Advanced Placement Specialty Conference TEACHING THE ......Advanced Placement Specialty Conference TEACHING THE IDEAS BEHIND POWER SERIES Presented by LIN McMULLIN Sequences and Series
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Adv
ance
d P
lace
men
t Spe
cial
ty C
onfe
renc
e
TE
AC
HIN
G T
HE
IDE
AS
BE
HIN
D
PO
WE
R S
ER
IES
Pre
sent
ed b
y
LIN
McM
ULL
IN
S
eque
nces
and
Ser
ies
in P
reca
lcul
us
P
ower
Ser
ies
In
terv
als
of C
onve
rgen
ce &
Con
verg
ence
Tes
ts
E
rror
Bou
nds
G
eom
etric
Ser
ies
N
ew S
erie
s fr
om O
ld
P
robl
em
s
Q &
A
Fro
m th
e C
ours
e D
escr
iptio
n
*IV
. Pol
ynom
ial A
ppro
xim
atio
ns a
nd S
erie
s *
Con
cept
of s
erie
s. A
ser
ies
is d
efin
ed a
s a
sequ
enc
e of
par
tial s
ums,
an
d co
nver
genc
e is
def
ined
in te
rms
of th
e lim
it of t
he s
eque
nce
of
part
ial s
ums.
Tec
hnol
ogy
can
be u
sed
to e
xplo
re c
onverg
ence
or
dive
rgen
ce.
* S
erie
s of
con
stan
ts.
+ M
otiv
atin
g ex
ampl
es, i
nclu
ding
dec
imal
exp
ansi
on.
+ G
eom
etric
ser
ies
with
app
licat
ions
. +
The
har
mon
ic s
erie
s.
+ A
ltern
atin
g se
ries
with
err
or b
ound
. +
Ter
ms
of s
erie
s as
are
as o
f rec
tang
les
and
thei
r re
latio
nshi
p to
im
prop
er in
tegr
als,
incl
udin
g th
e in
tegr
al te
st a
nd it
s us
e in
test
ing
the
conv
erge
nce
of p-se
ries.
+
The
rat
io te
st fo
r co
nver
genc
e an
d di
verg
ence
. +
Com
parin
g se
ries
to te
st fo
r co
nver
genc
e or
div
ergen
ce.
* T
aylo
r se
ries.
+
Tay
lor
poly
nom
ial a
ppro
xim
atio
n w
ith g
raph
ical
dem
onst
ratio
n of
co
nver
genc
e. (
For
exa
mpl
e, v
iew
ing
grap
hs o
f var
ious T
aylo
r po
lyno
mia
ls o
f the
sin
e fu
nctio
n ap
prox
imat
ing
the s
ine
curv
e.)
+ M
acla
urin
ser
ies
and
the
gene
ral T
aylo
r se
ries
cente
red
at x
= a
. +
Mac
laur
in s
erie
s fo
r th
e fu
nctio
ns,
x e, s
in x
, cos
x, a
nd
1
1x
− .
+ F
orm
al m
anip
ulat
ion
of T
aylo
r se
ries
and
shor
tcuts
to c
ompu
ting
Tay
lor
serie
s, in
clud
ing
subs
titut
ion,
diff
eren
tiatio
n, a
ntid
iffer
entia
tion,
an
d th
e fo
rmat
ion
of n
ew s
erie
s fr
om k
now
n se
ries.
+
Fun
ctio
ns d
efin
ed b
y po
wer
ser
ies.
+
Rad
ius
and
inte
rval
of c
onve
rgen
ce o
f pow
er s
eries.
+
Lag
rang
e er
ror
boun
d fo
r T
aylo
r po
lyno
mia
ls.
Pre
calc
ulus
Seq
uenc
es a
nd s
erie
s:
•
Sig
ma
nota
tion,
• R
ecur
sive
and
non
-rec
ursi
ve d
efin
ition
s of
seq
uenc
es,
• B
asic
fo
rmul
as
for
the
sum
s of
si
mpl
e se
quen
ces
(2
11
1
con
stan
t,,
nn
n
kk
k
kk
==
=∑
∑∑
, etc
.).
• G
iven
a s
eque
nce
they
sho
uld
be a
ble
to w
rite
the
form
ula
for
the
nth t
erm
; gi
ven
the
nth t
erm
the
y sh
ould
be
able
to
writ
e th
e te
rms
of th
e se
quen
ce.
• D
efin
ition
of
conv
erge
nce
of a
ser
ies
as t
he li
mit
of
the
asso
ciat
ed
sequ
ence
of p
artia
l sum
s.
Typ
es o
f Ser
ies
•
Arit
hmet
ic s
erie
s,
• G
eom
etric
ser
ies,
•
Alte
rnat
ing
serie
s,
• H
arm
onic
ser
ies,
•
Alte
rnat
ing
harm
onic
ser
ies,
•
p-se
ries
Dec
imal
s
( ((() )))
( ((() )))
( ((() )))
( ((() )))
01
23
0.33
33...
0.3
0.03
0.00
30.
003
0.3
100.
310
0.3
100.
310
0.3
0.3
11
0.9
31
10
−−
−−
−−
−−
−−
−−
=+
++
+=
++
++
=+
++
+=
++
++
=+
++
+=
++
++
=+
++
+=
++
++
==
==
==
==
==
==
− −−−
⋯ ⋯⋯⋯
⋯ ⋯⋯⋯
( (((
) )))( (((
) )))( (((
) )))( (((
) )))
( ((() )))
01
23
0.99
90.
910
0.9
100.
910
0.9
10
0.9
0.9
11
0.9
110
10.
999.
..3
0.33
3...
31
3
−−
−−
−−
−−
−−
−−
=+
++
+=
++
++
=+
++
+=
++
++
==
==
==
==
==
==
− −−−
==
==
==
==
==
==
⋯ ⋯⋯⋯
10.
999.
..1.
999.
..0.
999.
..1
22
+ +++=
==
==
==
==
==
=
To
grap
h ( (((
) )))1
0.7
t
na=
−−
=−
−=
−−
=−
− in
the
plan
e:
Mod
e: P
aram
etri
c G
raph
form
at:
Do
t E
quat
ion
Edi
tor:
xt1
(t)
= t
yt1
(t)
= 1
– (
–0.7
)t̂
Win
dow
: tm
in =
1,
tm
ax
= 3
0,
ts
tep
= 1
,
xmin
= 0
,
xma
x =
31
,
xscl
= 1
,
ymin
=
0,
ym
ax
= 1
.5,
ys
cl =
1.
To
grap
h ( (((
) )))1
0.7
t
na=
−−
=−
−=
−−
=−
− o
n a
num
ber
line:
M
ode:
Par
amet
ric
Gra
ph fo
rmat
: D
ot
Equ
atio
n E
dito
r:
xt
1(t
) =
1 –
(–
0.7
)t̂
and
yt
1(t)
= 1
Win
dow
: tm
in =
1,
tm
ax
= 3
0,
ts
tep
= 1
,
xmin
= 0
,
xmax
= 2
,
xscl
= 1
,
ymin
=
0,
ym
ax =
2,
yscl
= 1
.
The
n T
RA
CE
the
grap
h an
d w
atch
it c
onve
rge.
Tay
lor
Pol
ynom
ial
If
f ha
s n d
eriv
ativ
es a
t c th
en th
e nth
Tay
lor
poly
nom
ial
for
f a
t x =
c is
( ((() )))
( ((() )))
( ((() )))( (((
) )))( (((
) ))) ( ((() )))
( ((() ))) ( (((
) ))) ( ((() )))
2
2!
!
nn
n
fc
fc
Tx
fc
fc
xc
xc
xc
n
′′ ′′′′′′′ ′′′
=+
−+
−+
+−
=+
−+
−+
+−
=+
−+
−+
+−
=+
−+
−+
+−
⋯ ⋯⋯⋯
If
c =
0, t
he nth
Mac
laur
in p
olyn
omia
l fo
r f
is
( ((() )))
( ((() )))
( ((() )))
( ((() )))
( ((() ))) ( (((
) )))2
2!
!
nn
n
fc
fc
Tx
fc
fc
xx
xn
′′ ′′′′′′′ ′′′
=+
++
+=
++
++
=+
++
+=
++
++
⋯ ⋯⋯⋯
Let
( ((() )))
32
1424
fx
xx
x=
+−
−=
+−
−=
+−
−=
+−
−
a.
Writ
e th
e po
wer
ser
ies
for f
cen
tere
d at
x =
2.
b. E
xpan
d th
e te
rms
of th
e po
wer
ser
ies
and
sim
plify
.
c. I
s th
is a
n ac
cide
nt o
r w
ill th
is h
appe
n w
ith a
ny p
olyn
omia
l? E
xpla
in.
( ((() )))
( ((() )))
( ((() )))
( ((() )))
23
402
27
22
fx
xx
x=
−+
−+
−+
−=
−+
−+
−+
−=
−+
−+
−+
−=
−+
−+
−+
−
N
ow e
xpan
d th
e bi
nom
ial t
erm
s an
d se
e w
hat y
ou g
et!
1988
BC
4: D
eter
min
e al
l val
ues
of x fo
r w
hich
the
serie
s
( ((() )))
0
2ln
2
kk
k
x k
∞ ∞∞∞ = ===+ +++
∑ ∑∑∑ c
onve
rges
.
()
()
()
()
() (
)(
)
11
2 ln3
ln2
limlim
2ln
32
ln2
By
L'H
opita
l's r
ule,
1ln
23
2lim
limlim
11
ln3
23
ln2
lim2
2ln
3 12
12
kk
kk
kk
kk
k
k
x kk
xk
x
k
kk
kk
kk
kx
xk
xx
++
→∞
→∞
→∞
→∞
→∞
→∞
+
+
=
+
+
++
+=
==
++
++
∴=
+
<⇔
<
()
()
()
0
0
0
11
Con
verg
es fo
r 2
2
11
At
, se
ries
beco
mes
2
ln2
1di
verg
es, b
y co
mpa
riso
n w
ith h
arm
on
ic s
erie
s 2
11
At
, se
ries
bec
om
es
2ln
2
con
verg
es, b
y th
e al
tern
atin
g se
ries
test
.
Ser
ie
k
k
k
k
x
xk
k
xk
∞ =
∞ =
∞ =
∴−
<<
=+
+
−=
−+
∴
∑
∑
∑
11
s co
nve
rges
for
22
x−
≤<
Mem
oriz
e th
e M
acla
urin
Ser
ies
and
the
inte
rval
of co
nver
genc
e fo
r
23
4
35
7
24
6
23
4
1
for
all
2!
3!
4!
sin
fo
r al
l 3
!5
!7
!
cos
1
for
all
2!
4!
6!
11
-11
1xx
xx
ex
x
xx
xx
xx
xx
xx
x
xx
xx
xx=
++
++
+=
++
++
+=
++
++
+=
++
++
+
=−
+−
+=
−+
−+
=−
+−
+=
−+
−+
=−
+−
+=
−+
−+
=−
+−
+=
−+
−+
=+
++
++
<<
=+
++
++
<<
=+
++
++
<<
=+
++
++
<<
− −−−
⋯ ⋯⋯⋯
⋯ ⋯⋯⋯ ⋯ ⋯⋯⋯
⋯ ⋯⋯⋯
2000
BC
3: T
he T
aylo
r se
ries
abou
t x =
5 fo
r a
cert
ain
func
tion
f co
nver
ges
to
()
fx
for
all
x in
the
inte
rval
of c
onve
rgen
ce. T
he nth
deriv
ativ
e of
f at
5x
= === is
giv
en b
y ( (((
) )))( (((
) )))(
)(
1)!
52
2
nn
n
nf
n
− −−−= ===
+ +++, a
nd
1(5
)2
f= ===
.
(a
) W
rite
the
third
-deg
ree
Tay
lor
poly
nom
ial f
or f
abo
ut x
= 5
.
(b
) F
ind
the
radi
us o
f con
verg
ence
of t
he T
aylo
r se
ries fo
r f
abou
t x =
5.
R
atio
test
giv
es
51
25
2
Rad
ius
is 2
x x
− −−−< <<<
−<
−<
−<
−<
23
31
11
1(
,5)(
)(
5)
(5
)(
5)
26
1640
Pf
xx
xx
=−
−+
−−
−=
−−
+−
−−
=−
−+
−−
−=
−−
+−
−−
(c
) S
how
that
the
sixt
h-de
gree
Tay
lor
poly
nom
ial f
or f ab
out x
= 5
appr
oxim
ates
( (((
) )))6f
with
err
or le
ss th
an
1
1000
.
By
the
alte
rnat
ing
serie
s te
st t
he e
rro
r is
less
tha
n
( ((() )))
( ((() )))
( ((() )))
77
77
17
!6
51
11
2(7
2)7
!2
911
5210
00
−−
−−
−−
−−
==
<=
=<
==
<=
=<
+ +++
Is th
is tr
ue a
t f(4
)? W
hy, o
r w
hy n
ot?
Legr
ange
For
m o
f the
Rem
aind
er an
d th
e Le
gran
ge E
rror
Bou
nd
Tay
lor’s
The
orem
:
If f
has
deriv
ativ
es o
f all
orde
rs o
n an
inte
rval
con
tainin
g a,
then
for
any
posi
tive
inte
ger n
and
for
all x
in th
e in
terv
al, t
here
exi
st a
num
ber c
betw
een x
and
a s
uch
that
:
( ((() )))
11
1
() (
) ()
()!c
nf
nR
xx
an
n
+ ++++ +++
=−
=−
=−
=−
+ +++
Thi
s is
cal
led
the L
egra
nge
For
m o
f the
Rem
aind
er.
( ((() )))
22
()
()
()
()
()
()(
)(
)(
)!
!
nf
af
an
fx
fa
fa
Rx
nx
ax
ax
an
′′ ′′′′′′′ ′′′
+−
+−
++
−+
−+
+−
+−
++
−+
−+
= ===+
−+
+−
++
−+
+−
+⋯
Exa
mpl
e 1:
app
lyin
g th
e th
eore
m to
the
sine
func
tion c
ente
red
at th
e or
igin
,
ther
e ex
ists
a c
betw
een x
and
zer
o su
ch th
at
35
61
11
612
06
!
35
61
11
612
072
0
35
11
612
0
sin
sin(
)
sin
(0.2
)0
.2(0
.2)
(0.2
)si
n(
)(0.
2)
sin
(0.2
)0
.2(0
.2)
(0.2
)0
.19
866
93
xx
xx
cx
c
=−
++
=−
−+
≈−
−≈
Not
ice
that
the
rem
aind
er te
rm is
not c
alcu
late
d at
x =
0, b
ut a
t som
e x =
c
in th
e in
terv
al (
0, 0
.2),
so
the
sixt
h po
wer
term
is us
ed,
61
572
0si
n()(
0.2
)R
c=
is
not z
ero.
In th
e op
en in
terv
al (
0, 0
.2)
the
larg
est th
e si
n(c)
can
be
is 1
: (N
ote:
sin
(0.2
)0.
19
8669
33.
..≈
), s
o th
e la
rges
t the
err
or c
an b
e is
68
1(1
)(0
.2)
8.8
910
72
0−
≈×
(or
6
81 72
0(0
.198
669
3)(0
.2)
1.7
71
0−≈
×).
The
act
ual
erro
r is
con
side
rabl
y le
ss, a
bout
92
.54
10−
×.
E
xam
ple
2: a
pply
ing
the
theo
rem
to x e
cent
ered
at t
he o
rigin
, the
re e
xist
s a
c be
twee
n x a
nd z
ero
such
that
The
n at
, say
, x =
0.2
:
In t
he i
nter
val
00
2[
,.
] th
e la
rges
t th
at
c e c
an b
e is
0.
21.
2214
0e
≈ ≈≈≈,
so t
he
larg
est
the
erro
r ca
n be
is
( ((() )))( (((
) )))45
1 241.
2214
00.
28.
1427
10− −−−≈
×≈
×≈
×≈
×.
Thi
s is
the
Lagr
ange
Err
or B
ound
. The
act
ual e
rror
is a
bout
5
6.94
10− −−−
× ×××.
23
11
23
!4
1 4!
1x
cx
xe
xe
x=
+=
+=
+=
++ +++
++
++
++
++ ( (((
) )))( (((
) )))2
30.
25
11
26
0.2
10.
20.
20.
21.
2213
338.
1427
10
1.22
1402
7581
6...
e e
− −−−≈
++
+=
±×
≈+
++
=±
×≈
++
+=
±×
≈+
++
=±
×
= ===
( ((() )))
( ((() )))
( ((() )))
23
0.2
11
24
16
241
0.2
0.2
0.2
(0.2
)c e
e+
++
++
++
++ +++
=+
=+
=+
=+
1999
BC
4: T
he fu
nctio
n f h
as d
eriv
ativ
es o
f all
orde
rs fo
r all
real
num
bers
x.
Ass
ume f
(2)
= –
3, f
'(2)
= 5
, f ''
(2)
= 3
, and
f '''(
2) =
–8.
(a)
Writ
e th
e th
ird-d
egre
e T
aylo
r po
lyno
mia
l for
f a
bout
x =
2 a
nd u
se it
to
app
roxi
mat
e f(
1.5)
.
()
()
()
()
()
23
38
32
6,2
35
22
2
1.5
4.9
58
Tf
xx
x
f
=−
+−
+−
−−
≈−
(b
) T
he fo
urth
der
ivat
ive
of f
sat
isfie
s th
e in
equa
lity
(
) ()
43
fx
≤ fo
r al
l x
in th
e cl
osed
inte
rval
[1.5
, 2].
Use
the La
gran
ge e
rror
bou
nd o
n th
e ap
prox
imat
ion
to f
(1.5
) fo
und
in p
art (
a) to
exp
lain
why
(
)1
.55
f≠
−.
LEB
=
43
1.5
24
!−
=0.0
0781
25,
()
1.5
4.9
583
0.0
078
125
4.9
66
5f
>−
−=
−>
−
(c)
Writ
e th
e fo
urth
deg
ree
Tay
lor
poly
nom
ial,
P(x
), fo
r
()
()
22
gx
fx
=+
abou
t x =
0. U
se P to
exp
lain
why
g m
ust h
ave
a re
lativ
e m
inim
um
at x
= 0
.
()(
)(
)(
)(
)(
) ()
23
38
32
6
22
43
22
,23
52
22
,22
35
Tf
xx
xx
Tf
xx
x
=−
+−
+−
−−
+=
−+
+
an
d fr
om th
e co
effic
ient
s (
)(0
)0
an
d
00
gg
′′′
=>
ther
efor
e a
min
imu
m b
y th
e S
econ
d D
eriv
ativ
e T
est.
Geo
met
ric S
erie
s
Met
hod
1: T
he e
xpre
ssio
n 1
1x
− −−− is
sim
ilar
to 1
ar
− −−−.
If w
e ha
d a
geom
etric
serie
s w
ith a
firs
t ter
m o
f a =
1 a
nd a
com
mon
rat
io o
f x, t
hen
11
Sx
= ===− −−−
.
Tur
ning
this
aro
und,
the
pow
er s
erie
s fo
r 1
1x
− −−− m
ust b
e th
e ge
omet
ric
serie
s
and
the
inte
rval
of c
onve
rgen
ce m
ust b
e al
l x su
ch th
at
1x
< <<< o
r 1
1x
−<
<−
<<
−<
<−
<<
.
23
41
11
nx
xx
xx
x=
++
++
++
+=
++
++
++
+=
++
++
++
+=
++
++
++
+− −−−
⋯⋯
⋯⋯
⋯⋯
⋯⋯
Rat
iona
l Exp
ress
ions
Met
hod
2: L
ong
divi
sion
yie
lds
the
sam
e re
sult:
23 2 2 2
3
1(1
)1 1
xx
x
x x xx x x
x
++
++
++
++
++
++
− −−−− −−−
− −−−
− −−−
⋯ ⋯⋯⋯
⋯ ⋯⋯⋯
Bin
omia
l The
orem
M
etho
d 3:
Exp
and
ing ( (((
) )))11
x− −−−
− −−−b
y th
e B
ino
mia
l The
ore
m a
lso
giv
es t
he s
am
e re
sult.
( ((() )))
( ((() )))
( ((() )))( (((
) )))1
22
33
11
2et
c.2
23
nn
nn
nn
nn
nn
ab
ana
ba
ba
b−
−−
−−
−−
−−
−−
−−
−−
−−
−−
−−
−−
−+
=+
++
++
=+
++
++
=+
++
++
=+
++
+⋅ ⋅⋅⋅
( ((() )))
( ((() )))( (((
) ))) ( ((() )))
( ((() )))
( ((() )))( (((
) )))( ((() ))) ( (((
) )))( (((
) )))
23
21
11 4
3
23
12
(1)
1(
1)1
()
12
12
31
23
1
xx
x
x
xx
x−−
−−
−−
−−
−−
−−
−−
−−
− −−−
−−
−−
−−
−−
−=
+−
−+
−−
=+
−−
+−
−=
+−
−+
−−
=+
−−
+−
−−
−−
−−
−−
−−
−−
+−
++
−+
+−
++
−+
⋅ ⋅⋅⋅=
−+
−+
=−
+−
+=
−+
−+
=−
+−
+
⋯ ⋯⋯⋯
⋯ ⋯⋯⋯
Rat
iona
l Exp
ress
ions
Oth
er f
unc
tion
s ca
n b
e ha
ndle
d in
the
sam
e w
ays
. O
ne
wa
y to
find
the
Mac
laur
in S
erie
s fo
r an
y ra
tiona
l exp
ress
ion,
suc
h a
s 215
5xx
+ +++,
is to
arr
ange
the
term
s w
ith t
he lo
wes
t pow
er f
irst
and
per
form
a lo
ng
div
isio
n.
35
73
33
525
125
2
3 3 35
3 5
57
33
525
73 25
3
515 15
3
3
3
xx
xx
xx x
x x xx x
x x
−+
−+
−+
−+
−+
−+
−+
−+
+ ++++ +++ − −−− −
−−
−−
−−
−
+ +++ − −−−
⋯ ⋯⋯⋯
�����������
�����������
�����������
�����������
T
his
is a
geo
met
ric
seri
es w
ith a
firs
t ter
m o
f 3x
and
a r
atio
of
2 5xr
=−
=−
=−
=−
And
the
inte
rval
of c
onv
erge
nce
is
( ((() )))
( ((() )))
( ((() )))
( ((() )))
22
2
2
23
55
5
5
35
73
33
525
125
33
33
31
3
xx
x
x
xx
xx
x
xx
xx
=+
−+
−+
−+
=+
−+
−+
−+
=+
−+
−+
−+
=+
−+
−+
−+
−−
−−
−−
−−
=−
+−
+=
−+
−+
=−
+−
+=
−+
−+
⋯ ⋯⋯⋯
⋯ ⋯⋯⋯
2 51 5
55
x
x
x
−<
−<
−<
−<
< <<<
−<
<−
<<
−<
<−
<<( (((
) )))2
2
5
153
51
x
xx
x= ===
+ +++−
−−
−−
−−
−
A “
Mis
take
”
23
23
45
11
1
(2)
(2)
(2)
12
24
816
22
nn
nk
k
k
xx
xx
xx
xx
x
xx
xx
xx
x−
−−
−−
−−
−
= ===
=+
++
+=
++
++
=+
++
+=
++
++
− −−−
=+
++
++
+=
=+
++
++
+=
=+
++
++
+=
=+
++
++
+=∑ ∑∑∑
⋯ ⋯⋯⋯
⋯⋯
⋯⋯
⋯⋯
⋯⋯
The
ser
ies
is g
eom
etr
ic a
nd c
onv
erge
s w
hen
1
12
1 or
2
2x
x<
−<
<<
−<
<<
−<
<<
−<
<.
So
the
inte
rval
of c
on
verg
enc
e is
11
22
x−
<<
−<
<−
<<
−<
<.
−6−5
−4−3
−2−1
12
34
56
7
−5−4−3−2−112345
x
y
11
22
Inte
rval
of
conv
erge
nce:
x
−<
<
func
tion
Po
wer
Ser
ies
But
wha
t if
you
do
this
?
( ((() )))
23
23
41
11
11
11
11
11
2 11
22
22
22
22
12
11
11
11
1
2
48
1632
22
nn
nn
n
xx
xx
xx
xx
xx
xx
∞ ∞∞∞
− −−−− −−−
= ===
− −−−
=
=−
−−
−−
==
−−
−−
−=
=−
−−
−−
==
−−
−−
−
− −−−
− −−−
− −−−=
−−
−−
−−
−−
==
−−
−−
−−
−−
==
−−
−−
−−
−−
==
−−
−−
−−
−−
=∑ ∑∑∑
⋯ ⋯⋯⋯
⋯⋯
⋯⋯
⋯⋯
⋯⋯
The
ser
ies
is g
eom
etr
ic a
nd c
onv
erge
s w
hen
1
11
or
22
xx
<<
<<
<<
<<
.
So
the
inte
rval
of c
on
verg
enc
e is ( (((
) )))( (((
) )))1
12
2,
,−∞
−∪
∞−∞
−∪
∞−∞
−∪
∞−∞
−∪
∞.
y
x−6
−5−4
−3−2
−11
23
45
67
−5−4−3−2−1123
1 2
Inte
rval
of C
onve
rgen
ce
x<
−
1 2
Inte
rval
of C
onve
rgen
ce
x>
New
Ser
ies
from
Old
T
reat
ing
2
11
x+ +++
as
a ge
omet
ric s
erie
s w
ith
2r
x=
−=
−=
−=
− g
ives
:
2
46
22
0
11
(1)
1
nk
k
k
xx
xx
x= ===
=−
+−
+=
−=
−+
−+
=−
=−
+−
+=
−=
−+
−+
=−
+ +++∑ ∑∑∑
⋯ ⋯⋯⋯ fo
r 1
1x
−<
<−
<<
−<
<−
<<
But
1
2
1ta
n1
dx
dxx
− −−−= ===
+ +++.
The
refo
re
2
1(
1)
13
57
21
11
13
57
21
21
1ta
n1
kn
kk
k
xdx
Cx
xx
xC
xx
− −−−− −−−
−−
−−
−−
−−
− −−−= ===
==
+−
+−
+=
+=
=+
−+
−+
=+
==
+−
+−
+=
+=
=+
−+
−+
=+
+ +++∑ ∑∑∑
∫ ∫∫∫⋯ ⋯⋯⋯
The
initi
al c
ondi
tion
( ((() )))
1ta
n0
0− −−−
= === te
lls u
s th
at C
= 0
. So
the
pow
er s
erie
s fo
r
1ta
nx
− −−− is
2
1(
1)
21
21
1
kn
kk
k
x− −−−
− −−−− −−−
− −−−= ===∑ ∑∑∑
for
11
x−
<<
−<
<−
<<
−<
<.
1993
BC
3
Let
f b
e th
e fu
nctio
n gi
ven
by
2(
)x
fx
e= ===
. (a
) Writ
e th
e fir
st fo
ur n
onze
ro te
rms
and
the
gener
al te
rm fo
r th
e T
aylo
r se
ries
expa
nsio
n of
(
)f
x a
bout
x =
0.
()
()
()
23
23
/2
23
/22
3
12
!3!
!
/2/2
/21
22
!3!
!
12
22
!2
3!2
!
nx
n
x
nx
n
xx
xe
xn
xx
xx
en
xx
xx
en
=+
++
++
+
=+
++
++
+
=+
++
++
+
⋯⋯
⋯⋯
⋯⋯
(b
) Use
the
resu
lt fr
om p
art (
a) to
writ
e th
e fir
st th
ree
nonz
ero
term
s an
d th
e ge
nera
l ter
m o
f the
ser
ies
expa
nsio
n ab
out
for
x =
0 f
or
21
()
x
eg
xx− −−−
= ===.
2
3/2
23
23
23
/2 /22
1
23
12
22
!2
3!2
!
11
22
2!
23!
2!
1 11 2
22
!2
3!2
!nx
n
n
nx x
n n
xx
xx
en
xx
xx n
e
xx
ex
xx
xn−
=+
++
++
+
++
++
++
−
−
=
−=
++
++
+
⋯⋯
⋯⋯
⋯⋯
(c)
For
the
func
tion
g in
par
t (b)
, fin
d
( ((() )))2
g′ ′′′ a
nd u
se it
to s
how
that
1
14(
1)!
4n
nn
∞ ∞∞∞ = ===
= ===+ +++
∑ ∑∑∑ .
()
()
()
()
()
()
2
23
2
2
23
1
11
2
22
!2
3!2
!
11 8
24
2! 1
21
22
22
2!
23!
2!
11
1
81
24
!
41
!
n
n
n
n
n
n
n
nx
xg
xn
nx
x
n ng
nn
nn n
−
−
−
∞ =
−′
=+
++
+
−=
++
++
−⋅
′=
++
++
−=
++
++
=+
∑
⋯⋯
⋯⋯
⋯⋯
⋯⋯
()
()
()(
)
()
()
()
/2
/2/2
2
1
1A
lso;
11
12 1
21
12
24
41
41
!4
x
xx
n
eg
xx
xe
eg
xx
ee
g
n n
∞ =
−=
−−
′=
⋅−
−′
==
∴=
+∑
()
21
23
1 22
2!
23!
2!
n n
xx
xg
xn−
=+
++
++
⋯⋯
Eul
er’s
For
mul
a
23
45
23
45
12
!3
!4
!5
!(
)(
)(
)(
)1
2!
3!
4!
5!
x ix
xx
xx
ex
ixix
ixix
eix
=+
++
++
+=
++
++
++
=+
++
++
+=
++
++
++
=+
++
++
+=
++
++
++
=+
++
++
+=
++
++
++
⋯ ⋯⋯⋯
⋯ ⋯⋯⋯
Exp
and
, an
d s
imp
lify
the
term
s. G
roup
tho
se te
rms
with
out
i an
d th
ose
with
i.
22
33
44
55
24
63
57
12
!3
!4
!5
!
12
!4
!6
!3
!5
!7
!
ix ix
ix
ix
ix
ix
eix
xx
xx
xx
ex
i
=+
++
++
+=
++
++
++
=+
++
++
+=
++
++
++
=−
+−
++
−+
−+
=−
+−
++
−+
−+
=−
+−
++
−+
−+
=−
+−
++
−+
−+
⋯ ⋯⋯⋯
⋯⋯
⋯⋯
⋯⋯
⋯⋯
24
63
57
12
!4
!6
!3
!5
!7
!ix
xx
xx
xx
ex
i
=
−+
−+
+−
+−
+=
−+
−+
+−
+−
+=
−+
−+
+−
+−
+=
−+
−+
+−
+−
+
⋯
⋯⋯
⋯⋯
⋯⋯
⋯
Rec
ogni
ze th
e M
acla
urin
Ser
ies
for co
s a
nd s
inx
x.
co
ssi
nix e
xi
x=
+=
+=
+=
+
F
inal
ly s
ubst
itute
xπ πππ
= === a
nd s
impl
ify a
gain
.
cos
sin
10
10
i i
i
ei
ei
e
π πππ π πππ
π πππ
ππ
ππ
ππ
ππ
=+
=+
=+
=+
=−
+=
−+
=−
+=
−+
+=
+=
+=
+=
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