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4:00 PM
Right. Thanks again :)
Hi Ted
 
Hi Astyx
 
Comment ça va ?
 
Ça va bien, merci, et toi?
 
Hi Ted
 
Ça va ça va
 
4:04 PM
Hi skull
 
Je viens de recevoir mes notes pour l'ENS
C'est l'info qui a pêché
 
Ah oui?
 
J'ai eu 8/20 coefficient 25
Avec un point de plus (en maths ou en info) j'étais classé
J'ai demandé vérification de ma note (on sait jamais)
 
Hi.
 
4:28 PM
hi @Ted
 
4:38 PM
I was arguing with someone on an answer on a Desmos graph that was wrong, but the guy wouldn't accept that the calculator was wrong
even though Desmos says it has an issue that cannot be resolved :/
I gave up
1
Q: Floor Function ?!

Viraam Question: If $x \ge 0$ and $y \ge 0$, then the area bounded by the graph of $ \lfloor x \rfloor + \lfloor y \rfloor =2 $ with the $x$ and $y$-axis is? Answer provided: $3$ units$^2$. My doubt: How do we graph this relation in order to find the area?

it's a weird question to ask about an area "bounded by the graph of" a region, not a curve
makes me think the question designer was confused
 
[Random] Possibly open question: Find an area preserving transformation such that the following definite integral can be computed
$$\int_1^{2.5}\frac{e^{x^x-x^2-{\sin^2 x}}}{\ln x}dx$$
 
do you have any reason to believe that there's a "nice" answer secret?
because I got dizzy just looking at it
 
4:54 PM
I am not sure, but since we can always deform shapes and there are subsets of deformations that can deform it in an area preserving manner, it might be possible to somehow mold it into a square
 
Who even expects Desmos to be correct for anything but elementary functions :/
 
I know how to do do that with clay, but how we can do that mathematically seemed either an interesting question, or had already been explored and I am not aware about
 
(Actually typo, lower limit is 1.01 cause it appears 1 is a vertical asymptote)
 
That's the actual graph ^
 
4:57 PM
@Secret A vertical asymptote need not be an issue
 
@GFauxPas See the comments; Xpw and you seem to have an accord now
 
@AkivaWeinberger what function am I looking at, is it a line of squares?
 
The set of solutions to the equation $\lfloor x \rfloor + \lfloor y \rfloor = 2$
 
Ah I see
 
So it's a union of squares, more or less. The boundaries of said squares are special.
 
5:01 PM
Yeah, the intersection of the boundary with the set is actually a bunch of disconnected angles
 
NB Integrals that has the following form:

$$\int \frac{e^{-stuff}}{\text{very fat denominator}}dx$$ tend to for some reason give me an impression it should look like a spike. Guess I have been staring at $\frac{\sin x}{x}$ for too long
 
It's a set that's open in the Sorgenfrey plane, too.
 
what program is that Akiva
 
Is there a way to integrate equations of this form: partial of variable / variable = constant ?
 
you want to integrate an equation?
 
5:09 PM
@GFauxPas Desmos
 
k
 
It was essentially a hack
 
5:25 PM
 
5:55 PM
@Pythonista You get a log on one side...
 
6:06 PM
@AkivaWeinberger here is an interesting thing to try and prove. If a and b are integers show that there exist unique q and r such that a = qb and r and r has less prime factors counting multiplicity then b.
it is parallel to the polynomial division theorem which uses order to sorta count factors
i think it is true
might be useful
 
@Typhon There is some error in what you wrote
 
@TobiasKildetoft how can you say I made an error?
i wrote what I wished to write. no typos.
 
@Typhon Because what you wrote is trivially not true
 
oh ok
 
and also is not grammatical
 
6:08 PM
it is grammatical.
that I do know
and it is also irrelevant.
 
"and $r$ and $r$"?
 
Hi
You mean qb+r?
What the hell ^
Tried to autocorrect "qb" into "without boundary"
 
real manifolds have boundaries
 
@TedShifrin now I am a king :)
 
6:14 PM
Is the Indian edition way cheaper?
 
The physical version, in all its flammable gory
 
@AkivaWeinberger hello!
 
Hi!
@Typhon Assuming you did mean $a=qb+r$, that seems to be false. Take $a=19$ and $b=6$. Then $19=(2)6+7$ and $19=(1)6+13$, and yet $13$ and $7$ both have fewer factors than $6$
so it's not unique
 
LOL @Kirill ... I'm just supposed to get a copy of the corrected printing today. Make sure you look at typos on my webpage.
smacks DogAteMy
hi @PVAL
 
hi @ted
 
6:24 PM
Like a forest
Beautiful, majestic, flammable
 
and even inflammable
is your point that .pdfs are less inflammable?
 
Yeah
@TedShifrin Look the thing either flams or it doesn't
 
@TedShifrin with great pleasure! I am already wondered of the great number of images (I think I have had none during my study (except the graphc theory))!
@TedShifrin and yes. Theorem 3.2 Chain rule :)
 
Pictures are helpful for most people, @Kirill.
I always laugh that Rudin's most famous analysis text has not a single picture/diagram.
 
@TedShifrin as I $\in$ people, for me too
 
6:28 PM
Yeah you really need to be ready to draw your own diagrams at all times with that thing
 
@TedShifrin I understand Rudin if I have understood it somewhere else. But not vice versa
 
Remember the theorem about how if the partial sums of $\sum a_n$ are bounded and $b_n$ decreases to zero then $\sum a_nb_n$ is convergent?
 
@MikeM: I think this statement is wrong. You can't get necessarily a global $i$-form on $X$ wedged with a global ($k-1$)-form on $Y$.
 
@AkivaWeinberger sandwitch?
 
It had only recently occurred to me to look at it geometrically
and it's a lot more obvious
@Kirill I'm not quite sure how you would do that. Remember that the $a_n$ could be negative
so $\sum a_n$ could oscillate without converging
 
6:30 PM
That's a summation by parts argument, DogAteMy.
So I'm curious to see the picture you're thinking.
 
@TedShifrin I don't think that's what it's saying, though. I think it's saying they're finite sums of things of the form $f\omega_X \wedge \eta_Y$.
 
Lots of rectangles
 
That lets you work on compactly supported forms in charts.
 
sometimes they could go the wrong way and have negative area and cancel other bits of rectangle out
I guess I'd have to draw it
 
Note the subscript $\mathcal O_{X \times Y}$ - they're talking sheafy tensor products?
I never think about the sheafy formalism but I suspect if what they wrote isn't true something's not right.
 
6:31 PM
I think the person is intending global sections by $\Omega^i_X$.
I guess we should ask for clarification.
 
Sure, but hit it with a bump function to put it in a chart.
 
@TedShifrin btw, if I may: in the Spectral Theorem you say that $T: \mathbb{R}^n \to \mathbb{R}^n$ must be a symmetric linear map. In my lecture notes I see that $T$ has to be normal. Are these conditions equivalent?
 
OK, as long as the person understands. I added to my comment.
@Kirill: Your theorem is more general than mine. I didn't treat normal operators. That normally goes in a more advanced linear algebra course.
 
@TedShifrin ok, got it
 
But there's a similar inductive proof.
The linear algebra is a bit more sophisticated, however.
 
6:34 PM
@TedShifrin ... you mentioned the day before?
 
Huh?
Oh, right.
For symmetric operators, it's clear that different eigenspaces are orthogonal. What's true for normal operators?
 
They're stable under the operator
 
@TedShifrin there is an adjoint operator $\varphi^{\ast}$ to $\varphi$ with $\varphi \circ \varphi^{\ast} = \varphi^{\ast} \circ \varphi$.
 
Who are stable?
I know the definition. But what can you say about distinct eigenspaces?
 
@TedShifrin two eigenvectors from different eigenspaces are orthogonal
 
6:39 PM
OK, so that's what allows the induction to work nicely :)
 
@TedShifrin the sum of eigenspaces is anyway direct, no normality is required
 
You still want an orthonormal (unitary) basis of eigenvectors, so it is required.
 
@TedShifrin I have read another mystic statement, for that I need time though: "if $L \subseteq V$ is the span of all eigenvectors to all eigenvalues of a normal map $\varphi: V \to V$, then $\varphi(L^{\perp}) \subseteq L^{\perp}$ and $L^{\perp}$ contains none of the eigenvectors of $\varphi$.
@TedShifrin sure, I mean the the sum of eigenspaces is direct, and I was told about it two months before I explored what the normality is
(of a map) sure
 
I don't know the point of that statement. Are they talking infinite-dimensional? I have no idea.
Also, I don't know what you meant when you said "no normality is required."
 
@TedShifrin so, $V$ stands for an euclidian or a unitary space.
 
6:46 PM
possibly infinite-dimensional?
I'm only talking about finite-dimensional spectral theorems.
In which case $L=V$. So I don't understand.
 
@TedShifrin I mean that I have read "if ... diagonalizable, then the sum of the eigenspaces is direct" a long time before "... is normal, if". No, finite-dimensional. "He" wants to show that every normal map is diagonalizable and needs this lemma for that purpose.
 
hi chat
 
@Semiclassical hi
 
You can always take a direct sum of eigenspaces, @Kirill. The point is that the operator is diagonalizable if and only if that direct sum is the entire vector space.
hi @Semiclassic
 
hello..I'm new here and I don't know what types of questions are discussed here..Can I ask a question from measure theory.
 
6:51 PM
@Kirill hmm. I'm trying to figure out which Riemann surface this is.
 
hi chat
 
I want to say $z^(1/3)$? But I'm not sure I'm counting right.
 
@mathiu_lady: You may ask anything you want. Depending on who's here at the time, you may get a helpful answer ... :)
@Semiclassic: You mean the picture on my book?
 
right.
I count three sheets, but I'm not sure I'm right about that.
 
If a function from $R$ to $R$ is Lipschitz then how can I show that it is differentiable $L^1$ almost everywhere? Here L^1 means Lebesgue measure
 
6:55 PM
That's a famous hard theorem.
I have no idea how the proof goes.
@EricSilva: Do you know?
That's used a lot in GMT.
 
I actually just lectured on this theorem on tuesday!
 
LOL, well, perfect, then. Rademacher's Theorem, right?
 
yup
 
Yeah it's a part from the Rademacher's theorem
 
introduces mathiu_lady to Eric
 
6:57 PM
the usual proof is in two steps: First to show that monotone functions are differentiable almost everywhere, and then to show that functions that are locally of bounded variation are everywhere the difference of monotone functions
and it's pretty easy to show that Lipschitz functions are locally of bounded variation
Terry Tao has a blog post about differentiation theorems that introduces the ideas behind the proof that monotone functions are diff-a.e., it's also in like lots of introductory real analysis texts that use measure theory
 
@EricSilva thanks , I'm trying
 
^The link to the Tao blog post
 
okay, plotted the Riemann surface for $z^{1/3}$ and got the same picture as the front of the book (apart from coloring etc). So $z^{1/3}$ it is :)
 
Well, of course, it's only an $\Bbb R^3$ slice of it, @Semiclassic.
 
Sure.
 
7:00 PM
I couldn't remember if I'd done $w=z^3$ or $w^2=z^3$.
 
for Rademacher one can also use the characterization of the dual of Hilbert spaces + Lebesgue differentiation theorem to give a quick proof that Lipschitz implies diff-a.e. but this is maybe hitting the problem with a bigger stick than it needs
 
@AlessandroCodenotti hello!
 
Hi @Alessandro
 
7:01 PM
ohi
 
@mathiu_lady: If you have a specific question or two, perhaps that would help?
hi Steamy
 
high @SteamyRoot
 
Looks like I'm bad at combinatorics and programming.
 
Of course, it depends on which slice you take. I can't find my original Mathematica file with which I did that. It must have been on my office computer ... long gone.
 
7:03 PM
And this is z^(1/3): i.stack.imgur.com/gYBLr.png
 
Me too, @user2219896.
 
btw my Professor told today that the notation of an adjoint and of a dual space is similar because they are connected somehow. How, in basic terms?
 
You know what dual spaces are?
 
The way I did it was to write $w^3=z=re^{i\theta}\implies w=\sqrt[3]{r}e^{i\theta/3}$, and then plot the parametric surface $(r\cos\theta,r\sin \theta,\sqrt[3]{r}\sin (\theta/3))$ for $\theta\in [0,6\pi)$
 
@Kirill if $X, Y$ are vector spaces and $f:X \to Y$ is a linear map between them then the adjoint $f^{\ast}$ is a linear map from $Y^{\ast} \to X^{\ast}$
 
7:06 PM
@TedShifrin to be honest I knew that :) I have to refresh it for the exam. But I have my lecture notes to look in.
 
so that's the im(z^(1/3)) slice.
 
@EricSilva oh I have leart it like dot product equals dot product
 
@Semiclassic: I must have done something similar/identical, but fiddled with coloring.
 
If I do the real-part slice instead, then what I get has the same orientation as the cover
 
7:08 PM
Aha.
I think we discussed this before here years ago ... But I'm too old to remember.
 
Though I think that's equivalent to rotating the im-slice by a certain angle ?
 
so yes, the set of all linear forms on $V$ is called a dual space
 
@Kirill: And if you have an inner product, you can say $V^*\cong V$ because every linear functional on $V$ is given by inner product with some vector.
 
and, there is a dual basis, where one uses Kroenecker-Delta notation
 
When you connect that with what Eric said above, you have the adjoint with the definition you're familiar with.
 
7:10 PM
@TedShifrin whaaaaaaat
 
As for the diagram color, Mathematica changed the default coloring a while back. (was running into that earlier)
 
I was trying to be consistent with Kirill's language, Alessandro, but I messed up.
 
@EricSilva , again I have not studied about Hilbert space yet
 
@TedShifrin I have to meditate about that, it is too fast
 
@Semiclassic: I know our graduate secretary wanted to use a similar image for the cover of the graduate handbook, and so I had to recreate that and do a bunch of fiddling with lighting/colors.
 
7:11 PM
lol
 
@TedShifrin nono I didn't mean that's wrong, I was surprised by this fact
 
What? @Alessandro ... You didn't know this?
 
Wait what assumptions are there on $V$?
 
Inner product space.
 
@mathiu_lady that's why i said it may be a bit too much for the problem at hand, the outline of the proof in Tao's notes don't use Hilbert space stuff
 
7:12 PM
Algebraic dual, not analytic dual.
 
@TedShifrin will be happy to get a full portion of mathematical terms. I am used to learn a lot of things :)
 
Anyhow, @Kirill: I hope you find my book a useful reference for you.
I'll be back later. Ta ta ...
 
@TedShifrin yes, I will read it first
 
@TedShifrin we did very little on dual spaces in the linear algebra course, I guess functional analysis next year will have much more stuff about them
 
later @ted
 
7:14 PM
@Alessandro you'll definitely see that in a functional analysis class
 
@TedShifrin anyway, I am somehow happy to get it
 
@TedShifrin don't we have $V^*\cong V$ iff $V$ is finite dimensional when $V^*$ is the algebraic dual?
 
depends what you mean by $\cong$
 
isomorphism?
"as big as"?
 
Let $X,Y$ be random vectors such that $X, AX+Y$ are independent and normal with mean zero and covariances $\Sigma_1, \Sigma_2$. Can I conclude that $(Y|X=0) \sim \mathscr N (0, \Sigma_2)$?
 
7:17 PM
@Waiting Hey there, I was not around. I guess my avatar was napping ;-)
 
oh i guess Alessandro said algebraic
I missed that
 
@Emolga in random space?
 
then his statement is correct, the dual in the infinite case is always "bigger"
 
@Kirill, Sorry, I do not understand the question...
 
@EricSilva Is it true that the boundary of an open ball (e.g. $B(0,1)$) is separable i.e., it has countable dense subset?
 
7:19 PM
@mathiu_lady Yes.
Separability is inherited by all subsets of metric spaces.
 
I'm confused, me and Ted are not thinking about the same thing probably. That's what happens when I jump into a conversation halfway through
 
@Emolga vectors are the elements of the vectors space. Even though I do not understand your question, in what space are you operating?
 
@mathiu_lady what space is it the open ball of
@Alessandro, one point is that I've never in my life seen someone care about the algebraic dual of an infinite dim space
 
@Emolga as much as what is $A$.
 
you usually care about the continuous dual with respect to some topology on the space
 
7:21 PM
What's an explicit countable dense subset?
 
$X$ is a random variable, whose values are vectors of real numbers. The same is true for $Y$. I guess if it's true for scalar random variables I will be able to adapt the proof to this case that I have, so it can be ignored. $A$ is just some constant matrix.
 
@EricSilva probably because it's a mess :P
 
@EricSilva suppose in $R^n$
 
then yes, it is true there @mathiu_lady
@Alessandro ur damn right it is
 
which is probably the same reason nobody cares about a basis of $L^2$ even though AC guarantees there is one
 
7:25 PM
@AlessandroCodenotti More like because any Hamel basis of $L^2$ is uncountable.
 
@Kirill Ok, I proved it.
(The distribution of $X$ does not play a role, only the assumption of independency)
 
Hey there!
 
@Alessandro we do care about Schauder bases of $L^{2}$ like a lot
 
yeah I meant Hamel basis
 
7:32 PM
but these aren't "algebraic bases" i guess
yeah
@Daminark that was a meaty lecture today
 
@robjohn hehe, OK. I was working on some integral with Lerch transcendent. :-)
 
oh, nice
Lately I've been running into the weirdness of confluent hypergeometric functions
 
@Emolga oh nice to hear that! I am ex-musician and mathematician in the first year, so am not eligable to talk about covariances at the moment :)
@Emolga but stiil, the space we are in is important
 
Any chance someone here familiar with computability theory?
In particular im trying to prove that $L = \{<M_1, \dots , M_n> : \cap M_i$ does not contains words of length smaller then $n \} \in CoRE - R$.
 
@Kirill Cool! I think the probability space is implied by the use of normal random variables, as they carry with them the assumption of working with an appropriate power of $\Bbb R$.
 
7:43 PM
@Emolga probability spaces? I think I have to deal with them in the next year
oh sure, measures are going to come in October.
we have had Moore-Penrose inverse today. Looking at the Wiki-page in English I meet $M(m,n,K)$ notation. What is that? A $m \times n$ matrix with entires in $K$?
or the set of all that matrices? Or something else?
ok, I´ve found the explanation
 
@EricSilva yeah, it was real good
 
7:59 PM
i just like the functional stuff lurking in there
 
@Waiting I've referred to that in an answer or two, but it gets me feeling queasy.
 
@robjohn I have one already added to my project. It's so enjoyable! However, the difficulty level is a bit crazy.
:-)
@robjohn These days I'm involved in research in more directions, I investigate some problems that apparently are new in the mathematical literature (that kind of problems I use to play with). Some are already included there.
 
@Waiting great! Is your book getting printed soon?
 
8:14 PM
@robjohn I'm very fascinated these days by a class of alternating series, it's actually about a sum of series which when combined together produce very impressive closed-forms.
@robjohn Yes.
@robjohn I'm NOT able yet to treat those series by elementary means, but I suppose that some fascinating cancellations happen.
At the moment I cannot even imagine how that happens, perhaps I will need to ponder over those problems for some weeks, months, or even more.
@robjohn In general we are used to the series solutions since they are far more elegant than calculating complicated integrals. I don't see why here things won't be the same.
 
8:34 PM
$$\int_0^1 \int_0^1 \frac{(x y)^{u-1}}{1-x y z} (-\log(x y))^s \textrm{d}x \textrm{d} y$$
This is a pretty cool double integral, known in the mathematical literature, of course.
@robjohn I didn't post (in chat) problems for years ... :-)
Wait, I think I posted some time ago an elementary integral with logarithm, maybe some weeks ago, but that one was more for fun, not really a serious problem.
Let's make it harder
$$\int_0^1 \int_0^1 \frac{(x y)^{u-1}}{1-x y z} (-\log(x y))^s\log(1-xy) \textrm{d}x \textrm{d} y$$
or
$$\int_0^1 \int_0^1 \frac{(x y)^{u-1}}{1-x y z} (-\log(x y))^s\log(1+xy) \textrm{d}x \textrm{d} y$$
 
9:10 PM
I took this tool and put non-Latin text into it and I think it broke
 
Hey @Eric, sorry I was out for a while, and yeah functional is some good stuff
I probably should've done what David did in my lecture and just black box some of the results, he actually got through most of what he wanted mainly because of that
(His lecture notes are like, 8 pages long)
 
9:31 PM
@Waiting Gotta go for a while. See you later.
 
@robjohn Sure. Later.
I think I also leave to take some sleep.
 
9:46 PM
What is analytic Number Theory ?
 
You use integrals and infinite sums to study primes somehow
 
@Akiva can you elaborate please :>)
 
I don't think I know enough about it to speak
I know this thing is used
 
You've probably heard of the Riemann zeta function, right?
 
Just curious, for any given polynomial function $f:\Bbb N\to\Bbb Q$, is the function $g:\Bbb N\to\Bbb Q$ defined by $g(n)=\sum\limits_{k=1}^nf(k)$ also a polynomial function?
 

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