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20:08
@IanMateus those series are treated by Ramanujan in his papers.
@Chris'ssis links?
Ay can somebody help me with the dedekind cut for sqrt(2)
?
en.wikipedia.org/wiki/Dedekind_cut says that you can take (2x+2)/(x+1) But how would I even think of that choice?
@Anthony There is no question here.
@JasperLoy I don't understand what you mean.
20:13
@Anthony What is your question? LOL
Oh I suppose I left out a line?
I thought dedekind cuts were something that were off-handedly mentioned in an obscure portion of a first rigorous analysis text, and then never encountered or dealt with by anyone in any context ever.
It says that you can take (2x+2)/(x+1) as a number greater than any x in the lower portion of your cut.
But I don't see how I would have thought of (2x+2)/(x+1)
It says x+2 in the denominator, actually.
What you wrote just simplifies to 2
Agh I'm sorry.
Yes. (2x+2)/(x+2)
20:17
And it's just one of many possible choices that satisfies the desired properties. How you see it is to basically try hard enough long enough. It's kind of the ages-old question for every thing mathematical ever: How did they see this? Hard to answer, but mostly it's just trying really hard for a long time.
So how could I think of another one? What kind of process would I use? Really only staring at it?
Like 2x/(x+1)
Would that work?
No. Maybe?
@Chris'ssis this also contains what Ethan had found, very nice
It's writing out the relevant equations, and trying to look for a very simple expression which "fits" it. 'Seeing it' comes from long experience (and a bit of inexplicable brilliance where your brain just somehow sees strange stuff).
@Zibadawa Agreed.
As someone recently told me that someone else once said: "This may seem complicated and inobvious at first, but after doing it 9,546 times it becomes completely trivial."
20:23
@IanMateus yeah :-)
And that's basically what any (reputable) author means when he says something like "this is obvious".
He's done it a tremendous number of times, and once you do it a tremendous number of times, you will definitely agree that it is obvious.
Novices also say it is obvious when they actually got it wrong, lol.
@robjohn hey, take a look at the last problem in the first column here. It is basically this question. The problem contains a "AMM 6189" thing, what is this? Is it American Mathematical Monthly? If so, how can I find the original paper by AMM? Google didn't help me much
@JasperLoy Yes, that's why I threw in the "reputable" part.
Then shouldn't the objective be to not have to do it so many times to make it "obvious"?
20:28
@skull I answered 2 lhf today, lol.
Cool :-)
Not really. Tao has a wordpress blog concerning the stages of mathematical maturity that gives a rather pleasant description of mathematical learning and research. The basic idea is that you first go through the stages of rote repetition and unrelenting rigor so that you can then blossom into the stage of intuitively driven research.
I don't understand why Tao is so famous, he is just a Fields medallist, and there are many others, but the others are not as famous as he is.
Because he's better than them.
Youth?
20:31
"Just a fields medalist" is a rather bizarre thing to write, incidentally. I'm not sure the English language permits this sentence.
@Zibadawa I don't think Tao says you should go through rote, he only describes what happens. Not that you have said so, but it seems like
Yeah, I'm going off of pure memory. I'm fairly sure he doesn't mention rote at all; whether it's implicit is another matter.
@skullpatrol I suspect Tao is only 3 letters, so it is easy to read and type.
Perhaps
He did set some record for the youngest PhD
Ah, Fefferman was also very young.
20:36
@JasperLoy You know how many papers he has published right? You know how long he used on his master and phd? ...
Tao is rare combination of remarkable traits. He is a genius amongst geniuses. He had remarkable results and accomplishments at a very young age. His specialties (if one would actually claim him to be limited thusly) are in popular areas. He is an altogether awesome and eloquent dude. And he's pretty handsome. He is a perfectly good reason for everyone to hate being who they are: because they are not him. Okay, that's kind of a ludicrous fanboy thing to say, but he's a pretty rad guy.
@N3buchadnezzar Yup, read all those.
Exactly..
Being socially intelligent, and a prodigy is something many wants to promote to "discurrage" the stereotype of a mathematician.
Eg introvert, weird old etc. Tao fits in quite well in doing so.
He came from Hong Kong, right?
Australian
20:41
Originally
Hi @mike!
He was born in Australia
You can respect him, and I do, but one should never start worshipping mathematicians as I think people sometimes do Tao
His parents immigrated from Hong Kong
I'm not sure if it's worship, but yes.
(a few pages down, the paragraph about the special culture)
20:44
I agree with you @mike
@Mike It's okay to start worshipping a little when the people are dead a hundred years ;) Waiting to start w'ing Dedekind.
I agree with @Daniel
Math SE users worship Andre Nicolas, lol.
I say we follow the egalitarian ideal and ensure that all mathematicians are worshiped.
Worship yourself first.
Imo
20:48
@skullpatrol I just noticed your new pic.
@IanMateus I can't read that page
@robjohn sorry. It reads "AMM 6189 by Edward T. H. Wang. Prove or disprove that for each natural number $n\leq 2$, one can arrange the numbers $1, 2, \ldots, n$ in a sequence such that the sum of any adjacent numbers is a prime."
@robjohn abbreviations like JMM 566, CRUX 327 and others also appear throughout the paper, it is a big list of problems. I'd like to know whether you have seen the AMM thing somewhere yet
@mike which page?
@Ian As stated, the problem is trivial. Perhaps you meant n >= 2. Or perhaps I missed the context and the whole chat is about the misprint.
20:58
@PaulEpstein the chat is about the reference.
@PaulEpstein hehehe, $n\geq 2$. It is the third time I mix them today!
So n>=2 not n<=2?
Sounds like a weird question for a Putnam-type exercise.
Is there anyone here who knows both French and German?
If the assertion is true, it must be too hard.
So it must fail for some smallish counterexample.
@PaulEpstein it is true up to $10^{165}$
21:01
But "Prove or disprove " sounds weird, like it's some type of easy quiz question. Is this an open conjecture?
The book is supposed to contain only known problems at this page, open ones only at the end. I found some discussion on Google, but nothing too far from "Hamiltonian paths" and a guy who found a trick to prove it for any twin prime (as I did last week).
It was unsolved as of 1979
I heard Guy's book on open number theory questions has a discussion (page 105) but I don't have it handy
@robjohn I can't see it, sorry
That was in a section of unsolved problems as of 1979
@robjohn hm... I didn't see it. Nice, thank you
21:11
I made a cubic equation fitter
give 3 points and a y intercept and voila a cubic
@Alizter numerical example
Works for integers at the moment though because I am lazy
hmmm (0, 2), (1, 5) and (5, -6). With y intercept 6
This $y$ intercept is $d$ in $ax^3+bx^2+cx+d$?
yes
change first coordinate to (-1, 2)
0 is misbehaving in my program
@Alizter Is it faster than Lagrange?
21:16
No way. I did it by <strike>using</strike> abusing cramars rule
and it only works for cubic's with integers points
hmmm didn't work
oh well back to homework
using
These days I created a question that involves the use of the balanced negapolygamma function.
It seems pretty interesting.
> negapolygamma
awesome
In mathematics, the generalized polygamma function or balanced negapolygamma function is a function introduced by Olivier Espinosa Aldunate and Victor H. Moll. It generalizes the polygamma function to negative and fractional order, but remains equal to it for integer positive orders. Definition The generalized polygamma function is defined as follows: : \psi(z,q)=\frac{\zeta'(z+1,q)+(\psi(-z)+\gamma ) \zeta (z+1,q)}{\Gamma (-z)} \, or alternatively, : \psi(z,q)=e^{- \gamma z}\frac{\partial}{\partial z}\left(e^{\gamma z}\frac{\zeta(z+1,q)}{\Gamma(-z)}\right), where \psi(z) is ...
21:32
So the user with high rep whose answers I hate also writes English in a way I hate, using punctuation wrongly most of the time.
@JasperLoy Somewhat. I can't appreciate literature in French, but I get the gist of most newspaper articles. Deutsch kann ich ziemlich gut.
@DanielFischer Ah, so for an English speaker, is it easier to learn German than French?
A little easier, probably. Not much, though.
I now have 3 accepted answers at 0 votes, lol.
 
1 hour later…
22:39
Once again I am struggling. Any set-theorists around that mind taking a look at my ashaming attempts at some question?
(Or people who just happen to know the subject of course. Pretty any knowledge is better than mine I am afraid..)
I'll throw it in here in case someone comes and might have an idea..
Let A:=N X N, and let <' be the following relation on A: <k1,n1><'<k2,n2> iff k1<k2 AND n1>n2. Now, I need to show that every chain of A is finite. I went around showing that the relation is a partial, non-strict order on A, and to show that A have minimal and maximal members (where the minimal members are <0,n> and the maximal are <k,0>). Now I try to approach this -
I try to look at the maximal chain of A, and I can get my head around how it should look and how to show it is finite, yet I have no idea how to proof-write it.
@Chris'ssis Just my luck. I usually deal with the unbalanced function.
22:54
@robjohn I see you have 2 answers at -1.
@Studentmath You have to show every chain is finite. You are just proving other stuff
"the relation is a partial, non-strict order on A, and to show that A have minimal and maximal members ,I try to look at the maximal chain of A"
@Theta30 I had to go around proving the relation is partial so that the chain, as it is defined in my books, can be defined.. the maximal and minimal numbers are other related questions I had to answer, but thought throwing it in as it may be useful since in my mind I would pick one of each into the maximal chain of A
So I didn't prove anything that I didn't have to :P still, I am clueless as to how to prove the that every chain is finite. If I prove the maximal chain is finite, then every chain is finite (since every chain is a subset of the maximal chain).
Now I don't know if I can just state the maximal chain will have to have a member of the type <0,n> and <k,0>, and if so why can I state that..
You don't need to prove the chain is defined unless asked to
@JasperLoy Did you just vote for them? they seem to have disappeared.
@robjohn I realised you capped, so I removed my upvote and they are there again.
23:01
I needed it for other parts, too. But still, not the main issue here, over-proving.
It's the lack of proving I have problem with..
@Studentmath At each step, up or down, one of the components must decrease.
Yes, I thought about it.
Oh.
I don't even have to look at the maximal chain.
make up an example and see what happens to it
2
Q: Prove that $x^{−1}Hx $ is a subgroup of G

JewelssLet $H$ be a subgroup of a group $G$ and, for $x\in G$, let $x^{-1}ax$ denote the set $\{x^{−1}ax : a\in H\}$. Prove that $x^{−1}Hx$ is a subgroup of $G$.

@DanielFischer Awesome, thanks, just realized it.
23:02
It is disturbing that a badly written answer gets the most votes above.
@JasperLoy The answers are okay, but I think someone thought they were not distinguished enough from other answers.
hi @Jasper, @robjohn, @Daniel, @studentmath
@TedShifrin Hi!
@JasperLoy Maybe disturbing, but not uncommon.
Hey Ted!
23:04
@robjohn You seem to always be downvoted by the same guy, lol.
Hi @Ted.
@Jasper: It constantly disturbs me that certain people write garbage answers and insist they're right.
Hey @Ted (don't take that phonetically)
LOL, cute, @robjohn
@TedShifrin like you haven't heard that before.
23:05
I am in yet another battle with the idiot Mhenni
no, @robjohn, that's a first!
@TedShifrin You seem to have such a good time with him.
yeah, almost as much fun as with my students who don't want to make an effort to pass :P
Arrogance is a wonderful thing ... oh, well, I don't have the time to bother to get it right, but I'm posting my s*** anyhow.
I try to check my answers before posting, but many times I still make errors. Luckily there is the 5 minutes where edits are not recorded.
@Jasper, we all make errors.
But this site is good for practising some basic latex!
23:09
The worst errors are the ones you see but cannot remove. Like texting or emails.
And I've posted some phenomenally wrong answers to questions. But I'm not doing it as a pattern, insisting that it's up to the OP to sort out what's right and what's wrong.
Yes, @Alizter, I've done some of those, too :(
There was a funny story I heard from somewhere. A employee has to write some emails to some customers addressing some financial issues. As a template he begins the letter with "Dear Rich Bastard". You can only guess what happens next :P
Well, anyhow, I have a bunch of multivariable calc (w/proofs) tests to grade tonight.
@DanielFischer I went about saying that for every <a,b> that we take from a chain, if we go on to one side we reach <k,0> and to the other <0,n> - and if we don't reach there, the chain is certainly finite as it goes between two <a,b>,<c,d> where a<c and d>b, so it has at most the number of natural numbers between d and b or a and c, which is of course finite. If it does go all the way to there on both sides, it has at the most either the number of natural numbers between 0 and n or 0 and k..
@TedShifrin Yes, no chatting for you!
23:11
You get the point. Sounds about right, correct?
and @JasperLoy It is good for that for sure..
Glad @Daniel is on the set theory watch tonight :)
aren't you glad, @Jasper !
Yes, I only answer algebra-precalculus here, lol.
Someone has to take care of us poor lads (nah, I am loving it. Even though I constantly feel not good enough in it..)
ahh quadratic extensions of fields is doing my head in
@Studentmath Yes, about right. If the chain contains $(n,k)$, it can have at most $n$ other elements on the one side, and at most $k$ on the other.
23:13
@Jasper: If you want to go to grad school, you should start pushing yourself on your answering :P
well, @Alizter, move on to cubic and quartic, then. :D
Yes, alright. Thanks-a-ton!
Notation: When somebody writes $F[x]$ for some field $F$. They are referring to a polynomial over $F$. Right? So why does $F[\sqrt{n}]$ and $F[i]$ not become polynomials?
It is, @Alizter, but the proposition is that those are actually fields. Because you can find the multiplicative inverse of any nonzero $p(\alpha)$ for such $\alpha$.
Maybe if I read on it should become evident.
It's the Euclidean algorithm, @Alizter. If $\alpha$ is the root of an irreducible polynomial $f(x)\in F[x]$, then if $p(\alpha)\ne 0$, this means that there are $g(x)$ and $h(x)$ so that $g(x)f(x)+h(x)p(x)=1$, and so $h(\alpha)p(\alpha)=1$.
23:20
@TedShifrin I am fine with the concept of irreducables and what not but why are the notations so similar? That was what I was trying to say.
You mean $F[\alpha]$ versus $F(\alpha)$?
No $F[x]$ for polynomials and for example $\Bbb Z[i]$ for gi.
No, @Alizter, that notation is perfectly consistent. $\Bbb Z[i]$ denotes all polynomial expressions in $i$ with integer coefficients. It is the image of the homomorphism $\Bbb Z[x]\to \Bbb C$ given by evaluation at $i$.
I thought you were wondering about why when we start with a field we usually end up with a field but write it as $F[\alpha]$.
@TedShifrin But wouldn't $Z[i]$ contain things like .... Oh wait. Does that mean that .... Damn that is clever.
Care to repeat that, @Alizter? :D
23:25
Thanks @TedShifrin
Now figure out my field stuff above :)
Hi everyone
Hi/bye @Paul.
@Ted, I think I solved that problem I had with the Szemeredi problem.
for F[X], X is transcendental while for F[i], i it is algebrical
23:26
wow, cool @Paul.
Hopefully, I can correct Szemeredi's original proof.
Well, yes, @Theta30, but that's a bit beyond where we were :)
@TedShifrin So when fields are extended they are really talking about trivial examples where the degree doesn't exceed 2. Usually?
No, no, no @Alizter.
I don't think anyone's bothered to do that before.
23:27
I didn't know there was anything amiss with it, @Paul. Totally not my stuff.
But I gave you my reference people :)
Plenty have tried to simplify Szemeredi's exposition but I don't think anyone has worked with the original exposition, merely eliminating the errors.
@Ted I don't think there is anything amiss with it.
Papers of that length are expected to contain multiple errors.
I spent a lot of my life trying to get rid of all of Griffiths's errors, @Paul. There were lots, but the guy is nevertheless a brilliant mathematician and expositor.
But if I have $\Bbb Z[\sqrt{n}]$ does this not contain things with terms with $\sqrt{n}^3$?
@Alizter, but that's $n\sqrt n$.
@Ted Which Griffiths?
23:28
Phillip.
@TedShifrin :| This abstract algebra stuff is cool.
I need sleep
@Alizter: But consider $\Bbb Q[\root 3\of 2]$, for example. Not quadratic!
Sleep is important, especially at your age :D
@Alizter which course are you taking?
I am going to bed, goodnight @ted, lol.
G'night, @Jasper.
I'm going to eat dinner and then grade exams all night. Bye, all.
23:30
In fact, Szemeredi's paper is remarkably error-sparse but even Alaska has some people living there.
@TedShifrin Thanks and bye!
Well, @Paul, one or two.
@Ted The dinner isn't exactly a major surprise. Bye.
Dayum, my timing sucks.
@Ted, I thought that sleep was age-independent. Old and young alike need to sleep.

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