Analysis

6.6 The Taylor Formula

First recall the Taylor formula with the Lagrange form of the remainder. It will only be needed on

so that is what I will show.

Proof: Let K be a number chosen such that

( m ) h(1)− h (0) + ∑ h(k)(0)+ K = 0 k=1 k!

Now the idea is to find K. To do this, let

( m ) ∑ h(k)(t) k m+1 F (t) = h(1)− h(t) + k! (1− t) + K (1− t) k=1

Then F

= F = 0. Therefore, by Rolle’s theorem there exists t between 0 and 1 such that F^′ = 0. Thus,

And so

′ h(m+1)(t) m ′ m = h (t)+ ---m!----(1 − t) − h (t) − K (m + 1)(1− t)

and so

h(m+1-)(t)- K = (m + 1)! .

This proves the theorem. ■

Now let f : U → ℝ where U ⊆ X a normed vector space and suppose f ∈ C^m

and suppose D^m+1f(x) exists on U. Let x ∈ U and let r > 0 be such that

B (x,r) ⊆ U.

Then for

< r consider

f (x+tv) − f (x) ≡ h (t)

for t ∈

. Then by the chain rule,

h′(t) = Df (x+tv)(v),h′′(t) = D2f (x+tv) (v) (v)

and continuing in this way,

(k) (k) (k) k h (t) = D f (x+tv )(v)(v)⋅⋅⋅(v) ≡ D f (x+tv)v .

It follows from Taylor’s formula for a function of one variable given above that

∑m D(k)f (x)vk D(m+1)f (x+tv )vm+1 f (x +v ) = f (x)+ ----k!-----+ ------(m-+-1)!------. (6.21) k=1

(6.21)

This proves the following theorem.