05/16/2014, 07:27 AM

(05/15/2014, 09:49 PM)tommy1729 Wrote: [ -> ]....

As for f(z) and f(f(z))/exp(z) being entire ...

I repeat my questions/remarks :

What does the weierstrass product of those look like ??

.....

Notice f(z) = exp(g(z)) ( 1 + a_1 z)(1 + a_2 z) ...

implies f(z) has no positive real zero's but an infinite amount of negative zero's ( the a_n ).

Also f(z) = exp(g(z)) ( 1 + a_1 z)(1 + a_2 z) ... could be wrong !

The weierstrass product form could be more complicated !

Is it ?

Similar questions for f(f(z)/exp(z).

Since f(f(z))/exp(z) must be close to 1 for positive real z , it follows it must go to oo for other part of the complex plane.

This post is just some pretty pictures, with all graphs from from -20 real to +20 real, and -5 imag to +20 imag with grid lines every 5 units. The first graph is the graph or the ratio that I wanted to create, which looks like I expected it would. Let's call the asymptotic half iterate function f (it needs a better name). This graph is f(f(z))/exp(z)-1, showing the very sharp switch from the region of good convergence, for positive reals, where f(f(z))/exp(z)-1 is very nearly zero, which is black. Here, I used the best approximation I have so far, which turns out to match the exponential function by a little better than 99% in the region of interest where the image is black. The boundary is pretty sharp. The conjecture is that there is a well defined boundary is exactly where f(z) takes on negative values, with imag(f(z))=0. At that boundary, the conjecture is that |f(f(z))/exp(z)-1|~=1. The 99.2% accuracy in this region may be as good as f(f(z))/exp(z) can be. With a 300 term Taylor series, accuracy peaks at 99.95% at z=2800, beyond which I would need to generate more Taylor series terms for convergence. f(f(pi i)=-1.021 + 0.053i, vs the exp(Pi*I)=-1.

[attachment=1058]

The next graph is f(z), the asymptotic half iterate itself, using the same grid coordinates. You can see zeros for f on the real axis, as black dots, at -0.71, -4.26, and -15.21. The pattern goes on forever, as f grows at the negative real axis, oscillating between positive and negative.

[attachment=1055]

The next graph is f(f(z)), same coordinates. You can clearly see the parts of the complex plane where f(f(z)) mimics the exponential, and the parts of the complex plane where it doesn't mimic the exponential function, which explains the first graph. And you can see the zeros, of f(f(z)), which are black dots. I'm pretty sure all the zeros of f(z) are on the negative real axis, and exp(z) has no zeros, so if you connect the black dots not on the real axis, this is where f(f(z))/exp(z) has to start to diverge from its asymptotic value of 1, and this is where f(z) is at the negative real axis.

btw; thanks for your comments. The Weierstrass product form sounds interesting; I haven't worked with it before, so it might be a lot of learning before I could generate any useful results.

- Sheldon

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