(50g) Nth Fibonacci Number
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02-27-2015, 05:10 AM
Post: #12
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RE: (50g) Nth Fibonacci Number
(02-26-2015 01:42 PM)rprosperi Wrote: So, can you explain that? For \(x_{n+2}=x_n+x_{n+1}\) you can define a 2nd sequence \(y_n=x_{n+1}\). Thus the relation becomes: \(y_{n+1}=x_{n+2}=x_n+y_n\). Now you combine both sequences into a vector: \(z_n=\begin{bmatrix} x_n \\ y_n \end{bmatrix}\) This allows to merge both equations into a single one: \(z_{n+1}=\begin{bmatrix} x_{n+1} \\ y_{n+1} \end{bmatrix}=\begin{bmatrix} y_n \\ x_n+y_n \end{bmatrix}=\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}\cdot\begin{bmatrix} x_n \\ y_n \end{bmatrix}=M\cdot z_n\) But this is just the recursive definition of a geometric sequence leading to: \(z_n=M^n\cdot z_0\) with \(z_0=\begin{bmatrix} 1 \\ 0 \end{bmatrix}\) HTH Thomas |
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