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2.2
Let sn
E?=i l/ diverges.
16-
1
?=i 1/n.
1
S2N~Sn~N+~1
Thus, {SN} is
=
not
+
a
'"
+
1
Series
139
Then
1
2N~N2N "2
Cauchy
sequence.
positive numbers is particularly easy to work with.
sequence of positive numbers, then the series {Yj=i ck)
{cn}
is an increasing sequence, so by Theorem 2.1 (as rewritten in Problem 1) this
sequence converges if and only if it is bounded.
The
sum
of
a
is
For if
series of
a
Proposition 4. Let {c} be a
following assertions are equivalent.
(i)
(ii)
(iii)
of nonnegative
sequence
numbers.
The
ck converges.
is bounded.
i
{Yj= ck}
e>0, there
For each
k
is
an
N such that for all m>N,
Zcfc<
N
=
proof of the equivalence of (i) and (ii) is essentially given in the preced
ing paragraph. Part (iii) is just the Cauchy criterion restated for positive
series (see Problem 11).
The
Examples
17.
=i l/!
converges.
1
1
<
2"_1
n!
and thus for all N,
N
y
1
<
by (2.5).
N-1
1
y <2
For n!
>
2""1 for all
n,
so
140
2.
Notions
of Calculus
18.
i
ta
AVn
)
cos(l/n)/
+
n
For
converges.
11
1
1
<
cos(l/n)
+
n
n
+ 1
n
n
Thus, for all N
V
/I
1
/1
\
1_\
_
cos(l/n)j ,Mn
~
M\n
n
+
n
+ 1
)
1
1
1
_
2+2
'"
+
_
3
1
+
There is
complex
putation
no
such
simple
criterion
as
Proposition
4 for
1
]v-]v+-T<1
series of
arbitrary
(or real) numbers, and the question of convergence as well
of a limit can become extremely subtle. However, if for
as com
a
given
series the series formed of the absolute values converges, the situation is
considerably clarified. Ordinarily we shall discuss the convergence of a
series only in the happy circumstance that the
values converges.
Lef
Proposition 5.
converges,
Yjck
be
{ck}
a
sequence
of complex numbers.
If
\c\
also converges.
Proof. Let t be the sequence of partial
Notice, for m > n
2 ft
corresponding series of absolute
sums
of 2 I ft I and
s
the
partial
sums
of
.
\Sn
I
S
Thus, if {t} is
a
Definition 3.
Cauchy
Let
ft
sequence,
{ck}
absolutely convergent, if
converges, we say
c is
There
are
such
2
<,
things
be
lC*l =tm~t
so
a
also is
{s}.
sequence of complex numbers.
c is
converges. If
|c| diverges, but
c
|c|
conditionally convergent.
as
conditionally convergent
sequences.
In
fact,
Z"=i (-!)"/ converges. But as we have seen in Example 16 the series
Yj?m ! 1/n of absolute values is divergent. It is easy to see that L t (- l)"/n
2.2
Series
141
converges. Let {s} be the sequence of partial sums. Then the
subsequence
{s2n} is decreasing, and bounded below by su and the subsequence {s2n+1}
is increasing, and bounded above by
s2
Thus, both these subsequences
.
Since
converge.
l^n + l
s2n\
~
they have the
<
+ 1
n
same
limit.
It is easy to deduce that the full
sequence also
Here is the proof in a more general case
converges to that common limit.
(known as Leibniz's theorem).
6.
Proposition
that lim c
=
Let
Then
0.
{cn} be a decreasing sequence of positive
(- l)"c converges.
Proof. Let j=2*=i ( l)*ft. We consider the sequences
partial sums separately. The sequence {s2} is decreasing, since
*2(n + l) S2
Similarly,
C2n+1
are
<S2 + i
partial
bounded, for, given
=S2 C2 + i
of
even
and odd
^0
the sequence of odd
these sequences
Sl
C2n + 1
numbers such
sums
{s2+i} is increasing.
Furthermore
any n,
<s2<,s2
so {s2} is bounded below by Si and above by s2
The same is true for the sequence
s' both exist. Furthermore,
fen+i}. Thus, by Theorem 2.1 lim s2 s, lim s2+i
.
=
s'
lim s2 + 1
sequences, of odd
s
=
limit, the
lim
s2n
partial
=
s2)
lim(c2n +0=0, so .$' .$. Since both
and even partial sums converge and have the same
\\m(s2n
sums
=
n-.oo
n-eo
+1
=
=
whole sequence also converges to that limit.
Notice that this argument does not give any hint as to the value of
( l)"/n. Outside of the case of Proposition 4, there is no positive asser
-
tion that
can
be made about
tend to behave very
badly,
as
conditionally convergent series.
following illustrative example
the
Example
19. The sequence
111111
2+2+4+4+4+4+-
In fact,
shows.
they
142
2.
of Calculus
Notions
is the
same
and thus
+ 1-1
1 + 1 H
as
general term is decreasing
conditionally convergent:
i
1
-,0,-,0,-,0,
...,,
2n
4
4
2
partial
sums
'
"
2n
2n
2n
4
n
The sequence of
V
h
I----H
44
4
2
2
1
11
111111
c
Since the
diverges.
to zero, by Leibniz's theorem this series is
times
is
0,
However, we may now
converges to zero.
it
that
no
so
longer converges ! Consider
rearrange terms of the series
we
first add the positive terms
in
each
the same series where
group
and thus
obviously
and then the
negative
terms :
(2.7)
The
sequence of
corresponding
n-1
1111
>
2
0,
,
4
,
2
,
4
partial
u,
Thus, there is
.
a
.
.
,
,
2n
.
.
.
1_
,
2n
,
n
u,
sums
.
.
is
.
subsequence: {$, \, ...}
and another:
{0,0, ...}
so
We leave to the student
we cannot have convergence of (2.7).
it
to
show
that
can
be
further
(Exercise 9)
rearranged so that it once
again
converges, but this time to
one
!
absolutely convergent series. We may
please. If we arrive at a limit, it is
attempt
an
the sum.
In fact, if
is
absolutely convergent series we may sum first
c
the positive terms, and then the negative terms; and
c is the sum of these
two sums.
We conclude this section with the proof of these facts.
No such foolishness holds for
to sum the series in any way we
2.2
7. Let
Proposition
c be
absolutely convergent
series
143
of real numbers.
Let
(i)
\ck
+
k
\0
Then the
sums
if
ck
if
c,<0
ck
>
0
_
_
k
f-c
if
c
j
if
c,>0
0
ck~ converge and YJck
,
Let g be
(ii) (Rearrangement.)
onto
an
Series
a one-to-one
Then
the positive integers.
(iii) (Regrouping.)
=
cg(n)
Let h be any
=
YJck
<
0
Z CH
mapping of the positive integers
c
~
.
strictly increasing function from
P into P.
Let
Hn)
Z
d=
k=
Then
</
=
c
Proof.
(i) Since the
2
>
Iftl
*=i
{2*=i Iftl)
is bounded by absolute convergence, and
2ft+.2ft"
the sequences
*=i
2*=i
say s, t
.
.
sequence
*=i
verge to,
Let e > 0
ck
h(n-l)
ft+,
2*=i
c*~
respectively, by
Then there
2ft+-*
*=i
are
are
also increasing and bounded. Thus they con
s
t.
We have to show that 2 ft
Theorem 2.1.
M, N2 such that for
=
n
>
M,
<
2'
and for n>N2,
n
2ft"Then for
n
>
<-
max(7v*i, N2),
2ft-(j-0
(ii)
Let ^ be
P onto P into
a
a
2ft+- 2
ft"
-fr-0
<
Then g-- is defined and also maps
one-to-one map of P onto P.
For each n, let JV =max(^(l), ...,g(n)).
one-to-one fashion.
144
2.
Notions
of Calculus
Then
n
2ift(i^ 2
*=i
for all n,
iftl
*=i
the series
so
^2
2 ft
iftl
is
absolutely convergent.
Nn
n
2 ftV> < 2
Similarly,
it
c*+ ^ 2 c*+ and
2 ft-<*> ^ 2 c*-
for all , so we have 2 ft+c*> ^ 2 c*+> 2 c<*> ^ 2 ft"- Applying the same reasoning
but reversing the roles of the two series, we obtain the reverse inequalities so that
in fact, 2ft+(*)=2ft+ and 2ft-(*)=2ft"- Thus, by part (i) we obtained the
desired equality; that is, 2 ft<*>
2 ft
Part (iii) is actually true for any convergent series.
increasing function h be given. Notice that h(ri) ;>
=
is
an
N such that
for all
n
2ft-
c
> N.
Thus, for
n>N,
<
24.= 2 2
*=1
and
h(n)
>
N,
so
cj=
*=lj=h(*-l)
that
ll<n)
2dn-c
<
*=i
2CJ~C
<
J=l
EXERCISES
7. Show that
n+lf
=i\n
converges.
8. What is
f
"=i
(-D"
a
where
a2
=
2",
2+i =3"?
2
y=i
cj
Let
n
2 ft
for all
=
n.
c, and the
For
e
>
strictly
0, there
2.3
Tests for
145
Convergence
9. Rearrange the series
i_i+i-i+i-1+...+i-i+...-i+
so
4
2
2
it has the
4
4
4
2
2
2
sum one.
10. Can the series
( 1)"/ be rearranged
2
so as
to have
sum
10,000?
PROBLEMS
2 z-
9. Suppose
10.
=
z
Suppose
(a) 2 z and lim
(b) 2
z
and
w
w
exist.
exist-
2w
2 z converges
and
11. Prove that
2
=
Snow tnat
w-
2 (z +
w)
=
z
+
w.
2 z w exist ?
2 z" w" ex'st ?
Does
Does
if and only if for all
s
>0, there exists
an
N > 0 such that
2
*
<
Deduce that
2.3
Tests for
for all
e
n
>N
Proposition 4(iii)
is true.
Convergence
Since the theory of series is so
unwieldy, there has developed
convergence which
have already given
are more or
some
important and the definition of convergence
a large collection of tests (or criteria) for
less easy to apply in the relevant cases. We
criteria for convergence.
> 0,
if for every
c converges if and only
(1) Cauchy criterion:
>
>
N.
n
for all m
+ cm\ <
is an integer N such that |c+1 +
then
to
(-D"cn converges.
decreases
zero,
If
the
sequence {c}
(2)
if and only
converges
is
c
If
the
nonnegative,
sequence {c}
(3)
is bounded.
sequence {Yj=i ck} of partial sums
there
if the
for absolute
The last condition, which can be considered as a condition
the
basic one.
is
criterion which
convergence, gives rise to the following
one (if we
The idea is to compare a given series with a known convergent
we
one (if
suspect that it
suspect that it converges) or to a known divergent
diverges).
146
2.
Notions
of Calculus
Z I/"'
converges,
Examples
20.
noticed that 1/n!
1
/
have
seen
and since
<2~n+1,
in
Example
17.
There,
is convergent,
Z2_B+1
we
so
is
For
Zl/!.
JV
as we
N
\
1
a>
1
Z^I ^r<Zr
n=\l
=l\n!/
forallN
n=lZ.
21.
>
sin
-I
For if
diverges.
is small
x
enough,
sin
x >
x/2.
Thus, there is
an
N such that if n>N,
sin(-)
\nj
>:
2n
and thus for m>N,
fi
\n/
But
we can
enough.
2jv+in
\n/
=i
make the last
Thus,
sum as
Z=i sin(5/n)
large
as we
please by taking
is not bounded, and
so
m
it is not
large
con
vergent.
22.
z
*b (1 + 0"
is
absolutely convergent.
m
V
,tb|i
1
=
+
i|"
Y1
1
r=-
nh(^2y
The idea behind these
For
|1
+
i\
=
J2,
so
for any m,
i-
< oo
examples
since J
v 2
>
1
is contained in the
following
theorem.
2.3
Theorem 2.3.
If
there is
a
(Comparison Test) Let {c}
positive number
K and
Tests for
be
a
sequence of complex numbers.
N, and
an
147
Convergence
a
sequence
{p} of positive
numbers such that
forn>N,
(i)\cn\<KPn,
CO
(ii) Z
Pn < >
n=l
then
Zc
converges
If instead,
we
absolutely.
have
forn>N,
W \c\>KPn,
oo
Z Pn
11=1
(ii)'
then
Z k|
>
=
diverges.
In the first
Proof.
2
iftl
=
2
case
k
=
2
iftl
=
2
2 iftl
iftl +
^
t=w + i
s=i
which is unbounded
as n
->
+
K=l
W+l
In the second case, the sequence of
kti
partial
2 ifti<2ifti
iftl +
k=l
k=l
the sequence of
partial
2 iftl
t=i
sums
*2/><co
sums
+
is bounded.
I
is unbounded.
p-
t=w + i
oo.
Examples
23.
V^=0z7n!
an
integer
so
that
\z\N+"
(N
+
Since
a
N
so
z.
converges absolutely for any complex
all
+
for
n)\
n, (N
that JV> 2|z|. Then,
Choose
>
(2|z|)n,
Jz\N
2"
n)\~
converges, so does
result we obtain lim
1/2"
corollary
have been derived
directly).
|z|7" ! by the comparison test. As
z7n! 0 for all z (this however could
=
148
2.
of Calculus
Notions
24. Z nkz" converges absolutely for all z,
and otherwise diverges. If |z| > 1, then
|z|
<
1 and all
limn*z#0,
integers k;
so
the series
hardly converge. Now suppose |z| < 1. We want to prove the
convergence by comparison with the geometric series, so we must
can
account for the effect of the coefficients
as n
->
oo, thus also
greater than 1
(n
1)* <
+
r-^-
-
(n
.
(n
n
+
,.t
if
<
1
->
Then there is
.
or
s
l)/nk
+
an
nk.
Note that
(Exercise 13).
Let
N such that for all
s
(n
+
l)/n
->
1
be any number
n>N,
k
snk
Thus, by induction we can conclude that, for all n > 0, (N + ri)k < s"Nk.
Thus, (N + n)k\z\N+n <(s\z\)"Nk\z\N. We should choose J<l/|z|,
Z (s\ z I)" < - With the choice then
apply the comparison test to obtain the
that
so
we can
series
Z
25.
of
s:
1
< s <
l/\z\,
convergence of
our
nkz".
Zw'z" diverges
for all z#0. We have seen in Example 2
c, lim c7n!
0, or, replacing c by z~l,
complex number
that for any
=
n->oo
lim
l/n!z"
This
0.
=
precludes
the
possibility
that limn!z"
=
0,
B-*00
the
so
26.
given
series cannot converge.
Z 1/n2
converges.
proof of this,
at
.i\n
1/
n
+
present
\n
n
+
1/
converges to 1
1
n
But
.
1
n
1
+ 1
n(n
thus
00
I
1
A n(n + 1)
=
In
1
+
1)
a
rely
N + 1
Thus, the series
^
we
later section
on a
tricky
we
shall
observation.
give
another
2.3
Now, 2n2
1
n2
>
+
n
=
n(n
+
Tests for
Convergence
149
1), thus
1
s<
n2
n(n
+
1)
by comparison
so
Z llnil+l)
27.
that ke
large
1
m(l
>
2.
Z l/2
converges for any e > 0. Let k be
Then, for any n ; if m ^ nk,
1
+
:
e)
n*(l+<0
Between nk and
there is
an
an
integer
so
1
<
-
m'"
also converges.
<
=
(n
nk+2
+
If there
(n
+
if
,(n
+
If
are
-
nk integers.
Since
1
n0 such that for
n >
n0
<,
2nk, or (n
+
if
-
nk
<
nk.
Thus,
n*
1
(BV)k
1
m=ii+im<1+-nk+2~n2
Well,
now we can
show that the sequence of
partial
sums
Ui 7^]
is
bounded, for
Nk
Z
n=l
nok
J
(l+e)
n
^
11
Z1
=
N"
1
7(1+7)
"
JV
+
J
h
_
=nofc+l
(B+l)k
n=no m=n+ 1
"
(1+e)
1
*
*o' + Z r2^o'I + Z^2<00
n=no
"
"
Now a special kind of a series is a power series: the geometric series, and
the series in Examples 24 and 25 are such series. A power series is a series
150
Notions
2.
of Calculus
of the form
00
Zfl-z*
B=0
series has the property that if it converges for some z0 then it con
it diverges for some zu then it
verges for all z such that |z| < |z0|, and if
diverges for all z such that \z\ > IzJ. Thus, the geometric series diverges
for |z| < 1 ; the series
> 1 and
z"/n! converges for all z,
for
Such
a
\z\
and
,
Z
converges
This general property of power series is
converges for no z.
deduced from the comparison test. We make the following somewhat
Z!z"
easily
stronger statement.
Proposition
8.
Let
{c}
be
a
sequence
of complex numbers.
(i) If {\c \t"} is bounded for some positive number t, then Z cz" converges
absolutely for all z, \z\ < t.
(ii) If{\c \t"} is unbounded, then Z cnz" diverges for all z, \z\ > t.
Proof.
(i) Suppose
M>
|c|
t
"
for all
n.
Let
|ftz"l<:|ft|f"(Y<m(^\"
z
be such that |z| <
for all
t.
Then
n
|z|/f<l, 2(lzl/0"<> so by the comparison test the series 2CZ"
converges absolutely.
(ii) If {|c| ?"} is unbounded so is {cz"} for all z, |z| > t. Thus, we cannot have
and since
lim cnz"
0,
=
so
2
Definition 4.
ft z" cannot converge.
Let
series associated to
{c} be a sequence of complex numbers.
{c} is the series Zb*=o anz"- The radius of
The power
convergence
of the power series is the least upper bound R of all real numbers f such that
the sequence {| c |f"} is bounded.
According
to
Proposition 8 the series Zb=o anz" converges for z inside
jR(|z| < R), and diverges for z outside that disk (see
the disk of radius
Problem
12).
Examples
has radius of convergence one. For if t > 1, then
is unbounded, and if t < 1, t"/n -* 0. Notice that we can make
clear assertion for z on the unit circle, since Zb=o 0)7W diverges,
28.
Zb=o z7n
{f7n}
no
but
Zb=o ( 1)7"
converges.
2.3
29. If
{c}
Tests for
Convergence
151
is bounded, but does not tend to zero, Zb*=o cbz" nas
For clearly {cf"} is bounded for f < 1,
radius of convergence one.
and unbounded for t > 1
.
There
are
two final tests of
cn
I)1'"
fr
< r
SOme r <
Z cb converges absolutely.
then
(|
then
c
|)1/n
>
for
R
If there is
< r <
Z cb
are as
follows :
some
1
If there
R
>
are
infinitely
many
n
such that
1
R
>
an r <
1 such that
eventually
1
absolutely.
converges
>
then
These
Z cn diverges.
Ratio test.
then
importance.
If eventually
Root test.
(I
some
for
1
If
infinitely
many
n
Z c diverges.
by comparison with the geometric series. We leave
it to the student to derive these tests (Problem 13). Let us here indicate why
the convergence assertions are true. Suppose (| c |)1/n < r < 1, for n large
enough (say n > N). Then \cH\ < r" eventually, so the partial sums Z kl
These
are
are
both derived
bounded
by
1
n
Z0 Kl
=
+ 7
t
by comparison with the geometric series.
< r
for
n >
N
As for the ratio test, suppose
152
Then
2.
we
Notions
of Calculus
have
kjy+il <r\cN\
|Cjv+2l<r|cN+1|<r2|cN|
\cN+3\<r3\cN\
\cN+k\
r"\cN\
<
by induction.
Tj=0 Id
Thus,
<
Z"=o k.1
+
\cM\
Z rk <
since
oo
r <
1.
EXERCISES
following series converge?
11. Which of the
ns + 8
v
(a)
2sin(i).
(e)
(b)
2sin(i).
(f)
v
(0
2^(1).
(g)
2-^2"
(d)
2
(h)
27^r,x,,,x>0.
(2n)\
(i)
2
(j)
2(-D"sini.
tenf-)
sin(^
\n}
n)
-
.
^4n6 + n*'
^
n3 + n2 + n+l
n* + n5 + n6 + 7
'
n*
x"' k
a
positive integer,
0 <
(1)
2(-i)n
^
+ 1
1
.
+ -LJ\.
+
z(L
(n+1)2
(+2)2/
\n2 ^
(n)
2(-\
"
(-l>- n
<
(m)
n
(k)
x
n
^-7 + -Ul
+ 2/
+ 1
n
-
(+l)2
12.
Verify directly that
13.
Suppose lim
c
=
c.
lim
z"\n\
=
0 for every
Then for any
z.
integer k,
lim c*
=
c*.
2.4
14. Find the disk of convergence of the
2Z-
00
<b>
.?
following
(f)
=
Convergence
153
power series.
2>!z".
n
-
in R"
=
0
(g)
2
(h)
2d+n)z".
(2n)2
z"
(C)
Jo^-
(d)
.S,^
n\
/
00
z
2 z"
\"
?o(2nj-
(e)
(j>
2(1+*)"
PROBLEMS
12. Let {c} be a sequence of complex numbers, and let R be the radius
of convergence of the power series 2 ft*"- Show that
2 ft*" converges
absolutely for |z| < R, 2 ftz" diverges for |z| > R.
13. Derive the convergence and
divergence
assertions of the root and
ratio tests.
2.4
Convergence
in R"
The notion of convergence of a sequence of vectors is easy to conceive,
a vector in R" is
just an n-tuple of real numbers. Thus, a sequence of
vectors is an n-tuple of real sequences, and the question of convergence of
since
the vector sequence is just that of the simultaneous convergence of those n
real sequences. We might also directly paraphrase Definition 2 of conver
gence,
using the
possible notions
Definition 5.
notion of distance in R" discussed in
are
in fact the
Let
{yk}
be
Chapter
1.
These two
same.
a
sequence of vectors in R".
The sequence
converges if there is a vector v e R" such that to every positive number > 0
there corresponds an integer K such that || yk
v|| < e for k > K. We write
lim yk
v if {yk} converges to v.
-
=
ft-* oo
154
Thus, lim
general
in Section 2.1
=
\ck
c\
-
yk
can
;'
1,
=
we
.
when
=
said that
0,
->
Now, if
2.
we
precisely that lim || yk
term yk and
way it sounds like
put this
n
v means
=
yk
between the
when
of Calculus
Notions
2.
.
vkJ
,
v
-
zero as
||
=
0 ; that is, the distance
When
k becomes infinite.
just the notion we have in mind. Recalling that
a complex sequence {ck} converges to c precisely
that this coincides with the above definition when
write out the sequence yk of vectors in R" as an n-tuple
(2.8)
(vk\...,vk-)
We
n.
->
tends to
we see
we
view the
.
v
vJ for
given sequence as the n real sequences {vkJ}, where
v precisely
now verify the fact mentioned above, that yk
in
fact
the case
is
that
2
that
Notice
all j.
Proposition
->
ofR2.
Proposition 9. The
if and only if lim vkJ
=
k-*
If
Proof.
w
sequence (2.8) converges to the vector
v] for allj.
v
=
(v1, ...,v")
oo
=
(w1,
.
.
.
,
vc")
is
a
vector in
R", then by definition
I|W||=(2(H'')2)1'2
Then, in particular
|ftJ-tfJl<l|v*-v||
j
(2.9)
\,...,n
that v -> v. Then, given e > 0, there is a
vJ\
Thus, by Equation (2.9) for each j, \vkJ
precisely that lim vkJ v1.
Suppose
now
for k 5: K.
means
=
-
K such that ||v*
<
e
for k ^ K.
v
||
< e
But this
=
k->oo
v')2 ->0 for ally,
Conversely, if vkJ -+vJ for ally, then (vkJ
But then, by Definition 5, v*-^v.
v ||-^-0 as k -*oo.
llv*
so
[2 OV
02]"2
=
precisely the same way we can verify that if the sequence of vectors (2.8)
a Cauchy criterion so do each of the real sequences {vkJ}, and thus
are convergent.
Hence, by Proposition 9 the sequence of vectors {yk} also
so
we
have a Cauchy criterion for vector sequences also. This
converges,
fact, as well as some basic algebraic properties of convergence of vectors is
easily verifiable. Accordingly, we make these assertions, leaving the proofs
In
satisfies
to the reader.
Proposition 10. (Cauchy Criterion) Let {yk} be a sequence of vectors in R".
whenever
Suppose to every e > 0 there corresponds a K such that \\ vr
vs || <
both r,s>K. Then the sequence {yk} is convergent.
2.4
Proposition 11. Suppose lim yk
{wj are sequences of vectors in R",
y, lim wk
=
and
{Ck}
Convergence
w, lim ck
=
is
a
sequence
=
in R"
c, where
155
{vj,
of real numbers.
Then
(i)
(ii)
(iii)
lim(vfc + yyk)
lim<vt,wfc>
lim ckyk
=
=
+ w,
v
<v, w>,
cv.
Example
30. Let
find
point of a given plane in R3 which is closest to
plane is given by the equation <x, a> c for fixed
Let m =g.l.b. {||x||; <x, a>
a, c.
c). Choose a sequence {x}
on the plane such that ||x|| ->m.
We shall show that {x} actually
the
us
origin.
a
A
=
=
converges.
||x
We
xj|2
-
Now,
=
||x||2
+
||xj|2
2
-
(2.10)
<x, xm>
estimate the last term by using the fact that the midpoint
+ xm) between x and xm must also be on the given plane.
can
i(x
2(xb + 0
+
4
^^
4
+ ^
2
<x xm>
,
Thus,
-2<x,xm>< ||x||2+ ||xj|2-4m2
and
Combining (2.10)
l|x
-
xm||2
<
2(||x||2
Now, since ||x||
n,m>n0,
then
||x
-
we
+
->m,
have
(2.11),
we
||xj|2
if
>
||x||
(2.11)
find that
2m2)
-
0 is
(2.12)
given, there is an n0 such that for
||xj| <m + e. Inequality (2.12)
< m + s,
gives
xj|2
<
2((m
+
)2
+
(m
+
b)2
-
2m2)
<
4ms + 2e2
=
fi(4w
+
2e)
please by choosing e small. Thus if
n,m
large enough, ||x-xj| is small, so the sequence {x} is
lim x,then ||x||
m,
lim||x||
and
thus convergent. Ifx
Cauchy,
so x is the closest point on the plane to the origin.
This
can
be made
as
small
as we
are
=
=
=
156
2.
Let
above
Notions
pause for
example, for
us
of Calculus
a
moment to consider the reasons,
studying
lem of calculus is to find
an
as
illustrated by the
the convergence of vectors. The central
object, usually considered as a point in a
prob
given
specified properties (i.e., the maximum
given function, or a zero of a function). At least, the theoretical aspect
of the problem is to prove the existence of a point with such and such
properties. Our technique for doing this is to use the desired properties to
develop a sequence of approximations; our hope is that the approximations
will converge; and that the limit will have the desired properties. It is thus
essential to be able to discuss the question of convergence without already
knowing the limit. Hence, for example, we have- the Cauchy criterion.
Further, we will need techniques, or criteria, to apply to the given properties
in order to be able to extract the desired Cauchy sequence of approximation.
For example, we will want to know : (a) If we have a convergent sequence
of points having a property, does the limit have that property ? (b) If we
have a sequence of points having a property, does the sequence converge?
These questions lead
or, at least does it have a convergent subsequence?
collection of points, which has certain
of a
us
to the reconsideration of the closed sets introduced in Section 1.11.
Recall that
precisely,
a
closed set in R" is
S is closed if and
a
set whose
complement
only if corresponding
to every v
is open. More
$ S, there is an
0 such that any vector within e of v is also not in S. In particular, if S
closed set, and v ^ S, then v cannot be the limit of a sequence of vectors
in S. To put it positively, a closed set contains the limits of all convergent
>
is
a
sequences it contains.
Proposition 12.
equivalent:
This is in fact
Let S be
a
a
Suppose
v
e
=
v.
This is
is
a v
contained in S, then lim yk
nonsense
such that ||v
since
Then there is
vector in S which is within
l//j there
assertions
e
are
S.
Let {yk} be a sequence contained in S and suppose
$ S, since S is closed, there is an e > 0 such that no vector
Thus, we must have v e S.
Suppose now S is not closed.
a
following
S is closed.
it converges to v. If
in S gets within e of
there is
criterion for closedness:
The
in R".
set
(i) S is closed.
(ii) If {yk} is a convergent sequence
Proof.
defining
-
e
of
v
is the limit of
a v
v.
v|| <, l/ and
in
v
a
sequence in S.
S such that for every
e
>0
particular, for each n, taking
e S.
Thus, v ->v so (ii) does
not hold for S.
We are now in a position to state our last basic
consequence of the funda
mental existence axiom for the real number system. This is that
every
bounded sequence in R" has a convergent subsequence. It is
easy to derive
2.4
Convergence
in R"
157
this from the Cauchy criterion, itself an assertion of existence. Let us
illustrate the situation in R2
Suppose {ck} is a sequence of complex numbers
which is bounded ; that is, it remains in some fixed square S0 of side length K.
Cut that square into four equal squares. At least one of these new squares
.
has
equal pieces
of the {ck} ;
way
many of the
{ck} ; let St be one such square. Cut St into four
and let S2 be one of these new squares which has infinitely many
now do the same with S2 and so on (see Figure 2.4).
In this
infinitely
obtain
we
a
sequence of squares
{S}
with the
properties:
Sm=>S+u
(i)
(ii)
(iii)
length of S is K/2n,
S has infinitely many of the {ck}.
side
Now that this is done, we can, for each integer n, select a k(n) such that
and {ckW} forms a subsequence of {ck}. (For this we need to
ck(n)eS,
know that
than any
For let
have
S contains infinitely many {ck}, so that we can choose k(n) greater
previously chosen index.) Now, {ct(B)} is a Cauchy sequence.
e >
0, and choose N
cm,cHm)eSN,
IC*(b)
C*(Bl)l
<
so
that
>
Kyj2/2N.
Then, if
n,
m >
N,
we
so
(K\2
\2NJ
(K\2 Kj2 <
2N
\2N)
~
Since the sequence {ct(B)} is a Cauchy sequence, by Proposition 10 it con
idea of the
verges, and the argument for R2 is concluded. This is the basic
verification of
Figure 2.4
158
2.
of Calculus
Notions
Theorem 2.4.
subsequence
Every
sequence in
which converges to
a
closed and bounded set S in R" has
a
a
point ofS.
Suppose that S is closed and bounded and {yk} is a sequence in S. We
Cauchy subsequence. Since the sequence is bounded, it is contained in
some ball B(0, R).
This ball can be covered by finitely many balls of radius 1.
Since the {yk} are infinite, there is one such ball which contains infinitely many. Call
it Bi, and let ykil) e Bi. Bi can be covered by finitely many balls of radius i. Let
B2 be one such which contains infinitely many of the {vj and let v*(2) e B2 with
k(2) > k(l).
Continuing in this way we obtain a sequence {B} of balls, a subsequence {vt<)} of
{yk} such that (i) B has radius 1/n, (ii) vJ(n) e B, (iii) B => B+i. Then {vMn)} is a
Cauchy sequence, for if n, m> N, yk{n) and vt(m) e BN which has radius 1/N, so
Proof.
shall find
a
2
\\ykw
vt(m)||
for all n,
<
m
>N
N
By Proposition 10 there is a v such that vt() - v as n -^ oo. Since S is
and {v*()} 6 S, we also have v e S, so the theorem is proven.
closed set,
a
Example
unit
31. The
sphere
S
=
{xeRn: \\x\\
=
1} is closed.
For
if
Now suppose
x, then certainly ||x|| -> ||x||, so if x e S, so is x.
x
T is a linear transformation of R" to R".
We want to know if there
->
is
an
xeS at which
||rx|| is a maximum. First of all,
||7x|| with x e S is bounded. Let A
representing T, and M max \a/\. Then
numbers of the form
the matrix
Tx
=
T(x\
the set of
=
(a/)
be
=
.
.
.
,
x")
=
(Z a/xJ,
.
.
.
,
Z af-x*)
so
II Tx||
=
<
Thus,
[(Z *y V)2
\nM2 \\x\2
+
+
+
+
(Z flyV)2]1'2
nM2
nM is the desired bound.
||x||2]1/2
<
(2.13)
nM
||x||
By the least upper bound axiom then,
sup{||7x|| : xeS} exists, and there is a sequence {x} c S such
that ||7x|| ->w. According to the above theorem there is a sub
Since ||7x|| ->w, we also
sequence {y} which converges, say to y.
have ||7y|| ->m, and by (2.13), in fact ||7y||
m.
lim||Ty||
m
=
=
=
2.5
Continuity
159
PROBLEMS
14. Prove
Proposition 10.
Proposition 11.
16. Let n be a plane in R3, and suppose x0 is the
point on n which is
closest to the origin. Show that if x e Tl, then x0 is
orthogonal to x x0
(Hint : If not, then one of x
x0 x + x0 is closer to the origin than
x0.)
17. Find the point on the plane given by the
equation <x, (1, 1, 1)>
3
which is closest to the origin.
18. Find the point on the plane <x, (1,0, 1)>=2 which is closest to
-0, 1, 1).
15. Prove
-
.
,
=
19. Let L be
range of L
are
a
linear function from R" to Rm
20. Let L: RP-+ R be
also lim
Show that the kernel and
both closed.
linear function.
a
Show that if lim x
=
x, then
L(x) =L(x).
21. Let
v0
be
a
vector in
R", and n the
set of
x
such that <x, v0>
=
c.
Show that IT is closed.
22. Show that for any v0
e
R" and
r
>
0,
{vefl": ||v-v0||<r}
is closed.
23. Show that yk
max
v'\
\vk
->
v
in R" if and only if
->0
lslsn
2.5
Continuity
We turn now to the consideration of functions from subsets of R" to Rm.
The basic notion of analysis being that of convergence, the fundamental
class of functions will consist of those which respect convergence ; that is,
those which take convergent sequences into convergent sequences.
are continuous functions.
Definition 6.
values in Rm.
Let S be
/is
a
set
continuous
on
These
in R", and /a function defined on S, taking
S if whenever yk -> v with vk e S, all k,veS,
then/(v,)-/(v).
We shall be concerned most
a
given point.
usually
For this purpose
we
with the local
study
of
a
function
make this additional definition.
near
2.
160
of Calculus
Notions
said to be continuous at v0
v->v0
will be
/ from a set in R", taking values in Rm,
of
v0 and
e R" if /is defined in a neighborhood
A function
Definition 7.
implies /(v) ->/(v0).
Examples
|| v
v||
-
-+0
so
that
||v||
For
continuous.
is
f:R"-*R,f(y)= \\y\\
32.
if
v->-v,
then
||v|| since
->
| ||v||-||v|| |<||v-v||
f:C-*C, f(z)
0, so that
z|
|z
33.
=
34. A linear function
also z
on
continuous:
is
z
=
->
-
->
z^z
implies
|z-z|
z.
R" is continuous.
Let
i=l
Then, if yk
'
Z"= i
;f
V
have
v we
-?
since the limit of
->
a
u1,
sum
.
.
.
,
vk"
is the
->
v",
sum
so
that
Z?=i a'
of the limits.
-"
Thus,
/(VftWWthe idea of continuity of a function /is this: as a moving point
to
close
p0 the value /(p) off at p gets close to/(p0). That is, we can
p gets
ensure that /(p) is as close as we please to /(p0) by choosing p sufficiently
criterion for continuity,
s
8
This leads to the so-called
close to p0
Roughly,
,
"
"
-
.
which
we now
give.
Let S be
Proposition 13.
defined on S.
a
subset
ofR", and let f be
an
Rm valued function
Let x0 e X. f is continuous at x0 if and only if, to every e > 0, there
corresponds a5>0 such that \\x x0 1| < 8 implies ||/(x) -/(x0) || < .
(ii) If S is open, f is continuous on S if and only if f is continuous at every
(i)
-
point of S.
Proof,
there is
a
8
>
x0
.
Let x
^ N,
||/(x)
x0
.
0 such that whenever
Since x-*x0, there is
n
->
/(xo)||
an
N such
< e, as
S criterion is true,
we shall show that /
/(x) -*/(x0). Given e > 0,
x is within 8 of x0 we have ||/(x)
/(x0)|| < e.
that n>N implies ||x
x0|| <8. Thus, for
(i) Supposing first that the
is continuous at
e
We have to show
desired.
2.5
Conversely, if the
8
>
0 there is
8
1
an
e
xd
8 criterion is false, then there is
x II < 8 but ||/(x)
-
for which ||x
!
l
1
2
3
n
Continuity
an e0
-
-
161
such that for every
/(x0) ||
> e0
Selecting
.
obtain the corresponding sequence Xi, Xi/2, ..., Xi,, which converges to x0.
But/(xi/n)+>/(x0) since the/(xi/n) are always outside the ball of radius e0 centered
we
at x0.
Part
(ii) is
left
as an
exercise.
Examples
/: R2
35.
/(*>')
-+
R defined
by
=
FT?
is continuous at
For
(0, 0).
5x
1 +
<5|x|<5||(x,^)||
y2
Thus, if
is
given
we can
choose <5
=
e/5.
Then
||(x, y) \\
<
8
implies
5x
<
v2
1 +
58
=
36.
f(x,
y,
z)
y3z
=
1 + x2 + z2
is continuous at (0, 0, 0).
|/(x,
y,
z)
-
f(0, 0, 0)|
Thus for each
|/(x,
y,
>
z)\ <84
We have
y3z
1 +
0 choose 8
=
e.
<
=
x2
=
+
e4
z2
=
e.
\y3z\
Then
<
\\(x,
||(x,
y,
y,
z)
z)\\<8 implies
162
Notions
2.
of Calculus
37.
/(*>y)
(4^4-
=
/(0,0)
(x,y)*(0,0)
=
0
This function is not continuous, since
4x2
f(x,x)
If
we
=
2
=
redefine
2^0
/(0, 0)
=
2, this
new
function is still not continuous,
since
/(0,y)
^=1^2
=
38. We
by
the
can
easily verify
8 criterion.
s
l/(v) -/(w) |
=
|
the
of the linear function
continuity
Z (' ~w')\< ||(a\
by Schwarz's inequality. Thus,
8
IKa1, ...,a") || h. Then || v
if
.
.
.
-
>
application
concerning convergence
study of continuity,
to the
v0
1|
<
a") || || v
,
8
might
w||
given, we can take
implies |/(v) -/(v0)| < .
discussed in
as
-
0 is
"
=
The facts
(2.14)
For
be
previous sections have
expected. In particular,
the assertion that every sequence in a closed bounded set has a convergent
subsequence has profound significance for the behavior of continuous func
tions.
Here is
an
important illustration.
Proposition 14. (Intermediate Value Theorem) Let f be a
function on the interval {x e R: a < x < b}, and suppose that f (a)
Then there is
a c
in the interval such
thatf(c)
=
continuous
<
y
<f(b).
y.
We seek (as in Figure 2.5) not just a point at which the value of /is y,
precisely the first such point c. We must find a way to describe this
point which permits us to use the existence theorem. If x < c we must have
f(x) < y, otherwise the graph of /crosses the line y y between a and c. Thus, c
is a lower bound for the set of x such that f(x) > y. Since c is in that set, it must
be the greatest such lower bound. So if there exists a first c at which /(c)
y, it is
the greatest lower bound of {x e R : a <, x <, b, f(x) ^ y}. We now show that this
point (which exists by the least upper bound property) is the desired c.
Proof.
but
more
=
=
2.5
Let
Continuity
163
c
g.l.b.{x :a<x<b, f(x) > y}. Then c is a limit of a sequence {*} in this
Since y </(*) we must also have y < lim/( jc) =/(c) since
/is continuous.
Now, if /(c) # y, we must have f(c) > y. Again, by continuity, there is a 8 such
=
set.
that if \x
c\
<
8, then
l/W-/(c)|<^^
from which it follows that for all
x
between
c
f(c 8) > y, contradicting the definition of c as a
with/(x) > y. Hence /(c) > y is impossible, so we
Now, the
that
they
most
important
bounded
are
and
c
8, f(x)
> y.
Thus,
lower bound for the set of
must have /(c)
=
x
y.
fact about continuous real-valued functions is
closed and bounded sets.
This follows
easily
on the
/
set S, then, for every positive integer n, there is an x e S such that/(x) > n.
If S is closed and bounded, {x} has a convergent sequence {x(t)}. Let
lim xn{k)
Since / is continuous, /(x0)
x0
lim/(xw) > lim n(k). But
from Theorem 2.4.
=
If,
on
is continuous and not bounded above
say,
=
.
fc-*00
n(k)
->
oo as
k
-*
oo,
so
this is
impossible.
Thus/is
fc->00
bounded
on
S.
What is
it attains its least upper bound. For if m is this least upper bound,
but is not a value of/ then g(x)
(/(x) m)'1 is an unbounded function
more
=
on
S, again
function
on
a
a
contradiction.
To conclude:
such that
/(x1)
/(x2)
=
=
if/ is
a
continuous real-valued
closed and bounded set S in i?", then there
sup{/(x):xeS}
inf{/(x):x6S}
Figure
2.5
are
xux2e S
2.
164
Here
are
Notions
the
of Calculus
proofs in
=
more
general
context.
continuous Rm -valued function
a
Then the set
bounded set S in R".
f(S)
f be
Let
Theorem 2.5.
slightly
a
of values
off
on
the closed and
on
S,
{f(x):xeS}
is closed and bounded.
Suppose ye/(S) and y^yeRm. We must
show that y ef(S). But this is easy. Since y ef(S), there is for each n,xeS
Since S is closed and bounded there is a subsequence {zk} of
such that /(x)
y
Since / is continuous, f(zk) ->/(z). On the other
which
zk^-zeS.
converges,
{*}
lim/(z4) y and
hand, [f(zk)} is a subsequence of {y}, so f(zk) -> y. Thus /(z)
First, f(S) is closed.
Proof.
=
.
=
=
ye/(S).
If f(S) is not bounded, there is for each n an x e S such that ||/(x) || ^ n. But
{x} has a convergent subsequence {z}. Let lim zk z. Then lim/(zt) =/(z).
But {f(zk)} is a subsequence of {/(x)}, so ||/(zt)ll -+ o, which is impossible since
{/(z*)} is convergent. Thus, /(S) must be bounded.
=
In
particular,
suppose /is
a
real-valued function defined
on
the closed and
Then/(S) is bounded, so M= sup{f: tef(S)} exists,
closed, M e/(S). Thus there is an xt e S such that
bounded set S.
since /(S) is
/(Xl)
Similarly,
we
state
=
sup{/(x):xS}
there is
x2 such
an
that/(x2)
=
inf{/(x) :
x e
S}.
This basic fact
as
Theorem 2.6.
on a
and
A continuous
function
attains its maximum and minimum
closed bounded set.
PROBLEMS
24. Let x0
e
Rn.
25. Show that
26. Prove part
a
Show
that/(x)
=
<x, x0> is continuous
linear function L : R"
(ii)
of
27. Show that if / is
-*
on
R".
Rm is continuous.
Proposition 13.
a
continuous real-valued function
on a
closed and
bounded set S, there is an x2 such that/(x2)
g.l.b.{/(x): x e S}.
28. Suppose that /, g are J?m-valued functions continuous at p0 6 R".
If c e R, then also
Show that /+ g and </, gy are also continuous at p0
=
.
cf is continuous
at p0
.
2.6
2.6
Calculus
of One Variable
165
Calculus of One Variable
Theorem 2.6, which asserts that a continuous function attains its maximum
on a closed and bounded
set, is the fundamental theoretical
and minimum
tool of the calculus.
of
We shall
now
give
a
brief review of the fundamentals
to the student's memory.
calculus, leaving the recollection of techniques
We shall
give brief justifications of some of the more basic or special facts.
First of all, we studied in the calculus a limit concept which was more
general than the sequential limit we have been studying. We recall the
definition.
Definition 8.
Suppose / is
{x:0 <\x
We say lim f(x)
x0\ <8 implies
-
First of all, the
one:
lim/(x)
L
=
a
set
x0\ <80}
-
L if and
only if, for all > 0, there is <5 > 0 such that
\f(x) L\ < e.
relationship between the two concepts of limit is an easy
if and only if for every sequence x converging to x0 we
=
\x
real-valued function defined in
a
-
,
x-*xo
have
lim/(x)
=
L.
We
can
thus
rephrase
the notion of
continuity using
B~*00
Definition 8.
/is
continuous at x0 if and
only
if
lim/(x) =f(x0).
x-*xo
Proposition 15.
(i) Suppose f is defined
in I
{x:
=
0
<
|x
x0
Then
\ <8}.
lim/(x)
=
L
x-*xq
if and only if, for
limf(x)
(ii) /// is
every sequence
=
also
in I such that x
{xn}
-*
x0
we
have
L
defined
at
x0,f is
if and only if
continuous at x0
\imf(x) =f(x0)
X-*Xo
Proof. We will prove only (i). The proof of (ii) is the same and is left as a
problem. Suppose first that \im f(x)=L. Let {*} be a sequence such that
X-*XQ
x-+x0.
Given
e
that
|jc-x0|<8.
\x
x0
-
\< 8.
>
0, there is
a
8
> 0
Now since x-*x0,
Thus if
n
>
such that \f(x)
there is
N, \f(x) -L\<e.
an
-
L\
< e
for any
N such that for
Thus, f(xn) ^L.
x
such
n>N,
166
2.
Notions
Now suppose
of Calculus
lim/(x)
=L is false.
Then there is
such that for every 8
an e0
x-*xo
we can
such that |x,-x0| <8but \f(x)-L\ ^e. Consider the sequence
Then |c
xl < l/, so certainly c ->x0
1, J,
,
1/n,
is always outside the interval of width e and center L, so it cannot converge
find
an x
of x's for 8
{c}
But/(c)
-
=
...
.
....
toL.
Definition 9.
e
x0
R.
/is
Let /be
a
real-valued function defined in
an
interval about
differentiable at x0 if the limit
lim/(X +
"
/(Xo)
f
r->0
If it does the limit is called the derivative of /at x0 and is denoted
exists.
en
^
/ C*o)
^
/
^
~ (*o)
01"
If/is differentiable in an interval J and the derivative/' is also differentiable
there, then / is said to be twice differentiable on / and (/')' is the second
derivative off and is denoted by
/"
The
or
dx2
higher derivatives/'",
manner.
.
.
.
,/(n),
...
are
defined
successively
A function which has derivatives of all orders
on
in the obvious
the interval will
infinitely differentiable there. If/ g are n-times differentiable
I,
are/+ g,fg, and cfTor c a real number. If/ is differentiable in an
interval I it is continuous there. If/ is differentiable at a point x0 where it
0. This, together
attains a local maximum (or minimum), then f'(x0)
with Theorem 2.6 gives this basic existence theorem.
be said to be
on
so
=
Theorem 2.7.
interval
[a, b~].
(Mean Value Theorem) Let f be differentiable
a point t; e (a, b) such that
on
the closed
There is
mJ{b)-f(a)
b
(215)
a
Proof. This theorem has a nice geometric interpretation (Figure 2.6). There is a
point (|, /(f)) on the graph y =f(x) at which the tangent line is parallel to the line
through (a, f(a)) and (b, f(b)). Clearly (see Problem 30), we need only verify this
when the latter line is horizontal, that is, f(b) =f(a). In this case, let f0 e [a, b],
2.6
Figure
Calculus
of One Variable
167
2.6
[a, b] be the points at which / attains its maximum and minimum respectively
(Figure 2.7). If either f 0 or f i is interior, then/has a local maximum
so
there,
/'(f) 0 for the appropriate f. If this is false, then {f0 f 1} are the points
{a, b), so /(a) =f(b) is at once the maximum and minimum of/. Thus, /is constant
on [a, b], so /' is identically zero and we can choose any point for our f
fi
e
on
the interval
=
,
.
Now suppose that / is a differentiable function defined on the interval
[a, b], and g is a function defined on the range of/ and differentiable there.
Then the composed function h
g f, defined by
=
h(x)=g(f(x))
is also differentiable
h(x)-h(x0)
X
~
~
Xn
on
a, b.
For if x0
e
\a, b], then
g(f(x))-g(f(x0)) f(x)-f(x0)
f(x)-f(x0)
(W(*,.))
:2_
(.))
Figure 2.7
(2.16)
168
2.
Taking
the limit
Notions
Hm
X
at/(x0),
g(f(x))-g(f(x0))
f(x) f(x0)
Hm
f(x)-f(x0)
~
X0
the limit
implies /(x) ->f(x0)),
have (since x->x0
we
Hm
=
f(x)->f(xo)
the right exist
on
so
both sides,
h(x)-h(x0)
x->x0
The limits
on
of Calculus
since/is differentiable at x0
,
X
X0
and g is differentiable
Thus h is differentiable and
the left exists.
on
x-+x0
we
obtain
the chain rule:
h'(x0)
that
(Notice
(g of)'(x0)
=
if/(x) =f(x0),
However, that
If /is
exists
a
f(x)
fg(y)
g
we
say
us
a
case can
g'(f(x0))f'(x0)
(2.16) is invalid and the proof breaks down.
separately.)
interval [a, b] to the interval [a, j8] and there
[a, b~] such that
then
be treated
function from the
function g:
a
=
[a, /?]
->
=
x
for all
=
y
for all y
x e
e
that/is invertible and g is
condition under which
a
[a, b]
[a, 0]
its inverse.
The
mean
value theorem
differentiable function is invertible.
/has an inverse, it must be
this will be guaranteed if/' is never
for the invertibility of/
function
one-to-one.
zero.
From
(2.14)
we see
gives
If
a
that
This is the sufficient condition
Theorem 2.8.
Suppose thatf is a continuously differentiable function defined
[a, b], andfi is never zero. Letf(a) a andf(b) p\ There
is a continuously differentiable function g defined on the interval between
a and ft such that
on
the interval
3(f(x))
=
=
x
and
g'(f(x))
=
-
f(x)
Proof, f is one-to-one. For if a < ai
a f between ai and bi such that
<
for all
=
x
bi ^ b, there is, by the
mean
value
theorem
f(bi)-f(ai)=f'()(bi-ai)*0
Thus
by hypothesis.
between
a
and
j8
f(bi) ^/(ai). By the intermediate value theorem every v
by /. Now we can define g as follows : let g(y) be that
is attained
2.6
x
such that
f(x)=y.
Clearly, g(f(x))
=
Calculus of One Variable
and
x
f(g(y))=y.
169
Now g is differ
entiable:
g(y)-g(yo)
..
hm
x-x0
..
=
y-yo
-ro
lim
n-yo
-
f(x0)
l
hm
=
-
f(x)
x-.xo
[f(x)
-
A further fundamental fact to be drawn from the
determined,
this: A function is
Theorem 2.9.
and that f'(x)
=
Let h
by
=/ g.
h(c)
-
By hypothesis h'(x)
[a, b], there is a f a
=
c e
,
0 for all
<
f<c
This for all
on
0,
h(c)
so
a
x e
[a, b].
By the
mean
h(a)
Now, given any real-valued function / defined
=
value theorem is
such that
so
'(f )
differs from g by
f'(x0)
its derivative.
[a, b],
But
mean
X0
+ C
value theorem, for any
=
constant,
-
Suppose that f, g are differentiable on the interval [a, b]
g'(x) for all x e [a, b]. Then there is a constant C such that
f(x)=g(x)
Proof.
up to a
1
f(Xo)]lx
=
h(a).
constant,
as
c e
is constant and thus /
desired.
interval J,
we
consider
=/. By Theorem
2.9, any two such functions differ by a constant; thus by specifying the value
of such an/at any point it is completely determined. We denote by Jj /
F(x) that function (if it exists) such that F(a) 0 and F'(x) =f(x) for
all x e fa, b]. j* /is called the indefinite integral of/. Every continuous
function has an indefinite integral, which is given by the process of Riemann
integration which we now describe.
Let /be a bounded function defined on the interval J. A partition P of I
those differentiable functions F defined
on
I such that F'
=
=
< a such that
sequence of points a0 < ay <
the approxi
to
/
[a0 a]. We now construct two sums, corresponding
2.8
:
in
mations to the area under the graph off given
Figure
consists of
an
increasing
=
,
S(P,/)=
a(P,f)=
J>(fl|-f-i)
Zwi(fli-ai-i)
i=l
170
2.
Notions
of Calculus
Figure
where
Mt,mf
[fl,-!, a,].
/ is
the maximum and minimum values
are
Definition 10.
interval.
Let
=
be
off
on
the interval
bounded real-valued function defined
a
on
the
integrable if
(2.17)
suPff(F,/)
p
p
(i.e., if we
we
/
Riemann
infE(P,/)
2.8
can
please).
the interval
find
partitions
for which the two
sums
2 and
o are as
close
as
In this case the common value is called the definite integral of/ over
/, and denoted
j"7/.
If/ g are integrable on the interval J, then so is/ + g and cf, ceR. Further
lif+li9> \icf=c\if- If/is integrable on the interval J,
then/is integrable on every interval J c I. If/ is integrable on the intervals
[a, i] and [b, c] with a <b < c, then/is integrable on [a, c] and
liW+9)
=
f
he c]
/= f
'[a, 6]
/+ f
J[b, c]
/
Furthermore, if f>g and both functions
Finally, if/is integrable on [a, b], then
F(x)
is
a
=
f
Jin
/
vi
continuous function of x.
are
integrable,
then
J//^J/^.
2.6
Calculus of One Variable
171
The fundamental theorem of calculus says more:
if/ is continuous on
that
the
definite
and
the
is,
indefinite integrals of
j [a,6]/= \baf;
The proof of this is actually quite
easy to describe. Define
[a, b], then
/ coincide.
these functions
on
the interval
[a, b], corresponding
to the two sides of
Equation (2.17);
F(x)
F(x)
=
=
inf {KP, /) : P
a
partition
of
[a, x] }.
sup{(j(P,/) : P a partition of [a, *]}
that/is Riemann integrable on [a, U] is to prove F(b) F(b). We
show, using Theorem 2.9, that in factF(x) F(x) for all x e [a, b~\.
First of all F is differentiable in [a, b~\. Let x e [a, ft] and h > 0, then
To prove
=
=
F(x
+
n)
<
F(x)
+ Mh
(2.18)
F(x
+
h)
>
F(x)
+ mh
(2.19)
where M, m are the maximum and minimum of/ in the interval [x, x + ].
These inequalities can be routinely verified (see Problem 32); Figure 2.9 is
+ h) is just F(x) plus the infimum of all 2 (P,f) over parti
[x, x + h]. Any such sum lies between Mh and mh. Now
Equations (2.18) and (2.19) give
convincing: F(x
tions of
J(x +
m <
h)
-
F(x)
;
n
^
<
M
r
*
Figure
2.9
x
+ h
172
2.
Notions
of Calculus
Letting h 0, since /
F'(x) exists and is/(x).
-*
and has the
F(a)
=
is the
value.
F differ
Thus, F and
0 is obvious, we have that F(x)
defined for all x, is differentiable and has
F(a)
J [a>x]/is
same
is continuous, M and m both tend to f(x). Thus
Similarly, one verifies that F'(x) also exists for all x
by
a
constant.
Since
F(x) for all x. Thus
derivative/. This, then,
=
=
proof of
(Fundamental Theorem of Calculus) Suppose f is contin
Then the integral J*/ exists for all x e [a, b].
uous on the interval [_a, b}.
This is a differentiable function off, and
Theorem 2.10.
d
r*
-r\f
dx Ja
f(x)
=
PROBLEMS
29. Prove
Proposition 15(ii).
mean value theorem is proven in the case where
The way to do the general case is to compare the graph of/
with the line through f(b) and f(a). More precisely, let g be the function
30. In the text the
f(p) =f(a).
whose
graph is that line,
(a) Show that
h(x) =/(*) -f(a)
and consider h
-f(bl
~f(a)
b
(x
-
=/ g.
(2.20)
a)
a
(b) Show that h(a) h(b) 0.
(c) Now from the text there is a f between a and b such that h'(0
Differentiating (2.20), deduce that
=
=
=
0.
m=m-m
Suppose that /is differentiable on the interval [a, b], and/'(x)>0
x.
Show that /is strictly increasing, that is, f(x) </( y) if x <y.
32. Verify inequalities (2.18) and (2.19).
33. Give an example of a continuous function of a real variable which is
not differentiable.
Give an example of an integrable function which is not
31.
for all
continuous.
34. Find the real-valued function
/,
such that
f f(t ) dt
=
j
f(t) dt
for all
x e
[0, 1 ]
continuous
on
the interval [0,
1]
2.7
Multiple Integration
35. Suppose /is k times differentiable on R, and fw(x)
1.
Verify that /is a polynomial of degree at most k
173
=0 for all
x.
Multiple Integration
2.7
The calculus of many variables results from the attempt to study functions
of several variable quantities by generalizing to that context the calculus of a
Some notions
variable.
single
of linear
algebra to
closer to that of
be
properly
generalize easily,
The
understood.
require some ideas
integration theory is much
others
variable than is differentiation, hence
one
we
shall describe
it first.
A closed
rectangle
in R" is
a
set of the form
{(x1,...,xn)eR":ai<xi<bi}
for
some
case
fixed
points
of intervals,
in the
same
we
a
=
=
(a1, ...,an),
denote the
l(a1,...,d>>,(b1,...,b't)-\
b
=
(b1, ...,bn)
corresponding
open and
As in the
way:
(a,V)
lsL,b)
(*,K]
=
=
=
{xeRn:ai<xi<bi}
{xeRn:ai<xi<bi}
{xeR":ai<xi<bi}
rectangle will refer to any of these possibilities.
is
rectangle R determined by the vectors a and b
The term
the
in Rn.
half-open rectangles
Vol(K)
=
The volume of
(b1 -a1) (b"-dr)
or halfNotice that the volume of R is the same whether R is closed, open
it should be since the faces contribute no volume.
open. Of course, this is as
The characteristic function of S, denoted by Xs
set.
Now let S be any
on
S and identically zero off S. We should want
one
is the function which is
with the integral of Xsso that the volume of S coincides
to define
integral
have J Xr
Vol(R). The notion
particular, for a rectangle R we shall
turn out that way.
of integral will be built up piece by piece so that things
of characteristic functions
Now suppose that /is a finite linear combination
called
of rectangles: /= Z/=i
*(*) Such a function is
and identically
of
collection
rectangles,
finite
some
It is constant on each of
=
In
simPle/u"ctl0,n:
zero
off their union.
174
2.
Notions
Definition 11.
[/=
>
of Calculus
Let/be
a
simple
function.
If/=
Z*=i
ailRi->
Z^Vol(R;)
we
define
(2.21)
;=i
We
immediately have a problem. It may be possible to also write the
function in another way, /
Zy= i c,- Xsj fr some other collection
of rectangles.
For Definition 1 1 to make sense, we must be assured that the
same
sum
=
Zy=i
and the
Vol(Sj) coincides with (2.21). In case the at and c; are all
and {Sj} are nonoverlapping (intersect only in faces),
cj
{Rt}
amounts to the assertion that the volume of
a
set is the
sum
one
this
of the volumes
of its
rectangular pieces, no matter how it is so partitioned. The verification
that (2.21) is the same for all expressions of the function /as a combination
of characteristic functions is a long verification which is omitted. We now
make this general definition of the integral.
Definition 12.
zero
outside
Let/be a bounded real-valued function which is identically
rectangle R. The upper integral off is
some
J /= inf{|
The lower
simple
o: a a
function
on
R such that
a
>/}
integral off is
j /= sup{|
/is integrable
a: a a
simple
function
on
R such that
a
</}
if
j /= j /;
the
common
value is the
integral f /
This is the direct
given
to
generalization of the definition of the Riemann integral
On the plane and in space it bears the same relation
volume as does the Riemann integral to length.
in Section 2.6.
area
and
Definition 13.
Let S be
a
set in R".
If Xs is
integrable,
we
define the
volume of S to be
Vol(S)=J**s
Now there
are
sets for which /s is not
and shall not
occur in this text.
logical
overlapping rectangles contained in the
integrable;
these
are
highly patho
Notice that if Ru ...,Rn are nonset S, then the sum of the volumes
2.7
Multiple Integration
175
J (Z Xr) is less than J Xs since /s > Z X*, Thus the volume
of S is greater than the sum of the volumes of any collection of nonoverlapping rectangles contained in S. Similarly, if now Ru
Rn are nonZ Vol (Rt)
=
,
.
.
.
,
overlapping rectangles containing S, J *s Z Vol(i?;).
S is trapped between the volume of any union of rectangles containing S and
the volume of any union of rectangles contained in S. If we can make these
<
two volumes as close as we
J
Xs is
integrable (for
Theorem 2.11.
then
J
please by
=
Xs
Let R be
proper choices of the
then
rectangles,
]Xs), and its integral is the volume of S.
closed
a
Thus, the volume of
rectangle
in R".
Iff is
continuous
on
R
and zero offR, then f is integrable.
J
Proof. Given e > 0, we must find simple functions
< J t + e Vol(.R); for then it will follow that
a,
such that
r
a
>/>
t
and
<j
f /< f
<j
for any >0.
("
<
t
+
e
Vol(7?)
Thus, lf<~lf.
<
f/+
e
Vol(R)
In any case, since the
inequality, J/<J/is
obvious, /is integrable.
Such functions a, t are easily found using the basic property of uniform con
tinuity (discussed in miscellaneous Problem 80). According to that theorem, given
>0, there is a 8>0 such that, if |x-y|<8 then |/(x) -/(y)| < e. Now
partition R into a finite set S of rectangles each of which has the property that any
two points are within 8 of each other.
Thus, if for each such rectangle p, mp and
m<e.
M are respectively the maximum and minimum of/on p, we must have M
,
Let
a
T=2mX<>0
pes
2 MPXe
=
peS
where p0 is the open
L
=
J
2 Mp Vol(p)
peS
<
f
J
since S is
These
rectangle corresponding
a
t
+
e
Then
a
>/> t certainly, and
2 (rne+ e) Vol(/>)
peS
2 Vol(p)
peS
partition
following
<
to p.
<{t+e Vol(R)
J
of R into rectangles.
basic
properties
of the
integral
are
easily derived.
176
2.
Notions
of Calculus
Proposition 16. The collection of integrable functions
the integral is a linear function. That is:
is
a vector
(i) Iff is integrable and c e R, then cfis integrable and \cf=c
(ii) Iff, g are integrable, so isf+g and \(f+g) J/+ J g(iii) Furthermore, iff< g then [f<\g.
space and
J/.
=
(ii) is certainly true for simple
where
R,, Sj are rectangles, then
2*jXs,>
/=2'Xi>
+ 2ZbJXsj is also simple, and thus integrable. By Definition 1,
We leave the
Proof.
f+g=2a>Xi
|(/+ 9)
More
proof
of
For if
functions.
=
generally,
simple functions
(i)
0
to the reader,
=
2 ", Vol(R,) + 2 bj Vol(Sj)
now
let
/,
g be any
cti, a2, ru t2,
CTi>/^(72
=
J>+ \g
integrable functions.
If
>
0, there
are
such that
ti;>0>t2
and
J
<^i
<,
I
(T2
J
Ti
<,
> a2
+
T2
+
J
T2
+
Thus
CTl
+
Ti
>/+ g
so
{ (f+g) ji + jri <|((72 + t2) +
<
Since
> 0 was
e
arbitrary,
we
obtain
2e ^
J(/+^)
J (/+ g) <, J (/+ g),
Finally,
j(f+g)^jcr2 + jr2 + 2e<jf+jg+2e
so
letting e->-0,
+ 2e
so
/+ # is integrable.
2.7
Multiple Integration
111
Similarly,
j(f+g) 2e>jai + jr2>jf+jg
+
so
again letting
e
->0,
j(f+g)>jf+jg
(iii) Finally iff<g, then gf>0. But certainly the function which is identically
is a simple function. Thus J (g -/) > J (# -/) > 0. By (ii) it follows that
zero
Sg-Sf^0,or!g>jf.
We shall
give
the basic tool for
computing integrals: Fubini's theorem.
integrate by integrating one variable at a
time. For the purpose of showing this, write the variable (x1, ...,x") of
R" as (x, y) where x e R"'1 and y e R: x
(x1,
x"'1), y x". Let /
be a function defined on a rectangle R in R", and suppose for each y fixed,
f(x, y) is an integrable function of x. Define F(y) J/(x, y) dx. If F
is an integrable function of y, its integral
According
now
to that result
we can
=
=
...,
=
JF(y)dy j\jf(x,y)dx
=
dy
off. We shall now show that iff is integrable
generally (after applying this principle n times)
J/
functions appearing in the following formula are integrable, then the
is called the iterated integral
this is the
if all
More
same as
formula is valid.
jfix1
x") dx1
=
dx"
J J"'" jf(x1,...,xm)dxl
dx2
dx"
(2.22)
This follows from Fubini's theorem.
Theorem 2.12.
refer
to
(i)
(ii)
Let f be
the coordinates
These functions
an
ofR"
ofy,
These functions ofx,
integrable function on a rectangle
(x, y), where xeRk,ye R"~k
R in R".
as
|/(x, y) dx, |/(x, y) dx are integrable.
j/(x, y) dy, |/(x, y) dy are integrable.
We
178
2.
Notions
of Calculus
(iii) If is given by
any iterated
integral off; for example,
J7(x, y) dx dy j \]f(x, y) dx] dy j \jf(x, y) dy
=
Proof.
It is
=
dx
easily verified that the collection of functions for which the
asser
tions (i), (ii), and (iii) are true is a vector space. Furthermore, these assertions are
obvious for the characteristic function of a rectangle. Thus, Fubini's theorem
holds for simple functions.
Now, suppose /is a bounded, real-valued function on the given rectangle R, and
suppose that or is a simple function, and /> a.
By definition of the lower integral
with respect to the
x
coordinate,
jf(x,y)dx>jo(x,y)dx
Now this
inequality is maintained after taking
the lower
integrals with respect
to y,
thus
J"
J*/(x,y)rfxpy>j[ja(x,y)rfx
dy
=
j o(x, y)
dx
dy
(2.23)
since Theorem 2.12 is true for simple functions. Equation (2.23) being true for
any o <;/, we can take the least upper bound on the right, obtaining
j jf(x,y)dx
dy>
\f(x,y)dxdy
Now, by considering simple functions
kind of
reasoning
we
a
such that
o
>:f and applying the
same
obtain this inequality
j J7(x,y)rfx dy^jf(x,y)dxdy
As
a
result,
we
obtain this string of
real-valued function
KlM
on
inequalities,
which is valid for any bounded,
R:
V
J'l
W\fU>
(2.24)
(The second and third inequalities follow immediately from the fact that the upper
integral always dominates the lower integral.) Now, if /is indeed integrable, the
2.7
first and last terms of
(2.24)
are
the same,
so
all
are
Multiple Integration
the
179
That the second and
same.
top third are equal implies that J /(x, y) dx is integrable. That the bottom third
and fourth are equal says that J/(x, y) dx is integrable. The equation
jf(x,y)dxdy j jf(x,y)dx
dy
=
now
just
Now
we
states the
equality of the end
shall illustrate the
we
should remark that
defined
only
rectangle; more often such
given measurable domain D.
Definition 14.
Let D be
function / defined
R
on
/ will
of
a
be
domain contained in
a
we
function is defined
We make the
say /is
integrable
a
or
rectangle
if this is
considered
following definition.
so
Given
R.
a
function/
for the
by
=
a
but rather
D,
a
Before doing that,
integrate functions
xe
0
D
xeR,x$
D
JD/=J/
We define
If D is
on
=/(x)
/(*)
graph
have the occasion to
on a
on a
defined
of Fubini's theorem.
use
rarely
we
terms with the interior terms.
subdomain of
function,
or
has
a
rectangle
some
integrable if / is.
tacitly assume our
other
R bounded
by
surface which is the
a
redeeming property,
then the function
We shall not pursue this theoretical
domains
inquiry,
redeemable.
are
Example
39.
Define
{D
(x, y):
y) x2
0
=
f(x,
=
+
<
y2
y
if
<
x2,
f(x, y) x2 + y2.
0 otherwise.
D, and f(x, y)
0<x<l},
(x, y)
e
=
=
Then
\j-\j- uj>-Hjx-n.c"<x2+/)*
since, for fixed x, f(x, y) is
We thus obtain
x2 + y2
It-?
x2y
zero
r1/
+
if
y-^ dx=L\x
a
x <
0
x6\
or
J
y
>
1
dx
x2 and otherwise is
1
26
+yr*=5+2T= 105
180
2.
of Calculus
Notions
y
=
g(x)
Figure 2.10
Let
J7
=
us
do the
example, iterating this time
f_ [f_ /(x, y) dx\ dy j^ [f_(x2
=
+
1
~
3
2
1
+
~~
or
dy
_
~~
15
3
Ibl
7
general technique can be described
(Figure 2.10).
as
follows :
=
{(*, y) :
a < x <
b, g(x)
<
y
<f(x)}
(Figure 2.11)
D
dx\
26
2
~
in either of these forms
D
y2)
in the other order.
t
_
The
same
=
{(x, y):a<y^b, <b(y)
Then, given the function/defined
on
rbf rg<-x>
r
f=\
f(x,y)dy
dx
in the first case; and in the second
rf
C
c*W
j/ j J
=
<x<
f(x,y)dx dy
D,
xjj(y)}
we can
write
Try to write the domain
2.7
Multiple Integration
181
Of course, if neither case can be obtained, then D might have to be broken
up into pieces in each of which either representation is possible.
The
computation of integrals in more than two dimensions is done in pretty much
way, but with a certain amount of additional care. For example,
should try to pick out one of the coordinates, say z, so that the
given
domain takes the form g(y) < x </(y), where y represents all the other
the
same
one
coordinates and ranges through some domain D0
break down D0 in the same way.
Now
.
one
proceeds
to
Examples
D={(x,y, z): x2
z) xyz.
40.
f(x,
y,
Now z ranges between 0 and
D
y2
+
z2
<
1,
(x2
+
y2))1'2,
+
x >
0,
=
{(x, y, z): x2
+
y2
<
(1
D
0,
z >
0},
1, 0
0
< x,
<
y, 0
so
< z <
Thus, continuing the analysis of
A>
>
y
=
{(x, y): x2
=
=
{(x, y, z):
0
< z <
[1
0
-
+
y2
<
< x <
(x2
+
1, 0
1, 0
< x,
<
y
0
<
<
(1
y}
-
x2)1/2,
y2)]1/2}
p
x
=
Hy)J
x
J
\
a
L,y
1
Figure 2.11
=
<l>(y)
[1
-
(x2
+
y2)]1/2}
182
2.
Notions
of Calculus
and
"
.1
.
J
f=
D
J0
2
V
y2))rfy
dx
y(l-(x2
J0 L
(1
X
y,
0
y2
+
z2
+
z >
^^0
dx
J
4
z): x2
1
x2)2]
-
2
{(x,
+
<
~24
1, (x
0}
y,
z):
0
i)2
y,
We may rewrite this domain
{(x,y, z):(x-i)2
{(x,
-
f(x,
+
< x <
+
1, 0
<
y
<
[i
-
(x
0
z)
r
r[i(x-i)2]'/2
=
<
i
1
as
-
< z <
[1
Jl-(x7
+
< z <
[1
y2)11/2
dx
L^o
Figure
2.12
-
(x2
+
y2)]1/2}
iff12,
Thus
ri
y2
y2<i,x>0,y> 0,
0
=
dx
Jl-X2)'/2
(see Figure 2.12).
=
dy
So
firx(l-x-2)2
D=
dz
z
\_J0
x >
D
.[l-(x2+y2)ll/2
y
2J/[Jo
1
41.
x\
.1
1
=
r.(i-xJ)'/2
rfy
az
-
(x2
+
y2)]1/2}
2.7
Integration is clearly of value
study of mass. Suppose
in
Multiple Integration
183
computing volumes;
it also plays a role
domain in R3 filled with a certain fluid.
shall let (>) be the mass of the fluid contained
in the
is
a
If D is any subdomain in E, we
in D. What information do we need in order to compute mass (D), and how
do we compute it ? The answer is suggested by comparison of the
properties
fact, it is clear that the intuitive properties
of mass are the same as the properties of volume ; so we should also expect
to be able to compute masses by integration.
In fact, we introduce the
notion of density: for x0 e E, the density o(x0) of the fluid at x0 is the limit
of
mass
with those of volume.
In
mass(i?)
r
where
Vol(R)
we mean
by
R
->
the sides of R tend to
density
and
,
that x0 is in the rectangle R, and the lengths of
(we might call mass (R)/'Vo\ (R) the relative
zero
of the fluid in the
domain is
such
x0
rectangle R).
in terms of this
computable
{R J is a
almost filling D. Then
a
domain and
collection
Now, the
mass
of the fluid in any
D is
density function a. Suppose
of pairwise disjoint rectangles
in D
approximation to mass (D) and as the size of the rectangles gets smaller
smaller, the approximation gets better. On the other hand, this sum is
the integral of a simple function approximating a, and thus approximates
JB a. Taking the limit we obtain mass (>) JD a.
is
an
and
=
EXERCISES
15. Compute the volume of these domains:
(a) {(x,y)eR2:x2 + y2<\).
(b) {(x,y)eR2:x2<y<\}.
(c) {(x,y, z)eR3:0<x<l,0<y<l,0<z<x2 + y2}.
(d) {(x, y, z)eR3: -1 <x < 1, 0 <y<2, y<z< y + x2}.
16. Verify that the volume of a right circular cylinder of radius r and
height h is inr2h.
17. Integrate the function/on the unit rectangle [(0, 0), (1, 1)] in R2
x cos 2-ny.
(a) f(x, y)
(b) f(x,y)=\(x-\)(y-\)\.
(c) f(x, y)=xe*y + ye~x.
=
184
2.
Notions
,a\
of Calculus
tt
x
^
(a)
f(x, y)=
i
n
\
f(x,y)=
\
(e)
tfx^y
.
.c
if
fx + y
x
v <
+
ifx +
j
1
y>l.
f(x,y) (i+x2 + y2Y'2.
Integrate the function /on the domain D in R2.
(a) 7J {(x, y):0<x,0<y,x + y< l},f(x, y)=x2 + y2.
(b) D {(*, y): 0 < y <x < l},f(x, y) xy2
{(x, y):0<y<x< \},f(x, y)=x2y.
(c) D
(d) D {(x, y): x2 + y2 < l},/(x, y) (x2 y2)2.
19. Integrate the function /on the domain D in .R3.
(a) D is the intersection of the unit ball with the octant {x > 0, y > 0,
x + v + z.
z > 0} and f(x, y, z)
(b) Z> is as above and f(x, y, z) xyz.
(c) Z> is the unit cube in the first octant and/(x, v, z)=x2 + y2 + z2.
(A) D is the domain in the first octant bounded by the coordinate
z.
\ and f(x, y, z)
axes and the plane x + y + z
(f)
=
18.
=
=
=
=
=
=
-
=
=
=
=
PROBLEMS
Verify that the integral on R" as defined in this section coincides,
1, with the Riemann integral defined in the previous section.
37. Let /be a bounded, nonnegative, real-valued function defined on the
interval /, and let D
{(x,y) e R2; x el, 0 < v </(*)}. Verify this
assertion: /is integrable if and only if D is measurable, and U f= Vol(Z>).
Let D be a domain in R2 and suppose
38. Use Problem 37 to verify this.
36.
when
n
=
=
that D is of the form
{(x, y)eR2:a<x<b,g(x)^y <f(x)}
Then, if D is measurable, Vol (D)
JS [f(x) g(x)] dx.
39. Complete the proof of Fubini's theorem by verifying the second and
=
third
-
inequalities of Equation (2.24).
40. State and prove Fubini's theorem in three dimensions.
41 Suppose the unit ball is filled with a fluid whose density is
.
proportional
to the distance to the
boundary. Find the radius of the ball centered at
the origin which has precisely half the mass.
42. Suppose a cone of base radius r and height h is filled with mud
(Figure 2.13). Suppose the density of the mud is equal to the distance from
the base.
What is the
43. A beach B is
B
=
{(x,y):
1 <x2 +
and the human
mass
of the mud?
shaped in
the form of
a
crescent
(see Figure 2.14)
y2;(x- i)2 + v2<l}
density a increases with the distance from the water. More
precisely, a(x, y)
(x2 + v2)"1. What is the mass of humanity on that
=
beach ?
2.8
2.8
2.13
Figure
2.14
Differentiation
185
Partial Differentiation
the
Although
it is
Figure
Partial
integral
computed by
in R" is defined without reference to the
succession of integrations,
a
one
coordinate at
coordinates,
a
time.
The
notion of differentiation is, to begin with, generalized to R" one coordinate
at a time.
Later we shall see how to build out of this generalization an
invariant notion of derivation.
Let x0
e
R", and suppose that / is
neighborhood
x' given by
J
(X0
,
.
of x0
.
.
,
X
,
.
a
real-valued function defined in
For each i consider the function of the
x0 )
single
a
variable
186
2.
of Calculus
Notions
If this function is differentiable, we denote the derivative by df/dx1, and call
More precisely,
it the partial derivative of/in the x' direction.
Let /be
Definition 15.
of x0 in R".
df
OX
The
v
,
i
(x0)
=
real-valued function defined in
partial derivative of/ with respect
xp'
/(V
,.
hm
+ f,
Another way of
This restriction is
partial
function
a
as
the
on
are
partial
one
computed merely by considering
constant.
42.
=
xy
f(x,y)
=
dx
y
dy
(x, y)
=
x
43.
=
(x2y)
f
dy
2xy
=
x2
1
44.
/(*, y)
=
cos[x(l
f(x,y)= -(l
dx
dy
(x, y)
=
45.
/(*, y)
=
x>
-x
+
+
y)]
y)sin[x(l
sin[x(l
Consider the
through
Examples
fix, y)
derivative is this.
x0 and in the E; direction.
variable and df/dx' is its derivative.
the line
function of
derivatives
relevant variable
neighborhood
*
describing
as a
a
to x! at x0 is the limit
,x0")-/(xo1, ...,Xo")
(-.0
function / only
These
a
+
y)]
+
y)]
all but the
2.8
x~(x, y)
'
yxy
=
dx
dy
(x, y)
=
Partial
x In
Differentiation
187
x
Of course, if the functions
dx1'""
are
dx"
also defined in
neighborhood
a
of x0
,
partial differentiation, and keep going in this
we
may
way
as
subject them
as possible.
far
to further
We shall
refer to any such operation as a partial differentiation and call its order the
number of individual partial derivatives involved.
Thus, the order of
\dxJJ
dx'
is 2; the order of
dx2 \dy
is 6.
\dz3JJ
We introduce
dx2
notational convention which deletes
\dxj
dx
d2f
a
d
/df\
dx
\dyj
_
dx
dy
*2
l2f
d
ldf\
dx1
\dxJj
_
1
dx dxj
d3f
dx1 dxj dxk
d6f
dx2 dy dz
and
so
forth.
((K\\
dx' \dxJ
d
3
dx
\dxkjf
d5f
\dx dy dz3)
/
parentheses.
188
2.
Notions
Suppose
df/dx1,
that
now
...,
df/dx"
of Calculus
/is
open set N in R" and that
set all the variables constant except
function defined in
a
all exist in N.
If
we
an
just the derivative off along this line. Thus, if
df/dx' 0, / is constant along the line on which only x' varies. In such
circumstances we say that /is independent of x', since /does not vary as x'
alone varies. If, moreover, df/dx' is zero at all points of N for all i, then/
depends on none of the variables, so is constant. As this is an important
one, say x'
then
,
df/dx'
is
=
observation,
make it.
we
Proposition 17. Suppose that f is a real-valued function defined in a neigh
of x0 in R". f is constant near x0 if and only if all the derivatives
df/dx" exist and are zero near x0
df/dx1,
borhood
.
..
Proof.
.
,
If /is constant, it is obvious that
df/dx'
=
suppose that these conditions are valid in a ball
y
(y1, ...,y")e B(x0 r). We will show that/(>0
=
,
the
proof.
0 for all i.
On the other hand,
Let
B(x0 r) centered at x0
=/(x0). Figure 2.15 illustrates
.
,
Consider the function of x" :
f(x0
.
,
.
.
,
Xo-1, x")
This function has derivative
/(xo1,
.
.
.
,
zero
by hypothesis,
XV1, xo") =f(xo'
so
is constant.
xl~\ y")
Now, the function of xn_x,
f(x0\...,x"0-2,x-\y)
(y'.y2,/)
/ (y',-to2,jto
(xn\xir, Jfi>:! )
Figure
2.15
Thus,
2.8
Partial
also has derivative zero, and thus must be constant,
f(x0\
...,
Differentiation
189
so
xTl, v") =f(x0\ ...,y-\ y")
This together with the preceding equation gives
f(x0\
xTl, x0") =f(x0\
,
...,
xl'1, y-\ y")
Continuing in this way, we can replace each x0J by the corresponding y]
time, ending up with the desired equation f(x0) f(y).
As far
fact
only
on
the
as
should
we
on
higher order
verify.
they
dy dz
basic
performed.
For
example,
d5f
d5f
dz dx dy dx dz
dy dx dz dz dx
equation; it being clear that all others follow
applications of the first one. The verification of (2.25)
interesting application of Fubini's theorem.
verify only
the first
succession of
amounts to
an
Theorem 2.13.
Let
f
be
a
function defined in a neighborhood
that all first- and second-order partial deriva
real-valued
of(x0 y0) in R2 and suppose
off exist and are continuous
,
tives
d2f
d2f
dx dy
dy dx
throughout
R
one
(2.25)
d5f
We shall
are
dy dx
dy
dx
N
concerned, there is
a
This is that each
^
*'
dx
a
are
at
partial differentiation depends
the number of derivatives with respect to each coordinate, and not
now
the order in which
from
differentiations
one
on
N.
Then
N.
Proof. We apply Fubini's theorem to d2f/dx 8y in
((xa yo), (s, t)) contained in N (see Figure 2.16)
a
sufficiently small rectangle
=
,
rs IY
e2/
l
r'
[V
e2/
dy
(2.26)
190
2.
Notions
of Calculus
Figure
Now,
we can
rs
Jxo
Integrating
and (2.27)
easily evaluate the integral
dy
once
\
df
df
Jxo 8x\8y
J
dy
dy
8
again (this time with respect
ay
J*o L-'vo dx
dy
right-hand side.
the
on
l8f
f
82f
dx
2.16
dy
dx
=
c'
j
y) we obtain from Equations (2.26)
to
8
-[f(s,y)-f(x0,y)]dy
=f(s, 0 -f(x0 t)
-
,
Now,
we can
[f(s, y0) -f(x0 y0)]
,
differentiate this
'
d
8s
L-
d2f
8x8y
'
(x, y) dy dx
-i.
82f
8x
Then, from (2.28)
f'
(2.28)
equation with respect to 5 first, and then t. By the
calculus, we know how to differentiate the integral on the
the upper limit of integration :
fundamental theorem of
left with respect to
For fixed y,
a2/
=
axey(''y)dy ax(s't)-8x<s'y)
8y
(s, y) dy
2.8
Differentiating
this
equation
with respect to
t , we
Differentiation
191
obtain
82f
82f
dx 8y
as
now
Partial
8y
8x
desired.
Another
allows
important application
Proposition 18.
two
variables
F
function
of Fubini's theorem is this result, which
to differentiate under the
us
on
F(x)=
Then F is
Suppose that f
integral sign.
is
a
continuously differentiable function of
and y, a < x < b, and y
the interval [fl, b~\ by
x
e
D,
a
domain in R".
Define
the
|7(x,y)dy
differentiable
and
d-f(x) j8/(x,y)dr
=
Jd dx
dx
We shall show that F is the indefinite
Proof.
f
integral of
the function
J
8f.
\yx(x,y)dy
and thus
by
the fundamental theorem of calculus, the
follows.
proposition
By
Fubini's theorem
i[ll^y)dy\dx=l[V^{x'y)dx
by the fundamental
f(t,y)-f(a,y). Thus
But
theorem of
dy
calculus, the inner integral
on
the
right
is
[f(t,y)-f(a,y)]dy=F(t)-F(a)
f'Tf ^(x,y)df\dx=\
JD
J
J LJD
Let
are
us
8x
return
obtained
consideration of the first-order derivatives.
to lines
differentiating after restricting the function
now
by
to the
These
parallel
192
2.
of Calculus
Notions
to the coordinate axes.
That
along any line.
We
is,
Let x0
Definition 16.
we
e
neighborhood of x0
derivative df(x0 v) to be
in
a
.
generalize
this notion to allow differentiation
make this definition.
R" and suppose /is a real- valued function defined
If v is a vector in R", we define the directional
,
jff(*o + 'v)
This is
clearly
the
Um/(x0 + fv)
same as
/(x0)
-
t
i-o
We leave it
as an
g(x0)
=
exercise to
that
verify
(2.29)
d/(x0,Ej)
Now, in certain pathological
cases
the directional derivatives need not
together in any nice way, but typically we need only know the
vatives in order to find any directional derivative.
Proposition 19.
Proof.
vary
one
...
,
+
linearly
.
in
v.
looking
at the difference
fv)-/(x0)
variable at
argument with
a
a
In order to expose the idea without
we consider the two-variable
time.
pile of indices,
encumbering
case.
difference
f(x0 + t h, y0 + tk)
{f(x0
deri
a neighborhood ofx0 and the partial
Then the directional derivatives
df/dx" all exist near x0
The argument consists in
/(x0
hang
Suppose f is defined in
derivatives df/dx1,
df(x0, y)
partial
+
th,
y0 +
f(x0 y0)
,
tk) -f(x0 + th, y0)} + {f(x0 + th, y0) -f(x0 y)}
,
the
Write the
2.8
find
We
can
the
mean
x0
and
better
a
expression for the
+ th,
y0)
f(x0 y0)
-
8f
=
,
Similarly, by applying the
8f
8y
is, there is
by applying
f0 between
a
,
value theorem to the function f(x0 + th, s),
mean
we can
as
(x0 + th, t]a)tk
-q0 between y0 and y0 + tk.
some
rectangle [(x0 yo), (x0
,
f(x0
+
df
ty)-f(x0)
=
--0,
we
Thus,
we
have for suitable
(f0 v)
,
in the
+ th, y0 + tk)],
dx
t
(f
.,8f,
o
,
yo)h +
oy
v.
,.
(x0 + th,
r)0)k
,
obtain by continuity that
d((x0 y0), (h, k))
,
Thus the
That
(fo yo)th
rewrite the term in the first set of braces
t
y0) of s.
193
+ th such that
x0
Letting
Differentiation
term in the second set of braces
value theorem to the function f(s,
f(x0
for
Partial
proposition
=
+
j (*o y0)h
,
(x0 y0)k
,
(2.30)
is verified, at least in R2.
This linear function, df(x0 v) of the vector v in R" is called the differential
We will make a systematic study of this in a later chapter. The
of/ at x0
,
.
vector-valued function
K\
\dxl'""dx")
IK.
gradient off and is denoted by V/
generalization of (2.30) to n variables
is called the
19 that the
df(*o ,v)
The
in the
what
=
et W
more
<. V/(x0)>
Proposition
t2-31)
total derivative." It is not as powerful
one variable and it is some
kind of tool. For
cumbersome, but it does provide a similar
gradient behaves as a
analysis of a function
example,
=
It is clear from
is
sort of
as
"
the derivative in
194
2.
Notions
Proposition 20.
attains
a
of Calculus
The
maximum
gradient of a function
vanishes at any point at which it
minimum value.
or
(xo1,
x0") is (for instance) a maximum value of /, then
Proof. If x0
f(x0l,
x',
x0"), as a function of x', attains a maximum at x0'. Thus,
8f/8x' vanishes at x0'. Since this is true for all i, Vf(x0) 0.
=
.
.
.
.
,
.
.
.
.
.
,
,
=
Examples
46. Consider /(x, y,
V/=(2x
Thus
x
=
-
zero
x
-
is, only
x2
+ xy +
y2.
when
2y
=
2
that
=
2y)
+ y,x +
V/is
z)
at the
This is the
origin.
only critical point,
and
a
minimum at that.
47.
f(x,
y,
V/= (cos y,
is
never
48.
zero,
z)
=
y +
x
sin y,
1)
so
/has
no
f(x, y,z)
V/= (cos(yz),
x cos
=
xz
x cos
z
critical values.
(yz)
sin yz, xy sin
yz)
V/is zero only when x Oandyz n(n + ^foranyintegern. Clearly,
/ has both negative and positive values near any point on the line
{x 0}, so no such point is critical. Thus, /has no critical points.
=
=
=
EXERCISES
20. Find the first
partial derivatives of these functions.
sin(xy) (c) x'' (d) x2y + y2x
21. Differentiate x*". (Hint: This is the same as finding the directional
derivative of x"z at a point (x, x, x) in the direction of (1, 1, 1).)
(a)
xyz
(b)
2.9
Improper Integrals
195
22. If /is differentiable at x0, then
8f
,-
(x0)
for all
=
df(x0 Ei)
,
i.
23.
Suppose that/, g are differentiable at x0 in R". Show
differentiable and V(fg)(x0) =f(x0)Vg(x0) + g(x0)Vf(x0).
24. If /is differentiable at x0, and/(x0) ^0, then
v(i)(x0) ^
=
V/(x0)
25. What is the minimum of x2 +
26. What is the maximum of
x+3v
that fg is also
y2 + (2v + l)2 ?
n
l+x2+y2'
27. Compute the differentials of the functions in Exercise 20.
PROBLEMS
44.
Suppose /is
a
differentiable function of two variables and gi,g2 are
one variable so that the range of (gi,g2) is in the
differentiable functions of
domain
Find the derivative of
of/.
45. Let /be
function of
46.
x
a
y alone if and
Suppose that
47. Let T: R"
h(t ) =f(gi(t), g2(t)).
differentiable function of two variables.
->
L
:
R"
R" be
->
a
only if 8f/8x + 8f/8y
R is
a
linear function.
linear transformation.
=
Show that
/is
a
0.
What is V/_ ?
Define the function
on
Show that / is differentiable, and V/(x, y)
f(x, y)
<T'y, Tx} (recall that T' is the transpose of T: if T is represented by the
matrix (a/), then T' is represented by (b/) where b/
at').
R" x R":
=
(Tx, y>.
=
=
48. If T: R"-+R" is
a
<7x, x> is differentiable,
2.9
g(x)
=
=
Improper Integrals
We return
be
linear transformation, then the function
Tx + Tx.
and V#(x)
now
considering
focus
on
the
"
introduce the
to the
study of functions of
functions defined
on
"
one
variable; in fact,
the whole real line.
infinity of such functions.
notion of lim/(x) as x -* oo.
behavior at
we
will
Our interest will
For this purpose
we
196
2.
Notions
If/ is
Definition 17.
{x: x > fl}
lim/(x)
=
a
say that
we
if, for
L
of Calculus
real-valued function defined in
an
infinite interval
converges to L as x becomes infinite, written
> 0 there is an M > 0 such that x> M implies
f(x)
e
every
x-*oo
|/(x)
L|
Similarly, if/ is defined
< .
in
{x: x<b]
we
say
lim/(x)
=
L
x-*<x>
(the
limit of
there is
an
f(x)
M
is L
as
x
0 such that
>
becomes
M
x <
negatively infinite) if,
implies \f(x) L\ < e.
for every
>
0
Examples
lim
49.
M
x >
1/x
=
For
0.
implies |l/x
0|
-
given
>
0,
we can
take M
=
_1.
Then
< .
50.
4x2
,.
+ 3x + 5
lim
-
ox
jc-+qo
For,
4x2
2
long
so
1
-
=
7
2
0,
as x >
4 +
+ 3x + 5
3/x
~
8x2
Now,
8
7
-
we can
-
5/x2
7/x2
+
(2.32)
compute the desired limit by using the standard algebraic
rules (the limit of a sum is the sum of the limits, etc.). (See Exercise
28.) Since 1/x, 1/x2 tend to zero as x->oo, the limit of (2.32) as
x-> oo
is
4/8
=
1/2.
51.
x\x\
,.
hm
x-*co
1
T
X
I'm
5=1
X-*
oo
x\x\
t + X
2= 1
If
X2
x|x|
1
=
'
Y+x1
=
TTx1
l +
i/x2
if
x <
0,
x|x|
x2
1 + x2
1 + x2
-1
1 +
1/x2
2.9
lim arctan
52.
x
=
Improper Integrals
197
n/2.
Definition 17 is the analog for functions defined on an infinite interval
of the notion of convergence of a sequence (a function defined on the integers).
lust as we pass from sequences to series we can pass from infinite limits of
functions
to infinite sums; that
is, integrals
Let/be a continuous
/ is integrable if lim \xaf exists,
Definition 18.
We say
absolutely integrable if lim
J?/ /is
over
infinite intervals.
function
on
in which
case we
\xa \f\
the interval
{x:
a}.
x >
write the limit
as
exists.
Examples
53.
x-2 is integrable
dx
1
x
=
Ji
=
the interval
on
[1, oo).
For
--+1
m
i
so
Cx~2dx=
Jl
54.
x~y
(--+1^1)
lim
m-oo\
is not
cos x
=
1
m
absolutely integrable
on
the interval
[1, oo).
For
-2HB +
OO
dx
>
X
Jl
Between
2;tn
-
JI/3 QQg x
Z I
b=1 J2nn-n/
and
jt/3
dx
X
3
2rcn +
1
n/3,
x
cos x >
(2nn
+
n/3)
1
\.
Thus,
f
Ji
The
2n
1
cos X
dx>
Z
^i
=
o
2
(27tn
+
tt/3)
infinite intervals is entirely analogous to the
We have the following facts (whose counterparts
theory of integration
on
theory of infinite series.
theory of series are easily recognized).
in the
oo
3
2.
198
Proposition
of Calculus
Notions
Let f be continuous
21.
on
the interval
{x
: x >
a}.
(i) fis absolutely integrable if and only if the set {\xa |/|} is bounded.
(ii) Iff is absolutely integrable, then fis integrable.
(iii) (Comparison Test). // there exists a b > a and a constant K and an
integrable positive function g defined on {x: x> b} such that Kg > |/|, thenf
is absolutely integrable.
Proof.
(i) If |/| is integrable, clearly {Jj |/|} is bounded. On the other hand, if {J; |/|}
is bounded, let L
supfjs |/|}. Then for e > 0, L e is not an upper bound, so
e.
Then for all x>.x0,
there exists an x0 such that S |/| > L
-
=
-
L>:
j l/l>|
!/!>-
so
-flfl\
<B
(ii) Suppose j? |/|
Let
e
> 0.
f
Then for n,
=
L.
Then there is
l/l-
m
>: x0
|c-cm|
=
Let c
an x0
=
rS/.
{c} is
a
Cauchy
sequence.
,
f fzf l/l^ f l/l- f l/l
Thus
{c} is Cauchy,
Let
>
n
so
0, and find N
in the
;> x0
,
a
f \f\-L
as
x
<-
-
e
We show that
such that for
so
+
converges, say to
that |c
previous computation.
c|
<
a
c.
j
|/|-.
<
We shall show that in fact
e/2 for n^>N.
Then for
x
>
J/=
c.
max(x0 N),
,
2.9
(iii) Under
the
given hypothesis, if x
b
x
j l/I^J"
e
Thus
>
Improper Integrals
199
1, then
co
1/1+
AT
J
g<oo
b
a
by (i), /is absolutely integrable.
Here is
easily derived relationship between
integrals which provides yet another
an
series and
the absolute convergence of
test for the convergence of
series.
Proposition 22. (Integral Test) Let f be a positive, decreasing function
+
Then Jj f exists if and only ifY.n=if(n) < defined on R
.
Proof. For x,n<x<n+l we have f(n)>x>n+ 1. Thus/()>|i;+1 />
f(n +1). Thus, by comparison the series 2 j+1 fand 2/(") converge or diverge
together. But the convergence of the first series is the same as the existence of
J /, and conversely.
This proposition gives an easy proof that 2 l/<1+e) < < for e > 0. (Compare
For if we consider the integral j? dtjt1*', we have
to the work of Example 18.)
r* dt
-1
x
i
1
1
ex'
e
1
~
as x ->-ao.
Example
55.
OO
1
?2 n(log n)2
< 00
For
log
au
/'e* du
f"*-*
dt
J
2
No*
'log 2
f(l0g f)2
I
1
,og2
log
_,
=
-"
1
2
logx
Thus
r
/
dt
h f(logf)2
i
_
*-\log2
M=_L
logx/
log:
< oo
200
2.
of Calculus
Notions
EXERCISES
28.
Verify
these
algebraic properties of lim.
Suppose lim/(x), lim g(x)
JC-.00
X-.00
exist.
(a)
lim f(x) + g(x)
=
x-*o
(b)
lim f(x)g(x)
=
x-*co
(c)
lim
-.
lim f(x) + lim
lim f(x) lim
-
=
#(x)
g(x).
x-*<x>
x-*ao
f(x)
g(x).
x-*co
x-*oo
Jim-^
*?
if lim
hm^tx)
#(x) ?t 0.
x-.co
x-oo
29.
Compute these limits
(a)
sin
as x
->
oo.
1
x
.
x
(d)
tan-.
(C)
XSinx-
x
x2 + 3x + 1
1
** + l
x2-l
(c)
FT!"
30. Which of these series converge:
2
(a)
n=2
nlogn
n=2
1
"
(>)
1
W
2^r-T
(10g lOg ri)2
=
n(logn)2
ii
=
2
n__.
1
OO
Z
n
2
=
oo
2
x2
(log n)2(log log n)2
n3'2
n=2(logn)2
1
oo
-J7T
f
* IV
'
2
W
2
2
(d>
Cg)
^r-^i
1
CO
n
2
(f)
-j
n
=
2
(log rt),2
1
( sin n)2
X-.00
2.70
2.10
The
The
Space of Continuous
Functions
201
Space of Continuous Functions
The mathematician attacks his
problems with a certain store of techniques.
problem will require the development of a new technique;
more often the problem is solved by viewing it in one way, and then another
and then again another until a viewpoint is obtained which allows for the
application of one of those techniques. Sometimes if the viewpoint is clever
enough, or profound enough or naive enough the applicable technique
is quite elementary and surprising and leads to further deep discoveries.
This is the case with the contraction lemma (a fixed point theorem) which we
shall apply several times in this text to obtain some of the basic facts of
calculus. First, in this section, we shall develop the particular viewpoint in
the relevant context. It is simple enough instead of looking at continuous
Occasionally
functions
Let
in
us
a
one
at
a
time,
we
illustrate this with
finding
a
consider them all.
particular problem. Suppose
differentiable function with these properties:
f'(x)=f(x)
for all
a
x
and
/(0)
=
we are
interested
(2.33)
1
function means first of all to verify that a solution to our
and
secondly to establish some technique for computing it.
problem exists,
We already have enough experience with calculus to know that this second
of
objective will be hard to fulfill. What we in fact seek is a means effectively
This provides a clue : let us look for a sequence
our solution
To find such
a
approximating
of functions {/}
.
which converges to a function with the properties (2.33).
Such a sequence would be a sequence of differentiable function {/} such
If we
that the sequence {fn(x)} converges for all x, and f'n(x) =f-i(x).
had such a sequence, we could take the limit and deduce that
lim/'n(x)
=
lim/n_1(x)
lim/(x) will solve our problem.
itself provides the tech
a good idea, because Equation (2.33)
be
Let
any function, and define
/0
nique for generating such a sequence.
Will the sequence
so forth.
and
/,=/' Then let/2=A,/3=A,
that
Notice
7 i -/ o.
/2
{/} converge? Well, that is a problem.
we must be very
Thus,
/3=/'2=/"o, and more generally / =/<5n)is
careful to choose an infinitely differentiable function for/0. Suppose f0
so all
and
0,
Then
/ + 1 =/o"+1
chosen as a polynomial of degree n.
converges,
the rest of our functions are zero. Thus, the sequence certainly
so/(x)
=
Now this is
=
202
but
2.
hardly
Notions
of Calculus
solution, since the condition /(0)
to a
=
1 is not verified. In
fact,
this present approach has obviously petered out fruitlessly and it may be
1 in our
because we have not incorporated the initial condition /(0)
=
approach. Can we put all of (2.33) in one statement, and then proceed
with this technique of generating an approximating sequence ? The funda
mental theorem of calculus says yes; in fact, (2.33) can be rewritten as
f(x)=ff(t)dt
+ l
(2.34)
'o
This
is
operation involving integration rather than differentiation,
advantage of not having to choose a very wellbehaved function for the first approximant. Let us try again, with (2.34)
rather than (2.33). Letting/,
1, we find
and
now
an
have the added
so we
=
/i(x)
=
f2(x)
=
f
1 dt + 1
=
+ 1
x
"o
2
f (f + 1) dt +
1
+
=
x
+ 1
2
Jo
/3(x)=/o(y '+1)d'+1=^+5+x
+
+ 1
/b(x)=/o7-i(o^+i=5+(-^ ---+^+-+i
+
(2.35)
Now we're
getting
converges for any
/(x)
=
somewhere.
x.
lim/(x)=
recognized
we
sought
after function.
the solution of
need
the limit in
now
our
problem
see
=
that the series
(2.35)
(Of course the reader has long since
as
being
the
exponential
is the theoretical mathematics that will allow
Xn
J/(f)d-f+l=
Z-f
Jo
B=on!
function.
that it did in fact turn out that
(2.35) and correctly deduce
r*
/(x)
seen
0n!
Thus he should be reassured to
What
already
-.
=
this must be the
We have
Thus, letting
us
way.)
to take
The
2.10
Space of Continuous Functions
203
led to the
question of convergence in the space of continuous
proceed to that theory.
functions.
Let X be a closed bounded set in R", and let C(X) denote the space of all
continuous complex-valued functions on X. We know that iff and g are two
functions in C(X), then so are/+ g ana\fg and cf, for c, a complex number.
In particular, C(X) is a vector space on which multiplication is defined.
The vector space C(X) is quite different from the vector spaces C", R" : C(X)
is usually infinite dimensional (see Problem 49).
C(X) does not have any
Thus
we are
We
now
in fact, we wouldn't know how to choose one.
however,
C(X) is not very different. There is in this
particulars,
obvious "standard basis"
In other
space a reasonable notion of closeness. Two functions are close if their
values are everywhere close ; that is, if the maximum of their difference is
small.
This leads to
Definition 19.
a
notion of
Let X be
a
space of continuous functions
length
and distance in
C(X).
closed and bounded set in jR", and
X.
If /e C(X), the length of/is
C(X)
the
on
H/ll =max{|/(x)|:xeX}
If/ g
The
are
in
C(X), the
properties
of
distance
length
between/ and g is \\f-g\\.
and distance
are
just
those of the
corresponding
notions in R" :
\\cf\\
=
\c\ H/ll
ll/+<7ll< 11/11
+
11*11
the
ll/n =0, then /=0. What is important is what we can consider
notion of convergence of a sequence of continuous functions. We say that
term of the
/n ->/if ll/n -/II -> . that is> if the distance between the general
becomes arbitrarily small. This is the same as saying that
and
If
sequence
/
of/n at points
of X converge to the values of/in a uniform manner.
The value of these notions lies not only in their naturality, but in the now
the values
possibility of finding specific functions satisfying given properties
by techniques of approximation. Let us make this precise.
realizable
Definition 20.
C(X).
Let X be
We say that
{/}
that
lim
n-* 00
||/
-
/||
=
0
a
closed bounded set, and
{/}
is uniformly convergent if there is
an
sequence in
/e C(X) such
a
204
2.
of Calculus
Notions
We say that the sequence is
such that
II/b /mil
uniformly Cauchy if, for
e >
every
0 there is
an
N
whenever n,m> N
<
Examples
56. Let
x
uniformly
converges
/n'(x)
to
(1
x)x". This sequence
||/,||.
compute max|/(x)|
[0, 1], /(x)
Let
zero.
us
=
-
=
n(l-x)x"-1-x"
=
so/'(x)
11/.
be the interval
=
0 has the solutions
V
which tends to
57. On the
l/U+1/
+
n
x
=
n
0,
x
l\n
+
n/(n
=
+
+
Thus
1).
lj
zero.
same
interval the sequence fn(x)
=
sin
x/n
tends to zero,
for
||/J
sin
=
-?
-
0
asn->oo
58. Consider the convergence of the sequence {nx sin x/n} on the
[0,1]. Now we know that sin x/n ->0 as n - oo, but
interval
-
nx
oo,
so
cannot
we
We have to refine
our
n, it is very close to
nx
nxsin
about the
For
x/n.
large
product.
values of
Thus
x/n.
(2.36)
x
n
n
so we
=
-
make any deduction
information about sin
guess that
nx
sin
x/n
-*
x2.
Let
us
prove it
by computing
(2.37)
x
nx sin
n
In order to do
.
x
x
n
n
that, let
<
sin
r
n
us
provide
an
in the interval
estimate to
[0, 1]
our
guess
(2.36).
(2.38)
2.10
Then
The Space
of Continuous Functions
205
(2.37) becomes
X
nx sin
x
/
2
x\
1 +
nj
x
.
nxlsin
=
\
n
n
I
nx
=
.
x
x\
n
nj
sin
\
X
.
lliwll
<
sm
-x2
n
(2.39)
X
-
n
1
< n
X
nx
(2.40)
n
.
-^
=
n
n2
and since n_1 -*0
as n
->
co, we are thrc)ughL.
59. On the interval
It is not
does not
m
=
2n,
[0, 1] the sequence {sin nx} is not convergent.
Cauchy sequence. The distance ||sinnx sinmx||
become arbitrarily small as n, m->co. In particular, if
even a
we
||sin(nx)
have
sin(2nx)||
>
sinl
I
n
\
2n/
sinf 2n
\
|
=
1
2n/
The basic theorem about convergence of continuous functions is the
which plays the same role in C(X) as the least upper bound axiom
following,
does for R. It provides the assertion of existence of functions with prescribed
properties. In order to verify that a sequence of functions has a continuous
limit, we need only verify that it is a uniformly Cauchy sequence.
Theorem 2.14.
uniformly
X.
uniformly Cauchy
Suppose {/} is
Proof.
on
A
This
n,m~^.N.
means :
This
l/n(x)
Thus, for
sequence
of
continuous
-
uniformly Cauchy
e > 0, there is an
precisely
a
for every
x
is
means
fm(x) \<e
for all
x e
sequence of continuous functions
N > 0 such that 1 1/
/m 1 1 < e for
-
(2.41)
X
each x, {/(x)} is a uniformly Cauchy sequence of real numbers, and
Denote the limit, lim/(x) by f(x). We must show that this
thus converges.
function
functions
convergent.
->/(x) is continuous, and that/
converges
uniformly to/.
206
2.
Notions
First of all, if
e
>
of Calculus
0, choose N as above, and let
m
->
in
oo
(2.41).
obtain, for
We
n>N,
lim
|/(x) -/(x)|
|/(x) -/(x)|
=
<
for all
e
x e
X
m-* oo
Thus, if n^N, ||/-/||>e.
This
implies
that lim
||/-/|| =0,
desired.
as
n- oo
/ is continuous. Fix x0 e X. Let e > 0 and choose iV so large that
< e/3.
Since / is continuous, there is a S>0 such that ||x
x0||<3
/II
ll/
implies |/v(x) /(x0)| < e/3. Then if |x x0| < 8,
Now
| f(x) -/(xo)l
l/(x) -/(x)| + |/(x) -/(x)| + IMx) -/(x0)|
^
e
e
e
<3+3+3=,
desired.
as
seen one vector space of functions, we can easily see them every
The collection of bounded real-valued functions on a set X is a
Having
where.
vector space
taking
functions
same
over
the reals.
values in R" is also
taking
a
The collection of all bounded functions
on
X
the space of continuous
All the spaces here are endowed with the
vector space;
values in R".
similarly,
concept of length :
ll/H =sup{||/(x) ||: xeX}
Of
even more
interest
analytic operations.
are
For
the spaces of functions on which is defined some
example, if I is an interval, the space of all real-
valued functions which
C](/)
are differentiable on / is a vector space.
The space
of all functions whose derivative is continuous is also a vector space,
is the space C(n,(/) of all functions which have continuous nth derivatives.
The space R(I) of functions which are integrable on / is a vector space. These
(and other) examples are further elaborated in the exercises. Suffice it to
as
say here that the mathematical
theory which follows this point of view
(20th-century) development which has
only in foundations of mathematics, but in the
(called functional analysis) is
a
recent
had profound impact, not
practical application of mathematics
Let
us
return to the space
bounded set X in R".
in
we
a
space,
are
on
which
easily led
Once
are
in all branches of science.
C(X) of continuous functions
we
begin thinking
defined such notions
to consider functions
on
on
of these functions
a
closed
as
points
distance and convergence,
that space.
Naturally, such a
as
function is continuous if it takes convergent sequences into convergent
sequences.
2.10
The
Space of Continuous Functions
207
Examples
60. Let g e
/-/, that is,
C(X)
and define
\\fng-fg\\< II/.-/H ll^n
61. Define
II/2 -/2||
If
/-?/>
thusalso
=
<b(f) =fg.
$
||/-/||-*0,then
i/,: C(X)
ll(/ -/)(/ +/)||
the term
continuous, for
if
->o
C(X), <P(f) =f\
-
js
<
||/ -/||
,/, js also continuous,
-
||/ +/||
||/ +/|| remains bounded
for
(2.42)
while
\\f-f\\->o
||/2-/2||->0.
62. If P is any
polynomial, \/ip(f)
=
P(f) is continuous
on
(Problem 55).
63. Define M:
C(X)
-^
R, M(f)
=
C(X)
||/||.
This is continuous, since
\\M(f)-M(g)\\
=
\ ||/||-|l*ll l<ll/-<7ll
64. Let x0
AT and define F0 : C(X) ^ R, F0(f) =f(x0).
Certainly
is
continuous: for if/->/in C(X), then the maximum over X
F0
f
\L(x) -/W|
Fo(L) F0(f).
tends to zero; in
particular, |/(x0) -/(x0)| ->0,
so
^
65. The definite
/
=
[a, b~]
<=
R.
integral is
J/"-//|-|-f//n_/)
so
if/n ->/,
also
J//,
than that of
from
a
continuous function
on
C(I), where
For
->
\,f
<\\L-f\\(b-a)
A stronger and more important statement
the indefinite integral, as a function
Example 65 is that
C(I) to C(I) is continuous.
This is contained in the next pro
position.
Proposition
23. Let I
{x e R: a < x < b}. Suppose f is a sequence of
functions on I converging uniformly to f. Let F(x) J-* /
jafi Then F^F uniformly.
continuous
F(x)
=
=
=
,
208
2.
Notions
of Calculus
Proof.
x
I
F(x)-F(x)\=
J
Thus, taking the maximum
(/-/)
the
on
SS
11/. -/IK*
-
<
a)
11/. -/IK*
-
a)
left,
l|F-F||<||/-/||(6-a)
if /
so
->/ uniformly
so
also F^F.
Problem 56 is intended to demonstrate that
tion is
not a
continuous function
on
C(I).
on
the other hand, differentia
(It isn't
even
everywhere defined;
i.e., there are continuous functions that do not have a derivative.)
less, Proposition 23 has this consequence for differentiation.
Neverthe
Proposition 24. Let {/} be a sequence of continuously differentiable
functions on the interval [a, b~] and suppose that (i) {/'} is uniformly Cauchy,
(ii) f(a) 0 for all n. Then {/} is uniformly convergent to a differentiable
function f and f lim/'.
=
=
The
Proof.
proof
of this
proposition consists in a rereading
By that theorem
of
Proposition
23
via the fundamental theorem of calculus.
X
fn(x)=j
/'.
a
by Proposition 23, / is also convergent. If we let g
lim/' then lim/,
Thus, lim/, is indeed differentiable and its derivative is g
lim/'
so
=
,
=
Jj g.
=
.
Let
to the consideration of our
original problem. In fact,
generalize
slightly. Let c be a complex number, and let us seek a
differentiable complex-valued function / such that
let
us
return
now
it
us
f'(x)
This is,
tinuous
=
cf(x)
for all
x
by the fundamental
function/such that
f(x)
=
c\Xf(t)dt
+ l
and
/(0)
=
1
theorem of calculus the
(2.43)
same as
seeking
a con
(2.44)
2.10
Now that
follow
we
a more
and define the
Tf(x)
The
Space of Continuous Functions
have the necessary
theory
sophisticated approach.
function T on C(I):
c
=
ff(t) dt
and
point
of view available,
Let I be the interval I
+ 1
=
209
we
may
\_-R, R],
(2.45)
'o
/ such that/= Tfi that is, a fixed point of the transfor
technique is that of successive approximation. Let /0 be any
continuous function, and define f
Tf0, f2
Tf =T2f0, and in general
T"f0. We must show that the sequence {/} converges. If
/ Tfn_i
1 we can compute the sequence explicitly, and we find that
we choose /0
We seek
function
a
Our
mation.
=
=
=
=
=
(cx)"~l
(ex)"
Then if
m >
L(x)
n,
f(x)
-
=
(ex)'1
(cx)m
k
+
f + -b^-rv,
(m-1)!
[-R, R] the maximum
by |c|, and x by R. Thus,
On the interval
replacing
c
m
ii f
Um
All-
'
+
(|c|Kr
i
+
k=o
^R)m'1
of this
i
+
i
+
k\
k=o
+
1)!
expression
is dominated
by
(lcl*>"+1
(n
(m_i)i
m,
(cx)"+i
;
(n
ml
+
l)!
K\
Since the series
(f.
(kl R)k
is a Cauchy sequence, so by (2.46),
converges, its sequence of partial sums
Since T is
{/} is a Cauchy sequence and is thus uniformly convergent.
continuous
on
lim/,
=
C(I),
lim
we
have
T(/B_,)
=
Tflim/,-,)
=
r(lim/)
210
so
to
2.
Notions
lim/ solves
spend a few
the
The
number
/'(*)
given problem. This function is important enough for
paragraphs discussing it.
exponential function, denoted exp(cx),
is the solution of the differential
c
cf(x)
=
us
more
Definition 21.
complex
of Calculus
/(0)
=
or
ecx, for any
equation
1
First of all, this definition makes sense, because there is
only
one
solution.
If g also solves, then
d
[ecx~\
ecxg'
cecxg
cecxg
cecxg
=
92
dx
since
g'
ecx
or
the
Thus e^g'1 is constant. Since its value at 0 is
cg.
From
these discussions we have these additional
g.
=
=
exponential
=
b
ex+y
we
from
(ii)
1,
of
0
n!
eV.
never zero.
=
Part
(ii) follows
and
=(*)
must have
h(x)
=
e',
'(0)=_=1
so
(ii) is verified.
Part
(iii) follows immediately
:
^CXq-CX
(e")-1
properties
=
"'(*)
Thus
=
function
Part (i) follows directly from the argument above.
uniqueness. Fix v, and define h(x)
e'+yle*. Then
Proof.
so
=
ecx is
from the
1, e^g'1
25.
Proposition
(ii)
(iii)
0
92
_
gCX-CX
_
Q
_
1
=<r".
PROBLEMS
49. Let / be
dimensional.
a
nonempty interval in R.
Show that
C(I) is infinite
The Fixed Point Theorem
2.11
50. Show that the sequence of functions
on
211
the closed unit disk in C
defined by
n
X
ux)=Lk2
converges.
| 2 Z"/M converge
51. Does the sequence
52. Let {a} be
these facts:
a
sequence of
on
the closed unit disk?
complex numbers such
that
2 M
< co-
Verify
For every z, |z| <
(a)
i/Wi
2
<
n=
l,/(z)
2*='
=
ifl-i
1
(b) /is continuous
{z
on
e
C:
\z\
<
1}.
2i?=N+il,.l^0as/V^*>.
/
Show that
g be continuous functions
\\fg\\
<
54. Show that
x
x
n
n
Is
\\g\l
the interval
for all
<
sin
ll/ll
on
This is true because
polynomials f(z)=^Liaz\
uniform limit of the
53. Let
az" converges, and
\\fg\\
<
on
since
/is
the
||/-/v||<
the closed and bounded set X.
11/11
lltfll possible?
[0, 1],
n
n2
55. Let xi,
.
.
.
,
xk e X
and p be any
polynomial
in k variables.
Define
r-.C(X)^C
,F(/)=K/(xi), ...,/(*))
Show that T is continuous.
funct.ons which
sequence {/} of differentiable
is
not
convergent.
that
{/'(*)}
convergent, but such
56. Find
is
a
uniformly
2.11 The Fixed Point Theorem
of the
point theorem is a generalization
the discussion
approximations described above in
The fixed
exponential
function
root of
Newton as a technique for finding
First,
is this.
method
Newton's
technique was first used by
polynomial equations. Simply stated,
This
technique of successive
of the
212
2.
Notions
of Calculus
technique is described by means of which one can transform a given approxi
a root into a better approximation.
One then chooses a reasonable
approximation, applies this technique to it to find a better one. Having
this, one again applies the technique : if it's a good one, the result is an even
better approximation. Continuing in this way, one obtains a sequence of
approximations which should converge to the root. Now, having described
the procedure, let us turn to Newton's specific technique for bettering
approximations.
Let / be a given real polynomial. We want to find a point x0 such that
f(x0) 0. Choose a px so that/f^) is small. Now, replace the function by
its linear approximation at p1: L(x) =f(Pi) +f'(Pi)(x
Pi), and let p2 be
the root of L(x)
0. In other words, replace the graph off by its tangent
line and let p2 be the x intercept of that line (see Figure 2.17). Now apply
this procedure to p2
Let p3 be the root of the linear approximation to /
at p2
and so forth. We can describe Newton's technique abstractly as
follows : For any point p, let T(p) be the zero of the linear approximation of
/at p:T(p) solves the equation f(p) +f'(p)(T(p) p) 0. (We must
a
mation to
=
=
.
,
-
Figure 2.17
=
2.11
# 0 for T to be
that/'
T(p)
assume
have
we
p, and
=
The Fixed Point Theorem
well-defined
a
conversely, thus
213
function.) Clearly, if f(p) 0,
are in reality
seeking a fixed
=
we
point of T\
T has the property of contraction
Suppose
1 such that
on some
interval I.
There is
a
Ty\ < c\x y\, all x, y e I. Then Newton's method
0 (or/'(x)
works. There is a root of/(x)
0) on the interval /, and it is
the limit of the sequence x0 Tx0 T2x0
where x0 is any point of /.
This is the content of the fixed point theorem.
We now state and prove it explicitly for subsets of C(X). It will be clear
that the theorem is true for subsets of R", by virtue of the same argument.
c <
\Tx
-
-
=
,
,
Suppose S is
of sequences
Theorem 2.15.
a
,
closed
a
S contains all limits
which is
=
in S.
contraction, that is, there is
II T(f)
Then there is
-
a
T(g) ||
unique
< c
||/
-
g \\
...,
of functions in C(X) : that
Suppose T is a mapping of S onto S
a c <
set
1 such that
for allf,
g
e
S
continuous function f0 such that
T(/0) =/0
.
Proof. Certainly the fixed point is unique. For if T(f0) =/0 and T(f) =/, then
l!/o-/ill= lir(/o)-r(/)]|<c||/o-/i||<||/o-/ill unless ||/0-/i||=0, that
is, /o =/i.
Let the sequence {/} be defined as follows :fi=f,f2
Now let fe C(X).
Tf,
Tfn-l. {/} is a Cauchy sequence. For
f3
Tf2,...,fn
=
=
=
ll/.+i-/.ll
so we can
=
ll7y.-3/.-ill^c||/.-/.-il|
verify by induction that
ll/.+i-/.ll<c"ll/i-/oll
Thus, for
m
> n we
11/- -/.!!<
have
ZUi+i-fj)
<mf\\fj+l-fj\\
J=n
c"
#)"*
<
Since
c
lim Tfn
<
=
II/1-/0IKII/1-/0I
Since T is continuous, Tf0
{/} is Cauchy, so has a limit f0 e C(X).
function.
fixed
lim/+1 =/0 and thus f0 is the desired
1
-
,
,
n-eo
n-> oo
As
an
\IcJ)
x0
an
>
illustration
0 such that
on
x02
the real numbers let
=
a,
us
prove that if
by Newton's method.
First,
a >
we
0, there
is
describe the
214
2.
p).
2p(x
Tp
=
of Calculus
0, the linear approximation to x2
Thus, the zero of this linear polynomial is
Let p
T.
map
+
Notions
contraction
a
p2
a
+ p
^2p
-IH
Clearly, if T has a fixed point x0
that Tis
at p is
a
>
,
1
|Tx-Tv|=;
we
x02
must have
=
Thus,
a.
we
must show
closed interval:
on some
a
a
1
X
V
~2
y +
a
,
(y -X)
x-y +
xy
-\ x-y\ 1-^
XV
only ensure that
\.
{x: x2 >a/2}.
(a/xy)
Then for x, y e I, xy > a/2, so a/xy < 2, which is the desired inequality.
Thus, by the fixed point theorem there is an x0 with x02 > a/2 such that
Since a, x,y
1
x02
=
now
give
a
Sometimes
theorem.
j
a
(a/xy)
<
1,
so
contraction with
somewhat
a
c
need
we
Let/
=
=
function of
x:
value (0, 1), and near
of y. The relations
=
1
y
more
subtle
application
of the fixed
point
relation between two real variables determines
function of the other.
as a
are
1
positive,
1, for Tto be
a.
We shall
as a
all
axe
>
=
For
(1, 0)
we
sin(x(log y))
example,
x2
x;
+
y2
the relation
=
1
gives
should write
=
x
y
=
(1
=
x
+ v
=
one
0 determines
x2)1/2
y2)1/2 as a
(1
near
the
function
0
somewhat less transparent, nevertheless
as a function of x.
we can
ask whether
or
not
they
do determine y
Suppose
now, in
F(x, y)
defined in the
(2.47)
=
general
we
have
equation (see Figure 2.18)
an
0
(2.47)
plane. We ask : does
saying y g(x) ?
amounts to
=
there exist
More
a
function g of x such that
is there a function g
precisely,
such that
F(x, y)
=
0
if and
It is not hard to find
a
only
if
y
=
#(x)
necessary condition.
For there to be such
a
function
2.11
y is
a
function of
The Fixed Point Theorem
y is not
jc
a
215
function of j:
Figure 2.18
it must be the
that each line
constant intersects the set
F(x, y) 0
F(x, y), as a function
of y on lines x
The root of
constant must take the value 0 only once.
0 is then the value g(x). Now we recall from one-variable theory
F(x, y)
that a function H(y) will take all values once if H'(y) j= 0. Thus the reason
in
only
one
case
x
point (see Figure 2.19).
=
=
Thus the function
=
=
able condition to
impose
on
F is that it has
a
continuous
partial
derivative
with respect to y, and dF/dy = 0. This condition turns out to be enough.
More precisely, suppose that F is defined and has continuous partial
derivatives in the neighborhood of the origin in R2, and dF/dy(0, 0) ^ 0.
Figure 2.19
216
We seek
a
and
F(x,
F(x0 y)
function g defined in
g(x)) 0. If we fix
This
0.
=
function of y
of F(x0
y)
,
a
brings
neighborhood of x
x
=
,
as a
of Calculus
Notions
2.
=
right
T(y)
us
is the
Newton did:
as
at y ; that
+
dy
0 such that
a
zero
of the linear
#(0)
=
0
root of
Define T
approximation
is,
dF
F(xo,y)
=
0, then we seek
x0
back to Newton's method.
near
(xo,y)(Ty-y)
=
0
or
dF
Ty
Just
as
Thus,
=
y-
Jy
in Newton's
need
we
y for x0
that fixed
we now
case
the solution of
point.
that T is
a
that it will have
This
(2.48)
,
,
only verify
near x so
F(x0 y)
(x0 y)
application
F(x0 y)
=
,
contraction in
fixed
a
0 is the fixed
some
point
of T.
interval of values of
point ; and we define g(x0) to be
point theorem really works, as
of the fixed
shall prove.
Suppose that F has continuous partial derivatives in
and that F(0, 0)
0, dF/dy(0, 0) # 0. Then there is
0),
neighborhood of(0,
in
x
some
interval
6, e) such that
g
definedfor
(
function
Theorem 2.16.
=
F(x, y)
Proof.
=
0
if and only if
Instead of (2.48)
we
y
=
a
a
g(x)
consider something
slightly simpler.
For
x near
0,
define
Tx(y)=y-
d2L-(0,0)
Ty
F(x,y)
(2.49)
We want to find the fixed point, if it exists, of (2.49).
e
17 < y < 77 in which Tx is
< x < e,
a
Thus
we
seek suitable intervals,
contraction
\8F (0,0)
.
T,( yi)
By the
mean
-
Tx(y2)
=yi-y2-
value theorem there is
a
[F(x,yi)-F(x,y2)]
t, between y1 and y2 such that
dF
F(x, y,)
-
F(x, y2)
=
dy
(x, 0(yt
-
y2)
(2.50)
2.11
Equation (2.50) becomes,
Tx(yi)
Tx(y2)
-
=
-
(0, 0)"1 8-f (x, 0
y2)\lI 3-f
dy
dy
(0, 0).
we
may choose
s<yi<e,
of e,
e <
y2
e so
and
< e
(2.51)
-
Now the term in brackets is continuous in
Thus
217
substitution,
upon
(yi
The Fixed Point Theorem
(x, <!;) and has the value 0 at
\ if -e < x < e,
that that term is less than
is between yy and y2
.
With this choice
(2.51) gives
\Tx(yi)-Tx(y2)\<i\yi-y2\
Then, if
Tx is indeed a contraction. Define g(x) as the fixed point of Tx
0, then by (2.49) Tx(y)
y, so we must have y
g(x). On the
F(x, y)
other hand, if y
g(x), then Tx(y) y, so again by (2.49) we must have
The
theorem
is proved.
0.
F(x, y)
To say that the function g exists is already good enough, but much more is
We will leave the verifica
true: g is a continuously differentiable function.
tion of this fact to the interested reader (see Problem 58). In Section 7.2
we shall reconsider this theorem (known as the implicit function theorem)
in many more variables. The beauty of the fixed point theorem is that the
general context does not at all complicate the ideas, nor the verifications.
so
.
=
=
=
=
=
=
EXERCISES
31. Find, by Newton's method, a sequence of numbers converging to
the square root of a, for any a > 0. Now, do the cube root.
32. Find a sequence converging to a root of these polynomials :
(c) xb-2x2-3x + 2
(a) x3 + x2 + x+l
(b)
(a)
F(x, y)
33.
(d) x5 -x- I
F(x, y)=x sin(xy). For what values of (x, y) such that
0 defines v as a
0 is it true that nearby the equation F(x, v)
x2-x+l
Let
=
=
function of x?
(b) Same problem for
(ii) F(x, y) x" y,
(i) F(x,y)=xy2 + 2xy+\,
x2
v2+
(iii) F(x, y)
34. Let F(x, y) be differentiable in a domain D, and (x0 ,y0)e D such
that F(xo,yo)=0. Suppose g is differentiable and has the property
Show that
0.
vo F(x, g(xj)
<7(*0)
=
=
=
=
,
9{-Xo)-
8F\dx(x0,y0)
8F/dy(x0,yo)
-
218
Notions
2.
35. Find
of Calculus
g' where
g is defined
implicitly by
l
(c) exy
(d) e"=y
xsin(xy)=0
cos(x+v)=y
(a)
(b)
=
PROBLEMS
57. Prove the fixed
Theorem
on
IfS
is
then there is
S,
a
a
point theorem in
subset
unique
of
R" and T is
e
S such that
y0
F(x0 y0)
,
0,
differentiable
= 0.
dF/8y(x0 jo)
(F(x, y(x))=0
,
Theorem 2.15
as
defined on S and is
T(y0) y<>
and
contraction
.
be
Let g
a
=
partial derivatives
58. Let F have continuous
=
R":
(x0 y0)
near
the
,
We
g(x0)=y0).
and suppose
function described in
can
prove that g is
follows.
(a) First of all, by the mean value theorem, for any (x, y), there is
(f 7j) on the line between (x0 y0) and (x, y) such that
,
,
8F
8F
F(x, y)
F(x0 y0)
-
(I t/)(x ~Xo)
=
,
+
>
(, -q)(y
Why is the mean value theorem applicable?
(b) Now, if we substitute y=g(x), y0 =g(x0),
(|, rj)(x -Xo) +
=
-
we
y0)
have
dF
8F
0
a
Yy
(f V)(9(x)
0(Xo))
-
Thus
g (x)
-
x
g(x0)
Xo
-
dF/8x(lj, ri)
8F/8y(, tj)
Conclude that g is differentiable and
g'(x0)
2.12
8F/8x(xo,g(x0))
=
8F/8y(Xo,g(xo))
Summary
z,... of
A sequence zx
to C.
complex numbers is a function from
{z} converges to z if, for every e
The sequence
positive integers
there is an N such that \z
z\ < e for
A convergent sequence is
n >
the
>
0
N.
bounded, but
not
conversely. A monotonic
Cauchy criterion: a
bounded sequence of real numbers is convergent.
2.12
219
Summary
sequence {z} converges if, for every e > 0, there is an N such that \z
for both n,m>N.
The series formed of a sequence {zn} is the
sequence of sums
zm \
-
< e
(?=i zj-
If the sequence of sums converges, we
say that the series converges and denote
the limit by "= i z
If 2 z- converges, then z -> 0, but not
conversely.
If {ck} is a sequence of nonnegative numbers,
c*. converges if and only
if the sequence =1 ck is bounded. A series
z converges absolutely if
.
XlzJ
Absolutely convergent
< oo.
series may be summed in any convenient
way.
Tests for
Convergence
Suppose |z| < \wB\ for all but finitely
(0
E |w| converges, z is absolutely convergent, (ii) if
so does
2 |h>|.
comparison test.
if
If
root test.
is
|c|1/n
< r
for
some r <
1 and all but
many
2
finitely
Then
n.
|z| diverges,
many n,
2
c
absolutely convergent.
If \cn+1/cn\ < r for some r < 1 and all but finitely many n,
is
2 c absolutely convergent.
The sequence {yk} of vectors in R" is said to converge to v if, for every
e > 0, there is an N such that
v || < s for k > N.
A sequence of vectors
|| yk
ratio test.
-
converges if and only if it does so in each coordinate.
A set S is closed if and only if yk e S, lim yk
v implies
=
v e
S also.
Every
sequence contained in a closed and bounded set has a convergent subsequence.
An .Revalued function defined in R" is said to be continuous at v0 iff is
defined in
a
neighborhood
tion is continuous
on a
of v0 and yk
->
v0
implies f(yk) </(v0).
point of S.
set S if it is continuous at every
A func
If S is
a
closed and bounded set, and / is a continuous real-valued function defined
on S, then/is bounded and attains its maximum and minimum.
Sections 2.6 and 2.7
the definitions here;
are
only
mainly about integration.
major results.
fundamental theorem of calculus.
interval [a,
F(x)
exists for all
b}.
=
Then the
We shall not recollect
the
Suppose / is continuous
integral
ff
x e
[_a, b~\.
Fis differentiable
on
(a, b) and F' =/.
on
the
220
R
2.
Notions
fubini's
theorem.
=
Jj
Let /be
I in R".
x
x
of Calculus
an
J/can
integrable function defined
computed by iteration:
\j ([ \l /(*'.
*") dx" ^"-1
=
Let /be
a
v
is
v
is defined
a
real-valued function defined in
R", the directional derivative
vector in
on a
rectangle
be
'
dx1
neighborhood of x0 in R". If
df(x0 v) of/atx0 in the direction
a
,
by
lim/(X + ty)'
~
/(Xo)
f->0
(if it exists).
The
,(x0)
=
partial derivative
df(x0,Ei)
derivatives
If these
partial
df(x0 y)
is linear in
,
If the
partial
of/with respect to x' at x0 is
v.
are
We
derivatives
can
all defined and continuous
df/dx'
all exist in
an
compute the derivatives d(df/dx')/dxJ. These
derivatives. If all first and second derivatives
in
an
open set
d2f
d2f
dx1 dx'
throughout N.
Suppose that /
an
F(x)
=
has continuous
dx
partial
interval of reals, and D is
f
f(x, y) dy
Then F is differentiable and
d,..,
00
x0
,
then
open set
are
we
may be able to
the second-order
off exist
and
are
partial
continuous
N, then
dx' dx1
where I is
near
write
=
f
W
J g^0y)dy
a
derivatives in the domain I
domain in Rn.
Let
x
D,
2.12
Suppose /is
verges to L
a
as x
real-valued function defined
->
written
oo
lim/(x)
=
L if
R.
on
|/(x)
-
We say
L|
221
Summary
that/(x)
con
be made arbi-
can
x->ao
trarily small by taking
on
x
sufficiently large.
If now /is
a
continuous function
R such that
f/
lim
^0
x- oo
exists,
say that
we
/is integrable
on
R.
If lim
J* |/|
exists, /is absolutely
X~*QO
If/ is a positive, decreasing continuous function
if and only if "= i/(n) < oo.
/exists
jf
We denote by C(X) the collection
Let X be a closed and bounded set in R".
of all complex-valued continuous functions on X.
C(X) is a vector space.
If/is in C(X), the /en#fn of/is
Integral
integrable.
defined
on
test:
R, then
H/ll =max{|/(x)|:xX}
For/ # in C(X) the distance between/and gis \\f- g\\. If {/} is a sequence
in C(Z), and ||/ -/|| ->0 as n-> oo for some /e C(Z), we say that {/}
Cauchy criterion. Suppose {/} is a sequence
converges uniformly to /
in C(X) satisfying the following condition: for each > 0, there is an N such
that ||/ -/J| < 6 whenever n,m>N. Then there is an/e C(X) such that
/->/ uniformly.
integration.
also
J*/
The
c
-"
If X is
an
interval in R,
and/, -^/uniformly
J/ uniformly.
exponential function,
denoted
is the solution of the differential
exp(cx), or
equation y'
ecx for any
=
cy,
y(0)
in
C(X)
complex
1.
=
then
number
It has these
properties :
(cx)n
=
c(x+y)
ecx is
_
o
n!
ecxecy
never zero.
fixed point theorem.
onto S
mapping of S
II T(f)
Then there is
-
a
Let S be
which is
T(g) ||
< c
||/
-
a
a
closed
set
offunctions
in
contraction; that is, there is
g II
C(X) and
c <
for all figeS
unique continuous function f0 such that T(f0) =f0
T
a
1 such that
222
of Calculus
Notions
2.
implicit
theorem.
function
Suppose
(0, 0), and
derivatives in
a
neighborhood
Then there is
a
function g defined for
F(x, y)
=
if and
0
of
only if
y
x
=
in
that F has continuous
that
some
F(0, 0)
=
interval
(
partial
0, dF/dy(0, 0) # 0.
e,
e) such that
g(x)
FURTHER READING
M.
Spivak, Calculus, Benjamin,
New
This is
York, 1967.
It is
text in the one-variable calculus.
an
eloquent
an
excellent reference for
a
full
treatment of the material in this
chapter.
Lichtenberg, Mathematics for Scientists, Benjamin,
New York, 1966. This is a review of the theory of calculus from the point
of view of the physical scientist. It includes a chapter on numerical analysis.
T. A. Bak and J.
C. W. Burrill and J. R. Knudsen, Real Variables, Holt, Rinehart and
Winston, New York, 1969. An advanced text, going thoroughly through
the material of this
chapter and beyond to
MISCELLANEOUS
59. Let
So also is
the
a
of Lebesque
Then {x +
yn} is also
a
linear
subspace
Show that the collection C of convergent sequences is a linear subAlso C0 the collection of all sequences converging to zero is a
.
,
subspace of B.
These spaces are all infinite dimensional.
lim
on convergent sequences in the obvious
R : lim{x}
lim x
Show that lim is a linear function.
62. Define the function
way : lim : C
-*
"
"
=
.
63. What is the dimension of the space of linear functional
annihilate C0?
64. Let xi
x+i
sequence.
Show that it is not finite dimensional.
vector space.
space of B.
linear
a
r; thus the collection S of all real
60. Show that the collection B of bounded sequences is
of the vector space S of all sequences (Problem 59).
61
integration.
PROBLEMS
{x} and {yn} be sequences.
{rx} for any real number
sequences is
theory
=
=4,
i(xn + 3/x).
x2
=
on
C which
x are defined, let
i(4 + ), and once x2
{x} converges. Assuming that, find the
,
.
.
.
,
Prove that
limit.
65.
(a) Show that for
lim
n"/(n + 1)'
=
every
integer k,
1
limnV(n+l)t+1=0
lim
nk+i/(n
+
1)" does
not exist
(b) Let k be an integer, and 1 > h > 0.
(c) Show that lim n/h" does not exist.
Show that lim n"h"
=
0.
2.12
66. Let
x+i
xi
=
Summary
223
1, and in general
1+x,
_,
3
3 + x
Find lim x
67. Suppose lim
.
Let y
Let A: be
(a)
(b)
=
z
=
z.
J(z_, + z). Then
a positive
integer.
lim ,y
z.
Now let {y} be defined by
=
1
y-
=
+
~k~+\ ^Zn Zn+1
"'
^
z"+)
Then lim y=z also.
(c) This time take
1
y
=
-
(zi +
+ z)
Once
again lim > z.
Suppose that / is
lim/(c) =/(c).
=
68.
continuous
at
c,
and
lim
c
=
c.
Then
69. Let {c} be a sequence of complex numbers, and suppose (|c|)""
R.
Show that R'1 is the radius of convergence of 2 cz".
70. Let {.$}, {?} be two sequences of positive numbers such that lim s tn '
=
exists and is
nonzero.
Then 2 s converges if and only if 2 t converges.
{c} be a sequence of positive numbers. Suppose that for every
we have also
sequence of positive numbers {/?} such that 2 Pi <
2 C"P < - Prove that {c} is bounded.
72. Verify Schwarz's inequality:
71. Let
iiaAiV^iw2- !>i2
/
1=1
n=l
n=l
by virtue of the same fact for finite sums, which was dis
Chapter 1 .)
Is the reverse
73. Prove that if 2 kl2 < , then 2 (Un)\a\ < co.
It is true
(Hint:
cussed in Problem 74 of
implication
true?
74. Let S be
a
S} is a closed
75. Suppose that / is
for all
on a
76.
R".
Suppose
Let x0
there is
,
an x2 e
a
/is
{yeR": <v, s>=0
a
R" and
positive real-valued function defined
log /is also continuous.
continuous
Show that
that
Xi e
=
set.
s 6
set S in R".
Show that (S)
subset of Rn.
continuous real-valued function defined
c e
R" such that
R be such that
f(x2)
=
c.
/(x0)
<
c
</(x,).
on
all of
Show that
224
2.
Notions
of Calculus
11. Show that if /is
a
continuous function
on
the interval /
taking only
rational values, then /must be constant.
78. A set S in R" is called connected if every continuous real-valued func
tion has the intermediate value property. Show that this is equivalent to the
following definition:
A set S is not connected if there is
defined
79.
a
continuous real-valued function /
two values.
S which takes
precisely
Verify the following assertions :
(a) A ball in R" is connected.
(b) The set of integers is not connected.
1} is connected.
(c) The sphere {xe R3: ||x ||
(d) The union of two balls in R" is connected if and only if they
on
=
intersect.
(e) An open set is not connected if and only if it can be written as the
disjoint union of two nonempty open subsets.
(f ) A closed set is not connected if and only if it can be written as the
disjoint union of two nonempty closed sets.
80. Let / be a continuous function on the closed and bounded set X.
Then/is uniformly continuous; that is, given e > 0, there is a S > 0 such that
for all x, y e A'such that |x
y\ < S we have |/(x) f(y)\ < e. Supposing
not,
derive
we can
a
contradiction
as
"
follows.
There is
an e0
such that
"
for every 8,
is not true. Taking
|x
y | < S implies \f(x) f(y)\ < e0
S
1/n, there are x,yn with |x y\ <l/but \f(xn)f(y)\ >e0. Since
X is closed and bounded, these sequences have convergent subsequences:
=
Show that lim x'
{x\}, {y/}.
=
lim
/ but |/(lim x') /(lim y')|
> e0
,
a
contradiction.
81. Let L be
a
linear functional
on
R" and choose v0 such that
||w0||
=
1
and
L(v0)=max{L(v): N|
=
l}
Show that for every v e R", L(v)
L(v0) <y, v0~>.
82. Let / be an integrable function on the rectangle [a, b]. Let R, be
rectangle [a, b + <(b a)], for 0 <, t <, 1. Verify that / is integrable on
=
each rectangle R, and define F(t)
JRt /. Show that / is continuous. Is
/differentiable?
83. Let Q
{pjq : p, q integers with 0 <,p <l q). Q is a subset of the unit
interval [0,1] which is not measurable. For surely Jxo
0, and if
kj R => Q, then also Ri u
u R => [0, 1], so
Ri
J 2 Xi ^ !> and
u_=
,
=
=
thusj Xq
=
84. Let
1-
/ be
integrable nonnegative function defined on the domain
D
{(x, y,z)eR3; 0 <. z <>f(x, y) ; (x, y)eB}.
Verify that Vol(Z>) j /.
B
c
R2
and
an
consider
=
=
2.12
85.
225
Summary
Suppose that / is a continuous decreasing real-valued function of a
0. Then f<f f(x) sin x dx converges (compare
f(x)
real variable and lim
=
X-.CO
this with Leibniz's theorem for
86.
Suppose that /is
/(x)^ + oo
if, for
-s-
Hxll-*
as
every M there is
that if /is
|jx||
a
a
series).
real-valued function defined
We say that
R".
oo
if such that
a
f(x)
> M
real-valued continuous function
oo, then /attains
on
a
minimum at
some
on
whenever ||x || > K.
R" such
that/(x)
->
Show
+
oo as
point.
87. Define
/(x)->0
in
a
way
||x||-*oo
as
suggested by the definition in
continuous function
and
a
88.
minimum
on
on
problem.
Show that if
a
a
maximum
R".
Suppose / is
a
real-valued function which has continuous
derivatives in the ball {x
<7(x)=f
the above
R" has this property, then it attains both
e
R": ||x|| <
1}.
partial
Show that the function
f(tx)dt
same properties, and find Vg.
89. Let /2 be the space of sequences {c} of real numbers such that
has the
2ici2<
n
=
l
Because of the result in Problem 72
are
in
(Schwarz's inequality), if {c}
and
{dn}
I2, then
<{*}, ,}>
2
=
n
=
c*d
l
Show that I2 is a Euclidean vector space with that inner product.
90. The space of continuous functions on the unit interval can be made
into a Euclidean vector space in this way:
converges.
</<?>=[
f(t)g(t)dt
Corresponding
by ||
1 12
so as
product is a notion of length which we denote
distinguish it from the modulus || || introduced in the
to this inner
to
226
2.
Notions
of Calculus
Show that this length is deficient in these respects :
We can have ||/||2-*0 without having ||/||-^0.
text.
(a)
(b)
Cauchy
We
can
have
a
converge to
a
sense
of continuous functions which is
{/,}
sequence
sequence in the
of the
continuous function.
length ||
||2
On the other
(c) if ||/. ||. ^0, then ||/||2 -*0 also.
91. Suppose L: C[0, l]-*j? is a linear function.
tinuous if and only if there is an M > 0 such that
,
a
but which does not
hand, show that
Show that L is
con
\L(f)\^M\\fU
92. Show that there is
f'o(x)
=
a
for all
(/o(x))2
differentiable function
unique
x
and
Do it
/o(0)
by applying the fixed point theorem
ontheset{/<=C[i ]: ||/||<f}:
=
f0 such that
i
to the function T defined below
x
Tf(x)=\
93. We
of (n
f2(t)dt+i
talk of open and closed sets, and convergence in the space M "
n) matrices, merely by considering them as vectors in R"2. Doing so
these statements :
can
x
verify
(a)
(b)
(c)
(d)
The set G of invertible (n x n) matrices is open.
The set of triangular matrices is closed.
The function A->A2 is continuous.
lfp is
any
polynomial in
one
variable the function
T->p(T)
is continuous.
(e) lim (x/n!) 2=o (l/n!)TB
exists for all TeL(R",
Rm).
fl-00
94.
Suppose g is a continuous real-valued function
[a, a]. Show that the implication
f
J
g(t)f(t)dt
=
on
the interval
0
-0
for all
F
implies g 0 holds whenever F is any one of these classes :
(a) F=C([-a,a]).
(b)F=Ci([-a,a]).
(c) F is the collection of all polynomials.
(d) Fis the collection {xi'-Ia subinterval of [a, a]}.
(e) Fis the collection of all continuously differentiable functions such
that/(-a)=/(a) 0.
fe
=
=