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00:09
@user2236 Also, the vague direction of WB's anti-infinity stance has been covered quite meticulously by logicians who have charted the power of all the different axioms and logical systems, so that anybody can make choices about which ones are allowable, which then implies which subset of mathematics is then derivable. This is the point of Reverse Mathematics
I don't know if (but I seriously doubt that) WB has made a coherent choice there.
can we make "converge uniformly on compact subsets" a new norm?
I suppose not
but can we make this into a topology?
Did someone say topology?
$$\langle \chi_V, \chi_W \rangle = \dim \operatorname{Hom}_G(V,W)$$
lol hi @CaptainAmerica16
@CaptainAmerica16 how's it going
00:17
@LeakyNun Pretty well. Been livin' it up for the past few hours. I have cheesecake and ice cream.
Zee
Zee
@LeakyNun what r u looking to do ? Make a new topology such?
Is the proof given in the answer correct?
0
Q: Prove that $A=\{x\in X:$ there is a simple chain consisting of members of $\mathscr O$ that connects $a$ and $x\}$ is open

Math geekLet $(X,\mathscr T)$ be a topological space and let $a,b \in X$. A simple chain connecting $a$ and $b$ is a finite set $U_1,U_2,...,U_n$ of open sets such that $a\in U_1\setminus U_2$, $b\in U_n\setminus U_{n-1}$, and each $i,j=1,2,...,n, U_i\cap U_j \neq \emptyset$ iff $|i-j|\leq 1.$ Let $\m...

ah, more junk food, @CaptainAmerica.
I guess on your birthday I won't complain.
Can we give proof without cases? please help me to complete the proof.
why do we require cases here?
00:37
@TedShifrin I had a semi-healthy dinner. Now I'm about to go get some sweet and spicy beef jerky :D
I don't see why cases are needed, @Mathgeek, but I'm not thinking very hard.
@CaptainAmerica: That after ice cream and cheesecake?
@TedShifrin I can hold a lot of food.
That does sound like a weird combo tho.
LOL, and without getting sick, apparently ... although once you did get sick, as I recall.
@TedShifrin does there exist a proof cover all cases in single step?
00:39
It seems to me that showing $U_n\subset A$ does it, regardless of the value of $n$, @Mathgeek, but I haven't thought about it too hard.
hi @Semiclassic
so, some news
@TedShifrin Lol, yeah. I learned my limits. It ends after the jerky - hopefully.
@CaptainAmerica: You're still a kidlet :D
Don't ruin this for me Mr. Ted >:|
00:41
I do my best. LOL.
@TedShifrin you mean this proof is okay.right?
I don't know what you mean by "this proof."
solution given by quasi.
It's too complicated, for my taste.
Brb every one. Not that it's really necessary to say.
00:42
Just work it out yourself, @Mathgeek.
i've been working with another prof on some stuff outside of my thesis work; it's related to stuff I've rambled on about here. that's heading to a paper eventually, but i knew that already
Yes, @Semiclassic?
what i found out about today is that it's like that said prof is going to get invited to present it at a conference next year
and he wants me to come along if possible.
well, that's very cool ...
00:44
so you haven't totally given up on academia yet :P
especially since the conference is three days in Italy :)
is it accepted for publication?
not yet. it's not submitted yet
oh ...
even that's a ways off
the conference is in June
So, not for a while
00:45
I guess it's a question of whether said prof has extra grant money to pay for your travel
oh, in that case, no problem
where in Italy?
Viterbo. Not far from Rome I think?
there's also a number of big names in foundations of physics, so that's really cool
looks like halfway from Rome to Perugia
i have no sense of scale when it comes to Europe
00:47
relatively speaking, it's pretty close ... like 55 miles or somethin'.
makes sense
Looks like a torturous train ride with zillions of stops.
I wonder if you'll get to give half the talk :P
lol
probably not, but he does want me around for the Q&A
00:51
That's truly awesome, Semiclassic. Congrats.
yeah, it's pretty great
and it's far enough off that I can make plans around it
@TedShifrin I am not getting a proof without cases.:(
are you kidding me
that seems pretty lazy
it might be an accurate summary, but still
@Mathgeek: If $x\in U_n$ is connected to $a$ by a chain, why does it not follow immediately that every $y\in U_n$ is likewise connected to $a$ by a chain?
00:53
obscure numberings
I refuse to read crap like that, Leaky.
People like Federer's GMT book who refer to everything with sequences of 12 numbers should be shot.
Luckily, I have to go eat a quick dinner and leave for bridge. So ... farewell, y'all.
Ted are you willing to say that infront of a judge ? am building a case
2
damn it i just missed Ted ._.
00:56
"By 2.3.1, 2.3.2, 4.6.7(ii), we satisfy the conditions of 5.6.1, so the result follows"
:47417556I also think the same. quasi was saying that what if $y\in U_n \cap U_{n-1}$? then chain terminate at $U_{n-1}$
in the first comment, I write the same. but he told me that it is wrong @TedShifrin
@Leaky which book is that?
@Daminark I would rather not say
:shrug:
01:36
I would love for some help on this problem! math.stackexchange.com/questions/2978523/…
 
1 hour later…
02:44
Is there an identity that can simplify the computation of the Frobenius norm of the logarithm of a matrix $A$?
 
1 hour later…
03:47
Graph of a 4x4 Sudoku. (The edges are the constraints.)
 
2 hours later…
05:48
Good morning sirs and madams
And entities/non-entities with other (possibly collections of) pronouns
Happy Halloween! #math #mathchat
Oh my
A ghost
It's a c=-26 fermionic ghost CFT
Zee
Zee
06:06
Does anyone feel topology and geometry are much more fuzzy compared to analysis and algebra ? I like geometry btw
 
2 hours later…
07:36
I have a question about Euler Angles and Gimbal Lock
If you’re simply applying rotations one at a time to a vector in space, would you not eat gimbal lock? Is it only when you try to preserve the overall set of rotation matrices that gimbal lock occurs?
07:52
lol it's hilarious when see comment like this: math.stackexchange.com/q/215145/390226
 
3 hours later…
11:12
hi, sorry for using the wrong terminology, i don't know exactly how to correctly formulate the question: given a space in R with n dimensions, can someone explain the process needed to project one of the existing axis (A) on a new axis(B) orthogonal to another existing(C) ? i am referring to a sentence here(2/3 of the article) : https://blog.insightdatascience.com/generating-custom-photo-realistic-faces-using-ai-d170b1b59255

or: can anyone help me formulate this question correctly in order to post it on math.se ?
 
1 hour later…
12:19
Hey
My question has been obviously closed so I'll ask directly to you guys !
Here it is : https://math.stackexchange.com/questions/2968819/project-ideas-on-chaos-theory-cellular-automata-fractals-games-ia
TL;DR I'm looking for a project in computer science but mostly in math domain like fractals (terrain generation), chaos, chaos game, IA etc..
0
Q: reflection and bilinear forms

ninja hatori$r_{N}$ : N $\perp$ $N^{\perp}$ $\to$ N $\perp$ $N^{\perp}$ s.t $(x,y)$ $\to$ $(x,-y)$ preserves inner product of bilinear forms. I know that $r_{N}$ . $r_{N}$= id

12:38
@Ninjahatori might want to mention that you've since deleted the question
(though i guess that's superfluous now, since I just did)
12:55
Is there anyone familiar with QR-factorization? I want to know some overview about how they're useful in reducing the computation?
Is the following true (I think I might be mis-remembering something, and want to be certain) $f(z_0)=\frac{1}{2\pi i}\int_C\frac{f^n(z)}{(z-z_0)^{n+1}}dz$
@Rithaniel Are you thinking of Cauchy's integral formula?
That's the one I'm trying to remember, yes.
typically that'd be $\displaystyle f^{(n)}(z_0)=\frac{1}{2\pi i} \int_C \frac{f(z)}{(z-z_0)^{n+1}}\,dz$
Okay, I had the derivative in the wrong place.
Danke schon.
13:04
np
just take $f^{n}(z)$ as the nth antiderivative, lol
(don't do that, it's more confusing than helpful)
At that point would we also want $(z-z_0)^{1-n}$ below?
In the version I wrote, yes. In the version you wrote, no
the point is that you'd take the function in the numerator of your integrand, differentiate it $n$ times, and evaluate at $z=z_0$
so you'd have $f^{n}(z)\mapsto f(z_0)$ if you were to interpret your original formula in that way
by contrast, the usual expression is $f(z)\mapsto f^{(n)}(z_0)$
all of this assumes that $f(z)$ is analytic on the unit circle, of course.
Naturally. (Also, to check my intuition, given $f(z)=\frac{z^2}{z-3}$, it follows that $\int_C f(z)dz=3^2 2\pi i$, correct?)
looks right. As a check, we have $f(z)=\frac{(z-3)^2+6(z-3)+3^2}{z-3}=(z-3)+6+\frac{3^2}{z-3}$
the first two integrate to zero, so you're left with the pole at $z=3\implies \int_C f(z)\,dz = 2\pi i\cdot 3^2$
13:22
Well, then I have an issue, because my homework states that this should equal zero. :/
huh
oh, derp
Where's the pole?
unit circle around 0
that's the contour
where's the pole in your $f(z)$?
In that case, I'm not actually given a pole, I think.
uh
then how are you deducing that $\int_C \frac{dz}{z-3}=2\pi i$?
13:27
(I think this is an issue with my lack of volcabulary. Would the pole just be the origin?)
No.
I mean, why do we expect $\int_C \frac{dz}{z-z_0}\neq 0$?
Because $f'(z)\neq 0$?
then why would $\int_C z\,dz=0$?
Also note that the Cauchy integral formula implies that $\displaystyle \int_C \frac{dz}{(z-z_0)^{n+1}}=0$ for $n>0$
So what's special about $n=0$?
Okay, so the issue with $\frac{z^2}{z-3}$ is that we're trying to evaluate it at $3$ instead of $0$?
13:35
Darn
This is basic complex analysis stuff, so if you're not familiar with it you should review it
You mean the fact that the integral at $n=0$ is $2\pi i$?
Eh. even that's too strong.
But, what in the integrand $\frac{1}{z-z_0}$ is giving the possibility of the integral being nonzero?
Maybe there's a vocabulary/translation issue, but the terminology of poles etc. is elementary in complex analysis
I've googled the term and have been doing some reading. I think we were calling them "singular points?" We've not used the term "poles" at all, though.
ah, okay
the usual terminology is that an nth-order pole of $f(z)$ is an nth-order zero of $1/f(z)$
this differs from a more generic singular point, though, which could instead be removable or essential
e.g. the point $z=0$ is a removable singularity of $\frac{\sin z}{z}$ and an essential singularity of $e^{1/z}$
13:45
Okay, so zeroes and poles are types of singular points?
poles are. zeros aren't, since there's nothing singular about $f(z)\to 0$ as $z\to z_0$
Right, because a zero is not necessarily a discontinuity.
It typically won't be in complex analysis, though cases like $f(z)=\sqrt{z}$ are a bit weird
Because half of the closed contour centered on 0 is equal to 0?
Or, rather, no, is that even true?
well, the point is that if you take $z=e^{i\theta}$ and consider $\sqrt{z}$ from $\theta=0$ to $2\pi$, then $\sqrt{z}$ goes from $1$ to $-1$
so $\sqrt{z}$ doesn't return to itself if you analytically continue $z$ around the origin
to get back to $1$, you need $\theta$ to run all the way to $4\pi$
13:51
Okay, so my intuition was along the correct line.
by contrast, that won't happen if $z$ is an integer, positive or negative
a true statement would be something like this: If you integrate $\frac{1}{\sqrt{z}}=2\frac{d}{dz}\sqrt{z}$ with $z=e^{i\theta}$ from the $\theta=0$ to $4\pi$, then thte result is zero
basically because the antiderivative $\sqrt{z}$ is then identical at the endpoints. but that requires some care to understand correctly and you'll hopefully talk about this
anyways. the simpler classification is that if $f(z)=a(z-z_0)^n+g(z)$ with $g(z)$ analytic and integer $n$, then there are two possibilities.
1) $n\geq 0\implies f(z)$ s nonsingular at $z=z_0$; if $n>0$, it's a zero of order $n$
2) $n<0\implies f(z)$ is singular at $z=z_0$ and is a pole of order $-n$.
In the case of n=-1, you call it a simple pole or just a pole
In that case, the integrand in $\int_C \frac{dz}{(z-z_0)^{n+1}$ will have a pole of order $n+1$ at $z=z_0$
With that in mind, there's a very specific criterion for whether $\int_C\frac{dz}{(z-z_0)^{n+1}\neq 0$
one condition is that $n=0$, as you noted
but there's another condition which you have not indicated you know
We said earlier that $f(z)$ "of course" must be analytic inside the unit circle.
Why does that matter? For instance, what can we conclude about $\int_C f(z)\,dz$ in that case?
(Had to move to a different location. Reading)
Are we looking for the fact that there exists an antiderivative of $f(z)$?
14:08
eh, sorta
I mean, 1/z has ln(z) as antiderivative for instance
but what's weird about that case?
the last question I asked is the more pertinent tho
If $f(z)$ is analytic, what can we say about $\int_C f(z)\,dz$?
That it equals zero? Is that what we want?
that's a place to start.
Now, why doesn't this work if we do $\int_C dz/z$?
Because it's not analytic on all of $\mathbb{C}$?
That's a sufficient statement but not a necessary one.
I mean, $f(z)$ is analytic on most of CC
How much of the complex plane do we actually care about when doing a closed contour integral?
In any case, you should have an actual theorem for this
i.e. a theorem which tells you a sufficient condition for $\int_C f(z)\,dz=0$
Ah, that the value of $f(z)$ at a particular $z$ is the same regardless of direction of approach? Which is an issue for $ln(z)$ due to a value at the branch cut differing depending on direction of approach.
14:18
right. of course, you won't want to approach $z=0$ in any case
I feel like you're looking for something specific, and I'm just listing everything I know.
i'm looking for a theorem, yes
and if you've picked up any theorems from complex analysis, this should be among them
I guess here's another way to see the issue, though. Consider the following two examples
first, take $f(z)=\frac{z-1/2}$. What are the first few terms in the Laurent series at $z=0$?
Actually, we can handle both cases at once if we go to $f(z)=1/(z-z_0)$. So start from that instead
(Had to google what a Laurent series was)
So, Laurent series goes from negative infinity to infinity? The is no "first term," then?
In principle, yes. For instance, the Laurent series of $e^z+e^{1/z}$ about $z=0$ won't have a first term
But in most cases of interest you will have a first term.
Can you suggest some undergraduate project topics?
14:29
(I'm still unclear why my calculation of $\int_C\frac{z^2}{z-3}dz$ was incorrect. Because, as best I can tell, I applied Cauchy-Goursat correctly, but, I obviously got something wrong.)
Well, what's your statement of Cauchy-Goursat?
14:43
See, I think the issue is that I don't think in terms of theorems. I think in terms of formula. I took note of the Cauchy-Goursat formula, but not the theorem.
Then you're going to be entirely confused, because the formula is itself a theorem.
It's got premises, therefore it's a theorem. If you don't know the conditions under which the formula applies, then you don't know the formula.
(Confusingly, though, there is a difference between the Cauchy integral theorem and the Cauchy integral formula.)
Okay, yeah, I was about to post the integral formula.
Well, start with what you know, but include the premises
Given a simple, closed contour centered at a point $z_0$, and a function which is analytic on that curve, $f^{(n)}(z)=\frac{1}{2\pi i}\int_C\frac{f(z)}{(z-z_0)^{n+1}}dz$
There's one issue with that. Analytic on that curve?
14:50
inside that curve.
Right
Also, it's $f^{(n)}(z_0)$ on the LHS
Right
As a special case, we have $f^{(n)}(0)=\frac{1}{2\pi i}\int_C f(z)z^{-n-1}\,dz$
@LeakyNun: Oh! Thanks! I should've known that it was due to the extreme value theorem.
Okay, and the issue with $\frac{z^2}{z-3}$ was that my initial calculation was using a $z_0$ of $3$, when we're actually meant to be evaluating it at $z_0=0$, or is there some other miscalculation on my part?
14:53
Yeah, that'll work to explain your issue.
The more generic point, though, is that $\int_C f(z)\,dz=0$ if $f(z)$ is analytic inside the unit circle
$f(z)$ needn't be analytic everywhere; it only need be analytic inside the curve
With that in mind, where does $f(z)=\frac{z^2}{z-3}$ fail to be analytic?
Okay, then I do understand where I went wrong. That is a relief!
I guess I should go with an argument from the theorem, instead of showing it via arithmetic, though.
(It fails where $z=3$)
Right. Which lies outside the unit circle
So $f(z)=z^2/(z-3)$, while not analytic everywhere, is analytic inside the unit circle
and that's all you need to conclude that $\int_C f(z)\,dz=0$
Alright, thank you for your patience with helping me out.
the point $z=3$ is said to be a (first-order) pole of the integrand, and you're only interested in (first-order) poles which lie inside the unit circle
Biggest lesson I've learned today is that I need to pay more attention to theorems when they're stated.
14:58
Yep.
Perko pair
(I don't know if they're actually equivalent, or just each equivalent to the other's mirror image)
15:31
I have a friend who is a designer who is looking for some mathematical inspiration for designs.
Any I deas?
The Newton fractal is a boundary set in the complex plane which is characterized by Newton's method applied to a fixed polynomial p ( Z ) ∈ C [ Z ] {\displaystyle p(Z)\in \mathbb {C} [Z]} or transcendental function. It is the Julia set of the meromorphic function z ↦ z − p ( z ) ...
@PiyushDivyanakar Here are two^
This also looks neat:
You can also go through this, some of these are very pretty:
1081
Q: Visually stunning math concepts which are easy to explain

RBSSince I'm not that good at (as I like to call it) 'die-hard-mathematics', I've always liked concepts like the golden ratio or the dragon curve, which are easy to understand and explain, but are mathematically beautiful at the same time. Do you know of any other concepts like these?

I'm not quite sure what this is:
15:47
There's a nice picture here
11
Q: How to find a conformal map of the unit disk on a given simply-connected domain

Taras BanakhBy the classical Riemann Theorem, each bounded simply-connected domain in the complex plane is the image of the unit disk under a conformal transformation, which can be illustrated drawing images of circles and radii around the center of the disk, like on this image taken from this site: I am ...

Hyperbolic tilings are always nice
(Found on this Wikipedia page)
@MikeMiller Are such mappings unique in some way?
(Up to Möbius stuff on the left I guess)
Here's a limit I should know how to take (and don't): $\lim_{x\to \infty}\frac{g(x+\epsilon)-g(x)}{g'(x)}$
@AkivaWeinberger yes, and by the Mobius comment it seems you even see the proof
Given two you can do one and then the inverse of the other to get a biholomorphism of the disc
(I'm pretty sure I should be stating some conditions here but I don't quite know what they should be)
On the subject of tilings, there are aperiodic tilings such as the Penrose tiling and Wang tiles (the latter of which has applications in procedural generation) @PiyushDivyanakar
@Semiclassical What is it if $g(x)=e^x$?
(Or $e^{cx}$)
If it's $e^x$ it's $e^\varepsilon-1$, yeah?
15:56
Hmm, yeah
If it's $e^{cx}$ then it's $\frac1c(e^{c\varepsilon}-1)$, if I didn't mess up
If $g(x)=\log x$, it's the limit of $x\log(1+\epsilon/x)\to \epsilon$ as $x\to\infty$
@Semiclassical Which is $\varepsilon$, yeah?
yeah, I've been using \epsilon not \varepsilon
What if $g$ is $x^n$
@Semiclassical Right, yeah, what I meant is that the answer to that limit is epsilon
15:59
Yes, which is why I included $\to \epsilon$ :3
OH
I am blind
lol
tbf i wrote that in a weird way
@AkivaWeinberger thanks @AkivaWeinberger
If $g(x)=x^p$ you get $\frac{(x+\epsilon)^p-x^p}{p x^{p-1}}$
which should also be $\epsilon$, no?
16:01
yep
And it seems like there's fairly generic conditions under which it should converge to $\epsilon$
So we got $\epsilon$ for everything except $e^x$, and I'm guessing it's because that's too fast
I suspect the way to look at this is to write $g(x+\epsilon)=g(x(1+\epsilon/x))$
There's a geometric interpretation here
yeah
i mean, it's $\Delta g/g'(x)\to \Delta x$ as $x\to\infty$
which is intuitive enough
but it can't be true in general
Slope of secant over slope of tangent should tend to one
16:05
right
if the secant has constant width $\epsilon$
What if we substitute in $\ln g$
or $e^g$
Throwing random suggestions out there
Define $g=e^f$ and substitute that in
or $g=\ln f$
and see if the resulting expressions look simple
Guessing the latter'll be more fruitful than the former
but I dunno
If $g=e^f$ then it becomes $\dfrac{e^{f(x+\varepsilon)}-e^{f(x)}}{f'e^{f(x)}}$
or $\frac1{f'(x)}(e^{f(x+\varepsilon)-f(x)}-1)$
and to be honest that looks pretty unhelpful
Unless $f(x+\epsilon)-f(x)\to0$ (as $x\to\infty$) I guess
in which case, the limit is equivalent to the limit of $\dfrac{f(x+\varepsilon)-f(x)}{f'(x)}$, if it exists
in which case I guess our property is closed under taking logs?
16:19
I cannot seem to do arithmetic, today. Allow C to be the counterclockwise boundary of upper half disk $\{re^{i\theta}: 0\geq r\geq 1, o\geq\theta\geq\pi\}$ and pick a branch of $z^{\frac{1}{2}}$ such that $f(z)=\sqrt(r)e^{\frac{i\theta}{2}}$ if $r>0$ and $-\frac{\pi}{2}\geq\theta\geq\frac{3\pi}{2}$ and $f(z)=0$ if $r=0$. Apparently, the integral $\int_C f(z)dz=0$, but I'm getting a value of $\frac{8}{3}+\frac{8}{3}i$
@Semiclassical Conjecture: If $g(x+\varepsilon)/g(x)\to1$, then $\dfrac{g(x+\varepsilon)-g(x)}{g'(x)}\to\varepsilon$.
As $x\to\infty$, for both limits.
Hmm.
full disclosure: this is motivated by an AMM problem I'm working on. Lemme post the problem as best I can recall
Amendment: If the former limit is true for all epsilon, then the latter limit is true for all epsilon
Let $f:[1\infty)\to \mathbb{R}$ be a convex, continuously differentiable function with $f'(x)>1$ for all $x>1$. Show that $\int_1^\infty \frac{dx}{f'(x)}$ is convergent if and only if $\sum_{n=1}^\infty [f^{-1}(f(n)+\epsilon)-n]$ is convergent.
By the integral test, the integral converges if and only if $\sum_{n=1}^\infty \frac{1}{f'(n)}$ converges
If one can show that $\frac{f^{-1}(f(n)+\epsilon)-n}{1/f'(n)}$ converges to a nonzero constant as $n\to\infty$, the statement will follow from the limit comparison theorem
For convenient notation, now let $g(x)=f^{-1}(x)$ with $g:\mathbb{R}\to [1,\infty)$
Since $g(f(x))=x$, we have $g'(f(x))f'(x)=1\implies 1/f'(x)=g'(f(x))$. Writing $x=f(n)$, the ratio becomes $\frac{g(x+\epsilon)-g(x)}{f(x)}$.
Since $f(n)$ has positive derivative and is convex, I think we can say for sure that it increases without bound as $n\to\infty$?
in which case we're just down to this limit
oh, blah. My initial statement of the problem should have ended with "...is convergent for any $\epsilon>0$"
grrr, $f'(x)>0$ for all $x>1$
16:43
Is gershgorin's circle theorem the quickest/best way to check if a matrix is positive definite
I guess the problem with this however, you don't know for sure whether an eigenvalue is positive or negative
@AkivaWeinberger I think there is a more convincing illusion that does not have infinitely many hexagons. Just put 12 pentagons regularly around the circular edge that we cannot see properly. =)
If you can find a way to make that work, go for it
The more I stare at this problem, the more I am convinced that it's not actually true.

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