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CountVonCount · #22 10-26-2016, 03:32 PM

Quote:

Originally Posted by magdon

Suppose $e^{-\frac12\epsilon^2N}\ge\frac14$

Then, $e^{-\frac18\epsilon^2N}\ge e^{-\frac12\epsilon^2N}\ge\frac14$ .

In which case $4 m(2N) e^{-\frac18\epsilon^2N}\ge 4m(2N)/4\ge 1$ and the bound in Theorem A.1 is trivial.

Hello,

thanks for the answer. I understand this argument, however this holds also for

$e^{-\frac12\epsilon^2N}\ge\frac18$

or for

$e^{-\frac12\epsilon^2N}\ge\frac{1}{16}$

Thus the value 1/4 is somehow magic for me.

Edit: I think you choose 1/4 because it is so easy to see, that the RHS of Theorem A.1 gets 1. Nevertheless with a different value you would get a different outcome of the final formula.