Newton’s Method Calculator

This Newton’s method calculator finds a root of \( f(x) = 0 \) numerically. You give it a function and a starting guess, and it runs Newton’s iteration step by step, showing \( x_n \), \( f(x_n) \) and the next estimate \( x_{n+1} \) in a table, then reports the root it lands on and how close \( f \) is to zero there.

Use it to check a numerical analysis assignment, see how fast the method converges, or find where a guess goes wrong. Also, it’s free and needs no sign-up.

How to use the Newton’s method calculator

1. Enter \( f(x) \), written so that you are solving \( f(x) = 0 \). To solve \( \cos x = x \), type cos(x) - x. Use ^ for powers, sqrt(x), sin(x), ln(x), e^(x) and pi; implicit multiplication such as 2x works.
2. Enter the initial guess \( x_0 \). For example, a sketch of the graph (try the graphing calculator), or a sign change of \( f \) between two numbers, helps you choose it.
3. Set the number of iterations (1 to 50). Five or six is usually plenty.
4. Press Iterate. If two iterates agree to machine precision before the limit, the calculator stops early and labels the result “Root (converged)”; otherwise it shows “Root estimate”.

You don’t need to enter \( f'(x) \). The calculator differentiates your function exactly.

Formula

Newton’s method replaces the curve with its tangent line at \( x_n \) and moves to where that line crosses the x-axis:

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

Near a simple root, each step roughly doubles the number of correct digits (quadratic convergence), which is why a handful of iterations is often enough.

Worked example

Solve \( \cos x = x \), that is, find the root of \( f(x) = \cos x - x \), starting at \( x_0 = 1 \).

The derivative is \( f'(x) = -\sin x - 1 \), so the update is

$$x_{n+1} = x_n - \frac{\cos x_n - x_n}{-\sin x_n - 1}$$

\( n \)	\( x_n \)	\( f(x_n) \)	\( x_{n+1} \)
0	1	\( -0.45969769 \)	0.75036387
1	0.75036387	\( -0.018923074 \)	0.73911289
2	0.73911289	\( -0.000046455899 \)	0.73908513
3	0.73908513	\( -2.8472058 \times 10^{-10} \)	0.73908513

The root is \( x \approx 0.739085133215 \). Look at \( f(x_n) \). The error shrinks from about \( 10^{-2} \) to \( 10^{-5} \) to \( 10^{-10} \). This shows quadratic convergence. Enter cos(x) - x with \( x_0 = 1 \) above. A newton’s method calculator will reproduce the table. It adds one more row. This confirms the value no longer changes. Then it stops.

When Newton’s method fails (and how the calculator shows it)

• Zero derivative. If \( f'(x_n) = 0 \), the tangent is horizontal and never meets the x-axis. The calculator stops with a message asking you to choose another starting point. Example: x^2 - 4 with \( x_0 = 0 \).
• Cycling. For \( f(x) = x^3 - 2x + 2 \) starting at \( x_0 = 0 \): \( x_1 = 0 - \frac{2}{-2} = 1 \), then \( x_2 = 1 - \frac{1}{1} = 0 \). The iterates bounce between 0 and 1 forever. Run it with 6 iterations and the table makes this obvious, and the final \( f(\text{root}) \) is 2, nowhere near zero. Always check that value.
• Undefined values. If an iterate leaves the domain (for example a negative input to ln(x)), the calculator reports where \( f \) or \( f' \) is undefined.
• Wrong root. With several roots, Newton’s method finds the one its starting point leads to, which isn’t always the nearest. As a result, try different guesses to find the others.

For the derivation and more on convergence, read the Newton’s method guide. Because every step is a tangent line, the equation of a tangent line guide is useful background too, and the tangent line calculator shows the line at your starting guess.

Worked example: computing √2

Solve \( f(x) = x^2 - 2 = 0 \) with \( f'(x) = 2x \), starting at \( x_0 = 1 \):

Step	\( x_n \)
0	1
1	1.5
2	1.416667
3	1.414216

The true value is 1.414214. The number of correct digits roughly doubles each step, which is called quadratic convergence. This is close to how calculators compute square roots.

Choosing a starting point

• Sketch the graph or test a few values to find where the function changes sign, and start near there.
• Avoid points where \( f'(x) \approx 0 \): the tangent is nearly flat and sends the next guess far away.
• If the equation has several roots, different starting points find different roots.

Worked example: solving cos x = x

Rewrite as \( f(x) = \cos x - x = 0 \), with \( f'(x) = -\sin x - 1 \). Starting at \( x_0 = 1 \), a newton’s method calculator settles on \( x \approx 0.739085 \) quickly. However, no algebraic formula gives this number. Therefore, numerical methods are important.

Newton’s method and tangent lines

Each step replaces the curve by its tangent line at the current guess and jumps to where that line crosses zero:

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

Common mistakes

• Using the wrong derivative, which makes the method converge slowly or not at all.
• Stopping too early: keep going until two successive guesses agree to the accuracy you need.
• Forgetting to set the calculator to radians for trig equations.

Practice problems

1. Do one step of Newton’s method for \( x^3 - 5 = 0 \) from \( x_0 = 2 \).
2. Why does \( x_0 = 0 \) fail for \( x^2 - 2 = 0 \)?

Answers: 1. \( 2 - \frac{3}{12} = 1.75 \) (the true root is 1.70998). 2. \( f'(0) = 0 \), so the formula divides by zero.

Your Step-by-Step Guide to Newton's Method

First, identify the equation you want to solve. Rewrite it so one side is zero. For example, change 'x^2 = 2' into 'x^2 – 2 = 0'. This gives you the function f(x) to enter.

Next, input your function f(x) into the calculator's 'Function' field. Use standard mathematical notation. For instance, 'x squared' becomes 'x^2', and 'e to the x' is 'e^(x)'.

Then, provide an initial guess for the root. This is your starting point, x0. A visual estimate from a graph often works well. Alternatively, choose a value where f(x) is close to zero for a newton’s method calculator.

Set the number of iterations. Five to ten steps are usually sufficient for high precision. The newton's method calculator will perform the calculations sequentially. Finally, press the 'Iterate' button to see the results.

The calculator will display a table for each step. It lists x_n, f(x_n), and the next estimate x_(n+1). Observe how x_n converges to the root. Watch f(x_n) approach zero in this newton’s method calculator.

The final result shows the estimated root and the function's value at that point. A very small f(x) confirms a good approximation.

When Other Methods Might Be Better

Newton's method is powerful but not always the best choice. For instance, if you need to find *all* roots of a polynomial, methods like polynomial division or factoring might be more direct. Newton's method finds only one root at a time.

Consider the Bisection Method when you are sure a root exists within a specific interval. This method is slower but guarantees convergence if f(a) and f(b) have opposite signs. It is robust when Newton's method struggles with a poor initial guess.

If the derivative f'(x) is difficult or impossible to calculate, Newton's method cannot be applied. In such cases, the Secant Method offers an alternative. It approximates the derivative using previous function values. This avoids the need for explicit differentiation.

Related guides

• Newton’s Method: Formula, Worked Example and When It Fails
• How to Find the Equation of a Tangent Line
• Linear Approximation

Further reading

• Newton’s Method (Paul’s Online Math Notes) — the iteration formula with worked examples.
• Linear Approximations (Paul’s Online Math Notes) — the tangent-line idea that every Newton step is built on.

FAQ

How many iterations of Newton’s method do I need?

Near a simple root, usually 4 to 6 are enough for full calculator precision, because the number of correct digits roughly doubles each step. The calculator stops early once the iterates stop changing.

How do I choose the initial guess?

Pick a value close to the root, ideally between two points where \( f \) changes sign and away from places where \( f'(x) = 0 \).

Why doesn’t Newton’s method converge for my function?

The usual causes are a starting point near a horizontal tangent, a cycle between two values, or a root where \( f' \) is also zero, which slows convergence. Try another starting point and check that \( f(\text{root}) \) is close to 0.

Can it solve an equation like e^x = 3x?

Yes. Move everything to one side and enter e^x - 3x. The calculator finds a root of that function, which is a solution of the original equation.