Step by step
The question
Question: find the Maclaurin series of e to the -1x.
f(x) = e−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 1.
f(0) = 1
The derivative of e to the kx is k times e to the kx.
f′(x) = −1e−x
f′(0) = −1
f″(x) = 1e−x
f″(0) = 1
f‴(x) = −1e−x
f‴(0) = −1
f⁗(x) = 1e−x
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 Check
Let's check close to 0, at x = 0.1.
Our terms give about 0.90484.
A calculator gives 0.90484 for e to the -1x.
Almost the same.
Right!
x = 0.1: the terms ≈ 0.90484, e^(-1x) ≈ 0.90484
Let's look back.
First: the derivatives.
Then: divide by n factorial.
Then: write the series.
Last: check.
Answer
f(x) = 1 − x + x22 − x36 + x424 + ...