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=11x+1y? ("Reciprocal addition" common for parallel resistors)
and as you can see, the operation of interest is x⊕y=11x+1y=xyx+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!
x⊕y=x−y
11x+1y=x−y
xyx+y=x−y
xy=x2−y2
0=x2−xy−y2
and, using the quadratic formula,
x=y±√y2+4y22
x=y1±√52
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
11x+1y=x+y
xyx+y=x+y
xy=x2+2xy+y2
0=x2+xy+y2
x=−y±√y2−4y22
x=y1±√−32
x=y1±√3i2
and this is the "imaginary golden ratio"!
φi=1+√3i2
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.
φi=1−1φi
φ2i=φi−1
and generally
φni=φn−1i−φn−2i
This complex ratio also lies on the unit circle in the complex plane, and has a representation as a power of e!
φi=cos(π/3)+isin(π/3)=eiπ/3
|φi|=1
It is also a nested radical, because of the identity φ2i+1=φi
φi=√−1+√−1+√−1+√−1+...
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
fn=fn−1−fn−2
f0=0
f1=1
0,1,1,0,−1,−1,0,1,1,...
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.
Answer
fn=(ϕ−ni−ϕni)i√3
You can prove this by induction on n, using the fact that ϕi and ϕ−1i are the two solutions to the quadratic equation x2−x+1=0. (Alternatively, it's easy to see that the three sequences ⟨ϕni|n∈N⟩, ⟨ϕ−ni|n∈N⟩, and ⟨fn|n∈N⟩ 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 fn 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 u1=⟨ϕni|n∈N⟩ and u2=⟨ϕ−ni|n∈N⟩ satisfy the recurrence relation (because ϕi and ϕ−1i are roots of the quadratic equation above), so u1 and u2 belong to V.
You can check that {u1,u2} is a basis for V. The formula I gave for fn is just the particular linear combination of u1 and u2 that happens to yield the sequence ⟨fn|n∈N⟩.
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