+0  
 
0
88
1
avatar

The Fibonacci sequence, with \(F_0 = 0\), \(F_1 = 1\) and \(F_n = F_{n - 2} + F_{n - 1\), had a closed form \(F_n = \frac{1}{\sqrt{5}} \left( \phi^n - \widehat{\phi}^n \right),\) where \(\phi = \frac{1 + \sqrt{5}}{2} \; \text{and} \; \widehat{\phi} = \frac{1 - \sqrt{5}}{2}.\) The Lucas numbers are defined in a similar way. Let \(L_0\) be the zeroth Lucas number and \(L_1\) be the first. If \(\begin{align*} L_0 &= 2 \\ L_1 &= 1 \\ L_n &= L_{n - 1} + L_{n - 2} \; \text{for}\; n \geq 2 \end{align*}\) Find \((a,b)\) such that \(L_n = a\phi^n + b\widehat{\phi}^n.\)

Guest Mar 20, 2018
 #1
avatar+19482 
+2

The Fibonacci sequence, with \(F_0 = 0\), \(F_1 = 1\) and \(F_n = F_{n - 2} + F_{n - 1}\), had a closed form \(F_n = \frac{1}{\sqrt{5}} \left( \phi^n - \widehat{\phi}^n \right),\) where \(\phi = \frac{1 + \sqrt{5}}{2} \; \text{and} \; \widehat{\phi} = \frac{1 - \sqrt{5}}{2}.\) The Lucas numbers are defined in a similar way. Let \(L_0\) be the zeroth Lucas number and \(L_1\) be the first. If \(\begin{align*} L_0 &= 2 \\ L_1 &= 1 \\ L_n &= L_{n - 1} + L_{n - 2} \; \text{for}\; n \geq 2 \end{align*}\) Find \((a,b)\) such that \(L_n = a\phi^n + b\widehat{\phi}^n.\)

 

 

\(\text{Fibonacci sequence $0,1,1,2,3,5,8,13,21,34,55,\ldots$} \\ \begin{array}{|lcr|} \hline a_0 = a_0 &=& 0\cdot a_1 + 1 \cdot a_0 \\ a_1 = a_1 &=& 1\cdot a_1 + 0 \cdot a_0 \\ a_2 = a_1+a_0 &=& 1\cdot a_1 + 1 \cdot a_0 \\ a_3 = a_2+a_1 &=& 2\cdot a_1 + 1 \cdot a_0 \\ a_4 = a_3+a_2 &=& 3\cdot a_1 + 2 \cdot a_0 \\ a_5 = a_4+a_3 &=& 5\cdot a_1 + 3 \cdot a_0 \\ a_6 = a_5+a_4 &=& 8\cdot a_1 + 5 \cdot a_0 \\ a_7 = a_6+a_5 &=& 13\cdot a_1 + 8 \cdot a_0 \\ a_8 = a_7+a_6 &=& 21\cdot a_1 + 13 \cdot a_0 \\ \ldots \\ a_n = a_{n-1}+a_{n-2} &=& F_n\cdot a_1 + F_{n-1} \cdot a_0 \\ && \boxed{a_n = F_n\cdot a_1 + F_{n-1} \cdot a_0 } \\\\ \text{Lucas numbers:} \\ a_0 = L_0 = 2 \\ a_1 = L_1 = 1 \\ a_n = L_n =F_n\cdot 1 + F_{n-1} \cdot 2 \\ && \boxed{L_n =F_n + 2F_{n-1}} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline F_n &=& \frac{1}{\sqrt{5}} \left( \phi^n - \widehat{\phi}^n \right) \\ F_{n-1} &=& \frac{1}{\sqrt{5}} \left( \phi^{n-1} - \widehat{\phi}^{n-1} \right) \\ \boxed{L_n =F_n + 2F_{n-1}} \\ L_n &=& \frac{1}{\sqrt{5}} \left( \phi^n - \widehat{\phi}^n \right) + 2\frac{1}{\sqrt{5}} \left( \phi^{n-1} - \widehat{\phi}^{n-1} \right) \\ &=& \dfrac{ \left( \phi^n - \widehat{\phi}^n \right)+2\left( \phi^{n-1} - \widehat{\phi}^{n-1} \right) } { \sqrt{5} } \\ &=& \dfrac{ \phi^n - \widehat{\phi}^n+2 \phi^{n-1} - 2\widehat{\phi}^{n-1} } { \sqrt{5} } \\ &=& \dfrac{ \phi^n+2 \phi^{n-1} - \widehat{\phi}^n - 2\widehat{\phi}^{n-1} } { \sqrt{5} } \\ &=& \dfrac{ \phi^n\left( 1+\dfrac{2}{\phi} \right) - \widehat{\phi}^n\left( 1+\dfrac{2}{\widehat{\phi}} \right) }{ \sqrt{5} } \quad | \quad \frac{1}{\phi} = \phi -1 \quad \frac{1}{\widehat{\phi}} = \widehat{\phi} -1 \\ &=& \dfrac{ \phi^n\Big( 1+2(\phi -1 ) \Big) - \widehat{\phi}^n\left( 1+2(\widehat{\phi} -1) \right) }{ \sqrt{5} } \\ &=& \dfrac{ \phi^n( 2\phi -1 ) - \widehat{\phi}^n\left( 2\widehat{\phi} -1 \right) }{ \sqrt{5} } \quad | \quad 2\phi -1 = \sqrt{5} \quad 2\widehat{\phi} -1= -\sqrt{5} \\ &=& \dfrac{ \phi^n\sqrt{5} - \widehat{\phi}^n\left( -\sqrt{5} \right) }{ \sqrt{5} } \\ &=& \dfrac{ \phi^n\sqrt{5} + \widehat{\phi}^n \sqrt{5} } { \sqrt{5} } \\ &=& \phi^n + \widehat{\phi}^n \\ \boxed{L_n =\phi^n + \widehat{\phi}^n } \\ \hline \end{array}\)

 

(a,b) = (1,1)
 

laugh

heureka  Mar 21, 2018
edited by heureka  Mar 22, 2018

19 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.