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12:45 AM
hello. i am wondering if i am misinterpreting (4.70). For the $n=7$ case, I am getting that $\sum_i n_i^2 \neq 1$
I have defined n0 to n7 above and then the final line at the bottom is computing the sum of squares
 
1:05 AM
the only thing i can think of is maybe all $n_i$ s.t. $2 \leq i \leq n$ should have a $\cos$ as the first trig function
because there is certainly no way to obtain $1$ out of the way I interpreted the table
bleb i think that's it...
god
 
1:31 AM
I'm high def. just like my rep. 1080p, but we're still gettin' set. jet bundles in check. ramified covers'll wreck. just ask Xander, my questions are always 100% correct. weird flex. stay vexed. get vaxed. took automorphic L's so many times I lost my critical strip and couldn't find my way back. but that ain't that. $A.G.$ said count stacks. Mumford or Deligne. Triple ramified locus supreme. No I ain't talking taco bell but it's a wrap you know what I mean?
 
 
2 hours later…
3:31 AM
anon sent me here
 
3:58 AM
@Pancho_Squancho welcome
 
 
6 hours later…
9:41 AM
that feeling when author deletes their own question while you already wrote an answer
 
 
2 hours later…
11:39 AM
Hi everyone. A few years ago i asked this question: https://math.stackexchange.com/questions/3464491/is-this-contrived-sequence-eventually-periodic and I want to return to investigate this question. In short, we start with a tape full of zeroes, fix a number n and then look at the number $t_0$ in the 0th cell, then look at the number $t_1$ in the $t_0$th cell, and so on, and then increase the number in $t_{n-1}$th sell by $1$. Then do it again and again. My goal is to describe the possible evolutions of the tape depending on the number $n$.
 
@Jakobian brutal!
 
11:55 AM
@Nikita the message is okay but a message of this length will probably get lost under other messages, and its pretty long to digest all at once
 
 
2 hours later…
1:31 PM
The centered Hardy Littlewood function M_c f and the centered HLW w.r.t. cubes M_c f' are comparable, i.e. there exist positive constants a and b such that
a M_c ' f <= M_c f<= b M_c' f (A)
Can we say the same about the uncentered HLW functions M_u and M_u ' ?
' indicates w.r.t. cubes
I know that M_c is weak (1,1) so by (A), I can say that M_c' is weak (1,1) as well.
But how do I conclude from (A) that M_u and M_u' are weak (1,1) as well?
 
 
1 hour later…
2:43 PM
nvm, I got it now.
 
 
3 hours later…
5:24 PM
@Gwen I can't find the chat room where you were going to help me before, so I'm trying to make a new one. Sorry to invite you here, didn't know how else to contact you here.
And now I see why I can't make one, rep is less than half from the number it needs to be to make them.
 
5:56 PM
Looks like she's gone, darn it
 
6:12 PM
I'm not seeing how $G \cong N_1\times N_2\times \dots$
$f(a_1,a_2,...) = a_1\cdot a_2\cdot \dots $ doesn't seem injective
for example $\mathbb{Z}_{12}$, $(0,1)\mapsto 1$ and $(1,0)\mapsto 1$ for "addition"
 
that needs a lot more context to make sense
 
yeah and I just realized why it's injective
they're disjoint normal subgroups of $G$ so you can't write $(a,b) \mapsto c$ and $(b,a)\mapsto c$ because $b \notin A$, $a\notin B$ or whatever
i'm still not a fan of using $=$ to mean $\cong$
it's just like the addition near-ring thing.. I have a preconceived notion of $=$ and now they're using it and $\cong$ interchangeably T_T
 
6:30 PM
typically, "=" means "natural isomorphism" or at the very least "an implicit canonical choice of isomorphism", you typically wouldn't use it to denote an arbitrary isomorphism
 
natural isomorphism as in between groups?
I forgot what that means, it was in dummit & foote
I think it was $\pi$ to denote natural homomorphism
 
it's not something you should worry about
 
@Obliv Gross. $\iota$ is for isomorphisms, $\pi$ is for projections.
 
why ioughta
shakes fist
 
I was taught to pronounce $\iota$ as 'YO-tah', so it took me a second to cotton on to what you were punning there.
 
6:37 PM
what is this called again: $gcd(a,b) = 1 \implies d = au+bv$
does it have some fancy name
 
It's some guy's theorem.
 
ok I will credit some guy then
 
B something...
Not Binet... that's for the Fibonacci sequence...
BEZOUT!
That's the guy's name, I think.
In mathematics, Bézout's identity (also called Bézout's lemma), named after Étienne Bézout who proved it for polynomials, is the following theorem: Here the greatest common divisor of 0 and 0 is taken to be 0. The integers x and y are called Bézout coefficients for (a, b); they are not unique. A pair of Bézout coefficients can be computed by the extended Euclidean algorithm, and this pair is, in the case of integers one of the two pairs such that |x| ≤ |b/d| and |y| ≤ |a/d|; equality occurs only if one of a and b is a multiple of the other. As an example, the greatest common divisor of 15 and...
 
Thanks bezout
 
$\iota$ is for structure maps into a colimit, $\pi$ is for structure maps out of a limit
 
6:42 PM
:(
 
I keep wanting to switch between add. and mult. notation for groups even though the group is under addition D:
 
It's a group. What does it mean to be "additive" or "multiplicative"? There is one operation.
 
I gotta remember that half of having the stuffs in math is having swag & writing wicked nice
like Ted's blackboard writing
 
being clearly legible & aesthetic earns you 10 leslie coin
or it should
@XanderHenderson Wait isn't it possible to have a set be a group under diff. operations though
like ($\mathbb{R},+$) and ($\mathbb{R},\cdot$)
 
6:48 PM
@Obliv Sure. But unless you are specifically thinking about some set of numbers, the different operations are just abstractions, and would probably be denoted $+_1$ and $+_2$ (if you needed to distinguish them).
 
yeah I guess you're right. It's not like $\mathbb{Z}_n$ is a group under multiplication anyway. but having consistency in notation is important imo
everything is blurring together
is $U_{n}$ in general all the elements in $\mathbb{Z}_n\setminus \{0\}$ that are relatively prime with $n$
yesh
 
@Obliv I mean, maybe? I've not seen it in any generalized setting to mean that, but perhaps your book / professor uses it that way?
By the way, $\mathbb{Z}_n$ isn't great notation. Better to write $\mathbb{Z}/n\mathbb{Z}$ (or $\mathbb{Z}/n$, if'n yer lazy).
 
oh sorry I should have said $U_n$ to mean units
yeah that makes sense, to denote quotient group
 
For $p$ prime, $\mathbb{Z}_p$ can also denote the $p$-adic integers, so there is potential ambiguity.
 
p-adic integers? where have I heard that before
 
7:01 PM
@Obliv No idea.
I can't read your mind, nor see your memories.
 
not all bijections are isomorphisms right
since they must "respect" the operation
 
Correct.
 
a homomorphism is kinda like invariance in physics
the invariant is preserved under transformations similar to performing operations in the respective sets
we didn't cover group actions but is it relevant to how for a subgroup $N$ of $G$ we have $N$ being normal if $gN = Ng$ for all $g \in G$
normal means the subgroup has weak commutativity in this way
 
7:59 PM
what you have to understand is that semiotics do not subsume semantics
 
 
2 hours later…
10:00 PM
very interesting video
 
10:24 PM
> Lemma Let there be three measurable spaces $(E,\mathcal A)$, $(F_1,\mathcal B_1)$ and $(F_2,\mathcal B_2)$, and equip the product $F_1\times F_2$ with the product $\sigma$-algebro $\mathcal B_1\otimes \mathcal B_2$. Define $f: E\to F_1\times F_2$ by $f(x)=(f_1(x),f_2(x))$ for every $x\in E$, where $f_1: E\to F_1$ and $f_2: E\to F_2$. Then $f$ is measurable iff both $f_1$ and $f_2$ are measurable.
Apparently the "only if" part is very easy and left to the reader. I struggle with this; in other words, if $f$ is measurable, why are $f_1$ and $f_2$ measurable? Note here $\mathcal{B}_1$ and $\mathcal{B}_2$ are not necessarily Borel $\sigma$-algebros.
 
10:34 PM
Maybe it has something to do with $f^{-1}\left(B_1\times B_2\right)=f_1^{-1}\left(B_1\right)\cap f_2^{-1}\left(B_2\right)$, although this equality is already used in proving the "if" part.
 
11:02 PM
Holy crap......quiz complete
But these long division probs with polynomials are a nightmare
 
@psie $f_i=\pi_if$, where $\pi_i\colon F_1\times F_2\rightarrow F_i$ are the projections, $i=1,2$
 
ok, and is $\pi_i$ measurable? If it is, then one can say that the compositions of measurable functions is measurable.
 
measurability of pi_i ought to be very close to the definition of the product sigma algebra
 
indeed, I would argue it's the whole point of the definition
 
another win for the category nerds
 
11:27 PM
May I have some assistance with this one guys? It says, Add The Polynomials. $6a^4 b^4 +(8a^4 b^4 - 9b)$
 
11:38 PM
@YourLordJoyBoy That's like $6a^4b^4+1(8a^4b^4-9b)=6a^4b^4+8a^4b^4-9b$
the $1$ is implicit and you basically don't need the parentheses there
so from there u can just add the terms with the same base
 

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