The number $r:=e^{-e}$ is the smallest number for which the infinite power tower $r\uparrow r \uparrow r\uparrow \cdots$ converges. In other words, the iterarion $x_1=r$ , $x_{n+1}=r^{x_n}$ converges.

However, the convergence is rather slow.

Is there a tight asymptotic formula for the error we have if we take $n$ numbers $r$ ?

Experimentint with PARI/GP, it seems that the error is approximately $\frac{1}{\sqrt{n}}$


Yes, but depends upon the tightness you want? When you approach the fixed point $a=-1/\log r$ then you may compute the asymptotic behavior. Setting $f(x)=f_r(x)=r^x$ you have a pitchfork bifurcation for $g=f_r\circ f_r$ at $r=e^{-e}$ and $a=-1/\log r$. This leads to the asymptotic behavior: $$ g(a+z) = a+z - c z^3 + o(z^3)$$ with $c=-1.23...$ This translates into a differential equation for the error (for the second iterate): $\Delta(n) = g^n(a+z_0)-a$: $$ \frac{d\Delta}{dn} = -c\Delta^3 + ..$$ or solving by truncating to the leading term $$\Delta(n) = \Delta(0) \frac{1}{\sqrt{1+ c n^2 \Delta(0)^2}} \asymp \frac{1}{\sqrt{c}\;n}$$ but a rigorous bound for this error term is not simple.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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