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