Newton's method can double its mistake every step, and the answer was 0.01 away

By the 42Math team · 22 June 20265 min read

Start 0.01 from the answer. Twenty steps later you are at −5,242.88, and every step was computed correctly.

Solve x³ − x − 2 = 0. There is no neat formula worth remembering, so do what every piece of numerical software does: guess, then let Newton's method correct the guess.

Start at 1.5. Three iterations later you have 1.52138, and the function evaluated there is 5.89387 × 10⁻⁷, within a millionth of zero. Three steps. Each step roughly doubles the number of correct digits, which is why this method is inside your calculator, your CAD package and the solver in your spreadsheet.

Now ask it for something easier. Solve ∛x = 0.

You know the answer. It is zero, and it is the only one. Start at 0.01 — already within a hundredth — and watch:

0.01 → −0.02 → 0.04 → −0.08 → 0.16 → −0.32 → 0.64 → −1.28 → 2.56 → …

After twenty iterations the estimate is −5242.88. The residual, which began at 0.215, is now 17.37. The method has not stalled or wandered; it has moved away from the answer in a perfectly straight line of reasoning, doubling the error and flipping the sign every single step.

Nothing went wrong

The temptation is to look for a bug. There isn't one, and the arithmetic is worth doing because it is so short.

Newton's step is x − f(x)/f′(x). For f(x) = x^1/3, the derivative is (1/3)x^−2/3, so

f(x)/f′(x) = x^1/3 ÷ ⅓x^−2/3 = 3x

and the next estimate is x − 3x = −2x. Not approximately: exactly. Every iteration multiplies the estimate by −2, from any starting point, forever. That is where −5242.88 comes from: it is 0.01 × (−2)¹⁹, and you can read the whole geometric sequence off the iteration table in the Newton's Method Explorer under its diverge preset.

What the cube root has done is break the assumption the method rests on. Newton's method replaces the curve with its tangent line and jumps to where that line crosses zero. Near a root that is an excellent approximation: it is the first term of the Taylor expansion, and the ignored terms shrink quadratically, which is exactly why the digits double. But ∛x has a vertical tangent at zero. Its derivative doesn't merely get small there, it becomes infinite, and the curve is not locally straight at any magnification. Tangent lines near zero point almost straight up, and the axis crossing still lands at −2x: twice as far from the root as where you started, on the other side.

Watch the two columns beside each other in the table: f(x) grows while f′(x) shrinks toward 0.001. The step is their ratio, so it grows twice over.

The second failure is worse, because it looks like success

Divergence at least announces itself.

Take f(x) = x³ − x, whose roots are −1, 0 and +1. Start at 0.57. The method converges cleanly, thirteen iterations, residual 2.3 × 10⁻¹¹ — and the root it hands back is −1.

Look at where 0.57 sits. It is 0.43 from +1 and 1.57 from −1. It is also only 0.57 from the root at zero. Newton's method returns the one that is furthest away, having sailed straight past two nearer roots, and reports total success with eleven correct decimal places.

Now start at 0.58 instead. Fifteen iterations, and the answer is +1.

One hundredth of a change in the initial guess, and the solver lands at the opposite end of the function. Between those two starting points is 1/√3 ≈ 0.5774, where the derivative of x³ − x is zero. The tangent there is horizontal and crosses the axis nowhere near; approach that point from either side and the first step is enormous, in opposite directions. Every flat spot on a curve is a launch pad, and the region it throws you into has nothing to do with proximity.

This is the property that makes "which root will I get?" an unanswerable question in general. For polynomials over the complex numbers the sets of starting points leading to each root are called basins of attraction, and their boundaries are fractal: arbitrarily close to a point that converges to one root sits a point that converges to another. Hubbard, Schleicher and Sutherland showed in 2001 that you can nevertheless construct a finite set of starting points guaranteed to find every root of a given polynomial. A real result, and a measure of how much work "just use Newton's method" quietly hides.

What this changes about using it

Three practical consequences follow, and the first two are visible on the page.

First, a small residual is not a certificate. In the diverge run above the residual is 17.37 and rising, so that one is caught. But in the 0.57 run the residual is 2 × 10⁻¹¹ and the answer is a perfectly good root, just not the one anybody asking the question wanted. Convergence tests can tell you that you have found a solution. Nothing in the method can tell you it is the solution.

Second, iteration caps are load-bearing. The explorer stops at twenty by default, which is the only reason the divergent run has an ending rather than an overflow. Production solvers do the same thing, and the number is not arbitrary: a healthy Newton iteration on a well-behaved function reaches machine precision in five to ten steps, so anything still running at twenty is not converging slowly, it is not converging.

Third, the fix for a bad start is usually not a better algorithm but a bracket. Bisection cannot diverge — if the function changes sign across an interval, the midpoint rule cannot leave it — and it cannot pick the wrong root, because there is only one inside the bracket. It is merely slow, gaining one bit per step rather than doubling digits. Most real solvers therefore run a hybrid: bisect until the estimate is provably close, then let Newton finish in three steps. You get the speed where the assumptions hold and the guarantee where they don't.

There is a footnote about the name. Newton's own method, from around 1669, did not use derivatives and was written for polynomials as a sequence of substitutions; Raphson simplified it in 1690; the form taught today, with f′ and an arbitrary differentiable f, is Simpson's from 1740. Ypma traced the whole descent in 1995. The thing that fails so instructively on the cube root is a eighteenth-century tidy-up of a seventeenth-century recipe, and it has been in every numerical toolbox since, which is the case for knowing precisely where it breaks.

Published 22 June 2026 · corrections welcome via the corrections page.