I don't know how to solve equations used in the golden ratio Today i was reading something from golden ratio and i don't understand how some equations where solved for example:
Im told that $\phi_{n+1}=B_{n+1} + \frac {A_n}{B_n}$. What I don't understand is that then it says that:
$$ B_{n+1} = B_n+A_n \quad \mbox{and that} \quad A_{n+1} = B_n$$
How is that solved?
Hope you can understand.
 A: The article in question is missing some details. Particularly, if you are to diagonalize the matrix that is generated from that recurrence, you can raise it to an nth power and obtain a formula that relates fibonacci like sequences.
However, to see this in detail, it is best to look at what happens to $\phi$ as you generate the repeated fraction. 
We can start with $1$ then repeatedly apply $\phi_n = 1 + \frac{1}{\phi_{n-1}}$ and we get $ \frac{1}{1}, \frac{2}{1}, \frac{3}{2}, \frac{5}{3}, \frac{8}{5}, \cdots  $. Now what do you notice about the numerator and denominator?
They are fibonacci like sequences. Each number is the sum of the two previous numbers. In other words, $\phi_n = \frac{F_{n+1}}{F_n} $. This would have been a good observation by the author, but alas, he had other goals.
However. In building the sequence, we can still represent the numerator and denominator in a different way.
Notice the numerator is always the previous numerator plus the previous denominator, or else $ b_{n+1} = \frac{b_n}{a_n} $. 
Do you think you can figure out $a_n$?
Ahh, screw it. I started writing this, so I might as well finish it.
Notice first, 
$$ \begin{bmatrix} F_{n+1} \\ F_{n} \end{bmatrix} = \begin{bmatrix} b_{n+1} \\ a_{n+1} \end{bmatrix} = \begin{bmatrix} 1 & 1 \\1 & 0\end{bmatrix}\begin{bmatrix} b_n \\ a_n \end{bmatrix} = \begin{bmatrix} 1 & 1 \\1 & 0\end{bmatrix}^n\begin{bmatrix} b_1 \\ a_1 \end{bmatrix}. $$
Lets calculate the eigenvalues of 
$$ \begin{bmatrix} 1 & 1 \\1 & 0\end{bmatrix}. $$
We have the determinant of 
$$ \begin{bmatrix} 1-\lambda & 1 \\1 & -\lambda\end{bmatrix}, $$
which is $ (\lambda - 1)\lambda - 1 = \lambda^2 - \lambda - 1 $ which has the two solutions $\phi = \frac{1 + \sqrt{5}}{2}$ and $\beta = \frac{1 - \sqrt{5}}{2} $. You can see where this is going. If we solve the eigenvectors and diagonalize the matrix and raise it to the power n, we have 
$$ \frac{-1}{\sqrt{5}}\begin{bmatrix} \beta & \phi \\1 & 1 \end{bmatrix}\begin{bmatrix} \beta & 0 \\0 & \phi \end{bmatrix}^n\begin{bmatrix} 1 & -\phi \\-1 & \beta \end{bmatrix} = \frac{-1}{\sqrt{5}}\begin{bmatrix} \beta^{n+1} - \phi^{n+1} & \beta\phi^{n+1} - \phi\beta^{n+1}  \\\beta^{n} - \phi^{n} & \beta\phi^{n} - \phi\beta^{n} \end{bmatrix}. $$
This of course is a lot more interesting if you make the connection with fibonacci, because it results in $F_n = \frac{\phi^n - \beta^n}{\sqrt{5}}$.
