While playing with the results of defining a new operation, I came across a number of interesting properties with little literature surrounding it; the link to my original post is here: Finding properties of operation defined by $x⊕y=\frac{1}{\frac{1}{x}+\frac{1}{y}}$? ("Reciprocal addition" common for parallel resistors)

and as you can see, the operation of interest is $x⊕y = \frac{1}{\frac{1}{x}+\frac{1}{y}} = \frac{xy}{x+y}$.

In wanting to find a condition such that $x⊕y = x-y$, I found that the ratio between x and y mus be φ=1.618... the golden ratio, for this to work!


$\frac{1}{\frac{1}{x}+\frac{1}{y}} = x-y$

$\frac{xy}{x+y} = x-y$

$xy = x^2-y^2$

$0 = x^2-xy-y^2$

and, using the quadratic formula,

$x = \frac{y±\sqrt{y^2+4y^2}}{2}$

$x = y\frac{1±\sqrt{5}}{2}$

$x = φy$

This result is amazing in and of itself. Yet through the same basic setup, we find a new ratio pops out if we try $x⊕y = x+y$ and it is complex.

$x⊕y = x+y$

$\frac{1}{\frac{1}{x}+\frac{1}{y}} = x+y$

$\frac{xy}{x+y} = x+y$

$xy = x^2+2xy+y^2$

$0 = x^2+xy+y^2$

$x = \frac{-y±\sqrt{y^2-4y^2}}{2}$

$x = y\frac{1±\sqrt{-3}}{2}$

$x = y\frac{1±\sqrt{3}i}{2}$

and this is the "imaginary golden ratio"!

$φ_i = \frac{1+\sqrt{3}i}{2}$

It has many properties of the golden ratio, mirrored. This forum from 2011 is the only literature I could dig up on it, and it explains most of the properties I also found and more. http://mymathforum.com/number-theory/17605-imaginary-golden-ratio.html

This number is extremely cool, because its mathematical properties mirror φ but also have their own coolness.

$φ_i = 1-\frac{1}{φ_i}$

$φ_i^2 = φ_i - 1$

and generally

$φ_i^n = φ_i^{n-1} - φ_i^{n-2}$

This complex ratio also lies on the unit circle in the complex plane, and has a representation as a power of e!

$φ_i = cos(π/3)+ isin(π/3) = e^{iπ/3}$


It is also a nested radical, because of the identity $φ_i^2 + 1 = φ_i$


Since the only other forum which I could find that has acknowledged the existence of the imaginary golden ratio (other than the context of it as a special case imaginary power of e) I'd like to share my findings and ask if anybody has heard of this ratio before, and if anybody could offer more fine tuned ideas or explorations into the properties of this number. One specific qustion I have involves its supposed connection (according to the 2011 forum) to the sequence

$f_n = f_{n-1} - f_{n-2}$




could somebody explain to me how this sequence is connected to φ_i? The forum states there is a connection, but I can't figure out what it is based on the wording. What am I missing?

Thanks for your help with my question/exploration.

  • 1
    $\begingroup$ In general the recurrence $x_n=cx_{n-1}+dx_{n-2}$ has solutions of the form $x_n=a\alpha^n+b\beta^n$ for some constants $a,b$, where $\alpha$ and $\beta$ are the roots of the quadratic $x^2=cx+d$ (if distinct). $\endgroup$
    – stewbasic
    Jul 7 '16 at 3:57
  • $\begingroup$ It's worth noting that your operation is one half of the Harmonic Mean. en.wikipedia.org/wiki/Harmonic_mean $\endgroup$ Jul 8 '16 at 5:50

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

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 $\langle f_n \;\vert\;n \in \mathbb{N}\rangle.$

  • 1
    $\begingroup$ This isn't really part of the answer, but it might be of interest to note that you can expand $\phi_i^{\;\pm n}=\text{cos}\;n \pi \pm i \;\text{sin}\;n \pi$ in the formula for $\;f_n\;$in my answer to yield $$f_n = \frac{2}{\sqrt{3}}\text{sin}\frac{n \pi}{3}.$$ (Again, both sides have period 6, so, if you prefer, you can just verify this by calculating it for $n=\;$0, 1, 2, 3, 4, and 5.) $\endgroup$ Jul 8 '16 at 5:27
  • 1
    $\begingroup$ In my previous comment, I omitted a couple of $\frac{1}{3}\text{'s}$ in the derivation. I meant to say that we could expand $\phi_i^{\;\pm n}=\text{cos}\frac{n \pi}{3} \pm i \;\text{sin}\frac{n \pi}{3}$ to yield $$f_n = \frac{2}{\sqrt{3}}\text{sin}\frac{n \pi}{3}.$$ $\endgroup$ Jul 26 '16 at 22:47

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'!


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