Is there an infinite, countable connected $T_2$-space $(X,\tau)$ such that $(X,\tau)$ has the fixed point property? (This means that for every continuous map $f:X\to X$ there is $x\in X$ such that $f(x) = x$.)

Let $S_1,S_2,\dots,S_k$ be subsets of the set $S=\{1,2,\dots,n\}$, not necessarily distinct. We will color each element of $S$ red, green, or blue. From this coloring, each set $S_i$ will receive one or more color according to the following rule:

Let $r_i,g_i,b_i$ denote the number of red, green, and blue elements of $S_i$, respectively, and let $m_i=\max(r_i,g_i,b_i)$. If $r_i\geq m_i-1$, we give the color red to $S_i$. Similarly for green and blue.

What is the maximum constant $d$ for which we can always color the elements of $S$ in such a way that for any color, at least a fraction $d$ of the sets $S_i$ receive that color?

An algorithm that starts with a two-coloring and change the color of one element at a time to the third color achieves $d=1/5$, while an example shows that $d=1/3$ is the best one can hope for.

I recently heard the following statement in a talk:

A complex random process is stationary if its Fourier transform is band limited and the phases are delta correlated. In that case, the correlation length of the process is proportional to the bandwith of the Fourier transform.

Where, as I understand it:

phase delta correlated means that if the process Fourier transform is $\hat{\psi} (\omega) = r_{\omega}e^{i\varphi _{\omega}}$, then $$\left(\varphi_{\omega}, \varphi_{\nu}\right)_{L^2} = \delta (\omega -\nu) \, .$$

The $L^2$ product is over the probability space (i.e., all possible realizations).

- Stationary means that its correlation function $R(t,t')$ obeys $R(t,t') = r(|t-t'|)$.

Because it was stated very informally, and I have almost no background in stationary processes, I have three, basic **questions**:

- Is this true
*as stated*, or does it need further refinements to be exact? Could you refer to a proof? - Is this condition also necessary, or only sufficient?
- What assures us that the Fourier transform exists for
*every*realization?

Is there a real number $A$ such that $$\left \lfloor n^{A} \right \rfloor$$ is a prime number ($\forall n \in \mathbb{N})$? It is obvious that $A>1+\epsilon$ from the prime number theorem.

The ambient space if ${\bf M}_n({\mathbb R})$.

Let us begin with facts. 1- The cone of positive semi-definite symmetric matrices is convex. 2- It is a little subtler that the cone $K^+$ of matrices with non-negative determinant is rank-one convex; this is because $\det(M+xy^T)=\det M+x^T\hat My$, where $\hat M$ is the cofactor matrix. Therefore, whenever $\det A,\det B\ge0$ and $B-A$ is rank-one, then the determinant is $\ge0$ over the segment $(A,B)$.

Is there an interesting example (say, a non-convex one) of a "singularly convex" cone ? By singularly convex, I mean that if $A,B\in K$ and $\det(B-A)=0$, then $(A,B)\subset K$.

Of course, if $n=2$, singular convexity is just rank-one convexity.

Mind that I am specially interested in cones formed of symmetric matrices.

In "Black Swan" (Page 159 in the Penguin 2010 Edition) Taleb makes an interesting estimate of potential project overruns. As example, he takes a Project that is planned to terminate in 79 days. If the Project is still not finished on day 79, then he states it should be expected to take another 25. If still not finished on day 90, should take a further 58. 100 gives 89, 119 gives 149, and at 600, you will wait 1'590 days.

In the notes (Page 391 in my Penguin), he gives his basis which is a probability based on a power-law, where f=Kx^1.5

He also gives his background theory : this is two integrals which I have no idea how to write with this editor !

I guess I am just not smart enough to follow his Maths - and would really like to understand this in detail.

If anyone could take me through the steps, I would be immensely grateful.

Regards

Bill

Suppose $C$ is a small category with a monoidal structure. Then by the special case of the Day convolution theorem for presheaves, $\operatorname{Psh}(C)$ is equipped with a corresponding biclosed monoidal structure. If $C$ is equipped with a Grothendieck topology, is there any useful condition for when the biclosed monoidal structure on presheaves descends to a biclosed monoidal structure on the category of sheaves for that Grothendieck topology?

Let $S_n$ denote the symmetric group on $n$ symbols. Given a subgroup $H\le S_n$ and a particular cycle structure, is there a relatively efficient way to compute the number of elements of $H$ having that cycle structure?

Let $A,B$ finitely generated local $k$ algebra and $m$ be the maximal ideal of $A$. Suppose $B$ is finite type and flat over $A$. Let $G$ be a finitely generated $B$ module which is $A$ flat. Then given a B module homomorphism $\oplus_{i=1}^r B\xrightarrow{\phi}\oplus_{i=1}^r B$, consider $Hom_B(\oplus_{i=1}^r B,G)\xrightarrow{\phi}' Hom_B(\oplus_{i=1}^r B,G)$.

Now consider $\oplus_{i=1}^r B\otimes_A k\xrightarrow{\phi\otimes k}\oplus_{i=1}^r B\otimes_A k$ and $Hom_B(\oplus_{i=1}^r B\otimes_A k,G\otimes_A k)\xrightarrow{(\phi\otimes k)}' Hom_B(\oplus_{i=1}^r B\otimes_A k,G\otimes_A k)$.

Also consider $Hom_B(\oplus_{i=1}^r B,G)\otimes_A k\xrightarrow{\phi'\otimes k} Hom_B(\oplus_{i=1}^r B,G)\otimes_A k$.

Is it true that $\phi'\otimes k =(\phi\otimes k)'$ ?

It might happen because $G$ and $B$ are flat over $A$.

Hi i am currently playing a game, where you earn experience by crafting stuff. Now is their a feature where leveling in crafting is become faster. But to be more efficient with mine materials, I want to work out some formulas to caluculate the materials for an x amount of experience. But i can't find the exact formula.

So there are 3 variables: M (Material), P (Product) and exp (Experience). The case is, for every 2 materials you get 4 products and 8 experience, but to make leveling easy, for every 4 products you get 1 material and 8 experience. And i want to make a formula, with a input of amount of experience and output amount of material.

With this case in mind, I got this formulas: (m / 2) = 4p + 8exp and (p / 4) = m + 8exp.

My thought was to rewrite the second formula, so i can set that one in the first formula. So (p/4) = m + 8 exp, should become 4p = 16m + 128 exp, multiply both sides with 16. So now there are 2 the same variables: 4p. Now put the second in the first and i get: (m / 2) = (16m + 128exp) + 8exp; => (m / 2) - 16m = 136exp with m >= 2. I wasn't satisfied with this formula and i think i made a mistake somewhere...

Can someone help me with this case and maybe give some advise what i could do better.

Here are a couple of curious related results about a generalized 2-player 1-set tennis game: the winner of the set is the first player to win $n$ games, and the winner of each game is the first player to win $k$ points. Let $p>1/2$ be the probability that the stronger player wins a point, and let $P(n,k,p)$ be the probability that the stronger player wins the set. Then

- $P(n,k,p)>P(nk,1,p)=P(1,nk,p)$ if $n,k>1$, and
- $P(n,k,p)>P(k,n,p)$ if $n>k>1$.

The way the paper goes about proving this is pretty clever, but I'm still wondering if there is a more direct combinatorial proof. Say, for a rational $p$, after clearing denominators, we are counting more of something on the left than on the right because of such-and-such an injection.

Here I consider cuspidal automorphic representations $\pi$ over the similitude group $\mathrm{GSp}(4,\mathbb{A}_\mathbb{Q})$. Let $f$ be a non-zero vector in the representation $\pi$. I want to know if there is any reference/work on relating special values of the complex adjoint $L$-function $L(s,\pi,\mathrm{Ad})$ of $\pi$ to the Petersson norm $\langle f,f\rangle$.

I know there is an article of Atsushi Ichino ('On critical values of adjoint $L$-functions for $\mathrm{GSp}(4)$') on this topic. Yet this article assumes $\pi$ to be unramified over all the finite places and to be of a special type over the archimedean place. So can we remove or weaken these assumptions? Is there any work after this?

Any comment or suggestions would be welcome.

Let $f:X\longrightarrow Y$ be a map between CW-complexes $X$ and $Y$. By the Whitehead Theorems, if one of the conditions:

1- (homotopy version) $\pi_n (f):\pi_n (X)\longrightarrow \pi_n (Y)$ is an isomorphism for all $n\geq 1$,

or

2- (homology version) $\pi_1 (f):\pi_1 (X)\longrightarrow \pi_1 (Y)$ and $H_n (\tilde{f}):H_n (\tilde{X})\longrightarrow H_n (\tilde{Y})$ are isomorphisms for all $n\geq 2$,

hold, then there is a map $g:Y\longrightarrow X$ such that $g\circ f\simeq id_X$ and $f\circ g\simeq id_Y$.

Question: Is there any weaker condition (with respect to above conditions) under which there is a map $g:Y\longrightarrow X$ such that we have only $g\circ f\simeq id_X$?

We know that the Pitman efficacy of the RSS sign test is

$eff(S^{+}_{RSS})=\frac{meff(S^{+}_{SRS})}{4\sum_{i=1}^{m}\frac{B(0.5;i,m+1-i)B(0.5,m+1-i,i)}{B^{2}(i,m+1-i)}}$

But I do not know why the Pitman efficacy of the MRSS sign test is

$eff(S^{+}_{MRSS})=\frac{eff(S^{+}_{SRS})}{4^mB(0.5;c_i,m+1-c_i)B(0.5,m+1-c_i,c_i)}$

Can you help me? Thanks a lot

In this paper R.H. Bing has constructed his famous example of a countable connected Hausdorff space.

The Bing space $\mathbb B$ is the rational half-plane $\{(x,y)\in\mathbb Q\times \mathbb Q:y\ge 0\}$ endowed with the topology consisting of the sets $U\subset \mathbb B$ such that for any $(a,b)\in U$ there exists $\varepsilon>0$ such that each point $(x,0)$ with $\min\{|x-(a-b/\sqrt{3})|,|x-(a+b/\sqrt{3})|\}<\varepsilon$ belongs to $U$.

It is easy to see that the action of the homeomorphism group on $\mathbb B$ has at most two orbits.

**Problem.** Is the Bing space $\mathbb B$ topologically homogeneous?

**Remark.** It seems that the first example of a topologically homogeneous countable connected Haudorff space was constructed by Joseph Martin. A simple transparent example of such space is the rational projective space $\mathbb QP^\infty$ discussed in this MO post. It is interesting who first realized that the space $\mathbb QP^\infty$ is connected?

We know the following theorem

$\bullet$ For fixed m and k, $P(S^{+}_{RSSU}=y)=\sum_{r_1+...+r_m=y}\prod_{i=1}^{m}\left( \begin{array}{c} k \\ r_i \end{array} \right)\frac{B^{k-r_i}(F(\theta);i,i)B^{r_i}(\overline{F}(\theta);i,i)}{B^{k}(i,i)}$, $y=0,1,...,n,$ where $\overline{F}(\theta)=1-F(\theta)$.

For any given population distribution, the powers of the three sign tests can be computed. I can not prove for example when m=3 and n=24, the powers of $S^{+}_{RSS}$, $S^{+}_{MRSS}$ and $S^{+}_{RSSU}$ are

$\beta_{RSS}(\theta)=\sum_{y=0}^{8}\sum_{r_1+...+r_m=y}\prod_{i=1}^{m}\left( \begin{array}{c} k \\ r_i \end{array} \right)\frac{B^{k-r_i}(F(\theta);i,m-i)B^{r_i}(\overline{F}(\theta);i,m-i)}{B^{k}(i,m-i)}$,

$\beta_{RSSU}(\theta)=\sum_{y=0}^{7}\sum_{r_1+...+r_m=y}\prod_{i=1}^{m}\left( \begin{array}{c} k \\ r_i \end{array} \right)\frac{B^{k-r_i}(F(\theta);i,i)B^{r_i}(\overline{F}(\theta);i,i)}{B^{k}(i,i)}$,

$\beta_{MRSS}(\theta)=\sum_{y=0}^{7}\left( \begin{array}{c} 24 \\ y \end{array} \right)\frac{B^{24-y}(F(\theta);C_m,m-C_m)B^{y}(\overline{F}(\theta);C_m,m-C_m)}{B^{24}(C_m,m-C_m)}$,

Could you help me to prove the three above equations? Thanks alot

If $(X_n,\mathcal{F_n})_{n\in \mathbb{N}}$ is a martingale such that $\forall$ n $\in \mathbb{N}, \frac{X_{n+1}}{X_n}\in L^1$ How can be demonstrated that:

$\mathbb{E}[\frac{X_{n+1}}{X_n}]=1$ and that the random variables $\frac{X_{n+1}}{X_n}$ and $\frac{X_n}{X_{n-1}}$ are uncorrelated?

Although the MO question Limit of lights in rooms was quickly closed, it suggests a related question:

** Q0**. What is the probability that a random quadrilateral $Q$
is entirely illuminated from a random point $p \in Q$?

What is a "random quadrilateral"?
What constitutes a random simple polygon is not easily answered.
See Jeff Erickson's webpage on this topic.
But perhaps we can take this
as the following definition of a *random quadrilateral*:

Let $p_i$ be four points uniformaly randomly distributed in a unit-radius disk. Let $Q$ be the quadrilateral formed by connecting the points in order: $(p_1,p_2,p_3,p_4,p_1)$. A certain fraction of these will have crossing segments, and the others will be simple polygons. My simulations suggest that about half are simple, half have segment crossings.

** Q1**. What percentage of quadrilaterals (as described above) are simple (non-self-crossing)?
Near $\tfrac{1}{2}$? Exactly $\tfrac{1}{2}$?

The answer to ** Q1** is likely known, but I could neither
find it nor derive it.

Among the random simple quadrilaterals $Q$,

** Q2**. For what proportion does a random internal
point $x$ insides $Q$ illuminates all of $Q$?

Any point in the kernel (yellow) will illuminate the quad.
I expect the answer to ** Q2** is well above a half, nearer to (but less than) $1$.

If two parallel lines are cut by a transversal, the alternate interior angles are equal, and the corresponding angles are equal.

(Spherical Angle definition: A spherical angle is a particular dihedral angle; it is the angle between two intersecting arcs of circles on a sphere. It measured by the angle between the planes containing the arcs .)

Through a given point, only one line can be drawn parallel to a given line.

For example, two lines can intersect in no more than one point, intersecting lines have equal opposite angles, and adjacent angles of intersecting lines are supplementary.

For any line R and any point P which does not lie on R. This implies that there are through P an infinite number of coplanar lines that do not intersect R.

In On independence and large cardinal strength of physical statements we see that their are physical *statements* which are independent of ZFC, and even strong cardinal axioms. There were many answers, but unfortunately do not lead to finite experiments whose outcomes are independent of ZFC.

We can get close by running a machine that looks for contradictions in ZFC. If ZFC is inconsistent in our universe, then after (non-standard) finite amount of time, the machine will output a contradiction in ZFC. If ZFC is consistient in our universe though, this machine will never halt, and running the machine for a finite amount of time is not guaranteed to answer the question (indeed, assuming ZFC is consistient in our universe, it won't).

If we had a computer that could perform its first step in $1$ second, second step in $\frac12$ seconds, its third step in $\frac14$ steps, etc, we could perform the experiment in finite time. If ZFC is inconsistent, there will be some finite non-standard real number $r=\frac{2^{n}-1}{2^{n-1}}$ such that machine halts in the $r$ seconds (if the machine runs for $n$ steps). If ZFC is consistent, it will never halt after $r$ seconds for any $r \in [0,2)$. After $2$ seconds, we will know what the result is.

The problem is that, as far as we know, no such machine exists. Is there some other way we could devise an experiment whose result is independent of ZFC? I think it will have to involve the real numbers in some way, but I'm not sure.