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CALCULUS II MA 1123
Infinite Sequences andSeries
Telkom Institute of Technology
Bandung - Indonesia
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Infinite Se uencesInfinite Se uences
Definition
Sequences or infinite sequence is a function whosedomain is the set of positive integers.
Notation: f : N R
= nThis function is also sequence of real number. We
denote a sequence by a1, a2, a3, ...., by or{ }
=0nna n n .
Formula of sequence:
1. An explicit formula for the nth term2. The first few terms of the sequence
3. A recursion formula1 111 naaa == 1
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n n ...,
4,
3,
2,
n
n
a++
1
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Conver ence of Se uenceConver ence of Se uence
Definisi:e say a sequence an converges o or as m
L and write
La =lim
If for each positive number , there is acorres ondin ositive number N such that
n
A sequence that fails to converge to any finite
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Notes:Notes:
Here and is many sequence problem, it is convenient touse the followin almost obvious fact.
Lxfx
=
)(limIf Lnfn
=
)(lim, then
This is convenient because we can apply lHopitals Rule
to the continuous variable problem.
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Pro erties of Limit Se uencePro erties of Limit Se uence
Let {an} be convergent to L and {bn} be convergent toM, then
1. MLblimalimbalim ==nnn
2.( ) ( ) ( )
M.Lblim.alimb.alimnnnnnnn
==
3.( )
M
L
blim
alim
b
alim
n
nn
n
n
n==
, untuk M 0
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Monotone Se uenceMonotone Se uence
Definition:
A sequence {an} is called
(i) increasing if an < an+1
(ii) nondecreasing if an an+1
(iii) decreasing if an > an+1
(iv) nonincreasing if an an+1
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Exam leExam le
Determine if the sequence coverges or diverges :
1n2na n =
1.
Answer:
We has )( =xxf , in particular, I Hopital Rule,
2
1
12lim)(lim =
=
x
xxf
xx
that is, an converges to .
2
1
12lim =
nn
n
,
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Exam le Cont.Exam le Cont.
2.
n
n
11a
+=
Answer:We has
x
xxf
+= 11)( , In particular, IHopitals Rule,
+= x
xx
11ln.limexp
+=
x
x 1
1ln
limexpx
x x
+
11lim
ee
x
x
x==
+
=
1
1
limexp
+
=
2
2
1
1.
1
limexp x
x
xx
en
n
n=
+
11lim
Thus, x
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that is, an converges to e.
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Problem SetProblem Set
2
Determine if the sequence coverges or diverges :
3n2n
a2n
+
=
2n32 +
=
...4
,3
,2
,1
an+1 =
2
(an +
na
) , a1=21. 7.
1nn +
na n =
...9
5,
7
4,
5
3,
3
2,1
.
9.3.
( )n
n
n4
a
=
...
4
31
1,
3
21
1,
2
11
1,110.4.
na n =
...
15
4,
14
3,
13
2,
12
1an+1 = 1 +
2
1an , a1=1 11.6.
5.
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Se uence of Partial SumsSe uence of Partial Sums
Let Sn denote the sum of the first n terms of the series
, then
=1i
ia
S1 = a1
.
.
.Sn = a1 + a2 + a3 + a4 + + an =
=iia
1
n =1i
i
Sequence {Sn} is called the sequence of partial sums
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Definition: Convergent of Infinite SeriesDefinition: Convergent of Infinite Series
The infinite series
ia conver es and has sum S if the=1i
sequence of partial sums {Sn} converges to S, if {Sn}
diverges, then the series diverges.
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Geometric SeriesGeometric Series
A series of the form
=
1n
1nar =a +ar +a r
2
+ ... + a r
n-1
+ ...
w a .
The partial sum of geometric series isnSn =
=
1i
1iar = a +ar +a r + ... + a rn-
r1an
nr1
, .
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Pro erties ofPro erties of Geometric SeriesGeometric Series
n .
n
nrlim
= 0, then the series converges to
r1a
2. if then sequence {rn} diverges because
r > 1 nn
rlim
= ,
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Exam leExam le
...11111
+++++1. Show converges
Answer:Consider the partial sums
S1 =2
= 1 -2
S2 =42
+ =4
= 1 (2
)2
S3 = 111 ++ = 7 = 1 ( 1 )3
then,
S = 1 1 n
2and
nn
Slim
=n
lim (1 (2
1)n) = 1
so because the se uence of artial sums Sn
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converges to 1, then that infinite series converges to 1.
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Exam le 2Exam le 2
1
2.2. (Kolaps Series). Show the infinite series converges.=1i
Answer:Consider the partial sums
)1i(i +=
i-
1i +
then,
and
Sn =
+
/
++
/
/
+
/
/
+
/
1nn
...43322
1 =
+
1n
1
so because the se uence of artial sums Sn
nn
Slim
=n
lim
+
1n
1 = 1
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converges to 1, then that infinite series converges to 1.
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Exam le 3Exam le 3
3.3.
= i
1(Show that the harmonic series diverges)
Answer:
We show that Sn grows without bound. Imagine n to beLarge and write
Sn = 1 +n
...8765432
++++++++
Sn = 1 + 1...1111111 ++ ++++ ++n
1 +
n
1...
8
1
8
1
8
1
8
1
4
1
4
1
2
1++
++++
++
It is clear that by taking n sufficiently large, we can
= 1 +n
...2222
+++++
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.n
, ,so does the harmonic series.
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Dever ence tests ofDever ence tests ofnnth termth term
If the series
=0nna converges, then n
nalim
= 0, ekivalen if
nn
alim
0 then the series diverges.
2n
= ++1n 4n3n3
That series diverges sincen 2 1 1
4n3n3m
2n ++=
2
n
n
4
n
33
m
++ 3
= (Tidak Nol)
So proved that diverges.
= ++1n2
2
4n3n3
n
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The New ProblemThe New Problem
To prove that a series converges is does not suffice to
s ow t at nn
am
= , s nce t s property may o or
divergent as well as convergent series.
=1n n
1
=1 + n
1
...8
1
7
1
6
1
5
1
4
1
3
1
2
1
++++++++ + . . .It is clear that n
nalim
= 0, but the harmonic series diverges.
.
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Positive Series TestPositive Series Test
1. The integral test
Let function f be continuous, positive, nonincreasing
function the interval [1,)
.
1,
=1n
)n(f converges.
. ,
=1n
)n(f diverges.1
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Exam leExam le
1. Determine the series
n 2en converges or diverges=1n
Answer. We has2
xex)x(f = , thus2 2b
b1 2dxex
1
xexm1b
1bxem
2b
x 2elim1 11lim1 1
= =
= = =1 eebb
e
So because dxex2x
converges, then
n 2en
=n
also converges.
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Exam leExam le
2. Determine the series
1converges or diverges
Answer. We has , thus
=2n
xxxf
ln1)( =
=b
b
xx
dx
xx
dx
22
ln
lim
ln
=
2
ln
)(lnlim
x
xd
b
bb
dx 1,
also diverges.
2 lnxx=2n
nlnn
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Positive Series TestPositive Series Test
2. Convergence ofp-series
e ser es
=1
1i
piw ere p s a constant, s ca e p-ser es.
By the integral test, we have
1
tp1x
1t p1
xx
m1 pt
=1
t p1 =
p1t
Note the case
1. p = 1 gives the harmonic series, then the seriesdiverges
2. p > 1 givesp1tlim = 0, then the series converges
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Positive Series TestPositive Series Test
3. p < 1 givesp1tlim =, then the series diverges.
4. p < 0, the nth term of =n
i Pi1
1, namely,
doesnt evenPn
1
Tend toward 0. So the series diver es b nth term test.
So we has, namely:
-.2. The p-series diverges if 0 p 1
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Exam leExam le
Does the following series converge or diverge?
1.
=1
001,1
nn
-
1
=1,
n n
because p=1,001 > 1
1.
By the p-series test, the series diverges
=1 2n n
= 21
1
because p= < 1
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Exam leExam le
Does the following series converge or diverge:
n.
= 3n2 5n
Answer:
1
n
n n
n=
5n 2 we know that
,
=1n n
1
is harmonic series andn
1
5n2 n
, so ecause t e ser es=1n
n
2
n
iverges, t en
the series diverges.= 2n n
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Exam leExam le
1
Answer:
= +1n 5n
We will comparing the series with bn= and an=
we know that the series is p-series with p = 2 > 1
5n2 +2n
21
, so because the series
2n
1
5n
12 +
=1n
and
=1n2n
1converges
then the series converges.= +1n
2 5n
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Problem SetProblem Set
Does the following series converge or diverge
= +1n2
5n ( )= 3n22n
4.1.
= 3n2
5n
1 = 1n 1n2
12. 5.
= +1nn 12
13.
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4. Limit Comparison Test
Suppose an and bn positive series andn
n
n b
a
lim = L
.= `1n
n
= `1nn
diverge together
2. i L = 0 an= `1n
n= `1n
naconverges, t en converges.
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Exam leExam le
Does the following series converge or diverge :
= +1n23 7n5n
1.
Jawab:
If we note the series, the nth term is like bn=
Thus,
2n
32n +
n
n
n b
alim
= 2
1
2
23
175lim
n
nnn
+= 75
lim23 +
+=
nn
nn
n
+
23
3n2the series converges.
So because L=2 and the series=1
2n n
converges, then
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Exam leExam le
12.
Answer:
= +1n 2 4n
If we note the series, the nth term is like bn =
thus,
n1
n
n
n b
alim
= 1
n1
4nlim
2
n
+ 4n
nlim
2
2
n += =
the series diverges.
So because L=1 and the series diverges, then
= +1n 2 4n
1 =1nn
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Problem SetProblem Set
Does the following series converge or diverge:
= ++1n
2
3n2n
= 1n3
4n
4.1.
= +1n 1nn
=1n
2n2.5.
=
+
1n2n
3n23.
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5. Ratio Test
=1k kaa
Let be a series of positive term and
= kk a
m
=ka1. if < 1, the series converges.
suppose
=1kka diverges.2. if > 1, the series
= , ..
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Exam leExam le
Does the following series converge or diverge: 3n
.=1 !n n
n 1+nAnswer:
Suppose nth term is an =!n, then (n+1)th is an+1=
thus,( )!1+n
1+n31n+
3
n
( )1lim
+=
nn0=
( )!13lim
+=
nnn!
3lim
n
nn=
n
n
n a1lim +
o ecause = < , en e ser es=1 !n n
converges.
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Problem SetProblem Set
Does the following series converge or deverge:
=1n
n
n
=1 !n n 3
4.1.
( )
=1 !2n n
n
( )=1 !2n n2. 5.
=
+
1 !
4
n
n
n
n3.
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6. The Root Test
=1k kaLet be a series with a positive terms and suppose
ka1. if a < 1, the series converges
kk
=
=1kka diverges
=
2. if a > 1, then the series
3. if a
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Exam leExam le
Does the following series converge and diverge: + 22
n
n.
= 1 1n n
Answer: n
Suppose nth is an =n
1
, thus
22 +n1
=
= nn
nn
So because a = 2 (> 1), then the series
=
+
1
22n
n
n
diverges.
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Exam leExam le
+ 2n
n.
= 1 12n n
Answer: n
Suppose nth term is an =n
12
, thus
12+nn212 nnnn
n
So because a = (< 1), then the series =
1 1n n
converges.
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Problem SetProblem Set
Does the following series converge or diverge: n n
=
1 lnn n =
+1 2n n
3.1.
=
+
1 12
23
n
n
n
n
=
+1 23n n
n2. 4.
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Alternating SeriesAlternating Series
The series of the form
( ) ...aaaaa1 43211n
n1n ++=
=
+
with an > 0, for all n.
An important example is the alternating harmonic
series, namely;
( ) ...4
1
3
1
2
11
n
11
1n
1n ++=
=
+
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Exam leExam le
1. Answer (Alternating series test)
1
1
e ave an= n , an an+1 = 1+n
since
, e ser es converges,
1a. 11
11
1
>+==+
=+ nnn
n
an
n an >an+1
1.
nnn
n
So because a dan b proved, the alternating seriesconverges.
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Exam leExam le
2. Answer (alternating series test)
1
1
e ave an=!n, an an+1 =( )!1+n
since
, e ser es converges,
1
a.
( )11
!11
1
>+=+
=+
n
nan
n an >an+1
1.! nn
nn
So because a dan b proved, the series converges.
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Problem SetProblem Set
Does the following alternating series converge or deverge:
( )=
+
+
1
1
13
1n
n
n
)=
1 3
1
n
n
n
4.1.
( )
= ++
1
21
n
n
nn
n ( )= +
1 )1(
1n
n
nn2. 5.
( )
=
+1
1
!1
n
nn
n
n3.
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Absolute Convergence and ConditionalAbsolute Convergence and Conditional
If the series is said to converge adsolutely if the series
=1nnb
In the other words, a series
converges absolutely, then
=1nnb converges.
of absolute values converges.
nb
nb converges.
And the series is said converge conditionally if the series
diver es but=1n =1n
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Ratio Test for Absolute ConvergenceRatio Test for Absolute Convergence
Let
na with an 0 andna 1lim +
= r, then=1n n
1. if r < 1, the series converges absolutely2. if r > 1, the series diverges
. ,convergence can be drawn from this test.
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Exam leExam le
Does the following series conditional converge, absoluteconver e or diver es:
1.( )
=
+1
1
!
21
n
nn
n
We have an= ( )!
21 1
n
nn+ , and an+1 =( )
( )!1
21
12
+
+
+
n
nn
thus,
( )
( ) 21
!11
limlim1
1 n
a
ar
nn
n
nn
n
n +
+
+
+==
( )!12!2
lim1
+=
+
n
nn
n
n 1
2lim
+=
nn0=
We conclude from the Absolute Ratio Test, the seriesabsolute converges.
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Exam leExam le
2. ( )
+ 11
1 n
=n
Answer:
+ 11
1 n =1n n
diverges, since it is a p-series
=
=
=1 1
1
n n
n
n
a
However,
( )
+ 11
1n
So the series
with p = .
conditional converges.=n
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Problem SetProblem Set
Does the following series conditional converge,
( )
1n
n n ( )
11 n1. 4.
,
=1n
)4(n
=1n
+
1
)1(n
=12
n n
n
=1 lnn nn
+
1n
. .
= +1 23n n = +1 1n nn3. 6.
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Power SeriesPower Series
The power series has two form:
1. A power series in x has the form
nxa = 2
=0n
2. A power series in (x b) has the form
( )
=
0n
nn bxa = a0 + a1 (x-b) + a2 (x-b)
2 + . . .
For this time, we talk about convergence set / for what xsdoes the power series converges.
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Radius of Conver enceRadius of Conver ence
Radius of conver ence found with the ratio test forabsolutely converges :
Let ( )
nn bxa and nn
n bxaL)(
lim1
1 =+
+
=0n n
1. If L < 1, the series converges.. , ,tried.
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nx
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AnswerAnswer nx
=0n
. .
n
n
n
n
n
xxL :lim
1
1
= ++
x=
)1(lim +=
nx
So the series converges absolutely if L< 1, namelyIf x = 2 or x = -2 , the ratio fails. However, When x = 2
( ) ( )
=
= +=
+ 11 11
21
2
nnn
n
nn
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AnswerAnswer
When x = 2
The series is the alternatin harmonic series which
( ) ( )==
+
=+
11 121 nnn nn
We conclude that the convergence set is the interval
converges.
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!)1( n
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Answer 3Answer 3 + !)1(nxn
=
. e app y t e rat o test or a so ute convergence.
( ) nn xnxn
L!1
lim+
+=
( )xn
n2lim +=
==
0,
,
xjika
xa
= .
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Theorem 1Theorem 1
n
nxa is alwa s=0n
an interval of one of the following three types.
1. The single pointx= 0.2. An interval (-c, c), plus possibly one or both endpoints.
3. The whole real line.
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Theorem 2Theorem 2
nbxa is alwa s=0n
an interval of one of the following three types.=. .
2. An interval (-c + b, c + b), plus possibly one or bothendpoints.
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Problem SetProblem Set
Find the convergence set of the given power series: n
( )= +
02
1n n
1.
...
81.4
4ln2
27.3
3ln2
9.2
2ln2
3
2+
++
++
++
+ xxxx2.
( ) ...!3!2
2 ++
++
++xx
x3.
32n
x nn
nn
( )= +02
41nnn
4.( )= +
0
221n
nn5. ( )= +0
231n
nn6.
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O eration on Power SeriesO eration on Power Series
We have done this for one series, a geometric series.
11,
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Theorem 3Theorem 3
Suppose that S(x) is the sum of a power series oninterval I that is
S(x)=
=0n
n
nxa = a0 + a1x+ a2x2 + a3x3+ . . .
[ ]
=0n
nnxaD
Then,1. S(x) = = D[a0 + a1x+ a2x
2 + a3x3+ . . .]
2
=
1
1
n
nn xna
x x
1 1 1
1 2 3 . . .
=
dttS0
)(=0
0n
n tta
=
+
+1
1
nn xn
a
2 3 4=
=
= a0x+ a1x + a2x a3x+ . . ..
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Exam leExam le
Apply Theorem 3 to the geometric series
x1= 1 +x + x2 +x3 + . . . for -1
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ContohContoh
a. ln (1 x)
Integrating term by term givesx
++++== dttttdttx0
32
0
...11
)1ln(
111111 432432x
...432
...432 0
, -1
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Problem SetProblem Set
Find the power series representation for f(x):
x
xf
+
=
1
)(
=x
x1
ln1
1. 5. f(x)=tan-1(x)
x 122
+ x1( )21 x+..
1=xx ++ 11
.( )x32 +
.
2
1)(xf =4.
x
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Taylor and MaclaurinTaylor and Maclaurin SeriesSeries
Definition: Let f(x) be function with derivatives of all orderatx=b. then we define the Taylor series for fabout x = b
to be )( )( nn bf
)()('' 2bxbf
=
0
!n
nx = = - . . .
In the special case where b = 0, the taylor series for f
!2
)0(" 2xf
, ,
( ) )(
!
)0( nn
x
n
ff(x) = = f(0) + f(0)(x)+ + . . .
=n
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Exam leExam le
Find the Maclaurin Series for the given function:
=.
Answer:
=
f(x) = cosx
f(x) = - sinx f(0) = 0
f(0) = 1
f(x) = - cosx f(0) = -1
flV (x) = sinx flV(0) = 0
thus,753
xxx +12n
n x
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...
!7!5!3 ( )= +=
0 !12n n
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Exam leExam le
2. f(x)= ex
f(x) = ex
fx = ex
f(0) = 1
f(x) = ex f(0) = 1
f(x) = ex f(0) = 1
flV (x) = ex flV(0) = 1
...!4!3!2
1)(432
+++++==xxx
xexf x
=
=0 !n
n
n
x
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Exam leExam le
Find the Taylor series about x = 1 for f(x)= ex
f(x) = ex
f(x) = ex =
f(1) = e
f(x) = ex f(1) = e
f(x) = ex
f(1) = eflV (x) = ex flV(1) = e
thus,
( ) ( )...
!3
1
!2
1)1()(
32
+
+
++==x
ex
exeeexf x( )
=
=
0 !
1
n
n
n
xe
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