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00:40
https://p.migdal.pl/tagoverflow/?site=mathoverflow.net&size=64

An interactive graph of Mathoverflow and Mathematics tags, credits to Piotr Migdał. Thought you guys may find it interesting
 
2 hours later…
03:01
Would anyone happen to be able to point me in the right direction in solving this problem: math.stackexchange.com/questions/3907887/…
I've spent some time on it, and cannot seem to make any progress
Admittedly, I don't know much about the subject.
03:15
Hey does it always happen that when a big list question has many answers, an answer that is given very late is neglected?
For example, I answered a famous big list question very late, and it didn't get even a view, I think.
Here is my answer to that question:
0
A: Different methods to compute $\sum\limits_{k=1}^\infty \frac{1}{k^2}$ (Basel problem)

epic_mathHere's mine. I'm answering late, I know that, but I am still answering it. We'll use the expansion of $\tanh^{-1}$: $$\frac{1}{2}\log\frac{1+y}{1-y}=\sum_{n\geq0}\frac{y^{2n+1}}{2n+1},\quad|y|<1$$ We start with this inequality: $$\int_{-1}^{1}\int_{-1}^{1}\frac{1}{1+2xy+y^2}dy\,dx=\int_{-1}^{1}\f...

no, a question gets bumped to the homepage every time it gets a new answer
Oh
Hey it got an upvote
Why wasn't my reputation increased?
03:34
cause you made it community wiki
that means you don't get rep from upvotes
04:16
Hey can anyone answer this:
0
Q: Consequences of the Riemann Hypothesis

epic_mathNote:This is the same question, but it doesn't have any answers and it is very old and inactive, so it doesn't answer my question. This may also look similar, but doesn't answer my question, you can see the answers yourselves. While reading this section of the Wikipedia article on the Riemann hyp...

96
Q: Collection of equivalent forms of Riemann Hypothesis

Jon BannonThis forum brings together a broad enough base of mathematicians to collect a "big list" of equivalent forms of the Riemann Hypothesis...just for fun. Also, perhaps, this collection could include statements that imply RH or its negation. Here is what I am suggesting we do: Construct a more o...

Thanks but equivalents are not the same as consequences
if you prove RH, you prove every statement equivalent to it
Yeahh
But it is still different
For example, RH tells many things about the distribution of primes but those statements aren't 'equivalent' to RH
those are as genuine consequences as any other
of course, they aren't all of them
04:31
0
Q: Equivalent condition for twin primes starting with the modular arithmetic condition $4((n-1)! + 1) = -n \pmod{n(n+2)}$ (related to Wilson's theorem).

CommutativeAlgebraStudentSee characterisation of the twin primes for a modular arithmetic viewpoint. Shift $m$ by $1$ so consider twin prime pairs $(m-1, m+1)$ and the formula becomes: $$ 4((n-2)! + 1) = -(n-1) \pmod {n^2 - 1} $$ Since $\gcd(n \pm 1, n) = 1$ for all $n \geq 2$ we have that we can faithfully multiply bo...

then again, "consequence" and "equivalent" are but informal terms anyhow
in the end, any true statement implies any other true statement
Start with Wilson's characterisation of twin primes, then arrive at something related but looking way different
Hi, Thorgott. You helped me greatly the other day
on continuity of spectral map for rings
@Thorgott what do you think of that twin prime set up
05:09
Hello everyone. Sorry if this is just one more stupid question, but honestly, how much of intro analysis are reverse calculations? Does the rest of analysis go in this direction as well?
Hello, can anyone take a look at my post?
0
Q: A condition for a Toeplitz operator to be Fredholm and find its index if its Fredholm

MikeI have a question about showing that a Toeplitz operator is Fredholm: Show that the operator $T_{e^{inx}}:L_2^+\to L_2^+$ acting on $L_2^+=\sum_{k\ge 0}a_ke^{ikx}$ with $\sum_{k}|a_k|^2<\infty$ is Fredholm for every $n\in\mathbb{Z}$ and find its index. Denote $S^1$ to be the unit circle. The sp...

 
1 hour later…
06:30
Hey guys I want to inform you about something. In my analytic number theory room, I have decided to post my research (not my personal research, but from reliable articles and journals) on a topic which I call 'generalized partition function'. Please see it if you are interested. I will post some research everyday.
06:47
@AttractorNotStrangeAtAll I don't think I'm going to agree with you that the $\delta$-$\epsilon$ proofs we were discussing are "reverse calculations." If you call it "reverse" to say that I can make the sum of two terms $<\epsilon$ by making each term $<\epsilon/2$, then I'll grant a certain "reverse"-ness. But I don't consider that to be the case. I consider "reverse" to be doing starting with the result (say an equation) and doing if-and-only-if steps to get back to the hypothesis.
For gamma_1,gamma_2:I->R^3 be regular space curves with the same domain. For sigma(t,u) = u gamma_1(t)+(1-u) gamma_2(t), K(t,1/2) = 0 for any t where K is a Gaussian curvature.
In my calculation, sigma_uu=0 so what I need to show is <sigma_{tu},sigma_u \times sigma_v> =0. And after calculation, I need to show <gamma_2-gamma_1,gamma_2'\times gamma_1'> = 0. This is where I'm stuck
does anyone know some interesting research topics in number theory?
 
1 hour later…
07:59
is there any method to evaluate (or turn into a product, sum, etc.) a non-periodic continued fraction?
 
1 hour later…
09:19
3
A: Congratulations: the big thread!

skullpatrolCongratulations robjohn for getting into the 300K club.

I meant this, sir.
71
Q: Congratulations robjohn for getting into the 100k club.

skullpatrolWe truly appreciate your constant presence and insightful hints in the Mathematics chat room. Your dedication to learning is inspiring. Thank you also, for all your hard work as a moderator.

not this classic
:-)
10:09
@epic_math I'm sure there are many methods. Alas, the only one I know of, due to Euler and Hurwitz, works with sequences of certain power series satisfying a recurrence relation that can be turned into a continued fraction. It does lead to a few cool examples.
Like the formula for [1/x,3/x,5/x,...] in terms of the exponential function, another for [1/x,2/x,3/x,4/x,...] in terms of certain Bessel functions. And to e=[2,1,2,1,1,4,1,1,6,1,1,8,1,1,...]
My source is Joe Roberts' book Elementary Number Theory - A Problem Oriented Approach. Egad, I studied that book intensively 40 years ago, but have neglected it since (for purposes other than looking for topics for undergrad essays).
10:51
How can I get the formula about (f\circ\gamma)'' (t)?
Oh don't need to answer this
11:48
I just had a thought if $y(x)=f(x)=x+1$ then we normally write $f^{-1}(x)=x-1$. Shouldn't we actually write $f^{-1}(y)$ or something because the inverse of the function of course would take the variable $y$. How am I confused?
Hey I want to share a picture
I was zooming into a mandelbrot set and the deepest that dumb computer could zoom was this:
I zoomed verrryyyyy deep so it is not much clear
Can someone help me with this: Suppose you are given a coin which is not necessarily unbiased. Giving proper justification, explain how you would obtain the probability of getting“head” in a single toss of the coin.
12:05
Could someone please ping me ( @ ) when giving answer to Tapi, sounds like an interesting q
I don't know the answer but probably I will call you
Hey I have an interesting problem in my mind
I noticed that if $p_n$ is the nth prime number, then $$\sqrt{p_{n+1}}-\sqrt{p_n}<1$$
I couldn't prove it even after a lot of effort
I rewrote it as $$g_n<2\sqrt{p_n}+1$$ where $g_n$ is the prime gap
@geocalc33 hello!
@northerner It doesn't matter what you call the variable, the function is the same. That said, I've seen people write $f^{-1}(y)$ for precisely that reason. It's a matter of preference.
ok thanks
Let me give some graphs related to the problem
It would take time to make the plot
@Tapi Are you looking for something sophisticated? If something simple suffices, just throw the coin a lot of times and calculate the ratio of head tosses to total tosses. This is the unbiased ML estimator for that problem and converges a.s. to the actual probability.
12:21
what
Hey guys! I don't know if I am at the right forum or not! Let's talk about finding closed form of $\zeta(3)$.
I wasted my whole day
I do know that $\liminf_{n\rightarrow\infty}\sqrt{p_{n+1}}-\sqrt{p_n}=0$ is true, however
@T.H.Shehadi are you serious or joking?
I am computationally serious
12:23
okay
or at least let's talk about most accurate approximations you can think of :(
Okay I am interested
In mathematics, at the intersection of number theory and special functions, Apéry's constant is the sum of the reciprocals of the positive cubes. That is, it is defined as the number ζ ( 3 ) = ∑ n = 1 ∞...
If $\log \Lambda(s)\ \in \mathrm{SL}(2,\Bbb R)$ then $\Lambda(s) \in G$ s.t. $G \simeq \mathrm{SL}(2,\Bbb R)$ right?
is $\zeta(3)$ irrational ?
Yes it is, I think
But where can I get a proof?
12:41
@Thorgott I'm not looking for anything sophisticated, I've just started college (studying b.sc stats) and I've been taught classical probability and the definitions of conditional probability and frequency definition. So I should use the frequency definition here right? Since I've to write it in a formal way, should I like say that p(getting a head)= lim (n tends to infinity) number of throws in which I've got head/ total number of throws (n)?
I have solved the Riemann hypothesis,
Let me write that in my thesis.
Yes I have solved it using induction,
On the Riemann zeta function!
I will post it on researchgate,
and 1 million dollars will be paid!
looool
@Tapi I don't know what the frequency definition is. The formal version of what I said is just the law of large numbers.
@epic_math Yes, it's irrational. That's the famous issue that Apéry settled like 40 or so years ago. I'm sure you can find a proof on google.
lol*2^100
13:00
0
Q: Prove that the Guassian curvature of the given surface is zero

love_sodam Let $\gamma_1,\gamma_2:I\to\mathbb{R}^3$ be a pair of regular space curves with the same domain, $I$. Define $\sigma:I\times(0,1)\to\mathbb{R}^3$ as $\sigma(t,u) = u\cdot\gamma_1(t)+(1-u)\cdot\gamma_2(t)$. Assume that $\gamma_1,\gamma_2$ are chosen such that $\sigma$ is regular. Prove that $K(t,...

Isn't there anyone could handle this problem?
13:29
can anyone verify my answer?
When showing the existence of a splitting field of a polynomial over K[x] (K being any field), why is it not enough to simply adjoin all the roots?
i.e. Why isn't " K(a1,..., an) is a splitting field" a proof
13:57
define what that expression means
the subtlety is that, a priori, it doesn't necessarily make sense to talk about "the roots"
14:24
unless, you know, you constructed the algebraic closure
only a brute would construct algebraic closures before constructing splitting fields
15:24
@skullpatrol I figured, thanks.
That moment when I realised that conflict and bifrucations etc. are just weird shaped cusps when the higher dimensional object is plotted
 
1 hour later…
16:49
Why does (2) extend the definition of (1) to all real numbers except the non positive integers? By the way, for $\alpha=0$, $\Gamma(-1)$ in (2). Is this defined?
Because for any $-1<\alpha<0$, you can now define $\Gamma(\alpha):=\Gamma(\alpha+1)/\alpha$ (and $\Gamma(\alpha+1)$ is defined, because $\alpha+1>0$) and that satisfies the recursion by definition. Then, for $-2<\alpha<-1$, you again define $\Gamma(\alpha):=\Gamma(\alpha+1)/\alpha$ (and now $\Gamma(\alpha+1)$ is defined by the previous step, since $-1<\alpha+1<0$)$ and this again satisfies the recursion by definition. You continue this process recursively.
This doesn't work for negative integers, because you would define them via $\Gamma(0)$, which doesn't exist and also cannot defined recursively, because you'd have to set "$\Gamma(0)=\Gamma(1)/0$", but division by $0$ doesn't make sense
From a broader point of view, you can't possibly hope to extend the Gamma function to the non-positive integers, because they are poles of the function
@TedShifrin Thanks for your answer! Yes, that's what I meant. Starting with a $<\varepsilon$ inequality then arriving at a $<\varepsilon\cdot A+B$, with this $\varepsilon\cdot A+B$ that'll be my delta. This is what I think feel like "reverse calculation".
@Thorgott How does $\Gamma(\alpha):=\Gamma(\alpha+1)/\alpha$ satisfy the recursion, for say $-1<\alpha<0$?
cause that definition is precisely the recursion formula, but rearranged
17:05
Got it.
@Thorgott Thanks for the reply. In layman terms, what is the pole of a function?
it's a point where the function isn't defined, but near which the function is defined, and so that the functions values approach infinity (or minus infinity) as that point is approached
think of a function like 1/x and the point 0 for example
Great, thanks.
18:02
actually, there is slightly more technicalities to poles so as not to mix up with essential singularities
a pole can be said to obey some kind of reciprocal power behaviour in its neighbourhood, hence why they can be removed by multiplying $(z-a)^n$
@Thorgott have you ever heard about universally measurable sets? I just stumbled into those while reading a book
18:22
ㅤㅤㅤㅤㅤㅤㅤㅤ
ㅤHi
I believe I saw that term used once in a survey on the automatic continuity of homomorphisms between Polish groups that someone (I believe it was Lukas) linked here some time ago
@user2103480 Turns out I'm not the only one who was wondering about this question, they even cite the same paper you found a few days ago! mathoverflow.net/questions/376493/…
hi chat
@Thorgott Ah it would make sense in context. By the way do you happen to have a link to that survey? Automatic continuity is one of the things my supervisor studies so I might need to learn about it eventually
@AlessandroCodenotti I sent you a message on discord
Oh I missed it, let me see
universally measurable homomorphisms from locally compact polish groups into polish groups are continuous
Ah Rosendal, of course
beautiful
@Thorgott Neat, I knew that result for Baire measurable homs
18:32
@AlessandroCodenotti haha nice
what are maps that are Baire or universally measurable, but not just plain measurable
Measurable wrt which $\sigma$-algebra?
any one that a sane person would seriously consider
Baire sets are contained in universally measurable sets which are contained in Lebesgue measurable sets, so all baire/universally measurable functions are measurable
0
A: Writing $\sqrt[\large3]2+\sqrt[\large3]4$ with nested roots

QuantoNote $$(\sqrt[3]2+\sqrt[3]4 -1)^2= 5- \sqrt[3]{4} $$ which yields $$\sqrt[3]2+\sqrt[3]4 =\sqrt{5-\sqrt[3]{4}}+1$$

A new answer to a question I asked five years ago
On whether $\sqrt[3]2+\sqrt[3]4$ can be obtained through integer addition/subtraction, positive integer roots, and positive integer powers
Indeed, the powers are not necessary
$$\sqrt[3]2+\sqrt[3]4 =\sqrt{5-\sqrt[3]{4}}+1$$
18:39
en.wikipedia.org/wiki/Universally_measurable_set there is a nice example on wiki showing that Lebesgue measurable does not imply universally measurable
Wait actually I don't know if subtracting $\sqrt[3]4$ there is legal
Uh well no it should be fine: $\sqrt[3]{-4}$
huh, but the Lebesgue measure of the Cantor set is zero
coin flipping measure should be a singular measure, no
I'm not sure what's your objection
they say the Lebesgue measure of that subset of the cantor set is the coin flipping measure and I'm objecting that the Lebesgue measure of that is zero, but the coin flipping measure shouldn't always be zero
5
Q: Which explainable artificial intelligence techniques are there?

nbroExplainable artificial intelligence (XAI) is concerned with the development of techniques that can enhance the interpretability, accountability and transparency of artificial intelligence and, in particular, machine learning algorithms and models, especially black box ones, such as artificial neu...

18:52
in a graph n-hypercube, what is the vertex set and edge set of an n-1 hypercube?
@Thorgott No they are mapping the Cantor set onto $[0,1]$ and taking the measure of the image
That's how I read it
oh, duh, they're doing binary sequences
so it's surjective
I was thinking of base 3 Cantor set
ok, that's a neat construction, I like it
It really is
19:29
@AlessandroCodenotti that is neat
measure theory is such a mess
20:19
watching a physics lecture where the prof said the Levi-Civita symbol was named after two mathematicians
lol
That's like Ryll-Nardzewski
@BalarkaSen btw, my crusade of stochastic process pedantery hasn't ended yet, but it's (luckily) still the case that everything works out fine. My most recent question was the following:

Do processes stay independent if we switch sigma algebras? Example: Let $X,Y \colon \Omega \rightarrow \R^{[0,1]}$ be continuous processes that are measurable with respect to the sigma algebra $$\mathcal{G} = \bigotimes_{t \in [0,1]} \mathcal{B}(\R).$$
Further, assume they are independent, so that for $F, G \in \mathcal{G}$, we have that $\P(X \in F \text{ and } Y \in G) = \P(X \in F) \cdot \P(Y \in G)$. N
seems like something that is of course true
but i wouldnt be able to prove
The answer is rather chill. First off, $X^{-1}(F) = X^{-1}(F \cap C([0,1]))$
20:29
the answer is yes you mean :P
yes it is
why are you going down this incredibly dry measure theory rabbithole tho
and $F \cap C([0,1])$ is measurable in that second sigma algebra so the inverse images actually don't change
@BalarkaSen I cannot not check the stuff
at some point I accept things, but these basic facts I want to understand, especially since they're mentioned nowhere
It just bugs me that nobody mentions this stuff so I have to prove it myself
@BalarkaSen it's routine once you know these kinds of proofs. Principle of good sets
yeah so thats why you shouldnt do it over and over
Haven't done a principle of good sets proof in a year tbh
20:34
i dont actually think measure theory is an issue at all in probability
oh fine
It is when you construct new limit objects I think
Once you do normal stochastic analysis it's fine, but everytime you move to a different part all the basic stuff comes up again, then one verifies it and starts calculating without care
Hmm I see
is there a fast way to show that in a finite field of order $q^d$ where $q$ is the power of a prime, if $X^{q^d} - X$ divides $X^{q^s} - X$, then $d | s$?
People usually write "the Kuratowski and Ryll-Nardzewski selector theorem" instead of Kuratowski-Ryll-Nardzewski because it's very confusing
@AlessandroCodenotti who writes that
20:39
@AlessandroCodenotti like BSD conjecture
In mathematics, the Birch and Swinnerton-Dyer conjecture describes the set of rational solutions to equations defining an elliptic curve. It is an open problem in the field of number theory and is widely recognized as one of the most challenging mathematical problems. The conjecture was chosen as one of the seven Millennium Prize Problems listed by the Clay Mathematics Institute, which has offered a $1,000,000 prize for the first correct proof. It is named after mathematicians Bryan Birch and Peter Swinnerton-Dyer, who developed the conjecture during the first half of the 1960s with the help of...
sounds niche even to me
Descriptive set theorists usually
lmao Cantor-Bernstein theorem is abbreviated CBT
@porridgemathematics X^q^d-X divides X^q^s-X, iff X^q^d-X splits in GF(q^s), iff GF(q^d) <= GF(q^s), iff [GF(q^d):GF(q)] divides [GF(q^s):GF(q)], iff d divides s
this can be avoided by using em dashes
@AlessandroCodenotti You use em-dashes between distinct people's names
en-dash
Kuratowski--Ryll-Nardzewski
20:44
sniped
if I want an adjective to say that the base rings in $M \otimes_A N \otimes_B P$ are different, is there an adjective that is something like heterobasic?
call it a bimodule
as in, in contrast to $M \otimes_A N \otimes_A P$
I'm taking a "heterobasic" tensor product
@LeakyNun thanks, im trying to reformulate that in the language of the algebra 3 course (which honestly I find a bit too sparse, like the lecture notes prove things pretty tersely for me), so are we saying that in $$F_{q^d}$$, we have $$X^{q^d} - X | X^{q^s} - X$$, in other words all elements are zeros of $$X^{q^s} - X$$, now if $$X^{q^s}-X$$ does not split into linear factors, mod out by the ideal generated by a non-linear irreducible factor
, which gives you a field of order $$(q^d)^k_1 = q^{dk_1}$$, then at some point this procedure stops, and it necessarily stops when we reach $$q^s$$ so $$dk_1..k_t = s$$ and $$ d|s$$ as desired?
oh, it's algebra 3 again
do they teach finite fields there?
yeah but they don't go into enough detail
at least for someone with almost zero experience with this stuff
20:51
I don't understand why you need to start using contradiction
I mean
did they tell you that GF(q^s) is the splitting field of X^q^s-X?
and in fact GF(q^s) is nothing but all the roots of X^q^s-X
we haven't defined 'splitting field' ><
there is this 'tower of fields' construction that we are using
how did you define finite fields
yeah, what's GF(q^d) to you?
the idea is you start with $$F_p$$, which satisfies $$X^{p^r} - X = 0$$ for all $$x \in F_p$$, and then if that poly doesn't split into linear factors, you mod out by a non-linear irreducible factor, and get a strictly larger field that still satisfies $$X^{p^r}-X$$
i understand its the unique finite field with those many elements
yeah but when you write GF(q^d) you're already assuming that it exists and is unique
yeah but what you said is exactly what splitting field means
okay, actually yeah i think the construction were doing is defining that
you can just count the number of roots of X^q^d-X
just not explicitly naming it
which agrees with the degree of the polynomial, which is q^d
so I don't understand which step in my proof needs to be reformulated
14 mins ago, by Leaky Nun
@porridgemathematics X^q^d-X divides X^q^s-X, iff X^q^d-X splits in GF(q^s), iff GF(q^d) <= GF(q^s), iff [GF(q^d):GF(q)] divides [GF(q^s):GF(q)], iff d divides s
hi @TedShifrin
20:55
X^q^d-X splits in GF(q^s), iff GF(q^d) <= GF(q^s) is what im trying to prove, basically
reformulate only in the sense that we cant take that for granted
and one way you can do that is by incrementally increasing the size of your field starting with GF(q^d)
at least the way thats taught in algebra 3
=>: X^q^s-X splits in GF(q^s) so any polynomial that divides X^q^s-X must split into a subset of the linear factors of X^q^s-X
yes but you are assuming that GF(q^s) is the splitting field
<=: every root of X^q^d-X is in GF(q^s), and GF(q^s) consists of exactly the roots of X^q^s-X
@porridgemathematics no, I just said that X^q^s-X splits in GF(q^s)
yes but what is GF(q^s) ?
assuming you only have GF(q^d)
for d < s
or rather, 'a finite field of order q^d'
you need to construct it from the smaller field, right?
1. take a large field F in which X^q^s-X splits
2. consider the roots of that polynomial. they form a subfield of size q^s.
3. call it GF(q^s)
20:59
1. is the issue
its nitpicky, but what i was trying to do was construct the larger field so that the smaller one very obviously sits in it
because its a tower
F_{q^d} \subset .. \subset F_{q^s}
if you want everything to be subfields, just use an algebraic closure to begin with
it is near-impossible to extend upwards
we extend upwards at the very beginning of finite fields in algebra 3
why is it near impossible?
because it is much easier to redefined the smaller fields as the subfield
it's not near impossible, it's just baaaaaad
consider extending R to C
21:02
it probably is, but im just trying to follow the course structure thats al
let's say we have R, now let's define C to be { (x,y) | x,y in R } with some operations that make sense
oh no, then C does not contain R
sure, so just work with isomorphisms?
so we need to define C instead to be { (x,y) | x,y in R and y != 0 } U R
isnt that all of math
well if you're working with isomorphisms, then what's the difference between extending upwards and contracting downwards?
21:03
well i just wanted an explicit construction of the larger field
how do you construct it
@LeakyNun not having to know what an algebraic closure is
lets say we only have a field of size p
how do you show that it sits in a larger field?
without constructing the larger field
and showing an isomorphic copy of ther smaller one sits in it
what's the difference?
you said you're working with isomorphisms
all im saying is my 'reformulation' clarified 1.
i dont know what an algebraic closure is
it is a set-theoretic obstacle that you can't (easily) extend upwards without reidentifying the bottom field
21:05
okay but were working with finite fields
we don't need to invoke anything funky here
here's a semi-explicit construction of GF(p^d)
i wasn't attacking your solution, sorry if it came across that way, it was useful to me
i just wanted to put it in the language of the course im doing
take an irreducible polynomial P of degree d over GF(p), and then just take GF(p)[X]/P
yes thats what i did
huge caveat over "take an irreducible polynomial of degree d"
21:06
yes, of course, but once you show it exists (by contradiction) you can mod out by the ideal generated by it
show that exists by cOuNtInG
i dont take issue with contradiction there at least at this stage
or yeah by a counting argument
I don't understand why you're taking issue with this alternative approach that constructs all the GF(p^d) independently first
btw "the" embedding GF(p^d) -> GF(p^de) is not unique
so I'm not sure what you're asking is possible to start with
what do you mean?
it definitely is possible
there are 2 embeddings GF(7^2) -> GF(7^6)
21:08
you mean prove d|s via constructing the larger field?
so you need to make one choice
no, I'm defining what GF(p^d) means
I constructed every GF(p^d)
the approach i would take is define one, and then show they are all isomorphic, isn't that standard?
do you accept that GF(p^d) makes sense for any d?
i could after we prove uniqueness
lets continue this later, sorry, i gtg
21:11
thanks for your help, and sorry again
I don't need uniqueness to solve your problem, I don't think
I can prove that X^q^d-X splits in GF(p^d) for every choice of GF(p^d)
Howdy @Leaky. I didn’t realize I was here.
if it shows that your icon here but you don't think you are here, are you really here?
Nope. Page in background.
(I was parodying "if a tree falls in the forest but nobody hears it, does it make a sound?")
21:20
I was always of the opinion that it doesn't.
Spend the day in philosophical debate!
22:05
Hi @TedShifrin
Can I ask you guys something, sometimes you have a written exam to enter an academic job. I'd like to know the name of this exam, does anyone knows what its called? Is it called entrance exam? I'm writing here because this is the only chat I ask questions in stack exchange and I don't know if there is a better place to put it.
@LeakyNun If you turn on optimizations it doesn't make sound to save resources if it's in nobody's render range
@user777 I've never heard of any such thing.
Hi Karim.
@TedShifrin I have decided to do youtube videos in my free time covering everything needed to do research in Algebraic geometry, complex geometry, and arithmetic geometry in my free time. In essence covering algebra, analysis, and algebraic geometry.
This is really for me as it strengthen the connections in my brain
I am sure it did strengthen things when you taught your multivariable analysis class.
@TedShifrin In Saudi Arabia, If you're going to be appointed as a lecturer, you need to take a general exam in your field about what you have studied.
22:20
@user777: I have never heard of such a thing. Is this a lecturer without a doctorate degree?
@Karim: I dunno. I was already an expert and had taught the course 15 times when I made my videos.
I think you may not be qualified to make authoritative videos on these subjects. Although who knows if anyone thinks videos are expert authority.
Maybe I should wait
Give a few seminar talks for your fellow graduate students. I did that quite a bit when I was a graduate student. I learned things better and they got to see stuff they would not otherwise have learned.
Yeah that is good idea.
I remember even giving talks on currents in several complex variables to learn that stuff better. Now it's long gone. But informal seminar talks are very different from posting things world-wide as if you're authoritative.
right yeah and if you say something that you might be missing something then it becomes encoded in people's brain.
I don't want to have that on my conciousness for now for people mathematical progress.
22:30
Think about the difficulties people have when there are mistakes in textbooks. They go nuts about that.
Or they believe it's true and tell everyone else wrong things. A stupid case of that happened to me. I have a remark in my algebra book (which is correct) that you can have an algebraic number of degree 4 over $\Bbb Q$ that is not constructible by compass and straightedge. I got an angry email from someone who told me that this was crap because Fraleigh had a T/F question indicating that it was correct. ... That person wouldn't believe me when I said Fraleigh was wrong
and that I had a counterexample later in my book.
Yeah I don't want that to happen to me I am very sensitive.
I will take it to heart ahaha
I will just record things for myself to view, also do grad seminars. The best purchase I made recently was iPad
with a pen
Oh yeah, I've never done a pencil with the iPad, but you can have lots of fun.
If you want to be amused by incorrect terminology driving lots of MSE people nuts (and the OP calling me silly and judgmental), look at this.
Okay :D
LOL
22:44
I'm such a meanie.
Math should be like that in order to keep the rigour.
Strange language that google would tell him commuting = invertible.
I have learned a lot by putting that emphasis. Besides he will thank you in the long run.
Well, calling himself alpha male did strike me as prophetic.
22:48
Anyhow, keep me posted on your seminars/lectures :)
Sounds good. Things I really need to strengthen is multivariable complex analysis. I found this book called from holomorphic functions to complex manifolds.
Great book so far.
What does it mean for a complex valued function to be continuously differentiable?
Does it mean the real and imaginary parts are continuously differentiable functions?
What author(s), @Karim?
@TedShifrin Klaus Fritzsche
22:52
Oh yeah, rings a vague bell.
I obviously do not know it, but I've perhaps looked at it or heard of it.
there is couple of videos on youtube by Alan Huckleberry he is also nice but not very rigorous though
besides I already know this stuff just want rigorous book or videos to add to my studies
@user2103480 ah I just noticed, the version on wiki is way worse than the one I have in mind. The one I know says that if X is Polish with the Borel sigma algebra and you put the Vietoris topology on $F(X)$, the hyperspace of closed sets, then there is a Borel function $f:F(X)\to X$ with $f(C)\in C$ for all $C$
@AlessandroCodenotti that is better?
And you can actually get a sequence of such functions $f_n:F(X)\to X$ s.t. $\{f_n(C)\mid n\in\Bbb N\}$ is dense in $C$ for all $C$
That's way better
What is the vietoris topology and why is F(X) a hyperspace
22:55
It's saying that the closed sets have a super nice choice function
what's a hyperspace and why is the terminology so sci fi
Hyperspace is just a vague term for a space whose elements are subspaces of some other space
@AlessandroCodenotti literally everything
Borel means measurable?
So there's a measurable choice function=
Nice
Borel means measurable when both spaces have the Borel $\sigma$-algebra
that is a cool theorem indeed
22:58
Nuts theorem
You can almost get something even better
Say that $X\subset\Bbb R^2$ is universal for the closed sets if $X$ is closed, and every closed $C$ in $\Bbb R$ can be realized as a slice of $X$
@AlessandroCodenotti is there any chance you told me this construction before?
By the Jankov-Von Neumann uniformization theorem there is a function $f:\Bbb R\to\Bbb R$ such that $(x,f(x))$ is always in the slice of $X$ over $x$
And the theorem also tells you that there is a universally measurable such function, meaning that it is measurable with respect to any Borel measure
(but some sets $C$ will appear as slices of $X$ many times, so it's not quite a choice function)
@user2103480 yes but I was talking about Borel codes, because the uniformization can be constructed in ZF for all Borel sets even, but it's impossible to extract a choice function without AC
23:17
@TedShifrin Yes, when you get a PhD, you promoted to Assistant Professor.
@TedShifrin lol, that's even funnier when you consider that in fact almost all such numbers aren't constructible by compass and straightedge
23:33
Yup. But even textbook writers make errors. Shocking.
user486313
hey @TedShifrin

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