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00:01
Sorry, that ain't it.
Unfortunately I can't seem to find the blog entry where Jeff/Joel explain the whole bit... huh.
But that answer seems to be in agreement with per-answer rep cap.
It is interesting that it was Jon Skeet who answered...
(sadly it does not look to have been glanced at by any SE employee)
I don't know who JS is, but from looking around, he seems to be a popular guy.
He's one of the best SO answerers. If we have Arturo, they have Jon Skeet.
QED
QED
00:09
> I feel that perhaps "proper forcing" is a household concept in set theory, and the reader probably knows the basic facts about forcing and proper forcing. However we demand no prerequisites - projecteuclid.org/euclid.pl/1235419814
Whose words, @QED?
QED
QED
Shelah
It's irritating when you think something is going to be an introductory lecture on X, then they say "but you already know all about X!"
4
@QED Weren't you reassured by his "However we demand no prerequisites"? =)
QED
QED
no
I have heard that before
2
@QED but you do need to build from something, no?
00:13
@QED Er, it's a book on forcing, and he says that the reader perhaps knows about these household concepts in set theory?
Not looking good, dude! =)
I am pretty sure this question is a dupe, but I have to run now. :(
QED
QED
bye
I wonder how ZFC + a set that is a countable model of ZFC isn't contradictory
Er, @QED, that last comment of mine was unrelated to my other ones...
QED
QED
maybe I should study godels constructable sets first
Bye @QED and @J.M.
Bye!
QED
QED
00:18
why do I keep getting facebook when I google this stuff?
"click here if you Like 'Gödel's constructible universe'"
what are your exact search terms?
QED
QED
"godels constructable"
everyone gets different google search results though
they profile you by various means
Yeah, I don't seem to get any FB result after typing in "godel's constructible"...
QED
QED
I can't be bothered with this set theory stuff
Maybe try adding a -facebook to flush those out...
QED
QED
00:30
I don't really understand why people care about all those weird real numbers
aren't they just some weird side effect of set theory
e.g. almost all numbers are "normal" or "transcendental" or .. -- but why does it matter?
that's not exactly what I mean
QED
QED
what?
I'm agreeing, that's all.
QED
QED
I wonder why set theory is so popular
I guess it's probably just the first thing people got to work
Facebook for some reason is subsuming some Wikipedia pages so that people can "like" them. I've come across the phenom before while googling math.
00:39
@QED Having a countable model of ZFC is just a matter of not letting the model know it has too few set.
QED
QED
I am still strugging to find out what math I should study
@QED Well, what do you like?
QED
QED
@Henning, but is it okay for ZFC to know it's been given a set that is a model of ZFC?
The model's representation of R will be countable at the metalevel, so at the metalevel there's a bijection between the model's N and the model's R.
QED
QED
Maybe that side steps the 'can't prove your own consistency' thing because it's not proving that ZFC+model of ZFC is consistent
00:40
But as long as this bijection is not in the model, it will never notice the difference.
I think all you can do is have a relative consitency proof: if ZCF is consistent, then ZCF+inaccessible cardinal is consistent too.
QED
QED
@J.M. Well I like the algorithmic resolution of diophantine equations book you recommended but I got stuck on strassmans theorem and gave it up
And even so, I suspect that you need not only consistency of ZFC, but also that ZFC has a model whose omega is the true integers, which is not even formalizable.
QED
QED
@Henning, that makes sense!
thanks
@QED: I only suggested that book since you mentioned you like Diophantines...
Still, you should know what you want to do...
QED
QED
it's as I said, everything seems a few levels above me - I can't find anything that is just one level above me
I suppose it's because I don't have a maths library
00:44
Anyway, I need my bed. Just popping in to see what happened while I was away sulking.
QED
QED
good night
Have a good one, Henning.
00:57
Kind of annoying when a (in this case psychology) paper is double-spaced, and even triple-spaced between headers and sections. You're not in goddamn high school anymore, Mr. Author dude.
Maybe it's the house style for that journal...
(but I agree with that being annoying)
Well, then Mr. and Mrs. Journal Standards Deciders.
QED
QED
what are the topics in diophantine equations?
other than elliptic curves
It's pretty wide. The linear ones are the simplest.
quadratic forms. formal number theory.
QED
QED
01:01
never heard of formal number theory
You have polynomial ones like Pell's... and then even more complicated ones like Fermat's (which is one of the things that elliptic curves were useful for) and Catalan's
QED
QED
that seems to be stuff like arithmetization of syntax
as I understand it, modular forms are important to all of those
pell/quadratic forms, fermat, catalan
I am stuck :S
lol, in "Bayesian Statistics for Archaeologists" the author snuck in a picture from Indiana Jones
QED
QED
01:23
Maybe I should work through Tao's random matrix theory notes
seems a lot of difficult stuff I don't know from probability though
Hello everyone, I'm doing homework currently and needed a little bit of help.. Does anyone know how to convert y=(x=5)^2 + 10 to standard form? (Factoring)
This is for college level math class.
QED
QED
what is standard form?
by the way, did you mistype it? (x=5)^2 looks wrong
The ^ is to mark that the 2 is an exponent
It should be y = (x + 5)^2 + 10, I suppose
QED
QED
but x=5 squared??
01:30
rewrite as y = (x + 5)(x + 5) + 10... "FOIL" (distribute)
I meant + 5
y = (x^2 + 10x + 25) + 10
QED
QED
ah ok
y = x^2 + 10x + 35
a = 1; b = 10; c = 35
@TheChaz That clears up my question! Thanks a ton! :D
01:31
No problem. Good luck out there!
@TheChaz Enjoy a little rep there too ;)
I'm off. Toodles.
?!? It would take like 18k to get me to an enjoyable level, but thanks :)
@TheChaz I'm with you on that one ;)
QED
QED
bye JM
01:32
Sorry to just bust in like that, Hello Math.SE!
QED
QED
hi
@SimonSheehan Also, there are implements in the software that screen for "serial downvoting"... not sure if there is such a thing for "serial upvoting", but watch yourself :)
@TheChaz Oh I'm aware :) Been around a while
There is serial upvoting actually
Just so you can't give your friend a ton of rep each day.
01:35
Are you working on HW for algebra right now?
Nope... were doing a unit on factoring. I'm only in Grade 11 :P
Nice. When I was in high school, they taught us to factor quadratics by the "guess and check" method*... later on in life, I learned that not everyone is as good at guessing, and - more importantly! - that you never have to guess.
@TheChaz Guessing seems to take me a while.. That's exactly what were doing right now :P
I'm struggling a bit I feel, but getting better... just gotta be ready for the test friday and I wanted to confirm things.
If your method doesn't involve looking at the product "a*c", then you're probably doing it wrong (or guessing).
@TheChaz A C ?
01:38
ax^2 + bx + c
Ah you just use different variables than us... got it.
ac is the product of the coefficient of x^2 and of the constant term.
@TheChaz Awesome, thanks for telling me that... Well, time to finish this. thanks for all the help! hope to see you around again sometime
Drop by Super User if you're in need of computer problems ;)
Adios. I use a mac, so...
:P
QED
QED
haha
01:43
We speak Mac too! :P
QED
QED
I don't know what math to do..
that's my problem
I punish myself for mistakes from my past.
That's my problem.
o_O
The only n for which x^-1=x for all x in (Z/nZ)* are the divisors n|24. Interesting.
QED
QED
I'm not sure what that refers to
I've heard that one before anon
I think there was a whole paper about it but I cant remember the name
01:55
from "What's so special about the divisors of 24"
QED
QED
that's it!
I was being facetious.
QED
QED
oh
I've been desperately trying to find some topic to get into for the last few weeks
My standard response is "aperiodic tilings". They fascinate me.
temperley-lieb algebras. high tantalizing / difficulty ratio.
QED
QED
01:57
are you talking about the penrose dart-kite type or the wang/polyomino type?
Yes.
There are many aperiodic sets of tiles.
QED
QED
both?
"My only regret is not making love to a Mobius strip" "Would you be top or bottom?" "Yes."
7
QED
QED
hehe
source, anon?
01:59
Reddit, google it
ok. Thought it might have been xkcd
QED
QED
I think these tilings are very attractive but I don't know what maths there is for it
Creating them, proving that they are aperiodic ...?
QED
QED
looks like quantum physics applied to knot theory
 
1 hour later…
03:09
"there are as many kinds of love as there are hearts" - Leo Tolstoy, Anna Karenina. I think it'd be more like hearts*(hearts-1)/2, technically...
03:32
but who's counting
 
3 hours later…
06:49
@QED, @Henning: The theory ZFC+There is a model of ZFC is a stronger theory than ZFC. It proves ZFC is consistent, but it cannot prove its own consistency, this is similar to how ZFC+Inaccessible cardinal is a stronger theory. And as Dana Scott said once, if you want to prove more you need to assume more.
07:02
In response to my earlier question, this article in Annals of Mathematical statistics find the Laplace transform of the density.
 
1 hour later…
08:10
Heh. I spent way too much tme working on an answer and Gerry Myerson will probably get accepted for "See the discussion at MathOverflow".
Hah.
Yeah, some people really like just to post a link.
Ha!!, not only I was the first to answer. I also answered the longest and most detailed answer. I feel like mini-Arturo or mini-JDH when these things happen.
I remember seeing this problem and it is easy to see the solution, but it just takes a bit of bookkeeping to justify the answer.
@robjohn I upvoted your answer, FWIW. :) (I also liked Robert's answer at MO.)
08:31
I commented on (and upvoted) your answer.
Obviously, there can be no counterexample since the result is true, but I have seen proofs like Robert's look good, but fail because the diagram lies.
His answer leaves me worrying. Perhaps I am not thinking clearly.
09:27
@Rob: hi. why didn't you vote to close math.stackexchange.com/questions/80453/…
seems perfect coincidence with that question
09:55
@Gortaur I guess I could.
@robjohn I think now that maybe it's not so clear
Geez, I've been downvoted.
It's a perfectly valid argument, perhaps not what the OP wanted. Oh, well.
@robjohn the reason is that for a fixed k (say 2) we can put explicitly that 7!>2^7
while for arbitrary k it maybe more rigor to find explicitly N(k), that's why I decided to put my answer there
@robjohn ratio test is D'Alamber criteria?
09:59
The ratio test only need apply past a certain point and for n>2k the ratio of terms is < 1/2
@Rob icic. I've put a link there to ratio test if you don't mind
Perhaps I should ammend.
I don't mind. :-)
ooops, there was a typo in my answer
@anon so here you've lost egoists
So here I've lost egoists? What does that mean?
with egoists it would be (n+1)n/2
and also you've assumed that heart A loves heart B only in the same way as B loves A. naive )
10:09
oh, referring to the quote. I gotcha. actually, your second comment reminds me of a quasi-mathematical topic I brought up on efnet's #math: "closeness" between people might be a bit of a misnomer in that relational closeness isn't symmetric like with physical objects.
but it's easier to figure out the model behind h*(h-1)/2 than it is with more realistic assumptions
@anon I believe it's called a pseudometric which lead to anisotropy. my former supervisor wrote a book on differential geometry for that spaces, but left me do only boring triangulation (
10:23
@Gortaur I modified my answer to use the ratio by itself, and not use series.
@Rob: seems that your downvote has gone
I just noticed that :-)
I should check the answer to the similar problem to see if that was downvoted at the same time -_-
doesn't seem to have been.
10:39
@robjohn which que are you talking about?
I checked the similar problem to see if my answer there, which was very similar, was also downvoted.
10:59
@JM Perhaps I am misreading things, but I think that Robert's argument shows something else:
Oops. JM isn't here.
11:32
hello
12:23
@Ben: hi
@Srivatsan: good morning, sir
As always, I prefer Good morning, Respected Sir Narayanan! =)
@Srivatsan dunno, respected professors seem not to be the right salutation on MSE
@Srivatsan you're sir but not Sir; Sri as a maximum )
@Gortaur No no, I was just alluding to iyengar's style...
@Srivatsan he does not have it )
12:50
@AsafKaragila Yes, of course. Not thinking.
@Gortaur But his posts have seen a dramatic improvement in quality of writing. (content, I am no judge of that...) =)
By the way, 'Morning, @Gortaur. =)
@Srivatsan usually I fix punctuation like !!!->! or :-))))->:-) and I fight with myself to leave eyes for emoticon in the last case
@Srivatsan thank you. it's late morning here, like 2pm
my mouse has gone mad
I have kind of soft set-theoretic question
@Gortaur What kind of mouse?
In computing, a mouse is a pointing device that functions by detecting two-dimensional motion relative to its supporting surface. Physically, a mouse consists of an object held under one of the user's hands, with one or more buttons. It sometimes features other elements, such as "wheels", which allow the user to perform various system-dependent operations, or extra buttons or features that can add more control or dimensional input. The mouse's motion typically translates into the motion of a cursor on a display, which allows for fine control of a graphical user interface. Naming The fi...
The computer peripheral (I am yet to meet a mad peripheral, by the way =)), or the animal?
12:54
I don't respect any animals with a size less than a size of a kitten
and I do love cats. they're perfect neverminders
@Gortaur How did it go mad?
@AsafKaragila Um, but now that I read back, didn't I qualify it with something like "...also assume that ZFC is a true theory"? If ZFC is consistent, then there is a model of it at the metalevel, and if it's also true (that is, the theorems it can prove about its integers are also true about meta-N) then it cannot prove not-(ZFC has a model), thus under those assumptions ZFC+(ZFC has a model) is consistent.
Cool. I've received my employment letter :-).
13:34
@JonasTeuwen congrats! 1 year of go-no-go?
Say, there is a model of dynamics in a discrete time: X{n+1} = f(X{n},Y{n},Z{n}) and Y{n+1} = g(Y{n},Z{n}). Now, for example I have a model of the form X{n+1} = X{n}+Z{n}.
how should I say what is the state space for Y? is it empty, or it is a singleton
@Srivatsan I'm totally frustrated. If there is something incorrect in this answer that I am missing?
@Gortaur Yeah.
Actually, the first two months they can just kick me out, after that after a year and if they still keep me it will be hard to stop me from finishing the four years.
@JonasTeuwen yeah, I know it
The first downvoter removed their downvote, now there is another downvote on my amended answer. I thought I knew what the issue was; I guess not.
@JonasTeuwen Congratulations!!
Thanks! :D.
I have at least given you an upvote to cancel the downvote :D.
@JonasTeuwen Thanks. I wish I knew what they objected to now.
13:47
I don't know...
@Rob I cannot understand why my answer is downvoted as well. Maybe there is some mistake I fail to find?
Downvotes are free now, right? Maybe this is some quirk of that.
what do you mean they are free?
They don't subtract rep from you.
from me they do )
-1 if I downvote the answer, 0 when downvote the question
13:52
@Gortaur 5/8 of the answers were downvoted.
someone is being really picky.
Oh, maybe they have changed it to downvoting questions only, not to answers as well (I thought both).
becomes annoying
@Gortaur Seems correct.
@Srivatsan: thanks for checking
Except that "play the role of 7 in our case" should be "play the role of 2 in our case". =)
Imagine getting downvoted for that =)
14:02
@Srivatsan of 7 because N(3) = 7, 3=2+1
@Srivatsan I noticed that, but I couldn't imagine that being the reason.
@robjohn what are you talking about? with 7 I referred to the first number n for which n!>(2+1)^n. Then I give a lower bound for such number through j^2+j in the general case
@robjohn Oh no no, I misunderstood (and so did you, apparently =)). Gortaur is correct. 7 = N(3) is the N_0 such that every term n!/ 3^n is sufficiently large.
So, k plays the role of 2, and N plays the role of 7.
@Gortaur okay, I guess I misunderstood, but definitely upon a quick reading, it looks as if 2->k is the sub, but it's different.
14:06
@Gortaur Sorry, what does that mean? Am I wrong again?
@Srivatsan it means that I agree with you )
In any case, I do think that the answer wasn't too clear at that point. =)
@robjohn I guess that's because my formulation is unclear, so I will change it now to avoid misunderstanding. Cannot imagine as well that it can be a reason for downvote
But why do you need all this, @Gortaur?
The sequence j^n / n! is eventually decreasing. So it has a limit, say L.
Now it boils down to showing L=0.
So, to be frank =), I guess I wouldn't upvote your answer, but the downvote is unjustified.
14:11
It is also literally unjustified because the voter "forgot" to leave a comment.
wth )
I can speak TeX comfortably @Gortaur. You can even just paste the code =)
ah, it does not like spaces
@Gortaur Er, I don't follow this.
@Srivatsan: so how to show that $L=0$? Compare it with C\alpha^n where C is the product of some first terms
14:13
No, I would go like this. The sequence j^n/n! has a limit L.
@Gortaur Then L is also the limit of j^{n+1}/(n+1)! = j^n / n! \times j/n. You know what to do from now... =)
@Gortaur in a URL, you must use %20
@Srivatsan ok
@robjohn I've stolen this idea already from the link you've provided ))
@Srivatsan I was following the principle of half-full kettle to prove this fact
@Gortaur Other things should be escaped as well, but some browsers pass things through anyway.
most browsers?
@Gortaur \alpha < 1?
14:16
@robjohn I'm scared to ask whom are you talking to
@robjohn yes
@Gortaur I had said "some browsers" to you and then added as an afterthought, "most browsers?"
@Gortaur Of course, my proof (as written) is inferior in that I don't even get a modest upper estimate on j^n/n!, whereas yours shows that it goes down exponentially. But who cares, right?
there are already so many browsers for android so you cannot talk about most of them w.r.t. countnig measure
@Gortaur indeed :-D
@Srivatsan I mean - I wrote an answer once to similar problem. So I put a link to that answer and also show how to reduce new problem to the old one
14:20
@Gortaur The android code that Steve Jobs said was stolen from Apple contains some of my QuickDraw GX code.
Increasing success by lowering expectations!
@Gortaur Yes, yes. Makes sense.
@JonasTeuwen the keystone of American education.
@robjohn that will warm my soul each time I use it now )
I +1'ed your request for explanation. But this looks like a hit-and-run case. @Gortaur
14:21
@JonasTeuwen Actually, maintaining the status quo by lowering expectations.
thanks, but usually I don't allow people to hit me and run. Either they are strong enough not to afraid of me and hence they don't need to run, or they are not so strong so I can catch them. @Srivatsan
@Srivatsan both Gortaur and I had answers downvoted. Actually, 5 of the 8 answers were downvoted. I wonder if the downvoter was doing so because the question was an almost exact duplicate.
@robjohn oh sorry. that would be very stupid action of him
@robjohn How is the question being a near-dupe anyone's problems? I strongly recommend that they get a life. =)
14:25
@Rob personally I downvoted pedja's answer, but besides other two guys I've put a comment why
Unless they give some explanation, I can only speculate why they downvoted.
I checked the question; it's not downvoted. Perhaps that isn't the cause anyway...
@Gortaur I +1'd your comment. I don't quite think he deserves another downvote or I would have.
@Srivatsan Listen to my paranoid part: the reason may be that this guy thought that we are looking for easy rep answering a duplicate question %)
@Gortaur That's the only reason I mentioned it. However, I added more to my answer from the previous answer (before he downvoted).
14:29
@robjohn too much thought of us about this guy maybe. He should suffer now from the hiccup
@Gortaur a fitting punishment :-) I have to take the dog to the park. BBL
@robjohn have fun )
@robjohn Ciao
Update: David Mitra's answer is also downvoted.
It looks ok to me.
arggghhhhhhhh
how stupid. rather then solving (I-A)x = b
I solved Ax = b obtaining (of course) solution in the range of 10^8
@Gortaur Why do you need to solve (I-A)x=b?
14:39
@Srivatsan for the case study I am solving one integral equation and toolbox gives wrong results
the most simple way to imagine a solution is to simply discretize this equation into system of linear ones
replacing the integral operator by its discretized version (matrix A)
15:06
Are there any <2000 rep users present? If so, does this link work for you: math.stackexchange.com/suggested-edits/2966
Or, for that matter, for <10k users?
Works for me, a <10k user.
@Henning
15:29
@HenningMakholm works for me (9540)
I thought I was being smart picking zw <= 1/e^2 z^2 + e^2 w^2 in my integral so that everything nicely converges, but then sqrt(2) < e < 1/sqrt(2) =(.
@robjohn, @Srivatsan: thanks. So if there is a problem it would be under the 2000 rep cutoff for approving suggested edits.
@JonasTeuwen Whut? 1/sqrt(2) is much less than e on my calculator.
Oh e != exp(1).
Just \epsilon.
My idea just doesn't work.
15:59
@JonasTeuwen why? just because sqrt(2)>1/sqrt(2)? ;-)
@robjohn Exactly.
picky picky :-)
It is just some nasty estimate which I can't seem to finish for the moment. It is for my thesis, so I wonder if I can post it on MSE. I had an argument written down but I can't figure out anymore how it works 8-).
@robjohn and what is a value of 2 here? maybe 2<1 in your case
The integral "clearly" converges, but why...
16:04
@JonasTeuwen which integral?
Yes, maybe I should just do the argument and write: "For sufficiently small values of 2".
@robjohn I'll post it anyway. One moment.
\begin{align} does not work here?
@robjohn Here it is!
My maximal function is very maximal :-).
16:31
@JonasTeuwen You might want to include a link to the Ornstein-Uhlenbeck operator.
@robjohn Done that. But I don't think it is needed, the action of the Ornstein-Uhlenbeck semigroup can be written against the Mehler integral kernel which I have given. You need to estimate that one since it would be quite horrible to work with the semigroup when you don't have such a cute thingie :D.
@JonasTeuwen Well, since you mentioned it, the first thing people are going to think they need to do is to look it up. It is just nice to have a link there.
According to me (so it seems) it would hold that 4 exp(-t^2) |z||w| <= 2|w|. But I think monkey abused my brain when I did that.
My advisor wrote next to it: "Zie ik niet; waarom?" which translates to: "I don't see this; why?". So I suppose he asks me why that holds, not why he does not see it...
@Jonas: do you mind if I insert a link to the OUOperator?
@robjohn No, but pay attention. Some people give -L instead of L.
16:41
In mathematics, the Ornstein–Uhlenbeck operator can be thought of as a generalization of the Laplace operator to an infinite-dimensional setting. The Ornstein–Uhlenbeck operator plays a significant role in the Malliavin calculus. Introduction: the finite-dimensional picture The Laplacian Consider the gradient operator ∇ acting on scalar functions f : Rn → R; the gradient of a scalar function is a vector field v = ∇f : Rn → Rn. The divergence operator div, acting on vector fields to produce scalar fields, is the adj...
How about that link?
@robjohn That one is okay.
So edited :-)
Thanks!
I have 41 holiday days a year, cool!
Which you can exchange for a bike or stuff like that.
Exciting.
41 is quite a lot. Does that include Saturdays?
No.
40 hours a week and 41 holiday days in a year.
16:48
So eight weeks and a bit. Impressive.
Well, the salary is probably less impressive (2048 EUR).
(PhD "student", but according to the Dutch law a PhD student is formally not a student but an employee)
Same in Denmark. But the salary is a bit higher here (though by a factor less than 2).
Yes, but you get 8% holiday allowance and 8.3% at the end of the year. Plus the second year is 2379 EUR.
41 days paid vacation and holiday allowance?

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