10/08/2015, 12:26 PM
Sheldon , in your link you apparantly considered similar things.
But what is that about Laurent series ?
You mention Laurent and then you drop the negative terms ??
Or Maybe it related to the fake ln in the MSE threads.
I assume that.
Anyway.
As Said in the previous post , we seem to have a base problem.
So I reconsider.
I believe exp( ln ^ u ) ~ J is optimal for u = 2.
And i wonder about fake ( d(x) ) = J(x).
So i consider,
a > 1
Fa = exp( ln^a(x) ).
Ga = x^a
Ga ' = a x^(a-1)
Ga ' ^[-1] = (x/a)^(1/(a-1))
So a_n = exp( (n/a)^(a/(a-1)) - n (n/a)^(1/(a-1)) ).
So we get
1 + 1/(a-1) = 2
=> a = 2.
I assume t ' (x) = t(x/w) gives
t(x) ~ exp( ln(x) ^a(w) )
Where 1 + 1/(a(w) -1) = w.
Or t(x) ~ exp( ln(x) ^(1 + 1/(a-1)) )
( im running out of time to decide ).
So does a better estimate for J give a better fake !?
Or not ?
Regards
Tommy1729
But what is that about Laurent series ?
You mention Laurent and then you drop the negative terms ??
Or Maybe it related to the fake ln in the MSE threads.
I assume that.
Anyway.
As Said in the previous post , we seem to have a base problem.
So I reconsider.
I believe exp( ln ^ u ) ~ J is optimal for u = 2.
And i wonder about fake ( d(x) ) = J(x).
So i consider,
a > 1
Fa = exp( ln^a(x) ).
Ga = x^a
Ga ' = a x^(a-1)
Ga ' ^[-1] = (x/a)^(1/(a-1))
So a_n = exp( (n/a)^(a/(a-1)) - n (n/a)^(1/(a-1)) ).
So we get
1 + 1/(a-1) = 2
=> a = 2.
I assume t ' (x) = t(x/w) gives
t(x) ~ exp( ln(x) ^a(w) )
Where 1 + 1/(a(w) -1) = w.
Or t(x) ~ exp( ln(x) ^(1 + 1/(a-1)) )
( im running out of time to decide ).
So does a better estimate for J give a better fake !?
Or not ?
Regards
Tommy1729

