Let's solve the **Tent map** problem (parts a, b, c): ### Tent Map Definition \[ x_{n+1} = \begin{cases} 2x_n, & 0 \leq x_n \leq \frac{1}{2} \\ 2 - 2x_n, & \frac{1}{2} < x_n \leq 1 \end{cases} \] --- ## (a) Find all the fixed points, and classify their stability. **Fixed points:** \( x^* \) such that \( x_{n+1} = x_n = x^* \). ### Case 1: \( 0 \leq x^* \leq \frac{1}{2} \) \[ x^* = 2x^* \implies x^* = 0 \] ### Case 2: \( \frac{1}{2} < x^* \leq 1 \) \[ x^* = 2 - 2x^* \\ x^* + 2x^* = 2 \\ 3x^* = 2 \implies x^* = \frac{2}{3} \] Check if \( \frac{2}{3} > \frac{1}{2} \) (yes). So, **fixed points:** \( x^* = 0 \) and \( x^* = \frac{2}{3} \). --- **Stability:** - Calculate derivative at each fixed point. - Map derivative: - For \( 0 \leq x \leq \frac{1}{2} \): \( f'(x) = 2 \) - For \( \frac{1}{2} < x \leq 1 \): \( f'(x) = -2 \) - Fixed point is **stable** if \( |f'(x^*)| < 1 \). ### At \( x^* = 0 \): \( f'(0) = 2 \) ⟹ **Unstable** ### At \( x^* = \frac{2}{3} \): \( f'(\frac{2}{3}) = -2 \) ⟹ **Unstable** --- ## (b) Show that the map has a period-2 orbit. Is it stable or unstable? A period-2 orbit: \( x_1 \to x_2 \to x_1 \), \( x_2 \neq x_1 \). Set \( x_{n+2} = x_n \), \( x_{n+1} \neq x_n \). Let’s try \( x_1 \in [0, \frac{1}{2}] \), then \( x_2 = 2x_1 \). If \( 2x_1 > \frac{1}{2} \), so \( x_1 > \frac{1}{4} \). Let’s try \( x_1 \in (\frac{1}{4}, \frac{1}{2}] \). \[ x_2 = 2x_1 \in (\frac{1}{2}, 1] \] Now, \( x_3 = 2 - 2x_2 \) But for period-2, \( x_3 = x_1 \): \[ x_3 = 2 - 2x_2 = 2 - 2(2x_1) = 2 - 4x_1 \] Set \( x_3 = x_1 \): \[ x_1 = 2 - 4x_1 \\ 5x_1 = 2 \\ x_1 = \frac{2}{5} \] Check \( x_2 = 2x_1 = \frac{4}{5} \), which is in \( (\frac{1}{2}, 1] \). So, period-2 points: \( x_1 = \frac{2}{5} \), \( x_2 = \frac{4}{5} \). **Stability:** The stability is determined by the product of derivatives over one period: \[ |f'(x_1) f'(x_2)| = |2 \cdot (-2)| = 4 > 1 \] **So, the period-2 orbit is unstable.** --- ## (c) Find all period-3 and period-4 points. Are the corresponding periodic orbits stable or unstable? ### **Period-3 points:** In general, a period-3 point \( x \) satisfies \( f^3(x) = x \) and is not a fixed point or period-1/2. For the tent map, the period-\( n \) points can be constructed as follows: For period-\( n \), the tent map has \( 2^{n}-2 \) periodic points (excluding lower periods). For period-3: \( 2^3 - 2 = 6 \) period-3 points. **Explicit calculation:** They can be found by solving \( f^3(x) = x \), and excluding fixed and period-2 points. But, let's write the general form: Because the tent map is piecewise linear, the period-\( n \) points are at positions: \[ x = \sum_{k=1}^{n} \frac{a_k}{2^k}, \quad a_k = 0 \text{ or } 1 \] with the binary sequence \( a_1, a_2, ..., a_n \) not repeating with shorter period. For period-3, the sequences are: - 001 - 010 - 011 - 100 - 101 - 110 So, the points are at: \[ x_1 = 0.001_2 = \frac{1}{8} \\ x_2 = 0.010_2 = \frac{1}{4} \\ x_3 = 0.011_2 = \frac{3}{8} \\ x_4 = 0.100_2 = \frac{1}{2} \\ x_5 = 0.101_2 = \frac{5}{8} \\ x_6 = 0.110_2 = \frac{3}{4} \] But you must exclude points that are already period-1 or 2. **All period-3 orbits are unstable** for the tent map, since the derivative over one period is \(|(±2)^3| = 8 > 1\). --- ### Period-4 points Number: \( 2^4 - 2 = 14 \). Similarly, all period-4 orbits are unstable, since \(|(±2)^4| = 16 > 1\). --- ## **Final Answers** ### (a) - **Fixed points:** \( x^* = 0 \), \( x^* = \frac{2}{3} \) - **Stability:** Both are **unstable**. ### (b) - **Period-2 orbit:** \( x_1 = \frac{2}{5}, x_2 = \frac{4}{5} \) - **Stability:** **Unstable**. ### (c) - **Period-3 and period-4 points:** There are \( 6 \) period-3 points and \( 14 \) period-4 points, all are **unstable**. --- **If you need explicit values for all period-3 and period-4 points, let me know!**

Certainly! Let's approach each part systematically, providing detailed steps and reasoning. --- # **Part (a): Find all fixed points and classify their stability** ## Step 1: Recall the tent map: \[ f(x) = \begin{cases} 2x, & 0 \leq x \leq \frac{1}{2} \\ 2 - 2x, & \frac{1}{2} < x \leq 1 \end{cases} \] ## Step 2: Find fixed points, i.e., solutions to \( x = f(x) \). ### For \( 0 \leq x \leq \frac{1}{2} \): \[ x = 2x \implies x = 0 \] Check if \( x = 0 \) satisfies the domain: yes, it does. ### For \( \frac{1}{2} < x \leq 1 \): \[ x = 2 - 2x \implies x + 2x = 2 \implies 3x = 2 \implies x = \frac{2}{3} \] Check the domain: \( \frac{2}{3} \approx 0.666 \), which is greater than 0.5, so valid. **Thus, fixed points:** \[ \boxed{ x^* = 0 \quad \text{and} \quad x^* = \frac{2}{3} } \] ## Step 3: Classify their stability. The derivative \( f'(x) \) is: \[ f'(x) = \begin{cases} 2, & 0 \leq x \leq \frac{1}{2} \\ -2, & \frac{1}{2} < x \leq 1 \end{cases} \] **Stability criterion:** A fixed point is stable if \( |f'(x^*)| < 1 \). - At \( x=0 \): \[ |f'(0)| = 2 > 1 \implies \text{Unstable} \] - At \( x=\frac{2}{3} \): \[ |f'(\frac{2}{3})| = 2 > 1 \implies \text{Unstable} \] --- # **Part (b): Show that the map has a period-2 orbit. Is it stable or unstable?** ## Step 1: Find period-2 points. A period-2 orbit consists of points \( x_1 \) and \( x_2 \), with: \[ f(x_1) = x_2, \quad f(x_2) = x_1, \quad \text{and} \quad x_1 \neq x_2 \] ## Step 2: Assume \( x_1 \) in \( [0, \frac{1}{2}] \): \[ x_2 = f(x_1) \] - If \( x_1 \in [0, \frac{1}{2}] \): \[ x_2 = 2x_1 \] - Now, \( x_2 \) must be in \( (\frac{1}{2}, 1] \) for the second part of the map to apply (since \( f(x_2) \) must map back to \( x_1 \)): \[ x_2 = 2x_1 > \frac{1}{2} \implies x_1 > \frac{1}{4} \] - For \( x_2 \), applying \( f \): \[ f(x_2) = \begin{cases} 2x_2, & x_2 \leq \frac{1}{2} \\ 2 - 2x_2, & x_2 > \frac{1}{2} \end{cases} \] Since \( x_2 > 1/2 \), we have: \[ f(x_2) = 2 - 2x_2 \] Set \( f(x_2) = x_1 \): \[ x_1 = 2 - 2x_2 \] Recall that \( x_2 = 2x_1 \), so: \[ x_1 = 2 - 2 (2x_1) = 2 - 4x_1 \] Solve for \( x_1 \): \[ x_1 + 4x_1 = 2 \implies 5x_1 = 2 \implies x_1 = \frac{2}{5} \] Corresponding \( x_2 \): \[ x_2 = 2x_1 = 2 \times \frac{2}{5} = \frac{4}{5} \] **Check domain:** - \( x_1 = 0.4 \), which is in \( [0, 0.5] \). Good. - \( x_2 = 0.8 \), which is in \( (\frac{1}{2}, 1] \). Good. ## **Answer:** The **period-2 orbit** is: \[ \boxed{ \left( x_1, x_2 \right) = \left( \frac{2}{5}, \frac{4}{5} \right) } \] ## Step 3: Determine stability. The stability of a period-2 orbit: \[ \left| f'(x_1) \times f'(x_2) \right| = |2 \times (-2)| = 4 \] Since \( 4 > 1 \), the orbit is **unstable**. --- # **Part (c): Find all period-3 and period-4 points. Are the orbits stable?** ## Step 1: Period-3 points. ### General approach: - The tent map is piecewise linear. - Period-\( n \) points (excluding fixed points and lower period points) can be constructed via binary sequences. ### Step 2: Binary sequences for period-3 points. Sequences of length 3 (excluding repetitions of shorter periods): All sequences: - 001 - 010 - 011 - 100 - 101 - 110 **Note:** Sequences like 000, 111 are fixed points or period-1 points. ### Step 3: Convert binary sequences to \( x \). Each sequence corresponds to a ternary expansion: \[ x = \sum_{k=1}^{n} \frac{a_k}{2^k} \] - For 001: \[ x = \frac{0}{2} + \frac{0}{4} + \frac{1}{8} = \frac{1}{8} \] - For 010: \[ x = 0 + \frac{1}{4} + 0 = \frac{1}{4} \] - For 011: \[ x = 0 + \frac{1}{4} + \frac{1}{8} = \frac{3}{8} \] - For 100: \[ x = \frac{1}{2} + 0 + 0 = \frac{1}{2} \] - For 101: \[ x = \frac{1}{2} + 0 + \frac{1}{8} = \frac{4}{8} + \frac{1}{8} = \frac{5}{8} \] - For 110: \[ x = \frac{1}{2} + \frac{1}{4} + 0 = \frac{2}{4} + \frac{1}{4} = \frac{3}{4} \] ### Step 4: Period-3 points are at \[ x \in \left\{ \frac{1}{8}, \frac{1}{4}, \frac{3}{8}, \frac{1}{2}, \frac{5}{8}, \frac{3}{4} \right\} \] ### Step 5: Stability of period-3 orbits. - The derivative over one period: \[ |f'(x_1) \times f'(x_2) \times f'(x_3)| = |2 \times (-2) \times 2| = 8 \] which is **greater than 1**, indicating **unstable** periodic orbits. --- ## **Similarly for period-4:** - Number of sequences: \( 2^4 - 2 = 14 \) - These sequences are all binary strings of length 4, excluding repeats of smaller period. The points are: \[ x = \sum_{k=1}^4 \frac{a_k}{2^k} \] with \( a_k \in \{0, 1\} \), and excluding sequences with smaller periods. Explicitly, the points are: - 0001: \( \frac{1}{16} \) - 0010: \( \frac{2}{16} = \frac{1}{8} \) - 0011: \( \frac{3}{16} \) - 0100: \( \frac{4}{16} = \frac{1}{4} \) - 0101: \( \frac{5}{16} \) - 0110: \( \frac{6}{16} = \frac{3}{8} \) - 0111: \( \frac{7}{16} \) - 1000: \( \frac{8}{16} = \frac{1}{2} \) - 1001: \( \frac{9}{16} \) - 1010: \( \frac{10}{16} = \frac{5}{8} \) - 1011: \( \frac{11}{16} \) - 1100: \( \frac{12}{16} = \frac{3}{4} \) - 1101: \( \frac{13}{16} \) - 1110: \( \frac{14}{16} = \frac{7}{8} \) Again, these points are all **unstable** because the product of derivatives over the cycle: \[ |2 \times (-2) \times 2 \times (-2)| = 16 > 1 \] --- # **Final Summary:** | Part | Fixed/Periodic Points | Values | Stability | |---|---|---|---| | (a) | Fixed points | \( 0 \), \( \frac{2}{3} \) | Unstable | | (b) | Period-2 orbit | \( \left( \frac{2}{5}, \frac{4}{5} \right) \) | Unstable | | (c) | Period-3 points | \( \frac{1}{8}, \frac{1}{4}, \frac{3}{8}, \frac{1}{2}, \frac{5}{8}, \frac{3}{4} \) | Unstable | | (c) | Period-4 points | \( \frac{1}{16}, \frac{1}{8}, \frac{3}{16}, \frac{1}{4}, \frac{5}{16}, \frac{3}{8}, \frac{7}{16}, \frac{1}{2}, \frac{9}{16}, \frac{5}{8}, \frac{11}{16}, \frac{3}{4}, \frac{13}{16}, \frac{7}{8} \) | Unstable | --- **Note:** The instability is typical of the tent map's chaotic regime, with all periodic orbits of period ≥ 1 being unstable. --- Let me know if you'd like any further clarifications or detailed derivations!