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Cauchy sequence question

by Unassuming
Tags: cauchy, sequence
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Unassuming
#1
Oct16-08, 04:09 PM
P: 169
Let a_1 and a_2 be arbitrary real number that are not equal. For [tex]n \geq 3[/tex], define a_n inductively by,

[tex]a_n = \frac{1}{2} (a_{n-1}+a_{n+2} )[/tex]



I cannot get the result that the book gets. I proceed,

[tex]a_{n+1} - a_{n} = \frac{1}{2}(a_n + a_{n-1} ) - \frac{1}{2} (a_{n-1} + a_{n-2} ) = \frac{1}{2} ( a_n - a_{n-2}
)= \frac{1}{2}(a_n + a_{n-1} ) [/tex]

The book got the answer,

[tex] a_{n+1} - a_n = \frac{-1}{2} (a_n - a_{n-1} ) [/tex]

Any help for me?
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Dick
#2
Oct16-08, 04:28 PM
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You've got some typos in there that are making this really confusing. The original recursion is a_n=(1/2)*(a_n-1+a_n-2), right? That says 2*a_n-a_n-1=a_n-2. I agree with the book.
Unassuming
#3
Oct16-08, 05:19 PM
P: 169
I can't see it. I've done some algebra and I still can't get anywhere from your last step.

Dick
#4
Oct16-08, 05:24 PM
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Cauchy sequence question

You've got a_n+1-a_n=(1/2)(a_n-a_n-2), right? Put the expression for a_n-2 into that. If you still aren't getting it show your work.
Unassuming
#5
Oct16-08, 05:40 PM
P: 169
I got it. Thanks Dick
HallsofIvy
#6
Oct17-08, 07:12 AM
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PF Gold
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Here's a "cheat":
Since [itex]\sum x^n/n![/itex] is the Taylor's series for ex, 1+ 1/2!+ 1/3!+ ... converges to e. Since it converges, it is a Cauchy sequence!

Or you could just note that the numeric series converges by the ratio test. Again, since it converges, it is a Cauchy sequence.

But I doubt that either of those answers is what was wanted!
Unassuming
#7
Oct17-08, 03:35 PM
P: 169
Those are nice things to put the problem into perspective though. Sometimes I (and hopefully others) get lost in the algebra of the problem.


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