Step by step
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 + ...