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.

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?

share|improve this question
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

1 Answer 1

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 $\mathrm e=2.71828$).

share|improve this answer

Your Answer


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.