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12:08 AM
@TedShifrin do you get a discount with line bundles?
 
Hi @robjohn! Generally, in packages of 10 or more.
 
Is that a mathematician's dozen?
 
12:27 AM
What would we do with a baker’s?
 
@TedShifrin. Hello Prof, I found dealing with logical equivalences needs just practicing the identities!
The more I practice the fast I can solve though still not easy to get some in 1 MINUTE :/
I am not sure, truth tables also very fast to find out
The problem is since we have a multiple choice, then doing truth table might not be smart choice!
If we have 4 choices, then each needs 30 seconds at most to find out
Which of the following set of expressions is not equivalent to each other?
a) 𝑝 and (𝑝 ∧ 𝑞) ∨ (𝑝 ∧ ¬𝑞)
b) 𝑝 → (𝑞 → 𝑟) and (𝑝 ∧ 𝑞) → 𝑟
c) (𝑝 → 𝑞) → 𝑟 and (𝑝 ∧ 𝑞) → 𝑟
d) All of the above are correct
In this question, we have 2 on each side times 3, we need 6 truth tables!
 
i don't personally know of how to speed that up. except for the usual piece of advice, which also applies to getting to madison square garden.
 
1:14 AM
Can anyone give me a quick tip? About sets
 
you're gonna have to be slightly more specific than that
 
possibly? it isn't rude to just burst into the chat with a question about sets :)
 
smacks leslie
 
I need to solve a problem, but I don't know what ]1, 5[ means. All I know is it's an open set, but what are the elements of that set? How can I know?
 
well, it isn't as rude as smacking people, anyway
 
1:17 AM
@BahSoh think it's just R \ (1, 5)
 
B = ]1, 5[
@hyper-neutrino What does the R \ mean?
 
my guess is it's european-inflected notation for what in america we would write as (1,5) or the set of real numbers x satisfying 1 < x < 5.
 
real numbers minus (1, 5)
 
No way
 
that's actually a good guess
but it's $(1, 5)$
 
1:18 AM
@Avra truth tables in general are not very efficient; they're reliable, but take a long time (like you already observed). the approach I typically use is if I think they'll be equivalent, I try to use whatever rules/laws I am allowed to (for my current class, that would be the essential laws of propositional logic) to make them equivalent, and if I think they'll be different, I try to find a countercase
 
Yes, depending on context, either integers or reals between 1 and 5
 
people, and i think some of them are french, prefer the ] [ because it performs the same task of telling you whether the endpoints are in there, without also perhaps looking like an ordered pair.
 
Right, ordered pair conundrum
 
yeah its not a bad notation
I ran into that very issue just the other week
 
i don't know what kind of math these europeans are doing, that this always comes up.
 
1:19 AM
coprimality is written the same way, too
 
but if it makes them happy, they can have it.
 
Not to mention thinking 0 is natural.
 
what IS the consensus on 0?
 
don't get me started.
 
is it a natural number? isn't it?
 
1:20 AM
world war 3 will be fought over that, joe.
 
i don't think there is a consensus lol
 
reaches for popcorn
 
Given the sets A=[0,3] and B=]1,5[ find B−A. I found it is B = {4, 5}. Am I too far from the answer?
 
So your universe is integers?
Just the point I made earlier.
 
i think in one term i had two courses where 0 was natural in one but not the other
 
1:21 AM
i would almost be tempted to draw a picture.
 
You have gone to the dark side.
 
math departments should establish this department-wide, and post it on a statement of belief on their websites.
it should also be the #1 potential faculty interview question.
 
I’m too lazy to roll innumerate eyes.
 
@BahSoh well, if ]1, 5[ really is just (1, 5), then I don't think 5 should be in B
 
this is really really cute - what are the asymptotics (if he had just said "find the limit", then if I am reading this problem right, he would have just given it away to an extent) of $\int_{[0, 1]^n} f \Big( \frac{x_1 + \ldots + x_n}{n} \Big) dx_1 \cdots dx_n$ for $f$ continuous on $[0, 1]$
 
1:23 AM
ted they might make my wife be chair in the spring. she's going to get a lot of unsolicited advice from me about what her department ought to do.
 
My advice to you — butt out.
 
ok, fine, i'll enable chatjax to read that sorry integral.
 
But you never listen to reason.
 
@hyper-neutrino Thanks
 
it's worth it!
 
1:26 AM
Make an obvious change of variables?
 
don't make an obvious change of variables (or do, idk)
 
A=[0,3]. Do the square brackets mean that this set goes from 0 to 3? I ask this because I usually see braces
 
oh, you are asked to find the limit as $n \rightarrow \infty$, of course.
yeah no $x_i \mapsto n \cdot x_i$ won't do you any good here
 
Braces are totally different, @BahSoh. Learn the notation for your course. We’re not here for that.
No, dopey @JoeShmo
 
then whats the change of vars youre talking about?
you don't want to touch this with the solution I am thinking about
 
1:30 AM
Substitute for the mean, leave other variables alone.
I’m not giving it serious thought.
 
it's the strong law of large numbers
and then lets say dominated convergence
 
You didn’t specify context or tools.
 
probability theory
 
I’m thinking low level.
 
and observe that these are uniformly distributed on [0,1]
 
1:32 AM
Shrug.
 
apologies if I wasn't clear
so anyway, youre asked for the expected value here
 
Of course you weren’t .
 
so looks like after all is said and done, it's $f(\frac{1}{2})$ in the limit
yeah, no, I knew we had to use SLLN in this assignment, so that's almost cheating
I just thought this problem was really cute, wanted to share
but I guess that's not everything though. I guess he does want the rate of convergence. so perhaps its a bit more involved...
 
1:51 AM
Plz halp
0
Q: Probability density for a Markov process with Gaussian two point functions

DanielSankThe problem Consider a stochastic process with the following three properties: The process is Markov, meaning that $p(x_n,t_n|x_{n-1},t_{n-1},\ldots x_1, t_1) = p(x_n,t_n|x_{n-1},t_{n-1}).$ The conditional probability is $$p(x_n, t_n|x_{n-1},t_{n-1}) = \left[ 2 \pi \sigma^2 (1 - e^{-2 \gamma (t_...

 
2:05 AM
@hyper-neutrino. %100! Took me 3 days to realize that turth table won't work with multiple choices! We have to use equivalences otherwise we are lost.
 
2:50 AM
Hi @Ted
 
3:09 AM
Hi @Lukas
 
@Ted if I have a hyperbolic manifold and choose a point, is the isotropy subgroup of the isometry group at that point compact?
the motivation for that question is this question: math.stackexchange.com/questions/4283521
the proof I gave is complex-analytic
but I'm wondering if you can work hyperbolically as well
 
3:25 AM
Where did hyperbolic come from?
 
well, if you have a bounded domain in $\Bbb C$, you have a hyperbolic metric such that all holomorphic automorphism are isometries and conversely
or do I have that wrong
 
it came from the greek, hyper- (over) and ballein (to throw)
this is different from hyperballin' which is what i do in the club
 
oh wow, someone actually talking about ancient greek
I'm happy
τῶν πασῶν γλώττων τὴν Ἑλληνικὴν ἕτι προτιμῶ
 
3:45 AM
Good evening everyone. Is there a meta-algebra that describes, say, what the $n$th term of a polynomial is in terms of its coefficients and powers of coefficients? Consider, for example, that $(a+b)^n$ describes an $n$th order polynomial. What is the $(n-k)$th term going to look like for arbitrary integer $k$?
 
@AMDG Are you talking about the Binomial Theorem?
 
If what I searched on Binomial Theorem is correct, then I'm looking for a generalization of the binomial theorem.
i.e. to arbitrary sums $(\sum_{n}^{k} p(x_0, x_1, ..., x_n))^m$ where function or relation $p$ describes a polynomial with $x_n\in\mathbb{N}_0$ for my particular needs, though generalized to the reals is fine too.
 
multinomial theorem for a start
 
@AMDG are you talking about summing a polynomial of a variable number of variables? I don't know how the sum is being computed for various $n$.
 
@robjohn Well what I want, to get to the point, is a means of symbolically computing the first power of two greater than an arbitrary integer as $2^{\lfloor\log_2(x^y)\rfloor + 1}$, and I can represent the sums of powers of two as a polynomial expansion of an arbitrary number of terms to raise to an arbitrary integer power, in my case up to 64.
Perhaps the multinomial theorem as mentioned by Leslie can allow me to efficiently compute exponentiation without having to solve for optimal addition-subtraction chains since we would be operating on individual terms in parallel.
I could also just be looking at this the wrong way right now. :)
You see, I realized it's probably faster to compute reciprocals using the fact that $(0.1)_y = \sum_{n=2}^{\infty} (y-1) y^{-n}$.
Now the issue is no longer a matter of computing $2^x\bmod y$. It's been simplified to finding the first power of two greater than an arbitrary integer as $2^{\lfloor\log_2(y^n)\rfloor + 1}$ if I'm not mistaken.
And of course $\lfloor\log_2(y^n)\rfloor = \lfloor n\log_2(y)\rfloor$, so if I have a fast log2 function or a method of just computing $y^n$ efficiently or specifically the highest order bit of $y^n$ which is all I need, then I'll be good.
 
4:18 AM
What method would you guys suggest for doing what I want?
Also why is it called multinomial instead of polynomial? Is polynomial theorem something else?
 
4:33 AM
i answered a convex psq. i feel guilty now.
 
4:49 AM
hm, is it possible to find the rate of convergence in the problem I posted above?
we don't know anything about $f$ other than that it is continuous on [0,1]
 
i think there should be an unclose operation on questions.
not sure where the question is. probably need to hit the sack anyway.
 
its right there. scroll up
the question is asking for "asymptotics". but does he actually want the rate of convergence.. or just the limit
 
Meanwhile, it's 1 AM here.
 
same, twins
it's a miracle I have enough brain capacity to type
 
5:15 AM
What, we're both EST or are you saying something else?
Is that you, my long lost twin brother!?
 
6:00 AM
Am reading the Wikipedia page on the multinomial theorem. My one functioning brain cell is having a hard time. How would I compute the highest order term?
I think I'm starting to get it.
So if I understand correctly, then I can compute $2^{\lfloor\log_2(x)\rfloor}$, compute the coefficient of this term in the sum, multiply, then compute the floored log2 of the product and that will give me $\lfloor\log_2(x^y)\rfloor$?
So three would be a binomial as $(2 + 1)^n$. The highest order term is $2$. I compute the binomial coefficient here for $2$. Does the logarithm of this product give me what I'm looking for?
 
 
2 hours later…
7:54 AM
does anyone know why if $f : (a,b) \rightarrow [0,k]$ for some $k > 0$ is log-concave, i.e. $\log(f)$ is concave on $(a,b)$, then $f$ is necessarily unimodal?
i.e. there is some value $m \in (a,b)$ such that $f$ is monotonically increasing for $(a,m]$ and monotonically decreasing on $[m,b)$
 
 
3 hours later…
11:22 AM
if $z_1,...,z_n$ are points on the unit circle, and $p_j \geq 0$ are such that $p_1 + ... + p_n = n$, why is $\prod_{j} (1 + \frac{z}{z_j})^{p_j}$ analytic? I can see that its analytic in the unit disk say, but I don't know how to prove its analytic everywhere..
 
@Shaun hey, how is topos theory study going?
 
 
2 hours later…
1:24 PM
@AbstractSpacecraft
 
@geocalc33
what up mon?
I'm trying to answer: prove that $\Bbb{Z}^*$ is cyclic using only group theory (no rings or field theory) it has a 100 point bounty. It's difficult though :)
 
not too much, you?
 
$\Bbb{Z}_p^*$
For prime $p$ is always a cyclic group
 
good luck
 
1:27 PM
question
 
Sure
Hit me wid it
 
how would I learn SAS or Python
 
What's the definition of software a service?
Doesn't that just mean a server-side API?
That you can call from other apps?
 
SAS is a programming language for statisticians
 
oh cool
I know Python very well
but not sas
I can teach you
Want to zoom right now?
 
1:28 PM
sure if you have time
 
Zoom away
I'll switch it on
@geocalc33 emailed you a Zoom lnk
 
2:28 PM
In Spivak's intro, (Ch 1, Calculus) he asks a series of question ( Question 16) all in somewhat similar vein. From "Show that $(x+y)^{2} = x^{2} + y^{2}$ only when $x = 0 $ or $y = 0$", to "find out when $ (x+y)^{5} = x^{5} + y^{5}$", and finally to the generalization "when $(x+y)^{n} = x^{n} + y^{n}$" Is this pattern fundamental to calculus , or rather an exercise in understanding manipulations to come? I am trying to see the forest :-)
 
2:41 PM
does anyone know why $\Phi$ is log-concave in this answer? mathoverflow.net/questions/64099/…
 
3:07 PM
@user1115542 well, there's a general expression for $(x+y)^n$ that is generally fundamental
 
@Thorgott Often, one finds oneself immersed in the details of the proofs. And, the one pattern I see is him building on each previous proof. But, for his own reasons, he often never explains where this fits into the larger scheme of things....if that makes any sense. Hence the question
 
3:43 PM
@AMDG we're both EST
 
I figured
 
if $f$ $g,h$ are polynomials in a field $\mathbb{F}$, with $f=gh$ then $f$ has no roots if and only if $g$ and $h$ have no roots?
in the field $\mathbb{F}$
 
4:17 PM
yep
one direction is basically 0k = 0 for k in F, and the other is that if a, b in a field F satisfy ab = 0 then one of a, b must be zero
 
4:44 PM
this suggests possible generalizations of both directions
 
5:16 PM
Say you have two $k$-tensors $m$ and $n$ drawn as nodes on a piece of paper, and connected by $k$ lines (with labels for each of the lines). This defines an inner product between the two $k$-tensors. Could you generalize this inner product by considering $m$ and $n$ as sitting in some 3-space with $k$ lines connecting $m$ and $n$? I'm not sure if there's any difference between the 2d and 3d case however because in both cases you have $k$ lines connecting the tensors...
say $m$ and $n$ are connected by a single line. Then this is the inner product between the two vectors
 
5:43 PM
this rush to close really drives me mad.
 
copper would rather work with unclosable operators on incomplete normed spaces.
copper isn't there a vote to unclose? maybe i'm missing something. i rarely vote on anything.
despite being awarded my ten-year chip on math.SE the other day. getting that was the definition of mixed feelings.
 
@AbstractSpacecraft Really?
@copper.hat with blatant no-effort homework statements, I now just vote to close. I used to dither around, but I’m fed up.
 
i did think there was a reopen option. but the system won't do much if people don't cast those votes. i think it's easier for borderline posts to attract close votes. for whatever reason.
 
I’m not talking borderline.
 
you're not, but i think copper might be.
when i see a question with a bunch of close votes stacked up on it i usually agree with them.
 
5:53 PM
How can $\Bbb Z^*$ possibly be cyclic?
 
i.e. i do not think the majority of closed questions are borderline.
i'm wondering what $\Bbb Z^*$ is right now.
 
I assume nonzero integers as a multiplicative group. Otherwise, Spacecraft should elucidate.
 
@TedShifrin I am sympathetic, but I just ignore those questions.
 
Nonzero integers do not constitute a group.
 
Oh good grief.
Right. So is it just the units?
I didn’t write this. I just read it. So of course $\{\pm 1\}$ is cyclic.
 
6:02 PM
I don't know. I find this notation ambiguous, and I prefer notations such as $\operatorname{GL}_1(R)$.
 
$\mathbb Z^\times$ is another alternative notation.
 
Yes, agreed, the group of units.
 
I would probably only use GL(1,R) if I were in a context where I was also inclined to consider GL(n,R) for other n.
Otherwise seems kind of cumbersome.
 
i was thinking maybe units in integers mod n (which i denote by Z^n, n\Z, or n\Z/n\Z/n), but nope, many of those aren't cyclic either.
maybe the question is about units in integers mod a prime, which would explain the field theoretic aside.
maybe we'll never know.
 
Maybe one replaces $\mathbb Z$ by a ring of integers and asks about its torsion part.
 
6:17 PM
maybe the real cyclic object here was the friends we made along the way.
 
a polynomial of degree 5 over a field is irreducible iff it has no linear and no quadratic factors?
 
think of possible degrees of factorizations.
 
had to do a double take on this title "Example of an absolutely summable series that is not summable". the question makes sense, but just for a moment...
 
oh right, the result holds because if $p=fg$ then $degp = degf+deg g$ for polynomials over a field $\mathbb{F}$
@leslietownes
 
copper: ha, ditto. i secretly hate questions about incomplete normed spaces.
they're always in the most boring chapter of the functional analysis book.
 
7:18 PM
@hyper-neutrino. This is one example, most solved it with truth tables, but with 1 minute constraint what would you do!
 
1 minute constraint? easy, if p=1 q=0 r=0 then the first is true but the second is false
since and is stricter than or, you can get p∨q true with p∧q false, and then since → is basically <= you just imagine getting LHSs of 0 and 1, and then RHS of 0 will distinguish them
 
But you think proving taughtologies needs more than 1 minute though?
 
possibly. depends on the problem but writing out the rules might take a while
finding countercases is ofc usually going to be much easier (or at least faster)
 
@hyper-neutrinoy. That is the point, this is not fair to bring proving taughtologies in a minute constraint!
Not complaining, but...
you are right about the above one though
@hyper-neutrino. Wow! You are really smart. You guess we can replace implication statemenets with their $\land, \lor$ and solve accordingly as we can play on sticktiness gap!
Show that (p → q) → r and p → (q → r) are not logically equivalent.
This is indeed the question I was trying to solve not the one above. Concentration is at its lowest levels on my side :/
 
well i happen to already know off the top of my head that the latter is only false when p=q=1 r=0 :P
 
7:27 PM
@hyper-neutrino. Your trick would be to play on stricktness gap and replace implication statements with $\and$ or $\lor$ if possible
 
already knowing the answer is the most devious form of cheating.
very hard to detect.
one of the benefits of not starting chatjax is you get to see who uses land and lor vs. wedge and vee.
 
i used to use wedge and vee until i found out land lor existed
 
@leslietownes. :}
 
also used to type out rightarrow fully until i foud out \to works
@Avra well, here, it's a bit trickier since the other one you presented has the symmetry of both ending with →r
 
@hyper-neutrino. YEEEEEEEEAH
 
7:29 PM
whereas this one can't be broken down into segments that overlap between the two expressions
 
I told you
 
i mean here p=q=1 r=0 is the only truth valuation where the right side is 0
 
I love mathematicians, but sometimes they go very extreeeeeeeeme
 
so you just need to find any other case where the left side is 0 and that means the right side will be 1
in this case pretty easy to do p=q=r=0 lol
 
@hyper-neutrino. You think it's not easy to solve the last question in minute and half constraint though!
 
7:30 PM
@hyper-neutrino I have an idea for your Halloween costume on site!! Disguise your self as Chi
 
i don't like time contraints on problems like these in general; it really doesn't demonstrate understanding as opposed to memorizing and using hacky shortcuts
 
 
@amWhy. :0
 
like for example, if I need to do things in a short time constraint, I'm going to use the fact that → basically means ≤ for 0,1 which is a horrible way of understanding boolean logic but it's fast for my intuition
 
@hyper-neutrino. Appreciated. tnx
 
7:31 PM
@amWhy hmm. interesting idea; why chi of all things though, just curious? is there something in particular about it :P
 
I will practice as many as I can and see where it goes :'
 
i had changed my pfp to は at some point though I forget why
 
hyper: it kinda looks like a lambda that has been stabbed in the back. very halloweeny
 
@hyper-neutrino I was thinking of ease of transition. All it takes is a stab in the back, per @leslie!
 
haha
i see it now :p nice
 
7:35 PM
i would do zombie grandpa arising out of the kiddie pool but i'm not that good at photoshopping extremely low resolution images
 
@robjohn favor to humbly ask you. Could you work on my identicon to shift from blue, to orange for Halloween? Any other artistic contributions are welcome, but not necessary. :-)
 
Who needs Halloweeen when we have abstract algebra
 
@Avra Even abstract algebra likes to disguise itself in places one may not anticipate!
 
@amWhy. :(
 
8:13 PM
Excuse me, but is there any way I can improve this log2 approximation by removing multiplies while maintaining approximately this same level of accuracy? desmos.com/calculator/17e19dfjcs
As it stands, the multiplies alone would make this cost 12 cycles on Ryzen Family 17h. Better than Newton-Raphson, but it's not good enough, and it only needs to be just accurate enough for computing the first power of two greater than $x^y$ as I previously mentioned.
Getting rid of the 5th order term gives me sufficient accuracy for $x^{63}$ only. That brings me down to 9c. I want to push it the extra mile and get it accurate for $y\in [0, 64]$, though I am working on some ideas for reducing this requirement.
 
8:31 PM
Hello everybody
 
o/
 
I was practicing Fourier transformation and I'm not sure If I'm solving the problems correct.
Can you check if my solution to this problem is correct, please?
I tried to calculate the inverse Fourier transform of the F(k) I found but things got more complicated. I mean to check if my solution correct.
 
wolfram alpha comes pretty close to that for the real part of F(k). i don't think i have the memory to wade through all of the computations
 
8:47 PM
Can an insane person recognize their own insanity?
 
see e.g. tinyurl.com/3db7a6y5 for my wolfram alpha input and output. i omitted the factor of sqrt(2pi). i don't know how good wolfram alpha is at actually applying its assumption that k is real
 
from a legal perspective I'm guessing that the answer is no
 
in the US the test for insanity as a defense in criminal matters varies by jurisdiction. as far as i know, it nowhere really corresponds with any colloquial or clinical interpretation of concepts like that
 
I want to be a lawyer
 
how the standard is written matters less than what a judge lets in as evidence of a person's state of mind (to satisfy whatever standard). judges can be very lenient or strict about that depending on the cases, with no real way of making it uniform.
 
8:53 PM
hmm
that's crazy
 
Witness our former president.
 
yeah lol
 
tmw you paste a random directory from your PC.
 
ooh yeah, maybe delete that.
 
I can't now
 
9:04 PM
an admin could. bat signal
 
What will happen?
 
that has a name in it and refers to a photo that likely came from an iPhone. :)
 
Yep it's a photo
nobody hack
 
Not sure about other platforms, but HEIC I've only seen come out of an iPhone, and the serial name of the image implies iPhone in tandem with this format.
 
my user name is a string of nonsense characters on most of my windows pcs for this reason. pathnames wind up in all sorts of places you don't expect them to, and while i'm not paranoid i don't need random websites or whatever correlating my real name with other data.
 
9:06 PM
administrator
 
Downloads implies it came from the internet or was transferred directly to that folder. You can glean a lot of insight from a single directory string. :)
 
HEIC appeared a few iOS changes ago. I hadn't really paid attention, but there it was.
 
please delete that please
 
I think the only typical room residents who have that power are robjohn and Xander.
 
@robjohn if you see this maybe delete geocalc's pathname hiccup.
 
9:07 PM
ping all teh room moderators
 
Room owners like me have no power (of which to speak).
 
i do hope that facebook likes what it's getting about me from my daughter and cat's instagram accounts. instagram tries to sell me cat toys.
 
Sad
 
@leslie I'm waiting for the picture of munchkin in her crazy cat outfit next to the actual cat.
Private communication of said picture would be fine.
 
we do need to pair them. she's begun a campaign of wanting to be a 'scary strawberry' or a 'strawberry ghost.' we have enough stuff to make a cat costume. we don't know how to do strawberry ghost.
this isn't something they have at those seasonal halloween stores.
whatever she ends up being, i'm sure olivia's expression when they are paired together in a photo will be underwhelmed.
 
9:11 PM
I'll be sure to send you Screech's reaction.
 
Can we apply triangle inequality on more than 2 sides pls?
$|a+b+c| \le |a|+|b| + |c|$
 
Yes, as many as you like. Heard of mathematical induction?
 
Yeeeees prof
tnx
 
Sometimes you need to stop and think instead of pestering us.
3
 
okay :/
 
9:14 PM
By the way, has the message to be deleted been flagged for moderator?
 
Leslie pinged a request, but maybe the person who wants it removed should flag it himself.
 
:)
 
can I follow up about a question?
thank you moderator
0
Q: diagram of inner product - generalization?

geocalc33Consider a sort of graphical language of linear algebra...tensor network diagrams. So consider the space of $n$ non-intersecting paths connecting a node $A$ to a node $B.$ This "picture" is precisely and exactly an "inner product." See short blog post: Matrices as Tensor Network Diagrams. See lec...

Maybe it's not a proper generalization - but I don't really know
they don't really discuss it in the video
 
9:50 PM
i'm not getting a lot of content out of this (admittedly i am skimming it)
in particular i don't fully understand the formalism (i'm generally aware of it but this page is not helping) and don't see what paths being planar or nonintersecting would have to do with it
if this hogwash diverts math people and math dollars away from the defense industry, quantitative finance, or helping people sell advertising, i guess i'm OK with it
 
Hello!!
Let $(\Omega, p)$ be a discrete probability room with induced probability measure $P$ and let $A, B\subseteq \Omega$ be two events.
I want to show that $P(A\cap B)-P(A)P(B)=P(A^c)P(B)-P(A^c\cap B)$.
I have done the following :

We have that $(A\cap B)\cap (A^c\cap B)=\emptyset$ and so $$P((A\cap B)\cup (A^c\cap B))=P(A\cap B)+P (A^c\cap B) \\ \Rightarrow P((A\cup A^c)\cap B)=P(A\cap B)+P (A^c\cap B)$$
Now we have to show that $P((A\cup A^c)\cap B)=P(A)P(B)+P(A^c)P(B)=[P(A)+P(A^c)]P(B)$, right?
We get that result if $(A\cup A^c)$ and $B$ are independent, or not? How can we show that?

Or is there an other (better) way to show the desired expression?
 
@leslietownes I see that egreg handled that. Sorry, I have been away for a while.
@TedShifrin It has been deleted. It might still be in peoples' caches until they refresh.
 
Hello @robjohn !! Do you maybe have an idea bout my problem about? I got stuck and I don't really know how to continue. Is maybe my approach not the corret one?
 
10:09 PM
@robjohn I hope you are all hanging in there. I don't want to make things worse. Just know that we're thinking of you.
Hi, demonic @Alessandro.
 
10:24 PM
You very cleverly avoided weighing in when I was asking you to tell me I wasn't a total idiot, even though some "students of logic" suggested I might be :D
 
Oh yeah, I saw that but I was busy
 
Yeah, I wasn't meaning to interrupt you. I just wanted your professional opinion at some point :)
 
10:45 PM
I'm most definitely not a professional logician :P
I mostly think about Polish groups nowadays
 
You're more of one than the rest of us will ever be. :P
Not that I ever wanted to be a "formal proof" writing mathematician.
Are Polish groups like Polish spaces?
Instead of topological groups?
 
Are there good reasons that one should care about them?
 
General topological groups are too hard, but for Polish groups there is a lot of machinery from descriptive set theory
 
Hmm ... that sounds like professional logic.
I never got past Lie groups :)
 
10:49 PM
They appear naturally in many places, Lie groups, Iso(M) for M metric separable, unitary group of the separable Hilbert space, automorphism groups of countable structures (in the model theory sense)
 
Lie groups have manifold structure. So Polish spaces are nice enough that one can say a lot more like that?
 
I usually think about nonlocally compact groups though, there are some interesting phenomena that cannot happen for locally compact groups, they are too nice
@TedShifrin what do you mean?
 
I wouldn't have thought Polish spaces were sufficiently nice to gain you manifold-y type things.
 
my impression is that they don't, but in operator theory they get you better-than-nothing.
they might be the last safe stop on the train ride to hell that is general topology
 
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