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.
Theorem 6.6.1 Let h :
→ ℝ have m + 1
derivatives. Then there exists
t ∈ such that
m∑ h(k)(0) h(m+1)(t)
h (1) = h(0)+ --k!-- + (m-+-1)! .
k=1
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Proof: Let K be a number chosen such that
( m )
h(1)− h (0) + ∑ h(k)(0)+ K = 0
k=1 k!
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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
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Then F
=
F = 0
. Therefore, by Rolle’s theorem there exists
t between 0 and 1 such
that
F′ = 0
. Thus,
m
0 = − F ′(t) = h′(t)+ ∑ h(k+1)(t)(1− t)k
k=1 k!
∑m (k)
− h--(t)k(1− t)k− 1 − K (m + 1) (1 − t)m
k=1 k!
And so
∑m (k+1) m∑−1 (k+1)
= h′(t)+ h----(t)(1− t)k − h-----(t) (1 − t)k
k=1 k! k=0 k!
− K (m + 1)(1− t)m
′ h(m+1)(t) m ′ m
= h (t)+ ---m!----(1 − t) − h (t) − K (m + 1)(1− t)
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and so
h(m+1-)(t)-
K = (m + 1)! .
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This proves the theorem. ■
Now let f : U → ℝ where U ⊆ X a normed vector space and suppose f ∈ Cm
and
suppose
Dm+1f(
x) exists on
U. Let
x ∈ U and let
r > 0 be such that
Then for
< r consider
for t ∈
. Then by the chain rule,
h′(t) = Df (x+tv)(v),h′′(t) = D2f (x+tv) (v) (v)
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and continuing in this way,
(k) (k) (k) k
h (t) = D f (x+tv )(v)(v)⋅⋅⋅(v) ≡ D f (x+tv)v .
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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.
Theorem 6.6.2 Let f : U → ℝ and let f ∈ Cm+1
. Then if
and
< r, there exists t ∈ such that 6.21 holds.