Worked examples
The question
Question: find the Maclaurin series of e to the x.
f(x) = ex
The Maclaurin series: the nth derivative at 0, divided by n factorial, times x to the n.
Maclaurin means a = 0.
f(x) = Σ f⁽ⁿ⁾(a)n!(x − a)n
1 The derivatives
Then put in x = 0: f(0) is 1.
f(0) = 1
The derivative of e to the kx is k times e to the kx.
f′(x) = ex
f′(0) = 1
f″(x) = ex
f″(0) = 1
f‴(x) = ex
f‴(0) = 1
f⁗(x) = ex
f⁗(0) = 1
2 Divide by n factorial
Divide each value by n factorial: 0!
= 1, 1!
= 1, 2!
= 2, 3!
= 6, 4!
= 24.
numbers : 10!, 11!, 12! = 12, 13! = 16, 14! = 124
Careful!
Do not forget to divide by n factorial.
The x squared term is f″ over 2 factorial.
Careful: not the x² term = f″(0) × x² ✗
3 Write the series
Put them together: each number times x to its power.
The dots mean it goes on for ever.
f(x) = 1 + x + x22 + x36 + x424 + ...
4 The pattern
The pattern: x to the n/n factorial.
With Σ the whole series fits on one line.
f(x) = Σn = 0∞xnn!
5 Check
Let's check close to 0, at x = 0.1.
Our terms give about 1.10517.
A calculator gives 1.10517 for e to the x.
Almost the same.
Right!
x = 0.1: the terms ≈ 1.10517, e^x ≈ 1.10517
Let's look back.
First: the derivatives.
Then: divide by n factorial.
Then: write the series.
Then: the pattern.
Last: check.
Answer f(x) = Σn = 0∞xnn!
▶ Watch it solved
The question
Question: find the Maclaurin series of ln(1 + x).
f(x) = ln(1 + x)
The Maclaurin series: the nth derivative at 0, divided by n factorial, times x to the n.
Maclaurin means a = 0.
f(x) = Σ f⁽ⁿ⁾(a)n!(x − a)n
1 The derivatives
Then put in x = 0: f(0) is 0.
f(0) = 0
Take the derivative again and again.
f′(x) = 1(1 + x)
f′(0) = 1
f″(x) = −1(1 + x)2
f″(0) = −1
f‴(x) = 2(1 + x)3
f‴(0) = 2
f⁗(x) = −6(1 + x)4
f⁗(0) = −6
2 Divide by n factorial
Divide each value by n factorial: 0!
= 1, 1!
= 1, 2!
= 2, 3!
= 6, 4!
= 24.
numbers : 00!, 11!, −12! = −12, 23! = 13, −64! = −14
Careful!
Do not forget to divide by n factorial.
The x squared term is f″ over 2 factorial.
Careful: not the x² term = f″(0) × x² ✗
3 Write the series
Put them together: each number times x to its power.
The dots mean it goes on for ever.
f(x) = x − x22 + x33 − x44 + ...
4 Check
Let's check close to 0, at x = 0.1.
Our terms give about 0.09531.
A calculator gives 0.09531 for ln(1 + x).
Almost the same.
Right!
x = 0.1: the terms ≈ 0.09531, ln(1 + x) ≈ 0.09531
Let's look back.
First: the derivatives.
Then: divide by n factorial.
Then: write the series.
Last: check.
Answer f(x) = x − x22 + x33 − x44 + ...
▶ Watch it solved
The question
Question: find the Taylor polynomial of degree 3 of √x at x = 4.
f(x) = √x
The Taylor series: the nth derivative at the point a, divided by n factorial, times x minus a to the n.
f(x) = Σ f⁽ⁿ⁾(a)n!(x − a)n
1 The derivatives
Then put in x = 4: f(4) is 2.
f(4) = 2
Take the derivative again and again.
f′(x) = 12√x
f′(4) = 14
f″(x) = −14x^(3/2)
f″(4) = −132
f‴(x) = 38x^(5/2)
f‴(4) = 3256
2 Divide by n factorial
Divide each value by n factorial: 0!
= 1, 1!
= 1, 2!
= 2, 3!
= 6, 4!
= 24.
numbers : 20!, 14 ÷ 1!, −132 ÷ 2! = −164, 3256 ÷ 3! = 1512
Careful!
Do not forget to divide by n factorial.
The x squared term is f″ over 2 factorial.
Careful: not the (x − 4)² term = f″(4) × (x − 4)² ✗
3 Write the series
Put them together: each number times (x − 4) to its power.
Stop at the power 3.
P₃(x) = 2 + (x − 4)4 − (x − 4)264 + (x − 4)3512
4 Check
Let's check close to 4, at x = 4.1.
Our terms give about 2.02485.
A calculator gives 2.02485 for √x.
Almost the same.
Right!
x = 4.1: the terms ≈ 2.02485, √x ≈ 2.02485
Let's look back.
First: the derivatives.
Then: divide by n factorial.
Then: write the series.
Last: check.
Answer P₃(x) = 2 + (x − 4)4 − (x − 4)264 + (x − 4)3512
▶ Watch it solved
