Step by step
The question
Question: find the Taylor series of e to the x at x = 2.
f(x) = ex
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 = 2: f(2) is e to the 2.
f(2) = e2
The derivative of e to the kx is k times e to the kx.
f′(x) = ex
f′(2) = e2
f″(x) = ex
f″(2) = e2
f‴(x) = ex
f‴(2) = e2
f⁗(x) = ex
f⁗(2) = e2
2 Divide by n factorial
Divide each value by n factorial: 0!
= 1, 1!
= 1, 2!
= 2, 3!
= 6, 4!
= 24.
numbers : e0!, e1!, e2!, e3!, e4!
Careful!
Do not forget to divide by n factorial.
The x squared term is f″ over 2 factorial.
Careful: not the (x − 2)² term = f″(2) × (x − 2)² ✗
3 Write the series
Put them together: each number times (x − 2) to its power.
The dots mean it goes on for ever.
f(x) = e2 + e2(x − 2) + e22(x − 2)2 + e26(x − 2)3 + e224(x − 2)4 + ...
4 Check
Let's check close to 2, at x = 2.1.
Our terms give about 8.16617.
A calculator gives 8.16617 for e to the x.
Almost the same.
Right!
x = 2.1: the terms ≈ 8.16617, e^x ≈ 8.16617
Let's look back.
First: the derivatives.
Then: divide by n factorial.
Then: write the series.
Last: check.
Answer
f(x) = e2 + e2(x − 2) + e22(x − 2)2 + e26(x − 2)3 + e224(x − 2)4 + ...