Solve Newton's method for x^3 - 3x + 1 = 0 from x = 0 Step by Step

Step by step

The question

Question: estimate a root of x³ − 3x + 1 = 0 with Newton's method, starting at x = 0.

x³ − 3x + 1 = 0 =

1 The formula

Newton's idea: draw the tangent line at your guess.

Where it meets the x-axis is the next guess.

That gives this formula.

x_n+1 = x_n − f(x_n)f′(x_n)

See it on the graph.

Start at x₀ = 0.

Follow the tangent line down to the x-axis.

It lands at x₁: much closer to the root.

2 Find f′(x)

The slope: f′(x) = 3x² − 3, by the power rule.

f′(x) = 3x² − 3

Careful!

f goes on top, and f′ at the bottom.

Not the other way round.

Careful: not x₀ − f′(x₀)f(x₀) ✗

3 Round 1

f at 0 is 1.

f(0) = 0³ − 3(0) + 1 = 0 + 0 + 1 = 1

f′ at 0 is −3.

f′(0) = 3(0)² − 3 = 0 − 3 = −3

So x₁ is 0 minus 1 over −3: about 0.33333.

x₁ = 0 − 1−3 ≈ 0.33333

4 Round 2

f at 0.33333 is about 0.03704.

f(0.33333) ≈ 0.03704

f′ at 0.33333 is about −2.66667.

f′(0.33333) ≈ −2.66667

So x₂ is 0.33333 minus 0.03704 over −2.66667: about 0.34722.

x₂ = 0.33333 − 0.03704−2.66667 ≈ 0.34722

After 2 rounds, the estimate is 0.34722.

root ≈ 0.34722

5 Check

Let's check: put 0.34722 into f.

We get about 0.000196: almost 0.

Right!

f(0.34722) ≈ 0.000196

Let's look back.

First: the formula.

Then: find f′(x).

Then: round 1.

Then: round 2.

Last: check.

Answer