Climbing Powers

There are two definitions of $2^{3^4}$ . Definition 1 gives $(2^3)^4$ which is $2^{12}$ and definition 2 gives $2^{(3^4)}$ which is $2^{81}$.

Similarly the values of $(\sqrt{2}^{\sqrt{2}})^{\sqrt{2}}$ and $\sqrt{2}^{(\sqrt{2}^{\sqrt{2}})}$ are not equal. The first of these is $f(f(\sqrt{2}))$ where $f(x) = x^{\sqrt{2}}$ ; the second of these is $g(g(\sqrt{2}))$ where $g(x) =(\sqrt{2})^{x}$.

To see what happen if you iterate the functions many times you should now experiment, using your calculator or computer, by iterating both $f$ and $g$ in each case starting with the value $\sqrt{2}$.

Using these two definitions, we think of
\[ \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{.^{.^{.}}}}}}} \]

(where the powers of root 2 go on for ever) as the limit as $n$ to infinity of the sequence \[ x_1, x_2, x_3 , \dots x_n \]

where, according to the first definition, $x_{n+1}= f(x_n)$, or equivalently, \[ x_{n+1} = x_{n}^{\sqrt{2}} \]

and, according to the second definition, $x_{n+1}= g(x_n)$, or equivalently, \[ x_{n+1} = ( \sqrt{2})^{x_n} \] In both cases, if the limit exists, you will find it by putting $x_{n+1} = x_n = x$.