Find Limit of Fibonacci Sequence | Determine L

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Homework Help Overview

The discussion revolves around finding the limit of the ratio of consecutive Fibonacci numbers, denoted as \( x_n = \frac{f_{n+1}}{f_n} \), and determining the value of \( L \) where \( \lim(x_n) = L \). Participants express uncertainty about how to derive this limit without explicitly calculating numerous terms.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the assumption that if a limit exists, then \( x_n \) can be equated to \( x_{n-1} \). There are questions about how to derive a numerical value from the Fibonacci sequence and the role of the quadratic equation in this context. Some express confusion about how to manipulate the Fibonacci relationships to arrive at a solvable equation.

Discussion Status

The conversation is ongoing, with various participants exploring different interpretations of the Fibonacci sequence's properties. Some guidance has been offered regarding the formation of a quadratic equation, but there is still uncertainty about how to proceed with the calculations and the implications of the Fibonacci relationships.

Contextual Notes

Participants note the challenge of working with the Fibonacci sequence without specific numerical values and question the assumptions made in the derivation process. There is also mention of the theorem regarding subsequences converging to the same limit, which adds to the complexity of the discussion.

Punkyc7
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Let f[itex]_{n}[/itex] be the Fibonacci sequence and let [itex]x_{n}[/itex] = [itex]f_{n+1}[/itex]/[itex]f_{n}[/itex]. Given that lim[itex](x_{n}[/itex])=L exist determine L.

Ok so I know that the limit is [itex]\frac{1+\sqrt{5}}{2}[/itex] from previous experience with the sequence, but I am not sure how do you show that without writing out a lot of terms and then noticing what I all ready know it is. How do you find the limit of a sequence to a number if your not given any numbers to work with?
 
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Hi Punkyc7 ! :smile:

If there is a limit, then you can assume that xn = xn-1 :wink:
 
true, but how do you ever get a number when you are dealing with f[itex]_{n}[/itex] and f[itex]_{n+1}[/itex]. How can you just make a jump and say there is a [itex]\sqrt{5}[/itex] in there
 
Punkyc7 said:
How can you just make a jump and say there is a [itex]\sqrt{5}[/itex] in there

quadratic equation? :wink:
 
Ok I am not sure how you got there but this is what I have so farlet [itex]x_{n}[/itex] = [itex]f_{n+1}[/itex]/[itex]f_{n}[/itex] and let lim[itex](x_{n}[/itex])=L. From here is where I get stuck. I know that every sub sequence of a convergent sequence converges to the same number by some theorem, but I am not sure how that is at all helpful.

Would you do something like
[itex]x_{n}[/itex] = [itex]f_{n+1}[/itex]/[itex]f_{n}[/itex]
[itex]x_{n-1}[/itex] = [itex]f_{n}[/itex]/[itex]f_{n-1}[/itex]
 
put xn = xn-1
 
Ok so you get

= [itex]f_{n}[/itex]/[itex]f_{n-1}[/itex] = [itex]f_{n+1}[/itex]/[itex]f_{n}[/itex]

I think I see where you got the quadratic equation now

[itex]f_{n}[/itex] ^2= [itex]f_{n-1}[/itex] [itex]f_{n+1}[/itex]=[itex]f_{n}[/itex] ^2 - [itex]f_{n-1}([/itex] [itex]f_{n+1}[/itex])to use the quadratic equation is this [itex]f_{n-1}([/itex] [itex]f_{n+1}[/itex]) considered b or c?

and that doesn't look very pretty to solve...
 
Last edited:
erm :redface:

you'll also need fn+1 = … ? :wink:
 
Ok

[itex]f_{n+1}[/itex]=[itex]\frac{f_{n}}{f_{n-1}}[/itex]


[itex]f_{n}[/itex] ^2 - f[itex]_{n-1}[/itex]* [itex]\frac{f_{n}}{f_{n-1}}[/itex]

[itex]f_{n}[/itex] ^2 -[itex]f_{n}[/itex]=0

is that right?
 
  • #10
This is the Fibonacci sequence!

So fn+1 = ? :smile:
 
  • #11
oh so
f[itex]_{n+1}[/itex]= f[itex]_{n}[/itex] +f[itex]_{n-1}[/itex]


do I use that for f[itex]_{n+1}[/itex]
Also how do you know when to use what?
 
  • #12
Punkyc7 said:
do I use that for f[itex]_{n+1}[/itex]

yes … that should give you your quadratic equation :wink:
Also how do you know when to use what?

You're told it's a Fibonacci sequence, so you obviously have to use that information somewhere! :smile:

and now I'm off to bed :zzz:
 
  • #13
let [itex]x_{n}[/itex] = [itex]f_{n+1}[/itex]/[itex]f_{n}[/itex] and let lim[itex](x_{n}[/itex])=L. Since we know the sequence converges we can say

[itex]x_{n}[/itex] =[itex]x_{n-1}[/itex] Which Implies

= [itex]f_{n}[/itex]/[itex]f_{n-1}[/itex] = [itex]f_{n+1}[/itex]/[itex]f_{n}[/itex]

[itex]f_{n}[/itex] ^2= [itex]f_{n-1}[/itex] [itex]f_{n+1}[/itex]=[itex]f_{n}[/itex] ^2 - [itex]f_{n-1}([/itex] [itex]f_{n+1}[/itex])=0

[itex]f_{n}[/itex] ^2 - f[itex]_{n}[/itex]f[itex]_{n-1}[/itex]-f[itex]_{n-1}[/itex]^2=0How do you hammer this into the quadratic equation I am thinking the a=1 b=not sure c=not sure ? Also how do you get numbers from this when we don't have a single number?
 
Last edited:
  • #14
Punkyc7 said:
let [itex]x_{n}[/itex] = [itex]f_{n+1}[/itex]/[itex]f_{n}[/itex] and let lim[itex](x_{n}[/itex])=L. Since we know the sequence converges we can say

[itex]x_{n}[/itex] =[itex]x_{n-1}[/itex] Which Implies

= [itex]f_{n}[/itex]/[itex]f_{n-1}[/itex] = [itex]f_{n+1}[/itex]/[itex]f_{n}[/itex]
...which also equals L:

[itex]\frac{f_n}{f_{n-1}} = \frac{f_{n+1}}{f_n} = L[/itex]

Now take this portion:
[itex]\frac{f_{n+1}}{f_n} = L[/itex]

Replace the numerator with its equivalent, and then rewrite as a sum of two fractions. A substitution can be made, and you will end up with an expression on the left side with NO f's. Soon you will see a quadratic equation in terms of L. Solve for L.
 

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