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00:02
but total boundedness is preserved, right?
oh yes. Ofcourse, because its equivalent to compactness
wait, no. It isn't
lol.
Yeah, I just proved that total boundedness is preserved
00:17
@topologicalorientablesurface note that however metrics inducing the same topology don't even preserve boundedness in general
(total) boundedness is really a property of the metric rather than the topology
01:08
What does $\{$ $0,1$ $\}^n$ where $n$ is a natural number mean?
the cartesian product of $\{0,1\}$ with itself $n$ times, usually
or you can think of it as the set of functions from $n=\{0,...,n-1\}$ to $\{0,1\}$
which either is the same thing or just naturally identified with it, depending on your conventions
or, the corners of the $n$-dimensional unit hypercube
 
2 hours later…
03:30
0
Q: Symmetries of the Cayley group law table for composite $|G|$?

EnjoysMathHere is the group law for $\Bbb{Z}/6\Bbb{Z}$. As well as those other group law tables for small group order. I noticed that no matter how you order the elements for a group, you can always Block-decompose the square multiplication table for group of order $|G| = mn$ into $m\times m$ submatrices...

 
4 hours later…
07:26
@Semiclassical Sir are you there?
08:00
@BalarkaSen hi
@BalarkaSen have you gone back home?
Not yet, flight is at night
09:05
Hi @Balarka
how many people had the internet in 1964?
Hi @Alessandro
How long will you have to stay home because of the virus?
At least two weeks
Possibly three
@BalarkaSen did you know that any two smooth plane curves of the same degree are diffeomorphic?
09:20
I see, is your uni doing online classes in the meantime or are you free to do nice maths in the meantime?
At least one of my professors is doing online classes, but I think the other classes are postponed in the meantime
So yeah I would be able to do my stuff
@LeakyNun Like, curves in $\Bbb{CP}^2$, right? That's degree-genus.
@BalarkaSen but genus just gives you homeomorphic right
Orientable closed connected 2-manifolds are determined by genus in both TOP and DIFF
@BalarkaSen nice
TOP = DIFF in dimension 2
09:23
what
anyway the proof given to me seems too simple to be true
so might I ask you about it
let H be the set of homogeneous polynomials in 3 variable of a fixed degree
Then P(H\0) makes sense where P is projectivization
then consider S, the subset of P(H\0) consisting of the smooth curves
etc etc S is path-connected
so given two smooth plane curves f0, f1 in S
Yeah that works
let γ be a path and consider the pullback {(t,f,z) | γ(t)=f, f(z)=0}
now this is the part I don't understand
You get an I-bundle of curves which restrict to f0 and f1 at the ends
09:26
somehow the vector field d/dt lifts to a vector field in the pullback
But bundles over intervals are trivial
and then you can flow f0 to f1
so the part that I don't understand is why d/dt lifts to a vector field
I mean, why can't I do this to flow along the cobordism from S1 to S1 U S1
Right, I will tell you, give me a second to parse my thoughts.
ok thanks
Let $E = \{(t, f, z)| \gamma(t) = f, f(z) = 0\}$. Then you have a map $\pi : E \to [0, 1]$ where $\pi(t, f, z) = t$, agreed?
09:31
right
I claim $\pi$ is a submersion.
aha
so it's precisely because the curves are smooth
thanks
Right. In general if you have a cobordism $M$ between $W_0$ and $W_1$ which admits a submerison $\pi : M \to [0, 1]$ such that $\pi^{-1}(i) = W_i$ for $i = 0, 1$, $M$ is the trivial cobordism
The proof is the same thing you described, by flowing along the horizontal lift of $\partial_t$ by $\pi$
This is "baby Morse theory"
ok $\pi$ is a submersion, and then how do you pullback the vector field?
Consider $TM = \ker d\pi \oplus H$ where $H \subset TM$ is a $1$-dimensional subbundle
09:35
where did that come from?
That is always true, right? $d\pi : TM \to T[0, 1]$ is a fiberwise surjective map of bundles, so look at the kernel. You get a short exact $0 \to \ker d\pi \to TM \to T[0, 1] \to 0$ sequence of bundles.
yeah but why does it split?
Any SES of real bundles does. Here's a splitting explicitly; choose a Riemannian metric on $M$, and look at the orthocomplement of $\ker d\pi$. Call that $H$.
Then $H = (d\pi)^*(T[0, 1])$ by submersivity of $\pi$
2
Q: Split exact sequence of vector bundles

user7090I have read that a short exact sequence of differentiable vector bundles is always split. I was interpreting this as the splitting being fiber wise, i.e that the fibers of the middle component of the short exact sequence are isomorphic as a $\mathbb{C}$-module to the direct sum of the fibers of t...

hmm
it's a bit hard to believe that every SES of real bundles splits lol
Since $T[0, 1]$ is the trivial $\Bbb R$-bundle on $[0, 1]$, it's pullback to $M$, $H$, is also a trivial $\Bbb R$-bundle
So you can take a section, which is your lift of $\partial_t$, and flow along that
@LeakyNun Right, I imagine this is a bit confusing for algebraic geometers
I run into conversation difficulties a lot with my roommate, whose bundles are always complex/algebraic vector bundles
09:42
this is weird
Here's a different perspective. Once you choose a Riemannian metric on $M$, you can define gradient field $\nabla \pi$ of the map $\pi : M \to [0, 1] \subset \Bbb R$. Since $\pi$ has no critical points, $\nabla \pi$ is a nonzero vector field on $M$ which flows from $W_0$ to $W_1$
Inward pointing on $W_0$ and outward pointing on $W_1$
You simply flow along this gradient. That is Morse theory exactly; if you have a Morse function on your cobordism with no critical points then it must be the trivial cobordism
did you just generalize Morse theory or something
There's no global change of topology otherwise you'd have seen it manifested as a critical point of the height
@LeakyNun I am just explaining why this is baby Morse theory :)
brilliant
Here's a general result; if $\pi : M \to N$ is a smooth map which is surjective and submersive then $\pi$ is a smooth fiber bundle.
09:45
:o
why?
This is Ehresmann's fibration theorem, the proof is a little more work because you end up lifting $\partial/\partial x_1, \cdots, \partial/\partial x_n$ and flowing along them to construct local trivializations for $\pi$. But same idea.
brilliant
In mathematics, or specifically, in differential topology, Ehresmann's lemma or Ehresmann's fibration theorem states that a smooth mapping f : M → N {\displaystyle f\colon M\rightarrow N} where M {\displaystyle M} and N {\displaystyle N} are smooth manifolds such that f {\displaystyle f} is a surjective submersion, and a proper map, (in particular, this condition is always satisfied if M is compact...
this one?
Yeah.
Oh I forgot properness, my manifolds were compact. Sorry about that.
and locally trivial fibration means fiber bundle?
In this context, yeah.
09:50
wait doesn't locally trivial already mean fibre bundle
Problem 2 is a proof
interesting stuff
Yeah I never really understood why people use "locally trivial fibration" instead of "smooth fiber bundle"
Maybe I miss some subtlety. They're the same thing to me
@LeakyNun The cool fact is that there is a singular version of Ehressmann fibration theorem
Let's see if I can state it
ok go ahead
I will try to specialize to an interesting case than give the full blown general version. Suppose $X$ and $Y$ are projective algebraic varieties such that the singularities are "nice" in some sense (for example, singularities arising from non-flat families of curves eg Whitney umbrella where a sequence of nodes degenerate to a cusp isn't allowed).
Let $X_1 = \text{Sing}(X)$ and denote $X_n = \text{Sing}(X_{n-1})$ iteratively. Define $X_0 = X$. Denote $S_n = X_{n-1} \setminus X_n$, and similarly $T_n = Y_{n-1} \setminus Y_n$. These are the "smooth strata of depth $n$" of $X$ and $Y$ respectively.
$f : X \to Y$ be a regular map such that $f^{-1}(T_n) \subseteq S_n$.
Suppose $f|S_n : S_n \to T_n$ is proper surjective submersion for each $n$.
Then $f$ is an uh um
Ok I dunno how to say it in a super motivating way but $f$ will also be locally trivial in some appropriate and nontrivial sense
@BalarkaSen I dunno probably historical
Makes sense, @MikeMiller
10:03
@LukasHeger gibts nichts hier
OK, maybe even easier: Suppose $X$ is a projective algebraic variety with nice singularities, and $Y$ is smooth (necessarily quasiprojective, as you will see). Assume that $f : X \to Y$ is a map such that $f|S_n : S_n \to Y$ is a proper surjective submersion for all $n$.
@LeakyNun A surjective map of real vector bundles splits because of partitions of unity. Given p: E -> F, note that the set of sections of p: E_x -> F_x is (noncanonically) diffeomorphic to Map(F_x, ker(p)), which is contractible.
In particular one can find that there is a finer bundle S(p) with contractible/convex fibers whose fiber at x is the space of sections of p_x.
Then there is a homemorphism $\Phi : X \to E$ where $E$ is an $F$-bundle over $Y$ such that $\Phi$ takes fibers of $f$ to fibers of $E$. (Here $F$ will be some space with singularities)
Any fiber bundle with contractible fibers has a section. It's easiest when they are actually convex fibers, as here.
Then it's easy to give a formula. Pick a section over each chart. pick a partition of unity. Add them all up.
that's a lot of cool information
thanks
 
2 hours later…
ABC
ABC
11:49
If a differential form $A(x,y)dx+B(x,y)dy$ is exact I know that exist a potential function that has $\frac{d(U(x,y)}{dx}=A(x,y)$ AND $\frac{d(U(x,y)}{dy}=B(x,y)$.
So now I know that $\int_{\alpha} (\omega) = \int_a^b \frac{dU(x,y)}{dx} x'(t) + \frac{dU(x,y)}{dy} y'(t) dt = \int_a^b \frac{d(U(x(t),y(t))}{dt}dt = U(\alpha (b)) - U(\alpha (a))$

My problem is: I can't see that intuitively.

I see that all passages are correct and so the result is correct, but I could never have done it because I don't see the intuition behind this generalization!
Thanks in advance!
 
2 hours later…
13:52
Can someone tell me what is going wrong here? Say $V=\frac{4}3\pi r^3$ and $A=4\pi r^2$, so $V=\frac{1}3rA$. I want to find $\frac{dV}{dA}$ which then should be $\frac{1}3r$ however this obviously isn't the right answer. Which step is causing this discrapency? I assume it's got to do something with that lone $r$ and the chain rule but I can't figure out the details.
And what would I need to change if I were to correctly differentiate V with respect to A instead of differentiating both V and A with respect to r?
The area is a function of radius and vice versa
You can do the problem treating r and V separately but that requires the multi variable chain rule
You’re better of expressing V as a function of r alone or of A alone, and only then differentiating
I tried writing V completely in terms of A but that gives a different result than the expected one
Let me write it out
I have to find the maximum value of theta ($0 \leq \theta \leq \pi/2$ )given the expression $$ \tan \theta = \frac{5.6x}{x^2 + d^2 +5.6d}$$ I thought that tangent is an increasing function in that interval of theta, so max value of $\tan \theta$ will of course give me the max value of $\theta$
Well then, I just tried it again and it worked
And since the max value of $\tan \theta$ is $\infty$ therefore I tried setting up the denominator equal to zero
14:07
I must have made a mistake the first time because writing V completely in terms of A doesn't cause this problem
@Semiclassical Thanks for the help, I hadn't heard about the multivariable chain rule before, I'll check it out.
That is $$ x^2 + d^2 +5.6d = 0 \\ x = \sqrt{-d^2 -5.6d} $$
But you see it’s strange that I’m getting negative value under the root sign.
What’s the mistake? Why there is a discrepancy?
But now if I follow the calculus step, that is setting up the derivative equal to zero $$ 5.6(x^2 +d^2 +5.6 d) - 11.2x^2 =0 \\ x = \sqrt{d^2 + 5.6 d}$$
14:40
Why should the derivative of $\tan\theta$ be 0 at $\tan\theta=\inf$? I believe this rule is only used to find finite maximums.
Your first answer seems to be correct (I did some rudimentary graphing)
@Typo Well to find the maximum of any function we find it’s derivative and set it to zero. I didn’t set the derivative of $\tan theta$ to be zero, I set the derivative of $\frac{5.6x}{x^2 +d^2 +5.6d}$ to zero
14:56
Yes but that is if they have local, finite maximums.
$\tan\theta$ is strictly increasing so it doesn't have a slope of 0 as $\theta$ approaches $\frac{\pi}2$ (Sorry the reasoning might be a little off, I do not know how to phrase it in English)
@Typo Yes, I agree. But then how to find the $x$ for which that expression become maximum
?
Well as I said, your first answer seems to be correct
after I did some basic 3d graphing
I suggest you wait for someone more versed in this topic to reply, I'm just a learner myself
I got it now
$\tan \theta$ Would behave graphically as $$\frac{5.6x}{x^2 +d^2 +5.6d}$$
And hence my aim would be to find the max value of that rational function.
 
1 hour later…
16:14
@Thorgott Do you know statistics?
16:28
@AbhasKumarSinha Yes he knows it.
didn't reply anything :|
@AbhasKumarSinha He is not online
@AbhasKumarSinha You ask your question he will be here around 14:00 or 15:00
Okaiii.....
Second point there^
Is that law of large numbers?^
16:38
Ping him
Weak law of large numbers to be specific @Thorgott
@AbhasKumarSinha You know something about Maxima and Minima?
@Knight yap.
@AbhasKumarSinha All right! $\theta$ is one of the angles of a triangle, $x$ is positive variable and $d$ is a positive constant
16:43
We have the relation $$ \tan \theta = \frac{ 5.6x}{x^2 + d^2 + 5.6d}$$ and we want to find the maximum value of $\theta$
That is we want $x$ for which $\theta$ is maximum
How to go for it?
@Knight $\tan \theta = \dfrac p h$? right?
So, the fraction can be written as a function of $x$ as $f(x)=\frac{5.6x}{x^2 + d^2 + 5.6d}$
Okay!
But triangle need not be necessarily a right triangle
@Knight What you mean by Right triangle? it's not given in the question.
Find, $x_o \in \mathbb R$ such that $f'(x_o) = 0$ and $f''(x_o) < 0$
@AbhasKumarSinha Well you said $\tan \theta = \frac{p}{h}$ That’s the definition we use only for right triangles
@Knight no, $\tan \theta$ is defined for rightangles doesn't mean that other triangles don't have that.
then, $\theta = \tan^{-1} f(x_o)$
16:49
@AbhasKumarSinha Agree so far
Why a subscript 0?
Is it your style or something else?
@Knight $x_o$ is a particular number which satisfies the condition:
2 mins ago, by Abhas Kumar Sinha
Find, $x_o \in \mathbb R$ such that $f'(x_o) = 0$ and $f''(x_o) < 0$
Where as $x$ is a random variable.
Okay,
You will get $x_0 = \sqrt {d^2 + 5.6 d}$
3 mins ago, by Abhas Kumar Sinha
then, $\theta = \tan^{-1} f(x_o)$
$$\theta = \tan^{-1} \frac{ 5.6 \sqrt{d^2 +5.6d}}{2d^2 + 11.2 d}$$
Is this what you’re saying?
yap, I guess
16:56
@AbhasKumarSinha But you see $$ \tan \theta = \frac{5.6x}{x^2 +d^2+5.6d}$$
hmm... I don't check calculations.
At $\theta \approx \pi/2$ we would have $\tan \theta \approx \infty$
Means the maximum value of theta is pi /2
@Knight no, why $\theta/2$?
no
maximum value is
4 mins ago, by Knight
$$\theta = \tan^{-1} \frac{ 5.6 \sqrt{d^2 +5.6d}}{2d^2 + 11.2 d}$$
not theta/2
@AbhasKumarSinha When I wrote $\theta /2$ ?
@Knight sorry I mean $\pi/2$
17:01
Why? At maximum value of theta we should get the maximum of tangent function, so at theta close to $\pi/2$ we have a very large tangent function
@Knight no, maximum value is that^ not $\pi/2$ and no, it should get to $\pi/2$ as you are claiming, but the fractions of that form can't ever have the value of $\pi/2$
So, the maximum value is that^
See this expression $$ \frac{5.6x}{x^2 +d^2 + 5.6d}$$ we want to make it large, beacuse a large $\tan \theta$ will ensure a large $\theta$, Do you agree so far?
@Knight k, but we can't make an expression of that form to become as much large as we like, they've some limits (bounded fractions)
@AbhasKumarSinha But it is equal to $\tan \theta$
17:07
And we can make $\tan \theta$ as large as we want, eh?
Why?
take this fraction for example $\frac {1} {x^2} $, can you make it smaller than 0?
No, (if you mean the whole expression is squared)
We can make it as close to zero as we want
@TedShifrin If you happen to come here today then please resolve this doubt of mine (the doubt which we are discussing)
@Knight exactly, that's the case with your fraction, we can't make is bigger than specified number.
@Secret help him, I'm busy now...
I've to go out to buy medicines now.
So, sorry....
bye.
17:12
@AbhasKumarSinha Why? Are you sick?
@Knight No, I'm completely fine, I just need basic medicines,
@AbhasKumarSinha For corona virus?
uh... differentiate that by x and find stationary points?
@Knight acidity.... XD
heh
@Secret What is stationary point?
@AbhasKumarSinha I knew :)
17:15
what study level are you on. Do you learn calculus yet?
@Secret Yeah! Of course :-)
@Ante How old are you?
you ask for my age or?
@Knight I guess that you are confused because, of that tan thing, no trigonometry $\neq$ tright angled triangles...
$\tan \theta$ is just a real number, nothing related to triangles here.
@Ante Yes
almost 35
17:18
@AbhasKumarSinha Yes you’re right
@Ante Wow! When we got your birthday?
a little less than two months
@Ante From this day? I mean from March 17th ?
yes
almost exactly two months
18.5.
There we go! Book a ticket for me upto Croatia, I will attend your party
party can be expected only if i solve Beal´s in this month
17:21
@AbhasKumarSinha You’re also invited
@Ante Can you please help me with that problem?
you all are invited i just need that 1 000 000
which one, Beal´s?
@Ante It seems very close to Fermat’s last Theorem
yes it is
@Ante Please help me $$ \tan \theta = \frac{5.6 x}{x^2 + d^2 + 5.6d}$$ find the value of $x$ for which $\theta$ is maximum
d is some constant?
17:25
$\theta$ is an angle of any triangle, $x$ is a positive real variable and $d$ is also a positive real constant
Can we do it without Calculus?
i do not know, it seems easier to transform it to 1/tan and then to search minimum of 1/tan
@Semiclassical Hello
@Ante That is maximum of $5.6x$ ?
Sometimes, elementary real analysis weirds me out: If we sum 1/p over prime p, then the sum diverges. But if we sum 1/(p-1) instead, it converges to 1.
@Semiclassical Wow! That’s a great result
Yeah. These are classic series, to be clear—nothing novel on my part
17:30
Would you mind seeing my doubt? If you are not too busy
That problem seems gross
Not difficult but just kinda bleh
Especially without calculus
are there any other nontrivial series such that 1/m diverges but 1/(m-1) converges?
i mean ,there are, but an example of
@Semiclassical My main problem is $\tan \theta$ can go to $\infty$ but that rational fraction cannot.
without differentiation, what if you write tan(a)x^2-5.6x+tan(a)(d^2+5.6d)=0 and consider a family of quadratic equations, each for every a? is that an overkill?
@Knight I don't see why that is a problem. Just take arctan and do the normal second derivative test, keeping in mind that you are restricting the values of $\theta$ to a suitable domain
17:39
@Knight sure? So y=f(x) can’t be made arbitrarily large by changing x, and thus theta can’t be made arbitrarily close to pi/2
@Semiclassical I don’t know why I’m feeling that $\tan \theta$ can be made arbitrarily larger
If it were a function of theta, it would be
But it’s not in this case
it’s a function of x, because you have theta = arctan( rational function of x )
@Semiclassical OHO! It’s a function of $x$ that solve my problem
tan(a,-x,d)=-tan(a,x,d), so only positive x need to be considered?
I’m assuming the range of theta is restricted to (-pi/2,pi/2), yeah
17:44
Thank you everyone @Semiclassical @Ante @SayanChattopadhyay
But if you don’t do that, then theta isn’t unique
Yes
Depends on the problem statement ofc
Theta is in interval of 0 and 180 degrees, because it’s angle of a triangle
May I shoot something from Physics?
experimental of theoretical?
17:47
Theoretical, magnetostatics
:(
Why?
Be Happy your birthday is near :)
My doubt is how he could write $I = I_s \hat{s} + I_z \hat{z}$ as $$ I = \langle I_s \cos \phi, I_s \sin \phi, I_z \rangle$$
Because $\phi$ is the angle by which the vector $\mathbf r'$ has been rotated from the $x-$ axis, why the current got that angle in it's component?
you have two different angles i think $\phi$ and ${\phi}^´$
and they are somehow related
Let's consider the current at $\mathbf r'$ with figure, the blue arrow shows the current
@Ante No we got just one angle $\phi '$
and for simplicity I have written it without primes
@Semiclassical are you there?
i think that $\phi$ is the angle associated to $r$ and ${\phi}^´$ associated to $r´$
17:59
@Ante Where we got $\phi$ ?
by transforming the coordinates because $r$ and $r´$ are related somehow
find the relation between them
@Ante $\mathbf r$ is the point where want our field and $\mathbf r'$ is the point where we got the current
can $Li(x)$ be expressed as a differential equation? (logarithmic integral)
that yields insight on the primes?
logarithmic integral?
18:07
as differential equation with what-like coeeficients? polynomials?
well I was thinking of putting a flow on Li(x)
what kind of flow?
hyperbolic flow
but I don't know if it's possible
can you type more details of what you want to achieve?
yeah
start with the logarithmic integral and operate on it with a differential operator $d/dx(Li(x))$
then exponentiate, so $\exp(d/dx(Li(x)))$
and then do a change of variables to get $1/x$.
I'm not sure what this process preserves, if anything
but you can define a hyperbolic flow on $1/x$ so maybe you could work backwards to put the flow onto $Li(x)$?
18:21
there are series representations of Li(x), did you try to differentiate them and to observe if there is some pattern?
@Ante a little yeah
you think that Li(x),Li´(x),Li´´(x),...,Li^(k)(x),.. all have some information about primes?
no I just took $1$ derivative
well all of them should contain some information about primes
even up to a point of the reformulation of RH
yeah I think they have some information about primes
18:52
any ideas?
0
Q: Can some triangle be partitioned into the almost-disjoint union of infinite number of regular $n$-gons, one for each $n \geq 3$?

AnteThat is, does there exists triangle $T$ and regular $n$-gons (one for each $n \geq 3$) $R_3,...,R_{k+3},...$ such that $T=\bigcup_{n=3}^{+ \infty}R_n$ and $R_n \subset T$ for every $n \geq 3$ and if $m \neq n$ then $R_m \bigcap R_n$ is either empty or is a line segment or is a point? This is jus...

19:10
Am I right to assume the provided algorithm for modular exponentiation here has an error?
It would seem that it would be claiming that:
$2^{n+1}\bmod2\equiv2(2^n\bmod50)\bmod2$
which is obviously zero?
er no, I that should be:
$2^{n+1}\bmod2\equiv2(2^n\bmod50)\bmod100$
It would seem to me that this process can be sped up though, in the case of repeated factors
 
1 hour later…
20:35
Given $\mu(X)<\infty$ and $f$ be a bounded measurable function, then $\int\,f=\inf\{\sum_{i=1}^{m}\mu(A_i)\sup_{A_i}f\}$ where $A_i,\dots,A_m$ are partition.
$\int\,f\leq\sum\mu(A_i)\sup\,f$ by boudeness, for the left > right, we define $f>\sum\alpha_i\phi$ where $\phi$ is simple function
forgot to say that $f$ is non-negative
21:03
in base 3, what are the two numbers lying between $0.1 and 0.11$?
21:17
0.101 and 0.102
@topologicalorientablesurface "the" two numbers?
I think my idea for > is incorrect
22:04
@topologicalorientablesurface What about $0.1000212$?
@knight I got your ping. I am on iPad for the indeterminate future, so chat is more difficult. I don’t know what you wanted.
22:16
I'm trying to get back into algebra and this question is bugging me again. According to this question and answer: math.stackexchange.com/questions/73652/… if you have a ring that's an integral extension of some subring and you quotient by an ideal, the quotient is an is an integral extension of the quotient of the subring by the intersection of the ideal with the subring.
But what happens if, when you take the quotient of the subring, the polynomial that defines the integral extension already has a solution?
for example, Z[(sqrt(2)] is an integral extension of Z, and you can quotient the extended ring by the ideal 7Z[sqrt(2)], but if you intersect 7Z[sqrt(2)] with Z, you get 7Z, and in Z/7Z, x^2-2=0 has two solutions, 3 and 4
22:34
Suppose you have a ring $R$ with generators $\{r_i\}_{i\in I}$. Is there a name for the subring $\langle r_i^n+r_j^n-(r_i+r_j)^n\mid i\neq j,i,j\in I,n\in\mathbb{N}\rangle$? Is it clear whether this subring can even be a proper subring?
It's trivial when you have only one generator, so it can be proper
True enough. (So many times I forget to check the trivial case)
22:54
Hi @Ted
Hi, demonic Alessandro!
hi @Ted
Hi Leaky. Glad to see everyone's Ok so far!
Oh oh, it's CaptainA.
@KrullDimension $\Bbb Z[\sqrt2]/7\Bbb Z[\sqrt2] = ``(\Bbb Z/7\Bbb Z)[\sqrt2]" = (\Bbb Z/7\Bbb Z)[X]/(X^2-2) = (\Bbb Z/7\Bbb Z)[X]/((X-3)(X-4)) = (\Bbb Z/7\Bbb Z)^2$
@LeakyNun The squared means the direct sum of the ring with itself, right?
23:25
@TedShifrin I've started self-studying again, albeit a bit more organized. I have a "schedule" now lol
You mean now that school is on hiatus?
I started before, but I've added a couple more hours a day due to the free time.
23:43
Did you watch round one of the candidates? @Leaky
@AlessandroCodenotti I saw the Giri - Nepo match on agadmator and am now watching Ding Liren - Wang Hao on agadmator
Giri - Nepo was very nice
I saw the live coverage on chess24 of the first round
Sasha - Kirill was sad, Grischuk had a big advantage, but as he always is he was in huge time trouble and got a draw
I'm rooting for Grischuk
3 hours ago, by Simple
Given $\mu(X)<\infty$ and $f$ be a bounded measurable function, then $\int\,f=\inf\{\sum_{i=1}^{m}\mu(A_i)\sup_{A_i}f\}$ where $A_i,\dots,A_m$ are partition.
this is upper lebesgue integral, I need to define $f<g$ where is $g$ is a simple function. the forward direction can simple apply monotonicity, struggling on the reverse direction

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