I've had this idea about special exponential functions (which means special sine and cosine with *i*) for a while now. I have to clean this up later- you can read it, but I'm pretty sure since I've combined 2 different messages, with unique math in each (different ways of looking at the same thing), there will be errors until I fix it later (I'm posting before it's cleaned up).

some history and explanation (TLDR it): I tried (half heartedly) to enlist others in an attempt to map out certain rational components of the functions some months back, around the time of the latest Reimann furor. One of the main problems I have had is wording my questions in a way that makes the questions themselves easy to understand. Another problem is crossed math: when I combine math from several sources, and don't make sure the different symbols I use match up with the math from each source (each source being something I wrote at a different stage in the development of this idea).

Basically, I found a way to smoothly connect real valued iterative functions when they are constrained to certain value ranges.

Assume we are adding in c every time, and then performing a square root operation on c+ last result:

\[ z_0=\sqrt{c} \\ z_1=\sqrt{z_0+c} \\ z_2=\sqrt{z_1+c} \\ z_n= \sqrt{z_{n-1} +c} \]

so \( z_4 =\sqrt{c+\sqrt{c+\sqrt{c+\sqrt{c+\sqrt{c}}}}} \).

There is a way, using the following functions, to get \( z_{2.5235} \) (2.5235 iterations) or whatever sub iteration we want. We start out with iterations at infinity, and use the value we get for that to smoothly ramp up iterations.

This is for arbitrary n (square root being n=2, cube root n=3..., square n=1/2), and you can figure out what values n and x will not work. k is the number of radicals

\[ Q_{x,n,k} =\sqrt[n] { \left( nx^{n-1} \right)^k \, \times \left[x - \sqrt[n]{x^n-x + \sqrt[n]{x^n-x + \sqrt[n]{x^n-x+...}}} \right]} \]

The number generated when k-->infinity is used to create smooth iterations. I write it like:

\[ q_{x,n} = Q_{x,n,k \to \infty} \]

an example you might recognize (pi): \[ q_{2,2} = \lim_{k\to\infty} \,\, Q_{2,2,k} =\pi \]

\( a_n \) is the coefficients of specific terms

\( b= nx^{n-1} \) is the nx^(n-1)

\( \gamma= q_{x,n} \)

The cosine like extraction function is below:

\[ Q_{x,n,k} = \frac {\gamma^2}{a_2 \, b^{0k}} - \frac {\gamma^4}{a_4 \, b^{1k}} +\frac {\gamma^6}{a_6 \, b^{2k}}- \frac {\gamma^8}{a_8 \, b^{3k}} \,\, ... \]

The sine like extraction function is below:

\[ Q_{x,n,k} =\frac {(\frac{\gamma}{2})}{a_1 \, b^{0k}} - \frac {(\frac{\gamma}{2})^3}{a_3 \, b^{1k}} + \frac {(\frac{\gamma}{2})^5}{a_5 \, b^{2k}} -\frac {(\frac{\gamma}{2})^7}{a_7 \, b^{3k}}+ \frac {(\frac{\gamma}{2})^9}{a_9 \, b^{4k}} \,\, ... \]

Back to the pi example:

Using the cosine like function, one can increase k with non-integer values, and traverse smoothly between iterations of radicals:

\[ Q_{2,2,3} = \,\,\ \frac{ pi^2}{ 4^{0}} - \frac{pi^4}{12 \times 4^{3}} +\frac{ pi^6}{360 \times 4^{6}} - \frac{ pi^8}{20160 \times 4^{9}}\,\,\, \dots = \,\,\, 4^3 \, \times \left[2 - \sqrt{2 + \sqrt{2}} \right] \]

so

\[ Q_{2,2,3} = \,\,\ \frac{ pi^2}{ 4^{3}} - \frac{pi^4}{12 \times 4^{6}} +\frac{ pi^6}{360 \times 4^{9}} - \frac{ pi^8}{20160 \times 4^{12}}\,\,\, \dots = \,\,\,2 - \sqrt{2 + \sqrt{2}} \]

so

\[ 2 - \,\,\ \frac{ pi^2}{ 4^{3}} - \frac{pi^4}{12 \times 4^{6}} +\frac{ pi^6}{360 \times 4^{9}} - \frac{ pi^8}{20160 \times 4^{12}}\,\,\, \dots = \sqrt{2 + \sqrt{2}} \]

You can see how using smooth k in the cosine extraction equation will result in smooth iteration:

\[ Q_{x,n,k} = \frac {\gamma^2}{a_2 \, b^{0k}} - \frac {\gamma^4}{a_4 \, b^{1k}} +\frac {\gamma^6}{a_6 \, b^{2k}}- \frac {\gamma^8}{a_8 \, b^{3k}} \,\, ... \]

If we focus on n=2 cases, it simplifies everything. I'm going to post those cases later, after I visit my friend.

Basically, looking for the a_m factors. At 2,2, they are a_m = m! (or m!/2 for the cosine.. I forget). With an a_m generating function for various x and n, we'll be able to smoothly iterate. Might be an interesting path to take... continuously iterated functions.

Hi everyone! Nice to be back. Matt