08/10/2014, 06:08 AM
(This post was last modified: 08/16/2014, 12:25 AM by sheldonison.)

There are advantages to using Tommy's 2sinh^{0.5} since there is a simple accessible closed form for the formal half iterate, unlike the exp^{0.5} which requires Kneser's Riemann mapping. In pondering an exact integral for all the derivatives asymptotic for

We draw the cutpoints at the negative real axis, and using the h_n formulation, one can get approximations for the derivatives, but I ran into difficulties getting a converging solution for all of the derivatives as the magnitude of z gets arbitrarily large. The equation above is exactly equivalent to

we desire , a very good entire asymptotic

I came up with this formulation which leads to all derivatives converging by way of the method in post#70 and post#76, by using this approximation:

where f(z) is the desired asymptotic, as per post#70/#76

There is one other step since post#70/76 used , then

where the sqrt term goes to 0.5 and the ln term goes to 0 as z grows exponentially,

There is also an "exact" version of g(z), which can be derived with some algebra from the equations above. However, using g_exact doesn't effect numerical results much as long , but the sqrt term above is really important since we actually use fairly small values of z for the Taylor series numerical approximations of the nth derivatives.

g_exact can be trivially generated from g(z)

this is our friend h(n) which is used for the Gaussian approximation using g or g_exact(z), for the nth derivative.

2sinh(z) and exp(z) behave nearly the same for big positive numbers, but the the key difference is that exp(big negative number) is very small, which is good for the convergence of the numeric integral, and lets all of the derivatives converge, where as 2sinh(big negative number) is an even bigger negative number. This is all assuming we sum up multiple wrappings around the unit circle, until we hit the minimum magnitude, to define a converging integral for all of the derivatives as the radius gets arbitrarily large.

I haven't done the numeric integrals for the derivatives of 2sinh, defined in this way, but I certainly could. This would yield another entire f(z) where f(f(z)) is an asymptotic to exp(z). I would expect it to behave very similarly in terms of convergence and error terms, with f(f(z)) eventually oscillating between larger and smaller than exp(z) an infinite number of times. The oscillations would even have nearly the same magnitude. It would also have zeros at the negative real axis, in much the same way.

We draw the cutpoints at the negative real axis, and using the h_n formulation, one can get approximations for the derivatives, but I ran into difficulties getting a converging solution for all of the derivatives as the magnitude of z gets arbitrarily large. The equation above is exactly equivalent to

we desire , a very good entire asymptotic

I came up with this formulation which leads to all derivatives converging by way of the method in post#70 and post#76, by using this approximation:

where f(z) is the desired asymptotic, as per post#70/#76

There is one other step since post#70/76 used , then

where the sqrt term goes to 0.5 and the ln term goes to 0 as z grows exponentially,

There is also an "exact" version of g(z), which can be derived with some algebra from the equations above. However, using g_exact doesn't effect numerical results much as long , but the sqrt term above is really important since we actually use fairly small values of z for the Taylor series numerical approximations of the nth derivatives.

g_exact can be trivially generated from g(z)

this is our friend h(n) which is used for the Gaussian approximation using g or g_exact(z), for the nth derivative.

2sinh(z) and exp(z) behave nearly the same for big positive numbers, but the the key difference is that exp(big negative number) is very small, which is good for the convergence of the numeric integral, and lets all of the derivatives converge, where as 2sinh(big negative number) is an even bigger negative number. This is all assuming we sum up multiple wrappings around the unit circle, until we hit the minimum magnitude, to define a converging integral for all of the derivatives as the radius gets arbitrarily large.

I haven't done the numeric integrals for the derivatives of 2sinh, defined in this way, but I certainly could. This would yield another entire f(z) where f(f(z)) is an asymptotic to exp(z). I would expect it to behave very similarly in terms of convergence and error terms, with f(f(z)) eventually oscillating between larger and smaller than exp(z) an infinite number of times. The oscillations would even have nearly the same magnitude. It would also have zeros at the negative real axis, in much the same way.

- Sheldon