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Dec 11, 2012 at 22:09 history edited Gerry Myerson CC BY-SA 3.0
typo, tidied link
Dec 11, 2012 at 22:07 comment added Gerry Myerson Since everything is conjectural, I'm not sure that "worst case" has a well-defined meaning. In any event, all I know is what I wrote.
Dec 11, 2012 at 15:47 comment added Charles So the weaker version gives $a^x\ll n^{2+\varepsilon}$. But this applies only to the common case of exponents 2 and 3; is this the worst case?
Dec 11, 2012 at 11:47 history answered Gerry Myerson CC BY-SA 3.0