# Fibonacci Look-a-like

• Jan 15th 2006, 01:37 PM
ThePerfectHacker
Fibonacci Look-a-like
Define a fibonacci series generated by {a,b} such as,
$\displaystyle u_1=a$
$\displaystyle u_2=b$
$\displaystyle u_{n+2}=u_{n+1}+u_{n}$
(The original fibonacci series is generated by {1,1})
($\displaystyle a\not =0$)

Then there are a number of interesting properties:
1)If $\displaystyle K$ is the finite continued fraction for $\displaystyle b/a$
then $\displaystyle \frac{u_{n+1}}{u_n}=[1;1,1...,1,K]$
Where the $\displaystyle 1$ appears $\displaystyle n-2$ times.

2)Thus, from here we have that $\displaystyle \lim_{n\rightarrow \infty}\frac{u_{n+1}}{u_n}=\psi$
(Thus, any fibonacci series generated by any two real numbers converges to the divine proportion).

3)The formula for $\displaystyle u_n$ is given by
$\displaystyle u_n=F(n-2)a+F(n-1)b$
where $\displaystyle F(n)$ is the n-th fibonacci number.
But by Binet's formula we have that,
for the n-th fibonacci number we can find a formula for $\displaystyle u_n$ but it is rather messy and will be omitted. Giving a second method for proving statement 2.
• Jan 20th 2006, 03:57 AM
I think you mean $\displaystyle \phi$ and not $\displaystyle \psi$. Very interesting, though.
• May 28th 2006, 02:55 PM
Natasha1
Quote:

Originally Posted by Treadstone 71
I think you mean $\displaystyle \phi$ and not $\displaystyle \psi$. Very interesting, though.

$\displaystyle \lim_{n\rightarrow \infty}\frac{u_{n+1}}{u_n}=\phi$
• May 28th 2006, 04:10 PM
ThePerfectHacker
Quote:

Originally Posted by Natasha1
$\displaystyle \lim_{n\rightarrow \infty}\frac{u_{n+1}}{u_n}=\phi$

That is not important ;)
• Jun 3rd 2006, 08:45 PM
Soroban
The proof of the second statement is rather neat.

Quote:

$\displaystyle 2)\;\;\lim_{n\to\infty}\frac{u_{n+1}}{u_n}\:=\: \phi$

The limit of the ratio of consecutive Fibonacci numbers is already established.

. . . . . $\displaystyle [1]\;\lim_{n\to\infty}\frac{F_{n+1}}{F_n}\:=\:\phi \qquad\qquad[2]\;\lim_{n\to\infty}\frac{F_n}{F_{n+1}} \:=\:\frac{1}{\phi}$

We have: $\displaystyle u_n\:=\:a\!\cdot\!F(n-2) + b\!\cdot\!F(n-1)$

The ratio is: $\displaystyle R\;=\;\frac{u_{n+1}}{u_n}\,=\,\frac{a\!\cdot\!F(n-1) + b\!\cdot\!F(n)}{a\!\cdot\!F(n-2) + b\!\cdot\!F(n-1)}$

Divide top and bottom by $\displaystyle F_{n-1}:\;\;R\;=\;\frac{a + b\!\cdot\!\frac{F(n)}{F_{n-1)}}}{a\!\cdot\!\frac{F(n-2)}{F(n-1)} + b}$

Take the limit: $\displaystyle \lim_{n\to\infty}R\;=\;\lim_{n\to\infty}\left( \frac{a + b\!\cdot\!\frac{F(n)}{F_{n-1)}}}{a\!\cdot\!\frac{F(n-2)}{F(n-1)} + b}\right)$

From [1] amd [2], this becomes: $\displaystyle \frac{a + b\!\cdot\!\phi}{a\!\cdot\!\frac{1}{\phi} + b}$

Multiply top and bottom by $\displaystyle \phi:\;\;\frac{\phi(a + b\!\cdot\!\phi)}{a + b\!\cdot\!\phi}\;=\;\phi$

• Jun 4th 2006, 05:37 AM
ThePerfectHacker
Quote:

Originally Posted by Soroban
The proof of the second statement is rather neat.

[size=3]

My proof is simpler.
Because of condition 1), since this countinued fractions converges to $\displaystyle [1:1,1,...]=\phi$.
• Jun 4th 2006, 12:09 PM
Soroban
Hello, ThePerfectHacker!

Quote:

My proof is simpler.
Of course it is . . . I wasn't claiming otherwise.

I was demonstrating a straight algebraic approach to the limit
. . for those not familiar with continued fractions.

Please understand: I never ever compete with other posters here.
I may post an alternate approach or even an improvement
. . . but never with a snicker or a sneer, implied or otherwise.