Skip to main content

Babylonian Square Root Method – Iterations

Quick answer

Heron’s iteration x←(x+S/x)/2 nails √2 in three steps: 1 → 1.5 → 1.4167 → 1.41422.

  • Every iterate and its error shown.
  • Converges fast.

Figures are estimates from this calculator’s standard formula — adjust the inputs in the tool below for your exact number.

How the Babylonian Method Calculator Works

xₖ₊₁ = (xₖ + S/xₖ)/2

The Babylonian (Heron’s) method refines a square-root guess by averaging it with S divided by it. If the guess x is too high, S/x is too low, and their average is closer to √S than either. For S = 2 starting at x₀ = 1: x₁ = (1 + 2)/2 = 1.5; x₂ = (1.5 + 2/1.5)/2 = 1.41667; x₃ = 1.4142169 — already within 0.0000031 of √2 ≈ 1.4142136. The calculator lists every iterate, its square (to show the residual error), and the number of correct digits, which roughly doubles per step — quadratic convergence, the same engine inside Newton’s method applied to f(x) = x² − S. Any positive seed works; poor guesses just add a step or two (x₀ = 7 for √2 catches up by the fourth iterate). The method doubles as the classic divide-and-average algorithm taught before calculators.

  1. 1

    Enter S and a starting guess

    Any positive guess; 1 or S/2 are common defaults

  2. 2

    Average x with S/x

    Each iterate xₖ₊₁ = (xₖ + S/xₖ)/2 is printed

  3. 3

    Watch errors halve their digits

    Correct digits double per iteration

  4. 4

    Stop at tolerance

    The run halts when |xₖ² − S| falls below your threshold

Use Cases

Numerical Methods Class

Demonstrate quadratic convergence on √2

Mental Math Training

Two hand iterations get √10 to 3.1623

Embedded Systems

Integer square roots via divide-and-average loops

Tips

  • 1

    Averaging x and S/x always moves toward √S

  • 2

    Digits double each iteration — three steps beat six decimal places

  • 3

    The seed hardly matters; bad guesses cost one extra step

Common Mistakes

  • Averaging x with x/2 instead of S/x

  • Stopping after one iterate and expecting full precision

  • Using a negative seed, which sends the iteration complex

  • Confusing the method with digit-by-digit longhand square rooting

FAQs

How fast does the method find √2?

From x₀ = 1: 1.5, 1.41667, 1.41422 — three iterations reach six-digit accuracy.

Why does averaging work?

If x overshoots √S, S/x undershoots; their mean is always closer than the worse of the two.

Does the starting guess matter?

Only marginally — x₀ = 7 for √2 converges one iteration later than x₀ = 1.

Can it compute cube roots?

Not directly — the analogous iteration for cube roots averages in triplets: x ← (2x + S/x²)/3.