Steply

Use Newton's method on x^3 - 5 = 0 starting at x = 2

Step by step, with the reason for every step · Newton's method · Calculus 1

Step by step

The question

Question: estimate a root of x³ − 5 = 0 with Newton's method, starting at x = 2.

x3 − 5 = 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.

xn+1 = xn − f(xn)f′(xn)

See it on the graph.

Start at x₀ = 2.

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, by the power rule.

f′(x) = 3x2

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 2 is 3.

f(2) = 23 − 5 = 8 − 5 = 3

f′ at 2 is 12.

f′(2) = 3(2)2 = 12

So x₁ is 2 minus 3/12: 1.75.

x1 = 2 − 312 = 1.75

4 Round 2

f at 1.75 is about 0.35938.

f(1.75) ≈ 0.35938

f′ at 1.75 is 9.1875.

f′(1.75) = 9.1875

So x₂ is 1.75 minus 0.35938/9.1875: about 1.71088.

x2 = 1.75 − 0.359389.1875 ≈ 1.71088

After 2 rounds, the estimate is 1.71088.

root ≈ 1.71088

5 Check

Let's check: put 1.71088 into f.

We get about 0.007973: almost 0.

Right!

f(1.71088) ≈ 0.007973

Let's look back.

First: the formula.

Then: find f′(x).

Then: round 1.

Then: round 2.

Last: check.

Answer
root ≈ 1.71088

Your turn

Try this one yourself, then check:

Estimate a root of x3 − 7 = 0
Show the answer
x ≈ 1.91294

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