5.2 Contents

1. Proof of$\medspace\pi$ 🔧

2. The Natural Number$\medspace e$ 🔧

3. The Golden Ratio$\medspace\varphi$ 🔧

4. The Plastic Ratio 🔧

5.2.1 Proof of $\pi$ 🔧

In regular polygons, as the number of sides becomes infinite, the perimeter approaches a circumference, wherein $\pi$ can be manually calculated.

5.2.2 The Natural Number $e$ 🔧

Continually compounded growth with 100% (1) return at a continuous rate yields a limit at $e$ $$e=\lim_{x\to\infin} {\bigg( 1+\frac{1}{x} \bigg)}^x=\lim_{x\to 0} {(1+x)}^{1/x}\approx 2.7 \medspace 1828 \medspace 1828 \medspace 459$$

$e$ Limit

$$\lim_{x\to 0}\frac{e^x-1}{x}=1$$

Natural Logarithm

The inverse function of $e^x$ is $\log_e x$, represented as $\ln(x)$

Limits of Natural Exponents & Logarithms

$$\lim_{x\to\pm\infin}e^{\pm x}=\infin\qquad\lim_{x\to\pm\infin}e^{\mp x}=0$$ $$\lim\limits_{x\to\infin}\ln(x)=\infin\qquad\lim\limits_{x\to 0^+}=-\infin\thickspace$$

Proof of $e$ Limit

Given the second equality $$e=\lim_{x\to 0}{(1+x)}^{1/x}$$

5.2.3 The Golden Ratio $\varphi$ 🔧

Definition

Two quantities are in the golden ratio if their ratio is the same as the ratio of their sum to the larger of the two quantities $$\frac{x+a}{x}=\frac{x}{a}$$

Ratios

$$\varphi=\frac{1+\sqrt{5}}{2}\approx1.61803$$ $$\phi=\frac{1-\sqrt{5}}{2}\approx-0.61803$$

Proof of Ratios

Given the ratio equality, substitute 1 for $a$ $$\frac{x+1}{x}=x$$ Multiply by $x$ $$x+1=x^2$$ Rearrange to appear as a standard quadratic equation $$x^2-x-1=0$$ Use the quadratic formula to find the values of $x$ $$x=\frac{-(-1)\pm\sqrt{(-1)^2-4\cdot 1\cdot-1}}{2\cdot 1}$$ Simplify

Fibonacci Series & $\varphi$ Limit

A series of a recursive function whose numbers are the sum of its previous two numbers, starting with 1,1 $$F_n=F_{n+1}+F_{n+2}$$ $$\{1,1,2,3,5,8,13,21,34,55,89,144,...\}$$ It can be rearranged to determine negative values before 1,1 $$F_{n+2}=F_n-F_{n+1}$$ $$\{...,-144,+89,-55,+34,-21,+13,-8,+5,-3,+2,-1,1,0,1,1\}$$ As the numbers of the recursive sequence become larger, the number divided by its previous number approaches the limits $$\lim\limits_{x\to\pm\infin}\frac{F(x)}{F(x-1)}=\pm\varphi^{\pm 1}$$

Golden Ratio Powers

Since $x^2=x+1$ as stated in the proof of ratios, $\varphi^2=\varphi+1$, and the following holds for all powers of $\varphi$ $$\varphi^x=\varphi^{x-1}+\varphi^{x-2}$$ When calculated, the result simplifies to $$\varphi^x=F(x)\cdot\varphi+F(x-1)$$

Algebraic Functions for the Fibonacci Series

$$F(x)=\frac{\varphi^x-(1-\varphi)^x}{\sqrt{5}},\medspace x\isin\Z$$ $$F(x)=\frac{\varphi^x-\cos(\pi\cdot x)\cdot\varphi^{-x}}{\sqrt{5}},\medspace x\isin\R$$

5.2.4 The Plastic Ratio 🔧

$$\rho=\frac{a}{b}=\frac{b+c}{a}=\frac{b}{c}$$ $$a>b>c>0$$ The only real solution of $x^3=x+1$

$\approx1.32472$

(Need to insert cubic formula in 2.3)