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A recurrence relation

by asmani
Tags: recurrence, relation
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asmani
#1
Aug8-11, 02:24 AM
P: 86
Hi all

Suppose that [itex]a_1=\sqrt5[/itex], [itex]a_{n+1}=a_n^2-2[/itex] and [itex]g_n=\frac{a_1a_2...a_n}{a_{n+1}}[/itex].
Evaluate [itex]\lim_{n\rightarrow \infty } g_n[/itex].

I have seen some information in this link. Besides, the sequence gn seems as a good rational approximation for [itex]\sqrt5[/itex]. I know that the answer is 1, But I can't find any nice solution. Any hint is strongly appreciated.
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Mentallic
#2
Aug8-11, 03:07 AM
HW Helper
P: 3,531
I have yet to study recurrence relations for myself, but I played around with it for a bit and it seems like it was headed in the right direction.

[tex]g_n=\frac{a_1a_2...a_n}{a_n^2-2}[/tex]

[tex]=\frac{a_n^2\left(\frac{a_1a_2...a_{n-1}}{a_n}\right)-2\left(\frac{a_1a_2...a_{n-1}}{a_n}\right)+2\left(\frac{a_1a_2...a_{n-1}}{a_n}\right)}{a_n^2-2}[/tex]

[tex]=\frac{\left(a_n^2-2\right)\left(\frac{a_1a_2...a_{n-1}}{a_n}\right)+2\left(\frac{a_1a_2...a_{n-1}}{a_n}\right)}{a_n^2-2}[/tex]

[tex]=\frac{a_1a_2...a_{n-1}}{a_n}+2\frac{a_1a_2...a_{n-1}}{a_na_{n+1}}[/tex]

[tex]=g_{n-1}+\frac{2}{a_n^2}g_n[/tex]

Hence we can express gn in terms of gn-1, and the factor multiplying gn-1 seems to say a lot about what gn is as n gets very large.
asmani
#3
Aug8-11, 04:07 AM
P: 86
Thanks. I can't see what does it say about gn. Can you help me on this?

I found out how to show that:
[tex]a_n=\frac{\varphi ^{2^n}-\varphi ^{-2^n}}{\varphi ^{2^{n-1}}-\varphi ^{-2^{n-1}}}[/tex]
And then the rest is easy. But I'm still looking for a more elegant solution.

Mentallic
#4
Aug8-11, 06:05 AM
HW Helper
P: 3,531
A recurrence relation

Hmm, after looking at it a bit more I realize it's really not telling us as much as I originally thought.

After solving for gn in

[tex]g_n=g_{n-1}+\frac{2}{a^2_n-2}g_n[/tex] to obtain

[tex]g_n=\frac{a_n^2}{a_n^2-2}g_{n-1}[/tex]

I was simply arguing that the factor approaches 1 as n gets large, but this doesn't really tell us about the limit of gn...
tiny-tim
#5
Aug8-11, 07:03 AM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,148
hi asmani!
Quote Quote by asmani View Post
I found out how to show that:
[tex]a_n=\frac{\varphi ^{2^n}-\varphi ^{-2^n}}{\varphi ^{2^{n-1}}-\varphi ^{-2^{n-1}}}[/tex]
And then the rest is easy. But I'm still looking for a more elegant solution.
reverse-engineering that expression, we get
an = sinh(C2n)/sinh(C2n-1) = 2cosh(C2n-1)
sooo try defining an = 2coshbn
asmani
#6
Aug8-11, 08:45 AM
P: 86
Thanks. I think it's better not to simplify the fraction to get:
[tex]g_n=a_1a_2\cdots a_n\frac{1}{a_{n+1}}=\frac{\sinh(C2^1) }{\sinh(C2^0)}\frac{\sinh(C2^2) }{\sinh(C2^1)}\cdots \frac{\sinh(C2^n) }{\sinh(C2^{n-1})}\frac{\sinh(C2^n) }{\sinh(C2^{n+1})}[/tex]
[tex]=\frac{\sinh^2(C2^n) }{\sinh(C2^0)\sinh(C2^{n+1})}[/tex]
Where C=Log(φ).
And now calculating the limit is easy.
Mentallic
#7
Aug8-11, 04:23 PM
HW Helper
P: 3,531
Sorry about wasting your time asmani, I should've left it to the big guys
asmani
#8
Aug9-11, 03:44 AM
P: 86
No, I appreciate your participation in this thread.


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