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Example of a problem

Let’s say we have the recurrence relation : $$ x_{n+1}=\frac{2}{3}(2x_n -x_n^2). $$ We can easily find its fixed points, namely $x=0$ and $x=1/2$. I have found numerically that for $x_0 \in (0,1)$ we have $x_n \rightarrow 1/2$. I would like to prove that this is indeed the case using generally applicable methods.

General Question

I have read about Jacobians etc. to be used to study the stability of fixed points and to show that a sequence indeed converges to (or does not converge to) a given fixed point. However, I do not seem to find a decent reference on the internet where this theory is discussed in detail.

HolyMonk
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1 Answers1

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For $x\in(0,1/2)$ we have

$$0<x<\frac{4x-2x^2}3<\frac12$$

and for $x\in(1/2,1)$ we have

$$\frac12<\frac{4x-2x^2}3<x<1$$

which proves this is monotone and bounded, and hence convergent. We can then easily see that it converges to $1/2$ because we find equality in this interval only when $x=1/2$.

Simply Beautiful Art
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