Newton's algorithm is

Example 1

Newton's algorithm applied to a square root — Newton's algorithm is

Use Newton's algorithm on f(x) = x² − 4 with a starting value x₀ = 5 to obtain two improved estimates of √4, then report the error after the second iteration.

Given

$f(x) = x^{2} - 4$
$x_{0} = 5$

Find

x₁, x₂ and the absolute error |x₂ − √N|

Start with the thinking

Newton's algorithm is x_{k+1} = x_k − f(x_k)/f′(x_k); for f = x² − N it collapses to the averaging form.
Convergence is quadratic, so two iterations already give several correct digits.

Step-by-step solution

Formula
$x_{k+1} = x_k - f(x_k)/f'(x_k) = ½(x_k + N/x_k)$
Iteration 1
$x_{1} = ½(5 + 4/5) = 2.90000$
Iteration 2
$x_{2} = ½(2.90000 + 4/2.90000) = 2.13966$
Exact value — √4 = 2.00000
Error — |x₂ − √4| = 0.139655

Answer:

x_{2} = 2.13966 with error 0.139655

Why the other options are there

0.8000 (single division, no averaging)
-16.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Newton's algorithm is

Example 2

Newton's algorithm applied to a square root — Newton's algorithm is (2)

Use Newton's algorithm on f(x) = x² − 3 with a starting value x₀ = 3 to obtain two improved estimates of √3, then report the error after the second iteration.

Given

$f(x) = x^{2} - 3$
$x_{0} = 3$

Find

x₁, x₂ and the absolute error |x₂ − √N|

Start with the thinking

Newton's algorithm is x_{k+1} = x_k − f(x_k)/f′(x_k); for f = x² − N it collapses to the averaging form.
Convergence is quadratic, so two iterations already give several correct digits.

Step-by-step solution

Formula
$x_{k+1} = x_k - f(x_k)/f'(x_k) = ½(x_k + N/x_k)$
Iteration 1
$x_{1} = ½(3 + 3/3) = 2.00000$
Iteration 2
$x_{2} = ½(2.00000 + 3/2.00000) = 1.75000$
Exact value — √3 = 1.73205
Error — |x₂ − √3| = 0.017949

Answer:

x_{2} = 1.75000 with error 0.017949

Why the other options are there

1.0000 (single division, no averaging)
-3.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Newton's algorithm is

Example 3

Newton's algorithm applied to a square root — Newton's algorithm is (3)

Use Newton's algorithm on f(x) = x² − 8 with a starting value x₀ = 2 to obtain two improved estimates of √8, then report the error after the second iteration.

Given

$f(x) = x^{2} - 8$
$x_{0} = 2$

Find

x₁, x₂ and the absolute error |x₂ − √N|

Start with the thinking

Newton's algorithm is x_{k+1} = x_k − f(x_k)/f′(x_k); for f = x² − N it collapses to the averaging form.
Convergence is quadratic, so two iterations already give several correct digits.

Step-by-step solution

Formula
$x_{k+1} = x_k - f(x_k)/f'(x_k) = ½(x_k + N/x_k)$
Iteration 1
$x_{1} = ½(2 + 8/2) = 3.00000$
Iteration 2
$x_{2} = ½(3.00000 + 8/3.00000) = 2.83333$
Exact value — √8 = 2.82843
Error — |x₂ − √8| = 0.004906

Answer:

x_{2} = 2.83333 with error 0.004906

Why the other options are there

4.0000 (single division, no averaging)
6.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Newton's algorithm is

Example 4

Newton's algorithm applied to a square root — Newton's algorithm is (4)

Use Newton's algorithm on f(x) = x² − 9 with a starting value x₀ = 5 to obtain two improved estimates of √9, then report the error after the second iteration.

Given

$f(x) = x^{2} - 9$
$x_{0} = 5$

Find

x₁, x₂ and the absolute error |x₂ − √N|

Start with the thinking

Newton's algorithm is x_{k+1} = x_k − f(x_k)/f′(x_k); for f = x² − N it collapses to the averaging form.
Convergence is quadratic, so two iterations already give several correct digits.

Step-by-step solution

Formula
$x_{k+1} = x_k - f(x_k)/f'(x_k) = ½(x_k + N/x_k)$
Iteration 1
$x_{1} = ½(5 + 9/5) = 3.40000$
Iteration 2
$x_{2} = ½(3.40000 + 9/3.40000) = 3.02353$
Exact value — √9 = 3.00000
Error — |x₂ − √9| = 0.023529

Answer:

x_{2} = 3.02353 with error 0.023529

Why the other options are there

1.8000 (single division, no averaging)
-11.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Newton's algorithm is

Example 5

Newton's algorithm applied to a square root — Newton's algorithm is (5)

Use Newton's algorithm on f(x) = x² − 5 with a starting value x₀ = 5 to obtain two improved estimates of √5, then report the error after the second iteration.

Given

$f(x) = x^{2} - 5$
$x_{0} = 5$

Find

x₁, x₂ and the absolute error |x₂ − √N|

Start with the thinking

Newton's algorithm is x_{k+1} = x_k − f(x_k)/f′(x_k); for f = x² − N it collapses to the averaging form.
Convergence is quadratic, so two iterations already give several correct digits.

Step-by-step solution

Formula
$x_{k+1} = x_k - f(x_k)/f'(x_k) = ½(x_k + N/x_k)$
Iteration 1
$x_{1} = ½(5 + 5/5) = 3.00000$
Iteration 2
$x_{2} = ½(3.00000 + 5/3.00000) = 2.33333$
Exact value — √5 = 2.23607
Error — |x₂ − √5| = 0.097265

Answer:

x_{2} = 2.33333 with error 0.097265

Why the other options are there

1.0000 (single division, no averaging)
-15.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Newton's algorithm is

Example 6

Newton's algorithm applied to a square root — Newton's algorithm is (6)

Use Newton's algorithm on f(x) = x² − 9 with a starting value x₀ = 6 to obtain two improved estimates of √9, then report the error after the second iteration.

Given

$f(x) = x^{2} - 9$
$x_{0} = 6$

Find

x₁, x₂ and the absolute error |x₂ − √N|

Start with the thinking

Newton's algorithm is x_{k+1} = x_k − f(x_k)/f′(x_k); for f = x² − N it collapses to the averaging form.
Convergence is quadratic, so two iterations already give several correct digits.

Step-by-step solution

Formula
$x_{k+1} = x_k - f(x_k)/f'(x_k) = ½(x_k + N/x_k)$
Iteration 1
$x_{1} = ½(6 + 9/6) = 3.75000$
Iteration 2
$x_{2} = ½(3.75000 + 9/3.75000) = 3.07500$
Exact value — √9 = 3.00000
Error — |x₂ − √9| = 0.075000

Answer:

x_{2} = 3.07500 with error 0.075000

Why the other options are there

1.5000 (single division, no averaging)
-21.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Newton's algorithm is

Example 7

Newton's algorithm applied to a square root — Newton's algorithm is (7)

Use Newton's algorithm on f(x) = x² − 6 with a starting value x₀ = 2 to obtain two improved estimates of √6, then report the error after the second iteration.

Given

$f(x) = x^{2} - 6$
$x_{0} = 2$

Find

x₁, x₂ and the absolute error |x₂ − √N|

Start with the thinking

Newton's algorithm is x_{k+1} = x_k − f(x_k)/f′(x_k); for f = x² − N it collapses to the averaging form.
Convergence is quadratic, so two iterations already give several correct digits.

Step-by-step solution

Formula
$x_{k+1} = x_k - f(x_k)/f'(x_k) = ½(x_k + N/x_k)$
Iteration 1
$x_{1} = ½(2 + 6/2) = 2.50000$
Iteration 2
$x_{2} = ½(2.50000 + 6/2.50000) = 2.45000$
Exact value — √6 = 2.44949
Error — |x₂ − √6| = 0.000510

Answer:

x_{2} = 2.45000 with error 0.000510

Why the other options are there

3.0000 (single division, no averaging)
4.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Newton's algorithm is

Example 8

Newton's algorithm applied to a square root — Newton's algorithm is (8)

Use Newton's algorithm on f(x) = x² − 5 with a starting value x₀ = 5 to obtain two improved estimates of √5, then report the error after the second iteration.

Given

$f(x) = x^{2} - 5$
$x_{0} = 5$

Find

x₁, x₂ and the absolute error |x₂ − √N|

Start with the thinking

Newton's algorithm is x_{k+1} = x_k − f(x_k)/f′(x_k); for f = x² − N it collapses to the averaging form.
Convergence is quadratic, so two iterations already give several correct digits.

Step-by-step solution

Formula
$x_{k+1} = x_k - f(x_k)/f'(x_k) = ½(x_k + N/x_k)$
Iteration 1
$x_{1} = ½(5 + 5/5) = 3.00000$
Iteration 2
$x_{2} = ½(3.00000 + 5/3.00000) = 2.33333$
Exact value — √5 = 2.23607
Error — |x₂ − √5| = 0.097265

Answer:

x_{2} = 2.33333 with error 0.097265

Why the other options are there

1.0000 (single division, no averaging)
-15.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Newton's algorithm is

Example 9

Newton's algorithm applied to a square root — Newton's algorithm is (9)

Use Newton's algorithm on f(x) = x² − 4 with a starting value x₀ = 2 to obtain two improved estimates of √4, then report the error after the second iteration.

Given

$f(x) = x^{2} - 4$
$x_{0} = 2$

Find

x₁, x₂ and the absolute error |x₂ − √N|

Start with the thinking

Newton's algorithm is x_{k+1} = x_k − f(x_k)/f′(x_k); for f = x² − N it collapses to the averaging form.
Convergence is quadratic, so two iterations already give several correct digits.

Step-by-step solution

Formula
$x_{k+1} = x_k - f(x_k)/f'(x_k) = ½(x_k + N/x_k)$
Iteration 1
$x_{1} = ½(2 + 4/2) = 2.00000$
Iteration 2
$x_{2} = ½(2.00000 + 4/2.00000) = 2.00000$
Exact value — √4 = 2.00000
Error — |x₂ − √4| = 0.000000

Answer:

x_{2} = 2.00000 with error 0.000000

Why the other options are there

2.0000 (single division, no averaging)
2.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Newton's algorithm is

Example 10

Newton's algorithm applied to a square root — Newton's algorithm is (10)

Use Newton's algorithm on f(x) = x² − 5 with a starting value x₀ = 5 to obtain two improved estimates of √5, then report the error after the second iteration.

Given

$f(x) = x^{2} - 5$
$x_{0} = 5$

Find

x₁, x₂ and the absolute error |x₂ − √N|

Start with the thinking

Newton's algorithm is x_{k+1} = x_k − f(x_k)/f′(x_k); for f = x² − N it collapses to the averaging form.
Convergence is quadratic, so two iterations already give several correct digits.

Step-by-step solution

Formula
$x_{k+1} = x_k - f(x_k)/f'(x_k) = ½(x_k + N/x_k)$
Iteration 1
$x_{1} = ½(5 + 5/5) = 3.00000$
Iteration 2
$x_{2} = ½(3.00000 + 5/3.00000) = 2.33333$
Exact value — √5 = 2.23607
Error — |x₂ − √5| = 0.097265

Answer:

x_{2} = 2.33333 with error 0.097265

Why the other options are there

1.0000 (single division, no averaging)
-15.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Newton's algorithm is

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Worked exam-style examples