Worked examples
The question
Question: estimate a root of x² − 2 = 0 with Newton's method, starting at x = 1.
x2 − 2 = 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 = 1.
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) = 2x, by the power rule.
f′(x) = 2x
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 1 is −1.
f(1) = 12 − 2 = 1 − 2 = −1
f′ at 1 is 2.
f′(1) = 2(1) = 2
So x₁ is 1 minus −1/2: 1.5.
x1 = 1 − −12 = 1.5
4 Round 2
f at 1.5 is 0.25.
f(1.5) = 0.25
f′ at 1.5 is 3.
f′(1.5) = 3
So x₂ is 1.5 minus 0.25/3: about 1.41667.
x2 = 1.5 − 0.253 ≈ 1.41667
After 2 rounds, the estimate is 1.41667.
root ≈ 1.41667
5 Check
Let's check: put 1.41667 into f.
We get about 0.006944: almost 0.
Right!
f(1.41667) ≈ 0.006944
Let's look back.
First: the formula.
Then: find f′(x).
Then: round 1.
Then: round 2.
Last: check.
▶ Watch it solved
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.
▶ Watch it solved
The question
Question: estimate a root of x³ − 2x − 5 = 0 with Newton's method, starting at x = 2.
x3 − 2x − 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 − 2, by the power rule.
f′(x) = 3x2 − 2
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 −1.
f(2) = 23 − 2(2) − 5 = 8 − 4 − 5 = −1
f′ at 2 is 10.
f′(2) = 3(2)2 − 2 = 12 − 2 = 10
So x₁ is 2 minus −1/10: 2.1.
x1 = 2 − −110 = 2.1
4 Round 2
f at 2.1 is 0.061.
f(2.1) = 0.061
f′ at 2.1 is 11.23.
f′(2.1) = 11.23
So x₂ is 2.1 minus 0.061/11.23: about 2.09457.
x2 = 2.1 − 0.06111.23 ≈ 2.09457
After 2 rounds, the estimate is 2.09457.
root ≈ 2.09457
5 Check
Let's check: put 2.09457 into f.
We get about 0.000186: almost 0.
Right!
f(2.09457) ≈ 0.000186
Let's look back.
First: the formula.
Then: find f′(x).
Then: round 1.
Then: round 2.
Last: check.
▶ Watch it solved
