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12:03 AM
If $K=Z(G) \cap X(G)$ where $Z$ and $X$ are zonotopes and copies of each other but have the same center and just differ by a rotation of 90 degrees, then is $K$ guaranteed to be a zonotope?
2
Q: Intersection of Zonotopes

AAAA zonotope is specified by a set of generators $G=\{g_1,\dots,g_n\}\subseteq\mathbb{R}^d$, and is defined as $Z(G)=\{z:z=\sum_{i=1}^n x_i g_i, 0\leq x_i\leq 1 \forall i\}$. Let $v\in Z(G)$, and consider the (non-empty) set $I=Z(G)\cap (Z(G)-v)$. That is, $I$ i s the intersection of $Z(G)$ and a s...

 
let $R$ be a ring. in general, $R$-algebras are the structures where unknowns coexist with elements of $R$, right?
 
@TedShifrin what is false?
 
@noballpointpen Ah, it’s false. Make up a counterexample with $S$ a subset of $\Bbb R$.
 
@TedShifrin you mean, they put an exercise that asks you to prove a thing which is false?
 
Apparently, unless you’re missing something.
It makes you understand what is needed for the intermediate value theorem to hold.
 
12:17 AM
"Here it is", page 6, number 1.2:4.
 
I hate mornings
 
Oh, my counterexample is wrong. Good.
 
Is your hint still relevant?
 
I assume it’s the right way to go,
 
12:33 AM
Ok
Seems like this version of the document is very different from what I have. Not talking about this exercise, it's the same. Just noticed more adorable formatting and then that it has 122 pages instead of 89.
Don't think I am going to dump the old. I have PDF marks made which exercises I finished.
 
Hi!
Shouldn't $S_2$ be less than $S_1$ as it consist of one term less?
 
Maybe they count (n-1) terms from that 2nd?
 
$\frac12\ln2$ is the sum of S1 for n=infty
@noballpointpen Oh
 
Good day, leslie
 
So I have to compare 1st term and the last term of S2. Thanks :)
Hmm S2 is greater than S1. But I can't see how it is greater than $\frac12\ln2$
 
12:56 AM
The last term evaluates to $1/2$ for $S_2$. Did you see it? Just noticing, don't know if its helpful.
Look at the difference of $S_1$ and $S_2$, and try to somehow compare it with $1/2 \ln 2$. That's where I would start.
Ah, notice that $1/2 \ln 2 < 1/2$ the result is immediate.
 
1:30 AM
@noballpointpen Isnt the last term $\frac1{2n}$ ?
 
Oops, yes.
Thought the numerator had power.
Did you solve it?
 
What?
 
You wondered why $S_2 > 1/2 \ln 2$.
 
@noballpointpen But for all n except 1 , $1/2 \ln 2 >1/(2n)$ . So isnt the given answer wrong?
 
Ah, I see :)
We also could look at the inequality $\frac{1}{1+n^2} < \frac{1}{2n}$. From which we could reconstruct the desired inequality.
Or not, I was just working in backwards mode.
Nvm, Wolgwang.
 
 
2 hours later…
3:26 AM
How many distinct subspaces can be formed from A (subset of a topological space X) by taking only closure and interior? (i.e., find m such that the no. of 'such subsets' can never exceed m in a topological space.)
$m\ge 6$ as $A=[1,2)\cup [3,4)\cup (4,5)\cup (Q\cap (6,7))\subset R$ shows.
(as A, int A, cl (A), int (cl (A)), cl(int A), int (cl (int A)) are all distinct.)
It is also true that cl(int (cl (int A)))= cl (int A).
so if m>6, then the contribution will come from int (cl (A)).
 
3:52 AM
5
Q: Playing with closure and interior to get the maximum number of sets

mathcounterexamples.netCan you find $A \subset \mathbb R^2$ such that $A, \overline{A}, \overset{\circ}{A}, \overset{\circ}{\overline{A}}, \overline{\overset{\circ}{A}}$ are all different? Can we get even more sets be alternating again closure and interior?

 
The set B= $A\cup \{9\}$ gives me m=7.
 
Not quite the classic question.
 
int cl int cl A= int cl A (trying to prove this one).
cl int cl int A= cl int A (this is true.)
 
it kinda embeds in the classic question, though, as that answer observes.
 
what's a ring where a maximal ideal has more than 2 generators
 
3:55 AM
@leslietownes I searched a lot, how did you find it? :P
I searched Kuratowski problem.
The variant that I was getting involved complements too...
 
@shintuku $K[x,y,z]$
 
before grothendieck left functional analysis for some other field he proved a kuratowski-like theorem, that up to a certain notion of equivalence, there are at most 14 norms you can get on a tensor product of banach spaces by applying various natural norming operations. i think the 14 is a numerical coincidence, however. the norming operations satisfy some relations, but i don't think they generate the same abstract semigroup as you get with the closure/complement problem.
in some alternate universe timeline, grothendieck never leaves functional analysis and devotes his life to proving increasingly weird shit about banach spaces.
2
[picture of alternate timeline, which resembles a midcentury painting of a 21st century paradise in which everyone is zipping around in rocket cars and poverty no longer exists]
 
Did your alternative universe annihilate white supremacy, too?
 
yes. it is also the peaceable kingdom.
 
Sign me up.
 
4:07 AM
I think the claim above is true: cl int cl A$\subset $ cl int cl A (interior Z is contained in Z). The set on RHS is contained in in cl A (because int cl A is contained in cl A so cl int cl A is also contained in cl A so cl int cl A is contained in cl A. Taking int on both sides, we get the said inclusion). Inclusion in the other direction is of course true.
 
if you don't go up to "equivalence," and ask if the grothendieck operations still might generate only a finite number of tensor norms, i think that's an open problem (the semigroup relations that are known do not force finiteness as they do in the kuratowski case). an open problem from grothendieck which probably nobody is working on.
 
So we have the following lemma: Denoting int cl by f. $f^m(A)= f(A)$, where m is any integer $\ge 2$. f^m means f applied m times.
similarly for g= cl int.
 
 
1 hour later…
5:16 AM
An inner automorphism of a group G restricted to a normal subgroup H need not be an inner automorphism of H.
If such group G exists then G can't be abelian.
 
would the semidirect product construction help here?
i.e. given any automorphism f of a group G, can you do a semidirect product in which G is a normal subgroup, and in which conjugation by some element in the semidirect product, when restricted to G, gives you f back
 
My strategy: Pick a non abelian group $G$ and an inner automorphism such that the restriction to it's center ( non trivial center)or any abelian normal subgroup is non identity.
Let $G=S_3 $ and $H=\langle (123) \rangle$
$f(123) =(132) $ and $f(12) =12$
Since $H$ is an abelian group, the identity map is the only inner automorphism of $H$
Correct?
 
5:58 AM
@TedShifrin cute problem.
 
6:19 AM
@leslietownes Corollary I hadn’t realized. A symmetric polynomial must have even degree.
That has to be nonsense. Yet …
Ah, it is nonsense. Interesting that $1$ and $-1$ aren’t behaving the same.
 
oh hrm, the distinctiveness of 1 pops right out of the way i was doing it.
 
I used the derivative with $f(x)=x^n f(1/x)$.
 
writing R for that operation (on polynomials of degree n), writing f of degree n as f(x) = (x-1)^m q(x) with q(1) nonzero, if if f(x) = Rf(x) = x^n (1/x-1)^m q(1/x) = x^m (1/x-1)^m x^{n-m} q(1/x) = (1-x)^m Rq(x) = (-1)^m (x-1)^m Rq(x), deduce (-1)^m Rq(x) = q(x), evaluate at 1, note that Rq(1) = the sum of coefficients of q = q(1) is nonzero, deduce (-1)^m = 1.
while some of this algebra "only" uses that 1 satisfies 1/a = a, there's no reason for Rq(-1) and q(-1) to be the same.
 
6:51 AM
math.stackexchange.com/questions/4705703/… this is pretty wild. for people who like weird definite integrals only.
 
7:25 AM
Need to evaluate a line integral: $\int_{L} x^2-y^2 dz, \ L: \{(x^2+y^2)^2+z^4=a^3z, \ x^2y+y^3-z^2x=0\}$ from the point $(0,0,a)$ to the point $(0,0,0)$, where $x\geq 0$. How do I parametrize $L$ normally?
 
7:42 AM
Okay. Seems like I got something:
$x^2y+y^3=z^2x \rightarrow y(x^2+y^2)=z^2x \rightarrow x^2+y^2 = \dfrac{z^2x}{y} \rightarrow \dfrac{z^4x^2}{y^2} +z^4 = a^3z \rightarrow z^3=\dfrac{a^3y^2}{x^2+y^2}$, so I took $x=\cos t, y= \sin t \ \rightarrow z=a\cdot (\sin t)^{2/3}$. Therefore, we get the integral $2a/3 \int_{1}^{\pi/2} \cos t (\cos^2 t - \sin^2 t)/\sin^{1/3} t dt$. Is that right?
 
8:08 AM
0
Q: Understanding a proof of Riemman integrability iff not continuous on a set of measure $0$

KoroLet $f:[a,b]\to R$ be a bounded function such that $D:=$ the subset of $[a,b]$ such that $f$ is not continuous at any point of $D$. $f$ is Riemann integrable iff $\lambda (D)=0$. Proof: For $n\in \mathbb N$, let $P_n$ be the partition that divides $[a,b]$ in $2^n$ equal subintervals $I_j, j=1,2,...

 
 
4 hours later…
11:50 AM
0
Q: $g(x)=\int_{(-\infty,x)} f d\lambda,$ where $f:R\to R$ is integrable, is uniformly continuos on $\mathbb R$.

KoroI tried to prove it the following way but got stuck: Take any two sequences $x_n$ and $y_n$ such that $x_n-y_n\to 0$. Define $E_n:=[x_n, y_n]\cup [y_n, x_n]$. $|g(x_n)-g(y_n)|=|\int f\chi_{E_n}d\lambda|$ $|f\chi_{E_n}|\le |f|$ and $\int |f|d\lambda <\infty$. So by the Dominated Convergence Theore...

 
@Koro isn't it absolutely continuous?
huh, moduli problem of Riemann surfaces is quite well-known
 
@onepotatotwopotato I don't know.
but the answer there proves 1) continuity, 2) does not explain why $\int f\chi_{[x_n,x]}\to 0$.
 
12:22 PM
couldn't you use density of smooth compactly supported functions in L^1 to prove uniform continuity? For an individual such function uniform continuity should be plain, then use that a sequence of uniformly continuous functions on R converging uniformly to a function on R makes the limit UC too.
but really a better way is to prove its actually absolutely continuous
here is the idea: First prove the following "If $\mu << \nu$ and $\mu$ is a finite measure, then for every $\epsilon > 0$, there exists $\delta > 0$ such that if $\nu(E) < \delta$, the $\mu(E) < \epsilon$
then notice that $g(x)$ defines a measure that is absolutely continuous with respect to the lebesgue measure on the real line
and you get uniform continuity pretty much as an immediate side effect of the first quoted result I just stated.
so you get that as long as $E$ is a borel set with sufficiently small lebesgue measure, $\int_{E} f d \lambda$ can be made arbitrarily small
uniform continuity follows just by replacing $E$ by a closed interval.
no need to mess around with sequences really
 
 
3 hours later…
3:00 PM
Dead as a morgue
 
 
1 hour later…
4:06 PM
jesus christ does The Economist suck
it's like what michael from the office would read unironically
 
4:20 PM
Any one have any idea where can I find the solution to this question of the Kyiv City Olympiad 2021 for 11th graders i.e " Let $f_{n}(m)= 1^{2n}+2^{2n}+3^{2n}...+m^{2n} $.($m,n$ integers). Show, that there are only finite amount of pairs $(a;b)$ such that $f_n(a)+f_n(b)$ is a prime number." ? I solved the problem, but I am interested in seeing other approaches.
A different query :
I got a strange message while creating meta stack exchange account it says, "You cannot register at this time. If you continue to find this error, contact us"
 
spine-chilling
 
@shintuku what ?
 
blood-curdling
 
@shintuku I am not sure, but if you are intending these words to me, then surely, it makes no sense to me.
It feels weird you know, if you randomnly post words like these!
 
would you say it feels ominous or eerie
 
4:29 PM
@shintuku absolutely!
Can't hold the suspense, what's up?
 
5:00 PM
strange notification again...
what is this? @robjohn
 
5:17 PM
confused screaming
 
5:28 PM
Let $p(x) \in\Bbb{R}[x]$ where $p(x) =x^3+ax^2+bx+c$ . Let $C(p)$ be the matrix companion to the polynomial $p$ .Then the set $A=\{(a, b, c) : C(p) has a jordan normal form \}$ is path connected.
$C(p) $ has Jordan normal form if the characteristic polynomial split inside $\Bbb{R}$
The characteristic polynomial of $C(p)$ is exactly the polynomial $p$
The question is equivalent to : Is the set $A=\{(a, b, c) \in\Bbb{R}^3 : p(x) splits in \Bbb{R}[x] \}$ path connected?
@TedShifrin Please help me to solve :)
 
@Koro it looks as if someone raised a flag in another chatroom and you're being asked to evaluate it.
 
5:49 PM
@AlessandroCodenotti Any idea :)
 
6:02 PM
Oh, you mean Koro isn't being flagged? :)
@Sourav So don't be so fancy. You're looking at the set of $(a,b,c)$ so that all the roots are real. So you have one polynomial with real roots $\alpha_j$ and another with real roots $\beta_j$. Can't you find an obvious path?
 
6:16 PM
$f(t) =(1-t) x+ty$ ?
$x=(0, 0,0) ~y=(a, b, c) $ $f(t) =(1-t) x+ty=ty$
 
if F is a field, is writing F(x) for the field of quotients of F[x] standard?
i.e. ring of fractions
 
6:37 PM
@TedShifrin Should I change that?
 
@冥王Hades $\frac{\sqrt3}8\left(-1+\sqrt3\right)$ times the square of the side of the square
so we need to find the side from the $24$
 
7:01 PM
@robjohn the largest angle in the $24$ area triangle is $150°\circ$. That might be a hint
I actually found a solution. Wanna see? @robjohn. You don't need to find any sides for it
 
For what values of $z\in \mathbb{C}$ does the series $\sum_{k=1}^{\infty} \frac{2k+i}{k+2i} z^k $ converge? I have barely studied any power series, but I believe this is one. Maybe it's possible to solve without using terminology associated with power series? Anyway, grateful for some hints/help.
 
@schn Are you familiar with the various tests of convergence?
 
Yes.
 
E.g. the ratio test and the root test?
 
Yes.
 
7:11 PM
Okay, so apply one of those.
 
Ok, will try :)
 
Keep in mind that the usual presentation of these tests is in the context of real analysis---they are theorems about real numbers (in particular, you compute some limit, then compare it to $1$; is it bigger or smaller than $1$). The complex numbers are not ordered, so you can't directly apply these tests.
But you probably know that if a series converges absolutely, then is must converge.
So test for the absolute convergence of the series, using one of those tests.
In this case, my guess is that the ratio test will be the relevant tool.
 
Time to learn some Fake Analysis
 
You still are not supposed to be posting geometry other than affine/projective geometry.
 
@TedShifrin someone sent it to me because he couldn't do it
I can see why. Without rotating that little triangle at the top this problem is annoying difficult
 
7:44 PM
1 hour ago, by shintuku
if F is a field, is writing F(x) for the field of quotients of F[x] standard?
 
@shintuku Does it need to be standard?
 
yes or i won't use it
marginal notation is for marginals
 
@shintuku What an obnoxious thing to say.
In mathematics, we can use whatever notation we like to mean whatever we want, so long as we explain what that notation means.
 
no
 
Yes, and your refusal to believe this demonstrates a lack of mathematical maturity. We use notation to communicate. If it helps communication, we should use that notation. If it doesn't help, then use different notation.
 
7:48 PM
i am the math baby
 
That being said, I would understand $F[x]$ to denote the ring of polynomials with coefficients in $F$.
Is that what you mean by it?
And what motivates your question about $F(x)$? Are you trying to find notation, or are you reading this notation in some paper or set of notes?
 
yes, and Gallian denotes the field of quotients of F[x] by F(x) if F is a field, but Artin doesn't do this
 
Okay... who cares?
 
i do
 
If Gallian uses the notation, then you know that there is at least one author who uses it.
So you should feel free to use it, as well.
So long as you are clear about what it means.
You are way too hung up on notation. You should always use whatever notation is convenient and clear.
 
7:56 PM
no standard notation, no standards
 
What an asinine thing to say.
 
@shintuku Of course Artin does it. It's universally used notation.
 
perfect i shall use it
thanks
 
You really should actually read a good book and learn it. This is getting absurd.
You and your "cute" quips are annoying almost everyone here. I recommend you curtail this.
Or soon you'll just be talking to yourself.
 
thanks for the advice
 
8:14 PM
@TedShifrin correction: Artin uses it to denote the field extension of F by an element x. it then follows as a distinct fact this field is isomorphic to the field of quotients of F[x].
 
I should win a noble peace prize for the amount of restraint I am showing from saying something snarky
 
@冥王Hades Saying "I'm not saying anything snarky" is saying something snarky. Don't congratulate yourself too much.
2
 
didn't know you could think about something else than high school geometry
 
8:33 PM
maybe worth noticing that some books treat this kind of notation slightly differently for 'indeterminates' being added on top of a known thing, than they do for subobjects of something that is already known (where the definition might be simpler, but also not meaningful for indeterminates).
 
Ay yo may man Shif.
 
@shintuku Go on. Keep arguing. Then read p. 344 of Artin.
 
SHould I do a quick refresh of Series before I look at the implicit and inverse function theorems in CHap 6?
 
@TedShifrin you get a pass that is almost not free, he says as I do in the first edition at page 494 and you're citing the second edition
 
 
2 hours later…
10:47 PM
I'm requesting to have my rude-abusive flag on this post disputed, because I remember flagging it for being gibberish, but it seems now that the gibberish was just self-vandalism
 
0
Q: About nilpotent Jordan algebras, matrix representations and formally real algebras

mickGiven an non-commutative associative unital algebra A of characteristic $0$, one can construct a Jordan algebra $A+$ using the same underlying addition vector space. Notice first that an associative algebra is a Jordan algebra if and only if it is commutative. To make it commutative we can define...

any ideas ?
Or this one
1
Q: Generating residues with $ a^n + b^n \mod p $

mickSay there exist some non-zero distinct residues $a,b$ such that $$ a^n + b^n \mod p $$ generates all nonzero residues for some $n$. Does such a pair $a,b$ exist for every odd prime $p > 13$ ? Or for all sufficiently large prime $p$ ? And if so, how many pairs exist as a function of $p$ ? I consid...

 

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