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Fibonacci Number :

I try to demonstrate that
$\phi^{n-2} \leq F_n \leq \phi^{n-1} $ where $\phi = \frac{1+\sqrt 5}{2}$

I started by having : $\phi^{n-3} \leq F_{n-1} \leq \phi^{n-2} $. Then : $\phi^{n-2} +\phi^{n-3} \leq F_{n+1}\leq \phi^{n-1} +\phi^{n-2} $

$\phi^{n-3}(\phi +1) \leq F_{n+1}\leq \phi^{n-2}(\phi +1)$

and here I did know how to conclude for $F_{n+1}$

Xander Henderson
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4everLRG
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    Use the fact that $\phi + 1 = \phi^2$. Also you need to use two induction hypotheses, with two base cases; or a strong induction, with two base cases. – player3236 Oct 23 '20 at 13:10
  • I suppose her that p(n ) is True and p(n-1) is True too. But how can I simplify and get rid of $\Large \phi^2$ Once I get the inequalities : $\Large \phi^{n-3} \phi^2 \leq F_{n+1}\leq \phi^{n-2}\phi^2$. – 4everLRG Oct 23 '20 at 13:15
  • Got it , thanks you it's Obvious, – 4everLRG Oct 23 '20 at 13:18
  • Don't use \large indiscriminately, and remember to end math equations with a $ as well. $P(n)$ is true does not imply $P(n-1)$ is also true, unless you are using strong/complete induction, where they are both true. Remember to write the base cases. – player3236 Oct 23 '20 at 13:18

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Since $\phi + 1 = \phi^2$, $$\phi^{n-3}(\phi +1) \leq F_{n+1}\leq \phi^{n-2}(\phi +1)$$ is equivalent to $$\phi^{n-3} \phi^2 \leq F_{n+1}\leq \phi^{n-2}\phi^2$$ which simplifies to $$\phi^{n-1} \leq F_{n+1}\leq \phi^{n}$$ as required.

lhf
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