- #1

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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?...

- Thread starter eljose79
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- #1

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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

HallsofIvy

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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

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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).

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:

- #4

mathman

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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

Hurkyl

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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

Integral

Staff Emeritus

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

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