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The polynomial for the Fibonacci recurrence $F(n) = F(n-1) + F(n-2)$ is

$$x^{2} = x + 1.$$

The solutions are :
$ϕ = \frac{1+\sqrt 5}{2}$ and $ψ = \frac{1-\sqrt 5}{2}.$

So the Fibonacci sequence, for real number has the form $F(n) = u ϕ^{n} + v ψ^{n}$.

With the boundary conditions $F(0) = 0$ one can see that $u = -v$.

The second boundary condition $F(1) = 1$ leads to $1 = u ϕ^{1} + v ψ^{1}$.

Substituting $v$ and simplifying the equation leads to $1 = u\sqrt 5$.

Therefore $u = \frac{1}{\sqrt 5}$ and $v = -\frac{1}{\sqrt 5}$

As a result $$F(n) = \frac{ϕ^{n} - ψ^{n}}{\sqrt 5}.$$

Is there any other way to derive this formula ?

kipf
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    Once you have found the formula , you can prove it by induction. – Peter Jun 05 '23 at 09:34
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    You could also use generating functions, but the way you have sketched here is the standard method. – lulu Jun 05 '23 at 09:38
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    Here is what @lulu is referring to. You can split it into partial fractions, then expand them as geometric series. – J.G. Jun 05 '23 at 09:40
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    Please surround entire expressions with dollar signs rather than individual symbols. It looks a whole lot nicer: rather than F(n) = u * $ϕ^{n}$ + v * $ ψ^{n}$ use $F(n) = u \cdot \phi^n + v \cdot \psi^n$ to get $F(n) = u \cdot ϕ^n + v \cdot ψ^n$. – Arthur Jun 05 '23 at 10:07
  • https://artofproblemsolving.com/wiki/index.php/Binet%27s_Formula – Mostafa Jun 05 '23 at 14:08

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