# Imaginary Golden Ratio

#### Agno

We all know the Golden Ratio.

$$\displaystyle \Phi=\frac{a}{b}=\frac{a+b}{a}$$

$$\displaystyle \Phi^2-\Phi-1=0 \Leftrightarrow \Phi = \frac{1 \pm \sqrt{5}}{2}$$

The Golden Ratio has all sorts of interesting properties like:
* subsequent powers of the Golden Ratios are summable (so-called "recurrence relation");
* there exists a direct relationship with the Fibonacci series;
* it can be expressed as a nested radical only using the number 1
* it is the "worst" real number for rational approximation because of its continued fraction representation (only using 1's)

But suppose we'd define an imaginary equivalent of this Golden Ratio (let's call it $$\displaystyle \Phi_{i}$$) as follows:

$$\displaystyle \Phi_{i}=\frac{a}{b}=\frac{a-b}{a}$$

$$\displaystyle \Phi_{i}^2-\Phi+1=0 \Leftrightarrow \Phi_{i} = \frac{1 \pm i\sqrt{3}}{2}$$

To my surprise, this Imaginary equivalent possesses all the same atrributes as the standard Golden Ratio:

There is a direct link to $$\displaystyle \pi$$ and $$\displaystyle e$$:

$$\displaystyle \Phi_{i} = e^{\pm \frac{\pi i}{3}}$$

There is a nice link to the Fibonacci series but then for the negative equivalent:

$$\displaystyle F_0 = 0$$
$$\displaystyle F_1 = -1$$
$$\displaystyle F_n = F_{n-1} - F_{n-2}$$

And this induces the series:

0,-1,-1, 0, 1, 1, 0,-1,-1, 0, 1, 1, 0...

and this series also follows a simple formula:

$$\displaystyle F_n = \frac{(1-i \sqrt(3))^n - (1+i \sqrt(3))^n}{2^ni\sqrt(3)}$$

or can be written as:

$$\displaystyle F_n = -\frac23 \sqrt{3} \sin(\frac{\pi n}{3})$$

Also, subsequent powers of $$\displaystyle \Phi_{i}$$ can be subtracted (instead of summed):

$$\displaystyle \Phi_{i}^n =\Phi_{i}^{n-1} - \Phi_{i}^{n-2}$$

And for n = 1, the special relation occurs:

$$\displaystyle \Phi_{i} =1 - \frac{1}{\Phi_{i}}$$

And this can be used to establish the recursive fraction (by recursively substituting the full formula $$\displaystyle \Phi_{i}$$ into the $$\displaystyle \Phi_{i}$$ in the denoninator):

$$\displaystyle \Phi _{i}= 1- \frac{1}{1- \frac{1}{1- \frac{1}{1- \frac{1}{1- ...}}}}$$ (yes, there seems to be another "worst" number for rational approximation).

And there also exists a representation in terms of a nested radical:

$$\displaystyle \Phi _{i}= \sqrt{-1+\sqrt{-1+\sqrt{-1+\sqrt{-1+\sqrt{-1+...}}}}}$$

Note:
This series should (I think) be interpreted as:

$$\displaystyle 1,\ \ \ 1-(\frac12+\frac12 i \sqrt3), \ \ \ 1-\frac{1}{1-(\frac12+\frac12 i \sqrt3)},\ \ \, \ 1-\frac{1}{1-\frac{1}{1-(\frac12+\frac12 i \sqrt3)}}\ \ \ \cdots$$ and "converges" to $$\displaystyle \frac12-\frac12 i \sqrt3$$.

I also tested whether there exists a similar connection to Pascal's triangle and it that also works out very nicely:

It's appears also true that the alternating sum of all n 'Imaginary' Fibonacci numbers equals the (n+2)-th Fibonacci number minus 1.

$$\displaystyle \displaystyle\sum_{k=0}^n (-1)^k F_{i(k)} = F_{i(n+2)}-1$$

And it is also true that the sum of n alternating squares of the 'Imaginary' Fibonacci series is $$\displaystyle F_{i(n)} * F_{i(n+1)}$$:

$$\displaystyle \displaystyle\sum_{k=0}^n (-1)^k F_{i(k)}^2 = F_{i(n)}*F_{i(n+1)}$$

Another nice similarity is that by dividing each n-th (original) Fibonacci number by $$\displaystyle 10^n$$ and then summing the results, the fraction $$\displaystyle \frac{1}{89}$$ results. For the alternating series it is: $$\displaystyle \frac{1}{109}$$.

When applied to the 'Imaginary' Fibonacci series: $$\displaystyle 0,-1,-1, 0, 1, 1, 0,-1,-1, 0, 1, 1, 0...$$ the result is $$\displaystyle \frac{1}{91}$$ and for the alternating version $$\displaystyle \frac{1}{111}$$.

Just as for the normal Golden Ratio, the sine of certain complex numbers involving $$\displaystyle \Phi_{i}$$ gives particularly simple answers, for example:

$$\displaystyle \sin(i \ln(\Phi_{i})) = -\frac12 \sqrt{3}$$

$$\displaystyle \sin(\frac12 \pi - i \ln(\Phi_{i})) = \frac12$$

This is as far as I've got, but guess there are many more advanced connections to be made.

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#### al-mahed

That's cool, I was playing with some continued fractions some time ago and found:

$$\displaystyle \Huge\phi_i=\frac{1}{i-\frac{1}{i-\frac{1}{i-\frac{1}{i-...}}}}\large\ ==>\ \phi_i^2-i\phi_i+1=0$$

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#### Agno

al-mahed said:
That's cool, I was playing with some continued fractions some time ago and found:

$$\displaystyle \Huge\phi_i=\frac{1}{i-\frac{1}{i-\frac{1}{i-\frac{1}{i-...}}}}\large\ ==>\ \phi_i^2-i\phi_i+1=0$$
Nice!

Played a bit further with these fractions:

$$\displaystyle \frac{F_0}{10^1} + \frac{F_1}{10^2} + \frac{F_2}{10^3} + \cdots = \frac{1}{89}$$ (normal Fibonacci)

$$\displaystyle \frac{F_i0}{10^1} + \frac{F_i1}{10^2} + \frac{F_i2}{10^3} + \cdots = \frac{1}{91}$$ ('Imaginary' Fibonacci)

and wondered if other solid fractions would exist when the respective functions:

$$\displaystyle \Phi^2 - \Phi = 1$$ and $$\displaystyle \Phi_{i}^2 - \Phi_{i} = -1$$

would be generalized to $$\displaystyle \Phi_k^2 - \Phi_k = k$$

And the following table (yeah, I love math!) immediately shows a (probably pretty trivial) pattern:

$$\displaystyle \begin{array}{|c|ccc|r|} \hline k & (\frac12+\frac12\sqrt k) & (\frac12+\frac12\sqrt k)^2 & \Phi_{k}^2 - \Phi_{k} & F_{k0}, F_{k1} , F_{k2}... -> \sum_{n=0}^\infty \frac{F_{kn}} {10^{n+1}} \\ \hline -7 & (\frac12+\frac12\sqrt {-7}) & (\frac12+\frac12\sqrt {-7})^2 & -2.00 & 0, -2.00, -2.00, 0, 2.00, 2.00, 0, -2.00 -> -\frac{2}{91} \\ -6 & (\frac12+\frac12\sqrt {-6}) & (\frac12+\frac12\sqrt {-6})^2 & -1.75 & 0, -1.75, -1.75, 0, 1.75, 1.75, 0, -1.75 -> -\frac{1.75}{91} \\ -5 & (\frac12+\frac12\sqrt {-5}) & (\frac12+\frac12\sqrt {-5})^2 & -1.50 & 0, -1.50, -1.50, 0, 1.50, 1.50, 0, -1.50 -> -\frac{1.50}{91} \\ -4 & (\frac12+\frac12\sqrt {-4}) & (\frac12+\frac12\sqrt {-4})^2 & -1.25 & 0, -1.25, -1.25, 0, 1.25, 1.25, 0, -1.25 -> -\frac{1.25}{91} \\ -3 & (\frac12+\frac12\sqrt {-3}) & (\frac12+\frac12\sqrt {-3})^2 & -1.00 & 0, -1.00, -1.00, 0, 1.00, 1.00, 0, -1.00 -> -\frac{1}{91} \\ -2 & (\frac12+\frac12\sqrt {-2}) & (\frac12+\frac12\sqrt {-2})^2 & -0.75 & 0, -0.75, -0.75, 0, 0.75, 0.75, 0, -0.75 -> -\frac{0.75}{91} \\ -1 & (\frac12+\frac12\sqrt {-1}) & (\frac12+\frac12\sqrt {-1})^2 & -0.50 & 0, -0.50, -0.50, 0, 0.50, 0.50, 0, -0.50 -> -\frac{0.50}{91} \\ 0 & (\frac12+\frac12\sqrt{0}) & (\frac12+\frac12\sqrt {0})^2 & -0.25 & 0, -0.25, -0.25, 0, 0.25, 0.25, 0, -0.25 -> -\frac{0.25}{91} \\ 1 & (\frac12+\frac12\sqrt 1) & (\frac12+\frac12\sqrt 1)^2 & 0.00 & 0, 0, 0, 0, 0, 0, 0, 0 -> 0 \\ 2 & (\frac12+\frac12\sqrt 2) & (\frac12+\frac12\sqrt 2)^2 & 0.25 & 0, 0.25, 0.25, 0.50, 0.75, 1.25, 2.00, 3.25 -> \frac{0.25}{89} \\ 3 & (\frac12+\frac12\sqrt 3) & (\frac12+\frac12\sqrt 3)^2 & 0.50 & 0, 0.50, 0.50, 1.00, 1.50, 2.50, 4.00, 6.50 -> \frac{0.50}{89} \\ 4 & (\frac12+\frac12\sqrt 4) & (\frac12+\frac12\sqrt 4)^2 & 0.75 & 0, 0.75, 0.75, 1.50, 2.25, 3.75, 6.00, 9.75 -> \frac{0.75}{89} \\ 5 & (\frac12+\frac12\sqrt 5) & (\frac12+\frac12\sqrt 5)^2 & 1.00 & 0, 1.00, 1.00, 2.00, 3.00, 5.00, 8.00, 13.00 -> \frac{1}{89} \\ 6 & (\frac12+\frac12\sqrt 6) & (\frac12+\frac12\sqrt 6)^2 & 1.25 & 0, 1.25, 1.25, 2.50, 3.75, 6.25, 10.00, 16.25 -> \frac{1.25}{89} \\ 7 & (\frac12+\frac12\sqrt 7) & (\frac12+\frac12\sqrt 7)^2 & 1.50 & 0, 1.50, 1.50, 3.00, 4.50, 7.50, 12.00, 19.50 -> \frac{1.50}{89} \\ \hline \end{array}$$

When k increases by 1 then the difference between $$\displaystyle \Phi_k^2$$ and $$\displaystyle \Phi_k$$ increases by 0.25.

Generalized:

$$\displaystyle k \in \mathbb{R}$$

$$\displaystyle \Phi_k = \frac12 + \frac12 \sqrt{k}$$

$$\displaystyle \Phi_k^2 - \Phi_k = \frac {k-1}{4}$$

$$\displaystyle F_{k0} =0, F_{k1} = \frac {k-1}{4}, F_{k n} = F_{k n-1} + F_{k n-2}$$

$$\displaystyle \sum_{n=0}^\infty \frac{F_{kn}} {10^{(n+1)}} = \begin{cases}\displaystyle\frac{k-1}{4*91}&k<1 \\ \\ \displaystyle\frac{k-1}{4*89}&k\ge 1\end{cases}$$

I'm almost certain (no proof) that the connection to Pascal's triangle (see picture in previous post) is valid for all $$\displaystyle \Phi_k$$.

The value $$\displaystyle \frac {k-1}{4}$$ goes in the top of the triangle, apply the summation rules and all $$\displaystyle F_{kn}$$ emerge on the diagonals (use +++ for $$\displaystyle k \ge 1$$ and -+- for $$\displaystyle k \lt 1$$).

Should also work for $$\displaystyle k = \pi$$, but haven't tried for complex numbers like $$\displaystyle k = i$$.

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#### The Chaz

Forum Staff
This is very interesting. Thanks for showing your investigations, guys!

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#### al-mahed

Agno said:
There is a direct link to $$\displaystyle \pi$$ and $$\displaystyle e$$:

$$\displaystyle \Phi_{i} = e^{\pm \frac{\pi i}{3}}$$
That's the most cool thing; how did you derive it?

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#### CRGreathouse

Forum Staff
The number is quite well known, it's a sixth root of unity. (Or rather, your numbers are two of the sixth roots of unity; the others are 1, -1, and the Eisenstein units $$\displaystyle e^{\pm\frac{2\pi i}{3}}.$$)

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#### Agno

I did find a generic formula for calculating the n-th 'Fibonacci' number in the system $$\displaystyle \Phi_k^2 - \Phi_k = k$$

$$\displaystyle F_{kn} = \begin{cases}\displaystyle \frac{(1-k)}{4} \frac{\left((1-\sqrt{-3})^n - (1+\sqrt{-3})^n\right)}{2^n\sqrt{-3}} & k < 1 \\ \\ \displaystyle\frac{(k-1)}{4} \frac{\left((1+\sqrt{5})^n - (1-\sqrt{5})^n\right)}{2^n\sqrt{5}} & k \ge 1 \end{cases}$$

This formula can generate each cell in the right column of the table above for any value of $$\displaystyle k \in \mathbb{R}, n \in \mathbb{N}$$ .

It also works for $$\displaystyle k \in \mathbb{C}$$, but I think only for $$\displaystyle k = a + bi$$ when $$\displaystyle a \ne 0$$ (this is about ranking complex numbers against real numbers and not sure my reasoning is correct).

Very curious to learn what happens when $$\displaystyle a = 0$$ and whether the discontinued function could be continued in that point?

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#### fbrdata

I was playing with similar ideas. Made some videos - check my youtube f.b.r data science.

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