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find the the limit points of this sequence.. [duplicate]
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This question is an exact duplicate of:
Find the sub-sequential limits of a recursive sequence
1 answer
Given $a_1 in mathbbR$ , consider the sequence $a_n$ defined by
$ a_n+1 = begincases fraca_n2 &text for even n ,\ frac 1+ a_n2 & text for odd n endcases$.
find the the limit points of this sequence..
My attempt : For n odd i get $ a_n+1= frac 1+ a_n2$
nad putting $a_n+1 = a_n = l$
Now i get $l= frac 1+ l2$ as i get $l = 1$ for odd..
as I don't know how to find the $a_n+1$ when n will even
Pliz help me,,,,,
real-analysis
asked Jul 31 at 21:56
stupid
52418
marked as duplicate by Clement C., amWhy, Xander Henderson, max_zorn, Piyush Divyanakar Aug 1 at 6:42
This question was marked as an exact duplicate of an existing question.
 |Â
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up vote
0
down vote
favorite
This question is an exact duplicate of:
Find the sub-sequential limits of a recursive sequence
1 answer
Given $a_1 in mathbbR$ , consider the sequence $a_n$ defined by
$ a_n+1 = begincases fraca_n2 &text for even n ,\ frac 1+ a_n2 & text for odd n endcases$.
find the the limit points of this sequence..
My attempt : For n odd i get $ a_n+1= frac 1+ a_n2$
nad putting $a_n+1 = a_n = l$
Now i get $l= frac 1+ l2$ as i get $l = 1$ for odd..
as I don't know how to find the $a_n+1$ when n will even
Pliz help me,,,,,
real-analysis
asked Jul 31 at 21:56
stupid
52418
marked as duplicate by Clement C., amWhy, Xander Henderson, max_zorn, Piyush Divyanakar Aug 1 at 6:42
This question was marked as an exact duplicate of an existing question.
1
Hint: Look at the subsequences $a_2k$ and $a_2k+1$.
– Dark Malthorp
Jul 31 at 22:07
1
"Pliz help me,,,,," I don't think this is necessary, is it
– Rumpelstiltskin
Jul 31 at 22:10
im not getting @ Dark,,,,can u elaborate more
– stupid
Jul 31 at 22:12
1
Set $b_n = a_2n+1$ and $c_n = a_2n$. Derive the recurrence relations for $(b_n)_n$ and $(c_n)_n$, find their limits. Conclude.
– Clement C.
Jul 31 at 22:16
thanks u clement
– stupid
Jul 31 at 22:18
 |Â
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This question is an exact duplicate of:
Find the sub-sequential limits of a recursive sequence
1 answer
Given $a_1 in mathbbR$ , consider the sequence $a_n$ defined by
$ a_n+1 = begincases fraca_n2 &text for even n ,\ frac 1+ a_n2 & text for odd n endcases$.
find the the limit points of this sequence..
My attempt : For n odd i get $ a_n+1= frac 1+ a_n2$
nad putting $a_n+1 = a_n = l$
Now i get $l= frac 1+ l2$ as i get $l = 1$ for odd..
as I don't know how to find the $a_n+1$ when n will even
Pliz help me,,,,,
real-analysis
asked Jul 31 at 21:56
stupid
52418
This question is an exact duplicate of:
Find the sub-sequential limits of a recursive sequence
1 answer
Given $a_1 in mathbbR$ , consider the sequence $a_n$ defined by
$ a_n+1 = begincases fraca_n2 &text for even n ,\ frac 1+ a_n2 & text for odd n endcases$.
find the the limit points of this sequence..
My attempt : For n odd i get $ a_n+1= frac 1+ a_n2$
nad putting $a_n+1 = a_n = l$
Now i get $l= frac 1+ l2$ as i get $l = 1$ for odd..
as I don't know how to find the $a_n+1$ when n will even
Pliz help me,,,,,
This question is an exact duplicate of:
Find the sub-sequential limits of a recursive sequence
1 answer
real-analysis
asked Jul 31 at 21:56
stupid
52418
asked Jul 31 at 21:56
stupid
52418
asked Jul 31 at 21:56
stupid
52418
asked Jul 31 at 21:56
stupid
52418
52418
marked as duplicate by Clement C., amWhy, Xander Henderson, max_zorn, Piyush Divyanakar Aug 1 at 6:42
This question was marked as an exact duplicate of an existing question.
marked as duplicate by Clement C., amWhy, Xander Henderson, max_zorn, Piyush Divyanakar Aug 1 at 6:42
This question was marked as an exact duplicate of an existing question.
1
Hint: Look at the subsequences $a_2k$ and $a_2k+1$.
– Dark Malthorp
Jul 31 at 22:07
1
"Pliz help me,,,,," I don't think this is necessary, is it
– Rumpelstiltskin
Jul 31 at 22:10
im not getting @ Dark,,,,can u elaborate more
– stupid
Jul 31 at 22:12
1
Set $b_n = a_2n+1$ and $c_n = a_2n$. Derive the recurrence relations for $(b_n)_n$ and $(c_n)_n$, find their limits. Conclude.
– Clement C.
Jul 31 at 22:16
thanks u clement
– stupid
Jul 31 at 22:18
 |Â
show 1 more comment
1
Hint: Look at the subsequences $a_2k$ and $a_2k+1$.
– Dark Malthorp
Jul 31 at 22:07
1
"Pliz help me,,,,," I don't think this is necessary, is it
– Rumpelstiltskin
Jul 31 at 22:10
im not getting @ Dark,,,,can u elaborate more
– stupid
Jul 31 at 22:12
1
Set $b_n = a_2n+1$ and $c_n = a_2n$. Derive the recurrence relations for $(b_n)_n$ and $(c_n)_n$, find their limits. Conclude.
– Clement C.
Jul 31 at 22:16
thanks u clement
– stupid
Jul 31 at 22:18
1
1
Hint: Look at the subsequences $a_2k$ and $a_2k+1$.
– Dark Malthorp
Jul 31 at 22:07
Hint: Look at the subsequences $a_2k$ and $a_2k+1$.
– Dark Malthorp
Jul 31 at 22:07
1
1
"Pliz help me,,,,," I don't think this is necessary, is it
– Rumpelstiltskin
Jul 31 at 22:10
"Pliz help me,,,,," I don't think this is necessary, is it
– Rumpelstiltskin
Jul 31 at 22:10
im not getting @ Dark,,,,can u elaborate more
– stupid
Jul 31 at 22:12
im not getting @ Dark,,,,can u elaborate more
– stupid
Jul 31 at 22:12
1
1
Set $b_n = a_2n+1$ and $c_n = a_2n$. Derive the recurrence relations for $(b_n)_n$ and $(c_n)_n$, find their limits. Conclude.
– Clement C.
Jul 31 at 22:16
Set $b_n = a_2n+1$ and $c_n = a_2n$. Derive the recurrence relations for $(b_n)_n$ and $(c_n)_n$, find their limits. Conclude.
– Clement C.
Jul 31 at 22:16
thanks u clement
– stupid
Jul 31 at 22:18
thanks u clement
– stupid
Jul 31 at 22:18
 |Â
show 1 more comment
1 Answer
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We have $$a_n+1 = begincases fraca_n2 &text for even n ,\ frac a_n2+frac 12 & text for odd n endcases$$
Let us start with $a_1=a$ and write a few terms to see the pattern.
$$ a, a/2, a/4+1/2, a/8+1/4 , a/16+5/8, a/32+5/16, a/64+21/32,...$$ It is clear that the parts including $a$ will tend to zero.
The remaining parts $$1/2,1/4,5/8,5/16,21/32,21/64,...$$
has two subsequences namely $$1/2 ,5/8,21/32,...$$ and $$1/4 ,5/16,21/64,...$$ The first subsequence is $$1/2 ,1/2+1/8,1/2+1/8+1/32,...$$ which converges to $2/3$
The second subsequence converges to half of the first limit which is $1/3$
Thus we have two limit points $$ 1/3,2/3$$
answered Jul 31 at 22:32
Mohammad Riazi-Kermani
27.3k41851
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
We have $$a_n+1 = begincases fraca_n2 &text for even n ,\ frac a_n2+frac 12 & text for odd n endcases$$
Let us start with $a_1=a$ and write a few terms to see the pattern.
$$ a, a/2, a/4+1/2, a/8+1/4 , a/16+5/8, a/32+5/16, a/64+21/32,...$$ It is clear that the parts including $a$ will tend to zero.
The remaining parts $$1/2,1/4,5/8,5/16,21/32,21/64,...$$
has two subsequences namely $$1/2 ,5/8,21/32,...$$ and $$1/4 ,5/16,21/64,...$$ The first subsequence is $$1/2 ,1/2+1/8,1/2+1/8+1/32,...$$ which converges to $2/3$
The second subsequence converges to half of the first limit which is $1/3$
Thus we have two limit points $$ 1/3,2/3$$
answered Jul 31 at 22:32
Mohammad Riazi-Kermani
27.3k41851
add a comment |Â
up vote
1
down vote
accepted
We have $$a_n+1 = begincases fraca_n2 &text for even n ,\ frac a_n2+frac 12 & text for odd n endcases$$
Let us start with $a_1=a$ and write a few terms to see the pattern.
$$ a, a/2, a/4+1/2, a/8+1/4 , a/16+5/8, a/32+5/16, a/64+21/32,...$$ It is clear that the parts including $a$ will tend to zero.
The remaining parts $$1/2,1/4,5/8,5/16,21/32,21/64,...$$
has two subsequences namely $$1/2 ,5/8,21/32,...$$ and $$1/4 ,5/16,21/64,...$$ The first subsequence is $$1/2 ,1/2+1/8,1/2+1/8+1/32,...$$ which converges to $2/3$
The second subsequence converges to half of the first limit which is $1/3$
Thus we have two limit points $$ 1/3,2/3$$
answered Jul 31 at 22:32
Mohammad Riazi-Kermani
27.3k41851
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
We have $$a_n+1 = begincases fraca_n2 &text for even n ,\ frac a_n2+frac 12 & text for odd n endcases$$
Let us start with $a_1=a$ and write a few terms to see the pattern.
$$ a, a/2, a/4+1/2, a/8+1/4 , a/16+5/8, a/32+5/16, a/64+21/32,...$$ It is clear that the parts including $a$ will tend to zero.
The remaining parts $$1/2,1/4,5/8,5/16,21/32,21/64,...$$
has two subsequences namely $$1/2 ,5/8,21/32,...$$ and $$1/4 ,5/16,21/64,...$$ The first subsequence is $$1/2 ,1/2+1/8,1/2+1/8+1/32,...$$ which converges to $2/3$
The second subsequence converges to half of the first limit which is $1/3$
Thus we have two limit points $$ 1/3,2/3$$
answered Jul 31 at 22:32
Mohammad Riazi-Kermani
27.3k41851
We have $$a_n+1 = begincases fraca_n2 &text for even n ,\ frac a_n2+frac 12 & text for odd n endcases$$
Let us start with $a_1=a$ and write a few terms to see the pattern.
$$ a, a/2, a/4+1/2, a/8+1/4 , a/16+5/8, a/32+5/16, a/64+21/32,...$$ It is clear that the parts including $a$ will tend to zero.
The remaining parts $$1/2,1/4,5/8,5/16,21/32,21/64,...$$
has two subsequences namely $$1/2 ,5/8,21/32,...$$ and $$1/4 ,5/16,21/64,...$$ The first subsequence is $$1/2 ,1/2+1/8,1/2+1/8+1/32,...$$ which converges to $2/3$
The second subsequence converges to half of the first limit which is $1/3$
Thus we have two limit points $$ 1/3,2/3$$
answered Jul 31 at 22:32
Mohammad Riazi-Kermani
27.3k41851
answered Jul 31 at 22:32
Mohammad Riazi-Kermani
27.3k41851
answered Jul 31 at 22:32
Mohammad Riazi-Kermani
27.3k41851
answered Jul 31 at 22:32
Mohammad Riazi-Kermani
27.3k41851
27.3k41851
add a comment |Â
add a comment |Â
1
Hint: Look at the subsequences $a_2k$ and $a_2k+1$.
– Dark Malthorp
Jul 31 at 22:07
1
"Pliz help me,,,,," I don't think this is necessary, is it
– Rumpelstiltskin
Jul 31 at 22:10
im not getting @ Dark,,,,can u elaborate more
– stupid
Jul 31 at 22:12
1
Set $b_n = a_2n+1$ and $c_n = a_2n$. Derive the recurrence relations for $(b_n)_n$ and $(c_n)_n$, find their limits. Conclude.
– Clement C.
Jul 31 at 22:16
thanks u clement
– stupid
Jul 31 at 22:18