Mixing
Let $x_1:=a>0$, $x_{n+1} := x_n+frac{1}{x_n}$ for all $n in mathbb{N}$. Determine if $(x_n)$ is convergent...
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This question already has an answer here:
let $x_1= a > 0 , x_{n+1} = x_n+dfrac{1}{x_n}$ for n ∈ N. determine whether it converges or diverges
1 answer
Let $x_1:=a>0$, $x_{n+1} := x_n+frac{1}{x_n}$ for all $n in mathbb{N}$. Determine if $(x_n)$ is convergent or not.
Notice that $x_n>0$ for all $n in mathbb{N}$, and $x_n$ is a solution of the equation $t^2-x_{n+1}t+1=0$, and then $Delta=x^2_{n+1}-4 geq 0$. So, $x^2_{n+1} geq 4$.
Thus, $(x_n)$ is bounded below by $4$.
Also we have that
$x_{n+1}-x_n=frac{1}{x_n}geq 0$ for all $nin mathbb{N}$.
Hence, $(x_n)$ is increasing.
We have shown that $(x_n)$ is increasing and bounded below. It follows by the monotone convergence theorem that $(x_n)$ converges to a limit that is at least $4$, say $x$.
We have $lim x_{n+1}=lim (x_n+frac{1}{x_n})Rightarrow x=x+frac{1}{x}$
how can I complete it, please?
real-analysis
edited Feb 4 at 13:38
David C. Ullrich
61.8k44095
asked Feb 3 at 2:46
DimaDima
873616
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marked as duplicate by rtybase, Adrian Keister, Andrew, drhab, Community♦ Feb 5 at 20:04
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
let $x_1= a > 0 , x_{n+1} = x_n+dfrac{1}{x_n}$ for n ∈ N. determine whether it converges or diverges
1 answer
Let $x_1:=a>0$, $x_{n+1} := x_n+frac{1}{x_n}$ for all $n in mathbb{N}$. Determine if $(x_n)$ is convergent or not.
Notice that $x_n>0$ for all $n in mathbb{N}$, and $x_n$ is a solution of the equation $t^2-x_{n+1}t+1=0$, and then $Delta=x^2_{n+1}-4 geq 0$. So, $x^2_{n+1} geq 4$.
Thus, $(x_n)$ is bounded below by $4$.
Also we have that
$x_{n+1}-x_n=frac{1}{x_n}geq 0$ for all $nin mathbb{N}$.
Hence, $(x_n)$ is increasing.
We have shown that $(x_n)$ is increasing and bounded below. It follows by the monotone convergence theorem that $(x_n)$ converges to a limit that is at least $4$, say $x$.
We have $lim x_{n+1}=lim (x_n+frac{1}{x_n})Rightarrow x=x+frac{1}{x}$
how can I complete it, please?
real-analysis
edited Feb 4 at 13:38
David C. Ullrich
61.8k44095
asked Feb 3 at 2:46
DimaDima
873616
$endgroup$
marked as duplicate by rtybase, Adrian Keister, Andrew, drhab, Community♦ Feb 5 at 20:04
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
If $x = x + frac{1}{x}$, then $x^2 = x^2 + 1$, whence $1 = 0$; wassup with that? Something seems fishy here . . .
$endgroup$
– Robert Lewis
Feb 3 at 2:53
$begingroup$
Maybe there's a problem with the assertion that $x_n$ is decreasing . . .
$endgroup$
– Robert Lewis
Feb 3 at 2:56
2
$begingroup$
The monotone convergence theorem does not apply. You should probably re-check your statement of that theorem.
$endgroup$
– JonathanZ
Feb 3 at 3:34
$begingroup$
Additionally, the sequence is not necessarily bounded below by 4, though it is bounded below by $a$.
$endgroup$
– JonathanZ
Feb 3 at 3:36
$begingroup$
@rtybase No, it diverges for any $a>0$. Whether $a=1$ or not, if $L$ is the limit we have $L=L+1/L$. The recurrence in the question at that link is different.
$endgroup$
– David C. Ullrich
Feb 4 at 13:35
add a comment |
$begingroup$
This question already has an answer here:
let $x_1= a > 0 , x_{n+1} = x_n+dfrac{1}{x_n}$ for n ∈ N. determine whether it converges or diverges
1 answer
Let $x_1:=a>0$, $x_{n+1} := x_n+frac{1}{x_n}$ for all $n in mathbb{N}$. Determine if $(x_n)$ is convergent or not.
Notice that $x_n>0$ for all $n in mathbb{N}$, and $x_n$ is a solution of the equation $t^2-x_{n+1}t+1=0$, and then $Delta=x^2_{n+1}-4 geq 0$. So, $x^2_{n+1} geq 4$.
Thus, $(x_n)$ is bounded below by $4$.
Also we have that
$x_{n+1}-x_n=frac{1}{x_n}geq 0$ for all $nin mathbb{N}$.
Hence, $(x_n)$ is increasing.
We have shown that $(x_n)$ is increasing and bounded below. It follows by the monotone convergence theorem that $(x_n)$ converges to a limit that is at least $4$, say $x$.
We have $lim x_{n+1}=lim (x_n+frac{1}{x_n})Rightarrow x=x+frac{1}{x}$
how can I complete it, please?
real-analysis
edited Feb 4 at 13:38
David C. Ullrich
61.8k44095
asked Feb 3 at 2:46
DimaDima
873616
$endgroup$
This question already has an answer here:
let $x_1= a > 0 , x_{n+1} = x_n+dfrac{1}{x_n}$ for n ∈ N. determine whether it converges or diverges
1 answer
Let $x_1:=a>0$, $x_{n+1} := x_n+frac{1}{x_n}$ for all $n in mathbb{N}$. Determine if $(x_n)$ is convergent or not.
Notice that $x_n>0$ for all $n in mathbb{N}$, and $x_n$ is a solution of the equation $t^2-x_{n+1}t+1=0$, and then $Delta=x^2_{n+1}-4 geq 0$. So, $x^2_{n+1} geq 4$.
Thus, $(x_n)$ is bounded below by $4$.
Also we have that
$x_{n+1}-x_n=frac{1}{x_n}geq 0$ for all $nin mathbb{N}$.
Hence, $(x_n)$ is increasing.
We have shown that $(x_n)$ is increasing and bounded below. It follows by the monotone convergence theorem that $(x_n)$ converges to a limit that is at least $4$, say $x$.
We have $lim x_{n+1}=lim (x_n+frac{1}{x_n})Rightarrow x=x+frac{1}{x}$
how can I complete it, please?
This question already has an answer here:
let $x_1= a > 0 , x_{n+1} = x_n+dfrac{1}{x_n}$ for n ∈ N. determine whether it converges or diverges
1 answer
real-analysis
real-analysis
edited Feb 4 at 13:38
David C. Ullrich
61.8k44095
asked Feb 3 at 2:46
DimaDima
873616
edited Feb 4 at 13:38
David C. Ullrich
61.8k44095
asked Feb 3 at 2:46
DimaDima
873616
edited Feb 4 at 13:38
David C. Ullrich
61.8k44095
edited Feb 4 at 13:38
David C. Ullrich
61.8k44095
edited Feb 4 at 13:38
David C. Ullrich
61.8k44095
61.8k44095
asked Feb 3 at 2:46
DimaDima
873616
asked Feb 3 at 2:46
DimaDima
873616
asked Feb 3 at 2:46
DimaDima
873616
873616
marked as duplicate by rtybase, Adrian Keister, Andrew, drhab, Community♦ Feb 5 at 20:04
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by rtybase, Adrian Keister, Andrew, drhab, Community♦ Feb 5 at 20:04
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
If $x = x + frac{1}{x}$, then $x^2 = x^2 + 1$, whence $1 = 0$; wassup with that? Something seems fishy here . . .
$endgroup$
– Robert Lewis
Feb 3 at 2:53
$begingroup$
Maybe there's a problem with the assertion that $x_n$ is decreasing . . .
$endgroup$
– Robert Lewis
Feb 3 at 2:56
2
$begingroup$
The monotone convergence theorem does not apply. You should probably re-check your statement of that theorem.
$endgroup$
– JonathanZ
Feb 3 at 3:34
$begingroup$
Additionally, the sequence is not necessarily bounded below by 4, though it is bounded below by $a$.
$endgroup$
– JonathanZ
Feb 3 at 3:36
$begingroup$
@rtybase No, it diverges for any $a>0$. Whether $a=1$ or not, if $L$ is the limit we have $L=L+1/L$. The recurrence in the question at that link is different.
$endgroup$
– David C. Ullrich
Feb 4 at 13:35
add a comment |
1
$begingroup$
If $x = x + frac{1}{x}$, then $x^2 = x^2 + 1$, whence $1 = 0$; wassup with that? Something seems fishy here . . .
$endgroup$
– Robert Lewis
Feb 3 at 2:53
$begingroup$
Maybe there's a problem with the assertion that $x_n$ is decreasing . . .
$endgroup$
– Robert Lewis
Feb 3 at 2:56
2
$begingroup$
The monotone convergence theorem does not apply. You should probably re-check your statement of that theorem.
$endgroup$
– JonathanZ
Feb 3 at 3:34
$begingroup$
Additionally, the sequence is not necessarily bounded below by 4, though it is bounded below by $a$.
$endgroup$
– JonathanZ
Feb 3 at 3:36
$begingroup$
@rtybase No, it diverges for any $a>0$. Whether $a=1$ or not, if $L$ is the limit we have $L=L+1/L$. The recurrence in the question at that link is different.
$endgroup$
– David C. Ullrich
Feb 4 at 13:35
1
1
$begingroup$
If $x = x + frac{1}{x}$, then $x^2 = x^2 + 1$, whence $1 = 0$; wassup with that? Something seems fishy here . . .
$endgroup$
– Robert Lewis
Feb 3 at 2:53
$begingroup$
If $x = x + frac{1}{x}$, then $x^2 = x^2 + 1$, whence $1 = 0$; wassup with that? Something seems fishy here . . .
$endgroup$
– Robert Lewis
Feb 3 at 2:53
$begingroup$
Maybe there's a problem with the assertion that $x_n$ is decreasing . . .
$endgroup$
– Robert Lewis
Feb 3 at 2:56
$begingroup$
Maybe there's a problem with the assertion that $x_n$ is decreasing . . .
$endgroup$
– Robert Lewis
Feb 3 at 2:56
2
2
$begingroup$
The monotone convergence theorem does not apply. You should probably re-check your statement of that theorem.
$endgroup$
– JonathanZ
Feb 3 at 3:34
$begingroup$
The monotone convergence theorem does not apply. You should probably re-check your statement of that theorem.
$endgroup$
– JonathanZ
Feb 3 at 3:34
$begingroup$
Additionally, the sequence is not necessarily bounded below by 4, though it is bounded below by $a$.
$endgroup$
– JonathanZ
Feb 3 at 3:36
$begingroup$
Additionally, the sequence is not necessarily bounded below by 4, though it is bounded below by $a$.
$endgroup$
– JonathanZ
Feb 3 at 3:36
$begingroup$
@rtybase No, it diverges for any $a>0$. Whether $a=1$ or not, if $L$ is the limit we have $L=L+1/L$. The recurrence in the question at that link is different.
$endgroup$
– David C. Ullrich
Feb 4 at 13:35
$begingroup$
@rtybase No, it diverges for any $a>0$. Whether $a=1$ or not, if $L$ is the limit we have $L=L+1/L$. The recurrence in the question at that link is different.
$endgroup$
– David C. Ullrich
Feb 4 at 13:35
add a comment |
2 Answers
2
active
oldest
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$begingroup$
If there were an integer $N$ such that $x_nleq N$ for all $n$, then we would have $tfrac{1}{x_n}geq tfrac{1}{N}$ for all $n$. But then
$$x_{n+N^2}geq x_n + Ntext{.}$$
answered Feb 3 at 4:06
K B DaveK B Dave
3,696317
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add a comment |
$begingroup$
As you have observed the sequence is increasing. either it converges to a finite limit (which is necessarily $>0$) or it converges to $infty$. The first case is ruled out by taking limits in the given equation. [We cannot have $x=x+frac 1 x$ ]. Hence $x_n to infty$.
answered Feb 3 at 5:03
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
$begingroup$
Exactly the point I was trying to get at in my comments to the question itself! Cheers!
$endgroup$
– Robert Lewis
Feb 3 at 5:08
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If there were an integer $N$ such that $x_nleq N$ for all $n$, then we would have $tfrac{1}{x_n}geq tfrac{1}{N}$ for all $n$. But then
$$x_{n+N^2}geq x_n + Ntext{.}$$
answered Feb 3 at 4:06
K B DaveK B Dave
3,696317
$endgroup$
add a comment |
$begingroup$
If there were an integer $N$ such that $x_nleq N$ for all $n$, then we would have $tfrac{1}{x_n}geq tfrac{1}{N}$ for all $n$. But then
$$x_{n+N^2}geq x_n + Ntext{.}$$
answered Feb 3 at 4:06
K B DaveK B Dave
3,696317
$endgroup$
add a comment |
$begingroup$
If there were an integer $N$ such that $x_nleq N$ for all $n$, then we would have $tfrac{1}{x_n}geq tfrac{1}{N}$ for all $n$. But then
$$x_{n+N^2}geq x_n + Ntext{.}$$
answered Feb 3 at 4:06
K B DaveK B Dave
3,696317
$endgroup$
If there were an integer $N$ such that $x_nleq N$ for all $n$, then we would have $tfrac{1}{x_n}geq tfrac{1}{N}$ for all $n$. But then
$$x_{n+N^2}geq x_n + Ntext{.}$$
answered Feb 3 at 4:06
K B DaveK B Dave
3,696317
answered Feb 3 at 4:06
K B DaveK B Dave
3,696317
answered Feb 3 at 4:06
K B DaveK B Dave
3,696317
answered Feb 3 at 4:06
K B DaveK B Dave
3,696317
3,696317
add a comment |
add a comment |
$begingroup$
As you have observed the sequence is increasing. either it converges to a finite limit (which is necessarily $>0$) or it converges to $infty$. The first case is ruled out by taking limits in the given equation. [We cannot have $x=x+frac 1 x$ ]. Hence $x_n to infty$.
answered Feb 3 at 5:03
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
$begingroup$
Exactly the point I was trying to get at in my comments to the question itself! Cheers!
$endgroup$
– Robert Lewis
Feb 3 at 5:08
add a comment |
$begingroup$
As you have observed the sequence is increasing. either it converges to a finite limit (which is necessarily $>0$) or it converges to $infty$. The first case is ruled out by taking limits in the given equation. [We cannot have $x=x+frac 1 x$ ]. Hence $x_n to infty$.
answered Feb 3 at 5:03
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
$begingroup$
Exactly the point I was trying to get at in my comments to the question itself! Cheers!
$endgroup$
– Robert Lewis
Feb 3 at 5:08
add a comment |
$begingroup$
As you have observed the sequence is increasing. either it converges to a finite limit (which is necessarily $>0$) or it converges to $infty$. The first case is ruled out by taking limits in the given equation. [We cannot have $x=x+frac 1 x$ ]. Hence $x_n to infty$.
answered Feb 3 at 5:03
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
As you have observed the sequence is increasing. either it converges to a finite limit (which is necessarily $>0$) or it converges to $infty$. The first case is ruled out by taking limits in the given equation. [We cannot have $x=x+frac 1 x$ ]. Hence $x_n to infty$.
answered Feb 3 at 5:03
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
answered Feb 3 at 5:03
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
answered Feb 3 at 5:03
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
answered Feb 3 at 5:03
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
74.9k53270
$begingroup$
Exactly the point I was trying to get at in my comments to the question itself! Cheers!
$endgroup$
– Robert Lewis
Feb 3 at 5:08
add a comment |
$begingroup$
Exactly the point I was trying to get at in my comments to the question itself! Cheers!
$endgroup$
– Robert Lewis
Feb 3 at 5:08
$begingroup$
Exactly the point I was trying to get at in my comments to the question itself! Cheers!
$endgroup$
– Robert Lewis
Feb 3 at 5:08
$begingroup$
Exactly the point I was trying to get at in my comments to the question itself! Cheers!
$endgroup$
– Robert Lewis
Feb 3 at 5:08
add a comment |
1
$begingroup$
If $x = x + frac{1}{x}$, then $x^2 = x^2 + 1$, whence $1 = 0$; wassup with that? Something seems fishy here . . .
$endgroup$
– Robert Lewis
Feb 3 at 2:53
$begingroup$
Maybe there's a problem with the assertion that $x_n$ is decreasing . . .
$endgroup$
– Robert Lewis
Feb 3 at 2:56
2
$begingroup$
The monotone convergence theorem does not apply. You should probably re-check your statement of that theorem.
$endgroup$
– JonathanZ
Feb 3 at 3:34
$begingroup$
Additionally, the sequence is not necessarily bounded below by 4, though it is bounded below by $a$.
$endgroup$
– JonathanZ
Feb 3 at 3:36
$begingroup$
@rtybase No, it diverges for any $a>0$. Whether $a=1$ or not, if $L$ is the limit we have $L=L+1/L$. The recurrence in the question at that link is different.
$endgroup$
– David C. Ullrich
Feb 4 at 13:35