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Newton-Raphson Method Calculator – Roots Fast

Quick answer

Newton iteration x←x−f(x)/f′(x) solves x²−2 from x₀=1: 1 → 1.5 → 1.4167 → 1.41422.

  • Tangent-line steps and doubling digits shown.

Figures are estimates from this calculator’s standard formula — adjust the inputs in the tool below for your exact number.

How the Newton-Raphson Calculator Works

xₖ₊₁ = xₖ − f(xₖ)/f′(xₖ)

Newton’s method slides down the tangent line: at each guess xₖ it follows the tangent’s slope f′(xₖ) to where that tangent hits zero, which lands closer to the actual root. Solving f(x) = x² − 2 from x₀ = 1: the tangent update gives x₁ = 1 − (−1)/2 = 1.5, then x₂ = 1.41667, then x₃ = 1.41422 — √2 to five digits in three steps. Convergence is quadratic near a simple root: correct digits double each iteration (error 0.086 → 0.0025 → 0.0000031 in the run above). The calculator prints f and f′ at every iterate, the tangent step, and the residual, and warns about the failure modes — a zero derivative mid-run, or oscillation when the starting point sits far from the root on a curvy function (x₀ = −1 on this parabola bounces between −1 and 1; choosing x₀ = 2 fixes it). Compare slower but safer methods: bisection halves the interval every step.

  1. 1

    Enter f(x) and f′(x)

    The function and its derivative (or pick a preset)

  2. 2

    Choose the starting guess

    Close guesses converge fastest

  3. 3

    Iterate the tangent step

    Each xₖ, f(xₖ), f′(xₖ) and update is listed

  4. 4

    Stop at tolerance

    Run ends when |f(xₖ)| < your threshold

Use Cases

Root Finding

Solve x² − 2 = 0 to machine precision in 3–4 steps

Optimization

Set f′ = 0: Newton on the derivative finds maxima/minima

Financial Models

Iterate yield-to-maturity and implied volatility solutions

Tips

  • 1

    Quadratic convergence: digits double per step near a simple root

  • 2

    Each step costs one function plus one derivative evaluation

  • 3

    A derivative near zero stalls the method — pick another start

Common Mistakes

  • Using f′(x₀) for every step instead of recomputing f′(xₖ)

  • Starting far from the root on flat regions and diverging

  • Forgetting multiplicity: double roots converge only linearly

  • Skipping the residual check |f(x)| before accepting the answer

FAQs

How fast does Newton find √2?

From x₀ = 1: 1.5 → 1.41667 → 1.41422, with error shrinking 0.086 → 0.0025 → 0.0000031.

Why is it called the tangent method?

Each step follows the tangent line at xₖ down to its zero — geometry doing the algebra.

When does Newton fail?

When f′ vanishes near the iterates, or when a poor start cycles — x₀ = −1 on x²−2 oscillates between ±1.

How does it compare with bisection?

Newton doubles digits per step but needs f′ and a decent start; bisection halves the interval unconditionally.

Can Newton find maxima?

Yes — run it on f′ with f″ as the derivative to solve f′ = 0.