# Climbing Powers

$2\wedge 3\wedge 4$ could be $(2^3)^4$ or $2^{(3^4)}$. Does it make any difference? For both definitions, which is bigger: $r\wedge r\wedge r\wedge r\dots$ where the powers of $r$ go on for ever, or $(r^r)^r$, where $r$ is $\sqrt{2}$?

## Problem

We can define $2^{3^{4}}$ either as $(2^{3})^{4}$ or as $2^{(3^{4})}$ . Does it make any difference?

Now calculate $\left(\sqrt 2^{ \sqrt 2 }\right)^{ \sqrt 2 }$ and $\sqrt 2 ^{\left(\sqrt 2 ^{ \sqrt 2 }\right)}$ and answer the following question for the natural extension of both definitions.

Which number is the biggest \[ \sqrt 2 ^{\sqrt 2 ^{\sqrt 2 ^{\sqrt 2 ^{.^{.^{.}}}}}} \]

where the powers of root $2$ go on for ever, or \[ \left(\sqrt 2 ^{\sqrt 2 }\right)^{\sqrt 2} ? \]

## Getting Started

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 $x_{n+1} = x_n = x$.

## Student Solutions

Well done Paul Jefferys, you got close to a complete solution here. We have to consider two different values of these climbing powers depending on the order of operations which can be shown by putting in brackets. We can define $2^{3^4}$ either as $(2^3)^4 = 2^{12}$ or as $ 2^{(3^4)} = 2^{81}$. In the same way there are two interpretations of ${\sqrt 2}^{{\sqrt 2}^{\sqrt 2}}$ The first of these is $f(f(\sqrt 2))$ where $f(x)=x^{\sqrt 2}$ which gives:

$({\sqrt 2}^{\sqrt 2})^{\sqrt 2} =
{\sqrt 2}^{\sqrt 2 \times \sqrt 2} = {\sqrt 2}^2 =2$

In the second case we get $g(g(\sqrt 2))$ where $g(x) = (\sqrt 2)^x$, and using a calculator to get an approximate value gives:

${\sqrt 2}^{({\sqrt 2}^{\sqrt 2})}
= {\sqrt 2} ^{1.63...} = 1.76 $ to $2$ decimal places.

So

${\sqrt 2}^{({\sqrt 2}^{\sqrt
2})}< ({\sqrt 2}^{\sqrt 2})^{\sqrt 2}.$

Now consider

${\sqrt 2}^{{\sqrt 2}^{{\sqrt
2}^{{\sqrt 2}^{{\sqrt 2}^ {{\sqrt 2}^{..}}}}}} $

where the powers of $\sqrt 2$ go on forever. We have seen that we have two possibilities, namely

$X_1 = \lim x_n$ where $x_1 =
\sqrt 2,\ x_{n+1}= x_n^{\sqrt 2}$ or

$X_2 = \lim x_n$ where $x_1 =
\sqrt 2,\ x_{n+1} = (\sqrt 2)^{x_n}$.

N.B. Both iterations can be done on a calculator or computer: $X_1 = \lim x_n$ is equivalent to iterating $f(x) = x^{\sqrt 2}$ and $X_2 = \lim x_n$ is equivalent to iterating $g(x) = (\sqrt 2)^x$. If you do this experimentally, in each case starting with $x_1=\sqrt 2$, you will find that the first iteration appears to converge to infinity and the second appears to converge to $2$. We claim $X_1 = +\infty$.

Proof

We have $x_1=\sqrt{2}$ and $x_{n+1}=x_n^{\sqrt{2}}$; thus

$\log x_{n+1}=\sqrt{2}\,\log x_n,
\quad \log x_1 = \log \sqrt{2}.$

Thus

$\log x_{n+1} =
\big(\sqrt{2}\big)^n\log \sqrt{2},$

and as $\log x_n \to +\infty$ as $n\to \infty$, we see that $x_n\to +\infty$. We now claim that $X_2=2$.

Proof

First we show that $x_n < 2$ for all $n$, and the proof is by induction. Clearly $x_1 < 2$. Now suppose that $x_n < 2$ and consider $x_{n+1}$. We have

$x_{n+1} =
\big(\sqrt{2}\big)^{x_n} < \big(\sqrt{2}\big)^2 = 2$

as required. Hence (by induction) $x_n < 2$ for all $n$. Next, we show by induction that $x_n < x_{n+1}$. It is clear that

$x_1 = \sqrt{2} <
(\sqrt{2})^{\sqrt{2}} = x_2.$

Now suppose that $x_{n-1} < x_n$. Then

${x_{n+1}\over x_n} =
{\sqrt{2}^{x_n}\over \sqrt{2}^{x_{n-1}}}
=\sqrt{2}^{x_n-x_{n-1}}.$

Thus, as $x_n-x_{n-1}> 0$, we have $x_{n+1}/x_n > 1$ and hence $x_{n+1}> x_n$. This shows that $x_n$ is increasing with $n$, and that $x_n < 2$, and this is enough to see that $x_n$ converges to some number $X_2$, where $X_2\leq 2$. As $x_{n+1}=({\sqrt 2})^{x_n}$, if we let $n$ tend to infinity we see that $X_2$ is a solution of the equation $x=({\sqrt{2}})^x$. If we now plot the graphs of $y=x$ and $y=(\sqrt{2})^x$, we see that these two graphs meet at only two points, namely $(2,2)$ and $(4,4)$. Thus $X_2$ is either $2$ or $4$, and so it must be $2$.

## Teachers' Resources

Why do this problem?

The problem demonstrates that mathematical language needs to be very precise and it is important to avoid ambiguity.

If the learners are encouraged to do as much as they can without looking at the hint most of them will discover for themselves that $2^{3^4}$ is ambiguous because $2^{(3^4)}$ is not the same as $(2^3)^4$.

The problem gives opportunities for all the learners to have some success and for them to investigate the powers of $\sqrt 2$ using a calculator or computer and discover the behaviour of the two different sequences for themselves.

At the same time the problem provides a challenge for the most able students to prove the results which calls for an understanding of functions. One proof uses the logarithm function and the other mathematical induction.

Possible Approach

After the class has worked on indices this problem gives reinforcement material for everyone and some extension possibilities for the most able.

The hint gives the learners a good deal of 'scaffolding' thus enabling the teacher to set the problem for the learners to tackle independently without a lot of introduction and help from the teacher, at least initially.

The problem could be set as homework with the intention of going over the proofs in class the next lesson and having a whole class discussion.

Key Questions

What does $a^{b^c}$ mean?

What is the difference between $(a^a)^a $ and $a^{(a^a)}$?

If you repeat (or iterate) the function which maps $a^a$ to $(a^a)^a $what is the next value? What is this function?

If you repeat (or iterate) the function which maps $a^a$ to $a^{(a^a)} $ what is the next value? What is this function?

Possible support

Try the probem
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