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Let $\phi$ be the golden ratio. I'm tasked to prove by other means than induction that $x$ in the next equation $$\phi^n =\phi F_n +x,$$ is actually a Fibonacci number. I have tried to apply Binet's formula to $\phi^n -\phi F_n$: \begin{align} \phi^n -\phi F_n &= \phi^n -\phi \left(\dfrac{\phi^n -\left(-\frac{1}{\phi}\right)^n}{\sqrt{5}}\right)\\ &=\phi^n -\frac{\phi^{n+1} -\left(-\frac{1}{\phi}\right)^n \phi}{\sqrt{5}}\\ &=\frac{\sqrt{5} \phi^n -\phi^{n+1} +\left(-\frac{1}{\phi}\right)^n \phi}{\sqrt{5}}. \end{align} But then I got stuck. Could you please help me?


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closed as unclear what you're asking by Namaste, Xander Henderson, Leucippus, Wouter, A. Goodier Apr 9 '18 at 8:59

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    $\begingroup$ This is hard to read. Do you mean $\phi^n=\phi \times F_n+F_k$? If $\phi=\frac {1+\sqrt 5}2$ then how is your claim true when $n=2$? $\endgroup$ – lulu Apr 8 '18 at 12:38
  • $\begingroup$ Sorry about that still learning. Yes everything you wrote above is correct. except I have to prove for a general case. So, eventually i should be able to find out that k = n-1. $\endgroup$ – Hell Carrier Apr 8 '18 at 12:46
  • $\begingroup$ Have you tried using induction on $n$? $\endgroup$ – Gerry Myerson Apr 8 '18 at 12:51
  • $\begingroup$ @lulu, perhaps we're using $F_1=F_2=1$. $\endgroup$ – Gerry Myerson Apr 8 '18 at 12:51
  • $\begingroup$ @GerryMyerson Right. I worked that out (eventually). $\endgroup$ – lulu Apr 8 '18 at 12:52
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Starting from $$\phi^n -\phi F_n,$$ then plug $$F_n \equiv \frac{1}{\sqrt{5}} (\phi^n -\varphi^n)$$ where $$\phi =\frac{1+\sqrt{5}}{2}, \qquad \varphi =\frac{1-\sqrt{5}}{2},$$ and use properties of $\phi$ and $\varphi$, namely, that $$1-\frac{1}{\sqrt{5}}\phi = -\frac{1}{\sqrt{5}}\varphi, \qquad \phi \varphi=-1,$$ to compute another Fibonacci number.

$$\phi^n -\phi F_n =\phi^n -\phi \cdot \tfrac{1}{\sqrt{5}}(\phi^n -\varphi^n) =\phi^n -\tfrac{1}{\sqrt{5}} \phi^{n+1} +\tfrac{1}{\sqrt{5}} \phi \varphi^n \\ =\left(1-\tfrac{1}{\sqrt{5}} \phi\right)\phi^n +\tfrac{1}{\sqrt{5}}\phi \varphi^n =-\tfrac{1}{\sqrt{5}} \varphi \phi^n +\tfrac{1}{\sqrt{5}} \phi \varphi^n \\ =-\varphi \phi \cdot \tfrac{1}{\sqrt{5}} \left(\phi^{n-1} -\varphi^{n-1} \right) =\tfrac{1}{\sqrt{5}} \left( \phi^{n-1} -\varphi^{n-1} \right) =F_{n-1}.$$

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  • $\begingroup$ I would like to prove that ϕn - ϕFn = Fn−1. Using, the LHS only. Would that still be possible without induction? $\endgroup$ – Hell Carrier Apr 8 '18 at 13:10
  • $\begingroup$ If you can prove the RHS is equal to the LHS, then by reversibility, the LHS is equal to the RHS. Or write the computations backwards. $\endgroup$ – Rócherz Apr 8 '18 at 13:12
  • $\begingroup$ Yes that is completely true. But in the question i have to assume that ϕn - ϕFn = [Another Fibonacci number]. And that [Another Fibonacci number] is not allowed to be assumed to be Fn−1. So, practcially the question states that find the general case for the value of [Another Fibonacci number] for ϕn - ϕFn = [Another Fibonacci number] to be satisfied. So i have to prove it assuming i only know the LHS part of the equation which is: ϕn - ϕFn $\endgroup$ – Hell Carrier Apr 8 '18 at 13:14
  • $\begingroup$ Oh, I see. I edited my answer accordingly. Same reasoning. $\endgroup$ – Rócherz Apr 8 '18 at 13:27
  • $\begingroup$ Alright! Thanks alot $\endgroup$ – Hell Carrier Apr 8 '18 at 13:28
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$$ \phi F_n = \phi\frac{\phi^n - \bar{\phi}^n}{\phi - \bar{\phi}} =\frac{\phi^{n+1} - \phi\bar{\phi}^n}{\phi - \bar{\phi}} =\frac{\phi^{n+1} + \phi\bar{\phi}^{n-1}}{\phi - \bar{\phi}} $$ $$ \phi F_n +F_k =\frac{\phi^{n+1} + \phi\bar{\phi}^{n-1}+\phi^k - \bar{\phi}^k}{\phi - \bar{\phi}} $$ and $ \phi F_n +F_k = \phi^n$ if and only if $$ =\frac{\phi^{n+1} + \phi\bar{\phi}^{n-1}+\phi^k - \bar{\phi}^k}{\phi - \bar{\phi}} = \phi^n $$ or $$ \phi^{n+1} + \phi\bar{\phi}^{n-1}+\phi^k - \bar{\phi}^k = \phi^{n+1} - \phi^n\bar{\phi}= \phi^{n+1}+\phi^{n-1} $$

Setting $k=n-1$ gives the result you desire.

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