# " Solving :f ^ [f(x)] (x) = (22x^2-75x+64) / (10x^2 -34x +29) ; x , f real ,f monotonous."

9 views

### Alain Verghote

Dec 28, 2004, 4:00:04 PM12/28/04
to
Dear All,

Is this kind of equation known?
^[f(x)] means iteration f(x) times.
How do we 'deal' with such a thing?

Sincerely,Alain.

### Daniel Geisler

Dec 29, 2004, 2:00:04 PM12/29/04
to
Alain Verghote wrote:
> Dear All,
>
> Is this kind of equation known?
> ^[f(x)] means iteration f(x) times.
> How do we 'deal' with such a thing?

Alain, expressions like f^{f(x)}(x) have a clear meaning when
considered in the context of the Ackermann function.

f(x) = a+x
f^{n}(x) = a*n + x
f^{f(x)}(x) = a*(a + x) + x

Multiplication
f(x) = a*x
f^{n}(x) = a^n * x
f^{f(x)}(x) = a^{a*x} * x

See http://www.tetration.org/scimath/1 for an exported Mathematica
notebook. While both addition and multiplication result in solutions
for f(x), the plots of the solutions show they are not monotonic.
More generally, f^{f(x)}(x) is meaningful when using continuously
iterated functions, but then you loose the chance of arriving at a
closed form solution. I would be very surprised if anyone could find
a monotonic closed form solution.
Daniel

### Alain Verghote

Dec 30, 2004, 4:30:07 PM12/30/04
to

Dear Daniel,

I've got a different view and prefer standing heavily upon Abel
counting functions.
When phi(x) and f(x) are such as :phi(f(x))=phi(x)+1
phi(f^[r](x) = phi(x) + r ,r real ;with phi invertible
f^[r](x) = phi^[-1](phi(x) +r) ;and with r=f(x)
f^[f(x)](x) = phi^[-1](phi(x) +f(x)) .
With this tool you can ALSO solve for instance:
f((2x +1)^[f(x)])= 3*f(x) ...
I am sure there is still much work for us!.
COURAGE,
Alain.