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Here is the official theorem I'll use: enter image description here

Since the Fibonacci sequence is defined as $F_n=F_{n-1}+F_{n-2}$, we solve the equation $x^2-x-1=0$ to find that $r_1 = \frac{1+\sqrt 5}{2}$ and $r_2 = \frac{1-\sqrt 5}{2}$

So we have $F_n = c_1\left(\frac{1+\sqrt 5}{2}\right)^n + c_2\left(\frac{1-\sqrt 5}{2}\right)^n$

We know that $F_0 = F_1 = 1$. So we can solve the following system to find the values of $c_1$ and $c_2$:

$1 = c_1 + c_2$

$1 = c_1\left(\frac{1+\sqrt 5}{2}\right) + c_2\left(\frac{1-\sqrt 5}{2}\right)$

Solving this system does not give $c_1 = 1/\sqrt5, c_2 = -1/\sqrt 5$ , even though that is apparently the right answer, i.e. the closed form of the Fibonacci sequence is apparently $$\frac1{\sqrt 5}\left(\frac{1+\sqrt 5}{2}\right) -\frac1{\sqrt 5}\left(\frac{1-\sqrt 5}{2}\right)$$

Where did I go wrong? Why doesn't solving the system of equations give me $c_1 = 1/\sqrt5, c_2 = -1/\sqrt 5$?

RobPratt
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James Ronald
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    Well, with the latter values we have $c_1+c_2=0$. So perhaps they were using $F_0=0,F_1=1$ as initial conditions. That's standard anyway. – lulu Nov 18 '19 at 21:32
  • See https://math.stackexchange.com/questions/65011/how-to-prove-that-the-binet-formula-gives-the-terms-of-the-fibonacci-sequence – Arnaud D. Nov 18 '19 at 22:00
  • The initial values for the Fibonacci sequence are defined as either $F_0=0,F_1=1$ or $F_1=F_2=1$ (of course, they both produce the same sequence). Maybe you confused the two ways? It would be easier to substitute $n=0$ and $n=1$. – bjorn93 Nov 18 '19 at 23:13
  • @lulu Thank you for the help!, so does the closed form of the Fibonacci sequence depend on what you choose $F_0$ and $F_1$ to be? That would be weird, since no matter your choice of $F_0$ and $F_1$ you still end up with the same sequence... – James Ronald Nov 19 '19 at 16:14
  • @bjorn93 Thank you for the response! So why is it the choice of initial values changes the closed form of the sequence? As you said, the choice of initial values doesn't change the sequence itself, so how can the closed form change? – James Ronald Nov 19 '19 at 16:16
  • @bjorn93 Oh, but then I'm back to my original question: if my choice of initial values is irrelevant, why did I still arrive at the incorrect closed form? – James Ronald Nov 19 '19 at 16:19
  • @JamesRonald because you have a mistake in your calculations :) see the first comment: it's $c_1+c_2=0$, not $1$, as you have it in your solution. – bjorn93 Nov 19 '19 at 16:22
  • @bjorn93 $c_1 + c_2 = F_0$, right? And $F_0=1$? So why is it incorrect to say $c_1+c_2 = 1$? Sorry for getting confused, I really appreciate the help – James Ronald Nov 19 '19 at 16:25
  • Of course the initial conditions effect the closed form! You tried to solve for $F_0=1$ where someone else used $F_0=0$. Any given formula can, at best, yield one or the other, not both. Of course, it is a triviality to switch from one of those choices to the other, but the exact form depends on the choice. – lulu Nov 19 '19 at 16:29
  • Just to stress, if we let $F_n$ denote your choice (with $F_0=1$) and we let $\mathscr F_n$ denote the alternate choice (with $\mathscr F_0=0$) then we get $F_n=\mathscr F_{n+1}$, so it is very straightforward to translate between the two. – lulu Nov 19 '19 at 16:32
  • @JamesRonald Yes, the choice of initial values matters for the closed-form solution. I didn't make myself clear. If you change the indexing (as you have), the constants $c_1$ and $c_2$ change. However, between the two standard ways to define the initial values of the Fibonacci sequence (see my first comment), both ways will produce the same closed form. (That's what I meant). Your definition is non-standard, however. – bjorn93 Nov 19 '19 at 16:34

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Let's see...

$$f_n = \left\{ \begin{array}{ll} 0 & \text{ for } n = 0 \\ 1 & \text{ for } n = 1 \\ f_{n-1} + f_{n-2} & \text{ for } n>1 \end{array} \right.$$

Now, the recursion can be written as $$f_n - f_{n-1} - f_{n-2} = 0,$$ so characteristic equation is $$x^2-x-1=0.$$ Now, the roots of the equation are $$X_{1,2} = \frac{1 \pm \sqrt{5}}2,$$ so general solution is $$f_n = C_1\cdot\left(\frac{1 + \sqrt{5}}2\right)^n + C_2\left(\frac{1 - \sqrt{5}}2\right)^n$$ From the $f_1$ and $f_2$ we get \begin{eqnarray} 0 &=& C_1 + C_2 \\ 1 &=& C_1\left(\frac{1 + \sqrt{5}}2\right) + C_2\left(\frac{1 - \sqrt{5}}2\right) \end{eqnarray} From the first equation we get $$C_2 = -C_1,$$ so \begin{equation} 1 = C_1\left(\frac{1 + \sqrt{5}}2\right) -C_1\left(\frac{1 - \sqrt{5}}2\right) \end{equation} Now, we have $$C_1\left[\frac{1 + \sqrt{5}}2 - \frac{1 - \sqrt{5}}2\right] = 1$$ or $$C_1\cdot\sqrt{5} =1$$ So, $$C_1 = \frac{1}{\sqrt{5}}.$$ Now, $$C_2 = -\frac{1}{\sqrt{5}}.$$ The particular solution for the equation is therefore $$f_n = \frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}2\right)^n - \left(\frac{1-\sqrt{5}}2\right)^n\right]$$

noname
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