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
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
$$\varphi=\frac{1+\sqrt{5}}{2}\approx\phantom{+}1.61803$$
$$\phi=\frac{1-\sqrt{5}}{2}\approx-0.61803$$
$$\phi=1-\varphi=-1/\varphi$$
Fibonacci Series & $\varphi$ Limit
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 $\varphi^2=\varphi+1$, 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
$$F(x)=\frac{\varphi^x-(1-\varphi)^x}{\sqrt{5}},\medspace x\isin\Z$$
$$F(x)=\frac{\varphi^x-\cos(\pi\cdot x)\cdot\varphi^{-x}}{\sqrt{5}},\medspace x\isin\R$$
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
5.2.4 The Plastic Ratio $\rho$
Definition
The only real solution of $\rho^3=\rho+1$
$$\rho=\frac{a}{b}=\frac{b+c}{a}=\frac{b}{c}$$
$$a>b>c>0$$
Ratio
$$\rho=\sqrt[3]{\frac{9+\sqrt{69}}{18}}+\sqrt[3]{\frac{9-\sqrt{69}}{18}}\approx1.32472$$
Plastic Series & $\rho$ Limit
A recursive function whose numbers are the sum of the previous two numbers before the one before it, starting with $0,1,1$
$$F_n=F_{n-2}+F_{n-3}$$
$$\{0,1,1,1,2,2,3,4,5,7,9,12,...\}$$
As the numbers of the recursive sequence become larger, the number divided by its previous number approaches the limits
$$\lim\limits_{x\to\infin}\frac{F(x)}{F(x-1)}=\rho$$
Proof of ratio
Let $\rho=a+b$, and recall the
cubic binomial
$$(a+b)^3=3\cdot a\cdot b\cdot(a+b)+a^3+b^3$$
Substitute $\rho$ for $a+b$
$$\rho^3=3\cdot a\cdot b \cdot \rho+a^3+b^3$$
Equate the coefficients with the original equation
$$3\cdot a\cdot b=1\medspace\land\medspace a^3+b^3=1$$
Cube the first equation and isolate $b^3$
$$b^3=\frac{1}{27\cdot a^3}$$
Substitute it into the second equation
$$a^3+\frac{1}{27\cdot a^3}=1$$
Multiply by $a^3$ and isolate $0$
$$a^{3^2}-a^3+\frac{1}{27}=0$$
Apply the
quadratic formula to solve for $a^3$, and take the cube root
$$a=\sqrt[3]{\frac{9\pm\sqrt{69}}{18}}$$
Since $a$ and $b$ are interchangeable, one is $+$, the other is $-$. Add $a+b$.
$$a+b=\sqrt[3]{\frac{9+\sqrt{69}}{18}}+\sqrt[3]{\frac{9-\sqrt{69}}{18}}$$
Substitute $\rho$ for $a+b$
