9 views

Skip to first unread message

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?

Is this kind of equation known?

^[f(x)] means iteration f(x) times.

How do we 'deal' with such a thing?

Sincerely,Alain.

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?

> 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.

Addition

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

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

to

Dear Daniel,

Thank for your reply.

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.

Reply all

Reply to author

Forward

0 new messages

Search

Clear search

Close search

Google apps

Main menu