Interesting a Fibonacci quesiton. Need help.

Question

Alice claims that she knows another formula for the Fibonacci numbers: Fn = $e^{n/2−1}$ for $n = 1,2,\cdots$ (where $e = 2.718281828$... is, naturally, the base of the natural logarithm). Is she right? Why or Why not?

Answer 1

I suppose this is some sort of practice on proof techniques. So here is a disproof by counter example:

Using her conjecture, $F_3 = \sqrt{e}$, which obviously is not the 3rd Fibonacci number.

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Here is a disproof by contradiction:

Suppose her conjecture was true. Then,

$$F_n + F_{n+1} = e^{\frac{n}{2} - 1} + e^{\frac{n + 1}{2} - 1} = e^{\frac{n}{2}}(e^{-1} + e^{\frac{1}{2}})$$

On the other hand,

$$F_{n + 2} = e^{\frac{n + 2}{2} - 1} = e^{\frac{n}{2}}(e^0)$$

But $e^0 \not = e^{-1} + e^{\frac{1}{2}}$. Hence $F_{n+2} \not = F_n + F_{n + 1}$. So her conjecture must be false.

Answer 2

There is a simple formula for $F_n$, but it has nothing to do with powers of e.

Define

$$\phi := \frac{1+\sqrt{5}}{2}$$

Then, the number

$$\frac{\phi^n}{\sqrt{5}}$$

is very close to the n-th Fibonacci-number $F_n$

In fact, when the given number is correctly rounded to an integer, it is $F_n$.