Change of base formula for Tetration
#1
Branching this to a new topic, as I think it deserves one. Due to the length, I'll split into two posts, one to make my claim, the other to back it up.

This is in response to a comment made here:
http://math.eretrandre.org/tetrationforu...d=21#pid21
bo198214 Wrote:However because there is no base transform formula for tetration, it maybe that \( e^{1/e} \) is the only base with a certain uniqueness.

Depends on what you mean by base transform formula. For example, assume you have two bases a and b, each greater than \( \eta \) (the Greek letter eta). As a refresher, I've previously defined \( \eta=e^{1/e} \). I think this constant will turn out to be so important to tetration that it deserves its own name. The letter eta is basically an alternative version of "e", just as this constant serves a similar purpose to the constant e. And it also makes for an easily pronounced "cheta" function, written \( {}^x \check \eta \),which I've previously described.

Anyway, so long as you have an exact solution for base b, you can find the tetration of base a for any real exponent r > -2.

An exact formula for base conversion requires infinite iterations, but arbitrary precision can be achieved with very low iteration counts, so long as each base isn't too close to eta. I've seen this fact alluded to by many authors, in fact. Peter Walker discussed almost exactly this same principle.

It's the basis for my solution for base e, by extending my function \( ^{x} \check \eta \). I don't think any of what I'm posting is "new", just put together so that the significance is obvious.

First, we need the constant of base conversion. It's essenstially a form of superlogarithmic constant. Think of it as the equivalent of the constant \( log_b(a) \) used for converting \( a^z = b^{z \times log_b(a)} \), assuming a and b are positive real numbers. We can find \( log_b(a) \) by taking a and b to very high integer powers:

\( log_b(a)\ =\ \lim_{n \to \infty} \left(\frac{n}{k}\right),\ a^k \le b^n \le a^{k+1} \)

By analogy, for tetration, we're going to tetrate them each a large number of times. However, as you will see, tetration to integer powers won't work, not if we want to find the superlogarithmic constant. If the superlogarithmic constant isn't an integer, you can only approximate without an exact solution for one of the bases. In other words, in almost all cases, we must have an exact solution for fractional iteration for at least one of the bases. That doesn't mean the constant doesn't exist, only that we can't uniquely determine its value without an exact solution for some base.

\( \begin{array}{|ccccc|}& & & & \\
\hline
\\[10pt]

\\
\hspace{10} & {\Large ^{\normalsize x} a} & = & {\Large \lim_{n \to \infty} log_a^{\circ n}\left({}^{\left(\normalsize n+x+\mu_b(a)\right)} b\right)} & \hspace{10} \\
\\[10pt]

\\
\hline
\end{array} \)

In a twist of irony, the logarithmic constant for exponentiation (hyper-3) is multiplicative (hyper-2), but the superlogarithmic constant for tetration (hyper-4) is additive (hyper-1). And I say it's a "superlogarithmic" constant, but it should not be confused with \( slog_b(a) \). I think they're related, but I haven't pinned down the nature of the relationship yet. This will require more study.

Moving along... Because logarthmic constants are multiplicative, we have:

\( log_a({c}) = log_a(b)\ \times\ log_b({c}) \)

On the other hand, since superlogarithmic constants are additive, we have:

\( \mu_a({c}) = \mu_a(b)\ +\ \mu_b({c}) \)

Edit: Boxed my base conversion formula, so it's easier to pick out later one when I revisit this thread.
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Messages In This Thread
Change of base formula for Tetration - by jaydfox - 08/12/2007, 06:39 AM
RE: Parabolic Iteration - by jaydfox - 08/15/2007, 09:19 PM
RE: Parabolic Iteration - by bo198214 - 08/15/2007, 09:30 PM
RE: Parabolic Iteration - by jaydfox - 08/15/2007, 11:41 PM
RE: Parabolic Iteration - by bo198214 - 08/16/2007, 08:17 AM
RE: Parabolic Iteration - by jaydfox - 08/16/2007, 05:51 PM
RE: Parabolic Iteration - by bo198214 - 08/16/2007, 06:40 PM
RE: Parabolic Iteration - by jaydfox - 08/16/2007, 09:47 PM
RE: Parabolic Iteration - by bo198214 - 08/16/2007, 10:07 PM

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