Figure-It-Out Function! - Part 6: The Second Reciprocal

f(x)f(x-1) + f(x)f(x+1) = 1.

Define g(x) = f(x)(x+1).

Then g(x) + g(x-1) = 1.

g(1) = f(1)f(2) = 2.

g(2) = 1 - g(1) = -1.

g(x+2) = 1-g(x+1) = 1-(1-g(x)) = g(x)

So these values will just repeat from here onwards. We can observe that g(x) = -1 for even x.

We are asked to find $g(2014)^{g(2012)^...{g(2)^2}} = {-1^{-1^...{-1^2}}} = {-1 ^ {-1^ ... {-1^1}}} = {-1^{-1^{-1^{-1}}}} = -1$ as $-1^{-1} = -1$ .

Fact:

$\begin{cases}f(n)=a\\f(n+1)=-\frac{1}{a}\end{cases}\implies f(n+2)=-2a$ .

Proof:

Simply use the recursive formula:

$\begin{aligned}f(n+1)&=\frac{1}{f(n)+f(n+2)} \\ -\frac{1}{a}&=\frac{1}{a+f(n+2)} \\ f(n+2)&=-2a\square \end{aligned}$

Fact:

$\begin{cases}f(n)=a\\f(n+1)=-\frac{1}{a}\\f(n+2)=-2a=b\end{cases}\implies f(n+3)=-\frac{1}{b}$

Proof:

Simply use the recursive formula again. $\square$ .

We also have $n=6\implies \begin{cases}f(n)=a\\f(n+1)=-\frac{1}{a}\end{cases}$ .

We can see from our proofs (and from the observation that $f(2)f(3)=-1, f(4)f(5)=-1$ ) that $f(2k+1)$ is a negative multiplicative inverse of $f(2k)$ $\forall k\in\mathbb N_{>0}$ .

Thus our answer is (we also have $f(1)=2$ ):

$\underbrace{(-1)^{(-1)^{\cdots^{(-1)^2}}}}_{1007 (-1)\text{'s}}=\boxed{-1}$

f(x) = 1/(f(x-1)+f(x+1)); f(1)=2 ; f(2)=1. Thus, f(0) = -1/2. Thus, f(1) = 2, f(2)=1; f(3)=-1......... And thus, f(2n) = (2^(n-1))*((-1)^(n+1)). And, f(2n+1) = -(f(2n))^(-1). Thus, f(2n)f(2n+1)=-1. And, (-1)^((-1)^n) = -1.

From the first few values of $f(x)$ , we note that $f(x)f(x+1) = \begin{cases} 2 & \text{for }x \text{ is odd.} \\ -1 & \text{for }x \text{ is even.} \end{cases}$ . Let us prove by induction that the claim is true for all $x \ge 1$ .

Proof: From $f(x) = \dfrac 1{f(x-1) + f(x+1)}$ $\implies f(x-1)f(x) + f(x)f(x+1) = 1$ . For $x=1$ and $x=2$ , $\implies f(1)f(2) = 2 \times 1 = 2$ . Then $f(1)f(2) + f(2)f(3) = 1$ , $\implies 2 + f(2)f(3) = 1$ , $\implies f(2)f(3) = -1$ . Therefore, the claim is true for $x=1$ and $x=2$ respectively. Assuming the claim is true for $x$ and x+1). Then, when \(x is even and $x+1$ is odd, $f(x+1)(x+2) = 1 - f(x)f(x+1) = 1 - (-1) = 2$ , when $x$ is odd and $x+1$ is even, $f(x+1)(x+2) = 1 - f(x)f(x+1) = 1 - 2 = -1$ . Therefore the claim are also true for $x+2$ and $x+3$ , and hence true for all $x \ge 1$ .

Now we have:

$\begin{aligned} X & = (f(2014)f(2015))^{(f(2012)f(2013))^{\cdots^{(f(2)f(3))^{f(1)}}}} \\ & = (-1)^{(-1)^{(-1)^{\cdots^{(-1)^{(-1)^2}}}}} \\ & = (-1)^{(-1)^{(-1)^{\cdots^{(-1)^1}}}} & \small \color{#3D99F6} \text{Note that }(-1)^{-1} = - 1 \\ & = \boxed{-1} \end{aligned}$