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Newton's Iteration (Newton's Method) — Definition, Formula & Examples

Newton's Method is a technique for finding approximate solutions to equations of the form f(x)=0f(x) = 0 by repeatedly improving a guess using the function's derivative. Each iteration draws a tangent line at the current guess and uses its xx-intercept as the next, better approximation.

Given a differentiable function ff and an initial approximation x0x_0 near a root of f(x)=0f(x) = 0, Newton's iteration generates a sequence {xn}\{x_n\} defined by xn+1=xn−f(xn)f′(xn)x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}, provided f′(xn)≠0f'(x_n) \neq 0. Under suitable conditions (e.g., ff is twice continuously differentiable and x0x_0 is sufficiently close to a simple root), the sequence converges quadratically to the root.

Key Formula

xn+1=xn−f(xn)f′(xn)x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}
Where:
  • xnx_n = Current approximation of the root
  • xn+1x_{n+1} = Next (improved) approximation
  • f(xn)f(x_n) = Value of the function at the current approximation
  • f′(xn)f'(x_n) = Value of the derivative at the current approximation

How It Works

Start by choosing an initial guess x0x_0 reasonably close to the root you want. Compute f(x0)f(x_0) and f′(x0)f'(x_0), then apply the formula to get x1x_1. Repeat the process: plug x1x_1 back in to get x2x_2, and so on. Each step typically doubles the number of correct decimal digits, so convergence is fast when it works. Stop iterating when successive approximations agree to the desired accuracy or when ∣f(xn)∣|f(x_n)| is small enough.

Worked Example

Problem: Use Newton's Method with initial guess x0=2x_0 = 2 to approximate 3\sqrt{3} by solving f(x)=x2−3=0f(x) = x^2 - 3 = 0. Perform two iterations.
Setup: Here f(x)=x2−3f(x) = x^2 - 3 and f′(x)=2xf'(x) = 2x. The iteration formula becomes:
xn+1=xn−xn2−32xnx_{n+1} = x_n - \frac{x_n^2 - 3}{2x_n}
Iteration 1: Start with x0=2x_0 = 2. Compute f(2)=4−3=1f(2) = 4 - 3 = 1 and f′(2)=4f'(2) = 4.
x1=2−14=1.75x_1 = 2 - \frac{1}{4} = 1.75
Iteration 2: Now use x1=1.75x_1 = 1.75. Compute f(1.75)=3.0625−3=0.0625f(1.75) = 3.0625 - 3 = 0.0625 and f′(1.75)=3.5f'(1.75) = 3.5.
x2=1.75−0.06253.5≈1.732143x_2 = 1.75 - \frac{0.0625}{3.5} \approx 1.732143
Answer: After two iterations, x2≈1.732143x_2 \approx 1.732143, which is already accurate to four decimal places compared to 3≈1.732051\sqrt{3} \approx 1.732051.

Why It Matters

Newton's Method is one of the most widely used root-finding algorithms in engineering, physics, and computer science. Optimization routines in machine learning often rely on Newton-type iterations. In calculus courses, it connects the geometric meaning of the derivative (tangent line slope) to a powerful computational tool.

Common Mistakes

Mistake: Choosing an initial guess where f′(x0)=0f'(x_0) = 0 or is very close to zero.
Correction: The formula divides by f′(xn)f'(x_n), so a zero or near-zero derivative causes the method to fail or produce wildly inaccurate results. Choose x0x_0 where the tangent line is not horizontal.