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If I have a geometric progression where $$x =\sqrt[j]{a}$$ and then the next term is $$\sqrt[x]{a}.$$ How can I express this mathematically to find the $n$'th iteration of this?

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Do you mean $x_i = \sqrt[x_{i-1}]{a}$? That does not look like a geometric progression... – Karolis Juodelė Dec 5 '12 at 8:06
There is no closed formula, valid for general $n$ and using only usual functions. – Did Dec 5 '12 at 8:17
Yes! thats what I'm taking about. Sorry I'm still in school so my mathematical vocabulary is very limited and this was the closest thing I have ever heard of. What would you call this relationship? – Jordan Brown Dec 5 '12 at 8:17
Here's something crazy for you. If you take any number < 16, my theory is that this function will converge to a single number! At 16, it bounces between 2 and 4 because $2^4 = 4^2 = 16$ and these are the only 2 numbers this works for: $a^b =b^a$ where a and b are two different numbers. This also works for -2 and -4 but I consider that to be essentially the same. – Jordan Brown Dec 5 '12 at 8:20
Also, it doesn't matter what value of j you start with. Each value of (a) will converge to a single number if a<16 regardless of what staring value of j is used. – Jordan Brown Dec 5 '12 at 8:43
up vote 1 down vote accepted

The question is to determine the asymptotic behaviour of a sequence $(x_n)_{n\geqslant0}$ defined by $x_0\gt0$ and $x_{n+1}=u(x_n)$ for every $n\geqslant0$, where $u:x\mapsto a^{1/x}$ for some $a\gt1$.

The function $u$ is decreasing from $u(0^+)=+\infty$ to $u(+\infty)=1$. The function $v=u\circ u$ is increasing from $v(0)=1$ to $v(+\infty)=a$ hence $1\leqslant x_n\leqslant a$ for every $n\geqslant2$, for every $x_0\gt0$. Furthermore $(x_{2n})_{n\geqslant0}$ and $(x_{2n+1})_{n\geqslant0}$ are monotone hence both these sequences converge. If their limits $\ell$ and $\ell'$ coincide, then $\ell=\ell'$ is a fixed point of $u$ (note that $u$ always has a fixed point). Otherwise, $\ell'=u(\ell)$ for some fixed point $\ell$ of $v$ not a fixed point of $u$.

When $a=16$, both cases occur, namely $2.74537$ is a fixed point of $u$ and $(2,4)$ is a $2$-cycle. When $a=2$ for example, the $2$-cycle does not occur hence $x_n\to\ell=1.55961$. Likewise, when $a=4$, $x_n\to\ell=2$. But when $a=20$, the fixed point is $2.85531$ and the $2$-cycle is $(1.50907,7.28017)$.

One can show that $u$ and $v$ have the same unique fixed point when $a\leqslant a^*$ and that $v$ has two distinct fixed points additionally to the fixed point of $u$ when $a\gt a^*$, where $a^*=\mathrm e^\mathrm e=15.15426$ (and for $a=a^*$ the fixed point is $\mathrm e=2.71828$).

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