Multivariable Advanced Calculus

144 THE DERIVATIVENow the idea is to find K. To do this, let()m∑ h (k) (t)F (t) = h (1) − h (t) +(1 − t) k + K (1 − t) m+1k!k=1Then F (1) = F (0) = 0. Therefore, by Rolle’s theorem there exists t between 0 and 1such that F ′ (t) = 0. Thus,And soand so0 = −F ′ (t) = h ′ (t) +−= h ′ (t) +m∑k=1m∑k=1m∑k=1h (k+1) (t)k!(1 − t) kh (k) (t)k (1 − t) k−1 − K (m + 1) (1 − t) mk!h (k+1) (t)k!−K (m + 1) (1 − t) m= h ′ (t) + h(m+1) (t)m!m−1∑(1 − t) k −k=0h (k+1) (t)k!(1 − t) k(1 − t) m − h ′ (t) − K (m + 1) (1 − t) mK = h(m+1) (t)(m + 1)! .This proves the theorem. Now let f : U → R where U ⊆ X a normed vector space and suppose f ∈ C m (U).Let x ∈ U and let r > 0 be such thatThen for ||v|| < r considerfor t ∈ [0, 1]. Then by the chain rule,and continuing in this way,B (x,r) ⊆ U.f (x+tv) − f (x) ≡ h (t)h ′ (t) = Df (x+tv) (v) , h ′′ (t) = D 2 f (x+tv) (v) (v)h (k) (t) = D (k) f (x+tv) (v) (v) · · · (v) ≡ D (k) f (x+tv) v k .It follows from Taylor’s formula for a function of one variable given above thatm∑ D (k) f (x) v kf (x + v) = f (x) +k!k=1+ D(m+1) f (x+tv) v m+1. (6.38)(m + 1)!This proves the following theorem.Theorem 6.11.2 Let f : U → R and let f ∈ C m+1 (U). Then ifB (x,r) ⊆ U,and ||v|| < r, there exists t ∈ (0, 1) such that 6.38 holds.