Fibonacci Look-a-like

ThePerfectHacker · #1 $Old$ January 15th, 2006, 02:37 PM

Define a fibonacci series generated by {a,b} such as,
$u_1=a$
$u_2=b$
$u_{n+2}=u_{n+1}+u_{n}$
(The original fibonacci series is generated by {1,1})
( $a\not =0$ )

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

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

3)The formula for $u_n$ is given by
$u_n=F(n-2)a+F(n-1)b$
where $F(n)$ 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 $u_n$ but it is rather messy and will be omitted. Giving a second method for proving statement 2.

Treadstone 71 · #2 $Old$ January 20th, 2006, 04:57 AM

I think you mean $\phi$ and not $\psi$ . Very interesting, though.

Natasha1 · #3 $Old$ May 28th, 2006, 03:55 PM

Quote:

Originally Posted by Treadstone 71

I think you mean $\phi$ and not $\psi$ . Very interesting, though.

$\lim_{n\rightarrow \infty}\frac{u_{n+1}}{u_n}=\phi$

ThePerfectHacker · #4 $Old$ May 28th, 2006, 05:10 PM

Quote:

Originally Posted by Natasha1

$\lim_{n\rightarrow \infty}\frac{u_{n+1}}{u_n}=\phi$

That is not important

Soroban · #5 $Old$ June 3rd, 2006, 09:45 PM

The proof of the second statement is rather neat.

Quote:

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

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

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

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

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

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

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

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

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

ThePerfectHacker · #6 $Old$ June 4th, 2006, 06:37 AM

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 $[1:1,1,...]=\phi$ .

Soroban · #7 $Old$ June 4th, 2006, 01:09 PM

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.