Rigor only makes sense in context of the language math is spoken. We consider the language of mathematics during Renaissance primitive, and we will be seen as primitive in the future.
The foundations in Elements was criticized almost 2000 years after Euclid, by Gauss, when he discovered hyperbolic geometry. During that time foundations in geometry was nearly unchanged. Immediately afterwards Klein developed a new context for geometry which is used still to date.
It's taken a long time to break through the standards set by Euclid.
And as Fargle says, we still use that kind style of writing in mathematics.
The fact that it took about 2000 years to confirm Euclid's intuition that you needed the parallel postulate to do ordinary flat geometry speaks volumes for its longevity
So then was Elements more of the type of text that was ahead of its time, or was it just that everything after that and before the 18th century was a regression of sorts?
Of its time more than ahead---Greek mathematics, as far as I recollect, represented the latter half of a slow and subtle shift in Western mathematical traditions from concrete to abstract. Don't quote me on that though, I'm not a math historian
It was some geometric method, yeah. I don't recall specific either. It's probably true that contemporary mathematics in Islamic tradition already made advances when Europe was too busy with the crusades
The scientific revolution in 17th century was the big comeback
Starting with Galileo
It's important to note that there are adverse effects of rigor too. Grassmann developed an abstract theory of exterior algebras that a man of high stature such as Kummer didn't understand
Of course the Galois example also fits the bill but that's been reiterated too often
Ah Grassmann apparently wrote a second textbook on linear algebra which was completely rigorous but it was still ignored by and large by the scientific community
I liked his exposition of the proof of isoperimetric inequality without sharp constants using entropy
It did make sense that the Kolmogorov point of view was obscuring the main idea
But things like finite probability spaces, discretizing general probability measure spaces (what he calls the subcategory of finite prob spaces being dense in the topological category of probability measure spaces) are well-known to computer scientists
Mixed. Started specializing in logic, up to forcing and such, now shifted more towards probability theory & analysis, but taking a course on semantics of HoTT on the side which is giving me headaches
Understandable, it seems like the kind of topic. Trying different literature as side references is a mess since the expositions differ a lot in structure, notation, and approaches to proofs
I looked at that and was horrified. Mostly, I use Rezk's intro to quasicategories, which I find pretty understandable, but doesn't cover everything in the course
I just never ended up using any substantial simplicial homotopy theory. I think all I wanted to understand was if $G$ is a simplicial group acting freely on a simplicial set $X$ then $X \to X/G$ is a Kan fibration or some variant of this. This simplifies a lot of fact-checking when dealing with fibrations of space of immersions, embeddings, diffeomorphism groups of manifolds, ...
Having skipped straight from no algebraic topology to simplicial sets, which is more manageable than expected, books such as Goerss-Jardine expect too many prerequisites
Percolation is heavy. In my introductory course on martingales and ergodic theory, we were given a few exercises that scratched the surface of percolation theory, since that is what the professor and assistants specialized in
@BalarkaSen I was referring more on a conceptual level; I know Cal 1 and I can understand how to understand some stuff beyond that like Linear Algebra, Multivar Cal, and ODE, but then beyond that it just looks like alien language.
@BalarkaSen it's interesting once you're used to the theory. Most applications are in finance or physics though, which isn't so lucky for me. But I'm taking stochastic PDE soon from a prof that specializes in stochastic neuroscience, so that might be a cool topic to dive in
@BalarkaSen but it will always stay an alien language once you move away from your comfort zone
I was asking, primarily because I was looking into univariate polynomials for months and months, and it took me a while, but I eventually figured out how to invert a cubic ($y = ax^3 + bx^2 + cx + d$) with some substitutions.
Quadratic is just a linear substitution, cubic was a weirder rational one, and IDK quartic. How come you can't do some substitution magic with a quintic though?
I will say that in the end it is somewhat arbitrary, because (IIRC) Abel-Ruffini Theorem says the general quintic is unsolvable in terms of elementary operations and $\sqrt[n]$ roots.
What does "order matters" mean in 5. and 6. of the twentyfold way (en.wikipedia.org/wiki/Twelvefold_way#The_twentyfold_way)? My understanding is that the twelvefold way (which doesn't include these two) already covers ordering VS nonordering of both the domain and the codomain. What additional order is being considered?
Suppose we have a function $f : \mathbb{R}^m \rightarrow \mathbb{R}^n$ with the condition that the rows of the Jacobian matrix of $f$ are linearly independent on $B \subseteq \mathbb{R}^m$.
Let $B_p$ = $\{ t \in B : \forall s \in \mathbb{R}^m \; \; \lvert s - t \rvert < \frac{1}{p} \implies \fra...
$$\dfrac{x^2}{(x-1)^2}+\dfrac{y^2}{(y-1)^2}+\dfrac{z^2}{(z-1)^2}=\left(\dfrac{x}{x-1} +\dfrac{y}{y-1}+\dfrac{z}{z-1}-1\right)^2+1$$ how to think,to derive at this?! intuition?
@user69608 Don't you need some additional conditions for that to be true? Maybe $xyz=1$?
How is it humanly possible to derive this? How to derive the fact that $$\dfrac{x^2}{(x-1)^2}+\dfrac{y^2}{(y-1)^2}+\dfrac{z^2}{(z-1)^2}=\left(\frac{x}{x-1} +\dfrac{y}{y-1}+\dfrac{z}{z-1}-1\right)^2+1$$I can prove this by expanding, but there's no way on earth I would've found that equality myself. — user26486Dec 23 '13 at 21:38
i've done calc 1 & 2 only; can i learn abt the lagrange inversion theorem and the lambert W func? what more do i need to learn to be off with enough prerequisites?
"Let $\mathfrak{m} \in \text{Max}(R)$ and $\mathfrak{p} \in \text{Spec} R$. Show that $(R, \mathfrak{m})$ is a local ring iff $1 + \mathfrak{m} \subset R^{\times}$." this pset has lots of unnecessary junk in the problem statements
@Alessandro Do you know much about the geometry of Heisenberg group? A presentation seems to be $\langle a, b | [[a, b], a] = 1, [[a, b], b] = 1\rangle$
Here's a picture of the Cayley graph. $z = [x, y]$, I believe. Doing a commutator along $x$ and $y$ climbs you up the $z$-axis
Ah the geodesics literally swerve upward like Lagrangians in a contact manifold. This is what they mean by Carnot-Caratheodory metric
And this has to do with commutator length I am sure. Nutcases, Gromov and his student Pansu
"Let $\mathcal{A}_{\text{Zar}}$ be the set of closed sets of a topology $\mathcal{T}_{\text{Zar}}$ ... " I also don't get why there's a distinction being made between A_Zar and T_Zar
Even when you make some mistake it's because of a forward/backward issue. Maybe pushforward became pullback, or a left arrow because a right arrow in your diagram, ...
Sets of the form $U(f) := \lbrace \mathfrak{p} \in \operatorname{Spec} R : f \notin \mathfrak{p}\rbrace$ form a basis for the Zariski topology; if $\mathfrak{p} \in U(f) \cap U(f')$ then $\mathfrak{p} \in U(ff') \subset U(f) \cap U(f')$ and i think $\bigcup_{f \in R\setminus I} U(f) = V(I)$
@Naganite I don't have a full answer to this, but maybe a useful comment: If you have a quadratic polynomial $X^2+pX+q$, you can substitute $X\mapsto X-\frac{p}{2}$ to bring it to the form $X^2+(q-\frac{p^2}{4})$. This substitution is precisely done so that the $X$ terms cancel out. Similarly, if you have a cubic $X^3+aX^2+bX+c$, you can substitute $X\mapsto X-\frac{a}{3}$ to get the cubic $X^3+(b-\frac{a^2}{3})X+(\frac{2a^3}{27}-\frac{ab}{3}+c)$. Indeed, there is a pattern here. If you have a monic polynomial of degree $n$, substitute $X\mapsto X-\frac{a}{n}$, where $a$ is the coefficient …
@Thorgott just thought, my argument for the injectivity of $S^{-1}N$ given an injective $N$ was wrong yesterday, since $\operatorname{Hom}_{S^{-1}R}(S^{-1}(-), S^{-1}N)$ is contravariant, so you actually get a surjection $$\operatorname{Hom}_{S^{-1}R}(S^{-1}R, S^{-1}N) \to \operatorname{Hom}_{S^{-1}R}(S^{-1}I, S^{-1}N)$$
A while ago I was reading Frobenius theorem (manifold) and although I didn't understand it completely, a key takeaway was the existence of integral manifolds which was used to prove the existence of solution on 1st order homogenous linear PDE. This got me thinking that to prove the higher order inhomogenous case, we (if possible) can extend this theorem for a tensor field. Is there any research/result in this direction.
I have to find the sum of the roots of $tan^2(x) - 9tan(x) + 1 = 0$ between $0 <= x < 2\pi$. I've ended up with $\frac{n([(n-1)\pi + arcsin(\frac92)] + [(n-1)\pi + 2arctan(\sqrt{\frac{81}{4} - 1} + \frac{9}{2}])}{2}$ but apparently It's doable without a calculator as well. Could I please get any hints?
@VioAriton If you do a substitution you get a polynomial (so the term roots actually makes sense). Then use that the sum of the roots can be read directly from the coefficients.
@henceproved The generalizations of Frobenius were started by Cartan and Kähler a century ago. Look up the Cartan-Kähler theorem and exterior differential systems. I recommend the relatively recent book by Bryant, Chern, Gardner, Goldschmidt, Griffiths with the title Exterior Differential Systems. But it's quite advanced.
For topological spaces $X$ and $Y$, is it possible that $X \times X$ and $Y \times Y$ are homeomorphic, but $X$ and $Y$ are not homeomorphic?
(I poked around with finite spaces, and manifolds, and the Cantor set, without seeing any examples.)
This was inspired by Existence of topological space ...
Does there exist a topological space $X$ such that $X \ncong Y\times Y$ for every topological space $Y$ but
$$X\times X \times X \cong Z\times Z$$
for some topological space $Z$ ?
Here $\cong$ means homeomorphic.
True, I guess restricting yourself to a subcategory you can actually say something about gives a way of somewhat systematically ruling out which spaces will and which spaces won't work/what properties they must satisfy. When working in the general topological category, you don't really have many options other than guessing.
especially in the second case, I think "being a square" is probably a very hard property to detect in general
There's also positive dimensional spaces homeomorphic to their square though, they are used to show that the inequality $\dim(X\times Y)\leq\dim X+\dim Y$ can be strict in the product theorem for the covering dimension
The subspace of $\ell^2$ made of points with only rational coordinates (this is sometimes called the Erdos space) has dimension 1 and is homeomorphic to its square
And you can actually look at the subspace made of sequences all of whose coordinates are of the form 1 over an integer, the closure of that is still 1 dimensional, but now we even have a completely metrizable example
Those are totally disconnected spaces with positive dimension which is another weird combination of properties
Where $\pi(x,q;a)$ count the prime less than x that is congruent to a modulus q. $\phi(x)$ is the ruler totient function, and $Li(x)$ is the logarithmic integral
The prime number theorem implies $\pi(x) = Li(x) + exp(xe^{a\sqrt{\log{x}}})$ as $x \to \infty$, but i'm not sure how to use that to give a bound on $|\pi(x,q;a) - \frac{Li(X)}{\phi(q)}|$
Well, from what i understand, Dirichlet's theorem is about the infinitude of primes in arithmetic progressions. I've been told to look at Brun-Titchmarsh theorem but I don't see how to use B-T to bound this error
Numbers that are congruent to a modulo q is precisely what an arithmetic progression is, so, in your notation, Dirichlet's theorem is the assertion that $\pi(x,q;a)\rightarrow\infty$ as $x\rightarrow\infty$ (when $a,q$ coprime of course, nothing happens otherwise) and what you want is a quantitative result in that direction
I don't know any analytic number theory, for the record, but that's the direction in which I'd look
For topological spaces $X$ and $Y$, is it possible that $X \times X$ and $Y \times Y$ are homeomorphic, but $X$ and $Y$ are not homeomorphic?
(I poked around with finite spaces, and manifolds, and the Cantor set, without seeing any examples.)
This was inspired by Existence of topological space ...
Does there exist a topological space $X$ such that $X \ncong Y\times Y$ for every topological space $Y$ but
$$X\times X \times X \cong Z\times Z$$
for some topological space $Z$ ?
Here $\cong$ means homeomorphic.
