Saturday, February 4, 2017

The Golden Ratio: Reciprocals and Powers

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.

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