Preliminaries
Consider the function f (x) = cos(x), its derivative f
(x) = −sin(x), and its antiderivative
F(x) = sin(x) + C. These formulas were studied in calculus. The former
is used to determine the slope m = f
(x0) of the curve y = f (x) at a point (x0, f (x0)),
and the latter is used to compute the area under the curve for a ≤ x ≤ b.
The slope at the point (π/2, 0) is m = f
(π/2) = −1 and can be used to find the
tangent line at this point (see Figure 1.1(a)):
ytan = m
x − π
2
+ 0 = f
π
2
x − π
2
= −x + π
2
.
y
0.5
0.0
0.5
1.0 1.5 2.0
x
1.0
−0.5
y = cos(x)
Figure 1.1 (a) The tangent line to
the curve y = cos(x) at the point
(π/2, 0).
1
2 CHAP. 1 PRELIMINARIES
0.5
0.0
0.5
1.0 1.5 2.0
x
1.0
−0.5
y
y = cos(x)
Figure 1.1 (b) The area under the
curve y = cos(x) over the interval
[0, π/2].
The area under the curve for 0 ≤ x ≤ π/2 is computed using an integral (see Figure
1.1(b)):
area =
π/2
0
cos(x) dx = F
π
2
− F(0) = sin
π
2
− 0 = 1.
These are some of the results that we will need to use from calculus.
1.1 Review of Calculus
It is assumed that the reader is familiar with the notation and subject matter covered in
the undergraduate calculus sequence. This should have included the topics of limits,
continuity, differentiation, integration, sequences, and series. Throughout the book we
refer to the following results.
Limits and Continuity
Definition 1.1. Assume that f (x) is defined on an open interval containing x = x0,
except possibly at x = x0 itself. Then f is said to have the limit L at x = x0, and we
write
(1) lim
x→x0
f (x) = L,
if given any > 0 there exists a δ > 0 such that | f (x) − L| < whenever 0 <
|x − x0| < δ. When the h-increment notation x = x0 + h is used, equation (1)
becomes
(2) lim
h→0
f (x0 + h) = L.
SEC. 1.1 REVIEW OF CALCULUS 3
Definition 1.2. Assume that f (x) is defined on an open interval containing x = x0.
Then f is said to be continuous at x = x0 if
(3) lim
x→x0
f (x) = f (x0).
The function f is said to be continuous on a set S if it is continuous at each point
x ∈ S. The notation Cn(S) stands for the set of all functions f such that f and its
first n derivatives are continuous on S. When S is an interval, say [a, b], then the
notation Cn[a, b] is used. As an example, consider the function f (x) = x4/3 on the
interval [−1, 1]. Clearly, f (x) and f
(x) = (4/3)x1/3 are continuous on [−1, 1],
while f
(x) = (4/9)x−2/3 is not continuous at x = 0.
Definition 1.3. Suppose that {xn}∞
n=1 is an infinite sequence. Then the sequence is
said to have the limit L, and we write
(4) lim
n→∞
xn = L,
if given any > 0, there exists a positive integer N = N( ) such that n > N implies
that |xn − L| < .
When a sequence has a limit, we say that it is a convergent sequence. Another
commonly used notation is “xn → L as n→∞.” Equation (4) is equivalent to
(5) lim
n→∞
(xn − L) = 0.
Thus we can view the sequence { n}∞
n=1
= {xn − L}∞
n=1 as an error sequence. The
following theorem relates the concepts of continuity and convergent sequence.
Theorem 1.1. Assume that f (x) is defined on the set S and x0 ∈ S. The following
statements are equivalent:
(a) The function f is continuous at x0.
(b) If lim
n→∞
xn = x0, then lim
n→∞
f (xn) = f (x0) (6) .
Theorem 1.2 (Intermediate Value Theorem). Assume that f ∈ C[a, b] and L is
any number between f (a) and f (b). Then there exists a number c, with c ∈ (a, b),
such that f (c) = L.
Example 1.1. The function f (x) = cos(x −1) is continuous over [0, 1], and the constant
L = 0.8 ∈ (cos(0), cos(1)). The solution to f (x) = 0.8 over [0, 1] is c1 = 0.356499.
Similarly, f (x) is continuous over [1, 2.5], and L = 0.8 ∈ (cos(2.5), cos(1)). The solution
to f (x) = 0.8 over [1, 2.5] is c2 = 1.643502. These two cases are shown in Figure 1.2.
4 CHAP. 1 PRELIMINARIES
y
0.0 x
0.5 1.0 1.5 2.0 2.5
0.2
0.4
0.6
0.8
1.0
c1 c2
y = L
y = f (x)
Figure 1.2 The intermediate value
theorem applied to the function
f (x) = cos(x − 1) over [0, 1] and
over the interval [1, 2.5].
0.0
(a, f (a))
y
(x1, f (x1))
y = f (x)
(x2, f (x2))
x
(b, f (b))
0.5 1.0 1.5 2.0 2.5 3.0
10
20
30
40
50
60
Figure 1.3 The extreme value
theorem applied to the function
f (x) = 35 + 59.5x − 66.5x2 + 15x3
over the interval [0, 3].
Theorem 1.3 (Extreme Value Theorem for a Continuous Function). Assume that
f ∈ C[a, b]. Then there exists a lower bound M1, an upper bound M2, and two
numbers x1, x2 ∈ [a, b] such that
(7) M1 = f (x1) ≤ f (x) ≤ f (x2) = M2 whenever x ∈ [a, b].
We sometimes express this by writing
(8) M1 = f (x1) = min
a≤x≤b
{ f (x)} and M2 = f (x2) = max
a≤x≤b
{ f (x)}.
Differentiable Functions
Definition 1.4. Assume that f (x) is defined on an open interval containing x0. Then
f is said to be differentiable at x0 if
(9) lim
x→x0
f (x) − f (x0)
x − x0
SEC. 1.1 REVIEW OF CALCULUS 5
exists. When this limit exists, it is denoted by f
(x0) and is called the derivative of f
at x0. An equivalent way to express this limit is to use the h-increment notation:
(10) lim
h→0
f (x0 + h) − f (x0)
h
= f
(x0).
A function that has a derivative at each point in a set S is said to be differentiable
on S. Note that the number m = f
(x0) is the slope of the tangent line to the graph of
the function y = f (x) at the point (x0, f (x0)).
Theorem 1.4. If f (x) is differentiable at x = x0, then f (x) is continuous at x = x0.
It follows from Theorem 1.3 that if a function f is differentiable on a closed interval
[a, b], then its extreme values occur at the endpoints of the interval or at the critical
points (solutions of f
(x) = 0) in the open interval (a, b).
Example 1.2. The function f (x) = 15x3−66.5x2+59.5x+35 is differentiable on [0, 3].
The solutions to f
(x) = 45x2 − 123x + 59.5 = 0 are x1 = 0.54955 and x2 = 2.40601.
The maximum and minimum values of f on [0, 3] are:
min{ f (0), f (3), f (x1), f (x2)} = min{35, 20, 50.10438, 2.11850} = 2.11850
and
max{ f (0), f (3), f (x1), f (x2)} = max{35, 20, 50.10438, 2.11850} = 50.10438
(see Figure 1.3).
Theorem 1.5 (Rolle’s Theorem). Assume that f ∈ C[a, b] and that f
(x) exists for
all x ∈ (a, b). If f (a) = f (b) = 0, then there exists a number c, with c ∈ (a, b), such
that f
(c) = 0.
Theorem 1.6 (Mean Value Theorem). Assume that f ∈ C[a, b] and that f
(x)
exists for all x ∈ (a, b). Then there exists a number c, with c ∈ (a, b), such that
(11) f
(c) = f (b) − f (a)
b − a
.
Geometrically, the mean value theorem says that there is at least one number c ∈
(a, b) such that the slope of the tangent line to the graph of y = f (x) at the point
(c, f (c)) equals the slope of the secant line through the points (a, f (a)) and (b, f (b)).
Example 1.3. The function f (x) = sin(x) is continuous on the closed interval [0.1, 2.1]
and differentiable on the open interval (0.1, 2.1). Thus, by the mean value theorem, there
is a number c such that
f
(c) = f (2.1) − f (0.1)
2.1 − 0.1
= 0.863209 − 0.099833
2.1 − 0.1
= 0.381688.
The solution to f
(c) = cos(c) = 0.381688 in the interval (0.1, 2.1) is c = 1.179174.
The graphs of f (x), the secant line y = 0.381688x + 0.099833, and the tangent line
y = 0.381688x + 0.474215 are shown in Figure 1.4.
6 CHAP. 1 PRELIMINARIES
a 0.5 1.0 c 1.5 2.0 b
f (a)
f (b)
1.0 (c, f (c))
(a, f (a))
(b, f (b))
m = f ′(c)
0.5
y
x
Figure 1.4 The mean value theorem applied to f (x) =
sin(x) over the interval [0.1, 2.1].
Theorem 1.7 (Generalized Rolle’s Theorem). Assume that f ∈ C[a, b] and that
f
(x), f
(x), . . . , f (n)(x) exist over (a, b) and x0, x1, . . . , xn ∈ [a, b]. If f (x j ) = 0
for j = 0, 1, . . . , n, then there exists a number c, with c ∈ (a, b), such that f (n)(c) = 0.
Integrals
Theorem 1.8 (First Fundamental Theorem). If f is continuous over [a, b] and F
is any antiderivative of f on [a, b], then
(12)
b
a
f (x) dx = F(b) − F(a) where F
(x) = f (x).
Theorem 1.9 (Second Fundamental Theorem). If f is continuous over [a, b] and
x ∈ (a, b), then
(13)
d
dx
x
a
f (t) dt = f (x).
Example 1.4. The function f (x) = cos(x) satisfies the hypotheses of Theorem 1.9 over
the interval [0, π/2]; thus by the chain rule
d
dx
x2
0
cos(t) dt = cos(x2)(x2)
= 2x cos(x2).
Theorem 1.10 (Mean Value Theorem for Integrals). Assume that f ∈ C[a, b].
Then there exists a number c, with c ∈ (a, b), such that
1
b − a
b
a
f (x) dx = f (c).
The value f (c) is the average value of f over the interval [a, b].
SEC. 1.1 REVIEW OF CALCULUS 7
0.0
0.2
0.4
0.6
0.8
0.0 0.5 1.0 1.5 2.0 2.5
y
y = f (x)
x
Figure 1.5 The mean value
theorem for integrals applied to
f (x) = sin(x) + 13
sin(3x) over the
interval [0, 2.5].
Example 1.5. The function f (x) = sin(x) + 13
sin(3x) satisfies the hypotheses of Theorem
1.10 over the interval [0, 2.5]. An antiderivative of f (x) is F(x) = −cos(x) −
19
cos(3x). The average value of the function f (x) over the interval [0, 2.5] is
1
2.5 − 0
2.5
0
f (x) dx = F(2.5) − F(0)
2.5
= 0.762629 − (−1.111111)
2.5
= 1.873740
2.5
= 0.749496.
There are three solutions to the equation f (c) = 0.749496 over the interval [0, 2.5]:
c1 = 0.440566, c2 = 1.268010, and c3 = 1.873583. The area of the rectangle with
base b − a = 2.5 and height f (c j ) = 0.749496 is f (c j )(b − a) = 1.873740. The area
of the rectangle has the same numerical value as the integral of f (x) taken over the interval
[0, 2.5]. A comparison of the area under the curve y = f (x) and that of the rectangle
can be seen in Figure 1.5.
Theorem 1.11 (Weighted Integral Mean Value Theorem). Assume that f, g ∈
C[a, b] and g(x) ≥ 0 for x ∈ [a, b]. Then there exists a number c, with c ∈ (a, b),
such that
(14)
b
a
f (x)g(x) dx = f (c)
b
a
g(x) dx.
Example 1.6. The functions f (x) = sin(x) and g(x) = x2 satisfy the hypotheses of
Theorem 1.11 over the interval [0, π/2]. Thus there exists a number c such that
sin(c) =
π/2
0 x2 sin(x) dx
π/2
0 x2 dx
= 1.14159
1.29193
= 0.883631
or c = sin−1(0.883631) = 1.08356.
8 CHAP. 1 PRELIMINARIES
Series
Definition 1.5. Let {an}∞
n=1 be a sequence. Then
∞
n=1 an is an infinite series. The
nth partial sum is Sn = nk
=1 ak . The infinite series converges if and only if the
sequence {Sn}∞
n=1 converges to a limit S, that is,
(15) lim
n→∞
Sn = lim
n→∞
n
k=1
ak = S.
If a series does not converge, we say that it diverges.
Example 1.7. Consider the infinite sequence {an}∞
n=1
=
1
n(n + 1)
∞
n=1
. Then the nth
partial sum is
Sn =
n
k=1
1
k(k + 1)
=
n
k=1
1
k
− 1
k + 1
= 1 − 1
n + 1
.
Therefore, the sum of the infinite series is
S = lim
n→∞
Sn = lim
n→∞
1 − 1
n + 1
= 1.
Theorem 1.12 (Taylor’s Theorem). Assume that f ∈ Cn+1[a, b] and let x0 ∈
[a, b]. Then, for every x ∈ (a, b), there exists a number c = c(x) (the value of c
depends on the value of x) that lies between x0 and x such that
(16) f (x) = Pn(x) + Rn(x),
where
(17) Pn(x) =
n
k=0
f (k)(x0)
k! (x − x0)k
and
(18) Rn(x) = f (n+1)(c)
(n + 1)! (x − x0)n+1.
Example 1.8. The function f (x) = sin(x) satisfies the hypotheses of Theorem 1.12. The
Taylor polynomial Pn(x) of degree n = 9 expanded about x0 = 0 is obtained by evaluating
SEC. 1.1 REVIEW OF CALCULUS 9
−1.0
−0.5
0.0
0.5
1.0
1 2 3 4 5 6
y
y = P(x)
x
y = f (x)
Figure 1.6 The graph of f (x) = sin(x) and the Taylor
polynomial P9(x) = x − x3/3! + x5/5! − x7/7! + x9/9!.
the following derivatives at x = 0 and substituting the numerical values into formula (17).
f (x) = sin(x), f (0) = 0,
f
(x) = cos(x), f
(0) = 1,
f
(x) = −sin(x), f
(0) = 0,
f (3)(x) = −cos(x), f (3)(0) = −1,
...
...
f (9)(x) = cos(x), f (9)(0) = 1,
P9(x) = x − x3
3!
+ x5
5!
− x7
7!
+ x9
9! .
A graph of both f and P9 over the interval [0, 2π] is shown in Figure 1.6.
Corollary 1.1. If Pn(x) is the Taylor polynomial of degree n given in Theorem 1.12,
then
(19) P(k)
n (x0) = f (k)(x0) for k = 0, 1, . . . , n.
Evaluation of a Polynomial
Let the polynomial P(x) of degree n have the form
(20) P(x) = anxn + an−1xn−1 +· · ·+a2x2 + a1x + a0.
10 CHAP. 1 PRELIMINARIES
Horner’s method or synthetic division is a technique for evaluating polynomials. It
can be thought of as nested multiplication. For example, a fifth-degree polynomial can
be written in the nested multiplication form
P5(x) = ((((a5x + a4)x + a3)x + a2)x + a1)x + a0.
Theorem 1.13 (Horner’s Method for Polynomial Evaluation). Assume that P(x)
is the polynomial given in equation (20) and x = c is a number for which P(c) is to be
evaluated.
Set bn = an and compute
(21) bk = ak + cbk+1 for k = n − 1, n − 2, . . ., 1, 0;
then b0 = P(c). Moreover, if
(22) Q0(x) = bnxn−1 + bn−1xn−2 +· · ·+b3x2 + b2x + b1,
then
(23) P(x) = (x − c)Q0(x) + R0,
where Q0(x) is the quotient polynomial of degree n − 1 and R0 = b0 = P(c) is the
remainder.
Proof. Substituting the right side of equation (22) for Q0(x) and b0 for R0 in equation
(23) yields
P(x) = (x − c)(bnxn−1 + bn−1xn−2 +· · ·+b3x2 + b2x + b1) + b0
= bnxn + (bn−1 − cbn)xn−1 +· · ·+(b2 − cb3)x2
+ (b1 − cb2)x + (b0 − cb1).
(24)
The numbers bk are determined by comparing the coefficients of xk in equations (20)
and (24), as shown in Table 1.1.
The value P(c) = b0 is easily obtained by substituting x = c into equation (22)
and using the fact that R0 = b0:
(25) P(c) = (c − c)Q0(c) + R0 = b0. •
The recursive formula for bk given in (21) is easy to implement with a computer.
A simple algorithm is
b(n) = a(n);
for k = n − 1: −1: 0
b(k) = a(k) + c ∗ b(k + 1);
end
SEC. 1.1 REVIEW OF CALCULUS 11
Table 1.1 Coefficients bk for Horner’s Method
xk Comparing (20) and (24) Solving for bk
xn an =bn bn =an
xn−1 an−1 =bn−1−cbn bn−1 =an−1+cbn
...
...
...
xk ak =bk −cbk+1 bk =ak +cbk+1
...
...
...
x0 a0 =b0 − cb1 b0 =a0 + cb1
Table 1.2 Horner’s Table for the Synthetic Division Process
Input an an−1 an−2 · · · ak · · · a2 a1 a0
c xbn xbn−1 · · · xbk+1 · · · xb3 xb2 xb1
bn bn−1 bn−2 · · · bk · · · b2 b1 b0 = P(c)
Output
When Horner’s method is performed by hand, it is easier to write the coefficients of
P(x) on a line and perform the calculation bk = ak + cbk+1 below ak in a column.
The format for this procedure is illustrated in Table 1.2.
Example 1.9. Use synthetic division (Horner’s method) to find P(3) for the polynomial
P(x) = x5 − 6x4 + 8x3 + 8x2 + 4x − 40.
a5 a4 a3 a2 a1 a0
Input 1 −6 8 8 4 −40
c = 3 3 −9 −3 15 57
1 −3 −1 5 19 17 = P(3) = b0
b5 b4 b3 b2 b1 Output
Therefore, P(3) = 17.
Numerical Methods Using Matlab, 4th Edition, 2004
John H. Mathews and Kurtis K. Fink
ISBN: 0-13-065248-2
Prentice-Hall Inc.
Upper Saddle River, New Jersey, USA
http://vig.prenhall.com/
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