In his work Liber Abbaci, published in 1202, Leonardo Fibonacci of Pisa, Italy, speculated on the reproduction of rabbits: How many pairs of rabbits will be produced in a year beginning with a single pair, if every month each pair bears a new pair which become productive from the second month on? The answer to his question is contained in a sequence known as a Fibonacci sequence.
After Each Month Start n = 0 1 2 3 4 5 6 7 8 9 10 11 12 Adult pairs 1 1 2 3 5 8 13 21 … Baby pairs 0 1 1 2 3 5 8 13 … Total pairs 1 2 3 5 8 13 21 34 … \begin{matrix}
\text{} & \text{After Each Month}\\
\text{}{\text { Start } n=} & \text{}{ \begin{array}{llllllllllll}{0} & {1} & {2} & {3} & {4} & {5} & {6} & {7} & {8} & {9} & {10} & {11} & {12}\end{array}}\\
\text{Adult pairs } & \text{}{\begin{array}{llllllllll}{1} & {1} & {2} & {3} & {5} & {8} & {13} & {21} & {\dots}\end{array}}\\
\text{Baby pairs } & \text{}{0 \quad 1 \quad 1 \quad 2 \quad 3 \quad 5 \quad 8 \quad 13 \quad \ldots}\\
\text{Total pairs } & \text{}{\begin{array}{lllllllll}{1} & {2} & {3} & {5} & {8} & {13} & {21} & {34} & {\dots}\end{array}}\\
\end{matrix}
Start n = Adult pairs Baby pairs Total pairs After Each Month 0 1 2 3 4 5 6 7 8 9 10 11 12 1 1 2 3 5 8 13 21 … 0 1 1 2 3 5 8 13 … 1 2 3 5 8 13 21 34 …
Each of the three rows describing rabbit pairs is a Fibonacci sequence and can be defined recursively by a second-order difference equation x n = x n − 2 + x n − 1 , n = 2 , 3 , … x_{n}=x_{n-2}+x_{n-1}, n=2,3, \ldots x n = x n − 2 + x n − 1 , n = 2 , 3 , … x 0 x_0 x 0 x 1 x_1 x 1 x 0 = 1 , x 1 = 1 x_{0}=1, x_{1}=1 x 0 = 1 , x 1 = 1 y n − 1 = x n − 2 y_{n-1}=x_{n-2} y n − 1 = x n − 2 y n = x n − 1 y_{n}=x_{n-1} y n = x n − 1
x n = x n − 1 + y n − 1 y n = x n − 1 \begin{array}{l}{x_{n}=x_{n-1}+y_{n-1}} \\ {y_{n}=x_{n-1}}\end{array}
x n = x n − 1 + y n − 1 y n = x n − 1
. Write this system in the matrix form X n = A X n − 1 \mathbf{X}_{n}=\mathbf{A} \mathbf{X}_{n-1} X n = A X n − 1 n = 2 , 3 , … n=2,3, \ldots n = 2 , 3 , …
A m = ( λ 2 λ 1 m − λ 1 λ 2 m + λ 2 m − λ 1 m λ 2 − λ 1 λ 2 m − λ 1 m λ 2 − λ 1 λ 2 m − λ 1 m λ 2 − λ 1 λ 2 λ 1 m − λ 1 λ 2 m λ 2 − λ 1 ) \mathbf{A}^{m}=\left(\begin{array}{cc}{\frac{\lambda_{2} \lambda_{1}^{m}-\lambda_{1} \lambda_{2}^{m}+\lambda_{2}^{m}-\lambda_{1}^{m}}{\lambda_{2}-\lambda_{1}}} & {\frac{\lambda_{2}^{m}-\lambda_{1}^{m}}{\lambda_{2}-\lambda_{1}}} \\ {\frac{\lambda_{2}^{m}-\lambda_{1}^{m}}{\lambda_{2}-\lambda_{1}}} & {\frac{\lambda_{2} \lambda_{1}^{m}-\lambda_{1} \lambda_{2}^{m}}{\lambda_{2}-\lambda_{1}}}\end{array}\right)
A m = ( λ 2 − λ 1 λ 2 λ 1 m − λ 1 λ 2 m + λ 2 m − λ 1 m λ 2 − λ 1 λ 2 m − λ 1 m λ 2 − λ 1 λ 2 m − λ 1 m λ 2 − λ 1 λ 2 λ 1 m − λ 1 λ 2 m )
or
A m = 1 2 m + 1 5 ( ( 1 + 5 ) m + 1 − ( 1 − 5 ) m + 1 2 ( 1 + 5 ) m − 2 ( 1 − 5 ) m 2 ( 1 + 5 ) m − 2 ( 1 − 5 ) m ( 1 + 5 ) ( 1 − 5 ) m − ( 1 − 5 ) ( 1 + 5 ) m ) \mathbf{A}^{m}=\frac{1}{2^{m+1} \sqrt{5}}\left(\begin{array}{cc}{(1+\sqrt{5})^{m+1}-(1-\sqrt{5})^{m+1}} & {2(1+\sqrt{5})^{m}-2(1-\sqrt{5})^{m}} \\ {2(1+\sqrt{5})^{m}-2(1-\sqrt{5})^{m}} & {(1+\sqrt{5})(1-\sqrt{5})^{m}-(1-\sqrt{5})(1+\sqrt{5})^{m}}\end{array}\right)
A m = 2 m + 1 5 1 ( ( 1 + 5 ) m + 1 − ( 1 − 5 ) m + 1 2 ( 1 + 5 ) m − 2 ( 1 − 5 ) m 2 ( 1 + 5 ) m − 2 ( 1 − 5 ) m ( 1 + 5 ) ( 1 − 5 ) m − ( 1 − 5 ) ( 1 + 5 ) m )
, where λ 1 = 1 2 ( 1 − 5 ) \lambda_{1}=\frac{1}{2}(1-\sqrt{5}) λ 1 = 2 1 ( 1 − 5 ) λ 2 = 1 2 ( 1 + 5 ) \lambda_{2}=\frac{1}{2}(1+\sqrt{5}) λ 2 = 2 1 ( 1 + 5 ) X n = A n − 1 X 1 \mathbf{X}_{n}=\mathbf{A}^{n-1} \mathbf{X}_{1} X n = A n − 1 X 1
