# For what range of numbers the iteration x = x^2 -2 does not go to infinity?

1. Jun 27, 2011

### gursimran

This is a problem related to julia sets but its more of a mathematical problem so I posted it here.

x= x^2 - 2

For what values the iteration does not go to infinity. I can't figure out how to calculate that. I tried calculating a nth term of this in terns of initial term but all in vain.

2. Jun 27, 2011

### micromass

Staff Emeritus
Hi gursimran!

Before I'll explain it to you, take a look at this website:

http://aleph0.clarku.edu/~djoyce/julia/julia.html

There they explain graphically what happens. Such a graphic intuition is a very good thing to have in these matters.

So, what we do first is calculate the fixed points of f(x)=x2-2. Thus we look for the points such that x=x2-2. A little calculation shows us that the fixed points are -1 and 2.

The fixed point with largest absolute value is 2, so I claim that [-2,2] does not go to infinity. Indeed, it suffices to calculate

$$f([-2,2])=[-2,2]$$

So any point in [-2,2] will be sent to a point in [-2,2]. So such a point cannot go to infinity.

Now, what with the points $x>2$. For such a points, it is easy to show that f(x)>x (thus x2-2>x). Thus the sequence starting with an x>2 will always be increasing. Furthermore |f(x)-x| will increase as x gets larger. So the sequence is increasing and the distance between terms get larger, thus it will go to infinity.

For x>-2, we see easily that f(x)>2, so we are in previous situation.

3. Jun 27, 2011

### gursimran

Hello micromax
Thanks a lot for answering my so many questions. The link you gave really helped me a lot n I get a better feel of the whole thing. Again thanks a lot..