Calculus: Analytic Geometry and Calculus, with Vectors

5.4 Theorems about continuous and differentiable functions 319

and setting y = 0 gives

xn.F.I = xn AX-)

f,(xn)

When the method is applied, we start with a first approximation x1, set n = 1

to calculate a second approximation x2, set n = 2 to calculate a third approxima-

tion x3, and so on. To test the Newton method and understanding of it in situa-

tions where computations are not too onerous to be done with a pencil, calculate

x2 when

(a) f (x) = x2 - 2, x3 = 1.4

(b) Ax) = x3 - 20, x1 = 3

(c) f(x) = x4 + x - 20, x1 = 2

(d) f(x) = x4 + 5x - 50, x1 = 2

13 It is not always easy to tell what is obvious and what is not, the funda-
mental difficulty being that some things that have been thought to be "obviously
true" are false. Consider, for example, the "obvious" statement that "each
finite set of numbers contains a greatest element." If, as is usual, the empty
set is considered to be a finite set, the statement is false. Consider, then, the
revised statement "each nonempty finite set S of numbers contains a greatest
element." Is this obviously true? Let n be a positive integer and let the num-
bers be x1, x2, x3,- -. , xn. The only thing we know about these things is that
they are numbers. One possible method of attacking the problem starts with
a comparison of x1 and x2. If x1 < x2, we discard x1 and consider the remaining
set, but if x2 < x1, we discard x2. Instead of employing this "finite mathematics,"
we introduce some "infinite mathematics" that will make us think about least
upper bounds. The fact that xk S I xkl for each k implies that

xk s Ixil + IX2I +... + Ixnl

for each k. Hence the set S has an upper bound, and it follows from Theorem
5.46 that the set has a least upper bound M. If there is a k for which xk = M,
then this xk must be a (or perhaps the) greatest element of S. If we suppose
that there is no k for which xk = M, and hence that xk < M for each k, we can
obtain a contradiction of the hypothesis that S contains only n elements. To do
this, let y1 be an element of S. Then yI < M and there must be an element y2
of S for which yI < y2 < M. The same argument shows that there must be an
element y3 of S for which y2 < ys < M, and so on. We run into a contradiction
of the assumption after we have used n elements of S. This proves that the set
S does contain a greatest element and provides the possibility that schemes for
finding "it" might even work.
13a Remark: To put the following problems and their consequences upon
a rigorous base, we should have a definition of the set S, of positive integers.
This set S, can be defined to be the subset of the set S of positive real numbers
for which the number 1 is the least element in SI; a + b is in S1 whenever a and
b are in Si; b - a is in SI whenever a and b are in S1 and a < b. It follows from
this that if a and A are numbers for which 0 < a < 1, then the interval a 5 x <
a + A can contain at most one integer.