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Newton's Method — Formula, Examples & How to Find

Newton's Method

An iterative process using derivatives that can often (but not always) be used to find zeros of a differentiable function. The basic idea is to start with an approximate guess for the zero, then use the formula below to turn that guess into a better approximation. This process is repeated until, after only a few steps, the approximation is extremely close to the actual value of the zero.

Note: In some circumstances, Newton's method backfires and gives successively worse and worse approximations.

Newton's Method formula x_{n+1} = x_n - f(x_n)/f'(x_n), with example approximating √2 using f(x)=x²-2, starting at x₁=1.5,...

Key Formula

xn+1=xnf(xn)f(xn)x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}
Where:
  • xnx_n = The current approximation of the zero (the nth guess)
  • xn+1x_{n+1} = The next (improved) approximation of the zero
  • f(xn)f(x_n) = The value of the function evaluated at the current guess
  • f(xn)f'(x_n) = The value of the derivative of the function evaluated at the current guess

Worked Example

Problem: Use Newton's Method to approximate a zero of f(x) = x² − 9, starting with an initial guess of x₀ = 2. Perform two iterations.
Step 1: Identify the function and its derivative.
f(x)=x29,f(x)=2xf(x) = x^2 - 9, \quad f'(x) = 2x
Step 2: Apply the formula with x₀ = 2.
x1=2f(2)f(2)=2494=254=2+1.25=3.25x_1 = 2 - \frac{f(2)}{f'(2)} = 2 - \frac{4 - 9}{4} = 2 - \frac{-5}{4} = 2 + 1.25 = 3.25
Step 3: Apply the formula again with x₁ = 3.25.
x2=3.25(3.25)292(3.25)=3.2510.562596.5=3.251.56256.53.250.24043.0096x_2 = 3.25 - \frac{(3.25)^2 - 9}{2(3.25)} = 3.25 - \frac{10.5625 - 9}{6.5} = 3.25 - \frac{1.5625}{6.5} \approx 3.25 - 0.2404 \approx 3.0096
Step 4: Compare to the exact root. The true zero is x = 3. After just two iterations, our approximation of 3.0096 is already within 0.01 of the exact answer.
Answer: After two iterations, x₂ ≈ 3.0096, which is very close to the exact zero x = 3.

Another Example

This example shows how Newton's Method can compute nth roots — values that don't simplify to clean numbers. It demonstrates a practical application beyond solving polynomial equations with known integer roots.

Problem: Use Newton's Method to approximate the cube root of 10 by finding a zero of f(x) = x³ − 10, starting with x₀ = 2. Perform two iterations.
Step 1: Set up the function and derivative. Finding the cube root of 10 is the same as solving x³ = 10.
f(x)=x310,f(x)=3x2f(x) = x^3 - 10, \quad f'(x) = 3x^2
Step 2: Apply the formula with x₀ = 2.
x1=223103(22)=281012=2212=2+0.162.1667x_1 = 2 - \frac{2^3 - 10}{3(2^2)} = 2 - \frac{8 - 10}{12} = 2 - \frac{-2}{12} = 2 + 0.1\overline{6} \approx 2.1667
Step 3: Apply the formula with x₁ ≈ 2.1667.
x2=2.1667(2.1667)3103(2.1667)22.166710.17131014.08342.16670.012162.1545x_2 = 2.1667 - \frac{(2.1667)^3 - 10}{3(2.1667)^2} \approx 2.1667 - \frac{10.1713 - 10}{14.0834} \approx 2.1667 - 0.01216 \approx 2.1545
Step 4: The actual value of the cube root of 10 is approximately 2.15443. After two iterations, our estimate of 2.1545 is accurate to nearly four decimal places.
Answer: After two iterations, x₂ ≈ 2.1545, which approximates ∛10 ≈ 2.15443 to four decimal places.

Frequently Asked Questions

When does Newton's Method fail?
Newton's Method can fail if the derivative f'(xₙ) equals zero at any iteration, because the formula requires division by f'(xₙ). It can also fail if the initial guess is too far from the root, if the function has an inflection point near the root, or if the iterations cycle without converging. In such cases, the approximations may diverge or oscillate rather than approach a zero.
Why does Newton's Method work?
At each step, the method constructs the tangent line to the curve at the current point (xₙ, f(xₙ)). The x-intercept of that tangent line becomes the next approximation xₙ₊₁. Because the tangent line is a good local approximation of the curve, its x-intercept is usually closer to the actual root than xₙ was. This geometric reasoning is why the method converges so rapidly when it works.
How many iterations of Newton's Method do you need?
When Newton's Method converges, it typically does so very quickly — often 3 to 5 iterations are enough for high accuracy. The method exhibits quadratic convergence near a simple root, meaning the number of correct decimal digits roughly doubles with each step. In practice, you stop iterating when successive approximations differ by less than a desired tolerance.

Newton's Method vs. Bisection Method

Newton's MethodBisection Method
ApproachUses the tangent line (derivative) to find the next approximationRepeatedly halves an interval known to contain a root
Speed of convergenceQuadratic convergence — very fast when it worksLinear convergence — slower but steady
RequirementsNeeds the function to be differentiable; needs a good initial guessOnly needs continuity and two starting values where f changes sign
Guaranteed to converge?No — can diverge or cycle with a poor starting guessYes — always converges to a root within the interval
When to useWhen you can compute the derivative and have a reasonable initial guessWhen you need a guaranteed result or lack derivative information

Why It Matters

Newton's Method appears in AP Calculus and university calculus courses as a key application of derivatives. Engineers, scientists, and programmers use it routinely because many real-world equations — from orbital mechanics to financial modeling — cannot be solved algebraically, and Newton's Method provides fast numerical solutions. Understanding it also deepens your grasp of how tangent lines approximate curves, a central idea in differential calculus.

Common Mistakes

Mistake: Forgetting to use the derivative in the denominator, or accidentally using f(x) in both the numerator and denominator.
Correction: The formula is xₙ₊₁ = xₙ − f(xₙ)/f'(xₙ). The numerator is the original function, and the denominator is its derivative. Double-check that you have computed f'(x) correctly before substituting.
Mistake: Choosing an initial guess where f'(x₀) is zero or nearly zero.
Correction: If f'(x₀) = 0, the tangent line is horizontal and never crosses the x-axis, so the formula is undefined. Pick a starting value where the derivative is not close to zero. Sketching the function first can help you choose a good x₀.

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