Answer Key
QA #3 Series
QuestionsΒ andΒ AnswersΒ SheetΒ 3
Β
Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β SeriesΒ
Β
QuestionΒ #1Β
Tell whetherΒ theΒ series converge.Β IfΒ itΒ converges,Β findΒ theΒ sum.Β
Β
β(πΒ =Β 0)
π
π(ππ/2)
π
Β Β
Β
Answer:Β
Β
β(πΒ =Β 0)
π
π(ππ/2)
π
Β
Β
πΒ =Β 1Β =Β (ππ/2)
2
/(ππ/2)Β =Β ππ/2Β =Β 1.57Β >
Β Β
Thus,Β theΒ seriesΒ diverges.
Β
QuestionΒ #2Β
Tell whetherΒ theΒ series converge.Β IfΒ itΒ converges,Β findΒ theΒ sum.Β
Β
β(πΒ =Β 1)
π
π(3/7)
π
Β
Β
Answer:Β
Β
β(πΒ =Β 1)
π
π(3/7)
π
Β Β
Β
πΒ Β =Β Β 3/7Β Β <Β Β 1Β
,Β soΒ theΒ series converges.Β
Sum:Β Β
(3/7)/(1βΒ 3/7)Β =
Β Β
Β
=Β Β 3/7Β π₯π₯Β 7/4Β Β =Β Β 3/4
Β
Β
QuestionΒ #3Β
Tell whetherΒ theΒ seriesΒ convergesΒ orΒ diverges,Β andΒ findΒ sum,Β ifΒ itΒ converges.Β
Β
β(πΒ =Β 3)
π
π(1/(πβΒ 2)βΒ 1/π)
Β
Β
Answer:Β
Β
β(πΒ =Β 3)
π
π(1/(πβΒ 2)βΒ 1/π)
Β Β
Β
β(πΒ =Β 3)
π
π1/(ππ₯π₯(πΒ βΒ 2))
Β Β
ThisΒ is notΒ geometric seriesΒ
Β
π
π
=Β 1Β +Β 1/2Β =Β 3/2ππ πΒ βΒ ππ.
Β
Thus,Β seriesΒ converges.
Β Β
QuestionΒ #4Β
PleaseΒ justifyΒ whetherΒ theΒ seriesΒ convergesΒ orΒ diverges,Β andΒ findΒ sum,Β ifΒ itΒ converges.Β
Β
β(πΒ =Β 1)Β ππΒ 3/π
2
.Β
Β
Answer:Β
Β
β(πΒ =Β 1)Β ππΒ 3/π
2
Β Β
Β
β(πΒ =Β 1)Β ππΒ 1/π
π
Β Β
βΒ ifΒ Β
πΒ Β >Β Β 1π
Β Β
βΒ series converges,Β ifΒ Β
πΒ Β <Β Β 1
Β ,Β series diverges.Β
Therefore,Β series convergesΒ by p-test.Β
QuestionΒ #5Β
Tell whetherΒ theΒ seriesΒ convergesΒ orΒ diverges.Β
Β
β(πΒ =Β 1)
π
π5/(πΒ +Β 4)
4
Β
Β Β
Answer:Β
Β
β(πΒ =Β 1)
π
π5/(πΒ +Β 4)
4
Β Β
PutΒ Β
πΒ Β +Β Β 4Β Β βΒ Β π
Β Β
So,Β
β(πΒ =Β 5)
π
π5/π
4
Β Β
FromΒ p-seriesΒ test:Β
ForΒ series Β
β(πΒ =Β 1)
π
π1/π
π
,Β ifΒ
πΒ >Β Β 1
,Β series converges.Β
So,Β forΒ seriesΒ inΒ equationΒ 1Β Β
πΒ Β =Β Β 4
,Β soΒ this seriesΒ convergesΒ
TheΒ seriesΒ is p-series withΒ
πΒ Β =Β Β 4
Β andΒ itΒ converges.
Β
QuestionΒ #6Β
JustifyΒ ifΒ theΒ seriesΒ is convergent:Β
Β
β(πΒ =Β 1)
π
ππ
(βπ)
Β Β
Answer:Β
Β
β(πΒ =Β 1)
π
ππ
(βπ)
=
Β Β
Β
=Β β(πΒ =Β 1)
π
π1/π
π
Β
Here,Β commonΒ ratio,Β Β
πΒ Β =Β Β 1/π
Β Β
So,Β Β
πΒ =Β (1/π)/(1βΒ 1/π)Β =Β (1/π)/((πβΒ 1)/π)
Β Β
Β
πΒ =Β 1/(πβΒ 1)
.Β
This seriesΒ converges.
Β
QuestionΒ #7Β
ForΒ whichΒ zβsΒ does theΒ infiniteΒ series converge?Β ToΒ whatΒ functionΒ doesΒ itΒ converge?Β
Β
1Β +Β π§/2Β +Β π§
2
/4Β +Β π§
3
/8Β +Β β―
Β
Β
Answer:Β
Β
1Β +Β π§/2Β +Β π§
2
/4Β +Β π§
3
/8Β +Β πππΎΒ β¦
Β (geometric series)Β
So,Β Β
=Β 1/(1Β βΒ (π§/2))Β =Β 1/((2βΒ π§)/2)
Β Β
So,Β Β
=Β 2/(2βΒ π§)
Β Β
This series convergesΒ ifΒ Β
|π§|Β <Β 2
Β Β andΒ itΒ convergesΒ Β
2/(2Β βΒ π§)
Β .
Β
QuestionΒ #8Β
SimplifyΒ
(π ππΞΈΒ βΒ π‘ππΞΈ)(1Β +Β π ππΞΈ)
Β
Answer:Β
Β
(π ππΞΈΒ βΒ π‘ππΞΈ)(1Β +Β π ππΞΈ)Β =
Β Β
Β
=Β π ππ(ΞΈ)π ππ(ΞΈ)Β βΒ π‘ππ(ΞΈ)Β βΒ π‘ππ(ΞΈ)π ππ(ΞΈ)Β =
Β Β
Β
=
1
πππ ΞΈ
+
π ππΞΈ
πππ ΞΈ
βΒ π‘ππΞΈΒ β
π ππΞΈ
πππ ΞΈ
π ππΞΈΒ =
Β Β Β
Β
[:
β²
π ππΞΈΒ =
1
πππ ΞΈ
ππππ‘ππΞΈΒ =
π ππΞΈ
πππ ΞΈ
]
Β Β
Β
=
1
πππ ΞΈ
+Β π‘ππΞΈΒ βΒ π‘ππΞΈΒ β
π ππ
2
ΞΈ
πππ ΞΈ
=
Β Β
Β
=
1
πππ ΞΈ
β
π ππ
2
ΞΈ
πππ ΞΈ
=
Β Β Β
Β
=
1βπ ππ
2
ΞΈ
πππ ΞΈ
=
Β Β Β
Β
=
πππ
2
ΞΈ
πππ ΞΈ
=Β πππ
Β Β
Β
[:
β²
π ππ
2
ΞΈΒ +Β πππ
2
ΞΈΒ =Β 1]
Β Β
Answer:Β Β
(π ππΞΈΒ βΒ π‘ππΞΈ)(1Β +Β π ππΞΈ)Β =Β πππ
Β
Β Β
QuestionΒ #9
Β
ComputeΒ eachΒ ofΒ theseΒ doubleΒ sums:Β
Β
β
β
(πΒ +Β π)
3
π+1
2
π=1
Β
Β
Answer:
Β
ComputeΒ thisΒ doubleΒ sumΒ by firstΒ computingΒ theΒ innerΒ sumΒ followedΒ by theΒ outerΒ sum.Β
Β
β
β
(πΒ +Β π)
3
π=1
2
π=1
Β Β
Β
=Β β
(πΒ +Β 1Β +Β πΒ +Β 2Β +Β πΒ +Β 3)
2
π=1
Β Β
Β
=Β 3Β β
(πΒ +Β 2)
2
π=1
Β Β
Β
=Β 3(1Β +Β 2Β +Β 2Β +Β 2)
Β Β
Β
=Β 21
Β
Β
QuestionΒ #10Β
ComputeΒ eachΒ ofΒ theseΒ doubleΒ sums:Β
Β
=Β β
(πΒ +Β 1Β +Β πΒ +Β 2Β +Β πΒ +Β 3)
2
π=1
Β
Β
Answer:Β
ComputeΒ thisΒ doubleΒ sumΒ by firstΒ computingΒ theΒ innerΒ sumΒ followedΒ by theΒ outerΒ sum.Β
Β
=Β β
β
(2πΒ +Β 3π)
3
π=0
2
π=0
Β Β
Β
=Β β
(2πΒ +Β 0Β +Β 2πΒ +Β 3Β +Β 2πΒ +Β 6Β +Β 2πΒ +Β 9)
2
π=0
Β Β
Β
=Β β
(8πΒ +Β 18)
2
π=0
Β
Β
=Β 8(0)Β +Β 18Β +Β 8(1)Β +Β 18Β +Β 8(2)Β +Β 18
Β Β
Β
=Β 78
Β
Β
Β
QuestionΒ #11Β
WriteΒ anΒ equivalentΒ series withΒ theΒ indexΒ ofΒ summationΒ beginningΒ atΒ =Β 1.Β
Β
β
(β1)
π
β
π=0
+Β 1(πΒ +Β 1)π₯
π
Β
Β
Answer:Β Β
This equationΒ is givenΒ inΒ theΒ series.Β AddΒ 1Β toΒ bothΒ sides andΒ introduceΒ k.Β SubtractΒ 1Β fromΒ bothΒ sides.Β
Β
πΒ =Β 0
Β ,Β soΒ Β
πΒ +Β 1Β =Β 1Β =Β π
Β .Β Therefore,Β Β
πΒ =Β πΒ βΒ 1
Β Β
InΒ theΒ originalΒ series,Β plugΒ inΒ Β
πΒ βΒ 1
Β Β forΒ nΒ
Β
β
(β1)
[(πβ1)+1]
[(πΒ βΒ 1)Β +Β 1]
β
π=1
β Β π₯
πβ1
Β
Simplify.Β SinceΒ kΒ isΒ aΒ dummy variable,Β youΒ canΒ replaceΒ itΒ withΒ any otherΒ variable.Β LetΒ n=kΒ
Β
β
(β1)
π
π
β
π=1
β Β π₯
πβ1
=Β β
(β1)
π
π
β
π=1
β Β π₯
πβ1
Β
Result:Β Β
β
(β1)
π
π
β
π=1
β Β π₯
πβ1
Β
QuestionΒ #12Β
WriteΒ anΒ equivalentΒ series withΒ theΒ indexΒ ofΒ summationΒ beginningΒ atΒ =Β 1.Β
Β
β
(β1)
π
β
π=0
+Β 1(πΒ +Β 1)π₯
π
Β
Β
Answer:Β
TheΒ TaylorΒ series ofΒ Β
lnΒ π₯
Β aroundΒ c=1.Β WeΒ knowΒ fromΒ theΒ definitionΒ ofΒ TaylorΒ seriesΒ ofΒ theΒ functionΒ fΒ
Β
π(π₯)Β =Β π(π₯)|
π₯=π
+
π
β²
(π₯)|
π₯=π
1!
(π₯Β βΒ π)Β +
π
β²β²
(π₯)|
π₯=π
2!
(π₯Β βΒ π)
2
+
π
β²β²β²
(π₯)|
π₯=π
3!
(π₯Β βΒ π)
3
+Β β―
Β
NowΒ theΒ values ofΒ differentialΒ functions atΒ givenΒ pointΒ
Β
π(1)Β =Β ln(1)Β =Β 0,Β π
β²
(1)Β =
1
π₯
|
π₯=1
=Β 1
Β Β
Β
π
β²β²
(1)Β =Β β
1
π₯
2
π₯=1
=Β β1,Β π
β²β²β²
(1)Β =
2
π₯
3
|
π₯=1
=Β 2π
β²β²β²β²
(1)Β =Β β
6
π₯
4
|
π₯=1
=Β β6
Β Β
Β
π
π
(1)Β =Β (β1)
π+1Β (πβ1)!
π₯
π
|
π₯=1
=Β (β1)
π
((πΒ βΒ 1)!)
Β Β
Thus theΒ TaylorΒ seriesΒ expressionΒ becomeΒ
Β
π(π₯)Β =Β 0Β +Β (π₯Β βΒ 1)Β β
1
2!
(π₯Β βΒ 1)
2
+Β 2!
1
3!
(π₯Β βΒ 1)
3
βΒ 3!
1
4!
+Β (β1)
π+1
(πΒ βΒ 1)!
1
π!
(π₯Β βΒ 1)
π
Β Β
OrΒ weΒ canΒ writeΒ asΒ
Β f
(π₯)Β =Β (1Β βΒ π₯)Β β
1
2
(π₯Β βΒ 1)
2
+
1
3
(π₯Β βΒ 1)
3
β
1
4
(π₯Β βΒ 1)
4
+Β (β1)
π+1
(πΒ βΒ 1)!
1
π!
(π₯Β βΒ 1)
π
βΒ β―
Β
Β
QuestionΒ #13Β
UseΒ theΒ summationΒ formulas toΒ rewriteΒ theΒ expressionΒ withous summatiomΒ notation.Β
Β
β
4π+5
π
2
π
π=1
Β Β
UseΒ theΒ result toΒ findΒ theΒ sums forΒ n=10,Β 100.Β
Answer:Β
Β
β
π
π
π=1
=
π(π+1)
2
Β Β
Β
β
π
π
π=1
=Β π.Β π
,Β whereΒ k=constantΒ
forΒ Β
πΒ =Β 10
Β Β
Β
β
4π+5
(10)
2
10
π=1
=
1
100
[4Β β
π
10
π=1
+Β β
5
10
π=1
]
Β
1
10(10Β 1)
[]4
10Β 5]
2.7
100
2
+
=
+Β ο΄Β =
Β
forΒ n=100Β
Β
β
4π+5
(100)
2
100
π=1
=
1
(100)
2
[4Β β Β β
π
100
π=1
+Β βΒ πΒ =Β 1
100
5]
Β
Β
=
1
(100)
2
[4
100(100+1)
2
+Β (5Β ΓΒ 100)]
Β Β
Β
=Β 2.07
Β
Β
QuestionΒ #14Β
DetermineΒ theΒ sumΒ ofΒ theΒ followingΒ series.Β
β
(
1
π
+9
π
11
π
)
β
π=1
Β Β
Answer:Β
WeΒ haveΒ toΒ findΒ sumΒ ofΒ series.Β
SeriesΒ isΒ givenΒ as:Β
Β
βΒ β
(
1
π
+9
π
11
π
)
β
π=1
Β Β
WeΒ willΒ useΒ geometric seriesΒ testΒ toΒ findΒ sumΒ ofΒ series .Β
Geometric series testΒ is givenΒ below:Β
Β
βΒ β
ππ
πβ1
β
π=1
Β Β
Β
βΒ ππ|π|Β <Β 1
Β (series converges)Β
Β
βΒ β
=
π
1βπ
Β (infiniteΒ gp)Β
WithΒ theΒ helpΒ ofΒ geometric series weΒ willΒ findΒ sumΒ .Β
WorkΒ is shownΒ below:Β
Β
βΒ β
(
1
π
+9
π
11
π
)
β
π=1
Β
Β
βΒ β
(
1
11
)
π
β
π=1
+Β (
9
11
)
π
Β Β
Β
βΒ β
(
1
11
)
π
β
π=1
=
1
11
1β
1
11
Β Β
Β
βΒ β
(
1
11
)
π
β
π=1
=
1
10
Β Β
Β
βΒ β
(
9
11
)
π
β
π=1
=
9
11
1β
9
11
Β Β
Β
βΒ β
(
9
11
)
π
β
π=1
=
9
2
Β Β
Β
βΒ β
(
1
11
)
π
β
π=1
+Β (
9
11
)
π
=
1
10
+
9
2
Β Β
Β
βΒ β
(
1
11
)
π
β
π=1
+Β (
9
11
)
π
=
23
5
Β
QuestionΒ #15Β
WhichΒ equationΒ couldΒ beΒ usedΒ toΒ calculateΒ theΒ sumΒ ofΒ theΒ geometric series?Β
Β
1
3
+
2
9
+
4
27
+
8
21
+
16
243
Β
Β
Answer:Β
TheΒ formulaΒ forΒ theΒ nthΒ partialΒ sumΒ ofΒ aΒ geometric sequenceΒ is,Β
Β
π
π
=
π
1
(1βπ
π
)
1βπ
,Β πΒ β Β 1
Β Β
where,Β Β
π
π
Β Β is theΒ sumΒ ofΒ GPΒ withΒ nΒ termsΒ
Β
π
1
Β Β is theΒ firstΒ termΒ
rΒ isΒ theΒ commonΒ ratioΒ
nΒ is theΒ numberΒ ofΒ termsΒ
TheΒ sumΒ ofΒ theΒ geometric series givenΒ is,Β
Β
1
3
+
2
9
+
4
27
+
8
21
+
16
243
Β Β
ThereΒ areΒ 5Β terms inΒ theΒ series andΒ theΒ commonΒ ratioΒ ofΒ theΒ seriesΒ is foundΒ byΒ dividingΒ anyΒ termΒ by theΒ
previous term.Β Suppose,Β divideΒ theΒ secondΒ termΒ by theΒ firstΒ termΒ andΒ theΒ commonΒ ratioΒ isΒ obtainedΒ as,Β
Β
πΒ =
2
9
Γ·
1
3
Β Β
Β
=
2
3
Β Β
Therefore,Β rΒ is notΒ equalΒ toΒ 1Β andΒ theΒ sumΒ ofΒ theΒ givenΒ seriesΒ is,Β
Β
π
5
=
1
3
(1β(
2
3
)
5
)
(1β
2
3
)
Β
QuestionΒ #16Β
FindΒ whetherΒ theΒ seriesΒ divergesΒ andΒ itsΒ sum:Β
Β
β
(β1)
π+1Β 3
5
π
β
π=1
0
1
2
2
3
3
0
(1
)
(Β )1
(
)Β ][(Β )1Β (
)
(Β )1Β (
)Β (Β )1
(
)
(Β )1
(
)Β ...
0
1
2
3
n
n
nΒ k
k
n
n
n
n
k
n
n
n
n
n
x
x
x
x
x
x
k
β
β
β
β
=
β
=
β
β
+
βΒ +
β
+
β
ο₯
Β
Answer:Β
NoticeΒ thatΒ
Β
(β1)
π+1Β 3
5
π
=Β β3
(β1)
π
5
π
=Β β3Β (
β1
5
)
π
Β Β
SinceΒ Β
β
ππ
π
β
π=1
=
ππ
1βπ
Β Β (ifΒ Β
|π|Β <Β 1
Β )Β
Β
β
β
π=1
βΒ 3Β (
β1
5
)
π
=
β3β
β1
5
1β
β1
5
=
3
5
6
5
=
1
2
Β Β
andΒ theΒ sumΒ convergesΒ becauseΒ
1
1
|
|
1
5
5
β
=Β οΌ
Β
Β
QuestionΒ #17Β
UseΒ linearΒ approximationΒ (orΒ differentials)Β toΒ estimate:Β
β99.2
Β Β
WhatΒ amΒ IΒ supposedΒ toΒ doΒ withΒ this?Β IΒ amΒ notΒ givenΒ xΒ orΒ dx
Β
Answer:Β
UeΒ TaylorΒ series forΒ Β
βπ₯
Β Β aboutΒ Β
π₯Β =Β 100
Β .Β TheΒ reasonΒ toΒ expandΒ theΒ TaylorΒ seriesΒ aboutΒ 100Β is thatΒ 100Β isΒ
theΒ closestΒ squareΒ toΒ 99.2.Β
Β
π(π₯)Β =Β π(100)Β +Β π
β²
(100)(π₯Β βΒ 100)
Β Β +Β higherΒ orderΒ termsΒ
Hence,Β
Β
β99.2Β βΒ β100Β +
1
2
(99.2β100)
β100
=Β 10Β βΒ 0.04Β =Β 9.96
Β
Β
QuestionΒ #18Β
DetermineΒ whetherΒ theΒ geometric seriesΒ is convergentΒ orΒ divergent.Β
Β
10Β βΒ 4Β +Β 1.6Β βΒ 0.64Β +Β β―.
Β Β
IfΒ it's convergentΒ findΒ itsΒ sum.
Β
Answer:Β
StepΒ 1Β
LetΒ Β
πΒ =Β 10Β Β βΒ Β 4Β Β +Β 1.6Β 0.64Β Β +Β β―
Β Β
GivenΒ thatΒ SΒ is aΒ geometric series,Β Β
TheΒ knownΒ factΒ is thatΒ aΒ geometric series Β
πΒ +Β ππΒ +Β ππ
2
+Β β―
Β Β isΒ convergentΒ Β
ππΒ π|π|Β <Β 1
.Β Β
By comparingΒ theΒ givenΒ series withΒ itsΒ general form,Β
Β
πΒ Β =Β Β 10
Β ,Β
πΒ =
β4
10
=Β β0.4
Β .Β
Β
|π|Β =Β |β0.4|Β <Β 1
Β .Β
ThisΒ implies thatΒ theΒ series Β
10Β Β βΒ 4Β +Β 1.6Β βΒ 0.64Β +Β β―
Β Β isΒ convergent.Β
StepΒ 2Β
TheΒ knownΒ factΒ is thatΒ aΒ geometric series Β
πΒ +Β ππΒ +Β ππ
2
+Β β―
Β -Β isΒ equalΒ toΒ Β
π
1βπ
(πΒ <Β 1)
Β .Β Β
Β
10Β βΒ 4Β +Β 1.6Β βΒ 0.64Β Β±Β β―Β =Β (10Β +Β 1.6Β +Β 0.256Β +Β β―Β )Β βΒ (4Β +Β 0.64Β +Β β―Β )
Β Β
?
10,
0.16
10
4
(
)Β (
)
4,
0,16
1Β 0.16
1Β 0.16
a
r
a
r
=
=
ο©
οΉ
=
β
οͺ
οΊ
=
=
β
β
ο«
ο»
Β
Β
=Β (
10β4
1β0.16
)
Β Β
Β
=Β (
6
0.84
)
Β Β
Β
=Β 7.1428571428571429
Β
Β
QuestionΒ #19Β
BinomialΒ expansionΒ ofΒ Β
(1Β βΒ π₯)
π
Β Β
We'veΒ beenΒ givenΒ withΒ followingΒ binomialΒ expansionΒ
Β
(π₯Β +Β 1)
π
=Β 1Β +Β ππ₯Β +
π(πβ1)
2!
π₯
2
+
π(πβ1)(πβ2)
3!
π₯
3
β¦
Β Β
HowΒ canΒ weΒ getΒ theΒ formulaΒ ofΒ Β
(1Β βΒ π₯)
π
Β Β
Answer:Β
Well,Β as IΒ understandΒ it,Β weΒ couldΒ writeΒ theΒ binomialΒ expansionΒ as:Β
0
1
2
2
3
3
0
(1
)
(Β )1
(
)Β ][(Β )1Β (
)
(Β )1Β (
)Β (Β )1
(
)
(Β )1
(
)Β ...
0
1
2
3
n
n
nΒ k
k
n
n
n
n
k
n
n
n
n
n
x
x
x
x
x
x
k
β
β
β
β
=
β
=
β
β
+
βΒ +
β
+
β
ο₯
Β
whichΒ simplifiesΒ toΒ
Β
1Β βΒ ππ₯Β +
π(πβ1)
2!
β Β π₯
2
β
π(πβ1)(πβ2)
3!
β Β π₯
3
β¦
Β Β
WhichΒ is theΒ answerΒ everyoneΒ elseΒ has given.
Β
QuestionΒ #20Β
HowΒ toΒ findΒ theΒ MaclaurinΒ seriesΒ forΒ Β
π
π₯
Β ?Β
Answer:Β
TheΒ MaclaurinΒ seriesΒ is obtainedΒ by theΒ PowerΒ Series:Β
Β
π(π₯)Β =Β π(0)Β +Β π
β²
(0)
π₯
0!
+Β π
β²
(0)
π₯
2
2!
+Β π
β²
(0)
π₯
3
3!
+Β β―
Β Β
As weΒ haveΒ
π(π₯)Β =Β π
π₯
,
Β then:Β Β
π(π₯)Β =Β π
π₯
βΒ π(0)Β =Β 1
Β Β Β
Β
π
β²
(π₯)Β =Β π
π₯
βΒ π
β²
(0)Β =Β 1
Β Β Β
Β
π
β²β²
(π₯)Β =Β π
π₯
βΒ π
β²β²
(0)Β =Β 1
Β Β Β
Β
π
3
(π₯)Β =Β π
π₯
βΒ π
3
(0)Β =Β 1
Β Β Β
Β
π
π
(π₯)Β =Β π
π₯
βΒ π
π
(0)Β =Β 1
Β Β Β
So,Β theΒ MaclaurinΒ seriesΒ is:Β Β
Β
π
π₯
=Β 1Β +Β 1
π₯
0!
+Β 1
π₯
2
2!
+Β 1
π₯
3
3!
+Β 1
π₯
4
4!
+Β β―Β +Β 1
π₯
π
π!
Β Β
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