## FANDOM

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Tetration is a binary mathematical operator $^ba$ defined by the recurrence relation:

$^0a = 1$
$^{b + 1}a = a^{^ba}$

More intuitively, $^ba = a^{a^{a^{.^{.^{.^{a^a}}}}}}$ with b copies of a. $^ba$ is pronounced "a tetrated to b" or "to-the-b a."

Tetration leads to very large numbers, even with small inputs. For example, $^43 = 3^{3^{3^3}} = 3^{3^{27}} = 3^{7625597484987}$, which has 3638334640025 digits.

## Generalizing

The problem with the above definition is that $^ba$ works only for nonnegative integers b. What is $^\pi 2$, for example?

Generalizing b to the real numbers is a tricky and interesting problem. We present a solution proposed by Daniel Geisler of http://tetration.org/. Heavy differential calculus is ahead, so be warned. This is intended to be a gentle introduction; visit the original page if you don't need a tutorial.

Tetration is part of a class of general problems involving function iteration. Function iteration $f^t(z)$ is defined by the recurrence relation $f^0(z) = z$, $f^{t + 1}(z) = f(f^t(z))$, so $f^2(z) = f(f(z))$, $f^3(z) = f(f(f(z)))$, etc. Tetration could be defined as $^ba = f^b(1)$ where $f(z) = a^z$. So if we can define $f^t(z)$ for real $t$, we're all set for defining continuous tetration.

First, we translate f so that $f(0) = 0$. This simplifies a lot of the math. Then we consider the Maclaurin series (Taylor series around 0) of $f^t$:

$\sum_{i = 0}^\infty D^if^t(0)z^n/n! = f^t(0) + Df^t(0)z + \frac{D^2f^t(0)z^2}{2!} + \frac{D^3f^t(0)z^3}{3!} + \ldots$

This converges to $f^t(z)$ for $0 \leq |z| < R$ for some radius $R$. What we need to do is find $f^t(0),\,Df^t(0),\,D^2f^t(0),\,D^3f^t(0),\,Df^t(0)\ldots$.

### Constant term

First, $f^t(0) = 0$ because $f(0) = 0$.

### First derivative

To compute $Df^t(0)$, we need to find $Df^t(z)$. Define $g(z) = f^{t - 1}(z)$ for convenience:

$Df^t(z) = Df(g(z)) = f'(g(z))g'(z)$ (Chain Rule)
$= f'(f^{t - 1}(z))Df(f^{t - 2}(z)) = f'(f^{t - 1}(z))f'(f^{t - 2}(z))Df^{t - 3}(z)$
$= f'(f^{t - 1}(z)) \cdot f'(f^{t - 2}(z)) \cdots f'(f(z)) \cdot f'(z)$
$= \prod_{i = 0}^{t - 1} f'(f^i(z))$

Plugging in $z = 0$, we can take advantage of the fact that $f(0) = 0$:

$= \prod_{i = 0}^{t - 1} f'(f^i(0)) = \prod_{i = 0}^{t - 1} f'(0) = f'(0)^t$

Since we'll see them a lot, we'll define $\lambda_1 = f'(0)$, $\lambda_2 = f''(0)$, etc. So we can write our latest finding as $Df^t(0) = \lambda_1^t$.

### Second derivative

This one is somewhat nastier. Again define $g(z) = f^{t - 1}(z)$:

$D^2f^t(z) = D^2f(g(z)) = D[f'(g(z))g'(z)]$ (Chain Rule)
$= f''(g(z))g'(z)^2 + f'(g(z))g''(z)$ (Product Rule)
$= f''(f^{t - 1}(z))Df^{t - 1}(z)^2 + f'(f^{t - 1}(z))D^2f^{t - 1}(z)$ (Product Rule)
$= f''(f^{t - 1}(z))(\lambda_1^{t - 1})^2 + f'(f^{t - 1}(z))D^2f^{t - 1}(z)$

Setting $z = 0$:

$= \lambda_2\lambda_1^{2t - 2} + \lambda_1D^2f^{t - 1}(0)$
$= \lambda_2\lambda_1^{2t - 2} + \lambda_1(\lambda_2\lambda_1^{2(t - 1) - 2} + \lambda_1D^2f^{t - 2}(0))$
$= \lambda_2\lambda_1^{2t - 2} + \lambda_1\lambda_2\lambda_1^{2t - 4} + \lambda_1^2\lambda_2\lambda_1^{2t - 6} + \ldots + \lambda_1^{t - 1}\lambda_2\lambda_1^0$
$= \lambda_2 \sum_{i = 0}^{t - 1} \lambda_1^{2t - i - 2}$

### Third derivative

$D^3f^t(z) = D^3f(g(z)) = D[f''(g(z))g'(z)^2 + f'(g(z))g''(z)]$ (Chain Rule + Product Rule)
$= Df''(g(z))g'(z)^2 + Df'(g(z))g''(z)$
$= f'''(g(z))g'(z)^3 + f''(g(z))[2g''(z)g'(z)] + f''(g(z))g''(z)g'(z) + f'(g(z))g'''(z)$ (Product Rule, 2x)
$= f'''(g(z))g'(z)^3 + 3f''(g(z))g''(z)g'(z) + f'(g(z))g'''(z)$ (combining like terms)