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tetration base conversion, and sexp/slog limit equations
In continuing to ponder the base change properties of tetration, I have come across an interesting function, that is analytic, but seems like it must converge to a non-analytic function. Consider the function f(x), defined such that

I chose sexp_e, since Dimitrii's base e super exponential taylor series has been published. And super .upper is exactly the same as Jay's "cheta" function, which is a well defined analytic super exponential, with base , and which has previously been explored in many posts, especially Jay suggests choosing cheta(0)=2e. Since , this leads to cheta(1)=e*e, and cheta(2)=e^e. Cheta(x) has super exponential growth as x increases in the positive reals. For negative values of x, cheta(x) assymptotically approaches e. Continuing on, solving for f(x)

Here we use the inverse of Dimitrii's base(e) super exponential, which is an analytic function. Since cheta(x) is analytic, then f(x) must also be an analytic function. Here is the graph of f(x), for real values of x.

[Image: slog_e_sexp_cheta.gif]

F(x) is an interesting analytic function. On the left, it converges to the line f(x)=-x+1. On the right, it converges to a 1-cyclic function. By definition

So f(x) converts from to cheta(x) or . As x increases past 4, f(x) is converging to a one-cyclic function. You can see the detail of the one cyclic function in the graph. It averages around -0.5843. The convergence is super-exponential, with the value of f(5.0) converging to 42 decimal digits, and the value of f(6.0) converging to something like 10^42 decimal digits.

f(2.0) = 0.000000000000 = slog_e(sexp(2.0))-2.0 = slog_e(e^e)-2.0 = 0
f(2.5) = -0.267810330237 = slog_e(sexp(2.5))-2.0 = slog_e(35.32)-2.5
f(3.0) = -0.455343068531 = slog_e( 263.734)-3.0
f(3.5) = -0.554032644495 = slog_e( 439911)-3.5
f(4.0) = -0.583032469819 = slog_e( 1.3686e+042)-4.0
f(4.5) = -0.584037761409 = slog_e( (439911)/e)-3.5
f(5.0) = -0.584589240445 = slog_e((1.3686e+42/e)-4.0
f(5.5) = -0.584037958805 = slog_e(( 439911)/e-1)-5.5
f(6.0) = -0.584589240445 = slog_e((1.37e+42/e-1)-6.0
f(6.5) = -0.584037958805 = f(5.5)

Back to the base change function
Since we have , then we can convert from cheta(x) to the sexp_e_baseconv(x), by iterating ln of cheta(x). I'm using the equations for the sexp.base.change_e(x) from earlier in this post.

Define the small constant

But here, cheta(x) can be replaced with

We can substitute and simplify, getting rid of the iterated ln since

Next, we define , which is just the 1-cyclic limit of f(x), and plug it in, now with all limits removed.

Next question, is the 1-cyclic function analytic? Previous arguments made by Henryk and Jay suggest that it cannot be, since the the base.change version of sexp_e seems to be only defined at the real axis. But this new equation is the same as the base change version of sexp_e! We have succeeded in removing all the iterated logarithms from the base change function. But this time, the base change version of sexp_e is being generated by sexp_e combined with , which is the limiting behavior of the graph of f(x), an analytic function. The other version was generated by iterated logarithms of cheta(x). It makes sense that infinitely iterated logarithms might not be analytic. But f(x) has been shown to be a well defined analytic function. And since f(x) quickly converges to , it seems like should be analytic. Is it analytic?

For a quick review, the base change versions of super exponentials is defined at the real axis, and infinitly differentiable. But so far, it cannot be defined on the complex plane. The nice thing about the base change version of super exponentials is that the value of "k" converges to a base change constant for increasing values of x, for any two bases a, and b. converges to a base conversion constant as x grows.

then the value of k is a constant for increasing values of x.
sexp_a(x+k) = sexp_b(x)

Henryk and Jay showed that the base change version of sexp has an infinite number of singularities, arbitrarily close to any point on the real axis. So the radius of convergence of the taylor series would be zero, or, as I understand it, the function isn't analytic. And that means isn't analytic either, because if was analytic, then sexp_e(x+) would also be analytic. But is merely the limiting behavior of f(x), for values of x>=6, where f(x) is a 1-cyclic sinusoid. Can an analytic function converge in this way, to a non-analytic function?

For me, the base change is a source of continued fascination. Why is it that for any particular value of x, there is a value of k such that the equation sexp_a(x+n)=sexp_b(x+n+k) holds for integer values of n, but not for fractional or real values of n? For cheta(x+n), why are there some values of x where cheta(x) seems "larger" than it should be compared with cheta(x-0.5), at least as compared with other sexp bases? I'm also still attempting to make some sense of the base change function on the complex plane, and may eventually post some graphs of cheta, on the complex plane, along wth some thoughts on how the base change version of sexp might be defined on the complex plane. I've also made many graphs of patterns of the amplitude and phase of the function, that I may eventually post, which show a predictable amplitude and phase. I would also like to look at for different bases, and compare the fourier series for different versions of . Unfortunately, so far, any real progress I make seems to be made at a glacially slow pace...

Hopefully, others also find this subject interesting.
- Sheldon Levenstein

Messages In This Thread
Is it analytic? - by sheldonison - 12/22/2009, 11:39 PM

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