Fibonacci Sequence converge exercise

  • Thread starter Calabi_Yau
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  • #1
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Let Fn denote the Fibonacci sequence.
un is the sequence given by: un= Fn+1/Fn. Show that mod(un - [itex]\phi[/itex]) [itex]\leq[/itex][itex]\frac{1}{\phi}[/itex]mod(un-1-[itex]\phi[/itex]) and therefore mod(un - [itex]\phi[/itex]) [itex]\leq[/itex] [itex]\frac{1}{\phin-1}[/itex][/itex]mod(u1-[itex]\phi[/itex]) and then conclude un converges to [itex]\phi[/itex]


I have tried with the identity [itex]\phi[/itex] = 1+ [itex]\frac{1}{\phi}[/itex] if anything came to light... And I tried dividing the mods but it got even more complicated.

I can prove from the seocnd equation that un converges to [itex]\phi[/itex] as n-1 converges to infinity and thus 1/+inf =0, the right side becomes zero and we get mod(un - [itex]\phi[/itex]) [itex]\leq[/itex]0 which is the definiton of convergence. But I can't get from the first equation to the second. I don't know how to pass from un-1 to u1 and the part of the [itex]\phi[/itex]n-1. Can someone shed some light on this issue?
 

Answers and Replies

  • #2
pasmith
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Let Fn denote the Fibonacci sequence.
un is the sequence given by: un= Fn+1/Fn. Show that mod(un - [itex]\phi[/itex]) [itex]\leq[/itex][itex]\frac{1}{\phi}[/itex]mod(un-1-[itex]\phi[/itex]) and therefore mod(un - [itex]\phi[/itex]) [itex]\leq[/itex] [itex]\frac{1}{\phin-1}[/itex][/itex]mod(u1-[itex]\phi[/itex]) and then conclude un converges to [itex]\phi[/itex]


I have tried with the identity [itex]\phi[/itex] = 1+ [itex]\frac{1}{\phi}[/itex] if anything came to light... And I tried dividing the mods but it got even more complicated.

I can prove from the seocnd equation that un converges to [itex]\phi[/itex] as n-1 converges to infinity and thus 1/+inf =0, the right side becomes zero and we get mod(un - [itex]\phi[/itex]) [itex]\leq[/itex]0 which is the definiton of convergence. But I can't get from the first equation to the second. I don't know how to pass from un-1 to u1 and the part of the [itex]\phi[/itex]n-1. Can someone shed some light on this issue?
We have
[tex]
|u_n - \phi| \leq \frac{1}{\phi} |u_{n-1} - \phi|
[/tex]
But since this is true for all n, we also have
[tex]
|u_n - \phi| \leq \frac{1}{\phi} |u_{n-1} - \phi| \leq \frac{1}{\phi} \frac{1}{\phi} |u_{n-2} - \phi| = \frac{1}{\phi^2} |u_{n-2} - \phi|
[/tex]
and we can clearly keep going, picking up a factor of [itex]\phi^{-1}[/itex] each time, until we have a multiple of [itex]|u_1 - \phi|[/itex] on the right.
 
  • #3
35
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Hmm, I didn't see that pattern. That is the same as making n-1=p and then doing the same for up and up-1.

Nice, thank you :)
 

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