A calculus problem by Ash R

Calculus Level 2

$\LARGE {\sqrt2} ^{{\sqrt2}^{{\sqrt2}^{{\sqrt2}^{\cdot^{\cdot^\cdot}}} } } = \, ?$

2

\sqrt2

8

2\sqrt2

32

4

\infty

4\sqrt2

1 solution

Peter Macgregor
Apr 6, 2016

By substituting the equation into itself (!) we get

$\left(\sqrt{2}\right)^p=p$

and by inspection, (or by trying out the suggested solutions) $\boxed{p=2}$

Note that we can only get going on this problem by assuming that the power tower converges. Euler showed that this is so provided that the bricks of the tower lie between $e^{-e}$ and $e^{\frac{1}{e}}$ .

In our case the bricks are $\sqrt{2}$ which is just beneath the upper limit for convergence.

The equation $\left(\sqrt{2}\right)^p=p$ has the solution $p=4$ as well. Why is the answer 2 rather than 4?

Otto Bretscher - 5 years, 2 months ago

Oh! Thank you for raising this question.

I was lucky that the poser didn't head the multiple choices with 'four'!

Calculating the first few approximations of the power tower certainly suggests that it converges to 2, but I can't find an easy way to show this analytically.

I think four can be ruled out as a possible solution because the maximum possible value of a convergent power tower is e.

As it stands my solution is incomplete. Can someone with more analytical muscle fill in the the gaps?

Peter Macgregor - 5 years, 2 months ago

Consider the increasing exponential function $f(x)=(\sqrt{2})^x$ , with the fixed points 2 and 4, where $f(x)=x$ . We know that $f(x)>x$ when $x<2$ since $f(x)$ is convex.

Define a sequence recursively by $x_0=\sqrt{2}$ and $x_{n+1}=f(x_n)$ . Our task is to find the limit of $x_n$ as $n$ goes to infinity; this limit will be the value of the power tower, by definition.

We can show by induction that $x_n<2$ for all $n$ . Indeed, $x_0<2$ , and $x_n<2$ implies that $x_{n+1}=f(x_n)<f(2)=2$ . Since $x_n$ is an increasing function bounded by 2, it has a limit $L.$ Since the limit must be a fixed point, by continuity of $f(x)$ , we have $L=2$ .

This somewhat abstract proof can be illustrated nicely with a cobweb, as @Agnishom Chattopadhyay did here .

As Mr Macgregor points out, my countryman Leonhard P. Euler has done all this work for us a long time ago, but it is fun to figure it out by ourselves.

Otto Bretscher - 5 years, 2 months ago

It has ambigous solution :v

Resha Dwika Hefni Al-Fahsi - 5 years, 2 months ago

No, 4 isn't a solution, but one has to explain why.

Otto Bretscher - 5 years, 2 months ago

How about you, Sir? Don't you know?

Resha Dwika Hefni Al-Fahsi - 5 years, 2 months ago

@Resha Dwika Hefni Al-Fahsi – The issue is discussed here , for 3 instead of 4, but the reasoning is the same.

Otto Bretscher - 5 years, 2 months ago

A calculus problem by Ash R

1 solution

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