• Tetration is a binary mathematical operator ${\displaystyle ^ba}$ defined by the recurrence relation:

• ${\displaystyle ^{0}a=1}$
• ${\displaystyle ^{b+1}a=a^{^{b}a}}$

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

Tetration leads to very large numbers, even with small inputs. For example, ${\displaystyle ^{4}3=3^{3^{3^{3}}}=3^{3^{27}}=3^{7625597484987}}$, which has 3638334640025 digits.

Generalizing

The problem with the above definition is that ${\displaystyle ^ba}$ works only for nonnegative integers b. What is ${\displaystyle ^{\pi }2}$, 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 ${\displaystyle f^{t}(z)}$ is defined by the recurrence relation ${\displaystyle f^{0}(z)=z}$, ${\displaystyle f^{t+1}(z)=f(f^{t}(z))}$, so ${\displaystyle f^{2}(z)=f(f(z))}$, ${\displaystyle f^{3}(z)=f(f(f(z)))}$, etc. Tetration could be defined as ${\displaystyle ^{b}a=f^{b}(1)}$ where ${\displaystyle f(z)=a^{z}}$. So if we can define ${\displaystyle f^{t}(z)}$ for real ${\displaystyle t}$, we're all set for defining continuous tetration.

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

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

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

Constant term

First, ${\displaystyle f^{t}(0)=0}$ because ${\displaystyle f(0) = 0}$.

First derivative

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

${\displaystyle Df^{t}(z)=Df(g(z))=f'(g(z))g'(z)}$ (Chain Rule)

${\displaystyle =f'(f^{t-1}(z))Df(f^{t-2}(z))=f'(f^{t-1}(z))f'(f^{t-2}(z))Df^{t-3}(z)}$

${\displaystyle =f'(f^{t-1}(z))\cdot f'(f^{t-2}(z))\cdots f'(f(z))\cdot f'(z)}$

${\displaystyle =\prod _{i=0}^{t-1}f'(f^{i}(z))}$

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

${\displaystyle =\prod _{i=0}^{t-1}f'(f^{i}(0))=\prod _{i=0}^{t-1}f'(0)=f'(0)^{t}}$

Since we'll see them a lot, we'll define ${\displaystyle \lambda _{1}=f'(0)}$, ${\displaystyle \lambda _{2}=f''(0)}$, etc. So we can write our latest finding as ${\displaystyle Df^{t}(0)=\lambda _{1}^{t}}$.

Second derivative

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

${\displaystyle D^{2}f^{t}(z)=D^{2}f(g(z))=D[f'(g(z))g'(z)]}$ (Chain Rule)

${\displaystyle =f''(g(z))g'(z)^{2}+f'(g(z))g''(z)}$ (Product Rule)

${\displaystyle =f''(f^{t-1}(z))Df^{t-1}(z)^{2}+f'(f^{t-1}(z))D^{2}f^{t-1}(z)}$ (Product Rule)

${\displaystyle =f''(f^{t-1}(z))(\lambda _{1}^{t-1})^{2}+f'(f^{t-1}(z))D^{2}f^{t-1}(z)}$

Setting ${\displaystyle z=0}$:

${\displaystyle =\lambda _{2}\lambda _{1}^{2t-2}+\lambda _{1}D^{2}f^{t-1}(0)}$

${\displaystyle =\lambda _{2}\lambda _{1}^{2t-2}+\lambda _{1}(\lambda _{2}\lambda _{1}^{2(t-1)-2}+\lambda _{1}D^{2}f^{t-2}(0))}$

${\displaystyle =\lambda _{2}\lambda _{1}^{2t-2}+\lambda _{1}\lambda _{2}\lambda _{1}^{2t-4}+\lambda _{1}^{2}\lambda _{2}\lambda _{1}^{2t-6}+\ldots +\lambda _{1}^{t-1}\lambda _{2}\lambda _{1}^{0}}$

${\displaystyle =\lambda _{2}\sum _{i=0}^{t-1}\lambda _{1}^{2t-i-2}}$

Third derivative

${\displaystyle D^{3}f^{t}(z)=D^{3}f(g(z))=D[f''(g(z))g'(z)^{2}+f'(g(z))g''(z)]}$ (Chain Rule + Product Rule)

${\displaystyle =Df''(g(z))g'(z)^{2}+Df'(g(z))g''(z)}$

${\displaystyle =f'''(g(z))g'(z)^{3}+f''(g(z))[2g''(z)g'(z)]+f''(g(z))g''(z)g'(z)+f'(g(z))g'''(z)}$ (Product Rule, 2x)

${\displaystyle =f'''(g(z))g'(z)^{3}+3f''(g(z))g''(z)g'(z)+f'(g(z))g'''(z)}$ (combining like terms)

External links

• http://en.citizendium.org/wiki/Tetration
• Knuth's up-arrow notation