Fix $a > 0$ and let $x_1 > sqrt a$ [duplicate]
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This question already has an answer here:
Find the limit if it exists of $S_{n+1} = frac{1}{2}(S_n +frac{A}{S_n})$
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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
edited Nov 19 at 16:32
Tianlalu
2,9361935
asked Nov 19 at 16:21
RandomThinker
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marked as duplicate by Nosrati, rtybase, Trevor Gunn, Arthur, user90369 Nov 19 at 20:14
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.
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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
edited Nov 19 at 16:32
Tianlalu
2,9361935
asked Nov 19 at 16:21
RandomThinker
192
marked as duplicate by Nosrati, rtybase, Trevor Gunn, Arthur, user90369 Nov 19 at 20:14
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
Nov 19 at 16:30
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
Nov 19 at 16:31
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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
edited Nov 19 at 16:32
Tianlalu
2,9361935
asked Nov 19 at 16:21
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
edited Nov 19 at 16:32
Tianlalu
2,9361935
asked Nov 19 at 16:21
RandomThinker
192
edited Nov 19 at 16:32
Tianlalu
2,9361935
asked Nov 19 at 16:21
RandomThinker
192
edited Nov 19 at 16:32
Tianlalu
2,9361935
edited Nov 19 at 16:32
Tianlalu
2,9361935
edited Nov 19 at 16:32
Tianlalu
2,9361935
2,9361935
asked Nov 19 at 16:21
RandomThinker
192
asked Nov 19 at 16:21
RandomThinker
192
asked Nov 19 at 16:21
RandomThinker
192
192
marked as duplicate by Nosrati, rtybase, Trevor Gunn, Arthur, user90369 Nov 19 at 20:14
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 Nov 19 at 20:14
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
Nov 19 at 16:30
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
Nov 19 at 16:31
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
Nov 19 at 16:30
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
Nov 19 at 16:31
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
Nov 19 at 16:30
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
Nov 19 at 16:30
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
Nov 19 at 16:31
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
Nov 19 at 16:31
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.
answered Nov 19 at 16:32
Arthur
109k7103186
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.
answered Nov 19 at 16:32
Arthur
109k7103186
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.
answered Nov 19 at 16:32
Arthur
109k7103186
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.
answered Nov 19 at 16:32
Arthur
109k7103186
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.
answered Nov 19 at 16:32
Arthur
109k7103186
edited Nov 19 at 16:39
answered Nov 19 at 16:32
Arthur
109k7103186
answered Nov 19 at 16:32
Arthur
109k7103186
answered Nov 19 at 16:32
Arthur
109k7103186
109k7103186
add a comment |
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
Nov 19 at 16:30
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
Nov 19 at 16:31