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Starting from $F_n = F_{k-1} F_{n-k-1} + F_{k-1} F_{n-k} + F_{k-2} F_{n-k} $

and letting $F_k \approx \phi^k$, I am hoping to find the corresponding statement for the Golden ratio:

$\phi^n = 2 \phi^{n-2} + \phi^{n-1}$ or just $\phi^2 = 2 + \phi$. Where did I go wrong?

In fact, starting from an even simpler identity: Proof of identity $F_m F_n + F_{m−1} F_{n−1} = F_{m+n−1}$ for Fibonacci numbers you can get the same faulty asymptotics

$\phi^{m+n} + \phi^{m+n-2} = \phi^{m+n-1}$ or $\fbox{$\phi^2 + 1 = \phi$}$.

See also Showing that an equation holds true with a Fibonacci sequence: $F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$

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cactus314
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  • I tried writing $F_k F_{n-k} \approx \phi^n + \phi^{n-2k}$ still doesn't work. – cactus314 Feb 15 '14 at 20:37
  • What is your question exactly? What faulty asymptotics are you referring to? – Erick Wong Feb 15 '14 at 21:06
  • I took an existing Fibonacci identity and asked when $n,k >> 1$, what the asymptotics look like? It turns out $F_k \approx \frac{1}{\mathbf{\sqrt{5}}}\phi^k$ so and that $\frac{\phi^2 + 1}{\sqrt{5}}=\phi$. – cactus314 Feb 15 '14 at 21:27

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The "faulty" asymptotics is caused by omission of the multiplicative constant. Such constants are often unimportant (such as in the big-Oh notation, when saying $F_n=\Theta(\phi^n)$ is perfectly correct), but sometimes play very important role.

In fact, reasoning of this kind can sometimes be used to determine the constant! For example, consider the identity $$F_{n+m}=F_{n−1}F_m+F_nF_{m+1}$$ If we assume that $F_n\approx \alpha \phi^n$, we get $$\alpha \phi^{n+m} = \alpha^2 \phi^{n+m-1} + \alpha^2 \phi^{n+m+1}$$ which reduces to $$\alpha = \frac{\phi}{1 + \phi^2}=\frac{1}{\sqrt{5}}$$ ... exactly as expected.

Peter Košinár
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