Newton's Method — Step by Step with Examples

Q: How fast does Newton's method converge?

Quadratic convergence — the number of correct digits roughly doubles each iteration. Finding √2: x₀=1.5 (1 digit), x₁=1.4167 (3 digits), x₂=1.41421 (5 digits).

Q: When does Newton's method fail?

When f'(x) = 0 (horizontal tangent), when the initial guess is too far away, or when the function oscillates. A common example: f(x) = x^(1/3) near x=0.

Q: What is a good stopping criterion?

Stop when |xₙ₊₁ − xₙ| < ε (tolerance) or |f(xₙ)| < ε. Common tolerance: 10⁻⁶ to 10⁻¹⁰ depending on required precision.

Q: How do I choose the initial guess?

Graph the function to estimate where it crosses zero. Use the Intermediate Value Theorem: if f(a) and f(b) have different signs, a root lies between a and b. Start midway.

Newton's method (also called Newton-Raphson) is an iterative algorithm for finding roots of equations — values where f(x) = 0. Starting from an initial guess, it uses the tangent line at each point to jump closer to the root. It converges very fast (quadratically) when it works, roughly doubling the number of correct digits each iteration.

Newton's Method (Newton-Raphson) (xₙ₊₁)

xₙ₊₁ = xₙ − f(xₙ)/f'(xₙ)

Step-by-Step Method

Choose an Initial Guess x₀

Pick a starting value near the expected root. For f(x) = x² − 2 (finding √2), start with x₀ = 1.5 since √2 ≈ 1.4.

x₀ ≈ expected root

Compute f(x₀) and f'(x₀)

f(1.5) = 1.5² − 2 = 0.25. f'(x) = 2x, so f'(1.5) = 3.

f(x₀) = 0.25, f'(x₀) = 3

Apply the Formula

x₁ = 1.5 − 0.25/3 = 1.5 − 0.0833 = 1.4167. Already close to √2 = 1.4142!

x₁ = x₀ − f(x₀)/f'(x₀)

Repeat Until Convergence

x₂ = 1.4167 − (1.4167² − 2)/(2 × 1.4167) = 1.4167 − 0.00694/2.8334 = 1.4142. After just 2 iterations, we have √2 to 4 decimal places.

|xₙ₊₁ − xₙ| < tolerance → stop

When This Method Fails

Newton's method can fail when: f'(x) = 0 at some iteration (division by zero), the initial guess is too far from the root (may diverge or cycle), the function has multiple roots nearby, or the function is not differentiable. Always check that iterations are converging by monitoring |xₙ₊₁ − xₙ|.

Real-World Applications

Newton's method is used in: computer graphics (ray tracing, finding intersections), financial modeling (implied volatility in options pricing), physics simulations (solving nonlinear equations), optimization algorithms (finding where the gradient equals zero), and engineering (solving circuit equations).

Frequently Asked Questions

How fast does Newton's method converge?▼

Quadratic convergence — the number of correct digits roughly doubles each iteration. Finding √2: x₀=1.5 (1 digit), x₁=1.4167 (3 digits), x₂=1.41421 (5 digits).

When does Newton's method fail?▼

When f'(x) = 0 (horizontal tangent), when the initial guess is too far away, or when the function oscillates. A common example: f(x) = x^(1/3) near x=0.

What is a good stopping criterion?▼

Stop when |xₙ₊₁ − xₙ| < ε (tolerance) or |f(xₙ)| < ε. Common tolerance: 10⁻⁶ to 10⁻¹⁰ depending on required precision.

How do I choose the initial guess?▼

Graph the function to estimate where it crosses zero. Use the Intermediate Value Theorem: if f(a) and f(b) have different signs, a root lies between a and b. Start midway.

Try It Yourself

Skip the manual calculations. Use our free Newton's Method Calculator for instant results.

Open Newton's Method Calculator →

Newton's Method — Step by Step with Examples

Newton's Method (Newton-Raphson) (xₙ₊₁)

xₙ₊₁ = xₙ − f(xₙ)/f'(xₙ)

Step-by-Step Method

Choose an Initial Guess x₀

Pick a starting value near the expected root. For f(x) = x² − 2 (finding √2), start with x₀ = 1.5 since √2 ≈ 1.4.

x₀ ≈ expected root

Compute f(x₀) and f'(x₀)

f(1.5) = 1.5² − 2 = 0.25. f'(x) = 2x, so f'(1.5) = 3.

f(x₀) = 0.25, f'(x₀) = 3

Apply the Formula

x₁ = 1.5 − 0.25/3 = 1.5 − 0.0833 = 1.4167. Already close to √2 = 1.4142!

x₁ = x₀ − f(x₀)/f'(x₀)

Repeat Until Convergence

x₂ = 1.4167 − (1.4167² − 2)/(2 × 1.4167) = 1.4167 − 0.00694/2.8334 = 1.4142. After just 2 iterations, we have √2 to 4 decimal places.

|xₙ₊₁ − xₙ| < tolerance → stop

When This Method Fails

Real-World Applications

Frequently Asked Questions

How fast does Newton's method converge?▼

Quadratic convergence — the number of correct digits roughly doubles each iteration. Finding √2: x₀=1.5 (1 digit), x₁=1.4167 (3 digits), x₂=1.41421 (5 digits).

When does Newton's method fail?▼

When f'(x) = 0 (horizontal tangent), when the initial guess is too far away, or when the function oscillates. A common example: f(x) = x^(1/3) near x=0.

What is a good stopping criterion?▼

Stop when |xₙ₊₁ − xₙ| < ε (tolerance) or |f(xₙ)| < ε. Common tolerance: 10⁻⁶ to 10⁻¹⁰ depending on required precision.

How do I choose the initial guess?▼

Graph the function to estimate where it crosses zero. Use the Intermediate Value Theorem: if f(a) and f(b) have different signs, a root lies between a and b. Start midway.

Try It Yourself

Skip the manual calculations. Use our free Newton's Method Calculator for instant results.

Open Newton's Method Calculator →