Step by step
The question
Question: find the Taylor series of e to the x at x = 1.
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 = 1: f(1) is e.
f(1) = e
The derivative of e to the kx is k times e to the kx.
f′(x) = ex
f′(1) = e
f″(x) = ex
f″(1) = e
f‴(x) = ex
f‴(1) = e
f⁗(x) = ex
f⁗(1) = e
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 − 1)² term = f″(1) × (x − 1)² ✗
3 Write the series
Put them together: each number times (x − 1) to its power.
The dots mean it goes on for ever.
f(x) = e + e(x − 1) + e2(x − 1)2 + e6(x − 1)3 + e24(x − 1)4 + ...
4 Check
Let's check close to 1, at x = 1.1.
Our terms give about 3.00417.
A calculator gives 3.00417 for e to the x.
Almost the same.
Right!
x = 1.1: the terms ≈ 3.00417, e^x ≈ 3.00417
Let's look back.
First: the derivatives.
Then: divide by n factorial.
Then: write the series.
Last: check.
Answer
f(x) = e + e(x − 1) + e2(x − 1)2 + e6(x − 1)3 + e24(x − 1)4 + ...