Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A friend gave me the following problem:

Let $c$ be any positive real number. Define a sequence recursively by $$a_0=c,\;\text{and }\; a_n=c^{a_{n-1}}\;\text{for }\;n=1,2,\ldots$$ For what values of $c$ does this sequence converge?

The problem is trickier than it first seems, as I believe there are values $c>1$ for which the sequence converges, and also values $0<c<1$ for which the sequence diverges. This is supposed to be able to be solved by a first year calculus student, so elementary methods are preferred.

A followup question: Suppose the answer to the first question is some set $D\subset\mathbb{R}^+$. Then we have a well-defined function $f:D\rightarrow\mathbb{R}$ given by $f(c)=L_c$ where $L_c$ is the limit of the sequence defined above. Is $f$ continuous? Is $f$ differentiable?

Thanks!

share|improve this question

2 Answers 2

up vote 1 down vote accepted

Define $A=a_\infty$ Then $c^A=A$, with $A$ is a fixed point of $c^z$. Consider that for an infinitesimal $dz$

$c^{A+dz}=c^{dz} c^A =c^{dz} A = (1+ \ln(c){dz})A = A+ \ln(c^A){dz} = A+ \ln(A){dz}$.

So exponentiation maps ${A+dz}$ into $A+ \ln(A){dz}$ or more simply ${dz}$ into $\ln(A){dz}$. But this means that $\ln(A)$ is the multiplier or Lyapunov characteristic at $A$ of the fixed point of $c^z$.

The question now is where is the multiplier exactly on the unit circle. For multipliers inside the unit circle we have convergence and with the multiplier outside the unit circle we have divergence. Therefore $\ln(A) = exp(2 \pi i x)$ and $A = exp(exp(2 \pi i x))$ where $x$ ranges from zero to one.

Since $a^A=A$, $A^\frac{1}{A} = a $. So the curve $exp(exp(2 \pi i x))^{exp(exp(2 \pi i x))^{-1}}$ gives the boundary of convergence.

See Projective Fractals.

share|improve this answer
    
I see you've extended the domain to the complex numbers, but plugging in $x=0$ and $x=\frac{1}{2}$ give the same bounds that Shu Xiao Li posts. I'll have to do some work to fully understand your answer, but thanks for the help. Any chance you could provide a simplified solution for the original question (only consider $c\in\mathbb{R}^+$)? –  Jared Mar 19 '13 at 18:10
    
Sorry, as Shu Xiao Li stated, Euler proved the limits for the real numbers, but I am not familiar with the proof Euler used. From my work with the subject I suspect that this is the simplest approach to proving the domain of convergence. –  Daniel Geisler Mar 19 '13 at 23:37

This operation is known as Tetration.

As shown in wikipedia, Euler proved that it converges for $e^{-e}\leq c\leq e^{1/e}$ and diverges otherwise.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.