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:

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:

["Title: Proving the Sequence Converges to 0: Rigorous Verification with the Function ( G(t) = t\left(1 - \dfrac{t}{4}\right) )", "In mathematical sequences and analysis, convergence is a cornerstone concept. For many iterative sequences defined by recurrence relations, establishing convergence involves verifying key properties such as monotonicity, boundedness, and fixed points. In this article, we rigorously prove that a particular sequence defined by ( G(t) = t\left(1 - \dfrac{t}{4}\right) ) converges to 0 for ( t \in (0, 4) ), following the recurrence ( b_{n+1} = G(b_n) ) with initial value ( b_1 = 1 ).", "---", "### Sequence Setup: Defining ( G(t) ) and the Iteration", "We begin with the function\n[\nG(t) = t\left(1 - \dfrac{t}{4}\right) = t - \dfrac{t^2}{4}.\n]\nFor ( t \in (0, 4) ), observe that ( 1 - \dfrac{t}{4} \in (0,1) ), so ( G(t) ) maps ( (0,4) ) into ( (0,4) ) — specifically, ( G(t) < t ) for all ( t \in (0,4) ), provided ( t > 0 ). This technical detail is crucial for the next steps.", "Given the sequence defined by ( b_1 = 1 \in (0,4) ) and recurrence ( b_{n+1} = G(b_n) ), we analyze the behavior and limit.", "---", "### The Sequence is Positive and Decreasing", "Since ( b_1 = 1 > 0 ) and ( G(t) > 0 ) for all ( t \in (0,4) ), by induction, ( b_n > 0 ) for all ( n ).", "Next, consider the difference:\n[\nb_{n+1} - b_n = G(b_n) - b_n = \left(b_n - \dfrac{b_n^2}{4}\right) - b_n = -\dfrac{b_n^2}{4} < 0.\n]\nThus, ( b_{n+1} < b_n ) for all ( n ), so the sequence is strictly decreasing.", "A strictly decreasing sequence bounded below by 0 is convergent — this establishes the first major result.", "---", "### Applying the Monotone Convergence Theorem", "Let ( {b_n} ) be the monotonic decreasing sequence bounded below by 0. By the Monotone Convergence Theorem, it converges to some limit ( L \geq 0 ).", "Since ( G ) is continuous on ( [0,4] ) (in fact, on ( (0,4) )), and ( b_n \ o L ), we take limits on both sides of the recurrence:\n[\n\lim_{n\ o\infty} b_{n+1} = \lim_{n\ o\infty} G(b_n) \quad \Rightarrow \quad L = G(L).\n]\nSo, ( L ) must satisfy ( L = G(L) = L\left(1 - \dfrac{L}{4}\right) ).", "---", "### Finding Fixed Points in ( [0, 4) )", "Solve ( L = L\left(1 - \dfrac{L}{4}\right) ):\n[\nL = L - \dfrac{L^2}{4} \quad \Rightarrow \quad 0 = -\dfrac{L^2}{4} \quad \Rightarrow \quad L^2 = 0 \quad \Rightarrow \quad L = 0.\n]\nThus, the only fixed point in ( [0,4) ) satisfying the fixed-point equation is ( L = 0 ).", "Therefore, the sequence converges to\n[\n\boxed{0}.\n]", "---", "### Why ( t = 0 ) is the Unique Attractive Fixed Point", "The fixed point equation ( G(t) = t ) has only one solution in ( [0,4) ), namely ( t = 0 ). For any ( t > 0 ), iterating ( G(t) ) reduces the value strictly toward 0. Hence, despite starting in ( (0,4) ), all terms remain in this interval and converge unconditionally to 0.", "This convergence behavior makes ( G(t) ) a useful model in dynamical systems, demonstrating global asymptotic stability toward extinction or equilibrium at zero when the input is within the defined domain.", "---", "### Conclusion", "We have verified—through monotonicity, boundedness, fixed-point analysis, and continuity—that the sequence defined by ( b_{n+1} = t(1 - t/4) ) with ( b_1 = 1 ) and ( t \in (0,4) ) converges to 0. The stable long-term value is not just a limit but the only feasible fixed point in the interval, confirming both mathematical rigor and practical convergence for ( t > 0 ).", "This proof exemplifies how careful analysis of function properties and sequence behavior underpins convergence in iterative processes.", "---", "Keywords: sequence convergence, fixed point ( G(t) = t ), recurrence relation, ( G(t) = t(1 - t/4) ), monotonic sequence, bounded decreasing, mathematical proof, ( \lim_{n\ o\infty} b_n = 0 ), analysis of iterated functions."]

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