Modifying Binet's Formula for the Fibonacci Sequence with a Complex Offset

I am investigating non-integer, complex-valued extensions of the Fibonacci sequence.

Problem

How do I modify the function

$$Ft(n, θ) = \frac{ϕ^{n + m i} + e^{i θ} (−ψ)^{n + m i}} {ϕ − ψ}$$

so it stays in contact with the spiral of the function

$$F(n) = \frac{ϕ^{n + m i} − ψ^{n + m i}} {ϕ − ψ}$$

as we vary the value of the constant $$m$$?

Background

The Fibonacci sequence $$\{0, 1, 1, 2, 3, 5, 8, \,...\}$$ is initially defined as a positive integer sequence with the relation $$F_n = F_{n − 1} + F_{n − 2}$$ and the starting values $$F_0 = 0$$ and $$F_1 = 1$$. It can be extended to negative values easily enough with the relation $$F_n = F_{n + 2} - F_{n + 1}$$. But calculating $$F_n$$ for an arbitrary value of $$n$$ with these relations requires calculating every value between the starting values and the desired value, which becomes tedious as $$n$$ gets large.

Luckily, there is a better way: Binet's formula $$F(n) = \frac{ϕ^n − ψ^n} {ϕ − ψ}$$, using the golden ratio $$ϕ = \frac{1 + \sqrt{5}} {2} ≈ 1.618$$ and its conjugate $$ψ = \frac{1 − \sqrt{5}} {2} ≈ −0.618$$. This function gives the $$n$$th Fibonacci number directly, without needing to calculate any intervening values.

Notice that Binet's formula does not require whole-number inputs for the value of $$n$$. However, because $$ψ$$ is a negative number being raised to the power of $$n$$, non-integer values of $$n$$ produce complex outputs. We can plot this in a $$3\text{D}$$ graph, with a one-dimensional real input $$n$$ on the $$x$$-axis, the real part of the two-dimensional complex output $$F(n)$$ on the $$y$$-axis, and the imaginary part of the complex output on the $$z$$-axis.

For negative values of $$n$$, the $$ψ^n$$-term dominates as a wide spiral around the $$n$$-axis that contracts towards the value of $$ϕ^n$$ as $$n$$ increases towards zero.

For positive values of $$n$$, the $$ϕ^n$$-term dominates, with the $$ψ^n$$-term contributing only a tiny wobble.

The spiral crosses the $$n\text{Re}$$ ($$xy$$) plane at every integer value of $$n$$.

Binet's formula also has some noteworthy variations:

• $$Fa(n) = \text{round}\left(\frac{ϕ^n} {ϕ − ψ}\right)$$ gives all Fibonacci numbers for positive integer values of $$n$$
• $$Fe(n) = \frac{ϕ^n − (−ψ)^n} {ϕ − ψ}$$ gives even-index Fibonacci numbers for even integer values of $$n$$
• $$Fo(n) = \frac{ϕ^n + (−ψ)^n} {ϕ − ψ}$$ gives odd-index Fibonacci numbers for odd integer values of $$n$$

Plotting these functions alongside the main function, we can see that $$Fa(n)$$ is a standard exponential curve in the center of the spiral of Binet's formula, while $$Fe(n)$$ is a sigmoid curve along the lower-real side of the spiral that touches it at every even-integer $$n$$, and $$Fo(n)$$ is a U-shaped curve along the upper-real side of the spiral that touches it at every odd-integer $$n$$.

Upon seeing this I wondered if there was a way to modify the pattern of $$Fe(n)$$ and $$Fo(n)$$ to get curves that touched the spiral at other points, and it turns out there is.

The function

$$Ft(n, θ) = \frac{ϕ^n + e^{i θ} (−ψ)^n} {ϕ − ψ}$$

rotates around the spiral of Binet's formula by the angle $$θ$$, starting out equivalent to $$Fo(n)$$ when $$θ = 0°$$ and becoming equivalent to $$Fe(n)$$ when $$θ = 180°$$.

So far so good. What happens if we allow complex values of $$n$$? Unfortunately, properly displaying a $$2\text{D}$$ input and a $$2\text{D}$$ output would require a $$4\text{D}$$ plot, which humans aren't well equipped to comprehend.

Instead, we can modify our 3D plot of Binet's formula and variations with a constant imaginary offset $$m$$, giving us for example $$F(n + m i)$$.

For $$m = 0$$ we get the original purely-real-input setup. For positive values of $$m$$ the spiral contracts around the exponential curve of $$Fa(n + m i)$$, which rotates up around the $$x$$-axis. For negative values of $$m$$ the spiral expands away from the exponential curve of $$Fa(n + m i)$$, which rotates down around the $$x$$-axis. However, while the plots of $$Fe(n + m i)$$, $$Fo(n + m i)$$, and $$Ft(n + m i, θ)$$ rotate with the exponential curve of $$Fa(n + m i)$$, they do not expand and contract with the spiral.

I have been unable to figure out how to modify them to match the expanding/contracting behavior.

Does anyone know how to change the function

$$Ft(n, θ) = \frac{ϕ^{n + m i} + e^{i θ} (−ψ)^{n + m i}} {ϕ − ψ}$$

so it stays in contact with the spiral of the function

$$F(n) = \frac{ϕ^{n + m i} − ψ^{n + m i}} {ϕ − ψ}$$

as we vary the value of the constant $$m$$?

• I am upvoting for both the analysis and quality of the graphics. Jun 20, 2022 at 19:52
• Have you tried using a factor of $\,(-1)^{(m\,i)} = (e^{-\pi})^m$? Jun 20, 2022 at 20:23
• @Somos That's it! Updating the function to $Ft(n, θ) = \frac{ϕ^{n + m i} + e^{i θ - π m} (−ψ)^{n + m i}} {ϕ − ψ}$ works exactly as I wanted. If you make that an answer I'll accept it. Jun 21, 2022 at 1:56
• @martycohen The graphics were made in GeoGebra, which is surprisingly powerful free graphing software. It's designed mostly for geometry and algebra, but it can also handle a decent amount of calculus. Jun 21, 2022 at 14:15

In the context of extending the Fibonacci sequence to non-integer index value the natural formula is $$F(n+m\,i) = \frac{\phi^{n + m\,i} − \psi^{n + m\,i}} {\psi − \psi}.$$ Since $$\,\psi=-1/\phi<0,\,$$ there is no problem with $$\,\psi^n\,$$ but what about $$\,\psi^{m\,i}?\,$$ Use the definition of the power function as in DLMF to get $$\psi^{m\,i} = (-1)^{m\,i}(-\psi)^{m\,i} = (e^{-\pi})^m \phi^{-m\,i}.$$ Thus, a suitable formula is $$F(n+m\,i) = \frac{\phi^{n + m\,i} −\ \psi^n (e^{-\pi})^m \phi^{-m\,i}} {\psi − \psi}$$ or some simple variation as needed.