Fix $a > 0$ and let $x_1 > sqrt a$ [duplicate]

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This question already has an answer here:

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  Find the limit if it exists of $S_{n+1} = frac{1}{2}(S_n +frac{A}{S_n})$
  
  2 answers
  
  

My current thought is to take $x_{n+1}-x_n$ and to prove the difference is smaller than $sqrt a$. But it won't help me to prove $x_n$ converges to $sqrt a$.

real-analysis convergence recurrence-relations

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edited yesterday

Tianlalu

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RandomThinker

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marked as duplicate by Nosrati, rtybase, Trevor Gunn, Arthur, user90369 yesterday

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.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well, the question starts by asking you to show that the sequence is decreasing; so $x_{n + 1} - x_n < sqrt{a}$ is useless, because you actually need to show that $x_{n + 1} - x_n le 0$. Then you'll probably invoke a theorem about monotone convergence.
  – T. Bongers
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I answered here to this question sometimes ago, in a different context (it was required that the approximating succession ${x_n}_{ninmathbb{N}}$ was made of only rational numbers, but the method of proof is basically the same).
  – Daniele Tampieri
  yesterday
  
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

This question already has an answer here:

• 
  Find the limit if it exists of $S_{n+1} = frac{1}{2}(S_n +frac{A}{S_n})$
  
  2 answers
  
  

My current thought is to take $x_{n+1}-x_n$ and to prove the difference is smaller than $sqrt a$. But it won't help me to prove $x_n$ converges to $sqrt a$.

real-analysis convergence recurrence-relations

share|cite|improve this question

edited yesterday

Tianlalu

2,594632

asked yesterday

RandomThinker

192

marked as duplicate by Nosrati, rtybase, Trevor Gunn, Arthur, user90369 yesterday

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.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well, the question starts by asking you to show that the sequence is decreasing; so $x_{n + 1} - x_n < sqrt{a}$ is useless, because you actually need to show that $x_{n + 1} - x_n le 0$. Then you'll probably invoke a theorem about monotone convergence.
  – T. Bongers
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I answered here to this question sometimes ago, in a different context (it was required that the approximating succession ${x_n}_{ninmathbb{N}}$ was made of only rational numbers, but the method of proof is basically the same).
  – Daniele Tampieri
  yesterday
  
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

This question already has an answer here:

• 
  Find the limit if it exists of $S_{n+1} = frac{1}{2}(S_n +frac{A}{S_n})$
  
  2 answers
  
  

My current thought is to take $x_{n+1}-x_n$ and to prove the difference is smaller than $sqrt a$. But it won't help me to prove $x_n$ converges to $sqrt a$.

real-analysis convergence recurrence-relations

share|cite|improve this question

edited yesterday

Tianlalu

2,594632

asked yesterday

RandomThinker

192

This question already has an answer here:

• 
  Find the limit if it exists of $S_{n+1} = frac{1}{2}(S_n +frac{A}{S_n})$
  
  2 answers
  
  

My current thought is to take $x_{n+1}-x_n$ and to prove the difference is smaller than $sqrt a$. But it won't help me to prove $x_n$ converges to $sqrt a$.

This question already has an answer here:

• 
  Find the limit if it exists of $S_{n+1} = frac{1}{2}(S_n +frac{A}{S_n})$
  
  2 answers
  
  

real-analysis convergence recurrence-relations

real-analysis convergence recurrence-relations

share|cite|improve this question

edited yesterday

Tianlalu

2,594632

asked yesterday

RandomThinker

192

share|cite|improve this question

edited yesterday

Tianlalu

2,594632

asked yesterday

RandomThinker

192

share|cite|improve this question

share|cite|improve this question

edited yesterday

Tianlalu

2,594632

edited yesterday

Tianlalu

2,594632

edited yesterday

Tianlalu

2,594632

2,594632

asked yesterday

RandomThinker

192

asked yesterday

RandomThinker

192

asked yesterday

RandomThinker

192

192

marked as duplicate by Nosrati, rtybase, Trevor Gunn, Arthur, user90369 yesterday

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 Nosrati, rtybase, Trevor Gunn, Arthur, user90369 yesterday

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.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well, the question starts by asking you to show that the sequence is decreasing; so $x_{n + 1} - x_n < sqrt{a}$ is useless, because you actually need to show that $x_{n + 1} - x_n le 0$. Then you'll probably invoke a theorem about monotone convergence.
  – T. Bongers
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I answered here to this question sometimes ago, in a different context (it was required that the approximating succession ${x_n}_{ninmathbb{N}}$ was made of only rational numbers, but the method of proof is basically the same).
  – Daniele Tampieri
  yesterday
  
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well, the question starts by asking you to show that the sequence is decreasing; so $x_{n + 1} - x_n < sqrt{a}$ is useless, because you actually need to show that $x_{n + 1} - x_n le 0$. Then you'll probably invoke a theorem about monotone convergence.
  – T. Bongers
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I answered here to this question sometimes ago, in a different context (it was required that the approximating succession ${x_n}_{ninmathbb{N}}$ was made of only rational numbers, but the method of proof is basically the same).
  – Daniele Tampieri
  yesterday
  
  
  

  
  
  
  

Well, the question starts by asking you to show that the sequence is decreasing; so $x_{n + 1} - x_n < sqrt{a}$ is useless, because you actually need to show that $x_{n + 1} - x_n le 0$. Then you'll probably invoke a theorem about monotone convergence.
– T. Bongers
yesterday

Well, the question starts by asking you to show that the sequence is decreasing; so $x_{n + 1} - x_n < sqrt{a}$ is useless, because you actually need to show that $x_{n + 1} - x_n le 0$. Then you'll probably invoke a theorem about monotone convergence.
– T. Bongers
yesterday

I answered here to this question sometimes ago, in a different context (it was required that the approximating succession ${x_n}_{ninmathbb{N}}$ was made of only rational numbers, but the method of proof is basically the same).
– Daniele Tampieri
yesterday

I answered here to this question sometimes ago, in a different context (it was required that the approximating succession ${x_n}_{ninmathbb{N}}$ was made of only rational numbers, but the method of proof is basically the same).
– Daniele Tampieri
yesterday

add a comment | 

1 Answer
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Assuming $x_1>sqrtalpha$, show that $x_{n+1}>sqrtalpha$. Then use that to show $x_{n+1}-x_n<0$, which proves the sequence is monotonically decreasing.

Therefore (it is monotonically decreasing and bounded below) the sequence has a limit $L$. The limit must satisfy:
$$
L=frac12left(L+fracalpha Lright)
$$
which means you can figure out exactly what it is.

share|cite|improve this answer

edited yesterday

answered yesterday

Arthur

108k7103186

add a comment | 

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

Assuming $x_1>sqrtalpha$, show that $x_{n+1}>sqrtalpha$. Then use that to show $x_{n+1}-x_n<0$, which proves the sequence is monotonically decreasing.

Therefore (it is monotonically decreasing and bounded below) the sequence has a limit $L$. The limit must satisfy:
$$
L=frac12left(L+fracalpha Lright)
$$
which means you can figure out exactly what it is.

share|cite|improve this answer

edited yesterday

answered yesterday

Arthur

108k7103186

add a comment | 

up vote
2
down vote

Assuming $x_1>sqrtalpha$, show that $x_{n+1}>sqrtalpha$. Then use that to show $x_{n+1}-x_n<0$, which proves the sequence is monotonically decreasing.

Therefore (it is monotonically decreasing and bounded below) the sequence has a limit $L$. The limit must satisfy:
$$
L=frac12left(L+fracalpha Lright)
$$
which means you can figure out exactly what it is.

share|cite|improve this answer

edited yesterday

answered yesterday

Arthur

108k7103186

add a comment | 

up vote
2
down vote

up vote
2
down vote

Assuming $x_1>sqrtalpha$, show that $x_{n+1}>sqrtalpha$. Then use that to show $x_{n+1}-x_n<0$, which proves the sequence is monotonically decreasing.

Therefore (it is monotonically decreasing and bounded below) the sequence has a limit $L$. The limit must satisfy:
$$
L=frac12left(L+fracalpha Lright)
$$
which means you can figure out exactly what it is.

share|cite|improve this answer

edited yesterday

answered yesterday

Arthur

108k7103186

Assuming $x_1>sqrtalpha$, show that $x_{n+1}>sqrtalpha$. Then use that to show $x_{n+1}-x_n<0$, which proves the sequence is monotonically decreasing.

Therefore (it is monotonically decreasing and bounded below) the sequence has a limit $L$. The limit must satisfy:
$$
L=frac12left(L+fracalpha Lright)
$$
which means you can figure out exactly what it is.

share|cite|improve this answer

edited yesterday

answered yesterday

Arthur

108k7103186

share|cite|improve this answer

share|cite|improve this answer

edited yesterday

edited yesterday

edited yesterday

answered yesterday

Arthur

108k7103186

answered yesterday

Arthur

108k7103186

answered yesterday

Arthur

108k7103186

108k7103186

add a comment | 

add a comment | 

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