Newton's Method: Formula & Examples

Newton’s method approximates a zero of a function, a solution of f(x) = 0, by following tangent lines. Start with a guess x₀. Then repeat the formula $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$ . Each new value is where the tangent line at the old value crosses the x-axis, and the values usually close in on the zero very quickly.

Before this lesson: Linear Approximation, Equation of a Tangent Line and Optimization

The idea: slide down the tangent line

Near a guess x₀, the graph of f looks like its tangent line. A line is easy to solve: set $f(x_0) + f'(x_0)(x - x_0) = 0$ and solve for x. That gives a new guess, x₁, which is usually much closer to where the curve crosses the axis. Repeating the step gives x₂, x₃ and so on.

Each tangent line lands closer to the zero. Two steps already give √2 to three decimal places.

Newton’s method formula

x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}

Write f so the equation is f(x) = 0, and find f′(x).
Choose a starting guess x₀ close to the zero (a graph helps).
Use the formula to find x₁, then x₂, and so on.
Stop when two values agree to the number of decimal places you need.

Newton’s method can fail. It breaks down if f′(xₙ) = 0 (a horizontal tangent never reaches the axis), and a poor starting guess can jump to a different zero or bounce back and forth forever.

Newton’s method examples

Example 1: Approximating √2

Use Newton’s method with x₀ = 1 to approximate √2. Do three steps.

f(x) = x^2 - 2, \; f'(x) = 2x

√2 is the positive zero of x² − 2.

x_1 = 1 - \frac{1 - 2}{2(1)} = 1 + \frac{1}{2} = 1.5

Step 1: f(1) = −1 and f′(1) = 2.

x_2 = 1.5 - \frac{2.25 - 2}{3} = 1.5 - 0.083333 = 1.416667

Step 2: f(1.5) = 0.25 and f′(1.5) = 3.

x_3 = 1.416667 - \frac{0.006944}{2.833333} \approx 1.414216

Step 3: f(1.416667) ≈ 0.006944 and f′ ≈ 2.833333.

Answer: √2 ≈ 1.414216 after three steps. The true value is 1.4142136, so the error is about 0.000002.

Example 2: A cubic equation

Approximate the real solution of $x^3 - x - 1 = 0$ , starting at x₀ = 1.5.

f(x) = x^3 - x - 1, \; f'(x) = 3x^2 - 1

The function and its derivative.

x_1 = 1.5 - \frac{0.875}{5.75} \approx 1.347826

f(1.5) = 3.375 − 1.5 − 1 = 0.875 and f′(1.5) = 5.75.

x_2 = 1.347826 - \frac{0.100682}{4.449905} \approx 1.325200

Evaluate f and f′ at x₁.

x_3 = 1.325200 - \frac{0.002058}{4.268468} \approx 1.324718

Evaluate at x₂.

Answer: x ≈ 1.3247. The number of correct digits roughly doubles with each step.

Example 3: Solving cos x = x

Solve $\cos x = x$ (in radians), starting at x₀ = 1.

f(x) = \cos x - x, \; f'(x) = -\sin x - 1

Move everything to one side.

x_1 = 1 - \frac{-0.459698}{-1.841471} \approx 0.750364

f(1) = cos 1 − 1 and f′(1) = −sin 1 − 1.

x_2 = 0.750364 - \frac{-0.018923}{-1.681905} \approx 0.739113

Evaluate at x₁.

x_3 = 0.739113 - \frac{-0.000046}{-1.673633} \approx 0.739085

Evaluate at x₂.

Answer: x ≈ 0.7391. Algebra alone cannot solve this equation, but Newton’s method handles it easily.

Example 4: When Newton’s method fails

Try Newton’s method on $x^3 - 2x + 2 = 0$ starting at x₀ = 0.

f'(x) = 3x^2 - 2

Differentiate.

x_1 = 0 - \frac{2}{-2} = 1

f(0) = 2 and f′(0) = −2.

x_2 = 1 - \frac{1}{1} = 0

f(1) = 1 and f′(1) = 1.

x_3 = 1, \; x_4 = 0, \ldots

The values bounce between 0 and 1 forever.

Answer: the method cycles and never converges. A different start, like x₀ = −2, finds the real zero near −1.7693.

Example 5: Word problem with a bowl

A hemispherical bowl has radius 3 ft. When the water is h feet deep, the volume is $V = \frac{\pi h^2(9 - h)}{3}$ ft³. How deep is the water when the bowl holds 20 ft³? Start at h₀ = 1.5.

f(h) = \frac{\pi h^2(9 - h)}{3} - 20

Set the volume equal to 20 and move 20 to the left.

f'(h) = \frac{\pi(18h - 3h^2)}{3}

Expand to (9h² − h³)π/3, then differentiate.

h_1 = 1.5 - \frac{-2.328541}{21.205750} \approx 1.609807

Step 1.

h_2 = 1.609807 - \frac{0.055434}{22.202778} \approx 1.607310

Step 2.

Answer: the water is about 1.607 ft (19.3 in.) deep. A third step gives 1.607309, so the answer has settled.

Common mistakes

Mistake	Example of the error	Fix
Adding instead of subtracting	$x_1 = x_0 + \frac{f(x_0)}{f'(x_0)}$	The formula subtracts: $x_0 - \frac{f(x_0)}{f'(x_0)}$ .
Not setting the equation equal to 0	Using f(x) = cos x for cos x = x	Use f(x) = cos x − x.
Using degrees in trig problems	Calculator in degree mode for cos x = x	Use radians.
Rounding too early	Rounding x₁ to 1.3 in Example 2	Keep at least 6 decimal places until the end.
Starting where f′ = 0	x₀ = 1 for x³ − 3x + 1	f′(1) = 0, so the formula divides by 0. Pick another start.

Related methods

Newton’s method and linear approximation

Each step is a linear approximation: you replace f by its tangent line and solve the line exactly. Newton’s method is that idea repeated.

The divide-and-average square root

For f(x) = x² − a, the formula simplifies to $x_{n+1} = \frac{1}{2}\left(x_n + \frac{a}{x_n}\right)$ , an ancient method for square roots (Practice Problem 7).

Practice problems

Do one step of Newton’s method for $f(x) = x^2 - 5$ with x₀ = 2.
Approximate √10 with two steps from x₀ = 3.
Approximate $\sqrt[3]{5}$ with two steps from x₀ = 2.
Do two steps for $x^3 + 2x - 5 = 0$ from x₀ = 1.
Solve $e^x = 3 - x$ with two steps from x₀ = 1.
Why can’t you start Newton’s method at x₀ = 1 for $f(x) = x^3 - 3x + 1$ ?
Show that for $f(x) = x^2 - a$ , Newton’s formula becomes $x_{n+1} = \frac{1}{2}\left(x_n + \frac{a}{x_n}\right)$ .
Use one step from x₀ = 3 to approximate the solution of ln x = 1.
Use one step from x₀ = 2 to approximate the positive solution of $\sin x = \frac{x}{2}$ .
A cube-shaped tank must hold 40 ft³. Use two steps from x₀ = 3 to approximate its side length.

Show answers

x_1 = 2 - \frac{4 - 5}{4} = 2 + \frac{1}{4} = 2.25

f(2) = −1 and f′(2) = 4.

Answer: x₁ = 2.25

x_1 = 3 - \frac{-1}{6} \approx 3.166667

f(x) = x² − 10: f(3) = −1 and f′(3) = 6.

x_2 = 3.166667 - \frac{0.027778}{6.333333} \approx 3.162281

Evaluate at x₁.

Answer: √10 ≈ 3.1623

x_1 = 2 - \frac{3}{12} = 1.75

f(x) = x³ − 5: f(2) = 3 and f′(2) = 12.

x_2 = 1.75 - \frac{0.359375}{9.1875} \approx 1.710884

f(1.75) = 5.359375 − 5 and f′(1.75) = 9.1875.

Answer: ∛5 ≈ 1.7109 (true value 1.70998)

x_1 = 1 - \frac{-2}{5} = 1.4

f(1) = −2 and f′(x) = 3x² + 2 gives f′(1) = 5.

x_2 = 1.4 - \frac{0.544}{7.88} \approx 1.330964

f(1.4) = 0.544 and f′(1.4) = 7.88.

Answer: x ≈ 1.3310

x_1 = 1 - \frac{0.718282}{3.718282} \approx 0.806824

f(x) = eˣ + x − 3 and f′(x) = eˣ + 1.

x_2 = 0.806824 - \frac{0.047605}{3.240781} \approx 0.792135

Evaluate at x₁.

Answer: x ≈ 0.7921

f'(x) = 3x^2 - 3, \; f'(1) = 0

Differentiate and substitute.

Answer: the formula would divide by f′(1) = 0. The tangent line at x = 1 is horizontal and never crosses the x-axis.

x_{n+1} = x_n - \frac{x_n^2 - a}{2x_n}

f′(x) = 2x.

x_{n+1} = \frac{2x_n^2 - x_n^2 + a}{2x_n} = \frac{x_n^2 + a}{2x_n}

Use a common denominator.

x_{n+1} = \frac{1}{2}\left(x_n + \frac{a}{x_n}\right)

Split the fraction.

Answer: shown

x_1 = 3 - \frac{\ln 3 - 1}{\frac{1}{3}} = 3 - 3(0.098612)

f(x) = ln x − 1 and f′(x) = 1/x.

x_1 \approx 2.704163

Multiply and subtract.

Answer: about 2.7042 (the exact answer is e ≈ 2.7183)

x_1 = 2 - \frac{-0.090703}{-0.916147} \approx 1.900996

f(x) = sin x − x/2 and f′(x) = cos x − ½.

Answer: about 1.9010 (the true value is about 1.8955)

10.

x_1 = 3 - \frac{27 - 40}{27} \approx 3.481481

f(x) = x³ − 40 and f′(x) = 3x².

x_2 = 3.481481 - \frac{2.198039}{36.362140} \approx 3.421033

Evaluate at x₁.

Answer: about 3.421 ft (the true value is 3.41995)

Newton’s method FAQ

What is the formula for Newton’s method?

$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$ . Start with a guess x₀ and apply the formula again and again.

Why does Newton’s method work?

Near a point, a smooth curve looks like its tangent line. The zero of the tangent line is a good estimate of the zero of the curve, and each step uses a better tangent line.

When does Newton’s method fail?

When f′(xₙ) = 0, when the guess is far from the zero, or when the values cycle, as in Example 4. Try a different starting value.

How many steps should I do?

Stop when two values in a row agree to the accuracy you need. Near a zero, the number of correct digits roughly doubles each step.

Summary

Write the equation as f(x) = 0 and find f′(x).
Repeat $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$ from a good starting guess.
Each step is where the tangent line crosses the x-axis.
Keep plenty of decimals, and use radians for trig.
Watch for f′ = 0 and cycling.

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