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12:00 AM
It is starting to rain here! It's been a long time.
 
@Semiclassical in my course on stochastic processes in fluids, we handle a lot of correlation functions
and I can't quite to seem the physical importance
We started with defining the correlation functions, for some complex-valued process $X_t$, as $\Bbb E[X_t^\ast X_s]$ and with stationarity this is equal to $\Bbb E[X_{t-s}^\ast X_0]$
Then $C(t) = \Bbb E[X_t^\ast X_0]$ is our correlation function
Now first off, why complex-valued? What exactly are we measuring and how does that come up?
(Our professor's answers are always either too general or way too specific with examples such as this and that laser)
Then we use Bochner-Khintchin to show existence of real-valued random variables whose characteristic function is equal to correlation function and there again, I don't exactly know what this physically means or why we are interested in that
I will just describe further what we do since it's good to summarize this and you might see the context more
We also mention Wiener-Khinchin afterwards which says that the power spectral density of the process (what does this thing mean intuitively?) and the correlation function form a fourier-transform pair
For a stationary process
I can ofc give definitions if you need them, since these vary quite a bit under the same names
Ok so then we define the laplace transform (with an i factor in it, to make it a function defined on the upper complex half plane)
we relate it to the power spectral density although that's probably just a technicality
 
12:20 AM
that's a lot of words, but where's the algebra
 
So with all the fourier and laplace transforms I'm also confused why we need them to handle the processes. Is it the case that somehow, we only have the fourier transform at hand in practice, and need to conclude from this properties of the correlation function?
@Thorgott nowhere, thank god
Because afterwards we go all out fourier and define properties that ensure that a function from the upper half pflane to itself (+ the real line) is actually a the fourier transform of a correlation function, kinda similarly to what we did with Bochner-Khintchin
It's basically all about finding formulas like $i \hat{\phi}(z)=\frac{1}{\pi} \int_{\Bbb R} \frac{\operatorname{Re} \hat{\phi}(\omega+i 0)}{\omega-z} d \omega, \quad \Im z>0$
(It's amazing that automatic latex readers recognize my profs handwriting)
Here, this has got to do with the "Riesz-Herglotz representation of a Nevanlinna function"
Then we go more specific and consider processes that aren't just stationary, but also symmetric under time reversal, i.e. $\{-X_t\}$ has the same law as $\{X_t\}$
We say that such a process is in equilibrium
And have again formulas like $i \Phi(z)=\int_{R} \frac{z}{\omega^{2}-z^{2}} d F(\omega)$. Here $F$ is some distribution function so this is just a stieltjes integral. And again I wonder what this tells me
Apparently, Fourier inversion is easier in equilibrium, so that we get the correlation function
Our next concern then is finding out, from the fourier transform, so-called memory kernels
They satisfy $\frac{d}{d t} \phi(t)+\gamma \phi(t)+\int_{0}^{t} D\left(t-t^{\prime}\right) \phi\left(t^{\prime}\right) d t^{\prime}=0, \quad \phi(0)=1$
I should just be able to differentiate this and get a second order ODE right? Is this some spring type stuff?
Anyways it is apparent that from the fourier transform, and by approximating it right at short and high frequencies, we can say a lot about the process. I just don't see the natural setting
For example, we get a "harmonic oscillator with retarded friction" describing the differential equation with the memory kernel, in one example. What's that?
Afterwards it gets a little more hands-on again and we work with the "velocity auto-correlation function", look at integro-differential equations comparable to the one for memory kernels, only this time they're called "friction kernels", and the ODE is approx the same, but with $\gamma = 0$
the kernels we obtain again via the fourier transform, which makes a bit more sense now to me, since this directly comes from our knowledge of the process, and still purposefully uses the fourier transform
Does laplace-transforming just generally make handling the processes easier? What are the properties that we forget when we fourier-transform, which make it easier to handle the information? That might be a bit too general a question, but maybe there's a good answer
One of the thing's we're able to show then is that glassy dynamics are non ergodic, since the friction kernel does not go to zero. Pic incoming
Here $Z$ is the velocity auto-correlation function $Z(t) = \frac{d^2}{dt^2} \Bbb E[(R_t - R_0)^2]$ for some particle trajectory $R_t$
And the friction kernel is the $\zeta$ that shows up in the demoninator of the fourier-transform
 
12:57 AM
enters, looks, and faints
 
The part afterwards I can't summarize well yet. We start with scattering functions defined by $\begin{aligned} F_{s}(\vec{k}, t) &=\left\langle e^{-i \vec{k} \cdot \Delta \vec{R}(t)}\right\rangle &, \Delta \vec{R}(t)=\vec{R}(t)-\vec{R}(0) \\ &=\left\langle \rho_{s}(\vec{k}, t)^{*} \rho_{s}(\vec{k}, 0)\right\rangle, & \rho_{s}(\vec{k}, t):=e^{i \vec{k} \cdot \vec{R}(t)} \end{aligned}$

(I hope you can forgive the change of notation, the $\langle \cdot \rangle$ are expectations, I didn't want to type that all myself)
This is again $(\vec{k},t) \mapsto \varphi_{\Delta \vec{R}(t)}(\vec{k})$
So yeh fourier transforms and characteristic functions everywhere and I can handle the maths but its about as useful as an algebra class (@Thorgott) since I can't see how to throw these methods onto anything that might be interesting to me
I know from before that the scattering functions can help us distinguishing a process from a gaussian process with help of the cumulants. In this case not that surprising to me since ofc the scattering function is just a characteristic function and hence captures behavior of moments
@TedShifrin taking revenge on the geometers here
you kidnap the room regularly :P
anyways I'm finished, math room is now free again. Until semiclassical answers
Ah, one possible reason for all this might be that for a process at hand, with a given correlation function, we can take the fourier transform and approximate this at different frequency regimes. The approximation we can then, with the help of our machinery, transform back to see what kind of process the main driver of the trajectory is. But what do high/low frequencies correspond to, for the back-transformed process?
 
1:22 AM
@user2103480 If you look carefully, you'll see I discuss/help with analysis stuff and algebra stuff more than I do geometry. It was your page-long wall that stunned me.
 
1:33 AM
@TedShifrin and if it weren't for thors and your side-comments, the wall would be even mightier!
those were some well-needed line breaks though
To provide something more useful than math, here is an educational video:
 
So I am looking at my book right? and "the field of rational functions $\overline{k}(Z_f) \cong \overline{k}(x)[y]/(f)$ where $f$ is a polynomial not equal to $cx+d$ for $c,d \in \overline{k}$" shows up, and I'm here wondering "why"
many are asking this
It says things like $f(x,y) = a_n(x) y^n + ... + a_0(x)$ an irreducible polynomial in $\overline{k}[x,y]$. $\varphi : \overline{k}[x] \to \overline{k}[x,y]/(f)$ a natural map that is injective since $f \neq cx +d$ with $c,d \in \overline{k}$.
Since $\varphi$ injective we can extend it to an injection
$\overline{k}(x) \to \overline{k}(Z_f)$
s.t.
$g(x)/h(x) \mapsto \varphi (g(x))/ \varphi (h(x))$
and I'm sitting here, scratching my head, wondering "wow, that really makes it so $\overline{k}(Z_f) \cong \overline{k}(x)[y]/(f)$? that's nuts!"
 
1:50 AM
@user2103480 pure brilliance
id follow him on twitter, but he didnt give the handle
 
I totally searched for his frickin twitter I want to see that right now
"well it's good that no one completed it right"
"4 years later and this is still the most important video on Youtube."
"I like how they take him seriously when he’s talking about Thule and Atlantis but freak out when they think he’s a Nazi."
the comments are gold
"This kid just real life shitposted normies." I'm dying
 
"Trump is a Kantian" wew
@user2103480 he sounds like concetrated pol
 
@BigSocks the comments have an answer for this as well
"This is why /pol/ and /x/ shouldn't be allowed near each other"
- "This is why /pol/ and /x/ *should* be allowed near each other**"
 
lmao
 
but actually:
 
1:56 AM
youtube comments? you mean the ancient sacred texts
 
"This was 100% /lit/. Literally no one on /pol/ is redpilled on Schelling."
 
not nowadays anyway :x
 
hahahah yeh
 
lmao
 
 
2 hours later…
3:43 AM
Paragraph 2, end of the third sentence, why is $C_f[y/x] = \overline{k}[y/x]$ instead of $\overline{k}[y/x]/(f)$?
it seems odd bc the relation wouldn't hold otherwise
 
4:32 AM
the bases have different sizes. don't think it can be right, but I will consult the sandman and check back in tomorrow
 
4:54 AM
is it an equality of sets? sure. is it an isomorphism.? I doubt it.
 
@user2103480 Wow I am omegapilled
 
I am having problems with identifying wave and heat equations.
One is hyperbolic and the other parabolic.Can you explain which is which and why so?
 
5:18 AM
Why are they called so?What is so hyperbolic about a second order differential?
 
Think ellipse, hyperbola, parabola. Sum of squares, difference of squares, one square plus linear.
 
ok thought..
But what has this to do with differential equations?
Sum of squares and difference of squares would be equivalent in this regard @TedShifrin
I do not see a sign originating anywhere..
ok but @TedShifrin Why is wave a hyperbolic and heat parabolic?
 
5:47 AM
Somebody please respond to this...
 
I don't understand your confusion. The 1D heat equation is $u_{xx} = u_t$, the 1D wave equation is $u_{xx} = u_{tt}$. This formally resembles $x^2 - t$ (= const), $x^2 - t^2$ (= const) - a parabola and a hyperbola respectively.
 
@BalarkaSen My confusion is why is the one dimensional heat equation so...?
And one dimensional wave equation so..
 
What do you mean by "so"?
It's a differential equation which is called the heat equation. What is the question?
 
I understand why they are parabolic and hyperbolic to some extent...
But why "heat" and why "wave"..
 
Because they model how heat diffuses through a medium, and how a wave propagates through a medium, respectively.
 
5:54 AM
Because the physicists say so.
 
Why can someone give some intution?
 
This is not a mathematical question so you should ask physicists not us
 
Ok but you said that it resembles right..So in a sense you are saying it nothing to do with the actual meaning..
It is just an analogy....?
 
This is in regards to the terminology "elliptic, parabolic, hyperbolic"? Yes, this has no "meaning". It is a way to label a differential equation, but it is an important label because you need to appeal to different techniques to solve a PDE which is parabolic or hyperbolic rather than elliptic let's say.
 
6:39 AM
it has to do with the form of the pde. a sort of $b^2-4ac$ sort of thing.
 
6:51 AM
is 2021 over yet?
 
7:05 AM
@copper.hat my 'joke': 2020 is over, but the 2020s have just begun
 
7:20 AM
Hello! I have a quick question
Does it hold that $$e^{-\frac{1}{x}}$ is uniformly continuous on $(0,\+infty)$?
 
it looks quite uniform to me
 
@LeakyNun Let me tell you about mass-transport theory
 
what about it
 
Oh do you know it
 
I don't
 
7:33 AM
Let $G$ be a countable infinite group and $\Gamma$ be a Cayley graph of $G$ with respect to some generating set
 
@LeakyNun Ok, thank you!
 
A transportation scheme is a function $f : \Gamma \times \Gamma \to [0, \infty]$ such that $f(gx, gy) = f(x, y)$ for all $g \in G$. Think of $f(x, y)$ is a $G$-invariant way to transport a unit amount of mass from vertex $x$ to vertex $y$.
 
There's a PDE for that, isn't there
i know i've seen this discussed
 
My setting is discrete
 
hmm
In statistics, the earth mover's distance (EMD) is a measure of the distance between two probability distributions over a region D. In mathematics, this is known as the Wasserstein metric. Informally, if the distributions are interpreted as two different ways of piling up a certain amount of dirt over the region D, the EMD is the minimum cost of turning one pile into the other; where the cost is assumed to be amount of dirt moved times the distance by which it is moved.The above definition is valid only if the two distributions have the same integral (informally, if the two piles have the same...
is what i'm thinking of
 
7:36 AM
I don't know if this is what I am doing
You're thinking of optimal transport theory.
This is different
 
ah
suboptimal transport theory? :P
 
The mass-transport principle says for any transportation scheme $f$, $$\sum_{x \in \Gamma} f(o, x) = \sum_{x \in \Gamma} f(x, o)$$
 
Hello!! I am looking at the function $d(x,y)=|x^3+x-(y^3+y)|$ and I want to check if itdefines a metric.The first two conditions are satisfied. How can we checkthe triangle inequality? Or dowe consider this as a sum of the metric $d(x,y)\leq |x^3-y^3|+|x-y|$ and so we get agaithat $d$ defines also a metric?
 
Conservation of mass at $o$
@LeakyNun @Semiclassical
 
@MaryStar only clever thing i notice is that you can factor that polynomial: $x^3+x-y^3-y=(x^3-y^3)+(x-y)=(x-y)(x^2+xy +y^2)+(x-y)$
so $d(x,y)=|x-y|\cdot |x^2+xy+y^2+1|$
that's neat but i dunno if it tells you anything
 
7:42 AM
And then we show the inequality for $|x^2+xy+y^2+1|$ right? @Semiclassical
 
hmm, i don't see how that follows.
it's still d(x,y) which is the (proposed) metric
 
Ah yes.... The inequality holds, or not? (Or is it not a metric and so it cannot follows that it holds?) @Semiclassical
 
that's the question, yes
it does make me wonder more generally: under what conditions is $|f(x)-f(y)|$ a metric for polynomial $f$
probably nothing too nice one can say
i don't have much intution for this tho
 
So we have to check of $f(x)=x^3+x$ is injective,right? @Semiclassical
 
oh, nice
that'd seem to do it
 
7:49 AM
Which is injective and so $d$ defines a metric, right?@Semiclassical
 
alas, they don't show -why- it's true
well, $f'(x)=3x^2+1$ is positive everywhere, so it's monotonically increasing
so yeah, injective
of course, this is incomplete insofar as we haven't shown -why- $f$ being injective suffices
(and i don't know off the top of my head)
 
Ok! I will think about that if I can make the proof
 
sounds good
 
@Semiclassical no doubt 2020 is the hindsight year...
@MaryStar The map $p(x) = x+x^3$ is a bijection, hence $d(x,y)=|f(x)-f(y)|$ defines a distance.
the properties just come from the properties of absolute value.
$|f(x)-f(z)| \le |f(x)-f(y)|+|f(y)-f(z)|$.
injectivity is sufficient of course.
for example $f=\arctan$.
 
8:11 AM
Ahh! Thank you very much!! :-) @copper.hat
 
you are very welcome.
 
 
2 hours later…
10:30 AM
@BalarkaSen So f is in reality $\Gamma \to [0, \infty]$ because $f(x,y) = f(y^{-1}x, e)$ ?
Or do you only want the g's to be part of the generating set?
 
@Astyx Well yes but I do not advocate this way of thinking about it
 
Okay, sorry :)
So what is it you're doing with this graph?
 
The mass-transport principle is obvious but it has many applications. One of them is it gives a quick example of a transitive graph which is not a Cayley graph.
Should I say?
 
What's a transitive graph? A connected one?
Ah no, it's stronger than that, you want an automorphism
 
No, the automorphism group acts transitively on the graph
 
10:38 AM
Gotcha
 
Take a $3$-regular tree $T$. Fix an "end" $\xi$ of the tree; that is, consider a ray on the tree going off to infinity and look at the collection of all rays which differ from this ray at only finitely many vertices (can only differ at some initial segment, because tree).
This collection is what I call $\xi$
For any vertex $x \in T$, there is a unique ray $x_0 = x, x_1, x_2, \cdots$ starting at $x$, contained in $\xi$.
I define $x_2$ to be the grandchild of $x$. Join $x$ to $x_2$ by a new edge. Doing this for every $x$, I obtain a new graph $G$.
I claim this is transitive but not a Cayley graph
 
I'm not sure I follow. A 3-regular tree is just a binary one right? When you say ray, you don't only include paths that start at the root of the tree?
 
No, the binary tree has a root, which has degree 2. By 3-regular I mean a genuine 3-regular tree, there's no root.
Every vertex has degree 3.
 
Oh ok
My bad, yeah, I understand
 
Sorry for not emphasizing.
 
10:46 AM
No, it's just my graph terminology knowledge is very lacking
 
Same
 
If you don't add the grandchild edges, you get a cayley graph of the group of sequences $(\Bbb Z/2\Bbb Z)^{\Bbb Z}$ that stabilize at 0 right?
 
That will have many cycles regardless of what generating set you choose, so that cannot be right.
 
yup that doesn't work
 
$\Bbb Z_2 * \Bbb Z_2 * \Bbb Z_2$ will do the trick, where $*$ means free product.
 
10:52 AM
Hmmm yeah ok
 
Standard generators, one for each component.
 
So you start at the origin of a certain ray, and you move a finite path from there
And this gives you every possible ray in $\xi$
 
But aren't all the paths you're giving containing the original path?
They'll differ at some initial segment but one need not contain the other
These are rays which eventually agree
 
You mean with my sequence construction? Yeah, I had in mind your free product idea but failed to think it through
 
No I mean in your description of $\xi$
 
10:58 AM
If they eventually agree, they disagree on only a finite number of vertices. With that free product you can go "up the ray" at first to the vertex where both rays start agreeing, and construct from there
(I think this construction might rely on choice, because you need to color each edge in a way that no two adjacent edge have the same color?)
 
What do you mean by construct from there? I mean we fix a ray $x_0, x_1, x_2, \cdots$ and then look at all the rays $y_0, y_1, y_2, \cdots$ such that $x_k = y_k$ for all but finitely many $k$'s. That's $\xi$
I don't understand colors
Why is that relevant
 
Oh I thought we only wanted $y_k = x_{k+s}$ for some $s$ and $k$ large enough
 
Yeah this is not what I meant. Here's the picture: the "boundary" of the tree $T$ is a Cantor set. Fix a point $\xi$ there, and look at all rays which has "terminal point" at $\xi$
This is why I am using the suggestive terminology "end"
 
Yeah, this I got, but you don't want $x_{-1}, x_0, x_1, x_2,\dots$ to be in that ray right?
I cannot "lengthen" the ray
 
Why not?
No restriction on initial vertex of the ray
 
11:04 AM
Let me draw something
 
Green point is the end $\xi$, which is concretely the collection of all these rays ending at the green point
No restriction on initial vertex.
 
Ok so we had the same idea
Why do you think the free product construction doesn't work?
 
Which construction?
 
So in your drawing, say I chose red as a representative path of this ray
I label the edges with a,b, or c (each an element of $\Bbb Z_2$) such that no two adjacent edges have the same label
Now an element in $\Bbb Z_2*\Bbb Z_2*\Bbb Z_2$ is a (finite) word in a,b,c
Now I create a path from the origin of the red path like this: I read the word, and every time I meet an a, I go through the edge that is labeled a (same for b and c)
I can concatenate the path with the red path and reduce it to give me a new representative of the ray
I claim you get them all
 
Get all of what? I really am not able to understand what you want to do. If you want to get all rays in $\xi$ sure that is equivalent to all "infinite words" in $\Bbb Z_2 * \Bbb Z_2 * \Bbb Z_2$ which only differ from a representative word on an initial segment
Where an "infinite word" is a collection of words such that the $n$-th word is contained in the $n+1$-th word as a penultimate segment or whatever
But why? The picture I drew explains all of this rigmarole.
 
11:17 AM
Yes but you don't need to take infinite words. Instead, you can take one infinite word and all finite words
 
Sure, fine, but what does this accomplish?
I still don't understand what you want out of this
 
It shows it's a Cayley graph
 
Eh? The tree is already a Cayley graph.
We have established that
 
Ok nvm
 
Why would $G$ be a Cayley graph? This is obtained from the tree by adding new edges corresponding to (grandparent, grandchild) vertices along rays in $\xi$
$T$ is a Cayley graph by just drawing $Cay(\Bbb Z_2 * \Bbb Z_2 * \Bbb Z_2, \{a^{\pm 1}, b^{\pm 1}, c^{\pm 1}\})$. It's just the tree. This has nothing to do with rays, right?
 
11:20 AM
Yeah my bad
 
The point is grandparent-grandchild doesn't mean anything if you don't fix an end $\xi$.
It gives a notion of direction to the graph
Everything is flowing towards $\xi$, the green point, so to speak
 
So every vertex looks like this now
Where the new edges are the exterior ones now
 
Everything is valence 4 if you think about it carefully
 
Yup I missed edges where the circled vertex is a grandchild
Isn't everything valence 8?
 
@Astyx Nope, because of the direction associated with $\xi$.
 
11:31 AM
You're missing the edge where the one with 7 incoming edges is grandchild
 
Oh sorry I mean valence $4 + 3 = 7$.
Oh, touche.
$8$ it is.
 
So it's transitive because you can shift everything by "pulling" from $\xi$ (or pushing into it)
 
And I guess your idea is that we can make a transportation scheme that doesn't respect the mass transport principle or something
 
Yup
So the point is every vertex has a unique grandchild, but every vertex has 4 grandparents.
So incoming mass and outgoing mass are not the same, there's some storage of mass. This cannot happen for a Cayley graph
Let's write this precisely
Let $\Gamma$ be the graph (I wrote $G$ earlier, but $G$ should be a group not a graph). Let $f : \Gamma \times \Gamma \to [0, \infty]$, $f(x, y)$ = Indicator($y$ is a $\xi$-grandparent of $x$).
Then $\sum_x f(o, x) = 4$ but $\sum_x f(x, o) = 1$
I claim $f$ is $\text{Aut}(\Gamma)$-invariant. This will show $\Gamma$ is not a Cayley graph because if it is Cayley graph of some group $G$, then certainly $G$ is a subgroup of $\text{Aut}(\Gamma)$, so $f$ is also $G$-invariant.
But then it breaks the mass-transport principle.
 
11:49 AM
($Aut(\Gamma)$)
 
Lol, oh boy
 
Right yeah that works
 
yeah
 
That looks interesting
 
can anyone explain why the red line is ture?
I understand to rotation but why translation?
It seems translation can manipulate the gap between two parallel lines
 
12:22 PM
@BalarkaSen thulepilled
ah no wait thats varg
 
lol yeah
 
 
1 hour later…
1:50 PM
For people interested in the convolution and cross-correlation operations.
2
Q: What do the variables in the cross-correlation formula mean?

InvestingScientistI understand what cross-correlation does given a kernel and an input image, but the formula confuses me a little. Given here in Goodfellow's Deep Learning (page 329), I can't quite understand what $m$ and $n$ are. Are they the dimensions of the kernel along the height and width dimensions? $$S(i,...

 
@Semiclassical did the wall of text scare you off? :P
 
2:09 PM
@Astyx OK so there's an analogue of the mass-transport principle for transitive graphs in general as well
$\Gamma$ be a locally finite countable graph, $G$ be a group of automorphisms of $\Gamma$. If $f : V(\Gamma) \times V(\Gamma) \to [0, \infty]$ is a $G$-invariant transportation scheme then for all $x, y \in V(\Gamma)$, $$\sum_{z \in G\cdot y} f(x, z) = \sum_{w \in G\cdot x} f(w, y) \cdot \frac{|\text{Stab}(w) \cdot y|}{|\text{Stab}(y) \cdot w|}$$
This is the most general version.
If $G$ acts transitively on $\Gamma$ then you get the usual conservation of mass but with a factor: $$\sum_{x \in V(\Gamma)} f(o, x) = \sum_{x \in V(\Gamma)} f(x, o) \cdot \frac{|\text{Stab}(x) \cdot o|}{|\text{Stab}(o) \cdot x|}$$
In the grandparent graph, if $o$ is the $\xi$-grandparent of $x$, $|\text{Stab}(x) \cdot o| = 1$ whilst $|\text{Stab}(o) \cdot x| = 4$, so we're fuzzing each term in the left hand side out by $1/4$ (the rest of the terms don't matter if $f(a, b)$ is indicator of $b$ being the $\xi$-grandparent of $a$.
 
2:45 PM
How do you make sense of this?
 
@LHC2012 1/(1+x) = 1-x+x^2-x^3+...
 
@LHC2012 who tortures beginning students with these big o calculations?
 
Wait so I still don't know what to do given 1/(1+x) = 1-x+x^2-x^3+..., sorry I'm still very confused
Like it seems like you ignore the x^2 term, but why can you do that and not ignore the x term?
 
3:02 PM
@leaky
 
@LHC2012 O(y) means "constant times y + higher powers of y"
 
if possible can you elaborate? I'm. getting stuck on a simple proof of steffensen's method quadratic convergence for 2 days...
 
so 2y + y^3 is O(y)
 
you mean the case when y is small?
 
yeah
 
3:04 PM
ok thanks
Also do you know why newton's method doesn't require that $f$ is $C^3$? I thought since $g(x) = x - \frac{f(x)}{f'(x)}$ you need $g$ to be $C^2$ so $f$ is $C^3$ @leak
 
no idea
 
My thinking that $g$ needs to be $C^2$ is because I need the error term in taylor series of g(x) to be bounded
 
you need $f$ to be only $C^1$ for convergence sufficiently near the root
with $C^2$ you get quadratic rate of convergence
 
Oh I see
Can I check something then? If I have
It only converges cubically if $g$ is $C^3$, or $f$ is $C^5$??
@BalarkaSen
 
3:20 PM
I don't understand what is written
 
When you define $g(x) = x - \frac{f(x)}{f'(x)}$ I thought $g(x)$ needs to be $C^2$ to converge quadratically
 
That's not true.
 
Balarka when he's annoyed: "no hablas ingles"
 
But say if you taylor expand g(x) right, then if $g$ is not $C^2$, how do you bound the quadratic term?
 
You don't Taylor expand $g$
 
3:26 PM
Then I'm clueless how you would show quadratic convergence
 
One uses first order Taylor expansion on $f$ with remainders. You can find a proof in textbooks.
 
4:19 PM
@BalarkaSen Very cool!
Is it obvious that those cardinals are going to be finite (in the general case)
I guess that's what locally finite means
For each vertex $v$, $|\{v'\in V, d(v,v')\le n\}|\le \infty$ for all $n$
Then, given an automorphism preserves distance, $\text{Stab}(v).x \subset \{v'\in V, d(v, v')= d(v,x)\}\subset \{v'\in V, d(v, v')\le d(v,x)\} $
 
Yup!
That's it
 
 
2 hours later…
6:46 PM
Am I remembering something wrong and the MVP for harmonic functions only needs it to hold for small enough radii, as opposed to "all" radii (I think for all might follow from there exists)?
 
Your question doesn't make sense to me.
In general, a region will not contain balls of large radius centered at most points.
 
Say we got $f : \Bbb Z \hookrightarrow \Bbb Z[i]$ and consider the ideal in $\Bbb Z[i]$ generated by $2 +i$. Why is it the case that $[(2 + i)(2-i) = 5] \Rightarrow [f^{-1}((2+i)) = (5)]$?
 
Every ideal in $\Bbb Z$ is principal. What do you suggest, @BigSocks, if not $(5)$?
 
I mean, I could see that it was $(5)$ but probably arguing differently. Like taking some $g : \Bbb Z[x] \to \Bbb Z[i]$ with $x \mapsto i$ and thinking about how the ideals of one of them match up with the ideals of the other
I'm mostly confused as to how that product being $5$ means the preimage of the ideal is $5$
 
It's like rationalizing denominators. The smallest integer in the ideal has to divide $5$.
 
6:53 PM
$5=(2+i)(2-i)$ implies $5$ is contained in the ideal generated by $2+i$. $f^{-1}$ is just intersection with $\mathbb{Z}$, so $(2+i)\cap\mathbb{Z}$ contains $(5)$. $(5)$ is a maximal ideal ideal and $(2+i)\cap\mathbb{Z}$ is an ideal, so the only other option left is it being the entire ring, which is absurd (preimages of proper ideals are proper).
 
Ok I like both of these answers. Thank you @TedShifrin for the intuition and @Thorgott for the proof
 
LOL, Ted is worthless for proofs. :)
 
That's ok- me too. one day computers will help us with those tricky bits anyway. Intuition is important
 
I've known Thor to provide a little intuition once in a rare while.
 
thanks :P
 

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