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 $h \to 0^+$?
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
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
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
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.
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.
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.
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.
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.
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
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
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
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
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
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
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
"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
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.
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
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?
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 $v_0,...,v_k$?
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.
@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').
@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
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.
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 $i,j$
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
@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).
@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)
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
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
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?
I 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...
@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 $|G|=p^3$.
I 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...
