12/09/2015, 06:34 AM
I think super-roots are important.
Iterated exponentials (\( w = \exp_x^{y}(z) \)) are a function of three variables (trivariate? function), and so they have 3 inverse functions: negatively iterated exponentials (solving for z), trivariate super-logarithms (solving for y), and trivariate super-roots (solving for x). Trivariate super-logarithms can be expressed with bivariate super-logarithms, and so are not fundamental operations, but trivariate super-roots have no known expression in terms of bivariate super-roots, and so are, so far, a fundamental operation so far as I know.
My recent research into super-roots have convinced me that we know more about them than we think we know. We can calculate the derivatives of them to a rational number in some cases, and to any precision in other cases. Using a combination of power series and Lagrange inverse series, we can calculate many many things about them, but we still don't have a closed form for these apparently useful functions. I think that given enough time, effort, and insight, we can find at least a recurrence equation that expresses how to find super-root (n + 1) given complete knowledge of super-root (n).
I'm going to go out on a limb and make a notation for these trivariate super-roots:
If we could find a general way of expressing trivariate super-roots in terms of bivariate super-roots, then I think we would know much more about tetration than we do today. Perhaps along the way we will discover something new that will shed some light on super-logarithms, too, perhaps.
Regards,
Andrew Robbins
Iterated exponentials (\( w = \exp_x^{y}(z) \)) are a function of three variables (trivariate? function), and so they have 3 inverse functions: negatively iterated exponentials (solving for z), trivariate super-logarithms (solving for y), and trivariate super-roots (solving for x). Trivariate super-logarithms can be expressed with bivariate super-logarithms, and so are not fundamental operations, but trivariate super-roots have no known expression in terms of bivariate super-roots, and so are, so far, a fundamental operation so far as I know.
My recent research into super-roots have convinced me that we know more about them than we think we know. We can calculate the derivatives of them to a rational number in some cases, and to any precision in other cases. Using a combination of power series and Lagrange inverse series, we can calculate many many things about them, but we still don't have a closed form for these apparently useful functions. I think that given enough time, effort, and insight, we can find at least a recurrence equation that expresses how to find super-root (n + 1) given complete knowledge of super-root (n).
I'm going to go out on a limb and make a notation for these trivariate super-roots:
- \( \sqrt[y]{w}^{(z)}_{\mathrm{s}} = x \) iff. \( w = \exp_x^y(z) \)
- \( \sqrt[y]{w}^{(z)}_{\mathrm{s}}
= \sqrt[y]{x^w}^{(x^z)}_{\mathrm{s}}
= \sqrt[(y-1)]{w}^{(x^z)}_{\mathrm{s}}
= x \)
- \( \sqrt[3]{w}^{(z)}_{\mathrm{s}}
= \left(\sqrt[2]{w^z}^{\left(\frac{x^z}{z}\right)}_{\mathrm{s}}\right)^{1/z}
= x \)
- \( \sqrt[2]{w}^{(z)}_{\mathrm{s}}
= \left(\sqrt[2]{w^z}^{(1)}_{\mathrm{s}}\right)^{1/z}
= x \)
If we could find a general way of expressing trivariate super-roots in terms of bivariate super-roots, then I think we would know much more about tetration than we do today. Perhaps along the way we will discover something new that will shed some light on super-logarithms, too, perhaps.
Regards,
Andrew Robbins

