clc
clear all
%% Exercise 4 - Chapter 24
disp('Exercise 4 - Chapter 24')
disp('INPUTS')
% PURPOSE: Prompt the user to enter a vector of elements
% INPUTS/CODE
user_vec(1)=input('  enter the first element '); % first element
for i=2:1:inf                                    % for an infinte vector
    x=input('  enter the next element ');        % prompt the next element
    if x=='over'   % did the user type 'over' (as a string)?
        break      % if true, end loop
    end
    user_vec(i)=x; % to generate the vector
end
disp('OUTPUTS')
disp('  The vector, as entered')
disp(user_vec')     % to display the vector
%% Exercise 2 - Chapter 25
disp('Exercise 2 - Chapter 25')
% PURPOSE: Find the zeros of an entered function
% NOTE 1: This method only finds one zero of the function, other zeros may exist
% NOTE 2: This method may diverge to infinity 
% NOTE 3: This method only finds real zeros
% INPUTS
disp('INPUTS')
syms x        % symbolic variable
eqn ='x^2-6'; % function to solve x for when equal to zero - any function will work
g   =-34;     % inital guess - the closer the better 
fprintf('  The function is %s\n',char(eqn))
fprintf('  The inital guess is %g\n\n',g)
% CODE
deqn=diff(eqn,x,1); % finding deqn/dx 
deqn=char(deqn);    % inline function requires strings 
f  =inline(eqn);    % using inline to form f(x)
df =inline(deqn);   % using inline to form f'(x)
for i=1:1:50        % conducting 50 reps, which is probably over-done
    g=g-f(g)/df(g); % formula provided
end
% OUTPUTS
disp('OUTPUTS')
fprintf('  One estimated root of the function is %g\n\n',g)
disp('  The exact roots of the function are:')
disp(double(solve(eqn,x)))