Taylor's theorem

Taylor's theorem describes the margin of error for a Taylor series. It states that if $𝑓 : [𝑝, 𝑞] \to ℝ$ is $(𝑛 + 1)$ -differentiable, then for any $𝑎, 𝑥 \in (𝑝, 𝑞)$ there exists a $𝑧 \in (𝑎, 𝑥)$ such that #m/thm/anal

𝑓 (𝑥) = 𝑇 𝑓 𝑛, 𝑎 (𝑥) + 𝑅 𝑓, 𝑧 𝑛, 𝑎 (𝑥)

where $𝑇 𝑓 𝑛, 𝑎 (𝑥)$ is the Taylor series about $𝑎$ to order $𝑛$ and the Lagrange remainder $𝑅 𝑓, 𝑧 𝑛, 𝑎 (𝑥)$ is given by

𝑅 𝑓, 𝑧 𝑛, 𝑎 (𝑥) = 𝑓 (𝑛 + 1) (𝑧) (𝑛 + 1)! (𝑥 - 𝑎) 𝑛 + 1

Therefore a maximum error is given by the maximum of $𝑅 𝑓, 𝑧 𝑛, 𝑎 (𝑥)$ with respect to $𝑧$ .

Proof¹

Let

𝐴 = 𝑓 (𝑥) - 𝑇 𝑓 𝑛, 𝑎 (𝑥) (𝑥 - 𝑎) 𝑛 + 1 (𝑛 + 1)!

i.e. so that

𝑓 (𝑥) = 𝑇 𝑓 𝑛, 𝑎 (𝑥) + 𝐴 (𝑛 + 1)! (𝑥 - 𝑎) 𝑛 + 1

We want to show there exists a $𝑧 \in (𝑎, 𝑏)$ such that $𝐴 = 𝑓 (𝑛 + 1) (𝑧)$ . Now define

𝐹 (𝜉) = 𝑓 (𝑥) - 𝑇 𝑓 𝑛, 𝜉 (𝑥) - 𝐴 (𝑛 + 1)! (𝑥 - 𝜉) 𝑛 + 1

Then $𝐹 (𝑥) = 𝐹 (𝑎) = 0$ , so by Rolle's theorem there exists some $𝑧 \in (𝑥, 𝑎)$ such that $𝐹' (𝑧) = 0$ . Now

𝐹' (𝜉) = - 𝑑 𝑑 𝜉 (𝑓 (𝜉) + 𝑛 \sum 𝑖 = 1 𝑓 (𝑖) (𝜉) 𝑖! (𝑥 - 𝜉) 𝑖 + 𝐴 (𝑛 + 1)! (𝑥 - 𝜉) 𝑛 + 1) = - (𝑓' (𝜉) + 𝑛 \sum 𝑖 = 1 (𝑓 (𝑖 + 1) (𝜉) 𝑖! (𝑥 - 𝜉) 𝑖 - 𝑓 (𝑖) (𝜉) (𝑖 - 1)! (𝑥 - 𝜉) 𝑖 - 1) - 𝐴 𝑛! (𝑥 - 𝜉) 𝑛) = - (𝑓' (𝜉) + 𝑛 \sum 𝑖 = 1 𝑓 (𝑖 + 1) (𝜉) 𝑖! (𝑥 - 𝜉) 𝑖 - 𝑛 - 1 \sum 𝑖 = 0 𝑓 (𝑖 + 1) (𝜉) 𝑖! (𝑥 - 𝜉) 𝑖 - 𝐴 𝑛! (𝑥 - 𝜉) 𝑛) = - (𝑓' (𝜉) + 𝑓 (𝑛 + 1) (𝜉) 𝑛! - 𝑓' (𝜉) - 𝐴 𝑛! (𝑥 - 𝜉) 𝑛) = - 𝑓 (𝑛 + 1) (𝜉) + 𝐴 𝑛! (𝑥 - 𝜉) 𝑛

so $𝐹' (𝑧) = 0$ for $𝑧 \in (𝑥, 𝑎)$ implies $𝑓 (𝑛 + 1) (𝑧) = 𝐴$ , as required.

#state/tidy | #SemBr | #lang/en

1953. A Proof of Taylor's Formula ↩

Taylor's theorem

Footnotes