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We know:

$F_0 = 0$

$F_1 = 1$

$F_i = F_{i-1} + F_{i-2}$ for $i \geq 2$

Prove by induction:

$F_i = \dfrac{\phi^i-{\phi^{*}}^i}{\sqrt{5}}$

where $\phi = (1+\sqrt{5}) / 2$ and $\phi = (1-\sqrt{5}) / 2$.

My attempt:

Base case: $i = 0$: $F_0 = 0$ (easy to show)

Assume true for $i = k$ and $i = k + 1$.

We have:

$F_k = \dfrac{\phi^k-{\phi^{*}}^k}{\sqrt{5}}$

and

$F_{k+1} = \dfrac{\phi^{k+1}-{\phi^{*}}^{k+1}}{\sqrt{5}}$

Show it holds for $i = k + 2$ to complete induction. This is where I'm stuck. I have tried:

$F_{k+2} = F_{k+1} + F_{k} = \dfrac{\phi^{k+1}-{\phi^{*}}^{k+1}}{\sqrt{5}} + \dfrac{\phi^k-{\phi^{*}}^k}{\sqrt{5}}$

$F_{k+2} = \dfrac{\phi^k(1+\phi)-{\phi^{*}}^{k}(1+{\phi^*})}{\sqrt{5}}$

but I'm not sure where to go from here. Any help would be appreciated.

VividD
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Keir Simmons
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    You have a mistake in the last line, $\phi^{k+1} + \phi^k = \phi^k(1+\phi)$ and ditto for $\phi^\ast$. Now use $\phi^2 = 1+\phi$ (and ditto ...) – Daniel Fischer Mar 06 '14 at 15:31
  • Thanks Daniel - I wasn't aware of that property. Are there any other ways of proving this without use of that property? – Keir Simmons Mar 06 '14 at 15:39
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    @AliAmiri, that one was (rightly) closed for not showing any work/interest by who asked. This is different. – vonbrand Mar 06 '14 at 16:09
  • You need two base cases, $i = 0$ and $i = 1$ to get started. The property alluded to by @DanielFisher is since $\phi^2 - \phi - 1 = 0$ (solve the quadratic, the roots are your $\phi$ and $\phi^*$). – vonbrand Mar 06 '14 at 16:14

2 Answers2

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Hint $\rm\quad \phi^{\:n+1}\!-\:\bar\phi^{\:n+1} =\ (\phi+\bar\phi)\ (\phi^n\!-\:\bar\phi^n)\ -\ \phi\:\bar\phi\ (\phi^{\:n-1}\!-\:\bar\phi^{\:n-1})$

Hence, upon substituting $\rm\,\ \phi+\bar\phi\ =\ 1\ =\ -\phi\bar\phi\ $ and dividing by $\:\phi-\bar\phi = \sqrt 5\:$ we deduce $\rm\ \ldots$

Remark $\ $ To understand the essence of the matter it's worth emphasizing that such an inductive proof amounts precisely to showing that $\rm\:f_n\:$ and $\rm\: \bar{f}_n = (\phi^n-\bar\phi^n)/(\phi-\bar\phi)\:$ are both solutions of the difference equation (recurrence) $\rm\ f_{n+2} = f_{n+1} + f_n\:,\:$ with initial conditions $\rm\ f_0 = 0,\ f_1 = 1\:.\:$ The (trivial) induction simply proves the uniqueness of such solutions. It will prove quite instructive to structure the proof from this standpoint. It will also mean that you can later reuse this uniqueness theorem for recurrences.

Generally, just as above, uniqueness theorems provide very powerful tools for proving equalities - a point which I emphasize in many prior posts. For example, see my prior posts on telescopy and the fundamental theorem of difference calculus, esp. this one.

Community
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Bill Dubuque
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Use $1+\phi=\phi^2$, to get $$ F_{k+2} = \dfrac{\phi^k(1+\phi)-{\phi^{*}}^{k}(1+{\phi^{*}})}{\sqrt{5}}= \dfrac{\phi^{k+2}-{\phi^{*}}^{(k+2)}}{\sqrt{5}} $$

draks ...
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