Fibonacci Computation and Recurrence

Fibonacci Numbers

...linear algebra perspectives

$F_0 = 0, F_1 = 1$

$F_{n+2} = F_{n+1} + F_n$ for $n = 0,1,2,\ldots$

$F_0, F_1, F_2, \ldots$

#1. Fibonacci Numbers, Quickly

The Fibonacci numbers were first described in Indian mathematics, as early as 200 BC in work by Pingala on enumerating possible patterns of Sanskrit poetry formed from syllables of two lengths.

Hemachandra (c. 1150) is credited with knowledge of the sequence as well, writing that "the sum of the last and the one before the last is the number ... of the next mātrā-vṛtta."

$\underbrace{\left(\begin{array}{cc}{\color{IndianRed}1} & {\color{IndianRed}1} \\{\color{DodgerBlue}1} & {\color{DodgerBlue}0}\end{array}\right)}_{M}\left(\begin{array}{c}F_{n+1} \\F_n\end{array}\right) = \left(\begin{array}{c}F_{n+2} \\F_{n+1}\end{array}\right)$

$M\left(\begin{array}{c}F_{n+1} \\F_n\end{array}\right)=\left(\begin{array}{c}F_{n+2} \\F_{n+1}\end{array}\right)$

#1. Fibonacci Numbers, Quickly

$M\left(\begin{array}{c}1 \\0\end{array}\right)=\left(\begin{array}{c}F_{2} \\F_{1}\end{array}\right)$

$M\left(\begin{array}{c}F_{n+1} \\F_n\end{array}\right)=\left(\begin{array}{c}F_{n+2} \\F_{n+1}\end{array}\right)$

#1. Fibonacci Numbers, Quickly

$M\left(\begin{array}{c}1 \\0\end{array}\right)=\left(\begin{array}{c}F_{2} \\F_{1}\end{array}\right)$

#1. Fibonacci Numbers, Quickly

$M\left(\begin{array}{c}1 \\0\end{array}\right)=\left(\begin{array}{c}F_{2} \\F_{1}\end{array}\right)$

$M\left(\begin{array}{c}F_2 \\F_1\end{array}\right)=\left(\begin{array}{c}F_{3} \\F_{2}\end{array}\right)$

#1. Fibonacci Numbers, Quickly

$M\left(\begin{array}{c}1 \\0\end{array}\right)=\left(\begin{array}{c}F_{2} \\F_{1}\end{array}\right)$

$M{\color{Orange}\left(\begin{array}{c}F_2 \\F_1\end{array}\right)}=\left(\begin{array}{c}F_{3} \\F_{2}\end{array}\right)$

#1. Fibonacci Numbers, Quickly

$M\left(\begin{array}{c}1 \\0\end{array}\right)=\left(\begin{array}{c}F_{2} \\F_{1}\end{array}\right)$

$M \cdot {\color{Orange}M \left(\begin{array}{c}1 \\0\end{array}\right)}=\left(\begin{array}{c}F_{3} \\F_{2}\end{array}\right)$

#1. Fibonacci Numbers, Quickly

$M\left(\begin{array}{c}1 \\0\end{array}\right)=\left(\begin{array}{c}F_{2} \\F_{1}\end{array}\right)$

$M^2\left(\begin{array}{c}1 \\0\end{array}\right)=\left(\begin{array}{c}F_{3} \\F_{2}\end{array}\right)$

#1. Fibonacci Numbers, Quickly

$M\left(\begin{array}{c}1 \\0\end{array}\right)=\left(\begin{array}{c}F_{2} \\F_{1}\end{array}\right)$

$M^n\left(\begin{array}{c}1 \\0\end{array}\right)=\left(\begin{array}{c}F_{n+1} \\F_{n}\end{array}\right)$

#1. Fibonacci Numbers, Quickly

$\underbrace{M \longrightarrow M^{2} \longrightarrow M^{4} \longrightarrow M^{8} \longrightarrow M^{16} \longrightarrow M^{32} \longrightarrow \cdots}_{}$

repeated squaring

$\mathcal{O}(\log_2 n)$ multiplications of $2 \times 2$ matrices.

$M^n = M^{(2^{k_1} + 2^{k_2} + \cdots + 2^{k_\ell})} = M^{2^{k_1}} M^{2^{k_2}} \cdots M^{2^{k_\ell}}$

$\mathcal{O}(\log_2 n)$ multiplications of $2 \times 2$ matrices.

#1. Fibonacci Numbers, Quickly

#2. Fibonacci Numbers, The Formula

Consider the vector space of all infinite sequences:

$(u_0, u_1, u_2, \ldots )$

...with coordinate-wise addition and multiplication by real numbers.

In this space we define a subspace $\mathcal{W}$ of all sequences satisfying the equation:

$u_{n+2} = u_{n+1}+u_n$ for all $n = 0, 1, ...$

Easy to verify: $\mathcal{W}$ is a subspace.

Claim: $dim(\mathcal{W}) = 2$ .

Observe that $(0,1,1,2,3,\ldots)$ and $(1,0,1,1,2,\ldots)$ is a basis.

$(\tau^0,\tau^1,\tau^2,\tau^3,\cdots) \in \mathcal{W}$

$\tau^k = \tau^{k-1}+\tau^{k-2}$

$\tau^2 = \tau+1$

$\tau_{1,2}=(1 \pm \sqrt{5}) / 2$

$\alpha=(1 + \sqrt{5}) / 2$

$\beta=(1 - \sqrt{5}) / 2$

#2. Fibonacci Numbers, The Formula

$F = c \cdot (\alpha^0,\alpha^1,\alpha^2,\alpha^3,\cdots) + d \cdot (\beta^0,\beta^1,\beta^2,\beta^3,\cdots)$

$0 = c + d$

$1 = c\alpha + d\beta$

$d = 1/(\beta -\alpha)$

$c = 1/(\alpha -\beta)$

#2. Fibonacci Numbers, The Formula

$F_n = c \cdot \alpha^n + d \cdot \beta^n$

$d = 1/(\beta -\alpha)$

$c = 1/(\alpha -\beta)$

$\alpha=(1 + \sqrt{5}) / 2$

$\beta=(1 - \sqrt{5}) / 2$

$F_n = \frac{1}{\sqrt{5}} \cdot \left[ \left( \frac{1+\sqrt{5}}{2} \right)^n - \left( \frac{1-\sqrt{5}}{2} \right)^n \right]$

#2. Fibonacci Numbers, The Formula

$F_n = \left\lfloor \frac{1}{\sqrt{5}} \cdot \left( \frac{1+\sqrt{5}}{2} \right)^n \right\rfloor$

Exercise 1. Show that:

Exercise 2. Use this method to work out a closed form for:

$y_{n+2}=2 y_{n+1}-y_n$

(Source: generatingfunctionology, Wilf;
h/t Matthew Drescher and John Azariah for a fun Twitter discussion on this.)

Fibonacci Numbers ...linear algebra perspectives