# Bounding the product of exponents

I was reading de Weger's paper on bounding the cardinality of the Tate-Shafarevich group and in lemma 1 (pg 111), he claims that for any $n \in \mathbb{N}$, we have that $$c(n) << N^{((log \; 3 / 3)(1+\epsilon))/ log \; log \; n)}.$$ where $c(n)$ is the product of the exponents of the prime decomposition of $n$.

This is confusing however as in page 108, he defines $N$ as the conductor of an elliptic curve over $\mathbb{Q}$, and he doesn't seem to redefine it anywhere else.

Following his proof I computed that $$c(n) < n^\delta exp\left(-\dfrac{log \; 3}{3}+\delta \; log \; n\right)$$ where $$\delta = \dfrac{\frac{log \; 3}{3}(1 + \frac{\epsilon}{2})}{log \; log \; n}.$$

So my question is if this is indeed a typo or is there some connection here between $c(n)$ and $N$ that I am indeed missing here?

Also it is indeed a typo and if $$c(n) < n^\delta exp\left(-\dfrac{log \; 3}{3}+\delta \; log \; n\right)$$ then how does the result follow for $n$ sufficiently large?

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Looking at the proof, it's clearly a typo. And your estimate for $c(n)$ doesn't seem to be what he writes. Simply take his estimate for $c(n)$ (which is $\log n$ times a bunch of stuff) and exponentiate --- you get $n$ raised to the power of a bunch of stuff, which is exactly what he says it should be.
I continued from the point he said that $$log \; c(n) - \delta \; log \; n < (log \; 3 / 3)(\epsilon/2) (log \; n/log \; log \; n)$$ to derive the bound I mentioned above. Maybe I'm missing something. – Eugene May 20 '12 at 7:31
@Eugene: Dear Eugene, Rewrite this as $\log c(n) < (\log n)(\delta + (\log 3 /3)(\epsilon/2)/\log\log n)$ and exponentiate, to get $c(n) < n^{\delta + (\log3 / 3)(\epsilon/2)/\log\log n}$ and substitute in the given value of $\delta$, and I think everything should be clear. Perhaps I'm missing something? Cheers, – Matt E May 20 '12 at 12:25