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: = 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
 = 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(x))
) - and I ask: what does this function look like? What I'm doing then is this substitution:
 & = &x & \cdot f^{\circ 1}(x) & \cdot f^{\circ 2}(x) & \cdot f^{\circ 3}(x) & \cdots \\<br />
2.& \exp(f(x))&= && f^{\circ 1}(x) & \cdot f^{\circ 2}(x) & \cdot f^{\circ 3}(x) & \cdots \\ \\ \\<br />
3.& {\exp(f(x))\over \exp(x) } & = & \frac 1x \\ \\<br />
& \exp(f(x)) & = & &{ \exp(x) \over x} \\ \\ \\ <br />
4. & f(x)&=& x & - \log(x) \end{array} )
*(From 4. I know, that x is now restricted to 
)*
But if I do now the computation with some example *x* I get the result
\cdot f^{\circ 2}(x)\cdot f^{\circ 3}(x)\cdots <br />
\\ 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
 & (x_k-1) \\<br />
\hline<br />
1.50000000000 & 0.500000000000 \\<br />
1.09453489189 & 0.0945348918918 \\<br />
1.00420537512 & 0.00420537512103 \\<br />
1.00000881788 & 0.00000881787694501 \\<br />
1.00000000004 & 3.88772483656E-11 \\<br />
1.00000000000 & 7.55720220223E-22 \\<br />
1.00000000000 & 2.85556525627E-43 \\<br />
1.00000000000 & 4.07712646640E-86 \\<br />
1.00000000000 & 8.31148011150E-172 \\<br />
1.00000000000 & 1.020640763E-202 \\<br />
1.00000000000 & 1.020640763E-202 \\<br />
\cdots & \cdots<br />
\end{array}<br />
)
Problem with infinite product using iterating of a function:
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
(where the circle-notation means iteration, and
But if I do now the computation with some example *x* I get the result
***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