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