Question

Prove taylor's theorem??

Answer 1

Let f be a function having n+1 continuous derivatives on an interval
I. Let a ∈ I, x ∈ I. Then
(∗n) f(x) = f(a) + f
′
(a)
1! (x − a) + · · · +
f
(n)
(a)
n!
(x − a)
n + Rn(x, a)
where
(∗∗n) Rn(x, a) = Z x
a
(x − t)
n
n!
f
(n+1)(t)dt.
Proof. For n = 0 this just says that
f(x) = f(a) + Z x
a
f
′
(t) dt
which is the fundamental theorem of calculus.
For n = 1 we use the formula (∗0) and integrate by parts. That is we apply the
formula
Z x
a
u(t)v
′
(t)dt = u(x)v(x) − u(a)v(a) −
Z x
a
u
′
(t)v(t)dt
with u(t) = f
′
(t), v(t) = t − x.
We then get
Z x
a
f
′
(t) dt =
h
(t − x)f
′
(t)
ix
a
−
Z x
a
(t − x)f
′′(t)dt
= (x − a)f
′
(a) + Z x
a
(x − t)f
′′(t)dt.
Therefore by (∗0),(∗∗0)
f(x) = f(a) + Z x
a
f
′
(t) dt
= f(a) + (x − a)f
′
(a) + Z x
a
(x − t)f
′′(t)dt.
We prove the general case using induction. We show that the formula (∗n)
implies the formula (∗n+1). Suppose we have already proved the formula for a
certain number n ≥ 0. Then we integrate by parts in the remainder term Rn(x, a)
(cf. the above formula with u(t) = f
(n+1)(t), v(t) = (x−t)
n+1/(n+ 1)). We obtain
Z x
a
(x − t)
n
f
(n+1)(t) dt =
h−(x − t)
n+1
n + 1
f
(n+1)(t)
ix
a
−
Z x
a
−(x − t)
n+1
n + 1
f
(n+2)(t)dt
=
(x − a)
n+1
n + 1
f
(n+1)(a) + Z x
a
(x − t)
n+1
n + 1
f
(n+2)(t)dt.Deviding by n! yields
Rn(x, a) = (x − a)
n+1
(n + 1)! + Rn+1(x, a).
Assuming the correctness of (∗n), (∗∗)n we may deduce (∗)n+1, (∗∗n+1):
f(x) = f(a) + f
′
(a)
1! (x − a) + · · · +
f
n(a)
n!
(x − a)
n + Rn(x, a)
= f(a) + f
′
(a)
1! (x − a) + · · · +
f
n)
(a)
n!
(x − a)
n +
f
(n+1)(a)
(n + 1)! (x − a)
n+1 + Rn+1(x, a).
with Rn+1(x, a) = R x
a
(x−t)
n+1
(n+1)! f
(n+2)(t)dt.
We now want to estimate the remainder term Rn.
Theorem 2. Let f be as in Theorem 1 and Rn as in (∗∗n). Let
M = max{|f
(n+1)(t)| : t between a and x}.
Then
|Rn(x, a)| ≤ M
(n + 1)!|x − a|
n+1
.
Proof. Let I(a, x) be the interval with endpoints a and x
|Rn(x, a)| ≤ Z
I(a,x)



(x − t)
n
n)! f
(n+1)(t)



dt
≤
Z x
a
(x − t)
n
n!
M dt =
M
(n + 1)!(x − a)
n+1
.
The last theorem can be strengthened as follows.
Theorem 3. Let f be as in Theorem 1. There is a number γ between a and x such
that
Rn(x, a) = f
(n+1)(γ)
(n + 1)! (x − a)
n+1
Proof. Suppose first a < x.
Let k be the minimum of f
(n+1)(t) in the interval [a, x] (as above) and let K be
the maximum of f
(n+1) in this interval. Then
k
Z x
a
(x − t)
n
n!
dt ≤ Rn(x, a) ≤ K
Z x
a
(x − t)
n
n!
dt.
Evaluating the integral (as above) and deviding by the integral yields
k
(n + 1)! ≤
Rn(x, a)
(x − a)
n+1 ≤
K
(n + 1)!.