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Sep 26, 2013 at 22:26 history wiki removed François G. Dorais
Jul 27, 2011 at 1:55 comment added Noam D. Elkies You're welcome!$$ $$Getting an explicit upper bound on $\|\Delta\|$ is a calculus exercise. We may assume $\tau = x+iy$ with $y^2 \geq 3/4$ (fundamental domain). Then the factor $\prod_{n=1}^\infty (1-q^n)^{24}$ of $\|\Delta\|$ is at most $\prod_{n=1}^\infty (1+e^{-\sqrt{3}\pi n})^{24}$, while the rest is $y^6 e^{-2\pi y}$ which is maximized at $y = 3/\pi$, etc.$$ $$It seems (and is probably known) that in fact the max occurs at the sharper corner $\tau = (\pm 1 + \sqrt{3}i)/2$ of the fundamental domain, where $\|\Delta\| = .002+ = (2\pi^2/9)^6 / \Gamma(2/3)^{36}$.
Jul 26, 2011 at 13:00 comment added Ariyan Javanpeykar Thank you very much. This is a great answer. Do you think there's any hope in making the uniform upper bound on $\Vert \Delta \Vert$ explicit?
Jul 26, 2011 at 10:42 comment added Noam D. Elkies @Ariyan: As $j(\tau) \rightarrow \infty$ with $\tau$ in the usual fundamental domain, $q \rightarrow 0$ with $q \sim 1/j$. Thus $|q| \sim 1/|j|$. Also Im$(\tau) = \log(1/|q|)/(2\pi) \sim \log|j|/(2\pi)$. This accounts for two of the factors of $\|\Delta\|$, and the remaining factor $\prod_{n=1}^\infty (1-q^n)^{24}$ approaches $1$ as $q \rightarrow 0$. This gives the asymptotic formula for $\|\Delta\|$ in terms of $|j|$, and all the error estimates are readily seen to be effective.
Jul 26, 2011 at 9:10 comment added Ariyan Javanpeykar How does one obtain the effective lower bound for $\Vert \Delta\Vert$ in terms of the absolute value of $j$?
Jul 26, 2011 at 6:39 vote accept Ariyan Javanpeykar
Jul 26, 2011 at 1:34 history edited Noam D. Elkies CC BY-SA 3.0
Replaced "form of this bound" by a more precise formulation (and a paragraph break)
Jul 26, 2011 at 0:57 history answered Noam D. Elkies CC BY-SA 3.0