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-   -   When the growth function = 2^N (http://book.caltech.edu/bookforum/showthread.php?t=377)

timhndrxn 04-18-2012 06:54 PM

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? :confused:

jcatanz 04-19-2012 08:57 AM

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

Originally Posted by timhndrxn (Post 1436)
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? :confused:

Because exponential kills polynomial. I.e., for any real number a, in the limit N --> infinity N^a/2^N goes to zero.

htlin 04-19-2012 04:12 PM

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

Originally Posted by timhndrxn (Post 1436)
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? :confused:

They are both bad for small N. But for large N, the exponentially decreasing term in Hoeffding would kill N^{9999} and provides us some guarantee on learning. Hope this helps.

lucifirm 04-22-2012 10:58 AM

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!

htlin 04-22-2012 09:07 PM

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

Originally Posted by lucifirm (Post 1526)
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.

gersh 04-24-2012 03:10 PM

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

gah44 10-23-2012 12:54 AM

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

Originally Posted by htlin (Post 1484)
They are both bad for small N. But for large N, the exponentially decreasing term in Hoeffding would kill N^{9999} 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 N, 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 N) cases.


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