# Tetration Forum

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I http://math.stackexchange.com/questions/327995 I discuss the problem

Problem with infinite product using iterating of a function: $\exp(x) = x \cdot f^{\circ 1}(x)\cdot f^{\circ 2}(x) \cdot \ldots$
I think, because of the better latex-formatting it is easier to read there, but for completeness I'll copy&paste the problem here too.

Considering the iteration of functions, with focus on the iterated exponentiation, I'm looking, whether the function which I want to iterate can -hopefully with some advantage- itself be expressed by iterations of a -so to say- "more basic" function.

Now I assume a function f(x) such that
$\exp(x) = x \cdot f^{\circ 1}(x)\cdot f^{\circ 2}(x)\cdot f^{\circ 3}(x)\cdots$

(where the circle-notation means iteration, and $f^{\circ 0}=x, f^{\circ 1}(x)=f(x)$) - and I ask: what does this function look like? What I'm doing then is this substitution:

$\begin{array} {lrll}
1.& \exp(x) & = &x & \cdot f^{\circ 1}(x) & \cdot f^{\circ 2}(x) & \cdot f^{\circ 3}(x) & \cdots \\
2.& \exp(f(x))&= && f^{\circ 1}(x) & \cdot f^{\circ 2}(x) & \cdot f^{\circ 3}(x) & \cdots \\ \\ \\
3.& {\exp(f(x))\over \exp(x) } & = & \frac 1x \\ \\
& \exp(f(x)) & = & &{ \exp(x) \over x} \\ \\ \\
4. & f(x)&=& x & - \log(x) \end{array}$

$\qquad \qquad$ *(From 4. I know, that x is now restricted to $x \gt 0$)*

But if I do now the computation with some example *x* I get the result

$y = x \cdot f^{\circ 1}(x)\cdot f^{\circ 2}(x)\cdot f^{\circ 3}(x)\cdots
\\ y = \exp(x) / \exp(1)$

***Q:*** Where does this additional factor come from? Where have the above steps missed some crucial information?

<hr>
A code snippet using Pari/GP:
PHP Code:
f(x) = x-log(x)  // define the function         x0=1.5         //  = 1.50000000000    [tmp=x0,pr=1]              // initialize    for(k=1,64,pr *= tmp;tmp = f(tmp));   pr   // compute 64 terms, show result          // = 1.64872127070    exp(x0)        // show expected value           // = 4.48168907034         pr*exp(1)      // show, how it matches           //  = 4.48168907034

<hr>
Here is an example which shows the type of convergence; I use *x_0=1.5* and internal precision of 200 decimal digits. Then we get the terms of the partial product as
$\begin{array} {r|r}
x_k=f^{\circ k}(x) & (x_k-1) \\
\hline
1.50000000000 & 0.500000000000 \\
1.09453489189 & 0.0945348918918 \\
1.00420537512 & 0.00420537512103 \\
1.00000881788 & 0.00000881787694501 \\
1.00000000004 & 3.88772483656E-11 \\
1.00000000000 & 7.55720220223E-22 \\
1.00000000000 & 2.85556525627E-43 \\
1.00000000000 & 4.07712646640E-86 \\
1.00000000000 & 8.31148011150E-172 \\
1.00000000000 & 1.020640763E-202 \\
1.00000000000 & 1.020640763E-202 \\
\cdots & \cdots
\end{array}
$
Hi Mr. Helms,

If we consider the solution to the infinite product exp(x) = M * x * f(x) * f^[2](x) *..., it has the same solution as the one we get for M = 1 because of the fraction appears in your calculation which makes M/M = 1. So, it's likely that a factor appears but it seems your calculation doesn't counts it. It seems more like the constant of integration but except that it isn't integration - not very useful, but just my 2 cents.

Balarka
.
(03/12/2013, 10:01 AM)Balarka Sen Wrote: [ -> ]Hi Mr. Helms,

If we consider the solution to the infinite product exp(x) = M * x * f(x) * f^[2](x) *..., it has the same solution as the one we get for M = 1 because of the fraction appears in your calculation which makes M/M = 1. So, it's likely that a factor appears but it seems your calculation doesn't counts it. It seems more like the constant of integration but except that it isn't integration - not very useful, but just my 2 cents.

Balarka
.
Hi Balarka -

yes, that seems also to me the reason. But why is that factor just exp(1)? In the discussion in MSE I've tried to give an answer to this, but that what I got is merely an intuition yet, not yet a usable formalism.

I've one time also looked at the gamma-function in terms of functional iteration and arrived at the "incomplete gamma" where a seemingly similar effect appears. It might be interesting to compare this.(If you're interested, here's the link: http://go.helms-net.de/math/musings/Unco...gGamma.pdf , see specifically at pg 12)

What I'm after with this is to see, how iterates of exp(x) might look expressed with that function f(x). (or whether it might be more sensical to discuss iterates of exp(x-1) in this context... )

Gottfried
I was toying around with differentiating both sides of some of your equations. afterall exp is its own derivative and the derivative of x times y with respect to x is easy to rewrite.

No conclusion(s) yet though.

Nice post

regards

tommy1729
Another trivial proof.

exp(x)/M = x * f(x) * f^[2](x)*...

It was already shown that f was x - ln(x).

As usual we search for fixpoints.

1 is a fixpoint of x - ln(x) since 1 - ln(1) = 1.

We also know that f^[n](x) must approach 1 in the limit.

Hence if we plug in x=1 on the RHS we get 1 * 1 * 1 * ... = 1

Therefore we get

exp(x)/M = 1

=> exp(1)/M = 1

=> e/M = 1

=> M = e

Q.E.D.

This thread reminds me a bit of an idea I had not so long ago

http://math.eretrandre.org/tetrationforu...hp?tid=768

Im not sure where you want to go next with this. Like how to build tetration from it , or other intresting properties. Afterall we know it is analytic.

regards

tommy1729
I've updated the question in MSE: added some more information (general bases for the exponential) and added the question about the observation, that for bases b < exp(1) we might not find convergence and even for some unknown constant eta not even a set of fixed accumulation-points.

(I'll include that update here later today or tomorrow)
See

http://math.stackexchange.com/questions/...xpx-x-cdot