05/16/2023, 11:30 PM

Using f(x) = exp(x) - exp(-3/5 x) instead of 2sinh for the 2sinh method.

You might wonder why.

well the higher derivatives go into agreement with exp(x) for re(x) large unlike with 2sinh(x).

We also get the fixpoint at 0 and the function looks to go to exp in a more smooth way.

All the derivatives are positive at the origin and for x > 0.

This resembles exp(x) better thus in a way.

It also satisfies the semi-group isom property.

Notice that

exp(x) - exp( - 1/3 x) does NOT work since it has a fixpoint at some negative x.

This method is thus a consensus and compromise between exp(x) - 1 and 2sinh(x).

exp(x) - 1 is for base exp(1/e) however and is parabolic, so it is more 2sinh type.

regards

tommy1729

You might wonder why.

well the higher derivatives go into agreement with exp(x) for re(x) large unlike with 2sinh(x).

We also get the fixpoint at 0 and the function looks to go to exp in a more smooth way.

All the derivatives are positive at the origin and for x > 0.

This resembles exp(x) better thus in a way.

It also satisfies the semi-group isom property.

Notice that

exp(x) - exp( - 1/3 x) does NOT work since it has a fixpoint at some negative x.

This method is thus a consensus and compromise between exp(x) - 1 and 2sinh(x).

exp(x) - 1 is for base exp(1/e) however and is parabolic, so it is more 2sinh type.

regards

tommy1729