The Golden Ratio: Reciprocals and Powers
Determining the Golden Ratio
Take a line
segment with length of x + 1, and split it into two segments, one with width x,
the other width 1. (see the black lines on the left)
The golden
ratio is defined when given positive numbers a and b, where a > b, the
following ratio is true:
(a + b) / a = a
/ b
Applying this
to the above diagram:
(x + 1) / x = x
/ 1
Solving for x
yields:
(x + 1) * 1 = x
* x
x + 1 = x^2
x^2 – x – 1 = 0
By the
quadratic equation:
x = (1 ± √(1^2 –
4 * 1 * -1)) / 2
x = ( 1 ± √5 )
/ 2 (I)
Since x is a
measure, we will consider only the positive root and define the golden ratio (ϕ)
as:
ϕ = (1 + √5) / 2 (II)
ϕ ≈ 1.6180339
Reciprocals of ϕ
1 /
ϕ
1 / ϕ
= 2 / (1 + √5)
= 2 / (1 + √5)
* (1 - √5) / (1 - √5)
= 2 * (1 - √5)
/ -4
= (2 – 2 * √5)
/ -4
= (2 * √5 – 2)
/ 4
= (√5 – 1) / 2
Here is where a
trick comes in, we’re going to add and subtract 1:
= (√5 – 1 + 1 –
1) / 2
Now, note that:
= (√5 + 1 – 2)
/ 2
= (√5 + 1) / 2 –
2 / 2
= (√5 + 1) /
2 - 1
= ϕ – 1
Hence: 1 / ϕ = ϕ – 1
1 /
ϕ^2
1 / ϕ^2
= (2 / (1 + √5)
)^2
= 4 / (6 + 2 * √5)
= 4 / (6 + 2 * √5)
* (6 – 2 * √5) / (6 – 2 * √5)
= (24 – 8 * √5)
/ 16
= (3 - √5 ) / 2
Note that 4 – 1
= 3…
= (4 – 1 - √5)
/ 2
= 4 / 2 – (1 + √
5) / 2
= 2 - ϕ
Hence: 1 / ϕ^2 = 2 - ϕ
Powers of ϕ
ϕ^2
ϕ^2
= ( (1 + √5) /
2 )^2
= 1 / 4 * (1 +
2 * √5 + 5)
= 1 / 4 * (6 +
2 * √5)
= (3 + √5) / 2
= (2 + 1 + √5)
/ 2
= 1 + (1 + √5)
/ 2
= 1 + ϕ
Hence: ϕ^2 = 1 + ϕ
Take note, this
plays in determining further powers of ϕ.
ϕ^3
ϕ^3
= ϕ * ϕ^2
Using the fact
that ϕ^2 = 1 + ϕ
= ϕ * (1 + ϕ)
= ϕ + ϕ^2
= ϕ + 1 + ϕ
= 1 + 2 * ϕ
Hence: ϕ^3 = 1 + 2 * ϕ
ϕ^4
ϕ^4
= ϕ * ϕ^3
= ϕ * (1 + 2 * ϕ)
= ϕ + 2 * ϕ^2
= ϕ + 2 * (1 + ϕ)
= ϕ + 2 + 2 * ϕ
= 2 + 3 * ϕ
With ϕ^4 = 2 +
3 * ϕ. Moving on…
ϕ^5
ϕ^5
= ϕ * ϕ^4
= ϕ * (2 + 3 * ϕ)
= 2 * ϕ + 3 * ϕ^2
= 2 * ϕ + 3 *
(1 + ϕ)
= 2 * ϕ + 3 + 3
* ϕ
= 3 + 5 * ϕ
Hence: ϕ^5 = 3 + 5 * ϕ
ϕ^6
Quickly…
ϕ^6
= ϕ * ϕ^5
= ϕ * (3 + 5 * ϕ)
= 5 + 8 * ϕ
A
Connection to the Fibonacci Sequence
Sense a
pattern?
ϕ^2 = 1 + ϕ
ϕ^3 = 1 + 2 * ϕ
ϕ^4 = 2 + 3 * ϕ
ϕ^5 = 3 + 5 * ϕ
ϕ^6 = 5 + 8 * ϕ
Note the
coefficients come from the Fibonacci Sequence.
(1,1,2,3,5,8,13… ) The Fibonacci sequence is defined as:
F_n = F_n-1 +
F_n-2 where the first two initial terms
are 1 and 1.
Can we show
that ϕ^n = F_n+1 + F_n * ϕ?
Proof
Let assume that
the base case of:
ϕ^n = F_n-1 +
F_n * ϕ
Will this
relation hold up for ϕ^(n+1)? Then:
ϕ^(n + 1)
= ϕ * ϕ^n
= ϕ * (F_n-1 +
F_n * ϕ)
= ϕ * F_n-1 +
F_n * ϕ^2
= F_n-1 * ϕ +
F_n (1 + ϕ)
= F_n-1 * ϕ +
F_n + F_n * ϕ
= F_n + (F_n-1
+ F_n) * ϕ
Using the
definition of the sequence, F_n-1 + F_n = F_n+1. Hence:
= F_n + F_n+1 *
ϕ
Since ϕ^(n + 1)
= F_n + F_n+1 * ϕ, the relation holds.
Next time I
have a retro review of an early financial calculator from the 1980s, the TI
BA-II (without any plus or other annotation).
I have couple more retro calculator reviews coming (tentatively) later
this month: Casio CM-100 and Canon
FS-5.
That’s all for
now! Take care,
Eddie
This blog is property
of Edward Shore, 2017.
