Single-Variable-Calculus · Unit 11 · Video 3 · Interactive Practice

Newton's Method: Quadratic Convergence and How It Can Fail

IKey Formulas

Formula	Name	Video values
$x_{n+1} = x_n - \dfrac{f(x_n)}{f'(x_n)}$	Newton step	$\sqrt{5}$ : $x_0 = 2,\ x_1 = \tfrac{9}{4},\ x_2 = \tfrac{161}{72}$
$E_n = \lvert x - x_n \rvert$	Error of the guess $x_n$	$x$ = the root (no subscript)
$E_{n+1} \approx C\,E_n^2,\quad C = \dfrac{\lvert f'' \rvert}{2\lvert f' \rvert}$	Quadratic convergence	gap $\approx \tfrac{1}{2}\lvert f'' \rvert E_n^2$ , divided by $\lvert f' \rvert$
$E_{n+1} = \dfrac{E_n^2}{2x_n}$	Exact for $f(x) = x^2 - 5$	$C \to \tfrac{1}{2\sqrt{5}} \approx 0.22$

Key Insight: When $f'$ is not too small, $f''$ is not too big and $x_0$ starts near the root, each error is about a constant times the square of the last one, so the correct digits double. Without those conditions the method can converge to the wrong root, hit a horizontal tangent, or cycle.

IIWhy the Error Squares

At the root the tangent line sits $E_n^2$ below the curve, so the next error is $E_n^2 / f'(x_n)$ .

Guess xₙ 3.00

💡 For $f(x) = x^2 - 5$ , $\tfrac{1}{2}f'' = 1$ , so the gap is exactly $E_n^2$ ; for a general $f$ it is about $\tfrac{1}{2}\lvert f'' \rvert E_n^2$ , which gives $C = \dfrac{\lvert f'' \rvert}{2\lvert f' \rvert}$ .

IIICorrect Digits Double

For $\sqrt{5}$ , $E_{n+1} = E_n^2/(2x_n)$ exactly, so the number of correct digits roughly doubles each step.

Start x₀ Step n 0

IVThree Ways to Fail

The same rule can converge to the other root, meet a horizontal tangent, or cycle forever.

Function Start x₀ −2 Steps 3

💡 Challenge: find the one start on $x^2 - 5$ that has no $x_1$ , then a start on $x^3 - 5x$ that returns to itself after two steps.

VQuiz Questions

Problem 1 · Squaring the Error

Given: Near a root, an iteration's errors obey $E_{n+1} \approx E_n^2$ (constant $C = 1$ ), and $E_2 = 10^{-3}$ . Find the approximate size of $E_4$ .

✅ Correct! Two squarings:

10^{-3} \to 10^{-6} \to 10^{-12}

. The correct digits go

3 \to 6 \to 12

❌ That is $E_3$ . From

E_2

E_4

is two steps, so the error is squared twice.

❌ The digits do not grow by a fixed amount. Adding one digit per step gives

10^{-4}

, then

10^{-5}

; but squaring doubles the exponent:

(10^{-3})^2 = 10^{-6}

, then

10^{-12}

❌ One squaring too many.

10^{-24}

E_5

; from

E_2

E_4

there are only two steps.

Show solution

Each step squares the error:

E_3 \approx E_2^2 = (10^{-3})^2 = 10^{-6}, \qquad E_4 \approx E_3^2 = (10^{-6})^2 = 10^{-12}.

So $E_4 \approx 10^{-12}$ : about $12$ correct decimal places, double the $6$ of $x_3$ and four times the $3$ of $x_2$ .

Problem 2 · Which Root?

Given: Newton's method on $f(x) = x^2 - 5$ , started at $x_0 = -3$ . Find $x_1$ and the limit of the guesses.

What is $x_1$ ?

Where do the guesses converge?

✅ Correct!

x_1 = -\tfrac{7}{3} \approx -2.333

, and the guesses close in on

-\sqrt{5} \approx -2.236

. The method only knows where you started.

❌ Check the sign of $f'(-3)$ .

f'(x) = 2x

, so

f'(-3) = -6

, not

6

x_1 = -3 - \frac{4}{-6} = -3 + \frac{2}{3}

❌ A negative start cannot jump to the positive side here.

x_1 = \tfrac{1}{2}x_0 + \tfrac{5}{2x_0}

is negative whenever

x_0

is.

❌ The fraction is upside down. The step subtracts

\frac{f(x_0)}{f'(x_0)} = \frac{4}{-6}

, not

\frac{f'(x_0)}{f(x_0)}

❌ That is the root you might want, not the one you get. Every negative start for

x^2 - 5

stays negative and leads to

-\sqrt{5}

❌ No cycle here.

x_1 = -\tfrac{7}{3}

x_2 = -\tfrac{47}{21} \approx -2.238

: the guesses settle down, very fast.

❌ $0$ is not a root.

f(0) = -5

; the vertex is where the tangent is horizontal, the start with no

x_1

❌ Check $x_1$ . Use

x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}

with

f(-3) = 4

and

f'(-3) = -6

❌ Check the limit. Follow the sign of the guesses and which root lies on that side.

Show solution

Here $f(-3) = 9 - 5 = 4$ and $f'(x) = 2x$ , so $f'(-3) = -6$ :

x_1 = -3 - \frac{4}{-6} = -3 + \frac{2}{3} = -\frac{7}{3} \approx -2.333.

The next step gives $x_2 = \tfrac{1}{2}\big(-\tfrac{7}{3}\big) + \dfrac{5}{2(-7/3)} = -\tfrac{7}{6} - \tfrac{15}{14} = -\tfrac{47}{21} \approx -2.2381$ , then $x_3 \approx -2.2360689$ .

The guesses converge to $-\sqrt{5} \approx -2.2360680$ . Newton's method found a root, just not the positive one: which root it finds is decided by the start.

Problem 3 · The Constant for a Cube Root

Given: Newton's method on $f(x) = x^3 - 2$ , whose root is $x = \sqrt[3]{2} \approx 1.26$ . At some step $E_n = 10^{-3}$ . Estimate $E_{n+1}$ with $E_{n+1} \approx \dfrac{\lvert f'' \rvert}{2\lvert f' \rvert}E_n^2$ at the root.

✅ Correct!

C = \frac{6x}{2 \cdot 3x^2} = \frac{1}{x} = \frac{1}{\sqrt[3]{2}} \approx 0.794

, so

E_{n+1} \approx 0.794 \times 10^{-6}

. An actual step from

\sqrt[3]{2} + 10^{-3}

gives

7.93 \times 10^{-7}

❌ That sets $C = 1$ . The ladder with

C = 1

is an idealisation; here

C = \frac{\lvert f'' \rvert}{2\lvert f' \rvert}

must be computed.

❌ The $\tfrac{1}{2}$ is missing. The gap is

\tfrac{1}{2}\lvert f'' \rvert E_n^2

(quadratic approximation), so

C = \frac{\lvert f'' \rvert}{2\lvert f' \rvert}

, not

\frac{\lvert f'' \rvert}{\lvert f' \rvert}

❌ The constant is upside down.

f'

is in the denominator: a steep tangent shrinks the next error.

C = \frac{6x}{6x^2} = \frac{1}{x}

, not

x

Show solution

Derivatives: $f'(x) = 3x^2$ and $f''(x) = 6x$ . At the root $x = \sqrt[3]{2}$ :

C = \frac{\lvert f'' \rvert}{2\lvert f' \rvert} = \frac{6x}{2 \cdot 3x^2} = \frac{1}{x} = \frac{1}{\sqrt[3]{2}} \approx 0.794.

Then

E_{n+1} \approx 0.794 \times (10^{-3})^2 = 7.94 \times 10^{-7} \approx 7.9 \times 10^{-7}.

Check: one Newton step from $x_n = \sqrt[3]{2} + 10^{-3}$ lands $7.93 \times 10^{-7}$ above the root. The digits still double (about $3$ to about $6$ ); $C$ only shifts them slightly.

Problem 4 · A New Wiggly Curve

Given: Newton's method on $f(x) = x^3 - 2x + 2$ , started at $x_0 = 0$ . What happens to the guesses?

✅ Correct!

x_1 = 0 - \frac{2}{-2} = 1

and

x_2 = 1 - \frac{1}{1} = 0

: back to the start, a cycle like

\pm 1

x^3 - 5x

❌ Compute the first two steps.

x_1 = 1

and

x_2 = 0

return to the start, so the guesses never approach

-1.77

❌ Check $f'(0)$ .

f'(x) = 3x^2 - 2

, so

f'(0) = -2 \neq 0

; the tangent at

x_0

is not horizontal.

❌ $1$ is not a root.

f(1) = 1 - 2 + 2 = 1 \neq 0

, so the method takes another step, back to

0

Show solution

Here $f'(x) = 3x^2 - 2$ .

Step 1: $f(0) = 2$ , $f'(0) = -2$ , so $x_1 = 0 - \dfrac{2}{-2} = 1$ .

Step 2: $f(1) = 1$ , $f'(1) = 1$ , so $x_2 = 1 - \dfrac{1}{1} = 0$ .

$x_2 = x_0$ , so every later step repeats: $0 \to 1 \to 0 \to 1 \to \cdots$ . The guesses never settle on the only real root, $x \approx -1.769$ . This is the cycle failure; it needs a curve that wiggles, as $x^3 - 5x$ does in the video.

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