10/18/2015, 11:07 PM
Im considering to replace g'' in the gaussian method or the Tommy-Sheldon iterations by ln( e + exp(g'')).
Notice the condition f '' > 0 implies
For x > 1 :
g '' + g' (g'-1) > 0.
---
How good the Tommy-Sheldon iterations work are depending alot on how good the gaussian is.
Lets say the gaussian is off by a factor 5.
Then the next iteration gives
New g = ln( 5 f (*) ) = ln(5) + old g.
This change seems too weak.
Funny because if we consider gaussian( 5 f) then this weak change is exactly what we need ( to get S9 close to gaussian and both to correct )
, since min( 5 f/ x^m) = 5 min ( f/ x^m).
Reconsidering things seems necessary.
Fake function theory is tricky.
However , i assumed here the gaussian was off by a factor 5.
It is not certain this is possible.
What then gives back confidence to the Tommy-Sheldon iterations.
Or by the replacement suggested Above ,
" The exponential Tommy-Sheldon iterations ".
Shorthand ets.
Regards
Tommy1729
Notice the condition f '' > 0 implies
For x > 1 :
g '' + g' (g'-1) > 0.
---
How good the Tommy-Sheldon iterations work are depending alot on how good the gaussian is.
Lets say the gaussian is off by a factor 5.
Then the next iteration gives
New g = ln( 5 f (*) ) = ln(5) + old g.
This change seems too weak.
Funny because if we consider gaussian( 5 f) then this weak change is exactly what we need ( to get S9 close to gaussian and both to correct )
, since min( 5 f/ x^m) = 5 min ( f/ x^m).
Reconsidering things seems necessary.
Fake function theory is tricky.
However , i assumed here the gaussian was off by a factor 5.
It is not certain this is possible.
What then gives back confidence to the Tommy-Sheldon iterations.
Or by the replacement suggested Above ,
" The exponential Tommy-Sheldon iterations ".
Shorthand ets.
Regards
Tommy1729
