Daedalus' Thirteenth Challenge

[Daedalus]

It has definitely been a weekend for solving problems that I've been working on for a few years, and this challenge is about a discovery that I made today involving interpolation and iterated exponentials or nested exponentials as I like to call them. Some of you are aware of my thread, Discoveries by Daedalus, where I prove various relationships for iterated exponentials and the nested logarithm operation that I discovered.

Nested Exponentiation:

Most of us are aware of nested exponents and how to simplify the math that uses them:

[math](a^b)^c=a^{b\, c}[/math]

So I began to wonder about series of nested exponents and if I could derive any properties associated with them. Now I do realize that this next part may be known, but here is the relationship for a series of nested exponents along with how I denote nested exponentiation:

[math]a^{\left \langle n \right \rangle} = \underbrace{(((a)^a)^a)^{\cdot^{\cdot^{a}}}}_n = a^{a^{n-1}}[/math]

We can demonstrate this by the following:

[math]a^{\left \langle 1 \right \rangle} = a = a^{a^{(1-1)}}[/math]

[math]a^{\left \langle 2 \right \rangle} = (a)^a = a^{a^{(2-1)}}[/math]

[math]a^{\left \langle 3 \right \rangle} = ((a)^a)^a = a^{a^{(3-1)}}[/math]

[math]a^{\left \langle 4 \right \rangle} = (((a)^a)^a)^a = a^{a^{(4-1)}}[/math]

Now that we have the generalized form, [math]a^{a^{n-1}}[/math], that predicts the outcome of continued nested exponentiation, we can derive the derivatives / integrals using Calculus and define an algorithm for determining nested roots using the Newton-Rhapson method.

Ever since the work I did on nested exponentials, I've been trying to find a nested / iterated exponential interpolation formula, and today I have discovered exactly that. The discovery occurred to me after making a few posts in the thread, Finding a Polynomial, where I talk about mine, Newton's, and Lagrange's interpolation formulas.

This challenge is to use the equations that can be found in the above links and find at least one of the nested / iterated exponential interpolation formulas hidden within. I realize that this challenge is a tough one, but 90% of the work is already done for you. If you understand how those equations work, it's not very far leap to arrive at nested / iterated exponential interpolation.

 

The interpolation formula must produce an equation of the form

 

[math]f(x)=a^{\left \langle x \right \rangle} = a^{a^{x-1}}[/math]

 

and interpolates either a linear sequential set of points or any set of points. I'll give bonus points if you find both nested / iterated exponential interpolation formulas. While I wait to see if anyone else can solve this challenge, I'm going to start writing up a paper on it

 

Graphs of Interpolating [math]y=x[/math] Using Increasing Sets of Points

 

Here are some graphs of using iterated exponential interpolation on increasing values of [math]y=x[/math] where [math]x \ge 2[/math].

 

 

As we interpolate more points of [math]y=x[/math], the generated iterated exponential function matches the curve better. If the last value of [math]x[/math] in the set of points used to generate the function is odd, then the output goes to infinity.

 

 

However, if the last value of [math]x[/math] is even, then the equation is asymptotic to [math]y=1[/math] where it begins to follow [math]y=x[/math] along the interpolated points and then sharply approaches 1 at both ends.

 

 

The equation failed for slopes less than or equal to 1/2. When we approach the slope of 1/2 from the right, we can see that the iterated exponential equations begin to oscillate wildly around [math]y=\frac{1}{2}x[/math].

 

 

The oscillations grow larger as the slope gets closer to 1/2.

 

Edited May 14, 2016 by Daedalus

[Daedalus]

I know some of these challenges are super hard and crazy, but this one really is an oversight on my previous work. The solution is so simple that if you take a look at the links I posted, there's a good chance you'll see how to solve this challenge. Any takers?