Riding Tangents to the Root

Ride tangent lines down to a root. Each iteration doubles the number of correct digits.

Draw the tangent and take its root as the next guess. A straight line stands in for the curve. From a good start, the guesses converge astonishingly fast.

For example, guessing a number a 🧑 friend hid, each hint doubles the correct digits of your next guess. A wild first guess turns exact within a few rounds.

xn+1=xn−f(xn)f′(xn)x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}

Examples & Exercises

Prove algebraically that Newton's method for x2−2=0x^2 - 2 = 0 is the Babylonian method.

x−x2−22x=x+2/x2x - \frac{x^2-2}{2x} = \frac{x + 2/x}{2}
Starting at x0=1x_0 = 1 — 4 steps nail 2\sqrt{2} past 10 digits.
Show proof

  1. With f(x)=x2−2f(x) = x^2 - 2 and f′(x)=2xf'(x) = 2x, Newton's iteration is xn+1=xn−xn2−22xnx_{n+1} = x_n - \frac{x_n^2 - 2}{2x_n}.

  2. Putting over a common denominator gives 2xn2−xn2+22xn=xn+2/xn2\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 en+1≈f′′(x∗)2f′(x∗)en2e_{n+1} \approx \frac{f''(x^*)}{2f'(x^*)} e_n^2. (Idea level)

en+1≈C en2,C=f′′(x∗)2f′(x∗)e_{n+1} \approx C\,e_n^2, \quad C = \frac{f''(x^*)}{2f'(x^*)}
Check the error ratio ∣en+1∣/en2≈0.354=1/(22)|e_{n+1}|/e_n^2 \approx 0.354 = 1/(2\sqrt{2}).
Show hint

  1. Hypotheses: ff is C2C^2 near x∗x^* and x∗x^* is a simple root, i.e. f(x∗)=0f(x^*) = 0 with f′(x∗)≠0f'(x^*) \ne 0 (if f′(x∗)=0f'(x^*) = 0 the ratio f′′(x∗)2f′(x∗)\frac{f''(x^*)}{2f'(x^*)} below is meaningless).

    Under these hypotheses, near a root f(x∗)=0f(x^*) = 0, the second-order Taylor expansion gives 0=f(x∗)≈f(xn)+f′(xn)(x∗−xn)+f′′(x∗)2en20 = f(x^*) \approx f(x_n) + f'(x_n)(x^* - x_n) + \frac{f''(x^*)}{2}e_n^2 (with en=xn−x∗e_n = x_n - x^*).

  2. Substituting the Newton step xn+1=xn−f(xn)/f′(xn)x_{n+1} = x_n - f(x_n)/f'(x_n) and rearranging gives en+1≈f′′(x∗)2f′(x∗)en2e_{n+1} \approx \frac{f''(x^*)}{2f'(x^*)}e_n^2.

  3. 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)=x3−2x+2f(x) = x^3 - 2x + 2 from x0=0x_0 = 0 cycles 0→1→00 \to 1 \to 0 forever.

x1=0−2−2=1,x2=1−11=0x_1 = 0 - \frac{2}{-2} = 1, \quad x_2 = 1 - \frac{1}{1} = 0
The tangent bounces between 00 and 11 forever — no convergence.
Show proof

  1. f(0)=2f(0) = 2 and f′(0)=−2f'(0) = -2, so x1=0−2−2=1x_1 = 0 - \frac{2}{-2} = 1.

  2. f(1)=1f(1) = 1 and f′(1)=1f'(1) = 1, so x2=1−11=0x_2 = 1 - \frac{1}{1} = 0.

  3. Since x2=x0x_2 = x_0, the sequence repeats 0,1,0,1,…0, 1, 0, 1, \dots forever — Newton's method does not always converge.

10. Differentiation