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12:01 AM
@Asaf: seen that comment by Dave Renfro? If that's true that would answer my question of a few days ago...
 
I know, but money for conferences isn't the only thing that'll solve...
 
@tb I saw that a few days ago. I have a relevant paper by Arnold Miller, hang on.
 
Time to go pick up the kids. Catch you folks later...
 
See you!
 
Oh right, I left the print out at the office. Let me find the .pdf
It says in the introduction that indeed in the Feferman-Levy model (R as a countable union of countable sets) P(R)=Sigma^0_4
 
12:05 AM
@Asaf: great! thanks a lot for that link!
 
And that indeed this is the minimal number of steps.
 
lol, a calculation in Wikipedia's article on partitions is cited to WolframAlpha: en.wikipedia.org/wiki/Partition_(number_theory)#cite_note-2
 
There's a proof without the axiom of choice about Sigma_3\neq Pi_3
 
Was the citation necessary? I suppose he just put in what Alpha spat out... :)
 
@Asaf: sorry, I was having a peek at the paper.
 
12:13 AM
He writes well. I need to read this paper in-depth.
 
I don't need to, but my interest is sparked already.
 
:-)
I was already intrigued by the structure of the Feferman-Levy model, and was wondering what sort of sets hide there. This made me things worse, it is a paper I will have to read eventually. Arnold Miller is a good writer too, he writes with humor hidden between the words.
 
Yeah, it looks like a very nice paper at first glance. I also like that Specker is mentioned, great guy!
 
By the way, I am not sure it answers your question.
Since I am not sure that in the Feferman-Levy model there are non-separable sets of reals.
It does sound reasonable, though. I'm not 100% certain about it though.
 
Oh, right. Actually, I had a different question in mind that I apparently didn't spell out :)
 
12:23 AM
Spit it out, then :-)
 
I'm a bit reluctant because I completely lack intuition here... :)
Never mind.
 
Do you have intuition in emails? :P
C'mon, you know that this is how one develops intuition...
Also, if to quote some punk...
"I gots to know..."
 
On a completely different note: this paper reminds me a bit of \omega+\omega
I only see now that it has two reference lists and I was completely confused by that...
 
There's an appendix with its own references.
 
Yeah, that's what I just figured out now.
 
12:57 AM
Oh shoo, it's 3am again and I'm still awake :-)
Goodnight!
 
@Asaf: good night!
 
1:14 AM
it doesnt make any sense to want to check your downvotes on deleted questions or answers, right?
 
Nope.
(Unless they get automagically undeleted as zombies...)
 
@Jack: definitely not, unless you're of the curious kind. Deleted contributions don't count for anything.
 
cool. I wrote a little script to show the down votes you cast. I was worried because it misses deleted contributions, but since you can't "un vote" them, it seems fine.
at least until the zombie (post) apocalypse
fwiw: 50% of my downvotes are on deleted contributions, so i've got 2 permanent downvotes on my record
 
@Jack how did you find those downvotes on deleted contributions?
 
my profile reports 4 downvotes, but my reputation reports 2, and I vaguely remember 2 answers getting deleted after my vicious downvoting and namecalling
 
1:22 AM
Are you sure you didn't ever downvote a question? Those don't show as -1 for rep, I guess since the downvote penalty on questions was removed.
 
oh
could be that
 
If they're downvotes on answers that are now deleted, I presume the rep drop was undone when the answers were deleted. Maybe check math.stackexchange.com/reputation and compare with your displayed rep?
 
my rep is about 95 higher on the site versus on the rep calculator. i've already lost about 300 to rep recalc though, so i'm waiting for my 10k + 95 to recalc
(oh and my script just runs through the reputation page, that's where it gets the downvoted answers)
 
:D
 
@JackSchmidt I'd take the opportunity to recalc before hitting 10k+95, to lose and regain the privileges, or to see whether or not the privileges are given by the /reputation page. It's an interesting question.
Now I'm really off to bed :D
 
1:29 AM
between 10k and 10k+95?
 
After mulling over it for an entire evening, I have the feeling a book like "The Group Theory Activity Book" will sell decently...
Something with a lot of cardboard and acetate cutouts...
(Maybe the shorter title "Visual Group Theory" would suffice.)
 
oh, i took a picture of the laser-cut acrylic graph paper
(i think already a book called "visual group theory")
maybe touchy feely group theory
 
Aargh. Couldn't they just disable the displayed environments in titles?
 
It shouldn't be too hard to strip $$ or \displaystyle in titles, I reckon.
 
I was just expecting the curve, but the grid's a nice touch!
(though I think the grid makes things a bit more costly.)
 
i think it doesn't add too much to our cost, but it could be our weird accounting (we mostly only pay the human labor; electricity is on separate budgets, and I think the acrylic is cheap)
 
2:11 AM
@Jack: thanks for another nice and useful bookmarklet!
 
Indeed!
 
(However, I'd be much more interested in seeing the questions I downvoted...)
 
Okay, I need to step out. See you guys!
 
See you, @JM
 
 
2 hours later…
4:32 AM
woohoo 10k club. --partyhat--
 
@anon: congratulations!
(but you could have yelled a bit louder)
 
5:03 AM
@anon: congrats! That was fast.
Amazing! 19 in the chatroom.
 
It's getting crowded in here...
 
I wonder if we can fit 2 dozen in here.
 
5:21 AM
@robjohn: I see, you're adapting to Asaf's love of chaos.
 
@tb adapting in what way?
 
@robjohn: assimilatory?
 
@tb Oh, because I am willing to risk bringing down SE just to see if we can fit 24 people in a chatroom? For possibility, I wasn't really thinking about the crash, more about the availability of 24 people wanting to be in the chatroom at the same time.
 
what's so special about the number 24
 
@anon: looks like that's the number of free slots in the box on the top right of this window.
 
5:31 AM
I see 22 total slots. Guess my laptop isn't wide enough.
 
is it inappropriate that my favorite example of critical point is the height function of a pair of boobs?
boobs have a nice saddle point
 
i know i'm right, but whenever i try to explain this example to an audience i crack up
 
@anon: I'll add my congrats, too.
 
yea, congrats man 8)
 
5:37 AM
thanks !
 
6:08 AM
@tb I wonder if a third row appears after 24.
there are now 20
 
@robjohn Yes, probably. I already have three rows visible:
 
@tb I guess it is dependent on the screen real estate.
 
@robjohn: Looking at the page source, it seems as if there is an upper bound and then a link saying "more" is presented so that you can expand the box: last two entries of the source for the box are: <li id="present-user-25053" class="present-user user-container user-25053" style="opacity: 0.15; visibility: visible; width: 32px;"> <!-- some further stuff>
<li class="more" style="display: none;" title="show -24 more"></li>
 
6:24 AM
good morning everybody
@anon: so you're one of them, bravo )
 
now he's one of the... 1%!!!!!
 
we're unthinking majority
 
Speaking of witch... three witches watch three watches. Which witch watch which watch?
 
op-op-op, @tb
@AsafKaragila is it about permutations?
 
morning Asaf
 
6:30 AM
Top of the morrow to you Tuberculosis.
@Gortaur It is a tongue twister.
 
@AsafKaragila incredible, is it? does such simple stuff twist your tongue? ;p
 
Thanks :) but never present me as a present, because that's not part of my name
 
@tb I don't see any reason for OP to not even upvote your answer
 
@Gortaur Actually no. If would have known Hebrew I could have given you a real knotted tongue.
@tb How is the edited greet? :-)
 
@AsafKaragila shla @Sasha po shosse i sosala sushku
 
6:34 AM
thanks guys, that wasn't the intention...
 
I refuse to learn tongue twisters in languages I do not speak at some degree of fluency.
 
@AlexeiAverchenko: have you seen my 'sosala suchku'? dangerous typo
 
dude... how old are you?
 
me?
Asaf?
 
and there i was thinking my boob height function was juvenile :D
 
6:36 AM
@AlexeiAverchenko that's what I was thinking about you when reading this :p
 
@AlexeiAverchenko tell me, girls in Russia are still as cool as they were 3 years ago? (
 
i'm still officially allowed to obsess over breasts :P
your milage may vary lol
 
@Asaf: I think it should be "Which witch watches which watch?" if you want it to be grammatically correct.
 
math.stackexchange.com/questions/75656/… I cannot understand it. 5 minutes of work and +45. lol. very sad lol
 
6:39 AM
@anon And I think that "I gots to know" is a grammatically wrong, yet accurate quote. Sometimes you gotta let go of grammar for the fun.
 
@AsafKaragila 0_o7
@AlexeiAverchenko 0_07
 
?
I can see... with your apparent lack of capitalization and periods at the end of sentences.
 
Dammit, I did it again!
is 'dammit' a word btw?
or is it considered two words?
 
mda... see you, comrade Alexei, in my cabinet
 
in your what?
i'm afraid i wouldn't squeeze there lol
 
6:48 AM
anyone home?
 
i'm off to the obligatory accounting software classes. did i mention my uni is fucked up?
 
are you in actuarial/finance/applied program?
 
no, pure math!
we have function theory, kahler geometry, and accounting
if you ask why, your guess is as good as mine
 
i guess you uni has decided to equip their grads with mandatory safety helmets
for the rare and treacherous roads of the post-phd academic market
 
@yayu: This may be true. But do you think such safety helmets are actually effective? Anyone only slightly familiar with how students function will know that after their graduation/PhD. people will have forgotten 100% of such mandatory material.
So I wonder who thinks that it is worth the effort inflicting such things on unwilling students.
 
7:53 AM
@t.b true. Though I think these courses are intended to get marketable terms into a student's CV. So if someone chooses to apply to a bank etc, they can show that they have had such and such course. Ofcourse, having gained a degree in math they would be expected to get quickly familiar with such subjects if their employment required it.
It is a bit unfair to make these courses mandatory though, and I agree.
 
8:08 AM
@AlexeiAverchenko 1C? )
 
8:49 AM
21
 
@rob black jack?
 
No, the number of people in the chat room.
It's a sticky chatroom :-)
 
they're keeping silence on thoggh
 
Yeah, they're ghosts.
 
@rob I'm still 'hosting' BardDur )
so we can escape there in the case ghosts will wake up and flood the current room
btw, if you receive @rob messages?
 
8:57 AM
yes
I think anyone matching rob will get @rob messages.
 
could you try to send something to @G?
 
@G I don't think 1 character works
@gor I think it takes 3
 
that's right
 
@go not 1 or 2
 
@ro
bro ;)
 
9:00 AM
When others named @anon get pinged it goes in my inbox too.
 
yep
It's a blanket ping.... a ping in a blanket. now I'm hungry.
 
damn. secretary just written me that I should take British Airways
 
to where?
 
Orlando
 
Ah, for the conference.
 
9:03 AM
exactly
reading about adjunction spaces for the first time makes my brains boil a bit ( that does not say anything good about'em
 
Hmm.
 
Good morning guys.
 
@Jonas: it seems just a bit ago I said good night.
 
@robjohn Good night!
 
:-)
@JonasTeuwen Almost noon there.
 
9:19 AM
Here too... 11:19. Where is "here"?
 
9:42 AM
Weird. I was only in the other room for a few seconds, and that was almost a day ago, but the system still says I'm "in" that room. Must be a trap.
 
@anon perhaps the 22 showing in this room are trapped as well.
 
I was also thinking that when it was mentioned
 
You can check out, but you can never leave.
 
Excellent song :)
 
I agree.
 
9:47 AM
This is my kingdom. This is why you can never leave.
Welcome to Hell (Venom)
 
@mafutrct is that a hydra in your gravatar?
 
It's a sea anemone
If i recall correctly, that is.
 
@robjohn HYDRA was the organization led by The Red Skull.
 
took me many years to pronounce that correctly
 
@AsafKaragila Yes, but hydra are also sea animals that reproduce by budding and also a multi-headed mythical creature.
 
9:53 AM
@robjohn Of course. However The Red Skull was awesome.
 
@anon anonome :)
 
head asplode
 
@anon 22 years?
 
@Asaf: I haven't lived that long...
 
@anon That to say that you still cannot pronounce it properly :-)
 
9:54 AM
oh... sure I can. I just need to pause a moment first.
 
10:20 AM
@Gortaur what else? :)
 
programming language in Cyrillic alphabet is something really amazing )
 
@Gortaur you mean pushforwards in Top?
 
what?
 
no, about adjunction spaces
or do you mean something more elaborate?
 
ah, Top=topology
yeah, I beliebe them X \cup_f Y
 
10:23 AM
i had problems with them the first time i read about them
(about half a year ago)
but now they seem trivial
you have two maps f: A \to X and g: A \to Y
so you identify images of A in both
 
they're quite ok, but I need to get focused. Once I realized how to construct the quotient map and its fibers it became easier
 
and the rest of these spaces you can picture kind of 'flapping' on the sides
if A = \varnothing you just get a disjoint union
 
By Jove, it's a party here!
@anon: congrats; how does it feel? ;)
 
just like in case of pullbacks you get product when fibering over a point
 
sorta cool
 
10:26 AM
@JM it's a party, but Jove hasn't shown yet.
@anon: I'm jealous. I have 3K more to go.
 
@robjohn: You've been on the site half as long as I have.
 
@rob: You should have joined earlier! ;)
 
@JM Had I known it existed, I would have. :-p
 
@AlexeiAverchenko tell me, q(Y \ A) = Y \ A?
 
Oh wait, your profile now says 3 mo instead of 2 (now).
 
10:28 AM
what is q?
 
quotient map from disjoint union to X\cup_f Y
 
@anon In a few more days it will be 3 months. 7/28-10/25
 
10:41 AM
Hmm, peculiar: after deleting a comment, the software does not let me vote for comments until five seconds have passed.
 
It counts deletions as 'votes' even for comments; has always been that way for me.
I'm in the process of color-coding my latest answer on Dirichlet convolutions :)
 
@Gortaur afaik yes
 
@anon: I've upvoted already (matter of fact, I used up my last upvote for today on you), so I can't give you another one for coloring... :P
 
when A is the subset of X
wait
now i'm confused :D
what is being adjointed to what?
 
11:02 AM
Four colors this time. I consider myself a creative pioneer.
 
11:12 AM
@AlexeiAverchenko A is a subset of Y, f:A->X is countinuous. We say that a~f(a) and by this relation make a quotient space
 
@JM that's a lot of upvoting :-o
 
Well, sure. If I can read it, and it's genuinely nice, I upvote.
That a lot of posts satisfy both conditions is coincidental.
 
@Gortaur then yes, you are correct
 
ok, cool. thanks
I'm currently thinking how to prove that two 2-dim closed balls in attached by the boundaries are homemomrphic to the 2-dim sphere
I think I should consider them as quotient spaces of some universal space and then put quotient maps with the same identification
maybe simple, but sounds challenging to me )
@AlexeiAverchenko but in the case you have any advice - do not hesitate to tell it to me
 
11:40 AM
@Gortaur since when has anybody with advice ever bothered to withhold it? :-)
 
sometimes it's withheld due to the pedagogical reason
I would like to clarify that's it's not the case, I've stacked with my problem
 
would a quick conceptual linear algebra doubt be entertained this morning?
 
@Gortaur come again?
you have two 2-disks
 
@AlexeiAverchenko what?
aha
 
how do you attach their boundaries?
i mean to what?
are we talking a cw complex here?
 
11:55 AM
@AlexeiAverchenko: come here, ok? chat.stackexchange.com/rooms/1402/baraddur
 
12:11 PM
@Gortaur Runnin' away, eh?
Sorry for the multiple pings.
@yayu what's up?
24 in the room. It's a trap!
gotta go afk. bbl
Wow! 13 dropped from the list at once.
They were all trapped in our dimension.
 
What list?
 
@Jonas list of people in the room
 
The list of people in the chat room. There were 24 and then a whole row went away, now there are 11
 
Okay.
 
12:42 PM
now I returned
 
I've got a stochastics question that is probably very simple but I fail to figure it out, should I still create a post on the site?
 
@mafutrct say it here
 
I have two matrices that are possibly related by a similarity transform...
 
It's about a fairly complex issue from Mahjong but I tried to make it simpler:

Given is a number Q as well as b black and w white tiles. Repeatedly draw a tile. If it is white, toss it away and draw again until the first black tile is drawn. The result is Q plus the number of whites drawn (i.e. minimum Q+0 and maximum Q+w). What is the average result?
 
their trace is the same... a prelim check...
then , assuming that they are similar, i try to find the similarity trasnform...
 
12:55 PM
@mafutrct: first of all, you can get rid of Q
 
where T and J are matrices that are related ...
A = YX^-1
 
second, you have b and w tiles and the 0 step and then you need to know the probability p(k) that the first black one will appear on the k-th step
k<= w
 
now.. when I diagonalize T all eigenvalues are zero... so can we conclude that they are not related?
 
then sum (k-1) p(k) over k= 1,2,...,w
 
*related = similar
 
12:57 PM
if it does not give you any idea, then put the question on MSE
@yayu I'm not good in linear stuff, sorry
 
@Gortaur Ah, yes, that's the approach I thought of as well
But it appeared it should be simpler
If you read C# I could show you some code that implements the idea?
 
now, I read only LaTeX (
 
Haha okay
Well what I'm doing is:
1. Get the result for the 0 case (which is simple)
2. Sum over each probability multiplied with its respective result
(Just as you described)
 
@yayu: If all the eigenvalues are zero, then, the trace should also be zero, yes?
 
But the probability calculation I'm using is a little funny:
 
1:01 PM
yes
 
(Unfortunately, I need to step out for quite a while; just type out the rest of what you have and I'll get back to you later unless somebody does it for me in the interim.)
 
k.. but then the D is just the zero matrix... and it annihilates everything.. i.e J=T=0
 
It should be the product from 1 to n (n being the number of whites) over (w-i)/(n-i) for i misses, then multiplied by (n-w)/(n-i) for hitting
 
and what is the question?
 
I kept looking for a simple formula to get the result of this product right away instead of actually iterating over it
I think it's the same as the Product(i=0 to n) over (a+n)/(b+n)
Is there a way to get the result of this right away, or do I hate to actually multiply all values?
 
1:09 PM
hate=have?
)
 
yes, sorry
 
I've got that p(k) = (1- p(1) )(1-p(2) )...(1 - p(k-1) ) *b/(b+w+1-k)
 
The first part (left of the *) is what I had too, the right side I'm still trying to understand
Ah, I had (b+w-k) without the -1
Probably a mistake on my side
 
check it for k=2
 
1:30 PM
Ah, I see where it's from, I was thinking of k as the number of misses instead of the number of tiles taken
It was a confusion on my part, should be fine
 
@yayu: I'm back.
 
2:02 PM
@Gortaur So, is there any way to abbreviate that expression? I just had a look at various formularies but found nothing helpful
 
I'm sorry, I;m busy a bit
 
Sure, no worries
 
I guess there is - but if you would like to obtain the closed expression maybe it's better to put this question on MSE
interesting enough, and I;m not so good with summations
 
"closed expression" of?
"closed expression" of?
 
2:35 PM
@JM and one more time
 
I like summations
 
Yes, I accidentally sent it twice...
 
icic, but I spent 30 seconds trying to find the striking difference )
@JM closed expression of sum k p(k) with k=1,2,...,w
 
I think you need a Pochhammer symbol when finding striking differences.
 
Yeesh, that's one deep recursion. Not my expertise, then...
(\nabla, \Delta, what's the difference? :) )
 
2:44 PM
not mine as well. Though I thought that not as bad as Pochhammer
@JM the mirror map
 
Actually Pochhammer would be somewhat easier for me. ;)
Something sent to me: "But I do have good aesthetic sense! That's why I know that my solution is ugly!"
 
@JM soulless objects send you messages?
 
@Gortaur I assume p(0)=b/(b+w)?
 
@robjohn exactly
 
@Gortaur: No, a friend I am talking to in a different tab of my browser...
 
2:49 PM
@JM ah, ok. was confused by your ' something '
 
@Gortaur how high does k go?
 
from 1 to w
 
So dry... maybe I'll actually check my other SE accounts...
 
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