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2:47 AM
any one has a symbolic math software
1
Q: How can we solve the expression explicitly for $X$ in terms of $Y$?

BAYMAXI am thinking how to write $X$ explicitly in terms of $Y,A,B,C$? I have $AX^3 + X^2(B-1) + X(-C) + \alpha = Y$ I thought of using symbolic maths but could not find any. Any help is nice!

I am trying this but have no software, any help?
 
3:47 AM
Can there exist a continuous function $f:\mathbb R \to \mathbb R^2$: $f(\mathbb R)=\{(x,y) \in \mathbb R^2: x^2+y^2\}$?
Can there exist a continuous function $f:\mathbb R \to \mathbb R^2$: $f(\mathbb R)=\{(x,y) \in \mathbb R^2: x^2+y^2=1\}$?
Since, $R$ is not compact. But unit circle is compact.
 
4:04 AM
@N.Maneesh Why no simply $f(t)=(\cos t,\sin t)$?
BTW Math Meta Chat is not really a room for this type of question.
 
@MartinSleziak sorry!
Thank you.
 
4:27 AM
If anyone there, I think the answers are pretty obvious and they would be B and C but..
Well, Is anyone there?
 
5:08 AM
@user21820 That brings out the important point on how we really have no way to distinguish between 1) a potentially infinite process and 2) a process that is finite but looks never-ending because it is much larger than our personal tolerance before we start labelling something as going forever
> Anyway, we started getting really curious about this South Dakota Tractor Museum — it sounded sort of funny. It took 250 kilometers of driving before we passed it. We wouldn’t normally care about a tractor museum, but there was really nothing else to think about while we were driving. The only thing to see were fields of grain, and these signs, which kept building up the suspense, saying things like

ONLY 100 MILES TO THE SOUTH DAKOTA TRACTOR MUSEUM!
 
I don't know what you're talking about...
 
@user21820 I mean, I am just guessing, there must be a reason why when he is talking about ordinals, he keep going back to the South Dakota Tractor Museum and keep commenting about keep on seeing those road signs
To an ordinary person who is driving in a long road, it feels no different between seeing 100 signs vs seeing 10000000000 road signs along the way, it still will feel like the road is going forever
That said, I actually don't know what motivate you to talk about that link back above all of a sudden
so my guess may be way off
 
5:25 AM
@Secret It's just a joke. If you read through the series you will know why.
 
Ah...
 
That link was just part 1.
It's not often that a mathematically correct article about ordinals is actually amusing.
 
true
 
 
3 hours later…
8:39 AM
Hmm.. it does sounds like the joke is strongly correlated to which stage of the ordinal is in, for example "run out of gas" coincides with "ran out of ordinal notation when getting to $\epsilon_0$"
Also chapter 3 is not published back in 2017 when I first read this blog
 
 
1 hour later…
9:47 AM
---
(Unrelated)
https://math.stanford.edu/~feferman/papers/predicativity.pdf
Apr 23 at 4:31, by Rithaniel
Here's a question for you: We know that no set of axioms will ever decide all statements, from Gödel's Incompleteness Theorems. However, do there exist statements that cannot be decided by any set of axioms except ones which contain one or more axioms dealing directly with that particular statement?
Impredicativity thus mean: An object $M$ cannot be defined without using $\phi(V)=M$ where $\phi$ is some definable function and $V$ is the universe of the foundation in question
In other words, an object is impredicative if it cannot be defined "from below"
9
A: How can we know we're not accidentally talking about non-standard integers?

user21820 We know that any formal system cannot completely pin down the natural numbers. Incidentally, I said exactly this here. Besides what I said in that post, I wish to elaborate on the following points: A generalized version of the Godel-Rosser incompleteness theorem shows convincingly that the...

and it gets worse, you can have systems where the proof that demonstrate an inconsistency within the system is too long to fit within some specified finite number of words
Thus the question of "predicative infinity" can be rephrased as "Given any foundation F, is the following predicative "There exists a formula $\phi$ not involving $V$ such that $\phi \vdash m$ where $m$ is infinite" logically equivalent to Godel Rosser incompleteness theorems"
> If you want to read more about this, it helps to know that a function from ordinals to ordinals that’s continuous and strictly increasing is called normal. ‘Normal’ is an adjective that mathematicians use when they haven’t had enough coffee in the morning and aren’t feeling creative—it means a thousand different things. In this case, a better term would be ‘differentiable’.
Starting to wonder is the set of all normal objects not normal
 
10:07 AM
If $E$ is a subfield of $F(X)$, and $E$ contains some "rational polynomial" $f(X)/g(X)$, where $f(X),g(X) \neq 0$, does it follow that $E$ contains a "regular polynomial"; i.e., something in $F[X]$?
 
$g(X)$ can be constant polynomial or $g(X)$ divides $f(X)$?
both situations should give an element of $F[X]$
 
10:30 AM
@Secret Why must $g(X)$ divide $f(X)$?
 
Not necessary, it really depends on what $E$ is. But given so little info on what $E,F$ is (in particular, $E$ is not necessarily obtained as a quotient of $F(X)$), I don't see why you can rule out rational polynomials where $g(X)$ divides $f(X)$
 
Well, $F$ is a field of characteristic $0$ and $E = F(X^2) \cap F(X^2 - X)$; but I'm hoping what I claimed holds more generally.
 
10:50 AM
I am suppose to show that $E = F(X^2) \cap F(X^2 - X) = F$, but I'm beginning to wonder whether this is false.
 
11:08 AM
I don't think polynomials involving $X^2, X^2-X$ are likely to be linearly dependent, but to check whether $ab=c$ where $a,b$ are polynomials in $F(X^2)$ and $b$ are polynomials in $F(X^2-X)$ I don't know how it can be checked
typo: $c$ for the final $b$
 
11:28 AM
i.e. you need to check the algebraic independence of the elements in $F(X^2)$ and $F(X^2-X)$ to check if the intersection is the whole thing
which for polynomials this can be done by computing the determinant of a matrix containing the polynomials as its entries, but I am not sure if something different is needed for general polynomial fields
 
@Secret Not sure if you're making this mistake, but $F[X]$ and $F(X)$ are different things: the former is the polynomial ring over $F$, the latter is the fraction of field of $F(X)$.
 
you are right, I was talking about $F[X], F[X^2-X]$, in that case I am not sure as I am not really good with field of fractions
 
Sorry, "...the fraction field of $F[X]$"
I can show that the intersection doesn't contain any "proper" polynomials using a little bit of Galois theory, but I don't know how to show that it doesn't contain rational polynomials.
 
12:10 PM
Well...perhaps we can figure that out later...What does $\text{Aut }(E/F)$ denote?
Is it the group of all automorphisms of $E$ fixing the subfield $F$?
 
uh I thought $E$ is the subfield of $F$ according to what you said above?
 
Yeah, I'm working on something completely different now...sorry for the confusion in notation.
 
12:25 PM
it's ok
 
If $|E :F|$ is finite, do I know that $\text{Aut }(E/F)$ is finite?
 
12:39 PM
That is way beyond my knowledge field, goes you might have to ask other field theorists
 
Yes, |Aut(E/F)| is at most |E:F|. Let i : F -> E be the inclusion map. Fix an embedding x : F -> bar(F) and consider the set of all F-embeddings s : E -> bar(F) such that the diagram involving i, x and s commute. Call this set B. Aut(E/F) acts on an element s of B by precomposing an F-automorphism of E with s. The stabilizer of this action is trivial, so by orbit-stabilizer, size of every orbit is |Aut(E/F)|, so |Aut(E/F)| divides |B|.
E/F is an algebraic extension as |E : F| is finite (just take any element alpha in E and keep looking at 1, alpha, alpha^2, ... until you get a linear relation which you will as E is finite dimensional over F), so any embedding E -> bar(F) over a fixed embedding x : F -> bar(F) is determined by choice of an F-basis of E
This means |B| = |E:F|. So you get even better, |Aut(E/F)| divides |E:F|
 
12:58 PM
Ah, very nice! Thanks @BalarkaSen
 
1:18 PM
I suppose a slick argument for |B| = |E:F| is to choose a primitive element alpha for E/F, then E = F(alpha). The F-embeddings s : E -> bar(F) over x is determined by the image of alpha, which can be any of the deg(alpha)-many roots of the minimal polynomial of alpha over F, so there are deg(alpha) = |E:F| many embeddings.
Note to myself: Basically I was trying to do a covering space argument, i : F -> E gives a "covering map" Spec E -> Spec F whose degree is |E:F|. An embedding x : F -> bar(F) is really a basepoint Spec bar(F) -> Spec F, and s : E -> bar(F) over x is a lift of the basepoint to Spec bar(F) -> Spec E, one of the points in the preimage of x in the covering space upstairs
Aut(E/F) is the Deck transformation group of Spec(E) -> Spec(F), which acts faithfully on the set of preimages of x, of which there are |E:F| many
Someone should rewrite all of algebra from a topological perspective :3
 
1:40 PM
0
Q: Example of sigma algebra that is not a topology

user10293123There is a very nice explanation of an example of sigma algebra that is not a topology: here. I do not fully understand the answer. Apparently this is a basic question, but why do we want this sigma algebra to include sets which are uncountable with uncountable complements in order to be a top...

hmm... is there such thing as "non topologicalisable algebraic structures"?
3
Q: What is non-algebraic structure

Lance PollardIf an algebraic structure is a set of operations on a set of elements, what is a non-algebraic structure? https://en.wikipedia.org/wiki/Outline_of_algebraic_structures#Algebraic_structures_with_additional_non-algebraic_structure

Ah, that reminds me of that discussion a year ago...
Maths is roughly divided into the following fields:
Foundations, Number theory, Arithmetic, Algebra, Order theory, Combinitorics, Topology, Geometry, Analysis, Probability, Statistics, Numerical analysis, Computer science, etc.
It is relatively straightforward to work out what is:
not a foundation, not a number, not arithmetic, not algebraic, not an order, not combinatorics, not a topology, not analysis, not probability, not statistics, not numerical, not computer science
Except one
0
Q: Given axioms, how do we know it defines a geometry?

SecretIt is known that besides using coordinates and algebra, there are axiomisation of geometry such as Tarski, Hilbert and Euclid. However looking at the axioms of Tarski for example: Betweeness $B(\cdot,\cdot,\cdot)$ satisfy e.g.: \begin{align} Bxyz &\to x=y\\ (Bxuz \land Byvz) &\to \exists a(B...

There is no nice way to define what is not a geometry, because the concept is too flexible
 
2:08 PM
Problem: Let $(f_n)$ be a sequence of uniformly bounded measurable functions on $\Bbb{R}$ that converge pointwise to $f$. Prove that $f$ is measurable.
Question: What is the point uniform boundedness assumption? Isn't the statement true without that assumption?
"...point of the..."
 
3:08 PM
I think so. For any Borel subset $B \subset \Bbb R$, $f^{-1}(B) = \bigcap_{k \geq 1} \bigcup_{m \geq 1} \bigcap_{n \geq m} f_n^{-1}(B_{1/k})$ where $B_\epsilon$ denotes the union of $\epsilon$-balls around every point of $B$, if I didn't do some silly set theory mistake.
$f_n^{-1}(B_{1/k})$ are measurable for sure, as $B_{1/k}$ are open hence Borel, and $f_n$ are measurable. Now whatever algebra you do with these sets it's going to stay Borel.
The bounded hypothesis is unnecessary
 
3:40 PM
@BalarkaSen !hi
 
Hey
 
4:04 PM
Pretty sure that holds for arbitrary measure spaces without any assumptions apart from pointwise convergence
Hi @Balarka
 
can someone please check this out?
1
Q: If a sequence $(u_n)$ be such that its every subsequence has a subsequence that converges to $0$, then $\lim u_n= 0$

Subhasis BiswasSuppose that $(u_n)$ is unbounded above. Then, we pick an arbitrary monotone increasing subsequence $(v_n)$ of $(u_n)$. But by hypothesis, we can find a subsequence of $(v_n)$ that converges to $0$. Hence, $\lim v_n =0$. Therefore, $(u_n)$ must be bounded above. Again, let $(u_n)$ be unbounded b...

 
@SubhasisBiswas is $(u_n)$ a subsequence of itself?
Oh I misread the question!
 
well, ignoring that part
we cannot consider a sequence a subsequence of itself in this context
 
@AlessandroCodenotti I need the codomain to just be a metric space for my proof to work, I guess
Hi by the way
 
4:28 PM
[Random short]
Short ramble about ordinal sequences before going to sleep after a long day
Let $f$ be a normal function, then let $'$ be the fixed point of $f$
Then we have the following rapidly increasing sequence:
 
Howdy a @Balarka, @anakhro, demonic @Alessandro
 
My favourite Ted.
 
LOL, cuz you have so many Teds.
 
$0, 0'=0^0=1,1'=\omega,\omega'-\phi (2,0),\phi'(0,0)=LVO=\psi ({}^3\Omega),LVO' = BHO = \psi (\epsilon_{\Omega+1})$
 
Hey @Ted!
 
4:34 PM
@Balarka, good grief!.
 
LOL
I borrowed a monograph by Freedman on 4-manifold topology from the library for this summer. I think at the end he has a construction of exotic R^4's
Should read it some time
 
Yeah, I do not know this stuff at all.
 
My roommate asked me recently if Milnor's exotic S^7 has trivial tangent bundle, like the usual S^7. I had to think a lot on that one!
 
Doesn't it have some nontrivial characteristic class?
Hmm, maybe not.
 
There are no topological obstructions, no. Rank 7 bundles on S^7 are classified by $\pi_6 SO(7)$ by the clutching construction, which is indeed trivial. So every rank 7 bundle on S^7 is trivial.
 
4:41 PM
Right.
 
What's that curve that's wacky enough to make people doubt the Jordan curve theorem?
 
Osgood curve
 
Thanks!
 
5:03 PM
@BalarkaSen what would you think the def. of a $z$-invariant vector field would be?
 
Invariant under translation along the $z$-axis
Like you said
 
HMMMMMMMMMM
What is the best way to phrase this.
@BalarkaSen do you like knots?
 
I do but I don't know much about it!
 
Me either.
@BalarkaSen want to hear something cute?
 
Yeah sure
 
5:07 PM
Suppose you have a very small car in $\mathbb R^2$
It does not turn on a dime.
So you have to move forward as your turn.
So you have a direction associated to your car, $\theta\in S^1$.
And coordinates $(x,y)\in\mathbb R^2$
 
Okay
 
So the motion of your car is determined as curves in $\mathbb R^2\times S^1$.
 
Right
 
Because of the limitations of not being able to turn on a dime, these curves are directed by $\frac{\partial}{\partial\theta}$ and $\cos\theta\frac{\partial}{\partial x} + \sin\theta\frac{\partial}{\partial y}$.
 
This sounds like “holonomic constraints” business
 
5:11 PM
Reminds me of Bryant's talk
Ya
 
Of course, the obvious thing (considering it is me telling you about it) is that these vector fields define a plane field in $\mathbb R^2\times S^1$.
 
And the valid motion curves have to be tangent to this plane field for all time, right?
 
Yes.
In particular, the plane field $\xi = \ker(\sin\theta\,dx -\cos\theta\,dy)$.
 
The expenditure of a household on consumer goods, C, is related to the household's income, y, in the following way: When the household's income is $1000, the expenditure on consumer goods is $900, and whenever income increases by $100, the expenditure on consumer goods increases by $80. Express the expenditure on consumer goods as a function of income, assuming a linear relationship.
 
You can check this is a contact manifold we have just made.
 
5:13 PM
Coolio.
 
So your tiny car only travels along Legendrian curves
 
So any curve can be approximated by motion curves
 
Yes. That is, you can always park your little car.
 
By h-principle for Legendrian submanifolds, all motions are possible, more or less :3
Nice interpretation. I like this a lot.
 
There is actually a really nice paper on the proof.
The proof in like every book/set of notes that I have read has been EXTREMELY bad.
 
5:16 PM
Thanks a bunch, I need to read this.
 
"draw a zig zag on the curve, this lifts to a Legendrian curve"
Geiges's proof isn't even any good.
Though that probably shouldn't be a surprise.
 
Hey. Does anybody have a reference to read on the lattice of quotients of a group?
 
@anakhro Yeah I don't actually know the proof. The most general theorem along these lines is that holonomic approximation holds for any locally integrable microflexible Diff-invariant differential relation.
I should read it at some point
It's all in Eliashberg-Mishachev
We could try reading it togather if you want
 
I don't know how well I would follow.
What sort of background might one need?
 
This is Gromov's theorem, and if you're reading Gromov, I'm afraid no background is enough :D
 
5:30 PM
Heh
Well after I get my thesis done I will be free to do next to anything.
 
You could teach me the necessary contact geometry and I could teach you the necessary differential relations formalism maybe
@anakhro Cool!
 
You'd probably rapidly grasp more contact geometry than I do in like a week.
 
Approximately when should you be done with your thesis?
 
I am very slow.
Hopefully mid June at latest.
 
Nah.
@anakhro Oh fantastic. I'll be away in July as a visiting student to learn more about... h-principles! So that'd be a good time.
 
5:35 PM
Nice!
I also need to study for comprehensive exams for first week of september. >:(
 
That sucks.
 
Yeah. Kind of have to relearn all that I forgot.
Especially that algebraic topology nonsense. ;)
But mostly, algebra.
 
triggered
 
5:53 PM
@anakhro I was trying to remember this argument Ted told me some time ago to find paths between any two points in $\Bbb R^3$ tangent to the standard contact structure $dz = xdy$. Suppose I want to join $(0,0,0)$ with $(a,b,c)$, then my path will have to look like $(\gamma_1(t), \gamma_2(t), \int_0^t \gamma_2 \gamma_1')$.
Let $\gamma = (\gamma_1, \gamma_2)$ be the projection of the path downstairs, from $(0, 0)$ to $(a, b)$. I need to choose this so that $\int_0^1 \gamma_2 \gamma_1' = c$, equivalently, $\int_\gamma x dy = c$.
This is easy enough, because consider the straightline segment from $(0, 0)$ to $(a, b)$, call that $\sigma$. $\int_\sigma xdy = ab/2$ (check), so you need to choose $\gamma$ such that $\int_\gamma xdy - \int_\sigma xdy = c - ab/2$. This is the same thing as requiring $\gamma \cup \sigma$ bound a region of area $c - ab/2$ by Green's formula
I think you can modify this idea to get an approximation result
 
You can easily lift projections to Legendrian arcs.
 
Visually I can; it's just going zig-zags isn't it
 
6:08 PM
e.g. a projected curve on the $x,z$-plane, $\gamma(t) = (x(t),z(t))$ which doesn't have vertical tangencies just lifts to $(x(t),z'(t)/x'(t),z(t))$.
Then for $x,y$-plane, just set $z(t) := z(0) + \int_0^ty(s)x'(s)\,ds$.
But $z(0)$ is a parameter since the contact structure is vertically invariant.
 
@BalarkaSen My idea is to break the given curve you want to approximate into small bits, and then replace those small bits by the above procedure by Legendrian curves... instead of taking the straightline segment, just take the original curve as the "reference arc" with which you bound the disk
The resulting curve should be arbitrarily close to the given curve
Once you can do it for the standard contact structure you can do it for all contact structures because of Pfaff's theorem :3
 
Balarka has started talking to himself now.
 
Yeah, but I think the approximation of an arc continuously close by a Legendrian one needs more detail than that.
Because it just amounts to "draw the zigzag"
BUT CAN YOU DRAW THE ZIGZAG?
 
I see
 
You can always argue in the prototype space, but it would need a nice gluing argument, too.
That's how Geiges does it. But he gives neither the gluing nor the zigzag
 
6:14 PM
I understand the zigzag from a philosophical point of view, to be honest, rather than analytically. There's a very general holonomic approximation theorem, which says that, roughly speaking, every non-holonomic section can be approximated by holonomic sections. The zig-zags come up naturally in the same interpolation idea
@anakhro Got it
I think the nontriviality in the zigzags is that as you approximate better and better, the amplitudes of the zigzags go down, yet the frequencies go up
That's why it's only a C^0-approximation
@TobiasKildetoft Would you be surprised if I told you this is how I am in real life as well
 
not even a bit
 
Hi @Ted!
 
rehi, a @Balarka
 
You might like the discussion above
Starting from anakhro's cute story
 
@anakhro I don't like "coordinate-invariant." I prefer "independent of $z$" or "invariant under translation along the $z$-axis" or ...
I remember assigning car-parking exercises in beginning graduate differential geometry years ago (exercise in computing Lie brackets of drive and steer vector fields).
I originally saw that in some notes/book on relativity (I think).
 
6:40 PM
Hi, I am trying to find a function such that f(x)>0 for all x and f'(x),f''(x) < 0 for all x, any ideas?
 
Defined on all real numbers?
 
yes
 
Good luck :P
Can you do $f(x)>0$, $f'(x)<0$, and $f''(x)>0$ for all $x$? @JohnKeeper
 
@TedShifrin No, I can do f^2(x)<0 for all x, f(x)>0 for all x and f'(x)<0 for all x though. Maybe that is easier?
 
What do you mean by $f^2$?
 
6:48 PM
f(x)*f(x)
wait that can't be negative right?
 
Surely that can't be negative.
 
Oh I just realized f''(x) can be 0
 
Huh? Now I'm totally lost. You should have an easy example of the one I asked for. The one you asked for cannot possibly exist.
 
it's certain-e possible.
 
Which one?
 
6:55 PM
get outta here
hes making bad joke
 
I get the joke.
 
Ted, yours.
 
(I'm not as dumb as I look.)
Well, Demonark doesn't make as many horrid puns anymore, so I suppose someone has to fill the void.
 
D:
 
it's voiding the fill rather than filling the void
 
6:57 PM
Wow, a @Balarka got an apostrophe correct. Will wonders never cease.
 
Not again
Shall I always be ribbed by Ted for my apostrophe errors from the past
 
Yes.
Well, if they really are all in the past, it will cease.
But this is a more-than-frequent occurrence.
 
Sorry, Ted has syntactical pet peeves. And living in the time where ignorance rules the roost, and most people type "your" instead of "you're," my pet peeves are difficult to exorcize.
 
I dont know what your talking about, Balarkas accent's are alway's on point
 
7:00 PM
There you go.
 
Stoke's theorem gets Ted the hardest
Try that more often
 
Grrr.
Good thing it's lunchtime, so Ted can leave.
 
Enjoy!
 
Buon appetito!
 
7:38 PM
Hello
 
7:52 PM
Hey guys, let's say I have a smooth map $f: R^m \to R^n$ and a manifold $M= \{ (x,f(x)) \in R^{m+n}: x \in R^m, f(x) \in R^n\}$. Is there a quick way to implicitly define $M$ as the preimage of some regular value under some map $\phi$?
Oh it turns out I don't need to for what I'm doing . I'm still curious to know
Oh duh
It's obvious now
NEVERMIND
 
 
1 hour later…
9:19 PM
Evening all
 
I'll say the answer in case someone was curious and it's not obvious to them
The graph is a set of ordered pairs $(v,f(v))$ so if we define $\phi(z,f(v)) = z-f(v)$ or $f(v)-z$, then $\phi^{-1}(0)$ is the graph of $f$
Wait, this only works if $m=n$
Or does it
No it works anyway
I confused myself
 
10:16 PM
Why does plotting points of the form $(p_n,n)$
approximate the prime counting function
oh nevermind got it
 
10:44 PM
@GFauxPas You actually want $\phi(z,v)$ with things in appropriate domains.
 
Hello.
 
hello, @Earth.
 
I just want to verify, if anyone is around.

1) Over a local ring (or indeed a PID) a projective module is necessarily free
2) Thus for a local ring R, K_0(R) is Z, since they are classified by rank, and K_0(N)=Z
3) If R is a commutative ring, free modules on R correspond to free sheaves of O_X-modules on X = Spec(R)
4) ...
 
That looks right, but I haven't thought about commutative algebra in a very long time, so you should consult people like @MatheinBoulemos (who hasn't been around in a while) and @AlessandroCodenotti and others who are steeped in it now.
 
Sure thing, hopefully they turn up :-).
 
10:58 PM
Mathein hasn't been around in a while; I'm not sure where he disappeared to. I'm sure Alessandro is asleep, as it's late in Europe.
 
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