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most recent 30 from mathoverflow.net 2018-10-17T02:59:20Z

Relation between coefficients of expansions

Sat, 07/14/2018 - 16:08

Related to Relations between coefficients of expansions of a rational function at 0 and infinity

I commented at the linked question that the question seemed less about what happened "at infinity", and more about what happened "away from zero". And to some extent, the answer confirmed it by discussing the rank of (the biinfinite matrix corresponding to) the biinfinite sequence - which seems to be the degree of the rational function "away from zero and infinity". This is related to the functions $t$ and $t^{-1}$ on $\mathbb{P}^1$; they are the unique functions with a single zero and single pole at $0$ and $\infty$, or vice versa.

So that brings up a corresponding question:

Are there similar algebraic relations that could be made between the expressions of a rational function $f(t)$ when it is expressed as $A(t), B(t) \in F((t))$ such that $f(t) = A(t), f(t - c) = B(t)$ for some constant $c$?

This isn't quite a generalization of the original question, but gets into another interesting situation - when the zeroes don't match, but the poles do. More generally, we can separate the zeroes as well, leading to:

Are there similar algebraic relations that could be made between $A(t), B(t) \in F((t))$ such that $f(t) = A(t), B(t) = f(\frac{at + b}{ct + d})$?

As in the above question, there is no expectation of a finite algebraic relation. Instead, I'm hoping for a series of "loosening" conditions (similar to the conditions that the biinfinite matrix be rank $n$ for some $n \in \mathbb{Z})$, though I would expect that more than $1$ parameter would be necessary.

Description of (completely) bounded operator

Sat, 07/14/2018 - 16:04

I am somewhat a beginner in the field of operator algebras and was wondering about the following:

Let $T$ be a linear map between the space of bounded operators $B(H)$ on some Hilbert space and $S$ a map between the space of trace-class operators that we denote by $N(H)$ in the sequel.

Then, one defines maps $T_n: M_n(B(H)) \rightarrow M_n(B(H))$ by $$T_n((a_{ij})):=(T(a_{ij}))$$

Then, $T$ is completely bounded if and only if $\sup_n \left\lVert T_n \right\rVert< \infty.$

And analogously for $S.$

I would like to know whether a map $T$ is completely bounded if and only if $$\left\lVert T \otimes \operatorname{id}_{X} \right\rVert< \infty$$ or in case of $S$ whether complete boundedness is equivalent to $$\left\lVert S \otimes \operatorname{id}_{X} \right\rVert< \infty$$

for some $X$?

For $T$ I think I found a reference on math.stackexchange saying that $X=B(K)$ works for any infinite-dimensional Hilbert space $K$ (nothing on whether $K$ needs to be separable discussed in this thread.) This should imply that $X=N(H)$ works for $S$ as well.

Assuming this to be true, I am wondering about two things now:

1.)Assume I can take $X=B(H)$ for $T$ and $X=N(H)$ for $S$, then $T \otimes \operatorname{id}_{B(H)}$ is a map from $B(H) \otimes B(H)$ into itself. Can the space $B(H) \otimes B(H)$ be identified with $B(H \otimes H)$?- I assume that by a duality argument this would be equivalent to asking whether $N(H) \otimes N(H)$ is isomorphic to $N(H\otimes H)$.

2.) Are there any other choices of $X$ allowed in the above two examples?

Is this partition problem strongly NP-complete?

Sat, 07/14/2018 - 15:46

Some computational problems have variants that appear to be harder. For instance, Graph Automorphism (GA) problem has quasi-polynomial time algorithm ( by Babai's Graph Isomorphism result) while the fixed-point free GA problem is NP-complete.

Partition problem is weakly NP-complete problem since it has pseudo-polynomial time algorithm. I am interested in variants that are strongly NP-complete.

Here is a variant of partition problem:

Restricted partition problem

Input: Set $S$ of $2N$ integers, and a collection of pairs $P$ from $S$

Query: Is there a partition of $S$ into two equal cardinality parts $A$ and $S-A$ such that both parts have the same sum and no pair in $P$ has both elements in one side of the partition?

Is this variant of partition problem NP-complete in the strong sense?

Indecomposable ordinals and pseudointersection

Sat, 07/14/2018 - 14:38

Is the following claim correct (Chapter 13 before Theorem 87 of Todorcevic's book: Notes on forcing axioms): Let $\alpha$ be an infinite countable indecomposable ordinal and $U$ be an uniform ultrafilter on $\alpha$ (namely elements in $U$ have order type $\alpha$). Then for any collection $\{B_i\in U: i<\mathfrak{m}\}$ it is true that there exists $B\subset \alpha$ of order type $\alpha$ such that $B-B_i$ is finite for all $i<\mathfrak{m}$. Here $\mathfrak{m}$ is the least cardinal such that Martin's Axiom holds below $\mathfrak{m}$ (i.e. meeting any $\beta$ many dense sets for any $\beta<\mathfrak{m}$).

In fact, it will be interesting to know if this is true at all for countable collection $\{B_i: i<\omega\}$.

It is proved there that the statement is true if we relax the conclusion asking only $B-B_i$ to be bounded in $B$. Any thoughts?

Factoring x15−1 into irreducible polynomials over GF(3) [on hold]

Sat, 07/14/2018 - 14:19

Factoring x^15−1 into irreducible polynomials over GF(3) How can i find this? Thanks in advance.

Determinant as a Hamiltonian

Sat, 07/14/2018 - 13:40

Are there two symplectic structures $\omega_1, \omega_2$ on $M_{2n}(\mathbb{R})$ such that the function $Det:M_{2n}(\mathbb{R})\to \mathbb{R}$ is completely integrable with respect to $\omega_{1}$ but is not completely integrable with respect to $\omega_2$

Note that we do not limit the symplectic structures to structures with constant coefficients.

I had already asked a weaker version of this question in the following link but I confess that I did not understand the details(How does the action of orthonormal group guarantee's the integrability?)

http://mathforum.org/kb/thread.jspa?forumID=253&threadID=1653483&messageID=5990478#5990478

(In)stability of a two-dimensional dynamical system

Sat, 07/14/2018 - 13:04

Consider the following system of coupled differential equations $$ \dot{x}_1(t) = -x_1(t) - \cos(\omega t)x_1(t) + \cos(\omega t)x_2(t), \ x_1(0)\in\mathbb{R},\\ \dot{x}_2(t) = -\gamma x_2(t) - \cos(\omega t)x_2(t) + \cos(\omega t)x_1(t), \ x_2(0)\in\mathbb{R}, $$ where $\omega$ and $\gamma$ are positive real constants. Observe that $\bar{x}=(\bar{x}_1,\bar{x}_2)=(0,0)$ is an equilibrium of the above system.

It is almost trivial to see that if $\gamma=1$ then $\bar x$ is attractive. Indeed, in this case, we have that $x(t)=[x_1(t), x_2(t)]^\top$ can be explicitly computed as $$ x(t) = \exp\left(\begin{bmatrix}-t &0\\ 0 & -t\end{bmatrix} + \frac{1}{\omega}\sin(\omega t)\begin{bmatrix}-1 &1\\ 1 & -1\end{bmatrix}\right)x(0), $$ so that $x(t)\to 0$ for $t\to \infty$.

However in case $\gamma\ne 1$ proving the attractiveness of the origin is not obvious (and perhaps not even true!).

In particular, numerical simulations seem to suggest that for $\gamma$ and $\omega$ sufficiently small (e.g. $\gamma=0.001$ and $\omega=10$) the equilibrium $\bar{x}$ is not attractive.

I've struggled a lot to find a way of formally proving this, with no luck. So I decided to post the problem here hoping that some of you will provide some useful suggestions or tips. Thank you!

I post here the Mathematica code that I've used in my simulations:

(* nominal values for simulation *) values = {gamma -> 0.001, w -> 10}; equations = { {x1'[t], x2'[t]} == {-x1[t] - Cos[w*t]*x1[t] + Cos[w*t]*x2[t], -gamma*x2[t] - Cos[w*t]*x2[t] + Cos[w*t]*x1[t]}, {x1[0], x2[0]} == {0.1, 0.1}}; {x1t, x2t} = NDSolveValue[equations /. values, {x1[t], x2[t]}, {t, 0, 1000}]; Plot[x1t, {t, 0, 1000}, PlotRange -> {-0.2, 0.2}] Plot[x2t, {t, 0, 1000}, PlotRange -> {-0.2, 0.2}]

Existence of non-null-homotopic map from $M^n$ to $S^{n-1}$

Sat, 07/14/2018 - 09:27

Let $M^n$ be compact, connected, oriented $n$-dimensional smooth manifold without boundary, the Hopf degree theorem states that the homotopy class of continuous maps from $M^n$ to $S^n$ is classified by its degree. What about continuous map from $M^n$ to $S^{n-1}$ (assume $n>0$) ? I am mostly interested in the case $n>2$.

More specially, when does there exist a non null-homotopic map in the case $n=3$? If $H^2(M,\Bbb Z) \not=0$ then it's true as $[M,\Bbb{C}P^{\infty}]=H^2(M,\Bbb Z)$ and $\Bbb CP^1 \cong S^2$. This is not necessary as $M=S^3$ shows.

What is the minimal dimension of a complex realising a group representation?

Sat, 07/14/2018 - 02:42

This question is inspired by this one, which was about representations that can be realised homologically by an action on a graph (i.e., a 1-dimensional complex).

Many interesting integral representations of groups arise from a group acting on a simplicial complex that is homotopy equivalent to a wedge of spheres, by applying homology. A classical example is the action of groups of Lie type on spherical buildings. On homology this gives an integral form of the Steinberg representation.

One may ask if there exists a complex of lower dimension than the Tits building that realises the (integral) Steinberg representation in this way. I expect that the answer is No, but how to prove it?

More generally, given an integral $G$-representation that can be realised as the homology of a spherical complex with an action of $G$, is there an effective lower bound on the dimension of such a complex? One obvious lower bound is given by the minimal length of a resolution by permutation representations. Is this something that has been studied?

When every open cover admits a $\sigma$-disjoint subcover?

Sat, 07/14/2018 - 02:41

We say that a sequence $(\mathcal X_n)$ of families of subsets of a topological space $X$ is a $\sigma$-disjoint cover of $X$ if every family $\mathcal X_n$ consists of mutually disjoint sets and $\bigcup\limits_n\bigcup\mathcal X_n=X$.

Let us say that a space $X$ is weakly Lindelof, if every open cover of $X$ admits a $\sigma$-disjoint subcover. Clearly, every Lindelof space is weakly Lindelof.

Question 1. Is there any well-known in the literature name for the class of "weakly Lindelof" spaces?

The following question concerns weaker property than "weak Lindeloffness".

Question 2. Does there exist a $\sigma$-disjoint cover of a Banach space $X$ by open balls of diameters $\le 1$?

"Flat links", a reference request

Sat, 07/14/2018 - 01:21

A hyperbolic link is one whose complement admits a hyperbolic metric. Hyperbolic links, and especially hyperbolic knots, are quite popular these days. However, I am currently interested in links whose complement admits a flat (i.e. locally euclidean) metric. If I got it right, the major difference from the hyperbolic case is the existence (at least, I want for it to exist) of a natural compactification. This compactification makes the 3-sphere into an Alexandrov space with conical singularities along the components of the link, or something like this. Is there such a construction somewhere in the literature?

[EDIT] I think I need to clarify this a bit. What I want is a metric on ${\mathbb S}^3$ such that

1) The metric is flat outside of the link.

2) Each component of the link has a neighborhood isometric to a product of a conical point by ${\mathbb S}^1$.

(So, if it is a ``negative'' conical point, then it won't be an Alexandrov space.)

Moments of area of random triangle inscribed in a circle

Fri, 07/13/2018 - 14:47

The $2m$th moment of the (random) area of the triangle whose vertices are three independent, uniformly distributed random points on the unit circle appears to be $((3m)!/(m!)^3)/16^m$. Can anyone prove this? Better yet, can anyone give a conceptual explanation for why this moment should be rational? (If this observation is not new, references would be appreciated.)

This question was inspired by John Baez's posts https://johncarlosbaez.wordpress.com/2018/07/10/random-points-on-a-sphere-part-1/ and https://johncarlosbaez.wordpress.com/2018/07/12/random-points-on-a-sphere-part-2/.

A geometric proof of Krull's Principal ideal theorem

Fri, 07/13/2018 - 12:34

Krull's height theorem states that in a Noetherian, local ring $(A,\mathfrak m)$, for any $f \in \mathfrak m$, the minimal prime ideal containing $(f)$ is at most height $1$.

This is a very geometric statement and is essentially saying that hypersurfaces can cut down the dimension by at most one. However, I have never seen a geometrically motivated proof. All the proofs I have seen essentially muck around with symbolic powers of prime ideals and this seems very ad-hoc/unmotivated to me.

Surely a geometrical statement should have a geometric proof!

Does someone have a geometric way of seeing why this theorem should be true or what is going on? Or what is going on geometrically with the standard proof and symbolic powers?

Bezout theorem for germs of holomorphic functions

Fri, 07/13/2018 - 11:05

UPDATE.

It was pointed out by @Dmitri that two smooth curves given by $f=y$ and $g=y+x^k$ in $\mathbb C^2$ provide a simple counterexample.

Let $f_1, \ldots, f_p, g_1, \ldots, g_q$ be germs of holomorphic functions at $0$ in $\mathbb C^{p+q}$. Assume that the multiplicity ${\rm mult}(f_1, \ldots, f_p)$ is finite let $$ V:= \{ f_1 = \ldots = f_p =0\} $$ be (germ of) the zero variety of $f_1, \ldots, f_p$, and denote by ${\rm mult}_V(g_1, \ldots, g_q)$ the maximum number for points in the fiber of the restriction $g|_V$, where $g:=(g_1, \ldots, g_q)$.

Is the following true?

$$ {\rm mult}(f_1, \ldots, f_p, g_1, \ldots, g_q) \le {\rm mult}(f_1, \ldots, f_p) \cdot {\rm mult}_V(g_1, \ldots, g_q) $$

For example, if $f=y$ and $g=y+x^k$, we have ${\rm mult} (f) = {\rm mult} (g) = 1$ but ${\rm mult}(f, g) = k$. On the other hand, for $V=\{f=0\}$, we have ${\rm mult}_V (g) = k$, so the above relation is correct.

OLD QUESTION (the answer is no).

Is the following local holomorphic variant of Bezout theorem true:

Let $f_1, \ldots, f_p, g_1, \ldots, g_q$ be germs of holomorphic functions at $0$ in $\mathbb C^{p+q}$. Then $$ {\rm mult}(f_1, \ldots, f_p, g_1, \ldots, g_q) = {\rm mult}(f_1, \ldots, f_p) \cdot {\rm mult}(g_1, \ldots, g_q), \quad (*) $$ provided the first multiplicity is finite.

Here for $m$ holomorphic germs $(h_1, \dots, h_m)$ in $\mathbb C^n$, $m\le n$, the multiplicity ${\rm mult}(h_1, \ldots, h_m)$ is defined as the minimum dimension of the quotient $$ \cal O/\langle h_1, \ldots, h_m, l_1, \ldots, l_{n-m} \rangle $$ of the ring $\cal O = \cal O_{\mathbb C^n,0}$ of germs at $0$ of holomorphic functions in $\mathbb C^n$ modulo the ideal generated by the germs $h_1, \ldots, h_m, l_1, \ldots, l_{n-m}$, with $l_j$ being linear functions, and the minimum is taken over all choices of the $l_j$.

Equivalently, the multiplicity can be defined as the minimum number of points (over all choices of the linear functions $l_j$) in the preimage of a generic point in $\mathbb C^n$ near $0$ under the map $$ z\mapsto (h_1(z), \ldots, h_m(z), l_1(z), \ldots, l_{n-m}(z)), $$ where the same notation is used for the germ representative functions. (Instead of taking generic point, one can take the maximal number of points in a preimage.)

Orthogonal Grassmanians: cases where $\text{OG}( \mathbb{P}^1 , Q) \not \simeq \mathbb{P}^3$

Fri, 07/13/2018 - 08:26

Let $Q = \{ q(x_0, \dots, x_4) = 0 \}$ be a quadric-threefold over a field $k$. Are there cases where the orthogonal Grassmanian $\text{OG}( \mathbb{P}^1 , Q)$ is not a copy of $\mathbb{P}^3$?

Here's a Lemma (Kollar) that says it is a $\mathbb{P}^3$ iff it contains a line:

Lemma 61

(1) $\text{OG}(\mathbb{P}^1,Q)$ is a Severi-Brauer variety

(3) $\text{OG}(\mathbb{P}^1,Q) \simeq \mathbb{P}^3 \longleftrightarrow \text{OG}(\mathbb{P}^1,Q) \text{ contains a line }\longleftrightarrow q \sim y_0y_1 + y_2y_3 + a y_4^2$

but isn't this variety the Grassmanian of lines?

Here I believe $\simeq$ means "isomorphism of varieties" (instead of birational equivalence $\stackrel{bir}{\sim}$) and the symbol $\sim$ is "rational equivalence of quadratic forms".

I am still parsing the definition based on the comments I am wondering:

  • Are there other 3-dimensional Severi-Brauer varieties over $\mathbb{Q}$ other than $\{ 0\}$ and $\mathbb{P}^3$ ?
    I believe the answer is no because quadric sections themselves are varieties of this type.

However Wikipedia says this:

In mathematics, a Severi–Brauer variety over a field K is an algebraic variety V which becomes isomorphic to a projective space over an algebraic closure of K.

In dimension one, the Severi–Brauer varieties are conics. The corresponding central simple algebras are the quaternion algebras. The algebra $(a,b)_K$ corresponds to the conic C(a,b) with equation $$ C(a,b) = \big\{ z^{2} -ax^{2}-by^{2} = 0 \big\} $$ and the algebra $(a,b)_K$ splits, that is, $(a,b)_K$ is isomorphic to a matrix algebra over $K$, if and only if $C(a,b)$ has a point defined over $K$: this is in turn equivalent to $C(a,b)$ being isomorphic to the projective line over $K$.

So we get that quaternions play a role in SB varieties and that conics (and one could imagine quadrics) are Severi-Brauer varieties. Now I am confused again, that if the conic has a single point over $K$, then it is a projective line $\mathbb{P}^1(K)$.

Does this correspond to the trivial element of the Brauer group? There are even expositions of Severi-Brauer varieties that are more geometric and avoid the use of Galois cohomology.

In Neukirch's Cohomology of Number Fields, they give a defintion of the Brauer Group:

The Brauer group of a field $K$ is the set of similarity classes $[A]$ of central $K$-simple algebras $A$ endowed with the multiplication $$[A][B] = [A \otimes_K B] $$ Two central simple $K$-algebras $A$ and $B$ are similar if $$ A \otimes_K M_r(K) \simeq B \otimes_K M_s(K) $$ for some $r,s$. (Chapter 6.3)

The group $\mathbb{Q}/\mathbb{Z}$ keeps being mentions in regards to various exact sequences and the Hasse principle. It seems the challenge is to stay grounded and not to float away with these definitions, for what is a very concrete object.

Density on Hölder spaces whose elements vanish on the boundary

Fri, 07/13/2018 - 08:13

I would like to ask the following problem.

Let $\Omega$ be a $C^{r+1,\alpha}$ domain, $r\in \mathbb{N}, 0<\alpha<1.$ We denote $$C^{r,\alpha}_{0}(\overline{\Omega})=\{f\in C^{r,\alpha}(\overline{\Omega}): f=0 \mbox{ on }\partial \Omega\},$$ here $C^{r,\alpha}(\overline{\Omega})$ is Holder spaces. Is $C^{r+1,\alpha}_0(\overline{\Omega})$ dense in $C^{r,\alpha}_0(\overline{\Omega})?$

We see that when $r\geq 2$, the answer is positive. For any $u\in C^{r,\alpha}_{0}(\overline{\Omega}),$ we have $$\Delta u := f\in C^{r-2,\alpha}_{0}(\overline{\Omega}).$$ Then there exists a sequence $f_n\in C^{r-1,\alpha}_{0}(\overline{\Omega})$ such that $f_n\rightarrow f$ in $C^{r-2,\alpha}_{0}(\overline{\Omega}).$ With each $f_n,$ there exists unique $u_n\in C^{r+1,\alpha}_{0}(\overline{\Omega})$ such that

$$\left\{\begin{array}{ll}\Delta u_n=f_n &\mbox{ in }\Omega\\ u_n =0 &\mbox{ on }\partial \Omega. \end{array}\right.$$ Therefore, by eliptic regularity, we obtain that $||u_n-u||_{C^{2,\alpha}}\leq C||f_n-f||_{C^{r-2,\alpha}}.$ It implies the conclution. It seems that we can not apply the above method for the case $r=1.$

Examples of "miraculous" proofs

Fri, 07/13/2018 - 07:24

Concerning the proof that $\zeta(3)$ is irrational, Van der Poorten famously noted that

"Apéry's incredible proof appears to be a mixture of miracles and mysteries".

Indeed, many ideas introduced in Apéry's proof such as $\zeta(3)=\frac{5}{2} \sum_{k=1}^{\infty}{\frac{ (-1)^{k-1}} {\binom {2k}{k}k^{3}}}$ and the recurrence $n^3u_n + (n-1)^3 u_{n-2} = (34n^3-51n^2+27n-5)u_{n-1}, n\geq2$ amazed contemporary mathematicians (although the fast-converging series was already derived several years earlier by Hjortnaes).

What are other examples of "miraculous" proofs, whose ingredients amazed mathematicians of their time? In particular, proofs such that:

  • a large portion of the Theorems involved in the proof represent entirely new ideas,
  • these Theorems are applicable to a wide area of mathematics,
  • the general atmosphere among contemporary mathematicians was a mixture of surprise and awe ("where did this come from?").

Integral structures via lattices

Thu, 07/12/2018 - 18:52

I am looking at the paper "p-adic Groups" by Bruhat (in the Boulder Proceedings, 1965). I have a question about one of the statements. Let $k$ be the quotient field of a complete discrete valuation ring $\mathcal{O}$. Let $V$ be a vector space over $k$. Let $L$ be a lattice in $V$. Choose a basis for $L$.

Bruhat states the following in pp. 63-64:

(1) The algebra $\mathcal{O}[GL]:=\mathcal{O}[g_{ij},(det(g_{ij}))^{-1}]$ is an $\mathcal{O}$-structure for $GL(V)$.

(2) More generally, if $G$ is a linear algebraic group over $k$, then given a faithful rational representation $\rho:G \rightarrow GL(V)$, the image of $\mathcal{O}[GL(V)]$ in $k[G]$ is an $\mathcal{O}$-structure for $G$.

(3) Any $\mathcal{O}$-structure for $G$ may be obtained in this way.

How is (3) proven?

Solutions to a system of homogeneous equations (inequalities)

Thu, 07/12/2018 - 13:10

Let $f_1,\ldots,f_r \in \mathbb{R}[x_1,\ldots,x_n]$ be $r$ homogeneous polynomials of the same odd degree $d$, where $d \in \{3,5,7,\ldots\}$.

For which values of $r,n,d$ there exists a real solution $a=(a_1,\ldots,a_n) \in \mathbb{R}^n-\{(0,\ldots,0)\}$ to the system of $r$ inequalities $f_i \geq 0$?

For $r=2$ the answer is positive, namely, there exists such $a$, by the following argument that can be found in the answer to this question: By continuity there exists $t \in \mathbb{R}^n$ with $|t| = 1$ and $f_1(t) = f_1(-t) = 0$. Clearly, at least one of $\{f_2(t),f_2(-t)\}$ is $\geq 0$. Then for $s=t$ or $s=-t$, we have: $f_1(s)=0$ and $f_2(s) \geq 0$.

Actually, I am interested in the case $n=4$; I really apologize that I am not explaining why (it is quite complicated; hopefully later I will be able to explain).

Remarks:

(1) I wonder if the following idea may yield an answer to my question: For example, $n=4,d=3$, and order the monomials: $y_1:=x_1^3,y_2:=x_1^2x_2,y_3:=x_1^2x_3,y_4:=x_1^2x_4,x_1x_2x_1,\ldots,y_m:=x_4^3$. Then the system of $r$ inequalities $f_i \geq 0$ becomes a linear system in $m$ new variables $y_1,\ldots,y_m$. Now, $0=(0,\ldots,0) \in \mathbb{R}^m$ is a solution, and if the associated matrix is invertible, then $0$ is a unique solution, but this does not imply that the original system has $0 \in \mathbb{R}^n$ as a unique solution, so there is hope to get non-trivial solutions. On the other hand, if the associated matrix is not invertible, then there are non-trivial solutions, but I am not sure if it is possible to obtain from one of them a solution to the original system, probably not necessarily.

(2) This paper seems relevant.

(3) I have already asked the above question here, but have not received any comments, neither to it nor to another similar question.

Any hints and comments are welcome!

Edit: I have now found this question which almost answers my question, provided that I understood it correctly and if we take $k=0$ in that question notations, then we get that my question, if we replace $\mathbb{R}$ by $\mathbb{C}$, has a positive answer (= a common zero for all the $f_i$'s) for $n > r$.

intersection of sets

Thu, 07/12/2018 - 03:48

Let $A_n=\{d:d|n\}\cup \{kn|D(n)\leq k \leq n-1\}$ where $D(n)$ is the number of divisors of $n$.

Is it correct that
$$|A_i \cap A_j |=(i,j),$$ for each $i,j \in N$?

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