Many equations do not hand over their answers in a tidy formula. A calculator can still solve them by asking a more practical question: if the curve looks straight right here, where would that local straight line cross zero? Newton’s method repeats that question, shifting the tangent line after every answer.
It is a remarkably efficient bargain when the guess is near a simple root. It is also easy to overtrust. A tangent can point at the wrong root, leap across a boundary, stall on a flat patch, or cycle forever. The formula is short; the judgment around it is not.
f(x) = 0xn+1 = xn − f(xn) / f′(xn)The next estimate is the current estimate minus its residual divided by its local slope.
What every part of the update carries
The current x-coordinate, with the units of the unknown: metres, seconds, kelvin, or a dimensionless number.
How far the equation is from zero, expressed in the output units of f.
Output units per x-unit. It tells the tangent how rapidly the residual changes nearby.
A dimensionless counter that records repeated approximation, not elapsed time.
The units supply a quiet error check. The quotient f(xn) / f′(xn) has x-units, so it can be subtracted from xn. If that subtraction mixes unlike units, either the derivative or the model has been defined incorrectly.
NIST’s Digital Library of Mathematical Functions gives this rule for real twice-continuously differentiable functions and complex analytic functions. Its origin is only a first-order approximation.
Derive the rule from one tangent
At a current estimate xn, approximate the curve a short distance Δx away:
f(xn + Δx) ≈ f(xn) + f′(xn)Δx0 ≈ f(xn) + f′(xn)ΔxΔx ≈ −f(xn) / f′(xn)xn+1 = xn + ΔxThe approximation is the entire bargain. Newton’s method does not know the curve between samples; it temporarily replaces the curve with its tangent. The NIST phase-field numerical guide describes Newton-type methods in the same sequence: linearize, calculate a correction, iterate.
Worked example 1: finding √2 with no square-root key
Set f(x) = x² − 2; then f′(x) = 2x. The update simplifies to:
xn+1 = xn − (xn² − 2)/(2xn) = ½(xn + 2/xn)Start at x0 = 1.5, a value above √2.
x1 = 1.5 − 0.25/3 = 17/12 = 1.4166666667x2 = 17/12 − (1/144)/(17/6) = 577/408 = 1.4142156863x3 = 665857/470832 = 1.41421356237469x3² − 2 ≈ 4.51 × 10−12The reference value begins √2 = 1.41421356237310…. Three corrections have delivered eleven matching decimal places. The root is simple, its derivative is not zero, and the first guess is close enough for the tangent approximation to improve itself. DLMF calls this local behavior quadratic convergence: once in the right neighborhood, the number of correct digits can roughly double per step.
Worked example 2: a cubic whose root is not obvious
Solve x³ − 2x − 5 = 0 from x0 = 2. Here f′(x) = 3x² − 2.
| estimate | residual | slope | next estimate |
|---|---|---|---|
2 | −1 | 10 | 2 − (−1/10) = 2.1 |
2.1 | 0.061 | 11.23 | 2.094568121104185 |
2.094568121104185 | 0.000185723173272 | 11.161646841838 | 2.094551481698199 |
2.094551481698199 | 1.739761224 × 10−9 | 11.161437728452 | 2.094551481542327 |
Substitution at the final value leaves a residual of about 9 × 10−16 in IEEE double arithmetic. That is a strong numerical result, not a proof that a displayed decimal is exact. A serious result needs a tolerance, a stopping rule, and the function’s permitted domain.
When a fast tangent becomes a bad guide
If f′(xn) is zero or tiny, the division is undefined or launches a huge step. Rescale or use a safeguarded method.
The tangent may land near a different root or far outside the meaningful domain. A plausible first guess is not a guarantee.
For x³ − 2x + 2, starting at 0 gives 1, then 0 again. Track residual history and stop after a finite count.
For (x − 1)², the update becomes (xn + 1)/2: error halves rather than squares.
Stop on two kinds of evidence
|xn+1 − xn| ≤ x toleranceHas the estimate stopped moving materially in the units that matter?
|f(xn+1)| ≤ residual toleranceDoes the estimate satisfy the equation to the required output accuracy?
n < maximum iterationsReturn an explicit failure state rather than silently accepting a stall or cycle.
Neither numerical test is sufficient alone. A small step can mean the solver is trapped in a flat region; a small residual can mislead when the equation is badly scaled. Practical solvers often combine Newton’s fast local step with a bracket, damping, or a line search. NIST’s work on regulated Newton-Raphson methods makes the same broader point: global reliability needs extra structure.
Newton’s method can tell you how a smooth equation’s nearby tangent points toward zero. It cannot certify that the selected root is unique, physically permitted, globally reachable, or accurate beyond the precision of the supplied function and derivative. A fast update is evidence to check, not a conclusion to announce.