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1:09 PM
3 hours ago, by BAYMAX
$A_{1},A_{2},...,A_{m}$ are distinct $n \times n$ matrices such that $A_{i}A_{j} = 0$ for $i \neq j$ then relation between $m$ and $n$ is ?
any idea ?
 
1:44 PM
@BAYMAX let $A$ be a non-zero matrix with $A^2=0$... then you can get $m$ arbitrarily large by having $A, 2A, 3A, \cdots$...
 
$A^2 = 0$ is not given as a hypothesiss
 
@GFauxPas I'm describing an algorithm to make $m$ large
basically unlimited by $n$
 
but there may not be such an $A$
 
there must be such an $A$
 
why?
 
1:48 PM
just have $1$ on the top-right hand corner and $0$ elsewhere
well, except when $n=1$
 
how do I know that's a matrix in $\langle A_i \rangle$?
 
@GFauxPas no, I'm describing an algorithm to make $m$ large
essentially saying "there is no relation between $m$ and $n$"
 
but you don't know that there are any matrices there that have $A^2 = 0$?
maybe there are none?
 
@GFauxPas I just told you how to construct one
1 min ago, by Leaky Nun
just have $1$ on the top-right hand corner and $0$ elsewhere
 
you did, but you didn't show me its a matrix in the sequence he has
 
1:50 PM
@GFauxPas no, I'm describing an algorithm to make $m$ large, essentially saying "there is no relation between $m$ and $n$", or "you can't deduce the relationship from that constraint"
 
hmm
what if you assume $A_i A_j \ne 0$ for $i=j$?
 
@GFauxPas now this is more interesting
 
yup :)
 
and I'm not going to think about it
 
I like your algorithm though
lol
 
1:52 PM
Is this a correct answer to this question? : math.stackexchange.com/a/1054648/266135
 
thanks for looking into the problem, sorry i was out!
Also I was thinking how Weierstrass thought of constructing the Weierstrass function continuous everywhere but no where differentiable!
 
2:35 PM
There's also an existence proof somewhere
of, like, continuous but nowhere differentiable thingies
 
If I understood correctly, it is an example of a policy that is rolled out halfway, and then at the risk of being rolled back, and said rollback caused issues because of half of the stuff being completed (the classification of type 1 ISPs to type 2)
 
I realised that one defines a great circle as some circle whose centre is that of the sphere itself. How does one show that a great circle divides the sphere into two halves?
 
Or in symbolic terms:
Intended plan:
$\rightarrow\rightarrow$
Roll back plan
$.$
Half completed plan
$\rightarrow .$
The issue is that for the half complete plan, the harmful change stayed but the beneficial change does not
Or more accurately, part of the changes will not be harmful had it gone all the way through
 
2:50 PM
@AgnishomChattopadhyay Here's an attempt:
There are two sides to every plane.
So consider the plane containing the great circle.
We need to show that (1) there's a part of the sphere on either side of that plane, (2) those parts of the sphere are congruent to each other, and (3) that those parts of the sphere are connected
For (1), choose a point not on the plane on either side and project it radially onto the sphere
For (2), note that the sphere is symmetric by reflection across the plane
For (3), I suppose you could show that each hemisphere is the projected image of a disk, and use the fact that the projection of a connected space is connected
 
That's helpful, thanks!
 
3:07 PM
@LeakyNun The problem is I don't know how to analyze these when quantifiers get involved
Like consider the sentence $\exists a:(f(a,x)=0\implies g(a,x)=0)$ where $f$ and $g$ are some multivariate polynomials
I don't know how to figure out what sort of set that could define
 
multivariate polynomial = ?
Also, hi
 
ohi
@Astyx Polynomial in multiple variables
 
3:25 PM
@AkivaWeinberger $p \implies q \equiv \neg p \lor q$
$\exists a:(f(a,x)=0\implies g(a,x)=0)$
$\exists a:(f(a,x)\ne0\lor g(a,x)=0)$
 
Would another approach re great circles be to argue that you can always rotate the coordinate system such that the great circle becomes the equator?
 
3:57 PM
Hi all, I have a short calculus question which I think is too trivial to ask as an actual question
Could anyone point me towards some kind of cheat sheet for this integral? \int_{-\infty}^{\infty} e^{i(m-n)x} f(x) dx
 
Great to see you @robjohn back in the chat room.
We lost Daniel Fischer :(
 
@Justwinbaby where did you last put him?
 
4:23 PM
A general question: is the condition of a set being infinite equivalent to there being a surjective function from the set to the set of natural numbers? Thanks
 
@wilkersmon yes
because the natural numbers has the smallest infinite cardinality
 
Yeah, thanks. i thought there might be some subtleties in set theory, like some infinite sets might not have such functions, so thanks ;)
 
@LeakyNun Oh, so it's $(\exists a:f\ne0)\lor(\exists a:g=0)$ (I think you can distribute $\exists$ over $\lor$ like that)
And then it's just the union of the solution sets
It doesn't distribute over $\land$, though, so maybe something like $\exists a:(f(x,a)=0\land g(x,a)\ne0)$ would be more complicated
("There's a number that's even and perfect" is different from "there's a number even and there's a number that's perfect", so $\exists$ doesn't distribute over $\land$)
@wilkersmon Only if you assume the axiom of choice
Otherwise weird stuff can happen
 
4:59 PM
I think that, when choice is not assumed, the accepted definition is "there's no bijection between it and a set of the form $\{1,2,\dots,n\}$". There are various other notions of infinite that are equivalent when choice is taken, but not necessarily so without it.
 
@AkivaWeinberger I don't think that's equivalent
 
5:15 PM
@robjohn hey! How is it going?
 
$\int (e^{x\ln2})'dx=\int \frac{\ln 2}{(1+W(u))}du$
Need to find more nontrivial change of variables to present this integral, perhaps I should start considering using a series as a change of variable...
(The evaluation is of course straightforward, as one suitable change of variable will brought it into the inverse chain rule form as shown)
It is of course easy to write an arbitrary power series change of variables $u=\sum_{k=1}^{\infty}a_kx^k$, but without carefully considering what the sequence $(a_k)$ is and hence the functional identities that one can exploit with special functions, such a change of variable will be meaningless and does not make the integral look anymore sensible
 
@LeakyNun "I'm thinking of a number that is either even or a square" is the same as "I'm thinking of an even number, or I'm thinking of a square number"
 
@AkivaWeinberger it has to be the same $a$ satisfying both parts
 
5:31 PM
$\lor$ is "or"
 
hi chat
 
@AkivaWeinberger you're right
 
However, we can still proceed the general investigation without need to worry about that subtlety and see what happens. Suppose we have a dominant convergent series change of variable:

$$u=\sum_{k=1}^{\infty}a_kx^k$$

Then since it is convergent, we can interchange the limits and hence derivatives and sums to get:

$$du=\sum_{k=1}^{\infty}a_kkx^{k-1}dx$$

So a simple looking integral such as:

$$\int u du = \int \sum_{l=1}^{\infty}a_lx^l\left(\sum_{k=1}^{\infty}a_kkx^{k-1}\right) dx= \int \sum_{l=1}^{\infty}\sum_{k=1}^{\infty}a_la_kx^{l+k-1}dx$$
$$\int \sum_{l=1}^{\infty}\sum_{k=1}^{\infty}a_la_kx^{l+k-1}dx =\sum_{l=1}^{\infty}\sum_{k=1}^{\infty}\frac{a_la_k}{l+k}x^{l+k}+C$$
Now depending on what $a_l,a_k$ are, closed forms may be found for this subset of infinite series
 
@Secret In the realm of integrals involving the Lambert-W function, there's a rather interesting remark in a paper I know
Consider the function $\mathfrak{W}(z)=W_0(z)-W_{-1}(z)$ where $W_n(z)$ is the $n$th branch of the Lambert-W function.
The paper defines this as the 'principal branch difference'. (More generally it has $\mathfrak{W}_{nm}=W_n-W_m$.) It's the principal one because it's the only such difference which has some real values on the real line.
By comparison, if you just had the log function the branches would be $\text{Log}(z)+2\pi n i$, where $\text{Log}(z)$ is the principal logarithm, so the difference between two branches would always be an integer multiple of $2\pi i$. So it's not interesting for the log function, but it is interesting for Lambert-W.
Anyways, here's the remark
"It comes as a surprise to find that before Lambert W was named, people discovered the branch difference
independently. Thus in 1970, Jordan and Glasser, apparently working independently of each other, used a substitution equivalent to $\mathfrak{W}(xe^x)$ to evaluate an integral posed in SIAM Review by Logan, Mallows & Shepp. The integral was
$\int_0^\infty e^{-w/2}\sqrt{w}\,dw$ where $w=\frac{u}{1-e^{-u}}$.
[It includes the citation of Jordan and Glasser where they do this.] A further surprise comes when the difference used by Jordan and Glasser is extended to the complex plane. Both real and imaginary parts of the difference are separately always positive."
So that might be a fun integral to play with.
 
5:46 PM
Hmm... I wonder if the reason that substitution works can be traced deeper, so it must have something to do with the way the principal branch difference behave
 
Alas, that instance on springer is the only one I'm aware of.
 
(OMG I have a lot of papers I want to read. I have 2 QFT in CTC spacetime papers, 1 foundational quantum and a couple of chemistry paper, and now this and a couple of maths paper)

Seriously when we can ever stop time...
 
lol
would be damn convenient, yeah
 
Reading papers are fun, but the reading itself is SLLLLOOOOOWWWW
 
Eh, it's faster if you don't try to understand it :P
unfortunately, it usually takes time even just to understand if what you're reading has the potential to be relevant.
the Jordan and Glasser ref is this one: epubs.siam.org/doi/abs/10.1137/1012028
it's pretty short.
oh. and apparently that integral equals $\sqrt{2\pi}$.
 
5:53 PM
Well, that's the issue, things don't stick well in my mind if I don't understand what's going on, that's why in the long past I am some afraid of integrals and series

Wow, that paper *is* short, I think I can go through that one right now, and I will just return to one of y time travel papers and then read the rest later
 
yeah.
They don't talk in terms of Lambert-W, for historical reasons, but it's pretty obvious how it fits in there.
(What I'd actually like to find for my own interests is an instance of the principal branch sum)
 
Ah I think I see it now, $we^{-w}$ just shouts Lambert W so is $ve^{-v}$. Putting the two together gives the principal branch difference expression (or something similar)
 
Right.
One is the principal branch, the other is the -1 branch.
 
The integrand does have that Lambert W like symmetry once you pull the $e^{-\frac{w}{2}}$ inside the squareroot
 
Yeah.
 
6:06 PM
@Secret Where else do I clear my doubts from if I am not supposed to ask them here?
 
Your avatar looks so similar to Secret's
 
@AkivaWeinberger Mine? (I hate my avatar )
 
but maybe it's like his is a bland cracker and yours is a cheese flavored one
Yeah
 
@AkivaWeinberger First it was green, it suddenly changed into brown :"(
 
I guess that'd make mine the spicy one
 
6:10 PM
[Unrelated, continuation of previous posts] The above preliminary investigation suggest we can generate some infinite series representations that involve dominant convergence,from some functions and their integrals.

But can we push this further by starting with the constant function, integrate it, made a series change of variables, and then somehow remove the variable x that is initially introduced...?

Suppose we have a number of interest, e.g. $\frac{\pi}{2}$, we want to somehow lock it inside an integral with it, i.e. $\exists f(x)$
 
though the hole in the middle makes that interpretation a bit awkward.
 
Heya =)
Anyone familiar with Vim here?
 
I dunno I think yours is like a strawberry or something
Some sort of fruit/berry
 
hmm.
a fruit roll up, heh.
 
Hm yeah
 
6:12 PM
@Abcd If you are going to ask a question, don't just throw the screencap of the problem on here frequently, rather, process your problem and isolate the key area where you have issues with, and tell others what you have tried so their answers will be more relevant
 
What's the question
 
@Secret I will try to do that from now on. If that's done, then would the "help vampire tag" be removed from me?'
 
Akiva: well, it really start from this possibly too harsh comment of mine, because this is like the 5th user that I saw this homewokr pinging behaviour in the chat I frequented, after ramajaran, and many others in the past
2 days ago, by Secret
@Abcd From observing your interactions with other chat users throughout h bar, maths and chemistry over the past 5 months, You are really starting to sound like a help vampire. What exactly have you tried on that question?
Well, my issue is I am actually a very impatient person. I like to help people, but I don't like to felt like I am helping someone to do their homework
@Abcd If you have made you effort, I think you will be fine
chat is slightly more relaxed than the main when it comes to question types, and people are often happy to help, but just be a bit careful in the way a question is asked as you want to demonstrate to others that you have done some research and show effort before you get stuck on the problem and ask for help
In the past, I have encountered/saw people like koolman in h bar and periodic table, ramajaran in maths chat and periodic table, Mary star in some occassions, and at least 2 other users who tend to just dump the problem and wait for the people to solve for them. Ramajaran is particularly worse (he is much better now) he used to mass ping people, Balarka knows what it is like
 
"help vampire" is unfortuntely a pretty good description of that behavior
 
@Semiclassical But I didn't always behave like a help vampire, did I?
 
6:22 PM
Not always. But it's easy to fall into it.
 
This explains why, combined with my impatience, I get quite sensitive, and espeicallty I have a real life friend I sever contact with last year because I cannot handle his increasingly help vampire behaviour despite he has very interesting and unorthodox viewpoints, which the two tings combined result in me having an unstable psyche for that year as I struggle to attend his problems and hoping that some of his unique insights I can learnt from,
(but this is getting unprofitable as his help vampire behaviour intensified since his electromagentism course starts
That he also said that the textbook is wrong as the first step of nearly 90% of his questions is what really irritates me
 
next time when I see "l hopital rule is not working" I would outright skip the question
 
People not doing their due diligence is frustrating.
And not being able to do obvious sanity checks.
 
no, L hopital rule is working perfectly fine, it's your calculations which aren't working
 
And then there are users who are doing someone else's homework. Your downvote finger will get really trigger happy when you see those questions
 
6:29 PM
yeah.
 
in The h Bar, Mar 6 at 4:32, by Secret
Actually, I don't study physics, so that's why there's a lack of effort ^^'. I am trying to help my friend who has struggled with certain questions, I should have asked him how he tried to solve the question. — D.Winstein 43 secs ago
in The h Bar, Mar 6 at 4:32, by Secret
That's even worse than getting someone to do his homework
 
.......
ow
that gives me a headache just reading it
 
Hello!! We have a triangle ABC. The centroid S divides each of the medians in the ratio 2:1.
The centroid S is on MC, such that $|CS|:|SM| = 2$, where M is the midpoint of AB.

How can we write $$|CS|:|SM| = 2$$ using the coordinates? Do we use the formula of distance |CS| ?
 
6:45 PM
@Secret judging by recent activity, either the world has a sense of humorous irony or people just don't pay attention.
I suspect it's the latter, unfortunately.
 
As always, my first approach to these people is to ask them the question: "What have you tried?" and slowly let them to deduce their way closer to the answer. But for questions I have no background of, obviously I cannot answer
Most tend to be ok and actually show effort, only very few still continue to show no effort after that question
(For the latter, I tend to use koolman as a benchmark cause he was worse enough to get himself banned in periodic table in the long past)
(and thus my (now known to be an overeacted) harsh comment to abcd is because there are occassions I thought he exceeded koolman, and then that bad memory of that formal friend of mine flood in and, well you know the rest...)
Anyway, to prevent anymore derailing, my point is, always ask them what they have tried, and slowly hint their way to the solution, if one is short on time, dig some relevant MSE and let them read about it
 
eh, we all have limits to our patience.
 
7:01 PM
Hi Balarka
 
hi astyx
 
What's up ?
 
Hi, $$\text{ Let the function series }\sum \limits_{k=0}^n \cos(kx) \\
\text{Can you extract sub-sequence which converged to a function in } C^{\infty}(\mathbb R) ?$$
 
@LeakyNun That was rude. Whenever I said that, I kjew I was wrong. I said that in a light hearted, humorous way.
 
series*
 
7:05 PM
Not much. I wasted an afternoon on writing down my statistics practical
 
Merci
 
I think he was more commenting on how frustrating it is when people post L'Hopital problems and just say "L'H isn't working haaalp"
and not commenting on you specifically.
 
@Dattier I'm tempted to say "obviously not" because $x\mapsto\cos(kx)$ does not go to $0$ ?
 
@MaryStar please read the chat of the previous hour
 
@BalarkaSen At least that's done
 
7:07 PM
I'm still trying and failing to find a variational interpretation of these Lambert-W functions.
 
@Astyx I think yes, and there are an explanation
In first a subs-sequence to a continuous function ?
 
Just give your explanation
 
it is an enigma, which I propose here
 
I gave my answer
 
(For $z>0$, I can ask for which value of $w$ the function $zw-(w-1)e^w$ has a global maximum, and that'll be $w=W_0(z)$. But for $z<0$ this ceases to be a global maximum, and the other real branch $W_{-1}(z)$ only corresponds to a local minimum in this range.)
 
7:11 PM
Je viens de voir que je me trompe, désolé, en fait mon explication ne marche pas
 
Oh you even want the function to be infinitely differentiable ?
Ah, pas de quoi
 
Tes concours se sont bien passé ?
 
C'est pas encore fini malheureusement
Enfin jusqu'ici ça va
 
Bon courage pour la suite
 
We were talking in obscure English, and then suddenly we are talking in (obscure) French
 
7:13 PM
Bye.
 
Merci, bye
As long as we remain obscure @BalarkaSen
 
@Abcd What exactly do you mean? Which part?
 
@Astyx The best communication is in not communicating
Or rather, communicating in an unintelligible way
 
th bst com is not com
 
obviously you aren't big on postmodernism
I would say that as "best communism not"
 
7:15 PM
That "in" was pretty unnessecary
$\limsup \text{com} = \overline{\text{com}}$
Nice how communicate shifted to communism
 
Tomorrow is math exam day
Woohoo
 
You're done then?
modulo tomorrow i mean
 
Oh, no, it's a Balarka'.
 
Hi, Teds
 
7:20 PM
Bonne chance, bien sûr, @Astyx.
 
Would plural of " Balarka' " be "Balarka's"?
Something to ponder about I guess
 
@BalarkaSen Then I have physics on the afternoon, one last math exam the day after tomorrow in the morning then I'm FREE
(except from that thing on tuesday, but he)
@TedShifrin Merci
 
No, Balarka', since I'm making the rules, it'll be Balarkas' ... but one of you is already too many.
 
@Astyx wee
 
However these last two math exams are probably the most important I'll be taking this year (in my life ?)
So wee
 
7:23 PM
You'll do fine, @Astyx. Are these written or oral?
 
Oral
No preparation
And I can't do anything but feel helpless since I have no idea what they'll be on
 
So don't freak out if at some point they start asking you really hard stuff. That's a sign that they have already made a decision. That happened to me a few times on my Ph.D. exams.
 
I'm scared they might be on this part of the syllabus I forgot about
Thanks for reminding this to me, I tend to forget it
 
What part of the syllabus did you forget?
 
That's the point, I forgot
 
7:27 PM
lol
 
No, I think I probably know everything that has to be known, but there's always this feeling you're missing something important
Hi @AlessandroCodenotti
 
Hi @Alessandro (jumps out of the path of a careening automobile)
 
@Astyx I have a problem for you
 
Bah, @Astyx, no point thinking like that.
 
7:28 PM
When is a measure on a set a probability measure?
 
@Alessandro It's Arturo Bendeitto Giovanni Guiseppe Pietro Archengelo Alfredo Cortaffeli da Milano!
damn I remember it
 
I mean, I could try and prove all results there are to make sure, but I doubt that would be very productive
 
When $\mu(X)=1$, @Alessandro.
 
Giuseppe Arcangelo* @Balarka :P
 
@BalarkaSen I'm all ears (eyes)
 
7:29 PM
@TedShifrin that's enough? I don't know why but I was sure that was necessary but not sufficient
 
Aren't all measures countably additive?
 
Also, I've had only analysis in all my math exams so far
 
They are on their $\sigma$ algebra
 
@Astyx Prove the Riemann Hypothesis
 
There's that naughty DogAteMy again.
 
7:31 PM
DogAteMy posed a cute little problem yesterday that I was able to solve
 
@BalarkaSen Just biject it to a skew-symmetric topological space whose elements are countable homeomorphisms.
 
I think he saw it
 
Hm, ok, so there should be a random variable inducing the measure $\mu(E)=1$ whenever $E$ is nonempty on $\Bbb R$
 
@Astyx Here, take 1 million dollars
 
About describing positive integers from $\Bbb Z$, $1$ and operations ?
 
7:32 PM
yeah
 
Oh yeah I haven't had too much time to think about it
I think I'll give it more thoughts once my exams are finished
(in 38 hours)
 
I think the right approach takes a few minutes to solve it
 
But who's counting?
 
but sure
 
I definitely am
 
7:34 PM
I'll offer you crêpes when you're done, @Astyx :)
 
why would you offer him creeps
 
Through the mail ? I'm up for it
@BalarkaSen That's a french tradition
 
you offer each other the creeps for celebration?
duuuude
 
There's a new little French place (with a French owner) a few blocks from me that does crêpes, but they're really nothing impressive. I can do better.
 
Crêpes are better when you do them yourself anyway :p
 
7:39 PM
Well, for those of us who are good cooks, of course :)
 
Crêpes aren't that hard to cook
 
But making a good filling takes more skill.
 
Oh I actually have a maths question
 
In recent years, I've done more crêpes for dessert than for savory.
 
I had to prove that for a discrete random variable $X$ we have $P(X\ge E(X)) \gt 0$
 
7:42 PM
Hmm, I'm confused, wrt to that measure the only measurable sets are $\varnothing$ and $\Bbb R$ so it should be induced by a random variable with only 2 events
@Astyx by Markov's inequality you know that it it smaller than 1 :P
 
@AlessandroCodenotti Markov is too powerful
 
Markov is a funnt inequality, a lot of times it estimates a probability from above with something bigger than 1 and I'm like "wow, thanks a lot"
 
You have to fiddle with it sometimes to get useful things.
But it's been almost 2 years since I taught that stuff, so I forget.
 
All the probability you taught, you have forgotten?
I am shocked!!
 
Not all.
You're itching to meme, aren't you?
 
7:45 PM
lol
i don't get to speak though, because i don't know probability
 
So I proceeded by the abusrd : I supposed that $P(X\ge E(X)) =0$, deduced there existed a non-negligible event $\omega$ where $X(\omega) \lt E(X)$, and wrote $$E(X) = P(\omega)X(\omega) + \sum_{\omega'\in\Omega\setminus\{\omega\}}P(\omega')X(\omega') \le P(\omega)X(\omega) + (1-P(\omega))E(X) \lt E(X)$$
And the guy said that it was useless to take a specific event $\omega$
 
How can $E(X)$ be the expected value if everything of positive probability is less than it?
Ridiculous.
everything ... not some $\omega$.
 
Yeah, but then you'd have to deal with limits
 
Huh? I just want to write down the definition of expected value.
 
That's why I did that
 
7:48 PM
You need to assume $X$ not constant at some point
 
well, if $X$ is constant, then $X=E(X)$. Done.
 
The question is, if $u_n \lt v_n$, then prove $\sum_{n\in \Bbb N}u_n \lt \sum_{n\in \Bbb N}v_n$
 
Yeah, but $\Bbb P(X>\Bbb E[X])=0$ which could be less than $\Bbb E[X]$
 
Say what?
 
(it's $\ge$)
 
7:51 PM
Oh, I can't read
I was thinking about > (since that should work as well for nonconstant variables)
 
That could be deduced from the first by taking $X|X\ne E(X)$ instead of $X$ I guess
(Which has the same expectancy)
(and is legit cause $P(X\ne E(X))\ne 0$)
 
I still say that if $P(\omega) = 0$ for all $\omega\ge E(X)$, then you should write down the definition of $E(X)$ and get $E(X) = \sum \omega P(\omega) < \sum E(X) P(\omega) = E(X)$.
 
In $\Bbb Z_p$ is there any relation between $(1-x)^{-1}$ and the partial sums of $1+x+x^2+\dotsb$
Maybe it's the average of the partial sums
Like a Cesaro thing
 
@Ted My point is that these sums are infinite, so you need to justify the strict inequality for the limits
 
But everything is absolutely summable?
 
7:57 PM
Yes
 
Interesting query, DogAteMy.
 
Anyway, not that interresting
 

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