[yx+y2
<
which is 2
=
then3xy(x+y)=0,sox=0ory=0orx=-y.
(b)
The
first
equation
impliesthat
4x2
we
obtain
So
either
x+
y
=
(c)
Apply
part
(b)
with
y
on
k
(see
Chapter
true
for
a
k(k
=13+---+k3+(k+1)',
(n+1)n
,
part
(a).
(c)
Let
mi
1+Ê 1-Ä
so
Problem
1-18
<
+
(1)
for
all
i,
which
means
that
i<j
To check
The
=a·b+a
+
statement,
made
since
he
and
Prof.
Prof. B knows.
 
at
x
+-x+-
=a
x+
If
D
same as
imum
of
these
numbers
is
a
2
(-2md
(x
y2
x
y2
1=
0.
(e)
We
have
||v+w[2=(v+tv)•(v+w)=v•v+2v.w+w·w
||v-w||2=(v-w)•(v-w)=v•v-2v·w+w·w.
Subtracting
the
second
=4(v.w).
3.
(a)
We
have
Re
(v)
=a-(e·(b-w))=a-(b-(e-w))
w2
(and
7.
As
intersection
z
to
F2
is
the
length
of
the
vertical
line
L'
from
z
to
Ci
to
except
that
now
the
sum
will
always
be
symme ric with
(cos
curve
shown
below.
(Hence
the
name
 cardioid
heart-shaped).
(b)
The
point
with
polar
coordinates
(r,
0)
case
in
which
does
not
existatherwise
we
would
have
lim
f(x)
A
r
B
C
O
(b)
As
n
becomes
very
[a].
2.
12.
(a)
5
|¢(x)/xn|
on
R.
I
i
b
b
18.
(a)
Apply
Theorem
3
to
the
(continuous)
unction
d(z)
a
function
f
when
1/4
a
were
not
a
f
(x)
 
that there
sup{a,
is
uniformly
con-
tinuous
on
[0,
oo).
(The
argument
for
this
is
a
g(y)]
relation
between
f'
and
rational,
then
a
(x
made
very
origin
is
b(t)2
1
f'(x)
k=o
2a
4a2
2a
=--+c,
4a
or
b2
signs, then (x)
signs,
so
h(x)
can
be
incorporated
m
f(x)
On the
f
(x)
even.
may
consider
only
x
in
(a,
a
(It is
f (x)
f(x)
g(x)-g(a)
lim
To
prove
this,
h
62.
(a)
n;
that
f
is
convex
is
some
xo
in
(0,
x)
with
f'(xo)
that
k
n
obvious
way,
we
have
f(txi
-i
2
/(x)=0.
or
f'(-1)
19.
(ii)
the
a
reason-
of
to
the
tangent
line.
6.
(a)
Letting
u(Ð)
horizontal
axis.
For
to
the
tangent
line.
(d)
By
part
(a)
we
have
f
(9)
cos
Ð
oo
and
-oo]
8.
(a)
From
arch
of
the
cycloid
we
have
t
g¯'.
this
lb
mplies
that
xP
dx
the integral of
+-
---
+-
---
+-··
_<t;+c
stsct¿}=m¿'.
f(ct)dt.
ca
Ja
some t
the numerator
is
an
example.
P
which
contains
Pi
and
P2.
(b)
Part
(a)
showsthat
f
signs and consequently
i-1
5
(M¿
this is the Schwarz inequality.
(b)
First
proof:
If
g
[f(x)g(y)-
are
true
for
x
f°°
f.
we
have
-1
li
dt
lim
x
-
2x
1+
x2'
1--x2
are
within
x
of
each
other,
x
is
the
maximum
Then
sin
y
13-16,
a
step
function
s
5
with
Ï
b
f
on
[1
the
improper
that
cos
e2*7
to
have
a
pieced-together
solution
like
0
x
5
0
hm
the form
f
part
(a).
(c)
Let
P
=
2ue
du
4
(by
Problem
4(ii))).
becomes
1
2
1
f
(0)
i=1
-
m(bi
<e.
(and
shows
that
if
any
two
of
the
three
symbols
2n+1-1-3-5---(2n-1)
true
when
7
\
3t¿2t¿_i
(x,
a).
3
(b)
Now
by
Problem
19-4(ix)
3.
(ii)
9
x
Ï
x
e.
Choose
0<al'<a1<ai <a2'<a2<a2 <-··<an'<an<an <1.
Then
f
(a¿ )
f
(a¿')
mined
by
a
. . .
 
g(x)
by
b,
so
has
a
limit
l.
Clearly
l
a
convergent
subsequence,
say
f(x,
two
sequences
{f
(xx)}
and
{f
(ys)}.
(c)
Given
e
definitions.
a>1
a<1.
So, by
there
were
another
< <
6
ef
n=1
=
ai+2a+---+nan
terms,
and
E
q,
is
the
and0+a2+0+a4+0+as+--.,
whose
sum
is
Egg
an.
13.
For
every
N,
we
have
n
ak
k=1
i=1 i=1
Está mal... tomar las desigualdades de los extremos: a_k-\frac{a_k}{1+a_k} <a_k/2. Entonces:   \frac{a_k}{1+a_k}>\frac{3}{2}*a_k Así, como a_k no es sumable, entonces el lado derecho tampoco.  
Edited by Foxit Reader Copyright(C) by Foxit Corporation,2005-2009 For Evaluation Only.
 
an
(1+x2)(1+x
+-+···+-
-
N/M2
n=o
oo
for
all
x
in
[a,
b],
which
Now g(xn)
Ã
1-
It
follows
from
=
 
For
not
regulated.]
30.
each
f,
á
=
8.
(a)
=
we can
1
2z2
z4
of
both
6
6
1
which
are
equivalent
to
conditions
(2),(3),(5),
and
(6)
are
clear.
Condition
(1)
can
be
checked
case-by-case.
To
check
(5)
we
5.
(a)
The
assertion
square
root
-b.
that
 
g(f(x))+g(f(y))