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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)

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