Show that $sum_n=1^infty 2^2nsin^4frac a2^n=a^2-sin^2a$

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Show that $$sum_n=1^infty 2^2nsin^4frac a2^n=a^2-sin^2a$$

I am studying for an exam and I bumped into this question. It's really bothering me because I really don't have any clue what to do. Does it have anything to do with the Cauchy condensation? Can somebody help me I would really appreciate it

calculus sequences-and-series

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edited Aug 6 at 11:45

asked Jul 27 at 16:06

J.Doe

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Show that $$sum_n=1^infty 2^2nsin^4frac a2^n=a^2-sin^2a$$

I am studying for an exam and I bumped into this question. It's really bothering me because I really don't have any clue what to do. Does it have anything to do with the Cauchy condensation? Can somebody help me I would really appreciate it

calculus sequences-and-series

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edited Aug 6 at 11:45

asked Jul 27 at 16:06

J.Doe

84

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up vote
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up vote
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down vote

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2

Show that $$sum_n=1^infty 2^2nsin^4frac a2^n=a^2-sin^2a$$

I am studying for an exam and I bumped into this question. It's really bothering me because I really don't have any clue what to do. Does it have anything to do with the Cauchy condensation? Can somebody help me I would really appreciate it

calculus sequences-and-series

share|cite|improve this question

edited Aug 6 at 11:45

asked Jul 27 at 16:06

J.Doe

84

Show that $$sum_n=1^infty 2^2nsin^4frac a2^n=a^2-sin^2a$$

I am studying for an exam and I bumped into this question. It's really bothering me because I really don't have any clue what to do. Does it have anything to do with the Cauchy condensation? Can somebody help me I would really appreciate it

calculus sequences-and-series

share|cite|improve this question

edited Aug 6 at 11:45

asked Jul 27 at 16:06

J.Doe

84

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share|cite|improve this question

edited Aug 6 at 11:45

edited Aug 6 at 11:45

edited Aug 6 at 11:45

asked Jul 27 at 16:06

J.Doe

84

asked Jul 27 at 16:06

J.Doe

84

asked Jul 27 at 16:06

J.Doe

84

84

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add a comment | 

3 Answers
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accepted

Hint: Use the fact that $sin^4x=(sin^2 x)(sin^2 x)$ and a couple of trigonometric identities to reduce it to a telescopic series. Finally, use the identity $lim_xto0fracsin xx=1$ to obtain the desired value.

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edited Jul 27 at 16:24

answered Jul 27 at 16:16

Clayton

17.9k22882

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Using the hint that @Clayton gave this is what you get if you take the partial sum

$$2^2sin^2frac a2-sin^2a+2^4sin^2frac a2^2-2^2sin^2frac a2+ cdots+ 2^2nsin^2frac a2^n-2^2n-2sin^2frac a2^n-1=
2^2nsin^2frac a2^n-sin^2a$$

when you take the limit $$lim_nto inftyleft(2^2nsin^2frac a2^n-sin^2aright)=a^2-sin^2a$$

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answered Jul 27 at 16:58

J.Dane

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Solution

Notice that
beginalign*
2^2nsin^4frac a2^n&=2^2ncdotsin^2frac a2^ncdotsin^2frac a2^n\&=2^2ncdotsin^2frac a2^ncdotleft(1-cos^2frac a2^nright)\&=2^2ncdotsin^2frac a2^n-2^2ncdotsin^2frac a2^ncos^2frac a2^n\&=2^2ncdotsin^2frac a2^n-2^2n-2sin^2frac a2^n-1.
endalign*

Hence, the partial sum
beginalign*
&sum_n=1^mleft(2^2nsin^4frac a2^nright)\=&2^2sin^2frac a2-sin^2a+2^4sin^2frac a2^2-2^2sin^2frac a2+ cdots+ 2^2msin^2frac a2^m-2^2m-2sin^2frac a2^m-1\=&2^2msin^2frac a2^m-sin^2a.
endalign*

Let $m to infty$. We obtain $$sum_n=1^inftyleft(2^2nsin^4frac a2^nright)=lim_m to inftyleft(fracsindfrac a2^mdfraca2^mcdot aright)^2-sin^2 a=(1cdot a)^2-sin^2 a=a^2-sin^2 a.$$

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answered Jul 27 at 17:33

mengdie1982

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

Hint: Use the fact that $sin^4x=(sin^2 x)(sin^2 x)$ and a couple of trigonometric identities to reduce it to a telescopic series. Finally, use the identity $lim_xto0fracsin xx=1$ to obtain the desired value.

share|cite|improve this answer

edited Jul 27 at 16:24

answered Jul 27 at 16:16

Clayton

17.9k22882

add a comment | 

up vote
3
down vote



accepted

Hint: Use the fact that $sin^4x=(sin^2 x)(sin^2 x)$ and a couple of trigonometric identities to reduce it to a telescopic series. Finally, use the identity $lim_xto0fracsin xx=1$ to obtain the desired value.

share|cite|improve this answer

edited Jul 27 at 16:24

answered Jul 27 at 16:16

Clayton

17.9k22882

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

Hint: Use the fact that $sin^4x=(sin^2 x)(sin^2 x)$ and a couple of trigonometric identities to reduce it to a telescopic series. Finally, use the identity $lim_xto0fracsin xx=1$ to obtain the desired value.

share|cite|improve this answer

edited Jul 27 at 16:24

answered Jul 27 at 16:16

Clayton

17.9k22882

Hint: Use the fact that $sin^4x=(sin^2 x)(sin^2 x)$ and a couple of trigonometric identities to reduce it to a telescopic series. Finally, use the identity $lim_xto0fracsin xx=1$ to obtain the desired value.

share|cite|improve this answer

edited Jul 27 at 16:24

answered Jul 27 at 16:16

Clayton

17.9k22882

share|cite|improve this answer

share|cite|improve this answer

edited Jul 27 at 16:24

edited Jul 27 at 16:24

edited Jul 27 at 16:24

answered Jul 27 at 16:16

Clayton

17.9k22882

answered Jul 27 at 16:16

Clayton

17.9k22882

answered Jul 27 at 16:16

Clayton

17.9k22882

17.9k22882

add a comment | 

add a comment | 

up vote
2
down vote

Using the hint that @Clayton gave this is what you get if you take the partial sum

$$2^2sin^2frac a2-sin^2a+2^4sin^2frac a2^2-2^2sin^2frac a2+ cdots+ 2^2nsin^2frac a2^n-2^2n-2sin^2frac a2^n-1=
2^2nsin^2frac a2^n-sin^2a$$

when you take the limit $$lim_nto inftyleft(2^2nsin^2frac a2^n-sin^2aright)=a^2-sin^2a$$

share|cite|improve this answer

answered Jul 27 at 16:58

J.Dane

159112

add a comment | 

up vote
2
down vote

Using the hint that @Clayton gave this is what you get if you take the partial sum

$$2^2sin^2frac a2-sin^2a+2^4sin^2frac a2^2-2^2sin^2frac a2+ cdots+ 2^2nsin^2frac a2^n-2^2n-2sin^2frac a2^n-1=
2^2nsin^2frac a2^n-sin^2a$$

when you take the limit $$lim_nto inftyleft(2^2nsin^2frac a2^n-sin^2aright)=a^2-sin^2a$$

share|cite|improve this answer

answered Jul 27 at 16:58

J.Dane

159112

add a comment | 

up vote
2
down vote

up vote
2
down vote

Using the hint that @Clayton gave this is what you get if you take the partial sum

$$2^2sin^2frac a2-sin^2a+2^4sin^2frac a2^2-2^2sin^2frac a2+ cdots+ 2^2nsin^2frac a2^n-2^2n-2sin^2frac a2^n-1=
2^2nsin^2frac a2^n-sin^2a$$

when you take the limit $$lim_nto inftyleft(2^2nsin^2frac a2^n-sin^2aright)=a^2-sin^2a$$

share|cite|improve this answer

answered Jul 27 at 16:58

J.Dane

159112

Using the hint that @Clayton gave this is what you get if you take the partial sum

$$2^2sin^2frac a2-sin^2a+2^4sin^2frac a2^2-2^2sin^2frac a2+ cdots+ 2^2nsin^2frac a2^n-2^2n-2sin^2frac a2^n-1=
2^2nsin^2frac a2^n-sin^2a$$

when you take the limit $$lim_nto inftyleft(2^2nsin^2frac a2^n-sin^2aright)=a^2-sin^2a$$

share|cite|improve this answer

answered Jul 27 at 16:58

J.Dane

159112

share|cite|improve this answer

share|cite|improve this answer

answered Jul 27 at 16:58

J.Dane

159112

answered Jul 27 at 16:58

J.Dane

159112

answered Jul 27 at 16:58

J.Dane

159112

159112

add a comment | 

add a comment | 

up vote
1
down vote

Solution

Notice that
beginalign*
2^2nsin^4frac a2^n&=2^2ncdotsin^2frac a2^ncdotsin^2frac a2^n\&=2^2ncdotsin^2frac a2^ncdotleft(1-cos^2frac a2^nright)\&=2^2ncdotsin^2frac a2^n-2^2ncdotsin^2frac a2^ncos^2frac a2^n\&=2^2ncdotsin^2frac a2^n-2^2n-2sin^2frac a2^n-1.
endalign*

Hence, the partial sum
beginalign*
&sum_n=1^mleft(2^2nsin^4frac a2^nright)\=&2^2sin^2frac a2-sin^2a+2^4sin^2frac a2^2-2^2sin^2frac a2+ cdots+ 2^2msin^2frac a2^m-2^2m-2sin^2frac a2^m-1\=&2^2msin^2frac a2^m-sin^2a.
endalign*

Let $m to infty$. We obtain $$sum_n=1^inftyleft(2^2nsin^4frac a2^nright)=lim_m to inftyleft(fracsindfrac a2^mdfraca2^mcdot aright)^2-sin^2 a=(1cdot a)^2-sin^2 a=a^2-sin^2 a.$$

share|cite|improve this answer

answered Jul 27 at 17:33

mengdie1982

2,827216

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up vote
1
down vote

Solution

Notice that
beginalign*
2^2nsin^4frac a2^n&=2^2ncdotsin^2frac a2^ncdotsin^2frac a2^n\&=2^2ncdotsin^2frac a2^ncdotleft(1-cos^2frac a2^nright)\&=2^2ncdotsin^2frac a2^n-2^2ncdotsin^2frac a2^ncos^2frac a2^n\&=2^2ncdotsin^2frac a2^n-2^2n-2sin^2frac a2^n-1.
endalign*

Hence, the partial sum
beginalign*
&sum_n=1^mleft(2^2nsin^4frac a2^nright)\=&2^2sin^2frac a2-sin^2a+2^4sin^2frac a2^2-2^2sin^2frac a2+ cdots+ 2^2msin^2frac a2^m-2^2m-2sin^2frac a2^m-1\=&2^2msin^2frac a2^m-sin^2a.
endalign*

Let $m to infty$. We obtain $$sum_n=1^inftyleft(2^2nsin^4frac a2^nright)=lim_m to inftyleft(fracsindfrac a2^mdfraca2^mcdot aright)^2-sin^2 a=(1cdot a)^2-sin^2 a=a^2-sin^2 a.$$

share|cite|improve this answer

answered Jul 27 at 17:33

mengdie1982

2,827216

add a comment | 

up vote
1
down vote

up vote
1
down vote

Solution

Notice that
beginalign*
2^2nsin^4frac a2^n&=2^2ncdotsin^2frac a2^ncdotsin^2frac a2^n\&=2^2ncdotsin^2frac a2^ncdotleft(1-cos^2frac a2^nright)\&=2^2ncdotsin^2frac a2^n-2^2ncdotsin^2frac a2^ncos^2frac a2^n\&=2^2ncdotsin^2frac a2^n-2^2n-2sin^2frac a2^n-1.
endalign*

Hence, the partial sum
beginalign*
&sum_n=1^mleft(2^2nsin^4frac a2^nright)\=&2^2sin^2frac a2-sin^2a+2^4sin^2frac a2^2-2^2sin^2frac a2+ cdots+ 2^2msin^2frac a2^m-2^2m-2sin^2frac a2^m-1\=&2^2msin^2frac a2^m-sin^2a.
endalign*

Let $m to infty$. We obtain $$sum_n=1^inftyleft(2^2nsin^4frac a2^nright)=lim_m to inftyleft(fracsindfrac a2^mdfraca2^mcdot aright)^2-sin^2 a=(1cdot a)^2-sin^2 a=a^2-sin^2 a.$$

share|cite|improve this answer

answered Jul 27 at 17:33

mengdie1982

2,827216

Solution

Notice that
beginalign*
2^2nsin^4frac a2^n&=2^2ncdotsin^2frac a2^ncdotsin^2frac a2^n\&=2^2ncdotsin^2frac a2^ncdotleft(1-cos^2frac a2^nright)\&=2^2ncdotsin^2frac a2^n-2^2ncdotsin^2frac a2^ncos^2frac a2^n\&=2^2ncdotsin^2frac a2^n-2^2n-2sin^2frac a2^n-1.
endalign*

Hence, the partial sum
beginalign*
&sum_n=1^mleft(2^2nsin^4frac a2^nright)\=&2^2sin^2frac a2-sin^2a+2^4sin^2frac a2^2-2^2sin^2frac a2+ cdots+ 2^2msin^2frac a2^m-2^2m-2sin^2frac a2^m-1\=&2^2msin^2frac a2^m-sin^2a.
endalign*

Let $m to infty$. We obtain $$sum_n=1^inftyleft(2^2nsin^4frac a2^nright)=lim_m to inftyleft(fracsindfrac a2^mdfraca2^mcdot aright)^2-sin^2 a=(1cdot a)^2-sin^2 a=a^2-sin^2 a.$$

share|cite|improve this answer

answered Jul 27 at 17:33

mengdie1982

2,827216

share|cite|improve this answer

share|cite|improve this answer

answered Jul 27 at 17:33

mengdie1982

2,827216

answered Jul 27 at 17:33

mengdie1982

2,827216

answered Jul 27 at 17:33

mengdie1982

2,827216

2,827216

add a comment | 

add a comment | 

 

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