![]() |
#1
|
|||
|
|||
![]()
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
|
|||
|
|||
![]()
Because exponential kills polynomial. I.e., for any real number a, in the limit N --> infinity N^a/2^N goes to zero.
|
#3
|
||||
|
||||
![]() Quote:
![]() ![]() ![]()
__________________
When one teaches, two learn. |
#4
|
|||
|
|||
![]()
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
|
||||
|
||||
![]()
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.
__________________
When one teaches, two learn. |
#6
|
|||
|
|||
![]()
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
|
|||
|
|||
![]() |
![]() |
Thread Tools | |
Display Modes | |
|
|