Stefanica D. Solutions Manual: A Primer For The Mathematics Of Financial Engineering
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Vll
2
Numerical
integration.
Interest
Rates.
Bonds.
45
2.1
Solutions to
Chapter
2 Exercises. . . . . . . . . . . . . . .
..
45
2.2 Supplemental Exercises. . . . . . . . . . . . . . . . . . . .
..
57
2.3 Solutions
to
Supplemental Exercises. . . . . . . . . . . . .
..
58
1
Calculus
review.
Plain
vanilla
options.
17
1.
1 Solutions
to
Chapter
1 Exercises. . . . . . . . . . . . . . .
..
17
1.
2 Supplemental Exercises. . . . . . . . . . . . . . . . . . . .
..
32
1.3 Solutions
to
Supplemental Exercises . . . . . . . . . . . . .
..
33
Xl
IX
4
Lognormal
variables.
Risk-neutral
pricing.
91
4.1 Solutions
to
Chapter
4 Exercises. . . . . . . . . . . . . . .
..
91
4.2 Supplemental Exercises. . . . . . . . . . . . . . . . . . . . . . 106
4.3
Solutions
to
Supplemental Exercises. . . . . . . . . . . . . . . 107
3
Probability
concepts.
Black-Scholes
formula.
Greeks
and
Hedging.
63
3.1 Solutions to
Chapter
3 Exercises. . . . . . . . . . . . . . .
..
63
3.2 Supplemental Exercises. . . . . . . . . . . . . . . . . . . .
..
82
3.3 Solutions to Supplemental Exercises. . . . . . . . . . . . .
..
83
Acknowledgments
Contents
Preface
o.
Mathematical
preliminaries
1
0.1 Solutions
to
Chapter
0 Exercises. . . . . . . . . . . . . . .
..
1
0.2
Supplemental Exercises. . . . . . . . . . . . . . . . . . . .
..
11
0.3 Solutions to Supplemental Exercises. . . . . . . . . . . . .
..
11
V
l1l
CONTENTS
5
Taylor
￥
formula.
Taylor
series.
5.1 Solutions
to
Chapter
5 Exercises. .
5.2 Supplemental
Exercises.
. . .
5.3 Solutions
to
Supplemental Exercises.
3346
6
Finite
Differences.
Black-Scholes
PDE.
135
6.1 Solutions
to
Chapter
6 Exercises. . . . . . . . . . . . . . . . . 135
6.2 Supplemental Exercises . . . . . . . . . . . . . . . . . . . . . . 150
6.3 Solutions
to
Supplemental
Exercises.
. . . . . . . . . . . . . .
151
Preface
Bibliography
203
The
addition of this Solutions
Iv
lanual
to
"A
Primer
for
the
Mathematics of
Financial Engineering" offers
the
reader
the
opportunity
to
undertake rig-
orous self-study of
the
mathematical topics presented in
the
Iv
lath
Primer,
with
the
goal of achieving a deeper understanding of
the
financial applications
therein.
Every exercise from
the
Math
Primer is solved in detail in
the
Solutions
Manua
l.
Over
50
new exercises are included, and complete solutions
to
these supple-
mental exercises are
provided.
如
1any
of these exercises are quite challenging
and offer insight
that
promises
to
be most useful in further financial engi-
neering studies as well as job interviews.
Using
the
Solution Manual as a companion
to
the
Iv
lath
Primer,
the
reader
will
be
able
to
not
only bridge any gaps in knowledge
but
will also glean a
more advanced perspective on financial applications by studying
the
supple-
mental exercises
and
their solutions.
The
Solutions
Iv
lanual will be
an
important
resource for prospective financial
engineering
graduate
students. Studying
the
material from
the
Math
Primer
in
tandem
with
the
Solutions Manual would provide
the
solid mathematical
background required for successful
graduate
studies.
The
author
has been
the
Director of
the
Baruch College
MFE
Program
1
since
its inception in 2002. Over
90
percent of
the
graduates of
the
Baruch
Iv
IFE
Program are currently employed
i
口
the
financial industry.
"A Primer for
the
Iv
lathematics of Financial Engineering"
and
this Solutions
Manual are
the
first books
to
appear in
the
Financial Engineering Advanced
Backgrou
日
d
Series. Books on Numerical Linear Algebra,on Probability,and
on Differential Equations for financial engineering applications are forthcom-
mg.
7
孔
1ultivariable
calculus:
chain
rule
,
integration
by
substitu-
tion
,
and
extrema.
161
7.1 Solutions
to
Chapter
7 Exercises. . . . . . . . . . . . . . . . .
161
7.2 Supplemental
Exercises.
. . . . . . . . . . . . . . . . . . . . .
171
7.3 Solutions
to
Supplemental
Exercises.
. . . . . . . . . . . . . . 173
8
Lagrange
multipliers.
Newton's
method.
Implied
volatil-
ity.
Bootstrapping.
179
8.1 Solutions
to
Chapter
8 Exercises. . . . . . . . . . . . . . . . . 179
8.2 Supplemental
Exercises.
. . . . . . . . . . . . . . . . . . . . . 197
8.3 Solutions
to
Supplemental
Exercises.
. . . . . . . . . . . . . . 198
Dan
Stefanica
New York
, 2008
lBaruch
MFE
Program
web
page:
http://www.baruch.cuny.edu/math/master
山
tml
QuantN
etwork
student
forum
web
page:
httr
旷/
www.quantnet.org/forum/index
抖
lp
IX
Acknow
ledgments
"A Primer for
the
Mathematics of Financial Engineering" published in April
2008
,is based
on
material covered in a mathematics refresher course I
taught
to
students entering
the
Financial Engineering Masters
Program
at
Baruch
College.
Using
the
book as
the
primary
text
in
the
July 2008 refresher course was a
challenging
but
exceptionally rewarding experience.
The
students from
the
2008 cohort of
the
Baruch
WIFE
Program who
took
that
course were driven
to
master
the
material
and
creatively incorporate ideas
at
an
even deeper level
than
that
of
the
WIath
Primer
boo
k.
WIany
of
the
supplemental questions in
the
Solutions WIanual came
about
as a result of
the
remarkable interaction
I
had
with
也
is
very talented group. I
am
grateful
to
all of
them
for their
impressive efforts.
Special
thanks
go
to
those who supported
the
proofreading effort:
Barnett
Feingold,
Chuan
Yuan-Huang,Aditya Chitral,Hao He,Weidong Huang,Eu-
gene Krel
, Shlomo Ben Shoshan, Mark Su, Shuwen Zhao,
and
Stefan Zota.
The
art
for
the
book cover was again designed by Max Rumyantsev, and
Andy Nguyen continued
to
lend his tremendous
support
to
the
fepress.org
website
and
on QuantNet.org. I
am
indebted
to
them
for all their help.
This book is dedicated
to
my wonderful family. You give sense
to
my work
and make everything worthwhile.
Dan Stefanica
New York
, 2008
Xl
Xll
ACKN0
1i
VLEDGJVIENTS
Chapter
0
Mathematical
preliminaries.
0.1
Solutions
to
Chapter
0
Exercises
Problem
1: Let f :
IR
•
IR
be
an
odd
function.
(i) Show
that
xf(x)
is
a
丑
even
function and x
2
f (x) is
an
odd function.
(ii) Show
that
the
function
gl
:
IR
•
IR
given by gl(X) = f(x22 is
an
even
function
and
that
the
function
g2
:
IR
•
IR
given by
如何)
=
f(x
:J)
is
an
odd
function.
(iii) Let
h :
IR
•
IR
be
de
主
ned
as
h(x)
工
xtf(x
J
)
，
where i
and
j are positive
integers.
When
is h(x)
an
odd function?
Solutio
衍
Since
f(x)
is
an
odd
function,
it
follows
that
f(
-x)
= - f(x) ,
\j
x
εIR.
(1)
(i) Let h
(叫
=
xf(x)
and
f2
(x) = x
2
f(x). Using
(1)
,
we
自 nd
that
h(-x)
=
-xf(-x)
=
xf(x)
=
h(x)
,
\j
x
εIR;
(2)
f2
(-x)
=
(…
X)2
f(
-x)
= - x
2
f(x) = -
f2
(x) ,
\j
x
εIR.
(3)
We
co
卫
clude
from (2)
that
h(x) is
an
even function, and, from
(3)
,
f2
(x) is
an
odd function.
(ii) From
(1)
, it follows
that
gl(-X) = f ((_X)2) =
f(x
2
)
=
gl(X)
,
\j
x
εIR;
(4)
g2(-X) = f ((_x)3) =
f(-x
3
)
= -
f(x
3
)
= -
g2(X)
,
\j
x
εIR.
(5)
We conclude from (4)
that
gl(X)
is
an
even function,
and
,from (5),
that
如何)
is
an
odd function.
(iii)
If
j is a positive integer,
it
follows from
(1)
that
f ((- x
)J)
=
(-1
)J
f (x),
\j
x
εIR.
(6)
2
JVIATHE
1V
IATICAL
PRELIMINARIES
Problem
2: Let
S(n
,
2)
=
I:
~=1
k
2
and
S(
η
，
3)
=
I:
~=1
k
3
.
(i) Let
T(n
,2,
x)
=
I:
~=1
k
2
x
k
. Use
the
recursion formula
T(n
,2,
x)
-
吐
(T(n
，
1,x)),
(7)
0.1.
SOLUTIONS
TO
CHAPTER
0
EXERCISES
(
2
一
(η
十
1)3η
x
n
一
1
十
(2η2
十
2η-1)(η
十
2)(η
十
l)x
n
\
l\
-n
2
(
η+
3)(η
+
2)x
n
十
1
)
1
日
1
、，
6(1 -
x)
/一
(η
十
1)3η(η
-
1)x
n
-
2
十
(2η2
十
2η
-1)(n+2)(
η
十
1)η
x
n
一
1\
Ii\
-n
2
(
η
十
3)(η
十
2)(η
十
1)zn6/
1
日
1
、，
(一加川一川仇
2η
一阳仙
O
一
η2(η
十
3)(η
十
2
勾
)(η
十
1)
6
η(η+
1)
(一
(η
十
1)2(η-
1)
+
(2η2
十
2η-
1)(η
十
2)
…
η(η+
3)(η+
2))
6
3
Then
,
usi
吨
(6)
，
we
在 nd
that
h(-x)
=
(-x)if((-x)j)
=
(-1)ν
((-l)jf(x
j
))
工(
- 1)
i+
jxif (x
j
)
=
(-1)
忡
jh(x)
，
V x
εJR.
TTIerefore7if t
+j
is aIIeven iMeger7the
fmctionfz(z)is
an
eveIl function,
and, if
i
十
j
is
an
odd integer,
the
function
h(x)
is
a~
~dd
function.
口?
and
the
fact
that
η
(n+1)(2
η+
1)
6
T(
η
，
1
，
x)
Z
一
(η
十
1)x
n
+
1
十
nx
n
+
2
(1
-
X)2
(8)
Therefore,
T(
η
，
2
，
x)
η(η+
1)(2η+
1)
S(
η
，
2)
=
(iii) Finding
the
value of
T(
凡
3
，
x)
requires
usi
吨
Quotient
Rule
to
differen
t
杜
iate
T(
η
，
2
，
x).
(iv)
The
solution follows similarly
to
that
from
part
(ii), albeit with more
complicated
computations.
口
to
show
that
X
+x
2
一
(η
+
1
)2
x
n+1
十
(2η2
十
2η
-
1)x
n
+
2
_ n
2
x
n
+
3
(1
-
x)3
(ii) Note
that
S(
η
，
2)
=
T(
η
，
2"
l);"Ys~
yHopital's
rule
to
evaluate
T(
η
，
2
，
1)
，
and
conclude
that
S
(
川)=
7
加
1i(2
叫)
(iii) Compute
T(
η
，
3
，
x)
=
I:
~=1
k
3
x
k
using
the
recursion formula
Problem
3: Compute
S(
η
，
4)
=
I:
~=1
k
4
using
the
recursion formula
川
J)24mμ))
(iv) Note
that
S(
η
，
3)
=
T(
η
，
3
，
1).
U~e
I'Hopital's rule
to
evaluate
T(
η
，
3
，
1)
，
and
conclude
that
S(n
,
3)
=
(呼叫"
~.a
iution:
(i)
~he
res~lt
follows from
(7)
and
(8)
by
usi
鸣
Quotient
Rule
to
differentiate T
(
η
，
1
，
x).
(ii)
It
is
easy
to
see
that
T(
η
，
2
，
1)
=
I:
~=1
炉工
S(
η
，
2).
By using I'Hopital's
rule
we
盘时
that
T(
η
，
2
，
x)
is
equal to
lim
Z
十
x
2
一
(η
十
1)2
x
饥十
1
十
(2η2
十
2η
-
1)x
n
+
2
一
η
2
X
n
十
3
z• 1
(1-x)3
lim 1
+2x
一
(η
十
1)3
x
n
十
(2η2
十
2η-
1)(η
十
2)x
n
十
1
_
n2(
η
+3)x
n
+
2
-3(1
- x)2
\lit-/
-qJ
QU
+.3
.qb
/Il--\
MZH
+
n
(9)
for i = 4, given
that
=η(ηf1)
;
S(
川)
=
(叮
γ
n(
η+
1)(2η
十
1)
6 '
Solution: For i = 4,
the
recursion formula (9) becomes
仰，
4)
=
H
价十户一言
0)
s(n,
j))
4
l\!f
ATHENIATICAL
PRELIMINARIES
~(川
)5
_ 1 -
S(n
,
0)
一臼川一则川)一叫川))
η(η
十
1)(6η3
十
9η2
十
η-1)
口
Problem
4:
It
is
easy
to
see
that
the
sequence
(x
η)
吃
1
given by X
n
=
I::
~=l
k
2
satisfies
the
recursion
X
n
十
1
-
X
n
十
(η
+1?
，
V
η
三
1
，
nu
/Ftt\
with
Xl
=
1.
(i) By
s
由
stituti
吨
n
+ 1 for
ηin
(10),obtain
that
X
n
+2 -
X
n
十
1
+
(η
十
2?
Subtract
(10) from (11) to find
that
4
,
i
X
n
+2
2X
n
+l
- X
n
+
2η+3
，
V
η;
三 1
，
(12)
with
Xl
二
1
and
X2 =
5.
(ii) Similarly, show
that
X
n
十
3
3X
n
+2 -
3X
n
+l
+
X
n
十
2
，
V
η;
三
1
，
(13)
with
Xl
= 1, X2 = 5, and
句=
14.
(iii) Prove
that
the
sequence
(x
n
)
吃
1
satisfies
the
linear
rec
旧
s
lO
n
X
n
+4 -
4X
n
+3
+
6X
n
+2 -
4x
η+1
十
X
n
0, V n
二三
1.
Solve this recursion and show
that
X
n
二=
η(η
十
1)(2η+
1)
V
η>
1.
Conclude
that
S(
η
，
2)
三二
k
2
V
η
>
1.
η(η+
1)(2η+
1)
6
Solution: From (11),
we
obtain
that
the
first terms of
the
sequence
(xn)n
三
l
are
Xl
= 1, X2 = 5, X3 =
14
, X4 =
30.
(i)
The
recursion (12) follows immediately by subtracting (10) from (11).
0.1.
SOLUTIONS
TO
CHAPTER
0
EXERCISES
5
(ii) We substitute
η
十
1
for n in (12) and obtain
that
X
n
+3 -
2x
n
十
2
-
Xn+l
十
2(η+
1)
+
3.
By
subtracti
吨
(12)
from (14),
we
find
that
(14)
X
n
+3 =
3x
n
十
2
-
3X
n
+l
十
X
n
十
2
，
V n
~三1.
(iii) We
substitute
η
十
1
for n in (13) and obtain
that
X
n
+4 =
3x
n
十
1
一
3X
n
+2
十
Xn+l
十
2
，
V n
二三
1.
vhu
/iE\
By subtracting (13) from (15) ,
we
find
that
X
n
+4 -
4X
n
+3
十
6X
n
+2
-
4X
n
+l
+X
n
- 0, V n
~三1.
(16)
The
characteristic polynomial associated
to
the
recursion (16) is
P(z)
=
泸
-4
♂
+
6z
2
-
4z
十
1
=
(z-1)4.
The
polynomial P
(z)
has root
λ=
1 with multiplicity
4.
We conclude
that
the
there exist constants C
i
,i = 1 : 4,such
that
X
n
=
0
1
十
O
2η
十
0
3η2
十
04n3
，
V
η;
三1.
Since
Xl
= 1, X2 = 5, X3 =
14
, X4 =
30
,
it
follows
that
0
1,
O
2
, 0
3
and 04
satisfy
the
following linear system
(111
1)C1)-/1
1 2 4 8 \ I O
2
\
f 5
1 3 9
27
I I 0
3
I - I
14
1 4
16
64\0
4\
30
We obtain
that
0
1
= 0, O
2
=
~，
α=i
aBdq=i
and
therefore
2η3η(η
+
1)(2n
十
1)
n
-
…十一十一=
, V
n;
三1.口
Problem
5: Find
the
general form of
the
sequence
(x
n
)n
2:
0
satisfyi
吨
the
linear recursion
X
n
+3 =
2X
n
+l
十
X
n
, V n
二三
0,
with Xo = 1,
Xl
=
-1
, and X2 =
1.
First Solution: By direct computation,
we
obtain X3 =
-1
, X4 =
1
，圳工
-1
，
X6 =
1.
It
is
nat
盯
al
to
conjecture
that
X
n
=
(-l)n
for
a
町
positive
integer
η.
This can be easily checked by induction.
6
1V
IATHE
1V
IATICAL
PRELIMINARIES
0.1.
SOLUTIONS
TO
CHAPTER
0
EXERCISES
7
X
n
=
(-1)η
，
Vn
三
o.
口
Second
Sol
州
on:
Alternati
叫
y
，
we
note
that
the
seque
丑
ce (x
n
)
吃
o
satis
且
es
the
linear recursion
X
n
十
3
-
2x
η
刊
-
X
n
= 0,
with
characteristic polynomial
P(z)
=
z3
- 2z +1 =
(z
十
1)(z2
- Z - 1).
The
roots of
P(z)
are
一
-1
，与
E
眨，
a
时导L§.
Th
阳
e
臼
re£
白
f
扣
ore
，
t
出坠
h
阳
e
盯盯
re
exist
c
∞
con
臼∞
I
丑
nst
归
S
阳
t
C1
,O
2
,
and
0
3
such
that
/
严叩\
η
/
严-\
η
I
1
斗、
15
\
~
11
-
vi
月
1
X
n
=
0
1
(-1)
η
十
O
2
卜」二
I
+
0
3
←」二
I
, V
n
主
O.
\ 2 I
U\
2 I
By
solving
the
3 x 3 linear system for 0
1
, O
2
,
and
0
3
obtained by requiring
that
Xo
= 1, Xl =
-1
, and
X2
= 1
we
find
that
0
1
= 1, O
2
= 0, and 0
3
=
o.
We conclude
that
Since
Xo
= 1
and
Xl = 5,
we
obtain
that
C
1
=
-1
and
C2 = 2
and
therefore
X
n
=
2·
3
n
-
1, V
η
二三
O.
口
(24)
(23)
(22)
(20)
X
n
+2
-
4X
n
+1
-
3x
n
十
1
，
V
n
二三
o.
X
n
+3
-
5X
n
+2
- 7
X
n
十
1
+ 3x
n
, V n
~三
o.
Substitute
η+
1 for n in (22)
to
obtain
that
X
n
十
3
4x
η+2
-
3x
η
十
1
+1, V
η;
三
o.
Subtract
(22) from (23)
to
find
that
with
Xo
= 1, Xl = 5,
and
X2
=
18.
(ii) Find
the
general formula for X
n
,
n
主
o.
Solution: (i)
The
first three terms of
the
sequence can
be
computed from
(20) and are
Xo
= 1, Xl =
?'
and
X~_
~
18.
By
substituting
n + 1 for
ηi
口
(20)
we
obtain
that
X
n
+2
-
3X
n
+1
十
η+3
，
V
η
三
o.
(21)
Subtract
(20) from (21)
to
find
that
with
Xo
=
1.
(i) Show
that
the
sequence
(Xn)n:::::O
satisfies
the
1
且
i
丑
near
r
陀
ec
盯
S
Zη+3
-
5x
η+2
一
7X
n
+1
+
3x
r
阳
1
，
V
η
二兰三
0,
Problem
7:
The
sequence
(x
n
)
吨。
satisfies
the
recursion
X
n
+1
-
3x
n
十
η
+2
，
V n
二三
0,
(17)
X
n
十
1
3X
n
+ 2, V
η;
三
0
，
X
n
十
2
-
4x
n
十
1
- 3x
n
, V
η
二三
0,
with
Xo
= 1
and
Xl =
5.
(ii)
Find
the
general
forml
由
for
X
n
，
η
主
o.
Solution: (i) Let n = 0 in (17)
to
find
that
Xl = 5. By
Sl
郎
ti
扰
t1
巾
ηi
丑
(ο17η)
，
it follows
that
Problem
6:
The
sequence
(x
n
)
吃
o
satisfies
the
recursion
with
Xo
=
1.
(i) Show
that
the
sequence
(x
n
)
吨。
satisfies
the
linear recursion
(ii)
The
characteristic polynomial of
the
linear recursion (24) is
P (
x)
=
z3
- 5
z2
+7z - 3 = (z - 1)
2
(z - 3).
Therefore, there exist constants 0
1
,O
2
,0
3
such
that
X
n
=
0
1
3
η
十
O
2η
+ 0
3
, V
η
二三
o.
Since
Xo
=
1
，问=
5, and
X2
=
18
,
we
find
that
01 =
~，
O
2
=
-~，
α=-i
We
conclude
that
Zη+2
-
3x
n
十
1
十
2.
We
Sl
削
ract
(17) from (18)
and
obtain
that
X
n
+2
-
4x
η
十
1
+
3x
n
-
0, V n
~三
o.
(ii)
The
characteristic polynomial of
the
linear
rec
旧
sion
(19) is
P(z)
=
z2-4z
十
3
=
(z-1)(z-3)
,
which has roots 1
and
3.
Thus
,
X
n
二
0
1
+ 023n, V
η;
三
o.
(18)
(19)
X
n
-
3η+2
-
2η
4
~
Vη>
o.
口
8
JVIATHEJVIATICAL
PRELIMINARIES
0.1.
SOLUTIONS
TO
CHAPTER
0
EXERCISES
9
It
is easy
to
see
that
I
h(x)
+
h(x)
I /'
1~
YY>
"""J
h(x)
I
J..
lim
~;m
n
I
h(x)
I
<'
IiTftplznl:
三
liryhγ|
十
brylznl<
and
therefore
、
by
definition,
h(x)
+
h(x)
= O(x
n
) as x •
0.
Let
gl(X)
二可♂)
and
g2(X)
=
o(
俨
)asx
•
0.
The
凡
叫四
1=0
and
叫
92(~)
I= °
Z•
01
x
n
I x• oI x ,. I
Problem
8:
Let
P(z) =
2:立。
α
iZi
be
the
characteristic polynomial corre-
sponding
to
the
linear recursion
2
二
α
向十
i
= 0, V n
~
°
(25)
Assume
that
λis
a
root
of multiplicity 2 of P(z). Show
that
the
seque
丑
ce
(Yn)
吨。
given
by
Yn
=
c
ηλ
飞
η;
三
0
，
where C is
an
arbitrary
consta
风
satisfies
the
recursion
(25).
Sol
毗
on:
Note
thatλis
a
root
ofmultiplicity 2 ofP(z) if
and
only if
P(
λ)=0
and
P'(
λ)
= 0,where
k
P'(z)
=
汇
ω
ii-
1
lil--io
2-z
n3-
iil-lili--QU
-LZn
!li--io
-2
1i
l--iili--nud
z-B
oa--1
--nq
•
-znE
\l/-e
z-d
g-W
blzpk
dh
nt
mmm
Then
,for
anyη;
三
0
，
Problem
10:
Prove
that
汇
α
iYn+i
2
二
α
以
η+
i)λ
叫
Cωη2
汇二
α
川+刊
4
十
C2
汇二二
iω
叫
α
向
iλ
俨
η
肿十
Cη
队
λ
n
2
汇二
α
川十
C
队
λ
沪
ηω+
礼
1
2
汇二
ω
叫
ωtλ
沪俨
t
•
H
一」
1
C
η
λ
η
P(
λ
均)
+ C
λ
η
肿叫+礼
1
P'(
从川
λ
材
)
i.e., show
that
and
that
In
other
words,
the
sequence
(Yn)n?:.O
satisfies
the
rec
旧
sion
(25).
口
Similarly, prove
that
Problem
9:
Let
n >
0.
Show
that
nZ?ZM
。(♂)
+O(
♂)
o(x
n
)
+ o(x
n
)
。(♂)，
as x • 0;
o(x
n
)
, as x •
0.
(26)
(27)
Sol
州
on:
Let
h(x)
=
。但
n)
and
h(x)
=
O(
俨)
as x •
0.
Then
,
lims
叫
h(x)
I
∞
p
Ih(
川
DI
一一一
I
<
00
and
limsup
1
一一
~I
m•
aT
I
俨
I
-
x
→
0ι|
俨
i
Solution: Recall
that
。(的，
as
n
→∞;
?十州，
asη
→∞?
~n
1.2
lims
叩
μ17;1<
∞
n
一→。
C
10
~~_1
k
2
一手
lim
sup:::='
凡-J.
OJ
<∞-
n
一今汉二
'也，
。(的，
as
n
→∞;
Z
十州，
asη
→∞
2
二
k
2
中俨十
1)
删去
k
3
10
Then
,
lim
L:
~=1
k
2
k=l
-
n
→∞
ηU
""""η
7A2
η3
im
L...t
k=l
气一
τ
n
→∞
η"
""",,n
1
认
3
limι
~k=l
内
二
n
→∞
ητ
lim
L:~卢
η4
1
日
1
k=l
一
τ
n
→∞
ηu
We conclude
that
MATHEMATICAL
PRELI
]l.
ifINARIES
1
~…
η(η+
1)(2η+
1)
η
→∞
6η3
唱
3η2
十
η
..…二『
η
→∞
6η2
limd(η
十
1)2
一
η
→∞
4
,.
2η
十
1
..…-
n
→∞
4η
1
2
<
。(的?
as
η
→∞;
号
+
O(n
2
)
,
as
η
→∞;
。(的
7
as
η
→∞;
T
十
O(η3)
，
as
η
一→
00.
<
∞;
<
00.
∞;
口
0.2.
SUPPLEMENTAL
EXERCISES
11
0.2
Supplemental
Exercises
l
3
∞;
1.
Let
α>
0
be
a positive number. Compute
<
Vα
十卢+~
2.
Let
α>
0
be
a positive number. Compute
1
α
十一一--，
α
十注工
3.
(i)
Find
x > 0 such
that
Z
Z
Z
= 2.
(ii)
Find
the
largest possible value of x > 0
with
such
that
there exists
a number
b > 0
with
z
Z
b.
Also,
what
is
the
largest possible value of
b?
0.3
Solutions
to
Supplemental
Exercises
Pro
blem
1: Let
α>
0 be a positive number.
Compute
户十卢十日
τ
灿
t
剖仇
tω
ion
仰仰
η
t
出
ha
肮
t
limit by l,
ther
口
1
1
挝
t
follows
that
l must satisfy
jα
+j
α+
币二=
d
丰
l
，
(28)
which
can
be
solved for l
to
obtain
that
1+
VI+石
2
(29)
12
!\/f
ATHEMATICAL
PRELIMINARIES
0.3.
SOLUTIONS
TO
SUPPLE
1V
IENTAL
EXERCISES
13
X
n
<
X
叫字=}
X
n
<
va
τ
X
n
仁丰
zi-zn
一
α<
0
,
We now show
that
,for
anyα>
0,
the
limit
of
飞
/α+
、
/α
+va
τ-.
.-.
does
e
于风
whick
i?quimieM
to
provi
丑
g
that
the
叫
uen
配阳
C
臼
e (X
乌
ωηρ)η
哈三沪
o
i
坦
s
c
∞
on
口阴飞
whe
臼
re
X
均
0=
飞
vαand
since X
n
<
land
X
n
> 0 for all
n
三
O.
vVe
conclude
that
X
n
<
X
n
十
1
for all
n>
O.
We showed
that
the
seque
口
ce
(xn)n~O
is
increasi
吨
and
bounded from
above. We conclude
that
the
seque
口
ce
is convergent.
Therefore
、
the
limit
l=li
问→
00
X
n
satisfies
the
equation l =
vaτI
an-d
is given by (29),i.
e.
,
Xn
+1
-
va
丰百
<
va:+l = l,
since l is
the
positive
叫叫
O
∞
n
of
(闵
28
剑
8
创)，
a
肌削丑叫
d
t
山
he
臼盯
re
时
efl
伽
f
inductior
瓦
1
，
we
fi
且
II
卫
ld
tl
扭
lat
X
η
<
l for all η>
O.
(ii)
The
sequence
(x
r
山三
o
is
increasi
吨·
It
is easy
to
see
that
(33)
(34)
(32)
(31)
α+17V
n
三
O.
X
n
X
n
+1
-
l=
α+;
白
l2
-
al
- 1 = 0,
(α2
十
l)Yn
+
α
Yn+1
-
α
十一一-一一=
? V
n>07
α+
县
α
Yn
十
1
lin
with
Yo
=
α.
Note
that
Yn = X2n for all
η
主
O.
Assume
that
Yn < l, where l is given by (32). Recall from (31)
that
l2
-.al
- 1 = 0
Xo
<
X2
< l <
X3
<
Xl·
Based on this observation,
we
co
时
ecture
that
the
subsequence
(X2n)
吃
O
made of
the
even
terms
of
the
sequence
(xn)n~O
is
increasi
吨
and
has limit
equal
to
l,
and
that
the
subseq
旧
lce
(X2
附
1)
吃
o
made of
the
odd
terms of
(X
n
)
吨。
is
decreasing
and
has limit equal
to
l.
To show this
, let
(y
山
The
自
rst
few
terms
of
the
seque
丑
ce
(xn)n
三
o
are
α2
十
1
a
3
+
2αα4
十
3α2+1
Xo
=
α
，
Xl
=
一一一一
，
X2
=
一一一一-:
XQ
=
α2α2
十
l'
，1，
3α3
十
2α
We note
that
the
terms
of
the
sequence are alternatively larger
and
smaller
than
the
value of l given by (32),i.
e.
,
l=
α+va
可
2.
2
To show
that
the
continuous fraction (30) does have a limit,
we
must
prove
that
the
sequence
(xn)n
主
o
is
converge
风
where
Xo
=
αand
and therefore
Solution:
If
we
know
that
the
continuous fraction (30) has a limit l,
then
l
must satisfy
(x
n
_
1+
V;+
石
) (x
n
-
l-f
石)
川)
(x
n
十盯了一
1)
2:-zn
一
α
X
n
+1
-
va
百二
，
\I
η
主
O.
We will how
that
the
seq
阳
l
丑
creasm
丑
g.
Let l
be
given by
(羽)，
i.e., let l
=
与严
(i)
The
sequence
(x
n
)
吃
o
is bounded from above bv
l.
Note
that
Xo
=
va
< l.
If
we
assume
that
X
n
<
l
，
白
en
since X
n
>
O.
Note
that
lim x,..,
1
十飞
11
十
4α
h
1m
X
，..，
ι
川
η
→∞川
2
一
Pro
blem
2: Let
α>
0 be a positive number. Compute
1
α
十一一一寸
α
+
a~...
(30)
and
that
t
2
一
α
t
1
(t-
a+vt+
4
)
(t
一
α-F)
(35)