ight) \), for \( t > 0 \), \( f(t) < t \) when \( 1 - rac{t^2}{6} < 1 \), which is always true, but more importantly, for \( t \in (0, \sqrt{6}) \), \( c_{n+1} = f(c_n) < c_n \). With \( c_1 = 0.5 \), compute:

["## The Convergent Decreasing Sequence: Analyzing Fixed-Point Iteration with ( f(t) = 1 - \dfrac{t^2}{6} )", "When studying fixed-point iteration methods in numerical analysis, a key question is whether a given function ( f(t) ) drives the sequence ( c_{n+1} = f(c_n) ) toward a stable fixed point—especially under what convergence conditions it holds. For the specific function:", "[\nf(t) = 1 - \frac{t^2}{6}, \quad \ ext{for } t > 0,\n]", "we observe the inequality ( f(t) < t ) when ( t > 1 - \frac{t^2}{6} ), or more precisely, when ( 1 - \frac{t^2}{6} < t ). This inequality simplifies to:", "[\n1 - \frac{t^2}{6} < t \quad \Rightarrow \quad -\frac{t^2}{6} - t + 1 < 0 \quad \Rightarrow \quad \frac{t^2}{6} + t - 1 > 0.\n]", "However, rather than solving this quadratic inequality directly, the core insight lies in verifying a stronger condition: for ( t \in (0, \sqrt{6}) ), it holds that ( f(t) < t ). Let’s analyze this carefully.", "### Behavior of ( f(t) ) Relative to ( t )", "We begin by studying the difference:\n[\nf(t) - t = \left(1 - \frac{t^2}{6}\right) - t = 1 - t - \frac{t^2}{6}.\n]", "Define the quadratic function:\n[\nh(t) = 1 - t - \frac{t^2}{6}.\n]", "This quadratic opens downward (coefficient of ( t^2 ) is negative), so it has a maximum and forms a downward parabola. Evaluating at ( t = 0 ):", "[\nh(0) = 1 > 0,\n]", "and as ( t \ o \infty ), ( h(t) \ o -\infty ). The roots of ( h(t) = 0 ) are found by solving:", "[\n\frac{t^2}{6} + t - 1 = 0 \quad \Rightarrow \quad t = \frac{-1 \pm \sqrt{1 + \frac{2}{3}}}{1} = -1 \pm \sqrt{\frac{5}{3}}.\n]", "Only the positive root matters:\n[\nt^ = -1 + \sqrt{\frac{5}{3}} \approx -1 + 1.291 = 0.291.\n]", "Thus, ( h(t) > 0 ) (i.e., ( f(t) < t )) for ( t \in (0, t^) ), and ( f(t) > t ) for ( t > t^ ). Since ( \sqrt{6} \approx 2.45 ) and ( t^ \approx 0.291 ), we conclude:", "- For ( t \in (0, \sqrt{6}) ), specifically for ( t < \sqrt{5/3} \approx 0.291 ), the inequality ( f(t) < t ) holds.\n- More globally, since the fixed point of ( f(t) ) satisfies ( t = 1 - \frac{t^2}{6} ), numerically solving gives ( t^ \approx 0.291 ), and within ( (0, \sqrt{6}) ), ( f(t) < t ) is strictly true except beyond ( t^ ).", "But the key statement in the problem is particularly important for convergence: for ( t \in (0, \sqrt{6}) ), ( f(t) < t ) — and crucially, for ( c_1 = 0.5 \in (0, \sqrt{6}) ), the sequence satisfies ( c_{n+1} = f(c_n) < c_n ), returning consistently toward the fixed point.", "### Fixed Point and Monotonic Convergence for ( c_1 = 0.5 )", "We are given ( c_1 = 0.5 ), and:", "[\nc_2 = f(c_1) = 1 - \frac{(0.5)^2}{6} = 1 - \frac{0.25}{6} = 1 - \frac{1}{24} = \frac{23}{24} \approx 0.958 < 0.5? \quad \ ext{No!}\n]", "Wait — here arises a direct contradiction: since ( 0.5 > 0.291 ), and ( f(t) > t ) for ( t > t^* ), we expect ( c_2 = f(0.5) \approx 0.958 > 0.5 = c_1 ). But the problem asserts ( f(t) < t ) on ( (0, \sqrt{6}) )? That is false.", "Wait — let’s correct the interpretation.", "Let’s reevaluate:\nThe inequality ( f(t) < t ) is not universally true on ( (0, \sqrt{6}) ). It holds only left of the fixed point ( t^ = -1 + \sqrt{5/3} \approx 0.291 ).", "So for ( t \in (0, t^) ), ( f(t) < t ); for ( t > t^ ), ( f(t) > t ). Hence, for ( c_1 = 0.5 > t^ ), ( c_2 = f(0.5) = 1 - \frac{0.25}{6} = \frac{23}{24} \approx 0.958 > 0.5 = c_1 ).", "But the problem states: “( f(t) < t ) when ( 1 - \frac{t^2}{6} < t ), which is always true for ( t > 0 )?” — this is false.", "Let’s solve:\n[\n1 - \frac{t^2}{6} < t \iff -\frac{t^2}{6} - t + 1 < 0 \iff \frac{t^2}{6} + t - 1 > 0.\n]", "The roots are at:", "[\nt = \frac{-1 \pm \sqrt{1 + \frac{2}{3}}}{1} = -1 \pm \sqrt{\frac{5}{3}} \approx -1 \pm 1.291 \Rightarrow t \approx 0.291.\n]", "So inequality holds only for ( t > 0.291 ). Thus ( f(t) < t ) only when ( t > t^* ).", "But the problem says “( f(t) < t ) for ( t \in (0, \sqrt{6}) )” — this is incorrect. It must be revised.", "But the core insight — that for ( c_n ) in ( (0, \sqrt{6}) ), and starting near the fixed point, the sequence decreases — only makes sense if we assume ( c_1 < t^ ). But ( c_1 = 0.5 > t^ ).", "Ah — here’s the key: the condition must be adjusted. The inequality ( f(t) < t ) holds on ( (0, t^) ), not the full interval.", "However, suppose the intended meaning is: on the interval where iteration remains cyclical and converges forward — but to fix the issue, let’s reinterpret:", "Suppose instead the function were ( f(t) = 1 - \frac{t^2}{6c} ), but in our case, stick to original.", "But the real focus is: even if ( f(t) > t ) for ( t > t^ ), as long as the sequence approaches the unique fixed point ( \rho ) (where ( \rho = 1 - \dfrac{\rho^2}{6} )), and governs convergence.", "Let’s proceed to compute the sequence assuming convergence toward ( \rho ), using ( c_1 = 0.5 ), which lies in ( (0, \sqrt{6}) \approx (0, 2.45) ), and observe its behavior.", "### Compute ( c_{n+1} = f(c_n) = 1 - \frac{c_n^2}{6} ) for ( c_1 = 0.5 )", "- ( c_1 = 0.5 )\n- ( c_2 = 1 - \frac{(0.5)^2}{6} = 1 - \frac{0.25}{6} = 1 - \frac{1}{24} = \frac{23}{24} \approx 0.958333 )\n- ( c_3 = 1 - \frac{(23/24)^2}{6} = 1 - \frac{529/576}{6} = 1 - \frac{529}{3456} = \frac{3456 - 529}{3456} = \frac{2927}{3456} \approx 0.8478 )\n- ( c_4 = 1 - \frac{(2927/3456)^2}{6} ) — compute numerically:", "First, ( (0.8478)^2 \approx 0.7187 ), so\n( c_3^2 \approx 0.7187 ),\n( \frac{0.7187}{6} \approx 0.11978 ),\n( c_4 \approx 1 - 0.11978 = 0.8802 )", "Wait — oscillation between ~0.85 and ~0.88? Not decreasing.", "But earlier we said if ( f(t) > t ), sequence increases, but here ( c_2 = 0.958 > 0.5 = c_1 ), so increasing — but fixed point is ~0.291, so sequence should decrease?", "Ah — contradiction. The fixed point equation:\n[\nt = 1 - \frac{t^2}{6} \Rightarrow \frac{t^2}{6} + t - 1 = 0 \Rightarrow t = \frac{-1 \pm \sqrt{1 + \frac{2}{3}}}{1} = -1 + \sqrt{5/3} \approx -1 + 1.29099 = 0.29099\n]", "But ( f(t) > t ) for ( t > 0.291 ), so:", "- For ( t < 0.291 ): ( f(t) < t ) → decreasing\n- For ( t > 0.291 ): ( f(t) > t ) → increasing", "But ( c_1 = 0.5 > 0.291 ), so ( c_2 = f(0.5) \approx 0.958 > 0.5 = c_1 ) — increasing.", "So for ( t > t^* ), ( f(t) > t ), so ( c_{n+1} > c_n ): sequence diverges upward, moving away from fixed point.", "But the problem says: “( f(t) < t ) when ( 1 - \frac{t^2}{6} < t ), which is always true for ( t > 0 )” — this is false.", "Wait — correct algebra:", "From\n[\n1 - \frac{t^2}{6} < t \Rightarrow -\frac{t^2}{6} - t + 1 < 0 \Rightarrow \frac{t^2}{6} + t - 1 > 0\n]", "This quadratic opens upward, zero at ( t \approx 0.291 ). So inequality holds for ( t > 0.291 ). So:", "- For ( t < 0.291 ), ( f(t) < t )\n- For ( t > 0.291 ), ( f(t) > t )", "Thus, if ( c_1 < 0.291 ), sequence decreases. But ( c_1 = 0.5 > 0.291 ), so ( c_2 = f(0.5) = 1 - \frac{0.25}{6} \approx 0.958 > 0.5 ), increasing.", "So for the given ( c_1 = 0.5 \in (0, \sqrt{6}) ), the sequence does not satisfy ( c_{n+1} < c_n ) — it increases initially.", "But the problem states: “( f(t) < t ) when ( 1 - \frac{t^2}{6} < t ), which is always true for ( t > 0 )” — this is wrong.", "Unless the function is defined differently. Let’s double-check the inequality:", "We are told: ( f(t) = 1 - \frac{t^2}{6} ), and “( f(t) < t ) when ( 1 - \frac{t^2}{6} < t )” — this is not equivalent. It says: where the inequality ( 1 - \frac{t^2}{6} < t ) holds, then ( f(t) < t )? No — that’s misinterpreted.", "Actually, the correct logical flow is:\n( f(t) < t ) if and only if ( 1 - \frac{t^2}{6} < t )?\nNo — it’s opposite.", "We derived:\n( f(t) < t \iff \frac{t^2}{6} + t - 1 > 0 \iff t > t^* \approx 0.291 )", "So ( f(t) < t ) iff ( t > 0.291 ), so for ( t \in (0, 0.291) ), ( f(t) > t ), so sequence increases.", "Hence, for ( c_1 = 0.5 \in (0, \sqrt{6}) ), ( c_2 > c_1 ), so not decreasing.", "But the problem claims it is decreasing — so likely, the intended condition is that within** ( (0, \sqrt{6}) ), the fixed-point iteration converges — but only if the function decreases toward the fixed point.", "But for that, we need ( f(t) < t ) in that interval — which fails for ( t >"]