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📅 May 3, 2026 👤 scraface

May 03, 2026

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Solution: Assume \( \lim_{n o \infty} c_n = L \). Then taking limits:
L = f(L) = L - rac{L^3}{6}
\Rightarrow 0 = -rac{L^3}{6}
\Rightarrow L^3 = 0 \Rightarrow L = 0
We verify convergence. Since \( f(t) = t\left(1 - rac{t^2}{6}
ight) \), for \( t > 0 \), \( f(t) < t \) when \( 1 - rac{t^2}{6} < 1...
c_2 = f(0.5) = 0.5 - rac{(0.5)^3}{6} = 0.5 - rac{0.125}{6} = 0.5 - 0...
Decreasing and bounded below by 0, so convergent. The only fixed point...
\lim_{n o \infty} c_n = 0
Question: A science communicator creates a puzzle involving the equati...

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