# Writing about a function. Help me understand function iteration notation

• Nov 29th 2012, 11:06 PM
xaosman
Writing about a function. Help me understand function iteration notation
I have a kind of weird assignment. I have the following function:

g(x) = αg²(x/α)

Where g² apparently should be interpreted as the second iteration and not g squared. My assignment is to write as much as possible about this function and it's properties. The lecture was loosely connected to chaos, dynamical systems and the logistic function but I was told that it could be "solved" without knowing anything about dynamical systems or what was said during the lecture.

I have some trouble understanding the notation. Is there anyway to rewrite in a normal IMHO more understandable form? Is g²(x) = g(g(x)) ?

Where should I begin? Can someone help me by at least pointing me in the right direction by letting me know how and if it can be rewritten?
• Dec 9th 2012, 09:46 PM
Kiwi_Dave
Re: Writing about a function. Help me understand function iteration notation
Quote:

Originally Posted by xaosman
I have a kind of weird assignment. I have the following function:

g(x) = αg²(x/α)

Where g² apparently should be interpreted as the second iteration and not g squared. My assignment is to write as much as possible about this function and it's properties. The lecture was loosely connected to chaos, dynamical systems and the logistic function but I was told that it could be "solved" without knowing anything about dynamical systems or what was said during the lecture.

I have some trouble understanding the notation. Is there anyway to rewrite in a normal IMHO more understandable form? Is g²(x) = g(g(x)) ?

Where should I begin? Can someone help me by at least pointing me in the right direction by letting me know how and if it can be rewritten?

I think that your interpretation of the notation is correct.

$g(x)=ag(g(\frac xa))=ag^2(\frac xa)$ now replacing x with x / a

$g(\frac xa)=ag(g(\frac x{a^2}))$ substituting back gives $g(x)=ag(ag(g(\frac x{a^2})))$

And by the same reasoning:

$g(x)=ag(g(\frac x{a}))=ag(ag(g(\frac x{a^2})))=ag(ag(ag(g(\frac x{a^3}))))=ag(ag(ag(ag(g(\frac x{a^4})))))...$

Oops. Out of time.