- Christopher Stover added an answer:How to decrease the bounded gaps between primes?Christopher Stover · Florida State University
Terence Tao and his collaborators did tons of work on this in Polymath 8 (mentioned ambiguously in the above comment). You can read part of what was done here:
http://terrytao.wordpress.com/2014/09/30/the-bounded-gaps-between-primes-polymath-project-a-retrospective/
Tao's complete corpus of work on that project is:
http://terrytao.wordpress.com/tag/polymath8/
Following - Issofa Moyouwou added an answer:Up to now, what is known on (the maximal) domains that guarantee the transitivity of the majority rule?It is well known that the majority rule may not be transitive for some configurations of individual preferences. Domain restrictions are possible ways out. But what is known about maximal such domains (i) with respect to the cardinality? (ii) via set inclusion?Issofa Moyouwou · University of Yaounde I
Thank you very much. I also like the job done on mutiple issues.
Following - Patrick Solé added an answer:How can we express the recurrence formula about the labeled tree in Maple16?
In Maple 16, how can we with the software combstruct, to give the sentence about the recurrence formula,
A(x)=1+x[A(x)^{3}+3A(x)A(x^{2})+2A(x^{3})]/6
Patrick Solé · Institut Mines-TélécomVery strange, first the gf then the object to count!
Following - Peter T Breuer added an answer:Are there any statisticians to collaborate on a combinatorics challenge for a paper?
I and another three colleagues have an ongoing paper about the application of sociometry to multiple human resource allocation to multiple projects. The problem is mathematically quite challenging but we are on a dead end concerning one of the outcomes we are dealing with.
Particularly, we have been applying meta-heuristics and evolutionary algorithms to solve the problem of allocating groups of people so as to maximize cohesion among them. The research is quite interesting and we think it is going to open multiple and very interesting research and industry application in the near future.
However, we are stuck in one part. We are trying to calculate the number of viable combinations of people who can work either full-time, part-time or not work at all in several simultaneous projects and we need a person with advanced knowledge in combinatorics to give us a hand. We are willing to pay or to put his/her name as co-author on the paper.
Getting straight to the point, the problem statement is as follows:
There are N people (i=1...N) who can be selected to work in P simultaneous projects (j=1...P). Each person can have a dedication of work full-time (1), half-time (0.5) or not work (0), that is, three possible allocations (0, 0.5, 1).
Now, we know that each project j requires Rj people. How many different and viable combinations are there?
Numerical Example:
People available Project j=1 Project j=2 Project j=3=P
i=1 0 or 0.5 or 1 0 or 0.5 or 1 0 or 0.5 or 1 Row sum =<1
i=2 0 or 0.5 or 1 0 or 0.5 or 1 0 or 0.5 or 1 Row sum =<1
i=3 0 or 0.5 or 1 0 or 0.5 or 1 0 or 0.5 or 1 Row sum =<1
i=4 0 or 0.5 or 1 0 or 0.5 or 1 0 or 0.5 or 1 Row sum =<1
i=5 0 or 0.5 or 1 0 or 0.5 or 1 0 or 0.5 or 1 Row sum =<1
Requirements R_1=2 R_2=1 R_3=1 Row sum=4
,that is, we need 2+1+1 people working in these 3 projects and 1 out of the five available people is not used. We can use each person totally (1), half-time (0.5) or not use him/her, but the combinations must be feasible, i.e., each person cannot be assigned over 1 (full-time) and the requirements have to be fulfilled.
This problem is easier when the people can only work full-time or not work, but with half-times is far more complicated.
If anyone think that he/she is able to solve it, or just want further details, please contact me.
Thanks
Peter T Breuer · Birmingham City UniversityYes, notice that I added a few words in my answer saying effectively "well, I think you meant to be generous, and allow work to be shared among too many workers by letting many do only part-work, but ended up accidentally saying that you wanted parsimonious allocations, where as many workers as possible are fully employed; still never mind because the one reduces to the other". The reduction is
[generous] f_{g}(p;n_{1},n_{2})=f_{p}(p;n_{1},n_{2},p-n_{1}-n_{2}) [parsimonious]
There's an extra dummy project to absorb the efforts of under-employed workers.
So one can stick to the easier (parsimonious) calculation, which in the simplest picture is basically a 3-way recursion from p;n_{1},n_{2} to the sum of three smaller values in n_{1},n_{2} at p-1. The three ways are how to to distribute two half-person efforts among two projects. Those are 1+0, 0.5+0.5, 0+1. That leads to a recursion from p;n_{1},n_{2} to p-1;n_{1}-1,n_{2}-0 and p-1;n_{1}-0.5;n_{2}-0.5 and p-1;n_{1}-0;n_{2}-1. One sums those contributions.
If you are going to consider 1/3 allocations, and 1/4 allocations, etc, the same principle applies. For each person you need to consider how many ways there are of distributing his m/m of available effort among exactly n projects.
That's the number of partitions of a set of m objects into up to n labeled partitions (some of the partitions may be empty). Each partition represents the allocation of effort among the n projects by that person. That is the sum of Stirling numbers of the second kind ^{m}S_{0}+^{m}S_{1}+^{m}S_{2}+...+^{m}S_{n}. Unfortunately you don't so much as want that number as actually the partitions themselves. Each partition x_{1}+...+x_{n}=m is a point in the parameter space (p-1;n_{1}-x_{1}/m,...n_{n}-x_{n}/m) to recurse to and the sum of the Stirling numbers that I mentioned is merely a count of the number of terms in the recursion!
I should say that when n=m, i.e. you allow 1/nths of effort to be distributed among exactlu n projects, the count of terms above that I expressed as a sum of Stirling numbers is a well-known number, the Bell number B_{n}. It's big, as you might imagine. There are horrible recursions for it.
All that indicates that you don't actually want to calculate the numbers you say you want to calculate, because they're just nasty and meaningless. They're also unscaled, in that the number goes up if you allow each person to divide their efforts more finely. That should not happen! You should probably be scaling by 1/m^{n}, where n is the number of projects, and 1/m is the minimum unit of effort. Then your numbers would maybe stay on-scale.
I know that you want numbers, but I'm saying they aren't meaningful.
-----------
I'd rather look at approximations. In particular imagine that each person can give away a bit dp of their effort to any of n projects:
f(p;n_{1},n_{2},..)=f(p-dp;n_{1}-dp,n_{2},..)+f(p-dp;n_{1},n_{2}-dp,...)+... (*)
[This is an overestimate. In reality not every assignment of a dp to a project comprises a different overall assignment in the end, since if I assign 2dp to project 1 that is one choice, not two].That's not going to make sense unless I scale by n, the number of projects:
f(p;n_{1},n_{2},..)=(f(p-dp;n_{1}-dp,n_{2},..)+f(p-dp;n_{1},n_{2}-dp,...)+...)/n
in which case one gets
f(p;n_{1},n_{2},...)=f(p;n_{1},n_{2},...) -df/dp- 1/n df/dn_{1}- 1/n df/dn_{2}-...
which says
df/dp = -1/n(df/dn_{1}+...+df/dn_{n})
This is a partial differential equation. My PDEs are rusty, but I believe that is saying that if I were to move by k dp and k ndn_{1}, or k dp and k ndn_{2}, etc, the nett change in f would be zero. That means that this f is an arbitrary function of p-(n_{1}+n_{2}+...)/n . I.e. f(p;n_{1},n_{2},...)=F(p-(n_{1}+n_{2}+...)/n) for some F.
At least my scaling does "remove the infinities", but what am I counting? I scaled by 1/n when allocating a dp of a person, so if I had been allocating 1/mths of a person I would have scaled by 1/n^{m} when allocating all of one person. Therefore my number is 1/n^{mp} times too small for what you intend to count.
If you start with p=n_{1}+...+n_{n} exactly (supply equals demand), I conclude that your numbers grow as F(p-p/n) n^{mp} in the limit when you can allocate small amounts 1/m of each of p persons and distribute them among n projects.
The work n_{i} needed by each particular project does not seem to matter.
I don't know what the constant of proportionality F is, but the prediction is that it depends closely only on the number p of persons available (p-p/n, exactly speaking). I presume it is monotonic increasing, so you would be ok in taking it as a bound not an asymptote and writing F(p) instead of F(p-p/n). If one thinks about the one-project case (that needs p workers), the implication is that it is exactly 1.
[Note: this is an overestimate, deriving from the overestimate at (*) at the beginning of this approximation].
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There's a more accurate way of expressing the intention of yours that I expressed as (*) above, namely:
f(p;n_{1},n_{2},..)= Integral f(p-1;n_{1}-x_{1},n_{2}-x_{2},..) dA
where the integral is the sum that appeared in (*) extended to take account of the whole possible set of assignments of a unit of work from p to the projects 1 to n, so the integral is over the sums x_{1}+x_{2}+...=1 for x_{i} in the range [0,1]. That's the "diagonal" plane of the unit hypercube, so I suppose the area is something like 2^{n/2}, but you can tell me!
This would make sense as a stochastic expectation if one were to divide by that 2^{n/2}. I suggest the x_{i} are uniformly distributed over the range [0,1] taking into account the x_{1}+x_{2}+...=1 stipulation, which is to say that each equally sized subunit of the area of that hyperplane through the unit cube is equally probable. That's why dividing by the total area A=2^{n/2} make sense. It makes the distribution stochastic (total probability = 1).
The expectation is a kind of average count of the weighted number of ways of dividing up the work, counting a tiny fraction weight for each possible way.
If one extends this integral to an assignment of all of p, then
f(p;n1,n2,..)= Integral 1 dA
where the integral is over all x_{i} in the range [0,n_{i}] such that x_{1}+x_{2}+...=p. I believe one should weight all these possibiities equally again to get an expectation, which is to say that each subunit of equal area of the hyperplane x_{1}+x_{2}+...=p=n_{1}+n_{2}+... in the scaled cube [0,n_{1}]x[0,n_{2}]x... gets equal weight [but I'm willing to take suggestions!]. I imagine the total area of that plane segment is 2^{n/2} n_{1}n_{2}... or something very like that, so I have divided by that in getting an expectation, which will be E[1]=1 and the count represented by the unvarnished integral above is that multiplied by what I divided by, which is 2^{n/2} n_{1} n_{2} ....
However, the same scaling arguments as I applied for my earlier estimation applies again. Whereas I have been applying an integral in which each "way" of dividing up the effort counts for a tiny increment in the integral above, you have been counting it as "1" for each different way.... so I need to scale up by something more again to get to the kind of count you were doing. I'll work it out later, but I guess the extra scaling number is m^{n-1}. So the estimate here is something like 2^{n/2} n_{1} n_{2} ...n_{n} m^{n-1} for your numbers. This replaces the m^{np} over-estimate I had before [or will when I check my working and perfect the working .. watch this space].
Following - James F Peters added an answer:How do I use the Ptolemy inequality to study the geodesic angle?here, we try to use the ptolemy inequality to study the geodesic angle. and we introduce 2 preliminary results :
1. the ptolemy theorem in euclid spherical geometry
by studying the matrix of 4 quadruple points on the euclid spherical, we find the fact that, since there are 6 lines which connect these 4 points, so if 5 of the 6 lines are equal. then we will get that the sixth line's length is double as the other equally 5 lines;this result imply that the angle between the lines is included in $\frac{\sqrt{a}}{2r}$$\in(2k\pi+\pi/5,2k\pi+5\pi/6)$ ; then we substitute this result into the discriminant, and we get an inequality about the radius in $n+1$ dimension:
$op_{n+1}^2$$\ge$$\frac{r^2}{t^2}(1+\frac{r^2}{n})-\frac{r^2}{n}$
2. ptolemy inequality in minkowski geometry
here ,we use the centroid method to study the n-polygon problem in minkowski geometry. firstly, we introduce a well known problem such that:
if every angle of a polygon is equal, and the sidelines are :
$1^2,2^2,......,N^2$, then $\sum{n_{s}}e^{isa}=0$
by this theorem, we can factorize the mass on the vertex into pairs and the number of pairs is primes $N/2=\Pi{p_{i}^{a_{i}}}$ ,the weight of each pair is $\sum{4k-1}$,then we can divide these pairs into groups and every group has prime points too(page 21 in[1]); consequently we can divide the sidelines into 2 parts: $1^2,3^2,......$ and $2^2,4^2,......$;
the next step is to construct regular n-polygon and use the ptolemy inequality to make a regulation for the average of the sum of the mass in different group, and we can rearrange these groups of mass to ensure the first part $1^2,3^2,......$ is larger than the average and the second part $2^2,4^2,......$ is less than it.
therefore we can apply this average of sum to the distance formula in minkowski geometry in polar coordinate (page 24 in [1]).
here we also use combinatorics method (result we get in step 2) to study the natural logorithm in distance formula of minkowski geometry(page 25).
our goal is to represent the polar angle in minkowski geometry as the product of the mass lie on different vertex (page 26 in [1]).
so our question is how to apply the 2 results above to the geodesic angle?
by the inequality for the radius in $n+1$ dimension euclid spherical, we can ensure $v_{n+1}\ge0$; consequently we can substitute the representation of the polar angle in step 2 to spherical equation , which imply that we can also restrict the range of
$cos^2{\varphi}$ that is $[\sqrt{33}-4l,3]$
lastly,we apply the property of ptolemy space to get our estimate for geodesic angle,the bounde is :
$1-4e^2d/3+\frac{2}{3-4/3e^{2d}}$
is this method feasible? for more detail, you can refer to :
application of the ptolemy theorem (3) (page 15-27)
the analysis techniques for convexity: CAT-spaces (3)James F Peters · University of ManitobaThis is a good question with lots of interesting paths to consider in looking for answers. This question leads to a consideration of metric spaces that are Ptolemaic and Buseman convex spaces, stemming from
H. Busemann, The Geometry of Geodesics, 1955
A good place to start is
T. Foertsch, A. Lytchak, V. Schroeder, Non-positive curvature and the Ptolemy inequality:
http://carma.newcastle.edu.au/jon/Preprints/Papers/CAT(0)/Papers/fls-ptolemy.pdf
These authors prove that a metric space is CAT(0) if and only it is Ptolemy and Busemann convex, p. 11. Ultrarays assocated with geodesics and their enclosed angles are considered, starting on page 8.
Following - Lie Zhu added an answer:Can anyone suggest some good reference papers for beginners in the field of Combinatorial Design?It can either be related to Key Distribution or any other application, just an overview needed.Lie Zhu · Suzhou University
Design Theory, Zhe-Xian Wan, World Scientific Publishing Co Pte Ltd, 2009
Following - Ricky X. F. Chen added an answer:The maximum number of cycles of the product of a given permutation and a n-cycle?
Given a permutation A on the set [n], is there a way to determine the maximum number of disjoint cycles of AC where C ranges over all n-cycles on [n]? For which class of permutations A, this problem has been studied before? Thanks
Ricky X. F. Chen · University of Southern DenmarkThe titles of the papers I mentioned are something like:
Zagier, on the distribution of cycles....
Stanley, two enumerative results on product of cycles...
Following - Lie Zhu added an answer:Can anyone recommend ways to find a skew starter for a room square of side 667?Skew room squares exist for all odd values greater than 5. If n is prime it is a simple matter to generate a skew starter. But 667 is not prime. 667=23*29 which means a computer search has to be done in order to generate one. I would be satisfied with the skew room square of side 667, even though we can show it exists we can't seem to construct it. Any suggestions on this particular problem are appreciated.Lie Zhu · Suzhou University
The reference provided 10 minutes ago may not be appropriate.
But, one may try some direct product construction as follows.
Start from skew starter S_1 in Z_{23} and skew starter S_2 in Z_{29}.
Find a permutation P=(p_0,p_1,...,p_{28}) such that
P-I = (p_0-0, p_1-1, ..., p_{28}-28)
And
P+I = (p_0+0, p_1+1, ..., p_{28}+28)
Are both permutations in Z_{29}. For example, take p_s=2s.For each pair {x, y} in S_1, construct 29 pairs {(x,p_s), (y,s)}, s=0,1,…,28.
This generates 11x29 pairs.
Add 14 more pairs {(0,u),(0v)} such that {u,v} is from S_2.It may be easy to verify if the 11x29+14 pairs form a skew starter in Z_{23}xZ_{29}.
If it still does not work, please let me know.Following - Feng Qi added an answer:What is the power series expansion at zero of the secant to the power of three?
It is well known that the secant $\sec z$ may be expanded at $z=0$ into the power series
\begin{equation}\label{secant-Series}
\sec z=\sum_{n=0}^\infty(-1)^nE_{2n}\frac{z^{2n}}{(2n)!}
\end{equation}
for $|z|<\frac\pi2$, where $E_n$ for $n\ge0$ stand for the Euler numbers which are integers and may be defined by
\begin{equation}
\frac{2e^z}{e^{2z}+1}=\sum_{n=0}^\infty\frac{E_n}{n!}z^n =\sum_{n=0}^\infty E_{2n}\frac{z^{2n}}{(2n)!}, \quad |z|<\pi.
\end{equation}
What is the power series expansion at $0$ of the secant to the power of $3$? In other words, what are coefficients in the following power series?
\begin{equation}
\sec^3z=\sum_{n=0}^\infty A_{2n}\frac{z^{2n}}{(2n)!}, \quad |z|<\frac\pi2.
\end{equation}
It is clear that the secant to the third power $\sec^3z$ is even on the interval $\bigl(-\frac\pi2,\frac\pi2\bigr)$.Feng Qi · Tianjin Polytechnic UniversityDear All, How are you going? I have a good news to tell you: Several days ago I found a method to give a closed form for coefficients of MacLaurin series of the function $(sec z)^{a}$, where $a$ may be any given number. When I complete themanuscript, I would announce here.
Following - Jerry Rhee added an answer:What is your favourite combinatorial object, or mathematical object you find the most fascinating?I always am curious of what types of combinatorial or mathematical objects (in general) interest other researchers. For myself I have a pretty keen interest in types of restricted weak integer compositions (weak implies 0 is applied in integer sequence whereas the lack of this description means you can't use 0 in integer sequences, and restricted means finding subsets of these objects).
What combinatorial or mathematical object (preferably more elusive ones) fascinate you the most, or are your favouriteNot sure if the following qualifies but this phi spiral?
Following - Peter T Breuer added an answer:Which operations can be constructed using only xy and x+y+z?
What method can I use to charcterize which operations repeated multiplication xy and 3-input addition x+y+z are able to construct mod 4, or mod 8, or .. or mod 2^{32}?
The reason I ask is because one can prove that it is impossible to construct a known constant from unknown inputs using these operators (no formula f(x,y,..)=K for all x,y,..), and one can also characterise those other operators that one can add into the mix with them that preserve the property.
But I don't know what the totality of operations one can construct with xy and x+y+z is. Indeed, I haven't figured it out for plain old xy and x+y ! (yes, I can describe those: "all polynomials"; but how many are there and what operations cannot be formed as polynomials?)
Any clever ways to go about this? I am trying symbolic computation to generate the orbits of "x" and "y" under substitution in the binary operators xy and x+y+z mod 4.
Peter T Breuer · Birmingham City UniversityStrangely, I see 1436 operator tables constructed by xy and x+y mod 4. That's multiplication and the ordinary addition, not the more rarified three-input addition.
Shouldn't these plain old operations construct all polynomials in x,y with 0 constant term? Going by the coefficients, there should be 4^{15} of them (15 coefficients taking the values 0,1,2,3). But I suppose there are polynomials with different coefficients that are functionally equal, since there are divisors of zero mod 4? The difference might be zero mod 4, but have nonzero coefficients?
Can anyone count the number of functionally distinct multinomials in x,y mod 4?
How many polynomials are there? 64, according to the computer:
[0,1]x^{3}+[0,1]x^{2}+[0,1,2,3]x+[0,1,2,3]
That's 16 that are zero at zero.
That was due to reductions from 2x^{3}=2x^{2}=2x, and x^{4}=x^{2}. I'll try counting the multinomials. There are 4 possible coefficients of x, y and xy. There are 2 possible coefficients of x^{2}, y^{2}, x^{2}y,xy^{2},x^{2}y^{2}. There are 2 possible coefficients of x^{3},y^{3},x^{3}y,xy^{3},x^{3}y^{3}, x^{3}y^{2},x^{2}y^{3}. That is all. That's 4^{3} 2^{12}, or 2^{18}. It's no good .. I'll have to count them computationally.
What if a polynomial (zero at zero) is zero on all odd numbers? The coefficients satisfy a_{3}+a_{2}+a_{1}=0=a_{3}-a_{2}+a_{1}, so 2a_{2}=0 and a_{2}=0 or a_{2}=2. In one case a_{3}=-a_{1}, and the other a_{3}=2-a_{1}. so one has -a_{1}x^{3}+a_{1}x and (2-a_{1})x^{3}+2x^{2}+a_{1}x. If the polynomial is equal at even numbers too, then 2a_{1}=0 so a_{1}=0 or a_{1}=2. The polynomial is then 0, 2x^{3}+2x, 2x^{3}+2x^{2}, 2x^{2}+2x.
That gives four versions of each polynomial with the same functionality. This is the kernel of the embedding of polynomials into functions. It's a simple count .. there are 4^{4}/4=4^{3}=64 polynomials by functionality, 16 with zero constant term.
What if a multinomial x^{3}p_{3}(y)+x^{2}p_{2}(y)+xp_{1}(y)+p_{0}(y) is functionally zero?
p_{3}(y)+p_{2}(y)+p_{1}(y)+p_{0}(y)=0
-p_{3}(y)+p_{2}(y)-p_{1}(y)+p_{0}(y)=0
2p_{1}(y)+p_{0}(y)=0
p_{0}(y)=0
So 2p_{3}(y)=2p_{2}(y)=2p_{1}(y)=p_{0}(y)=0.
So p_{3},p_{2},p_{1} take only even values (0 or 2) and p_{0}(y)=0.
So p_{0} is one of the four functionally zero polynomials (that's 4 options).
That p takes only even values means it is b_{3}x^{3}+b_{2}x^{2}+b_{1}x+[0,2], where one or three of b_{1},b_{2},b_{3} are even: [0,2]x^{3}+[1,3]x^{2}+[1,3]x+[0,2], [1,3]x^{3}+[0,2]x^{2}+[1,3]x+[0,2], [1,3]x^{3}+[1,3]x^{2}+[0,2]x+[0,2], [0,2]x^{3}+[0,2]x^{2}+[0,2]x+[0,2], for 2^{4}4=2^{6}=64 options.
That is a total of 64^{3}4=2^{20} functions.
That would bring the number of functionally distinct multinomials down to about 4^{16}/2^{20} =2^{12}=4096! I don't seem to be able to get a stable estimate for this quantity :-(. My computer says the answer is 3686, 906 of which are zero at zero. That can't be right.
Following - Feng Qi added an answer:What are the values of the special Bell polynomials of the second kind?
Bell polynomials of the second kind B_{n,k}(x_{1,}x_{2},...,x_{n-k+1}) are also called the partial Bell polynomials, where n and k are positive integers. It is known that B_{n,k}(1,1,...,1) equals Stirling numbers of the second kind S(n,k).
What are the values of the special Bell polynomials of the second kind B_{n,k}(0,1,0,1,0,1,0,...) and B_{n,k}(1,0,1,0,1,0,...)? Where can I find answers to B_{n,k}(0,1,0,1,0,1,0,...) and B_{n,k}(1,0,1,0,1,0,...)? Do they exist somewhere?
Feng Qi · Tianjin Polytechnic UniversityBai-Ni Guo and Feng Qi, Explicit formulas for special values of Bell polynomials of the second kind and Euler numbers, ResearchGate Technical Report, available online at http://dx.doi.org/10.13140/2.1.3794.8808.
Following - Brian S. Blais added an answer:Can anyone give me an interpretation or link about improper uniform prior as a prior distribution in Bayesian estimation?
I need a good reason why improper uniform prior could be use as a prior in Serial Numbered Population (SNP) problem. Or maybe someone can tell me about improper uniform prior itself. An pdf link might be very helpful for me.
I attached the journal.
Brian S. Blais · Bryant UniversityI'm not sure if Jochen's answer is entirely correct; "An improper prior can be used when the resulting posterior is proper." I'd modify it to say that an improper prior can be used when the resulting posterior can be achieved with a *proper* prior and the limits of the distribution, after the posterior is calculated, is taken to infinity and the result is shown to be the same. For example, you can use a proper, bounded uniform prior and then, after you calculate the posterior, take the limits as the bounds grow to infinity to show that if you used the improper prior in the first place it would give the same answer.
Seen this way, the improper prior is a short-cut to make the analysis cleaner. For example, I personally find E T Jaynes' analyses very clear because he uses improper priors (when he can!), and warns that one has to be careful to do it properly if there is even a hint of trouble. I find a paper like Bretthorst's "Difference of Means" paper to be a much harder read because he uses proper priors throughout, even in cases where I think the improper prior would work.
Following - Feng Qi added an answer:Can anyone help me with the inequality for the ratio of two Bernoulli numbers?
Dear Colleagues
I need an inequality for the ratio of two Bernoulli numbers, see attached picture. Could you please help me to find it? Thank you very much.
Best regards
Feng Qi (F. Qi)
Feng Qi · Tianjin Polytechnic University[1] Feng Qi, A double inequality for ratios of Bernoulli numbers, ResearchGate Dataset, available online at http://dx.doi.org/10.13140/2.1.2367.2962.
[2] Feng Qi, A double inequality for ratios of Bernoulli numbers, RGMIA Research Report Collection 17 (2014), Article 103, 4 pages; Available online at http://rgmia.org/v17.php.
Following - Miroslav Rypka added an answer:Can anyone suggest references for Iterated Function Systems and combinatorics: uniqueness of addresses?
Could anyone suggest me some references about the uniqueness of the addresses of an Iterated Function System? E.g. points of Cantor set can be coded by a unique address, how general is this property?
More precisely, I am interested in the sufficient conditions to have a one-to-one relation between the shift space of the Iterated Function Systems and its attractor.
Thank you in advance!
Miroslav Rypka · Palacký University of OlomoucDear Anna,
There are three kinds of attractors of iterated function systems, totally disconnected, just touching and overlapping. Totally disconnected attractors have metrically equivalent structure to the Cantor set. This can be found in Barnley's book Fractals Everywhere. The remaining cases may be treated with the help of lifted IFS, which is also explained in the book.
Best regards
Miroslav
Following - Prasanth G. Narasimha-Shenoi added an answer:Let A & B be two square matrices such that A^2 is not equal to B^2, A is not equal to B but A^3=B^3 & A^2B=AB^2. What is determinant of A^2-B^2?If A and B Are non-singular then we have determinant of (A^2-B^2)=-determinant of AB , Am I correct? If so, then the problem is If A and B are singular, how can we prove?Prasanth G. Narasimha-Shenoi · Government College Chittur
@ Samuli Yeah the facts are correct. Sorry that I am not able to answer
Following - Sergey Perepechko added an answer:What is the formula to find the number of simple cycles in a graph? Is this problem NP Complete?I would like to list out all simple cycles in a connected graph.Sergey Perepechko · Petrozavodsk State University
You need to give additional details about your problem. If you know adjacency matrix of your graph and want to count number of simple cycles of fixed length look at slides and papers in my profile.
Sergey Perepechko
Following - Mostafa Eidiani added an answer:Which Software Package is the best for computations in Codes over Rings?What is the best software package available for carrying out computations related to algebraic coding theory, especially codes over rings.Following
- Albert Manfredi added an answer:Can anyone help me find amortized splay tree operation derivation?Please help me with a derivation of "amortized cost of a splay tree operation" . I am waiting for that please consider it fast and let me know.Albert Manfredi · The Boeing CompanyBoth of these articles include proofs. Not sure whether this is what you're asking. Hope they help.
http://www.cs.cornell.edu/Courses/cs312/2006fa/recitations/rec20.html
http://www.bowdoin.edu/~ltoma/teaching/cs231/fall10/Lectures/13-amortized/splay.pdfFollowing - Alina Abraham added an answer:In integral theory, how can you integrate all four perspectives in human development into a model?According to Wikipedia, http://en.wikipedia.org/wiki/Permutation, if you have 3 objects [a,b,c] combinatorics there are 6 possible permutations. But positioning in space and time are not considered. What if a 'c' is placed 0,5xx behind [a,b] or 'a' and 'b' separately either horizontal/vertical, or 'c' and 'b' are isolated back to back and 'a' spins around and around for 30 seconds more until reaching a stable position?
I am looking at Integral theory and trying to integrate all four perspectives in human development (internal-individual; internal-collective; external-individual; external-collective) into a model. I am not satisfied with those similar to the above ''max 6 possible permutation of the three elements". I believe that for 3 objects there are more than 1+2+3 permutations; and for 4 objects (my case) more than 10.Alina Abraham · ICL Business SchoolAdapting Ken Wilber/Terri O'Fallon model, using a spiraling trajectoryFollowing - Christopher Landauer added an answer:In terms of combinatorics, could (6677,333,166) be cyclical on the torus?Would like to know if (6677,333,166) could be cyclical on the torus.Christopher Landauer · The Aerospace CorporationThe notation is ambiguous - what combinatorial object are these the parameters for? Is it some kind of design? If so, what kind, and what parameters are indicated? The problem is that there are hundreds of different parameterized combinatorial designs that use the same kind of notationFollowing
- Mohan Shrikhande added an answer:Is there a database available on the net of symmetric designs?Symmetric designs=combinatorial 2-designs with as many points as blocks.Mohan Shrikhande · Central Michigan UniversityNice to hear from you, Patrick!Following
- Daniel Crespin added an answer:Spin - could you provide some clarification?I always thought that the sum of angular momentum should be zero. I am wondering if this applies on a sub-atomic scale as well as a macroscopic scale.Daniel Crespin · Central University of VenezuelaHello Donald,
A reasoning similar to yours implies that the total linear momentum of the Universe is zero. And furthermore, there should be a motionless center of mass of the Universe.
But how many technical details regarding these arguments can be worked out is not so clear. It may be worth a trial.
The term "spin" in Physics usually refers to angular momentum of atomic scale objects. Technically, for n >= 3 the special orthogonal group SO(n) is connected and has first homotopy group (also called fundamental group) isomorphic to Z_2, the integers modulo 2. Therefore its universal covering space, denoted Spin(n), is a well defined connected topological group (and a Lie group as well) with fibers consisting of two points, equivalently, it is a connected double cover.
If you consider instead the orthogonal group O(n), this is non-connected, has two connected components and the connected component of the identity is SO(n). The universal cover of O(n) is a topological space known as Pin(n), necessarily a non-connected double cover of O(n). But the group structure of Pin(n) is not unique. This subtlety is mentioned in http://en.wikipedia.org/wiki/Pin_group See references there.
In Classical Physics, the natural formalism for rotations in ordinary three dimensional space is based on SO(3), its tangent bundle, and the Lie algebra so(3) = tangent space to SO(3) at the identity. A good reference is the book Classical Mechanics by V. Arnold.
Then comes Quantum Mechanics and descendants. Schrödinger time dependent equation involves complex numbers. This forces the introduction of complex valued wave functions \psi. To go beyond the mathematical formalism of QM, the mathematical object \psi requires physical interpretation. The wave "amplitude" |\psi|^2 is interpreted "physically" as a probability distribution. This makes the "phase" disappear physically, because |\psi|^2 and |\exp(-i n t) \psi|^2 are one and the same physical state. But the phase is the natural way to consider rotations. Thus, rotations disappear in QM.
The quantistic way to recover something resembling rotation is to use Pauli matrices, equivalently, to use spin(3). I have been unable to make sense of these as rotations. Here the quantum dictum "Shut up and calculate" is acutely present. A nearby neon sign says "Abandon hope all ye who enter here".
In my opinion the unfortunate and mistaken choice of a unitary evolution equation ---that is, of Schrödinger evolution equation--- for the hydrogen atom made it impossible to understand microscopic rotational phenomena. Even worse, transitions themselves are contradicted by this equation.
On the other hand, Schrödinger eigenvalue equation is one of the most impressive wonders of science.
Most cordially and with best regards,
Daniel CrespinFollowing - Feng Qi added an answer:Can anyone help me with a combinatorial interpretation?I am asking for a combinatorial interpretation of a formula for Bell numbers in terms of Kummer confluent hypergeometric functions and Stirling numbers of the second kind. See the formula (8) and Theorem 1 in the attached PDF file or http://arxiv.org/abs/1402.2361．Could you please help me? Thank a lot.Feng Qi · Tianjin Polytechnic UniversityYes, th formula (8) is not the only such formula. Some mathematician asked me to provide a combinatorial interpretation, but I do not know the combinatorial meanings of the formula (8). I think that "the formula (8) just represents one more expression for those numbers, using Kummer confluent hypergeometric function" may be not a combinatorial interpretation of the formula (8).Following
- Marcel Van de Vel added an answer:Does the discrete n-circle (n even) admit a partition into n/2 pairs, all with a distinct diameter?A (discrete) n-circle is the set of complex n-th roots of unity, or: the vertices of a regular n-gon. The above question arose as part of a (nearly finished) research project on a method to produce unpredictable number sequences. Although my partial answers are no longer needed for the project, the simple-looking and still unsettled problem keeps intriguing me. I proved that if a partition exists into pairs of distinct diameters, then n must be of type 8k or 8k+2 (k>0 integer). Computer generated examples confirm that for n <= 112, these types are *exactly* the sizes that work. The computer was stopped after running for two days on the case n=114 (having inspected nearly 0.000...001% (about 300 zeros) of the total search space). The only hope on further information must come from construction methods other than brute-force search with back-tracking and from proofs. Specifically, the problem becomes this: Design an algorithm that is guaranteed to produce a partition (as desired) whenever there exists one and reports failure otherwise. Unlike the current backtracking brute-force search, the algorithm should provide answers in a reasonable time. [Added 09-12-2013: solved] The problem is certainly NP (Nondeterministic Polynomial), but chances are that it is NP-complete. [Added 09-12-2013: not NP complete] A weaker problem is to find a number b <= n/2 such that *any* b vertex pairs with different diameters can be rotated apart in the n-circle for *any* (even) n. It might be "(n/2)-1", I haven't checked on this yet. Ultimately, one should be able to determine the best b for each individual n (including the odd case). [Added 09-12-2013: this is still wide open. Exhaustive computer search is getting quite demanding, even for fairly low n]Marcel Van de Vel · VU University AmsterdamI have finished my paper containing the questions that I collected in my original posts. There are several more questions (and partial answers) in it, which all arose with the development of one major result on a theoretical method to produce unpredictable numbers.
As this paper is intended for publication in a regular journal, I cannot place it it here in public. If anyone is interested in receiving a copy (25 pp), please let me know. I'll send a pdf file by e-mail. The paper classifies mainly as combinatorial mathematics.Following - Henning Meyerhenke added an answer:How close is spectral partitioning to the solution of the min-cut problem?There are many approaches using different matrices and eigenvectors to solve the min-cut problem. What is the best theoretical result providing a good approximation from a spectral cut to the solution?Following
- Donald beverly Giles asked a question:Is there any interest in a very different spreadsheet algorithm for generating incidence matrices of projective planes?I have devised an algorithm for generating the incidence matrices of projective geometries not shown in traditional texts. Will include as an attachment. This design is self dual and relies on the 4 mols of order 5.Rather than use M*Mt can use M^2.Following
- Christopher Landauer added an answer:Square Root of a Symmetric MatricesThe square root of a 31 by 31 matrix with 6"s down the main diagonal and 1"s elsewhere is a symmetric binary matrix with six 1's in each row and column. If someone has an algorithm for this square root then perhaps they can apply it to a larger matix that Iam presently working on. This larger matrix is an 81 by 81 matrix with 16's down the main diagonal and 3's eveywhere else. The square root of this will be a binary symmetric martix with sixteen 1's in each row and column. For me this is not a simple problem. My knowledge of matrices does not extend to taking square roots of symmetric matrices and getting symmetric binary matrices as the answer.Christopher Landauer · The Aerospace Corporationsince the given matrices are linear combinations of the identify and the matrix commonly denoted J (the all 1 matrix), the eigenvalues are simple to compute (the eigenvalues of the k x k matrix J are one k and the rest 0), but only one of them is a square (36 in the first 31 x 31 example, 256 in the 81 x 81 example), and the rest are all equal (5 in the first example and 13 in the second) - also, since J . J = kJ, a square root of J is J/sqrt(k) - therefore, it is easily shown that a square root of rI + sJ in k x k matrices can be found having the same form aI+bJ, with
r = a^2, s = 2*a+b^2*k, which is easily solved for a and b (need r >= 0 for a to be real, and s-2*a >= 0 for b to be real) - in the two given examples, r=5, s=1 and r=13, s=3, then second condition does not hold and b will be imaginary
there are, of course, other square roots, as other people pointed outFollowing - Sanjib Kuila added an answer:Why does the four colorability of planar graphs not ensure the non-biplanarity of K_9?I hope the proof of the four color theorem is sufficient to explain the answer.Sanjib Kuila · Panskura Banamali CollegeI have already studied, that paper of Beineke. I also gone through the original works of J Battle et. al., and that of Tutte. In this topic, Harary wrote in his book, "No elegant or even reasonable proof is known."Following
- Donald beverly Giles added an answer:Proof of the existence of balanced tournament designs?I'm writing my senior research paper on balanced tournament designs, and I am looking for the proof of the existence of them which is in this paper.Donald beverly Giles · Board of Governors of the Federal Reserve SystemJennarose: Ron Mullin, Waterloo's first graduate student, proved the existence of room squares for all odd v greater than 5. The last one that needed proving was 257 which was constructed in the orient in the mid 70's. Further to this skew room squares exist for all odd v greater than 5. A skew room square is a room square where exactly one the cells (i,j) or (j,i) is occupied and the other cell is empty.
Don Giles is converting skew room squares of order 4n+3 into symmetric block designs with parameters (4n+3,2n+1,n) which in turn are being converted into Hadamard matrices of order (4n+4). All of these combinatorial structures are interrelated.
Paul Schellenberg developed the room square of order 25 as part of his PhD. He too was a student and then professor at Waterloo in the late 60's and early 70's
Addition and multiplication theorems were used after a collection of smaller room squares had been established by a host of researchers.
Hope this helps you with BALANCED TOURNAMENT DESIGNS>
I can elaborate more on their usage if you wish.Following
About Combinatorics
Combinatorics is a branch of mathematics concerning the study of finite or countable discrete structures. Aspects of combinatorics include counting the structures of a given kind and size (enumerative combinatorics), deciding when certain criteria can be met, and constructing and analyzing objects meeting the criteria (as in combinatorial designs and matroid theory), finding "largest", "smallest", or "optimal" objects (extremal combinatorics and combinatorial optimization), and studying combinatorial structures arising in an algebraic context, or applying algebraic techniques to combinatorial problems (algebraic combinatorics).
Last year, Dr. Yitang Zhang has published a paper for the upper bound of twin primes, which is 7*10^7. Does anyone has idea to achieve lower bounded gap?