# Where does Newton’s method lead?

Runs Newton’s method xₙ₊₁ = xₙ − f(xₙ)/f′(xₙ) from a first guess x₀ and shows every step.

- Page: https://www.acalculator.org/math/newtons-method-calculator
- JSON spec: https://www.acalculator.org/math/newtons-method-calculator.json
- Version: 10d74d48a9f2

## Default answer

Example with the default inputs (Function f(x) x^3 - 3x + 1, First guess x₀ 2, Steps 6): After the steps from x₀ = 2, Newton’s method for x^3 - 3x + 1 reaches 1.532088886.

## Inputs

| Key | Label | Description |
| --- | --- | --- |
| f | Function f(x) | The function whose zero you want. |
| x0 | First guess x₀ | Where to start. |
| n | Steps | How many steps to take (it stops early when xₙ₊₁ = xₙ). |

## Outputs

| Key | Label | Description |
| --- | --- | --- |
| root | Last estimate xₙ | The last value of x Newton’s method reached. |
| settled | Settled? | Whether the last step changed x by at most 10⁻¹⁰ of its size. |
| steps | Steps | Each estimate xₖ. |
| derivative | f′(x) = | The derivative used in each step. |

## Method

xₖ₊₁ = xₖ − f(xₖ)/f′(xₖ), with f′ from a computer algebra system, checked numerically.

## Assumptions

- Steps run in double precision; angles in radians.
- It stops when f′(xₖ) = 0 or f(xₖ) = 0.

## Worked examples

1. f = x^3 - 3x + 1, x0 = 2, n = 6 gives root = 1.532089, settled = yes. Source: OpenStax, Calculus Volume 1, section 4.9 Newton’s Method (https://openstax.org/books/calculus-volume-1/pages/4-9-newtons-method), Example 4.46: x₆ ≈ 1.532088886.
2. f = x^2 - 2, x0 = 2, n = 3 gives root = 1.414216, steps = x₀ = 2; x₁ = 1.5; x₂ = 1.416666667; x₃ = 1.414215686. Source: OpenStax, Calculus Volume 1, section 4.9 Newton’s Method (https://openstax.org/books/calculus-volume-1/pages/4-9-newtons-method), Example 4.47: x₁ = 1.5, x₂ ≈ 1.416666667, x₃ ≈ 1.414215686 (exactly 577/408).
3. f = x^3 - 2x + 2, x0 = 0, n = 4 gives root = 0, settled = not yet: take more steps or try another x₀. Source: OpenStax, Calculus Volume 1, section 4.9 Newton’s Method (https://openstax.org/books/calculus-volume-1/pages/4-9-newtons-method), Example 4.48: x₀ = 0, x₁ = 1, x₂ = 0, … the estimates swing between 0 and 1.

## FAQ

### What is Newton’s method?

A way to estimate a zero of a differentiable function f. From a guess xₙ, follow the tangent line at (xₙ, f(xₙ)) to where it crosses the x-axis: xₙ₊₁ = xₙ − f(xₙ)/f′(xₙ). Repeat until the estimates stop changing.

### How do I use Newton’s method to find a square root?

Find the positive zero of f(x) = x² − a. For √2, use f(x) = x² − 2 and x₀ = 2: x₁ = 1.5, x₂ ≈ 1.416666667, x₃ ≈ 1.414215686, x₄ ≈ 1.414213562. The number of correct digits roughly doubles each step.

### How do I choose x₀?

Pick a value close to the root, for example from a graph or from a sign change of f (f(1) < 0 < f(2) means a root between 1 and 2). Different first guesses can lead to different roots.

### When does Newton’s method fail?

When f′(xₙ) = 0 (the tangent is flat and never meets the x-axis), when the estimates swing back and forth (f(x) = x³ − 2x + 2 from x₀ = 0 gives 0, 1, 0, 1, …), or when they run away from the root. The page stops with a message in the first case and says "not yet" when the last step still changed x.

### How many steps do I need?

Near a simple root, a few. The page stops early when a step no longer changes x; "Settled?" says yes when the last step changed x by at most 10⁻¹⁰ of its size. At a double root, such as x² at 0, the method is slower: each step only halves the distance.

### Is Newton’s method the same as the Newton–Raphson method?

Yes. The method is named after Isaac Newton and Joseph Raphson; both names are used for the same formula.

## Sources

- OpenStax, Calculus Volume 1, section 4.9 Newton’s Method (retrieved 2026-10-03): https://openstax.org/books/calculus-volume-1/pages/4-9-newtons-method