09/03/2015, 10:31 PM
Ok trying to clarity the previous post.
G (x) ~ g(0) + f(x)^2 = g0 + g1 x + ...
(Taylor expansion)
F(x) = f1 x + f2 x^2 + ...
From this we can say gn = f1 f#n-1# + f2 f#n-2# + ... + (f#n/2#)^2.
However since (min [f / x^(n/2) ]) ^2 = min[f^2 / x^n] ,
gn is wrongly estimated as gn ~ (f#n/2#)^2.
Assuming
Ass 1
f#k# f#n-k# < ( f#n/2# )^2
( notice this depends on the convergeance speed of fn ! )
we thus get the improved estimate
g'n ~< (n/2) (f#n/2#)^2.
Hence a correcting factor upper bound : n/2.
---
Well that is the idea.
Handwaving informal and sketchy ... Yes i admit.
But still.
Issues ??
1) n is not even.
2) repeating the argument ... As in (q^2)^2. Leading to arbitrary correcting factors ??
3) similar to 2); replacing ^2 with other functions such as ^5 , also leading to arbitrary correcting factors.
Solutions to 1) 2) and 3) are considered but not formal.
I hope 1) 2) and 3) make clear what i meant with intuition failure.
Also the correcting factor n/2 is far from sqrt ( ln(n) n ).
---
Here i considered the n th Taylor polynomials with sqrt.
Clearly i Cannot meaningfully take anything beyond ^(2/n) since
( X^2 )^(n/2) = x^n.
---
Hope this clarifies a bit.
Sorry for the late reply , but this subject is tricky !
Since Ass 1 depends on converg speed im not even sure that TPID 17 is correct.
On the other hand it holds for semi-exp and exp and sheldon believes in a very similar variant.
I must be missing something.
The importance of the decending chain condition for the derivatives perhaps ??
---
Maybe i know someone who could help us here.
---
Pls inform me if its much simpler then i think but i guess it is not.
Regards
Tommy1729
G (x) ~ g(0) + f(x)^2 = g0 + g1 x + ...
(Taylor expansion)
F(x) = f1 x + f2 x^2 + ...
From this we can say gn = f1 f#n-1# + f2 f#n-2# + ... + (f#n/2#)^2.
However since (min [f / x^(n/2) ]) ^2 = min[f^2 / x^n] ,
gn is wrongly estimated as gn ~ (f#n/2#)^2.
Assuming
Ass 1
f#k# f#n-k# < ( f#n/2# )^2
( notice this depends on the convergeance speed of fn ! )
we thus get the improved estimate
g'n ~< (n/2) (f#n/2#)^2.
Hence a correcting factor upper bound : n/2.
---
Well that is the idea.
Handwaving informal and sketchy ... Yes i admit.
But still.
Issues ??
1) n is not even.
2) repeating the argument ... As in (q^2)^2. Leading to arbitrary correcting factors ??
3) similar to 2); replacing ^2 with other functions such as ^5 , also leading to arbitrary correcting factors.
Solutions to 1) 2) and 3) are considered but not formal.
I hope 1) 2) and 3) make clear what i meant with intuition failure.
Also the correcting factor n/2 is far from sqrt ( ln(n) n ).
---
Here i considered the n th Taylor polynomials with sqrt.
Clearly i Cannot meaningfully take anything beyond ^(2/n) since
( X^2 )^(n/2) = x^n.
---
Hope this clarifies a bit.
Sorry for the late reply , but this subject is tricky !
Since Ass 1 depends on converg speed im not even sure that TPID 17 is correct.
On the other hand it holds for semi-exp and exp and sheldon believes in a very similar variant.
I must be missing something.
The importance of the decending chain condition for the derivatives perhaps ??
---
Maybe i know someone who could help us here.
---
Pls inform me if its much simpler then i think but i guess it is not.
Regards
Tommy1729
