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

In summary, the problem is related to Julia sets and its more of a mathematical problem. However, by using a bit of graph analysis, it can be shown that the sequence will go to infinity.
  • #1
gursimran
36
0
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.
 
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  • #2
Hi gursimran! :smile:

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

[tex]f([-2,2])=[-2,2][/tex]

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 [itex]x>2[/itex]. 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
micromass said:
Hi gursimran! :smile:

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

[tex]f([-2,2])=[-2,2][/tex]

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 [itex]x>2[/itex]. 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.

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..
 

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

1. What is the purpose of the iteration x = x^2 -2?

The purpose of this iteration is to determine the behavior of the sequence x, where each term is generated by the formula x = x^2 -2. This can help us understand the convergence or divergence of the sequence.

2. Why does the iteration x = x^2 -2 sometimes go to infinity?

The iteration x = x^2 -2 can go to infinity if the starting value of x is greater than or equal to 2. This is because when a value greater than or equal to 2 is squared and then subtracted by 2, the resulting value is always greater than the original value. Therefore, the sequence will continue to increase and approach infinity.

3. What is the significance of the number 2 in the iteration x = x^2 -2?

The number 2 is significant because it is the critical point that separates the behavior of the sequence. When the starting value of x is less than 2, the sequence will eventually approach a specific value (known as a limit). However, when the starting value is greater than or equal to 2, the sequence will go to infinity.

4. Can the iteration x = x^2 -2 ever reach a negative value?

No, the iteration x = x^2 -2 can never produce a negative value. This is because the square of any number is always positive, and when 2 is subtracted, it will always result in a non-negative value.

5. How does the starting value of x affect the behavior of the sequence in the iteration x = x^2 -2?

The starting value of x greatly affects the behavior of the sequence in the iteration x = x^2 -2. As mentioned before, if the starting value is less than 2, the sequence will approach a limit. However, if the starting value is greater than or equal to 2, the sequence will go to infinity. This illustrates the sensitivity of the iteration to small changes in the starting value.

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