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#1
04-18-2012, 06:54 PM
 timhndrxn Junior Member Join Date: Apr 2012 Posts: 9
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
04-19-2012, 08:57 AM
 jcatanz Member Join Date: Apr 2012 Posts: 41
Re: When the growth function = 2^N

Quote:
 Originally Posted by timhndrxn 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?
Because exponential kills polynomial. I.e., for any real number a, in the limit N --> infinity N^a/2^N goes to zero.
#3
04-19-2012, 04:12 PM
 htlin NTU Join Date: Aug 2009 Location: Taipei, Taiwan Posts: 601
Re: When the growth function = 2^N

Quote:
 Originally Posted by timhndrxn 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?
They are both bad for small . But for large , the exponentially decreasing term in Hoeffding would kill and provides us some guarantee on learning. Hope this helps.
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#4
04-22-2012, 10:58 AM
 lucifirm Member Join Date: Apr 2012 Posts: 20
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
04-22-2012, 09:07 PM
 htlin NTU Join Date: Aug 2009 Location: Taipei, Taiwan Posts: 601
Re: When the growth function = 2^N

Quote:
 Originally Posted by lucifirm 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!
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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#6
04-24-2012, 03:10 PM
 gersh Junior Member Join Date: Apr 2012 Posts: 2
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
10-23-2012, 12:54 AM
 gah44 Invited Guest Join Date: Jul 2012 Location: Seattle, WA Posts: 153
Re: When the growth function = 2^N

Quote:
 Originally Posted by htlin They are both bad for small . But for large , the exponentially decreasing term in Hoeffding would kill and provides us some guarantee on learning. Hope this helps.
That is the way it always goes in theory.

In practice, we often don't have infinite , and so it might be that in some cases exponential isn't so bad. There are plenty of NP hard problems that can, in fact, be solved for practical (small ) cases.

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