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00:20
@AlessandroCodenotti play?
It's the middle of the night for Alessandro.
00:36
Ted is correct
 
2 hours later…
02:40
> The way to treat continua is to discretize them in such a way that you can refine the discretisation as many times as you want.
This is related to the "quantifier swapping" properties of nonstandard analysis - continuum is ∃∀, discretization-as-needed is ∀∃.
This is also why homology is so much easier than homotopy. Triangles can be broken up into triangles; spheres can't be broken up into spheres
03:04
you're just observing that homology is a homology theory?
there's plenty of things you can "break up" (things with excision) that are still incomprehensible
Stable homotopy for instance
03:56
@LeakyNun you could have easily brought up "nets".
 
1 hour later…
05:03
hello
05:32
In mathematics, the Dehn–Sommerville equations are a complete set of linear relations between the numbers of faces of different dimension of a simplicial polytope. For polytopes of dimension 4 and 5, they were found by Max Dehn in 1905. Their general form was established by Duncan Sommerville in 1927. The Dehn–Sommerville equations can be restated as a symmetry condition for the h-vector of the simplicial polytope and this has become the standard formulation in recent combinatorics literature. By duality, analogous equations hold for simple polytopes. == Statement == Let P be a d-dimensi...
Didn't know about this - this is the generalization of V-E+F=2
06:01
Ok, I think I'm correct with this notation, but here's where I'm at with it.

Does $\sum\limits_{i=1}^{2n^2} i^2$ mean the same thing as $1^2 + 2^2 + 3^2 + \ldots + (2n)^2$?
${}\dotsb+(2n^2)^2$
The LaTeX doesn't seem to be rendering in my browser.
Question: If a ring $R$ has an ideal $I$ such that $R/I\cong\mathbb{Z}$ can we say that $R$ is isomorphic to a direct product of rings such that one coordinate of that product is $\mathbb{Z}$?
@zacts latex isn't enabled by default. use the latex in chat link in the room description
@Semiclassical Cool, it works now. Thanks!
@AkivaWeinberger I don't understand your answer.
oh
I see. Let me think about this
06:07
If it says $2n^2$ on the top of the Sigma, then you should add until $i=2n^2$, which means the last term should be $i^2=(2n^2)^2$
So, $1^2 + 2^2 + 3^2 + \ldots + (2n)^2$ means $\sum\limits_{i=1}^{2n} i^2$?
If so then I think I might understand why.
06:29
Alright, more general question: given any ring homomorphism $\phi: R\to Q$, is it true that $\text{ker}(\phi)\times\text{Im}(\phi)\cong R$? I know there is an analogous statement that's true for vector spaces, but I can't find details about if this is true in a general ring
@Rithaniel no
Hmmmm, now I will seek a counterexample
it's not even true for groups
Huh, $\sum\limits_{i=1}^{n} (2i-1) = 1 + 3 + 5 + \ldots + (2n-1)$. On the other hand $\sum\limits_{i=1}^{2n-1} i = 1 + 2 + 3 + \ldots + (2n-1)$?
Note: I'm working through an exercise in this text Calculus by Spivak.
I'm just very confused by this notation.
@zacts yes
06:43
@LeakyNun so both of my above expressions are accurate?
Cool, thanks!
It's very helpful notation in the long term. It allows otherwise complex expressions to be written efficiently and relatively clearly
Sure, I just want to make sure that I understand how it works.
it's a very concise notation.
06:45
I initially thought that the (2n-1) was a drop in replacement for the top of the Σ. So whenever you would see (2n-1) you just put that at the top. But, that's not always the case.
it's $\displaystyle \sum_{i=1}^n f(i) = f(1) + f(2) + \cdots + f(n)$
so the last term is the function inside applied to the expression above Σ
otherwise $\displaystyle \sum_{i=1}^4 (2i+1)$ would make no sense in your notation
because 2i+1 cannot be 4
@AkivaWeinberger Funny enough there is a version of excision that works out in higher homotopy groups, under restrictive situations
It's very surprising to me that it does to be honest.
@BalarkaSen is the condition that the lower homotopy groups are trivial :P
You might also see something like $\sum\limits_{i\in I}f(i)$, which means "I have this bunch of things that I call $I$, and I want to apply this function $f$ to each thing in $I$ and then add together the resultant values" (and clearly, $\sum\limits_{i\in I}f(i)$ is easier to write)
06:51
@zacts and if the things inside are not integers, it would even make no sense
It's a complicated condition on the relative connectivity of the subset you are excising from the pair
@BalarkaSen It's precisely that which lets us get excision stably yeah
So, to summarize. The $(2n-1)$ in the $\ldots$ notation isn't necessarily what goes directly above the Greek letter Sigma in the summation notation.
Right, because the connectivity conditions become trivial if you suspend a lot I guess
Maybe not
06:54
cool
@zacts right
I think that's true
Stable homotopy is an extraordinary homology theory, or something, right?
I just call them homology theories
But yeah it satisfies excision and the easier axioms
extraordinary means that the point is not supported at zero?
yeah wiki confirms
but then shouldn't the point have trivial stable homotopy or something
its suspensions are just bigger points?
06:58
What?
according to wiki, extraordinary homology theory is $H_n$ not necessarily satisfying the axiom that $H_n(pt) = 0$ for $n > 0$
but it seems like stable homotopy satisfies that axiom
The value group is the stable homotopy group of the sphere dude
It's certainly not 0 lol
wait what
Wouldn't that be for $S^0=\{p,q\}$
yeah that's what I think
07:07
Homotopy groups require a basepoint
The homology theory here is $\pi_k^s(X) = \pi_{n+k} \Sigma^n X_+,$ where the plus denotes adding a disjoint basepoint
For n large
oh
nice
And let me also point out that suspension for based spaces is usually understood as the reduced suspension
So $\Sigma^k * = *$ on the nose
And if you didn't define it the way I did above the disjoint union axiom would be wrong
@MikeMiller Does any functor representable by a spectrum satisfy Mayer-Vietoris? Surely not, right? I suppose you can always get a long exact sequence by dualizing the Puppe sequence
(as far as I recall Mayer-Vietoris is part of the hypothesis when stating Brown representability)
@BalarkaSen I don't see why that shouldn't hold if it comes from Puppe
I would have to think to recover the proof though
I have actually forgotten how to verify [-, K(G, n)] satisfies the M-V, but I knew I checked it once
I guess the same proof pushes through
So I am less skeptical now
07:21
Yeah seems like you just need to know that every spectrum has the weak type of an infinite loop space
that's just H^n(G) so it satisfies the MV
circular @leaky
@MikeMiller ah, every spectrum is weak equivalent to a omega-spectrum? i didn't know that
@LeakyNun you don't know two homology theories are the same unless you verify the axioms for either of the theories lol
Omega Spectra are the "fibrant objects" in Spectra, the one for which the terminal map is a fibration
And suspension Spectra are the cofibrant objects
Ahh
so there should be some kind of fibrant replacement for everything i suppose
07:24
The fact that you can replace a spectrum with an Omega spectrum is the simplest part of the spectrum version of the path space construction
Yeah
actually now I realize you can avoid Mayer-Vietoris altogether by demanding the long exact sequence goes HA -> HX -> HX/A
it still determines the homology theory uniquely by the value group
and that is exactly what Puppe gives us, $A \to X \to X \cup CA \to SA \to SX \to \cdots$, as $X \cup CA \simeq X/A$
so there's actually no need to check Mayer-Vietoris to prove [-, K] = H as long as this version of LES is satisfied
you only use this to induct on the skeleta of CW complexes, collapsing each lower dimensional skeleton as you move up the ladder
07:43
@BalarkaSen I think of this as being the same as MV as being the same as excision
yeah me too
I have never bothered with the formal crap relating these all
 
3 hours later…
10:59
Given an odd number n, you can add 2*k to n and the result will be odd. Is there a way to extend this concept to skip multiples of 2 and 3? For example if you know n does not have factors 2 nor 3, what can you add to it to ensure the result also doesn't have multiples of two or 3?
11:33
Well, you just need to look at the number modulo 2 and modulo 3. (a + b) modulo p $\equiv$ (a modulo p) + (b modulo p)
So, if a $\equiv$ 1 mod 2, then you need to add something that is not equivalent to 1 mod 2. So it must be equivalent to 0 mod 2. In other words, it's of the form $2k$ for some $k\in\mathbb{Z}$. If a $\equiv$ 1 mod 3, then you need something that isn't equivalent to 2 mod 3, and so must be of the form $3k$ or $3k+1$ for some $k\in\mathbb{Z}$. If a $\equiv$ 2 mod 3, you need something that isn't equivalent to 1 mod 3, and so must be of the form $3k$ or $3k+2$ for some $k\in\mathbb{Z}$
So, $6k$ always works if you want to ensure that adding it does not introduce any factors of 2 or 3.
If a $\equiv$ 1 mod 3, then $6k+4$ also works. If a $\equiv$ to 2 mod 3, then $6k+2$ also works
11:48
@Rithaniel wanna explore variants?
Variants of chess? Sure. I should make an account on lichess, too
@Rithaniel tell me when you're ready
I'm good to go (That was a super short registration process)
12:41
@CowperKettle interesting
(After looking at the analysis, it is apparent that neither of us know how to play racing kings)
correct analysis
How do I invite you to a game?
go to the main page
click "play with a friend"
12:43
you can invite me by handle
hi chat, i'm studying on rudin's book an example of continuous function nowhere differentiabel
on fomrula (38), he defines $\delta_m$, which will become later the increment in the incremental rapport
he defines it with a $\pm$ saying we choose the sign in such a way there is no integer btween $4^mx$ and $4^m(x+\delta_m)$
now, why ho did this is obscure to me
any hint or explanation would be much appreciate, thanks in advance
 
1 hour later…
13:56
0
Q: $a_n = a_{n-1}^3 - a_{n-2} $ conjectured inf and sup

mickMy mentor tommy1729 told me : Let $b_1 = 1.$ Let $ b_2 = x.$ Also $ \frac{1}{2} < x < 2 $ Define for $ n>2 $: $$ b_n = \frac{\exp(\ln(b_{n-1})^3)}{b_{n-2}} $$ Then $$ \sup_{n>2} b_n = x , \inf_{n>2} b_n = \frac{1}{x} $$ Is this true ? How to show that ? We can rewrite the sequence $b_n$ ...

 
2 hours later…
15:36
hi
15:53
Hi, would it be right to say that a tangent to a curve $x=g(y)$ at the point $(x_0,y_0)$ is given by $x-x_0=g'(y)(y-y_0)$?
16:10
If i have some equation which equates to null and then another equation with different variables and different topic which also equates to null ( think of physics for example where you can use different subjects with different variables ) can then those two equations be equated to eachother_
A stupid example would be i dont know lets say that the velocity minus elephant equals zero and then say that the resistance plus voltage equals zero can then we say veloctiy minus elephant equals resistance plus voltage
stupid example but i think it clears the point
I think that the 0 output should be of the same "unit" in both cases in order to equal the two equations, but let's hear others
in your example, g' is the tangent if what you are trying to do is a derivitave.
Or actually that is nevertheless a tangent. Derivitave or not, but as far as I know is that the definition of a tangent is defined newly in Analysis. Considering what you mean I guess you do not know that.
But I wouldn't belive what I say I am just a noob afterall :D
Should start hunting for finite time singularities by using rougher fluids
16:28
@Shootforthemoon if $x_0=g(y_0)$, then that seems fine
(i imagine that was intended but it wasn't stated explicitly)
16:54
@Semiclassical Yes, thanks very much!
17:35
0
Q: Preservation of Second Countability

topologicalmagicianLet $f: X\rightarrow Y$ be a continuous open map. Show that if $X$ satisfies the second countability axiom, then so does $f(X)$. My attempt:Let $B_1,B_2,B_3....$ be an enumeration of the basis $\mathbb{B}$ the topology on $X$ admits. Then clearly, $A=$ $\{$ $f(B_i)$ $:$ $B_i \in \mathbb{B}$ $\}$...

any feedback would be greatly appreciated
18:21
@Ultradark
Hi, I have a question about the wave equation for pde.
u_tt-4u_xx=0 t>0
u(x,0)=1-x^2 when |x|<=1 and u(x,0)=0 when |x|>1
u_t(x,0)=4 when 1<=x<=2 and u_t(x,0)=0 else.
I'm looking for all points when u is not C^2 (twice differentiable)
18:37
@shi
18:50
                                                                                                                                                                                                                                                                                                         hello
@loch I answered my own question. Have at my answer https://stats.stackexchange.com/a/442135/82135.
 
3 hours later…
nice
do you know how to find
Find what? Prime numbers, no
a transformation that sends a point on a function along that function
for example f(x)=1/x
Along the graph of the function?
21:35
You have to be specific
A monoid hom between groups is a group hom, no?
@shi well I have verified $T_s = \begin{bmatrix} e^{s} & 0 \\\ 0 & e^{-s} \end{bmatrix}.$
this is the correct matrix
then why don't you just apply the first isomorphism theorem?
it's a squeeze mapping
21:39
@Thorgott because $\Bbb{Z}^{\times}$ is a monoid and the kernel of say $f(x) = x, x^2$ is a monoid not a group
but I am not sure how to find the correct one for $y=1/\ln(x)$
I thought for years that i had the right one
Thanks for the upvote, whoever did that
turns out I've been wrong for years...
the kernel of a group homomorphism is always a subgroup
@Ultradark my only suggestion is study algebra, lol. I hate analysis
@Thorgott in this case we have a monoid hom between monoids one of which is a group, so it's not a group hom
21:43
the units of a ring are a group under multiplication
Oh, crap, I meant the monoid not the units
edited post, thanks @Thorgott
Monoid problems seem harder than group problem from shear lack of textbooks about monoids. That is why if they were groups we'd already know the answer.
It's a trap. The question is really about twin primes, lol
J/k
@Thorgott sogar ein Normalteiler! lol
Ich spreche somme Deutsch!
22:00
@ShineOnYouCrazyDiamond isn't the map sending everything to 0 except 1 which is sent to 1 a monoid hom?
this question is rather demoralizing. not the question itself, but the comment to the deleted answer: math.stackexchange.com/questions/3486685/…
@AlessandroCodenotti yes
@Semiclassical look to the higher maths
@ShineOnYouCrazyDiamond So this answers your question
$\ker f = \{1\}$.
22:11
And so $\Bbb{Z}/\ker f \approx \Bbb{Z}$.
Still not sure though
What if $f$ wasn't contrived
For example $f(x) = x^{k}$ is a natural choice
It's definitely a monoid hom
So we can limit $f$ to polynomial maps and that rules out your counterexample
I broke maths
^_^
@ÍgjøgnumMeg huh? wovon?
What does it even mean to quotient by the kernel in general?
I guess the correct relation to quotient by is actually $x\sim y\iff f(x)=f(y)$
And with this relation the first isomorphism theorem looks true
22:26
using that relation you should even be able to get the first isomorphism theorem in the category of sets lol
It works for any algebra (as defined in universal algebra) and I guess a set is an algebra without operations
huh, and pointed sets are algebras with a single unary operation
bizarre
Why unary rather than 0-ary?
oh
0-ary then
words are hard
@Thorgott hab nur gemeint, dass der Kern eines Gruppenhoms ist immer ein Normalteiler lol
22:36
null-ary
immer ein Normalteiler ist* was ist bitte aus meinem Deutsch geworden
Makes sense then (that's what is often done in model theory too, to have 0-ary function symbols instead of constant symbols)
achso, dachte es geht um die Einheitengruppe und war verwirrt
Achso nein lol
This chat is getting conquered by the Germans
22:41
hehehe
I'm going back to snake mit in 1985
to learn about time travel
@ÍgjøgnumMeg actually maybe I should say people studying in Germany
Hmmm but then that includes me too
Merry Christmas to all of you!!
22:57
@manooooh merry xmas to you, but not to all of you since I don't know your holidays
You've got to be exact in math
Hahaha well that's true. A permission please. Here is 19:57 waiting for Santa Claus!
That's why I'm thinking if $x^{2^k} = 1$ modulo no more than $k+1$ odd primes, then there's some homological algebraic way of looking at the problem
Of twin primes
looks like Alessandro has played himself
Since $f(x) = x^{2^k}$ kernels
and stuff
qed
If $f(x) = x^2 = 1 \pmod {pq}$ but no other primes (using CRT here) can only be done finitely often, then $\ker f \subset \bigcup_{r \neq p,q} \ker \pi_{r}(x)$ for any $p, q \gt N$ for some $N \in \Bbb{N}$.
Since for $p,q$ past all the twin primes, we have $x^2 = 1 \pmod {pq} \implies x^2 = 1 \pmod r$ for some $r \neq p,q$.
Assuming twin primes false
Actually where $\pi_r : x \mapsto x^2 \pmod r$
math.stackexchange.com/questions/3484084/… Does anyone have any thoughts?
...concerns, trivialities, comments...
what are the primes if the twins are neglected?
23:10
Can you teach us more about the problem @Ultradark
I can't understand it
Not that it's your writing
It's me, I have no knowledge of that area
@Ultradark
What I'd like to see
is the actual mapping
In the form
What is the actual mapping?
In what form?
23:14
Show us how it's put together
So we can work with it
Also add in minimal background links
like wikipedia article
Show us here
I can help you edit
okay
what should I do
So $\ln(y) = 1/\ln(x)$
Can we solve for $y$ here?
what do you think
23:17
$y = e^{1/\ln x}$
In mathematics, a hyperbolic angle is a geometric figure that defines a hyperbolic sector. The relationship of a hyperbolic angle to a hyperbola parallels the relationship of an "ordinary" angle to a circle. The magnitude of the hyperbolic angle is the area of the corresponding sector of the hyperbola xy = 1. This hyperbola is rectangular with a semi-major axis of 2 {\displaystyle {\sqrt {2}}} , analogous to the magnitude of a circular angle corresponding to the area of a circular sector in a circle with radius...
So, the graph of $f(x) = y$
added this link to the post cause i think it's relevant
Is the set of all $(x, f(x))$
in physics, i think the hyperbolic angle would be equivalent to 'rapidity' in special relativity?
23:18
So $M$ a $2\times 2$ matrix takes $2\times 1$ column vectors to themselves
yeah I'm just doing the hyperbolic angle like method for a different function
So let $M = T_s$
oh hey, yeah
In relativity, rapidity is commonly used as a measure for relativistic velocity. Mathematically, rapidity can be defined as the hyperbolic angle that differentiates two frames of reference in relative motion, each frame being associated with distance and time coordinates. For one-dimensional motion, rapidities are additive whereas velocities must be combined by Einstein's velocity-addition formula. For low speeds, rapidity and velocity are proportional, but for higher velocities, rapidity takes a larger value, the rapidity of light being infinite. Using the inverse hyperbolic function artanh, the...
second sentence
Then $M \cdot (x, e^{1/ \ln x}) = (e^{e^s} \cdot x, e^{e^{-s}} \cdot e^{1/\ln x})$
So how is the right also a point in the graph?
To show that it is a point in the graph, we need the left component $e^{e^s}$ plugged into $f(x) = e^{1/\ln x}$ as an argument and the result has to be the right component $e^{e^{-s}} e^{1/\ln x}$
Is it though?
I don't think so
I think I just found the area of the sector between the rays and plugged that in for $s$
into the matrix
23:28
So it's not true for any $s$?
Just when $s$ is the area?
rereads post
I see
now
What did you just see?
That $s$ is the area
I am not comfortable working with such integrals, perhaps include small proof sketch?
it's just finding the area of the region
that big sector
$x,y \ne 0,1$
How does the golden ratio come into play?
it just does
I can show you though
$\exp(\frac{1}{\ln(x)})=x/e$
What do you define as hyperbolic rotation?
$\dfrac{1}{\ln x} = \ln \left ( \dfrac{x}{e} \right ) = \ln x - \ln e = \ln x - 1; \tag 2$
$1 = (\ln x)^2 - \ln x; \tag 3$
$(\ln x)^2 - \ln x - 1 = 0; \tag 4$
quadratic formula gives the roots
$x = \exp \left ( \dfrac{1 \pm \sqrt 5}{2} \right ). \tag 6$
What do I define as hyperbolic rotation?
A hyperbolic rotation is what we get when we slide all the points on the hyperbola along by some angle.
How does $\exp (1/\ln x) = x/e$?
The rest looks good
that's what you can start with
the function is bounded by the ray x/e
and also the ray y=x
and then it's a matter of calc 1 to find the area of the bounded region
@shi do you understand
23:48
exp(1/ln x) is definitely not x/e
for one, the left-hand side is undefined when x=1
nor does it work when x=e
It's actually $x^{(1/\ln x)^2}$, I believe.
it's an equation
believe nothing, not even what your eyes can see
Greetings, @MikeM.
they intersect
and then you solve for $x$
23:52
@MikeM: You have a moment for an embarrassing query?
If it's about math, please, and embarrassment is unnecessary. If it's about your insides, probably not.
and then you get the answer
LOL ... So let's take $E=TS^2$. Can you give me an argument that $\text{Hom}(E,E)$ is nontrivial? I don't think we can use characteristic classes. Doing $E$ the universal bundle on $G(2,4)$ (oriented) might give an easier example, but I can't do it either, yet.
I do much better in the complex world, I realize.
We are OK because the base is a small-dimensional sphere. The small-dimensionality reduces the problem to stable bundles, and the sphere lets us classify bundles as homotopy classes in the understood space BSO
Well, this is not my bailiwick, of course, but ...
23:56
Writing more detail give me a minute
$\text{Bun}^{or}_n(S^2) = [S^2, BSO(n)] = \pi_2 BSO(n) = \pi_1 SO(n)$.
In the $\tilde G(2,4)\cong S^2\times S^2$, I think I can do it with a curvature computation.
For large $n$ ---- $n \geq 3$ --- there are only two oriented rank n bundles over the 2-sphere
One of them has $w_2 = 0$, one of them has $w_2 \neq 0$
So we indeed can reduce to characteristic classes.
Well, this one clearly has a rank 2 trivial subbundle.
$\Bbb R$
23:58
Yes, clearly. My bad
Otherwise it would be C
Yes, I know how to do it with holomorphic line bundles :)
I wanted to make a direct argument that if it were trivial, we could get a nowhere-zero vector field.
I was close, but not quite a cigar.
I need a formula 1s

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