# Iterative functional equation

1. Mar 27, 2003

### eljose79

let be the equation f(g(t))=g(t) where f is known but g(t) not.... i think that perhaps we could solve it by iteration so g(t) would be

g(t)=fofofofofofofo...... where fof means the composition of f with itself...is that solution right? i do not even konw if my process to solve the functional equation is right and will converge to the solution.. has any other solution?...

2. Mar 27, 2003

### HallsofIvy

Staff Emeritus
Could you give a precise DEFINITION of fofofo...? That's a lot easier to write than to define! If for example f(x)= x^2 how would you find fofofo...(3)? Until you can answer that you don't have a function to BE the solution!

Taking the example f(x)= x^2, you are asking for a function
g(x) such that (g(x))^2= g(x) for each x. Since, for a specific x,
g(x) is a number, g(x) must always satisfy the numerical equation
u^2= u. Of course, 0 and 1 are the only numbers that satisfy that so g(x) must be always either 0 or 1. That gives an (uncountably) infinite number of functions g that satisfy this equation. If you require that g be countinuous then there are exactly two solutions:
g(x)= 0 for all x and g(x)= 1 for all x.

3. Mar 27, 2003

### arcnets

I think it works in some special cases.
For instance, you have a linear operator F which obeys |F*v| = |v| for any vector v, and you look for an eigenvector g so that F*g = g.
(I used your letters).
Then g = F^infinity * v. (At least that's what I think)

Of course, v and g could also be functions.

EDIT: Oops, I forgot some more restrictions. || is the 1-Norm (sum of components). Plus, all entries must be non-negative.

Here's a simple example:
F=
| 2/3 1/3 |
| 1/3 2/3 |
Let v = (1;0)
Then:
F*v = (2/3 ; 1/3)
F*F*v = (5/9 ; 4/9)
F*F*F*v = (14/27 ; 13/27)
...
This converges to:
g = (1/2 ; 1/2).

Last edited: Mar 27, 2003
4. Mar 27, 2003

### mathman

red herring?

Let x=g(t). Then the equation is f(x)=x. Since f(x) is supposed to be known, you just simply solve for x. t is irrelevant.

5. Mar 27, 2003

### Hurkyl

Staff Emeritus
Repeated iteration will only converge to a solution if your initial point is in the basin of an attracting fixed point (I don't know if basin is the right word).

Repeated iteration is not generally a nice thing. It's very easy for it to blow up to infinity, approach a periodic orbit, or simply bounce around an interval chaotically.

Iterative algorithms, such as Newton's Method or Conjugate Gradients, are specifically designed so that the desired solution is an attracting fixed point (though not always with 100% success).

Hurkyl

6. Mar 27, 2003

### Integral

Staff Emeritus
I believe to guarentee convergence of a interive scheme you need |f'|<1 at least in a neighborhood of the fixed point.

You might to a search on "fixed point" numerical schemes, there is a significant amount of literature about them.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?