08/26/2015, 07:36 PM
In the context of TPID 17
1) does this conjecture even hold ??
We have waay to little supporting examples.
We have the sqrt(n) factor in semi-exp and exp but for instance double exponential is not confirmed.
For instance for SUM x^n / n^(n^n).
Im Hoping you guys can help.
--- assuming its true ---
I noticed
Min ( f^2) = min ( f )^2
Yet (1 + f)^2 = 1 + 2f + f^2.
therefore
A_n = min ( f(x)/x^n )
Is improved by b_n = A_n + 2 sqrt A_n.
That is only a minor improvement.
I tried to use the same arguments and as you might have guessed
Replacing ^2 and sqrt by say ^3 and cuberoot.
However it seems to violate ...
I even hesitated to post this because intuition can get you in trouble here.
Clearly there is something missing here.
Yes i know , we approximate the LHS in
P(x) < min f
With a polynomial degree n , so if we take ^(n/2) we end Up with the largest naively possible root.
Because O(x^2)^(n/2) = O(x^n)
We cant ALLOW growth smaller then O (x^2) since this violates the conditions f>0,f ' > 0 , f '' > 0.
But that does not explain enough.
It does show the upperbound factor
< O (n/2)
But that is very close from O ( ln(n) sqrt(n) ).
All intuïtieve logic fails.
However i believe we can repeat the b_n argument and thus arrive at the improved
C_n = a_n + 2 sqrt a_n + 2 sqrt(a_n + 2 sqrt a_n) + ...
~~ a_n + 2 ln(n) sqrt(a_n).
But that is still far from the desired
C ( a_n ln(n+1) sqrt(n) )
Its getting weird , I know.
Regards
Tommy1729
1) does this conjecture even hold ??
We have waay to little supporting examples.
We have the sqrt(n) factor in semi-exp and exp but for instance double exponential is not confirmed.
For instance for SUM x^n / n^(n^n).
Im Hoping you guys can help.
--- assuming its true ---
I noticed
Min ( f^2) = min ( f )^2
Yet (1 + f)^2 = 1 + 2f + f^2.
therefore
A_n = min ( f(x)/x^n )
Is improved by b_n = A_n + 2 sqrt A_n.
That is only a minor improvement.
I tried to use the same arguments and as you might have guessed
Replacing ^2 and sqrt by say ^3 and cuberoot.
However it seems to violate ...
I even hesitated to post this because intuition can get you in trouble here.
Clearly there is something missing here.
Yes i know , we approximate the LHS in
P(x) < min f
With a polynomial degree n , so if we take ^(n/2) we end Up with the largest naively possible root.
Because O(x^2)^(n/2) = O(x^n)
We cant ALLOW growth smaller then O (x^2) since this violates the conditions f>0,f ' > 0 , f '' > 0.
But that does not explain enough.
It does show the upperbound factor
< O (n/2)
But that is very close from O ( ln(n) sqrt(n) ).
All intuïtieve logic fails.
However i believe we can repeat the b_n argument and thus arrive at the improved
C_n = a_n + 2 sqrt a_n + 2 sqrt(a_n + 2 sqrt a_n) + ...
~~ a_n + 2 ln(n) sqrt(a_n).
But that is still far from the desired
C ( a_n ln(n+1) sqrt(n) )
Its getting weird , I know.
Regards
Tommy1729

