# This note has been used to help create the Fast Fibonacci Transform wiki

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 $$f_{n}, f_{n+1}$$.

Note by Calvin Lin
4 years ago

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This 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.

- 4 years ago

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

Staff - 4 years ago

For Fast Fibonacci Transform, here's a python code for calculating fibonacci numbers:

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 # (Public) Returns F(n). def fibonacci(n): if n < 0: raise ValueError("Negative arguments not implemented") return _fib(n)[0] # (Private) Returns the tuple (F(n), F(n+1)). def _fib(n): if n == 0: return (0, 1) else: a, b = _fib(n // 2) c = a * (b * 2 - a) d = a * a + b * b if n % 2 == 0: return (c, d) else: return (d, c + d) 

- 3 years, 2 months ago

I wish I saw this earlier

- 3 years, 2 months ago

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

Staff - 3 years, 2 months ago

Okay! Will go through this

- 3 years, 2 months ago

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 $$\begin{pmatrix} 1&1\\1&0 \end{pmatrix} = ADA^{-1}$$.

- 4 years ago

(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.

- 4 years ago

(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.

- 4 years ago

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

- 4 years ago