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doodlenoodle
 Post subject: Contraction sequence
PostPosted: Mon, 20 Feb 2012 06:31:38 UTC 
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I am suppose to be using the contraction principle to prove that a sequence converges. I am trying to find the a_n+2 term so I can do

|an+2-an+1|<=k|an+1-an|

I am given a1 and an+1, how do I figure out an+2?

My sequence is 1+1/an = an+1

Would my an+2 = 1+1/(an+1) ?
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Shadow
 Post subject: Re: Contraction sequence
PostPosted: Mon, 20 Feb 2012 06:34:36 UTC 
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doodlenoodle wrote:
I am suppose to be using the contraction principle to prove that a sequence converges. I am trying to find the a_n+2 term so I can do

|an+2-an+1|<=k|an+1-an|

I am given a1 and an+1, how do I figure out an+2?

My sequence is 1+1/an = an+1

Would my an+2 = 1+1/(an+1) ?


By definitions $a_{n+2}=1+{1\over a_{n+1}}, so iterate:

$a_{n+2}=1+{1\over 1+{1\over a_n}}, then look at how the pattern evolves as n\to\infty and you should be able to see how this all terminates.

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doodlenoodle
 Post subject: Re: Contraction sequence
PostPosted: Mon, 20 Feb 2012 06:56:11 UTC 
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For using the contraction principle to prove a_n converges tho, I just get something like

|an+2-an+1| = |(1+1/an+1) - (1+1/an)|= |1/an+1-1/an| <= k|(1+1/an - an|

How can I prove that k is between 0 and 1?
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outermeasure
 Post subject: Re: Contraction sequence
PostPosted: Mon, 20 Feb 2012 12:35:32 UTC 
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doodlenoodle wrote:
For using the contraction principle to prove a_n converges tho, I just get something like

|an+2-an+1| = |(1+1/an+1) - (1+1/an)|= |1/an+1-1/an| <= k|(1+1/an - an|

How can I prove that k is between 0 and 1?


You can't, unless you put some restriction on a_n.

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\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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Justin
 Post subject: Re: Contraction sequence
PostPosted: Mon, 20 Feb 2012 16:40:03 UTC 
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doodlenoodle wrote:
For using the contraction principle to prove a_n converges tho, I just get something like

|an+2-an+1| = |(1+1/an+1) - (1+1/an)|= |1/an+1-1/an| <= k|(1+1/an - an|

How can I prove that k is between 0 and 1?


Would showing that the sequence is Cauchy be enough to show it satisfies the Contraction Principle?
This sequence clearly converges.

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"Mathematicians are like lovers. Grant a mathematician the least principle, and he will draw from it a consequence which you must also grant him, and from this consequence another." Bernard Le Bovier Fontenelle (1657-1757)

"In great mathematics there is a very high degree of unexpectedness, combined with inevitability and economy."
G.H. Hardy (1877-1947)
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Shadow
 Post subject: Re: Contraction sequence
PostPosted: Mon, 20 Feb 2012 16:59:17 UTC 
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Justin wrote:
doodlenoodle wrote:
For using the contraction principle to prove a_n converges tho, I just get something like

|an+2-an+1| = |(1+1/an+1) - (1+1/an)|= |1/an+1-1/an| <= k|(1+1/an - an|

How can I prove that k is between 0 and 1?


Would showing that the sequence is Cauchy be enough to show it satisfies the Contraction Principle?
This sequence clearly converges.


Why should it?

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Justin
 Post subject: Re: Contraction sequence
PostPosted: Mon, 20 Feb 2012 17:44:11 UTC 
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Shadow wrote:
Justin wrote:
doodlenoodle wrote:
For using the contraction principle to prove a_n converges tho, I just get something like

|an+2-an+1| = |(1+1/an+1) - (1+1/an)|= |1/an+1-1/an| <= k|(1+1/an - an|

How can I prove that k is between 0 and 1?


Would showing that the sequence is Cauchy be enough to show it satisfies the Contraction Principle?
This sequence clearly converges.


Why should it?


Just a thought...although I see your point. If showing the sequence Cauchy was that easy, you wouldn't need to use the
Contraction Principle to begin with. I'm guessing a triangle inequality estimate is required here.

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"Mathematicians are like lovers. Grant a mathematician the least principle, and he will draw from it a consequence which you must also grant him, and from this consequence another." Bernard Le Bovier Fontenelle (1657-1757)

"In great mathematics there is a very high degree of unexpectedness, combined with inevitability and economy."
G.H. Hardy (1877-1947)
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