When Newton Slips Past the Root
A student is finding the root of f(x) = x³ − 2x + 2, which has a single real root near x = −1.8. She starts Newton–Raphson from x₀ = 0 and is puzzled: the iterates jump to 1, then to 0, then back to 1, and never settle. She tries x₀ = 0.5 and again sees oscillation. She then tries the bisection method on [−2, −1] and finds it converges steadily but slowly, needing many iterations for three decimal places. She wants to understand why Newton fails here, whether the failure is her arithmetic or the method, and how to get an accurate root efficiently.
1. Using f(x) = x³ − 2x + 2 and f'(x) = 3x² − 2, show what happens to the Newton iterates from x₀ = 0, and explain the cause.
Read the worked answer
Newton's formula is x_{n+1} = x_n − f(x_n)/f'(x_n).
At x₀ = 0: f(0) = 2, f'(0) = −2, so x₁ = 0 − 2/(−2) = 1. At x₁ = 1: f(1) = 1 − 2 + 2 = 1, f'(1) = 1, so x₂ = 1 − 1/1 = 0.
So the iterates cycle 0 → 1 → 0 → 1 … exactly, a 2-cycle. This is not an arithmetic slip: it is a genuine failure of Newton's method. The cause is that f'(0) = −2 is small in magnitude relative to f(0) = 2, so the first step overshoots far to the right; at x = 1 the tangent line sends the iterate straight back. The starting point lies where the tangent geometry does not funnel iterates toward the root. Newton converges only when the start is sufficiently close to the root and f' does not nearly vanish.
2. Explain why bisection on [−2, −1] is guaranteed to work here, estimate the number of iterations needed for the root to within 0.001, and suggest a better practical strategy.
Read the worked answer
Check the sign change: f(−2) = −8 + 4 + 2 = −2 < 0 and f(−1) = −1 + 2 + 2 = 3 > 0. Since f is continuous (a polynomial), the intermediate value theorem guarantees a root in (−2, −1), so bisection is valid.
After n halvings the interval length is 1/2ⁿ. We need 1/2ⁿ ≤ 0.001, i.e. 2ⁿ ≥ 1000. Since 2⁹ = 512 and 2¹⁰ = 1024, we need n = 10 iterations.
Better strategy: use a few bisection steps to get close to the root (say to within 0.1), then switch to Newton–Raphson from that point. Near the root f' is comfortably non-zero and Newton converges quadratically, so the hybrid reaches high accuracy in far fewer evaluations than bisection alone. Bisection gives guaranteed convergence; Newton gives speed but needs a good start.