6.7. SECOND DERIVATIVE TEST 171

Therefore, F (c) = 0 by the way K was chosen and also F (x) = 0. By the mean valuetheorem or Rolle’s theorem, there exists ξ between x and c such that F ′ (ξ) = 0. Therefore,

0 = f ′ (ξ) +

n∑k=1

f (k+1) (ξ)

k!(x− ξ)

k −n∑

k=1

f (k) (ξ)

(k − 1)!(x− ξ)

k−1 −K (n+ 1) (x− ξ)n

= f ′ (ξ) +

n∑k=1

f (k+1) (ξ)

k!(x− ξ)

k −n−1∑k=0

f (k+1) (ξ)

k!(x− ξ)

k −K (n+ 1) (x− ξ)n

= f ′ (ξ) +f (n+1) (ξ)

n!(x− ξ)

n − f ′ (ξ)−K (n+ 1) (x− ξ)n

=f (n+1) (ξ)

n!(x− ξ)

n −K (n+ 1) (x− ξ)n

Then therefore,

K =f (n+1) (ξ)

(n+ 1)!■

The following is a special case and is what will be used.

Theorem 6.7.4 Let h : (−δ, 1 + δ) → R have m+1 derivatives. Then there exists t ∈ [0, 1]such that

h (1) = h (0) +

m∑k=1

h(k) (0)

k!+h(m+1) (t)

(m+ 1)!.

Now let f : U → R where U ⊆ Rn and suppose f ∈ Cm (U) . Let x ∈ U and let r > 0 besuch that

B (x,r) ⊆ U.

Then for ||v|| < r, considerf (x+tv)− f (x) ≡ h (t)

for t ∈ [0, 1] . Then by the chain rule,

h′ (t) =

n∑k=1

∂f

∂xk(x+ tv) vk, h

′′ (t) =

n∑k=1

n∑j=1

∂2f

∂xj∂xk(x+ tv) vkvj ■

Then from the Taylor formula stopping at the second derivative, the following theorem canbe obtained.

Theorem 6.7.5 Let f : U → R and let f ∈ C2 (U) . Then if

B (x,r) ⊆ U,

and ||v|| < r, there exists t ∈ (0, 1) such that.

f (x+ v) = f (x) +

n∑k=1

∂f

∂xk(x) vk +

1

2

n∑k=1

n∑j=1

∂2f

∂xj∂xk(x+ tv) vkvj (6.19)

Definition 6.7.6 Define the following matrix.

Hij (x+tv) ≡∂2f (x+tv)

∂xj∂xi.

It is called the Hessian matrix. From Corollary 6.7.2, this is a symmetric matrix. Then interms of this matrix, 6.19 can be written as

f (x+ v) = f (x) +

n∑j=1

∂f

∂xj(x) vk+

1

2vTH (x+tv)v