bio | website | mathoverflow.net/users/19885/… |
---|---|---|
location | Between Two Moments | |
age | ||
visits | member for | 2 years, 7 months |
seen | Jul 15 at 6:41 | |
stats | profile views | 2,068 |
$T$. $S$. $Eliot$, 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. "
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 |
Jan 11 |
revised |
Letting $S(m)$ be the digit sum of $m$, then $\lim_{n\to\infty}S(3^n)=\infty$?
added 60 characters in body |
Jan 11 |
revised |
Letting $S(m)$ be the digit sum of $m$, then $\lim_{n\to\infty}S(3^n)=\infty$?
correction |
Jan 11 |
revised |
Letting $S(m)$ be the digit sum of $m$, then $\lim_{n\to\infty}S(3^n)=\infty$?
corrected some symbols |
Jan 11 |
answered | Letting $S(m)$ be the digit sum of $m$, then $\lim_{n\to\infty}S(3^n)=\infty$? |
Jan 9 |
answered | Properties of Graphs with an eigenvalue of -1 (adjacency matrix)? |
Jan 9 |
answered | Reflexive (hyperbolic) graphs |
Dec 11 |
awarded | Yearling |
Nov 27 |
comment |
Combinatorial identities
I just came back again for one up vote to this nice answer. |
Nov 27 |
comment |
Combinatorial identities
It is just a point of view. Suppose we want to construct binary words with length $4n+1$ that one half of these words has weight $n$ and the total weight of these words are greater or equal than $n$. Also, we need the half of these words with this property. This number can be obtained with the left hand side. For the right hand side, we choose $k$ positions from $4n+1$ positions and then from the last selected position (that is 1), we move $n+1$ positions forward (one way for obtaining a word by one half weight greater than $n$) and then choose $n-k$ positions among $3n-k$ remaining positions. |
Nov 10 |
revised |
Does the Alternating group of degree $n>7$ have exactly one irreducible character of degree $n-1$?
spelling edited |
Nov 10 |
comment |
Does the Alternating group of degree $n>7$ have exactly one irreducible character of degree $n-1$?
Dear Jim, thanks for your good spelling comment. |
Nov 10 |
comment |
Does the Alternating group of degree $n>7$ have exactly one irreducible character of degree $n-1$?
Dear Abdollahi, it seems that you are right. When I studied these papers, I thought that it is not difficult to prove your question. But now, I am reading these papers more carefully. |
Nov 10 |
revised |
Does the Alternating group of degree $n>7$ have exactly one irreducible character of degree $n-1$?
deleted 13 characters in body |
Nov 9 |
awarded | Custodian |
Nov 9 |
reviewed | Edit suggested edit on Does the Alternating group of degree $n>7$ have exactly one irreducible character of degree $n-1$? |