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Newton's Method Calculator

Find roots using Newton's method iteratively.

Reviewed for accuracy by the Math Ora X team Last updated

Result

About Newton's Method Calculator

Newton's method converges quickly to roots. Starting from an initial guess, each iteration improves the approximation quadratically.

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

How to use this calculator

  1. Enter the function f(x) whose root you want to find.
  2. Give an initial guess x_0 that is close to the root.
  3. Compute the next approximation with Newton's formula, then repeat with the new value.
  4. Stop when two successive values are close enough for the accuracy you need.

The formula explained

The formula updates a guess by moving along the tangent line to the graph of f(x). Each new value x_{n+1} is usually closer to a root than x_n.

  • \(x_n\) = the current approximation after n iterations
  • \(x_{n+1}\) = the next approximation
  • \(f(x_n)\) = the value of the function at the current approximation
  • \(f'(x_n)\) = the derivative of the function at the current approximation

Step by step method

  1. Choose a starting value x_0 near the root.
  2. Evaluate f(x_0) and f'(x_0), then compute x_1 = x_0 - f(x_0)/f'(x_0).
  3. Repeat the same process with x_1, x_2, and so on until the values stop changing much.

Worked example

Problem. Use Newton's method to approximate a root of f(x)=x^2-2 starting from x_0=1.

  1. Compute f(1)=1^2-2=-1 and f'(x)=2x, so f'(1)=2.
  2. Apply Newton's formula, x_1 = 1 - (-1)/2 = 1.5.
  3. Repeat: f(1.5)=0.25 and f'(1.5)=3, so x_2 = 1.5 - 0.25/3 = 1.416666..., which is already close to sqrt{2}.

Answer. A root is approximately 1.4167 after two iterations, and it continues toward 1.4142.

Tips and common mistakes

  • Your starting guess should be close to the root, otherwise the method may fail or jump away.
  • If f'(x_n)=0 or very close to 0, the formula breaks down or becomes unstable.

Frequently asked questions

How do I use the Newton's Method Calculator to find a root?+

Enter the function f(x), its derivative f'(x), and a starting guess x0. The calculator applies x_{n+1} = x_n - f(x_n)/f'(x_n) repeatedly until it reaches a close enough approximation to a root.

What does the Newton's method formula mean?+

The formula updates your current guess by subtracting the function value divided by the derivative at that point. If the graph is smooth and your guess is good, each step usually moves you closer to where f(x) = 0.

What should I choose as the initial guess?+

Pick a value near the root if possible, often by looking at where the graph crosses the x-axis or where the function changes sign. A poor starting guess can make the method converge slowly, jump to a different root, or fail if the derivative is zero or very small.

What happens if the derivative is 0 or very small?+

If f'(x_n) = 0, the formula cannot be used because you would divide by zero. If the derivative is close to 0, the next step can become extremely large, so Newton's method may behave badly or diverge.

How is Newton's method different from just solving f(x) = 0 algebraically?+

Algebraic solving gives an exact root when a closed-form solution is available, while Newton's method gives a numerical approximation. It is especially useful for equations that are hard or impossible to solve exactly, as long as you can compute both f(x) and f'(x).

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