In How Many Transformations, Daniel came to the realization that linear recurrence of the form

$\left\{\begin{array}{l}x_{n+1}=a_1x_n+b_1y_n,\\ y_{n+1}=a_2x_n+b_2y_n,\end{array}\right.$

could easily be solved by setting up the matrix interpretation, and then diagonalize the matrix (assuming that's possible) which would allow us quick exponentiation, and hence obtain the Nth term directly.

We've seen this in the context of the Fast Fibonacci Transform. Specifically, set $x_n = f_{n+1}$ and $y_n = f_n$, and you get the system of equations

$\begin{cases} x_{n+1} = 1x_n + 1y_n \\ y_{n+1} = 1x_n + 0 y_n \\ \end{cases}$

Hence, $\begin{pmatrix} x_n \\ y_n \\ \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 0 \\ \end{pmatrix}^n \begin{pmatrix} 1 \\ 1 \\ \end{pmatrix}$

Using this, show the following:

1. Performing the eigenvalue decomposition, prove Binet's formula.

2. Using only matrix properties, conclude that

$f_{2n+1} = f_{n+1} ^2 + f_n ^2.$

Hint: $A^{2n} = A^n \times A^n$.

3. Find a similar formula for $f_{2n}$.

4. Express $f_{3n}$ in terms of $f_{n}, f_{n+1}$.

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## Comments

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TopNewestThis kinda gave away the method one uses to solve my problem (if you don't use eigenvalue decomposition). However, I would still like to see how eigenvalue decomposition works, so it would be nice if a brilliant user could post a solution to the problems suggested in this note. Thanks.

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Added a solution. It's pretty basic matrix manipulation, and would have been the approach that I thought you used.

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Let $\rho =\frac{1+\sqrt{5}}2$ and ${\overline \rho} = \frac{1-\sqrt{5}}2$. Let $D = \begin{pmatrix} \rho&0 \\ 0&{\overline \rho} \end{pmatrix}$. Let $A = \begin{pmatrix} \rho&{\overline \rho} \\ 1&1 \end{pmatrix}$. Then $\begin{pmatrix} 1&1\\1&0 \end{pmatrix} = ADA^{-1}$.

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(1) So $\begin{pmatrix} f_{n+1} \\ f_n \end{pmatrix} = (ADA^{-1})^n \begin{pmatrix} 1\\0 \end{pmatrix} = AD^nA^{-1} \begin{pmatrix} 1\\0 \end{pmatrix}$. Some painful computations yield $\begin{pmatrix} f_{n+1}\\f_n \end{pmatrix} = \frac1{\sqrt{5}} \begin{pmatrix} \rho^{n+1}-{\overline \rho}^{n+1} \\ \rho^n - {\overline \rho}^n \end{pmatrix}$; Binet's formula falls out of this.

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(2) and (3): let $M = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} = ADA^{-1}$. Then $M^n \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} f_{n+1} \\ f_n \end{pmatrix}$, and $M^n \begin{pmatrix} 0 \\ 1 \end{pmatrix} = M^{n-1} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} f_n \\ f_{n-1} \end{pmatrix}$. So $M^n = \begin{pmatrix} f_{n+1} & f_n \\ f_n & f_{n-1} \end{pmatrix}$.

Now then, $\begin{pmatrix} f_{2n+1} \\ f_{2n} \end{pmatrix} = M^{2n} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = M^n \begin{pmatrix} f_{n+1} \\ f_n \end{pmatrix} = \begin{pmatrix} f_{n+1} & f_n \\ f_n & f_{n-1} \end{pmatrix} \begin{pmatrix} f_{n+1} \\ f_n \end{pmatrix}$.

So we get $\begin{pmatrix} f_{2n+1} \\ f_{2n} \end{pmatrix} = \begin{pmatrix} f_{n+1}^2 + f_n^2 \\ f_{n+1}f_n + f_n f_{n-1} \end{pmatrix}$, and the formulas we want can be read off from there.

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$f_{3n} = f_n(3f_{n+1}^2-3f_nf_{n+1}+2f_n^2)$. (Basically the same process as the previous, but I had to substitute $f_{n-1} = f_{n+1}-f_n$ in some places.)

(4) Similar computations giveLog in to reply

I wish I saw this earlier

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@Trevor B. @Trevor Arashiro @Agnishom Chattopadhyay @Michael Mendrin @Sharky Kesa

Could you guys help fill out the Fast Fibonacci Transform Wiki Page? Thanks!

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Okay! Will go through this

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For

Fast Fibonacci Transform, here's a python code for calculating fibonacci numbers:Log in to reply