Imaginary Golden Ratio

Solution 1:

$$f_n = (\phi_i^{\;-n}-\phi_i^{\;n})\frac{i}{\sqrt{3}}$$

You can prove this by induction on $n,$ using the fact that $\phi_i$ and $\phi_i^{\;-1}$ are the two solutions to the quadratic equation $x^2-x+1=0.$ (Alternatively, it's easy to see that the three sequences $\langle \phi_i^{\;n}\;\vert\;n \in \mathbb{N}\rangle,$ $\langle \phi_i^{\;-n}\;\vert\;n \in \mathbb{N}\rangle,$ and $\langle f_n \;\vert\;n \in \mathbb{N}\rangle$ are all periodic with period 6, so it's actually sufficient to check the formula above for $n=\;$0, 1, 2, 3, 4, and 5.)

However, a little linear algebra shows what's really going on. The set of all complex-valued sequences satisfying the same recurrence relation as your $f_n$ is closed under pointwise addition and multiplication by a constant, so it forms a vector space $V$ over the field of complex numbers. The two sequences $u_1=\langle \phi_i^{\;n}\;\vert\;n \in \mathbb{N}\rangle$ and $u_2=\langle \phi_i^{\;-n}\;\vert\;n \in \mathbb{N}\rangle$ satisfy the recurrence relation (because $\phi_i$ and $\phi_i^{\;-1}$ are roots of the quadratic equation above), so $u_1$ and $u_2$ belong to $V.$

You can check that $\lbrace u_1,u_2 \rbrace$ is a basis for $V.$ The formula I gave for $f_n$ is just the particular linear combination of $u_1$ and $u_2$ that happens to yield the sequence $\langle f_n \;\vert\;n \in \mathbb{N}\rangle.$

Solution 2:

I found this number in an unrelated context. Consider the differential equation:

$$f'(x) = f(f(x))$$

The function that is the solution to this has the property that its derivative is the same as composing the function with itself. Assume the function takes the form of $f(x) = A x^r$ where $A$ and $r$ are constants. Then:

$$r A x^{r - 1} = A^{r + 1} x^{r^2}$$

Assuming $x$ is nonzero:

$$r A x^{ (r - 1) - r^2} = A^{r + 1}$$

The RHS is a constant and equivalent to the LHS, so the exponent must be $0$, otherwise the LHS would vary with $x$, so:

$$(r - 1) - r^2 = 0$$

The solutions to this are $(1 \pm i \sqrt{3})/2$.

So a function $f(x)$ that has the property of its derivative equaling $f(f(x))$ is of the form $f(x) = A x^r$ where $A$ is some constant and $r$ is the 'imaginary golden ratio'!