This exercise is a continuation of the previous one.

The Lucas numbers \(L_n\) are defined by the same relationship as the Fibonacci numbers: \(L_{n+2}=L_{n+1}+L_n\text{.}\) However, we begin with \(L_0=2\) and \(L_1=1\text{,}\) which leads to the sequence \(2,1,3,4,7,11,\ldots\text{.}\)

  1. As before, form the vector \(\xvec_n=\twovec{L_{n+1}}{L_n}\) so that \(\xvec_{n+1}=A\xvec_n\text{.}\) Express \(\xvec_0\) as a linear combination of \(\vvec_1\) and \(\vvec_2\text{,}\) eigenvectors of \(A\text{.}\)

  2. Explain why

    \begin{equation*} L_n = \left(\frac{1+\sqrt{5}}{2}\right)^n + \left(\frac{1-\sqrt{5}}{2}\right)^n\text{.} \end{equation*}
  3. Explain why \(L_n\) is the closest integer to \(\phi^n\) when \(n\) is large.

  4. Use this observation to find \(L_{20}\text{.}\)