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04:00
If I recall the notes I have read from you more than 5 months ago, a piecewise constant function C(x) is one where the implied derivative is zero everywhere, right?
LOL
burn that entire thing out of your mind
unfortunately that is no longer the implied derivative
it never was
it was a deluded half attempt i tricked myself into believing was equivalent
it's not
no
piecewise constant functions are functions for which the left and right hand derivatives are 0 everywhere
(one can just consider the left and right derivatives for discontinuous functions by using the left or right limit as the point to construct the tangent line)
wouldn't just left or right derivative be everywhere zero enough for piecewise constant, why do we need both sets to be zero?
@Secret "a piecewise constant function C(x) is one where the implied derivative is zero everywhere, right?" likely so, but now I have to actually prove that rigorously
@Secret was not sure if one implied the other
safer to just say both
just in case
I mean, a typical step function looks like (a,b],(b,c],... where each interval has different function values that is some constant function. You may also have something like [a,b),[c,d],(d,e],... but just one of the left or right derivatives is enough to capture the constancy?
shrugs
in math it is sometimes best to state all requirements you know are sufficient and excluding
even if redundant
worst is that im just being redundant
i know im not excluding things from the set by accident
@Secret as a question of curiosity though
if I wanted to say the partial derivative of g(x,y) at the point g(x,C(x)) what would be a good way to say that?
just (partial fraction symbol) g(x,C(x))?
(i dont remember the latex for the partial derivative symbol so bear with me)
04:10
$\frac{\partial g (x,y)}{\partial y}|_{(x,y)=(x,C(x))}$?
eeeh
you mean \partial x?
but yeah
i like that notation
thanks!
04:26
there's more than one notation
$\partial_y g(x,C(x))$ also works
umm
\partial_x
right
do you have mathjax enabled
i have it in my browser
but i dont like to turn it on
it makes my pc overwork
tends to overheat when i do it
anyway great feel free to explain what you said in the morning or some other time, when im not tired
04:40
or $\partial_y g(x,y=C(x))$ to be more explicit
04:58
@GFauxPas it's just the partial derivative of g at that point
nothing particularly special
(well one might argue it's actually a very important partial derivative but I've never actually been able to find anything discussing it)
0
Q: A difficult theorem regarding piecewise constant functions.

The Great DuckFirst we will define the left and right derivatives as follows. Left derivative of $f(x)$: $f(x)^+ = \lim_{h \to 0^+} \frac {f(x+h) - \lim_{a \to x^+} f(a)}{h}$ Right derivative of $f(x)$: $f(x)^- = \lim_{h \to 0^-} \frac {f(x+h) - \lim_{a \to x^-} f(a)}{h}$ Now let us define the set of piecew...

@GFauxPas feel free to read that for further clarification
I think all is good if i think abt characteristic eqn
$A^k=I$ then is $A^2+A^3+...+A^k-1 = O $ necessarily true, sorry for typo its thru my phone, and due travel..any link to this question or explanatikn would beuch helpful
.
I xould have searchd but approach0 not in othr device except PC or TAblet


Abpve is A +A^2 +...+A^k-1 =0
.
Next is tr(A)+tr(A^2)+tr(A^3)+...+tr(A^k-1)=-n true?
.
Is A^-1 + A^-2 + ... +A^-(k-1)=-I
True?
@BAYMAX please proofread your post
@Baymax not necessarily for the first thing, here's the problem
Do you know any algebra?
In particular, do you know what an integral domain is?
Well there's a chance that you won't be back before I have to go so I'll just explain
So basically, if you have a set with a notion of addition and of multiplication (called a ring), you can ask whether $ab = 0$ implies $a=0$ or $b=0$
Such a set is called an integral domain
Now, if $R$ is a ring, you can always form the set of polynomials with coefficients in $R$, and multiplication/addition is defined in the only way that makes sense. If $R$ is an integral domain, then this "polynomial ring" $R[x]$ is also an integral domain, this isn't hard to prove
The point is, $\mathbb{R}$ is an integral domain, so if you're dealing with numbers, then it is true that polynomials $\mathbb{R} \to \mathbb{R}$ allow you to factorize and say that if $f = gh$ and $x$ is a root of $f$, then it's a root of $g$ or $h$
Linear transformations don't form an integral domain is the problem
05:19
Thanks daminark, perhaps u forgot me, i know a bit of abstract algebra but i was trying to prove this using linear algebra, still thanks a lot for looking and explaining..
For example, $A(x,y) = (x,0)$ and $B(x,y) = (0,y)$, then $AB = 0$
Yip
So you know that $0 = A^k - I = (A-I)(A^{k-1} + \ldots + A + I)$
The problem is that what you have on the right are matrices, so it's possible that both matrices are non-zero and yet their product is $0$
(Matrices, transformations, I'm kinda jumping between them back and forth carelessly but the same principle applies)
But A cant be I as eigenvalue of A is not 1
@Daminark yes , that is ok
So that's why it's not necessarily true that $A^{k-1} + \ldots + A + I = 0$
05:27
So A-I is not zero and still we have the product of matricws zero so can we say that $A^k-1+...+I$=O
Is not equal to O i meant typo sorry
assuming characteristic $0$, then $A^{k-1} + \dots + A + I= 0$ iff $1$ not an eigenvalue of $A$
And here we are dealing with C Field so it has char 0, so it is tje geometric series of A is zero right?
If $1$ is an eigenvalue, then if we let $v$ be an eigenvector for $1$, we get that $(A^{k-1} + \dots + A + I)v = kv \neq 0$
If $1$ is not an eigenvalue, then $A-I$ is invertible, so we can multiply the equation $0=(A-I)(A^{k-1} + \dots + A + I)$ with $(A-I)^{-1}$
but $A$ might have $1$ as an eigenvalue without being equal to $1$. Consider for example $\begin{pmatrix} 1 && 0 \\ 0 && e^{2\pi i/k} \end{pmatrix}$
@BAYMAX does this answer your question? In general you can't say whether $A^{k-1} + \dots + A + I$ is zero or not, it depends on $1$ being an eigenvalue or not (see above)
I meant $A$ might have $1$ as an eigenvalue without being equal to $I$
If 1 is not an eigenvalue then we have I+A+...+A^k-1 =O
05:40
Or A+..+A^k-1 = -I
Thats nice
But wht about the trace equation?
is tr(A)+tr(A^2)+tr(A^3)+...+tr(A^k-1)=-n true?
.
Is A^-1 + A^-2 + ... +A^-(k-1)=-I
True?
ah, you might need to use the fact that $A$ is diagonizable for that
@MatheinBoulomenos
do you know about the relation of $A$ being diagonizable and the minimal polynomial of $A$?
We know A diagonalizable right as k >1
yes, it's true that $A$ is diagonizable, but that's not so easy, depending on what you know
05:45
Yup it has distinct roots abt char polynomial which implies diagonalizability
ah okay, you know that, good
Bit how do i relate that to trace?
powers of $A$ are much easier to compute when $A$ is in diagonal form
changing the basis doesn't change the trace
Yes, but why it doesnot change the trace when we power the matrix
Ah nevermind, we don't even need that diagonazabilty stuff
05:47
The eleme ts in diagonal raised to power and get added
@BAYMAX that does change the trace
note that the trace is linear
so tr(A)+ tr(A^2) + ....+ tr(A^k-1)=tr(A+A^2+ \dots + A^k-1)
Yes but if trace of A is x+y the A^m trace is x^m+y^m
If A is a diagonal matrix
But you know what A+A^2+ ....+ A^k-1 looks like when 1 is not an eigenvalue
that's just -I
@MatheinBoulomenos yes
@MatheinBoulomenos yes
So the teace of whole thing is -1
careful, the trace of the identity matrix is not 1
05:50
Oh my yes its -n
Nice
but you need that assumption on 1 not being an eigenvalue, else it won't work
Yes we have tht 1 is not an eigenvalue
Nice
So now how abt this Is A^-1 + A^-2 + ... +A^-(k-1)=-I
yes
A^-1 satisfies the same conditions as A
Hmm why
I think you can figure that out
05:55
Yes they too donot have eigenvalue 1
Hey @Mathein and @Alessandro!
Hi @Daminark @AlessandroCodenotti
Good morning!
Thank u so much@MatheinBoulomenos Any hints on my last qn of complex analysis?
@BAYMAX what was the question?
06:02
Considwr a set in which z^60 = -1 and ^k is not equal to 1 for 0<k<60? , Number of elements in the set is?
Sorry for my typos
I m on mobile
Is it clear?
Depends on how much you know about roots of unity
I wouldn't see this as a complex analysis question
So how can we approach this, kind looks like puzzle
Okay so if this has slowed down a little bit I'll finally stop procrastinating and do some p-adics
06:10
@Daminark yes, p-adics are really cool
You're looking for elements $\zeta \in \Bbb C$ such that $\zeta^{60}=-1$, but $\zeta^k \neq 1$ for $0 < k < 60$. All such elements will have multiplicative order $120$ and conversely, every element in $\Bbb C^\times$ with order $120$ will satisfy $\zeta^{60} = -1$
so the answer is $\varphi(120)=32$
So okay, on the one hand a p-adic integer is a formal power series $a_0 + a_1p + \ldots $ where $0\le a_i < p$, where I'm guessing addition and multiplication is just the same as for formal power series? The set of p-adics is $\mathbb{Z}_p$. Now, we have a map $\mathbb{Z}_p \to \mathbb{Z}/(p^k)$
@Daminark no, addition is not the same as power series
you have no carrying for power series
Err, yeah I was thinking multiplication, addition is carrying
What a coincidence, I am reading about $p$-adic cohomology right now
(or rather, $l$-adic, since this is always used for a prime $l\neq p$ when working in char $p$)
multiplication should be different as well, but I don't see why at the moment
I know that the structure of the group $\Bbb Z_p^\times$ is quite different from $\Bbb{F}_p[[T]]^\times$, so it can't be the same
06:18
It's possible that calling it that wasn't right but the point is you just define the stuff the way you expect, you just expand out
yeah
@TobiasKildetoft really cool. So you start with the whole sheaves on a site stuff?
@MatheinBoulomenos any motivation of how u could relate that to multiplicatice order?
Okay so if $\alpha = a_0 + a_1p + \ldots \in \mathbb{Z}_p$, let $\alpha_k = a_0 + \ldots + a_{k-1}p^{k-1}$. Then the map $\alpha\mapsto \alpha_k$ is gonna give you a map $\mathbb{Z}_p \to \mathbb{Z}/(p^k)$.
Like they are complex mumbers
And this map is of course compatible with the projection $\mathbb{Z}/(p^j) \to \mathbb{Z}/(p^i)$ since $\alpha_k \equiv_{p^{k'}} \alpha_{k'}$
Inconsistent notation but yeah anyway, now if we go backwards, we have a sequence $\alpha_k$ such that $\alpha_k \equiv_{p^{k'}} \alpha_{k'}$ for $k' < k$, we can inductively construct a p-adic, so this gives an inverse limit
Hmm, would there be any good example computations to get a grip on this stuff?
06:35
@Daminark what's $-1$ in $\Bbb Z_p$?
@MatheinBoulomenos No, I am just reading some down-to-earth accounts of what it is about, since it shows up in the Deligne-Lusztig construction of complex characters of finite groups of Lie type
@TobiasKildetoft ah I see
the actual construction is quite technical
Yeah, I am reading the appendix in Carter, which does not do that. It just gives the basic properties and consequences.
one of the profs here is an expert on étale stuff and it took him almost a whole semester to even define étale cohomology
@TobiasKildetoft do you remember the question about characteristic $p$ represenations of $\operatorname{GL}_n(\Bbb F_q)$? (basically that the composition factors are determined by the restriction to a maximal torus)
@MatheinBoulomenos Yeah
06:39
I have an idea on how you can maybe bypass the algebraic groups stuff
Hmm, so there's a map $\mathbb{Z}\to \mathbb{Z}/(p^k)$ which is just projecting, and presumably when you say "$-1 \in \mathbb{Z}_p$" you mean the induced map $\mathbb{Z}\to \mathbb{Z}_p$ which factors through all that?
@TobiasKildetoft The idea is that you base change your representation (just the vector space, not the group) just to $\Bbb F_{q^n}$, that's enough to give you Jordan–Chevalley decompositions of elements in $\operatorname{GL}_n(\Bbb F_q)$, I thought maybe that's enough to make things work and then you probably have to Galois descent back
just a vague idea
@Daminark yeah, that's one way to describe it
If so, then you know $a_i = p-1$
@Daminark do you see why this is actually an inverse of $1$ if you think of that as a "power series"?
06:46
yo
wassup guys?
@MatheinBoulomenos Hmm, not sure how Jordan-Chevalley decomposition gives what we need here.
So if you add 1, $p-1 + 1 = p$, so that carries over and we have $p(p-1) + p = p^2$, and etc
Hmm, along those lines, probably should evalute $(p-1 + p(p-1) + p^2(p-1) + \ldots )(p-1 + p(p-1) + p^2(p-1) + \ldots)$
@Daminark yes and the reason that this is actually zero is becauser $p^n \to 0$ for $n \to \infty$
@TobiasKildetoft yeah, it's quite possible this is useless
@MatheinBoulomenos Also, I was assuming the representation was over the algebraic closure to start with. I have never thought much about what happens over the finite field itself or how much that changes
First term is clear, second is because we have a $-2p$ and a $(-p)(-1)$ twice, and I'll just say it's clear from there since this is gonna take a while
Oh I mean
06:54
@TobiasKildetoft I thought maybe it's useful since we can make the elements from a $p$-Sylow act by unipotent matrices and then it looks like there might be a relation to Jordan-Chevalley
If we're gonna abuse some analysis here, this is a geometric series whose limit is $\frac{p-1}{1-p} = -1$
exactly, the geometric series works out the same way
@MatheinBoulomenos Yeah, that might work actually. So we first show that it suffices that they are isomorphic as reps for the Borel, when show that the claim for reps of the Borel
I wonder if this is really a thing one should be dealing with in terms of this being a group with a BN pair
@TobiasKildetoft if you restrict to the Borel and induct back, do you get the thing back you started with?
(that would probably to good to be true)
@MatheinBoulomenos No, you get something much larger
06:58
ah, of course
Unless you get fancy with how you define induction
but maybe it decomposes in a useful way?
no idea
The idea is that the Borel is generated by the maximal torus and the $p$-Sylow
I can see how it might work for the Borel. How do we get from the Borel to the whole group?
Probably by mimicking the argument for algebraic groups
07:00
hmm, I don't know much about algerbaic groups
is this stuff in Springer somewhere?
a finite dimensional rep of the Borel will have a vector which is fixed by the $p$-Sylow
probably
@TobiasKildetoft okay I can see that. I hope that we can avoid talking about weights, though
@MatheinBoulomenos Well, the weight will be there, but hidden
The idea is that this vector will span a subrepresentation as a $B$-module
Which means that we can induce and get a hom from the module induced from this $1$-dimensional module to the module we started with
If we do this for some other rep which is isomorphic to the first as a $B$-module, then this gives a submodule of that isomorphic to the one we just got, if we are a bit careful
the being careful is just making sure that these hom's share some irreducible in their image
And this is where we need the "weight", which is really just the eigenvalue of $T$ on this $1$-dimensional rep
hmm, can you elaborate a bit on that?
hmm, trying to unravel the algebraic group stuff down to this
So the idea is that really we are picking the weight of the vector to be maximal, and this ensures that the image of the hom will in each case still have a vector of this weight, and since this was maximal, it is easy to show that they must have an irreducible in common
the issue is with this order stuff
07:10
in what sense can we pick elements in a characteristic p field to be maximal?
We are not picking elements in the field
oh I thought it was an eigenvalue
We are picking characters of the maximal torus
which has the same sort of issue of course
Since really the ordering is coming from a root system, which we were trying to avoid
Ahh, but being a fixed-point for the action of the $p$-Sylow should be a substitute for this maximality
since this is really what it means to be maximal
07:13
ahh!
The picture to have in mind is that applying elements from the $p$-Sylow gives something that is at least as large as what you had before
(this is all much more clear if one is familiar with the analogous stuff for Lie algebras)
I don't know Lie algebras either :/
But an example might be good to work out
07:16
you gave me a lot of pointers, maybe that's enough to make it work
what if the $p$-Sylow fixes a more than 1-dimensional subspace?
does it matter what vector we pick then?
@MatheinBoulomenos it should not, as long as we pick the same ones for each rep (meaning we pick ones on which $T$ acts the same)
since these should correspond to different irreducibles in the original rep
okay
Thanks a lot, I'll try to write down the details for this, maybe that can at least solve going from the maximal torus to the borel subgroup, which seems like a big step
07:29
Is there a simple way to see that if we extend scalars to a larger field in characteristic $p$ (say from one finite field to a larger one), then semisimple representations get mapped to semisimple representations?
Hmm, no idea actually
Is it even true?
I have a proof for finite groups
For finite fields, I would guess this follows by some Galois theory
Ohh, I read the statement the wrong way around.
In general, it is easier to be semisimple over a larger field I guess
oh yeah, I just proved it for finite fields
but at least perfectness seemed to be important
Yeah, the field not being perfect might make the intuition above wrong
07:34
How would I go about figuring out $\langle f,g,h \rangle / \langle f + h, g + h \rangle$?
@Fargle Is the top one freely generated?
free abelian, yes
@TobiasKildetoft Here is the proof: overleaf.com/read/qvnwssyxgxbd#/65011491
note that I used that the Brauer group of finite fields is trivial
I don't like that proof much, I thought maybe there is some more abstract/general stuff
@MatheinBoulomenos Quite possibly. I have never dealt much outside algebraically closed fields
@Fargle So you can rewrite it as modding out by $(1,0,1)$ and $(0,1,1)$
Alright.
I have $f = g = -h$.
07:42
@Fargle Ok, so now you can rewrite in terms of two generators
Rewrite which thing?
@TobiasKildetoft I can reduce much of the proof to the noncommutative algebra question: if $A$ is, say an Artinian $K$-algebra and $L/K$ is separable, is the Jacobson radical of $L \otimes_K A$ equal to $L \otimes_K \operatorname{J}(A)$? If that holds, then much of the proof is unnecessary and the result would follow in more generality
@MatheinBoulomenos The reason I brought up Lie algebras is that some of these things look a lot nicer, given that there we replace "is a fixed point for" with "is killed by". This means that we get very nice stuff like "if $v\in V_{\lambda}$ then $x_{\alpha}v\in V_{\lambda + \alpha}$" rather than "... then $x_{\alpha}v\in V_{\lambda} + V_{\lambda+\alpha}$"
@MatheinBoulomenos Hmm, that is a good question, which must surely be known to people working with artin algebras
@TobiasKildetoft I see
@MatheinBoulomenos Note in those statements what happens if $\lambda$ is "maximal" and $\alpha$ is "positive".
07:47
@TobiasKildetoft I don't really understand where those indices are coming from (root systems, I guess)
@MatheinBoulomenos ahh, the indices are characters of $T$ and $V_{\lambda}$ is the set of elements of $V$ on which $T$ acts via $\lambda$.
ah, so then $x_\alpha v=0$?
right, in the Lie algebra version
08:04
@TobiasKildetoft ah, the thing I mentioned about how the Jacobson radical of a $k$-algebra behaves under scalar extensions along separable field extension is actually a theorem in Lam, in full generality (no Artinian assumption), so that applied to group algebra gives that scalar extensions of semisimple representations of any group along a separable field extension is again semisimple
08:18
@MatheinBoulomenos Seems like I was wrong. Carter does apparently define all of the ingredients for $l$-adic cohomology. He just does so very briefly and with no proofs.
08:33
another piece of purity tainted with reality lol
Pretty decent for a Quanta article. Still required reading most of it before it was clear what it really was that the researchers had done
I don't read Quanta much, but are they usually bad?
@Daminark They tend to either exaggerate what has been done, or sometimes just make it sound like something else entirely
The reason is that it is written for people without any real knowledge of math
And sometimes the imprecision of the explanations become very misleading
Ah that's no good
08:59
I remember hearing someone saying 'number theory becomes hard when your primes are denoted by $l$ instead of $p$'
09:53
Hey @Alessandro
How's it going?
10:05
@Alessandro whats a good post rock record for me
10:24
@MatheinBoulomenos Ok, now I have read through the definition of $l$-adic cohomology. It does not seem as complicated as I had feared
@BalarkaSen Lift Your Skinny Fists Like Antennas To Heaven
That's a classic
I went through the entire discography of GY!BE the past week unfortunately :)
Though I had heard Lift Yr. before
lel
@BalarkaSen GY!BE is going to play in Milan in July but it's right before my graduation so I can't go :/
Oh :(
That sucks
10:32
Did I already suggest you "not for want of trying" by maybeshewill?
No!
I have not heard of that
That's my suggestion then!
Noted. Will listen today
In topology and related areas of mathematics, a subset A of a topological space X is called dense (in X) if every point x in X either belongs to A or is a limit point of A, that is the closure of A is constituting the whole set X. Informally, for every point in X, the point is either in A or arbitrarily "close" to a member of A — for instance, every real number either is a rational number or has a rational number arbitrarily close to it (see Diophantine approximation). Formally, a subset A of a topological space X is dense in X if for any point x in X, any neighborhood of x contains at least one...
One reason I like general topology more than metric spaces: No need to have more than 3 absolute signs and inequalities in one line, again and again just from reading one proof
In mathematics, a subset A {\displaystyle A} of a topological space is said to be dense-in-itself if A {\displaystyle A} contains no isolated points. Every dense-in-itself closed set is perfect. Conversely, every perfect set is dense-in-itself. A simple example of a set which is dense-in-itself but not closed (and hence not a perfect set) is the subset of irrational numbers (considered as a subset of the real numbers). This set is dense-in-itself because every neighborhood of an irrational number ...
and that is one reason why it is so hard to plot the rationals without it look like a straight line
Now, I just need to think of or find a nice topology on the reals such that a given point $a$ is not a limit point even though it satisfy the following:
> for all $x \in \Bbb{R}$ and some fixed $a \in \Bbb{R}$ such that $x < a$, there exists $y \in \Bbb{R}$ such that $x < y < a$
The above statement, pictorically speaking, defines a net of nested intervals in the reals whose length get arbtrarily small as it approaches $a$
4
Q: a different nested intervals theorem

James ProppIs there any literature on (and a standard name for) the proposition that for any arbitrary-cardinality collection of closed intervals in the reals that is nested (in the sense that, given any two of the intervals, one of them is a subset of the other), the intersection of all the intervals is no...

12:06
@Secret interestingly, I’ve run into that sum-of-squares business before
Namely because certain problems in QI can be formulated in terms of semidefinite programming
@Semiclassical can i borrow your eye sockets?
Eg finding the maximum violation of Bell inequalities by a quantum system
I'll take that as a no, then.
Sorry, phone ran out of battery @TheGreatDuck
What’s up?
ah no problem
i assumed you had to leave for some reason
These two questions
-1
Q: A question regarding certain partial derivatives of $f(x) = 0$.

The Great DuckFirst we will define the left and right derivatives as follows. Left derivative of $f(x)$: $f(x)^+ = \lim_{h \to 0^+} \frac {f(x+h) - \lim_{a \to x^+} f(a)}{h}$ Right derivative of $f(x)$: $f(x)^- = \lim_{h \to 0^-} \frac {f(x+h) - \lim_{a \to x^-} f(a)}{h}$ Now let us define the set of piecew...

-1
Q: A difficult theorem regarding piecewise constant functions.

The Great DuckFirst we will define the left and right derivatives as follows. Left derivative of $f(x)$: $f(x)^+ = \lim_{h \to 0^+} \frac {f(x+h) - \lim_{a \to x^+} f(a)}{h}$ Right derivative of $f(x)$: $f(x)^- = \lim_{h \to 0^-} \frac {f(x+h) - \lim_{a \to x^-} f(a)}{h}$ Now let us define the set of piecew...

Is it actually true that they are exact duplicates
I feel like people are just not reading the whole post when voting or something....
:-/
12:15
I’ll tajr a look once I’m on my laptop
Though, it looks from the thumbnail that the opening sentences are identical. That predisposes people to view them as being identical
@Semiclassical People shouldn't issue close votes based on tumbnails. By that reasoning someone cant ask two questions about differentiation within a few minuts of each other and reuse the definition of the derivative because they don't feel like rewriting background stuff.
Presumably the two questions have the same background but proceed in different ways
Did you read them
12:21
Not yet. But before I say more than that I should
indeed
i feel like people are voting by simply skimming
that's not right
Well, it’s a matter of degree. If two papers are identical in the first ten pages and only differ after that, I don’t think it’s unreasonable to miss the difference. (That is obvious hyperbole compared to yours, but the principle is the same)
One way to avoid that impression would be to arrange the text so that the question appears first, and only then give the (identical) background
Alternatively, you could replace the background in the second question with a link/reference to the first question and s statement to the effect of “it’s the same background so I won’t reiterate it”
12:37
Help for a clearer solution: $k$-form $\omega$ is smooth if only if it is smooth as a map $\omega \colon M\rightarrow \Lambda ^k(M)$. link: math.stackexchange.com/questions/2818346/…
As it stands, you’ve given two questions with identical opening paragraphs and similar format past that. This makes it really easy for people to file them away as duplicates and miss the relevant differences
That’s lazy on their part, but you’re not doing yourself any favors with that approach
13:11
@TheGreatDuck I would also suggest that question titles such as "A difficult problem in [BLANK]" or "A hard theorem about [BLANK]" typically don't impress people.
Your question title should clearly describe what is being asked in the question.
Otherwise, they look clickbaity, which is annoying.
For this question, is that a theorem that you know is true (i.e. from a text), and you just don't know how to prove it; or is it a conjecture that you have which currently lacks a proof?
If it is the former, a citation to the book might be helpful. If it is the latter, I would refer to it as a conjecture, rather than a theorem. In either case, a better title might be something like "If $f(t) = g(t,C(t))$ and $g_x(t,C(t)) = 0$ for all $t$, is $f$ piecewise continuous?"
with the question itself clarifying what all of that notation means.
Something similar can be said for the other question, and I would also note that any time that a mathematician says that something is "obvious," they are lying. It is not obvious that the two questions are asking about different results; it actually does require a fairly close reading of both. Rather than just asserting to the commenter that it is obvious, perhaps it would be more politic to try to explain without being condescending.
13:49
How to show $\sum _{n=1}^{\infty \:}\frac{\sqrt{n+1}-\sqrt{n-1}}{n}$ converges without using integral?
you use the integral but you disguise it as genius wisdom
or not even that
hmm
try to bound it with something of the form $f(n+1) - f(n)$
hmmm
oh wait i misread a bit
aw I thought there would be something easy but my idea can't work
what is the "using integral" proof that you are thinking about ?

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