Use tangent information without promising global success

Try

For f(x)=x²−2, start at x₀=1. The tangent line has zero x₁=3/2. A second tangent gives x₂=(3/2+2/(3/2))/2=17/12. The method replaces a nonlinear equation by a local linear model and solves that simpler model exactly. Draw the curve and tangent intersection, then derive the recurrence from the line equation. This makes the origin of the step visible rather than treating it as a formula to memorize.

Predict

Predict why a locally accurate tangent can be a bad global guide. Far from a root, its intercept can leave the useful region, encounter a singularity or enter a cycle. A tiny derivative can produce an enormous step. Newton’s rapid convergence theorem is local and concerns a simple root. It does not say that every start finds a root, or that the closest root necessarily attracts the iteration.

Experiment

Trace the square-root calculation exactly.

n	xₙ	f(xₙ)	Next value
0	1	−1	3/2
1	3/2	1/4	17/12
2	17/12	1/144	577/408

Check the final fraction with x/2+1/x. Compare with f(x)=(x−1)², whose error is merely halved at each nonroot step. The residual can still decrease quickly for a flat function.

Understand

Newton’s update is xₙ₊₁=xₙ−f(xₙ)/f′(xₙ). Suppose f is C² near a root r with f′(r)≠0. On a small closed neighborhood choose |f′|≥m>0 and |f″|≤M. Taylor expansion of f(r) about xₙ gives eₙ₊₁=f″(ξ)eₙ²/(2f′(xₙ)), hence |eₙ₊₁|≤C|eₙ|² with C=M/(2m). Choose the start close enough that C|e₀|<1 and the error remains inside that neighborhood. Induction proves convergence and the quadratic bound. This invariant-neighborhood argument matters: a local formula alone cannot guarantee that later iterates satisfy its hypotheses.

Apply

For a multiplicity-m root, write f(x)=(x−r)^m h(x), where h(r)≠0 and h is sufficiently smooth. Then f/f′ behaves like (x−r)/m, giving asymptotic error factor 1−1/m rather than quadratic convergence. Known multiplicity can motivate the modified correction m f/f′. Numerical evaluation near the root can still lose significant digits, and a small residual may conceal a substantial root error when f is flat. Derivative size, root conditioning and a maintained bracket are useful additional evidence. A repeated root can also split under perturbations, so exact multiplicity itself may be fragile.

A residual-to-root bound is available if |f′|≥m>0 throughout an interval joining x and a root r: the mean value theorem gives |x−r|≤|f(x)|/m. Without such a lower derivative bound, a residual tolerance has no direct distance meaning. Combining this certificate with a bracket or known monotonicity is stronger evidence than simply stopping because consecutive printed iterates look identical. Near machine precision, distinct mathematical steps may round to the same stored value.

A finite run needs an honest stopping contract

Consider f(x)=x³−2x+2. Its derivative is 3x²−2. Newton started at zero produces 1; started at 1 it produces zero. This exact two-cycle has nonzero residuals 2 and 1. It disproves the universal-start claim without contradicting the local simple-root theorem above.

A separate bracket [−2,−1] has f(−2)=−2 and f(−1)=3. On this interval f′≥1, so the function is strictly increasing and its root is unique. At each stage keep endpoints a,b with f(a)≤0≤f(b). Accept a Newton proposal only if it lies in the middle half [a+(b−a)/4,b−(b−a)/4]; otherwise use the midpoint. Evaluate the accepted point and retain the subinterval with opposite signs. If its residual is exactly zero, stop at that root. Otherwise continuity preserves existence, and each retained interval has width at most three quarters of its predecessor. Induction gives width at most (3/4)^N times the original width. Its midpoint is within half that width of the unique root. The middle-half rule is deliberately conservative: mere membership in the bracket would not give this uniform contraction factor.

Worked first step. A failed proposal outside [−2,−1] is replaced by −3/2. Since f(−3/2)=13/8, the new bracket is [−2,−3/2]. On the original interval the derivative lower bound 1 also gives the residual certificate |x−r|≤|f(x)|. The width certificate is often sharper early in the run.

Changed written task. Replace the start by x=−1. Its Newton proposal is −4, so reject it. Prove the invariant and calculate how many accepted steps suffice to make bracket width at most 1/100, using the conservative factor. Feedback: choose N with (3/4)^N≤1/100; N=17 suffices, while 16 does not. This bound counts accepted evaluations, not the extra derivative/proposal work. Floating-point sign certification and a maximum evaluation budget are implementation obligations; when the budget ends, return the retained interval and “budget exhausted,” never a fabricated convergence claim.

Predict, change the case, and repair the claim

Case Outside the bracketA root cycle

f(x)=x³−2x+2; bracket [−2,−1]; proposal from x=0 is 1.

Which next step preserves the stated bracket contract?

Accept 1 because it is a Newton step Use midpoint −3/2 and update by sign Declare no real root

Check this claim Reset practice

Before checking, write the hypothesis or calculation that decides your answer. Change cases and explain which assumption changed. These fixed cases are public guided practice, not a saved score or independent proof assessment.

Continue

Newton is a local model-based solver, not an unconditional root detector. Next, secant iteration replaces derivatives by slopes and safeguarded methods retain a bracket while trying faster steps. Record each evaluated point, derivative, proposed step, accepted step and stopping reason. The written implementation prompts are design exercises; executed code needs independent tests. A residual trace demonstrates what happened for those inputs, whereas a convergence theorem explains a specified class of inputs under assumptions that must be checked.

Review calculus and linear algebra when a prerequisite needs practice. Continue to Combine fast proposals with a reliable bracket.