03/12/2013, 08:58 AM

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:

<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}

\)

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}

\)

Gottfried Helms, Kassel