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zmudz
 one year ago
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 \(\begin{align*} L_0 &= 2 \\ L_1 &= 1 \\ L_n &= L_{n  1} + L_{n  2} \; \text{for}\; n \geq 2 \end{align*}\)
zmudz
 one year ago
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 \(\begin{align*} L_0 &= 2 \\ L_1 &= 1 \\ L_n &= L_{n  1} + L_{n  2} \; \text{for}\; n \geq 2 \end{align*}\)

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zmudz
 one year ago
Best ResponseYou've already chosen the best response.0Let \(\begin{align*} A_0 &= 6 \\ A_1 &= 5 \\ A_n &= A_{n  1} + A_{n  2} \; \textrm{for} \; n \geq 2. \end{align*}\) There is a unique ordered pair \((c,d)\) such that \(c\phi^n + d\widehat{\phi}^n\) is the closed form for sequence \(A_n\). Find \(c\) using the Fibonacci and Lucas number sequences.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1It seems \[\large A_n=c\phi^n+d\widehat{\phi}^n = 2F_n+3L_n\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't know why the problem is arranged too complicated. To my course, it is easy to find it out by characteristics equation \(r^2r1=0\), which gives us \(r_1= \dfrac{1+\sqrt5}{2},r_2=\dfrac{1\sqrt5}{2}\), Hence the closed form of \(A_n =C(\dfrac{1+\sqrt 5}{2})^n+D(\dfrac{1\sqrt5}{2})^n\) If \(A_0=6\rightarrow C+D=6\) If\(A_1= 5\rightarrow C(\dfrac{1+\sqrt5}{2}) +D(\dfrac{1\sqrt5}{2})=5\) Solve for C, D, we have \(C = 9\dfrac{2}{\sqrt5}, D= \dfrac{2}{\sqrt5}3\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0just a correction but i believe the solution to the above system is: \[C = 3 + \frac{2}{\sqrt{5}}\] \[D = 3  \frac{2}{\sqrt{5}}\]
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