Step by step
The question
Question: find the Maclaurin series of e to the 2x.
f(x) = e2x
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) = 2e2x
f′(0) = 2
f″(x) = 4e2x
f″(0) = 4
f‴(x) = 8e2x
f‴(0) = 8
f⁗(x) = 16e2x
f⁗(0) = 16
2 Divide by n factorial
Divide each value by n factorial: 0!
= 1, 1!
= 1, 2!
= 2, 3!
= 6, 4!
= 24.
numbers : 10!, 21!, 42! = 2, 83! = 43, 164! = 23
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 + 2x + 2x2 + 4x33 + 2x43 + ...
4 Check
Let's check close to 0, at x = 0.1.
Our terms give about 1.2214.
A calculator gives 1.2214 for e to the 2x.
Almost the same.
Right!
x = 0.1: the terms ≈ 1.2214, e^(2x) ≈ 1.2214
Let's look back.
First: the derivatives.
Then: divide by n factorial.
Then: write the series.
Last: check.
Answer
f(x) = 1 + 2x + 2x2 + 4x33 + 2x43 + ...