How do I use the Newton-Raphson method?

Differentiate f, pick a starting value x₀ close to the root, then repeat xₙ₊₁ = xₙ − f(xₙ)/f'(xₙ) until two successive values agree to the accuracy you need. For x² − 2 from x₀ = 1 the values are 1.5, 1.41667, 1.414216, 1.41421356.

When does Newton-Raphson fail?

When f'(xₙ) = 0 the tangent is horizontal and never meets the x-axis, so the method stops; this happens if you start at a stationary point. Starting near one sends the next value far away. It can also cycle: for x³ − 2x + 2 from x₀ = 0 the values go 0, 1, 0, 1 for ever. A different starting value usually fixes it.

How can I tell whether x = g(x) will converge?

Work out g'(x) near the root. If |g'(α)| 1 it moves away from the root, however close you start. The smaller |g'(α)|, the faster it converges.

How many steps does bisection need?

Each step halves the interval, so after n steps it has width (b − a)/2ⁿ. To get the root to within 10⁻⁶ from an interval of width 1 you need 2ⁿ > 10⁶, so n = 20 steps. That is about 3.3 steps for each extra decimal place, much slower than Newton but guaranteed whenever there is a change of sign.

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Numerical methods: Newton-Raphson, iteration, bisection

Most equations have no formula for their roots, so we close in on them step by step. Compare three ways of doing it, see how fast each one gets there, and see how each one can fail.

Error against step (log scale)

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What's happening

Newton-Raphson starts at a guess x₀, draws the tangent there and takes the point where it meets the x-axis as the next guess: xₙ₊₁ = xₙ − f(xₙ)/f'(xₙ). Near a simple root it converges quadratically: each error is roughly the square of the one before, so the number of correct digits roughly doubles. It fails if a tangent is horizontal (f'(xₙ) = 0) and it can fall into a cycle. Fixed-point iteration rearranges f(x) = 0 as x = g(x) and repeats xₙ₊₁ = g(xₙ). On the graph that is up to the curve, across to y = x, and again: a staircase when g' is positive and a cobweb when it is negative. Each step multiplies the error by about |g'(α)| at the root α, so it converges only if |g'(α)| < 1. Bisection needs only a change of sign: if f(a) and f(b) have opposite signs, a continuous f has a root between them, and testing the midpoint halves the interval every step. It never fails, but it gains just one digit every 3.3 steps.

xₙ₊₁ = xₙ − f(xₙ) / f'(xₙ)xₙ₊₁ = g(xₙ), converges if |g'(α)| < 1Newton: eₙ₊₁ ≈ C eₙ² iteration: eₙ₊₁ ≈ |g'(α)| eₙ

A-Level Maths (AQA, Edexcel, OCR): locating roots by change of sign, fixed-point iteration with staircase and cobweb diagrams, the Newton-Raphson method and how each one fails. First-year numerical analysis: order of convergence.

Work through the numbers with Equation Solver and Graphing Calculator.

Challenge

Predict first: Newton-Raphson on x² − 2 = 0 from x₀ = 1. What are x₁ and x₂, and how many correct digits of √2 does x₄ have? Type your x₁ into the box, check it, then step on.

FAQ

How do I use the Newton-Raphson method?: Differentiate f, pick a starting value x₀ close to the root, then repeat xₙ₊₁ = xₙ − f(xₙ)/f'(xₙ) until two successive values agree to the accuracy you need. For x² − 2 from x₀ = 1 the values are 1.5, 1.41667, 1.414216, 1.41421356.
When does Newton-Raphson fail?: When f'(xₙ) = 0 the tangent is horizontal and never meets the x-axis, so the method stops; this happens if you start at a stationary point. Starting near one sends the next value far away. It can also cycle: for x³ − 2x + 2 from x₀ = 0 the values go 0, 1, 0, 1 for ever. A different starting value usually fixes it.
How can I tell whether x = g(x) will converge?: Work out g'(x) near the root. If |g'(α)| < 1 the iteration converges for starting values close enough, a staircase if g'(α) is positive and a cobweb if it is negative. If |g'(α)| > 1 it moves away from the root, however close you start. The smaller |g'(α)|, the faster it converges.
How many steps does bisection need?: Each step halves the interval, so after n steps it has width (b − a)/2ⁿ. To get the root to within 10⁻⁶ from an interval of width 1 you need 2ⁿ > 10⁶, so n = 20 steps. That is about 3.3 steps for each extra decimal place, much slower than Newton but guaranteed whenever there is a change of sign.