exponential.m

%% Exponentiation and Compounding Interest Rate
% *back to* <https://fanwangecon.github.io *Fan*>*'s* <https://fanwangecon.github.io/Math4Econ/ 
% *Intro Math for Econ*>*,*  <https://fanwangecon.github.io/M4Econ/ *Matlab Examples*>*, 
% or* <https://fanwangecon.github.io/MEconTools/ *MEconTools*> *Repositories*
% 
% _See also_: <https://fanwangecon.github.io/Math4Econ/explog/exolog.html Exponential 
% Function and Log Function>.
%% *Exponential Function*
%% 
% * *Exopential Function:* Functions where the variable $x$ appears as an _exponent:_ 
% $a^x$
% * $a$ is the base of Exponential function.
%% 
% Remember that
%% 
% * $a^0=1$
% * $a^{\frac{1}{2}}=\sqrt{a}$
% * if  $a^b = c$, we can also write, $a = c^{\frac{1}{b}}$, for example, $2^3=8$, 
% and $2=8^{\frac{1}{3}}$
% * $a^{-b} = \frac{1}{a^b}$
% * $x^a\cdot x^b = x^{a+b}$
% * $x^{a\cdot b} = (x^a)^b$
%% Exponential Function Graphs?
%% 
% * Note that the domain of exponential function includes positive and negative 
% $x$, and the exponential function will always be positive.
% * If base is below $1$, then the curve is monotonically downward sloping
% * If base is above $1$, then the curve is monotonically upwards sloping
% * If base is above $1$, higher base leads to steeper curvature. 

syms x
a1 = 0.5;
f_a1 = a1^(x);
a2 = 1.5;
f_a2 = a2^(x);
a3 = 2.5;
f_a3 = a3^(x);
figure();
hold on;
fplot(f_a1, [-2, 2]);
fplot(f_a2, [-2, 2]);
fplot(f_a3, [-2, 2]);
line([0,0],ylim);
line(xlim, [0,0]);
title('Exponential Function Graph with different bases')
legend(['base=',num2str(a1)], ['base=',num2str(a2)],['base=',num2str(a3)]);
grid on;
%% Infinitely Compounding Interest rate
% with 100 percent interest rate (APR), if we compound $N$ times within a year, 
% interest we pay at the end of the year is
%% 
% * $(1 + \frac{1}{N})^N-1$
%% 
% Suppose $N=5$ (You can also think of this as a loan with interest rate of 
% $20$% for every $73$ days), then we pay $159$% interest rate by the end of the 
% year. 

r = 1.05;
N = 5;
(1 + r/N)^N - 1
%% 
% What if we do more and more compounding, if we say interest rate compounds 
% $10$, $50$, $100$ times over the year, what happens? With APR at 100%, the total 
% interest rate you pay at the end of the year does not go to infinity, rather, 
% it converges to this special number $e$, Euler's number, $2.7182818$. This means 
% if every second the interest rate is compounding, with an APR of 100%, you end 
% up paying 272% of what you borrowed by the end of the year, which is 172% interest 
% rate. 
%% 
% * $\lim_{N \rightarrow \inf} (1 + \frac{1}{N})^{N} = e \approx 2.7182818$
%% 
% We can visualize this limit below

r = 1;
syms N
f_compoundR = (1 + r/N)^N;
figure();
fplot(f_compoundR, [1,100])
ylabel({'Principle and Interests at End of Year Given 100% APR' 'for 1 dollar Borrowed, given infinite compounding'})
xlabel('Number of Evenly-divided Times to Compound Interest Rate in a Year')
grid on;
grid minor
double(subs(f_compoundR,[1,2,3,4,5,6,7,8,9,10]))
%% Infinitely compounding Interest rate, different $r$ (APR $r$)
% Given:
%% 
% * $\lim_{N \rightarrow \inf} (1 + \frac{1}{N})^{N} = e \approx 2.7182818$
%% 
% What is
%% 
% * $\lim_{N \rightarrow \inf} (1 + \frac{r}{N})^{N}$?
%% 
% We can replace $N$ by $N=r \cdot M$
%% 
% * $\lim_{N \rightarrow \inf} (1 + \frac{r}{N})^{N} = \lim_{M \rightarrow \inf} 
% (1 + \frac{r}{r \cdot M})^{r \cdot M} = \left( \lim_{M \rightarrow \inf} (1 
% + \frac{1}{M})^{M} \right)^r = e^r$
%% 
% This gives the base $e$ exponential function a financial interpretation. 

syms x
f_e = exp(x);
figure();
hold on;
fplot(f_e, [-3, 3]);
line([0,0],ylim);
line(xlim, [0,0]);
title('Exponential Function Graph with base e')
xlabel('r = interest rate');
ylabel({'Principle and Interests at End of Year' 'for 1 dollar Borrowed, given infinite compounding'})
grid on;