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