« first day (468 days earlier)      last day (4552 days later) » 

10:00 PM
Ah. I like writing fancy symbols irl so I'd never do that.
Uh-oh, time to relogon in T minus something.
 
It's just the first Hebrew letter you'd learn.
 
Literally. :p
 
@AsafKaragila That's how it is in the rest of the world('s math departments) too.
 
What is?
 
That aleph is the first Hebrew letter you'd learn.
 
10:05 PM
Ah, yes.
Some people get to Beth and rarely folks get to Gimel too.
 
I got to beth but not gimel.
 
Gimel(kappa)=kappa^cf(kappa), if I recall correctly.
 
It's a bit strange that beth_1 isn't more used than it is, given the popularity of aleph_0. I mean people will rather supposed that their readers remember a special meaning for blackletter lowercase c...
 
Last week one of the guys here proved some theorem about partitions and whatnot, it was about finite subsets. He said that if we want to go to infinite sets then we need to talk about beth numbers instead of aleph (or assume GCH, of course)
 
(@Srivatsan, your comment is duly noted)
 
10:14 PM
Thanks, @Henning.
 
My problem with GCH is that it implies AC.
 
what was the question on partitions?
 
I do not recall. It was something from a paper by some guy, the same title was used by Erdos and someone else (Rado?) 30 years later when they extended the results.
 
"On a property of families of sets"
He wrote it with Hajnal, not Rado.
@anon What about this question?
I remember it was originally tagged with which is not a term for "cardinalities which are uncountable".
 
10:22 PM
Can W|A calculate the gcd of gaussian integers?
Also, How?
 
W|A can preform induction up to epsilon_0, though. :P
 
Dunno, came up with a google search for beth and partitiosn on site:math.se
@TheChaz: the WA syntax is GCD[a+b*I, c+d*I]
Also, it appears the '@' sign and " sign are switched on my keyboard's circuits or the computers receipt of my keystrokes... weird.
 
Ok thanks. Maybe I can slip away during my test to check my answers... O_o
(jk)
 
just remember to capitalize GCD and I
 
I'll make sure Siri knows the syntax :)
 
10:29 PM
@TheChaz (disappointed) I read it as Sri at first glance. =)
 
Err, WA says both 2 and 1+I are primes. It should make up its mind...
 
Sleepy time here, good night everyone!
 
night Matt
 
@Srivatsan Wish I had THAT number during test time!
 
10:30 PM
woooooooooooooot
 
lol, Chaz can't see it..
 
Wait, what is $I$?
 
I (capital) is how Mathematica/WA understand Sqrt[-1]
 
I mean $I_A$ in the low-hanging fruit question
 
@TheChaz Indicator functions, I think. At least the question makes sense that way.
 
10:32 PM
(oh, yes. I'm used to it being the indicator/characteristic function.)
 
@tb Is the question as trivial as I think it is? Surely, I am missing something...
 
So @t.b., intuitively, that "A and B are necessarily Borel measurable" seems like it could be true, but the pessimistic analyst thinks that this is not the case.
 
@TheChaz a and b could be zero right?
 
I was thinking that if A was unmeasurable and B = A' ... but that might be misguided
 
according to the, uh, deleted answer, you should split into cases a=b and a=/=b
 
10:36 PM
@TheChaz: yes if a < b: consider {f \geq b}, {f \leq a}. You're right on track and you should follow Srivatsan's hint as well.
 
This is the highest hanging instance of low-hanging fruit that I have ever attempted to eat :)
 
Well. I am going to sleep.
 
@TheChaz: you can do it!
 
Also, there's a new data dump on data.SE, fresh from three days ago!
 
Think of the starving children in Africa and reach high.
 
10:41 PM
@t.b. If I may request a slight rewind... Are you saying that I_A:= {f \geq b}? If so, what is f ?
 
@TheChaz f = a I_A + b I_B
 
(doh)
 
@tb Are A and B disjoint?
 
@Srivatsan: sure, sorry
OTOH: doesn't matter, then you need to consider some more cases...
@TheChaz: assume 0 < a < b. Then A \cap B = {f \geq a+b}. B = {f \geq b} and A \cup B = {f \geq a}
 
Ok. I was still trying to work with A \notin (scriptM) and B = A' .... to find a counterexample.
 
10:49 PM
So these three are Borel. Finally A can be written using these three sets, so is Borel, too. The other cases are similar.
 
@tb You can also say A = {f=a} union {f=a+b}, similarly for B.
 
@Srivatsan: yes. even better.
This also makes it easier to see why it fails for a = b
 
Apparently, the devil has entered the mix. (cf Henning's comment on the L-H fruit)
 
"The adversary" - is there another??
 
10:53 PM
Oh, I thought you were referring to a chat line. Move along, nothing to see here.
 
:)
 
@tb Hmm, I wonder what happens if there is a larger number of indicators mixed. Like a1, a_2, ..., a_n. Seems like there could be elaborate failure modes...
 
@Srivatsan: that's just disgusting, but if you want...
 
I guess this will work if all 2^n subsets have distinct sums.
 
Would have said something similar.
 
10:55 PM
Ok, got to go. (Got called for dinner.) // See you!
 
bye
@TheChaz Finally I know what user lhf's letter mix stands for...
2
 
Oh you can't be serious!
 
ohohoho. I get it now. I always wondered what that was.
 
The ... two... questions that LHF has asked don't really hang that low :)
I'm still scratching my head about the Borel. It's not true, RIGHT?!?!
 
@TheChaz: It's true if a \neq b and both are non-zero
 
11:03 PM
Ok. My counterexample relied on a = b, and lots of handwaving.
Off to answer the call of the frisbee. If anyone wants to talk about Bounded Variation and Absolute Continuity later, I'll be here!
 
@TheChaz if a = b = 0 it's trivially wrong. If a = b you can take two non-measurable sets whose union is measurable (that's probably the one you did). If a = 0 then you can take A non-measurable.
 
Alright! I can see the light, at least a little :) ciao
 
@TheChaz: have fun!
 
@tb What if a=-b?
 
QED
hello
 
11:11 PM
@HenningMakholm Ooops, I forgot about that case too... Take two disjoint Borel sets A_1 and A_2 and split the complement into two non-measurable parts add one of them to both A_1 and A_2.
 
For example A_1=A_2=Ø.
 
QED
Does anyone know just how much mathematics from high school people in the various university courses use?
 
Yeah, it was a long day.
 
@QED: Shortest answer: "Most of it."
 
Anyone up for a quick conspiracy to make Michael Hardy hit 10,000 precisely?
 
QED
11:17 PM
@anon, that surprises me a lot
 
Well, I suppose a lot of it is implicit, and it really depends in which direction you go.
 
@HenningMakholm But that's too easy upvote one of his questions, downvote another one.
 
Um... long day for me too. I got into my mind somehow that there had to be one person upvoting and another downvoting :-/
 
And what if, just what if, there were an upvoting saboteur to this diabolical plan?
 
Then the saboteur shall be remembered in infamy for having ruined a Nice Number.
 
11:21 PM
Remind's me a bit of this. But I thought I had waited longer...
 
@anon (Sorry for replying to such an old chat message.) I'm surprised that someone with your math skills has never been to college. I presume you've taught yourself a lot since leaving high school? (Or was your high school particularly strong?)
 
(Whistles innocently)
 
@HenningMakholm You should take a screen shot and email it to person in question. :)
 
@Mike: I took calc 3 / lin alg / diff equ in 11th grade from someone who should have really been teaching in uni - few liked him because he was rarely cognizant how over-everyone's-heads he was speaking (contra a few other math teachers, who were probably below a number of students' heads..) But most math I taught myself before seeing it in class - e.g. my dad gave me his college calc book when I was 12ish - so I served as sort of a bridge between teacher and classmates, educationally.
But yes, I read math in my free time. Nowadays I'm at the point I don't have enough time/concentration/motivation to do actually challenging study and non-challenging things don't really feel worth thinking about at all.
 
11:37 PM
"my dad gave me his college calc book when I was 12-ish" - dude, that's awesome. :)
 
I know. I beat my best buddy in a math competition in 8th grade because a joke-bonus question came up asking the derivative of 3x^2 or some such. :)
 
QED
I was reading this thing about math education in high school (guy saying we don't need it) on reddit and everyone seemed to think it's was necessary to put everyone through the math classes
 
What, Lockhart?
 
QED
no but sort of along those lines, it was a school teacher giving a talk
 
@anon Thanks for satisfying my curiosity. I'm impressed that you are as good as you are given that you are largely self-taught. I suspect you would match up well with some of the best senior math majors at the university where I teach.
 
11:42 PM
@HenningMakholm Reminds me... I've seen some people feel uncomfortable with Iverson brackets for similar reasons.
 
@MikeSpivey Not sure he would appreciate it. He thinks I'm a dishonest idiot.
 
@Mike: Thanks! @Henning: oh wow.
 
@HenningMakholm Wow. Double wow. How is it running into the same people years later on a different math forum?
 
Grudge-carrying 'R' Us!
 
QED
remarks like that aren't appropriate on wikipedia edit summaries
 
11:45 PM
@QED That's not the first one like that on a Wikipedia edit summary I've seen from the person Henning is referring to.
 
I think I now remember what put me off editing wiki pages years before...
 
QED
someone better tell him off so he can realize and stop doing it
 
He sure has his pet-peeves. Latex apart, how often have you guys seen the Morera thing, the cumulative distribution thing, and don't dare proving the infinitude of primes starting with the sentence let p_1,...,p_n be the n first primes and derive a contradiction.
 
QED
I guess the whole education reform stuff is just pointless because nothing is going to happen
 
@QED What I linked to is probably not the best educational case -- like it's 5 years old!
 
QED
11:48 PM
I do think much better is possible
 
(But I have been uncomfortable around him ever since, so there will be no emailing of the screenshot).
 
QED
well if it's such a long time ago he might not still think that
he probably has a short temper
 
I know we get frustrated sometimes with OP's, but I seem to have run into a string of nice ones recently. (See comments by dhz, fog, Carolus, and han.)
 
QED
0
Q: How can I calculate $\int_0^{\pi/2}\frac{\sin^3 t}{\sin^3 t+\cos^3 t}dt$?

JackCalculating with Mathematica, one can have $$\int_0^{\pi/2}\frac{\sin^3 t}{\sin^3 t+\cos^3 t}dt=\frac{\pi}{4}.$$ How can I get this formula by hand? Is there any simpler idea than using $u = \sin t$? Is there a simple way to calculate $$ \int_0^{\pi/2}\frac{\sin^n t}{\sin^n t+\cos^n t}dt ...

I don't understand that answer at all
 
@Mike: The "responses" tab in your profile only shows up for you (same with everyone else). :)
 
QED
11:59 PM
he just shows how to turn it into a different problem
 

« first day (468 days earlier)      last day (4552 days later) »