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📅 March 15, 2026 👤 scraface

Mar 15, 2026

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📌 = -rac{L^2}{4}

📌 rac{L^2}{4} = 0 \Rightarrow L^2 = 0 \Rightarrow L = 0

📌 We now verify that the sequence converges to 0. Note that \( G(t) = t(1 - rac{t}{4}) \). For \( t \in (0, 4) \), \( 1 - rac{t}{4} \in (0,1) \), so \( G(t) < t \) as long as \( t > 0 \). Since \( b_1 = 1 \in (0,4) \), and \( b_{n+1} = G(b_n) < b_n \), the sequence is positive and strictly decreasing. A bounded decreasing sequence converges. The only fixed point in \( [0,4) \) satisfying \( G(t) = t \) is \( t = 0 \). Hence:

📌 \lim_{n o \infty} b_n = 0

📌 Question: A science communicator is designing an interactive exhibit on symmetry and defines a function \( h(x) = x^3 + px + q \) to model visual patterns. If \( h(1) = 4 \), \( h(2) = 10 \), and \( h(3) = 24 \), find the value of \( h(0) \).

📌 Solution: We are given a cubic polynomial \( h(x) = x^3 + px + q \), and values:

📌 h(1) = 1^3 + p(1) + q = 1 + p + q = 4 \quad ext{(1)}

📌 h(2) = 8 + 2p + q = 10 \quad ext{(2)}

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