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Claudio
Claudio
00:00
user image
basically my professor gets the transpose of my result :p This is weird since I'm pretty sure the result is correct, as the answer I've linked shows...
oh ok $A^T A$ is invertible since I've assumed $A \ne [0]$
ok I found the same solution hahahah
still I wonder how did they get the transpose of my result after computing $g$
Sine of the Time
Sine of the Time
00:19
@Claudio what do you get?
Ah, do you mean $\nabla f$ and $\nabla g$?
Claudio
Claudio
if you look at my previous messages, where the $f$ of the photo is my $g$
yeah exactly
but obviously the same solution
Thorgott
Thorgott
@ILikeMathematics the "thus" does not follow
Claudio
Claudio
@Jakobian I was thinking about what you said yesterday: are you sure that $f = \lVert Ax-b\rVert$ is not differentiable at $x^{\ast} = A^{-1}b$
Thorgott
Thorgott
you need to show additionally that the multiplicity with which each Eigenvalue appears as a zero of the minimal polynomial is smaller than the multiplicity with which it appears as a zero of the characteristic polynomial
Claudio
Claudio
i get the analogy with $x \mapsto |x|$ but maybe if one computes $$\mathbf{h} \in \mathbb{R}^n, \lim_{ \mathbf{h} \to \mathbf{0} } \frac{\lVert Ax+Ah-b \rVert -\nabla f(A^{-1} b) \dot \mathbf{h} }{ \lVert \mathbf{h}\rVert}$$
ILikeMathematics
ILikeMathematics
00:33
@Thorgott True, thanks... For the matrix splitting case, this directly follows by the Jordan canonical form.
What would you do for the non-splitting case?
Can we maybe do something with Jordan too, somehow
Claudio
Claudio
bruh
$$ \mathbf{h} \in \mathbb{R}^n, \lim_{ \mathbf{h} \to \mathbf{0} } \frac{\vert Ax+Ah-b \vert -\nabla f(A^{-1} b) \cdot \mathbf{h} }{ \lVert \mathbf{h}\rVert} $$
Thorgott
Thorgott
the Jordan canonical form is a deeper theorem than Cayley-Hamilton, I would not deduce the latter from the former
(there's a good chance that argument would be circular, too)
ILikeMathematics
ILikeMathematics
@Thorgott I've proven the former and I don't think I've usen the latter anywhere. Anyways, could we just extend our field to $\mathbb C$ without loss of generality?
Then it would directly follow. Otherwise not
Ben Steffan
Ben Steffan
The minimal polynomial may change when passing from $\mathbb{R}$ to $\mathbb{C}$
also, are you not interested in arbitrary fields
Thorgott
Thorgott
sure, you can always pass to an algebraic closure
I'm just not morally convinced by this order of operations
@BenSteffan but the characteristic polynomial does not
Ben Steffan
Ben Steffan
00:41
@Thorgott indeed, but I was under the impression that the critical point of the argument is to show that $\mu_A \mid \chi_A$
a priori this might be true over $\mathbb{C}$ but not over $\mathbb{R}$
Thorgott
Thorgott
yeah, but you can just do that argument over $\overline{K}$
ILikeMathematics
ILikeMathematics
overline K is the algebraic closure of K?
Ben Steffan
Ben Steffan
yes
Thorgott
Thorgott
so $\chi_A$ (w.r.t. $\overline{K}$) vanishes on $A$ (interpreted as matrix with entries in $\overline{K}$), but this implies $\chi_A$ (w.r.t. $K$) vanishes on $A$ since the former scenario is obtained from the latter by composing with the embedding $K\hookrightarrow\overline{K}$
Ben Steffan
Ben Steffan
ah, so you don't want to prove that $\mu_A \mid \chi_A$ over $K$ at all :)
Thorgott
Thorgott
00:45
right, though it is true
it would follow post hoc lol
ILikeMathematics
ILikeMathematics
Why not consider $\overline K$, prove $mu_A \mid chi_A$ over it, thus $\chi_A(A) = 0_{n \times n}$ over $\overline K$ and now continue with Thorgott's argument
Ben Steffan
Ben Steffan
yes that is the idea
ILikeMathematics
ILikeMathematics
Ok that sounds good
Thanks!
Thorgott
Thorgott
I don't like this argument, but yeah, this is the suggestion
the determinant trick is a crucial technique in commutative algebra, which, in its general form, also implies the Nakayama lemma among other things. I don't think it's avoidable, let alone that avoiding it is desirable
Ben Steffan
Ben Steffan
in that sense it is the most important statement in commutative algebra :^)
Claudio
Claudio
00:57
is it always true that if $f: \mathbb{R}^n \to \mathbb{R}$ is strictly increasing and $g: \mathbb{R}^{n} \to \mathbb{R}$ has a global minimum in $\mathbf{x}_0$ then given $h: \mathbb{R}^n \to \mathbb{R}, h:= f \circ g (\mathbf{x})$ we have that $$\min_{\mathbb{R}^n}f(\mathbf{x}) = \min_{ \mathbb{R}^n }g(\mathbf{x}) = f\left( \min_{ \mathbb{R}^n }g \right) $$
the middle equality is wrong, I just realized, just ignore it
Xander Henderson
Xander Henderson
@Claudio What does it mean for a function from $\mathbb{R}^n$ to $\mathbb{R}$ to be strictly increasing?
Claudio
Claudio
no ok wait, $f: \mathbb{R} \to \mathbb{R}$
in my case $f = \sqrt{\cdot}$
maybe I have to add $f$ of class $C^{0}$
@XanderHenderson thanks for pointing that out
ILikeMathematics
ILikeMathematics
@BenSteffan By the way, how would you pronounce $\overline K$ in German? 'K Linie' or 'K Strich'?
Xander Henderson
Xander Henderson
If $f$ is increasing, then $f$ is minimized by plugging in the smallest possible thing. If $g(x)$ is the smallest value that $g$ can possibly attain, then $(f\circ g)(x)$ is the smallest value the composition can attain.
Thorgott
Thorgott
@ILikeMathematics 'K quer'
'K Strich' would be $K^{\prime}$
Claudio
Claudio
01:09
@XanderHenderson now that I think about it tho, why does my solution write $\min_{R^n}f = \sqrt{\min_{R^n}g(\mathbf{x})} = \sqrt{g(\mathbf{x}_0)}$
Xander Henderson
Xander Henderson
More precisely, $f$ increasing (strictly, even) means that $x < y$ implies that $f(x) < f(y)$. If $x = \min_u g(u)$, then for any $v \in \mathbb{R}^n$, it must be the case that $g(u) \le g(v)$. But then $f(g(u)) \le f(g(v))$. Therefore $f(g(u))$ is the minimum.
ILikeMathematics
ILikeMathematics
@Thorgott Thanks
Claudio
Claudio
it should be $\min_{R} f$ and not $R^n$
Xander Henderson
Xander Henderson
No... because in that notation, $f$ is not the square root function. $f$ is the composition.
(At least, if that notation is going to make any sense).
ILikeMathematics
ILikeMathematics
@Thorgott Yeah I know, just thought one could use it for both
Thorgott
Thorgott
01:11
surely us mathematicians would never overload terminology like that :)
Xander Henderson
Xander Henderson
As I read the notation, they are saying that $f(x) = \sqrt{g(x)}$, and so $\min f(x) = \min \sqrt{g(x)} = \sqrt{\min g(x)}$.
It ain't great notation, but I suppose that is what it means...
Claudio
Claudio
wait what do you mean? in my case the original function is $\lVert A\mathbf{x}-\mathbf{b} \rVert$ so I consider its square
and then I revert it
yeah, but $f$ eats scalars so writing $\min_{R^n}f $ doesn't make sense right?
Xander Henderson
Xander Henderson
@Claudio I can only respond to what you actually write... if the notation $\min f = \sqrt{ \min g(x) }$, then the only reasonable interpretation is the one I gave.
Unless you mean $\min_{x\in \operatorname{Im} g} f(x)$.
(i.e. minimizing over the image of $g$).
leslie townes
leslie townes
where's copper, he can make a joke about "minge"
Claudio
Claudio
I'll be more explicit: $$ f = \lVert Ax-b \rVert ^2, g = \lVert Ax-b\rVert$$
then the solution writes this
user image
Xander Henderson
Xander Henderson
01:15
Yes, that is correct.
6 mins ago, by Xander Henderson
More precisely, $f$ increasing (strictly, even) means that $x < y$ implies that $f(x) < f(y)$. If $x = \min_u g(u)$, then for any $v \in \mathbb{R}^n$, it must be the case that $g(u) \le g(v)$. But then $f(g(u)) \le f(g(v))$. Therefore $f(g(u))$ is the minimum.
It is the same idea.
Claudio
Claudio
yeah I wrote down something similar as well in fact, I was just perplexed by the domain of f
maybe I shouldn't think of it as a composition
Xander Henderson
Xander Henderson
If $x_0$ minimizes $f$, then $x_0$ minimizes $\sqrt{f(\cdot)}$.
Claudio
Claudio
I was a bit confused, it all makes sense now, thanks
I would call this self-sabotage :p
well, it's 2:20 am for me as well
 
2 hours later…
nickbros123
nickbros123
03:20
Did anyone here study numerical linear algebra? How was it?
Although no one studies anything numerical cuz they like it
 
1 hour later…
leslie townes
leslie townes
04:27
nick like anything else it really depends on the instructor. something that hampers a lot of 'numerical' offerings from math departments is that an instructor cannot assume that students are familiar with any kind of programming, so there tends to be a lot of toy algorithms, made up data (or no data), modifying examples the instructor gave, and "how to talk to computers so computers will listen" in place of whatever the course is supposed to be about.
if a course has a beefy list of prerequisites that might actually mean something, maybe this is less of an issue. but it is often an issue. in designing major and minor programs, "let students opt into taking programming, or not" or "let students opt into numerical stuff, or not" doesn't mix all that well with giving meaningful work in numerical classes
and of course if you go outside the math department you quickly get into the land of "this isn't actually the singular value decomposition, but its a lot faster, and the results are visually close enough without affecting the frame rate" or even "if it runs without error messages then it's perfect"
nickbros123
nickbros123
04:45
@leslietownes yeah, this is kinda also how the "in-department" course in numerical analysis is for me at the moment. I liked it when we were discussing root finding methods, where everything was rigorous to some level. After a point it became spamming taylor expansions and abusing the $\approx$ sign. integration and differentiation was fine, Now we are discussing linear systems, and things seem to be a bit more rigorous. we'll have to see- things like operator norm identities etc are assumed though
leslie townes
leslie townes
it's generally just a tough balance to strike. a class like that can basically be a pure math class, or it can be really programming/numerics oriented, or something in between. i think the "average" class at a US university is closer to the former because of curricular obstacles preventing anything from going too deep on the programming side
which can be fun and fine, its just not always the most, like, "applied" or applicable or useful thing in the world. its just another math class
and it can be kind of annoying if the class mixes, like, okay we're just going to write "approx" without saying what we mean, and then on the other hand there's proofs as homework, and you're like, proofs of what? from what?
haha
when i was an undergrad there was a sequence of numerical analysis classes that had a deeper set of prerequisites than most undergrad math classes, and was pretty good if sparsely attended, and then there was something like "numerical methods for the physical sciences" which was a cook's tour of algorithms, with minimal programming and a lot of proofs as homework, where depending on who was teaching it, it could be a disaster.
it all comes back to the instructor, is it someone who has ever dealt with numerical methods (if not the exact content of the course) in "real life," whether in their own work or in dialogue with others, or is it just someone with a math phd and a pulse tasked with running through chapters 1-6 of the department's chosen numerical linear algebra book
and/or do they care about teaching or are they just punching a clock
nickbros123
nickbros123
05:18
I agree, depends largely on the professor
for us most of the classes are about proving convergence, order of convergence and error estimating for various things. Homework is like 2-3 proofs and the rest are algorithm implementations
Soumik Mukherjee
Soumik Mukherjee
05:31
@ThomasFinley yes
Thomas Finley
Thomas Finley
05:53
@SoumikMukherjee Thanks! I get it.
@copper.hat I see. But earlier you used to come here more frequently. Remember? We used to talk about problems and all. Ted Shifrin also joined the conversation but things went really quiet in here, after Ted stopped visiting this room. I wonder what the reason might be! But anyways, wish you all the best. You should visit here more often, hehe ... I kinda feel nostalgic fr!!! LMAO
 
2 hours later…
copper.hat
copper.hat
08:19
Hi @ThomasFinley, Ted was the life of the party
 
2 hours later…
PM 2Ring
PM 2Ring
10:08
From mathstodon.xyz/@[email protected]/…
> Turns out that LLM summaries are actually useful.

Not for *summarizing* text -- they're horrible for that. They're weighted statistical models and by their very nature they'll drop the least common or most unusual bits of things. Y'know, the parts of a message that are actually important.

No, where they're great is as a writing check. If an LLM summary of your work is accurate that indicates what you wrote doesn't really have much interesting information in it and maybe you should try harder.
> I suspect, in the most cynical way I can muster (which, tbh, is fantastically and enthusiastically cynical) this is why upper management loves them so much and thinks they're awesome for summarizing emails -- the overwhelming majority of C-suite/VP+ level communication is performative and essentially information-free.
nickbros123
nickbros123
is the following statement true?- If $A$ is a dense set in X, then for any $q>0$, the collection $\{B_{q}(x):x \in A\}$ of $q-$ radius open balls cover $X$. I think this may be wrong but im not able to find a counter example
Ben Steffan
Ben Steffan
@nickbros123 It is true :)
the proof is a good exercise in the definition of "dense"
 
2 hours later…
Xander Henderson
Xander Henderson
11:58
Not the Yankees won. Huzzah!
Jakobian
Jakobian
12:12
@Claudio its not just an analogy, its literally your function when for example $A = I, b = 0$ and $n = 1$
minimizing $x\mapsto \|Ax-b\|$ is a classic problem and its called linear least squares
its when you don't solve a linear system of equations $Ax = b$, but you give the best possible approximation to it
psie
psie
12:32
I'm looking at various definitions of operator norms. For instance, $$\sup \left\{\| Tx\|_{\beta} :x\in V,\|x\|_{\alpha} \le 1\right\}=\sup \left\{\| Tx\|_{\beta} :x\in V,\|x\|_{\alpha} = 1\right\}.$$Apparently this only holds if $V\neq\{0\}$. I'm wondering why only under this condition? Couldn't there be norms such that the supremum over the set where the domain is $V=\{0,1/2\}$ would also be empty?
By $1/2$ I meant any nonzero vector. I don't know why I wrote $1/2$. Just replace it with $v\neq0$.
Jakobian
Jakobian
@psie how can a vector space have two points
over $\mathbb{R}$ or $\mathbb{C}$ that is
psie
psie
@Jakobian well, how can it have one point?
Jakobian
Jakobian
uh... by being the trivial vector space?
But if there is a non-zero vector $x$ then there's also $\lambda x$ for $\lambda\in \mathbb{K} = \mathbb{R}$ or $\mathbb{C}$ which is already uncountably many points
and all of them are different
psie
psie
ok, I see. So we only need to exclude the trivial vector space
Jakobian
Jakobian
yes because for the trivial vector space there won't be vectors $x$ such that $\|x\|_\alpha = 1$
that's the reason
so its supremum over the empty set
but lets put this under more scrutiny...
if we are taking supremum over $[0, \infty)$ then that's $0$ in this case, the two are then equal anyway
so do we really need to exclude $V = \{0\}$? It holds anyway if the suprema are interpreted to be taken in $[0, \infty)$
psie
psie
12:48
@Jakobian if we weren't taking supremum over $[0, \infty)$, what would the alternative be?
Jakobian
Jakobian
I don't know, taking supremum over $\overline{\mathbb{R}} = [-\infty, \infty]$? Leaving supremum of empty set undefined?
those are possible choices - that lead to wrong results
but non-negative reals seems like a natural choice given that we only deal with non-negative numbers here
in this sense, the two equality holds for any choice of $V$, trivial or non-trivial
psie
psie
ok, then I think it makes sense to take the supremum over $[0,\infty)$
Ryder Rude
Ryder Rude
hi
Child's Play or Hellraiser?
Don't spoil any
i will pick one
original movies
nickbros123
nickbros123
13:19
@BenSteffan oh damn, i think i proved it, but using limit points
if $\cap_{x \in A} B_{\delta}(x)^C \neq \emptyset$ then $z$ is in this intersection, and also in $X$. since z is supposed to be a limit point of $A$, and it definitely is not a point of $A$, we must have that delta ball around $z$ contains a point of $A$, which means for this point, z is in the delta ball, contradicting the fact that z is in the complement of said ball
this theorem gets used in proving $\ell_{\infty}$ is non-separable metric space
ModularMindset
ModularMindset
14:00
Is there a known example of a stratified space that is topologically a torus but geometrically constructed from positively curved regions?
"positively curved regions" here refers to the decomposition operator isolating the highest dimensional strata each of which admit metrics of positive curvature
Ben Steffan
Ben Steffan
@nickbros123 sounds about right
ModularMindset
ModularMindset
I can put individual metrics of positive curvature on each highest dim. strata but once one fixes the geometry like this the space is not a torus anymore
Ben Steffan
Ben Steffan
I would just have observed that the condition implies that if $z$ is in no element of the cover, that means it's at least a distance of $q$ away from any point of $A$, so $B_{q / 2}(z)$ is an open ball that doesn't contain a member of $A$, a direct contradiction to the definition of dense set (your mileage may vary depending on what exact definition you use)
@nickbros123 follow up exercise: is this still true if you allow $q$ to depend on $x$? :)
nickbros123
nickbros123
14:50
@BenSteffan yeah got it, A must intersect every non empty open set is an equivalent characterisation, so it directly contradicts
@BenSteffan will try soon, it's diwali in my place so not much room for maths ;)
That reminds me fellas, happy diwali
Ben Steffan
Ben Steffan
Happy Diwali!
Jakobian
Jakobian
15:39
@nickbros123 Interestingly, its not true if we let $q$ vary as a function of $x$
For example, if you enumerate rational numbers as $q_n$ and consider the balls $B_n = B(q_n, 2^{-n})$ then $B_n$ don't cover $\mathbb{R}$ since $\mu(\bigcup_n B_n) < \infty$ where $\mu$ is the Lebesgue measure
Ben Steffan
Ben Steffan
...
nickbros123
nickbros123
@BenSteffan lol
dw I did not read that solution ;)
Ben Steffan
Ben Steffan
so much for that
ah, ok :)
Ben Steffan
Ben Steffan
16:30
why has my profile picture disappeared from my SE account? 🤔
ModularMindset
ModularMindset
Not sure - it hasn't disappeared in chat
Sine of the Time
Sine of the Time
@BenSteffan what do you mean?
Ben Steffan
Ben Steffan
when I visit my profile it shows a default profile pic
not the one I use, and which chat still shows correctly
Sine of the Time
Sine of the Time
In all your accounts, I see the same profile photo
Ben Steffan
Ben Steffan
ok good, then maybe it's just on my end, somehow
Sine of the Time
Sine of the Time
16:37
For example in my case if you go to my profile and click on "network profile" you'll see the default profile pic
ModularMindset
ModularMindset
my network profile also shows a diffferent pic I used a while back
Ben Steffan
Ben Steffan
yeah but mine is not just the network profile
it's the site profile on MSE, and every other site I have an account on
Sine of the Time
Sine of the Time
@ModularMindset I think that happens because you can choose to update you pic on a specific site (e.g. Math.SE) or on all your accounts
Ben: I tried to go to network profile and in all the communities you joined and I see the same pic
Ben Steffan
Ben Steffan
@SineoftheTime ty
Sine of the Time
Sine of the Time
np
ModularMindset
ModularMindset
16:46
Is it possible to apply a decomposition operator on the torus and break it up into a finite set of constant positive curvature smooth surfaces?
surfaces $L_i$ which when properly attached yield a topological torus with same euler char. same cohomology and same fund. group?
It seems do-able - tradeoffs are that when you fix the geometry a la piecewise metrics - on $L_i$ you lose access to a global metric
This means topological torus but geometrically distinct
Jakobian
Jakobian
@BenSteffan yes, it seems to be
ILikeMathematics
ILikeMathematics
If someone gave a presentation on the Jordan canonical form and you had to ask questions, off the top of your head, what would you ask?
Ben Steffan
Ben Steffan
@ILikeMathematics impossible to say without seeing the presentation
I guess a generic question that people may or may not ask is "why do we care" or "what can you do with it"
of course your talk may already cover this so
(in fact should)
ILikeMathematics
ILikeMathematics
@BenSteffan So you would tailor your questions towards the presentation and not just ask some general stuff
@BenSteffan True
Ben Steffan
Ben Steffan
no, it's just impossible to say without knowing what your presentation says
there may be no general stuff to ask after the presentation is over
or there may be gaps which people will ask you about
ILikeMathematics
ILikeMathematics
16:58
Alright, thanks
Ben Steffan
Ben Steffan
applications is one such general question
also whether something has analogues/can be extended to other settings, but I guess this will be less relevant in your context
unless you decide to impose restrictions on the generality you treat the subject with
say, restricting to a certain field or so
Soumik Mukherjee
Soumik Mukherjee
@SineoftheTime Come from real account, Mr. leslie
Sine of the Time
Sine of the Time
:O
SE trying to censor Ben because he's too handsome
ModularMindset
ModularMindset
I would ask: Given that any full rank matrix can be reduced to the Jordan canonical form,
Ben Steffan
Ben Steffan
aww, ty
tbh the pic is kind of not great anyways
grumpy >:(
Jakobian
Jakobian
17:06
Tell me you're European without telling me you're European:
psie
psie
Consider a linear operator $T:\mathbb R^n\to\mathbb R^n$, where $\mathbb R^n$ is equipped with the $\infty$-norm. The operator norm of $T$ is in this case $\|T\|_\infty=\sup\{\|Tx\|_\infty:x\in X,\|x\|_\infty\leq 1\}$, and I know we can take the $\sup$ also over just $\|x\|_\infty=1$. Now let $T=(T_{ij})\in GL(n,\mathbb R)$ (the general linear group). Apparently the operator norm I gave is the same as $$\|T\|_\infty=\max\limits_{1\leq i\leq n}\sum_{j=1}^n |T_{ij}|.$$
I'm trying to deduce this from the first formula I gave, but I'm getting stuck quite early. We have \begin{align*}\|T\|_{\infty}&=\max_{\|x\|_{\infty}=1}\|Tx\|_{\infty}=\max_{\|x\|_{\infty}=1}\max_{i=1,\ldots,n}\left|\sum_{j=1}^{n}T_{ij}x_j\right|\\&=\max_{i=1,\ldots,n}\max_{\|x\|_{\infty}=1}\left|\sum_{j=1}^{n}T_{ij}x_j\right|=\max_{i=1,\ldots,n}\sum_{j=1}^{n}\left|T_{ij}\right|.\end{align*}
I wonder;
(a) why can we write $\max$ instead of $\sup$?
(b) why can we swap the two $\max$ in the third equality?
ILikeMathematics
ILikeMathematics
@ModularMindset ...yes?
@BenSteffan Alright, thank you
Jakobian
Jakobian
@psie a) compactness, b) max is commutative
psie
psie
ah, ok
Obliv
Obliv
I'm working through some functional limit questions for this analysis course. How come we choose the minimum for $\delta$ like for this problem: $\lim_{x\to 2}(x^2+x-1)=5$ I got $\delta < \frac{\varepsilon}{6}$ but it should be the minimum of $1,\varepsilon/6$.
ModularMindset
ModularMindset
17:09
...is the Jordan can. form a useful tool for understanding the local linear behavior of mappings, flows, and transformations in differential geometry and topology. And how so? @ILikeMathematics
Obliv
Obliv
heres what I wrote down: we have $$\forall x\in A,\forall \varepsilon>0,\exists \delta>0,(0<|x-c|<\delta) \implies (|f(x)-L|<\varepsilon)$$ which for this particular limit is $$\forall x\in A,\forall \varepsilon>0,\exists \delta>0,(0<|x-2|<\delta) \implies (|x^2+x-6|<\varepsilon)$$ since $x^2+x-1$ can be factored into $(x+3)(x-2)$, we can rewrite the right side as $|(x+3)(x-2)|<\varepsilon$ or equivalently $|x+3||x-2|<\varepsilon$.
Ben Steffan
Ben Steffan
somehow I doubt this question is in scope for a talk about linear algebra :)
Obliv
Obliv
We want to choose $\delta$ such that this statement follows. Suppose $\delta < 1$. Then $0<|x-2|<1$ means $x\in(1,3)$ then $|x+3|\in(4,6)$ so $|x-2||x+3|\in (0,6)$ so $\delta < 1 \implies |x-2||x+3|<6\delta$ so $\delta < \epsilon/6$ means the statement follows.
ILikeMathematics
ILikeMathematics
Haha, yeah
The questions will only be on JNF as a topic in LA
psie
psie
@Jakobian max being commutative...is that something you can prove or is it a definition?
Ben Steffan
Ben Steffan
17:12
does the definition of max say anything about commutativity?
ILikeMathematics
ILikeMathematics
@BenSteffan Do you happen to have discord? If yes, I could send you the slides (there are 14)
Ben Steffan
Ben Steffan
sure
Jakobian
Jakobian
@psie you can prove that
Ben Steffan
Ben Steffan
@ILikeMathematics I think you should be able to find me just using my name (?)
I haven't used discord in so long that I've forgotten how it works
ILikeMathematics
ILikeMathematics
@BenSteffan If you click on settings on the left bottom it will say 'Username', could you tell me what that is please so I can add you?
Ben Steffan
Ben Steffan
17:18
ah
it's bensteffan
ILikeMathematics
ILikeMathematics
Alright
Jakobian
Jakobian
@psie really lets replace max by sup
$\sup_{(i, j)\in I\times J} a_{ij} = \sup_{i\in I} \sup_{j\in J} a_{ij}$ is what you can prove
psie
psie
ok
ah, yes, this I have heard of
Jakobian
Jakobian
here this can be a supremum in a ordered set not necessarily in $\mathbb{R}$
How to prove it, note that supremum is characterized by two things
psie
psie
in one max, I'm taking the max over $\|x\|_\infty=1$. What's the index set here?
Jakobian
Jakobian
17:29
supremum is characterized by two things
Its an element $s$ such that
1. $a_{ij}\leq s$ for all $i, j$
2. If $a_{ij}\leq x$ for all $i, j$ then $s\leq x$
Now to prove this formula you need to prove that $\sup_{i\in I}\sup_{j\in J} a_{ij}$ satisfies those two things
psie
psie
I need to prove both inequalities $\sup_{(i, j)\in I\times J} a_{ij} \leq \sup_{i\in I} \sup_{j\in J} a_{ij}$ and $\sup_{(i, j)\in I\times J} a_{ij} \geq \sup_{i\in I} \sup_{j\in J} a_{ij}$.
Jakobian
Jakobian
so first, take $a_{i_0j_0}$. Then $a_{i_0j_0}\leq \sup_{j\in J} a_{i_0j}$
and moreover $\sup_{j\in J} a_{i_0j} \leq \sup_i \sup_j a_{ij}$
so 1) clearly holds
@psie uh. No
did you listen to me
Now for 2)
Let $x$ be such that $a_{ij}\leq x$ for all $i, j$
then for a fixed $i$, $a_{ij}\leq x$ for all $j$, so $\sup_j a_{ij} \leq x$ for all $i$
and so $\sup_i \sup_j a_{ij} \leq x$
Thomas Finley
Thomas Finley
@copper.hat You know what happened to Ted? Is he alright? Or is he simply not visiting the chat no longer.
Sine of the Time
Sine of the Time
@ThomasFinley Ted is still active on the site, he's taking a break from chat
I think
Thomas Finley
Thomas Finley
@SineoftheTime Oh I see...
Obliv
Obliv
18:16
Guys I'm having a crisis, I don't think it's possible to prove $$\forall x\in A,\forall \varepsilon>0,\exists \delta>0,(0<|x-c|<\delta) \implies (|f(x)-L|<\varepsilon)$$
i.e., the definition of functional limit
Let $\delta = k$ for some $k \in \mathbb{R}_{>0}$ and let it be true that $$\forall x\in A,\delta=k,(0<|x-2|<\delta)$$ Then $x\in (2-\delta,2+\delta)$ meaning $|x+3| \in (5-\delta,5+\delta)$. It then follows that $0<|x-2||x+3|<k(5+k)$.
ModularMindset
ModularMindset
The twist map $\tau: X \to X$ permutes the pieces of $X$ and is defined as follows:

1. On each $k$-stratum $X_i$: The map $\tau_{X_i}: X_i \to X_{\sigma(i)}$ permutes the strata according to a permutation $\sigma$ on $\{ 0, 1, \ldots, n \}$, such that each $X_i$ maps to a new stratum $X_{\sigma(i)}$.

2. On $\Gamma$-set Components: For each $\gamma \subset \Gamma$, $\tau$ induces a twist around $\gamma$, permuting orientations or positions to match the permutation $\sigma$. This map respects the vector bundles $\mathcal V_\gamma$ along each $\gamma$ and re-aligns them with the permuted $\G…
Obliv
Obliv
input whatever $f(x)-L$ you want, but doesn't $\varepsilon$ ultimately get limited by what $\delta$ we chose? So we can't have the conclusion hold $\forall \varepsilon>0$ when we've chosen some specific $\delta\in\mathbb{R}$...
ModularMindset
ModularMindset
Does my "twist map" go by a different more common name?
And is it intuitive what I defined here?
$\gamma$ is a 1-strata by the way
Gistingly I feel that I have successfully driven a point in such a way that one simply looks at this as perhaps a permutation of strata (grouped in terms of dimension) so only same dim. strata can be pre-muted.
Damn that sentence sucks
burn it!!
pre-muted should be permuted.
Vladimir Lysikov
Vladimir Lysikov
18:34
@Obliv Order of quantifiers is important
This is a very important idea you learn on these definitions
$\forall \varepsilon > 0 \exists \delta > 0$ means that for every $\varepsilon$ there exists $\delta$ (so for every $\varepsilon$ we choose its own $\delta$, which can depend on $\varepsilon$)
If it was $\exists \delta > 0 \forall \varepsilon > 0$, then it would mean that there exists $\delta$ which works for all $\varepsilon$
@ModularMindset Maybe you can introduce an action of the symmetric group on whatever structures you have
And say that a twist map is a map from $X$ to $\sigma.X$
But I don't understand anything in your area, so maybe that's not it
 
1 hour later…
Pizza
Pizza
20:07
Hi
Sine of the Time
Sine of the Time
@Pizza Ciao :)
Pizza
Pizza
$\int_{\gamma} \frac{\cos(kz)}{(z-2)^2} dz = 0$
I had to see for which parameters $k \in \Bbb R$ , was zero, where $\gamma$ is a closed, simple and regular curve
wait
$Res = \frac{d}{dz} \left[ (z - z_0)^2 \frac{f(z)}{(z - z_0)^2} \right] \Big|_{z = z_0}$
This one here
$= \frac{d}{dz} \left[ \cos(kz) \right] \Big|_{z = 2} = -k \sin(kz) \Big|_{z = 2} = -k \sin(2k)$
So, the integral on the closed curve $\gamma$ is proportional to this residue
$-k\sin(2k) = 0$ when $k = 0$ or $k = \frac{n\pi}{2}$ with $n \in \Bbb Z$
Is that so?
Sebastiano
Sebastiano
20:25
I express my solidarity with all Spaniards and residents of other nationalities affected by the devastating flood disaster.
Jakobian
Jakobian
Oh. What happened?
I know Poland had floods recently. Seems like Bosnia and Herzegovina got too. And it seems like France got floods after that, and then Spain
and maybe Germany somewhere in between
yeah I guess the floods were pretty bad lately
in Europe in general
Sine of the Time
Sine of the Time
20:40
@Pizza do you have other conditions on $\gamma$?
Pizza
Pizza
@SineoftheTime No, it just says being $\gamma$, a regular, simple and closed curve
Sine of the Time
Sine of the Time
so if $z=2$ is not "inside" the curve, the function is analytic
Pizza
Pizza
@SineoftheTime Maybe I need to do both?
If z Is inside or outside
Sine of the Time
Sine of the Time
where did you find this exercise?
Pizza
Pizza
I took it from an exam test
and it says the same as I told you
Sine of the Time
Sine of the Time
20:50
if $z=2$ is not inside, then the function is analytic for all $k$, isn't it?
Pizza
Pizza
Yes
there is no need for any special conditions on k, because the integral will always be zero i think
Sine of the Time
Sine of the Time
yes
if $z=2$ is inside, then your method is fine
Ben Steffan
Ben Steffan
The $A_\infty$-$\infty$-operad can be described as the Dedekind cut functor $\Delta^\mathrm{op} \to \mathrm{Fin}_*$. Didn't expect to ever see a Dedekind cut again in my life but here we are.
Pizza
Pizza
@SineoftheTime ah ok, however I didn't understand very well why in the formula there are (z-z0)² in the numerator and denominator
That is, can't be simplified?
Sine of the Time
Sine of the Time
Is what you wrote the solution of the exercise?
Pizza
Pizza
21:02
@SineoftheTime It should be like this, because I checked someone's resolution and he did so
Sine of the Time
Sine of the Time
I assume here $f(z)=\cos (kz)$
So yes, you can simpify
Pizza
Pizza
yes anyway I saw that in this exam that 3 exercises came out + 2 theoretical questions, but sometimes more exercise come out
Claudio
Claudio
21:17
@Jakobian I see, thanks
 
1 hour later…
Thorgott
Thorgott
22:31
@BenSteffan lol what
Ben Steffan
Ben Steffan
exactly
leslie townes
leslie townes
ben: just wait until you see how they define the multiplication in terms of 6-8 different cases
 
1 hour later…
Ben Steffan
Ben Steffan
23:57
this reminds me that there's a seminar on Grothendieck's resume in functional analysis going here this term
something something 14 topological tensor products something
14 is a normal and healthy amount of topological tensor products to have

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