# Limit of a recursively defined bivariate function.

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 ?

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Are you just making this up, or do you have a source? –  Pedro Tamaroff Sep 7 '12 at 20:14
What does it have to do with tetration or dynamical systems? –  tomasz Sep 7 '12 at 20:23
repeatedly doing exp or ln is clearly tetration related. If you dont agree , plz give a definition of tetration and show the evidence of that definition. –  mick Sep 7 '12 at 20:28
You still haven't given us even the remotest reason why you think this is ture. –  Thomas Andrews Sep 7 '12 at 20:48
@mick What does this new sentence even mean? "The limit is taken in the sense that ln(x) > n" does not mean anything, as far as my mathematical experience goes. –  Thomas Andrews Sep 7 '12 at 21:26

Let $h(x)=e\ln x$. If $x>e$ then $h(x)>e$. On the other hand, $h(x)\ge x$ iff $\frac{\ln x}x\ge\frac1e$, and $\frac{d}{dx}\left(\frac{\ln x}x\right)=\frac{1-\ln x}{x^2}<0$ for $x>e$, so $h(x)<x$ for $x>e$. Thus, for any $x>e$ the sequence $\langle h^n(x):n\in\Bbb N\rangle$ is decreasing and bounded below by $e$, so it converges. (Here $h^n$ is the $n$-th iteration of $h$.) Its limit $g(x)$ must be a fixed point of $h$.

Now $\frac{d}{dx}(x-e\ln x)=1-\frac{e}x$, so $x-f(x)$ has a minimum at $x=e$, where it is $0$. In other words, $x=e$ is the only fixed point of $h$. In other words, $g(x)=e$ for all $x\ge e$, and $\lim_{x\to\infty}\frac1{g(x)-e}$ does not exist.

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This is not true. For any $x$, eventually $f(n)$ will fail to be defined (and for any reasonable sized $x$, rather quickly). For $x=10^{23}$ it only takes five steps. $\lim_{n \to \infty} f(x)$ does not exist.

Added: Your notation is confusing. It would be better to give $f$ arguments of $n$ and $x$, so $f(0,x)=x, f(n,x)=e \ln (f(n-1,x)$, which I will use. I still don't believe the result with the $e$ multiplying. If $f(n,x)=e+\delta$, (thinking of $\delta$ as small), $f(n+1,x)\approx e+\delta-\frac {\delta^2}{2e^2}$ so $\lim_{n \to \infty}f(n,x)$ must be the fixed point of your function, which is $e$.

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removed a typo , now its true. sorry –  mick Sep 7 '12 at 20:21
@mick: It certainly is. The successive values of $f$ are approximately $10^{23},19.48269549699,1.092427808658,0.03252148686231$, and $-1.260301355808$. –  Brian M. Scott Sep 7 '12 at 20:21
@mick I think that is because you changed the question from divided by $e$ to multiplied by $e$. –  William Sep 7 '12 at 20:36
@mick: but it still isn't –  Ross Millikan Sep 7 '12 at 21:08
It's not now true. It's clear that $g(x)=e$ for all $x$, so your limit not only doesn't exist, it actually is undefined for all $x$. –  Thomas Andrews Sep 7 '12 at 21:12

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.

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Thanks for your plot. It seems different programs and precisions have quite an influance. Considering your data as correct , i get about 1/3 value higher ( overal quite equidist ). So i would dare to replace 163^(1/3) to 163^(1/3)-1/3 as the limit. The plot suggests that there is a limit > 0 i feel , not ? Anyone dare to conjecture 5 :) What version of maple are you using ? –  mick Sep 8 '12 at 22:33
It should be $k(m) = \ln(m)\big(f(m,\lfloor \ln m\rfloor)-e\big)$ –  Fabian Sep 9 '12 at 4:01
@Fabian: fixed, thanks. @ mich: Maple 16. –  GEdgar Sep 9 '12 at 13:16

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.

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In case this helps (too long for comment): Letting $h_n(x) = f_n(x)/e$ we get $h_1(x) = 1 +\log(x)$, $h_2(x) = 1 + \log(1 + \log(x))$, etc-

So $h_n(x) = L^{[n]}(x)$ where $L^{[n]}$ is the function $1 +\log(\cdot)$ composed with itself $n$ times. Clearly $L^{\infty}(x)= 1$ for any $x$.

But what the OP wants (I'm starting to suspect) is to find

$$\lim_{x\to \infty} L ^{\lfloor \log(x) \rfloor} (x)$$

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