In the original post, I put the function $$a(n+1) = m \cdot \exp(-K \cdot a(n) / m),\ n \geq 2$$ which is not the function I wanted to study. The correct one is the one given below

I came up on a recursive definition of a function, given by $$a(n+1) = m \cdot \exp\left(\frac{-K \cdot (m - a(n))}{m}\right),\ n \geq 1$$ with $m$ and $K$ being fixed integers ($m$ large).
The first terms of the recursion are $$a(0) = m-1$$ $$a(1) = m \exp \left( -K / m \right)$$ $$a(2) = m\exp\left(-K \left( 1 - \exp \left( \frac{-K}{m} \right) \right) \right)$$ $$a(3) = m \exp\left( -K\left( 1 - \exp\left(-K \left( 1 - \exp \left( \frac{-K}{m} \right) \right) \right) \right) \right) $$ and so on... It looks like a kind of tetration to me. The plot of this recursion looks like this: Plot

So far I've proved that this recursion converges (decreasing + bounded below), but as you can see, this function seem to converge rather fast. So my question is, how would you find either a closed form of this function, or find a bound/function such that $a(n) \leq b(n)$ for all $n$?

I also think that we can determine the limit of the sequence by setting $a(n+1) = a(n)$ and expressing the answer with Lambert's $W$ function, what do you think?

  • $\begingroup$ It may help if you first multiply through by $K$, you get $K \, a(n+1) = K m \, \exp(-K \, a(n) / m),\ n \geq 2$ and you can then put $b(n+1) = Km \, \exp(-\, b(n) / m),\ n \geq 2$. You can then divide through by $m$ giving $\frac{b(n+1)}{m} = K \, \exp(- \frac{b(n)}{m})$ which is of the form $c(n+1) = K \, \exp(-c(n))$. $\endgroup$ – jim May 29 '16 at 13:00
  • $\begingroup$ Should be for $n \ge 1$, with $c(0) =K(1- \frac{1}{m}), c(1) = K (1-\frac{K}{m})$ and if $m$ "large" only depends (in first approximation) upon $K$? Do you, in fact, need $c(0)$ and just put $c(1) = K'$? $\endgroup$ – jim May 29 '16 at 13:55
  • $\begingroup$ Have a look at math.stackexchange.com/questions/1045707/… $\endgroup$ – jim May 29 '16 at 14:01
  • $\begingroup$ $c(0) = m -1$ actually. I think it could be for $n \geq 1$ yes. $m$ and $K$ are fixed integers, typically, $m = 1000000$ and $K = 4$. I tried what you said in your first comment, but I don't seem to go anywhere better. Also I looked for some means to approximate tetrations, but I could not find anything simple. $\endgroup$ – Laurent Hayez May 29 '16 at 14:16
  • $\begingroup$ $a(0) = m -1, \, b(0) = K \, a(0), c(0) = b(0)/m = K \, a(0)/m$? What about this link math.stackexchange.com/questions/409505/tetration-limit?rq=1? $\endgroup$ – jim May 29 '16 at 15:13

The corrected version of the question gives an easier result; the fixpoint to which the iteration proceeds is now expressible using the Lambert-W function only.

Let $a(0)=m-1$ , define always $b(k) = a(k)/m$, let $J=\exp(K)$ and iterate using the $b(k)$-version by

$\qquad b(k+1) = J^{b(k)-1} $

The fixpoint $\small t=\lim_{n \to \infty}b(n) $, if one exists, can be computed by

$ \qquad \begin{array}{rrlrl} &&& t &= J^{t-1} \\ &&&-t J^{-t} &= - \frac1J \\ &&&-t K \cdot e^{-t K} &= -\frac KJ \\ &&&-t K &= W(- \frac KJ) \\ \to& t &= {W( -K e^{-K}) \over -K} \end{array}$

and the fixpoint for $\small u=\lim_{n \to \infty}a(n) $ is then also $\small u=t \cdot m$ . For $\small m=1000$ and $\small K=4$ I get

$\qquad \small \begin{array}{} t \approx 0.01982740128177841 \\ u = t\cdot m \approx 19.82740128177841 \end{array}$

* Searching the interpolation-function*

1) First steps

First of all, when looking for an interpolating function, it is even better to define $c(n)$ as $c(n)=b(n)-1=a(n)/m-1$ and to rewrite the recursion on the $c(n)$ because this is a well known form: $$ c(n+1) = J^{c(n)} - 1 \qquad \text{ where always } a(n) = (c(n)+1)\cdot m\tag 1$$

To define a function, which interpolates the $c(n)$ it is useful to remember, that such a function shall have different values wenn started at different values $c(0)$. So such a function should have two parameters: the initial value (which I'll call here formally $x$ or $x_0$) and the recursion-depth $h$ (which I denote according to the practice to call it the iteration-height of the power-tower , or better exponential-tower elsewhere). So we'll get a function $f(h,x)$ to denote the interpolated $c(h)$ when started at $c(0)=x = x_0$.

It is also convenient to denote $f(h,x_0)$ shorter as $x_h$ in the following formulae.

Then the problem of interpolating $x_h$ to fractional $h$ can be solved by creating exponential power series (and iterates of them) for a specific $h$.
Remember that $J=\exp(K)$ or $K=\log(J)$ by my first proposal above. Then the iteration height $h=1$ gives the power series in $x$ as $$ x_1 = J^{x}-1 = Kx + (Kx)^2/2! + (Kx)^3/3! + ... \tag 2$$ For the iteration height $h=2$ we had $$x_2 = Kx_1 + (K x_1)^2/2! + (K x_1)^3/3! + ... $$ where if we expand $x_1$ into its power series (and collect $x$ of like powers) get the series $$\begin{array} {} x_2 &= K^2 x \\ &+ (1/2 K^4 + 1/2 K^3) x^2 \\& + (1/6 K^6 + 1/2 K^5 + 1/6 K^4) x^3 \\ & + O(x^4) \end{array}$$

If we try to continue for higher $h$ this we'll see that we get very complicated expressions for the coefficients at the powers of $x$. Of course we can try to find some formula for that coefficients depending on $h$ - and actually it is well known, that that coefficients are polynomials in $K$ depending on $h$.

But to arrive at a more handy notation we can observe, that at the terms of the series for $x_2$ we have that of powers of $x_1$ - respectively of $x_1$'s formal power series. This leads to the idea to introduce that whole mechanism in terms of matrix-notation, namely "Carleman-matrices":

1.a) Explanation of Carlemanmatrices with simpler function $\exp(x)-1$

We denote an infinitely sized vector of powers of $x$ as "Vandermonde-vector" $$ V(x) = [1,x, x^2,x^3,...]$$ Then, with a vector $\small E_1 = [0,1,1/2!,1/3!,..]$ we get $\small V(x) \cdot E = \exp(x)-1 $ . Again with a certain vector $\small E_2 $ we get $\small V(x) \cdot E_2 = (\exp(x)-1)^2$. This can even be done by hand, and a software like Pari/GP can easily determine that coefficients in $\small E_2$ and even for higher powers of $\small \exp(x)-1$. Of course a vector $\small E_0 = [1,0,0,...]$ gives the formula $\small V(x) \cdot E_0 = (\exp(x)-1)^0 = 1$.
The idea is now to concat all those vectors $\small [E_0,E_1,E_2,...]$ to a matrix $E$ such that $$ \begin{array}{} V(x) \cdot E & = [(\exp(x)-1)^0, (\exp(x)-1),(\exp(x)-1)^2,...] \\ V(x) \cdot E & = V(\exp(x)-1) \end{array} $$ The matrix $E$ is then said to be "the Carleman-matrix" for the map of $x$ to $\exp(x)-1$ and of course we see immediately that we can implement iterations to any iteration-"height" $h$ easily by $$ V(x_0) \cdot E = V(\exp(x)-1) \\ V(\exp(x)-1) \cdot E = V(\exp(\exp(x)-1)-1) $$ or $$ \begin{array} {}V(x_0) \cdot E^2 &= V(\exp(\exp(x)-1)-1) &= V((\exp(x)-1)^{°2}) \\ V(x_0) \cdot E^h &= V(...) &= V((\exp(x)-1)^{°h}) \end{array} $$ where the notation $ ()°h$ means the hth iteration. This shall then be generalized to fractional heights h and which can then be seen as a problem of fractional powers of the Carleman-matrix - and for cases, where the matrix is triangular like here this can be done by diagonalization.

2) Carleman-approach applied to $f(h,x)$

Of course in the above derivation for $\small E_1$, $\small E_2$ we do not want the function $\small \exp(x)-1$ but $\small f(1,x) = J^x-1 = \exp(Kx)-1$ and so we have to rewrite the above formulae where the logarithm $\small K=\log(J)$ is included to get $\small F_1$, $\small F_2$, $\small F_0$ and so on to arrive at the Carleman-matrix $F$ which provides $$ V(x) \cdot F = V(J^x-1) = V(x_1) \\ V(x) \cdot F^h = V(x_h) \tag 4$$

Fractional powers of the matrix $F$ can now be found (if $\small K \ne 1$) by diagonalization: we solve for matrices $\small M,D,W$ such that $$ F = M \cdot D \cdot W \qquad M \text{ triangular }, D \text{ diagonal}, W=M^{-1} \tag 5$$ Then also $$ F^h = M \cdot D^h \cdot W \tag 6$$ and because $D$ is diagonal, fractional powers of $D$ are definable by fractional powers of the (scalar) values in the diagonal; so $$ D^h = diag (\{(d_{k,k})^h\} \tag 7$$ is well defined for positive $d_{k,k}$ (which is the case here: $\small d_{k,k}=K^k$).

In my exercises I also normalize $M$ columnwise to have ones in the diagonal (which is always possible in diagonalization) so this becomes a Carlemanmatrix too. To have now the sought interpolation for the $c(h)=x_h$ we write $$ V(x) \cdot (M \cdot D^h \cdot W) = V(x_h) $$ and because $x_h$ is in the second column of $V(x_h)$ we need only the second column of $W$ and can write $$ x_h = V(x) \cdot M \cdot D^h \cdot W[,1] \tag 8$$ This gives expanded as power series in $\small x$ and coefficients in terms of polynomials in $\small K^h$ the final form:
$$ \tag 9 $$ bild_coefficients

3) Power series expansion of the interpolating function $f(h,x)$ and results

Of course it is useless to write down this expanded symbolic representation; one would only use a software system to compute this numerically, for instance Pari/GP which has also a diagonalization procedure.

What I've got for $\small x_{0.5}$ when $\small m=1000,K=4$, $\small a(0)=m-1$ and $\small x_0 = c(0)= a(0)/m-1=-1/1000$ is

a(0)    = 999
a(0.5)  = 998.0013329778173
a(1)    = 996.0079893439915

By the construction using the diagonalization and powers of $D$ we get a series, where the parameter $h$ is in the exponent of $K$. Since we have an example, where the initial value $a(0)$ resp. $x_0$ is a constant and $k$ has a known numerical value, we can even convert that series into a power series in $h$, which makes the implementation for this function $f(h,x_0)$ in a software even simpler.

Such a (well approximated(!)) powerseries for the interpolation of $a(n)$ when $a(0)=m-1=999$ $$a(h) = ( f(h,m-1)+1 ) \cdot m = 1000 \cdot f(h,999)+1000 $$


   a(h)=       999.0000000000000                        (10)
        + O(h^32)

This gives the smooth curve

Original answer, referring to and based on a wrong formula is now deleted

  • $\begingroup$ Hey, I am so sorry, I just see know that I did not copy the right function on my post. It is in fact $a(n+1) = m \cdot \exp \left( \frac{-K(m - a(n))}{m} \right)$. I will edit it. If you plot it now, you should obtain a plot similar to mine. Also, I think we can determine the asymptotic value of this sequence by solving $a(n+1) = a(n)$ and expressing the anwer with Lambert's $W$ function. What do you think about it? $\endgroup$ – Laurent Hayez May 31 '16 at 10:54
  • $\begingroup$ Thank you for the update. I found this fixed point to. However, I have absolutely no idea how to determine for which $n$ it is reached other than numerically. Do you have any idea on that? Also, do you have any idea how to approximate this function? Looks like an inversed sigmoid or logistical function to me, so maybe get a differential equation similar to the logistic equation? $\endgroup$ – Laurent Hayez May 31 '16 at 11:54
  • $\begingroup$ @LaurentHayez : ?? The fixpoint is reached for no n; it is the limit when n is assumed to grow without bound. And what do you mean with "approximate this function" ? Your function is an exact expression in terms of the exponential function - I doubt there is another function which converges faster/better than the exp()-function itself. Or do I misunderstand something? $\endgroup$ – Gottfried Helms May 31 '16 at 18:23
  • $\begingroup$ Regarding your computation of the limit value, when I compute it with python, I get (for $m = 100$ and $K = 4$) $\sim 1.9827$. When I let the recursion go, from round 16, I get always get the same value $1.94071233282$. That's why I thought the fixed point was reached for a $N \in \mathbb{N}$. What I mean by approximating by the function, is that instead of computing it recursively, I would like to find a formula to directly compute $a(2345)$ for example, or a formula that says $a(2345) \leq b(2345, m, K)$ where $b$ is a function depending on $m$ and $K$ (maybe). $\endgroup$ – Laurent Hayez May 31 '16 at 19:32
  • $\begingroup$ @LaurentHayez : Ah, I see. I'll have time for this tomorrow. There is a method using matrices (namely the Stirling numbers 2nd kind) which provides such a direct computation feasability (keyword "Carleman-matrix" for function $f(x)=J^x - 1$ ) when moreover instead of $b(n)=a(n)/m$ we use $ x(n) = a(n)/m-1 $ If this helps/is of interest so far, then you may possibly look at my website go.helms-net.de/tetdocs for some discussion; there I call that matrices $U_t$ (for the base-parameter $t$, here I should call them $U_J$) (...) $\endgroup$ – Gottfried Helms May 31 '16 at 21:27

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