#1




When the growth function = 2^N
Please elaborate a little bit more in the text why this is an issue. It is crucial that the bound be polynomial. So why is 2^N bad, but N^9999 good?

#2




Re: When the growth function = 2^N
Because exponential kills polynomial. I.e., for any real number a, in the limit N > infinity N^a/2^N goes to zero.

#3




Re: When the growth function = 2^N
Quote:
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#4




Re: When the growth function = 2^N
Dear Professor,
Could you explain why you have chosen a polynomial? Could you have chosen another type of function, or series, for example? A sine or cosine? Thanks! 
#5




Re: When the growth function = 2^N
I don't think we have chosen anything. Any function that can be killed by the exponentially decreasing term in Hoeffding is of course welcomed, but technically, polynomial is what has been proved. Hope this helps.
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When one teaches, two learn. 
#6




Re: When the growth function = 2^N
The polynomial is multiplied by a negative exponential. For any k>0, alpha, Lim_{x>\infinity} \alpha*p(x)*e^(kx) =0. On other for l>k, k>0, lim_{x>\infinity) e^(k*x)*e^(l*x) = \infinity

#7




Re: When the growth function = 2^N

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