}$
$$\sum_{k = 1}^p a^{p-k} b^{k} = (a - b)^{p-1} + \sum_{k = 1}^{p-1} \left [ 1 + (-1)^k \binom{p-1}{k}\right ] a^{p -1-k} b^k $$
The problem is that some stuffs quickly gets ugly.
r9m
r9m
@robjohn for $f \in C^n[a,b]$, with $f(a) = f(b)$, is the following inequality true ? :o
$$\left|\int_a^{\frac{a+b}{2}}f(x)dx-\int_{\frac{a+b}{2}}^bf(x)dx\right|\leq\frac{(b-a)^{n+1}}{(n+1)!2^n}\sup_{x\in [a, b]}|f^{(n)}(x)|$$
12:44
@r9m Now what new hell is this
@r9m That looks very familiar...
@robjohn Is there any possible mathematics that doesn't look familiar to you and Chris'ssis?!
=D
r9m
r9m
@robjohn hehe .. Takashima Senko's Q for the case $n=3$ :D .. here
@blue I do not know what ordered pairs of matrices means. Why pairs?
r9m
r9m
@robjohn That solution you posted there was truly Ingenious Wizardry !! :D
@BalarkaSen Jesus Christ !! what is that ? :O
12:58
@r9m Ah, that's exactly what I was looking for...
r9m
r9m
@robjohn I have it nicely arranged in my bookmarks :D
@r9m How about if we use a cubic which matches at the endpoints and maximizes the difference of integrals. That has $0$ on the right side, for $n\ge4$
@r9m or am I missing something?
r9m
r9m
@robjohn but LHS won't vanish for such a cubic ? :o
(scratches head) I must be missing sth basic :| .. wait
Prove the polynomial FLT : There are no polynomials $X(t), Y(t), Z(t)$ in $\Bbb Z[t]$ satisfying $X(t)^n + Y(t)^n = Z(t)^n$ for $n \geq 3$.
Ack... in another question, another answerer, who answered one minute before I did, but with a wrong answer, deleted his answer and fixed it after seeing mine, gets upvoted. Perhaps my answer looks too simple. I don't understand people.
@r9m the fourth derivative of a cubic is $0$, is it not?
13:07
@robjohn One of the reason I never answered even medium hanging fruits.
r9m
r9m
@robjohn ya ... but is the LHS $$\left|\int_a^{\frac{a+b}{2}}f(x)dx-\int_{\frac{a+b}{2}}^bf(x)dx\right|$$, non vanishing for such a cubic ?
@r9m That was the point of my answer to the question that you found. The cubic maximizes it.
@DanielFischer, so I can say then that $\lVert A\mathbf{x} \rVert^2 \leq \lVert A\rVert^2 \lVert \mathbf{x} \rVert^2$ ?
r9m
r9m
@robjohn ow !! okay .. right !! .. right !!
@r9m Just try $x^3-x$ on $[-1,1]$
r9m
r9m
13:14
@robjohn yes !! .. I got that :-) .. you proved that in your answer :) .. I was looking at $\sin x$ in $[0,2\pi]$, the LHS = $2$, but the RHS $\to 0$ for large $n$ :D
Is somebody willing to test whether this link works for someone who does not have an account on Quora. (Whether all answers are visible without having to register/log in.) And whether there are some problems with other links from my post on meta: meta.math.stackexchange.com/questions/16243/…
@r9m ah, and looking at the bounds that that gives gives that constant...
"...that that gives gives..." almost stuttering :-)
@MartinSleziak looks good to me...
r9m
r9m
@robjohn ? what do you mean ? ;)
@DanielFischer, is that matrix norm, $\lVert A \lVert$ the Frobenius norm or the Hilbert–Schmidt norm (p=2)? Thanks again!
13:25
@r9m well, I used two words repeatedly... the grammar is correct, but if you pull those words out, it sounds like stuttering.
r9m
r9m
@robjohn ah .. thats interesting :D (grins)
@MartinSleziak I can't see who asked that question on Quora. Is that a feature of Quora?
r9m
r9m
@robjohn Senko accepted your answer(meaning he atleast saw the answer) and didn't upvote mine(the only upvote was yours) .. (possibly didn't even read it at all :P lol), I should have removed my initial calculations and the mechanism of getting the polynomials and presented the two cubics from the start and showed how clever I was :P XD
I'm incorrigible :P lol
r9m
r9m
13:46
@G.T.R are you active on AoPS ?
@r9m no, are you ?
r9m
r9m
@G.T.R nope :| .. but I am thinking of starting answering stuff there :D
@r9m all questions there are similar to what math110 and chinamath post ?
r9m
r9m
@G.T.R olympiad and competitive exam oriented Q's .. yes mostly :)
13:58
@r9m it must be really cold in that room facebook.com/…
@Hippalectryon hello
@robjohn To be honest, I do not now. I do not have too much experience with Quora, either.
r9m
r9m
@G.T.R LOL XD maybe
I kid you not, some people are wearing wool gloves on other photos
r9m
r9m
ic
Hmm, probably transcendence of $\pi/\sqrt{3} + \log(3)$ is a breakthrough.
I am not familiar with any other transcendence result on the set $\{\pi, \log (3)\}$
14:11
@r9m Yeah, that seems to be the way to go. As I remember, I showed too much of the thinking behind my answer on another question and it worked against me, there.
@r9m Heh... I just read your deleted comment, which I think is talking about a similar episode. :-)
@MartinSleziak This is the first I've heard of it.
Well, then at least my post was not completely useless. A few people learned about potentially useful Q&A site.
r9m
r9m
@robjohn XD .. can't be helped .. :-)
@G.T.R AoPS = Art of Pretending Smart :P lol
I think I have seen a question of algebraic independence of $\pi$ and $\log(2)$ in either here or MO, but I can;t find it anymore.
14:28
@nullgeppetto $\lVert A\mathbf{x}\rVert \leqslant \lVert A\rVert\cdot \lVert \mathbf{x}\rVert$ holds for many matrix norms (generally, one only considers matrix norms satisfying that). I have no idea which one is used in that context.
@BalarkaSen Almost all reals are transcendent, but they are hidden really well.
@robjohn I am not talking in a general probabilistic sense.
Transcendentals are uncountably infinite and is dense in $\Bbb R$
But that doesn't stop certain constants from being algebraic =)
@BalarkaSen Neither am I. In a set theoretic and measure theoretic sense, that is true as well.
I am not familiar with measure theory, so I am not sure.
@BalarkaSen non-transcendentals (algebraic numbers) are countable.
14:32
@robjohn I never contradicted that =D
That's why probabilistically you get a greater chance of stepping into a transcendental than algebraic.
@BalarkaSen That means that the set of transcendentals have full measure.
But my particular concern is about whether $\pi$ is $\mathcal{EL}$
And I have found much less reference than I should have.
@BalarkaSen you have $100$ percent chance of hitting a transcendental.
@robjohn Probability is 1, yes.
@DanielFischer Could you please help me with some Latex ?
14:36
@G.T.R it's uncomfortable when it is humid...
@G.T.R Maybe. What's the problem?
@robjohn Actually I think it is far more interesting to prove transcendence of certain constant than to state that it is likely to be transcendental.
As the former is trivially true but the latter is not
@BalarkaSen I am not contradicting that... I was just saying that so many reals are transcentdent (almost all reals are transcendent), but it is so hard to show which ones are (they are hidden really well)
you mean transcendetal
@BalarkaSen As a matter of fact, I think that statement is confirming what you are saying
14:39
@robjohn Yes, indeed it does.
Ah, I think this might be what I was looking for : mathoverflow.net/questions/118523/….
@BalarkaSen Ah $a^\pi=b$?
@DanielFischer I started writing a file here writelatex.com/1262185tdqxpr#/3065150 Don't worry about the fontsize (20). (1)how can I add some blank space between "Introduction"and the theorem that follows ? (adding \\ returns an error). (2) I want to have the "Il existe" aligned with the "Soit $f$"
@robjohn I don't think that's ever seriously possible, though (As Timothy's answer suggests : it'd contradict Schanuel's conjecture)
But algebraic independence of $\{\pi, \log(3)\}$ might be close
@G.T.R My connection is acting up, it loads really slowly. I can't see anything yet. Is it vertical whitespace that you want? How about simply adding a blank line to start a new paragraph?
14:49
@G.T.R I see now. In the preamble, add \setlength{\parindent}{0pt} and \setlength{\parskip}{0.5ex plus 0.5ex minus 0.2ex} (or whatever your preferred spacing is) to avoid the ugly habit of $\LaTeX$ to start paragraphs with indentation.
@DanielFischer it works, thanks. Do you know how to separate Introduction from what follows ?
That's a helpful latex hint. I usually indent each line separately.
@DanielFischer Do you know how to control line spacing in the preamble?
@G.T.R Usually, you'd use sectioning commands, \section*{Introduction} if you don't want a section number, \section{Introduction} if you do. And consider using a theorem environment for theorems (and lemmata, corollaries, ...).
This is interesting. I typed "Yitang Zhnag" in math genology project and it gave me that his advisor's advisor is S. S. Abhyankar himself. genealogy.ams.org/id.php?id=16848
@JohnDoe No, I'm typically happy with the default there. Would have to look up what name the pertinent length has. Is it important?
14:57
No not important, just out of interest. Might want to cover more pages with less work at some stage.

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