algorithms – Find maximal difference between two consecutive numbers

Determine the minimum $a$ and maximum $b$ in a first pass.

Create $n+1$ buckets, each of width $frac{b-a}{n+1}$, with the first beginning at $a$. We have $n+1$ buckets and only $n$ points, so by the pigeonhole principle, there must be at least one empty bucket. Since we know that the first and last buckets contain at least one point, we know furthermore that there must be an empty bucket with a nonempty bucket somewhere to its left and another nonempty bucket somewhere to its right, implying that the solution must be at least $frac{b-a}{n+1}$.

This means that no two points within the same bucket can provide the solution, since they must be strictly less than $frac{b-a}{n+1}$ apart from each other. So, any solution will consist of a pair of points from different buckets.

A valid solution must consist of the rightmost point from some bucket $i$ and the leftmost point from some bucket $j > i$ with all buckets in between (which may be no buckets at all, when $j=i+1$) empty. So it is sufficient to know, for each bucket, its leftmost and rightmost points.

To compute these, iterate through all the points again, updating the leftmost and rightmost positions of the current point’s bucket whenever necessary. (For the first point in a bucket, both of these positions will need to be updated.) A point’s bucket can be computed by subtracting $a$ and dividing by $frac{b-a}{n+1}$.

Finally, iterate through all the buckets, keeping track of the rightmost point in the last nonempty bucket seen, and comparing it with the leftmost point in the current bucket: Update the maximum whenever this exceeds the current maximum.

ca.classical analysis and odes – Maximal solution and its domain of $x'(t) = x(t)^{x(t)}$

How can I proof that the maximal solution to

x'(t) = x(t)^{x(t)}, quad x(0) = x_0,

where $x_0 > 0$ and $t geq 0$, is not (globally) defined on $mathbb{R}_{0}^{+}$?

I am given the hint that it might help to first look at $x_0 > 1$. Unfortunately, that does not help. I thought about maybe taking into account the derivative of $x^x$ which we can calculate and kind of substitute it to this problem, which at least up to now has not helped either. I am close to giving up on this one so any help is appreciated.

maximal detail in extreme low light: longer lens or shorter and faster lens?

Have you considered flash? A cobra-head zooming flash going to 200mm tends to have a guide number of about 70 at that length. Its reach at F2.8 and ISO25600 would be 400m. For full exposure.

When working with a 200mm lens, 400m is actually quite a distance. Though you would want to work with a dark background and a lit subject, so you’ll likely reduce exposure time to your flash duration (typically 1/200s) and possibly dial down ISO a bit.

Birds of prey can sit still a lot better than photographers so you could use longer exposure times, too, but particularly with the longer lenses that would require use of a tripod.

As you can see, with 1/50s and the fastest lens, your best is a washed-out silhouette against a noisy backdrop.

There are of course a number of disadvantages for a flash shot, too: probably the worst is that with a shy subject at a good distance you get to have exactly one shot per opportunity. Another is that structures such as pylons might be equipped with reflectors or reflective signs that basically halve the distance to the camera. While you can edit them out in post, they will mess with any metering (like TTL).

Also autofocus is not going to do an overly convincing job.

And there will be the red-eye equivalent of birds’ eyes, and particularly eyes with good nocturnal vision will be affected.

harmonic analysis – Hardy Littlewood maximal function bounds

Let $u in W^{1,p}(mathbb{R}^n) in L^{infty}(mathbb{R}^n)$ be a given function for some $1<p< infty$ and let $k in mathbb{R}$ be any number and consider the following maximal function
mathcal{M}_{leq k}(|nabla u|)(x) = sup_{R>0}frac{1}{|{B_R(x):u leq k}|} int_{{B_R(x):u leq k}} |nabla u| dx.

Question: Is it true that $mathcal{M}_{leq k}(|nabla u|)(x)$ is finite almost everywhere?

Analogous question regarding $mathcal{M}_{geq k}(|nabla u|)(x)$ can also be asked.

reference request – Finite simple groups with three conjugacy classes of maximal local subgroups

In (1) it was proved that

A finite nonsolvable group $G$ has three conjugacy classes of maximal subgroups if and only if $G/Phi(G)$ is isomorphic to ${rm PSL}(2,7)$ or ${rm PSL}(2,2^q)$ for some prime $p$.

This implies that, among simple groups, only only ${rm PSL}(2,7)$ and ${rm PSL}(2,2^q)$ have three conjugacy classes of maximal subgroups.

My question: I wonder if we can also find all simple groups with three maximal local subgroups.

A subgroup is a local subgroup if it is the normalizer of some nontrivial subgroup of prime power order. A proper local subgroup is a maximal local subgroup if it is maximal among proper local subgroups.

Maximal subgroups are not necessarily local and maximal local subgroups are not necessarily maximal subgroups. I know that the three non-conjugate maximal subgroups of ${rm PSL(2,4)}=A_5$ and ${rm PSL(2,7)}$ are local respectively, but is it true that ${rm PSL}(2,2^q)$ has three conjugacy classes of maximal local subgroups for each prime $p$? And how can I find all simple groups with such property?

Any help is appreciated!


(1) Belonogov, V. A.: Finite groups with three classes of maximal subgroups. Math. Sb., 131, 225–239 (1986)

time complexity – Maximal subsets of a point set which fit in a unit disk

Suppose that there are a set $P$ of $n$ points on the plane, and let $P_1, dots, P_k$ be not necessarily disjoint subsets of $P$ such that every point in $P_i| 1 leq i leq k$ fits inside a unit disk $D_i$.

Moreover, each $P_i$ is maximal. This means that if the corresponding unit disk $D_i$ moves to cover another point, then one point which was inside the disk will be uncovered.

Here is an example:
enter image description here

In the above figure, there are three maximal subsets.

I don’t know whether this problem has a name or was studied before, but my question is:

  1. Can $k$ be $O(1)^n$?
  2. If not, then can we find those subsets in polynomial time w.r.t. $n$?

reference request – Maximal order of $x^n-d$ and its dependence on $d$

It’s well known that the structure of the maximal order of $mathbb{Q}(sqrt{d})$ depends on $d$ modulo $4$: (assuming $d$ is squarefree), the maximal order is $mathbb{Z}left(frac{1+sqrt{d}}{2}right)$ if $d equiv 1 mod{4}$ and $mathbb{Z}(sqrt{d})$ otherwise. The structure of the biquadratic extension $mathbb{Q}(sqrt{d},sqrt{e})$ has a similar but more complicated description.

Now let’s consider $K=mathbb{Q}(sqrt(n){d})$. For simplicity, let’s consider only $d$ prime. Based on some SAGE computation, it seems that there is a degree $n$ extension $M$ of $mathbb{Q}(mu_{n^2})$ such that $mathcal{O}_K$ contains more than just $mathbb{Z}(sqrt(n){d})$ iff $d$ splits in $M$. (Note that if $n$ is prime, then $M$ is unique.) For $n=2$, one sees that $M=mathbb{Q}(i)$.

Is this known? It doesn’t seem too hard to prove that if $d$ is split, then there are more algebraic integers in $K$. But does this appear anywhere?

ag.algebraic geometry – In $mathbb{C}[x,y]$: If $langle u,v rangle$ is a maximal ideal, then $langle u-lambda,v-mu rangle$ is a maximal ideal?

I have asked the following question at MSE and got one answer. It would be nice to have more elaboration on it, please:

Let $u=u(x,y), v=v(x,y) in mathbb{C}(x,y)$, with $deg(u) geq 2$ and $deg(v) geq 2$.
Let $lambda, mu in mathbb{C}$.

Assume that the ideal generated by $u$ and $v$, $langle u,v rangle$, is a maximal ideal of $mathbb{C}(x,y)$.

Is it true that $langle u-lambda, v-mu rangle$ is a maximal ideal of $mathbb{C}(x,y)$?

My attempts to answer my question are:

(1) By Hilbert’s Nullstellensatz, $langle u,v rangle= langle x-a,y-b rangle$, for some $a,b in mathbb{C}$, so
$x-a=F_1u+G_1v$ and $y-b=F_2u+G_2v$, for some $F_1,G_1,F_2,G_2 in mathbb{C}(x,y)$.
Then, $x=F_1u+G_1v+a$ and $y=F_2u+G_2v+b$.

(2) $frac{mathbb{C}(x,y)}{langle u,v rangle}$ is a field (since $langle u,v rangle$ is maximal); actually, $frac{mathbb{C}(x,y)}{langle u,v rangle}$ is isomorphic to $mathbb{C}$. Is it true that $frac{mathbb{C}(x,y)}{langle u,v rangle}$ is isomorphic to $frac{mathbb{C}(x,y)}{langle u-lambda,v-mu rangle}$? In other words, is it true that $frac{mathbb{C}(x,y)}{langle u-lambda,v-mu rangle}$ is isomorphic to $mathbb{C}$?
See this question.

(3) If $langle u-lambda,v-mu rangle$ is not maximal, then it is contained in some maximal ideal: $langle u-lambda,v-mu rangle subsetneq langle x-c,y-d rangle$, $c,d in mathbb{C}$.
It is not difficult to see that $(u-lambda)(c,d)=0$ and $(v-mu)(c,d)=0$,
so $u(c,d)-lambda=0$ and $v(c,d)-mu=0$, namely,
$u(c,d)=lambda$ and $v(c,d)=mu$.

Remark: Is it possible that $langle u-lambda,v-mu rangle = mathbb{C}(x,y)$. If so, then there exist $F,G in mathbb{C}(x,y)$ such that
$F(u-lambda)+G(v-mu)=1$. Then at $(a,b)$ we get:
$F(a,b)(-lambda)+G(a,b)(-mu)=1$ (since, by (1), $u(a,b)=0$ and $v(a,b)=0$).

Thank you very much!

Maximal Result

All trader actually want to get maximum trading result, but in forex trading, this is not a simple way, loss also as part in trading which as a trader must considering with this risk, learn how to determine best entry point and exit point is a crucial part to make better skill, forex trading will have more fun if we can make money regularly