# Could someone check the attached document for mistakes?

• May 27th 2006, 02:58 PM
Natasha1
Could someone check the attached document for mistakes?
I would love someone to please have a look at the Fibonacci document attached and check the proof of the golden ratio and tell me if there are any mistakes.

Many thanks

Nat
• May 27th 2006, 05:39 PM
ThePerfectHacker
That document is not working on my computer.
• May 28th 2006, 01:47 AM
Natasha1
Quote:

Originally Posted by ThePerfectHacker
That document is not working on my computer.

Here is the document (as screenshots) could this be corrected please :-)
• May 28th 2006, 04:01 AM
rgep
After a quick read-through I think it's OK. There are many many ways of showing that the ratio of successive Fibonacci numbers tends to the Golden Ratio.

Let's establish some notation. Let $\phi = \frac{1+\sqrt5}2$ be the Golden Ratio and $\bar\phi = \frac{1-\sqrt5}2$ the conjugate, so that $\phi,\bar\phi$ are the roots of $\Phi^2 - \Phi - 1$ with $\phi+\bar\phi=1$ and $\phi\bar\phi = -1$.

One way is to find an explicit expression for F_n. A recurrence relation of the form $x_{n+1} = Px_n + Qx_{n-1}$ has a general solution of the form $A\alpha^n + B\beta^n$ where $\alpha,\beta$ are the roots of the auxiliary polynomial $f(t) = t^2 - Pt - Q$, and A,B are determined by two initial values. There is an exception when the auxiliary has repeated roots, in which case the general solution is $(An+B)\alpha^n$: that case doesn't apply here.

From this is is easy to derive $F_n = \frac{\phi^n-\bar\phi^n}{\phi-\bar\phi}$, and if you don't want to rely on the assertions just made about the theory of recurrence relations, you can prove it directly by induction (true for cases n=0 and 1, and check that the expression given on the RHS of that formula has the same recurrence relation as F_n). In turn, it's now easy to check that $F_{n+1}/F_n \rightarrow \phi$.

Other ways are based on the observation you already made, that if the sequence of ratios tends to a limit at all, that limit must be the Golden Ratio.
So you could directly consider $e_n = F_{n+1}/F_n - \phi$. It's not too hard to show that $e_{n+1} = F_{n+2}/F_{n+1} - \phi$ = $1 + F_n/F_{n+1} - \phi = \frac{1}{e_n+\phi} + 1 - \phi$ = $\frac{1}{e_n+\phi} - \frac{1}{\phi}$ = $\frac{\phi - e_n - \phi}{(e_n+\phi)\phi} = \frac{-e_n}{(e_n+\phi)\phi}$. In magnitude this is less than $\frac{-e_n}{2}$ at least when n is larger than 6, say. That is, the difference is alternating in sign and diminishing in magnitude by a factor of at least 2 each step, which is easily enough for convergence.
• May 28th 2006, 09:34 AM
ThePerfectHacker
The most difficult part of the proof is showing that the sequence is indeed convergent. Then once you have that you can manipulate it to get your result.

This is why I asked you in your other post if your studied countinued fractions, because all countinued fractions are convergent, thus that messy part is already proven.
---------
This is perhaps an easier way, if you would like. There is a formula for the fibonacci sequence in terms of an exponention function due to Binet. I can provide a proof if you would like.
• May 28th 2006, 09:42 AM
ThePerfectHacker
Since you are so interesting in fibonacci sequence. I once was playing around with this sequence and found something interesting Look Here
• May 28th 2006, 02:41 PM
Natasha1
Quote:

Originally Posted by ThePerfectHacker
Since you are so interesting in fibonacci sequence. I once was playing around with this sequence and found something interesting Look Here

Thanks for the offer but as I have never studied Binet, therefore I don't think I should use it. I'm trying hard on the other thread though the one on 'Proof by induction' to get an answer. To fill the gaps basically and it is due in tomorrow :-(
• May 28th 2006, 02:44 PM
ThePerfectHacker
Quote:

Originally Posted by Natasha1
...never studied Binet, ....

I am assuming you think that 'Binet' means some kind of theory/branch in math. No it is just a formula involving fibonacci numbers, (and its proof is based on induction). It is very simple nothing complicated if that was what frightened you.
• May 28th 2006, 03:07 PM
Natasha1
Well in that case, and if you don't mind I would love to see it :-). Thank you!
• May 28th 2006, 03:53 PM
ThePerfectHacker
Quote:

Originally Posted by Natasha1
Well in that case, and if you don't mind I would love to see it :-). Thank you!

As you know the 'divine proprtion' satisfies,
$x^2=x+1$
Now this equation has two roots,
$\phi=\frac{1+\sqrt{5}}{2}$ called 'the divine proprtion'
and,
$\psi=\frac{1-\sqrt{5}}{2}$ no special interest.
----
Therefore,
$\phi^2=\phi+1$
Multiply both sides by $\phi$ thus,
$\phi^3=\phi^2+\phi$
but, $\phi^2=\phi+1$ thus,
$\phi^3=(\phi+1)+\phi=2\phi+1$
Multiply both sides by $\phi$ thus,
$\phi^4=2\phi^2+\phi$
but, $\phi^2=\phi+1$ thus,
$\phi^4=2(\phi+1)+\phi=3\phi+2$
Multiply both sides by $\phi$ thus,
$\phi^5=3\phi^2+2\phi$
but, $\phi^2=\phi+1$ thus,
$\phi^5=3(\phi+1)+2\phi=5\phi+3$ thus,
-----
Did you notice something?
Firstly that,
$\phi^n=A\phi+B$ for some integers $A,B$, that you probably saw. But that you see that,
$A=F(n)\mbox{ and }B=F(n-1)$?
I will leave that as an excerise to you that prove that this fibonacci pattern still holds (this is induction type proof).

Similarly we have that,
$\psi^n=F(n)\psi+F(n-1)$
because it is developed in the same manner as for $\phi$.
---------
As a result we have two equations, for $n>0$.
$\left\{ \begin{array}{c}\phi^n=F(n)\phi+F(n-1) \mbox{ (1)}\\\psi^n=F(n)\psi+F(n-1)\mbox{ (2)}$

From (1) subtract (2) to get,
$\phi^n-\psi^n=F(n)(\phi-\psi)$
Thus,
$F(n)=\frac{\phi^n-\psi^n}{\phi-\psi}$
Thus,
$F(n)=\frac{ \left(\frac{1+\sqrt{5}}{2}\right)^n - \left( \frac{1-\sqrt{5}}{2}\right)^n}{\sqrt{5}}$-this is called binet's formula.

Anyways, you wish to demonstrate that the fibonaccis converge to $\phi$ you can do this by showing that,
$\lim_{n\to\infty}\frac{F(n+1)}{F(n)}=\frac{\phi^{n +1}-\psi^{n+1}}{\phi^n-\psi^n}=\phi$

Though this looks like a complicated limit it is not. Divide the numerator and denominator by $\phi^n$ thus,
$\lim_{n\to\infty}\frac{\phi-\psi\left(\frac{\psi}{\phi}\right)^n}{1-\left(\frac{\psi}{\phi}\right)^n}$
But,
$-1<\frac{\psi}{\phi}<1$ therefore as $n\to\infty}$ we have that, $\left(\frac{\psi}{\phi}\right)^n=0$
Thus,
$\lim_{n\to\infty}\frac{\phi-\psi\left(\frac{\psi}{\phi}\right)^n}{1-\left(\frac{\psi}{\phi}\right)^n}=\frac{\phi-\psi\cdot 0}{1-0}=\phi$

$\mathbb{Q}.\mathbb{E}.\mathbb{D}$
• May 28th 2006, 04:01 PM
Natasha1
Quote:

Originally Posted by ThePerfectHacker
As you know the 'divine proprtion' satisfies,
$x^2=x+1$
Now this equation has two roots,
$\phi=\frac{1+\sqrt{5}}{2}$ called 'the divine proprtion'
and,
$\psi=\frac{1-\sqrt{5}}{2}$ no special interest.
----
Therefore,
$\phi^2=\phi+1$
Multiply both sides by $\phi$ thus,
$\phi^3=\phi^2+\phi$
but, $\phi^2=\phi+1$ thus,
$\phi^3=(\phi+1)+\phi=2\phi+1$
Multiply both sides by $\phi$ thus,
$\phi^4=2\phi^2+\phi$
but, $\phi^2=\phi+1$ thus,
$\phi^4=2(\phi+1)+\phi=3\phi+2$
Multiply both sides by $\phi$ thus,
$\phi^5=3\phi^2+2\phi$
but, $\phi^2=\phi+1$ thus,
$\phi^5=3(\phi+1)+2\phi=5\phi+3$ thus,
-----
Did you notice something?
Firstly that,
$\phi^n=A\phi+B$ for some integers $A,B$, that you probably saw. But that you see that,
$A=F(n)\mbox{ and }B=F(n-1)$?
I will leave that as an excerise to you that prove that this fibonacci pattern still holds (this is induction type proof).

Similarly we have that,
$\psi^n=F(n)\psi+F(n-1)$
because it is developed in the same manner as for $\phi$.
---------
As a result we have two equations, for $n>0$.
$\left\{ \begin{array}{c}\phi^n=F(n)\phi+F(n-1) \mbox{ (1)}\\\psi^n=F(n)\psi+F(n-1)\mbox{ (2)}$

From (1) subtract (2) to get,
$\phi^n-\psi^n=F(n)(\phi-\psi)$
Thus,
$F(n)=\frac{\phi^n-\psi^n}{\phi-\psi}$
Thus,
$F(n)=\frac{ \left(\frac{1+\sqrt{5}}{2}\right)^n - \left( \frac{1-\sqrt{5}}{2}\right)^n}{\sqrt{5}}$-this is called binet's formula.

Anyways, you wish to demonstrate that the fibonaccis converge to $\phi$ you can do this by showing that,
$\lim_{n\to\infty}\frac{F(n+1)}{F(n)}=\frac{\phi^{n +1}-\psi^{n+1}}{\phi^n-\psi^n}=\phi$

Though this looks like a complicated limit it is not. Divide the numerator and denominator by $\phi^n$ thus,
$\lim_{n\to\infty}\frac{\phi-\psi\left(\frac{\psi}{\phi}\right)^n}{1-\left(\frac{\psi}{\phi}\right)^n}$
But,
$-1<\frac{\psi}{\phi}<1$ therefore as $n\to\infty}$ we have that, $\left(\frac{\psi}{\phi}\right)^n=0$
Thus,
$\lim_{n\to\infty}\frac{\phi-\psi\left(\frac{\psi}{\phi}\right)^n}{1-\left(\frac{\psi}{\phi}\right)^n}=\frac{\phi-\psi\cdot 0}{1-0}=\phi$

$\mathbb{Q}.\mathbb{E}.\mathbb{D}$

That's very neat. Thanks so much :-)