Infinite Range of Golden Ratios
Infinite PHI – ∞ φ

PHI series of numbers, 1^st order: 1

Since this is a 1^st order PHI, it has no further iterations. Once is all unity needs to authenticate itself.

It also has no relationship with any other value, so there is no need to approximate its solution by dividing some other number into unity!

Hence, all we get is... X = 1

Subtracting 1 from both sides, we get...
A linear polynomial in one unknown: X − 1 = 0

PHI series of numbers, 2^nd order:

After 35 iterations, it's pretty obvious that the approximation of the 2^nd order of PHI accurate to eleven decimal places is... 1.6180339887499

And it's reciprocal is... 0.6180339887499

To be able to make a 2^nd order polynomial in one unknown out of these two numbers will require that one of them is given a negative sign value... − 0.6180339887499

Now we can form this polynomial by multiplying these two values together. But first, we have to turn them into linear expressions in one unknown...

X₁ = 1.6180339887499

Subtract 1.6180339887499 from both sides of the equal sign...

X₁ − 1.6180339887499 = 1.6180339887499 − 1.6180339887499

Yields...
(X − 1.6180339887499) = 0

X₂ = − 0.6180339887499

Add 0.6180339887499 to both sides of the equal sign...

X₂ + 0.6180339887499 = − 0.6180339887499 + 0.6180339887499

Yields...
(X + 0.6180339887499) = 0

Multiplying these two roots together...
(X − 1.6180339887499) × (X + 0.6180339887499) = 0

Yields...
X² + 0.6180339887499X¹ − 1.6180339887499X¹ − 1 = 0

Simplifying further, yields a quadratic polynomial in one unknown...
X² − X¹ − 1 = 0

PHI series of numbers, 3^rd order:

After 54 iterations, the approximation of the three roots of the 2^nd order of PHI accurate to eleven decimal places are...

X₁ = a/c = 5263322654587402449 ÷ 2342398945624433584 = 2.2469796037175
X₂ = b/a = 4220857052430132287 ÷ 5263322654587402449 = 0.80193773580484
X₃ = c/b = 2342398945624433584 ÷ 4220857052430132287 = 0.55495813208737

To be able to make a 3^rd order polynomial in one unknown out of these three numbers will require that one of them is given a negative sign value... − 0.80193773580484

Now we can form this polynomial by multiplying these three values together. But first, we have to turn them into linear expressions in one unknown...

X₁ = 2.2469796037175

Subtract 2.2469796037175 from both sides of the equal sign...

X₁ − 2.2469796037175 = 2.2469796037175 − 2.2469796037175

Yields...
(X − 2.2469796037175) = 0

X₂ = − 0.80193773580484

Add 0.80193773580484 to both sides of the equal sign...

X₂ + 0.80193773580484 = − 0.80193773580484 + 0.80193773580484

Yields...
(X + 0.80193773580484) = 0

X₃ = 0.55495813208737

Subtract 0.55495813208737 from both sides of the equal sign...

X₃ − 0.55495813208737 = 0.55495813208737 − 0.55495813208737

Yields...
(X − 0.55495813208737) = 0

Multiplying these three roots together...
(X − 2.2469796037175) × (X + 0.80193773580484) × (X − 0.55495813208737) = 0

Yields...
X³ − 2.2469796037175X² + 0.80193773580484X² − 0.55495813208737X² − 1.8019377358048X¹ + 1.2469796037175X¹ − 0.44504186791263X¹ + 1 = 0

Simplifying further, yields a 3^rd order polynomial in one unknown...
X³ -2X² -1X¹ + 1 = 0

PHI series of numbers, 4^th order:

After 42 iterations, the approximation of the four roots of the 4^th order of PHI accurate to eleven decimal places are...

X₁ = a/d = 8416100665310424765 ÷ 2922881087185190704 = 2.8793852415718
X₂ = b/b = 7400994716656735704 ÷ 7400994716656735704 = 1
X₃ = c/a = 5493219578125234060 ÷ 8416100665310424765 = 0.65270364466614
X₄ = d/c = 2922881087185190704 ÷ 5493219578125234060 = 0.53208888623796

To be able to make a 4^th order polynomial in one unknown out of these four numbers will require that two of them are given a negative sign value... − 1 and − 0.53208888623796

Now we can form this polynomial by multiplying these four values together. But first, we have to turn them into linear expressions in one unknown...

X₁ = 2.8793852415718

Subtract 2.8793852415718 from both sides of the equal sign...

X₁ − 2.8793852415718 = 2.8793852415718 − 2.8793852415718

Yields...
(X − 2.8793852415718) = 0

X₂ = − 1

Add 1 to both sides of the equal sign...

X₂ + 1 = − 1 + 1

Yields...
(X + 1) = 0

X₃ = 0.65270364466614

Subtract 0.65270364466614 from both sides of the equal sign...

X₃ − 0.65270364466614 = 0.65270364466614 − 0.65270364466614

Yields...
(X − 0.65270364466614) = 0

X₄ = − 0.53208888623796

Add 0.53208888623796 to both sides of the equal sign...

X₄ + 0.53208888623796 = − 0.53208888623796 + 0.53208888623796

Yields...
(X + 0.53208888623796) = 0

Multiplying these four roots together...
(X − 2.8793852415718) × (X + 1) × (X − 0.65270364466614) × (X + 0.53208888623796) = 0

Yields...
X⁴ − 2.8793852415718X³ + 1X³ − 0.65270364466614X³ + 0.53208888623796X³ Ê
Ä − 2.8793852415718X² + 1.8793852415718X² − 1.532088886238X² Ê
Ä − 0.65270364466614X² + 0.53208888623796X² − 0.34729635533386X² Ê
Ä + 1.8793852415718X¹ − 1.532088886238X¹ + 1X¹ − 0.34729635533386X¹ + 1 = 0

Simplifying further, yields a 4^th order polynomial in one unknown...
X⁴ -2X³ -3X² + 1X¹ + 1 = 0

The Expanded Euclidean Algorithm arises from the Infinite Range of Golden Ratios...