Calculating limit of second-order recursive sequence(limit of) a linear second order recurrence relation with variable coefficientsProving the limit of a recursive sequenceRecursive equation with limitRecursive sequence with square rootCalculating limit of sequence by Euler $e$Finding the limits to sequences given by recursive transformation of a vector: $a_n = a_n-1M$How can I find an explicit expression for this recursively defined sequence?Calculate a limit of recursive sequenceLimit of a sequence with real and natural number variablesFinding the limit of a recursive complex sequence
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Calculating limit of second-order recursive sequence
(limit of) a linear second order recurrence relation with variable coefficientsProving the limit of a recursive sequenceRecursive equation with limitRecursive sequence with square rootCalculating limit of sequence by Euler $e$Finding the limits to sequences given by recursive transformation of a vector: $a_n = a_n-1M$How can I find an explicit expression for this recursively defined sequence?Calculate a limit of recursive sequenceLimit of a sequence with real and natural number variablesFinding the limit of a recursive complex sequence
$begingroup$
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
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
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
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.
$endgroup$
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
sequences-and-series limits recurrence-relations sequent-calculus
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.
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
asked Mar 10 at 19:35
F.LF.L
112
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.
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.
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Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
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 $a_2-a_3<6$.
answered 2 days ago
HK LeeHK Lee
14.1k52360
$endgroup$
add a comment |
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1 Answer
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1 Answer
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oldest
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votes
$begingroup$
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 $a_2-a_3<6$.
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 $a_2-a_3<6$.
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 $a_2-a_3<6$.
answered 2 days ago
HK LeeHK Lee
14.1k52360
$endgroup$
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 $a_2-a_3<6$.
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
answered 2 days ago
HK LeeHK Lee
14.1k52360
14.1k52360
add a comment |
add a comment |
F.L is a new contributor. Be nice, and check out our Code of Conduct.
F.L is a new contributor. Be nice, and check out our Code of Conduct.
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F.L is a new contributor. Be nice, and check out our Code of Conduct.
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