Newton's algorithm is

Mathematics · FE Reference Handbook section

Mathematics

4 formulas

10 exam-style examples

~53 min

All Mathematics lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Newton's algorithm is within Mathematics. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what newton's algorithm is describes physically and when it applies.
State every one of the 4 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: radians vs degrees — set the calculator before the first trig entry.

Lecture

Why this section exists. Newton's algorithm is is the part of Mathematics that lets you connect an algebraic or calculus expression that must be evaluated exactly to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as short symbolic manipulations with one numeric evaluation at the end. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. radians vs degrees — set the calculator before the first trig entry. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Three engineers in hard hats and safety vests reviewing drawings on a truck tailgate. — Photo 1. Where this shows up in practice: newton's algorithm is.
Capstone Studio instructional photograph

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes an algebraic or calculus expression that must be evaluated exactly. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 4 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

xk + 1	Quantity produced by "xk + 1 = xk - K 2 h2 O 2h , where" — read its definition and unit from the handbook line directly above the equation.
L x	Quantity produced by "L x = xk P x = xk" — read its definition and unit from the handbook line directly above the equation.
2h	Quantity produced by "2h = S 2 W" — read its definition and unit from the handbook line directly above the equation.
2 2h	Quantity produced by "2 2h = S 1 2 2" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

J N- 1
K 2 O
K 2x O 2x
S 2h W
S2x1 W
S 2h W
S2x W
2x Sg W
Sg W
S 2h W
S2x W
R 2 V
S2 h 2 2h g g 2 2h W
S2x12 2x1 2x2 2x1 2xn W
S W
S 2 2h 2 2h 2 2
h W
S g g W
2 x 2 x 2x2 2 2 x 2 xn
2x 2 Sg g g g g W
S W
Sg g g g g W
S 2 W
S 2h 2 2h g g 2 2h W

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

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) = x² − 4
x₀ = 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: x₂ = 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) = x² − 3
x₀ = 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: x₂ = 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) = x² − 8
x₀ = 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: x₂ = 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) = x² − 9
x₀ = 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: x₂ = 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) = x² − 5
x₀ = 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: x₂ = 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) = x² − 9
x₀ = 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: x₂ = 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) = x² − 6
x₀ = 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: x₂ = 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) = x² − 5
x₀ = 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: x₂ = 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) = x² − 4
x₀ = 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: x₂ = 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) = x² − 5
x₀ = 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: x₂ = 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

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given an algebraic or calculus expression that must be evaluated exactly, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Newton's algorithm is contains 4 relations; you must be able to find this page in under 15 seconds.
Exam style: short symbolic manipulations with one numeric evaluation at the end.
Unit rule: radians vs degrees — set the calculator before the first trig entry.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

radians vs degrees — set the calculator before the first trig entry
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.