Newton’s method

Newton’s method finds a root by replacing the function with its tangent line, solving that instead, and repeating from the answer. Each step is one formula, so the marks are in setting the equation up as $f(x) = 0$ , carrying the arithmetic exactly, and recognising a starting point where the method stalls or runs away.

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In this topic

Taught: Before the final, flavours A and B
Questions: 13
Total marks: 40
Suggested time: 67 min
Difficulty: Introductory to Capstone
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Key results

Newton’s method: $x_{n+1} = x_{n} - \frac{f(x_{n})}{f'(x_{n})}$
When $f'(x_{n}) \neq 0$ ; $x_{n+1}$ is where the tangent line at $x_{n}$ crosses the $x$ -axis.
When it fails: $f(x) = x^{3} - 2x + 2, \quad x_{0} = 0 \implies x_{1} = 1, \quad x_{2} = 0, \quad x_{3} = 1, \ \ldots$
A horizontal tangent, $f'(x_{n}) = 0$ , gives no next estimate; a nearly horizontal one sends it far away, possibly to another root; and the estimates can cycle, as here, and never settle.
An equation as a root: $g(x) = h(x) \iff f(x) = 0, \qquad f(x) = g(x) - h(x)$

Read the theoryUBC Math 100 §10 Newton’s methodUBC Math 100 §10.1 Newton’s method

Worked example

Standard3 marks · 4 min

Use Newton’s method on $f(x) = x^{2} - 5$ with $x_{0} = 2$ to approximate $\sqrt{5}$ . Find $x_{2}$ .

Strategy

Take two steps, each with the value and the slope at the latest estimate. Keep the fractions exact; the denominators stay small.

Working

$f(2) = -1, \quad f'(2) = 4, \qquad x_{1} = 2 - \frac{-1}{4} = \frac{9}{4}$
The first step, from x₀ = 2.
$f\!\left(\tfrac{9}{4}\right) = \frac{81}{16} - 5 = \frac{1}{16}, \qquad f'\!\left(\tfrac{9}{4}\right) = \frac{9}{2}$
Value and slope at the new estimate.
$x_{2} = \frac{9}{4} - \frac{1/16}{9/2} = \frac{9}{4} - \frac{1}{72} = \frac{162 - 1}{72} = \frac{161}{72}$
Dividing by 9/2 is multiplying by 2/9: (1/16)(2/9) = 1/72.
$\frac{161}{72} = 2.236111\ldots, \qquad \sqrt{5} = 2.236068\ldots$
Two steps already agree with √5 to four decimal places.

Answer

x_{2} = \frac{161}{72}

Check

(161/72)² ≈ 5.0002, so x₂ squares to within 0.0002 of 5, as an estimate of √5 should.

Where this goes wrong

Reusing the slope f′(2) = 4 in the second step, which gives 143/64. Each step uses the tangent line at the latest estimate, so the slope must be recomputed at 9/4.

On the paper

Write the iteration $x_{n+1} = x_n - f(x_n)/f'(x_n)$ with your $f$ and $f'$ , then show each iterate to the precision asked.

Leans on

Linear approximation, from Linear approximation

Your turn

Standard3 marks · 5 min

The equation $\cos x = x$ has exactly one solution, the zero of $f(x) = \cos x - x$ . Use one step of Newton’s method on $f$ from $x_{0} = \dfrac{\pi}{4}$ to estimate the solution. Give $x_{1}$ exactly.

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Practice questions

5 of the 13, across every level. Open a solution to read the working, or answer the question yourself and it counts towards your progress.

Introductory1 mark · 2 min
Newton’s method improves an estimate $x_{n}$ of a root of $f(x) = 0$ by following the tangent line at $x_{n}$ to the point where it crosses the $x$ -axis. Which formula gives the next estimate $x_{n+1}$ ?
Strategy
The next estimate is the x-intercept of the tangent line at the current one. Write that line down and solve for where it is zero.
Working
$y = f(x_{n}) + f'(x_{n})\,(x - x_{n})$
The tangent line at xₙ matches the value and the slope of f there.
$0 = f(x_{n}) + f'(x_{n})\,(x_{n+1} - x_{n})$
The next estimate is where this line meets the x-axis.
$x_{n+1} = x_{n} - \frac{f(x_{n})}{f'(x_{n})}$
Solve for xₙ₊₁. This needs f′(xₙ) ≠ 0.
Answer
$x_{n+1} = x_{n} - \dfrac{f(x_{n})}{f'(x_{n})}$
The answer is x sub n plus 1 equals x sub n minus f of x sub n over f prime of x sub n.
Check
On f(x) = x − 2, a line and so its own tangent, the formula takes x₀ = 5 to 5 − 3/1 = 2, the root, in one step.
Where this goes wrong
Adding the ratio instead of subtracting it. With a plus sign the estimate moves the right distance in the wrong direction, away from the root.
Leans on
Linear approximation, from Linear approximation
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Let $f(x) = x^{3} + x - 3$ . Starting from $x_{0} = 1$ , use one step of Newton’s method to find $x_{1}$ .
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Introductory2 marks · 3 min
Let $f(x) = x^{3} + x - 3$ . Starting from $x_{0} = 1$ , use one step of Newton’s method to find $x_{1}$ .
Strategy
One Newton step needs two numbers at the starting point, the value and the slope, and then the update formula.
Working
$f(1) = 1 + 1 - 3 = -1$
The value at x₀. It is not zero, so 1 is not the root.
$f'(x) = 3x^{2} + 1, \qquad f'(1) = 4$
The slope at x₀.
$x_{1} = 1 - \frac{-1}{4} = 1 + \frac{1}{4} = \frac{5}{4}$
f is negative and rising at 1, so the root is to the right, and the step moves right.
Answer
$x_{1} = \frac{5}{4}$
The answer is five quarters, or 1.25.
Check
f(1.25) ≈ 0.20 is far closer to 0 than f(1) = −1, and the sign change puts the root between 1 and 1.25.
Where this goes wrong
Sending the step the wrong way, by adding f(1)/f′(1) or by losing the minus sign of f(1), which gives 3/4. Since f is negative and increasing at 1, x₁ has to be bigger than 1.
Leans on
The power rule, from Differentiation rules
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Use Newton’s method on $f(x) = x^{2} - 5$ with $x_{0} = 2$ to approximate $\sqrt{5}$ . Find $x_{2}$ .
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Standard3 marks · 5 min
After a dose, the concentration of a drug in the blood is $C(t) = 5t\,e^{-0.4t}$ mg/L, where $t$ is in hours. The drug takes effect once the concentration reaches 3 mg/L. Starting from $t_{0} = 1$ , use one step of Newton’s method on $f(t) = C(t) - 3$ to estimate the time at which this first happens.
Strategy
Rewrite “C reaches 3” as a zero of f(t) = C(t) − 3, differentiate with the product rule, and take one step from t₀ = 1. Leave e^0.4 as it is: the answer is calculator-ready.
Working
$f(t) = 5t\,e^{-0.4t} - 3$
Move the target to the left side, so the time wanted is a zero of f.
$f'(t) = 5e^{-0.4t} + 5t\cdot\left(-0.4e^{-0.4t}\right) = e^{-0.4t}(5 - 2t)$
Product rule, with the chain rule on the exponential. The constant 3 differentiates to 0.
$f(1) = 5e^{-0.4} - 3, \qquad f'(1) = 3e^{-0.4}$
Value and slope at t₀ = 1.
$t_{1} = 1 - \frac{5e^{-0.4} - 3}{3e^{-0.4}} = 1 - \frac{5}{3} + e^{0.4} = e^{0.4} - \frac{2}{3}$
Split the fraction: 3 divided by 3e^−0.4 is e^0.4.
$t_{1} = e^{0.4} - \tfrac{2}{3} \approx 0.825\ \text{h}$
About 50 minutes. The true crossing time is 0.839 h, so one step lands within a minute of it.
Answer
$t_{1} = e^{0.4} - \frac{2}{3} \approx 0.825\ \text{h}$
The answer is e to the 0.4, minus two thirds: about 0.825 hours.
Check
C(0.825) ≈ 2.97 mg/L, just short of 3, and C is still rising there, so the true time is only a little later.
Where this goes wrong
Applying Newton’s method to C(t) itself instead of C(t) − 3, which gives t₁ = −2/3, a time before the dose. Newton’s method finds zeros, so the target value has to be moved to the left side first.
Leans on
The product rule, from Differentiation rules
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The equation $\cos x = x$ has exactly one solution, the zero of $f(x) = \cos x - x$ . Use one step of Newton’s method on $f$ from $x_{0} = \dfrac{\pi}{4}$ to estimate the solution. Give $x_{1}$ exactly.
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Challenging3 marks · 5 min
The function $f(x) = x^{3} - x$ has roots $-1$ , $0$ and $1$ . Newton’s method is started at $x_{0} = \tfrac{1}{2}$ . Which root do the estimates approach, and why?
Strategy
Take the first step exactly. The slope at 1/2 is small, which is the warning that the step will be long.
Working
$f\!\left(\tfrac{1}{2}\right) = \tfrac{1}{8} - \tfrac{1}{2} = -\tfrac{3}{8}, \qquad f'(x) = 3x^{2} - 1, \quad f'\!\left(\tfrac{1}{2}\right) = -\tfrac{1}{4}$
Value and slope at the start.
$x_{1} = \tfrac{1}{2} - \frac{-3/8}{-1/4} = \tfrac{1}{2} - \tfrac{3}{2} = -1$
The step is 3/2 long, three times the distance from 1/2 to either neighbouring root.
$f(-1) = 0 \implies x_{2} = -1 - \frac{0}{f'(-1)} = -1$
f′(−1) = 2 is not zero, so the step is zero and every estimate from x₁ on is −1.
Near 1/√3 ≈ 0.58, where f′ = 0, the tangent lines are nearly flat, so a start there is thrown a long way: here past the root 0 to the root −1.
Answer
$x_{1} = -1, \quad \text{so the estimates converge to the root } -1$
The answer is The first step lands exactly on −1, so the estimates converge to the root −1..
Check
The tangent at x = 1/2 simplifies to y = −(x + 1)/4, which crosses the axis at −1, itself a root of x³ − x.
Where this goes wrong
Assuming Newton’s method finds the root nearest the start. The step is set by the tangent line, and a shallow tangent can carry the estimate past a nearby root to a distant one.
Leans on
Slope of a tangent line, from The definition of the derivative
Try a similar question
Newton’s method is applied to $f(x) = x^{3} - 3x^{2} + x + 3$ , whose only real root is near $x = -0.77$ , starting from $x_{0} = 1$ . What happens to the estimates?
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Capstone6 marks · 10 min
The equation $e^{x} = 3 - x$ has exactly one solution, $r$ . Write it as $f(x) = 0$ and find $x_{1}$ and $x_{2}$ exactly by Newton’s method from $x_{0} = 0$ . Then use the concavity of $f$ to decide, without finding $r$ , whether $x_{2}$ is above or below $r$ .
Explain also why every later estimate lies on the same side of r, and why each is closer than the one before. Compare your working with the model solution.
Strategy
Compute the two steps exactly. Then compare each tangent line with the curve: concavity says which is higher, which fixes the sign of f at the next estimate, and the sign of f fixes the side of r.
Working
$f(x) = e^{x} + x - 3, \qquad f'(x) = e^{x} + 1, \qquad f''(x) = e^{x}$
Every term on one side; r is the zero of f.
$x_{1} = 0 - \frac{f(0)}{f'(0)} = 0 - \frac{-2}{2} = 1$
f(0) = 1 − 3 = −2 and f′(0) = 2.
$x_{2} = 1 - \frac{e - 2}{e + 1} = \frac{(e + 1) - (e - 2)}{e + 1} = \frac{3}{e + 1} \approx 0.807$
f(1) = e − 2 and f′(1) = e + 1; over a common denominator the e terms cancel.
$f(x_{n+1}) > f(x_{n}) + f'(x_{n})\,(x_{n+1} - x_{n}) = 0$
f″ = eˣ > 0, so the curve lies above the tangent at xₙ, and xₙ₊₁ is where that tangent is zero.
$f'(x) = e^{x} + 1 > 0 \implies \bigl(f(x) > 0 \iff x > r\bigr)$
f is increasing and zero only at r, so it is positive exactly to the right of r.
$x_{1},\ x_{2},\ x_{3},\ \ldots > r$
Every estimate from x₁ on overshoots. In particular x₂ = 3/(e + 1) is above r, which is 0.7921 to four places.
$x_{n} > r \implies x_{n+1} = x_{n} - \frac{f(x_{n})}{f'(x_{n})} < x_{n}$
Above r, f and f′ are both positive, so each step goes left: the estimates decrease while staying above r.
Answer
$x_{1} = 1, \quad x_{2} = \frac{3}{e + 1} \approx 0.807 > r; \quad \text{later estimates stay above } r \text{ and decrease}$
The answer is x₁ = 1 and x₂ = 3/(e + 1), about 0.807. Both overshoot r, and every later estimate is above r and smaller than the one before..
What earns the marks
Writes f(x) = eˣ + x − 3, or an equivalent with every term on one side, and f′(x) = eˣ + 1.
Finds x₁ = 0 − (−2)/2 = 1 and x₂ = 1 − (e − 2)/(e + 1) = 3/(e + 1) ≈ 0.807.
Notes that f″(x) = eˣ > 0, so f is concave up and its graph lies above every tangent line; at the point where a tangent line crosses the x-axis, f is therefore positive.
Uses f′ > 0 (f is increasing, so it is positive only to the right of r) to conclude that x₁, x₂ and every later estimate are larger than r.
Explains that at an estimate above r, f and f′ are both positive, so the step f/f′ is positive and the next estimate is smaller: the estimates decrease while staying above r.
Check
f(x₂) ≈ 0.048 is positive and f is increasing, so x₂ ≈ 0.807 does lie above r, as concavity predicted.
Where this goes wrong
Concluding from x₀ < r that the estimates approach r from below. The first step jumps past r, because a concave-up curve lies above its tangent lines, and from then on every estimate overshoots.
Leans on
Concavity and the second derivative, from Curve sketching
Try a similar question
The equation $\cos x = x$ has exactly one solution, the zero of $f(x) = \cos x - x$ . Use one step of Newton’s method on $f$ from $x_{0} = \dfrac{\pi}{4}$ to estimate the solution. Give $x_{1}$ exactly.
Answer this question

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Linear approximation
Tangent-line estimates, whether they run high or low, and approximating a value without a calculator.
14 questions

Key results

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Strategy

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Answer

Check

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Linear approximation