Recent MathOverflow Questions

Matrix operations preserving the middle coefficients of characteristic polynomial

Math Overflow Recent Questions - Fri, 01/18/2019 - 13:22

Let $n$ be a positive integer. For a ring $A$ and matrix $M \in \mathrm{Mat}_{n \times n}(A)$, let $\chi_{M}(t) = \det(M-t \operatorname{id}_{n}) = (-1)^{n}(t^{n} - \sigma_{M,1}t^{n-1} + \dotsb + (-1)^{n}\sigma_{M,n}) \in A[t]$ be the characteristic polynomial. Here $\sigma_{M,1}$ is the trace of $M$ and $\sigma_{M,n}$ is the determinant of $M$. Recall that $\sigma_{M,1}$ is additive and $\sigma_{M,n}$ is multiplicative, i.e. $\sigma_{M_{1}+M_{2},1} = \sigma_{M_{1},1} + \sigma_{M_{2},1}$ and $\sigma_{M_{1}M_{2},1} = \sigma_{M_{1},1}\sigma_{M_{2},1}$.

Are there any matrix operations that preserve the middle coefficients $\sigma_{M,2},\dotsc,\sigma_{M,n-1}$ in any reasonable sense?

This is related to the question Geometric interpretation of characteristic polynomial but I don't yet understand whether it answers my question.

Dimensions of Lie algebra powers of irreducible representations

Math Overflow Recent Questions - Fri, 01/18/2019 - 13:21

Consider the following plethysms. For semisimple Lie algebras $L$, if $A$ is the adjoint irrep, as far as I know, $d_1=|A^{\bigotimes 1}|=0$ (no lollipops), $d_2=|A^{\bigotimes 2}|=1$ (Schur's Lemma), $d_3=|A^{\bigotimes 3}|=1$ (structure constant tensor, 6j symbols).
And if $A$ is the adjoint of a non-semisimple $L$? I also read something like that the Killing form $K$ can be brought to $\mathrm{diag}(\{0,1\})$ in the right basis, implying $K^2=K$ and thus $d_2=2$ (unless $L$ is nilpotent, $K=0$ and thus $d_2=1$). But I'm confused - rescale the basis and the Killing form might have more than two different eigenvalues for more than 4 generators? (Does it?)
Assuming it miraculously hasn't, this means $d_3\leq 8$ (one or no Killing form pasted on the structure constant tensor). I already once jumped to conclusions :-) and thought $d_3=1$, but I found a $L$ with $d_3=4$. So, can you name a $L$ with $d_3=8$?.
(In graphic form the answers to my question seem obvious :-)

Generating random 6 digit numbers

Math Overflow Recent Questions - Fri, 01/18/2019 - 12:48

We can generate a random 6 digit number $M$ by first generating a random distribution over the 10 digits and then sampling from that distribution.

Formally, let $x = (x_0, x_1, x_2, \ldots, x_9)$ be a point sampled from the uniform distribution on the unit sphere in 10 dimensions.

Then $x$ can be used to define a new RV $D_x$ on the set $D = [0, 1, 2, \ldots, 9]$ such that $P[D_x=i] = {x_i}^2$.

Now consider the RV $M_{D_x}$ which is defined by first sampling $D_x$ 6 times (with replacement) to get a set $\{s_0, s_1, s_2, \ldots, s_5\}$ and then taking the sum $\sum_{j = 0}^5 10^j s_j$ (that is, use the 6 numbers generated from $D_x$ as digits of a number between 0 and 999999).

My question is: what is $P[M = n]$ for $n$ between 0 and 999999? Is this a uniform distribution over 6 digit integers?

How do we introduce a signed finite measure on the space of curves confined into the box $[0,1]^{n}$?

Math Overflow Recent Questions - Fri, 01/18/2019 - 12:48

Given $\Omega_{n} = \{\alpha:[0,1]\rightarrow[0,1]^{n}\,|\,\alpha\,\,\text{is smooth}\}$, consider the equivalence relation: \begin{align*} & \alpha_{1} \sim \alpha_{2} \Leftrightarrow \int_{0}^{1}\langle G(\alpha_{1}(t)),\alpha^{\prime}_{1}(t)\rangle\,\mathrm{d}t = \int_{0}^{1}\langle G(\alpha_{2}(t)),\alpha^{\prime}_{2}(t)\rangle\,\mathrm{d}t,\\ \end{align*}

Where $\alpha_{1}(0) = \alpha_{2}(0)$ and $\alpha_{1}(1) = \alpha_{2}(1)$. It is assumed that $G:[0,1]^{n}\rightarrow[0,1]^{n}$ is known. Such set can naturally be considered as a metric space (thus a topological space) in accordance to the norm:

\begin{align*} \lVert\alpha\rVert = \max_{0 \leq t \leq 1}\lVert\alpha(t)\rVert_{2} \end{align*}

Let us define $\Omega := \Omega_{n}/G$ as the quotient space according to the above-mentioned equivalence relation. Thus we can introduce a topology on $\Omega$. Precisely speaking, the quotient topology: \begin{align*} \tau_{\Omega} := \{O\subset\Omega\,|\,\pi^{-1}(O)\in\tau_{\Omega_{n}}\} \end{align*}

Where $\pi$ is the map which associates each $\alpha\in\Omega_{n}$ to $[\alpha]\in\Omega$: $\pi(\alpha) = [\alpha]$. Finally, given the topological space $(\Omega,\tau_{\Omega})$, we can construct its associated Borel $\sigma$-algebra $\Sigma$.

Here is my question: how do we introduce a signed finite measure on $(\Omega,\Sigma)$? Precisely, I would like to define a triple $(\Omega,\Sigma,\mathbb{P})$ such that $\mathbb{P}([\alpha]) = -\mathbb{P}([\alpha^{-}]) \geq 0$, where $\alpha^{-}(t) = \alpha(1-t)$.

Such problem makes part of my research project on negative probabilities. I apologize if the question does not fit into Math Overflow context. Any help is appreciated. Thanks in advance.

Is the metric completion of a Riemannian manifold always a geodesic space?

Math Overflow Recent Questions - Fri, 01/18/2019 - 11:55

A length space is a metric space $X$, where the distance between two points is the infimum of the lengths of curves joining them. The length of a curve $c: [0,1] \rightarrow X$ is the sup of $$ d(c(0), c(t_1)) + d(c(t_1), d(t_2)) + \cdots + d(c(t_{N-1}), c(1)) $$ over all $0 < t_1 < t_2\cdots < t_{N-1} < 1$ and $N > 0$.

A geodesic space is a length space, where for each $x,y \in X$, there is a curve $c$ connecting $x$ to $y$ whose length is equal to $d(x,y)$.

A Riemannian manifold $M$ and its metric completion $\overline{M}$ are length spaces. If the Riemannian manifold is complete, then it is a geodesic space.

But is $\overline{M}$ necessarily a geodesic space? If not, what is a counterexample?

This was motivated by my flawed answer to Minimizing geodesics in incomplete Riemannian manifolds

Also, note that if $\overline{M}$ is locally compact, then it is a geodesic space by the usual proof. One example of $M$, where $\overline{M}$ is not locally compact is the universal cover of the punctured flat plane. However, this is still a geodesic space.

Under which conditions do ellipsoids have a focal property?

Math Overflow Recent Questions - Fri, 01/18/2019 - 10:56

Given an ellipsoid $E$, we consider the trajectory that light would do inside if the surface of $E$ would be a mirror. In other words, we consider trajectories $\gamma:[0,\infty)\rightarrow E$ such that that $\gamma'$ is locally constant for any $t$ such that $\gamma(t)\in \mathrm{int}\,E$ and such that $\gamma'(t)$ becomes its reflection with respect the hyperplane $T_{\gamma(t)}\partial E$ for $t$ such that $\gamma(t)\in\partial E$.

When $E$ is an ellipse, any such trajectory that start at a foci will return to a foci after a reflection. In one generates a 3-ellipsoid by rotating the ellipse around the axis containing the foci, the same property continue to be true.

Now, imagine you rotate the ellipse with respect the other axis of symmetry. Then the image of pair of foci becomes a circle $C$. The question is the following: Do any light trajectory starting at $C$ come back to $C$ after a reflection? If this is not true, does this hold for the convex hull of $C$?

More generally, one can we said about ellipsoids, light trajectories and subsets with the property that light trajectories become back to them after one reflection?

[This was a question made to me by an experimental physicist, but after thinking about it I have been unable to find any reference about it despite its elementary looking form].

Quadrics over the univariate function field with discriminant of minimal degree

Math Overflow Recent Questions - Fri, 01/18/2019 - 10:54

Consider a non-degenerate quadric $Q(x,y,z) \subset \mathrm{P}^2$ over the univariate function field $\mathbb{F}_p(t)$, where $\mathbb{F}_p$ is a prime finite field, $p > 2$. For simplicity assume that coefficients of $Q$ lie in $\mathbb{F}_p[t]$.

Let $\Delta_Q(t) \in \mathbb{F}_p[t]$ be the discriminant of $Q$, i.e., the determinant of the corresponding symmetric matrix.

How to understand is there or not (at least in some cases) a quadric $Q^\prime$ isomorphic over $\mathbb{F}_p(t)$ to $Q$ such that the polynomial $\Delta_{Q^\prime}(t)$ has strictly smaller degree than $\Delta_Q(t)$. And how can I explicitly construct a linear transformation (over $\mathbb{F}_p(t)$) between $Q$ and $Q^\prime$?

On permanents involving trigonometric functions

Math Overflow Recent Questions - Fri, 01/18/2019 - 10:19

Here I pose my conjectures on permanents involving trigonometric functions.

Conjecture 1. Let $p$ be an odd prime. Then both $$c_p:=\sqrt p\,\text{per}\left[\cot\pi\frac{jk}p\right]_{1\le j,k\le (p-1)/2}$$ and $$t_p:=\frac1{\sqrt p}\text{per}\left[\tan\pi\frac{jk}p\right]_{1\le j,k\le(p-1)/2}$$ are integers. Moreover, $$c_p\equiv1\pmod p\ \ \text{and}\ \ t_p\equiv (-1)^{(p+1)/2}\pmod p.$$

Remark 1. Via Mathematica I find that \begin{align}&c_3=1,\ c_5=-4,\ c_7=22,\ c_{11}=1816,\ c_{13}=-5056, \\&c_{17}=-2676224,\ c_{19}=58473280. \end{align}

Conjecture 2. Let $p$ be an odd prime. Then both $$s_p:=\frac{2^{(p-1)/2}}{\sqrt p}\text{per}\left[\sin2\pi\frac{jk}p\right]_{1\le j,k\le(p-1)/2}$$ and $$r_p:=\frac{\sqrt p}{2^{(p-1)/2}}\text{per}\left[\csc2\pi\frac{jk}p\right]_{1\le j,k\le(p-1)/2}$$ are integers. Moreover, $$s_p\equiv(-1)^{(p+1)/2}\pmod p\ \ \text{and}\ \ r_p\equiv1\pmod p.$$

Conjecture 3. Let $p$ be an odd prime. Then $$a_p:=\frac1{2^{(p-1)/2}}\text{per}\left[\sec2\pi\frac{jk}p\right]_{1\le j,k\le(p-1)/2}\in\mathbb Z.$$ If $p>3$ and $p\equiv3\pmod 4$, then

$$\text{per}\left[\sec2\pi\frac{jk}p\right]_{1\le j,k\le (p-1)/2}\equiv \text{per}\left[\cos2\pi\frac{jk}p\right]_{1\le j,k\le (p-1)/2} \equiv-1\pmod p.$$

Remark 2. Using Galois theory, I have shown that $a_p,c_p,r_p,s_p,t_p$ are rational numbers for any odd prime $p$.

Your comments are welcome!

Enumerating all partitions induced by Voronoi diagrams for clustering

Math Overflow Recent Questions - Fri, 01/18/2019 - 10:05

a classical results by M. Inaba et al. in "Applications of Weighted Voronoi Diagrams and Randomization to Variance-Based k-CLustering" (Theorem 3) says

The number of Voronoi partitions of $n$ points by the Euclidean Voronoi diagram generated by $k$ points in $d$-dimensional space is $\mathcal{O}(n^{dk})$, and all the Voronoi partitions can be enumerated in $\mathcal{O}(n^{dk+1})$.

They basically divide the $d$-dimensional space into equivalence classes where two sets of center $\mu^1$ and $\mu^2$ are equivalent if they lead to the same Voronoi Diagram. Then they show that the arrangement of the $nk(k-1)/2$ surfaces

$$ ||x_i-\mu_j||^2- ||x_i-\mu_{j'}||^2 = 0 $$

for each point $x_i$ and two cluster center $\mu_j$ and $\mu_{j'}$ coincides with the equivalence relation from Voronoi partitions.

Next they argue that the combinatorial complexity of arrangements of $nk(k-1)/2$ constant-degree algebraic surfaces is bounded and that this implies and algorithm with running time $\mathcal{O}(n^{dk+1})$. Unfortunately, the cited source (Evaluation of combinatorial complexity for hypothesis spaces in learning theory with application, Master's Thesis, Department of Information Science, University of Toko, 1994) I cannot find anywhere. More precisely I cannot see the two following things.

  1. Where can I find a bound for the combinatorial complexity of the arrangement of $nk(k-1)/2$ constant degree algebraic surfaces and
  2. How does this help me to compute the arrangement?

For 2. I found the Bentley–Ottmann algorithm, however that only works for line segments and not degree 2 polynomials. How can this algorithm be generalized?

Thanks so much!

False constantes in first Hardy-Littlewood conjecture?

Math Overflow Recent Questions - Fri, 01/18/2019 - 10:04

Let $q$ be a prime number, and $m$ an even number.

Let $\displaystyle\mathcal{B}_q=\{b \in \mathbb{N}^{*} \, | \, b \wedge {\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a} \right)}=1 \text{ and } b \leq {\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a} \right)} \}$

Let $I_{q,m}(n)$ denote the number of elements $(b,b+m)$ less than $n$ and coprime to $\displaystyle{\small \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a}}$

Let $q(n)$ be the small prime verify $n+m < \displaystyle{\small \prod_{\substack{a \leq q(n) \\ \text{a prime}}} {\normalsize a}}$

Let $b_1$ be the prime that the gap $(b,b+m)$ is appear the first time in $\mathcal{B}_{b_1}$ and $s = \#\{(b,b+m)\in\mathcal{B}_{b_1}^2\}$

I proove that $I_{q(n),m}(n) \sim \dfrac{\theta_m}{\displaystyle{\small \prod_{\substack{3 \leq a < b_1 \\ \text{a prime}}} {\normalsize (a-2)}}} \; 2 \, C_2 \; \dfrac{n}{\ln(\ln(n))^2} e^{-2 \gamma}$

With $\theta_m$ is a constante verify $\dfrac{s}{b_1 - 2} \leq \theta_m \leq \displaystyle{\small \left( \prod_{\substack{a | m \\ \text{a prime}}} {\normalsize \frac{a-1}{a-2}} \right)}$ , and $C_2 = \displaystyle{\small \left( \prod_{\substack{3 \leq a \\ \text{a prime}}} {\normalsize \frac{a (a-2)}{(a-1)^2}} \right)}$

Show that $\theta_2 = \theta_4 = 1$

If we denote $I_n$ the number of elements less than $n$ and coprime to $\displaystyle{\small \left( \prod_{\substack{a \leq q(n) \\ \text{a prime}}} {\normalsize a} \right)}$ ($q(n)$ the small prime verify $n < \displaystyle{\small \left( \prod_{\substack{a \leq q(n) \\ \text{a prime}}} {\normalsize a} \right)}$)

I prove that $I_n \sim \dfrac{n}{\ln(\ln(n))} \, e^{-\gamma}$

Show that : $\dfrac{n}{\ln(\ln(n))} \, e^{-\gamma} = \dfrac{n}{\ln(n)} \dfrac{\ln(n)}{\ln(\ln(n))} \, e^{-\gamma} \sim \pi(n) \big( \pi(q(n)) e^{-\gamma} \big)$

Then if we use the same analogy with the case $I_{q(n),m}(n)$ :

$I_{q(n), m}(n) \sim \pi_m(n) \big( \pi(q(n)) \, e^{- \gamma} \big)^2$

Then : $\pi_m(n) \sim \dfrac{\theta_m}{\displaystyle{\small \prod_{\substack{3 \leq a < b_1 \\ \text{a prime}}} {\normalsize (a-2)}}} \; 2 \, C_2 \; \dfrac{n}{\ln(n)^2}$

Well that's just a conjecture (i don't proove $I_{q(n), m}(n) \sim \pi_m(n) \big( \pi(q(n)) \, e^{- \gamma} \big)^2$)

But the Hardy-Littlewood conjecture state that $\pi_m(n) \sim \displaystyle{\small \left( \prod_{\substack{a | m \\ \text{a prime}}} {\normalsize \frac{a-1}{a-2}} \right)} 2 \, C_2 \; \dfrac{n}{\ln(n)^2}$

http://lagrida.com/prime_numbers_construction.html

http://lagrida.com/Calculate_Gaps_Cardinality.html

http://lagrida.com/Fondamentale_Conjonctures_Prime_Numbers.html

Algebra parameter [on hold]

Math Overflow Recent Questions - Fri, 01/18/2019 - 10:00

I got 4 equasions, where are x,y,z,u variables , and a,b parameters. So the solution from the book is like this:

x+(a+1)y-(a+1)z-au=1

ax+(a+1)y+az-2u=2

ax+(a+1)y-2z+au=b

(a-1)x+3(a+1)z-4u=3-b

So when I do all transformations it gets this form:

(a+1)y+x-(a+1)z-au=1

(a-1)x+(2a+1)z+(a-2)u=1

(-a-2)z+(a+2)u=b-2

Now, next steps are these:

1) If a#-1 , a#1 , a#-2

Equasion is indefinite

2)If a=-1

Equasion is indefinite

3)If a=1

Equasion is indefinite

4)If a=-2

Equasion can be either indefinite or impossible

My question is how did we got that first part where a can't be 1, -1 or -2. Why it can be 2 or 0 or some other number?

Proving that the exterior algebra is symmetric via the polynomial ring

Math Overflow Recent Questions - Fri, 01/18/2019 - 09:23

Recall that a finite dimensional algebra $A$ over a field $K$ is called Frobenius in case $A \cong D(A)$ as right modules, and it is called symmetric in case $A \cong D(A)$ as bimodules (where $D=Hom_K(-,K)$).

It is known that the Koszul dual algebra $A^{!}$ of an Artin-Schelter regular Koszulalgebra $A$ is a finite dimensional Frobenius algebra.

Question 1: In case $A=K[x_1,...,x_n]$ is the polnyomial ring in $n$ variables over a field $K$ of characteristic 0, $A^{!}$ is an exterior algebra and in case I made no mistake this should be symmetric if and only if $n$ is odd. My question is whether one can prove this just by looking at $K[x_1,...,x_n]$. Or asked differently: What property corresponds to being symmetric in $K[x_1,...,x_n]$, if there is any? (so $K[x_1,...,x_n]$ should have this property only for odd $n$)

Question 2 (more general): Is there a condition on $A$ that describes when $A^{!}$ is a symmetric algebra?

Algebra parameters [on hold]

Math Overflow Recent Questions - Fri, 01/18/2019 - 09:20

I got 4 equasions, where are x,y,z,u variables , and a,b parameters. So the solution from the book is like this:

x+(a+1)y-(a+1)z-au=1 ax+(a+1)y+az-2u=2 ax+(a+1)y-2z+au=b (a-1)x+3(a+1)z-4u=3-b

So when I do all transformations it gets this form:

(a+1)y+x-(a+1)z-au=1 (a-1)x+(2a+1)z+(a-2)u=1 (-a-2)z+(a+2)u=b-2

Now, next steps are these:

1) If a#-1 , a#1 , a#-2
Equasion is indefinite

2)If a=-1 Equasion is indefinite

3)If a=1 Equasion is indefinite

4)If a=-2

Equasion can be either indefinite or impossible

My question is how did we got that first part where a can't be 1, -1 or -2. Why it can be 2 or 0 or some other number?

Is being of general type stable under generization

Math Overflow Recent Questions - Fri, 01/18/2019 - 08:59

This question is about how varieties of general type over an algebraically closed field of characteristic zero $k$ behave under generization in families.

Definition. An integral projective scheme X over k is of general type if some desingularization of X is of general type. A reduced projective scheme X over k is of general type if every (reduced) irreducible component of $X$ is of general type. Finally, a projective scheme X is of general type if X_{red} is of general type.

My question is as follows.

Let S be an integral variety over $k$ with function field $K=K(S)$. Let $X\to S$ be a projective flat morphism. Suppose that there is a closed point $s$ in $S(k)$ such that $X_s$ is of general type over $k$. Then the generic fibre $X_K$ of $X\to S$ is of general type over $K$.

Please note that I do not make any assumptions on the singularities of $X_s$, nor do I assume $X_s$ to be irreducible.

A partial (positive) answer follows from deep theorems of Siu, Kawamata, and Nakayama on the constancy of plurigenera. But, as far as I know, these theorems require some conditions on the singularities of $X_s$ (e.g., canonical singularities). Can one reduce to this situation using semi-stable reduction, maybe?

Probably, it suffices (and is more natural?) to assume only that at least one irreducible component of $X_s$ is of general type (instead of the entire fibre being of general type). This would at least allow one to reduce to the case that $X$ is smooth over $k$.

Looking for example of integral transformations that preserve number of zeros

Math Overflow Recent Questions - Fri, 01/18/2019 - 08:49

Let $f:\mathbb{R} \to \mathbb{R} $ have $n<\infty$ zeros.
I am looking for non-trivial examples of integral transformation \begin{align} g(x)= \int f(t) h(t,x) dt \end{align} such that $f$ and $g$ have the same number of zeros. Note that positions of zeros are allowed to change.

Edit: Suppose we assume the following regularity conditions on $f$

  1. $k$ times differentiable
  2. $f$ is absolutely integrable (i.e., $\int |f(x)| dx <\infty$)

Note: please feel free to add or change these assumptions. The point here is to see some meaningful examples.

Replacing the initial conditions for a PDE

Math Overflow Recent Questions - Fri, 01/18/2019 - 08:27
The problem

The PDE I am working with is given by $\left(\partial_a^b \leftrightarrow\frac{\partial^b}{\partial a^b}\right)$

$$\partial_t \psi = i \partial_x^2 \psi$$ $$\psi(x,t=0) = \psi_0(x)$$ $$\psi:\mathbb{R}^2\rightarrow \mathbb{C}$$

Let $F:\mathbb{R}^2 \rightarrow \mathbb{R}_{\ge 0}$ be a function such that for a solution $\psi$ of the above mentioned PDE, then $F=|\psi|^2$. The solutions $\psi$ that interest me have to be square-integrable.

At this point I am interested in computing $\psi$ for a known $F$, which from my point of view, should be equivalent to replacing the initial condition $\psi_0$ with the now known function $F$.

I have not been able to prove this, but I consider this approach to not hold on for a general case, in the sense that there might not be solutions of the PDE that are compatible with the choice of $F$. This is why I have changed the condition. However this is only an assumption, so if I can be proven wrong it'll be great.

The second attempt is as follows. For a solution $\psi$ of the PDE consider the curves given by $|\psi|^2 = a$, where $a \in [0,\max(|\psi|^2)]$ is a constant. I label this curves as $\Gamma(s)=\left(g(s), h(s)\right)$ where $g,h:\mathbb{R} \rightarrow \mathbb{R}$. Since $|\psi|^2$ is constant on a curve $\Gamma$, then it is true that

$$\nabla_{x,t}|\psi|^2 \perp T_{\Gamma}$$

where $\nabla_{x,t}|\psi|^2=\left( \partial_x |\psi|^2, \partial_t |\psi|^2\right)$ and $T_{\Gamma}(s)=\left(\partial_s g, \partial_s h\right)$ is the tangent of the curve $\Gamma$.

This condition becomes

$$\nabla_{x,t}|\psi|^2 \cdot T_{\Gamma}=0$$

or explicit

$$\partial_s g \cdot \partial_x |\psi|^2 + \partial_s h \cdot \partial_t |\psi|^2 = 0.$$

This is a transport equation which defines, by the choice on $g$ and $h$ the curves on which $|\psi|^2$ is constant, independent of the actual value of $|\psi|^2$ on that curve. Unlike the first attempt where I enforce $F$ and then look for the solution (if possible), here I enforce the curves on which the solution has the same amplitude. This should not be as restrictive as the first approach, but it might also be too little to allow for computing $\psi$.

Context

For the eventual questions about why do I want to formulate the problem in such a manner, or what is the purpose of it, I will try to give some pre-emptive answers.

There is a solution for this PDE, which although not square-integrable, satisfies the condition $F(x,t) = F(x-t^2)$. More about this here. Following this result, square-integrable versions of this solution have been developed that qualitatively satisfy the same condition, but only for $t<t_{max}\in \mathbb{R}$.

These results do not give an algorithm for generating solutions that have a particular behavior, it is more of a guessing game, where one pluggs an initial condition in the PDE and then hopes to get an oddly interesting solution. This is the reason why I want to replace the initial condition with something similar to what I have presented above.

My attempts

Following the article I have mentioned in the Context section, the choice I made for $g$ and $h$ has been

$$g(s) = x_0 + \frac{s^n}{n!}$$ $$h(s) = s$$

where $n\in \mathbb{N}$. For $n = 2$ a rescaled solution similar to the the one in the article should be obtained. Because of the choice on $g$ and $h$, I can replace $s$ with $t$. The condition becomes

$$t \partial_x F + \partial_t F = 0$$

At this point I considered writing $g$ as

$$g(t) = \sum_{n=0}^{\infty} \frac{t^n}{n!} \partial_t^n g(t=0)$$

I know the values of $\partial_t^n g(t=0), \forall n \in \mathbb{N}$ since I chose $g(t) = x_0 + t^2/2$, so I can derivate the condition with respect to $t$ in order to get to each $\partial_t^n g(t=0)$ term after I evaluate the expresion at $t=0$.

By evaluating the condition as it is, it gives

$$\partial_t|\psi(x,t =0)|^2 = 0$$

Derivating once and evaluating it gives

$$\partial_x|\psi|^2(x, t=0) + \partial_t^2 |\psi(x,t=0)|=0$$

and

$$\partial_t^n |\psi|^2(x,t=0) = 0, \forall n>2.$$

These conditions, I thought, could be used in order to compute an initial condition. Since there are an infinite number of such conditions, I thought that in the worst case scenario, if this approach works, I should be able to use a finite number of them in order to get an approximate initial condition that satisfies qualitatively the condition I emposed by choosing $g$ and $h$. As it turns out, the computation is not as simple as would have hoped, since the evaluation of the temporal derivatives can only make use of

$$\partial_t |\psi|^2 = i\left(\partial_x^2\psi \cdot \psi^* - \psi \cdot \partial_x^2 \psi^*\right)$$.

Although I wasn't able to actually compute an initial conditions from conditions for $n \le 2$, I did manage to validate the result by plugging in the initial condition from the article. I did not try for $n>2$ since even using Wolfram Alpha it still was difficult to follow the expressions.

Conclusion and questions

I am interested in validating this approach by answering to these questions:

  • Can I solve the PDE (numerical methods will do) without imposing other conditions, and if not, why is that? I looked for methods of proving that a problem is well-posed, but my background is in engineering and physics, not mathematics so I did not really know where to look for.
  • Does this approach already exist and I just didn't look where I had to in order to find it?

Terminology about trees

Math Overflow Recent Questions - Fri, 01/18/2019 - 05:42

In set theory, a tree is usually defined as a partial order such that the set of elements below any given one is well-ordered. I am interested in the class of partial orders $P$ such that for every $p \in P$, the set of $q \leq p$ is just linearly ordered. Does this have a name?

One-Sided Analyticity Condition Guarantees Analytic Function?

Math Overflow Recent Questions - Fri, 01/18/2019 - 00:07

Let $f \ \colon \ [0,\infty) \to \mathbb{R}$ be a function satisfying:

  • $f$ is differentiable infinitely many times in $(0,\infty)$, and has a right-derivative of any order at $0$.
  • $f$ satifsfies the condition (condition 3 here) for analyticity: for every compact $K \subset [0,\infty)$ there exists a constant $C_K$ such that $$\forall x \in K:\forall n \geq 0:|f^{(n)}(x)|\leq C_K^{n+1}n!$$ where in the last formula, if $K$ contains $0$ and $x=0$, then the $n$-th derivative in the formula is the $n$-th right derivative in $0$.

Is it true in this case that $f$ is analytic in $[0,\infty)$ and that for some $\epsilon > 0$,

$$\forall x \in [0,\epsilon) \ \colon \ f(x) = \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n$$

?

If in addition we have a constant $C$ such that $$\forall x \in [0,\infty):\forall n \geq 0:|f^{(n)}(x)|\leq C^{n+1}n!$$

does the following hold: $$\forall x \in [0,\infty) \ \colon \ f(x) = \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n$$

?

Littlewood’s three precepts of refereeing in mathematics: is it (1) new, (2) correct, (3) interesting?

Math Overflow Recent Questions - Thu, 01/17/2019 - 16:37

I have a question regarding Littlewood’s three precepts of refereeing a mathematical paper, namely whether it is (1) new, (2) correct, and (3) interesting.

I have found these mentioned in the literature on refereeing, e.g.:

  • “you should address Littlewoods’s three precepts: (1) Is it new? (2) Is it correct? Is it surprising?” (Krantz, 1997, p. 125); or
  • “the fundamental precepts ‘Is it true?’, ‘Is it new?’, and ‘Is it interesting?’ to which, Littlewood believed, a referee should always respond.” (Moslehian, 2010: 1245)

Unfortunately, I haven’t been able to track down the original source. Does anyone know where Littlewood might have formulated these three precepts?

Thank you!

REFERENCES

Krantz, S. G. (1997). A Primer of Mathematical Writing: Being a Disquisition on Having Your Ideas Recorded, Typeset, Published, Read, and Appreciated. Providence, RI: American Mathematical Society.

Moslehian, M. S. (2010). Attributes of an ideal referee. Notices of the American Mathematical Society, 57 (10), 1245. (pdf)

Are fully general Frobenioids necessary?

Math Overflow Recent Questions - Thu, 01/17/2019 - 10:59

Mochizuki's notion of a Frobenioid introduced in The geometry of Frobenioids I is rather elaborate. However, he also introduces a myriad of further properties that a Frobenioid may satisfy, and his main results in that paper are always under considerable additional assumptions. This leads to me to wonder

Question: To what extent is it necessary to think about Frobenioids in their full generality to understand Mochizuki's theory? For applications is it safe to assume that all Frobenioids one will encounter are (for example)

  • of isotropic type?

  • of standard type?

  • obtained as "model Frobenioids" (Thm 5.2)?

  • Frobenius-normalized?

  • ...

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