Limit of a recursively defined bivariate function.

Question

Let m and n be positive integers.

Let $f(m,0)=m$

Let $f(m,n)= e \ln(f(m,n-1))$

$$\lim_{m\to\infty} \ln(m)\Big(f(m,\lfloor\ln m\rfloor)) - e\Big) = 163^{1/3}+C$$

Where $C$ is a constant.

It seems $0.005 > C > 0$

Is this true ? Why is this so ? Is $C = 0$ ?

Is this an analogue to the computation of the Paris constant ?

Can we give a closed form for $C$ ?

EDIT :

Conjecture :

$$\lim_{m\to\infty} \ln(m)\Big(f(m,\lfloor\ln m\rfloor)) - e\Big) = A$$

$A > 0$

Is this true ? How to prove this ?

Answer 1

Write $k(m) = \ln(m)\big(f(m,\lfloor \ln(m)\rfloor)-e\big)$. Here is a graph of $k\big(10^{\displaystyle 10^x}\big)$ according to Maple. The horizontal line is $163^{1/3}$.

I see no reason to think $163^{1/3}$ has anything to do with the limit.

Answer 2

It is intresting to note that if we replace (in the limit) $m$ by $m+O(m^{\frac{1}{2}})$ we would arrive at the same value. This leads to the simplification in the limit : replace both $log(m)$ by $m$.

This is probably the first step.

The second step is probably finding a good taylor series (converging for the desired interval) and error term for $e$ $ log(m)$ so that the rate of $f(m,m)$ can be expressed.

This problem them resembles many other problems concerning iterations and limits and should thus be solvable. I note that $exp(\dfrac{x}{e})$ has a parabolic fixpoint.

Not a solution yet but worth it imho.