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12:20 AM
By First order Taylor's formula, we can write for a differentiable real-valued function with real domain, at $x\in \mathbb R$ as $f(x+h)-f(x)=f'(x)h+E(h)$, E(h) is an error function. Usually derivative at $x$ means slope of the tangent at that point. But according to the derivative of functions of several variables. Here Derivative of $f(x)$ is $T$: T(y)=f'(x)y$. Right? Is in't a tangent line?
But derivative at $x$ is the slope of the tangent at $x$ :(
 
Let $Av_h f(x) = \frac{1}{h} \int_{x}^{x+h} f$. If $f$ is continuous, then why does it follow that $Av_h f(x) - Av_h f(y) \to f(x) - f(y)$ as $h \to 0^+$?
 
@user193319: Easier question: Why is $\lim\limits_{h\to 0^+} Av_h f(x) = f(x)$?
@Mathgeek: Even in the single-variable situation, you have a linear function $L(y) = my$ where $m=f'(x)$.
 
Actually which one is derivative? $L(y)$ or $m$?
@TedShifrin
 
The linear function $L=f'(x)$ is the derivative at $x$.
 
@TedShifrin Well, I would say this has something to do with the fund. thm. of calc., but I am dealing with the Lebesgue integral.
 
12:34 AM
Just use continuity, @user193319. I don't care which sort of integral :P
 
howdy yall
 
@Mathgeek: In single-variable calculus we call the derivative $f'(x)$ a number, but the right viewpoint is that of a linear function.
heya @JoeShmo
 
hows everything going
 
Isn't that $L(h)$ the Tangent line?
 
The graph of the linear function (appropriately translated) is the tangent line.
 
12:38 AM
is there a well known special case best rank-1 approximation to an arbitrary $m \times n$ matrix $A$?
Eckart–Young–Mirsky theorem will give me rank-k
 
Not that I've heard of.
 
for any k, granted..
 
I have actually never heard of this question before.
 
the guy this semester is..
peculiar..
 
I suppose it's somewhat related to the spectral theorem in the square case.
But that gives a sum of rank-1's.
 
12:41 AM
All aboard the train to nerdville! How's everything going?
 
Oh hell. It's Demonark.
Spring break about over?
 
mathematics is cool!
 
How do I write $T(x)=5x^2+4x$ as $T$? For $T(x)=sin(x)$, I can write $T=\sin$.right? I lack this this knowledge. This might create the problem. @TedShifrin
 
Barely started actually
 
Oh, in general there's no way to write it unless you say $T = 5s + 4I$ where $s(x)=x^2$ and $I(x)=x$, @Mathgeek.
Oh, Demonark.
@Mathgeek, so you could write $T = 5I\cdot I + 4I$ for $I(x)=x$.
 
12:45 AM
I'm in South Bend, yesterday and today was the ND visit
 
Oh ... I was gonna say to say hi to Nancy Stanton if she hasn't retired.
 
@Daminark are you going to grad school at ND?
 
She seems to be Emeritus based on her website
 
I figured. I probably know some other people there, but I knew her quite well in college and grad school.
 
@JoeShmo not likely, chances are I'm heading to Madison
 
12:48 AM
very nice! any idea what youll be studying?
(dont say mathematics)
 
Lmao, got me
I'm not 100% sure yet. Tentatively something under the algebra and/or topology umbrellas
 
Wow, I hadn't known Karsten Grove moved to ND.
 
A number of people have been coming here, they're really trying to grow their algebra and geometry/topology groups
The school in general has had a lot of money that they're not sure what to do with. They gave the math department enough a few years ago to open 5 new tenure/tenure-track spots, and also they have a nice geometry/topology RTG
 
They used to be a real presence in complex geometry and several complex variables, but those people are all retired.
Oh, the RTG is great for a prospective grad student.
 
Yeah, also their people are really nice, especially for stuff like TQFT
But yeah it was a pretty good visit, got to meet some really nice folk over the past few days, and tbh even if I'm likely heading to Madison I hope to keep contact with these guys
 
12:58 AM
Contacts and collaborators are always important :)
 
For sure. I'll be heading to the Madison visit Tuesday so I'll report back soon on that as well
 
Cool, Demonark. I hope it goes well.
 
Thanks! In any event, how've things been going on your end?
 
Well, my midterm exam for my AoPS kids was an unmitigated disaster. I think I'm gonna hang up my teaching hat.
No fun teaching kids who don't have time to dedicate to what I want to be a good course.
 
what are you teaching them?
maybe reduce the size of the group?
quality over quantity?
 
1:05 AM
Ha ... It's already just 5 kids.
 
Find out which teacher has the best students and swap :P
But really that's sad
 
All but one are enrolled in a high school AP calculus course. I'm trying to do a better job (in about an hour a week) than their high school teachers of teaching concepts and even some mathematics.
 
count your chickens..
 
AoPS is really focused on the young kids. By high school age they're too busy with stuff.
I volunteered to teach an advanced course for some really good kids, but I doubt they have the time to put into it (and nor will the powers that be want it anyhow, I'm sure).
Nah, @JoeShmo. I missed teaching, but this isn't really the teaching I take pride in.
 
yeah :\
 
1:06 AM
Hmm, actually I wonder, when it comes to K-12 math education, do you think there's a country that just seems to get it right?
 
Certainly not us.
Look at this latest college admission bribery scandal. Typical entitledness shit in this country.
 
i can tell you my math education in israel was fairly OK. Certainly we did real mathematics
 
I think our chat friends from France had a superior "high school" experience. But they specialize so early and turn into grad students so young.
I am a big fan of general, broad education.
 
and although i was never educated in russia, all my math teachers were, and i understand that the ruskis did a good job too.
french huh?
 
Well, that's for the superior students. I'm sure most people in Russia fall by the wayside in education.
 
1:09 AM
oh yes of course
we did real -- albeit trivial -- mathematics
 
But we abandoned vocational education, which I thought was an important piece of the story in the US. When I went to high school, there were serious classes for students who wanted to work with their hands, etc., and had no interest in college. We need to go back to more of this.
 
as opposed to plug and chug (non-)mathematics in highschool in the land of the free
 
I realize this is not the point of the discussion here.
 
I feel like Germany takes an interesting gamble, if, around the equivalent of middle school I think, students don't seem like they're particularly academically inclined, they sorta redirect then to more technical training
 
Such tracking shouldn't be too young, though, Demonark. That's my point, that kids in Europe have to almost declare a graduate school dedication in high school, and it would be virtually impossible to major in math and some foreign literature in college there, which I loved doing here.
 
1:11 AM
well.. manual labor isn't so hot anymore..
 
It needn't be so manual, @JoeShmo.
Car repair requires more technical/computer skills these days.
Plumbers and electricians still make a great living.
 
and i realize its going to take several decades before that skill becomes completely obsolete..
sure, say for the next 30 years.
 
You think we're going to make plumbing and electrical work in houses/buildings obsolete?
 
@TedShifrin Is that a challenge?
And do you really want to make it a challenge? :P
 
i think the maintenance thereof isn't going to be a human job in.. idk. certainly 100 years
 
1:13 AM
Automating of that sort of thing requires a bit of an infrastructure overhaul and America doesn't strike me as very good at it
 
Interesting ... but this will be long after I'm dead :P
 
i certainly cant offer a timeline for that shift, but that's obviously where were going
 
I take your point, @JoeShmo, but we have real people right now who are being failed by our system.
 
@Dair I mean it's not a bad outcome by any means so I don't think too many people aside from the electricians and plumbers would mind if you accept
 
well, i can support the allocation of public funds for retraining programs.
 
1:15 AM
But I agree that everyone needs more basic computer skills, and basic accounting skills, etc. This should be part of high school for those who want/need it, just like typing used to be a thing when I was in high school. And shop.
 
Barack Obama actually tried in the Midwest, but it turns out coal mine workers like their jobs
and the cancer that it brings
 
They're just scared to tackle something new.
Being scared is real.
 
well..
how do you tell an illiterate 40 year old southern gentleman that perhaps computers are a better fit for him if coal is all he's been doing since age 5
 
Not all problems have a good solution
 
Be careful with your quantifiers and English :P
I guess that's OK.
 
1:17 AM
Doesn't like Gowers or somebody of similar callibre think mathematicians will be replaced some time in the (relatively speaking) soon future?
 
mathematicians are the last skilled workers to be replaced
 
@JoeShmo How much money are you willing to bet on that? :P
 
in which direction?
for or against?
 
Gotta get tenure before then
 
and time works in my favor..
 
1:19 AM
Tenured faculty are going to disappear ...
 
How much money are you willing to bet that mathematicians, indeed, will be the last skilled workers to be replaced?
 
well, hold on. not all is lost
hyper automation also means things are gonna get real cheap, real quick
well, careful with the phrasal there, @Dair
what am I signing here?
that mathematicians are, in fact, going to be replaced by the machine
 
But I mean, I dunno there's automated theorem proving in its seeming infancy (to me) but I feel like for math folk to be replaced fully we'd need computers to also have a way of determining what the interesting questions to ask are
 
or that they are the last to be so replaced?
 
the later.
@Daminark I wanna specialize in this at some point lol.
 
1:21 AM
oh. i am willing to put down a whopping $5
which is a lot, given that i am not a gambling man..
 
@Dair: Clearly English grammar/syntax experts will be needed.
 
So efficiency at however they prove things or creativity, and having the sense to ask the right questions? I feel like there's gonna be some time before AI gets on that, and chances are if it's that capable we may be in post-scarcity
 
in fact I can prove that thats true
(yes I did!)
(come on.. call me out)
 
@JoeShmo I literally don't know what you're trying prove.
 
I can prove that mathematicians are the last professionals to be so replaced
 
1:25 AM
Exercise: insert commas so my statement isn't horrendous to read. But in any event yeah, I think the more likely threat to mathematicians' careers is the public deciding we're not worth keeping around
 
@JoeShmo Wait don't tell me, I want some time to think about this exercise.
nah i got nothing go ahead.
 
If the machine can come up with an arbitrary $\lambda$-expression (otherwise known as a computer program, otherwise known as a mathematical proof) to solve an arbitrary mathematical problem,
then the same machine can solve any other problem as well (which are easier)
it can solve physics, then chemsitry, then biology
it can solve mechanical and electrical problems
 
"which are easier" big if true
 
it can solve problems in philosophy
well, if every (technical) discipline is a derivative discipline of mathematics
then thats easy to see
 
@JoeShmo What if the problem is undecidable? :P
 
1:33 AM
then the implication holds vacuously :)
 
well, see that's the theory, but I don't think it works in practice that way. :P
 
of course it doesn't
 
I mean they use tools from math but math can't replace their empirical elements. Thing is, if machines could reliably replace mathematicians, it'd have to be able to solve math problems in bounded time
 
no well you could write a program that operates within higher reliability parameters in an empirical setting than a human investigator does
yes. problems that are solvable in unbounded time are undecidable by definition (not solvable at all)
 
If it were capable of that it'd need to do so without aid of heuristics or making educated guesses or anything, for instance, since there's no guarantee that those methods of attack would lead to solutions
 
1:37 AM
perhaps there are stronger machines out there that i dont know of that exceed the class of all turing computable problems
(in fact i think there are things like co-turing languages, etc.)
but im not really familiar with that theory
 
rip my gaming computer died. Turns on but doesn't display anything on the monitor. :(
 
correct. so the particular theories i am alluding to are proof theory and model theory
and in particular proof mining in programming languages
to be clear, there's a long road ahead before any of these programs are "complete"
but hypothetically, the natural conclusion is everything we talked about
alright gents, homework won't write itself up unfortunately (yet!)
 
However, solving problems in many professional disciplines requires effectively being able to use heuristics and such
 
I feel like i've derailed chat yet again...
 
so i claim that this requirement is necessary today
but in the future it wont be
and, indeed, any heuristic begs a formalization
either there's a good reason for why it works, or it hasn't ever worked at all to start out with..
and, if heuristics bother you, check out Deepmind's AlphaGo!
 
1:41 AM
"any heuristics begs a formalization" even for math that's a ballsy claim which I wouldn't get behind with much confidence, and tackling IRL problems is likely a completely different story
 
Go (apparently) is intrinsically so based more on intuition than other games like chess
well, either a heuristic works, or it is a premature generalization of the present facts
would you agree?
if it works, there ought to be a good reason why
heuristic reasoning = intuition for the purposes of the discussion yeah?
 
I mean, the issue is that heuristics may not work reliably but may still give insight
 
i think we're getting at the same thing. if, in fact, a heuristic offers reliable insight, then it follows that there's an underlying reason as to why (dig in! go for the proof)
if it doesn't work reliably, then it doesn't work.
alright. this was fun. i have homework, and chores.
neither of which will do itself, as mentioned..
cheers!
 
See you!
 
 
2 hours later…
3:28 AM
peeks in I don't suppose anyone's around who is familiar with the Priestley topology for the prime spectrum of a distributive lattice?
 
 
6 hours later…
9:06 AM
@N.S.JOHN You only need basic number theory to prove this. — Kenny Lau Oct 7 '17 at 20:03
@LeakyNun Randomly spotting you on the main
I gave him a hint for a proof
 
hi
 
'lo
 
9:24 AM
I'm stuck at combinatorics ;__;
the sadness is unreal ;__;
 
 
1 hour later…
10:47 AM
just checking in to make sure everything is properly censored, looks all good team, no evidence of important people getting their feelings hurt so it can only mean everyone is being diligent
 
11:36 AM
Is it true that if $K$ is field extension of $F$ then basis of vector space $K$ over $F$ contains one and only one element of $F$?
 
@AkivaWeinberger that moment when I'm randomly spotted on the main
@Silent it can contain no elements of $F$
but $F$ is a subspace of $K$ with $\{1\}$ being a basis, so use dimension theorem
 
12:03 PM
@LeakyNun I don't understand what do you want to say by this: Do you say that $F$ has only one element in its base?
 
$\{1\}$ is a basis of $F$ as a subspace of $K$
 
i see that. so vector space $K$ over $F$ has at most one element of $F$ in basis?
 
12:50 PM
you sound like a vector space can only has one basis... I don't know how to interpret your questions
 
1:15 PM
i am sorry if i was ambiguous. I am asking, in any basis of K over F, we can have at most one element of F, right?
 
yes
 
thank u
 
1:38 PM
A vector field with less information is called a weak vector field
A vector field with complete information is called a smooth vector field
As you can see there are a lot more vectors in the smooth vector field
Hope that helps
 
@Ultradark, was that for me? I was not talking about vector field from multivariable calculus!
 
@Silent that was for everyone because not a lot of people are familiar with this concept
 
ok!
 
2:40 PM
If $Y$ is a contractible subspace of $X$, and $X$ deformation retracts to $Y$, does it follow that $X$ is contractible? If not, what "niceness" assumptions need to be imposed on $X$?
 
The identity on $X$ is homotopic to a contraction $X\to Y$, which restricts to the identity to $Y$. The latter is homotopic to a constant map, can't you just compose the homotopies?
 
So, if $f_t : Y \to Y$ denotes the null-homotopy, $g_t : X \to X$ the deformation retraction, are you saying consider $f_t \circ g_t$?
If $x \in X \setminus Y$, then $f_0(g_0(x)) = f_0(x)$ doesn't make sense.
 
@user193319 Nah there are issues with the domains, it needs to be fixed a bit
 
Well, $f_t$ induces a retraction $r : X \to Y$, so should it be $f_t \circ r \circ g_t$?
Wait, I think I mean $g_t$ induces a retraction.
 
2:56 PM
I think something like $F:X\times [0,1]\to X$ with $F(x,t)=g_{2t}(x)$ for $t\in[0,1/2]$ and $F(x,t)=f_{2t-1}(g_1(x))$ for $t\in[1/2,1]$ could work
I'm a bit worried by continuity but it might be fine
 
Why would continuity be an issue? The two definitions agree for $t=1/2$.
 
it's fine I think
 
3:53 PM
Let $X_n(\Bbb{Z})$ be the simplicial complex whose vertex set is $\Bbb{Z}$ such that the vertices $v_0,...,v_k$ span a $k$-simplex if and only if $|v_i - v_j| \le n$ for all $i,j$.....Question: what are the elements of $X_n(\Bbb{Z})$? Also, what is meant by "span"? Does it mean all convex combinations of the integers $v_0,...,v_k$?
 
Yo nerds
 
sup yo
 
Hi @Dami @Ígjøg
 
Hey @Alessandro
 
given a vector field is there an easy way to find the corresponding differential equation
 
4:11 PM
A vector field is really a function of the form:
$$\begin{pmatrix}f_0 (\Bbb{x})\\ \vdots \\ f_n (\Bbb{x}) \end{pmatrix}$$
where $\Bbb{x}$ are the spacial coordinates
if the vector field is a velocity field then:
$$\dot{\Bbb{x}}=\begin{pmatrix}f_0 (\Bbb{x})\\ \vdots \\ f_n (\Bbb{x}) \end{pmatrix}$$
thus your job is to find the $f_i$ along each basis direction
 
Say I have: $\vec F(x,y)=(x,y)$
 
I am trying to show that $X_n(\Bbb{Z})$ is contractible, but I am having difficulty even visualizing it.
...in the sense of not knowing what the elements of $X_n(\Bbb{Z})$ are.
 
4:30 PM
Fail
Matrices that look like this (matrices whose column vectors sum to zero) always have an eigenvector of eigenvalue 0 regardless of basis
 
Hi there , I'm taking this class called: "Models of geometry". For now we are dealing with lines in $\mathbb{E}^2$ ( set $\mathbb{R}x\mathbb{R}$ with dot product) , mirroring , isometries , translations ... if anyone knows about any good literature ( that is available [free] online )that could help me with that , it would be much appreciated.
 
5:03 PM
@AlessandroCodenotti Just saw this; I am compelled to point out that it is worse than that. Given any nice formal system S (i.e. S has a proof verifier program and can reason about finite program runs), if S proves Con(S) then S also proves "0=1". On the other hand, S might prove ¬Con(S) without proving "0=1". @LeakyNun: For example, PA' = PA+¬Con(PA) proves ¬Con(PA) and hence also proves ¬Con(PA').
 
 
1 hour later…
6:25 PM
What is required condition for a sequence to have a derangement?
For example let sequence be 1 1 2 and there is no rearrangement such that numbers on same position are different
but for 1 1 2 2 there is a derangement 2 2 1 1
I think No number must occur greater than floor(N/2) times to begin with
Other than that?
 
6:43 PM
@user193319 I think it's not going to be practical to visualize X_n for arbitrary n. I mean, once you get to n=4, you're dealing with 4-simplices which are 4-dimensional polytopes
you can certainly project a 4-simplex into 2- or 3-dimensional space to help visualize them, but you can't visualize a four-dimensional object as such
 
Okay, I see. So what are the elements of $X_n(\Bbb{Z})$, from a formal set theory perspective?
E.g., is $X_1(\Bbb{Z})$ is subset of $\Bbb{R}$ or $\mathcal{P}(\Bbb{R})$?
 
Dunno. I didn't go as far as point-set topology
Just saying that some neat 'visualization' of X_n seems like too much to hope for
 
I figured as much. I'm now just curious to know what the elements of $X_n(\Bbb{Z})$; I feel like that might help in showing that $X_n(\Bbb{Z})$ is contractible.
 
What's the definition of X_n again?
 
Let $X_n(\Bbb{Z})$ be the simplicial complex whose vertex set is $\Bbb{Z}$ such that the vertices $v_0,...,v_k$ span a $k$-simplex if and only if $|v_i - v_j| \le n$ for all $i,j$
 
6:54 PM
okay. So the question is partly just what the elements of a simplicial complex are
looking at the definition of simplicial complex, I get that it's a set of 0-simplices (vertices), 1-simplices (edges), etc
so in the case of X1 it'll just be the vertex set (ZZ) and the 1-simplices i.e. edges
insofar as you think of the vertices as singleton sets and the 1-simplices as 2-element sets from ZZ, I guess you can think of the simplicial complex as being a subset of P(ZZ)
specifically, X_n will be a subset of P(ZZ) where none of the subsets of ZZ contain more than n elements
In doing so, though, I'm specifically interpreting a set like {-2,0,3,6} as representing a 4-simplex
 
7:10 PM
So {-2,0,3,6} is in X_4(Z) (and X_n(Z) for n \ge 4), and a 4-simplex is something that lives in R^4?
 
well, {-2,0,3,6} is possible as an element of X_4(Z)
but I don't think it actually would be in your case, since |6-(-2)|=8 is bigger than 4
 
Ah, of course.
 
so I think it'd only show up once you got all the way up to X_8, lol
 
That's true...This stuff is kind of strange.
 
but yeah, you can only realize a 4-simplex by being in at least 4 dimensions
you can have a 4-simplex living in a higher dimensional space, of course, just as you can have a triangle living in 3 dimensional space
but you can't have a triangle in 1-dimensional space
I feel like the simplest way to understand X_n might be to start with the n-simplices
so stuff like {0,1,2,3,...,n}
(along with {1,2,3,...,n+1} etc)
the remaining k-simplices with k<n will just be subsets of these
so P({0,1,2,3,...,n}) + P({1,2,3,4,...,n+1}) + P({-1,0,1,2,...,n-1}) + ...
 
7:17 PM
Ah, okay. I think this might help me (not entirely certain, though) show that X_n(Z) deformation retracts to X_{n-1}(Z)
 
That's part of the hint to show that X_n(Z) contractible.
 
@jeea I'm fairly certain it's: if you have one number that's in more than half of the positions, it's undoable
 
X_1(Z) is contractible because evidently it's just a graph.
 
right. just edges and vertices
 
7:18 PM
(like with 1 1 2 there are three positions and the 1 occurs in two of them)
and otherwise, it's doable @jeea
 
well, it's moreover a connected graph
 
For example: 1 2 2 3 3 has no number that's more than half of the positions (and it has the derangement 3 3 1 2 2)
and 1 2 3 3 has no number that's more than half of the positions (and it has the derangement 3 3 1 2)
but 1 2 2 2 has more than half of them being 2, so there's no derangement
 
but phrases like "X deformation retract Y" are beyond my knowledge
so I can't really say any more
 
I had never considered derangements with repeated/indistinguishable objects. That was interesting, thanks @jeea
 
7:31 PM
@Semiclassical So, is X_n(Z) obtained from X_{n-1}(Z) by adding in more vertices and connecting them to X_{n-1}(Z) with edges/line segments?
If so, these line segments are homeomorphic to [0,1] which, e.g., deformation retract to $0$. So deformation retract all these added edges and you'll get X_{n-1}(Z).
 
I don't think that's quite right. I doubt you're just adding edges
 
Shoot...
 
at least not once n is large enough
probably you should look at how X2 goes to X3?
 
7:45 PM
@Semiclassical Hey I have a numerical experiment question
that you may or may not be interested in
Let $S_n$ be the set of pairs of nonnegative integers adding to $2n$ (so $\{(0,2n), (1,2n-1), (2,2n-2), \dotsb, (n,n)\}$)
(not caring about the orders of the elements in the pairs)
Let $S_{n,k}$ be $\{(a,b)\in S_n\mid p_i{\not\mid}a,b,~1\le i\le k\}$
so, like, the set of pairs where neither of the entries are a multiple of any of the first $k$ primes
This is like the sieve of Eratosthenes, where at each step we're sifting out the things that are multiples of the $k$th prime
(ex: $S_{49,3}$ is the set of pairs of things adding to $98$ where neither is a multiple of $2$, $3$, or $5$)
(which is $\{(1,97),(7,91),(19,79),(31,67),(37,61),(49,49)\}$)
Also as $k\to\infty$, the set $S_{n,k}$ becomes the set of primes adding to $2n$
(plus possibly $(1,2n-1)$, if $2n-1$ is prime)
so Goldbach's conjecture says that that isn't empty, right? (And that it's still not empty when we delete $(1,2n-1)$, which is an annoying edge case)
So, the question
Consider $|S_{n,k}|/|S_{n,k-1}|$
Intuitively that should be around $(p_k-2)/p_k$ or $(p_k-1)/p_k$
(About $1/p_k$ of the elements of $S_{n,k-1}$ will have the first entry be a multiple of $p_k$, and about $1/p_k$ of them will have the second entry be a multiple of $p_k$)
(and those might overlap if $2n$ itself is a multiple of $p_k$)
So the question is, how accurate is that guess
Like, if you compute $|S_{n,6}|/|S_{n,7}|$ for lots of $n$, how closely is it approximated by $15/17$ and $16/17$
The reason I'm interested in this is that, if the lower bound is reasonably close to $(p_k-2)/p_k$, this implies the Goldbach conjecture.
The reason is that $n(\frac{3-2}3)(\frac{5-2}5)(\frac{7-2}7)\dotsb(\frac{p-2}p)$ where $p$ is the largest prime less than $\sqrt{2n}$, goes to infinity
and basically is never near zero except for small $n$
so if that's a good approximate lower bound for $|S_{n,\infty}|$, then we get that there's lots of pairs of primes adding to $2n$
Apparently if we graph that approximate lower bound we get this
(Orange is $n(\frac{3-2}3)(\frac{5-2}5)(\frac{7-2}7)\dotsb(\frac{p-2}p)$ where $p$ is the largest prime less than $\sqrt{2n}$, blue is the number of pairs of primes adding to $2n$)
You'll notice that this isn't actually a perfect lower bound - for example, there are 13 pairs of primes adding to 992 and the formula predicts a lower bound of 15
(well, 14 if we ignore the pair (1,991) — 991 is prime)
so this approximation is quite rough, and this is not an actual proof of the Goldbach conjecture
(I wish!)
but it's a good heuristic support I think
(Credit for the idea goes to a certain Reddit user who is convinced that this is a valid proof)
(but I don't really think it'll go anywhere)
 
8:11 PM
I didn't follow all of that, to be honest, but what I'm always curious about when it comes to heuristic arguments (like this) is how far out you'd have to go before said heuristic could reasonably be expected to break down
 
@Semiclassical Can I just do a quick example here
 
So let's say we want pairs that add up to 68
and we've sieved out 2 and 3
What we're left with is
1,67
3,65
7,61
13,55
19,49
25,43
31,37
Now let's sift out 5s
Intuitively, we should expect that about a fifth of these should have a multiple of 5 in the first entry, and about a fifth should have a multiple of 5 in the second entry
We end up with
1,67
---3,65---
7,61
---13,55---
19,49
---25,43---
31,37
Wait how do I do strikethrough
Oh why did it work there and not there
Whatever, we end up with
1,67
7,61
19,49
31,37
Yeah? @Semiclassical
We started with 7 and ended up with 4
We would expect that we remove around 2/5 of these, and 7*2/5 is 2.8
and we actually removed 3
So our estimate that we'd remove roughly $2/p$ is roughly correct
Random question
What percent of Earth's surface is within the Arctic Circle?
 
Shouldn't 3,65 have been seived out initially since 3 is divisible by 3?
 
8:27 PM
Oh right whoops
Actually no it should stay in @Semiclassical
I wrote the definition wrong above
You shouldn't sieve out the prime itself
because otherwise, at the end, nothing will be left
whereas I just want to remove composite numbers
Like the usual sieve of Eratosthenes
 
whereas 2,66 gets kicked because 66 is divisible by 2?
 
Mhm
and 3,65 does get removed when we're doing 5s
By the way, Goldbach originally formulated his conjecture with 1 defined as prime, right?
Is that equivalent to the modern version?
 
dunno
 
The only way it wouldn't be is if there's an $n$ for which $2n-1$ is prime, and there's no pair of primes adding to $2n$
That doesn't sound unreasonable though
 
9:15 PM
Can anyone give me a few tips on how to render a Mandelbulb using the OpenGL (or a similar) API? I'm just learning OpenGL and I can render a Mandelbrot to a quad surface with parallel processing via the Fragment Shader (I test if scaled pixel coordinates on the quad are in the Mandelbrot set) but I'm not sure how I would go about generating points (or an appropriate visual representatoin) in 3D space... Do I have to resort to learning the Compute Shader first?
I'm thinking that maybe I could do ray tracing -- I'll draw a quad that covers the view port completely (like I did for the Mandelbrot), and for each pixel on the quad I'll mimic light rays shooting out into space (by using the Fragment Shader). I'll then determine they hit (or miss) the surface of the Mandelbulb and color the pixel from which the ray emanated accordingly... thoughts?
 
9:48 PM
Also tell me if this is not the right place to ask haha
 
10:18 PM
2
Q: Non-abelian Groups of Order $p^3$

user193319I am trying to show that a non-abelian group $G$ of order $p^3$ is isomorphic to one of two groups constructed on page 48 of these group theory notes (see examples 3.14 and 3.15 on that page). Here is some of my work so far: Since $G$ is a $p$-group, it must have non-trivial center; but, a...

 
Classifying $p$-groups in general is a difficult endeavour
 
Would someone mind taking a look at the comment I made on Beginner's answer?
There's just one small part of the proof I don't understand.
Whoops, I forgot I actually made two comments. The first one is more important than the second.
BRB--off to make dinner.
 
10:51 PM
@user193319 For $Z(G)=G'$, note that both are nontrivial. (One is an important fact about $p$-groups; the other is from the definition.) We have $G'\subseteq Z(G)$ because $G/Z(G)$ is abelian. By considering orders, you get $G'=Z(G)$. All we needed was the fact that $G$ is non-abelian and $|G|=p^3$.
 
the other is from the definition [of non-abelian]
 
@KarlKronenfeld Ah, dang it! Of course. Thanks!
 
Case 2.1 can be ruled out in a different way if you know about the connection between the Frattini subgroup and generators.
I'm still thinking about the wlog assumption you asked about.
 
hi @Karl and @Leaky
 
hey
 
10:55 PM
@LeakyNun Yes, I said that at the end (it's implied in the context).
hey @Ted
 
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