1
Review of Calculus
1.1
Limits and Continuity
Definition 1 We say that L is the limit of f (x) as x approaches x0 , written
as lim f (x) = L, if f (x) becomes arbitrarily close to L when x is sufficienlty
x→x0
close, but not equal, to x0 .
Definition 2 Let x0 be an interior point of domain of f. Then we say that f
is continuous at x0 if lim f (x) = f (x0 ).
x→x0
Remark
The set of all functions that are continuous on the set X is
denoted C (X) . For example, the set of all functions continuous on the closed
interval [a, b] is denoted C [a, b] . The set of all functions that are continuous at
every real number is denoted by C (R) or by C (−∞, ∞) .
1.2
The Derivative
Definition 3 Let x0 be an interior point of domain of f. The derivative of f
df
, is defined by
at x0 , denoted by f 0 (x0 ) or dx
x=x0
f (x) − f (x0 )
x→x0
x − x0
f 0 (x0 ) = lim
provided that the limit exists. If f 0 (x0 ) exists then f is said to be differentiable
at x0 .
Theorem
If the function f is differentiable at x0 , then f is continuous at
x0 .
Remark The set of all functions that have n continuous derivatives on X
is denoted C n (X) and the set of functions that have derivatives of all orders
on X is denoted C ∞ (X) . For example, polynomial, rational, trigonometric,
exponential, and logarithmic functions are in C ∞ (X), where X consists of all
numbers for which the functions are defined.
1.3
Some Important Theorems
Theorem (The Intermediate Value Theorem) If f is continuous on a closed
interval [a, b] and y0 is any number between f (a) and f (b), then there exists a
number x0 in (a, b) for which f (x0 ) = y0 .
Example 1 Show that x3 − 15x + 1 = 0 has three distinct solutions in the
interval [−4, 4] .
Solution Let f (x) = x3 − 15x + 1. Note that f is continuous everywhere.
1
We have f (0) = 1 > 0 and f (1) = −13 < 0. So, there is at least one solution
of f (x) = 0 on (0, 1) .
Also, we have f (4) = 5 > 0. This implies that a number x exists, with
1 < x < 4, for which x3 − 15x + 1 = 0.
Finally, we have f (−4) = −3 < 0.This implies also that a number x exists,
with −4 < x < 0, for which x3 − 15x + 1 = 0.
Theorem (Rolle’s Theorem) Suppose f is continuous on a closed interval
[a, b] and f is differentiable on (a, b) . If f (a) = f (b), then there exists at least
one number c ∈ (a, b) such that f 0 (c) = 0.
Example 2 Let f (x) = 1 − ex + (e − 1) sin( π2 x). Show that there is at least
one c in [0, 1] with f 0 (c) = 0.
Solution We have f is continuous on [0, 1] and differentiable on (0, 1).
Also, f (0) = f (1) = 0. Hence, according to the Rolle’s Theorem there exists at
least one number c ∈ (0, 1) such that f 0 (c) = 0.
Theorem (Mean Value Theorem) If f ∈ C [a, b] and f is differentiable on
(a, b), then a number c in (a, b) exists with
f 0 (c) =
f (b) − f (a)
b−a
Theorem (Extreme Value Theorem) If f ∈ C [a, b] , then c1 , c2 ∈ [a, b] exist
with f (c1 ) < f (x) < f (c2 ), for all x ∈ [a, b] . In addition, if f is differentiable
on (a, b) , then the numbers c1 and c2 occur either at the endpoints of [a, b] or
where f 0 is zero.
Example 3 Find the absolute extreme values of f (x) = x4 − 2x2 − 3 on
the interval [−2, 2] .
Solution
We have
f 0 (x)
=
4x3 − 4x
=
4x(x − 1)(x + 1)
The critical points are 0,-1 and 1. We have
f (−2)
f (0)
f (−1)
= f (2) = 5
= −3
= f (1) = −4
Hence, absolute maximum value is 5 (at x = ±2) and absolute minimum
value is -4 (at ±1).
2
1.4
Taylor Polynomials and Series
Taylor polynomials are used extensively in numerical analysis.
Theorem (Taylor’s Theorem) Suppose f ∈ C n [a, b], that f (n+1) exists on
[a, b], and x0 ∈ [a, b]. For every x ∈ [a, b], there exists a number ξ(x) between
x0 and x with
f (x) = Pn (x) + Rn (x),
where
Pn (x) = f (x0 ) + f 0 (x0 )(x − x0 ) +
and
Rn (x) =
f (n) (x0 )
f 00 (x0 )
(x − x0 )2 + ... +
(x − x0 )n
2!
n!
f (n+1) (ξ)
(x − x0 )n+1 .
(n + 1)!
Here Pn (x) is called the nth Taylor polynomial for f about x0 , and Rn (x)
is called the remainder term (or truncation error) associated with Pn (x).
The value of ξ(x) lies between x and x0 .
The infinite series obtained by taking the limit of Pn (x) as n → ∞ is called
the Taylor series for f about x0 . In the case x0 = 0, the Taylor polynomial is
often called a Maclaurin polynomial, and the Taylor series is often called a
Maclaurin series.
Example 4 Let f (x) = (1 + x)1/2 .
a) Find the third Taylor polynomial for f√about x0 = 0.
b) Use the polynomial in (a) to evaluate 1.1 and find an upper bound for
the error.
0.1
R
c) Use the polynomial in (a) to evaluate (1 + x)1/2 dx and find an upper
0
bound for the error.
Solution:
a) We have
f (x)
=
f 0 (x)
=
f 00 (x)
=
f 000 (x)
=
f (4) (x)
=
(1 + x)1/2 ⇒ f (0) = 1
1
1
(1 + x)−1/2 ⇒ f 0 (0) =
2
2
1
1
−3/2
00
− (1 + x)
⇒ f (0) = −
4
4
3
3
−5/2
000
(1 + x)
⇒ f (0) =
8
8
15
15
−7/2
(4)
− (1 + x)
⇒ f (ξ) = − (1 + ξ)−7/2
16
16
3
So, the third Taylor polynomial for f about x0 = 0 can be written as
P3 (x)
=
f (0) + f 0 (0)(x − 0) +
=
1+
f 00 (0)
f 000 (0)
(x − 0)2 +
(x − 0)3
2!
3!
x x2
x3
−
+
2
8
16
with the remainder term
R3 (x) = −
5
(1 + ξ)−7/2 x4 ,
128
0<ξ<x
b) When x = 0.1, this becomes
√
1.1 = f (0.1) ≈ P3 (0.1) = 1 +
0.1 (0.1)2
(0.1)3
−
+
= 1.0488125
2
8
16
√
The approximation to 1.1 given by the Taylor polynomial is therefore 1.0488125.
√ The error occurring if we use the approximation 1.0488125 for the value of
1.1 is bounded by
√
1.1 − 1.0488125
5
4
(1 + ξ)−7/2 (0.1)
128
=
−
≤
5 (0.1)
max (1 + ξ)−7/2 ≤ 3.91 × 10−6 ,
128 0≤ξ≤0.1
4
max (1 + ξ)−7/2 = 1
0≤ξ≤0.1
Hence
1.0488125 − 3.91 × 10−6
≤
1.04880859
≤
√
√
1.1 ≤ 1.0488125 + 3.91 × 10−6
1.1 ≤ 1.04881641
c)
0.1
Z
1/2
(1 + x)
0.1
Z
dx
≈
0
0.1
Z
P3 (x)dx =
0
=
x x2
x3
−
+ )dx
2
8
16
0
2
=
(1 +
3
0.1
x
x
x4
−
+
4
24 64 0
0.1024598958
x+
The error:
0.1
Z
1/2
(1 + x)
0
0.1
Z
dx −
P3 (x)dx
0.1
Z
R3 (x)dx = −
=
0
0.1
Z
0
≤
5
128
0
0.1
Z
x4 dx =
0
4
5
(1 + ξ)−7/2 x4 dx
128
0.1
5 x5
= 7.81 × 10−8
128 5 0
Hence
0.1024598958 − 7.81 × 10
−8
0.1
Z
(1 + x)1/2 dx ≤ 0.1024598958 + 7.81 × 10−8
≤
0
0.1
Z
0.10245981
(1 + x)1/2 dx ≤ 0.1024599739
≤
0
5