Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**yttrium88****Member**- Registered: 2005-12-01
- Posts: 20

Ok I admit it, this is more of a ponder than a discovery.

First, by ^n(x), I mean the function iterated on x, n times.

Now, if P(x) is a polynomial, is P^∞(x) a polynomial?

Is it continous?

Is it a step (also called discrete, I think) function?

Is it just a fractal set?

Hmmm....

Offline

**God****Member**- Registered: 2005-08-25
- Posts: 59

I would say it is not a polynomial. Apart from the fact that you cannot iterate a polynomial infinitely many times, consider this:

(x) = x^2

a(0) will always be 0, and a(1) and a(-1) will always be 1. For all x not equal to 0 such that |x| < 1, a(x) approaches 0 as a approaches ∞, and for all other numbers, a(x) tends to ∞ as a tends to ∞ - that is, it does not exist. So what you're left with in a limiting case is a discontinuous (and nonexistent) function defined only for the domain [-1, 1].

Offline

Pages: **1**