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QA #3 Series

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Questionsย andย Answersย Sheetย 3

ย 

ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย Seriesย 

ย 

Questionย #1ย 

Tell whetherย theย series converge.ย Ifย itย converges,ย findย theย sum.ย 
ย 

โˆ‘(๐‘›ย =ย 0)

๐‘œ

๐‘œ(๐‘๐‘–/2)

๐‘›

ย ย 

ย 

Answer:ย 

ย 

โˆ‘(๐‘›ย =ย 0)

๐‘œ

๐‘œ(๐‘๐‘–/2)

๐‘›

ย 

ย 

๐‘Žย =ย 1ย =ย (๐‘๐‘–/2)

2

/(๐‘๐‘–/2)ย =ย ๐‘๐‘–/2ย =ย 1.57ย >

ย ย 

Thus,ย theย seriesย diverges.

ย 

Questionย #2ย 

Tell whetherย theย series converge.ย Ifย itย converges,ย findย theย sum.ย 

ย 

โˆ‘(๐‘›ย =ย 1)

๐‘œ

๐‘œ(3/7)

๐‘›

ย 

ย 

Answer:ย 

ย 

โˆ‘(๐‘›ย =ย 1)

๐‘œ

๐‘œ(3/7)

๐‘›

ย ย 

ย 

๐‘Žย ย =ย ย 3/7ย ย <ย ย 1ย 

,ย soย theย series converges.ย 

Sum:ย ย 

(3/7)/(1โ€“ย 3/7)ย =

ย ย 

ย 

=ย ย 3/7ย ๐‘ฅ๐‘ฅย 7/4ย ย =ย ย 3/4

ย 

ย 

Questionย #3ย 

Tell whetherย theย seriesย convergesย orย diverges,ย andย findย sum,ย ifย itย converges.ย 

ย 

โˆ‘(๐‘›ย =ย 3)

๐‘œ

๐‘œ(1/(๐‘›โ€“ย 2)โ€“ย 1/๐‘›)

ย 

ย 

Answer:ย 

ย 

โˆ‘(๐‘›ย =ย 3)

๐‘œ

๐‘œ(1/(๐‘›โ€“ย 2)โ€“ย 1/๐‘›)

ย ย 

ย 

โˆ‘(๐‘›ย =ย 3)

๐‘œ

๐‘œ1/(๐‘›๐‘ฅ๐‘ฅ(๐‘›ย โˆ’ย 2))

ย ย 

Thisย is notย geometric seriesย 

ย 

๐‘†

๐‘›

=ย 1ย +ย 1/2ย =ย 3/2๐‘Ž๐‘ ๐‘›ย โ†’ย ๐‘œ๐‘œ.

ย 

Thus,ย seriesย converges.

ย ย 

Questionย #4ย 

Pleaseย justifyย whetherย theย seriesย convergesย orย diverges,ย andย findย sum,ย ifย itย converges.ย 

ย 

โˆ‘(๐‘›ย =ย 1)ย ๐‘œ๐‘œย 3/๐‘›

2

.ย 

ย 

Answer:ย 

ย 

โˆ‘(๐‘›ย =ย 1)ย ๐‘œ๐‘œย 3/๐‘›

2

ย ย 

ย 

โˆ‘(๐‘›ย =ย 1)ย ๐‘œ๐‘œย 1/๐‘›

๐‘

ย ย 

โ€“ย ifย ย 

๐‘ย ย >ย ย 1๐‘

ย ย 

โ€“ย series converges,ย ifย ย 

๐‘ย ย <ย ย 1

ย ,ย series diverges.ย 

Therefore,ย series convergesย by p-test.ย 

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Questionย #5ย 

Tell whetherย theย seriesย convergesย orย diverges.ย 

ย 

โˆ‘(๐‘˜ย =ย 1)

๐‘œ

๐‘œ5/(๐‘˜ย +ย 4)

4

ย 

ย ย 

Answer:ย 

ย 

โˆ‘(๐‘˜ย =ย 1)

๐‘œ

๐‘œ5/(๐‘˜ย +ย 4)

4

ย ย 

Putย ย 

๐‘˜ย ย +ย ย 4ย ย โ‡’ย ย ๐‘›

ย ย 

So,ย 

โˆ‘(๐‘›ย =ย 5)

๐‘œ

๐‘œ5/๐‘›

4

ย ย 

Fromย p-seriesย test:ย 

Forย series ย 

โˆ‘(๐‘›ย =ย 1)

๐‘œ

๐‘œ1/๐‘›

๐‘

,ย ifย 

๐‘ย >ย ย 1

,ย series converges.ย 

So,ย forย seriesย inย equationย 1ย ย 

๐‘ƒย ย =ย ย 4

,ย soย this seriesย convergesย 

Theย seriesย is p-series withย 

๐‘ย ย =ย ย 4

ย andย itย converges.

ย 

Questionย #6ย 

Justifyย ifย theย seriesย is convergent:ย 

ย 

โˆ‘(๐‘›ย =ย 1)

๐‘œ

๐‘œ๐‘’

(โˆ’๐‘›)

ย ย 

Answer:ย 

ย 

โˆ‘(๐‘›ย =ย 1)

๐‘œ

๐‘œ๐‘’

(โˆ’๐‘›)

=

ย ย 

ย 

=ย โˆ‘(๐‘›ย =ย 1)

๐‘œ

๐‘œ1/๐‘’

๐‘›

ย 

Here,ย commonย ratio,ย ย 

๐‘Žย ย =ย ย 1/๐‘’

ย ย 

So,ย ย 

๐‘†ย =ย (1/๐‘’)/(1โ€“ย 1/๐‘’)ย =ย (1/๐‘’)/((๐‘’โ€“ย 1)/๐‘’)

ย ย 

ย 

๐‘†ย =ย 1/(๐‘’โ€“ย 1)

.ย 

This seriesย converges.

ย 

Questionย #7ย 

Forย whichย zโ€™sย does theย infiniteย series converge?ย Toย whatย functionย doesย itย converge?ย 

ย 

1ย +ย ๐‘ง/2ย +ย ๐‘ง

2

/4ย +ย ๐‘ง

3

/8ย +ย โ‹ฏ

ย 

ย 

Answer:ย 

ย 

1ย +ย ๐‘ง/2ย +ย ๐‘ง

2

/4ย +ย ๐‘ง

3

/8ย +ย ๐‘๐‘†๐พย โ€ฆ

ย (geometric series)ย 

So,ย ย 

=ย 1/(1ย โˆ’ย (๐‘ง/2))ย =ย 1/((2โ€“ย ๐‘ง)/2)

ย ย 

So,ย ย 

=ย 2/(2โ€“ย ๐‘ง)

ย ย 

This series convergesย ifย ย 

|๐‘ง|ย <ย 2

ย ย andย itย convergesย ย 

2/(2ย โˆ’ย ๐‘ง)

ย .

ย 

Questionย #8ย 

Simplifyย 

(๐‘ ๐‘’๐‘ฮธย โˆ’ย ๐‘ก๐‘Ž๐‘›ฮธ)(1ย +ย ๐‘ ๐‘–๐‘›ฮธ)

ย 

Answer:ย 

ย 

(๐‘ ๐‘’๐‘ฮธย โˆ’ย ๐‘ก๐‘Ž๐‘›ฮธ)(1ย +ย ๐‘ ๐‘–๐‘›ฮธ)ย =

ย ย 

ย 

=ย ๐‘ ๐‘’๐‘(ฮธ)๐‘ ๐‘–๐‘›(ฮธ)ย โˆ’ย ๐‘ก๐‘Ž๐‘›(ฮธ)ย โˆ’ย ๐‘ก๐‘Ž๐‘›(ฮธ)๐‘ ๐‘–๐‘›(ฮธ)ย =

ย ย 

background image

ย 

=

1

๐‘๐‘œ๐‘ ฮธ

+

๐‘ ๐‘–๐‘›ฮธ

๐‘๐‘œ๐‘ ฮธ

โˆ’ย ๐‘ก๐‘Ž๐‘›ฮธย โˆ’

๐‘ ๐‘–๐‘›ฮธ

๐‘๐‘œ๐‘ ฮธ

๐‘ ๐‘–๐‘›ฮธย =

ย ย ย 

ย 

[:

โ€ฒ

๐‘ ๐‘’๐‘ฮธย =

1

๐‘๐‘œ๐‘ ฮธ

๐‘Ž๐‘›๐‘‘๐‘ก๐‘Ž๐‘›ฮธย =

๐‘ ๐‘–๐‘›ฮธ

๐‘๐‘œ๐‘ ฮธ

]

ย ย 

ย 

=

1

๐‘๐‘œ๐‘ ฮธ

+ย ๐‘ก๐‘Ž๐‘›ฮธย โˆ’ย ๐‘ก๐‘Ž๐‘›ฮธย โˆ’

๐‘ ๐‘–๐‘›

2

ฮธ

๐‘๐‘œ๐‘ ฮธ

=

ย ย 

ย 

=

1

๐‘๐‘œ๐‘ ฮธ

โˆ’

๐‘ ๐‘–๐‘›

2

ฮธ

๐‘๐‘œ๐‘ ฮธ

=

ย ย ย 

ย 

=

1โˆ’๐‘ ๐‘–๐‘›

2

ฮธ

๐‘๐‘œ๐‘ ฮธ

=

ย ย ย 

ย 

=

๐‘๐‘œ๐‘ 

2

ฮธ

๐‘๐‘œ๐‘ ฮธ

=ย ๐‘๐‘œ๐‘ 

ย ย 

ย 

[:

โ€ฒ

๐‘ ๐‘–๐‘›

2

ฮธย +ย ๐‘๐‘œ๐‘ 

2

ฮธย =ย 1]

ย ย 

Answer:ย ย 

(๐‘ ๐‘’๐‘ฮธย โˆ’ย ๐‘ก๐‘Ž๐‘›ฮธ)(1ย +ย ๐‘ ๐‘–๐‘›ฮธ)ย =ย ๐‘๐‘œ๐‘ 

ย 

ย ย 

Questionย #9

ย 

Computeย eachย ofย theseย doubleย sums:ย 

ย 

โˆ‘

โˆ‘

(๐‘–ย +ย ๐‘—)

3

๐‘–+1

2

๐‘–=1

ย 

ย 

Answer:

ย 

Computeย thisย doubleย sumย by firstย computingย theย innerย sumย followedย by theย outerย sum.ย 

ย 

โˆ‘

โˆ‘

(๐‘–ย +ย ๐‘—)

3

๐‘—=1

2

๐‘–=1

ย ย 

ย 

=ย โˆ‘

(๐‘–ย +ย 1ย +ย ๐‘–ย +ย 2ย +ย ๐‘–ย +ย 3)

2

๐‘–=1

ย ย 

ย 

=ย 3ย โˆ‘

(๐‘–ย +ย 2)

2

๐‘–=1

ย ย 

ย 

=ย 3(1ย +ย 2ย +ย 2ย +ย 2)

ย ย 

ย 

=ย 21

ย 

ย 

Questionย #10ย 

Computeย eachย ofย theseย doubleย sums:ย 

ย 

=ย โˆ‘

(๐‘–ย +ย 1ย +ย ๐‘–ย +ย 2ย +ย ๐‘—ย +ย 3)

2

๐‘–=1

ย 

ย 

Answer:ย 

Computeย thisย doubleย sumย by firstย computingย theย innerย sumย followedย by theย outerย sum.ย 

ย 

=ย โˆ‘

โˆ‘

(2๐‘–ย +ย 3๐‘—)

3

๐‘—=0

2

๐‘–=0

ย ย 

ย 

=ย โˆ‘

(2๐‘–ย +ย 0ย +ย 2๐‘–ย +ย 3ย +ย 2๐‘–ย +ย 6ย +ย 2๐‘–ย +ย 9)

2

๐‘–=0

ย ย 

ย 

=ย โˆ‘

(8๐‘–ย +ย 18)

2

๐‘–=0

ย 

ย 

=ย 8(0)ย +ย 18ย +ย 8(1)ย +ย 18ย +ย 8(2)ย +ย 18

ย ย 

ย 

=ย 78

ย 

ย 

ย 

Questionย #11ย 

Writeย anย equivalentย series withย theย indexย ofย summationย beginningย atย =ย 1.ย 

ย 

โˆ‘

(โˆ’1)

๐‘›

โˆž

๐‘›=0

+ย 1(๐‘›ย +ย 1)๐‘ฅ

๐‘›

ย 

ย 

background image

Answer:ย ย 

This equationย is givenย inย theย series.ย Addย 1ย toย bothย sides andย introduceย k.ย Subtractย 1ย fromย bothย sides.ย 

ย 

๐‘›ย =ย 0

ย ,ย soย ย 

๐‘›ย +ย 1ย =ย 1ย =ย ๐‘˜

ย .ย Therefore,ย ย 

๐‘›ย =ย ๐‘˜ย โˆ’ย 1

ย ย 

Inย theย originalย series,ย plugย inย ย 

๐‘˜ย โˆ’ย 1

ย ย forย nย 

ย 

โˆ‘

(โˆ’1)

[(๐‘˜โˆ’1)+1]

[(๐‘˜ย โˆ’ย 1)ย +ย 1]

โˆž

๐‘˜=1

โ‹…ย ๐‘ฅ

๐‘˜โˆ’1

ย 

Simplify.ย Sinceย kย isย aย dummy variable,ย youย canย replaceย itย withย any otherย variable.ย Letย n=kย 

ย 

โˆ‘

(โˆ’1)

๐‘˜

๐‘˜

โˆž

๐‘˜=1

โ‹…ย ๐‘ฅ

๐‘˜โˆ’1

=ย โˆ‘

(โˆ’1)

๐‘›

๐‘›

โˆž

๐‘›=1

โ‹…ย ๐‘ฅ

๐‘›โˆ’1

ย 

Result:ย ย 

โˆ‘

(โˆ’1)

๐‘›

๐‘›

โˆž

๐‘›=1

โ‹…ย ๐‘ฅ

๐‘›โˆ’1

ย 

Questionย #12ย 

Writeย anย equivalentย series withย theย indexย ofย summationย beginningย atย =ย 1.ย 

ย 

โˆ‘

(โˆ’1)

๐‘›

โˆž

๐‘›=0

+ย 1(๐‘›ย +ย 1)๐‘ฅ

๐‘›

ย 

ย 

Answer:ย 

Theย Taylorย series ofย ย 

lnย ๐‘ฅ

ย aroundย c=1.ย Weย knowย fromย theย definitionย ofย Taylorย seriesย ofย theย functionย fย 

ย 

๐‘“(๐‘ฅ)ย =ย ๐‘“(๐‘ฅ)|

๐‘ฅ=๐‘

+

๐‘“

โ€ฒ

(๐‘ฅ)|

๐‘ฅ=๐‘

1!

(๐‘ฅย โˆ’ย ๐‘)ย +

๐‘“

โ€ฒโ€ฒ

(๐‘ฅ)|

๐‘ฅ=๐‘

2!

(๐‘ฅย โˆ’ย ๐‘)

2

+

๐‘“

โ€ฒโ€ฒโ€ฒ

(๐‘ฅ)|

๐‘ฅ=๐‘

3!

(๐‘ฅย โˆ’ย ๐‘)

3

+ย โ‹ฏ

ย 

Nowย theย values ofย differentialย functions atย givenย pointย 

ย 

๐‘“(1)ย =ย ln(1)ย =ย 0,ย ๐‘“

โ€ฒ

(1)ย =

1

๐‘ฅ

|

๐‘ฅ=1

=ย 1

ย ย 

ย 

๐‘“

โ€ฒโ€ฒ

(1)ย =ย โˆ’

1

๐‘ฅ

2

๐‘ฅ=1

=ย โˆ’1,ย ๐‘“

โ€ฒโ€ฒโ€ฒ

(1)ย =

2

๐‘ฅ

3

|

๐‘ฅ=1

=ย 2๐‘“

โ€ฒโ€ฒโ€ฒโ€ฒ

(1)ย =ย โˆ’

6

๐‘ฅ

4

|

๐‘ฅ=1

=ย โˆ’6

ย ย 

ย 

๐‘“

๐‘›

(1)ย =ย (โˆ’1)

๐‘›+1ย (๐‘›โˆ’1)!

๐‘ฅ

๐‘›

|

๐‘ฅ=1

=ย (โˆ’1)

๐‘›

((๐‘›ย โˆ’ย 1)!)

ย ย 

Thus theย Taylorย seriesย expressionย becomeย 

ย 

๐‘“(๐‘ฅ)ย =ย 0ย +ย (๐‘ฅย โˆ’ย 1)ย โˆ’

1

2!

(๐‘ฅย โˆ’ย 1)

2

+ย 2!

1

3!

(๐‘ฅย โˆ’ย 1)

3

โˆ’ย 3!

1

4!

+ย (โˆ’1)

๐‘›+1

(๐‘›ย โˆ’ย 1)!

1

๐‘›!

(๐‘ฅย โˆ’ย 1)

๐‘›

ย ย 

Orย weย canย writeย asย 

ย f

(๐‘ฅ)ย =ย (1ย โˆ’ย ๐‘ฅ)ย โˆ’

1

2

(๐‘ฅย โˆ’ย 1)

2

+

1

3

(๐‘ฅย โˆ’ย 1)

3

โˆ’

1

4

(๐‘ฅย โˆ’ย 1)

4

+ย (โˆ’1)

๐‘›+1

(๐‘›ย โˆ’ย 1)!

1

๐‘›!

(๐‘ฅย โˆ’ย 1)

๐‘›

โˆ’ย โ‹ฏ

ย 

ย 

Questionย #13ย 

Useย theย summationย formulas toย rewriteย theย expressionย withous summatiomย notation.ย 

ย 

โˆ‘

4๐‘–+5

๐‘›

2

๐‘›

๐‘–=1

ย ย 

Useย theย result toย findย theย sums forย n=10,ย 100.ย 

Answer:ย 

ย 

โˆ‘

๐‘–

๐‘›

๐‘–=1

=

๐‘›(๐‘›+1)

2

ย ย 

ย 

โˆ‘

๐‘˜

๐‘›

๐‘–=1

=ย ๐‘›.ย ๐‘˜

,ย whereย k=constantย 

forย ย 

๐‘›ย =ย 10

ย ย 

ย 

โˆ‘

4๐‘–+5

(10)

2

10

๐‘–=1

=

1

100

[4ย โˆ‘

๐‘–

10

๐‘–=1

+ย โˆ‘

5

10

๐‘–=1

]

ย 

1

10(10ย 1)

[]4

10ย 5]

2.7

100

2

+

=

+ย ๏‚ดย =

ย 

background image

forย n=100ย 

ย 

โˆ‘

4๐‘–+5

(100)

2

100

๐‘–=1

=

1

(100)

2

[4ย โ‹…ย โˆ‘

๐‘–

100

๐‘–=1

+ย โˆ‘ย ๐‘–ย =ย 1

100

5]

ย 

ย 

=

1

(100)

2

[4

100(100+1)

2

+ย (5ย ร—ย 100)]

ย ย 

ย 

=ย 2.07

ย 

ย 

Questionย #14ย 

Determineย theย sumย ofย theย followingย series.ย 

โˆ‘

(

1

๐‘›

+9

๐‘›

11

๐‘›

)

โˆž

๐‘›=1

ย ย 

Answer:ย 

Weย haveย toย findย sumย ofย series.ย 

Seriesย isย givenย as:ย 

ย 

โ‡’ย โˆ‘

(

1

๐‘›

+9

๐‘›

11

๐‘›

)

โˆž

๐‘›=1

ย ย 

Weย willย useย geometric seriesย testย toย findย sumย ofย series .ย 

Geometric series testย is givenย below:ย 

ย 

โ‡’ย โˆ‘

๐‘Ž๐‘Ÿ

๐‘›โˆ’1

โˆž

๐‘›=1

ย ย 

ย 

โ‡’ย ๐‘–๐‘“|๐‘Ÿ|ย <ย 1

ย (series converges)ย 

ย 

โ‡’ย โˆ‘

=

๐‘Ž

1โˆ’๐‘Ÿ

ย (infiniteย gp)ย 

Withย theย helpย ofย geometric series weย willย findย sumย .ย 

Workย is shownย below:ย 

ย 

โ‡’ย โˆ‘

(

1

๐‘›

+9

๐‘›

11

๐‘›

)

โˆž

๐‘›=1

ย 

ย 

โ‡’ย โˆ‘

(

1

11

)

๐‘›

โˆž

๐‘›=1

+ย (

9

11

)

๐‘›

ย ย 

ย 

โ‡’ย โˆ‘

(

1

11

)

๐‘›

โˆž

๐‘›=1

=

1

11

1โˆ’

1

11

ย ย 

ย 

โ‡’ย โˆ‘

(

1

11

)

๐‘›

โˆž

๐‘›=1

=

1

10

ย ย 

ย 

โ‡’ย โˆ‘

(

9

11

)

๐‘›

โˆž

๐‘›=1

=

9

11

1โˆ’

9

11

ย ย 

ย 

โ‡’ย โˆ‘

(

9

11

)

๐‘›

โˆž

๐‘›=1

=

9

2

ย ย 

ย 

โ‡’ย โˆ‘

(

1

11

)

๐‘›

โˆž

๐‘›=1

+ย (

9

11

)

๐‘›

=

1

10

+

9

2

ย ย 

ย 

โ‡’ย โˆ‘

(

1

11

)

๐‘›

โˆž

๐‘›=1

+ย (

9

11

)

๐‘›

=

23

5

ย 

Questionย #15ย 

Whichย equationย couldย beย usedย toย calculateย theย sumย ofย theย geometric series?ย 

ย 

1

3

+

2

9

+

4

27

+

8

21

+

16

243

ย 

ย 

Answer:ย 

Theย formulaย forย theย nthย partialย sumย ofย aย geometric sequenceย is,ย 

background image

ย 

๐‘†

๐‘›

=

๐‘Ž

1

(1โˆ’๐‘Ÿ

๐‘›

)

1โˆ’๐‘Ÿ

,ย ๐‘Ÿย โ‰ ย 1

ย ย 

where,ย ย 

๐‘†

๐‘›

ย ย is theย sumย ofย GPย withย nย termsย 

ย 

๐‘Ž

1

ย ย is theย firstย termย 

rย isย theย commonย ratioย 

nย is theย numberย ofย termsย 

Theย sumย ofย theย geometric series givenย is,ย 

ย 

1

3

+

2

9

+

4

27

+

8

21

+

16

243

ย ย 

Thereย areย 5ย terms inย theย series andย theย commonย ratioย ofย theย seriesย is foundย byย dividingย anyย termย by theย 

previous term.ย Suppose,ย divideย theย secondย termย by theย firstย termย andย theย commonย ratioย isย obtainedย as,ย 

ย 

๐‘Ÿย =

2

9

รท

1

3

ย ย 

ย 

=

2

3

ย ย 

Therefore,ย rย is notย equalย toย 1ย andย theย sumย ofย theย givenย seriesย is,ย 

ย 

๐‘†

5

=

1
3

(1โˆ’(

2
3

)

5

)

(1โˆ’

2
3

)

ย 

Questionย #16ย 

Findย whetherย theย seriesย divergesย andย itsย sum:ย 

ย 

โˆ‘

(โˆ’1)

๐‘›+1ย 3

5

๐‘›

โˆž

๐‘›=1

0

1

2

2

3

3

0

(1

)

(ย )1

(

)ย ][(ย )1ย (

)

(ย )1ย (

)ย (ย )1

(

)

(ย )1

(

)ย ...

0

1

2

3

n

n

nย k

k

n

n

n

n

k

n

n

n

n

n

x

x

x

x

x

x

k

โˆ’

โˆ’

โˆ’

โˆ’

=

โˆ’

=

โˆ’

โˆ’

+

โˆ’ย +

โˆ’

+

โˆ’

๏ƒฅ

ย 

Answer:ย 

Noticeย thatย 

ย 

(โˆ’1)

๐‘›+1ย 3

5

๐‘›

=ย โˆ’3

(โˆ’1)

๐‘›

5

๐‘›

=ย โˆ’3ย (

โˆ’1

5

)

๐‘›

ย ย 

Sinceย ย 

โˆ‘

๐‘Ž๐‘Ÿ

๐‘˜

โˆž

๐‘˜=1

=

๐‘Ž๐‘Ÿ

1โˆ’๐‘Ÿ

ย ย (ifย ย 

|๐‘Ÿ|ย <ย 1

ย )ย 

ย 

โˆ‘

โˆž

๐‘›=1

โˆ’ย 3ย (

โˆ’1

5

)

๐‘›

=

โˆ’3โ‹…

โˆ’1

5

1โˆ’

โˆ’1

5

=

3
5

6
5

=

1

2

ย ย 

andย theย sumย convergesย becauseย 

1

1

|

|

1

5

5

โˆ’

=ย ๏€ผ

ย 

ย 

Questionย #17ย 

Useย linearย approximationย (orย differentials)ย toย estimate:ย 

โˆš99.2

ย ย 

Whatย amย Iย supposedย toย doย withย this?ย Iย amย notย givenย xย orย dx

ย 

Answer:ย 

Ueย Taylorย series forย ย 

โˆš๐‘ฅ

ย ย aboutย ย 

๐‘ฅย =ย 100

ย .ย Theย reasonย toย expandย theย Taylorย seriesย aboutย 100ย is thatย 100ย isย 

theย closestย squareย toย 99.2.ย 

ย 

๐‘“(๐‘ฅ)ย =ย ๐‘“(100)ย +ย ๐‘“

โ€ฒ

(100)(๐‘ฅย โˆ’ย 100)

ย ย +ย higherย orderย termsย 

background image

Hence,ย 

ย 

โˆš99.2ย โ‰ˆย โˆš100ย +

1

2

(99.2โˆ’100)

โˆš100

=ย 10ย โˆ’ย 0.04ย =ย 9.96

ย 

ย 

Questionย #18ย 

Determineย whetherย theย geometric seriesย is convergentย orย divergent.ย 

ย 

10ย โˆ’ย 4ย +ย 1.6ย โˆ’ย 0.64ย +ย โ‹ฏ.

ย ย 

Ifย it's convergentย findย itsย sum.

ย 

Answer:ย 

Stepย 1ย 

Letย ย 

๐‘†ย =ย 10ย ย โˆ’ย ย 4ย ย +ย 1.6ย 0.64ย ย +ย โ‹ฏ

ย ย 

Givenย thatย Sย is aย geometric series,ย ย 

Theย knownย factย is thatย aย geometric series ย 

๐‘Žย +ย ๐‘Ž๐‘Ÿย +ย ๐‘Ž๐‘Ÿ

2

+ย โ‹ฏ

ย ย isย convergentย ย 

๐‘–๐‘“ย ๐‘“|๐‘Ÿ|ย <ย 1

.ย ย 

By comparingย theย givenย series withย itsย general form,ย 

ย 

๐‘Žย ย =ย ย 10

ย ,ย 

๐‘Ÿย =

โˆ’4

10

=ย โˆ’0.4

ย .ย 

ย 

|๐‘Ÿ|ย =ย |โˆ’0.4|ย <ย 1

ย .ย 

Thisย implies thatย theย series ย 

10ย ย โˆ’ย 4ย +ย 1.6ย โˆ’ย 0.64ย +ย โ‹ฏ

ย ย isย convergent.ย 

Stepย 2ย 

Theย knownย factย is thatย aย geometric series ย 

๐‘Žย +ย ๐‘Ž๐‘Ÿย +ย ๐‘Ž๐‘Ÿ

2

+ย โ‹ฏ

ย -ย isย equalย toย ย 

๐‘Ž

1โˆ’๐‘Ÿ

(๐‘Ÿย <ย 1)

ย .ย ย 

ย 

10ย โˆ’ย 4ย +ย 1.6ย โˆ’ย 0.64ย ยฑย โ‹ฏย =ย (10ย +ย 1.6ย +ย 0.256ย +ย โ‹ฏย )ย โˆ’ย (4ย +ย 0.64ย +ย โ‹ฏย )

ย ย 

?

10,

0.16

10

4

(

)ย (

)

4,

0,16

1ย 0.16

1ย 0.16

a

r

a

r

=

=

๏ƒฉ

๏ƒน

=

โˆ’

๏ƒช

๏ƒบ

=

=

โˆ’

โˆ’

๏ƒซ

๏ƒป

ย 

ย 

=ย (

10โˆ’4

1โˆ’0.16

)

ย ย 

ย 

=ย (

6

0.84

)

ย ย 

ย 

=ย 7.1428571428571429

ย 

ย 

Questionย #19ย 

Binomialย expansionย ofย ย 

(1ย โˆ’ย ๐‘ฅ)

๐‘›

ย ย 

We'veย beenย givenย withย followingย binomialย expansionย 

ย 

(๐‘ฅย +ย 1)

๐‘›

=ย 1ย +ย ๐‘›๐‘ฅย +

๐‘›(๐‘›โˆ’1)

2!

๐‘ฅ

2

+

๐‘›(๐‘›โˆ’1)(๐‘›โˆ’2)

3!

๐‘ฅ

3

โ€ฆ

ย ย 

Howย canย weย getย theย formulaย ofย ย 

(1ย โˆ’ย ๐‘ฅ)

๐‘›

ย ย 

Answer:ย 

Well,ย as Iย understandย it,ย weย couldย writeย theย binomialย expansionย as:ย 

0

1

2

2

3

3

0

(1

)

(ย )1

(

)ย ][(ย )1ย (

)

(ย )1ย (

)ย (ย )1

(

)

(ย )1

(

)ย ...

0

1

2

3

n

n

nย k

k

n

n

n

n

k

n

n

n

n

n

x

x

x

x

x

x

k

โˆ’

โˆ’

โˆ’

โˆ’

=

โˆ’

=

โˆ’

โˆ’

+

โˆ’ย +

โˆ’

+

โˆ’

๏ƒฅ

ย 

whichย simplifiesย toย 

ย 

1ย โˆ’ย ๐‘›๐‘ฅย +

๐‘›(๐‘›โˆ’1)

2!

โ‹…ย ๐‘ฅ

2

โˆ’

๐‘›(๐‘›โˆ’1)(๐‘›โˆ’2)

3!

โ‹…ย ๐‘ฅ

3

โ€ฆ

ย ย 

Whichย is theย answerย everyoneย elseย has given.

ย 

background image

Questionย #20ย 

Howย toย findย theย Maclaurinย seriesย forย ย 

๐‘’

๐‘ฅ

ย ?ย 

Answer:ย 

Theย Maclaurinย seriesย is obtainedย by theย Powerย Series:ย 

ย 

๐‘“(๐‘ฅ)ย =ย ๐‘“(0)ย +ย ๐‘“

โ€ฒ

(0)

๐‘ฅ

0!

+ย ๐‘“

โ€ฒ

(0)

๐‘ฅ

2

2!

+ย ๐‘“

โ€ฒ

(0)

๐‘ฅ

3

3!

+ย โ‹ฏ

ย ย 

As weย haveย 

๐‘“(๐‘ฅ)ย =ย ๐‘’

๐‘ฅ

,

ย then:ย ย 

๐‘“(๐‘ฅ)ย =ย ๐‘’

๐‘ฅ

โ‡’ย ๐‘“(0)ย =ย 1

ย ย ย 

ย 

๐‘“

โ€ฒ

(๐‘ฅ)ย =ย ๐‘’

๐‘ฅ

โ‡’ย ๐‘“

โ€ฒ

(0)ย =ย 1

ย ย ย 

ย 

๐‘“

โ€ฒโ€ฒ

(๐‘ฅ)ย =ย ๐‘’

๐‘ฅ

โ‡’ย ๐‘“

โ€ฒโ€ฒ

(0)ย =ย 1

ย ย ย 

ย 

๐‘“

3

(๐‘ฅ)ย =ย ๐‘’

๐‘ฅ

โ‡’ย ๐‘“

3

(0)ย =ย 1

ย ย ย 

ย 

๐‘“

๐‘›

(๐‘ฅ)ย =ย ๐‘’

๐‘ฅ

โ‡’ย ๐‘“

๐‘›

(0)ย =ย 1

ย ย ย 

So,ย theย Maclaurinย seriesย is:ย ย 

ย 

๐‘’

๐‘ฅ

=ย 1ย +ย 1

๐‘ฅ

0!

+ย 1

๐‘ฅ

2

2!

+ย 1

๐‘ฅ

3

3!

+ย 1

๐‘ฅ

4

4!

+ย โ‹ฏย +ย 1

๐‘ฅ

๐‘›

๐‘›!

ย ย 

of 9