power series, again in greater detail than in
most comparable textbooks. The instructor who
chooses not to cover these sections completely
can omit the less standard topics without loss in
subsequent sections. Chapter 5 is devoted to
real-valued functions
2, a2 D 0, a3 D 14, and anC1 D 9an 23an1 C
15an2; n 3: Show by induction that an D 3 n1 5
n1 C 2, n 1. 17. The Fibonacci numbers fFng1
nD1 are defined by F1 D F2 D 1 and FnC1 D Fn C
Fn1; n 2: Prove by induction that Fn D .1 C p
5/n .1 p 5/n 2 n p 5 ; n 1:
surgically Familial Familial polyposis polyposis
or Gardner or Gardner s syndrome s syndrome
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55,0
or b < a: (G) If a < b and b < c, then a < c. (The
relation < is transitive.) (H) If a < b, then a C c <
b C c for any c, and if 0 < c, then ac < bc. A field
with an order relation satisfying (F)(H) is an
ordered field. Thus, the real numbers form an
orde
thrombosis, ulcer, gangrene Pain in the Pain in
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abscesses abscessespain, fever, pain, fever,
swelling swelling Fistula Fistula-in-ano
connection between anus and connection
between anus and skinchronically
and b1 b2 bn. Let f`1; `2; : : : `ng be a
permutation of f1; 2; : : :; ng, and define Q.`1;
`2; : : : ; `n/ D Xn iD1 .ai b`i / 2 : Show that Q.`1;
`2; : : : ; `n/ Q.1; 2; : : :; n/: 1.3 THE REAL LINE
One of our objectives is to develop rigorously
the conc
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propositions; one for each integer n n0; such
that (a) Pn0 is true I (b) for each integer n n0;
Pn implies PnC1: Then Pn is true for every
integer n n0: Proof For m 1, let Qm be the
proposition defined by Qm D PmCn01. Then Q1
D Pn0 is true by (a). If m 1
we wish to show that D . Since by definition
of F , it suffices to rule out the possibility that <
. We consider two cases. 26 Chapter 1 The Real
Numbers CASE 1. Suppose that < and 62 S.
Then, since S is closed, is not a limit point of S
(Theorem 1.3.5).
Induction) Let P1; P2;. . .; Pn; . . . be
propositions; one for each positive integer; such
that (a) P1 is trueI (b) for each positive integer
n; Pn implies PnC1: Then Pn is true for each
positive integer n: Proof Let M D n n 2 N
and Pn is true : From (a)
numbers r1 and r2 such that a < r1 < r2 < b:
(1.1.10) Let t D r1 C 1 p 2 .r2 r1/: Then t is
irrational (why?) and r1 < t < r2, so a < t < b,
from (1.1.10). Infimum of a Set A set S of real
numbers is bounded below if there is a real
number a such that x a
infinite sequence fSkg1 kD1 of sets are written
as S1 kD1 Sk and T1 kD1 Sk. Example 1.3.2 If F is
the collection of sets S D x < x 1 C ; 0 <
1=2; then [ S S 2 F D x 0 < x 3=2 and \ S
S 2 F D x 1=2 < x 1 : Example 1.3.3 If, for
each positive integer k, th
x x 2 7 (d) S D x j2x C 1j < 5 (e) S D x
.x2 C 1/1 > 1 2 (f) S D x x D rational and x 2
7 6. Prove Theorem 1.1.8. HINT: The set T D x
x 2 S is bounded above if S is bounded below:
Apply property (I) and Theorem 1.1.3 to T: 7. (a)
Show that inf S sup S
y1/ are in f , then y D y1. The set of xs that
occur as first members of f is the of f . If x is in
the domain of f , then the unique y in Y such
that .x; y/ 2 f is the value of f at x, and we write
y D f .x/. The set of all such values, a subset of
Y , i
integer n, by Theorem 1.2.1. The major effort in
an induction proof (after P1, P2, . . . , Pn, . . .
have been formulated) is usually directed
toward showing that Pn implies PnC1. However,
it is important to verify P1, since Pn may imply
PnC1 even if some
functions, the intermediate value theorem,
uniform continuity, and additional properties of
monotonic functions. SECTION 2.3 introduces
the derivative and its geometric interpretation.
Topics covered include the interchange of
differentiation and arithmet
the requestors faculty status. Although this
book is subject to a Creative Commons license,
the solutions manual is not. The author
reserves all rights to the manual. TO BEVERLY
Contents Preface vi Chapter 1 The Real
Numbers 1 1.1 The Real Number System 1
1, xn > xnC1 > p R and xn p R 1 2 n .x0 p R/2
x0 : 9. Find and prove by induction an explicit
formula for an if a1 D 1 and, for n 1, (a) anC1 D
an .n C 1/.2n C 1/ (b) anC1 D 3an .2n C 2/.2n C
3/ (c) anC1 D 2n C 1 n C 1 an (d) anC1 D 1 C 1 n
n an 10. Let a
that there really is an object that satisfies the
definition. (For example, does it make sense to
define the smallest positive real number?) This
observation is particularly appropriate in
connection with the definition of the
supremum of a set. For examp
test CEA blood test Treatment Treatment
surgery to remove primary, surgery to remove
primary, evaluate extent of spread, allow
staging evaluate extent of spread, allow staging
and plan further therapy and plan further
therapy Colon Cancer Colon Cancer-Sta
11 D .1/.1/ D 1 and 1 1 D 1.1/ D .1/1 D 1: Finally,
we define j1j D j 1j D 1: The introduction of 1
and 1, along with the arithmetic and order
relationships defined above, leads to
simplifications in the statements of theorems.
For example, the inequality
properties (a) and (b). Suppose that 1 < 2 and
2 has property (b); thus, if > 0, there is an x0
in S such that x0 > 2 . Then, by taking D 2 1,
we see that there is an x0 in S such that x0 >
2 .2 1/ D 1; so 1 cannot have property (a).
Therefore, there cann
Series 178 4.1 Sequences of Real Numbers 179
4.2 Earlier Topics Revisited With Sequences 195
4.3 Infinite Series of Constants 200 iv Contents
v 4.4 Sequences and Series of Functions 234 4.5
Power Series 257 Chapter 5 Real-Valued
Functions of Several Varia
pelvic pressure pregnancy, increased pelvic
pressure (ascites ascites, tumor), portal
hypertension, , tumor), portal hypertension,
diarrhea diarrhea Found in 3 positions: left
lateral, right Found in 3 positions: left lateral,
right anterior, right poster
integer (d) S D x x D 1=n; n D 1; 2; 3; : : : 12.
Prove: A limit point of a set S is either an
interior point or a boundary point of S. 13.
Prove: An isolated point of S is a boundary point
of S c . 14. Prove: (a) A boundary point of a set S
is either a l
unions of pairs of these sets are S [ T D S; S [ U
D S; and T [ U D S; and their intersections are S \
T D T; S \ U D U; and T \ U D ;: Section 1.3 The
Real Line 21 Also, S U D T and S T D U: Every set
S contains the empty set ;, for to say that ; is
not
an D 1 n : (1.2.5) This is given for n D 1. If we
assume it is true for some n, substituting it into
(1.2.4) yields anC1 D 1 n C 1 1 n D 1 .n C 1/ ;
which is (1.2.5) with n replaced by n C 1.
Therefore, (1.2.5) is true for every positive
integer n, by The
of b; D sup S if no point of S is to the right of ,
but there is at least one point of S to the right
of any number less than (Figure 1.1.1). (S =
dark line segments) b Figure 1.1.1 Example
1.1.1 If S is the set of negative numbers, then
any nonnegative n
prove (1.2.3) by induction once it has been
conjectured, induction is not the most efficient
way to find sn, which can be obtained quickly
by rewriting (1.2.2) as sn D n C .n 1/ C C 1 and
adding this to (1.2.2) to obtain 2sn D n C 1 C
.n 1/ C 2 C C 1 C n: