Golden Ratio

We say two quantities are in the golden ratio if their ratio is equivalent to the ratio of the sum of the quantities to the larger of the two quantities. It is obtained by dividing a line into two parts such that the longer part divided by the smaller part is also equal to the whole length divided by the longer part:

The golden ratio

The golden ratio, also called the golden number, divine proportion, etc., has a very close association with the Fibonacci sequence. It is often represented by the Greek letter phi, $\varphi$ or $\phi$. For the sake of uniformity, we shall denote the golden ratio by $\phi$.

We say two quantities $a$ and $b$, where $a>b$, are in the golden ratio, if

$$\frac{a}{b}=\frac{a+b}{a}.$$

In this section, we will try to evaluate the numerical value of the golden number or the golden ratio. We know that two quantities $a$ and $b$, where $a>b$, are in the golden ratio, if

$$\phi=\frac{a}{b}=\frac{a+b}{a}.$$

Then, we have

$$\begin{aligned} \phi=\frac{a}{b} & =\frac{a+b}{a} \\ & =1+\frac{b}{a} \\ & =1+\frac{1}{\phi} \qquad \left(\text{since } \frac{a}{b}=\phi\right) \\ \Rightarrow \phi &=\frac{\phi+1}{\phi} \\ \Rightarrow \phi^2-\phi -1 &= 0. \end{aligned}$$

The last equation, being a quadratic equation, has its roots given by the quadratic formula as

$$\begin{aligned} \phi & =\frac{1\pm\sqrt{(-1)^2-4\cdot(-1)\cdot1}}{2} \\ & = \frac{1\pm\sqrt{5}}{2}. \end{aligned}$$

But since the ratio is always positive,

$$\phi=\frac{\sqrt5+1}{2}\approx 1.61803.$$

The golden ratio can be expressed as a beautiful infinite continued fraction:

$$\displaystyle \phi = 1+ \cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots}}}$$

The golden ratio, or simply the golden number, is a very special number and can be found in some very beautiful nested functions. Some of them are discussed below.

$$\phi=\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}$$

Let

$$\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}} =x.$$

Then

$$\begin{aligned} \sqrt{1+x} &= x \\ 1+x &=x^2 \\ x^2-x-1 &= 0. \end{aligned}$$

Note that this is the equation which has roots $\frac{1\pm\sqrt{5}}{2}$. But since the radical is positive,

$$x=\frac{\sqrt{5}+1}{2}=\phi. \ _\square$$

$$\phi = \sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2+ \cdots }}}}}$$

Let $x=\sqrt{a-\sqrt{a+\sqrt{a-\sqrt{a+\cdots}}}}$ and $y=\sqrt{a+\sqrt{a-\sqrt{a+\sqrt{a-\sqrt{a+\cdots}}}}},$ then

$$\begin{aligned} x^2&=a-y &\qquad (1)\\ y^2&=a+x. &\qquad (2) \end{aligned}$$

Eliminating $a$ from (1) and (2), we get $x=y-1$. Now, putting this in (1) gives

$$y^2-y+(1-a)=0 \implies y=\frac{1+\sqrt{4a-3}}{2}.$$

Now, putting $a$ = 2, we get the given nested radical. $_\square$