**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.