Taylor series

Taylor’s theorem

Taylor’s theorem describes the margin of error for a Taylor series. It states that if 𝑓 :[𝑝,𝑞] →ℝ is (𝑛 +1)-differentiable, then for any 𝑎,𝑥 ∈(𝑝,𝑞) there exists a 𝑧 ∈(𝑎,𝑥) such that anal

𝑓(𝑥)=𝑇𝑓𝑛,𝑎(𝑥)+𝑅𝑓,𝑧𝑛,𝑎(𝑥)

where 𝑇𝑓𝑛,𝑎(𝑥) is the Taylor series about 𝑎 to order 𝑛 and the Lagrange remainder 𝑅𝑓,𝑧𝑛,𝑎(𝑥) is given by

𝑅𝑓,𝑧𝑛,𝑎(𝑥)=𝑓(𝑛+1)(𝑧)(𝑛+1)!(𝑥−𝑎)𝑛+1

Therefore a maximum error is given by the maximum of 𝑅𝑓,𝑧𝑛,𝑎(𝑥) with respect to 𝑧.

Proof ¹

Let

𝐴=𝑓(𝑥)−𝑇𝑓𝑛,𝑎(𝑥)(𝑥−𝑎)𝑛+1(𝑛+1)!

i.e. so that

𝑓(𝑥)=𝑇𝑓𝑛,𝑎(𝑥)+𝐴(𝑛+1)!(𝑥−𝑎)𝑛+1

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

𝐹(𝜉)=𝑓(𝑥)−𝑇𝑓𝑛,𝜉(𝑥)−𝐴(𝑛+1)!(𝑥−𝜉)𝑛+1

Then 𝐹(𝑥) =𝐹(𝑎) =0, so by Rolle’s theorem there exists some 𝑧 ∈(𝑥,𝑎) such that 𝐹′(𝑧) =0. Now

𝐹′(𝜉)=−𝑑𝑑𝜉(𝑓(𝜉)+𝑛∑𝑖=1𝑓(𝑖)(𝜉)𝑖!(𝑥−𝜉)𝑖+𝐴(𝑛+1)!(𝑥−𝜉)𝑛+1)=−(𝑓′(𝜉)+𝑛∑𝑖=1(𝑓(𝑖+1)(𝜉)𝑖!(𝑥−𝜉)𝑖−𝑓(𝑖)(𝜉)(𝑖−1)!(𝑥−𝜉)𝑖−1)−𝐴𝑛!(𝑥−𝜉)𝑛)=−(𝑓′(𝜉)+𝑛∑𝑖=1𝑓(𝑖+1)(𝜉)𝑖!(𝑥−𝜉)𝑖−𝑛−1∑𝑖=0𝑓(𝑖+1)(𝜉)𝑖!(𝑥−𝜉)𝑖−𝐴𝑛!(𝑥−𝜉)𝑛)=−(𝑓′(𝜉)+𝑓(𝑛+1)(𝜉)𝑛!−𝑓′(𝜉)−𝐴𝑛!(𝑥−𝜉)𝑛)=−𝑓(𝑛+1)(𝜉)+𝐴𝑛!(𝑥−𝜉)𝑛

so 𝐹′(𝑧) =0 for 𝑧 ∈(𝑥,𝑎) implies 𝑓(𝑛+1)(𝑧) =𝐴, as required.

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