Newton-Raphson Method Calculator: Root Finder with Bisection and Secant Methods and the Iteration Table

Solve any equation numerically and watch the iterations.

Calculated from standard textbook formulas, with every step shown. The formulas are fixed, so the working does not need checking: what matters is the numbers you type in. Use these calculators only if you are competent to choose the inputs (loads, material strength, dimensions, supports) and check your units and assumptions. For learning and checking your own working: design and sign-off are for a qualified engineer. Drafting Central does not store the numbers you type.

How this calculator works

Type the function with the equation arranged so that one side is zero.

Choose Newton, secant or bisection and enter the starting guesses or interval.

The calculator lists the iterations and gives the root with the size of f at that root.

Formulas used

  • x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}
  • x_{n+1} = x_n - f(x_n)\frac{x_n - x_{n-1}}{f(x_n) - f(x_{n-1})}
  • m = \frac{a + b}{2}

Worked example

With the default numbers in the calculator above:

  1. Root found: 2.094551482 (after 5 iterations, f(x) = -8.88e-16)
  2. Method: Newton-Raphson: x_new = x - f(x) / f'(x)
  3. Iteration 1: from x = 2 (f = -1.000e+0) to x = 2.1
  4. Iteration 2: from x = 2.1 (f = 6.100e-2) to x = 2.094568121
  5. Iteration 3: from x = 2.09456812 (f = 1.857e-4) to x = 2.094551482
  6. Iteration 4: from x = 2.09455148 (f = 1.740e-9) to x = 2.094551482
  7. Iteration 5: from x = 2.09455148 (f = -8.882e-16) to x = 2.094551482

Change the numbers in the calculator to see the full working, every step, for your own case.

A second worked example

A different case, with these numbers: Method Secant (two starting guesses).

  1. Root found: 2.094551482 (after 6 iterations, f(x) = 3.55e-15)
  2. Method: Secant method: the next guess comes from the line through the last two points
  3. Iteration 1: from x = 3 (f = 1.600e+1) to x = 2.058823529
  4. Iteration 2: from x = 2.05882353 (f = -3.908e-1) to x = 2.08126366
  5. Iteration 3: from x = 2.08126366 (f = -1.472e-1) to x = 2.094824146
  6. Iteration 4: from x = 2.09482415 (f = 3.044e-3) to x = 2.094549431
  7. Iteration 5: from x = 2.09454943 (f = -2.289e-5) to x = 2.094551481
  8. Iteration 6: from x = 2.09455148 (f = -3.513e-9) to x = 2.094551482

Common questions

How does Newton's method work?

It follows the tangent line at the current guess down to where it crosses the axis, and uses that as the next guess.

When does Newton's method fail?

When the slope is near zero, or the starting guess is far from the root, the guesses can jump away or oscillate.

Which method is safest?

Bisection always converges if the function changes sign in the interval, but it is the slowest.

Related calculators

Got a question about this? Ask an engineering question and get a plain-language answer with references.

Further reading: Any numerical methods or first-year calculus text.

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