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user193319
user193319
1:50 AM
0
Q: Showing a Certain Set is an Inner-product $A$ Module

user193319 Proposition 1.1 If $E$ is a semi-inner-product $A$-module and $x,y \in E$, then $\langle y,x \rangle \langle x,y \rangle \le ||\langle x,x \rangle || \langle y,y \rangle$ I am trying to verify that $N := \{x \in E \mid \langle x,x \rangle = 0 \}$ is an inner-product $A$-module (i.e., a sub-$...

functional-analysis modules operator-algebras c-star-algebras
 
anakhro
anakhro
2:20 AM
@Ultradark $\mathbb R^3$ is not compact?
 
Ultradark
Ultradark
2:46 AM
@anakhro yeah $\Bbb R^3$ is not compact usually, but it's implied in this question that it's compact because it's inside a compact space
 
Ted Shifrin
Ted Shifrin
3:44 AM
@Ultra: That's absurd. You don't get to declare a noncompact space compact. "Usually"?
 
Ultradark
Ultradark
4:19 AM
@TedShifrin To map $\Bbb R^3$ inside $S^2$ you can do: $(x,y,z)\mapsto(x,y,z)/\sqrt{1+r^2}$ so, $\left(\dfrac x{\sqrt{1+x^2+y^2+z^2}},\dfrac y{\sqrt{1+x^2+y^2+z^2}},\dfrac z{\sqrt{1+x^2+y^2+z^2}}\right)$. Then you can deform $S^2$ to a cube because they are homeomorphic, which also deforms what's inside $S^2.$ And what I'm asking is if you can re-create this from the information of six copies of square models of $\Bbb R^2$ on the boundary of a cube
like a hologram
 
anakhro
anakhro
4:54 AM
@TedShifrin Ultradark operates from within the context of modal logic. He's 1-upped us all.
 
 
2 hours later…
user76284
user76284
7:24 AM
Does it make sense to extend the definitions of Hessenberg addition and multiplication to exponentiation? That is, can the exponentiation of two ordinals $\alpha$ and $\beta$ be defined as as the maximum order type of a total order extending $\alpha \rightarrow \beta$? How is the latter to be interpreted?
 
 
1 hour later…
Adam
Adam
8:27 AM
Has anyone seen a movie called experimenter?
 
skullpetrol
skullpetrol
not me, is it any good?
 
Adam
Adam
Well content wise yeah I mean it depends how much of a film Nazi you are but the subjects it involves is good to think about in the context of modern day, particularly the way social media works
 
skullpetrol
skullpetrol
yeah, "fake news" and all :P
 
Adam
Adam
The general gist of it is that a psychologist conducted a series of tests and concluded that no matter how cruel the expectations, or how ridiculous etc 2/3 of the population with cooperate with authorities and 1/3 will question the morality of what is happening and object
I mean I was basically asleep when it came on the other night so I wanna re-watch it but yeah it is very cool
 
skullpetrol
skullpetrol
8:44 AM
hmmm...
reminds me of the "take a knee" protest in the NFL...
...did you know that there are still 3 players left who continue to take a knee during the anthem?
 
Adam
Adam
8:58 AM
what is the reason for it? I'm not a sports enthusiast
 
skullpetrol
skullpetrol
the originator of the protest wanted to draw attention to young black men being assaulted by the police
but it got generalized to all forms of social inequality
 
Adam
Adam
yeah and that will totally make a difference lmfao no when celebrities and athletes do anything social justice warrior it is always for their career
 
skullpetrol
skullpetrol
whelp, trump ended the originators football career and threatened to have anybody who continued to be fired
(this was three years ago)
 
Adam
Adam
come on how did he do that
 
skullpetrol
skullpetrol
he tried to get the team owners and the league to mandate that all players must stand
that's when the players union stepped in and said it is unconstitutional to do so
as I mentioned, there are still 3 players left who take a knee during the anthem to this day
they have reported getting many death threats through social media for their actions
as you said, that's the way social media works
 
Adam
Adam
9:38 AM
so how did he end the career of the first footballer you mentioned?
 
skullpetrol
skullpetrol
he never got hired by any team to play
 
Adam
Adam
well that's unfortunate I'm afraid my opinion doesn't change there is only one effective form of protest in my opinion
 
skullpetrol
skullpetrol
passive resistance?
 
Adam
Adam
Thích Quảng Đức
Thích Quảng Đức (Vietnamese: [tʰǐk̟ kʷâːŋ ɗɨ̌k] (listen); 1897 – 11 June 1963; born Lâm Văn Túc) was a Vietnamese Mahayana Buddhist monk who burned himself to death at a busy Saigon road intersection on 11 June 1963. Quảng Đức was protesting the persecution of Buddhists by the South Vietnamese government led by Ngô Đình Diệm. Photographs of his self-immolation were circulated widely across the world and brought attention to the policies of the Diệm government. John F. Kennedy said in reference to a photograph of Đức on fire, "No news picture in history has generated so much emotion around the world...
 
skullpetrol
skullpetrol
yeah, that works also
the first footballer essentially set his football career on fire
team fans have said (on social media) that they will boycott the game if he plays again
 
Adam
Adam
9:57 AM
Oh there is one other thing that works, vandalism of property vital to the operations of law enforcement
but yeah unless it's a stolen APC you are probably going to die doing that and even with the APC you are looking at a long prison sentence
but that's life anyway the Experimenter is a good film historically significant indeed
 
skullpetrol
skullpetrol
yup it looks pretty good, I'll it on my "to watch" list, thanks :-)
I'll put it on
mate
 
 
4 hours later…
Alessandro Codenotti
Alessandro Codenotti
1:53 PM
I found a paper where they use $r<\infty$ to introduce positive constants that should be thought of as being "very big" and $0<r$ to introduce positive constants that should be thought of as being "very small" and that's just genius
 
Leaky Nun
Leaky Nun
2:10 PM
@AlessandroCodenotti how about $r \gg 0$ and $r \ll \infty$
 
Alessandro Codenotti
Alessandro Codenotti
But that's boring
 
Ultradark
Ultradark
hi
if $\frac{x-1}{3}=k$ and $k=3$ what is the value of $x$?
When do the isotopic subgroups of two non-isomorphic abelian varieties commute?
can it be explained in terms of $\Bbb{C}^n/\Lambda,\Bbb{R}^{2n}/\Lambda?$
 
 
1 hour later…
Semiclassical
Semiclassical
3:38 PM
@LeakyNun plus, everything finite is much less than infinite
 
Semiclassical
Semiclassical
3:49 PM
I guess my impulse would be $0<r\ll 1$
 
Balarka Sen
Balarka Sen
4:23 PM
Whew, semester's over, back home
 
ÍgjøgnumMeg
ÍgjøgnumMeg
Waddup
 
Emanuel Frátrik
Emanuel Frátrik
hi could anybody say me how to determine about convergence of integral int_0^oo x*cos(x^4) dx
 
Alessandro Codenotti
Alessandro Codenotti
Finally there's time to do math? :P @Balarka
 
Balarka Sen
Balarka Sen
Yup, guess so
 
Emanuel Frátrik
Emanuel Frátrik
pls guys help me
 
MadSpaceMemer
MadSpaceMemer
4:35 PM
Good evening,
Does $\frac{d^0.f(x)}{d.x^0} = 0$ ?
If yes can someone please explain why as well or what does this notation represent.
thanks.
 
ÍgjøgnumMeg
ÍgjøgnumMeg
$0$th derivative is just the function itself
 
MadSpaceMemer
MadSpaceMemer
So it is like multiplying with 1?
 
ÍgjøgnumMeg
ÍgjøgnumMeg
well
it's just differentiating a function 0 times
which is.. doing nothing
at least that makes sense
 
MadSpaceMemer
MadSpaceMemer
Ah I see I am trying to make a formel for a derivation n times and for the first time the term appears null but for the seond n it is the first dervitave and such on
i am not sure how such formel looks like that's why i thought if i write the fraction as above it will work out.
 
ÍgjøgnumMeg
ÍgjøgnumMeg
if you differentiate a function 0 times and get 0 then you've got $f(x) \equiv 0$ no?
 
Alessandro Codenotti
Alessandro Codenotti
4:40 PM
@BalarkaSen Teach me Cech cohomology if you don't know what to think about :P
 
Balarka Sen
Balarka Sen
Aight what do you want to know
 
MadSpaceMemer
MadSpaceMemer
@ÍgjøgnumMeg can i say that in my case is the fraction of the null dervitave = 0 then applies the formel ?
 
Balarka Sen
Balarka Sen
I plan to read Galois theory this winter, alongside other things
 
Alessandro Codenotti
Alessandro Codenotti
I guess I should start by understanding the nerve of a cover
 
ÍgjøgnumMeg
ÍgjøgnumMeg
@MadSpaceMemer erklär mal auf Deutsch was du meinst
 
Balarka Sen
Balarka Sen
4:43 PM
Oh that's a cool construction. If you have a locally finite open cover $\{U_\alpha\}$, which means every point is contained in finitely many of these open sets, you add a $k$-simplex corresponding to every $k$-fold intersection of open sets from your cover
This makes a big simplicial complex which is what the nerve (or rather, it's geometric realization) is
 
MadSpaceMemer
MadSpaceMemer
@ÍgjøgnumMeg Meine Formel scheint zu funktionieren nur unter der Vorraussetzung dass die 0te Ableitung null ergibt, ansonsten stosse ich an probleme.
@ÍgjøgnumMeg Koennte ich die annahme treffen, dass meine 0 ableitung immer null ist
 
ÍgjøgnumMeg
ÍgjøgnumMeg
@MadSpaceMemer kannst du deine Formel hier hinschreiben? :) Weil für mich hört das sich ein bisschen komisch an
 
MadSpaceMemer
MadSpaceMemer
@ÍgjøgnumMeg wie waeres wenn wir in einem privaten raum weiterchatten?
 
ÍgjøgnumMeg
ÍgjøgnumMeg
Ja können wir :P
 
Alessandro Codenotti
Alessandro Codenotti
Hmm ok I think my issue is that I don't know how to go from an abstract simplicial complex to an honest topological space
 
Balarka Sen
Balarka Sen
4:48 PM
Most of the time I never think of an abstract simplicial complex in the first place
 
Alessandro Codenotti
Alessandro Codenotti
But that's what you get out of the cover
 
Balarka Sen
Balarka Sen
Yes, but why not just think of creating a disjoint union of many many simplices corresponding to every *-fold intersections, and then pasting them togather appropriately, giving the resulting space the quotient topology?
You paste two simplices togather whenever they share a common face
 
Alessandro Codenotti
Alessandro Codenotti
That seems more reasonable
So if I cover $S^1$ with two open (slightly more than)half circles I get two one-simplices joined together in a space homeomorphic to $S^1$
 
Balarka Sen
Balarka Sen
If you want me to be super-precise, let $X$ be your abstract simplicial complex, so it's subset of some power set $P(V)$ on an abstract set $V$ of vertices, such that for any element $F \in X$, every subset of $F$ also belongs to $X$ (it's closed under inclusions)
 
Alessandro Codenotti
Alessandro Codenotti
So this construction is the same as having a vertex for every set in the cover, joining two vertices with a 1-simplex if the corresponding two sets intersect, put 2-simplices for triple intersections and so on
Ah no wait that's too simple, there could be many components in every intersections
 
Balarka Sen
Balarka Sen
4:54 PM
Totally order $V$ in a completely arbitrary way for now. Then consider the directed system: for every $n$-order subset $\Omega$ of $V$, associate an $n$-simplex $\Delta_\Omega$. Whenever there is an inclusion $\Omega \subseteq \Omega'$ you get a natural injective map $\Delta_{\Omega} \hookrightarrow \Delta_{\Omega'}$
 
Alessandro Codenotti
Alessandro Codenotti
Wait should I actually keep track of that? Or do I just get a standard 1-simplex in my $S^1$ example above?
 
Balarka Sen
Balarka Sen
This gives a directed system of topological system. The direct limit is your geometric realization. And sorry let me get back to your concrete example instead of rambling abstract nonsense
 
Alessandro Codenotti
Alessandro Codenotti
@BalarkaSen I see, thanks
 
Balarka Sen
Balarka Sen
@Alessandro In your example, you have two vertices corresponding to the two open arcs, two edges corresponding to the two intersections, and nothing else
 
Alessandro Codenotti
Alessandro Codenotti
Hmm I'm getting confused, I think there should be a single edge here
(and then wikipedia suggests that we need to look at "good covers" where the finite intersections are all contractible to avoid issues)
 
Balarka Sen
Balarka Sen
4:58 PM
Ah, yes, I was thinking of good covers in mind. I usually add edges for every connected component of intersections in the cover
 
ÍgjøgnumMeg
ÍgjøgnumMeg
Let $K$ be a field and $A\subset K$ an integral domain lying in $K$. Let $I$ be a finitely generated $A$-submodule of $K$... then is it obvious that $K \otimes_A I \cong K$?
 
Alessandro Codenotti
Alessandro Codenotti
Ah ok so apparently you should either take simplicial cohomology of the nerve of a good cover or just don't worry about it and take a direct limit over all possible covers
 
Balarka Sen
Balarka Sen
@Alessandro Yes, you're right, your geometric realization is just $[0, 1]$
Yes, if you take nerve of a good cover, it's going to be homotopy equivalent to the original space
 
Alessandro Codenotti
Alessandro Codenotti
Makes sense
So I guess that the point is that every cover can be refined to a good one
(Is that even true? It sounds true for sure)
 
Balarka Sen
Balarka Sen
Uh, what would refining a cover mean?
A larger cover containing the original one?
 
Alessandro Codenotti
Alessandro Codenotti
5:03 PM
A cover $V$ refines a cover $U$ if every set in $V$ is contained in a set in $U$
 
Balarka Sen
Balarka Sen
What about a Hawaiian earring? Any open set containing the earring point contains a lot of homotopy
Your space has to necessarily have a contractible neighborhood around every point for it to admit a good cover, right?
 
Alessandro Codenotti
Alessandro Codenotti
Hmmm looks like it
 
Balarka Sen
Balarka Sen
My locally contractible comment is not correct, because you can take cone on the Hawaiian earring. That has a good cover: the whole space itself
Yet it is not locally contractible :)
 
Alessandro Codenotti
Alessandro Codenotti
So the definition of Cech cohomology taking the direct limit over all possible covers ordered by refinement is actually more general
 
Balarka Sen
Balarka Sen
Yeah, I think so, although I must admit I have never really thought about this subtlety
They are equivalent on good enough spaces, eg for a manifold, because you can always refine to a good cover, like you said
 
Alessandro Codenotti
Alessandro Codenotti
5:08 PM
True but isn't the whole point of Cech cohomology to deal with ugly spaces :P?
 
Balarka Sen
Balarka Sen
For me it's always been a gateway drug to sheaf cohomology
 
Alessandro Codenotti
Alessandro Codenotti
That's fair
But I like ugly spaces :P
 
Balarka Sen
Balarka Sen
Me too but I don't really know much about them
@ÍgjøgnumMeg Consider the map $K \to K \otimes_A I$, $\alpha \mapsto \alpha \otimes 1$. Note that any basic element $a/b \otimes i$ of $K \otimes_A I$ can be written as $(ai)/b \otimes 1$, so every element can be written as $\alpha \otimes 1$ by multilinearity. This proves your map is surjective.
Injectivity is direct, yes?
Consider the $A$-billinear map $K \times I \to A$, $(x, j) \mapsto jx$. If $\alpha \otimes 1 = 0$, then image of $(\alpha, 1)$ must be zero by this billinear map, because it factors through the tensor product by universal property. But then $\alpha = 0$
Whoops, what do I mean by $1$ in the second component there lol
Er
For some reason I was thinking about $K \otimes_A A$
OK, fix some nonzero $i \in I$ and do the same argument with $K \to K \otimes_A I$, $\alpha \mapsto \alpha \otimes i$. Every element $a/b \otimes j$ can be written as $(ja)/(ib) \otimes i$, so it's surjective. If $\alpha \otimes i = 0$ then image of $(\alpha, i)$ by the billinear map $K \times I \to K$ I wrote down must be zero so $\alpha i = 0$ but then $\alpha = 0$
You want $I$ to be nonzero as well, right? Otherwise $K \otimes_A (0) = (0)$, not $K$
Basically when you tensor with $K$ you can invert nonzero elements - remember $K = \text{Frac} A = (A \setminus (0))^{-1} A$, so it's the same thing as localizing away from $0$. But doing that in an ideal would imply it's all of $K$
Unless there's nothing to invert, in which case it's zero and you have the zero ideal
 
nbro
nbro
5:30 PM
Hi. I will appreciate any feedback on this given answer https://ai.stackexchange.com/a/16732/2444.
 
PreSchooler
PreSchooler
If I have a linear space of functions, and I want to prove that it is complete....then I have to prove that a cauchy sequence is convergent in that space....but pointwise convergent or uniform convergent?
 
Thorgott
Thorgott
that depends on your definition of the space
 
Alessandro Codenotti
Alessandro Codenotti
@PreSchooler Your space is not just a linear space, it is also a topological space (most likely with a topology coming from a norm), that's the topology you want to check completeness in
 
Balarka Sen
Balarka Sen
@Alessandro Did you listen to the new We Lost The Sea album
 
Alessandro Codenotti
Alessandro Codenotti
WAIT IS IT OUT ALREADY?
 
Balarka Sen
Balarka Sen
5:42 PM
Yeah
 
Alessandro Codenotti
Alessandro Codenotti
Holy crap I missed it
 
Balarka Sen
Balarka Sen
I got a notification a month ago but didn't get the chance to listen to it, I'm on it now
 
Alessandro Codenotti
Alessandro Codenotti
I'm going to listen to it later tonight
Lately I've been listening a lot to Hælos, they have a KEXP session on youtube which is fantastic
 
Balarka Sen
Balarka Sen
Oh I didn't know of that band
 
Alessandro Codenotti
Alessandro Codenotti
check them out when you have time
Also Vildhjarta announced a new album out of nowhere (they are a really good prog metal band but definitely on the heavy side, i.e. rather meshuggah inspired)
 
Adam
Adam
5:46 PM
The fermi paradox isn't really a paradox
a principle of least interference applies I think, for example you don't see cows or pigs evolving at a faster rate due to our domestication, in fact we caused them to be more retarded than wildlife
 
Balarka Sen
Balarka Sen
@PreSchooler To add on to Alessandro's note, the topology of pointwise convergence doesn't come from a norm in general. Say, functions $\Bbb R \to \Bbb R$ with pointwise convergence, which is really the product topology on $\Bbb R^{\Bbb R}$, is not first countable.
So in particular it isn't a metric space!
 
user76284
user76284
Anyone familiar with this?
1
Q: Hessenberg power of ordinals

user76284According to these notes on ordinal arithmetic: The Hessenberg sum $\alpha + \beta$ is the supremum of ordinals that are isomorphic to some well-order on $\alpha \sqcup \beta = (\{0\} \times \alpha) \cup (\{1\} \times \beta)$ extending the union partial-order: $$x \leq y \Longleftrightarrow (x_...

set-theory order-theory ordinals well-orders
 
Balarka Sen
Balarka Sen
I forget the argument that it's not first countable though, so I'll rewrite it. Suppose some countably many basic neighborhoods $V_n$ forms a base around the constant zero function $(0)_{\alpha \in \Bbb R}$. Then $V_n$ is a product of intervals which are all but finitely many $(-\infty, \infty)$'s for every $n \in \Bbb N$. These exceptional intervals forms a countable family of intervals in $\Bbb R$.
 
Adam
Adam
in the same manner the wealthy provide for their children to every extent they can, but the ability of the child is completely independent of this, with the additional hinderance of the temptation to be a self entitled trust fund kid that's proud of their mediocrity, sheltered from necessary challenges to achieve their actual optimal capability
 
user76284
user76284
@AlessandroCodenotti What about $\log r \gg 0$ and $\log r \ll 0$?
 
Balarka Sen
Balarka Sen
5:53 PM
Hm, how to proceed
 
user76284
user76284
Or $\log r \approx \infty$ and $\log r \approx -\infty$ :-)
 
Alessandro Codenotti
Alessandro Codenotti
@BalarkaSen Can't you now take the union over $n$ of the coordinates in which the product defining $V_n$ is a proper interval and then produce a set not contained in any $V_n$ by a diagonal argument?
 
Balarka Sen
Balarka Sen
Ah
 
Alessandro Codenotti
Alessandro Codenotti
(Because countable union of countable sets is countable)
Ah wait no, I got confused
You already have the countable set
 
Balarka Sen
Balarka Sen
Take some index which doesn't appear as a coordinate of a proper interval in any of the $V_n$'s, right?
 
Alessandro Codenotti
Alessandro Codenotti
5:56 PM
Yes
I guess that my argument shows that not even the "countable support" product is first countable
 
Balarka Sen
Balarka Sen
Say that's $\alpha \in \Bbb R$. Then just take a product of $\Bbb R$'s indexed over $\Bbb R$ and replace the $\alpha$-th $\Bbb R$ by some proper interval
@AlessandroCodenotti Oh ok
Funnily $\Bbb R^{\Bbb R}$ is second countable. You can take $\Bbb Q^{\Bbb Q} \times \{0\}^{\Bbb R \setminus \Bbb Q}$ (horrible notation) and that's a countable dense subset, I believe.
So it's even separable
 
Alessandro Codenotti
Alessandro Codenotti
No that can't be
 
Balarka Sen
Balarka Sen
I always forget the relation between second countability and separability. Second countable means countable basis which implies separability, yeah? I can take a point from each base
That's a countable dense subset
 
Alessandro Codenotti
Alessandro Codenotti
The product is Hausdorff and regular (because $\Bbb R$ is) so if it were second countable it'd be metrizable by Urysohn's theorem
 
Balarka Sen
Balarka Sen
Ah my bad then
Separability does not imply second countability
This would be a counterexample?
 
Alessandro Codenotti
Alessandro Codenotti
6:03 PM
Oh, right, I got confused
Separable and second countable are equivalent for metric spaces, not in general
 
Balarka Sen
Balarka Sen
Nah, yeah, you were spot on
I need to brush up my point set topology!
Next sem I have point set topology, so what better opportunity than this?
I'll start by reading Steen and Seebach :P
 
Alessandro Codenotti
Alessandro Codenotti
lol
Why not Engelking?
 
Balarka Sen
Balarka Sen
Oh yeah that would be good
How thick is Engelking?
I am intimidated by massive books on dry subjects
 
Alessandro Codenotti
Alessandro Codenotti
Not sure, but at some point I will actually work through it
There's a lot of super cool stuff on dimension theory
And generally a lot of things you don't see in Munkres or any other introductory text
 
Balarka Sen
Balarka Sen
Fuck 540 pages
 
Alessandro Codenotti
Alessandro Codenotti
6:07 PM
Damn that's thick
 
Brandon
Brandon
why $R_l$ (Lower limit topology ) is not second countable . We can choose basis elements [a,b) where a,b are rationals ?
 
Balarka Sen
Balarka Sen
Speaking of point set topology!
@Brandon What about $[\sqrt{2}, 2)$?
 
Alessandro Codenotti
Alessandro Codenotti
That's another example of separable but not first countable
Also a normal space whose square isn't
 
Brandon
Brandon
But then why $R_k$ is second countable ? web.math.ku.dk/~moller/e02/3gt/opg/S34.pdf
(just read first para)
 
Balarka Sen
Balarka Sen
Union of $[a_n, 2)$ where $a_n$ is a sequence of rational numbers converging to $\sqrt{2}$ from above, say, would give you $(\sqrt{2}, 2)$ but not $[\sqrt{2}, 2)$.
What is $\Bbb R_k$?
@AlessandroCodenotti Oof
 
Alessandro Codenotti
Alessandro Codenotti
6:12 PM
 
Brandon
Brandon
YES
can any one tell me why we cannot use same reasoning to say $R_l$ and $R$ (with usual topology ) are second countable ?
Oh yeah @BalarkaSen you are right , that is counter example
@BalarkaSen .
But we can say $R$ with usual topology is second countable
 
Alessandro Codenotti
Alessandro Codenotti
That doesn't prove that the space is not second countable though, just that your suggested countable basis doesn't work
 
Brandon
Brandon
since open sets are of form $(a,b)$ we can use same reasoning
 
Balarka Sen
Balarka Sen
I am not familiar with K-topology but $\Bbb R_\ell$ is not second countable because for every $\alpha \in \Bbb R$, $[\alpha, \infty)$ is an open set containing $\alpha$. But this forces that there is a basic set in whatever basis you choose which contains $\alpha$
That's a cardinality contradiction right there
 
Brandon
Brandon
@AlessandroCodenotti right
 
Balarka Sen
Balarka Sen
6:17 PM
Because you can write $[\alpha, \infty)$ as a union of half-open intervals only if one of them contains $\alpha$
 
nbro
nbro
There's a bounty on the following question. Maybe someone of you is interested in answering it.
6
Q: What are the state-of-the-art results on the generalization ability of deep learning methods?

Shirish KulhariI've read a few classic papers on different architectures of deep CNNs used to solve varied image-related problems. I'm aware there's some paradox in how deep networks generalize well despite seemingly overfitting training data. A lot of people in the data science field that I've interacted with ...

deep-learning research reference-request generalization
 
skullpetrol
skullpetrol
Hi professor @TedShifrin
 
Ted Shifrin
Ted Shifrin
hi skull
 
Balarka Sen
Balarka Sen
Hi @Ted!
 
Ted Shifrin
Ted Shifrin
Hi a @Balarka!
 
Brandon
Brandon
6:23 PM
Hi Ted
 
Ted Shifrin
Ted Shifrin
Hi Brandon
 
Brandon
Brandon
Lol @
 
Ted Shifrin
Ted Shifrin
(exhausted from all the greetings)
 
Brandon
Brandon
lmao
 
skullpetrol
skullpetrol
how goes it?
 
Ted Shifrin
Ted Shifrin
6:24 PM
It's still early, but just came back from food shopping
How are you, skull?
 
skullpetrol
skullpetrol
fine thanks
 
johnny09
johnny09
0
Q: By restricting the domain of a non-bijective function, can we still find the inverse function?

johnny09Suppose we have a cubic polynomial function $f:\mathbb{R}\to\mathbb{R}$ given by $$y=f(x)=ax^3+bx^2+cx+d,$$ where $a,b,c,d\in\mathbb{R}$ and $a\neq 0$. As far as I know there are two ways to find the inverse function of $f$, namely the Cardano method (where we use the depressed cubic equation) an...

inverse-function
any help would be appreciated!!! :)
 
Ted Shifrin
Ted Shifrin
In general, you cannot do this explicitly. Cardano will take you through complex numbers, no matter what you do, anyhow.
 
Balarka Sen
Balarka Sen
Casus irreducibilis, right?
 
Ted Shifrin
Ted Shifrin
Yup.
@johnny09: Why do you want an explicit formula?
 
Alessandro Codenotti
Alessandro Codenotti
6:40 PM
Hi @Ted
I'm confused by forms again
 
Ted Shifrin
Ted Shifrin
Oh dear, demonic @Alessandro.
 
skullpetrol
skullpetrol
@TedShifrin did you hear that the instigator of the "take a knee movement" during the anthem is trying to get back in the league?
 
Ted Shifrin
Ted Shifrin
No, I don't exactly follow football.
 
Alessandro Codenotti
Alessandro Codenotti
Rather I'm confused by how to actually compute stuff with them
 
Ted Shifrin
Ted Shifrin
How so, @Alessandro?
Maybe you need to watch some of my lectures :P
 
skullpetrol
skullpetrol
6:45 PM
it doesn't look too hopeful that they will ever let him play again
 
Ultradark
Ultradark
I’m trying to get into the league too
 
skullpetrol
skullpetrol
srsly?
 
Alessandro Codenotti
Alessandro Codenotti
Let's say I have $i:S^2\hookrightarrow\Bbb R^3$ and $\alpha=xz\, dy\wedge dz$ a $2$-form on $\Bbb R^3$. I want to compute $\int_{S^2}i^\ast\alpha$ (full disclaimer: I made up the form and manifold, I have a similar homework with a different form on a different manifold)
 
Ultradark
Ultradark
Yeah as a punter
 
skullpetrol
skullpetrol
coolio
 
Ted Shifrin
Ted Shifrin
6:47 PM
@Alessandro: The best way (ignoring Stokes's Theorem so far) is the parametrize (all but a set of measure zero of) the sphere by spherical coordinates and pull back and integrate.
 
Alessandro Codenotti
Alessandro Codenotti
So I'm lucky here because I can look at the "spherical coordinate" map $\phi$ from $D=(0,\pi)\times(0,2\pi)$ and integrate $\phi^\ast i^\ast\alpha$ on $D$
Right
 
Ted Shifrin
Ted Shifrin
Ayup.
I even have some examples in 4D in one of my lectures (after Stokes's Theorem), in case you're intrigued.
The definition using partitions of unity is made up to have a definition, but is never used.
In general, you can use one or maybe two parametrizations and cover all but a set of measure 0.
 
johnny09
johnny09
@TedShifrin it should be possible, right? I want to understand the method of finding an inverse for cubic polynomials
 
Alessandro Codenotti
Alessandro Codenotti
So $\phi^\ast i^\ast\alpha=(i\circ\phi)^\ast\alpha=(xy\circ i\circ \phi)(i\circ\phi)^\ast(dy)\wedge(i\circ\phi)^\ast(dz)$ so I need to understand $(i\circ\phi)^\ast(dy)$ for example, which is where I start running into issues
 
Ted Shifrin
Ted Shifrin
No, it's true that the inverse function theorem guarantees you an inverse function, but almost never can you find an explicit formula for the inverse.
$\phi^*dy = d(\phi^*y)$, @Alessandro.
 
Alessandro Codenotti
Alessandro Codenotti
6:51 PM
Oh right, I forgot about that
 
Ted Shifrin
Ted Shifrin
You can get it out of the second row of the derivative matrix of $\phi$, @Alessandro. Those are equivalent.
 
Ultradark
Ultradark
I answered a question on math overflow 😅
 
skullpetrol
skullpetrol
::applause::
 
Ted Shifrin
Ted Shifrin
The question is whether it's a good answer.
 
Ultradark
Ultradark
We’ll see if it gets some upvotes
 
skullpetrol
skullpetrol
6:55 PM
link please
 
Alessandro Codenotti
Alessandro Codenotti
I'm confused by what the pullback of a zero form means now though, the $n=0$ doesn't seem to work well with my definition of pullback
 
Ted Shifrin
Ted Shifrin
It sure should. Composition of functions.
 
Alessandro Codenotti
Alessandro Codenotti
Ah ok I should just get $\phi^\ast y=(y\circ\phi)$, right? so $\phi^\ast dy=d(y\circ\phi)$
 
Ted Shifrin
Ted Shifrin
Yuppers.
Whence my comment about the derivative matrix of $\phi$ :P
 
Alessandro Codenotti
Alessandro Codenotti
Right, because if $f$ is a smooth function then $df=\sum_j\partial/\partial x^jfdx^j$
 
MadSpaceMemer
MadSpaceMemer
7:04 PM
$f^{-1}{'}(y)= \frac{1}{f^{'}(f^{-1}(y))}$ and this is expressed in the Leibniz notation as follows: $\frac{dx}{dy}|_{f(x)} = \frac{1}{\frac{dy}{dx}|_x}$

What I do not understand is the following:

Why is it $\frac{dx}{dy}|_{f(x)}$ ? I did the derivation from the start and I do not get it flipped, thus I get $\frac{dy}{dx}|_{f(x)}$

Can you explain this step to me?
 
Ted Shifrin
Ted Shifrin
Ugh @ your notation, @Alessandro. But, yes.
@MadSpaceMemer: Write $f(x)=y$ and then it matches.
 
Alessandro Codenotti
Alessandro Codenotti
or $df=\partial_jfdx^j$, as a physicist would say? What's wrong with my notation?
 
MadSpaceMemer
MadSpaceMemer
@TedShifrin that is not my question.
 
Ted Shifrin
Ted Shifrin
$\partial/\partial x^jf$ is abhorrent. Either put in parentheses around $f$ or write, as we all do, $\partial f/\partial x^j$.
 
Alessandro Codenotti
Alessandro Codenotti
Oh right, I didn't think about that
 
Ultradark
Ultradark
7:06 PM
@skullpetrol can you tile a space with another space
 
Ted Shifrin
Ted Shifrin
@MadSpaceMemer: Well, your "derivation" is obviously wrong. Can you check it with the chain rule?
 
Ultradark
Ultradark
Paros examplis
That means examples of unemployment in Catalan actually
Partos excomplos
 
Alessandro Codenotti
Alessandro Codenotti
Ok I think my confusion is cleared now, thanks @Ted!
 
MadSpaceMemer
MadSpaceMemer
@TedShifrin can you check?
 
Ted Shifrin
Ted Shifrin
OK, @Alessandro. And, quite seriously, I have lots of explicit examples in my lectures, if you need them.
 
Ultradark
Ultradark
7:14 PM
If one enacts R^2 such that copies are produced and therein lies a ruling of space, can such tiling tile a space?
 
Alessandro Codenotti
Alessandro Codenotti
The lectures on youtube you mean? I might check them out then
 
Ted Shifrin
Ted Shifrin
Yup, @Alessandro.
 
Alessandro Codenotti
Alessandro Codenotti
The course I'm doing has no practical examples basically but then all the exercises are actual computations
 
Thorgott
Thorgott
I don't think you want differentiation to be an equivalence
 
MadSpaceMemer
MadSpaceMemer
Yea right. I just wanted to show that the next step is the one that follows...….
I am a newb excuse my blunders. :D
 
Ted Shifrin
Ted Shifrin
7:15 PM
@MadSpace: So you need to write $\dfrac d{dx}(f^{-1})(f(x))$, not the way you wrote it. But you have $\dfrac{df^{-1}}{dx}(y) =\dfrac 1{df/dx (x)}$, which is correct, where $y=f(x)$.
 
Ultradark
Ultradark
Mad space is my favorite spacer
In all of outer space
 
Ted Shifrin
Ted Shifrin
So where are you getting what you wrote above, @MadSpace?
 
MadSpaceMemer
MadSpaceMemer
@TedShifrin It is my derivation.
 
Ted Shifrin
Ted Shifrin
No, I mean what you typed above with $\frac{dy}{dx}\Big|_{f(x)}$.
That doesn't appear in your derivation.
 
MadSpaceMemer
MadSpaceMemer
Aha. It is copied from a book I am using, this is supposed to be the correct answer.
 
Ted Shifrin
Ted Shifrin
7:18 PM
The other question is whether you understand what this is talking about geometrically, in terms of the graph of $y=f(x)$ and the graph of $y=f^{-1}(x)$.
Well, the book is wrong.
Let me explain the geometry to you. Do you agree that the graph of $f^{-1}$ is obtained by reflecting the graph of $f$ across the line $y=x$?
 
MadSpaceMemer
MadSpaceMemer
I do agree.
 
Ted Shifrin
Ted Shifrin
So the tangent line at $(x,y)$ on the graph of $y=f(x)$ reflects to the tangent line at the point $(y,x)$ on the graph of the inverse.
This says that the slope of $f^{-1}$ at $y$ is the reciprocal of the slope of $f$ at $x$.
 
MadSpaceMemer
MadSpaceMemer
Yes
 
Ted Shifrin
Ted Shifrin
And $y=f(x)$ or $x=f^{-1}(y)$.
That's it. If the book wrote what you put, be very careful with the book, because it may have lots of serious errors.
 
MadSpaceMemer
MadSpaceMemer
@TedShifrin Can you please explain this step to me? i find it hard to understand.
TThe first one you mentioned.
 
Thorgott
Thorgott
7:24 PM
This is but a side comment, but when you write $f^{-1}(f(x))=x$, you want to say that this equality holds for all $x$ in a certain interval; you are talking about an equality of functions, not an equality of just a particular value. This is important, because it makes no sense to take the derivative of a number $f(x)$, but only to take the derivative of a function $f$, defined in a neighborhood of $x$, at $x$.
So, in particular, what you wrote in your second line is dangerous. You don't differentiate an equation, both notationally and substantially, you differentiate two functions that agree locally. Sorry in case you are aware, but I feel this is noteworthy.
 
Ted Shifrin
Ted Shifrin
@MadSpace: There's a huge difference between $\dfrac{df}{dx}(g(x))$ and $\dfrac d{dx}(f(g(x))$. That's the whole point of the chain rule.
 
MadSpaceMemer
MadSpaceMemer
I see thorrgott
 
Ted Shifrin
Ted Shifrin
@Thorgott: That is indeed an important point.
 
MadSpaceMemer
MadSpaceMemer
Aha and the first one means that f is derived and the second one is merly a chain derivation?
 
Ted Shifrin
Ted Shifrin
The second is differentiating the composition of functions. The first is differentiating $f$ and evaluating at ...
 
MadSpaceMemer
MadSpaceMemer
7:26 PM
aha!
Yes truly i agree i have missed this point.
How do you make the last transformation
 
Ted Shifrin
Ted Shifrin
What are you referring to, precisely?
 
MadSpaceMemer
MadSpaceMemer
$\dfrac{df^{-1}}{dx}(y) =\dfrac 1{df/dx (x)}$
 
Ted Shifrin
Ted Shifrin
That's what you had. The chain rule tells you the product of the two derivatives is $dx/dx = 1$, and then you divide.
 
MadSpaceMemer
MadSpaceMemer
Ah thats what i had i thought you rewrote the left side
 
Thorgott
Thorgott
You can also compute the derivative of $f^{-1}$ directly using the difference quotient in this case. It's a different approach, but it might illuminate the situation, as it, in a way, captures both the specific application of the chain rule and the geometric argument Ted gave above.
 
Ted Shifrin
Ted Shifrin
7:38 PM
The chain rule argument we've been discussing does not give a proof that the inverse function is differentiable. But, if we assume that fact, then implicit differentiation (the chain rule) is the easiest way to proceed for most students.
 
MadSpaceMemer
MadSpaceMemer
8:35 PM
Thank you very much. I understand =)
 
Jaakko Seppälä
Jaakko Seppälä
8:51 PM
Has someone mathematician idea for this programming problem? math.stackexchange.com/questions/3341706/…
 
Shine On You Crazy Diamond
Shine On You Crazy Diamond
@JaakkoSeppälä that is interesting
Are you looking for a Python 3.x answer?
2, 1, 3
 
Jaakko Seppälä
Jaakko Seppälä
@ShineOnYouCrazyDiamond I think the language does not matter but I know Python the best. I think I understand some other languages as well but it is hard for me to develop a suitable algorithm.
 
Shine On You Crazy Diamond
Shine On You Crazy Diamond
What if a number is already in position
then do you "collect on it"
?
For example, 1,2,3
then what?
You're not being very communicative right now
@JaakkoSeppälä
So it's hard to help you
 
Jaakko Seppälä
Jaakko Seppälä
You have to do one swap, like 2, 1, 3. Then one the first round you collect 1, and on the second round 2 and 3. If you swap 3 and 1 at the beginning, you collect firs 1, then 2 and on the last round 3 so it is not a useful swap.
I'm a slow to type, sorry.
 
Shine On You Crazy Diamond
Shine On You Crazy Diamond
What if there are zero changes to 1,2,3,... ?
Shouldn't your algorithm handle that first?
 
Jaakko Seppälä
Jaakko Seppälä
9:04 PM
You have to do one swap at the beginning and the compute how many round it takes to collect all numbers.
 
Shine On You Crazy Diamond
Shine On You Crazy Diamond
It seems like bad code if it doesn't handle the default case
Okay, so you're saying that a test case will not be empty
In swaps
 
Jaakko Seppälä
Jaakko Seppälä
Yes.
 
Shine On You Crazy Diamond
Shine On You Crazy Diamond
Can you collect more than two elements at a time?
 
Jaakko Seppälä
Jaakko Seppälä
No. I want to compute only number of rounds. On every round you can collect many numbers. All you need is that those appears in ascending order.
 
Shine On You Crazy Diamond
Shine On You Crazy Diamond
K
So, measure first for one swap:
2,1,3
It's one round
now for two swaps:
I honestly don't know how to solve this one
but I think it has something to do with inversions
if $i \lt j$
 
Jaakko Seppälä
Jaakko Seppälä
9:07 PM
Me neither.
But thanks for trying.
 
Shine On You Crazy Diamond
Shine On You Crazy Diamond
It's probably inversions
since there is a formula for it
@JaakkoSeppälä some problems are notoriously difficult to solve :D
I would just tell your prof to take their own permutations :)
First make them provide you with an application
:>
 
 
2 hours later…
Rithaniel
Rithaniel
11:37 PM
I always misspell "identity" as "indentity" and I don't know why
 
Ted Shifrin
Ted Shifrin
You had a strict typing teacher who stressed the importance of proper indentation.
 
Rithaniel
Rithaniel
That must be it
 
anakhro
anakhro
Mario teaches typing, si.
 
Ted Shifrin
Ted Shifrin
11:53 PM
Mario Andretti?
 
anakhro
anakhro
No, the plumber.
But close.
 
Ted Shifrin
Ted Shifrin
Oh, right, the plumber.
 
anakhro
anakhro
@TedShifrin read anything interesting lately?
 
Ted Shifrin
Ted Shifrin
Nope.
Too busy being impeached.
 
anakhro
anakhro
Impeached from what?
 
Ted Shifrin
Ted Shifrin
11:56 PM
Watching.
 
anakhro
anakhro
Oh, that.
Do you find it interesting?
 
Sha Vuklia
Sha Vuklia
heya guys
 
Ted Shifrin
Ted Shifrin
Somewhat, yes. I followed most of the Watergate hearings back in 1974.
Heya @Sha
 
Sha Vuklia
Sha Vuklia
oh wows, also, hi Ted
 

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