00:00 - 17:0019:00 - 00:00

12:05 AM
Well, assuming we're talking about individual pixels, you have a discrete number of points to work with. There isn't an automatic way to translate it into an infinitely differentiable (presumably continuous) surface
Arbitrary ways, on the other hand, might be more possible

12:50 AM
@MikeMiller I meant $(p)$ is the unique maximal ideal of $k[x]/(p^n)$, period. There are no other ideals with quotient field $k[x]/(p)$ because such an ideal would have to be maximal, and hence in turn would be $(p)$
It's a local ring
Maybe I misunderstand you

1:07 AM
I slept for over 12 hours and woke up the next day fuck me

my life
Havin' lots o' fun with $y^2 + 5 = x^3$

That's a good one.

it's basically exactly the same proof as Fermat's last theorem for regular primes
rofl
assume a solution exists, consider the equation as ideals, use considerations on the class group to derive a shitload of contradictions

Wait, really? Isn't it reduction mod 4?

I mean, you can do it with reduction mod 4 I think
but it's cooler if you write $(y + \sqrt{-5})(y - \sqrt{-5}) = (x)^3$ and show that the ideals on the LHS are coprime, so they are both equal to an ideal cubed, but the fact that $\Bbb Z[\sqrt{-5}]$ has class number $2$ means that they must be the cube of a principal ideal, so you get $y + \sqrt{-5} = \pm \alpha^3$ (unit group is $\pm 1$)
and then
do some shit from there
one reduces mod 4 to show those ideals are coprime though so that's probably the connection

1:19 AM

Interesting, let me try to understand. Why can't $(y + \sqrt{-5})$ be cube of a non-principal ideal? Because class number $2$ implies square of a non-principal ideal is always principal in $\Bbb Z[\sqrt{-5}]$, so cube has to be non-principal as well?
Which is not the case here

I did not use any code to make that golden thing

Well $\operatorname{Cl}_K \cong \Bbb Z/(2)$ and $a^3 \equiv 0 \bmod 2$ means $a \equiv 0 \bmod 2$ right?

That's what I said, right?

Urr

1:22 AM
Class group is group of fractional ideals modulo principal ideals, so there is a single non-principal ideal class, which squares to 0

yes

Aka squares to a principal ideal.
Nice.

@BalarkaSen Your comment's quite right, missed that it was a local ring

so if you have a principal ideal that is the cube of an ideal then the ideal of which it is the cube must itself be a principal ideal by class number 2
hahaha
Ahhh I see what you said
So then from that you get $y + \sqrt{-5} = \pm\alpha^3$ for some $\alpha \in \Bbb Z[\sqrt{-5}]$
and $y - \sqrt{-5} = \pm \beta^3$
Ah nice, if you set $\alpha = a + b\sqrt{-5}$ then I think you can just compare coefficients and get that $a = b = 0$ which is a contradiction

@ÍgjøgnumMeg Here's what I had in mind earlier: $y^2 = x^3 - 1$ mod $4$ implies $y$ is even and $x \equiv 1 \pmod{4}$. So write it as $y^2 + 4 = x^3 - 1 = (x - 1)(x^2 + x + 1)$. But $x^2 + x + 1$ is $3 \pmod{4}$ so has to have a prime factor which is $3 \pmod{4}$ as well, which hits $y^2 + 4$, so that $-4$ is a quadratic residue mod $p$ - so $-1$ is a quadratic residue mod $p$.

1:29 AM
Yeah this is the proof I've seen before

But $(-1|p) = 1$ implies $p \equiv 1\pmod{4}$

Indeed

Your proof is way cooler. I wonder if there's a concrete link between these passing to various integral extensions and doing factorization of ideals magic there and quadratic reciprocity
I have seen this happen before, you can do most of these problems in two ways

There definitely is by class field theory I think
but I don't understand class field theory
lmfao

Yeah I had that in mind, but I don't know what it means
Someday

1:31 AM
idk there are various reciprocity theorems in class field theory
I'll have that next semester so I'll tell you then rofl

I'll try to read Galois theory before semester starts and look into class field theory instead of Galois theory next semester

so $(a+b\sqrt{-5})^3 = a^3 + 3a^2b\sqrt{-5} - 15ab^2 - 5b^2\sqrt{-5} = y + \sqrt{-5}$, if the sign is +
Looool if you assume this then you get $a^2 = 2$ out
how FUN
I LIVE for this shit

it seems busywork after the unique factorization argument, thats my least favorite part lol

Yeah it's pretty quick though, comparing coeffs gives $3a^2b - 5b^3 = 1$ so $b = \pm 1$ and $3a^2 - 5b^2 = \pm 1$, i.e. $3a^2 = 6$ or $3a^2 = 4$

Aha nice.

1:42 AM
soz I got a bit overexcited
ah well, that was fun
gotta work out why poincaré series are a basis for the space of cusp forms now
which.. I do not know how to do

rip

need to remember some linear algebra
something about orthogonal spanning set is basis or smth
the problem is that my inner product is a bit ... hard to work with

1:58 AM
hi @ÍgjøgnumMeg @BalarkaSen

hiya @Leaky

I was gonna calculate the unit group of $\Bbb Z[\sqrt2]$ but then I took a nap and had dinner
but here I am now

nise

$|a+b\sqrt2|_\Bbb R = |(a+b\sqrt2, a-b\sqrt2)|_{\Bbb R^2} = \sqrt{(a+b\sqrt2)^2 + (a-b\sqrt2)^2} = \sqrt{2a^2+4b^2}$
the roots of unity satisfy $\le \sqrt2$
i.e. $a^2+2b^2 \le 1$
so there's only $\pm 1$
now the Dirichlet's unit theorem tells us to take the trace-zero space of the log
so $(\log|a+b\sqrt2|, \log|a-b\sqrt2|)$
to find a fundamental unit is to find a small enough $\varepsilon$ such that $0 < \log|a+b\sqrt2| < \varepsilon$ has exactly one solution
so $1 \le |a+b\sqrt2| \le e^\varepsilon$
no that isn't finite
how do I look at the trace-zero space

2:14 AM
I'm gonna go huff some cancer smoke into my lungs on the roof

I'll just pretend that $1+\sqrt2$ is small enough
graphical proof ^
so the units are $\pm(1+\sqrt2)^n$
if $b^2-2c^2=1$ and $c$ is even then $n = \frac{-1+|b|}2$ satisfies $n(n+1)/2 = (c/2)^2$
i.e. $(b+c\sqrt2)$ is a unit
lemme Python this
a,b = 1,0
for n in range(20):
print("$(1+\sqrt2)^{%d} = (%d)+(%d)\sqrt2$"%(n,a,b))
a,b = a+2*b,a+b
$(1+\sqrt2)^{0} = (1)+(0)\sqrt2$
$(1+\sqrt2)^{1} = (1)+(1)\sqrt2$
$(1+\sqrt2)^{2} = (3)+(2)\sqrt2$
$(1+\sqrt2)^{3} = (7)+(5)\sqrt2$
$(1+\sqrt2)^{4} = (17)+(12)\sqrt2$
$(1+\sqrt2)^{5} = (41)+(29)\sqrt2$
$(1+\sqrt2)^{6} = (99)+(70)\sqrt2$
$(1+\sqrt2)^{7} = (239)+(169)\sqrt2$
$(1+\sqrt2)^{8} = (577)+(408)\sqrt2$
$(1+\sqrt2)^{9} = (1393)+(985)\sqrt2$
$(1+\sqrt2)^{10} = (3363)+(2378)\sqrt2$
$(1+\sqrt2)^{11} = (8119)+(5741)\sqrt2$
$(1+\sqrt2)^{12} = (19601)+(13860)\sqrt2$
$(1+\sqrt2)^{13} = (47321)+(33461)\sqrt2$
7 hours ago, by Ted Shifrin
Here are the first 7. Still computing the 8th. 8,

49,

288,

1681,

9800,

57121,

332928,
7 hours ago, by Ted Shifrin
The eighth is 1940449
a,b = 1,0
for n in range(20):
res = "$(1+\sqrt2)^{%d} = %d+%d\sqrt2$"%(n,a,b)
if b%2 == 0: res += "; $1+2+...+%d = %d^2$"%((a-1)//2, b//2)
print(res)
a,b = a+2*b,a+b
$(1+\sqrt2)^{0} = 1+0\sqrt2$; $1+2+...+0 = 0^2$
$(1+\sqrt2)^{1} = 1+1\sqrt2$
$(1+\sqrt2)^{2} = 3+2\sqrt2$; $1+2+...+1 = 1^2$
$(1+\sqrt2)^{3} = 7+5\sqrt2$
$(1+\sqrt2)^{4} = 17+12\sqrt2$; $1+2+...+8 = 6^2$
$(1+\sqrt2)^{5} = 41+29\sqrt2$
$(1+\sqrt2)^{6} = 99+70\sqrt2$; $1+2+...+49 = 35^2$
$(1+\sqrt2)^{7} = 239+169\sqrt2$
$(1+\sqrt2)^{8} = 577+408\sqrt2$; $1+2+...+288 = 204^2$
$(1+\sqrt2)^{9} = 1393+985\sqrt2$
$(1+\sqrt2)^{10} = 3363+2378\sqrt2$; $1+2+...+1681 = 1189^2$
$(1+\sqrt2)^{11} = 8119+5741\sqrt2$
@ÍgjøgnumMeg big spoil time

2:32 AM
lol nice

$(1+\sqrt2)^{0} = 1+0\sqrt2$; $1+2+...+0 = 0^2$
$(1+\sqrt2)^{1} = 1+1\sqrt2$
$(1+\sqrt2)^{2} = 3+2\sqrt2$; $1+2+...+1 = 1^2$
$(1+\sqrt2)^{3} = 7+5\sqrt2$
$(1+\sqrt2)^{4} = 17+12\sqrt2$; $1+2+...+8 = 6^2$
$(1+\sqrt2)^{5} = 41+29\sqrt2$
$(1+\sqrt2)^{6} = 99+70\sqrt2$; $1+2+...+49 = 35^2$
$(1+\sqrt2)^{7} = 239+169\sqrt2$
$(1+\sqrt2)^{8} = 577+408\sqrt2$; $1+2+...+288 = 204^2$
$(1+\sqrt2)^{9} = 1393+985\sqrt2$
$(1+\sqrt2)^{10} = 3363+2378\sqrt2$; $1+2+...+1681 = 1189^2$
$(1+\sqrt2)^{11} = 8119+5741\sqrt2$

sick spoil

2:44 AM
This doesn't look like my list!
Oh, I see how it's my list now.

yay Ted agrees with me

lol
i should really get some sleep
but
I am not tired :(

3:46 AM
@TedShifrin @ÉricoMeloSilva are you here?

3:56 AM
There is a dimension N beyond which a sphere of radius 1000km is smaller than a cube of sidelength 0.001cm. When this first happens, what is the length of the cube's diagonal?
(Hypersphere, hypercube, "smaller" means smaller hypervolume)

lol I like how you casually slap in real world units

NB: I haven't actually done this problem

I'll just python it

but clearly the diagonal is larger than 1000km 'cause otherwise it'd be contained in the sphere
which gives a lower bound on the dimension of 10^22 I think?
So at least ten sextillion dimensions

then maybe I won't just python it

4:01 AM
Try it with more reasonable constants maybe

let's say it's even dimensional, then we want $V_{2n}(10^6) = \frac{\pi^n}{n!} 10^{12n} \le 10^{-5n}$
i.e. $\pi^n 10^{17n} \le n!$

At this point, maybe bring in Sterling's formula?

$n! \sim \sqrt{2\pi n} (n/e)^n$
LHS is $(10^{17}\pi)^n$ so approximately we need $10^{17} \pi = n/e$
so $n \sim 10^{17} \pi e \sim 10^{18}$ lol

What about my lower bound of $10^{22}$

idk

4:09 AM
I think you can't throw away the $\sqrt{2\pi n}$ like that
wait
@LeakyNun You compared $V_{2n}(S)$ with $V_n(C)$

oh lol

You want${}\le10^{-10n}$
and then double $n$ for the dimension
I think you get an upper bound of $20\cdot 10^{22}$ for the dimension in the end?
Meaning the diagonal of the cube should be 4000km-ish

@AkivaWeinberger

Wow
In any case, let this be a lesson to you, that, uh, higher-dimensional cubes are pointy and wide I guess
and big
Arbitrarily bigger than the set of points within walking distance

4:44 AM
lol

Most of the volume of the cube $[-1, 1]^n$ can be equated to volume of an extremely thin annulus of radius $\sqrt{n/3}$ in $\Bbb R^n$.
The thickness going to $0$ as $n$ grows huge
One way to understand this is to set up iid random variables $X_1, \cdots, X_n \sim \text{Unif}([-1, 1])$. Then by law of large numbers, $(X_1^2 + \cdots + X_n^2)/n$ converges to $\Bbb E X_1^2 = 1/3$.
So with high probability a random point from $[-1, 1]^n$ will have norm in $(\sqrt{n/3} - \varepsilon, \sqrt{n/3} + \varepsilon)$
Probability going to $1$ as $n$ going to $\infty$, for any fixed $\varepsilon > 0$.

5:02 AM
you ph***cist

It's math, there's no physics here
Pure probability :)
I was thinking about some probability a while back. It's very hot topic nowadays to understand expected Betti numbers of random simplicial complexes. Me, being a pea-brain, wanted to do the calculation for a random graph.
So take an Erdos-Reyni random graph $G(n, p)$ with $n$ vertices and edge probability $p$. I want to compute the expected $b_1$, or the expected number of cycles.
The expected number of $k$-cycles is clearly $\binom{n}{k} p^k$, which falls out if you associate an indicator variable $T_{ij}$ for every edge $(i, j)$ in $K_n$, and basically count $T = \sum T_{i_1 i_2} \cdots T_{i_k i_1}$, the sum running over all edges $(i_1, i_2), \cdots, (i_k, i_1)$ in a $k$-cycle in $K_n$.
$\Bbb E T$ counts the expected number of $k$-cycles, which is $\binom{n}{k} p^k$ by additivity of expectation.
So the expected number of cycles is $\sum_{k = 3}^n \binom{n}{k} p^k$.
There are two questions you can ask now: (1) What is the scaling limit? Say you let $p = \lambda/n$ and $n \to \infty$. Then the expected number of cycles become something like $e^\lambda - 1 - \lambda - \lambda^2/2$.
(2) What is the rate of convergence? How fast does expected number of cycles concentrate around $\sum_{k = 3}^n \binom{n}{k} p^k$, where $p$ is fixed, and you're letting $n$ grow? If you just use plain vanilla estimations I think you get an $O(1/n^2)$ error term, i.e., polynomial concentration around the central tendency.
But I believe you can do some trick like Chernoff bounds and get exponential concentration as well.
It's just all a bit hard because the indicator random variables corresponding to each $k$-cycle are not independent; they are "weakly" independent, so you can somehow make the proof of law of large number still work because the covariance terms are so small.

5:20 AM
@BalarkaSen Cycles forming a vector space over F_2, doesn't the number have to be 2^n?
Well algebraic cycles (sunsets of edges with an even number of edges per vertex)
Yeah never mind it's not the same thing

I think I still miscounted
Number of $k$-cycles is $n!/(n-k)! \cdot 1/{2k}$.
Womp womp
I did it for a triangle and forgot you can have multiple cycles through the same set of vertices if the number of vertices is more than $3$ :P
So the expected number of cycles is $\sum_{k = 3}^n \frac{n!}{(n-k)! 2k} p^k$.
What does that sum to lmao

you can factor out p^3 from that, but beyond that I'm not getting anything immediately useful from mathematica
it's a polynomial of degree n, of course, but that's not too illuminating

That's disappointing. I would like to understand the limit as $n \to \infty$ and $p = \lambda/n$ if it helps
I don't know how to estimate that

5:36 AM
as in, holding $\lambda$ fixed?

Yeah

5:47 AM
i've found one thing via OEIS, though I"m not sure how helpful it is:
the first few cases are p^3, 4p^3+3p^4, 10p^3+15p^4+12p^5,...
oh, even better: oeis.org/A284947
@BalarkaSen so your expected number of cycles is known as the cycle polynomial of the complete graph K_n

$n!/(n-k)! \leq n^k$. If you use that you get that the sum is bounded by $\sum_{k = 3}^n (np)^k/2k$

oh ffs mathworld

And $np = \lambda$ for us, so it's $1/2 \cdot \sum_{k = 3}^n \lambda^k/k$

mathworld.wolfram.com/CyclePolynomial.html: "The cycle polynomial, perhaps defined here for the first time, is therefore..."
if you're defining it yourself, wtf are you doing making an entry about it

MathWorld is baffling sometimes
You know their picture of Perko pair is wrong?

5:52 AM
That's a crime

what

Drawing knot theory pictures wrong is a horrible crime

isn't the whole point of the Perko pair that it was wrong for the longest time

Ken Perko jokingly calls it the "Weisstein pair" (after Weisstein, the founder(?) of MathWorld)
@Semiclassical Yeah. The deal was, knots #161 and #162 in the table were the same

sheesh

5:53 AM
or $10_{161}$ and $10_{162}$ I guess

@AkivaWeinberger So many places have drawn the Siefert surface of the trefoil wrong

That's what Perko found
so they revised the tables

They draw some nonorientable surface, the triple Moebius strip

and what used to be 10_163 got moved to 10_162
MathWorld has a picture of 10_161 and what-used-to-be-10_163-but-is-now-10_162
@BalarkaSen Makes sense, it's the obvious first guess

Yeah

5:56 AM
So in any case my impression is MathWorld was poorly put together and isn't maintained or updated
or, if it is, it's done poorly

@Semiclassical Ok at least it's clear that if $\lambda > 1$ then the expected number of cycles of $G(n, \lambda/n)$ go to infinity as $n \to \infty$
The sum doesn't converge

TIL the three levels of sensitive government information are top secret, secret, and confidential (in descending order)
I'm disappointed it's not top secret, secret, and bottom secret

It stands for "surveillance", "scandal" and "gossip"

Where's "surprises"

Can anyone help me in verifying this proof of Graph theory

6:26 AM
@LeakyNun Final answer (I think): the diagonal is $\sqrt{2\pi e}$ times 1000km, which is about 4133km

cool

and the dimension is $2\pi e$ times 10^22, which is about 1.70795*10^23
By "final answer" I just mean "I did the calculation again looking for more accuracy"
I really think that, since n has 24 digits, the error from the ~ term is essentially 0
Maybe it'll effect the last few digits

Random observation: I noticed that some finite sets can behave as if they are infinite as follows: {a,b,c} U {a} = {a,b,c} due to no multisets

Secret, you'll like this
3 hours ago, by Akiva Weinberger
There is a dimension N beyond which a sphere of radius 1000km is smaller than a cube of sidelength 0.001cm. When this first happens, what is the length of the cube's diagonal?

N=4?

6:29 AM
…?

O wait I forgot the units

Yeah but even so, the diagonal of a 4-cube of sidelength 0.001 is 0.002
and it has to be larger than 1000 otherwise it'd fit inside the sphere
@LeakyNun I shared the problem on Twitter. At least one person solved it

Solve for n: $\frac{\pi^{\frac{n}{2}}}{\Gamma (\frac{\pi}{2}+1)}10^{5n}<10^{-3n}$ then compute $\sqrt{n10^{-6n}}$ which I cannot do in mobile
Units will be in cm

Hint: Sterling, take nth roots

Stirling you mean

6:45 AM
Really?
Oh
Huh
Thanks
In terms of 3B1B's Numberphile appearance, ‪we're comparing a shrinking-target dart game to a fixed-target dart game
where the shrinking target starts with a radius of 1000km, and the fixed target has area (0.001cm)^2

Well wolfram give me ridiculous n=-218. Will work on that later in the pc

The number of dimensions has a ridiculous number of digits, so it might break computers

Also typo: $\frac{\pi^{\frac{n}{2}}}{\Gamma (\frac{n}{2}+1)}10^{8n}<10^{-3n}$
ah...

@Akiva Do you know about the accept-reject algorithm for sampling

6:56 AM
It gives a method to sample points from a probability distribution with an un-normalized pdf $f$
So $\int f \neq 1$
The idea is suppose you have another distribution with pdf $g$ such that $f \leq c g$ for some constant $c$.
You know how to sample from $g$, say
Sample a point $x$ from $g$ and $u$ from $\text{Unif}(0, 1)$
If $u \leq f(x)/cg(x)$, return $x$, or otherwise go back and pick new sample points
It's fun to figure out why this works: it relates to the dart on the target stuff

The area under the graph of g is the dartboard?

And the area under the graph of f is the target
and if it hits the board but not the target you go again

Precisely
Well

cg, not g

7:01 AM
area under the graph of $f$ is not the target because $f$ is not normalized

or I guess f/c
whichever

Oh but $c$ is not the normalizing constant of $f$ note
You dont know the normalizing constant

Right, but it's at least as big
'cause $\int g=1$ and $\int f\le\int cg=c$
This is the same way you pick a number from 1 to 5 using dice
Roll, if it's a six, roll again

Yeah

And at some point I vaguely remember us trying to work out how to optimize that for efficiency

7:04 AM
Basically in the process of picking a point under the graph of $f$ from the points under the graph of $cg$ the distribution following $f$ is already being normalized
Because proportions magic

The numerical instability in solving that volume seemed to stabilise at the nonsensical n=-193 only if I pre divided the n by at least $10^80$, but clearly that is not enough

Secret, take the nth root
n^(1/n) approaches 1
Most of the content of Stirling's approximation becomes very simple
I think you end up with $(n!)^{1/n}$ approx $n/e$, with an error rate quantumly tiny for the n we want
Also, ignoring that all - see if you can get an easy lower bound

Is there any version of the structure theorem for modules which are not finitely generated?
It would have been nice to expect something like that for operators. Like I could look at the C star algebra associated with the operator and do something like the structure theorem

Doubtful

Yeah seems too easy

7:21 AM
Think about how much breaks for infinitely generated modules over PIDs. Torsion free does not imply free, for example.
So even the torsion free part is complicated.

Q reaccs only

Fun fact: wolfram alpha give up very quickly with unable to parse, if if ask to solve for n expressions of the form $s \Gamma (n+1)^\frac{1}{n} < 1$ where s has a string length >2
It feels so dissatisfying we can only get bounds for that question when it is structurally speaking, not a complicated problem

Yeah torsion free implied free was crucial in breaking the structure theorem

One version of classification theorem that I like, which might or might not apply to your context, is classification of injective modules over PIDs.

What is an injective module?

7:32 AM
OK, nevermind then.

It's probably not relevant.

7:51 AM
Hi, can I ask what is the difference between a circle and a square nghd in R^2
In particular, I found thiis while studying The implicit function theorem: i.stack.imgur.com/v7ty3.png
Is this "rectangular"?

8:09 AM
whats an nghd

It doesn't matter what you take, inside any circular neighborhood you can find a square neighborhood and vice versa

8:30 AM
ah, thank you very much @SayanChattopadhyay

1 hour later…
9:55 AM
@Secret what is string length?

The number of characters in a string
e.g. "ABC" has a string length = 3

is that worth speaking of when you are talking about an element of $\mathbb C$?

Depends on whether you are talking about the decimal expansion, the symbol or something else entirely

well why don't you explain to me why its significant in each case then

2 hours later…
11:50 AM
Hey, does anyone know if the implicit function theorem (applied on a function F(x,y) that satisfies the assumptions) gives us the formula for f'(x), or just its value at (x_0,y_0)? Thanks in advance

0

So given picture I have to check whether divergence and curl is positive, negative or zero. So I check vector field at some point ; for figure b I can see that vector field goes outward so divergence is positive but for figure a I am not able to predict. Also how to conclude about curl?

In fact, f'(x) is $-F_x/F_y$, and by assumption $F_y$ is different form zero, but only at $(x_0,y_0)$

It will tell you the existence of such an $f$ (the implicit solution). From that the derivative at the point of interest is clear from composition rule.

No, the implicit function theorem does not give you any formulas.
As Balarka says, you need to do some additional (not too hard) work.
You might know what he calls the composition rule as the chain rule.

whoops i forgot chain rule was called chain rule
apostol calls it "function of a function rule" so i blame him

11:55 AM
By the chain rule I would get $-F_x/F_y$, right?

Tell me your argument in detail and I'll tell you if it's right.

Actually I got it watching this video, that is a simplified proof, the idea of the proof: youtube.com/watch?v=UYSwBx38cpQ
Aound minute 14.40
around*

I'm not going to watch a video, sorry.

We're definitely not watching a video. Tell us what you are confused about.

isnt the primitive of the function $f(z)= \frac{1}{z-z_0}$ the function $F(z)=log(z-z_0)$ the principal branch of log ??
i know this must be wrong cauase if it was right then the integral would be zero

11:58 AM
@MikeMiller no no sure

on closed curver around $z_0$
but why the derivative of $F$ is not $f$

@ManolisLyviakis Why must the integral be zero? Give me an argument.
It is indeed true that where defined $F' = f$.

i.stack.imgur.com/sfbvZ.png This pic summarizes my doubt
Is this application of the chain rule fine?

That's the sort of thing you're trying to train yourself to be able to tell --- not something you're trying to learn how to ask someone else to tell you :)

the integrl of $f$ around $z_0$ a closed curve is not zero. you can calculate the curvlinear integral easy

12:01 PM
What is the statement of the chain rule?
@ManolisLyviakis Yes, that's true.

@Shootforthemoon $F_y$ is different from zero in a neighborhood of $(x_0,y_0)$ so you get that formula in a neighborhood

What are you trying to use to argue that $F' = f$ implies that the integral is zero?

if the primitive of f existed => integral of $f$ around closed curves is zero

You're just asserting a statement, that's not an argument.

@RyanUnger ah, got it, thanks

12:03 PM
primitive of f is defined as the function that its derivative gives $f$

@MikeMiller You're right, the fact is that in my course we do not even use a book :( This is a topic I'm trying to delve into alone..

my argument is. $F(z)=log(z-z_0)$ is the primitive of $f$ therefore integral around closed curves is zero therefore around $z_0$ is zero . contradiction
haha

@RyanUnger If we integrate the formula in that neighbourhood, can we find the antiderivative of $f$, its primitive, if it does exist?

that might be hard considering the RHS possibly has a $y$ in it
the formula is only useful if you're given points already

0

So given picture I have to check whether divergence and curl is positive, negative or zero. So I check vector field at some point ; for figure b I can see that vector field goes outward so divergence is positive but for figure a I am not able to predict. Also how to conclude about curl?

12:20 PM
@RyanUnger we should have at least one point in general, so do you think this "boundary condition" would be enough to calculate the primitive?

the point you have is $(x_0,y_0)$

@RyanUnger Actually y=f(x) should not have y inside, so neither its derivative. no?
@RyanUnger yep

so what is the question

@Shootforthemoon Sorry, actually I did not understand exactly the answer to this

12:52 PM
well so far it's looking a little more clear, no thoughts? it's hard to stay on topic when on topic gets null response $$\mathcal H={\{h_k}\}_{k=1..N} \subset \mathbb N \land N \gt 1$$

$$f(n,N)=\sum _{k=1}^{N} \operatorname{irem} \left( h_{{k}}-n+1,n \right)\tag{1a}$$

$$g(n,N)=\sum _{k=1}^{N}\operatorname{irem} \left( h_{{k}}-1,n \right)\tag{1b}$$

where $\operatorname{irem}(n,m)$ is the integer remainder of the division of $n$ by $m$.

$$\max(f(n,N),g(n,N))\equiv\min(f(n,N),g(n,N))\pmod 2\tag{2}$$

Let $G$ be a topological group, and let $H$ be the closure of the set $\{1\}$. I am trying to prove that if $f : G \to Y$ is a continuous map, where $Y$ is a $T_1$-space (i.e., all singletons are closed), then $f(gh) = f(g)$ for every $g \in G$ and $h \in H$.
Here's what the authors state but I don't follow their reasoning: "For $y \in Y$, the singleton $\{y\}$ is closed, so $f^{-1}(\{y\})$, hence of the form $AH$ for some set $A \subset G$. This implies that $f(gh) = f(g)$ for every $g \in G$ and $h \in H$."

1:23 PM

1 hour later…
2:25 PM
@LeakyNun There's an exercise in Dummit-Foote which is asking me to prove that if $K/F$ is a finite extension, then if $K$ is a splitting field of some polynomial over $F$, then for every irreducible $f \in F[x]$, $f$ has a root in $K$ implies it splits completely in $K$.
I don't see a super-easy way to do this, but this line is coherent, right? Say $K$ is splitting field of the polynomial $g \in F[x]$. Then if the roots of $g$ are $\theta_1, \cdots, \theta_n$ in $\overline{F}$, $K = F(\theta_1, \cdots, \theta_n)$. Suppose $f \in F[x]$ is irreducible and $\alpha$ is a root of $f$ which belongs to $K$; then $\alpha$ is some polynomial on $\theta_1, \cdots, \theta_n$ with coefficients in $F$.
For any element $\sigma \in \text{Aut}(\overline{F}/F)$, $\sigma$ must permute $\{\theta_1, \cdots, \theta_n\}$, therefore $\sigma(\alpha) \in K$ as well, being a polynomial on $\theta_1, \cdots, \theta_n$ with coefficients in $F$.
I claim $\text{Aut}(\overline{F}/F)$ acts transitively on the roots of $f$. This, along with the previous conclusion, will prove the result.
This is because if $\alpha$ and $\beta$ are two roots of $f$ over $F$, then there is an isomorphism $F(\alpha) \to F(\beta)$. Compose with the inclusion $F(\beta) \to \overline{F}$ to get an embedding $F(\alpha) \to \overline{F}$, and extend that to an embedding $\overline{F} \to \overline{F}$ which will be an isomorphism.
This is an automorphism of $\overline{F}$ which fixes $F$ and sends $\alpha$ to $\beta$.
Therefore all the roots of $f$ are contained in $K$, hence $f$ splits completely in $K$.
For the sake of completeness: If $E$ is an algebraic extension of $F$, any embedding $i : E \to \overline{F}$ can be extended to an isomorphism $\overline{F} \to \overline{F}$ as follows. Take the set of tuples $(L, j)$ consisting of an extension $L/F$ and an embedding $j : L \to \overline{F}$ extending $i$.
Then this is nonempty, every chain has an upper bound, so Zorn's lemma applies and you get a maximal element which has to be $(\overline{F}, \phi)$ otherwise we can just add another algebraic. The final embedding $\phi : \overline{F} \to \overline{F}$ is surjective because $\overline{F}$ is an algebraic extension of $\phi(\overline{F})$, which being isomorphic to $\overline{F}$ is already algebraically closed, so cannot be proper.
I feel like I cheated and used Galois theory but I also don't see an easier way to solve this problem. I don't know what Dummit-Foote has in mind.

It’s decidedly not easy.
See Thm 9.9 in Ian Stewart @BalarkaSen

Checkin'
my proof's fine yeah?

2:43 PM
yeah it’s great

Thanks man

I like this proof
You don’t need the entirety of $\overline F$, just a finite subextension (this is like a philosophy now)
In particular, split f over K
Then L is a splitting field of fg and thus enjoys the extension property

Yeah I was initially working with the splitting field of $f$ but figured it's better to parse everything in terms of the universal extension

yeah so your proof is more “conceptual”

I suppose the key insight out of this mess is that if $K$ is splitting field of some polynomial (I suppose these are called normal extensions?) over $F$ then $\text{Gal}(K/F)$ acts transitively on the geometric points of $\text{Spec}\, K$, that is, every $F$-embedding $K \to \overline{F}$ must preserve $K$.

2:51 PM
they are finite normal extensions.

Ah, yes.

the geometric points of Spec K, ie, the only point of Spec K?

LOL
Geometric points are maps from Spec \bar K

Makes sense
Assumed it meant something like closed point

Geometry means something very different to them mutants

2:54 PM
@BalarkaSen and so an extension is normal iff it is a direct limit of finite normal extensions iff every minpoly splits

It doesn't mean "what you can see"
@LeakyNun Right, that every algebraic extension is direct limit of finite extensions is such a useful fact/philosophy

yeah

Thereby normal extensions are normal covering spaces and hence we can argue Poincare > Galois
flexes muscles

3:11 PM
@BalarkaSen have you looked at the ref?

1 hour later…
4:12 PM
hello guys, do we have $(|f'|)^2=(f')^2$?