Verification of a sequence with exactly two limit points.
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I was not allowed to edit the previous question because people spent their time and energy constructing an answer, therefore I have asked it again, with my initial error in formulating my answers corrected.
Is it true that this example I came up with, namely the sequence $$a_n=cosleft(frac{pi}{3} + frac{pi n}{2}right)$$
Has exactly $4$ limit points/accumulation points? What I tried to do was symmetrically pick an infinite amount of points that lie on the unit circle and all hop between four exact points $pm frac{1}{2}$ and $pm frac{1}{2} sqrt{3}$.
Is equality fine for accumulation points is basically what I am asking.
I also was thinking of an example with infinitely many accumulation points and wondered if the same example would cut it:
$$b_n = n cos(n) $$
Since it oscillates back and forth it comes back to every single point eventually.
As an alternative way of getting four accumulation points I could define unions of the sets $s_a= { a+ frac{1}{n}| n in mathbb{N} }$ for values $a=0, 1,2,3$ but this example feels very contrived and I do not know how to formulate this in terms of a sequence other than just listing the elements one by one in set notation: ${ frac{1}{1}, 1+frac{1}{1}, 2+frac{1}{1} dots, 1+ frac{1}{n}, 2+ frac{1}{n} dots }$
real-analysis
add a comment |
up vote
-2
down vote
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I was not allowed to edit the previous question because people spent their time and energy constructing an answer, therefore I have asked it again, with my initial error in formulating my answers corrected.
Is it true that this example I came up with, namely the sequence $$a_n=cosleft(frac{pi}{3} + frac{pi n}{2}right)$$
Has exactly $4$ limit points/accumulation points? What I tried to do was symmetrically pick an infinite amount of points that lie on the unit circle and all hop between four exact points $pm frac{1}{2}$ and $pm frac{1}{2} sqrt{3}$.
Is equality fine for accumulation points is basically what I am asking.
I also was thinking of an example with infinitely many accumulation points and wondered if the same example would cut it:
$$b_n = n cos(n) $$
Since it oscillates back and forth it comes back to every single point eventually.
As an alternative way of getting four accumulation points I could define unions of the sets $s_a= { a+ frac{1}{n}| n in mathbb{N} }$ for values $a=0, 1,2,3$ but this example feels very contrived and I do not know how to formulate this in terms of a sequence other than just listing the elements one by one in set notation: ${ frac{1}{1}, 1+frac{1}{1}, 2+frac{1}{1} dots, 1+ frac{1}{n}, 2+ frac{1}{n} dots }$
real-analysis
1
You have asked a question with the same title less than one hour ago, moreover without mentionning it. This is not a good practice. You should have modified the text of the initial question instead of asking a new one.
– Jean Marie
Nov 23 at 23:00
add a comment |
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
I was not allowed to edit the previous question because people spent their time and energy constructing an answer, therefore I have asked it again, with my initial error in formulating my answers corrected.
Is it true that this example I came up with, namely the sequence $$a_n=cosleft(frac{pi}{3} + frac{pi n}{2}right)$$
Has exactly $4$ limit points/accumulation points? What I tried to do was symmetrically pick an infinite amount of points that lie on the unit circle and all hop between four exact points $pm frac{1}{2}$ and $pm frac{1}{2} sqrt{3}$.
Is equality fine for accumulation points is basically what I am asking.
I also was thinking of an example with infinitely many accumulation points and wondered if the same example would cut it:
$$b_n = n cos(n) $$
Since it oscillates back and forth it comes back to every single point eventually.
As an alternative way of getting four accumulation points I could define unions of the sets $s_a= { a+ frac{1}{n}| n in mathbb{N} }$ for values $a=0, 1,2,3$ but this example feels very contrived and I do not know how to formulate this in terms of a sequence other than just listing the elements one by one in set notation: ${ frac{1}{1}, 1+frac{1}{1}, 2+frac{1}{1} dots, 1+ frac{1}{n}, 2+ frac{1}{n} dots }$
real-analysis
I was not allowed to edit the previous question because people spent their time and energy constructing an answer, therefore I have asked it again, with my initial error in formulating my answers corrected.
Is it true that this example I came up with, namely the sequence $$a_n=cosleft(frac{pi}{3} + frac{pi n}{2}right)$$
Has exactly $4$ limit points/accumulation points? What I tried to do was symmetrically pick an infinite amount of points that lie on the unit circle and all hop between four exact points $pm frac{1}{2}$ and $pm frac{1}{2} sqrt{3}$.
Is equality fine for accumulation points is basically what I am asking.
I also was thinking of an example with infinitely many accumulation points and wondered if the same example would cut it:
$$b_n = n cos(n) $$
Since it oscillates back and forth it comes back to every single point eventually.
As an alternative way of getting four accumulation points I could define unions of the sets $s_a= { a+ frac{1}{n}| n in mathbb{N} }$ for values $a=0, 1,2,3$ but this example feels very contrived and I do not know how to formulate this in terms of a sequence other than just listing the elements one by one in set notation: ${ frac{1}{1}, 1+frac{1}{1}, 2+frac{1}{1} dots, 1+ frac{1}{n}, 2+ frac{1}{n} dots }$
real-analysis
real-analysis
edited Nov 26 at 18:12
asked Nov 23 at 21:59
WesleyGroupshaveFeelingsToo
1,115321
1,115321
1
You have asked a question with the same title less than one hour ago, moreover without mentionning it. This is not a good practice. You should have modified the text of the initial question instead of asking a new one.
– Jean Marie
Nov 23 at 23:00
add a comment |
1
You have asked a question with the same title less than one hour ago, moreover without mentionning it. This is not a good practice. You should have modified the text of the initial question instead of asking a new one.
– Jean Marie
Nov 23 at 23:00
1
1
You have asked a question with the same title less than one hour ago, moreover without mentionning it. This is not a good practice. You should have modified the text of the initial question instead of asking a new one.
– Jean Marie
Nov 23 at 23:00
You have asked a question with the same title less than one hour ago, moreover without mentionning it. This is not a good practice. You should have modified the text of the initial question instead of asking a new one.
– Jean Marie
Nov 23 at 23:00
add a comment |
1 Answer
1
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votes
up vote
2
down vote
accepted
We have that
- $n=1 implies a_1=cosleft(frac{pi}{3} + frac{pi }{2}right)=-frac{sqrt 3}2$
- $n=2 implies a_2=cosleft(frac{pi}{3} + piright)=-frac12$
- $n=3 implies a_3=cosleft(frac{pi}{3} + frac{3pi }{2}right)=frac{sqrt 3}2$
- $n=4 implies a_4=cosleft(frac{pi}{3} + 2piright)=frac12$
- $n=5 implies a_5=a_1$
For the second question, yes of curse $b_n$ oscillates diverging assuming values $in mathbb{R}$ and $cos n$ is dense in $[-1,1]$. From that to conlcude that $ncos n$ is dense on the real line is not so trivial, see the related
- Is $n sin n$ dense on the real line?
- Does this sequence $a(n) = frac{1}{n^3sin(n)}$ converge
- Is it true that $varliminf_{n rightarrow +infty} |n sin n| = 0$
- Convergence of $sum(n^3sin^2n)^{-1}$
Could I simply take $cos(n)$ and that would give me an infinite amount of points because of density?
– WesleyGroupshaveFeelingsToo
Nov 23 at 22:14
@WesleyGroupshaveFeelingsToo Yes, refer to the following OP
– gimusi
Nov 23 at 22:17
I mean one sort of needs to prove that you will never start repeating a finite set of points.
– WesleyGroupshaveFeelingsToo
Nov 23 at 22:18
ah, very nice, thank you.
– WesleyGroupshaveFeelingsToo
Nov 23 at 22:19
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
We have that
- $n=1 implies a_1=cosleft(frac{pi}{3} + frac{pi }{2}right)=-frac{sqrt 3}2$
- $n=2 implies a_2=cosleft(frac{pi}{3} + piright)=-frac12$
- $n=3 implies a_3=cosleft(frac{pi}{3} + frac{3pi }{2}right)=frac{sqrt 3}2$
- $n=4 implies a_4=cosleft(frac{pi}{3} + 2piright)=frac12$
- $n=5 implies a_5=a_1$
For the second question, yes of curse $b_n$ oscillates diverging assuming values $in mathbb{R}$ and $cos n$ is dense in $[-1,1]$. From that to conlcude that $ncos n$ is dense on the real line is not so trivial, see the related
- Is $n sin n$ dense on the real line?
- Does this sequence $a(n) = frac{1}{n^3sin(n)}$ converge
- Is it true that $varliminf_{n rightarrow +infty} |n sin n| = 0$
- Convergence of $sum(n^3sin^2n)^{-1}$
Could I simply take $cos(n)$ and that would give me an infinite amount of points because of density?
– WesleyGroupshaveFeelingsToo
Nov 23 at 22:14
@WesleyGroupshaveFeelingsToo Yes, refer to the following OP
– gimusi
Nov 23 at 22:17
I mean one sort of needs to prove that you will never start repeating a finite set of points.
– WesleyGroupshaveFeelingsToo
Nov 23 at 22:18
ah, very nice, thank you.
– WesleyGroupshaveFeelingsToo
Nov 23 at 22:19
add a comment |
up vote
2
down vote
accepted
We have that
- $n=1 implies a_1=cosleft(frac{pi}{3} + frac{pi }{2}right)=-frac{sqrt 3}2$
- $n=2 implies a_2=cosleft(frac{pi}{3} + piright)=-frac12$
- $n=3 implies a_3=cosleft(frac{pi}{3} + frac{3pi }{2}right)=frac{sqrt 3}2$
- $n=4 implies a_4=cosleft(frac{pi}{3} + 2piright)=frac12$
- $n=5 implies a_5=a_1$
For the second question, yes of curse $b_n$ oscillates diverging assuming values $in mathbb{R}$ and $cos n$ is dense in $[-1,1]$. From that to conlcude that $ncos n$ is dense on the real line is not so trivial, see the related
- Is $n sin n$ dense on the real line?
- Does this sequence $a(n) = frac{1}{n^3sin(n)}$ converge
- Is it true that $varliminf_{n rightarrow +infty} |n sin n| = 0$
- Convergence of $sum(n^3sin^2n)^{-1}$
Could I simply take $cos(n)$ and that would give me an infinite amount of points because of density?
– WesleyGroupshaveFeelingsToo
Nov 23 at 22:14
@WesleyGroupshaveFeelingsToo Yes, refer to the following OP
– gimusi
Nov 23 at 22:17
I mean one sort of needs to prove that you will never start repeating a finite set of points.
– WesleyGroupshaveFeelingsToo
Nov 23 at 22:18
ah, very nice, thank you.
– WesleyGroupshaveFeelingsToo
Nov 23 at 22:19
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
We have that
- $n=1 implies a_1=cosleft(frac{pi}{3} + frac{pi }{2}right)=-frac{sqrt 3}2$
- $n=2 implies a_2=cosleft(frac{pi}{3} + piright)=-frac12$
- $n=3 implies a_3=cosleft(frac{pi}{3} + frac{3pi }{2}right)=frac{sqrt 3}2$
- $n=4 implies a_4=cosleft(frac{pi}{3} + 2piright)=frac12$
- $n=5 implies a_5=a_1$
For the second question, yes of curse $b_n$ oscillates diverging assuming values $in mathbb{R}$ and $cos n$ is dense in $[-1,1]$. From that to conlcude that $ncos n$ is dense on the real line is not so trivial, see the related
- Is $n sin n$ dense on the real line?
- Does this sequence $a(n) = frac{1}{n^3sin(n)}$ converge
- Is it true that $varliminf_{n rightarrow +infty} |n sin n| = 0$
- Convergence of $sum(n^3sin^2n)^{-1}$
We have that
- $n=1 implies a_1=cosleft(frac{pi}{3} + frac{pi }{2}right)=-frac{sqrt 3}2$
- $n=2 implies a_2=cosleft(frac{pi}{3} + piright)=-frac12$
- $n=3 implies a_3=cosleft(frac{pi}{3} + frac{3pi }{2}right)=frac{sqrt 3}2$
- $n=4 implies a_4=cosleft(frac{pi}{3} + 2piright)=frac12$
- $n=5 implies a_5=a_1$
For the second question, yes of curse $b_n$ oscillates diverging assuming values $in mathbb{R}$ and $cos n$ is dense in $[-1,1]$. From that to conlcude that $ncos n$ is dense on the real line is not so trivial, see the related
- Is $n sin n$ dense on the real line?
- Does this sequence $a(n) = frac{1}{n^3sin(n)}$ converge
- Is it true that $varliminf_{n rightarrow +infty} |n sin n| = 0$
- Convergence of $sum(n^3sin^2n)^{-1}$
edited Nov 23 at 22:36
answered Nov 23 at 22:04
gimusi
88.8k74394
88.8k74394
Could I simply take $cos(n)$ and that would give me an infinite amount of points because of density?
– WesleyGroupshaveFeelingsToo
Nov 23 at 22:14
@WesleyGroupshaveFeelingsToo Yes, refer to the following OP
– gimusi
Nov 23 at 22:17
I mean one sort of needs to prove that you will never start repeating a finite set of points.
– WesleyGroupshaveFeelingsToo
Nov 23 at 22:18
ah, very nice, thank you.
– WesleyGroupshaveFeelingsToo
Nov 23 at 22:19
add a comment |
Could I simply take $cos(n)$ and that would give me an infinite amount of points because of density?
– WesleyGroupshaveFeelingsToo
Nov 23 at 22:14
@WesleyGroupshaveFeelingsToo Yes, refer to the following OP
– gimusi
Nov 23 at 22:17
I mean one sort of needs to prove that you will never start repeating a finite set of points.
– WesleyGroupshaveFeelingsToo
Nov 23 at 22:18
ah, very nice, thank you.
– WesleyGroupshaveFeelingsToo
Nov 23 at 22:19
Could I simply take $cos(n)$ and that would give me an infinite amount of points because of density?
– WesleyGroupshaveFeelingsToo
Nov 23 at 22:14
Could I simply take $cos(n)$ and that would give me an infinite amount of points because of density?
– WesleyGroupshaveFeelingsToo
Nov 23 at 22:14
@WesleyGroupshaveFeelingsToo Yes, refer to the following OP
– gimusi
Nov 23 at 22:17
@WesleyGroupshaveFeelingsToo Yes, refer to the following OP
– gimusi
Nov 23 at 22:17
I mean one sort of needs to prove that you will never start repeating a finite set of points.
– WesleyGroupshaveFeelingsToo
Nov 23 at 22:18
I mean one sort of needs to prove that you will never start repeating a finite set of points.
– WesleyGroupshaveFeelingsToo
Nov 23 at 22:18
ah, very nice, thank you.
– WesleyGroupshaveFeelingsToo
Nov 23 at 22:19
ah, very nice, thank you.
– WesleyGroupshaveFeelingsToo
Nov 23 at 22:19
add a comment |
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You have asked a question with the same title less than one hour ago, moreover without mentionning it. This is not a good practice. You should have modified the text of the initial question instead of asking a new one.
– Jean Marie
Nov 23 at 23:00