12:00 AM
and plot the parameter space
you get many points
of course these points don't form an algebraic structure like a ring or field or group
but the $x$-axis itself is comprised of a subset of $\Bbb A$ (algebraic numbers)
$x$ is necessarily algebraic if $s,t \in \Bbb N$
and the algebraic numbers are closed

12:15 AM
@Ultradark When RHS = LHS implies p = q where p and q are polynomials (when s and t are natural).

what's p

I think I get what you're saying

okay yeah I just wanted to explore that a little
thanks for listening

But when you say the x-axis is itself comprised of a subset of $A$ what does that mean

@J.Doe I just mean since we're working in $x \in (0,1)$
a subset of the x-axis

12:28 AM
I mean there are all kinds of numbers in $(0,1)$. That the the solutions are algebraic comes from that they satisfy polynomials with coefficients in Q or Z, right?

yeah and natural number powers
if you just start with $x^t-(1-x)^s=0$
and plot the zero set
for the parameter space
I think you'd have $P(\Bbb A,\Bbb A)$
points with algebraic numbers as coordinates

0

I have rotations that are in Euler form only. All I would ever have to do to them mathematically is interpolate two rotations. Nothing else. The person providing them refuses to put them in a convenient form other than Euler angles because they believe that two interpolate two rotations you just ...

Help

@J.Doe but if you were to start with $2^{\frac{s}{\log_2(x)}}=2^{\frac{t}{\log_2(1-x)}}$
you'd get points of the form $P(\Bbb A, ?)$
so if they are equivalent equations
why can't I determine the nature of the y-coordinate for that one equation?

12:44 AM
But couldn't you actually solve this
Since it's over (0,1)
Use trig parametrisation
cos^2m = sin^2n etc

solve for the ?

1:09 AM
I'm thinking that the two representations are different
and there's a map between the two in the form of algebraic manipulations
What does simplifying really mean

When you say say $p(A, ?)$ what is $?$ referring to? The $y$ coordinate? Well define y well first.
Give me $y=...$ I'll give you your $y$ coordinate. In one sense you're talking about the $y$ coordinate of a single side of the equation $f(x)= g(x)$ as in $y=g(x)$ or $y=f(x)$; but you're also talking about $y$ in the sense of $y := f(x)-g(x)=0$.
Anyway, I'll stop writing nonsense since I really don't know what I'm talking about.

1:24 AM
Me too :)

:]

1 hour later…
2:51 AM
Can you always use Reduction of Order if you know a solution to a linear ODE, or do you have to get lucky for the right variable to cancel so that the order reduces?

3:06 AM
I was trying to reduce $y'' - y' + y = e^x, y_1 = e^x$ but when I plug in $ve^x$ I get $v'' + v' + v = 1$, which can't be reduced. I was under the impression that there should never be a $v$ term after simplifying.

3:19 AM
@user10478 when doing reduction of order, you let y= v y_1 where y_1 solves the homogeneous ODE

Ohhh
So I'm going to end up with $v'' + v' + v = 0$ I think, which still isn't reducible?

No, you’ll get a different ode for v(x)
One which shouldn’t have a v(x) term

Okay, thank you for clearing that up.

But I’m not sure reduction of order is the right approach here. y’’-y’+y=0 has two linearly independent solutions, and e^x is a particular solution to the inhomogeneous ode
So you should just have y=e^x + c1 y1 + c2 y2 as the general solution

So reduction of order is only for homogeneous linear equations.

3:26 AM
It’s not, but it’s more useful in cases where the second linearly independent solution isn’t obvious
Linear DEs with constant coefficients are easy enough
But when you start getting Bessel function solutions, then reduction is handy
It’s a useful tool for inhomogeneous ODEs, but in this case it’s not needed due to how simple the homogenous ode’s solutions are

Well you need reduction of order implicitly when solving a constant coefficient linear equation where the characteristic equation's discriminant is $0$, right? It's just that the answer is always $y_2 = ty_1$.

That’s true, I was being careless

When you say reduction of order is useful for inhomogeneous ODEs, do you mean just for solving the associated homogenous ODE?

No, I don’t. Judging from the wiki page, at least: they apply reduction of order to an inhomogeneous second-order linear DE
Okay, take a look at this chapter for more on reduction of order in general: howellkb.uah.edu/public_html/DEtext/Part3/Reduction_of_Order
The upshot being: if you have an nth-order linear nonhomogenous DE, and you know one solution to the homogeneous DE, then you can do reduction of order to get down to a new (n-1)th order linear nonhomogeneous DE

3:45 AM
Okay, so maybe $y_1$ and $y_2$ are poor choices of variables.
Since the two aren't parallel to each other.

Which is neat, but then you’re left with a nonhomogeneous DE for which it may be pretty hard to guess solutions
So reduction of order is most effective in the context of homogenous second-order odes, where the resulting homogeneous first-order ode can be solved by separation of variables
It’s not inapplicable outside that context, but it’s not as helpful

Okay, thanks for the help

2 hours later…

On physical infinity
Suppose I have a box which has volume 3 cm3 and inside, there is another box which measured to have volume 3 cm3
This will be an example of something that has a self injection that is not a surjection
We can go further and test whether the space is eulidean by checking how parallel lines behave near the boxes. If the space is euclidean, then it is more plausible that the box behave like infinity at least for objects of size less than 3cm3
Non-well-founded set theories are variants of axiomatic set theory that allow sets to contain themselves and otherwise violate the rule of well-foundedness. In non-well-founded set theories, the foundation axiom of ZFC is replaced by axioms implying its negation. The study of non-well-founded sets was initiated by Dmitry Mirimanoff in a series of papers between 1917 and 1920, in which he formulated the distinction between well-founded and non-well-founded sets; he did not regard well-foundedness as an axiom. Although a number of axiomatic systems of non-well-founded sets were proposed afterwards...

6:43 AM
I really don't like algebraists.
So I go to this talk, where the dude is supposed to talk on the Yang Baxter equation and it's set theoretical analogue with its connections to Braid groups and knot invariants. It turned out to be just a result on the solvability and nilpotency of a slew brace using the characterization of finite simple groups.
Algebraists draw out geometers and other mathematicians by talking about applications and then feed them finite group theory.

0

Question: Can we have a model of $ZF-\text {Regularity}$ where there exist an ordinal $\kappa$ such that $H_{\kappa}$ exists and $H_{\kappa}$ is not equinumerous to any well founded set? The motivation for this question comes in connection with defining Cardinality under some situations beyo...

Hmm... so quine atoms may look infinite when it is unravelled, but they actually do not have a well defined notion of cardinality
Aug 15 at 3:00, by Secret
More reflections on infinity as Chapter 5 of the book is just reached: Unreachability is not a unique trait of infinity. Dedekind finite sets can be indefinitely reduced in cardinality but the process will not complete in finite steps
So to update on this:
Let $X$ be some object with some properties $x$
Let $P$ be a process which outputs some objects or properties when given $x$
If there exists some $P$ such that given a property $x$ in $X$, it can be guarentee that $P$ continues and each output is distinct, then $X$ has some notion of being actual infinity as it contains what is effectively the output of a nonrepeating and neverending process
Example:
Let $U$ be the unravelling process which takes in a set equation $S$ and output the outcome when everything on the left of the equality is replaced by the corresponding thing on the right:
For example the quine atom: $X=\{X\}$
Unravelling gives: $U(X=\{X\})\implies U(X=\{\{X\}\})\implies U(X=\{\{\{X\}\}\})\implies,\cdots$
This process is nonrepeating and nonterminating, and yet it is clearly contained in $X$. Therefore $X$ is actual infinite wrt the unravel operator
Meanwhile let $A$ be the operator that takes a set and convert it into an accessible point graph
$A(X)$ terminates in one step to give a 1-cycle, and hence $X$ is finite wrt $A$
Thus, a process P that is guarentee to continue for each step and give distinct outputs at each step necessarily generates a potential infinity, and some object that contains all the output of this process is an actual infinity for P
In other words, one type of actual infinity is the completion of P
Apr 7 at 17:04, by user193319
I'm just talking about a universe which "goes on" forever and the states of the universe repeat, in accordance with Poincare' theorem.
Apr 7 at 17:14, by Secret
Anyway, I don't know if I will call a process that produce a sequence like 1,2,3,1,2,3,1,2,3,... as potentially infinite. Because you can produce the same sequence by cycling through the elements of the set {1,2,3} indefinitely, or you actually have an ordered tuple (1,2,3,1,2,3,1,2,3,...) and you move to the next element for each step. The latter case will be your suggestion of the Poncaire' theorem scenario. Otherwise, going to spent some time to read that book before thinking about this
Trying to capture sequences that repeat themselves at some point remains a challenge, as well the behavior of lie paradox and Yablo paradox
It is clear what the necessary condition for something to be actual infinite is now, but it is not clear what the sufficient condition is
That a notion of completion or closure over a potential infinity is necessary to define actual infinity, may open the door to proving there cannot be a predicative actual infinite object, because Godel's incompleteness theorem may appear in some form
We knew that a formal system that is sound and consistent, cannot prove nor disprove its own consistency, hence incomplete
Perhaps, a special case of this is given any potentially infinite process, we cannot construct its completion with the underlying formal system

7:44 AM
Let $P$ be a process which may or may not be terminating. If $P$ does not terminate (Given any step of execution of $P$, $P$ can continue its execution) then $P$ represents a potential infinity. An object $A$ such that given any output of a potentially infinite $P$ it is found in $A$ is an actual infinity wrt $P$
Thus $A$ is a completion of $P$

Morning all

Actually, to take account of things like successor ordinals such as $\omega +1$, maybe it needs to be defined this way:
Nontermination: A process $P$ is nonterminating if given a sufficiently large part of its execution history, there is a portion of its history such that given any step of execution of $P$, $P$ can continue its execution
For example a $P$ that generates $1,2,3,4,...,\omega$ is nonterminating since the subsequence $1,2,3,4,...$, $P$ can continue its execution at any point in this sequence without interruptions
That should capture everything from cyclic sequences, amorphous sets, infinite dedekind finite sets, alephs, other cardinals, uncomputability and so on
Thus a concrete example is the set of naturals, there exists no terminating process that can enumerate all naturals, even though there exists a finite representation for such a process, namely the rules of the successor operator given by peano axioms
> This can be very shocking to those people who are first introduced to the technical term “actual infinity.” It seems not to be the kind of infinity they are thinking about. The crux of the problem is that these people really are using a different concept of infinity. The sense of infinity in ordinary discourse these days is either the Aristotelian one of potential infinity or the medieval one that requires infinity to be endless, immeasurable, and perhaps to have connotations of perfection or inconceivability. This article uses the name transcendental infinity for the medieval concept alt
For those who think I have not spent my last 5 months reading the philosophy of infinity
> Dedekind’s new definition of "infinite" is defining an actually infinite set, not a potentially infinite set because Dedekind appealed to no continuing operation over time. The concept of a potentially infinite set is then given a new technical definition by saying a potentially infinite set is a growing, finite subset of an actually infinite set.
The problem of this in the context of predicative actual infinity is that we don't know if there exists formal systems which dedekind's definition of infinite is a theorem rather than an axiom, without employing stronger axioms such as there exists a universal set, as it is done in new foundations

8:32 AM
Math people
Given the Euclidian axioms, is there a unique Riemannian manifold that obeys them?
2
ie $(\mathbb{R^2}, \delta_{ij})$
I'm pretty sure that's true though it's a bit tricky to show because Euclidian axioms are a bit vague sometimes
I thought $(D^2, \delta_{ij})$ might also obey them but I think that's not actually true because there are not "circles of any sizes", ie you can find two points separated by a length superior to the radius of any possible circles
I'm pretty sure the first two axioms imply that it's geodesically complete and the last one that it's flat, but it's a bit tricky
Kind of depends on how one defines the elements
ie is a "line" just an inextendible geodesic or does it need to be infinite
and does it change what manifold can obey them

3 hours later…
11:34 AM
@AlessandroCodenotti I think I just found an inconsistency in Italian orthography
but idk if you would count it as such, since it's in a name
the "i" in "Lucia" is pronounced

hey
so i have a question
if you square both sides of an equation, is it still the same equation
for example we are studying about order and degree of a diff. equation
and the definition os degree is : Power od highest order derivative in a diff. equation without fraction or radical
and below that there is this example
here, to get rid of the radical sign we squared the equation
but will it still be the same equation? If the new equation we made by squaring original one have degree 2 then how can we say the original equatiin has degree 2?

1 hour later…
1:03 PM
Can you guys help me in this question?math.stackexchange.com/questions/3329862/…

1:42 PM
Sure no-one cares except maybe @LeakyNun :P but I made another ordinal collapsing function:
$\displaystyle S(A)\Leftrightarrow\forall f:\sup A\mapsto\sup A,\exists\alpha\in A,\forall\eta\in\alpha(f(\eta)\in\alpha)$

$\displaystyle\mathrm B(\alpha,\kappa)_0=\kappa\cup\{0,K\}$

$\displaystyle\mathrm B(\alpha,\kappa)_{n+1}=\{\gamma+\delta~|~\gamma,\delta\in\mathrm B(\alpha,\kappa)_n\}$

$\displaystyle\hphantom{\mathrm B(\alpha,\kappa)_{n+1}={}}{}\cup\{\Psi_\eta(\mu)~|~\mu\in\mathrm B(\alpha,\kappa)_n\land\eta\in\alpha\cap\mathrm B(\alpha,\kappa)_n\}$

$\displaystyle\mathrm B(\alpha,\kappa)=\bigcup_{n\in\mathbb N}\mathrm B(\alpha,\kappa)_n$
where K is a weakly compact cardinal and K' > K is closed under γ ↦ M(γ), the first γ-Mahlo.

@SimplyBeautifulArt what's $S$?

It's a proposition defined right there :P
($A$ is a set of ordinals)
It's basically whether or not $A$ is a stationary set, though I prefer this definition opposed to the usual.
You can read it on an intuitive level as "every function has an $\alpha\in A$ where it gets stuck at", which has its immediate implications as far as ordinal collapsing functions are concerned.

is there an example?

Oops there is a minor typo in my definition of $\Xi$
It should have $$S\bigg(\bigcap_{\eta\in\mathrm B(\alpha,\kappa)\cap\alpha}\Xi(\eta)\color{red}{{}\cap\kappa}\bigg)$$
@LeakyNun consider $A$, the set of countable additive principles $\{\omega^\gamma~|~\gamma\in\omega_1\}$.
Every function $f$ mapping countable ordinals to countable ordinals has an element $\alpha$ of the above set where $\eta\in\alpha\Rightarrow f(\eta)\in\alpha$
This result follows from the fact that $\omega_1$ is regular.

2:04 PM
hi some know how to choose a sequence to prove that this space is not a Banach space $(\mathcal{C}([0,\frac12],\mathbb{R}),||.||_1)$

3 hours later…
5:33 PM
Hi @Ted
Hi @ÉricoMeloSilva!

Henlo @Balarka @Ted

Hi @ÍgjøgnumMeg

If a Riemannian metric on a manifold $M$ admits a complete global orthonormal frame field, then is manifold complete?

@BalarkaSen yo

5:54 PM
Hi, a @Balarka and @Eric.
@s.harp: What is a "complete global" frame field?
So you're assuming that the manifold has trivial tangent bundle?
Oops, hi @ÍgjøgnumMeg

@TedShifrin yep, there are complete vectorfields $e_1,... , e_n$ defined on all of $M$ with $g(e_i, e_j)=\delta_{ij}$

So it's redundant. You only need to say global orthonormal frame.

which part is redundant? not every global orthonormal frame is complete (as an exampe: $(0,1)$ with $\partial_t$ )

I see.

i thought about it before, but I never really got anywhere :P

5:59 PM
The problem with that will come with the map's having rank 0 at some points.

I can't believe that the statement of the question is true, but I don't understand what he means with the first and second derivatives being independent

It's confusing. The first-order partial derivatives are everywhere linearly dependent.

oh dependent
now i can start believing

If the map has rank 1 everywhere, then I'm sure we can get it out of the rank theorem. But if the rank drops to 0 along some curves ...

6:17 PM
the progression of answers in this question amuses me: math.stackexchange.com/questions/3330146/…
Question: Can I find an example with n=2?
Answer 1: No, and here's why, but an example with n=3 is...
Comment to A1: That's not an example.
Answer 2: You can't do it for n<=3, but here's an example for n=4.
Comment to A2: That's not an example.
Alas, there is a trivial example for n=5 and therefore the pattern can't proceed beyond that.

6:58 PM
Since the chat has been talking about polynomials, here are some exercises that tell a fun story:
1. show that there is no injective polynomial $\Bbb C\times\Bbb C\to\Bbb C$ in $\Bbb C[x,y]$
2. show there is no injective polynomial $\Bbb R\times\Bbb R\to\Bbb R$ in $\Bbb R[x,y]$
however:
3. there exist injective polynomials $\Bbb Q\times \Bbb Q\to \Bbb Q$ in $\Bbb Q[x,y]$
4. it is unknown whether or not there exists a polynomial bijection $\Bbb Q\times \Bbb Q\to\Bbb Q$
(point 3 is conditioned on a conjecture though, see arxiv.org/abs/0902.3961 )

7:21 PM
@SimplyBeautifulArt Oh derp, I almost forgot, @Secret is probably interested in the above ordinal collapsing function.

1 hour later…
8:31 PM
@s.harp do you know of any transformations that map linear functions to vertical lines?
for example $y=x$ would map to $x=1/2$
i need it to be an isometric embedding, and compactification
but specifically lines of the form $y=x+b$

9:18 PM
Was Eisenstein An atheïst ?

9:47 PM
@mick indeterminable. Usually he gets placed on the "impersonal" deist side of things rather than a/theist.
@TedShifrin do you know anything about "generalized" complex and Kahler geometry?

I have never heard the term.

Me either. Some buzzwords from the course description: Dirac structures, sigma models, Gerbes, Courant algebroids, Gray-Hervella classification, Hitchin functional, Neveu-Schwarz fluxes, D-branes, etc.
Only D-branes I have previously heard of. And "algebroids", but not Courant.

This sounds like physics ...

"This is an introductory (i.e. first year graduate students are welcome and expected) course in generalized geometry, with a special emphasis on Dirac geometry, as developed by Courant, Weinstein, and Severa, as well as generalized complex geometry, as introduced by Hitchin. Dirac geometry is based on the idea of unifying the geometry of a Poisson structure with that of a closed 2-form, whereas generalized complex geometry unifies complex and symplectic geometry."
Yeah this sounds physicsy

Interesting. Hitchin's been around quite a while, but I've never heard any mention of this.

9:57 PM
ocw.mit.edu/courses/mathematics/18-969-topics-in-geometry-dirac-geometry-fall-2006/lecture-notes/18_969_geometry.pdf if you were interested in seeing the notes for this course.
I just stumbled upon it.
Maybe I will have to actually look at the notes.
;)
Anyways I am going to get some food, hope you are well, Ted!

It's not physics
In my experience I wouldn't call it very geometric
I doubt you'd like it

Whose lectures notes are these? A first-year graduate student, even at MIT, wouldn't have a clue in such a course. When I taught a topics in geometry course at MIT, I had all advanced grad students in it (some from Harvard). Ridiculous to say first-years are "expected."
Hmm, never heard of Gualtieri.
His webpage at Toronto says he works in math physics (and differential geometry).