Recent MathOverflow Questions

How to count Isomorphism Types of arbitrary structures?

Math Overflow Recent Questions - Fri, 08/04/2017 - 16:43

For all relational signatures $\sigma$ and nonnegative integers $n$, I want to count the number of isomorphism types of structures of order $n$ of the signature $\sigma$.

What I mean by structure is a pair $\mathfrak{A}=(A,( R(\mathfrak{A})_{R\in\sigma} )) $ for some relational signature $\sigma$ of predicates of arity $\geq 1$ and some finite set A. I think, I figured it out for the case when all predicates in $\sigma$ have arity $1$. Say, it is $\mathfrak{A}=(A, P_1(\mathfrak{A}),\ldots,P_m(\mathfrak{A}) ) $ for some nonnegative integer $m$. Then each vertex is coloured in certain ways: It could be that it is in $ P_1(\mathfrak{A}),P_2(\mathfrak{A}) $ but not in $P_3(\mathfrak{A})$ (for $m=3$); so we can assign to each vertex a subset of all colours it possesses. Then two structures of this signature are isomorphic if and only if for each subset of $\sigma=\{P_1,\ldots,P_m\}$, there are exactly the same number of vertices coloured this way in both structures.

But there are exactly $2^m$ subsets of the above $\sigma$, and so what we look for is the number of all $2^m$-tuples $ (a_1,\ldots,a_{2^m}) $ s.t. $\sum_{i\in [1,2^m]}a_i = n $ where n is the order of structures we want to compute the number of isomorphism types for. Well and this is a simple combinatorial fact this number is equal to $ \binom{n+2^m-1}{n} $.

I want to do the same thing without the restriction that $\sigma$ only contains predicates of arity $1$. I would be glad about ideas how to do this for structures of a signature with exactly one $r$-ary predicate for some arity $r\geq 2$, and also about ideas how to combine the numbers of isomorphism types of two such signatures (for some fixed $n$).

NBA Finals 2017 Live Stream, Schedule, Score

Math Overflow Recent Questions - Fri, 08/04/2017 - 16:34

NBA Finals 2017 Live Stream, Schedule, Score NBA Finals 2017 Live Stream, Schedule, Score NBA Finals 2017 Live Stream, Schedule, Score NBA Finals 2017 Live Stream, Schedule, Score

If a $\otimes$-idempotent object has a dual, must it be self-dual?

Math Overflow Recent Questions - Fri, 08/04/2017 - 15:27

Let $C$ be a symmetric monoidal category.

  • Recall that a dual for $X \in C$ is an object $X^\vee$ and maps $\eta: I \to X \otimes X^\vee$ and $\varepsilon: X^\vee \otimes X \to I$ (where $I$ is the monoidal unit) satisfying the triangle identities.

  • Let's say that $X$ is self-dual if there is a dual $X^\vee$ for $X$ with $X \cong X^\vee$.

  • Let's say that $X$ is idempotent if $X \cong X \otimes X$.

  • Let's say that $X$ is well-idempotent if there is a map $i: I \to X$ such that $X \otimes i: X \to X \otimes X$ is an isomorphism.

  • Dually, $X$ is co-well-idempotent if there is a map $p: X \to I$ such that $X \otimes p: X \otimes X \to X$ is an isomorphism.

Clearly, if $X$ is self-dual and idempotent, then $X$ is both well-idempotent and co-well-idempotent. Conversely, I ask

Questions:

  1. If $X$ is idempotent and dualizable, then is $X$ self-dual?

  2. If $X$ is well-idempotent and dualizable, then is $X$ self-dual?

  3. If $X$ is well-idempotent and co-well-idempotent and dualizable, then is $X$ self-dual?

I suspect the answer to all these questions is "no", but I don't know any examples.

Integration over arbitrary domains

Math Overflow Recent Questions - Fri, 08/04/2017 - 15:11

In mathematical physics, we sometimes encounter situations where we have integrals of the forms:

$$\text{(case 1):}\ \ \ \ \int\limits_{D} f(x,y,z) dx dy dz =k$$ $$\text{(case 2):}\ \ \ \ \int\limits_{D} f(x,y,z) dx dy dz =\int\limits_{D} g(x,y,z) dx dy dz$$ where $D$ is an arbitrary domain (e.g. volume) of intergation over the $(x,y,z)$ variables and $k$ is some constant.

Because of $D$'s arbitrariness (i.e. can be taken small enough so that the integrand is taken outside integral sign), we usually proceed by taking $f,g$ outside the integral and ending up with the corresponding results: $$\text{(case 1):}\ \ \ \ f=k/D$$ $$\text{(case 2):}\ \ \ \ f=g$$

Questions:

(a) Is there any formal/rigorous theorems in math (e.g. in analysis) that specialise in such situations or explain its conditions in general? What branch/subbranch of math would cover such problems?

(b) If we are faced with a problem of case (2) above, for example, and we could find some kind of domains $D$ of a certain shape but arbitrary size (e.g. say $D$ as a sphere of arbitrary radius) for which both sides (integrals) are known to be equal (and hence we reach equality of integrands), will that be enough to conclude that equality of integrals will reduce to equality of integrands in general (or do we need to first show that it is also true for all other arbitrary shapes of $D$)?

(c) Finally, could the same discussion/conclusions of case (2) above be extended to the case of integrals over closed surfaces (via Green's theorem for instance?): $\oint\limits_{A} \boldsymbol{F}\cdot \boldsymbol{dA}=\oint\limits_{A} \boldsymbol{G}\cdot \boldsymbol{dA}$?

(Note - I have asked this question on math SE, but got no answers.)

Is this algebra Gorenstein?

Math Overflow Recent Questions - Fri, 08/04/2017 - 13:28

Let $A$ be a finite dimensional algebra with the property that an indecomposable module $M$ is projective or $\Omega$-periodic iff $Ext^{i}(M,A)=0$ for all i>0. Is $A$ Gorenstein?

Galois descent for schemes over fields

Math Overflow Recent Questions - Fri, 08/04/2017 - 12:44

Let $K\subset L$ be a finite galois extension of fields (the case I have in mind is $K=\mathbb{R},L=\mathbb{C}$). Given a scheme $X$ over $K$ by pulling back to $L$ we get a scheme $Y=X\times _K L$ over $L$ with a $K$ - action of $G=\mathrm{Gal}(L/K)$. It is easy to see that this action satisfies the following: There exists an affine open cover $X=\bigcup_{i\in I}U_i $ such that any $\sigma \in G$ takes each $U_i$ to itself. Now let $Y$ be an arbitrary scheme over $L$ with a $K$ - action of $G$ that satisfies the above property (for all of this you can assume whatever finiteness and niceness conditions you want on everything, the main case I have in mind is that of finite type separated schemes over the base fields, not necessarily reduced or affine and if possible not necessarily quasi projective). My questions are the following:

1) What other necessary conditions on the action of $G$ on $Y$ (other then the one I stated) must I assume to have a chance of $Y$ coming from a scheme $X$ over $K$?

2) Are there any simple sufficient conditions for $Y$ to come from such an $X$?

3)What are important examples of schemes $Y$ over $\mathbb{C}$ with a $\mathrm{Gal}(\mathbb{C}/\mathbb{R})$ action not coming from a scheme over $\mathbb{R}$ I should keep in mind? If possible, I want all actions to be anti-holomorphic i.e. inducing anti-holomorphic homomorphisms on the stalks of the analytification. Interesting examples for analytic spaces which don't come from schemes will also be very nice.

Identifying poisoned wines, with a twist

Math Overflow Recent Questions - Fri, 08/04/2017 - 12:26

(This is a joint musing with Andrew Gordon and Wyatt Mackey)

There is a classic, elementary riddle, discussed before on MO and math.SE: suppose you have 1000 bottles of wine, and one is poisoned. The poison is slow-acting, and will not cause any negative effects until 24 hours after consumed. How many taste-testers (for the sake of humanity, let's say they are rats) do you need to determine which bottle is poisoned by the end of the 24 hours? There is a natural generalization to this problem, which is discussed here: what if there are two poisoned bottles? What about three?

The accepted answer of Sergey Norin cites a paper of Frankl and Furedi that develops theory to give a constructible solution of 43 taste-testers for the case of 1000 bottles, 2 of which are poisoned.

Apparently, there are also some sources out there that give the exact lower bound at 19 testers.

Our version of the problem extends the original one in a different direction: suppose there is one poisoned bottle and one bottle with antidote, so that any taste-tester who consumes wine from both the poison and antidote bottles will show no adverse effects. Then what is the fewest number of testers needed to locate the poisoned bottle?

This question as well has two further related questions:

  1. Assume that the antidote, when consumed without the poison, is also poisonous (this is not so ridiculous a situation: according to this article, the chemical atropine, while poisonous when consumed on its own, counteracts anthrax)

  2. Locate the antidote in addition to the poison.

We have had quite a bit of fun coming up with various solutions to this problem, but have been unable to determine an optimal one. For completeness and (perhaps) amusement, I'll present two of ours here:

First, we can find the poisoned bottle in a cellar of $n$ wines with $3\lceil\sqrt{n}\rceil - 1$ testers. The $-1$ is because we can leave out one bottle from the testing and still find all the information we need: if that bottle is poisoned, no one dies. If it isn't poisoned, we will have already found the poisoned one. We'll divvy up the tasting responsibilities as follows: arrange the bottles in a square (hence the $\lceil \sqrt{n}\rceil$); assign one taste-tester to all the bottles in each row, one to all the bottles in each column, and one to all the bottles along a diagonal, where the diagonals "wrap around":

Another construction is significantly less efficient and pointless, but sort of fun. If we consider testers as points in a finite projective space $\mathbb{P}_q^n : = \mathbb{P}\mathbb{F}_q^n$ for $q = p^e$ a prime power, and bottles of wine as lines in such a projective space, then a tester will drink from a bottle if and only if the corresponding point lies on the corresponding line. The problem can thus be transformed to the following, which would give an upper bound:

Question 1 Fix an integer $n$. For what prime power $q$ and what dimension $m$ does $\mathbb{P}_q^m$ contain at least $n$ lines and have the minimum number of points?

Perhaps "the problem can be reduced to" was not a great choice of words; I mean, any solution to Question 1 would give an upper bound to our problem, because lines in finite projective spaces contain at least three points, and those data are enough to distinguish any two points. The numbers of points and lines in finite projective spaces are known, so you can write a quick script to check for a minimum within some reasonable range of prime powers.

Anyway, our ideas were clearly insufficient, because Noam Elkies knocked them out of the park (literally) overnight, the following two algorithms being due to him:

Solution from Noam #1: We can solve this with $O(\log_2(n)^2)$ testers as follows. The solution to the 1-poison, no antidote problem comes from assigning a tester to each bit when the bottles are numbered base-2; this solution comes from assigning four testers to each pair of bits. That is, each pair of bits can be 00, 01, 10, or 11, and we assign a tester to each of those possibilities. This requires $2\ell(\ell - 1)$ testers, where $\ell = \lceil \log_2(n) \rceil$.

Solution from Noam #2: This one is $O(\log_2(n))$, the best we could possibly get via information-theoretic considerations, and it is non-constructive. So that I don't risk botching the construction, I'll quote directly Noam's solution:

Choose $k$ subsets of the $n$ bottles at random. (It will make the calculation easier if they're chosen independently "with replacement" so there's a positive albeit small possibly of stupidly choosing the same set twice.) Once $k >> \log(n)$, the probability that such a random choice works should be positive, so in particular some choice must work.

A collection of subsets fails if there's some two (poison, antidote) pairs $(x,y)$ and $(x',y')$ that they can't distinguish; that is, such that for each set $S$ the events ($S$ contains $x$ but not $y$) and ($S$ contains $x'$ but not $y'$) either both happen or both don't happen. We want to show that once $k >> \log(n)$ the expected number of such pairs of pairs is less than 1, which makes the success probability positive. Well, the expected number is just the sum over $\{(x,y), (x',y')\}$ of the probability that none of our sets distinguishes between them; and that probability is the $k$-th power of the probability that one set does not distinguish them. This one-set probability depends on which if any of the four coincidences $x=x', y=y', x=y', x'=y$ occur, but that's a finite number of possibilities, and for each of them we should get some probability $c<1$. So we can just take the largest of these and multiply its $k$-th power by the number of $\{(x,y), (x',y')\}$, which is less than $n^4$. Once $k$ exceeds some multiple of $\log(n)$, the resulting expected number $c^k n^4$ of failures falls below 1 and we're done.

Further refinements would include a more precise accounting of how many times each subset of the four coincidences occurs (e.g. there are only $O(n^3)$ ways to have $x=x'$; the full $n^4$ happens only when $x,x',y,y'$ are all distinct), or choosing randomly not among all subsets but only from subsets of size $\sim f \cdot n$ for some fixed $f$ in $(0,1)$ which would then be optimized by some possibly annoying calculus exercise.

For the generalization to $p$ poison bottles we should likewise get $O_p(\log(n))$: there are $n^{2p}$ possible failures, so $k$ must be large enough to make $c^k n^{2p} < 1$, but that's still a constant multiple of $\log(n)$.

This asymptotic improvement is fantastic, but we still don't know the answer in any specific cases. Might anyone have leads?

Calculating Govt Revenue on Fuel per anum [on hold]

Math Overflow Recent Questions - Fri, 08/04/2017 - 11:16

If a country consumes 2000 L of fuel per year, per capita and has a population of 4.8 million

What is the total amount of fuel consumed in litres per year by that country?

If the fuel is sold at 1.60c per litre. What is the total fuel sales in dollars per year?

What is 47% (tax) of this?

If less than 90 billion dollars, what is the percentage of 90 billion dollars?

Does the clique-coclique bound hold for all walk-regular graphs?

Math Overflow Recent Questions - Fri, 08/04/2017 - 11:10

The clique-coclique bound is said to hold for a simple graph $G$ on $n$ vertices if $\lvert \omega(G) \rvert \lvert \alpha(G) \lvert \leq n$, letting $\omega(G)$ and $\alpha(G)$ denote its clique and coclique (independent set) numbers respectively.

It is known, in particular, that the clique-coclique bound holds for all vertex-transitive graphs and all distance-regular graphs - two families of walk-regular graphs.

The clique-coclique also appears to hold for all of the examples of walk-regular graphs that I know of that are neither vertex-transitive nor distance-regular - which includes walk-regular graphs that are not semi-symmetric either.

Could it be possible that the clique-coclique bound actually holds for all (connected) walk-regular graphs?

By informal reasoning in head, it feels plausible to me that this could be the case? I wonder what would might be a good approach to take to try to prove or disprove this?

If the clique-coclique bound does not hold for all walk-regular graphs in general, might it at least hold for e.g. all semi-symmetric graphs, in addition to the known families of walk-regular graphs that it holds for?

invariant subspace [on hold]

Math Overflow Recent Questions - Fri, 08/04/2017 - 10:46

Let $\pi : \mathbb{R}^{n} \rightarrow \mathbb{R}^{n} $ be defined by $\pi (x_{1}, x_{2}, ...,x_{n})=(x_{2},x_{3},...,x_{n},x_{1})$:

What is a complete description of all $\pi \_$Invariant subspaces of $\mathbb{R}^n$?

Bounding exponential sum of the form $\sum_{\mathbf{x} \in (\mathbb{Z}/q \mathbb{Z})^n } \chi_1(x_1)\cdots \chi_n (x_n) e(a F(\mathbf{x})/q)$

Math Overflow Recent Questions - Fri, 08/04/2017 - 10:36

I have encountered the following exponential sum and I would like to obtain a non-trivial upper bound for it. I am not quite sure where to start, and I would greatly appreciate any suggestions on how I can go about to obtain an upper bound or where to look.

Let $q \in \mathbb{N}$, and $\chi_i$ be a Dirichlet character modulo $q$ (not necessarily primitive), and $F(x_1, \ldots, x_n)$ is a homogeneous polynomial of degree $d$ with integer coefficients which defines a smooth hypersurface over $\mathbb{P}^{n-1}$.

The exponential sum in question is $$ \sum_{\mathbf{x} \in (\mathbb{Z}/q \mathbb{Z})^n } \chi_1(x_1)\cdots \chi_n(x_n) \ e \left(F(\mathbf{x}) \frac{a}{q} \right), $$ where $(a,q)=1$.

I would greatly appreciate any suggestions or comments. Thank you very much for your time.

Yet another proof of Riemannian conjecture [on hold]

Math Overflow Recent Questions - Fri, 08/04/2017 - 10:20

Just published yesterday and the paper is astonishingly short.

Proof of the Riemann hypothesis by Frank Stenger https://arxiv.org/abs/1708.01209

Is this a valid proof for the long-standing conjecture? Have anyone review this paper could tell?

Squaring operation in KO theory

Math Overflow Recent Questions - Fri, 08/04/2017 - 10:17

There's an operation $\Lambda^2: KO^0(X) \to KO^0(X)$ such that $\Lambda^2(x+y)-\Lambda^2(x)-\Lambda^2(y)=xy$, which comes from the antisymmetric power. Similarly, there's a $\Lambda^2: KO^4(X) \to KO^8(X)$ with the same property, since the antisymmetric square of an $Sp$ bundle is an $O$ bundle.

Do they extend to the entire $KO^*$ so that $\Lambda^2: KO^d(X) \to KO^{2d}(X)$ and $\Lambda^2(x+y)-\Lambda^2(x)-\Lambda^2(y)=xy$?

If so, is there a corresponding operation in $KO(-,\mathbb{Z}/n\mathbb{Z})$?

At what point does exponential integral coincide with exponential?

Math Overflow Recent Questions - Fri, 08/04/2017 - 07:59

I'm looking for the solution to the equation $E_1(x)=e^{-x}$, where $E_1(x)$ is the exponential integral function. The solution is approximately 0.434818204, but is this constant well known/studied?

The reason I think this constant should be important is that exponential integral, which is upper bounded by $e^{-x}/x$, behaves like $e^{-x}$ for larger $x$ and like $-\log(x)$ for smaller $x$, so roughly this point (and, for that matter, when $E_1(x)=-\log(x)$, around $x \approx 0.676355077$) are natural "transition points" to study. Furthermore, the function $x E_1(x)$ is a "sharpened" version of $x e^{-x}$ and $x \log(x)$ and its maximum is attained at the point I'm interested in (whereas maximums of the other two functions are trivially attained at $x=1$ and $x=e$).

In general the elementary functions $x e^{-x}$ and $x \log(x)$ are well understood (e.g., Lambert W function tries to capture many complexities of the former), whereas there seems to be not much known about $x E_1(x)$ which exhibits a mixed behavior.

Why are Artin algebras so important? [on hold]

Math Overflow Recent Questions - Fri, 08/04/2017 - 07:27

Why are Artin algebras so important? Why Auslander and Reiten were interested in Artin algebras? What properties do they have that Noetherian algebras or Artinian rings do not in general, and that make them important?

Pairs of roots of unity whose real part satisfies a polynomial identity

Math Overflow Recent Questions - Fri, 08/04/2017 - 06:50

Some motivation: The matrix $M$ is Butson Hadamard if the entries of $M$ are $k^{\textrm{th}}$ roots of unity (for some $k$), and distinct pairs of rows are orthogonal under the usual Hermitian inner product. I am interested in classifying the Butson Hadamard matrices for which some power is a real scalar matrix. (This is equivalent to the corresponding unitary matrix having finite multiplicative order.)

In the case of $2\times 2$ matrices, after some reductions, the problem is equivalent to classifying the pairs of roots of unity $(\zeta_{1}, \zeta_{2})$ for which

$$\Re( \zeta_{1} ) = \sqrt{2} \Re(\zeta_{2}) $$

This can be restated as $2\Re(\zeta_{2})^{2} - \Re(\zeta_{1})^{2} = 0$. The solutions we have found, up to negation and complex conjugation, are $(i, i)$, $(1, \omega_{8})$ and $(\omega_{8}, \omega_{6})$. We ran computer searches which suggest that any further solution involves a $k^{\textrm{th}}$ root of unity for some $k \geq 33$.

Q1: Is this list of solutions to the displayed equation complete?

Q2: More generally, what techniques exist for finding all solutions to a polynomial equation in real parts of roots of unity?

Is the Lascar group over $A$ trivial when $T=T^{eq}$ and $A = acl(A)$?

Math Overflow Recent Questions - Thu, 08/03/2017 - 22:04

Let $T$ be a first-order theory which eliminates imaginaries, and let $A$ be an algebraically closed set in a model of $T$. Let $Gal_L(T[A])$ be the Lascar group of the theory $T[A]$, which is $T$ with constants added for $A$. Is $Gal_L(T[A])$ trivial? I would think not, but I don't know a counterexample.

Upper bound for $\|\textbf{A}^\dagger\|$, where $\textbf{A}$ is a matrix with specific sparse structure

Math Overflow Recent Questions - Thu, 08/03/2017 - 19:14

Consider the block matrix $$\textbf{A} = \left[ \begin{array}{ccc} \textbf{X}_{11} & \textbf{X}_{12}\\ \textbf{X}_{21} & \textbf{D}\\ \end{array} \right]$$ where $\textbf{X}_{11} \in \mathbb{C}^{r \times (r+1)}, \textbf{X}_{12} \in \mathbb{C}^{r \times 3nr}, \textbf{X}_{21} \in \mathbb{C}^{3nr \times (r+1)}$. These blocks are supposed to be dense matrices. We have some sparse structure on the block $\textbf{D}$. More precisely, we have that

$$\textbf{D} = \left[ \begin{array}{ccc} \left[ \begin{array}{ccc} D & \ldots & D\\ \vdots & \ddots & \vdots\\ D & \ldots & D\\ \end{array}\right] & \textbf{X} & \textbf{X}\\ \textbf{X} & \left[ \begin{array}{ccc} D & \ldots & D\\ \vdots & \ddots & \vdots\\ D & \ldots & D\\ \end{array}\right] & \textbf{X}\\ \textbf{X} & \textbf{X} & \left[ \begin{array}{ccc} D & \ldots & D\\ \vdots & \ddots & \vdots\\ D & \ldots & D\\ \end{array}\right]\\ \end{array}\right]$$

where each $\textbf{X}$ represents a dense matrix (but possible different from each other) in $\mathbb{C}^{nr \times nr}$. Each matrix in the form

$$\left[ \begin{array}{ccc} D & \ldots & D\\ \vdots & \ddots & \vdots\\ D & \ldots & D\\ \end{array}\right]\\$$ is composed by $r \times r$ of these matrices $D$, which are diagonal matrices (but possible different from each other) in $\mathbb{C}^{n \times n}$. So each of its $r^2$ entries is a diagonal matrix $D \in \mathbb{C}^{n \times n}$.

We have that $\textbf{A} \in \mathbb{C}^{(r+3nr) \times (1+r+3nr)}$ and what I'm trying to do is to obtain some upper bound for the norm of its pseudo inverse. Concretely, I want an upper bound for

$$\|\textbf{A}^\dagger\|$$ where the norm is the spectral norm or the Frobenius norm.

I was looking for some bound in terms of these blocks, in a way it take the advantage of its sparse structure.

I already looked through several papers and searched on the internet, but couldn't find anything helpful. All my ideas also didn't work, so I hope someone here may have a good idea. This looks to be a very specific problem, and since I'm not so familiar with results about sparse matrices, the best option is to share with you in the hope someone knows something about it. Thank you.

What does the group of automorphisms corresponding to $\mathfrak{g}$

Math Overflow Recent Questions - Thu, 08/03/2017 - 12:05

I am reading a book titled "Lectures on An Introduction to Grothendieck's Theory of the Fundamental Group" by J.P. Murre. I am in the chapter 4 titled "Fundamental groups". Here he fixes a base locally noetherian scheme $S$ and defines a category $\mathscr{C}$ taking the objects to be locally noetherian schemes with finite etale maps to $S$. Morphisms with in the objects are finite etale maps between them.

At the beginning of this chapter he is stating the basic properties of this category, like existence of fibre product and disjoint union. I am reading the proof of the existence of the quotient by a finite subgroup $\mathfrak{g}$ of the automorphisms of an object $X$. (I am unable to write proper notation for $\mathfrak{g}$, because I do not know what it is. What is the notation?) He states that the problem is enough to be solved locally and goes on to define a finite subgroup of automorphisms of $\operatorname{Spec}(A)$ corresponding to $\mathfrak{g}$. I am unable to understand what this means. It is not even clear to me why the problem is enough to be solved for affine open subschemes. I have attached the snapshot below.

An Optimization Problem for Exponential Polynomials

Math Overflow Recent Questions - Thu, 08/03/2017 - 04:47

Let $\omega$ be a primitive complex $n^{th}$ root of unity. I am interested in the following quantity $$ \max_{f(n)\leq \ell \leq g(n)} \quad \min_{0<k\leq n-1} \left| 1+\omega^k+\omega^{2k}+\cdots+\omega^{(\ell-1)k}\right|^2, $$ as $n$ goes to infinity. Note that the optimization simplifies to $$ \max_{f(n)\leq \ell \leq g(n)} \quad \min_{0<k\leq n-1} \left| \frac{\sin (\pi \ell k/n)}{\sin (\pi k/n)}\right|^2, $$ and for concreteness case I don't mind taking $f(n)=\log^2 n, g(n)=\sqrt{n}-$not the iterated log but square of log.

I know the minimum will in general can be very small (at least for arbitrary sums of roots of unity as in question by Terry Tao) but what if we are free to look at multiple lengths for the sum, and take consecutive powers, things must improve, but how much?

Edit: One can take $n$ prime, if it helps.

Pages

Subscribe to curious little things aggregator