Surprising continued fractions of numbers in the form $\sum_{n=0}^\infty \frac{1}{a^{2^n}}$, including the same pattern for every $a>2$

I've been interested in the numbers of this form because it can be proved that for integer $a \geq 2$ all of them are irrational: $$x_a=\sum_{n=0}^\infty \frac{1}{a^{2^n}}$$
They satisfy the conditions listed in this paper: The Approximation of Numbers as Sums of Reciprocals. This is related to my other question.
Now I decided to look at simple continued fractions of such numbers and noticed a surprising thing. For each $a$ I checked the CF entries consist of only three numbers (except the first entry, which is why I show $1/x$ instead of $x$):
$$x_2=\sum_{n=0}^\infty \frac{1}{2^{2^n}}=0.8164215090218931437080797375305252217$$
$$\frac{1}{x_2}=1+\cfrac{1}{4+\cfrac{1}{2+\cfrac{1}{4+\cfrac{1}{4+\cfrac{1}{6+\dots}}}}}$$
Writing the CF in the more convenient form we obtain for $200$ digits:
$1/x_2=$[1; 4, 2, 4, 4, 6, 4, 2, 4, 6, 2, 4, 6, 4, 4, 2, 4, 6, 2, 4, 4, 6, 4, 2, 6, 4, 2, 4, 6, 4, 4, 2, 4, 6, 2, 4, 4, 6, 4, 2, 4, 6, 2, 4, 6, 4, 4, 2, 6, 4, 2, 4, 4, 6, 4, 2, 6, 4, 2, 4, 6, 4, 4, 2, 4, 6, 2, 4, 4, 6, 4, 2, 4, 6, 2, 4, 6, 4, 4, 2, 4, 6, 2, 4, 4, 6, 4, 2, 6, 4, 2, 4, 6, 4, 4, 2, 6, 4, 2, 4, 4, 6, 4, 2, 4, 6, 2, 4, 6, 4, 4, 2, 6, 4, 2, 4, 4, 6, 4, 2, 6, 4, 2, 4, 6, 4, 4, 2, 4, 6, 2, 4, 4, 6, 4, 2, 4, 6, 2, 4, 6, 4, 4, 2, 4, 6, 2, 4, 4, 6, 4, 2, 6, 4, 2, 4, 6, 4, 4, 2, 4,...]
Clearly, all of the entries are $2,4$ or $6$.
The same goes for other examples:
$1/x_3=$[2; 5, 3, 3, 1, 3, 5, 3, 1, 5, 3, 1, 3, 3, 5, 3, 1, 5, 3, 3, 1, 3, 5, 1, 3, 5, 3, 1, 3, 3, 5, 3, 1, 5, 3, 3, 1, 3, 5, 3, 1, 5, 3, 1, 3, 3, 5, 1, 3, 5, 3, 3, 1, 3, 5, 1, 3, 5, 3, 1, 3, 3, 5, 3, 1, 5, 3, 3, 1, 3, 5, 3, 1, 5, 3, 1, 3, 3, 5, 3, 1, 5, 3, 3, 1, 3, 5, 1, 3, 5, 3, 1, 3, 3, 5, 1, 3, 5, 3, 3, 1, 3, 5, 3, 1, 5, 3, 1, 3, 3, 5, 1, 3, 5, 3, 3, 1, 3, 5, 1, 3, 5, 3, 1, 3, 3, 5, 3, 1, 5, 3, 3, 1, 3, 5, 3, 1, 5, 3, 1, 3, 3, 5, 3, 1, 5, 3, 3, 1, 3, 5, 1, 3, 5, 3, 1, 3, 3, 5, ,...]
$1/x_5=$[4; 7, 5, 5, 3, 5, 7, 5, 3, 7, 5, 3, 5, 5, 7, 5, 3, 7, 5, 5, 3, 5, 7, 3, 5, 7, 5, 3, 5, 5, 7, 5, 3, 7, 5, 5, 3, 5, 7, 5, 3, 7, 5, 3, 5, 5, 7, 3, 5, 7, 5, 5, 3, 5, 7, 3, 5, 7, 5, 3, 5, 5, 7, 5, 3, 7, 5, 5, 3, 5, 7, 5, 3, 7, 5, 3, 5, 5, 7, 5, 3, 7, 5, 5, 3, 5, 7, 3, 5, 7, 5, 3, 5, 5, 7, 3, 5, 7, 5, 5, 3, 5, 7, 5, 3, 7, 5, 3, 5, 5, 7, 3, 5, 7, 5, 5, 3, 5, 7, 3, 5, 7, 5, 3, 5, 5, 7, 5, 3, 7, 5, 5, 3, 5, 7, 5, 3, 7, 5,...]

How can this phenomenon be explained?

The further questions are:

Can we prove that all the CF entries for these numbers belong to a fixed set of three integers?
If so, can we make any conclusions about transcendentality of these numbers?

The implications are interesting. It is conjectured that algebraic numbers of degree $>2$ should have arbitrarily large CF entries at some point. See this paper. Meanwhile we know, that for degree $2$ the CF is (eventually) periodic.

Notice also the same 'pattern' which goes for the examples with odd $a$ here. We have a list of CF entries going the same way.
If we subtract the list of entries for $a=3$ from the list of entries for $a=5$, we obtain:
$$L_5-L_3=[0;2,2,2,2,2,2,2,2,2,...]$$
$$L_7-L_3=[0;4,4,4,4,4,4,4,4,4,...]$$
$$L_{113}-L_3=[0;110,110,110,110,110,110,...]$$
For $6$ and $2$ it goes the same way, but not for $4$ and $2$, there is some 'scrambling' there.

How can we prove/explain this facts? The same pattern for different $a$ seems very strange to me, especially if the numbers are transcendental.

Basically, if this turns out to be true, then from the CF for $x_3$ we will immediately obtain all the CFs for every $x_{2n+1}$

Important update. See http://oeis.org/A004200 for the case $a=3$, it seems like these continued fractions have pattern. And the following paper is linked: Simple continued fractions for some irrational numbers
The pattern is the same for every $a$ except $2$, so not only for the odd $a$.
Morevoer, look at the continued fractions for:
$$y_{ap}=a^p x_a$$
For integer $p$ you will notice a very apparent pattern.

My questions are largely answered by the linked paper. I will try to write up a short summary and post it as an answer, but anyone is three to do it before me.

Turns out a very close question was asked before Continued fraction for $c= \sum_{k=0}^\infty \frac 1{2^{2^k}} $ - is there a systematic expression?

@RobertIsrael, I have corrected my question, thank you

@RobArthan, yes, thank you. I get it now

@YuriyS: thanks for the link. I think you misunderstood the conjecture (which is far from trivial). the conjecture was that the simple continued fractions for algebraic numbers is either periodic or contains arbitrarily large coefficients. Read the abstract of the paper in your link (and then the rest of the paper) for clarification.

What is conjectured is that the continued fraction of any real algebraic number of degree $> 2$ has arbitrarily large entries.

@YuriyS: I am intrigued by the conjecture. Can you give a pointer to the paper you found about this conjecture, please.

Two simple CFs are equal iff their sequences of coefficients are equal. If the set of allowed coefficients includes two numbers $c_0$ and $c_1$, then you get a distinct CF for every real number $\alpha \in (0, 1)$, by mapping the binary representation of $\alpha$, a sequence of $0$s and $1$s, to the corresponding sequence of $c_0$s and $c_1$s.

@YuriyS: There is a bijection between them and the interval $[0, 1]$ expressed in base $b$, where $b$ is the number of integers in the set, right?

@RobArthan, how do you prove there are uncountably many such CFs?

There are uncountably many simple continued fractions with coefficients drawn from any given finite set of $2$ or more positive integers. So I don't think the conjecture you have in mind in your last two paragraphs can be correct without some further qualification.


Which digit occurs most often?

Is there any method to calculate, which digit occurs most often in the number
$$4 \uparrow \uparrow \uparrow \uparrow 4\  ,$$
the fourth Ackermann-number ?
Or would it be necessary to calculate the number digit by digit ?
I only know that the last n digits can be calculated, but this does not help much.

According to Wolfram Alpha, the number's too large to even represent, so that's saying something. It wouldn't be practical to calculate the 4th Ackermann number much less count its digits, so I can see why the urgency to have a simpler method is here.

I did not offer this bounty. And this is a very old question, I just wondered whether there is a trick that allows us to answer this question, which is apparently not the case. And I would find it very interesting, if it would be possible.

Even if we know the answer to this question, this is a dead end, unstructural information that needs a lot of work to obtain. (Despite of the bounty, please provide context and motivation, e.g. show the own effort to solve the issue. Why this function invented by humans for other reasons, why the digits in the very peculiar base ten used by humans for other reasons, and why $4\uparrow^44$?) As a parallel, let us consider some smaller "similar" number, $2\uparrow\uparrow 4=65536$, which is the profit to know a part of the statistics of the occurence of the digits?

In general, we know nothing about the digits of large powers, other than their last ones.


Do Hopf bundles give all relations between these "composition factors"?

Write a fiber bundle $F\to E\to B$ in short as $E=B\ltimes F$ (in analogy with groups).
(This is not necessary, but: given another bundle $X\to B\to Y$, we can write $E=(Y\ltimes X)\ltimes F$, but we may also compose $E\to B$ with $B\to Y$ to get fibrations $E\to Y$, whose fibers $Z$ fit inside fibrations $F\to Z\to X$ hence $Z=X\ltimes F$ and $E=Y\ltimes(X\ltimes F)$ as well. Thus we have a kind of associativity property $(Y\ltimes X)\ltimes F= Y\ltimes(X\ltimes F)$.)
I noticed something about sequences involving $\mathrm{Spin}(7)$.
First of all, unit imaginary octonions are square roots of negative one, so $L:\mathrm{Im}(\mathbb{O})\to\mathrm{End}_{\mathbb{R}}(\mathbb{O})$ which sends $u$ to left-multiplication-by-$u$ is "Clifford" and so extends to a representation of the Clifford algebra $\mathrm{Cliff}(\mathrm{Im}(\mathbb{O}))\to\mathrm{End}_{\mathbb{R}}(\mathbb{O})$, which restricts to a map $\mathrm{Spin}(7)\to\mathrm{SO}(8)$.
The point-stabilizer of any point in $S^7\subset\mathbb{O}\cong\mathbb{R}^8$ is $G_2=\mathrm{Aut}(\mathbb{O})$, and the point-stabilizer of any point in $S^6\subset\mathrm{Im}(\mathbb{O})$ is $\mathrm{SU}(3)$. Thus, we have some bundles
$$ \begin{array}{ccccc} G_2 & \to & \mathrm{Spin}(7) & \to & S^7 \\ \mathrm{SU}(3) & \to & G_2 & \to & S^6 \\ \mathrm{SU}(2) & \to & \mathrm{SU}(3) & \to & S^5 \end{array}$$
which combined with $\mathrm{SU}(2)\simeq S^3$ gives
$$ \mathrm{Spin}(7)=S^7\ltimes (S^6\ltimes (S^5\ltimes S^3)), $$
and when combined with the Hopf bundles $S^7=S^4\ltimes S^3$ and $ S^3=S^2\ltimes S^1$ becomes
$$ \begin{array}{lcl} \mathrm{Spin}(7) & = & (S^4\ltimes (S^2\ltimes S^1))\ltimes (S^6\ltimes(S^5\ltimes S^3)) \\ & \mathrm{or} & (S^4\ltimes S^3)\ltimes(S^6\ltimes(S^5\ltimes(S^2\ltimes S^1))). \end{array} $$
On the other hand, we have $\mathrm{Spin}(n)=S^{n-1}\ltimes\mathrm{Spin}(n-1)$ (where $\mathrm{Spin}(n)\to\mathrm{SO}(n)$ gives an action on $S^{n-1}\subset\mathbb{R}^n$ and we invoke orbit-stabilizer again), which means
$$ \mathrm{Spin}(7)=S^6\ltimes(S^5\ltimes(S^4\ltimes (S^3\ltimes (S^2\ltimes S^1)))). $$
So it looks like given any two sets of "composition factors" for $\mathrm{Spin}(7)$, they can made equivalent using the Hopf bundles. Is this a general principle or is there an obvious counterexample? That is, if something can be written as $S^{n_1}\ltimes\cdots\ltimes S^{n_k}$ in two different ways, can the (multisets of) composition factors be equated after using the Hopf bundles to replace $S^7$ with $S^4$ and $S^3$ etc.?
(Hopefully this is not too vague and sloppy so as to be cryptic, and not trivial.)


Gauss-Lucas Theorem (roots of derivatives)

Gauss-Lucas Theorem states:
"Let f be a polynomial and $f'$ the derivative of $f$. Then the theorem states that the $n-1$ roots of $f'$ all lie within the convex hull of the $n$ roots $\alpha_1,\ldots,\alpha_n$ of $f$."
My Question is:
Is there a theorem which states that there exists a permutation $\sigma \in S_n$ that the inner area of the polygon which edges go through the roots of $f$
$$\alpha_{\sigma(1)}\longrightarrow\alpha_{\sigma(2)}\longrightarrow\ldots\longrightarrow\alpha_{\sigma(n)}\longrightarrow\alpha_{\sigma(1)}$$ contains all roots of $f'$?

EDIT (OB) It is not completely clear from the original question wether the OP allowed for self intersections of the polygonal curve with vertices the roots of $f$. The question that has a bounty on its head asks for a polygonal Jordan curve with vertices the roots of $f$ containing the roots of $f'$ further assuming $f$ has simple roots. Roots of $f'$ are allowed to lie on the edges of the polygonal Jordan curve. We further assume $n\geq 3$ and that the roots of $f$ are not all aligned (i.e. not all contained in a real affine line.)

I found an interesting paper: arxiv.org/pdf/1405.0689

@HansEngler Indeed! The same happens if you consider $(X^2+1)X(X^2-1)(X^2-4)$ for instance.

It certainly can happen that some roots of $f'$ must lie on edges of any such polygonal Jordan curve. For example, take $f(z) = z^4 - z$ with roots $0$ and $a_i$, where the $a_i$ are roots of unity. Then the roots of $f'$ are $2^{-2/3}a_i$.

@Maesumi the OP certainly meant to write "whose edges". I might consider rewriting the question and including an example, the only problem is that I don't know how to draw plots. What the OP means by inner area I can explain in a few words: you have a closed polygonal path (in my question it is a Jordan path, so no self intersections) that links all roots of $f$, and there is a compact "inner area" and a non compact "outer area", the same way when you draw a circle, it defines a compact disk $\lbrace |z|\leq 1\rbrace$ and the non compact outer area $\lbrace |z|>1\rbrace$.

I think it might be better to re-write the question  from scratch and add an example or further explanation. what is "inner area" and what do you mean by "which edges"?

@LeonidKovalev you are right. I'll edit my edit ^^ thanks for your help.

@OlivierBégassat If the roots all lie on the same line, we can't have a polygonal Jordan curve through them. Maybe you wanted to say that the roots are in generic position: no three on the same line. And $n\ge 3$.

I started a bounty on this question. To get the reward, either give a counter-example, or produce a proof! Assume all roots to be distinct (this is not very restrictive, at least when it comes to producing a counter example, since we can modify the constant term of $f$ slightly, so that all its roots are now distinct, without affecting the position of its roots too much, and obviously not the roots of its derivative), and the polygon should be a polygonal Jordan curve.

Is the permutation required to be such that the edges do not cross? Otherwise it's hard to interpret the inner area. In any case, you have to allow the roots on the boundary of the polygon, as the case $n=2$ shows.


Number as the sum of digits of some degree

We will say that the measure of a number is equal to the maximum degree in which it is possible to represent a number in the form of a sum of digits copied (You can not rearrange the numbers). For example, for $55$ this will be $5$, because $$ 55^1 = 55, \quad 55 = 55$$ $$ 55^2 = 3025, \quad 30+25 = 55 $$ $$ 55^3 = 166375, \quad 1+6+6+37+5 = 55$$ $$ 55^4 = 9150625, \quad 9+15+0+6+25 = 55 $$$$55 ^ 5 = 503284375, \quad 5 + 0 + 3 + 28 + 4 + 3 + 7 + 5 = 55.$$
  Let $a_n$ be a sequence of numbers such that all smaller ones have a measure less than. What is the asymptotics of this sequence? Is it possible to somehow build numbers with a given measure? If not, what measures can not be built?

The task was put in a Russian forum, I put it a little different question, I will be glad to any help in its solution :)

We shouldn't have an immediate obstacle $\mod 9$, so interesting numbers are $0$ or $1$ modulo $9$. Assuming that, one would expect that the sum of digits of $n^k$ is about $\frac 9{2\log 10} k\log n$, so the power may be as large as $\frac {2n\log 10}{9\log n}$ and, if you are lucky, you can go up from here for numbers divisible by $10$ because the growth rate of the sum of the digits for them is slower. For the numbers in the table that do not end with $0$, this is an almost exact match. However, the third (and last) term in your sequence is formally $10$, whose measure is $+\infty$.

For Maximum, do you mean that all of the previous degrees have to have the same property? Or just the biggest degree with that property?

Yes, it is obvious that the problem for numbers ending with zero is uninteresting.

What is most interesting is that in the examples above the measure value coincides with the exact upper bound of the measure.

The Russian forum also presents some results, it seems to me amazing such large numbers on such relatively small results. $ n   \quad b(n) \\ 675 \quad  50 \\ 945 \quad 68 \\ 964 \quad 71 \\ 990 \quad 107 \\ 991 \quad 71 \\ 1296 \quad 84 \\  1702 \quad 114 \\ 2728 \quad 173 \\ 4879  \quad 285 \\ 5050 \quad 403 \\ 5149 \quad 300  \\ 5292 \quad 309 \\ $

I ask the answer to one of the questions. Do you think it's bad when there are so many of them?

All natural large ones are considered, the first term in the sequence is obviously $2$ with measure $1$, and the second term is 9, since $$ 9 ^ 2 = 81, \quad 8 + 1 = 9 $$


Does every finitely generated group have finitely many retracts up to isomorphism?

The infinite dihedral group $D_\infty = \langle a,b \mid a^2 = b^2 = \text{Id}\rangle $ is a finitely generated group with infinitely many cyclic subgroups of order 2, every one of which is a retract.

For the group $\mathbb{Z}\oplus\mathbb{Z}$, take $H_n=\langle (1,n)\rangle$ for any integer $n$ (with $K=\langle (0,1)\rangle$). Then we have  $\mathbb{Z}\oplus \mathbb{Z}=H_n \oplus K$ which shows that $\mathbb{Z}\oplus\mathbb{Z}$ (hence every finitely generated abelian group) has infinitely many different retracts.

Every free group $F_n$ of finite rank $n$ has infinitely many retracts. In fact, each free factor of $F_n$ is a retract and there are infinitely many free factors.

These are examples of finitely generated groups with infinitely many retracts. If we look at them, we'll find that they have finitely many retracts up to isomorphism. My question is that does every finitely generated group have finitely many retracts up to isomorphism?

@CheerfulParsnip You are right. I'll add free groups of finite rank to my list. Thanks very much for the comment.

You can add finitely generated free groups to your list of examples. Any retract of $F_n$ has rank $\leq n$. Maybe look for a counterexample in groups like Grigorchuk's Group.


A difficult integral for the Chern number

The integral
$$
I(m)=\frac{1}{4\pi}\int_{-\pi}^{\pi}\mathrm{d}x\int_{-\pi}^\pi\mathrm{d}y \frac{m\cos(x)\cos(y)-\cos x-\cos y}{\left( \sin^2x+\sin^2y +(m-\cos x-\cos y)^2\right)^{3/2}}
$$
gives the Chern number of a certain vector bundle [1] over a torus. It can be shown using the theory of characteristic classes that
$$
I(m) = \frac{\mathrm{sign}(m-2)+\mathrm{sign}(m+2)}{2}-\mathrm{sign}(m) = \begin{cases}1 & -2< m < 0 \\ -1 & 0 < m < 2 \\0 & \text{otherwise}\end{cases}.
$$
Is there any way to evaluate this integral directly (i.e. without making use of methods from differential geometry) to obtain the above result?
I should mention that the above integral can be written as ($1/4\pi$ times) the solid angle subtended from the origin of the unit vector $\hat{\mathbf{n}}$,
$$
I(m)=\frac{1}{4\pi}\int_{-\pi}^{\pi}\mathrm{d}x\int_{-\pi}^\pi\mathrm{d}y\, \hat{\mathbf{n}}\cdot\left(\partial_x \hat{\mathbf{n}}\times \partial_y \hat{\mathbf{n}}\right),
$$
where $\mathbf{n}(m)=(\sin x, \sin y, m- \cos x-\cos y)$. While this form makes it very straightforward to evaluate $I(m)$, I am interested in whether there is a way to compute this integral using more standard techniques.

[1] B. Bernevig Topological Insulators and Topological Superconductors Chapter 8

@TheSimpliFire done :) mathoverflow.net/questions/453343/…

@xzd209 I would suggest you crosspost this question on MathOverflow, ensuring the post includes a link to this MathStackExchange post.

@SetnessRamesory That case can probably be done with elliptic integrals but I still believe there must be an easier way to prove $I(m)=0\quad\forall|m|>2$, for instance. Due to the symmetry of the integrand and the exponent of $3/2$, an application of Green's theorem may be useful, similar to this approach.

@TheSimpliFire do you know$$\int_{0}^{\pi} \int_{0}^{\pi}  \frac{\text{d}x\text{d}y}{\sqrt{\sin(x)^2+\sin(y)^2+\left ( 1-\cos(x)-\cos(y) \right )^2 }  }=\frac{\Gamma\left ( \frac14 \right )^4 }{8\pi}?$$

The case $m=0$ is easy because we can rewrite as $-\pi I(0)=\int_0^\pi f(x)\,dx$ where $$f(x)=\int_0^\pi\frac{\cos x+\cos y}{(2+2\cos x\cos y)^{3/2}}\,dy$$ and we find that $f(x)=-f(\pi-x)$ for $x\in[0,\pi]$ after substituting $u=\pi-y$. So $I(0)=0$.

It appears there are two cases missing: $I(\pm2)=\mp1/2$.


Conjecture: No positive integer can be written as $a^b+b^a$ in more than one way

Today, I came up with the following problem when trying to solve this.

Are there distinct integers $a,b,m,n>1$ such that the equation $$a^b+b^a=m^n+n^m$$ holds? That is, is there ever an integer that can be written as $a^b+b^a$ in more than one way?

I claim that the answer is No, but I think solving this is beyond my knowledge. For a very preliminary observation, the simplest case is to consider the powers of $1,5,6,0$, since they end in those same digits. For example, $$\begin{cases}a\equiv5\pmod{10}\\b\equiv6\pmod{10}\end{cases}\implies a^b+b^a\equiv1\pmod{10}.$$ However, this brings about an issue, since there is hardly any indication as to what values $x$ and $y$ can take other than them having opposite parity.

PARI/GP code is
intfun(a,b,m,n)={for(i=2,a,for(j=2,b,for(k=2,m,for(l=2,n,if(i<>k && i<>l && j<>k && j<>l && i^j+j^i-k^l-l^k==0,print(i," ",j," ",k," ",l))))));}

No solutions have been found up to $a,b,m,n\le100$.

@ZachHunter I ran a code that checks for collisions as you suggested, up to $a, b \le 10000$, and I also ran it again for $a<70000$ and $b<500$. Did someone calculate asymptotics for this question? Can we say something along the lines "the function $f$ gets so big so fact that intuitively it's very unlikely that we'll find such number considering the bruteforces we ran for small numbers"?

Code could use updates:  like parfor, or using the fact that if $a,b$ are same parity, $m,n$ are going to need to be same parity as well. Or that The code as written checks $2^5+5^2$ and $5^2+2^5$ you're doing more than $4$ times as many checks as needed.

Maybe this could  help: $a^b+b^a=b^{\frac{b\ln(a)}{\ln(b)}}+b^a=b^a(b^M+1)$ where $M=\ln\left(\dfrac{\sqrt[\ln(b)]{b\ln(a)}}{e^a}\right)$

you can do a much broader search if you simply check if $f(a,b) = a^b+b^a$ is an injection for $(a,b) \in \mathbb{Z}^2, a \geq b \geq 2$. I was able to confirm this up for $a,b \leq 1300$ here repl.it/repls/OutstandingEarlyAutomaticvectorization. You would easily be able to search through higher values if you ran this locally on your computer instead of through repl, but I don't have access to that at the moment.

from my reconfiguring it, we get if any pair of them share a factor n, we can get a difference of nth powers on one side.

If $a | (b-1)$ and $a |(x-1)$ and $ a |( y-1)$ then there are no solutions (obviously).

$a^b-x^y=b^a-y^x$ etc mean they either pump out factors on both sides, or are all coprime.

Still no solution in the range $a,b,x,y\le 200$

We can accelerate the search by assuming $a\le b, x\le y, a\le x$


If two convex polygons tile the plane, how many sides can one of them have?

The set of convex polygons which tile the plane is, as of $2017$, known: it consists of all triangles, all quadrilaterals, $15$ families of pentagons, and three families of hexagons. Euler's formula rules out strictly convex $n$-gons with $n\ge 7$. (The pentagonal case is by far the most difficult one.)
I am interested in pairs of convex polygons that can collectively tile the plane. Specifically, I am curious how many sides can be in a polygon which is part of such a tiling.
Here are some conditions to impose on such a tiling, from weakest to strongest:

There is at least one copy of each tile. (Without this condition, one can trivially take a pair consisting of a tiling polygon and any other convex polygon, and just never use the latter shape.)

There are at least $k$ copies of each tile.

There are infinitely many of each tile.

Every tile borders a tile of the other type.

The tiling is $2$-isohedral, i.e., every tile can be carried to any other tile of the same shape by a symmetry of the tiling.

Each of these conditions implies those above it.
In the weakest case, the number of sides is unbounded, as exhibited by the following example:

(The tiling is constructed by decomposing "wedges" of central angle $2\pi/N$ into congruent isosceles triangles, and then combining the central triangles to yield an $N$-gon in the center.)
Requiring at least $k$ of each tile still yields arbitrarily high numbers of sides, by taking the above construction for $N=M\cdot k$ and subdividing the $N$-gon into $k$ "wedges" which are $(M+2)$-gonal.
On the other end of the spectrum, I have found a $2$-isohedral tiling using regular $18$-gons, shown below:

After consulting this paper, it seems that the tiling pictured above is of type $4_2 18_{12}-1\text{a}\ \text{MN}\ \text{p}6\text{m}$ in their classification scheme (shown at the bottom of page 109); there are no $2$-isohedral tilings which allow for any higher number of contacts between different shapes, although type $3_1 18_{12}-1\text{a}\ \text{MN}\ \text{p}6\text{m}$ also works (and can be obtained from the above construction by cutting each kite-shaped tile in two). Thus, it is maximal among $2$-isohedral tilings.
What are the maximal tilings under weaker conditions? The maximal number of sides under each successively stronger restriction is a weakly decreasing sequence which goes $\infty, \infty, ?, ?, 18$. So far, I have no bounds on the missing two terms except that they are each at least $18$.
Some notes on this problem:

It is not necessarily the case that one of the tiles may tile the plane on its own; see this math.SE question for a counterexample.

If convexity is relaxed for either piece, the number of sides is unbounded even in the $2$-isohedral case (in fact, both pieces can simultaneously have arbitrarily many sides).

Edit: Crossposted to Math Overflow here.


Progressive Dice Game

$(2019.)$ Edit: Rewriting the question to make it clear.

The progressive dice game

At the start, you have a fair, regular six sided dice $D=(1,2,3,4,5,6)$.
The game is played for $n$ turns. Each turn you make a roll, which will be $r\in D$.
Then to complete the turn, you make one of the following choices:

Bank the rolled number: You gain $r$ points (score).
Invest in (upgrade) the dice "evenly": If $r\lt 6$, choose $r$ sides and increase them by $1$ each. If $r\ge 6$, that is, $r=6k+k_0,k_0\lt 6$, then increase each of the six sides by $k$, and then choose $k_0$ sides and increase them by $1$ each.
Reroll the dice, effectively restarting this turn. But before rerolling, you must apply the penalty to the dice: "Evenly" downgrade the dice: If $S_0$ is the number of sides $\gt 0$ on the dice, then: If $r\lt S_0$, choose $r$ sides and decrease them by $1$ each. If $r\ge S_0$, that is, $r=S_0k+k_0,k_0\lt S_0$, then decrease each of the $S_0$ sides by $k$, and then choose $k_0$ sides and decrease them by $1$ each.

What is the optimal way to play to maximize your expected score at the end of the game?

If the dice was allowed to be upgraded/downgraded arbitrarily (not "evenly"), then one could downgrade the first five sides until they reach $0$. These sides now act as free rerolls. Then, keep investing the remaining points into the sixth side, which is now guaranteed to be rolled on each turn, after some amount of rerolls of that turn. Finally, bank that sixth rolled side in the last couple turns to maximize the expected value of the score.
But since we must upgrade/downgrade evenly, I'm not sure what is the optimal strategy.

If we ignore the "reroll" move:
If you upgrade the first $t$ turns, then bank the rest of the turns, you will expect the following amount of points on average:
$$ f(t) = 3.5\times\left(\frac{7}{6}\right)^{t}\times(n-t)$$
Which boils down to, that if you want to maximize your expected score, you should upgrade until the last $6$ (or $7$) turns and then bank those turns.
But this approach completely ignores the third action; the rerolls.

Can we do better than this strategy, if we use the rerolls somehow?

Rerolls?
I haven't worked out the strategy if the rerolls are considered.
A reroll will on average decrease the average value of the dice, and allow you to either improve or worsen your current turn, with equal probability on average?
But there seem to be exceptions? For example, rerolling a $1$ seems useful if used early (as later, if we had upgraded a lot, all sides will be much greater than $1$). Simply downgrade the rolled $1$ side when downgrading (rerolling). If you roll that side again, it will be $0$, and this allows you to again reroll the dice for free (downgrading $0$ points is a free reroll). Which means, choosing to downgrade when you roll a $1$, can only increase your expected score in that turn. But there is still a (small?) drawback: Lets say some other side is a number $\ge 6$. Then when upgrading later, you will have to put back at least one of those upgrade points into that downgraded $0$ side.

Seems to me that rerolls will decrease the expected score on average
  (as they decrease the average value of the dice), so it is always
  better not to use them (except in that early scenario of the game, if $n$ is small)? Is this true?

For example, for small $n$, rerolls can be useful to force larger values. For $n=1$ specifically, it seems we can always force the first turn (the only turn) to end up banking a $6$, the maximal possible score for $n=1$, by rerolling and downgrading strategically the rest of the sides if $6$ was not rolled.
But for large $n$, the rerolls seem to lower the average expected score at the end, if used anytime in previous turns, as larger upgrades will need to replenish those downgraded points inevetably at some point, as they are carried out "evenly".

As he described you could decrement 1 to 0 but the next time you decrement you have to decrement any other number before you can decrement the 1 which is 0 now

I think the increment/decrement rule can be described thusly:  If you roll a $k$, then you add/subtract $1$ to/from the smallest/largest number currently on the die, and repeat this action $k$ times.  Or do you want the player to be able to choose the sides whose numbers get changed?  (If so, then any time you roll a $1$, you should decrement that side to a $0$, which effectively gives you a free roll.)

@joriki Yes, a reroll is a new turn, more precisely it is starting your turn over, but with a bit downgraded dice. Yes, by maximise I tought of maximising the expected value, or in other words the average value over repeated games. Yes it "starts over", downgrade points work the same way as upgrades, but the zero field is simply not being looked at. The point is that the points are evenly distributed, not for example stacked at one side, and then being taken from the rest of the sides leading to a $(0,0,0,0,0,x)$ dice with $x$ being doubled each turn, and zeroes used as free rerolls.

Your formula seems right, but would be a bit easier to understand if you write it as $3.5\cdot\left(\frac76\right)^t(n-t)$.

And for the downgrade: Does this also start over once you've taken a point off all sides? If there are sides with $0$, does the downgrade start over once you've taken a point off all non-zero sides, or do the zero sides count towards the points despite not being further downgraded?

You can't maximise your score, since this is a random variable. I suspect what you mean is to maximise the expected value of the score?

And in the downgrade you remove as many points as you'd rolled, and then effectively get a new turn, in which you have a new choice whether to bank, upgrade or downgrade?

@mjqxxxx You do one side by one point, then repeat for another side, till you use up all your points that were rolled, if you do all sides and have points left to spent, you do the same with the remaining points again. (Example; rolled $4$ = do $4$ sides, each by $1$ point, Example; $8$ = do all sides by one point, then do two sides by one point)

Can you be more specific about the meaning of "[increment] its sides distributing the points one by one" and "downgrade the dice evenly taking one by one point"?  Are you incrementing/decrementing all sides by one?  Just the smallest/largest side?


Is $\Phi(q)$ rational for some $q \in \mathbb{Q}^*$, where $\Phi$ is the standard normal cumulative distribution function?

Suppose that we have rational numbers $q_1$, $q_2$ such that
$$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{q_1}e^{-\frac{t^2}{2}} \,\mathrm{d}t=q_2.$$
Does this imply that $q_1=0$ and $q_2=\dfrac{1}{2}$?

We know that $ q_{2} = \frac{1}{2} \Big( 1 + \text {erf} \Big( \frac{ q_{1} }{ \sqrt{2} } \Big) \Big)$ and  $ \text{erf} (x) = 1 - \frac{1}{ \sqrt{ \pi }} \Gamma \Big( \frac{1}{2} , x^2 \Big)$, so for $ q_{2} $ to be rational $ \text {erf} \Big( \frac{ q_{1} }{ \sqrt{2} } \Big) $ must be zero. Therefore $ q_{1} = 0$ and $q_{2} = \frac{1}{2}$.

I suspect $q_1$, $q_2$ can be even algebraic instead of rational (like in Lindemann-Weierstrass theorem showing that $\sin x$ is transcendental)

Two different questions here: (a) $\Phi(\Phi^{-1}(1/4)) = 1/4$ and $q = \Phi^{-1}(1/4) \approx -0.6744898$ is a number with rational $\Phi(q),$ with $q \ne 0$, so Yes to the title. (b) However, I would not want to claim that $q$ is rational, so the main question remains unanswered.


Open problems in Federer's Geometric Measure Theory

I wanted to know if the problems mentionned in this book are solved. More specifically, at some places, the author says that he doesn't know the answer, for example :"I do not know whether this equations are always true" p.361 4.1.8, or "I do not know..." p.189 2.10.26. Are there counterexamples or proofs of these asumptions?
I add some details : in theorem 2.10.25, if $f:X\rightarrow Y$ is a lipschitzian map of metric spaces, $A\subset X$, $0\leq k<\infty$, and $0\leq m<\infty$, then
$$
\int_Y^*\mathscr{H}^k(A\cap f^{-1}\{y\})d\mathscr{H}^m(y)\leq (\mathrm{Lip} f)^m \frac{\alpha(k)\alpha(m)}{\alpha(k+m)}\mathscr{H}^{k+m}(A).
$$ (where $\mathscr{H}^n$ is the Hausdorff measure associated to the metrics on $X$ and $Y$, and $\int^*$ is the upper integral) provided either $\{y\in Y, \mathscr{H}^k(A\cap f^{-1})>0\}$
 is the union of countable family of sets with finite $\mathscr{H}^m$ measure, or $Y$ is boundedly compact (each close bounded subset is compact). The question Federer asks is to determine if everything after "provided either..." is necessary.
The other question is about currents : let $S$, $T$ be two currents on open subsets $A$,$B$ of euclidean spaces, of degree $i$ and $j$ : if $S$, $T$ are representable by integration, do we always have $\Vert S\times T\Vert=\Vert S\Vert\times\Vert T\Vert$, and $\overrightarrow{S\times T}(a,b)=(\wedge_i p)\overrightarrow{S}(a)\wedge(\wedge_j q)\overrightarrow{T}(b)$ for $\Vert S\times T\Vert$ almost all $(a,b)\in A\times B$, where $p:A\rightarrow A\times B$ and $q:B\rightarrow A\times B$ are the canonical injections (it is indeed the case if either $\overrightarrow{S}$ or $\overrightarrow{T}$ is simple $\Vert S\Vert\times\Vert T\Vert$ almost everywhere).

It seems it has been proved in Measure theory and fine properties of functions, revised edition, by Evans.

Thanks, it would surprise me if there were no counter example or proof already available.

I'm a bit surprised, I have seen some discussion of Federer's text here before. Keep an eye on this from time to time, it may get an answer later.

Thank you anyway, I guess I'll have to find by myself. I really appreciate your help (I had no result either on math overflow).

I'll put a bounty on your question here once it's possible.

Thank you Mr Cook, I asked the question on math overflow and added the same détails here.

I wonder if you might get better input from math overflow on this one.


(Weil divisors : Cartier divisors) = (p-Cycles : ? )

Suppose $X$ verifies the suitable conditions in which Weil (resp. Cartier) divisors make sense.
The group of Weil divisors $\mathrm{Div}(X)$ on a scheme $X$ is the free abelian group generated by codimention $1$ irreducible subvarieties.
The sheaf of Cartier divisors is $\mathrm{Cart}_X:=\mathcal{K}_X^{\times}/\mathcal{O}_X^{\times}$.
The group of Cartier divisors is $\mathrm{Cart}(X):=\Gamma(X,\mathrm{Cart}_X)$.
The group $Z^p(X)$ of $p$-cycles is the free abelian group generated by irreducible subvarieties of codimension $p$, so in particular $\mathrm{Div}(X)=Z^1(X)$. So the notion of $p$-cycle is a direct generalization of the notion of Weil divisor.
My question:

Is there an analogous notion of group of "Cartier $p$-cycles" $\mathrm{Cart}^p(X)$? If yes, is there a sheaf $\mathrm{Cart}^p_X$ such that (naturally in $X$) we have $\mathrm{Cart}^p(X)=\Gamma(X,\mathrm{Cart}^p_X)$?

As Matt said we can consider ideal sheaves locally generated by $p$ element. A variety of codimension $p$ satisfies this is called local complete intersection, so we don't get all $p$-cycles, but certainly we get the smooth ones since by the Jacobian criterion they smooth (sub)varieties are lci.

One naive attempt would be to say that Cartier divisors are "locally principal." To get a codimension p object we'd need something like "locally the zero set of p things" which we could take to mean, lci or something. I have no idea how to make a sheaf that describes that, though.


Free medial magmas

A medial magma is a set $M$ with a binary operation $*$ satisfying $$(a*b)*(c*d) = (a*c)*(b*d)$$ for all $a,b,c,d \in M$. Medial magmas constitute a finitary algebraic category $\mathsf{Med}$, therefore there is a functor $M : \mathsf{Set} \to \mathsf{Med}$ which sends a set $X$ to the free medial magma $M(X)$ over $X$. Elements of $M(X)$ can be seen as equivalence classes of oriented non-empty binary trees whose leaves are marked with elements of $X$, where two such trees are equivalent if the one can be reached from the other by a finite number of steps, where each steps looks as follows:

Now I wonder if there is a more explicit description of the underlying set of $M(X)$, or a specific system of representatives. Can we simplify it? For example, when $X=\{\star\}$, we have no markings; what is a specific system of representatives? I also wonder if these free medial magmas are studied or used anywhere in the literature.

The commutative version has been studied under the name of "level algebras", see for example arxiv.org/pdf/math/0209363v3.pdf.

Me neither. In the introduction to the second paper I linked there are a few remarks on free medial groupoids. There is something to the effect that "It seems that there is no very nice description of the equational theory of medial groupoids and free medial groupoids." I can only see the first two pages, so I don't know if there is more in later sections.

Thank you. In fact this gives more search results. But unfortunately I couldn't find a paper which discusses free medial groupoids.

It seems that searching for "medial groupoid" instead of "medial magma" yields more results (in universal algebra groupoid is synonymous with magma). For instance, you can find the work of J. Ježek, T. Kepka and others who produced a number of papers on related structures and their representations, e.g. here (where the free cancellative and the free commutative medial magma are constructed) or here.


Is the Frog game solvable in the root of a full binary tree?

Frog game
The Frog game is the generalization of the Frog Jumping (see it on Numberphile) that can be played on any graph, but by convention, we restrict the game to Tree graphs (see wikipedia).
The game is simple to play, but it can be hard to determine if it is solvable in a given vertex.

In short, given a graph, the goal of the game is for all frogs to host a party on a single vertex of that graph. Initially, every vertex has one frog. All of the $f$ frogs on some vertex can "jump" to some other vertex if they both have at least one frog on them and if they are exactly $f$ edges apart from each other. If a "sequence of jumps" exists such that all frogs end up on a single vertex, then the party is successfully hosted on that vertex and we call that vertex a "lazy toad" because the frog that started there never jumped.

Formal rules
Let $v,w,u\in V$ be vertices of a graph $G=(V,E)$ and $n=|V|$ the number of vertices.
Let $f_m:V\to\mathbb N$ count the number of frogs on a given vertex where $m$ is the current declared move. A game takes $(n-1)$ moves to solve (if possible) and is played as follows:

We start the game on move zero $m=0$ where each vertex has one frog $f_0(v)=1,\forall v\in V$.

If it is move $m\lt n-1$ and the following conditions are met:

There exist two vertices $v,w\in V$ that have frogs on them $f_m(v)\ge 1$, $f_m(w)\ge 1$.
There exists a path from $v$ to $w$ containing exactly $f_m(v)$ unique edges.

Then, a legal move (or jump) can be made and is denoted as $(v\to w)$. If the move is made, then the following transitions occur:

All frogs jump from $v$ to $w$, that is, $f_{m+1}(v)=0,f_{m+1}(w)=f_m(v)+f_m(w)$.

The remaining frogs don't move, that is, $f_{m+1}(u)=f_{m}(u),\forall u\not\in\{v,w\}$.

We declare the next move $m+1$. (The game ends if we can't move.)

We say that the game is solvable in some vertex $w$ if there exists a sequence of legal moves such that we end up moving all frogs to the vertex $w$, which means $f_{n-1}(w)=n$. Consequently, if the game is solvable in $w$ then $f_{n-1}(v)=0,\forall v\ne w$. The solvable vertex $w$ is called a "lazy toad" because the frog on the vertex $w$ never jumped during the game.

For some examples, you can visit a post about this game on mathpickle. There is also an included link to a google drive where all trees with less than $15$ vertices have been computationally solved and categorized by the number of vertices then by the number of non-solvable vertices.
Remark. I've asked a general question about this game before Frogs jumping on trees, which asks if we can characterize solvable and non-solvable vertices of certain sets of graphs, but that appears to be a hard problem. Here, I've decided to restrict the problem to binary trees and only observe the root vertex.

Question
Let $T_h=(V,E)$ be a full binary tree of height $h$. This means that it has layers $0,1,2,\dots,h$ where the root $r\in V$ is on layer $0$. That is, we have $|V|=2^{h+1}-1$. Let $h\in \mathbb N$ because $h=0$ is a single vertex which is trivial.

I would conjecture that for all $h\ge 4$, every vertex of $T_h$ is a "lazy toad" (is solvable).

However, for simplicity, my question is just about the root vertex:

Let $h\in\mathbb N, h\ne 2$. Can we prove that the root $r$ of every such $T_h$ is a "lazy toad" (is solvable) ?

Notation
Let $v(i,j)\in V$ be the $j$th non-root vertex on the layer $i=1,2,\dots,h$ where $j=1,2,\dots,2^i$.
Lets declare a shorthand notation for moves:

"$(v\to w)$ then $(w\to u)$" to be equivalent to "$((v\to w)\to u)$".

"$(v\to w)$ and $(u\to w)$" to be equivalent to "$(v\to w \leftarrow u)$".

For clarity, let  $f=f_m(v)$ in "$\left(v\xrightarrow{f} w\right)$" stand for the number of frogs being moved.
Alternative suggestions for notation are welcome.

Examples (Reducing $T_3,T_4,T_5$ to $T_1$)
The fact that we can reduce examples to smaller examples makes me think that maybe induction is possible. Notice that $T_1$ was solved in the original version of the game where all vertices are on a line and that $T_2$ is not solvable, so $T_3$ is the first "real" case.
$$(h=1)$$ The trivial case $T_1$ is trivially solvable in all vertices:

[1.1] In $r$ using legal moves $v(1,1)\xrightarrow{1} r$ and $v(1,2)\xrightarrow{1} r$ in any order.
[1.2] In $v(1,1)$ by using moves $\left(r\xrightarrow{1} v(1,2)\right)\xrightarrow{2} v(1,1)$.
[1.3] In $v(1,2)$ by using moves $\left(r\xrightarrow{1} v(1,1)\right)\xrightarrow{2} v(1,2)$.

$$(h\ne 2)$$ The second case $T_2$ is not solvable in the root. This can be checked by listing all move sequences.
$$(h=3)$$ Lets break down the $T_3$ case. This case can be looked as a "base" trivial $T_1$ case that has two trivial $T_1$ cases "connected" to every leaf vertex. Let the vertices of those four "connected" trivial cases $T^{(1)}_1,T^{(2)}_1,T^{(3)}_1,T^{(4)}_1$ be indexed as: $v_1,v_2,v_3,v_4$ and let the vertices of the "base" trivial $T^{(0)}_1$ case be indexed as $v_0$.
That is, we can observe the $T_3$ as "composition" of $T_1$'s:

Now we can see that the solution in the root of $T_3$ follows from the solution of $T_1$. That is:

solve the "connected" $(T^{(i)}_1,i\in\{1,2,3,4\})$'s in a leaf vertex using [1.3]: $$\left(r_i\xrightarrow{1} v_i(1,1)\right)\xrightarrow{2} v_i(1,2),$$

then do the moves $v_i(1,2)\xrightarrow{3} r_0$ for $i\in\{1,2,3,4\}$ to transfer those frogs to root $r_0$,

then finally just solve the base $T^{(0)}_1$ with [1.1]: $v_0(1,1)\xrightarrow{1} r_0$ and $v_0(1,2)\xrightarrow{1} r_0$.

That is, the $T_3$ case is solved using the solutions of the trivial $T_1$ case.
In other words, if we know the solutions of $T_1$, we can reduce the $T_3$ to $T_1$ again.
$$(h=4)$$ Similarly as in the previous case, we can observe $T_4$ as "composition" of two connected $T_2$'s to each of the two leafs of one "base" $T_1$. That is, we "reduce" the $T_4$ to $T_1$ by "solving" the four $T_2$'s by moving all frogs from their vertices to the root of the "base".
We can "reduce" ("solve") $(T^{(i)}_2,i\in\{1,2,3,4\})$'s using following two move sequences:

$v_i(2,1)\xrightarrow{1} v_i(1,1)$ and $v_i(2,2)\xrightarrow{1} v_i(1,1)$, then $\left(v_i(1,1)\xrightarrow{3} v(2,4)\right)\xrightarrow{4} r_0$.

$r_i\xrightarrow{1} v_i(1,2)$ and $v_i(2,3)\xrightarrow{1} v_i(1,2)$, then $v_i(1,2)\xrightarrow{3} r_0$.

The vertices from the first (second) move sequence are colored blue (green) here in the $T_2^{(i)}$:

After the reduction, we are left with the "base" $T_1$ which is solved with [1.1] from the trivial case. In other words, we essentially moved all frogs from layers $2,3,4$ of $T_4$ to the root, leaving us only with layers $0,1$ which is equivalent to $T_1$.
$$(h=5)$$ Similarly to previous two cases, this $T_5$ case can be reduced to $T_1$ by solving four $T_3$'s. That is, we can reduce the layers $2,3,4,5$ by moving all their frogs to the base root. To do this, there are three separate move sequences highlighted in the following image (each puts $5$ frogs in a leaf vertex before moving the frogs to the base root):

$\left(\left(v_i(2,1)\xrightarrow{1} v(3,1)\right)\xrightarrow{2} v_i(1,1)\xleftarrow{1} r_i\right)$, then $\left(v_i(1,1)\xrightarrow{4}v_i(3,8)\right)\xrightarrow{5}r_0.$

$\left(v_i(3,3)\xrightarrow{1} v_i(2,2)\leftarrow{1} v_i(1,2)\right)$, then $\left(\left(v_i(2,2)\xrightarrow{3} v_i(1,2)\right)\xrightarrow{4}v_i(3,2)\right)\xrightarrow{5}r_0.$

$\left(v_i(3,5)\xrightarrow{1} v_i(2,3)\xleftarrow{1} v_i(3,6)\right)\xrightarrow{3} v_i(3,7)$, then $\left(v_i(2,4)\xrightarrow{1}v_i(3,7)\right)\xrightarrow{5}r_0.$

$$(h\ge 6)$$ These examples so far motivate the following question:

Is the reduction from $T_h$ to $T_1$ by solving four $T_{h-2}$'s always possible?

This could be one way to solve the problem. On the other hand, are there any other reductions?

Reduction pattern based on $T_6,T_7$
Since reductions from $T_h$ to $T_1$ over $T_{h-2}$'s seem to get more complex as $h$ grows, this begs the question if there are any other simpler reductions.
It is sometimes possible to reduce $T_h$ to $T_{h-2}$ over $T_1$'s. Take a look at the following examples.
Let $k=0,1,2,\dots,2^{h-2}-1$. If we look at the $T_6$, we can reduce its top two layers (reuducing it to $T_{4}$) using the following:

$v(6,4k+1)\xrightarrow{1} v(5,2k+1)$ and $v(6,4k+2)\xrightarrow{1} v(5,2k+1)$,

$\left(v(5,2k+2)\xrightarrow{1} v(6,4k+3)\right)\xrightarrow{2} v(6,4k+4)$,

$\left(v(5,2k+1)\xrightarrow{3} v(6,4k+4)\right)\xrightarrow{6} r$.

Similarly, we can reduce the $T_7$ to $T_5$ by using steps 1., 2. from above but modifying step 3. as (Note that you also modify the layers from $5$ to $6$ and $6$ to $7$ in all steps, of course.)

$\left(v(7,4k+4)\xrightarrow{3} v(6,2k+1)\right)\xrightarrow{6} r$.

We can ask if there are any other trees $T_h$ whose top two layers can be reduced.
Notice that we are collecting $T_1$'s in this reduction, that they contain $3$ vertices (frogs) each and that the total number of used $T_1$'s in our move sequence(s) must be a power of $2$ to completely reduce the top two layers, where we have $2$ possible layers $\{h,h-1\}$ to put the frogs on before jumping to root. Because of this, it is meaningful to try to extend this pattern to the full binary trees of the following kind:

If $h\in\{3\cdot2^t,3\cdot2^t+1\},t\in\mathbb N$, for which $t$'s is such $T_h$ reducible to $T_{h-2}$?

This now does smell like something that would be used in an inductive proof. But, even if this pattern holds for all $t$, it alone is not enough. The question is, can we find a sufficient set of reduction patterns?

Let $s\lt h$ and $h\gt 2$. Is it true that every $T_h$ can be reduced to some $T_s$ ?

Trivial reduction pattern

If $h=2^t+t-2,t\ge 2$ and $T_{t-1}$ is solvable in root, then $T_h$ can be reduced to $T_{h-t}$.

This is trivially true because the distance of the root of "$T_{t-1}$ that spans the top $t$ layers" to the "(base) root $r=r_0$", is precisely equal to $2^t-1$, which is the total number of frogs in $T_{t-1}$.
For example, if $t=4$ we have $h=18$. Solving top $t=4$ layers as a $(T_{t-1}=T_{3})$'s in root, puts $15$ frogs on a vertices that are on layer $15$. These frogs can jump to the root of $T_h$ which leaves us with $T_{h-t}=T_{14}$.
Both this and the previous reduction pattern are exponential and are not enough to cover all values of $h$ in the inductive argument. That is, we need more statements like this to cover all values.

Other examples
I've managed to find reductions for $T_h$ for $h$ up to $20$ by hand to show they are all solvable in the root, but I still don't see how I could complete the inductive argument or solve this problem in some other way for all $h\ne 2$.
For example, I can reduce $T_8,T_9,T_{10},T_{11}$ to $T_5,T_5,T_6,T_7$ respectively.
Then, $T_{12},T_{13}$ belong to $t=2$ from the above reduction pattern for $h\in\{3\cdot2^t,3\cdot2^t+1\}$ and can be reduced to $T_{10},T_{11}$. It is not as easy as $t=1$, but the move sequences do exist.
After those, I can reduce $T_{14},T_{15},T_{16},T_{17}$ to $T_{11},T_{11},T_{12},T_{13}$.
Then, $T_{18}$ belongs to $t=4$ of the trivial reduction pattern which means it can be reduced to $T_{14}$.
For $T_{19}$ I found a reduction to $T_{14}$,... and so on.
Note that these reductions are not the only ways to solve a tree in the root.

Necessary reduction condition
Notice that to reduce $T_h$ to $T_{h-a}$ for some $a\ge 2$, we need to move all frogs from top $a$ layers to the root. The number of frogs we are moving to the root in such reduction is $(2^a-1)2^{h-a+1}$. That is, we are shaving off the $T_{a-1}$'s at the top layers. For such a reduction to be possible, there must exist a set of moves that transfers the frogs to the root $r$
$$M_r=\{v(i,j)\xrightarrow{f_{i,j}}r\},\text{ where } i\in\{h,h-1,\dots,h-a+1\}.$$
Assuming such moves exist at some point in the game, they are legal only if $f_{i,j}=i$ where $i$ is the layer on which the vertex $v(i,j)$ is on. On the other hand, the sum of $f_{i,j}$'s must be equal to the total number of frogs that we are reducing (moving to root). That is, for $M_r$ to be able to exist, we necessarily need to be able to partition the number of frogs we are moving onto the corresponding layers, and this is essentially what we state in the necessary condition.
We also care if our move sequences repeat or have patterns (for example, see the use of $k$ in the $T_6,T_7$ reduction pattern), because then instead of considering all frogs at once, we can consider only $(2^a-1)2^{b},b\in[0,h-a+1]$ frogs for some $b$, and repeat the same move pattern $(2^{h-a+1}/2^b)$ times.
Based on all of this, we can state the following necessary condition:

(Necessary condition for reduction). Let $h\gt2,a\ge2$. If reduction from $T_h$ to $T_{h-a}$ is possible, then there exists $b\in[0,h-a+1]$ and a partition of the number $(2^a-1)2^b$ into the layers $H=\{h,h-1,h-2,\dots,h-a+1\}$ where $(h-l)\in H$ is used at most $(2^{a-l-1})2^b$ times.

Of course, the converse does not hold because this is only a necessary (not a sufficient) condition.
For an example, if we want a $T_h$ to $T_{h-2}$ reduction then we need to partition $3\cdot 2^b$ into $\{h,h-1\}$. If we assume that the solution is "simple", i.e. that all frogs are either on layer $h$ or layer $h-1$ (but not mixed between both layers), we get the following pattern:

$h$ must either be of form $3\cdot 2^t$ or $3\cdot 2^t+1$. This was essentially our assumption on $h$ when we talked about extending the $T_6,T_7$ pattern. But now we also need a sufficient condition. That is, if we want to prove this is possible for every $t$, then we need to for every $t$ find a sequence of corresponding legal moves or prove it exists.

Solvable partitions?
Surprisingly, the above condition is already very useful (efficient). I have that for all $h\in(2,20)$ I've solved so far, I needed to consider only $31$ or less vertices ($a\le 5,b\le 3$) to show that all vertices can be reduced to the root (that we can send all frogs to the root).
For example, the $T_{19}$ has over a million vertices! But, to solve it, one needs to consider just one $T_4$ (which has only $31$ vertices) at the top layers to find a solution! (Consider $a=5,b=0$ and the corresponding partition $31=15+16$ is solvable.) That is, the reduction pattern of that $T_4$ can simply be repeated on all adjacent $T_4$'s in the top five layers, which gives us a reduction to $T_{14}$ which we have already solved by reduction to $T_{11}$ then to $T_5$ then to $T_1$.
Essentially, we need to prove that at least one of the partitions given by the "(Necessary condition for reduction)" always allows the reduction to a smaller binary tree (that the converse holds for at least one partition). Then by induction, we have that they are all reducible to $T_1$.
I don't yet see how to construct move sequences for all cases of $h$. Alternatively, I do not know if we can maybe prove it without explicit construction of solutions?

Since the bounty didn't lead to anything new, its now crossposted to mathoverflow.net.


Is there an irrational number containing only $0$'s and $1$'s with continued fraction entries less than $10$?

The number $$0.10111001001000000000001$$ has continued fraction $$[0, 9, 1, 8, 9, 5, 1, 1, 5, 3, 1, 3, 1, 1, 4, 6, 1, 1, 8, 2, 5, 8, 1, 9, 9, 5, 2
, 8, 1, 1, 6, 4, 1, 1, 3, 1, 3, 5, 1, 1, 5, 9, 8, 1, 9]$$
So, the maximum values is $9$. But we are only at $23$ digits. Can we produce larger decimal expansions with the required property ? Perhaps arbitary large ones ?

Is there an irrational number, such that the decimal expansion contains only ones and zeros and the continued fraction contains no entry larger than $9$ ?

A shorter example without long sequences is $$0.1011010011100100111101101110101$$

But I would prefer sequences without such long $0$-sequences, if possible :)

(+1) An interesting question on the intersection of Cantor-like sets.


Increasing derivatives of recursively defined polynomials

Consider recursively defined polynomials $f_0(x) = x$ and $f_{n+1}(x) = f_n(x) - f_n'(x) x (1-x)$.
These polynomials have some special properties, for example $f_n(0) = 0$, $f_n(1) = 1$, and all $n+1$ roots of $f_n$ are in $[0,1)$.
Let $x_n$ denote the largest root of $f_n$. Then $f_n(x_n) = 0$ and $f_n'(x_n)>0$. Moreover, $x_n > x_{n-1}$ for all $n$.
I want to prove the following claim: $f_{n}'(x_{n+1}) > f_{n-1}'(x_{n+1})$ for all $n \geq 2$.
Note that the claim does not hold for arbitrary $x$. The derivatives are polynomials themselves, by Gauss-Lucas theorem all their roots are in $[0,x_n)$ and there are many points where $f_{n}'(x) < 0 < f_{n-1}'(x)$. However, I am quite sure that at $x \geq x_{n+1}$, the derivatives are ordered: $f_1'(x) < \dots < f_n'(x)$.
Some of the first polynomials are:

$f_1(x) = x^2$, $f_1'(x) = 2x$, $x_1 = 0$
$f_2(x) = 2x^3 - x^2$, $f_2'(x) = 6 x^2 -2x$, $x_2 = \frac{1}{2}$
$f_3(x) = 6 x^4 - 6x^3 + x^2$, $f_3'(x) = 24x^3-18x^2+ 2x$, $x_3 \approx 0.7887$
$f_4(x) = 24 x^5 - 36 x^4 + 14 x^3 -x^2$, $f_4'(x) =120x^4-144x^3 +42x^2 -2x$, $x_4 \approx 0.9082$

Therefore $f_2'(x)-f_1'(x) = 6x^2-4x \geq 0$ for all $x \geq \frac{2}{3}$. Note that, $x_2 < \frac{2}{3} < x_3$.
For $f_3'(x) - f_2'(x) = 24 x^3-24 x^2 + 4 x \geq 0$ for all $x \geq 0.7887$. Here it turns out that at $x_3$ the inequality holds as an equality (coincidence perhaps?), but of course then for $x_4 > x_3$ it holds as a strict inequality.

That's right, the claim does not hold for $n=1$. What I wanted was for all $n \geq 2$ (in my application $f_0$ is irrelevant, I'm just using it for construction). Sorry for the confusion. I'm updating the question.

The claim is false since $f'_1(x_2)=f'_0(x_2)$.

Just a conjecture, coming from looking at the plots for the first 5-6 cases.

Why are you quite sure the derivatives are ordered at $x \geq x_{n+1}$?


Calculate using residues $\int_0^\infty\int_0^\infty{\cos\frac{\pi}2(nx^2-\frac{y^2}n)\cos\pi xy\over\cosh\pi x\cosh\pi y}dxdy,n\in\mathbb{N}$

Q: Is it possible to calculate the integral
  $$
\int\limits_0^\infty \int\limits_0^\infty\frac{\cos\frac{\pi}2
\left(nx^2-\frac{y^2}n\right)\cos \pi xy}{\cosh \pi x\cosh \pi y}dxdy,~n\in\mathbb{N}\tag{1}
$$
  using residue theory?

For example, when $n=3$
$$
\int\limits_0^\infty \int\limits_0^\infty\frac{\cos\frac{\pi}{2}
\left(3x^2-\frac{y^2}{3}\right)\cos \pi xy}{\cosh \pi x\cosh \pi y}dxdy=\frac{\sqrt{3}-1}{2\sqrt{6}}.
$$
There is a closed form formula to calculate (1) for arbitrary natural $n$, but I don't know how to do it by residue theory. Maybe it is possible in principle, but is residue theory practical in this particular case? It seems such an approach would lead to a sum with $O(n^2)$ terms. Any hints would be appreciated.

You mentioned that there is a closed form formula. Would you care to provide it?

@AlfredYerger I know $I_{1,2}$ can be evaluated with residue theory. It has been done by Ron Gordon, for exampe, here on MSE. I believe his method can be applied to the case $n=3$ by calculating the integral over $x$ first, then similarly the remaining integral over $y$.

This is very interesting. So you would be content if the integrals $I_1$ and $I_2$ just below (26) in this paper would be evaluated with residue theory? arxiv.org/pdf/1712.10324.pdf

@AlfredYerger please see my paper at arxiv regarding 2D Mordell integrals.

It is also possible the LM estimates one typically sees are too coarse for this integral, and something better is needed to show that the integral along such an arc goes to $0$ as $R \to \infty$, but I don't have any ideas right now.

In particular, the place I got stuck is at estimating $| \int_\gamma \frac{\cos \frac{\pi}{2} u^2 \cos \pi u v}{\operatorname{cosh} \pi u/\sqrt{n}}|$ which I obtained by making the change of variables $u = \sqrt{n}x$ and $v = y/\sqrt{n}$, using the angle subtraction formula, distributing, and then pulling out some terms depending only on $v$ from the first integral. This integral is of an even function, so we can extend to an integral over the whole real line. If the LM estimates for a semi-circular arc in the upper half plane could be made to $\to 0$, we could try again to calculate residues.

Has the case $n=3$ been done with Residue theory? I spent some time thinking about this evening and I don't see how it can be possible. There is no residue theory to my knowledge for double integrals, so you must do a residue calculation on one or both of the iterated integrals, but after some algebra, I see no way to make the LM estimates work, so I don't see how it can be possible.  If the LM estimates could be worked out, I would be willing to try again.

@ Nemo:Based on residue theory.Unfortunately my browser can't display the formulas in your blog

@ Nemo: would you mind if I can offer a bounty for this question?

thx, that is cool stuff...are do you doing this on a recreational basis or is this for professional purposes?

can you give a reference for the amazing formula $n=3$? i bet it is due to ramanujan


When does $x^{x^{x^{...^x}}}$ diverge but $x^{x^{x^{...^c}}}$ converge?

Let us define these two sequences as follows:
$a_0=1$, $b_0=c$
$a_{n+1}=x^{a_n}$, $b_{n+1}=x^{b_n}$
$b_{n+1}\ne b_n$ for any $n$.
$x,c\in\mathbb C$
Is it possible for $a_n$ to diverge but $b_n$ to converge under these conditions?  For example, if $x=2$ and $c=i$, we have
$b_1=2^i=\operatorname{cis}(\ln2)$
$b_2=2^{\operatorname{cis}(\ln2)}=2^{\cos(\ln2)}\operatorname{cis}(\ln2\sin(\ln2))$
etc.
I haven't much clue as to whether it is the case that $b_n$ can converge when $a_n$ diverges, and I can hardly work out if $b_n$ converges with $x=2$ and $c=i$.

@miloBrandt b=exp(1/e) has a neutral fixed point=e, and iterating b^^n converges towards e.

One can work out that the only times that $x^{\ldots^c}$ can converge without hitting a fixed point is whenever $W(-\log(x))$ has absolute value at most $1$, where $W$ is the product log (and there are definitely converging $c$ when this absolute value is strictly less than $1$). I would maybe bet that there's an example of this happening where the absolute value is exactly $1$ and that where it's less than $1$, the tower converges starting at $1$.

@SheldonL If you still wish to I would be interested in seeing your numerical results.

@SheldonL: if you'll get something substantial - I'd like to see a workout also in our tetrationforum! (It seems to be an interesting facet in context with the D. Shell article)

Even considering complex values of x, I believe the surprising result is that there are no solutions!    I wrote a pari-gp loop that verified this for over 300,000 real and complex bases with attracting fixed points ....  I think this is the case since I think the domain of the basin of attraction of the attracting fixed point can be extended to f'=0; but I will need to do some reading to brush up on complex dynamics, and iterated functions.

@SheldonL Oddly, though I am asking such questions, such vocabulary is new to me.  But it's ok, I get the gist of what you are saying.  If you can think up a worthy answer, go for it and I'll ask clarifications later.  And do complex fixed points behave like real ones?  It seems quite hard to discern

@SimpleArt ouch!, but I think.  $0<b<e^{-e}$ has repelling fixed points.... negative bases?  complex bases? it might be easier to work with iterating $z \mapsto \exp(z)+k;\;\;\;k=\ln(\ln(b))$ which is congruent to $y \mapsto b^y;\;\;\;y=(z-k)/ln(b)$

@SheldonL Hm, I'm not entirely sure.  I'm also wondering about $x<e^{-e}$.

hmmm, for real values of x, the first requirement is that x>exp(1/e) so that the first power tower diverges.  But all complex fixed points for bases>exp(1/e) are repelling.  So doesn't that mean the 2nd power tower won't converge?  And therefore there are no real valued solutions to the Op's problem?

Hmm, the clause $c \ne x^c$ intends to exclude the "trivial" case of $c$ being the fixpoint. But consider cases , where the exponentialtower to base $x$ has a set of cyclic fixpoints (which might also be accumulation points such that the height-increasing exponentialtower of $b_n$ "converges" to a cycle over such a set of points) then you might want to add an analogue clause/to extend the given clause. (I've not yet a true answer to your question so far)

@Rohan I'm not so sure, both papers cover the topic of my sequence $a_n$, which I already know about, I'm more interested in $b_n$.


Consecutive prime numerators of harmonic numbers?

Let
$$\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}=\frac{a}{b}$$
and let $a$ and $b$ are coprime, $h_{n}=a$.
$h_{n}$ is prime for
$$n=2,3,5,8,9,21,26,41,56,62,69,79,89,91,122,127,143,167,201,230,247,252,290,349,376,459,489,492,516,662,687,714,771,932,944,1061,1281,1352,1489,1730, 1969,2012,2116,2457,2663,2955,3083,3130,3204,3359,3494,3572,3995,4155,4231,4250,4496,4616,5069,5988,6656,6883,8067,8156,8661,9097,\ldots$$
I guess proving that there are infinitely (or finitely) many primes of the form $h_n$ is very hard. But can we prove both of the $h_n$ and $h_{n+1}$ cannot be prime for $n>8$?

@Mr. Brooks, no, thank you. That is be some very useful information

@vrugtehagel Did you look in oeis.org/A056903/b056903.txt ? They call it a "b-file" and it goes up to $h_{97} = 78128$.

I haven't looked much into it before. The only thing I've done on this subject is prove that the harmonic series is never an integer for $n>1$. But this is interesting, so I'll definitely look into it

@vrugtehagel  : what do you already know about the numerators of the harmonic numbers ?

I edited the question to include some more values for $n$, more than oeis.org, which for some reason only includes values up to $3572$

Just for reference, the first values of $h_n$ are listed here: oeis.org/A001008

What sort of question is this? I think it isn't trivial. Where is the conjecture from, just looking on the numbers or are there reasons why it should be right?


An elegant description for graded-module morphisms with non-zero zero component

In an example I have worked out for my work, I have constructed a category whose objects are graded $R$-modules (where $R$ is a graded ring), and with morphisms the usual morphisms quotient the following class of morphisms:
$\Sigma=\left\lbrace f\in \hom_{\text{gr}R\text{-mod}}\left(A,B\right) \ | \ \ker\left(f\right)_0\neq 0, \ \mathrm{coker}\left(f\right)_0\neq 0\right\rbrace$
(by quotient I mean simply that this class of morphisms are isomorphisms, thus creating an equivalence relation) I am wondering if this category has a better (more canonical) description, or if I can show it is equivalent to some other interesting category.
Thanks!

Wow this is the first non-deleted question! Good job.

Well, I don't remember. I do remember that I was localizing by these morphisms and that was what I meant by quotienting. Judging by the date, I was probably trying to work out a ncag example. I don't have a clue what it was!

@MarianoSuárez-Alvarez I was trying to get the altruist badge, so I found the oldest unanswered question and put a bounty on it. It makes sense that this has been unanswered for three years...

This is very weird, really: you are making all modules with non-zero zero component isomorphic... Does this really make sense in some context?

Do you mean to make the maps in $\Sigma$ all the zero maps (which is quotienting) or do you mean to make the maps in $\Sigma$ isomorphisms (which is localizing)?  As $\Sigma$ is not an ideal the first isn't well defined, and as $\Sigma$ can contain the zero map between modules the second doesn't seem to make much sense either.

Since $\Sigma$ isn't an ideal in the category of graded $R$-modules, I don't think that quotienting by it makes much sense.


Hopf-like monoid in $(\Bbb{Set}, \times)$

I am looking for a nontrivial example of the following:

Let a monoid $A$ be given with unit $e$, and two of its distinguished disjoint submonoids $B_1$ and $B_2$ (s.t. $B_1\cap B_2=\{e\}$), endowed with monoid homomorphisms $\Delta_i:A\to B_i$ such that $\Delta_i|_{B_i}={\rm id}_{B_i}\ $  ($i=1,2$) .
  Moreover, also let a mapping $\sigma:A\to A$ be given, such that for all $a\in A$:
  $$\sigma(\Delta_1(a))\cdot\Delta_2(a)=e= \Delta_1(a)\cdot\sigma(\Delta_2(a)) $$
  In particular, each element of $B_1$ is right invertible, and each element of $B_2$ is left invertible in $A$.

Such an example is trivial if $B_1=B_2=A$.
Motivation: I am looking for Hopf-like monoids in cartesian categories, in which the counit axioms for the unique $A\to 1$ morphism are not posed. Then, the comultiplication map $\Delta:A\to A\times A$ has to be of the form $\Delta=(\Delta_1,\Delta_2)$ and coassociativity means ${\Delta_i}^2=\Delta_i$ and $\Delta_1\Delta_2=\Delta_2\Delta_1$. $B_1\cap B_2=\{e\}$ ensures this latter equality.
Of course, $\sigma$ would be the antipode.


Integral involving Complete Elliptic Integral of the First Kind K(k)

I have run into an integral involving the complete elliptic integral, which can be put into the following form after changing integration variables to the modulus:
$$\int_0^{\sqrt{\frac{\alpha}{1+\beta}}} dk\, \frac{ k^{11} K(k) } {\sqrt{(\alpha-\beta k^2)^2 - k^4} (\alpha - \beta k^2)^{11/2}}$$
$K(k)$ is the complete elliptic integral of the first kind, where $k$ is the modulus.  We can assume that $\alpha$ and $\beta$ are such the maximum value for $k$ is less than or equal to $1$.  Are there any ways to get a closed form solution out of this?  The indefinite integrals in G&R are not much help.

+1 for remembering to mention your argument convention for elliptic integrals. It looks a bit gnarly as it stands, but I'll see what I can do.


Expected number of operations on a vector until one of the coordinates becomes zero.

Let's say we have a vector $v = (x_1, ..., x_n) \in \mathbb{N}^n$ where $x_1 = x_2 = ... = x_n$. Next we choose an ordered pair of coordinates at random $(i, j)$ where $i, j \in \\{1, ..., n\\}$ and $i \neq j$. Finally we substitute the vector $v$ with a new vector $v' = (x_1, ..., x_i + 1, ..., x_j - 1, ..., x_n)$. Now we choose again an ordered pair of coordinates at random and substitute the vector $v'$ with a new vector doing the same we did for $v$. We continue doing this until one of the coordinates becomes zero.
What is the expected number of operations we are going to make?

I know the answer for $n = 2$ because you can model this process with a random walk. If $v = (x, x)$, then the expected number of operations is the same as the expected number of steps it will take to hit $x$ or $-x$ doing a random walk starting at zero. In this case the expected number of step starting at $y$ satisfies the recurrence relation $$E_y = 1 + \frac{1}{2} E_{y - 1} + \frac{1}{2} E_{y + 1}. $$ Then one can solve this linear recurrence.
I tried to do the same for the original problem but the recurrence relation is more difficult. Let $F_{(x_1, ..., x_n)}$ be the expected number of operations one can make to vector $v= (x_1, ..., x_n)$ before one of the coordinates becomes zero (in this case we allow $x_1, ..., x_n$ to be different). If I'm not wrong $F$ satisfies the following relation $$F_{(x_1, ..., x_n)} = 1 + \sum_{i, j} \frac{1}{n (n - 1)} F_{(x_1, ..., x_n) + e_{i, j}}, $$ where the $i$-th coordiante of $e_{i, j}$ is $1$, the $j$-th is $-1$ and the rest are all zero (the sum runs through all possible operations).

I think we can project the movement to hyperplane $x_1+\cdots+x_n=nN$, where $N=x_1=\cdots=x_n$ is the initial coordinate of $v$. With the stopping condition(i.e. stop when some coordinate becomes zero) we are on a bounded domain, with each transition $(0,\cdots,1,\cdots,-1,\cdots,0)$ parallel to some boundary line $x_k=0,k\not\in\{i,j\},x_i+x_j=nN$. Seem like we can reduce the dimension...will this help?

Note that raising operators do not modify the total weight of vectors. Moreover, whenever $a$ and $b$ are of equal weight and $a\leqslant b$, there is a raising operator $R$ such that $Ra =b$. It is not clear if this should really help here, though, but perhaps knowing these operators have a name might help in searching for references. Buena suerte! ;)

The operation you are considering is called an (elementary?) raising operator (a general raising operator is a product of perhaps repeated elementary ones). There is a partial order on $\mathbb Z^n$ given by $a \geqslant b$ iff for each $i$ we have $a_1+\cdots+a_i\geqslant b_1+\cdots+b_i$, and raising operators are nondecreasing for this order.


Reversal of an Autoregressive Cauchy Markov Chain

Let $\mu_0 (dx)$ be the standard one-dimensional Cauchy distribution, i.e.

\begin{align} \mu_0 (dx) = \frac{1}{\pi} \frac{1}{1+x^2} dx. \end{align}

Suppose I fix $h \in [0, 1]$, and form a Markov chain $\\{X_n\\}_{n \geqslant 0}$ as follows:

  1. At step $n$, I sample $Y_n \sim \mu_0$.
  2. I then set $X_{n+1} = (1 - h) X_n + h Y_n$

It is not so hard to show that this chain admits $\mu_0$ as a stationary measure, as this essentially comes from the fact that Cauchy distribution is a stable distribution.

What I'm interested in is the reversal of this Markov chain. More precisely, if the chain I describe above uses the Markov kernel $q (x \to dy)$, I want to understand the Markov kernel $r$ such that

\begin{align} \mu_0 (dx) q (x \to dy) = \mu_0 (dy) r (y \to dx). \end{align}

Fortunately, all of the quantities involved have densities with respect to Lebesgue measure, and as such, I can write down what $r (y \to dx)$ is:

\begin{align} r (y \to dx) = \frac{1 + y^2}{\pi} \frac{1}{1 + x^2} \frac{h}{ h^2 + (y - (1 - h) x)^2} dx. \end{align}

My question is then: is there a simple, elegant way to draw exact samples from $r$?

I would highlight that this is not a purely algorithmic question; I'd _really_ like to understand what this reversal kernel $r$ is doing. A nice byproduct of that would then be that I could simulate from it easily.

For completeness, some of the `purely algorithmic' solutions I had considered were the following.

  * I could try rejection sampling, and in principle this would work, but it wouldn't really give me insight into the nature of the Markov chain.
  * I could try something like the inverse CDF method, but it seems to me that the CDF of $r$ is not particularly nice to work with. As such, I'd have to use e.g. Newton iterations to use this method, and I'd prefer to not have to do this.

The general Cauchy distribution $c_w(dx)=\frac{b dx}{\pi(b^2+(x-a)^2)}$ where $b>0$ and $w=a+ib$ has Fourier transform $e^{iwt}$ for $t>0.$ Therefore your Markov chain defined by  $X_{n+1}=(1-h)X_n+hY_n$ where $Y_n\sim c_i=\mu_0$ is such that $X_n\sim c_{w_n}$ where $w_{n+1}=(1-h)w_n+ih$, hence $w_n=(1-h)^nw_0+ih$. I will try to understand your question in terms of these $w$ and $c_w.$


Extracting an (almost) independent large subset from a pairwise independent set of Bernoulli variables

Let $n>1$, and let $X_1,X_2, \ldots ,X_n$ be non-constant random variables with values in $\lbrace 0,1 \rbrace$. Let us say that a subset of variables $X_{i_1},X_{i_2}, \ldots,X_{i_d}$ is **complete** if the vector $\overrightarrow{X}=(X_{i_1},\ldots,X_{i_d})$ satisfies $P(\overrightarrow{X}=\overrightarrow{c})>0$ for any $\overrightarrow{c}\in \lbrace 0,1 \rbrace^d$.

Prove or find a counterexample : if $X_1,X_2, \ldots ,X_n$ are pairwise independent Bernoulli variables, then we may extract a complete subset of cardinality at least $t+1$, where $t$ is the largest integer satisfying $2^{t} \leq n$.

This is true for $n=3$ (and hence also true for $n$ between $3$ and $7$), as is shown in the main answer to that [MathOverflow question](https://mathoverflow.net/questions/66738/is-there-a-good-explanation-for-this-fact-on-pairwise-independent-variables). (That other [MathOverflow question](https://mathoverflow.net/questions/64973/sufficiently-random-sample) is also related, and provides several links)

If true, this result is sharp, as can be seen by the classical example of taking all arbitrary sums modulo 2 of an initial set of fully independent $t+1$ Bernoulli variables. This produces a set of pairwise independent $2^{t+1}-1$ variables, and where the maximal cardinality of a complete subset is $t+1$.

**Update 10/10/2012** : By induction, it would suffice to show the following : if $X_1, \ldots ,X_t$ is a fully independent set of $t$ Bernoulli variables and $X$ is another Bernoulli variable, such that the pair $(X_i,X)$ is independent for each $i$, then there are coefficients $\varepsilon_0,\varepsilon_1, \ldots ,\varepsilon_t$ in $\lbrace 0,1 \rbrace$ such that, if we put

$$ H=\Bigg\lbrace (x_1,\ldots,x_t,x) \in \lbrace 0,1 \rbrace ^{t+1} \Bigg| x=\varepsilon_0+\sum_{k=1}^{t}\varepsilon_kx_k \ {\sf mod} \ 2\Bigg\rbrace, \ \overrightarrow{X}=(X_{1},\ldots,X_{t},X) $$ then $P(\overrightarrow{X}=h)>0$ for any $h\in H$.


How to prove this lemma related to Rolle's theorem

For any function $f$ denote by $Z(f)$ and $Z_o(f)$ the cardinalities of $f^{-1}(0)\cap[0,1]$ and $f^{-1}(0)\cap(0,1)$, respectively. Let $H=\\{f\in C^\infty(\mathbb{R}): \text{supp}(f) = [0,1]\\}$

From [this question](https://math.stackexchange.com/questions/664741) we have
**Lemma 1**
Let $q:x\mapsto(x-r)p(x)$ where $p\in H$ and $r\in\mathbb{R}$. Then $Z(q^{(n)})\geq Z(p^{(n-1)})+1$ for all $n\in\mathbb{N}$.

**Proof**
Note that $q^{(n)}(x) = n p^{(n-1)}(x) + (x-r)p^{(n)}(x)$. Hence $r$ is a root of $q^{(n)}$ if and only if it is a root of $p^{(n-1)}$. Moreover we have

$$\underbrace{(x-r)^{n-1}q^{(n)}(x)}_{\text{LHS}} = n (x-r)^{n-1} p^{(n-1)}(x) + (x-r)^np^{(n)}(x) = \underbrace{\frac{d}{dx}(x-r)^n p^{(n-1)}(x)}_{\text{RHS}}$$

If $q^{(n)}(r) = 0$ then $Z(q^{(n)}) - 2 = Z_o(\text{LHS}) = Z_o(\text{RHS}) \geq Z(p^{(n-1)}) - 1$.
If $q^{(n)}(r) \neq 0$ then $Z(q^{(n)}) - 1 = Z_o(\text{LHS}) = Z_o(\text{RHS}) \geq Z(p^{(n-1)})\quad\quad\quad\quad\text{q.e.d.}$
$\text{ }$

The following modification
**Lemma 2**
Let $q:x\mapsto(x-r)(x-\bar{r})p(x)$ where $p\in H$ and $r\in\mathbb{C}\setminus\mathbb{R}$. Then $Z(q^{(n)})\geq Z(p^{(n-2)})+2$ for all $n\in\mathbb{N}\setminus\\{1\\}$.

I tried to prove the same way by writing

$$\frac{j(x)k(x)}{(x-r)(x-\bar{r})}q^{(n)}(x) = \frac{d}{dx}\left(j(x)^2 k(x)\frac{d}{dx}j(x)^{-1}p^{(n-2)}(x)\right)$$ where $\frac{d}{dx}$ is differentiation along the real axis and
$j(x)=c_1(x-r)^{1-n}+c_2(x-\bar{r})^{1-n}$
$k(x) = c_3 ((x-r)(x-\bar{r}))^n$
$c_1,c_2,c_3\in\mathbb{C}$

The real and imaginary part of $j(x)^{-1}$ are proportional, and $j(x)k(x)\in \mathbb{R}$ when choosing $c_1 = e^{i d_1}\\\c_2=e^{i d_2}\\\c_3=e^{-i\cdot (d_1+d_2)/2}\\\d_1,d_2\in\mathbb{R}$

but even then I couldn't get Rolle's theorem to work unless $j(x)$ has no roots on $(0,1)$. I have looked at many functions/different $n$s and even when choosing $r$ such that $j(x)$ has such roots, it seems impossible to find a counterexample to the lemma.

How to prove the lemma?

Or maybe this simpler version which remains when removing some of the assumptions:
If $p$ is smooth with $p^{(n-2)}$ having $m$ roots on $[0,1]$, then how to prove that the $n$'th derivative of $x \mapsto p(x)(x-r)(x-\bar{r})$ has at least $m-2$ roots on $(0,1)$.

In the last (simpler) lemma still $r$ purely imaginary?

@DavidSpeyer I'm interested in how to prove lemma 2 further than for those choices of $r$, $n$ and $p$ that implies that $j$ has no roots on $(0,1)$

@GerryMyerson Any function whose domain contains $[0,1]$ and whose codomain contains $\{0\}$

I take it "for any function $f$" means "for any real-valued function $f$ of a real variable"?

@WillFisher I mean the space of bump functions, but the other conditions implies the $c$ so it is redundant

What does the subscript $c$ signify in your notation $C^{\infty}_c$?

In the linked question, $r$ is assumed not in $(0,1)$. Do you want some analogous hypothesis in Lemma 2?


Existence of function satisfying $f(f'(x))=x$ almost everywhere

**My project is to Study the existence of a continuous function $f : \mathbb{R} \rightarrow \mathbb{R}$ differentiable almost everywhere satisfying $ f\circ f'(x)=x$ almost everywhere $x \in \mathbb{R}$**

I began the study by supposing $f\in C ^ 1(\mathbb{R}) $, I have shown that f does not exist.

After, I found some difficulties when we assume only f differentiable on $\mathbb{R}$, I had an answer using Darboux's theorem [Questions about the existence of a function](https://math.stackexchange.com/questions/3312572/questions-about-the-existence-of-a-function?noredirect=1#comment6815760_3312572).

Now, I want to attack the initial problem. Previous arguments do not work!

Do you have any suggestions for me?

Why this question is different from?$$f'(x)=f^{-1}(x)$$ which is solvable, see video

Convert your equation into a differential equation and seek out for a solution

@Jack D'Aurizio   the link math.stackexchange.com/questions/3312572/…  does not answer the question you asked me?

The question is not yet solved in mathoverflow.net/questions/337607/…

@Jack D'Aurizio   in the above link,  I explained this case   thank you for saving my question.

@Kavi Rama Murthy Can u explain me why u voted against this question?


Is there any example of a real-analytic approach to evaluate a definite integral (with an elementary integrand) whose value involves Lambert W?

I have never seen a real-analytic approach before to evaluate integrals of the form below $$\int_a^b\text{elementary function}(x)\,dx=\text{constant involving}\,W(\cdot)\,\text{in its simplest form}\tag1.$$

For instance, on MSE, all use the residue theorem:

  * [$\int_{-\infty}^{\infty}{e^x+1\over (e^x-x+1)^2+\pi^2}\mathrm dx=\int_{-\infty}^{\infty}{e^x+1\over (e^x+x+1)^2+\pi^2}\mathrm dx=1$](https://math.stackexchange.com/questions/2113205/int-infty-inftyex1-over-ex-x12-pi2-mathrm-dx-int-infty?noredirect=1&lq=1)

  * [Interesting integral related to the Omega Constant/Lambert W Function](https://math.stackexchange.com/questions/45745/interesting-integral-related-to-the-omega-constant-lambert-w-function?noredirect=1&lq=1)

  * [Prove that $\int_0^\infty \frac{1+2\cos x+x\sin x}{1+2x\sin x +x^2}dx=\frac{\pi}{1+\Omega}$ where $\Omega e^\Omega=1$](https://math.stackexchange.com/questions/2331600/prove-that-int-0-infty-frac12-cos-xx-sin-x12x-sin-x-x2dx-frac-pi?noredirect=1&lq=1)

  * [Proof for integral representation of Lambert W function](https://math.stackexchange.com/questions/3347447/proof-for-integral-representation-of-lambert-w-function)

  * [Evaluate $\int_{0}^{\infty} \ln(1+\frac{2\cos x}{x^2} +\frac{1}{x^4}) \, dx$](https://math.stackexchange.com/questions/4433252/evaluate-int-0-infty-ln1-frac2-cos-xx2-frac1x4-dx)

And the same applies to some of the wider literature I have come across:

  * [Stieltjes, Poisson and other integral representations for functions of Lambert W](https://arxiv.org/pdf/1103.5640.pdf)

  * [An Integral Representation of the Lambert $W$ Function](https://arxiv.org/pdf/2012.02480.pdf) **Note:** the proof is real-analytic, but the very first line assumes the validity of an integral identity which was only proven using complex analysis (Hankel contour).

So, my question is this:

Does anyone know of a proof of an identity of the form in $(1)$ that involves only real analysis (i.e. does not assume the existence of $\sqrt{-1}$)?

@ТymaGaidash No, the proof must not use the fact that $\sqrt{-1}$ exists.

@StevenClark No (I'm looking for an analytical proof rather than numerical). In that example, I'm actually not referring to the end result $\int_{-\infty}^{\infty}{e^x+1\over (e^x-x+1)^2+\pi^2}\,dx=1$ but rather the lemma $\int_{-\infty}^{\infty}\frac{a^2\,dx}{(e^x-ax-b)^2+(a\pi)^2}=\frac{1}{1+W\left(\frac{1}{a}e^{-b/a}\right)}$ in Jack's answer, which he said was evaluated only through the residue theorem. To qualify as an answer here, an example would be to prove the latter identity using real analysis when, say, $a=2$ and $b=3$, since $W(e^{-3/2}/2)$ doesn't have an obvious simplification.

Do you consider numerical integration a real-analytic approach? Some of your examples can be evaluated via numerical integration, but some of your examples also simplify which seems inconsistent with your desire to find an example that doesn't simplify. Excluding results that simplify seems rather arbitrary assuming one can show the example is inherently related to the Lambert W function. Trivial cases such as $\frac{W(10)}{W(10)}$ don't generally meet this criteria, but it seems to me $\int\limits_{-\infty}^{\infty}{e^x+1\over (e^x-x+1)^2+\pi^2}\mathrm dx=1$ does meet this criteria.

@StevenClark No, I think as long as your $x$ does not obviously simplify to $ye^y$ for some elementary $y$ then it's accepted (like obvious manipulations of $W((\sqrt2+1)e^{\sqrt2+1})$. Since otherwise we'd be running into much deeper holes involving Schanuel's conjecture.

Does one also have to prove a result involving $W(x)$ has no closed form that does't involve the $W(x)$ function?

@StevenClark I think there is a bit of misunderstanding -- I'm trying to find a valid integral representation (elementary integrand) for values of a constant involving $W(\cdot)$ such that it cannot be simplified to a constant that doesn't involve $W$. So $1/(1+W(2))$ would be valid, but constants like $W(-1/e)=-1$ or $\sin W(3e^3)^2=\sin9$ would not.

My example is less trivial than your example, and is $\int\limits_{\infty }^1 e^{1-x}\,dx$ not a valid integral representation of $W\left(-\frac{1}{e}\right)$?

@StevenClark No, which is the reason I added "in its simplest form" (or "in most simplified form" before this edit) so that we don't have trivial cases like $\int_0^12x\,dx=1=W(10)/W(10)$.

Does $\int\limits_{\infty }^1 e^{1-x}\,dx=W\left(-\frac{1}{e}\right)=-1$ count?


Minimum area contained between measurable set and translate by $\lambda$: A strengthening of 2018 USA TSTST #9

Question
Given $\lambda\in\mathbb{R}^+$, what is the smallest possible $c$ for which, given any measurable region $\mathcal{P}$ in the plane with measure $1$, there always exists a vector $\mathbf{v}$ with magnitude $\lambda$ so that the area shared between $\mathcal{P}$ and its translate by $\mathbf{v}$ is at most $c$?

Background
On this year's USA TSTST (a test that determines a group of about 30 people to take selection tests for the following year's USA team to the International Math Olympiad), there was an algebra/geometry/combinatorics hybrid problem that I found interesting:

Show that there is an absolute constant $c < 1$ with the following property: whenever $\mathcal P$ is a polygon with area $1$ in the plane, one can translate it by a distance of $\frac{1}{100}$ in some direction to obtain a polygon $\mathcal Q$, for which the intersection of the interiors of $\mathcal P$ and $\mathcal Q$  has total area at most $c$.

Essentially every solution can be boiled down to the following:
Step 1. For a given vector $\mathbf{v}\in \mathbb{R}^2$, define $f(\mathbf{v})=\mu\big(\mathcal{P}\cap(\mathcal{P}+\mathbf{v})\big)$, where $\mu(\cdot)$ is the area of a region, and $\mathcal{P}+\mathbf{v}$ consists of the points in $\mathcal{P}$ translated by $\mathbf{v}$.
Step 2. Prove $f(\mathbf{u}+\mathbf{v})\geq f(\mathbf{u})+f(\mathbf{v})-1$ and generalize it to
$$1-f\left(\sum_{i=1}^N \mathbf{v}_i\right)\leq \sum_{i=1}^N \left(1-f\left(\mathbf{v}_i\right)\right).$$
Step 3. Define, for real $t$,
$$I_t=\int_{0\leq ||\mathbf{v}||\leq t} f(\mathbf{v})\ d^2\mathbf{v} = \int_{0\leq ||\mathbf{v}||\leq t}\int_{\mathbf{x}\in \mathcal{P}} \mathbf{1}_{\mathcal{P}}(\mathbf{x}+\mathbf{v})\ d^2\mathbf{x}\ d^2\mathbf{v}.$$
We have
\begin{align}
I_t&=\int_{\mathbf{x}\in \mathcal{P}} \int_{0\leq ||\mathbf{v}||\leq t}\mathbf{1}_{\mathcal{P}}(\mathbf{x}+\mathbf{v})\ d^2\mathbf{v}\ d^2\mathbf{x}\\
&=\int_{\mathbf{x}\in \mathcal{P}} \mu\left(\mathcal{P}\cap\big\{\mathbf{x}+\mathbf{v}\big|0\leq ||\mathbf{v}||\leq t\big\}\right)\ d^2\mathbf{x}\\
&\leq\int_{\mathbf{x}\in \mathcal{P}} 1\ d^2\mathbf{x} = 1.
\end{align}
So, there must exist some $\mathbf{v}$ with $0\leq ||\mathbf{v}||\leq t$ that satisfies
$$f(\mathbf{v})\leq \frac{1}{\pi t^2}.$$
Step 4. Write this vector as
$$\mathbf{v}=\sum_{i=1}^{\lceil 100t\rceil}\mathbf{u}_i$$
where $||\mathbf{u}_i||=1/100$.
Step 5. We now have
$$1-\frac{1}{\pi t^2}\leq 1-f(\mathbf{v})\leq \sum_{i=1}^N (1-f(\mathbf{u}_i)),$$
so for some $i$ we must have
$$1-f(\mathbf{u}_i) \geq \frac{1}{\lceil 100t\rceil}\left(1-\frac{1}{\pi t^2}\right)$$
$$f(\mathbf{u}_i) \leq 1-\frac{1}{\lceil 100t\rceil}\left(1-\frac{1}{\pi t^2}\right).$$
As long as $t^2>1/\pi$, this gives us a working value of $c<1$. It is minimized at $t=0.98$, which gives
$$c\approx 0.99318.$$
More generally, if $\lambda$ is the length of our translate, this is minimized at the nearest integer multiple of $\lambda$ to $\sqrt{3/\pi}$, which gives
$$c\approx 1-2\lambda\sqrt{\frac{\pi}{27}}.$$
In particular, for large enough $\lambda$, this bound does nothing.

Progress
I've been able to improve the bound on $c$ given slightly in the following manner:
Assume, for all $||\mathbf{u}||=\lambda$, that $f(\mathbf{u})>1-\epsilon$. Then, for all vectors $\mathbf{v}$ of length $\leq n\lambda$, by writing $\mathbf{v}=\sum_{i=1}^n \mathbf{u}_i$ with $||\mathbf{u}_i||=\lambda$, we must have
$$1-f(\mathbf{v})\leq n\epsilon\implies f(\mathbf{v})\geq 1-n\epsilon.$$
Using (almost) our same integral as earlier and setting $t=N\lambda$, we have
$$\int_{\lambda< ||\mathbf{v}||\leq t} f(\mathbf{v})\ d^2\mathbf{v}\leq 1.$$
(we cannot start the integral at $0$ as vectors with magnitude $<\lambda$ cannot be represented as the sum of one vector with magnitude $\lambda$). However,
\begin{align}
\int_{\lambda \leq ||\mathbf{v}||\leq N\lambda} f(\mathbf{v})\ d^2\mathbf{v}
&=
\sum_{k=2}^N \int_{(k-1)\lambda< ||\mathbf{v}|| \leq k\lambda} f(\mathbf{v})\ d^2\mathbf{v}\\
&\geq
\sum_{k=2}^N \int_{(k-1)\lambda< ||\mathbf{v}|| \leq k\lambda} 1-k\epsilon\ d^2\mathbf{v}\\
&=
\pi\lambda^2\sum_{k=2}^N (2k-1)(1-k\epsilon)\\
&=
\pi\lambda^2\frac{N-1}{6}\left(6N+6-\epsilon\left(4N^2 + 7N  + 6\right)\right),
\end{align}
so
$$\pi\lambda^2\frac{N-1}{6}\left(6N+6-\epsilon\left(4N^2 + 7N  + 6\right)\right)\leq 1$$
$$N+1-\epsilon\left(\frac{4N^2 + 7N  + 6}{6}\right)\leq \frac{1}{\pi(N-1)\lambda^2}$$
$$N+1- \frac{1}{\pi(N-1)\lambda^2}\leq \epsilon\left(\frac{4N^2 + 7N  + 6}{6}\right)$$
$$\frac{6\left(\pi\left(N^2-1\right)\lambda^2- 1\right)}{\left(4N^2 + 7N  + 6\right)\left(\pi(N-1)\lambda^2\right)}\leq \epsilon$$
$$1-\frac{6\left(\pi\left(N^2-1\right)\lambda^2- 1\right)}{\left(4N^2 + 7N  + 6\right)\left(\pi(N-1)\lambda^2\right)}\geq 1-\epsilon.$$
Minimizing $N$ gives you about $0.9898$ for $\lambda=1/100$.
On the other hand, I've also been trying to find constructions that give a large value of $c$. I haven't come up with any great ones - the best I have is a circle of radius $\sqrt{\frac{1}{\pi}}$ which gives you a $c$ of about 0.9887. For larger $\lambda$ the best I can think of is something like a "star" with large radius and very many thin "prongs," but I haven't calculated the asymptotics on it yet.
I believe the bound given in step 2 of the solution is sharp iff
$$(\mathcal{P}+\mathbf{u})\cap(\mathcal{P}+\mathbf{v})\subseteq\mathcal{P}\subseteq(\mathcal{P}+\mathbf{u})\cup(\mathcal{P}+\mathbf{v}),$$
but all attempts I've made to find measurable sets for which this is true or nearly true for small vectors have failed. So I'm stuck.
Anyone have any ideas, either on sharper upper bounds on $c$ or a sharper $\mathcal{P}$?

Original problem can be solved, the value that half of circumference multiple 1/100 is upperbound. Thinned shape give a small c, maximum case would be about round and small shape, that's circle.

@mathworker21 You're right; I missed that. I've edited the question accordingly.

The question at the start is phrased poorly. You should reorder like "Given lambda, what is the smallest possible c>o so that for any P ....". The point is that c depends on lambda and not P.


For $n\in\mathbb{N}^+$, $\Re(s)>0$, evaluate $\int_{0}^{\infty}\sin\left(2\pi ne^{x}\right)\left[\frac{s}{e^{sx}-1}-\frac{1}{x}\right]dx$

Let $n$ be a positive integer, and $s\in \mathbb{C}\;,\Re(s)>0$. I want to compute the integral : $$\int_{0}^{\infty}\sin\left(2\pi ne^{x}\right)\left[\frac{s}{e^{sx}-1}-\frac{1}{x}\right]dx$$

I tried using the integral presentation found [here](http://functions.wolfram.com/ElementaryFunctions/Sin/07/02/) :

$$\sin(2\pi ne^{x})=\frac{\sqrt{\pi}}{2\pi i }\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{\Gamma(z)}{\Gamma\left(\frac{3}{2}-z \right )}\left(\pi n \right )^{-2z+1}e^{-(2z-1)x}dz\;\;\;\;\;0<\gamma<1$$ in conjunction with : $$r+\log(r)+\psi\left(\frac{1}{r}\right)=-\int_{0}^{\infty}e^{-y}\left(\frac{r}{e^{ry}-1}-\frac{1}{y}\right)dy$$ Where $\psi\left(\cdot\right)$ is the digamma function. But that made the problem even more difficult. Any help is highly appreciated.

**EDIT** :

Using the Fourier reciprocity : $$\frac{s}{e^{sx}-1}-\frac{1}{x}=2\int_{0}^{\infty}\left(\frac{1}{e^{2\pi \theta/s}-1}-\frac{s}{2\pi \theta} \right )\sin(x\theta)d\theta$$ Our integral reads : $$2\int_{0}^{\infty} g(\theta)\left(\frac{1}{e^{2\pi \theta/s}-1}-\frac{s}{2\pi \theta} \right )d\theta$$ Where : $$g(\theta)=\int_{0}^{\infty}\sin(2\pi ne^{x})\sin(x\theta)dx$$ $$=\frac{\sqrt{\pi}}{2\pi i }\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{\Gamma(z)}{\Gamma\left(\frac{3}{2}-z \right )}\left(\pi n \right )^{-2z+1}\frac{\theta}{\theta^{2}+(2z-1)^{2}}dz$$

Using change of variable $y=e^x$ plus noting that $\left[\frac{s}{e^{s x}-1}-\frac{1}{x}\right]=\left[\frac{s \coth \left(\frac{s x}{2}\right)}{2}-\frac{1}{x}-\frac{s}{2}\right]$ I think your integral can be written as $$I=\int_1^{\infty } \frac{\sin (2 \pi  n y)}{y} \left(\frac{s \coth \left(\frac{1}{2} s \log (y)\right)}{2 }-\frac{1}{ \log (y)}-\frac{s}{2 }\right) \, dy$$ The integral  of the $-\frac{s}{2}$ term involving just the sine integral is the largest or principal term. When integrated the term is $\left[ -\frac{1}{2} s \left(\frac{\pi }{2}-\text{Si}(2 n \pi )\right) \right]$.

Are you interested in any sort of asymptotics of $s$? Or do you need an exact closed form?

Are you sure you evaluated the integral with $\sin(2\pi n e^{x})$ not $\sin(2\pi nx)$ ?

A quick and dirty conjectural result found using Mathematica and tested for a few values of $n$ and real $s$ is $$I=\frac{1}{2} \pi  \left(\coth \left(\frac{2 \pi ^2 n}{s}\right)-1\right)-\frac{s}{4 \pi  n}$$

Did you consider numeric approximation (for fixed values of $s,n$) to get an idea? If you evaluate the integral from $0$ to some $n$ this integral should be well approximated even for relatively small $n$ because for large $x$ the sin term will just average out to zero.

It is esoteric indeed. This integral came up in the study of a certain number-theoretic function. The function between the parentheses is self reciprocal wrt the Fourier sine transform, and can be used to prove the transformation formula of the Dedekind eta function, when the sine factor is $\sin(2\pi n x)$. The problem above is a little bit different, but concerns a closely related function.

What context did this come up in? This seems like a very esoteric problem.

$\int_0^\infty\sin\left(2\pi ne^{x}\right)\left[\frac{s}{e^{sx}-1}-\frac{1}{x}\right]dx$ is the incomplete version of $\int_{-\infty}^\infty\sin\left(2\pi ne^{x}\right)\left[\frac{s}{e^{sx}-1}-\frac{1}{x}\right]dx$ which has way more chances to be evaluated, why do you need the former

You meant $\int_{-\infty}^\infty\sin\left(2\pi ne^{x}\right)\left[\frac{s}{e^{sx}-1}-\frac{1}{x}\right]dx$


The sum of eigenvalues of integral operator $S(f)(x)=\int_{\mathcal{X}} k(x,y)f(y)d\mu(y)$ is given by $\int_{\mathcal{X}} k(x,x) d\mu(x)$?

**Setup:** Let $(\mathcal{X},d_{\mathcal{X}})$ and $(\mathcal{Y},d_{\mathcal{Y}})$ be two separable metric spaces. Let $M^1(\mathcal{X})$ be the space of Borel probability measures on $\mathcal{X}$ with finite first moment, i.e. a Borel probability measure $\mu$ on $\mathcal{X}$ is in $M^1(\mathcal{X})$ if $\int d_{\mathcal{X}}(x,o) d\mu(x)<\infty$ for any $o\in\mathcal{X}$. The space $M^1(\mathcal{Y})$ is defined in similar fashion.

Fix $\mu\in M^1(\mathcal{X})$ and $\nu\in M^1(\mathcal{Y})$ and define $$ d_\mu(x_1,x_2)= d_{\mathcal{X}}(x_1,x_2) -\int d_{\mathcal{X}}(x_1,x)\, d\mu(x) - \int d_{\mathcal{X}}(x_2,x)\, d\mu(x) + \int d_{\mathcal{X}}(x,x')\, d\mu^2(x,x'), $$ and a similar definition of $d_\nu:\mathcal{Y}\times \mathcal{Y}\to\mathbb{R}$.

Now let $S:L^2(\mathcal{X}\times \mathcal{Y},\mathcal{B}(\mathcal{X})\otimes \mathcal{B}(\mathcal{Y}),\mu\times \nu) \to L^2(\mathcal{X}\times \mathcal{Y},\mathcal{B}(\mathcal{X})\otimes \mathcal{B}(\mathcal{Y}),\mu\times \nu),$ be a Hilbert-Schmidt operator given by $$ S(f)(x,y) = \int d_\mu(x,x')d_\nu(y,y') f(x',y') d\mu\times \nu(x',y'). $$ and let $\\{\lambda_i\\}_{i\geq 1}$ denote the non-zero eigenvalues of $S$ repeated according to multiplicity.

**Question:** How do I prove the following identity: $$\sum_{i=1}^\infty\lambda_i=\int d_\mu(x,x)d_\nu(y,y) \, d\mu\times \nu(x,y).$$ I tried but failed to show that $S$ is of trace class, since they under certain conditions (which I also can't verify in this setup) satisfy that $$ Trace(S)=\int d_\mu(x,x)d_\nu(y,y) \, d\mu\times \nu(x,y), $$ which yields the result since $Trace(S)=\sum_{i=1}^\infty\lambda_i$ (if it were of trace class).

_The identity $\sum_{i=1}^\infty\lambda_i=\int d_\mu(x,x)d_\nu(y,y) \, d\mu\times \nu(x,y)$ is stated in [Distance covariance in metric spaces by Russell Lyons (2013) Theorem 2.7](https://projecteuclid.org/euclid.aop/1378991840), without proof, so my approach with traces is only an idea._

_If another way of proving the identity appears or if there is a counterexample, that would more than satisfy my needs. In the case of a counterexample i would very much appreciate stronger initial conditions rendering the identity true._

Please bear in mind that i am a novice in the theory of operators and trace class operators (only went down this road to explain the equality above), so references would be much appriciated.

**Update:** A counterexample to the operator being of trace-class is presented by Russell Lyons in the errata to the mentioned paper. Furthermore a proof that the formula holds and that the operator is of traceclass, whenever the marginal spaces posses additional nice proporties (isometric embeddability into hilbert spaces), is also presented in this errata.

@Renart I think there are more subtle problems. For example $y\mapsto K(x,y) = \sum_{n}e_n(y) \int e_n(z) K(x,z)dz,$ only in $L^2(\mathbb{R})$ for each fixed $x$, which by definition means that for any $x$ $\|K(x,\cdot)-  \sum_{n=1}^k e_n(\cdot) \int e_n(z) K(x,z)dz\|_2\to_k 0.$ But in order to say that $\int \sum_{n=1}^\infty e_n(x)\int K(x,y)e_n(y)dy dx= \int K(x,x) dx$ we need stronger convergence. Anyways, all this assumes that the operator is of trace class, which i haven't even verified.

Well i don't have tried it myself but... From what i read the answer of jonhatan doesn't really use the fact that it's on $\mathbf R$. The only thing that need justification is the permutation of a sum and an integral at the very end. Fubini works pretty independently of the space you're concidering.

@Renart Yes that is indeed one of the questions that i have read, suggesting the trace approach. Though this is a integral kernel over $\mathbb{R}$, and i'm quite sure that problems arise when integrating over arbitrary metric spaces.

If not possible in the above setup, can the equality be shown for when $\mathcal{X}$ and $\mathcal{Y}$ are separable Hilbert spaces?