Imaginary Golden Ratio

Jan 2011
120
2
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.
 
Last edited by a moderator:
Dec 2007
687
47
That's cool, I was playing with some continued fractions some time ago and found:

\(\displaystyle \Huge\phi_i=\frac{1}{i-\frac{1}{i-\frac{1}{i-\frac{1}{i-...}}}}\large\ ==>\ \phi_i^2-i\phi_i+1=0\)
 
Last edited by a moderator:
Jan 2011
120
2
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 \(\displaystyle k = i\).
 
Last edited by a moderator:

The Chaz

Forum Staff
Nov 2009
2,767
5
Northwest Arkansas
This is very interesting. Thanks for showing your investigations, guys!
 
Last edited by a moderator:
Dec 2007
687
47
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?
 
Last edited by a moderator:

CRGreathouse

Forum Staff
Nov 2006
16,046
936
UTC -5
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 \(\displaystyle e^{\pm\frac{2\pi i}{3}}.\))
 
Last edited by a moderator:
Jan 2011
120
2
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?
 
Last edited by a moderator:
Nov 2019
1
0
Ibiza
I was playing with similar ideas. Made some videos - check my youtube f.b.r data science.
 
Last edited by a moderator: