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Newton's Method — Definition, Formula & Examples

Newton's Method is an iterative technique that uses the tangent line at a current guess to find successively better approximations to a root (zero) of a function. Starting from an initial estimate, each iteration refines the guess by following the tangent line to where it crosses the x-axis.

Given a differentiable function ff and an initial approximation x0x_0 to a root of f(x)=0f(x) = 0, Newton's Method generates a sequence {xn}\{x_n\} defined by the recurrence 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 to 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

You start with a guess x0x_0 near the root you want to find. At each step, you compute the tangent line to the curve y=f(x)y = f(x) at the point (xn,f(xn))(x_n, f(x_n)) and find where that tangent line hits the x-axis — that intersection becomes your next guess xn+1x_{n+1}. You repeat this process until two consecutive approximations agree to the desired number of decimal places. The method converges very quickly when it works — roughly doubling the number of correct digits with each iteration — but it can fail if the derivative is zero at an iterate, if the initial guess is too far from the root, or if the function has features like inflection points near the root that send iterates away.

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: Identify the function and its derivative.
f(x)=x2−3,f′(x)=2xf(x) = x^2 - 3, \quad f'(x) = 2x
Iteration 1: Apply the formula 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: Apply the formula with 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.

Another Example

Problem: Use Newton's Method with x0=1x_0 = 1 to find a root of f(x)=x3−x−1f(x) = x^3 - x - 1. Perform two iterations.
Setup: The derivative is f′(x)=3x2−1f'(x) = 3x^2 - 1.
f(x)=x3−x−1,f′(x)=3x2−1f(x) = x^3 - x - 1, \quad f'(x) = 3x^2 - 1
Iteration 1: Compute f(1)=1−1−1=−1f(1) = 1 - 1 - 1 = -1 and f′(1)=3−1=2f'(1) = 3 - 1 = 2.
x1=1−−12=1.5x_1 = 1 - \frac{-1}{2} = 1.5
Iteration 2: Compute f(1.5)=3.375−1.5−1=0.875f(1.5) = 3.375 - 1.5 - 1 = 0.875 and f′(1.5)=6.75−1=5.75f'(1.5) = 6.75 - 1 = 5.75.
x2=1.5−0.8755.75≈1.347826x_2 = 1.5 - \frac{0.875}{5.75} \approx 1.347826
Answer: After two iterations, x2≈1.3478x_2 \approx 1.3478. The true root is approximately 1.32471.3247, so further iterations would continue to improve accuracy.

Visualization

Why It Matters

Newton's Method is a core topic in Calculus I and II courses and appears frequently on AP Calculus exams. Engineers and scientists use it constantly — for instance, to solve equations arising in structural analysis, circuit design, and orbital mechanics that have no closed-form solution. It also forms the foundation for more advanced root-finding algorithms studied in numerical analysis courses.

Common Mistakes

Mistake: Forgetting the minus sign in the formula and computing xn+1=xn+f(xn)f′(xn)x_{n+1} = x_n + \frac{f(x_n)}{f'(x_n)}.
Correction: The formula subtracts the ratio: xn+1=xn−f(xn)f′(xn)x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}. The subtraction is what moves you toward the root along the tangent line.
Mistake: Using f′(x)f'(x) evaluated at xn+1x_{n+1} or at the root instead of at xnx_n.
Correction: Both ff and f′f' must be evaluated at the current approximation xnx_n to compute the next value xn+1x_{n+1}.

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