Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For an asymptotic estimate for the count of $y$-smooth numbers between $0$ and $x$, with the Dickman-de Bruijn function $\Psi(x, y) \approx x \rho(u)$ where $u = log( x)/log( y)$, Hildebrand and Tenenbaum provides the simple bounds for $\rho(u)$:

$0 < \rho(u) < 1/ \Gamma(u + 1)$.

Does an upper bound for $\rho(u)$ necessarily imply that $x/\Gamma(a+1)$ is an upper bound for $\Psi(x, y)$? My preliminary data suggests that this is true. If so, is it because $x\rho(u)$ is asymptotic that makes this true?

If not, I think I have some brute force checks to do to compare this function versus known upper bound formulas.

References: http://archive.numdam.org/ARCHIVE/JTNB/JTNB_1993__5_2/JTNB_1993__5_2_411_0/JTNB_1993__5_2_411_0.pdf

Page 426.

share|improve this question

1 Answer 1

An upper bound for $\rho(u)$ should not necessarily imply an upper bound for $\Psi(x,y)$. Pomerance asked whether it is true that $\Psi(x,y)\geq x\rho(u)$ for all $x\geq 2y\geq 2$. Thus a non-trivial lower bound for $\rho(u)$ might imply a non-trivial lower bound for $\Psi(x,y)$. The upper bound for $\rho(u)$ is rather weak, and thus in a rather large $xy$-region your upper bound for $\Psi(x,y)$ might be true, but this requires proof...

share|improve this answer
2  
Pieter: I've gone ahead and TeXified your answer; have a look and let me know if I've made any errors. In general, TeX in answers works much like math-mode TeX everwhere else, including the \$ and \$\$ delimiters; if you have any other math-mode questions, check out the guideline at meta.math.stackexchange.com/questions/107/… . –  Steven Stadnicki Oct 30 '12 at 17:31

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.