Golden simultaneous equations

Posted on Posted in Mathematics

Abstract

In this paper, we investigate the simultaneous equations derived from the Golden Ratio, and use differentiation to find the coordinates where Φ in the x-axis meets Φ in the y-axis in a golden quadratic graph. We demonstrate a proof for the solutions to the five simultaneous equations below by using algebraic and graphic solutions.

Introduction

The Golden Ratio is a number represented by the Greek letter Φ and is approximately equal to 1.6180339887…; it is irrational and is infinitely long, similar to π. Nevertheless, Φ can be represented by a fraction (which contains √5, an irrational number). The fraction which represents Φ is 1 +2 5 . Φ is an extremely interesting number; it appears everywhere in nature. The number is a ratio, the Golden Ratio, and it represents the “perfect cut” of nature. Everything in nature, from the number of pollen grains on a flower to the number of bonds in the compounds of uranium oxide, to the spiral shape of galaxies can be accurately modeled by Φ. Φ is a ratio, and it is often shown in nature as 1:Φ, such as the ratio of the distance between the shoulder and the elbow, and between the elbow and the fingertips. That is yet another Golden Ratio. However, if the larger section is considered to be 1, the smaller ratio can be Φ, known as the Lesser Golden Ratio.

In this paper, we aim to prove that the only possible solution to the simultaneous equations below is to use the Greater and Lesser Golden Ratios, therefore making the equations mathematically “beautiful.”

Most natural feats which contain the Golden Ratio are considered to be esthetically “beautiful.”

Simultaneous Equations

First, we introduce a set of simultaneous equations, which are at the center of this paper.

1. $x -1 = y$
2. $\frac{1}{x} = y\$
3. $n = x$
4. $n – 1 = x$
5. $2 + y = n$

Obviously, it is possible to rearrange these equations to create a longer string of equations such as: x -1=n – 2 = n − 1 = y, or n=2+y=x2=x-1. However, these cannot give any numerical value for the constants x, y, and z. This is why, the quadratic formula must be used. The final answer that we want to reach is

$x = \frac{1+\sqrt{5}}{2}$

therefore, we know that there must be a way of using ax2+bx+c=0. We shall then find the values of a, b,

and c which will give us the value of x by using the quadratic formula,

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

The values of a, b, and c must eventually make

$x = \frac{1+\sqrt{5}}{2}$

which is Φ ≈ 1.6180339887… Therefore, we use the formula

$\phi = \frac{1+\sqrt{5}}{2}$

and substitution to find the numerical values which would go into the formula as a, b, and c.

Algebraic Proof

Since we know that $\phi = \frac{1+\sqrt{5}}{2}$, 1=–b ∴ b=–1. 2

Then, $\sqrt{5} = \sqrt{b^2-4ac}$, but we know that

b = -1 ∴ $sqrt{5} = \sqrt{1-4ac}$ ∴ -12 = 1 ∴ -x2 = x

At this stage, it is much simpler to work out a because of the 2 in the Φ formula. It is evident that a=1 because 2=2a ∴ a=1. At this point, it becomes clear what the missing constant c is:

$\sqrt{5} = \sqrt{1-4c} \therefore \sqrt{5}$ -1∴ 4c = −4 ∴ c = −1

Another way to determine these constants is to use some of the simultaneous equations. Initially, we calculate:

$x – 1 = \frac{1}{x} \therefore x(x-1) = 1 = x^{2} – x = 1$

Therefore, we have obtained values for a, b, and c to use in the quadratic formula.

Now that we have all three of the constants needed for the quadratic formula to work, we must prove that it does;

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x = \frac{1\pm \sqrt{5}}{2}a = 1, b = -1, c=-1$

We now have x = ϕ ≈ 1 . 6180339887 … ∴ y = x – 1 ≈ 0.6180339887… ∴ n = x + 1 ≈ x + 1 ≈ 2.6180339887 and x = – φ, y = – ϕ, z = 0.382

Graphic Solution

Now that we have sorted out the equations Figure 1: Quadratic graph of the function y = x2 – x – 1 Figure 2: x-intercepts of the function y = x2 – x – 1; crossing at Φ and –ϕ

algebraically, we can also use graphs and differentiation to find the lowest possible point of the curve y = x2 – x – 1, which is derived from our a, b, and c from the quadratic formula. On a graph we can sketch this curve to find the x-intersects and the minimum point (not maximum because it is quadratic and continues increasing to infinity).

From Figure 1, it is possible to say that the x-intersects of the curve are Φ and –ϕ, the Golden Ratios.

Differentiation: y = x2 − x − 1

$\therefore (\frac{dy}{dx}(x^{n})=nx^{n-1}) \rightarrow \frac{dy}{dx} =\frac{d(x^{2}-x-1)}{dx}=2x-1=0$

$\therefore x=\frac{1}{2} \therefore \frac{1^{2}}{2}-\frac{1}{2}-1=\frac{-5}{4}=y$

Therefore, we can see that the minimum point of this curve has the coordinates

$(\frac{1}{2}, -\frac{5}{4})$

Figure 2 shows a magnification of the graph at the point (ϕ, 0), showing the exact point of the Golden Ratio.

References

1. Olsen S. The Golden Section: Nature’s Greatest Secret. Wooden Books. Walker & Company; 2006.
2. International Conference on Data Engineering. 16th International Conference on Data Engineering: 29 February-3 March 2000. San Diego, California: Institute of Electrical & Electronics Engineers; 2000.