Skip to main content
Search type Search syntax
Tags [tag]
Exact "words here"
Author user:1234
user:me (yours)
Score score:3 (3+)
score:0 (none)
Answers answers:3 (3+)
answers:0 (none)
isaccepted:yes
hasaccepted:no
inquestion:1234
Views views:250
Code code:"if (foo != bar)"
Sections title:apples
body:"apples oranges"
URL url:"*.example.com"
Saves in:saves
Status closed:yes
duplicate:no
migrated:no
wiki:no
Types is:question
is:answer
Exclude -[tag]
-apples
For more details on advanced search visit our help page
Results tagged with
Search options not deleted user 46744

Questions in which polynomials (single or several variables) play a key role. It is typically important that this tag is combined with other tags; polynomials appear in very different contexts. Please, use at least one of the top-level tags, such as nt.number-theory, co.combinatorics, ac.commutative-algebra, in addition to it. Also, note the more specific tags for some special types of polynomials, e.g., orthogonal-polynomials, symmetric-polynomials.

1 vote
Accepted

Integral with Bessel function and hypergeometric function ${}_2F_2$: explicit expression for...

polynomials can be expressed using analogous of the Lommel polynomials for the modified Bessel functions. … $r_k^0(x)$ and $s_k^0(x)$ are given either in terms of the Lommel polynomials or explicitly. …
Paul Enta's user avatar
  • 791
9 votes

Are these two new ways of representing odd zeta values as integrals known?

One may characterize more precisely the family of these polynomials. … cos\pi y/2}Q(y)\,dy=\left( -1 \right)^p8\left( 2^{2p+3}-1 \right)\frac{\zeta(2p+3)}{\pi^{2p+3}} \end{equation} The condition reads \begin{equation} Q(y)+Q(-y)=2g_p(y) \end{equation} The first polynomials
Paul Enta's user avatar
  • 791
5 votes

Are these two new ways of representing odd zeta values as integrals known?

{\pi^{2n-1}}\left( 2-2^{-2n} \right)\zeta(2n+1) \end{equation} where $E_n(x)$ are the Euler polynomials. … This result gives an explicit representation of the polynomials derived in my previous answer. …
Paul Enta's user avatar
  • 791