Calculate $lim_{ntoinfty}x_n$ when $x_{n+1} = frac{1}{4}x_n(5-x_n)$
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Sequence $x_n$ such that $x_0 in [0, 2]$ is defined by $x_{n+1} = frac{1}{4}x_n(5-x_n)$. For what values of $x_0 in [0, 2]$ does $x_n$ converges and to what limit?
For $x_0 = 0$, $lim_{ntoinfty}x_n = 0$. I guess that in other cases $lim_{ntoinfty}x_n = 1$. The sequence is bounded by $max(2, frac{25}{16})$ from AM-GM. How can I prove my assertion that $x_n$ converges to $1$?
I've tried to prove that for $x_0 in (0, 1)$ sequence increases and for $x_0 in (1, 2)$ decreases, but when proving first part I got stuck, because I don't know how to prove inductive step that $x_n < 1$ implies $frac{x_n(5-x_n)}{4} < 1$
sequences-and-series limits
edited 2 days ago
Chinnapparaj R
4,4351725
asked Nov 13 at 10:06
user4201961
593211
add a comment |
up vote
1
down vote
favorite
Sequence $x_n$ such that $x_0 in [0, 2]$ is defined by $x_{n+1} = frac{1}{4}x_n(5-x_n)$. For what values of $x_0 in [0, 2]$ does $x_n$ converges and to what limit?
For $x_0 = 0$, $lim_{ntoinfty}x_n = 0$. I guess that in other cases $lim_{ntoinfty}x_n = 1$. The sequence is bounded by $max(2, frac{25}{16})$ from AM-GM. How can I prove my assertion that $x_n$ converges to $1$?
I've tried to prove that for $x_0 in (0, 1)$ sequence increases and for $x_0 in (1, 2)$ decreases, but when proving first part I got stuck, because I don't know how to prove inductive step that $x_n < 1$ implies $frac{x_n(5-x_n)}{4} < 1$
sequences-and-series limits
edited 2 days ago
Chinnapparaj R
4,4351725
asked Nov 13 at 10:06
user4201961
593211
1
Assume the sequence converges to some $l$, and use the fact that ${x_{n+1}}$ and ${x_n}$ converge to the same number. You can then calculate precisely the possible limits of the sequence. Note that you still need to proof that the sequence coverges (usually done with arguments about monotonicity of the sequence).
– Donjim
Nov 13 at 10:15
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Sequence $x_n$ such that $x_0 in [0, 2]$ is defined by $x_{n+1} = frac{1}{4}x_n(5-x_n)$. For what values of $x_0 in [0, 2]$ does $x_n$ converges and to what limit?
For $x_0 = 0$, $lim_{ntoinfty}x_n = 0$. I guess that in other cases $lim_{ntoinfty}x_n = 1$. The sequence is bounded by $max(2, frac{25}{16})$ from AM-GM. How can I prove my assertion that $x_n$ converges to $1$?
I've tried to prove that for $x_0 in (0, 1)$ sequence increases and for $x_0 in (1, 2)$ decreases, but when proving first part I got stuck, because I don't know how to prove inductive step that $x_n < 1$ implies $frac{x_n(5-x_n)}{4} < 1$
sequences-and-series limits
edited 2 days ago
Chinnapparaj R
4,4351725
asked Nov 13 at 10:06
user4201961
593211
Sequence $x_n$ such that $x_0 in [0, 2]$ is defined by $x_{n+1} = frac{1}{4}x_n(5-x_n)$. For what values of $x_0 in [0, 2]$ does $x_n$ converges and to what limit?
For $x_0 = 0$, $lim_{ntoinfty}x_n = 0$. I guess that in other cases $lim_{ntoinfty}x_n = 1$. The sequence is bounded by $max(2, frac{25}{16})$ from AM-GM. How can I prove my assertion that $x_n$ converges to $1$?
I've tried to prove that for $x_0 in (0, 1)$ sequence increases and for $x_0 in (1, 2)$ decreases, but when proving first part I got stuck, because I don't know how to prove inductive step that $x_n < 1$ implies $frac{x_n(5-x_n)}{4} < 1$
sequences-and-series limits
sequences-and-series limits
edited 2 days ago
Chinnapparaj R
4,4351725
asked Nov 13 at 10:06
user4201961
593211
edited 2 days ago
Chinnapparaj R
4,4351725
asked Nov 13 at 10:06
user4201961
593211
edited 2 days ago
Chinnapparaj R
4,4351725
edited 2 days ago
Chinnapparaj R
4,4351725
edited 2 days ago
Chinnapparaj R
4,4351725
4,4351725
asked Nov 13 at 10:06
user4201961
593211
asked Nov 13 at 10:06
user4201961
593211
asked Nov 13 at 10:06
user4201961
593211
593211
1
Assume the sequence converges to some $l$, and use the fact that ${x_{n+1}}$ and ${x_n}$ converge to the same number. You can then calculate precisely the possible limits of the sequence. Note that you still need to proof that the sequence coverges (usually done with arguments about monotonicity of the sequence).
– Donjim
Nov 13 at 10:15
add a comment |
1
Assume the sequence converges to some $l$, and use the fact that ${x_{n+1}}$ and ${x_n}$ converge to the same number. You can then calculate precisely the possible limits of the sequence. Note that you still need to proof that the sequence coverges (usually done with arguments about monotonicity of the sequence).
– Donjim
Nov 13 at 10:15
1
1
Assume the sequence converges to some $l$, and use the fact that ${x_{n+1}}$ and ${x_n}$ converge to the same number. You can then calculate precisely the possible limits of the sequence. Note that you still need to proof that the sequence coverges (usually done with arguments about monotonicity of the sequence).
– Donjim
Nov 13 at 10:15
Assume the sequence converges to some $l$, and use the fact that ${x_{n+1}}$ and ${x_n}$ converge to the same number. You can then calculate precisely the possible limits of the sequence. Note that you still need to proof that the sequence coverges (usually done with arguments about monotonicity of the sequence).
– Donjim
Nov 13 at 10:15
add a comment |
1 Answer
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up vote
6
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The function $f(x)=frac {x(5-x)} 4$ maps $[0,2]$ into itself and it is increasing there. From this it follows by induction that $x_n in [0,1]$ for all $n$ and that $x_{n+1} geq x_n$. Hence $aequiv lim x_n$ exists and since $a=frac {a(5-a)} 4$ we get $a=0$ or $a=1$. Since $(a_n)$ is increasing it follows that $a=0$ if $x_0=0$ and $a=1$ otherwise.
edited Nov 13 at 10:38
Bernard
115k637108
answered Nov 13 at 10:16
Kavi Rama Murthy
39.3k31748
I understood my mistake. +1 by the brief and objective answer.
– MathOverview
Nov 13 at 10:29
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
The function $f(x)=frac {x(5-x)} 4$ maps $[0,2]$ into itself and it is increasing there. From this it follows by induction that $x_n in [0,1]$ for all $n$ and that $x_{n+1} geq x_n$. Hence $aequiv lim x_n$ exists and since $a=frac {a(5-a)} 4$ we get $a=0$ or $a=1$. Since $(a_n)$ is increasing it follows that $a=0$ if $x_0=0$ and $a=1$ otherwise.
edited Nov 13 at 10:38
Bernard
115k637108
answered Nov 13 at 10:16
Kavi Rama Murthy
39.3k31748
I understood my mistake. +1 by the brief and objective answer.
– MathOverview
Nov 13 at 10:29
add a comment |
up vote
6
down vote
The function $f(x)=frac {x(5-x)} 4$ maps $[0,2]$ into itself and it is increasing there. From this it follows by induction that $x_n in [0,1]$ for all $n$ and that $x_{n+1} geq x_n$. Hence $aequiv lim x_n$ exists and since $a=frac {a(5-a)} 4$ we get $a=0$ or $a=1$. Since $(a_n)$ is increasing it follows that $a=0$ if $x_0=0$ and $a=1$ otherwise.
edited Nov 13 at 10:38
Bernard
115k637108
answered Nov 13 at 10:16
Kavi Rama Murthy
39.3k31748
I understood my mistake. +1 by the brief and objective answer.
– MathOverview
Nov 13 at 10:29
add a comment |
up vote
6
down vote
up vote
6
down vote
The function $f(x)=frac {x(5-x)} 4$ maps $[0,2]$ into itself and it is increasing there. From this it follows by induction that $x_n in [0,1]$ for all $n$ and that $x_{n+1} geq x_n$. Hence $aequiv lim x_n$ exists and since $a=frac {a(5-a)} 4$ we get $a=0$ or $a=1$. Since $(a_n)$ is increasing it follows that $a=0$ if $x_0=0$ and $a=1$ otherwise.
edited Nov 13 at 10:38
Bernard
115k637108
answered Nov 13 at 10:16
Kavi Rama Murthy
39.3k31748
The function $f(x)=frac {x(5-x)} 4$ maps $[0,2]$ into itself and it is increasing there. From this it follows by induction that $x_n in [0,1]$ for all $n$ and that $x_{n+1} geq x_n$. Hence $aequiv lim x_n$ exists and since $a=frac {a(5-a)} 4$ we get $a=0$ or $a=1$. Since $(a_n)$ is increasing it follows that $a=0$ if $x_0=0$ and $a=1$ otherwise.
edited Nov 13 at 10:38
Bernard
115k637108
answered Nov 13 at 10:16
Kavi Rama Murthy
39.3k31748
edited Nov 13 at 10:38
Bernard
115k637108
edited Nov 13 at 10:38
Bernard
115k637108
edited Nov 13 at 10:38
Bernard
115k637108
115k637108
answered Nov 13 at 10:16
Kavi Rama Murthy
39.3k31748
answered Nov 13 at 10:16
Kavi Rama Murthy
39.3k31748
answered Nov 13 at 10:16
Kavi Rama Murthy
39.3k31748
39.3k31748
I understood my mistake. +1 by the brief and objective answer.
– MathOverview
Nov 13 at 10:29
add a comment |
I understood my mistake. +1 by the brief and objective answer.
– MathOverview
Nov 13 at 10:29
I understood my mistake. +1 by the brief and objective answer.
– MathOverview
Nov 13 at 10:29
I understood my mistake. +1 by the brief and objective answer.
– MathOverview
Nov 13 at 10:29
add a comment |
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Assume the sequence converges to some $l$, and use the fact that ${x_{n+1}}$ and ${x_n}$ converge to the same number. You can then calculate precisely the possible limits of the sequence. Note that you still need to proof that the sequence coverges (usually done with arguments about monotonicity of the sequence).
– Donjim
Nov 13 at 10:15