Single-Variable-Calculus · Unit 11 · Video 2 · Interactive Practice

Newton's Method: Tangent Lines That Find the Square Root of Five

IKey Formulas

Formula	Name	Role
$f(x) = 0$	Standard form	Every equation can be rewritten this way; $x^2 = 5$ becomes $x^2 - 5 = 0$
$y - y_0 = m(x - x_0)$ , with $y_0 = f(x_0)$ , $m = f'(x_0)$	Tangent line at the guess	The linear stand-in for the curve
$x_{n+1} = x_n - \dfrac{f(x_n)}{f'(x_n)}$	Newton's method	The tangent line's $x$ -intercept is the next guess
$x_{n+1} = \tfrac{1}{2}x_n + \dfrac{5}{2x_n}$	Newton's method for $\sqrt{5}$	From $f(x) = x^2 - 5$ , $f'(x) = 2x$

Key Insight: At the current guess, pretend the function is linear: replace the curve by its tangent line and solve the line instead. Starting from $x_0 = 2$ , the guesses $\tfrac{9}{4}$ , $\tfrac{161}{72}$ , $\tfrac{51841}{23184}$ close in on $\sqrt{5}$ with the number of correct digits roughly doubling each step.

IIVisualization 1 — Where the Tangent Line Crosses

The tangent line at $\big(x_0, f(x_0)\big)$ on $y = x^2 - 5$ crosses the $x$ -axis at the next guess $x_1$ .

Guess x₀ 2.00

IIIVisualization 2 — From the Picture to the Formula

Setting $y = 0$ in the tangent-line equation turns the crossing point into Newton's formula.

Step 1

1 · Tangent line through $(x_0, y_0)$ with slope $m$

y - y_0 = m(x - x_0)

2 · It crosses the axis at $(x_1, 0)$ : set $y = 0$ , $x = x_1$

-y_0 = m(x_1 - x_0)

3 · Divide by $m$

-\frac{y_0}{m} = x_1 - x_0 \quad\Longrightarrow\quad x_1 = x_0 - \frac{y_0}{m}

4 · The point is on the graph: $y_0 = f(x_0)$ , $m = f'(x_0)$

x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 2 - \frac{-1}{4} = \frac{9}{4}

5 · Repeat from the new guess

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

x_2 = \frac{9}{4} - \frac{1/16}{9/2} = \frac{9}{4} - \frac{1}{72} = \frac{161}{72}

IVVisualization 3 — Iterating Toward √5

Each pass of $x_{n+1} = \tfrac{1}{2}x_n + 5/(2x_n)$ roughly doubles the number of correct digits of $\sqrt{5} = 2.2360679774\ldots$

Step n 0 Start x₀

n	xₙ	decimal	\|√5 − xₙ\|

💡 Starting from $x_0 = 2$ , every later guess $x_n$ ( $n \ge 1$ ) satisfies $p^2 - 5q^2 = 1$ when written as $p/q$ (for example $161^2 - 5 \cdot 72^2 = 1$ ), so its error is exactly $\dfrac{1}{q\,(p + \sqrt{5}\,q)}$ ; why the digits double is the subject of the next video.

VQuiz Questions

Problem 1 · One Newton Step

Given: $f(x) = x^2 - 7$ and the starting guess $x_0 = 3$ — find the next guess $x_1$ .

✅ Correct!

x_1 = 3 - \tfrac{2}{6} = \tfrac{8}{3} \approx 2.667

, much closer to

\sqrt{7} \approx 2.646

than

3

was.

❌ Wrong sign. Newton's method subtracts the ratio:

x_1 = x_0 - f(x_0)/f'(x_0)

. Here

f(3) = 2

is positive and the curve rises, so the tangent line meets the axis to the left of

3

❌ The ratio is upside down. You computed

3 - f'(3)/f(3) = 3 - 6/2

. The step is

f(x_0)/f'(x_0) = 2/6

: height divided by slope.

❌ The slope is missing.

x_0 - f(x_0) = 3 - 2

forgets to divide by

f'(x_0) = 6

; the run of the tangent triangle is height over slope.

❌ Not quite. Compute

f(3)

and

f'(3)

, then apply

x_1 = x_0 - f(x_0)/f'(x_0)

Show solution

With $f(x) = x^2 - 7$ we have $f'(x) = 2x$ , so $f(3) = 9 - 7 = 2$ and $f'(3) = 6$ .

x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 3 - \frac{2}{6} = 3 - \frac{1}{3} = \frac{8}{3} \approx 2.667.

Equivalently, the simplified rule $x_1 = \tfrac{1}{2}x_0 + \tfrac{7}{2x_0} = \tfrac{3}{2} + \tfrac{7}{6} = \tfrac{16}{6} = \tfrac{8}{3}$ gives the same value. Since $\sqrt{7} \approx 2.6458$ , the error fell from about $0.35$ to about $0.02$ .

Problem 2 · A Cubic With No Practical Formula

Given: $f(x) = x^3 - 2x - 5$ and $x_0 = 2$ — find the Newton step $x_1$ .

✅ Correct!

f(2) = -1

and

f'(2) = 10

, so

x_1 = 2 - \tfrac{-1}{10} = 2.1

❌ Watch the double minus.

f(2) = -1

is negative, so

x_0 - f(x_0)/f'(x_0) = 2 - (-\tfrac{1}{10}) = 2 + 0.1

. The point lies below the axis on a rising curve, so the crossing is to the right.

❌ Check the derivative.

\dfrac{d}{dx}(x^3 - 2x - 5) = 3x^2 - 2

, so

f'(2) = 12 - 2 = 10

, not

12

❌ The ratio is upside down.

2 - f'(2)/f(2) = 2 - 10/(-1) = 12

. The step is

f(x_0)/f'(x_0)

, height over slope.

❌ Not quite. Evaluate

f(2)

and

f'(2)

carefully, then apply

x_1 = x_0 - f(x_0)/f'(x_0)

Show solution

$f(2) = 8 - 4 - 5 = -1$ and $f'(x) = 3x^2 - 2$ , so $f'(2) = 10$ .

x_1 = 2 - \frac{f(2)}{f'(2)} = 2 - \frac{-1}{10} = 2 + 0.1 = 2.1.

The two minus signs combine into a plus: the point $(2, -1)$ is below the axis and the tangent line rises with slope $10$ , so it reaches the axis $1/10$ to the right.

Problem 3 · Newton's Method for √10

Given: $f(x) = x^2 - 10$ and $x_0 = 3$ — find the simplified Newton recurrence and the second guess $x_2$ .

What is the simplified recurrence?

What is $x_2$ ?

✅ Correct!

x_1 = \tfrac{19}{6}

and

x_2 = \tfrac{721}{228} \approx 3.162281

, while

\sqrt{10} \approx 3.162278

❌ Only half of the constant survives. The term is

\dfrac{10}{2x_n} = \dfrac{5}{x_n}

: the

2

from

f'(x_n) = 2x_n

divides the

10

as well.

❌ The derivative is $2x_n$ , not $x_n$ . With the correct slope the ratio is

\dfrac{x_n^2 - 10}{2x_n}

❌ Sign slip.

-\dfrac{x_n^2 - 10}{2x_n} = -\dfrac{x_n}{2} + \dfrac{10}{2x_n}

: the minus in front of the fraction and the minus in front of

10

combine into a plus.

❌ That is $x_1$ . Apply the recurrence once more, starting from

\tfrac{19}{6}

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

\tfrac{19}{6} + \tfrac{60}{19} = \tfrac{721}{114}

is twice

x_2

; the recurrence averages

x_1

and

10/x_1

❌ This comes from $\tfrac{1}{2}x_n + 10/x_n$ . That rule sends

3

\tfrac{29}{6}

, farther from

\sqrt{10}

; the correct term is

10/(2x_n)

❌ Check the recurrence. Start from

x_n - \dfrac{x_n^2 - 10}{2x_n}

and split the fraction into two terms.

❌ Check $x_2$ . Compute

x_1

first, then apply the recurrence to

x_1

Show solution

With $f'(x) = 2x$ :

x_{n+1} = x_n - \frac{x_n^2 - 10}{2x_n} = x_n - \frac{x_n}{2} + \frac{10}{2x_n} = \frac{1}{2}x_n + \frac{5}{x_n}.

From $x_0 = 3$ :

x_1 = \frac{3}{2} + \frac{5}{3} = \frac{9 + 10}{6} = \frac{19}{6},

x_2 = \frac{19}{12} + \frac{30}{19} = \frac{361 + 360}{228} = \frac{721}{228} \approx 3.1622807.

Since $\sqrt{10} \approx 3.1622777$ , the error of $x_2$ is only about $3 \times 10^{-6}$ .

Problem 4 · Transfer: the Cube Root of 2

Given: we want $\sqrt[3]{2}$ by Newton's method starting from $x_0 = 1$ — choose the function to solve and find $x_2$ .

Which equation $f(x) = 0$ should Newton's method solve?

What is $x_2$ ?

✅ Correct!

x_2 = \tfrac{91}{72} \approx 1.2639

, against

\sqrt[3]{2} \approx 1.2599

❌ That equation's root is $\sqrt{2}$ . The cube root solves

x^3 = 2

❌ This needs the answer to set it up. The point is a function you can evaluate without knowing

\sqrt[3]{2}

x^3 - 2

❌ Its root is $-\sqrt[3]{2}$ . Move the

2

across:

x^3 = 2

becomes

x^3 - 2 = 0

❌ That is $x_1$ . One more step: apply the rule to

x_1 = \tfrac{4}{3}

❌ The derivative of $x^3$ is $3x^2$ . Using

f'(x) = x^2

sends

1 \to 2 \to \tfrac{1}{2}

; the factor

3

matters.

❌ Check the power. The second term is

\dfrac{2}{3x_1^2} = \dfrac{2}{3 \cdot 16/9} = \dfrac{3}{8}

; using

\dfrac{2}{3x_1} = \dfrac{1}{2}

gives

\tfrac{25}{18}

❌ Check the equation. Write

x = \sqrt[3]{2}

as a polynomial equation set equal to zero.

❌ Check $x_2$ . Simplify the recurrence first, then apply it twice.

Show solution

Solve $f(x) = x^3 - 2 = 0$ , with $f'(x) = 3x^2$ :

x_{n+1} = x_n - \frac{x_n^3 - 2}{3x_n^2} = \frac{2}{3}x_n + \frac{2}{3x_n^2}.

From $x_0 = 1$ :

x_1 = \frac{2}{3} + \frac{2}{3} = \frac{4}{3},

x_2 = \frac{2}{3}\cdot\frac{4}{3} + \frac{2}{3 \cdot \frac{16}{9}} = \frac{8}{9} + \frac{3}{8} = \frac{64 + 27}{72} = \frac{91}{72} \approx 1.2639.

Since $\sqrt[3]{2} \approx 1.2599$ , the error dropped from about $0.26$ at $x_0$ to about $0.07$ at $x_1$ and $0.004$ at $x_2$ .

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