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Exercise: Use $I_2$ to prove that $\alpha^{n-2}\le f_n\le \alpha^{n-1}$ for every positive integer $n$. Note that it is again necessary to make the first part of the proof for $n=1$ and $n=2$.

$I_2$ here means that we have to use strong induction. The variable $\alpha =\frac {1+\sqrt 5} {2}$

Attempt: I showed it holds for $n=1,2$ already and then assumed it held for up to $k$.

Im struggling on how to show it holds for $k+1$. I attempted to solve the left inequality first using the definition of fibonacci sequence $F_{k+1}= F_{k-1} + F_k$ and then using the induction hypothesis.

$F_k= F_{k-1}+F_{k}$

$x^{k-3}\le F_{k-1}\le x^{k-2}$ and $x^{k-2}\le F_k\le x^{k-1}$ (by our inductive hypothesis) (I replaced $x$ with $\alpha$)

therefore,

$$F_{k-1}\ge x^{k-3} + x^{k-2}\\ \ge x^{k-2}(x^{-1}+1)\\ \ge x^{k-2}(x^{-2})\\ \ge x^{k-4}$$

This is obviously wrong not sure why though!

$F_{k+1}\ge x^{k-4}$ (I replaced $\alpha$ with $x$). obviously not what I wanted which is $x^{k-1}\le F_{k+1}$

Seeker
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cheesewiz
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1 Answers1

1

You are attempting to use strong induction. After having checked for the case of $n=1$ and $n=2$, you should say:

Let $m>2$ be a natural number such that for all natural numbers $m'<m$, the statement in question is true.

This ensures that the statement is true for $m-1$ and $m-2$. I am assuming here they want you to use the recurrence relation $F_n=F_{n-1}+F_{n-2}$.

As it is currently written, you induction hypothesis does not imply the inequalities you say it does.

These are the flaws so far. I am not too sure on how to prove it yet.

Seeker
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