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:

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

Re: When the growth function = 2^N
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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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