Wikipedia:Reference desk/Archives/Mathematics/2012 April 8

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April 8

Polynomials modulo p

Given p, a prime, I am concerned with finding the roots of a polynomial f(x) mod p. Can anyone suggest where I might find information regarding algorithms I might use to investigate this? Thanks. meromorphic [talk to me] 10:45, 8 April 2012 (UTC)

See our articles on factorization of polynomials over a finite field and irreducibility tests, Berlekamp's algorithm, and Cantor–Zassenhaus algorithm. Gandalf61 (talk) 09:18, 9 April 2012 (UTC)

"Free" Nonassociative ring?

I'm certain I'm reinventing the wheel here (albeit with perhaps a somewhat distasteful choice of terminology and notation, since aesthetics has never been my strong suit), so my prompt is simply: please point me to the right literature to find out more about this.

Let ${\displaystyle R}$ be a Nonassociative ring, in which we recall that multiplication is not required to be invertible, have an identity, be commutative, or even (hence the name) be associative. For the most basic instance of this construction, we could take ${\displaystyle R=\{0\}}$. Let ${\displaystyle S}$ be a set of symbols, chosen to be disjoint from ${\displaystyle R}$. We may define the set of all "monomials" ${\displaystyle M}$ recursively as follows:

• If ${\displaystyle r\in R\backslash \{0\}}$, then the 1-tuple ${\displaystyle (r)\in M}$ is a monomial. If ${\displaystyle s\in S}$, then ${\displaystyle (s)\in M}$.
• If ${\displaystyle m_{1},m_{2}\in M}$ and at least one of the two is not equal to ${\displaystyle (r)}$ for some ${\displaystyle r\in R}$, then the 2-tuple ${\displaystyle (m_{1},M_{2})\in M}$.

This definition is carefully constructed to avoid any accidental collision of elements, as long as a 1-tuple cannot be interpretted as a 2-tuple and vice-versa. A monomial is built to encode the tree of a fully parenthesized multiplicative expression in elements of ${\displaystyle R}$ and ${\displaystyle S}$, with the caveat that adjacent multiplied elements of ${\displaystyle R}$ be collapsed into a single element of ${\displaystyle R}$.

Define the ring ${\displaystyle R[S]}$ as the set of all functions ${\displaystyle \sigma :M\to \mathbb {Z} }$ such that ${\displaystyle \sigma (m)=0}$ for all but finitely many ${\displaystyle m\in M}$. We think of this intuitively as ${\displaystyle \displaystyle \sigma =\sum _{m\in M}\sigma (m)m}$. Addition is pointwise, and multiplication is fun to define (actually, it almost ends up being easier to define than it usually is, as a consequence of the fact that it's usually possible to extract the original factors from a product of two monomials, but that's a bit of a tangent). It trivially satisfies the requirement that it form a group under addition, since as an additive set, it is nothing more than a direct sum of ${\displaystyle R}$ with many copies of ${\displaystyle \mathbb {Z} }$, and distributivity takes a bit more work, but it again follows as essentially a direct consequence of distributivity of multiplication over addition in ${\displaystyle \mathbb {Z} }$.

Then ${\displaystyle R[S]}$ satisfies a universal property: for any nonassociative ring ${\displaystyle R'}$ with a map ${\displaystyle \phi :R\to R'}$, and a function ${\displaystyle f:S\to R'}$, there is a unique extension ${\displaystyle {\hat {\phi }}:R[S]\to R'}$, with ${\displaystyle {\hat {\phi }}((r))=\phi (r)}$ and ${\displaystyle {\hat {\phi }}((s))=f(s)}$ for ${\displaystyle r\in R,s\in S}$.

For nonassociative rings, the definition of a two-sided ideal needed to define a quotient ring is identical to that for a noncommutative ring (although the ideal generated by an element or subset of a nonassociative ring is a much worse monster). --COVIZAPIBETEFOKY (talk) 17:01, 8 April 2012 (UTC)

On Hypergeometric Functions

Given the integral formula for the hypergeometric function ${\displaystyle F(a,b;c;z)={\frac {\Gamma (c)}{{\Gamma (b)}{\Gamma (c-b)}}}\int _{0}^{1}t^{b-1}(1-t)^{c-b-1}(1-tz)^{-a}\,dt}$, I have to express the integral ${\displaystyle \int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\!{\frac {e^{it(u+1)}}{e^{it}+iA}}\,dt\,}$ in terms of a combination of hypergeometric functions. I am keen to do as much of the legwork by myself as possible, so no answers or hints that would remove all challenge from the question please, but I have no clue how to express functions, however they may be represented, in terms of hypergeometric functions. Could someone please give me some help? Thanks. meromorphic [talk to me] 17:13, 8 April 2012 (UTC)

Hilbert's second problem

There is reference to : "Dawson (2006:sec. 2)" but no other identifying information. I would like to pursue reading his discussion; please help completing that reference; the citation itself & more on who Dawson is. Thank you. — Preceding unsigned comment added by 75.141.237.29 (talk) 18:20, 8 April 2012 (UTC)

The Hilbert's second problem article has Harvard-style citations, which means that you need to look down in the References section to see which Dawson is being referred to and what he wrote in 2006. One advantage of the Harvard style is that footnotes can then be used for notes rather than for citations. Hopefully this is enough information for you to find the Dawson ref; if not, please feel free to ask more. --Trovatore (talk) 19:18, 8 April 2012 (UTC)

yes, thank you. This is my first time using this forum - apologies for any format/procedure errors. But mainly, thank you for re-directing me to the references section again - it was inverted from the sequence I was used to; that's what caught me up (looking at the bottom of the list rather than the top. CeptualInstitute. — Preceding unsigned comment added by 75.141.237.29 (talk) 20:03, 8 April 2012 (UTC)

Showing an integral is well defined

Given ${\displaystyle \int _{0}^{\infty }\left({\frac {e^{-bx}}{e^{ia}e^{x}-1}}-{\frac {e^{bx}}{e^{-ia}e^{x}-1}}\right)\,dx\,}$ where a>0, ${\displaystyle a\not =2n\pi }$, 0<b<1, I am supposed to show that this integral is well defined. I'm not entirely sure what this process constitutes though. Is it essentially just showing that this integral exists and is finite? If so, how do you do this without directly evaluating it? Thanks. meromorphic [talk to me] 19:51, 8 April 2012 (UTC)

An improper integral is defined as a limit, so proving that the integral is well-defined means proving that the limit exists. You don't necessarily need to evaluate it to do that, although you'll probably need to get most of the way to evaluating it (if you know a particular term is finite you don't need to actually plug the numbers in and work it out, you can just move on to the next term). --Tango (talk) 21:46, 8 April 2012 (UTC)
Can't you use Cauchy's residue theorem and a suitable choice of contour? I get the integrand having poles at z = ±ia, while the residues are ∓eiab respectively. Alternatively, try multiplying the numerator and denominator of the first quotient by the denominator of the second and vice versa. I got a cosine and a hyperbolic sine when I tried. 02:29, 9 April 2012 (UTC)