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Calculating limit of second-order recursive sequence

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2

1

$begingroup$

Can someone give me a hint on how to calculate the limit for the following second-order recursive sequence:

beginalign
a_0 &= c_0, \
a_1 &= c_1, \
a_n &=fraca_n-1+a_n-22-frac(a_n-1)^2+(a_n-2)^22+(fraca_n-1+a_n-22)^2
endalign

I simulated it in Excel and I know that it converges to a constant or goes to negative infinity depending on $c_0,c_1$.

Can I separate the limit into 3 parts and calculate them one by one? Is it correct that I can not, because not all limits for each part exist?

I am mostly interested in the formula for the limit dependent on $c_0, c_1$ for an unrelated theoretical argument.

sequences-and-series limits recurrence-relations sequent-calculus

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edited Mar 10 at 20:59

F.L

asked Mar 10 at 19:35

F.LF.L

112

New contributor

F.L is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

2

1

$begingroup$

Can someone give me a hint on how to calculate the limit for the following second-order recursive sequence:

beginalign
a_0 &= c_0, \
a_1 &= c_1, \
a_n &=fraca_n-1+a_n-22-frac(a_n-1)^2+(a_n-2)^22+(fraca_n-1+a_n-22)^2
endalign

I simulated it in Excel and I know that it converges to a constant or goes to negative infinity depending on $c_0,c_1$.

Can I separate the limit into 3 parts and calculate them one by one? Is it correct that I can not, because not all limits for each part exist?

I am mostly interested in the formula for the limit dependent on $c_0, c_1$ for an unrelated theoretical argument.

sequences-and-series limits recurrence-relations sequent-calculus

share|cite|improve this question

edited Mar 10 at 20:59

F.L

asked Mar 10 at 19:35

F.LF.L

112

New contributor

F.L is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

Can someone give me a hint on how to calculate the limit for the following second-order recursive sequence:

beginalign
a_0 &= c_0, \
a_1 &= c_1, \
a_n &=fraca_n-1+a_n-22-frac(a_n-1)^2+(a_n-2)^22+(fraca_n-1+a_n-22)^2
endalign

I simulated it in Excel and I know that it converges to a constant or goes to negative infinity depending on $c_0,c_1$.

Can I separate the limit into 3 parts and calculate them one by one? Is it correct that I can not, because not all limits for each part exist?

I am mostly interested in the formula for the limit dependent on $c_0, c_1$ for an unrelated theoretical argument.

sequences-and-series limits recurrence-relations sequent-calculus

share|cite|improve this question

edited Mar 10 at 20:59

F.L

asked Mar 10 at 19:35

F.LF.L

112

New contributor

F.L is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Mar 10 at 20:59

F.L

asked Mar 10 at 19:35

F.LF.L

112

New contributor

F.L is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Mar 10 at 20:59

F.L

asked Mar 10 at 19:35

F.LF.L

112

New contributor

F.L is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 10 at 19:35

F.LF.L

112

New contributor

F.L is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 10 at 19:35

F.LF.L

112

1 Answer
1

active

oldest

votes

0

$begingroup$

When $a_1-a_2geq 6$ or $ a_2-a_1geq 4 $, it diverges. Other cases
are convergent.

Proof : i) $$a_n
= fraca_n-1+a_n-22 - (fraca_n-1-a_n-2 2)^2 $$

When $|a_1-a_2|geq 6$, then $|a_n-a_n-1|geq 6$. It does not converge.

ii) $a_2<a_1$ : If $a_1-a_2<6$ is close to $6$, then
$$ a_2-a_3 < a_1-a_2 $$

Hence there is large $N$ s.t. $a_i>
a_i+1$ for $ileq N-2$ and $a_N-1<a_N$.

iii) $a_1<a_2$ : If $a_2-a_1geq 4
$, then $a_2-a_3geq 6$.

If $a_2-a_1<4$, then $a_2-a_3<6$.

share|cite|improve this answer

answered 2 days ago

HK LeeHK Lee

14.1k52360

$endgroup$

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0

$begingroup$

When $a_1-a_2geq 6$ or $ a_2-a_1geq 4 $, it diverges. Other cases
are convergent.

Proof : i) $$a_n
= fraca_n-1+a_n-22 - (fraca_n-1-a_n-2 2)^2 $$

When $|a_1-a_2|geq 6$, then $|a_n-a_n-1|geq 6$. It does not converge.

ii) $a_2<a_1$ : If $a_1-a_2<6$ is close to $6$, then
$$ a_2-a_3 < a_1-a_2 $$

Hence there is large $N$ s.t. $a_i>
a_i+1$ for $ileq N-2$ and $a_N-1<a_N$.

iii) $a_1<a_2$ : If $a_2-a_1geq 4
$, then $a_2-a_3geq 6$.

If $a_2-a_1<4$, then $a_2-a_3<6$.

share|cite|improve this answer

answered 2 days ago

HK LeeHK Lee

14.1k52360

$endgroup$

add a comment | 

0

$begingroup$

When $a_1-a_2geq 6$ or $ a_2-a_1geq 4 $, it diverges. Other cases
are convergent.

Proof : i) $$a_n
= fraca_n-1+a_n-22 - (fraca_n-1-a_n-2 2)^2 $$

When $|a_1-a_2|geq 6$, then $|a_n-a_n-1|geq 6$. It does not converge.

ii) $a_2<a_1$ : If $a_1-a_2<6$ is close to $6$, then
$$ a_2-a_3 < a_1-a_2 $$

Hence there is large $N$ s.t. $a_i>
a_i+1$ for $ileq N-2$ and $a_N-1<a_N$.

iii) $a_1<a_2$ : If $a_2-a_1geq 4
$, then $a_2-a_3geq 6$.

If $a_2-a_1<4$, then $a_2-a_3<6$.

share|cite|improve this answer

answered 2 days ago

HK LeeHK Lee

14.1k52360

$endgroup$

add a comment | 

$begingroup$

When $a_1-a_2geq 6$ or $ a_2-a_1geq 4 $, it diverges. Other cases
are convergent.

Proof : i) $$a_n
= fraca_n-1+a_n-22 - (fraca_n-1-a_n-2 2)^2 $$

When $|a_1-a_2|geq 6$, then $|a_n-a_n-1|geq 6$. It does not converge.

ii) $a_2<a_1$ : If $a_1-a_2<6$ is close to $6$, then
$$ a_2-a_3 < a_1-a_2 $$

Hence there is large $N$ s.t. $a_i>
a_i+1$ for $ileq N-2$ and $a_N-1<a_N$.

iii) $a_1<a_2$ : If $a_2-a_1geq 4
$, then $a_2-a_3geq 6$.

If $a_2-a_1<4$, then $a_2-a_3<6$.

share|cite|improve this answer

answered 2 days ago

HK LeeHK Lee

14.1k52360

$endgroup$

share|cite|improve this answer

answered 2 days ago

HK LeeHK Lee

14.1k52360

answered 2 days ago

HK LeeHK Lee

14.1k52360

answered 2 days ago

HK LeeHK Lee

14.1k52360