bio | website | mathoverflow.net/users/19885/… |
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location | Between Two Moments | |
age | ||
visits | member for | 3 years, 1 month |
seen | Dec 18 '14 at 10:37 | |
stats | profile views | 2,330 |
$T$. $S$. $Eliot$, Introduction to Dante's Inferno: " Hell is a place where nothing connects with nothing. "
$Johann$ $von$ $Neumann$: " In mathematics you don't understand things, You just get used to them. "
$A$. $A$. $Zinoviev$: " Where there are problems, there is life. "
$P$. $R$. $Halmos$: "I do believe that problems are the heart of mathematics, and I hope that as teachers, in the classroom, in seminars, and in the books and articles we write, we will emphasize them more and more, and that we will train our students to be better problem-posers and problem-solvers than we are. "
$Shahrooz$: " Physics is mirror front of universe, but mathematics is the rules of reflection. "
Dec 18 |
comment |
Open problems with monetary rewards
Dear Voloch, I think you are right. But it is a little strange for me, why mathematicians should forget the heritage of the great mathematician like MacWiliams? It is just $10$ dollar, and I think someone must accept it and pay it in the future. If I can pay, I will accept it, maybe a good news. |
Dec 17 |
answered | Open problems with monetary rewards |
Dec 11 |
awarded | Yearling |
Nov 14 |
awarded | Nice Answer |
Nov 5 |
answered | Math books for advanced high school students |
Sep 16 |
revised |
Non-DS circulant graphs
added 337 characters in body |
Sep 16 |
comment |
Non-DS circulant graphs
In general, the answer is no, even for special type of graphs. |
Sep 16 |
answered | Non-DS circulant graphs |
Aug 25 |
answered | Largest eigenvalue adjacency matrix-link deletion |
Aug 14 |
comment |
For integers $a \ge b > 1$ is $f(a,b) = a^b + b^a$ injective?
Since nobody answered the question until now (after 8 vote), it seems that it is hard. But where am I wrong? Let $a=b+x$, so for every fixed $b$, the function $f_b(x)=(b+x)^b+b^{b+x}$ is increasing in variable $x$. So induction on $b$ must solve the problem. |
Aug 13 |
revised |
Cospectrality and dimension of graphs
Corrected a mistake |
Aug 12 |
answered | Cospectrality and dimension of graphs |
Aug 10 |
comment |
Do runs of every length occur in this sequence?
@r.e.s- About existence of the limit, you are right. I just talked approximately, but I believe that this limit is so close to $\frac{1}{2}$, even if it is irrational. For your second comment, it was just an optimistic view to convert the problem to bit-wise case (against half-block adding). In that case, the probability that $i$-th position be $1$ must increase for sufficiently large $i$ and it increases the probability of seeing a repeated $1$ sequence. This problem is like a phylogeny in genetics. What is the application of this nice problem? |
Aug 9 |
comment |
Do runs of every length occur in this sequence?
One vote and also favorating question. It seems that the behaviour of consequtive 1 in this squence is more chaotic. This is just a view: let $a_n$ and $L_n$ denote the number of ones and total length of sequnces in the $n$-th iteration, respectively. So (if I do not be wrong), $a_n\thickapprox \frac{3}{2}a_{n-1}$ and $L_n \thickapprox 2(3/2)^n$. Now, $\frac{a_n}{L_n}$ tends to $\frac{1}{2}$, when $n$ goes to infinity. So, I think the conjecture is true, but its proof must be difficult, because of $\frac{1}{2}$. |
Jul 27 |
accepted | irreducible polynomials on the polynomial sequence |
Jul 15 |
comment |
irreducible polynomials on the polynomial sequence
@David, the second question let us to choose carefully the $g_i(x)$, such that $P(x,y)$ generates infinite irreducible polynomial. It is a special case such that we know the conjecture is true for it. Anyway, thanks for your answer. |
Jul 14 |
asked | irreducible polynomials on the polynomial sequence |
Jun 11 |
awarded | Popular Question |
May 22 |
answered | Examples of graph properties characterized by forbidden (not necessarily induced) subgraphs |
Jan 31 |
awarded | Popular Question |