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Secret
Secret
3:27 AM
5
Q: What is the empty relation?

GabuI was reading the Wikipedia article on equivalence relations and one section says that "the empty relation R on a non-empty set X is vacuosly symmetric and transitive but not reflexive." What is the empty relation? And what is vacuosly symmetric? Thank you very much.

equivalence-relations
Probably that it is not reflexive is the most surprising to me
 
user10478
user10478
3:59 AM
Am I correct in inferring that the form of the forcing function times $t$ or $t^2$ is only a particular solution to a second order linear ODE if the particular solution would be linearly dependent with the homogeneous solution otherwise? If so, is there an intuition for why this occurs? How does the particular solution with the extra $t$ or $t^2$ know to "go away when it's not needed"?
 
Akiva Weinberger
Akiva Weinberger
4:14 AM
@Secret You can't have $aRa$ if nothingRanything
 
Semiclassical
Semiclassical
4:46 AM
@user10478 Here is an attempt. Suppose we have an $n$th order differential operator $L$ and solutions $\{y_k\}_{k=1}^n$ to the homogeneous ode $Ly=0$. One can now check that $L(xy)=xL y+L_1 y$ for some $(n-1)$th order differential operator $L_1$. (Formally, $L_1$ is what you get when you treat $L$ as a polynomial in $D_x$ and 'differentiate' with respect to $D_x$.)
Therefore, for any solution homogeneous $y_h$, we have $Ly_h=0\implies L(xy_h)=L_1 y_h$.
not so sure about this next point: It seems like one has $L(L_1 y_h)=L_1(Ly_h)=0$. This is trivial to justify if $L$ just has constant coefficients, but i'm not so sure otherwise.
In that case, however, $L_1 y_h$ is itself a solution to the homogeneous equation and thus can be expressed in the $\{y_k\}$ basis.
So $L_1$ then maps homogeneous solutions to homogeneous solutions. If the desired forcing function $f$ is in the image of $L_1$, you can thus find $y_h$ such that $L(xy_h)=L_1 y_h = f$ and solve the ODE.
So for constant-coefficient ODEs, at least, the $y_h\to x y_h$ tactic seems legit
(actually, does that method for getting a satisfactory particular solution work if it's not a constant-coefficient ode? haven't thought about that knid of thing in a while)
wolfram doesn't seem to think so: wolframalpha.com/input/…
 
 
2 hours later…
Mike
Mike
7:26 AM
Let us assume that we have N lines in a three dimensional space. How can we figure out if any of these lines intersects any other line?

It would be great, if the problem was solved in terms of linear algebra. If it is not possible, why?
 
 
1 hour later…
user55789
user55789
8:34 AM
Hi all, what's the closed form of the notion of a "signed sum" of summands
So suppose I have two elements, a_1, a_2, then the two signed sums possible would be a_1 + a_2 and a_1 - a_2. The best I could come up with was $\sum_{i=1}^2 \sigma_{ij} a_i$
And then constrain $\sigma_{1j}=1$
This shows up for example when considering partial fractions
 
 
1 hour later…
Alessandro Codenotti
Alessandro Codenotti
10:07 AM
In this notes I'm reading they consider a Borel measure $\mu$ on $G^{(0)}$ (this is the unit space of a topological groupoid, but this should be irrelevant, it is just some topological space with the Borel $\sigma$-algebra) and $H=\bigsqcup_{x\in G^{(0)}} H_x$ a $\mu$-measurable field of Hilbert spaces.
They don't define this term though. What exactly is a $\mu$-measurable field of Hilbert spaces?
 
Slereah
Slereah
Math folks
I must know
Can every diffeomorphism on a manifold be represented by the flow of a vector field
 
s.harp
s.harp
no
then every diffeomorphism would be homotopic to identity, no?
 
Mike Miller
Mike Miller
10:38 AM
Not even every diffeomorphism which is close to the identity is the time-1 flow of a vector field
Try constructing an example on the circle
The best you can do is time-1 map of a time-dependent vector field and that's only for diffeos isotopic to the identity
 
Slereah
Slereah
10:59 AM
What diffeomorphisms aren't isotopic to identity
Dehn twists?
 
Mike Miller
Mike Miller
11:10 AM
Diffeomorphisms of what? Lol
Look up mapping class groups
 
Leaky Nun
Leaky Nun
orientation not preserving ones arent isotopic to identity clearly
(right)
 
Slereah
Slereah
Manifolds
Also true
 
Mike Miller
Mike Miller
11:23 AM
"Dehn twists" are things for surfaces
 
Slereah
Slereah
Mapping class groups does indeed seem to be what I should look into
Dang it
I just wanted to have a decent notion of how an active coordinate transformation could be defined
Without resorting to Killing vectors immediatly
 
Slereah
Slereah
11:49 AM
Is there a proper notion of active coordinate transformation in mathematics?
ie : you have an original field as some section of a vector bundle $\phi \in \Gamma(E)$
And then you have some function $f : \Gamma(E) \to \Gamma(E)$ that maps it to a new field such that $f(\phi)(p) = \phi(\psi(p))$
Where $\psi$ is some diffeomorphism on the manifold
So that the coordinates stay the same but the field is moved along, so to speak
 
Mike Miller
Mike Miller
I have no idea what you're trying to say
 
Slereah
Slereah
Yeah I'm afraid I am speaking in physicist
Consider some real function $\phi$ on $\mathbb{R}$
With values $\phi(x)$
Now consider a mapping $\Phi$ between such functions such that $$\phi'(x) = \Phi(\phi)(x) = \phi(x + a)$$
In the case of a translation, for instance
The coordinates do not change here but simply we get a new field that has all its values translated compared to the original function
What is the corresponding proper mathematical notion here?
In such a way that it would work with fairly arbitrary sections of a vector bundle and diffeomorphism on the coordinates?
In physics there is some subtlety here between passive coordinate transformations (which is basically just picking a new coordinate patch) and active coordinate transformation as here, where you just move the field values around
I'm wondering if there is a corresponding mathematical notion
@BalarkaSen you speak physics
Do you get my meaning
 
s.harp
s.harp
12:25 PM
what you are writing looks like $\psi^*(\phi)$, the pullback of the $\phi$ by a diffeomorphism $\psi$
 
Slereah
Slereah
Hm, could be
Pullback by automorphism sort of thing
 
s.harp
s.harp
but what do you care about, do you just want to know what its called?
 
Slereah
Slereah
Well I'm trying to work out symmetries on the action functional in a proper sort of way
But first I must define what it means for a field to be transformed
I guess overall a field transformation is going to be some function $f : \Gamma(E) \to \Gamma(E)$, for $E$ the bundle of all the spacetimes fields, and in the case of a coordinate transform, that would be a pullback automorphism
so that things can be treated on roughly the same footing
Does that sound like it makes sense
 
s.harp
s.harp
whats a coordinate transform
 
Slereah
Slereah
Hard to be sure due to the whole physics $\leftrightarrow$ math sort of thing, but roughly a diffeomorphism on the manifold
 
s.harp
s.harp
12:38 PM
you should never have the entire diffeomorphism group as a group of symmetries
 
Slereah
Slereah
and indeed it isn't
But here I'm talking about transforms just
What will be symmetries will be for every $f$ such that $$S[\phi] = S[f\phi]$$
For the action functional $S$
I'm just trying to make it broad because not all transformations of the field will be symmetries
 
s.harp
s.harp
ok, so in physics you know that if you change the coordinates you get a different coordinate expression for $\phi$ and you have $S[\phi]=S[\phi']$, is that what you are trying to describe?
 
Slereah
Slereah
Yes
 
s.harp
s.harp
someting like $\int_A f(x) d^nx = \int_{y^{-1}A} f(y(x)) \det( \partial_i y^j) d^n y$
 
Slereah
Slereah
that sort of thing, yes
Although in this case it's not the coordinates that change but the field itself, if that makes sense
 
s.harp
s.harp
12:43 PM
thats part of the well-definedness of $S$ and $\phi$
 
Slereah
Slereah
(Although IIRC the two process are mathematically equivalent)
well, at least as far as this is concerned
But since $S$ is a function of the field configurations and not of the coordinates directly, I try to keep things focused on the field configurations
 
s.harp
s.harp
in physics you hvae some coordinate chart and then for this coordinate chart an expansion of $\phi$ in components, you have a rule for how this expansion changes when you switch coordinate charts
 
Slereah
Slereah
So that I treat translations on a field as a new field translated on the same manifold
 
s.harp
s.harp
the stuff in my previous message is just the physics way of saying that $\phi$ is a section of a vector bundle (or some other kindf of bundle)
 
Slereah
Slereah
Well fortunately as far as I know all fields are just vector bundles
 
s.harp
s.harp
12:46 PM
in physics the functional $S$ is usually written down using the way $\phi$ looks like in a given coordinate chart
 
Slereah
Slereah
Don't want to get any of those weird sphere bundles in my action
 
s.harp
s.harp
the statement that $S$ doesn'T change when you swap coordinates is nothing else than $S$ being well-defined, that is something where you put in sections of the bundle and not something that needs a coordinate chart to work
so your coordinate invariance thing is just the statement that the domain and codomain of $S$ are $\Gamma(E)$ and $\Bbb C$
 
Slereah
Slereah
Well as I said
This isn't quite the same as just changing the coordinates
 
s.harp
s.harp
there are also symmetries
 
Slereah
Slereah
Because of course, there are cases where the action isn't invariant under some coordinates transformation
 
s.harp
s.harp
12:49 PM
@Slereah no, under coordinate transformations the action is invariant unless its not well defined
 
Slereah
Slereah
Well
Perhaps I am using the wrong word here
There is the whole confusion in physics about "coordinate transform" meaning rather different things
 
s.harp
s.harp
let $E=TM$ the tangent bundle of $M$, then if you have a diffeomorphism $F:M\to M$ this induces a map $DF: TM\to TM$. If you have an action $S:\Gamma(TM)\to\Bbb C$ it need not be the case that $S[DF(X)]=S[X]$ for all $X\in \Gamma(TM)$
 
Slereah
Slereah
hence the whole passive v. active coordinate transformation
Yes that's basically what I'm doing I think
 
s.harp
s.harp
continuing wit hthe previous message if $S[DF(X)]=S[X]$ for all $X$, then $F$ is called a symmetry of $S$. Generally you should expect taht there are not so many symmetries
as an example consider on $M=S^1$, $E=TS^1$, the action $S[X]= \int_{S^1} \langle X, \partial_t \rangle dt$, then the rotations are symmetries of this action
 
Slereah
Slereah
Alright
Can I broaden the set of transformations of the field to $\text{Diff}(E,E)$, so as to include all possible transformations
ie, both the ones induced by diffeomorphisms on the manifolds as well as... I guess they are called internal symmetries in physics
 
s.harp
s.harp
12:55 PM
Lets stay with $E=TM$ for the time being, you can certainly ask the question for which $F\in\mathrm{Diffeo}(M)$ does $S[X]=S[DF(X)]$ for all $X\in\Gamma(E)$, thats a question that makes sense
 
Slereah
Slereah
ie the $U(1)$ symmetry $\phi \to \phi' = e^{i\alpha} \phi$ in $$S[\phi] = \int_\Omega \phi \phi^* d\mu[g]$$
 
s.harp
s.harp
ok, now we are broadening the context, we have now bundle-automorphisms $E\to E$ and asking when does $S[F(\phi)]=S[\phi]$ for $F\in \mathrm{Aut}(E)$
 
Slereah
Slereah
Yes
That's the fairly broad context I'd like to try to use
So as to get Noether's theorem properly
Although I guess I also need Lie group structures in there, but one step at a time
 
ALannister
ALannister
Anybody in here proficient in Matlab?
Or even if you’re not proficient in Matlab
Anybody in here familiar with Gauss-Newton and why it might not converge?
I have two open bounties open.
One has to do with why Gauss-Newton isn’t converging for the function I am trying to minimize.
The other is regarding my Matlab code.
Open bounties open hehheh redundancy at its finest
 
Slereah
Slereah
Even worse IIRC there are possible mix of internal and coordinate symmetries in some systems, I recall
Although they are considered pathological in physics
 
nitsua60
nitsua60
1:07 PM
@ALannister links?
 
Slereah
Slereah
@s.harp Is the automorphism on the bundle itself or the space of sections of the bundle, btw?
 
s.harp
s.harp
@Slereah a bundle automorphism is a map $E\to E$ such that 1. if $\pi(x)=\pi(y)$ then $\pi(F(x))=\pi(F(y))$ (ie its a bundle map) 2. the induced map $f:M\to M$, $\pi(x)\mapsto \pi(F(x))$ is a diffeomorphism (one says $F$ is a bundle-map over $f$) 3. any relevant structure of the fibres is preserved under $F$ (ie for vector bundles if $v,w\in\pi^{-1}(x)$ then $F(v+w) =F(v)+F(w)$ and $F(\lambda\, v)=\lambda\,F(v)$ for $\lambda\in \Bbb C$)
oh and 4. $F$ admits an inverse of that type
 
Slereah
Slereah
I assume a bundle automorphism induces a transformation on the sections
 
s.harp
s.harp
(note that point 2. follows from 1. + existence of an inverse satifying 1.)
certainly: if $s: M\to E$ is a section then $F\circ s\circ f^{-1}$ is also a section
 
Slereah
Slereah
true
I guess that's good enough for me
thx
Though I do wonder if there is a proper way to split that space
Like within the space of bundle automorphism, what is the subset of those automorphisms induced by a diffeomorphism on the base space?
 
s.harp
s.harp
1:18 PM
automorphisms of the bundle induce diffeomorphisms of the base, only in some cases do you have a method of getting automorphisms from diffeomorphisms of the base (like with the tangent bundle)
 
Slereah
Slereah
Hm
How to explain this
I mean for instance, if I have some bundle automorphism $\phi \to 0$, for any field $\phi$, we can't really say that this is just the same field under active coordinate change
While $\phi(x) \to \phi(x + a)$ is perfectly fine
 
s.harp
s.harp
what do you mean $\phi\to0$? No bunlde automorphism will do that
 
Slereah
Slereah
Yes I'm not sure if this is a good point
Trying to sort out things in my mind
because there's the whole Coleman-Mandula theorem in physics where basically spacetime and internal symmetries cannot mix in a non-trivial way
ie if you have some internal symmetry group $G$, then the invariant group for the action is something like $\mathbb{R}^4 \rtimes O(3,1) \times G$
The Poincaré group times the internal symmetries
 
ALannister
ALannister
 
Slereah
Slereah
So I'm trying to work out how all this work with that idea
 
ALannister
ALannister
1:24 PM
And
 
Slereah
Slereah
Let's say for instance you have a complex constant scalar field $\phi(x) = 1$
 
ALannister
ALannister
0
Q: Matlab: Gradient and Hessian of a function with vector input of user specified size

ALannisterI need to write a matlab m file that takes the following function of $x=(x_{1},x_{2},\cdots, x_{2n})$, and for $n=10$, $n=100$, $n=500$, $$f(x) = \frac{1}{2}\sum_{i=1}^{2n}i(x_{i})^{2}-\sum_{i=1}^{2n}x_{i}+\sum_{i=2}^{2n}\left[\frac{1}{4}\left(x_{i}+x_{i-1} \right)^{2} +\left(x_{i}-x_{i-1}\right)...

optimization vector-analysis matlab nonlinear-optimization hessian-matrix
 
Slereah
Slereah
Then any such coordinate transformation will leave it as a constant field $= 1$
But a transformation of the type $\phi \to e^{i\alpha} \phi$ will not
 
s.harp
s.harp
@Slereah id imagine the group of symmetries is the subgroup of $\mathrm{Aut}(E)$ preserving some kind of structure on $E$, like for example an action. An "internal" symmetry could be an automorphism over the identity diffeomorphism
 
Slereah
Slereah
Yeah I guess internal stuff might be something that has like... some property with respect to the projection of the bundle
Argh
It ain't easy
 
s.harp
s.harp
1:26 PM
additionally, like in the case $E=TM$ you should have some way of lifting diffeomorphisms to bundle-automorphisms, then the statement of that theorem could be understood as $\mathrm{Symm}(S) = \mathrm{Lifts of Poincare}\times \mathrm{Internal}$ (in particular the lifts of poinc commute with any internal symmetries)
 
Slereah
Slereah
I'm sure nLab has an article on that exact topic but I think nLab was written by some manner of aliens from another galaxy
All mathematical truth ever is contained in nLab if you can actually read it
Well I suppose there could be symmetries such as let's say $\phi(x) = e^{i\alpha(x)} \phi(\psi(x))$, for instance
And those would be the one forbidden by such a theorem
(It makes quantum field theories go bad)
 
s.harp
s.harp
@Slereah indeed $\phi\to (x\mapsto e^{i\alpha(x)} \phi(x))$ is an internal symmetry that does not commute with $\phi\mapsto \psi^*(\phi)$ if $\alpha$ is not constnat
 
Slereah
Slereah
Do you have any idea if there's a ressource on that kind of thing, btw?
Action functional and bundle automorphisms
 
s.harp
s.harp
i dont know wanything about qft
 
Slereah
Slereah
Well it's not QFT
 
s.harp
s.harp
1:32 PM
i would try books about principal bundles and gauge theory (with subtitles like "physics for mathematicians" or "mathematics for physicits")
 
Slereah
Slereah
This seems sufficiently awful to possibly be about that
That's always the problem, really
 
s.harp
s.harp
are chromomorphism groups related to genetics
 
Slereah
Slereah
When physicists do it it's usually insufficiently rigorous and when mathematicians do it it's impossible to comprehend
Also it tends to be very focused on specific gauge theories, usually
Oh man
you can tell this paper is going to be horrendous because it mixes three scripts in one symbol
$$\text{EP}\mathcal{A}ut$$
 
Slereah
Slereah
1:48 PM
Ah yes, I think what physicists call an internal symmetry is what is called a vertical automorphism in math
Well, an internal transformation anyway
it remains to be seen if it's a symmetry
I should write a math to physics dictionary one day
There does seem to be plenty on the topic although it is usually within the purview of immediatly using it for gauge theory which is slightly annoying
But I suppose it's doable to work out how it works pre-using it for symmetries
 
 
2 hours later…
Ultradark
Ultradark
3:46 PM
I figured out how to visualise the reciprocals of numbers
that is, numbers of the form $1/t$
 
Semiclassical
Semiclassical
4:01 PM
one way is circle inversion, which has a compass-and-straightedge construction: en.wikipedia.org/wiki/Inversive_geometry
 
Ultradark
Ultradark
oh that is neat
 
Semiclassical
Semiclassical
(you'd want to use a unit circle to get $x'=r^2/x=1/x$)
hi ted
 
Ultradark
Ultradark
@Semiclassical that's really cool
 
Semiclassical
Semiclassical
yeah
 
Ted Shifrin
Ted Shifrin
Hi @Semiclassic
 
Semiclassical
Semiclassical
4:05 PM
and as that article suggests, inversive geometry leads to some pretty interesting math beyond just "2D euclidean geometry"
 
Ted Shifrin
Ted Shifrin
It's really interesting what happens to the triangle inequality when you rewrite it with inversive geometry.
It gives you a famous fact about lengths in a quadrilateral.
 
Semiclassical
Semiclassical
oh, nice
and apparently inversive geometry is related to why this figure works:
[en.wikipedia.org/wiki/Peaucellier–Lipkin_linkage](Peaucellier–Lipkin linkage)
(note that none of the line segments change in length at all)
not sure how to fix that link
(I got a bit tired of having that animated gif in my eyes)
 
Ultradark
Ultradark
semi, is there a study based on a transformation that preserves the order of elements in a set
like you can shift all elements to the right one unit
 
Semiclassical
Semiclassical
As in, mappings $f$ such that $a<b\implies f(a)<f(b)$?
 
Ultradark
Ultradark
yeah i guess so
the transformation only needs to preserve the order of the elements (dilation is allowed)
 
Semiclassical
Semiclassical
4:14 PM
the terminology would regardless probably be monotonic: en.wikipedia.org/wiki/Monotonic_function
 
ALannister
ALannister
I asked why Gauss-Newton doesn’t converge for my function also stating I was fully aware that Newtob’s Method did. I had a bounty up. So what does this guy do?? He posts an answer with code for Newton’s method in it. Uggggh
 
Semiclassical
Semiclassical
note the bit on "monotonicity in order theory"
 
ALannister
ALannister
I don’t think I’m asking too much to have answers to my questions be on f-ing topic
 
Ultradark
Ultradark
yep I noted that :)
I wonder if there's a nice functional equation where the fixed points are $(N,N)$ and $(1/N,1/N).$
but then you can perform some good monotonic mappings on the fixed points
and measure the expansion and contraction
 
 
1 hour later…
MadSpaceMemer
MadSpaceMemer
5:38 PM
Hello is anyone with a masters degree or higher online who knows his stuff with set theory?
I just got an amazing idea.
Would like to run it by someone
 
Alessandro Codenotti
Alessandro Codenotti
5:50 PM
I know some set theory, depending on what you need
 
MadSpaceMemer
MadSpaceMemer
So i just was thinking. If two sets have the same cardinality, follows their powersets have the same cardinality. but that means if natural numbers and the positive integers have the same cardinlity thus also their powersets. So we have 2^N = 2^Z+ however if I want to add the begative integers, then the monotony of the addition doesn't allow the equation 2^N = 2^Z+ + 2^Z-
I know the main proof for why N and Z are same cardinality, I just got this idea and It seemed wierd.
 
Alessandro Codenotti
Alessandro Codenotti
Why not? 2^N=2^Z+ +2^Z- is true
Note that 2^Z+ +2^Z-=2^Z
 
MadSpaceMemer
MadSpaceMemer
that doesnt seem quit right?\
 
Semiclassical
Semiclassical
Infinity * 2 = Infinity isn’t absurd when it comes to set theory
 
MadSpaceMemer
MadSpaceMemer
if I ignore the concept of infinity and the bijective functions, the monotonie doesn't allow this addition
 
Alessandro Codenotti
Alessandro Codenotti
5:54 PM
Why not? For infinite cardinals A and B we have A+B=max(A,B)
 
Rithaniel
Rithaniel
Cardinal arithmetic is a bit different.
 
MadSpaceMemer
MadSpaceMemer
Oh really? didnt know that.i am newb student
Got excited about a wild idea, well nvm :D
 
Semiclassical
Semiclassical
cardinals are weird, so cardinal arithmetic is also weird
so, here's something weird
 
MadSpaceMemer
MadSpaceMemer
I am not convinced of these proofs I have. That R is uncountable and Z and Z are . It's logical and also illogical. Since you can say "stupidly thought" Z is like double N right? but they are still same size, however, R is bigger. since no bijection exists. but my mind keep saying, who cares? since N is anyway unlimitedly large and R as well
 
Rithaniel
Rithaniel
The weirdness with infinities arises from the fact that the most concrete definition of two infinite sets having the same "number of elements" relies on the existence of a bijective function between the sets.
 
Semiclassical
Semiclassical
5:58 PM
I'm looking at a linear algebra chapter "Chapter 5: Rank of the Partitioned Matrix and the Matrix Product". pretty boring
but here's the quotation that the author supplied for the start of the chapter: "More than any other time in history, mankind faces a crossroads. One path leads to despair and utter hopelessness. The other, to total extinction. Let us pray we have wisdom to choose correctly. -Woody Allen: My Speech to the Graduates"
wtf?
 
MadSpaceMemer
MadSpaceMemer
lol
Rithaniel more like not one that we can construct, what is the proof for R being overcountable? kantor diagonal right?
Well I always argued that that special number you construct to disprove the bijection I assign the number sixty trillion million to! huh!
Not quite mathematical, but somehow I keep thinking that they should be same size.
However my mind is simple.
To be honest I dont even know why I bother so much with these stupid futile questions, since I dont even study mathematics. :D
 
Slereah
Slereah
quick question
If I have a line bundle over $\mathbb{R}$, and a section $x(t)$, what would the action of an automorphism of the bundle $\mathbb{R}^2$ be?
I want to say something like $f(x(g(x,t), t)$ but it feels a bit wonky
 
Ultradark
Ultradark
what if every number had a velocity
 
Mike Miller
Mike Miller
6:16 PM
$\text{Diff}(\Bbb R) \times C^\infty(\Bbb R, \Bbb R^\times)$, with $(\varphi, f)$ acting on $\Bbb R^2$ by $(\varphi,f)(t,x) = (\varphi(t), f(t) \cdot x)$. The fact that you can define an automorphism as a map to $\Bbb R^\times$, instead of as a section of some bundle, is because the automorphism bundle of a line bundle is canonically trivial --- the only automorphisms of a line are scaling.
 
Ultradark
Ultradark
like 2 would have velocity 2
and 3 would have velocity 3
 
Slereah
Slereah
@MikeMiller thx
 
Mike Miller
Mike Miller
Your section is irrelevant, but if you want to know how this acts on a section $(t, x(t))$, you just plug into the formula for the action on the bundle: the new section is $(\varphi, f)(t, x(t)) = (\varphi(t), f(x(t)) \cdot x(t))$. This is not written in the correct form (the right should be written as a function of $s = \varphi(t)$, probably), but you may rewrite it as $$(s, f(x(\varphi^{-1}(s)) \cdot x(\varphi^{-1}(s))$$ if you want
And I misspoke above for what it's worth, it's not the product, it's the semidirect product of $\text{Diff}(\Bbb R)$ and $C^\infty$, with Diff(R) acting by pushforward
Anyway this is probably all worthless bullshit language to you
 
Slereah
Slereah
Well it's not trivial but I'll try to work through it
 
Slereah
Slereah
6:51 PM
Just got that dumbass idea to try to work out the symmetries of an action from its most general transformation
So I'm trying to work it out for the 1D free particle
Hopefully that's the 1+1D Galilean group!
 
anakhro
anakhro
7:20 PM
@RyanUnger what does it mean for some condition to be an "open condition" in the space of, say, 1-forms (with C^1 topology).
Does it mean that for any $\eta$ satisfying the condition, there exists an open neighbourhood in the space of 1-forms satisfying the condition?
 
Ultradark
Ultradark
7:38 PM
mathematics is just a way to supply more mathematics to mathematicians
corollaries?
 
anakhro
anakhro
The study of $x$ is just a way to supply more of the study of $x$ to people who study $x$.
I don't think the statement holds any more meaning than can be assigned from the fact "mathematic" occurs in it three times.
 
Semiclassical
Semiclassical
I'm trying to remember old analysis stuff. Suppose one is trying to determine continuity of $f(x,y)$ at $(x,y)=(0,0)$. It's easy to come up with examples where $f(x,ax)\to g(a)$ as $x\to 0$ (and therefore $f$ not continuous). It's also not so hard to come up with examples where $f(x,ax)\to 0$ for all $a$ as $x\to 0$ but $f(x,a x^2)\to g(a)$ and therefore blah blah blah
and more generally it's not hard to come up with stuff like $f(x,ax^p)\to g(a)$ for some power $p$
How much weirder can one get, though? That's all I remember but it feels like only scratching the surface
 
Tobias Kildetoft
Tobias Kildetoft
As far as I recall, it can work along any curve and still not be continuous
 
Semiclassical
Semiclassical
i feel like i've seen that before too
this answer suggests $$f(x,y)= \frac{(e^{-y^{-2}}-x)^2}{e^{-2y^{-2}}+x^2}$$ for $(x,y)\neq 0$
 
Ultradark
Ultradark
can anyone help me with intersection theory?
nevermind I figured it out
 
ALannister
ALannister
8:15 PM
Anybody here know anything about the Gauss-Newton method who could tell me why it doesn’t seem to be converging for a particular function?
 
Semiclassical
Semiclassical
@ALannister link, for reference?
I probably can't help either but I'm curious
 
Ted Shifrin
Ted Shifrin
You can make up functions @Semiclassic where everything looks fine on all curves $y=cx^k$ (all $c$ and $k$) but the function is still not continuous at $0$.
 
Semiclassical
Semiclassical
Yeah, the example in the answer I linked purports to work for any algebraic function $y=p(x)$
 
Ted Shifrin
Ted Shifrin
@Tobias: No, if it works along all continuous curves, then the function must be continuous.
I assigned that exercise to my multivariable math class as extra credit. :)
 
Semiclassical
Semiclassical
the point being, presumably, that "converges along all continuous curves" is pretty bloody strong
 
Ted Shifrin
Ted Shifrin
8:28 PM
I can write down something easier, @Semiclassic, if you'll allow a piecewise definition.
 
Semiclassical
Semiclassical
nah, it's fine.
 
Ted Shifrin
Ted Shifrin
Hint: Make it be $1$ for $0<y\le e^{-1/x^2}$, $0$ elsewhere.
 
Semiclassical
Semiclassical
@ALannister Here's what I notice about your function, which may be related to what's been noticed in comments: The Hessian has a zero eigenvalue at $\hat{x}=(1,1,1)$, with eigenvector $(4,5,3)$. That leads to the following: $f(1+4\epsilon,1+5\epsilon,1+3\epsilon)\approx f(1,1,1)+C\epsilon^4$ where $C$ is some positive constant
 
Ted Shifrin
Ted Shifrin
@anakhro Yes.
 
Semiclassical
Semiclassical
Which means that, along the $(4,5,3)$ direction, the function is very flat.
So probably an instructive example is to consider how Newton or Gauss-Newton would work for something like $f(x)=x^4$
 
Ted Shifrin
Ted Shifrin
8:35 PM
Demonic @Alessandro: This should be easy for you to answer. ;)
 
Semiclassical
Semiclassical
actually, this is sorta cute: Newton's method applied to $f(x)=x^p=0$ (assuming $p>1$ and $x_0>0$) gives $x_n=(1-p^{-1})^{n}x_0$
That still converges, but a good deal more slowly than Newton's method near a single root
upshot being that this kind of problem is one where Newton's method should do badly (and so Gauss-Newton as well, to the extent that those two methods are related)
blah, typo in the above: the relevant direction is (1,1,1) not (4,5,3). (had a small error in my mathematica)
 
 
3 hours later…
johnny09
johnny09
11:23 PM
if a polynomial function is increasing, then is it also strictly increasing?
 
Leaky Nun
Leaky Nun
@johnny09 sure
 
johnny09
johnny09
Cool! Thanks for your reply!
 
ALannister
ALannister
11:40 PM
@Semiclassical so do you think I should try again but with more iterations?
@johnny09 the only time something that is increasing is not also strictly increasing is if it's constant anywhere, because then it's $\leq$ instead of $<$.
 

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