Riding Tangents to the Root

Ride the tangent down to the root. The number of correct digits doubles each step.

Examples & Exercises

Prove algebraically that Newton's method for $x^2 - 2 = 0$ is the Babylonian method.

x - \frac{x^2-2}{2x} = \frac{x + 2/x}{2}

Starting at

x_0 = 1

— 4 steps nail

\sqrt{2}

past 10 digits.

Show proof

With $f(x) = x^2 - 2$ and $f'(x) = 2x$ , Newton's iteration is $x_{n+1} = x_n - \frac{x_n^2 - 2}{2x_n}$ .
Putting over a common denominator gives $\frac{2x_n^2 - x_n^2 + 2}{2x_n} = \frac{x_n + 2/x_n}{2}$ — exactly the Babylonian iteration.

Explain why the error follows $e_{n+1} \approx \frac{f''(x^*)}{2f'(x^*)} e_n^2$ . (Idea level)

e_{n+1} \approx C\,e_n^2, \quad C = \frac{f''(x^*)}{2f'(x^*)}

Check the error ratio

|e_{n+1}|/e_n^2 \approx 0.354 = 1/(2\sqrt{2})

Show proof

Near a root $f(x^*) = 0$ , the second-order Taylor expansion gives $0 = f(x^*) \approx f(x_n) + f'(x_n)(x^* - x_n) + \frac{f''(x^*)}{2}e_n^2$ (with $e_n = x_n - x^*$ ).
Substituting the Newton step $x_{n+1} = x_n - f(x_n)/f'(x_n)$ and rearranging gives $e_{n+1} \approx \frac{f''(x^*)}{2f'(x^*)}e_n^2$ .
So the error squares, doubling the correct digits each step — quadratic convergence. A rigorous proof needs the Taylor remainder estimate, so this stays at the idea level.

Show by direct computation that Newton's method for $f(x) = x^3 - 2x + 2$ from $x_0 = 0$ cycles $0 \to 1 \to 0$ forever.

x_1 = 0 - \frac{2}{-2} = 1, \quad x_2 = 1 - \frac{1}{1} = 0

The tangent bounces between

0

and

1

forever — no convergence.

Show proof

$f(0) = 2$ and $f'(0) = -2$ , so $x_1 = 0 - \frac{2}{-2} = 1$ .
$f(1) = 1$ and $f'(1) = 1$ , so $x_2 = 1 - \frac{1}{1} = 0$ .
Since $x_2 = x_0$ , the sequence repeats $0, 1, 0, 1, \dots$ forever — Newton's method does not always converge.