# Thread: Fixed Point Iteration and Logistic Map

1. ## Fixed Point Iteration and Logistic Map

Consider the logistic map $\displaystyle g : [0,1]\implies[0,1], g(x) = rx(1 - x)$, and the sequence of iterates $\displaystyle {x_0}, {x_1} = g({x_0}), ... , {x_{k+1}} = g({x_k})$. Here $\displaystyle r$ is a parameter $\displaystyle 0 < r \le 4$.
So my question is how can I (computationally) show that the map is well defined, that is, the range of g is included in $\displaystyle [0,1]$ for all $\displaystyle 0 < r\le 4$. Also, how can I find the fixed points of $\displaystyle g$ as $\displaystyle 0 < r < 1$ and $\displaystyle 1 < r < 3$ and determine whether the FPI method is convergent. That is, use the FPI Theorem and cobweb analysis to study the evolution of the iterates $\displaystyle {x_k}$ starting with a given initial value $\displaystyle {x_0}\in{[0,1]}$.

2. Originally Posted by sarahh
Consider the logistic map $\displaystyle g : [0,1]\implies[0,1], g(x) = rx(1 - x)$, and the sequence of iterates $\displaystyle {x_0}, {x_1} = g({x_0}), ... , {x_{k+1}} = g({x_k})$. Here $\displaystyle r$ is a parameter $\displaystyle 0 < r \le 4$.
So my question is how can I (computationally) show that the map is well defined, that is, the range of g is included in $\displaystyle [0,1]$ for all $\displaystyle 0 < r\le 4$. Also, how can I find the fixed points of $\displaystyle g$ as $\displaystyle 0 < r < 1$ and $\displaystyle 1 < r < 3$.
The function $\displaystyle g(x) = rx(1 - x)$ represents a parabola which crosses the x-axis at x=0 and x=1, and has its maximum value r/4 when x=1/2 (as you can verify by elementary calculus). So for $\displaystyle 0\leqslant x\leqslant 1$, g(x) lies between 0 and r/4. It follows that if $\displaystyle r\leqslant4$ then g maps the unit interval into itself.

The fixed points of g are those for which g(x) = x, or $\displaystyle x = rx(1 - x)$. That is a quadratic equation, which you should be able to solve.

3. I'm still confused Opalg--so the only fixed point on the interval $\displaystyle 0 < r < 1$ is 0, right?? I believe the fixed points on the interval $\displaystyle 1 < r < 3$ are just $\displaystyle 0$ and $\displaystyle 1 - \frac{1}{r}$ right? Or are there more fixed points? How do I then show convergence/divergence using the FPI method. I'm having trouble showing this--can you show me step my step how to show it? Also how do I use the FPI Theorem and cobweb analysis the to look at the evolution of the iterates $\displaystyle x_k$ starting with a given initial value $\displaystyle {x_0}\in{[0,1]}$.

4. Originally Posted by sarahh
I'm still confused Opalg--so the only fixed point on the interval $\displaystyle 0 < r < 1$ is 0, right?? I believe the fixed points on the interval $\displaystyle 1 < r < 3$ are just $\displaystyle 0$ and $\displaystyle 1 - \frac{1}{r}$ right? Or are there more fixed points?
You will always find it helpful in problems like this to draw a picture.

The red curve shows the graph of $\displaystyle g(x) = rx(1-x)$ with r=1. The blue curve is g(x) with r=3. The green line is the line y=x. The fixed points of the function are those where the curve meets the green line. You can see that if the parabola is shallow, like the red one, then the only fixed point will be at the origin. But for larger values of r, as in the blue curve, there is also another fixed point, at $\displaystyle x = 1-\tfrac1r$.

I don't know anything about the FPI Theorem or cobweb analysis, so I can't help you there. But I suspect that the significant thing about the condition 1<r<3 is that in that case $\displaystyle |g'(x)|<1$ when $\displaystyle x = 1-\tfrac1r$.

5. It makes a lot more sense from the diagram Opalg. Thanks! However you drew that (with your math software), what does the graph show for the intervals of $\displaystyle 1 < r < 3$ and also for $\displaystyle 3 < r \le4$. They will of course have the same 2 fixed points right? Namely $\displaystyle 0$ and $\displaystyle 1 - \frac{1}{r}$, so it's just a matter of convergence or divergence/chaotic? And also, I believe your right about testing for convergence using $\displaystyle |g'(x)|<1$, since $\displaystyle |g'(x)|<1$ implies convergence and $\displaystyle |g'(x)|>1$ implies divergence. But how would we calculate this using the intervals of r? That's what's messing me up--I'm not sure what to substitute in the derivate to make it work for all cases within the interval (usually we just use on number which is the fixed point I believe). I have to discuss what is going on and also show it mathematically through computation for this problem. And since we're working in the interval for $\displaystyle x_0\in[0,1]$ I'm not sure if it's still chaotic where it usually is or does the interval not matter?