91Ó°ÊÓ

Recall that a sequence of the form \(x_{n+1}=f\left(x_{n}\right)\) is called a dynamical system. (a) Using \(f(x)=x^{2}\) with \(x_{1}=a\), determine if \(x_{n+1}=f\left(x_{n}\right)\) has a limit if \(a=1, a=1.05\), and \(a=0.95\). This dynamical system is said to have a fixed point at \(x\) if \(f(x)=x\). To find the fixed points \(x_{n+1}=f\left(x_{n}\right)\) with \(f(x)=x^{2}\), we solve \(x^{2}=x\) or \(x^{2}-x=0\) with solutions and \(x=0\) and \(x=1\). In simple terms, a fixed point is called stable if a sequence that starts close to the fixed point has the fixed point as a limit. Otherwise, the fixed point is called unstable. (b) Would you classify \(x=1\) as stable or unstable? Would you classify \(x=0\) as stable or unstable? Briefly explain. (c) Consider \(x_{n+1}=f\left(x_{n}\right)\) with \(f(x)=2 x(1-x)\). i. Find the two fixed points. ii. Let \(x_{1}=0.25\). Does the sequence \(x_{n+1}=f\left(x_{n}\right)\) converge in this case? If so, what is the limit? iii. Let \(x_{1}=0.75\). Does the sequence \(x_{n+1}=f\left(x_{n}\right)\) converge in this case? If so, what is the limit? iv. Select any value of \(x_{1}\) between 0 and 1 . Does this choice affect the limit? v. Classify the two fixed points as stable or unstable. (d) Sometimes, unusual behavior can be observed when working with dynamical systems. For example, consider the dynamical system with \(f(x)=x+2.5 x(1-x)\) and \(x_{1}=1.2\). We see that the sequence oscillates between \(0.6\) and \(1.2\). We say that the dynamical system has a 2 -cycle because the values of the sequence oscillate between two numbers. (e) Describe the behavior of \(x_{n+1}=f\left(x_{n}\right)\) if \(f(x)=x+2.5 x(1-x)\) and \(x_{1}=1.201\). Do you see a cycle? If so, how many numbers. What are these numbers? Does a small change in the initial value of the sequence affect the resulting values of the sequence based on the results of this problem and the previous example? (f) Describe the behavion of \(x_{n+1}=f\left(x_{n}\right)\) if \(f(x)=x+2.5 x(1-x)\) and \(x_{1}=1.3\). Do you see a cycle? If so, how many numbers. What are these numbers? (g) Describe the behavior of \(x_{n+1}=f\left(x_{n}\right)\) if \(f(x)=x+2.5 x(1-x)\) and \(x_{1}=1.2\). If the values do not seem to approach a single value or a cycle of several values, we say that the dynamical system is chaotic. Does this system appear to be chaotic? In addition to your explanations, turn in the graphs obtained with plot for each problem.

Step by step solution

01

- Determine the Limit for a = 1

Given the function \(f(x) = x^2\) and the initial value \(x_1 = 1\), calculate the sequence \(x_{n+1} = f(x_n)\). Here, \(x_{n+1} = 1^2 = 1\). Since \(x_n\) remains 1 for all \(n\), the sequence has a limit of 1.

02

- Determine the Limit for a = 1.05

Using \(f(x) = x^2\) and initial value \(x_1 = 1.05\), calculate the sequence. \(x_2 = (1.05)^2 = 1.1025\). As \(n\) increases, calculate successive values: \(x_3 = (1.1025)^2 = 1.21550625\), and so on. Observe that the sequence diverges as \(n\) approaches infinity. Thus, no limit exists.

03

- Determine the Limit for a = 0.95

Using \(f(x) = x^2\) and initial value \(x_1 = 0.95\), calculate the sequence. \(x_2 = (0.95)^2 = 0.9025\). Continue the process: \(x_3 = (0.9025)^2 = 0.81450625\), and so forth. The sequence approaches 0 as \(n\) increases. Therefore, the sequence converges to the limit of 0.

04

- Classify Fixed Points as Stable or Unstable

Fixed points are solutions to \(f(x) = x\). For \(f(x) = x^2\), solve \(x^2 = x\), resulting in \(x = 0\) and \(x = 1\). To check stability, observe sequences near these fixed points. Sequences near \(x=1\) (e.g., starting with 1.05) diverge, indicating \(x=1\) is unstable. Sequences near \(x=0\) (e.g., starting with 0.95) converge to 0, indicating \(x=0\) is stable.

05

- Find Fixed Points for f(x) = 2x(1-x)

Solve \(f(x) = x\) for \(f(x) = 2x(1-x)\). Setting \(2x(1-x) = x\), yields \(2x - 2x^2 = x\). Rearrange to \(2x^2 - x = 0\), or \(x(2x - 1) = 0\). Solutions: \(x = 0\) and \(x = 0.5\). Thus, the fixed points are \(x = 0\) and \(x = 0.5\).

06

- Test Sequence Convergence for x1 = 0.25

With \(x_1 = 0.25\) and \(f(x) = 2x(1-x)\), calculate: \(x_{2} = 2(0.25)(1-0.25) = 0.375\). Continue to find \(x_3 = 2(0.375)(1-0.375) = 0.46875\), and so on. The sequence converges to 0.5.

07

- Test Sequence Convergence for x1 = 0.75

With \(x_1 = 0.75\), calculate: \(x_{2} = 2(0.75)(1-0.75) = 0.375\). Continue as in previous step, the sequence eventually converges to 0.5.

08

- Effects of Initial Value on Limit

For any initial value \(x_1\) between 0 and 1, the sequence \(x_{n+1} = 2x_n(1 - x_n)\) converges to the fixed point 0.5.

09

- Classify Fixed Points for f(x) = 2x(1-x)

Since sequences close to 0 approach 0 and sequences close to 0.5 approach 0.5, \(x = 0\) is stable, and \(x = 0.5\) is stable.

10

- Behavior of f(x) = x + 2.5x(1-x) with x1 = 1.201

With \(x_1 = 1.201\), calculate the sequence using \(f(x) = x + 2.5x(1-x)\). Observe the results: sequence oscillates without a fixed pattern, suggesting a chaotic system. No clear cycle is visible.

11

- Behavior of f(x) = x + 2.5x(1-x) with x1 = 1.3

With \(x_1 = 1.3\), calculate the sequence using \(f(x) = x + 2.5x(1-x)\). The sequence shows irregular behavior, illustrating chaos without an identifiable cycle.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

fixed points

A fixed point in a dynamical system is a value where the function evaluating at that point returns the point itself, meaning it remains unchanged through iterations. Mathematically, for a function \((f(x) = x)\), the fixed points are solutions to the equation \((f(x) = x)\). In the example provided with \((f(x) = x^2)\), solving \((x^2 = x)\) yields the fixed points \((x = 0)\) and \((x = 1)\). Fixed points are crucial because they represent states where the system could potentially stabilize or remain constant over time.

stability analysis

Stability analysis helps us understand whether small disturbances or variations in initial conditions will converge back to a fixed point or not. We classify fixed points as 'stable' or 'unstable'. A fixed point is stable if sequences starting close to it remain near and if they tend to this fixed point over iterations. Conversely, a fixed point is deemed unstable if sequences starting slightly away from it diverge.

For \((f(x) = x^2)\), analysis showed that \((x=1)\) is unstable since it diverges if the starting value is slightly above or below 1. On the other hand, \((x=0)\) is stable because sequences starting near 0 tend to converge to 0.

chaos theory

Chaos theory deals with systems that appear disordered but actually follow deterministic rules based on initial conditions. Tiny changes in starting values can lead to vastly different trajectories, making long-term prediction nearly impossible.

• In the given example with \((f(x)=x + 2.5 x(1-x))\), chaotic behavior is observed. Starting with initial values like \((x_1=1.201)\) causes the sequence to oscillate without a clear pattern, indicating chaos.
• As seen, slight variations in the initial value can lead to unpredictable and wild changes in the sequence.

The system is highly sensitive to initial conditions – a hallmark of chaos.

sequence convergence

Sequence convergence refers to a sequence approaching a fixed value as the number of iterations \((n)\) increases. In our example:

• For \((f(x)=x^2)\), with \((x_1=0.95)\), the sequence converges to 0.
• For \((f(x)=2x(1-x))\), sequences starting from any value between 0 and 1 converge to the fixed point \((x=0.5)\).

Identifying whether and where sequences converge is fundamental. It can indicate whether the system's long-term behavior settles into a pattern or remains unpredictable.