# Taylor’s Theorem ```Taylor’s Theorem
Theorem 1 (Taylor’s Theorem) Let a &lt; b, n ∈ IN ∪ {0}, and f : [a, b] → IR. Assume
that f (n) exists and is continuous on [a, b] and f (n+1) exists on (a, b). Let α ∈ [a, b] and
define the Taylor polynomial of degree n with expansion point α to be
Pn (x) =
n
X
1 (k)
f (α) (x
k!
− α)k
k=0
Then, for all x ∈ [a, b],
f (x) = Pn (x) + Rn (α, x)
where the error term Rn (α, x) is given by
R x 1 (n+1)
f
(t) (x − t)n dt, if f (n+1) is integrable.
(a) (integral form) Rn (α, x) = α n!
(b) (Lagrange form) Rn (α, x) =
and x.
(c) (Cauchy form) Rn (α, x) =
and x.
(d) If r ∈ IN, then Rn (α, x) =
α and x.
(e) Rn (α, x) =
1
Q(n) (α)(x
n!
1
(n+1)
(c)(x
(n+1)! f
1 (n+1)
f
(c) (x
n!
− c)n (x − α) for some c strictly between α
1
(n+1)
(c) (x−c)n−r+1 (x−α)r
rn! f
− α)n+1 where
(
Q(t) =
Proof:
by
− α)n+1 for some c strictly between α
f (t)−f (x)
t−x
′
f (x)
for some c strictly between
if t 6= x
if t = x
Fix any evaluation and expansion points x, α ∈ [a, b]. Define the function S(t)
f (x) = f (t) + (x − t)f ′ (t) + 12 (x − t)2 f ′′ (t) + &middot; &middot; &middot; +
1
(x
n!
− t)n f (n) (t) + S(t)
(1)
Observe that substituting t = α into (1) gives f (x) = Pn (x) + S(α). So we wish to find
Rn (α, x) = S(α)
The function S(t) is determined by its derivative and its value at one point. Finding a
value of S(t) for one value of t is easy. Substitute t = x into (1) to yield S(x) = 0. To find
S ′ (t), apply ddt to both sides of (1). Recalling that x is just a constant parameter,
0 = f ′ (t) + − f ′ (t) + (x − t)f ′′ (t) + − (x − t)f ′′ (t) + 12 (x − t)2 f (3) (t)
1
1
+ &middot; &middot; &middot; + − (n−1)!
(x − t)n−1 f (n) (t) + n!
(x − t)n f (n+1) (t) + S ′ (t)
=
1
n! (x
− t)n f (n+1) (t) + S ′ (t)
c Joel Feldman.
November 28, 2014
Taylor’s Theorem
1
so that
1 (n+1)
f
(t) (x − t)n
S ′ (t) = − n!
(a) By the fundamental theorem of calculus
Z
Z x
′
S (t) dt =
S(α) = − S(x) − S(α) = −
x
α
α
1 (n+1)
f
(t) (x
n!
− t)n dt
(c) By the mean value theorem, there is a c strictly between α and x such that
1 (n+1)
f
(c) (x − c)n (α − x)
S(α) = S(α) − S(x) = S ′ (c) (α − x) = − n!
=
1 (n+1)
(c) (x
n! f
− c)n (x − α)
(b) By the generalized mean value theorem (see the notes entitled “The Mean Value
Theorem”) with g(t) = (x − t)n+1 , there is a c strictly between α and x such that
S(α) = S(α) − S(x) =
S ′ (c)
g ′ (c)
g(α) − g(x)
n+1
1 (n+1)
1
= − n!
f
(c) (x − c)n −(n+1)(x−c)
n (x − α)
=
(n+1)
1
(c)(x
(n+1)! f
− α)n+1
Don’t forget, when computing g ′ (c), that g is a function of t with x just a fixed parameter.
(d) By the generalized mean value theorem with g(t) = (x−t)r , there is a c strictly between
α and x such that
′
(c)
S(α) = S(α) − S(x) = Sg ′ (c)
g(α) − g(x)
r
1 (n+1)
1
= − n!
f
(c) (x − c)n −r(x−c)
r−1 (x − α)
=
(n+1)
1
(c)(x
rn! f
− c)n−r+1 (x − α)r
(e) We’ll only consider the case that x 6= α. For x = α, the error Rn (α, x) is obviously
1
Q(n) (α)(x − α)n+1 = 0 even if Q(n) (α) is
zero and we’ll just take it as a convention that n!
not defined. Since f is n times differentiable, so is Q(t), at least for all t 6= x. In particular
Q(t) is n times differentiable at t = α. From the definition of Q we have that
f (t) = f (x) + (t − x) Q(t)
=⇒
f (α) = f (x) − (x − α) Q(α)
f ′ (t) = Q(t) + (t − x) Q′ (t)
=⇒
f ′ (α) = Q(α) − (x − α) Q′ (α)
f (2) (t) = 2Q′ (t) + (t − x) Q(2) (t)
..
.
=⇒
f (2) (α) = 2Q′ (α) − (x − α) Q(2) (α)
..
.
f (k) (t) = kQ(k−1) (t) + (t − x) Q(k) (t)
=⇒
f (k) (α) = kQ(k−1) (α) − (x − α) Q(k) (α)
c Joel Feldman.
November 28, 2014
Taylor’s Theorem
2
for k ≤ n. So
f (x) − P (x) = f (x) − f (α) −
n
X
1 (k)
(α) (x
k! f
k=1
n n
X
= (x − α)Q(α) −
− α)k
(k−1)
1
(α) (x
(k−1)! Q
− α)k −
1 (k)
(α) (x
k! Q
− α)k+1
k=1
o
The sum telescopes leaving
f (x) − P (x) =
(n)
1
(α) (x
n! Q
− α)n+1
as desired.
Remark 2 It is rarely necessary, or even possible, to evaluate Rn (α, x) exactly. It is
usually sufficient to find a number M such that f (n+1) (c) ≤ M for all c between the
expansion point α and the x of interest. Both parts (a) and (b) of Taylor’s Theorem then
imply that
|Rn (α, x)| ≤
1
(n+1)! M |x
− x0 |n+1
Example 3 (Sine and Cosine Series) The trigonometric functions sin x and cos x
have widely used Taylor expansions about α = 0. Every derivative of sin x and cos x is
one of &plusmn; sin x and &plusmn; cos x. Consequently, when we apply Theorem 1.b we always have
(n+1) |x|n+1
f
(c) ≤ 1 and hence |Rn (α, x)| ≤ (n+1)!
. This converges to zero as n → ∞. Consequently, for both f (x) = sin x and f (x) = cos x, we have lim Rn (α = 0, x) = 0 and
n→∞
h
f (x) = lim f (0) + f ′ (0) x + &middot; &middot; &middot; +
n→∞
1 (n)
f (0) xn
n!
i
First, compute all derivatives at general x.
f (x) = sin x
f ′ (x) = cos x
f ′′ (x) = − sin x
f (x) = cos x
f ′ (x) = − sin x f ′′ (x) = − cos x
f (3) (x) = − cos x
f (4) (x) = sin x
f (3) (x) = sin x
f (4) (x) = cos x &middot; &middot; &middot;
&middot;&middot;&middot;
The pattern starts over again with the fourth derivative being the same as the original
function. Now set x = α = 0.
f (x) = sin x
f (0) = 0 f ′ (0) = 1 f ′′ (0) = 0
f (3) (0) = −1
f (4) (0) = 0 &middot; &middot; &middot;
f (x) = cos x
f (0) = 1 f ′ (0) = 0 f ′′ (0) = −1
f (3) (0) = 0
f (4) (0) = 1 &middot; &middot; &middot;
c Joel Feldman.
November 28, 2014
Taylor’s Theorem
3
For sin x, all even numbered derivatives are zero. The odd numbered derivatives alternate
between 1 and −1. For cos x, all odd numbered derivatives are zero. The even numbered
derivatives alternate between 1 and −1. We conclude that
sin x = x −
cos x = 1 −
1 3
x
3!
1 2
2! x
+
+
1 5
x
5!
1 4
4! x
−&middot;&middot;&middot; =
−&middot;&middot;&middot; =
∞
X
n=0
∞
X
1
(−1)n (2n+1)!
x2n+1
(1)
(−1)
n
2n
1
(2n)! x
n=0
If we wish to evaluate sin x or cos x for some specific value of x, we may always use trig
identities like sin x = sin(x &plusmn; 2π) = − sin(x &plusmn; π) = cos π2 − x to reduce consideration
to 0 ≤ x ≤ π4 &lt; 1. Then the alternating series test tells us that the error introduced by
truncating the series in (1) is between 0 and the first term dropped. That is, if 0 ≤ x ≤ 1,
1 3
x +
sin x − x − 3!
1 2
cos x − 1 − 2!
x +
1 5
5! x
−&middot;&middot;&middot;+
1 4
4! x
−&middot;&middot;&middot;+
(−1)n−1 2n−1 (2n−1)! x
(−1)n−1 2n−2 (2n−2)! x
2n+1
x
is between 0 and (−1)n (2n+1)!
1
is between 0 and (−1)n (2n)!
x2n
This error decreases spectacularly quickly as n increases. For example
1
5!
1
9!
≈ 0.0083
≈ 0.000003
1
6!
1
10!
≈ 0.0014
≈ 0.00000028
1
7!
1
11!
≈ 0.0002
≈ 0.000000025
1
8!
1
12!
≈ 0.000025
≈ 0.000000002
Example 4 (Exponential Series) A similar phenomenon happens with the exponential
function f (x) = ex . By Theorem 1.b, for all natural numbers n and all real numbers x,
ex = 1 + x + 21 x2 +
1 3
3! x
+&middot;&middot;&middot;+
1 n
n! x
+
n+1
ec
(n+1)! x
for some c strictly between 0 and x. Now consider any fixed real number x. As c runs
from 0 to x, ec runs from e0 = 1 to ex . In particular, ec is always between 1 and ex and
so is smaller than 1 + ex . Thus the error term
|Rn (0, x)| = |x|n+1
ec
n+1 x
≤ [ex + 1]
(n + 1)!
(n + 1)!
|x|
Let’s call en (x) = (n+1)!
. I claim that as n increases towards infinity, en (x) decreases
(quickly) towards zero. To see this, let’s compare en (x) and en+1 (x).
n+1
en+1 (x)
=
en (x)
c Joel Feldman.
|x|n+2
(n+2)!
|x|
(n+1)!
n+1
=
|x|
n+2
November 28, 2014
Taylor’s Theorem
4
(x)
&lt; 12 . That is, increasing the
So, when n is bigger than, for example 2|x|, we have en+1
en (x)
index on en (x) by one decreases the size of en (x) by a factor of at least two. As a result
en (x) must tend to zero as n → ∞. Consequently lim Rn (0, x) = 0 and
n→∞
h
e = lim 1 + x + 21 x2 +
x
n→∞
c Joel Feldman.
1 3
x
3!
+&middot;&middot;&middot;+
1 n
x
n!
November 28, 2014
i
=
∞
X
1 n
x
n!
n=0
Taylor’s Theorem
5
```