Recent MathOverflow Questions

Variation on the definition of the uniform distribution mod 1 [on hold]

Math Overflow Recent Questions - Mon, 10/09/2017 - 16:04

A sequence $x_{n}$ is said to be uniformly distributed mod 1 if $\forall a,b$ with $0\leq a<b<1$, $$\lim_{n\rightarrow \infty}\frac{1}{n}|\lbrace j=1,...,n :\lbrace x_{j}\rbrace\in [a,b]\rbrace|=(b-a).$$ (here $\lbrace x_{j}\rbrace$ is the fractional part of $x_{j}$). But what about the following: $$\lim_{m\rightarrow \infty}\frac{1}{m}|\lbrace j=n,...,n+m :\lbrace x_{j}\rbrace\in [a,b]\rbrace|=(b-a),$$ when $n$ is a function of $m$. Does the former imply the latter?

In particular, I'm asking this in order to obtain the following bound: $$|\lbrace j=n,...,n+m :\lbrace f(j)\rbrace\in [a,b]\rbrace|\ll (b-a)m,$$ with $f(x)$ a polynomial with at least one irrational coefficient and $n=m^{3}$. A sort of short interval variation on the definition of uniform distribution mod 1.

Can anyone help me?

Thanks very much in advance.

EDIT: Previously, I had written that $n$ could be fixed. In this case, the answer is obvious, as GH from MO noted. I'm interested in the more complicated case when $n$ varies with $m$. So I edited the question.

Converse of the Archimedean property of the sphere

Math Overflow Recent Questions - Mon, 10/09/2017 - 16:01

In his remarkable book On the Sphere and Cylinder, where he came tantalizingly close to discovering calculus, Archimedes showed that the area of the portion of the sphere contained between a pair of parallel planes cutting the surface depends only on the distance between the planes (see p. 625 of this paper of King). This fact, which has been dubbed Archimedes hatbox theorem, is now a standard exercise in many calculus texts, and is even commemorated on the back of the Fields medal (if you look closely there, you can see the figure below in the background).

                                                     

Conversely, Blaschke showed that the only convex surface with this slab area property is the sphere. Indeed it is a simple exercise in classical differential geometry to check that any smooth convex surface with the slab area property must have constant curvature (Let $A(h)$ be the area trapped between a tangent plane and a prallel plane at distance $h$, and compute the limit of $A(h)/h$ as $h\to 0$). But can one still characterize the sphere if we fix the distance between the planes:

Question: Let $S\subset R^3$ be a convex surface with diameter $d$. Suppose that for some constant $0<h<d$, the area of the portion of $S$ trapped between every pair of parallel planes separated by the distance $d$ is constant, whenever both planes intersect $S$. Does it follow then that $S$ is a sphere?

A convex surface is the boundary of a compact convex set with interior points, and its diameter is the distance between its farthest points in the ambient space. Although this is known to be an open problem, I am not aware if it has been explicitly mentioned anywhere.

Addendum: See this question for a related problem.

A provably infinite infinitary aliquot sequence?

Math Overflow Recent Questions - Mon, 10/09/2017 - 15:40

A divisor of n is called infinitary if it is a product of divisors of the form p^{y_a 2^a}, where p^y is a prime power dividing n and sum_a y_a 2^a is the binary representation of y. [from OEIS A049417]

An infinitary aliquot sequence is defined by the map x -> A049417(x)-x. [from OEIS A127661] As with ordinary aliquot sequences, the map may end with a zero, enter a cycle, or wander haphazardly to increasingly large values with unknown outcome.

In a recent attempt to categorize some unknown-outcome infinitary aliquot sequences, I came across a new possibility. Starting at 706200, OEIS A293355(22), the number's evolution increases monotonically for at least 247 terms! If the condition of each term being greater than its predecessor is provably so, then of course we have a decidely infinite infinitary aliquot sequence. Can it be proven?

Algebraic model for the abelian category of descent data for modules in the non-affine case

Math Overflow Recent Questions - Mon, 10/09/2017 - 15:12

Let $f: X \to Y$ be a morphism of schemes. I'd like to have a completely algebraic description of the belian category of descent data for modules along $f$. Here's my attempt:

The category of quasi-coherent $\mathcal{O}_{X \times_Y X}$ modules caries a monoidal structure for which $\mathcal{O}_X$ is the unit (given by tensor product and pulling and pushing along obvious diagrams). Inside this category $\mathcal{O}_{X \times_Y X}$ has a canonical structure of a counital coalgebra over $\Delta_*\mathcal{O}_X$ with counit given by multiplication

$$\mathcal{O}_{X \times_Y X} \to \Delta_*\mathcal{O}_X$$

And comultiplication given by the insertion of the unit in the middle.

Denote by $LCoMod(\mathcal{O}_{X \times_Y X})$ the category of left quasi-coherent $\mathcal{O}_{X \times_Y X}$-comodules.

In the affine case this category is equivalently coalgebras over the comonad $f^*f_* \to Id$ which is the category of descent data by monadic descent.

Question: Is $LCoMod(\mathcal{O}_{X \times_Y X})$ a model for the category of descent data in the non-affine case?

What is known about stability of number theoretic statements for Beurling systems based on small perturbations

Math Overflow Recent Questions - Mon, 10/09/2017 - 13:35

Beurling considered a sequence of reals $1<x_1<x_2<\cdots <$ as "primes" and then the ordered sequence of all products of these "primes" as "integers". Let us consider Beurling primes which are small perturbations of the ordinary primes. My question is:

1) How small perturbation guarantees that the Beurling primes/integers satisfy the prime number theorem.

2) How small perturbation guarantees that the system of Beurling primes/integers satisfies the RH given the "ordinary" RH.

3) How small perturbation guarantees that the the system of Beurling primes/integers satisfies Cramér's conjecture ("the number of integers between the $n$th prime ($p_n$) and the $n+1$ prime is $O(\log^2 p_n)$) given the "ordonary Cramér's conjecture".

Remark: I left the notion "small perturbation" vague and there is some other vagueness in the questions. But I will be interested in answers for any variant. Other basic results/conjectures can also be considered, and I will be interested also in information about them.

Motivation: This is motivated by some questions raised during the discussion of polymath 4 that I recalled in the context of the recent question The enigmatic complexity of number theory .

Distribution relation in the Euler system of Heegner points

Math Overflow Recent Questions - Mon, 10/09/2017 - 10:27

I am trying to understand the details behind the so-called "distribution relations" between Heegner points on the modular curve $X_0(N)$, as given (for instance) in Gross's paper Kolyvagin's work on modular elliptic curves, [Proposition 3.7, (i)]. More precisely, the relation between Hecke and Galois actions on CM-points is still not clear for me. First of all let me recall in details the general settings, which the familiar reader can skip and go directly to the question.

Settings: Let $N$ be a positive integer, let $K$ be an imaginary quadratic field of discriminant $D<0$ such that every rational prime divisor of $N$ splits in $K$ (the so-called Heegner hypothesis). The Heegner hypothesis implies that one can find a (non-unique) ideal $\mathcal{N}$ of $\mathcal{O}_K$ such that $\mathcal{O}_N/\mathcal{N}\simeq \mathbb{Z}/N\mathbb{Z}$.

For a general $n$, denote by $\mathcal{O}_n=\mathbb{Z}+n\mathcal{O}_K\subset\mathcal{O}_K$ "the" order of $K$ of conductor $n$, and let $Pic(\mathcal{O}_n)$ be the ideal class group of $\mathcal{O}_n$ (which is isomorphic to $I(n)/P(n)$, with $I(n)$ the group of ideals in $\mathcal{O}_K$ that are prime to $n$; and $P(n)$ the subgroup of principal ideals in $\mathcal{O}_K$ generated by an element congruent mod $n\mathcal{O}_K$ to some $r\in\mathbb{Z}$, $(r,n)=1$). Class-field theory provides us with an abelian extension $K_n$ of $K$ such that $\mathrm{Gal}(K_n/K)\simeq Pic(\mathcal{O}_n)$.

Let now $n$ be an integer prime to $ND$. Setting $\mathcal{N}_n:=\mathcal{N}\cap\mathcal{O}_n$, we get that $\mathcal{O}_n/\mathcal{N}_n\simeq \mathcal{O}_K/\mathcal{N}\simeq \mathbb{Z}/N\mathbb{Z}$. One can thus define a point $x_n\in X_0(N)$ by setting $$x_n=[\mathbb{C}/\mathcal{O}_n\rightarrow \mathbb{C}/\mathcal{N}_n^{-1}]$$ (here $[E\rightarrow E']$ denotes the isomorphism class of the pair $(E,E')$ of elliptic curves, with $E\rightarrow E'$ a cyclic $N$-isogeny).

If $l$ is a rational prime not dividing $ND$ nor $n$, we can mimic the previous discussion and set $$x_{nl}=[\mathbb{C}/\mathcal{O}_{nl}\rightarrow \mathbb{C}/\mathcal{N}_{nl}^{-1}]$$ The theory of complex multiplication ensures that $x_n\in X_0(N)(K_n)$ (resp. $x_{nl}\in X_0(N)(K_{nl})$ )

Here comes my question: assuming $l$ is inert in $K/\mathbb{Q}$, why do we have $$\mathrm{Tr}_{K_{nl}/K_n}x_{nl}:=\sum_{\sigma\in\mathrm{Gal}(K_{nl}/K_n)}\sigma x_{nl} = T_l x_n$$ as divisors on $X_0(N)$ ?

What I "understood" is that:

  • For $\sigma\in \mathrm{Gal}(K_{nl}/K)$, the action of $\sigma$ on $x_{nl}$ is given by $\sigma x_{nl}=[\mathbb{C}/\mathfrak{a}_{\sigma}^{-1}\rightarrow \mathbb{C}/\mathfrak{a}_{\sigma}^{-1}\mathcal{N}_{nl}^{-1}]$
    where $\mathfrak{a}_{\sigma}$ is any (proper) ideal of $\mathcal{O}_{nl}$ such that $[\mathfrak{a}_{\sigma}]\in Pic(\mathcal{O}_{nl})$ corresponds to $\sigma$ via the isomorphism $\mathrm{Gal}(K_n/K)\simeq Pic(\mathcal{O}_n)$. Thus the sum $\mathrm{Tr}_{K_{nl}/K_n}x_{nl}$ rewrites as

$$\sum_{\sigma\in\mathrm{Gal}(K_{nl}/K_n)}[\mathbb{C}/\mathfrak{a}_{\sigma}^{-1}\rightarrow \mathbb{C}/\mathfrak{a}_{\sigma}^{-1}\mathcal{N}_{nl}^{-1}]$$ Here the condition $\sigma\in\mathrm{Gal}(K_{nl}/K_n)$ implies, I think, that $\mathfrak{a}_{\sigma}\mathcal{O}_n$ is a principal $\mathcal{O}_n$-ideal.

Here $\mathrm{Gal}(K_{nl}/K_n)\simeq \mathbb{F}_{l^2}^{\times}/\mathbb{F}_{l}^{\times}$, so the sum has $l+1$ terms.

  • The action of the Hecke operator $T_l$ ($l$ not dividing $N$) on divisors of the modular curve $X_0(N)$ can be described (at least in characteristic $0$) as $$T_l [E\xrightarrow{\phi} E']=\sum_{C\subset E[l], \#C=l} [E/C\rightarrow E'/\phi(C)]$$ (there are $l+1$ such $C$) In my situation, this can be rewritten (following Gross, Heegner points on $X_0(N)$, §6) as $$T_l x_n=\sum_{\mathfrak{b}\subset\mathcal{O}_n\text{lattice of index } l}[\mathbb{C}/\mathfrak{b}\rightarrow \mathbb{C}/\mathfrak{b}(\mathcal{N}_n\cap End(\mathfrak{b}))^{-1}] $$

What I don't understand is why the exactly the same terms should appear in both sums. I get that $(id)x_{nl}=x_{nl}$ appears in $T_l x_n$, as $\mathcal{O}_{nl}$ is a sub-lattice of order $l$ in $\mathcal{O}_n$ with $End(\mathcal{O}_{nl})=\mathcal{O}_{nl}$, but I don't see why $\mathfrak{a}_{\sigma}^{-1}$ is a sublattice of order $l$ in $\mathcal{O}_n$ if $\mathfrak{a}_{\sigma}$ fixes $K_n$. I think this involves properties about fractional ideals of orders which I don't quite understand.

I thank everyone taking the time to read this question and trying to provide me with any help me !

Can monoids of "continuous words" be realized as initial monoid objects?

Math Overflow Recent Questions - Mon, 10/09/2017 - 08:52

Whenever $X$ is a set, write $X^*$ for the monoid freely generated by $X$. The elements of $X$ are, of course, words in the letters $X$. When $X$ is finite, there also seems to be a great many sensible notions of continuous words in $X$. For instance, we could define that a continuous word in $X$ is a piecewise-linear curve $\gamma : [0,T] \rightarrow [0,\infty)^X$ satisfying $\gamma(0)=0$, such that each $\gamma_x : [0,T] \rightarrow [0,\infty)$ has non-negative derivative and such that $$\sum_{x \in X} (\gamma_x)'(t) = 1$$ for all $t \in (0,T)$. And this could be weakened considerably by replacing piecewise linear with piecewise differentiable, for example. In both cases, we get a monoid by concatenation of paths, and in both cases, $X^*$ can be embedded rather naturally in this monoid.

What I'd like to know is whether these kinds of monoids can be defined, in a non-artificial way, as initial objects in the category of monoids of an appropriately chosen monoidal category.

$R/I\otimes$ pure exact sequence

Math Overflow Recent Questions - Mon, 10/09/2017 - 04:27

A right $R$-module short exact sequence $\xi:0\rightarrow A \rightarrow B \rightarrow C\rightarrow 0$ is called pure if $\xi \otimes M$ is also a short exact sequence for arbitrary left $R$-module $M$.

Question: if $\xi \otimes R/I$ is exact for arbitrary left ideal $I$, is $\xi$ pure exact sequence?

I found this question here, but the question hasn't been solved.

Maybe this need the property: $\xi $ is pure exact if and only if $\mathrm{Hom}_{\mathbb Z}(\xi,\mathbb {Q/Z})$ is split. I don't know how to do. Thank you in advance!

Teichmuller groupoids in Grothendieck's esquisse d'un programme

Math Overflow Recent Questions - Mon, 10/09/2017 - 03:54

Grothendieck in his Esquisse d'un programme mentioned without any precise definition and construction that the absolute Galois group $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$ acts on the whole "tower" of the Teichmuller groupoids $\hat T_{g,\nu}$ (see p.5 in the above reference).

I do understand what does that mean in a very special case $g=0,\nu=4$.

What does it mean in general? What are these Teichmuller groupoids? Is there a more detailed exposition of the above general remark of Grothendieck?

ADD: If I understand correctly, $\hat T_{g,\nu}$ is very close to be a profinite completion of the fundamental group of the moduli space $M_{g,\nu}$ of smooth Riemann surfaces of genus $g$ with $\nu$ marked points (am I wrong?). Grothendieck claims that there are various natural morphisms between $\hat T_{g,\nu}$'s. Frankly I do not see any morphisms except $\hat T_{g,\nu}\to \hat T_{g,\mu}$ for $\mu<\nu$ induced by the map $M_{g,\nu}\to M_{g,\mu}$ which is just forgetting several marked points. Are there any other morphisms?

History of the abstract method in mathematics

Math Overflow Recent Questions - Sun, 10/08/2017 - 19:13

Recently I have "finished" a 13-year on and off research on the history of the mathematical notion of equivalence. At the end of which, I learned that we owe the nowadays rather elementary process of "Definition by abstraction" (taking equivalence classes as new objects") not to a person, a mathematical theory, a domain of study, but to the acceptance of what Timothy Gowers (2002, p, 18) calls “the abstract method in mathematics”, that " a mathematical object is what it does.” The following comment on the definition of cardinal numbers, written by Hausdorff about a century ago, shows the influence of this attitude on our understanding of equivalence.

This formal explanation says what the cardinal numbers are supposed to do, not what they are. More precise definitions have been attempted but they are unsatisfactory and unnecessary. Relations between cardinal number are merely a more convenient way of expressing relations between sets; we must leave the determination of the “essence” of the cardinal number to philosophy. (Hausdorff 1914, pp. 28-29)

I am not sure "the abstract method" to be a common name for this attitude/philosophy. To be honest, I've believed in it since my undergraduate days without having a name for it. Has it ever had any name? What is its story? Why and how did it start and spread? Who were the main advocates? The main antagonists? Asking a single MO-style question: what is the history or a history of the abstract method in mathematics? I would be more than happy to have some references for study.

Existence of a 1-parametric family of solutions to an ODE using fixed point theorems [on hold]

Math Overflow Recent Questions - Sun, 10/08/2017 - 06:46

This is an attempt to simplify my previous question. The Weierstraß equation $$y''=6y^2-\frac{1}{2} g_2 \tag{1} $$ with $g_2>0$ has two equilibria, at $y_{\text{eq}} = \pm \sqrt{g_2/12}$. In order to avoid radicals in what follows we substitute $$g_2=\frac{\alpha^4}{12},\;\alpha=(12g_2)^{1/4}>0$$

The ODE is thus $$y''=6y^2-\alpha^4/24, \tag{2}$$ and the equilibria are $y_{\text{eq}}= \pm \alpha^2/12$. For the purposes of this question, we will only consider the positive equilibrium. One can see that near the positive equilibrium, the ODE (2) has solutions of the form $$y=-\frac{\alpha^2}{6}+\frac{\alpha^2}{4} f^2 \left( \frac{\alpha}{2} \left(x-x_0 \right) \right) $$ where $x_0$ is an arbitrary constant, and $f \in \{\tanh,\coth\}$. In fact, the solutions have the following asymptotic behavior as $x \to +\infty$

$$\begin{align} -\frac{\alpha^2}{6}+\frac{\alpha^2}{4} \tanh^2 \left( \frac{\alpha}{2} \left(x-x_0 \right) \right)& \sim \frac{\alpha^2}{12}-\alpha^2 \mathrm{e}^{\alpha x_0} \mathrm{e}^{-\alpha x}+\mathcal{O} \left(\mathrm{e}^{-2\alpha(x-x_0)} \right), \\ -\frac{\alpha^2}{6}+\frac{\alpha^2}{4} \coth^2 \left( \frac{\alpha}{2} \left(x-x_0 \right) \right)& \sim \frac{\alpha^2}{12}+\alpha^2 \mathrm{e}^{\alpha x_0} \mathrm{e}^{-\alpha x}+\mathcal{O} \left(\mathrm{e}^{-2\alpha(x-x_0)} \right). \end{align} $$ We therefore have:

The ODE (2) has solutions whose deviations from the equilibrium $\alpha^2/12$ as $x \to +\infty$ behave like $k \mathrm{e}^{-\alpha x}$ for any real $k$ ($k=0$ corresponds to the equilibrium itself).

I'm interested in proving the aforementioned boxed statement, without solving the ODE (2). In order to do this, first use the translated variable $u=y-\alpha^2/12$ to get the DE $$u''=\alpha^2 u+6u^2 \tag{3} .$$ The relevant formulation of (3) as an integral equation is then $$u_k(x)=k \mathrm{e}^{-\alpha x}-\frac{1}{\alpha} \int_x^{+\infty} 6u_k^2(t) \sinh \left( \alpha (x-t) \right) \mathrm{d}t \tag{4}.$$ In order to show that (4) is solvable, I've considered the nonlinear integral operators $$\Phi_k:u(x) \mapsto k \mathrm{e}^{-\alpha x}-\frac{1}{\alpha} \int_x^{+\infty} 6u^2(t) \sinh \left( \alpha (x-t) \right) \mathrm{d}t $$ for $k \in \mathbb{R}$. I want to show that one can choose domains $X_k$ of $\Phi_k$ in such a way that the Banach fixed-point theorem may be applied. "Starting from the end", I looked at the following difference $$|\Phi_k[v_1](x)-\Phi_k[v_2](x)| \leq \frac{6}{\alpha} \int_x^{+\infty} |v_1(t)-v_2(t)| w(t) |v_1(t)+v_2(t)| \frac{|\sinh \left( \alpha(x-t) \right)|}{w(t)} \mathrm{d}t,$$ where $w(t)$ is some nonnegative weight function. Using Hölder's inequality I got $$|\Delta \Phi_k(x)| \leq \frac{6}{\alpha} \|u_1(t)-u_2(t) \|_{\mathcal{L}^p \left([x,+\infty),w^p(t) \mathrm{d}t \right)} \left\| \left( v_1(t)+v_2(t)\right)\sinh \left( \alpha(x-t) \right) \right\|_{\mathcal{L}^q \left([x,+\infty),\frac{\mathrm{d}t}{w^q(t)} \right)}. \tag{5}$$

I'm lost here: I don't know which $p,q$ nor $w$ to choose. In addition I need to choose metrics $d_k$ for the spaces $X_k$ (which I also haven't defined yet). I believe that one of the properties defining $X_k$ should be that its elements are continuously defined over an interval of the form $[x_0(k),+\infty)$ for some $x_0(k)$. Also, I suspect that $d_k$ could be taken as weighted $p$-norms on $[x_0(k),+\infty)$.

Any help with proving the integral equations have solutions would be greatly appreciated, such as advice on which metric spaces to choose. Moreover, tips on proving that the family $u_k$ depends continuously over $k$ in some sense are also welcome.

Generate non-negative linear combinations of non-negative vectors with different supports

Math Overflow Recent Questions - Sun, 10/08/2017 - 04:01

(I will not be surprised if this problem has been solved and/or has a trivial solution – I just do not know the right terminology to google for it.)

So the problem is as follows. I have an $m \times n$ matrix consisting of zeroes and ones. I treat the rows $\mathbf r_1, \mathbf r_2, \dotsc, \mathbf r_m$ of the matrix as vectors in $\mathbb R^n$ (or, to be more precise, in $\mathbb R_{\geq 0}^n$) and want to generate linear combinations that give vectors with non-negative entries:

$$ \alpha_1 \mathbf r_1 + \alpha_2 \mathbf r_2 + \dotsb \alpha_m \mathbf r_m = \mathbf h = (h_1, h_2, \dotsc, h_n) \in \mathbb R_{\geq 0}^n \,, $$ where the coefficients of the combination can be bothe positive, zeroes, and negative: $\alpha_1, \alpha_2, \dotsc, \alpha_m \in \mathbb R$.

In other words, I want to find many vectors from $\operatorname{span} \{\mathbf r_1, \mathbf r_2, \dotsc, \mathbf r_m\}$ that have non-negative entries. It goes without saying, that number of such vectors is infinite. But I am paricularly interested in generating vectors with different supports — this requires that some of $\alpha_1, \alpha_2, \dotsc, \alpha_m$ are sometimes negative. The routine can be (or should be?) randomised, and for my real task $n$ is usually around 100–500.

P.S. The support of the vector is a set of non-zero positions: $$ \operatorname{supp} (\mathbf h) = \{ i : h_i \neq 0 \} \,. $$

On fixed point probability in discrete logarithm?

Math Overflow Recent Questions - Sun, 10/08/2017 - 01:07

Given integer $n>2$ what is the probability that for a given $h\in\Bbb Z_n$ there is no $x\in[0,\varphi(n)-1]\cap\Bbb Z$ such that $h^{x\bmod\varphi(n)}\equiv x\bmod n$?

What is wrong with my argument about passing the limit inside the Laplacian? [on hold]

Math Overflow Recent Questions - Sat, 10/07/2017 - 20:43

Let $f$ be a locally integrable function on a bounded domain $D\subset \mathbb{R}^{n}$. Let $(\theta_{m})_{m}$ be a sequence of $C^{\infty}$ functions (without compact support). I will prove the following absurd result that I will call (1): \begin{equation} \lim_{j\rightarrow\infty}\int_{B}f(x)\Delta\theta_{j}(x)dx=\int_{B}f(x)\Delta\left(\lim_{j\rightarrow\infty} \theta_{j}(x)\right) dx. \end{equation} ($B$ is a relatively compact ball in $D$) I take $f_{\epsilon}$, the smooth regularization of $ f $ obtained by the convolution product of $f$ with $ \phi_{\epsilon}:=\epsilon^{-n}\phi(x/n) $, where $\phi$ is a nonnegative test function with support in $\lbrace |x|\leq 1\rbrace $ and $ \int\phi(t)dt=1. $ I also take an increasing sequence of compacts $(K_{k})$ such that $$\cup_{k=1}^{\infty}K_{k}=B.$$ For each $k$, I choose a test function $\rho_{k}$ that is =1 on a neighborhood of $K_{k}$ and its compact support is in the interior of $K_{j+1}$. Then for all $k$, $\rho_{k} f_{\epsilon}$ has compact support in $B$ and I can write \begin{align} \int_{B}\rho_{k}(x)f_{\epsilon}(x)\Delta\theta_{j}(x)dx&=\int_{B}\Delta\left( \rho_{k}(x)f_{\epsilon}(x)\right) \theta_{j}(x)dx. \end{align} Then I let $j\rightarrow\infty$ and slip the laplacian over $\theta$ in the right-side integral again. I obtain (1) with $ \rho_{k}(x)f_{\epsilon} $ instead of $f$. Finally I let successively $k\rightarrow\infty$ and $\epsilon\rightarrow0$ and I obtain (1)!!??

Could someone please tell me what's wrong with my argument? Thanks.

Large Cardinals about critical points of $j:V\rightarrow M$

Math Overflow Recent Questions - Sat, 10/07/2017 - 01:14

First Question: What is the largest transitive inner model $M$ (up to $\subset V$) known such that ZFC is consistent with $\mathrm{cp}(j)$ existing for some $j:V\rightarrow M$?

For example, huge cardinals (and all of their variants) make $M$ large by showing that every sequence of $M$ of length $j(\mathrm{cp}(j))$ is a subset of $M$ itself, meaning that it contains most of its own sequences.

However, one could possibly create an even larger $M$ by constructing similar models.

Generally, when one has some critical point of a $j:V\rightarrow M$ where $M$ is a large model, one could construct a larger cardinal axiom by suggesting an $M_\alpha$ with $\{x\in M:x\subset M_0\}\subset M_0$, $\{x\in M_\alpha: x\subset M_{\alpha+1}\}\subset M_{\alpha+1}$ and $\{x\in\bigcup_{\beta>\alpha}M_\beta:x\in M_\alpha\}\subset M_\alpha$ for limits $\alpha$. For any $M_\alpha$, the critical point of $j:V\rightarrow M_\alpha$ should be at least as large $j:V\rightarrow M$. One could use this tactic on any large cardinal axiom defined with critical points; like huge, strong, supercompact, and other cardinals.

Second Question: Is this method for creating large cardinal axioms inconsistent with ZFC?

Feynman-Kac and the classical heat equation

Math Overflow Recent Questions - Sat, 10/07/2017 - 00:40

Consider that for bounded functions $f(x)$ and $q(x)$, the Feynman-Kac formula gives us a representation for the bounded solutions of the initial value problem $$\begin{cases} u_t(t,x) = \Delta u(t,x) + q(x)u(x,t), & x \in \mathbb{R}, t >0 ,\\ u(t,x) = f(x), & x \in \mathbb{R}, t =0. \end{cases}$$

The formula is given by $$u(t,x) = \mathbb{E} \left( f(x+B_t) \exp \left( \int_0^t q(x+B_s) ds \right) \right).$$

If we take $q(x) =0$, then we get the heat equation.

The Feynman-Kac formula does not seem to revert back to the typical solution of the heat equation however. Can anyone illuminate the discrepancy?

Minimal non-klt center of asymptotic linear system

Math Overflow Recent Questions - Fri, 10/06/2017 - 23:41

Let $(X,\Delta)$ be a klt pair and $D $ a $Q $-Cartier divisor on $X $ such that the ring of sections of $D $ is finitely generated. Let $c$ be the log canonical threshold of the asymptotic linear system $||D||$, and denote by $V=V (\mathcal{J}(c \cdot ||D||))$, consider $E $ a general $Q $-divisor on $||D||$ and $c'$ its lc threshold with respect to $(X,\Delta) $. Is $V $ a minimal lc center of $(X,\Delta+c'E) $?

The distribution of Prime numbers [on hold]

Math Overflow Recent Questions - Fri, 10/06/2017 - 22:23

I have developed a Function P(x,t) for approximation of functions, inverse functions Pi() distribution of Prime numbers. Thus 1. t lies in the interval [0.95; 1]

  1. Lim P(x,t) = Pi^-1 (x) t->1enter image description here

Truncations of a series for the inverse tangent

Math Overflow Recent Questions - Fri, 10/06/2017 - 22:13

I am looking for a rapidly convergent series of the inverse tangent function. It is an interesting paper by Sofo and Villacorta (see the article here: https://www.researchgate.net/publication/273126245_NEW_IDENTITIES_FOR_THE_ARCTAN_FUNCTION) where the inverse tangent is given by this series $$\small\tan^{-1} z=z\sum\limits_{i=0}^{m-1}{2{{\left( -1 \right)}^{i}}{{z}^{2i}}}\sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{nm}}{{z}^{2n\left( m+1 \right)}}{{4}^{n}}\left( \begin{aligned} & 2nm+2i \\ & nm+i \\ \end{aligned} \right)}{\left( n+1 \right){{\left( 1+{{z}^{2}} \right)}^{n}}\left( \begin{aligned} & 2\left( nm+i+n+1 \right) \\ & nm+i+n+1 \\ \end{aligned} \right)\left( \begin{aligned} & nm+i+n+1 \\ & nm+1 \\ \end{aligned} \right)}}$$ When I truncated this equation (with respect to index $n$) I observed that accuracy is dependant on $m$. Suppose I take $N$ terms in truncation. How to choose the integer $m$ to get the most accurate approximation in truncation at given $N$?

About prime numbers and the sums of three distinct cubes

Math Overflow Recent Questions - Fri, 10/06/2017 - 22:00

I am not a professional mathematician, but I believe this question is not something unprofessional (for example, like solving for $x$ in the following: $\sqrt{x} + \sqrt{15 + x} = 15$) if you know what I mean. The reason being is because I cannot find solutions to the following equation through a not so exhausting algebraic method (for example, trial and error): $$a^3 + b^3 + c^3 = \left\{d^3 : a, b, c, d \in \mathbf{P}\right\}$$ $\mathbf{P}$ denotes the set of prime numbers, and I can only try to find solutions through some computer power using Ramanujan's Formula for certain integer solutions to the equation. This is where I found only one solution for which $0 < a < b < c \leqslant 2000$. $$193^3 + 461^3 + 631^3 = 709^3$$ However I grew curious to see whether there are other solutions. I have found many other near-solutions where $a, b, c, d$ are all prime except for one of the variables. Let the non-prime equal $k$ where $k = a, b, c$ or $d$, then it seems like $k$ is always the product of two primes, thus $k$ is semi-prime. And in my search, I have only come across one solution where $k$ is the product of two neighbouring primes. $$109^3 + 293^3 + 437^3 = \left\{479^3 : k = 437 = 19\times 23\right\}$$ I have only found eight-near solutions so far, but I did overlook solutions where $a, b, c, d$ are odd but $5$ divides one of them, due to a speculation that I had made.

If $5$ divides $a, b, c$ or $d$, then there exists (at least) another integer amongst $a, b, c, d$ that is (also) not prime. For example, $3^3 + 4^3 + 5^3 = 6^3$.

However, funnily enough, I found a counter-example (also a near-solution) right near the end of my search just by pure coincidence. $$415^3 + 1627^3 + 1997^3 = 2311^3$$ Because of this, I also tried looking for solutions for which $a, b, c, d$ were all odd, but none of them were prime, and I found two solutions, although there could be many more. $$147^3 + 471^3 + 1403^3 = 1421^3$$ $$255^3 + 539^3 + 1477^3 = 1503^3$$ And then during this particular search, I made two more speculations.

If $3$ divides $a, b, c$ or $d$, then $3$ also divides another integer amongst $a, b, c, d$.

Every perfect number cubed can be expressed as the sum of three cubes.

I am certain that $709$ is the smallest prime to be written as the sum of three cubes $a^3 + b^3 + c^3$ for which $a, b, c$ are prime, but that is assuming there exists other solutions. So my question is, does there exist other solutions? Is there an example of a near-solution for which $k$ is not semi-prime? Does there exist a counter-example to my two most recent speculations?

Thank you in advance.

Edit: I found another solution to $a^3 + b^3 + c^3 = d^3$ for which $a, b, c, d$ are prime. $$103^3 + 2179^3 + 2213^3 = 2767^3$$ so there you go, and there exists a near-solution where $k$ is not semi-prime. Here, $k = 605 = 5\times 11^2$. $$277^2 + 605^2 + 2111^3 = 2129^3$$ I also discovered other interesting properties with perfect numbers. $$3^3 + 4^3 + 5^3 = 6^3$$ $$18^3 + 19^3 + 21^3 = 28^3$$ $$57^3 + 82^3 + 495^3 = 496^3$$ $$2979^3 + 4005^3 + 7642^3 = 8128^3$$ $3$ divides all values of $a$, and it appears that: $$\frac{a + b + c}{2} = \left\{x^3 + y^3 + z^3 : x, y, z \in \mathbb{N}\right\}$$ Let $a = 3, b = 4, c = 5$ then $x = 3, y = 4, z = 5$.

Let $a = 18, b = 19, c = 21$ then $x = 11, y = 15, z = 27$.

Let $a = 57, b = 82, c = 495$ then $x = 15, y = 213, z = 281$ or $x = 162, y = 173, z = 282$.

And there must exist at least one prime amongst $x, y, z$ in each equation, but this raises another question; can $x, y, z$ all be prime?

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