zmudz
  • zmudz
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 \(\begin{align*} L_0 &= 2 \\ L_1 &= 1 \\ L_n &= L_{n - 1} + L_{n - 2} \; \text{for}\; n \geq 2 \end{align*}\)
Mathematics
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SOLVED
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katieb
  • katieb
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zmudz
  • zmudz
Let \(\begin{align*} A_0 &= 6 \\ A_1 &= 5 \\ A_n &= A_{n - 1} + A_{n - 2} \; \textrm{for} \; n \geq 2. \end{align*}\) 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
  • ganeshie8
It seems \[\large A_n=c\phi^n+d\widehat{\phi}^n = 2F_n+3L_n\]
anonymous
  • anonymous
I don't know why the problem is arranged too complicated. To my course, it is easy to find it out by characteristics equation \(r^2-r-1=0\), 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 \(C = 9-\dfrac{2}{\sqrt5}, D= \dfrac{2}{\sqrt5}-3\)

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dumbcow
  • dumbcow
just 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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