%% Lecture 6
%
% Last week.  
% a) Taylor Series expansion, and linear interpolation.
% b) Simulating time varying functions?
% c) Functions with multiple outputs.
%
% This week.
% a) Estimating Derivatives from functions
% b) Newton's method.
% c) Stopping conditions.

% One key question is "Finding the roots of an equation", finding the
% values x such that f(x) = 0.

% You may have seen this in math classes in the past, but one key
% question that is rarely addressed in a math class is:
%  How is the function f represented?

% What are reasonable choices for this?
% 1-- Maybe you have an array of the function values evaluated at many
%        points?  (e.g. you have the stock market closing prices for the last
%        10 years)
% 2-- Maybe you know the function completely, e.g. you know it is a
%        polynomial and you know the ceofficients?
% 3-- Maybe you have a "matlab function" that you can plug in values and
%        get results?

% How could we compute the zeros of a function for which we have samples of
%   the function for all values x from 1 to 1000?


% If we know the function completely?  If the function is something
% like y = x.^2 + 2x - 14
%
%   Maybe you can solve this directly (completing the square or
%   something, but if you can't, you can at least compute the derivative
%   and use newton's method).  What is the key to newtons method?
%
%  The Fundamental Theorem of Engineering, things are linear.
%
%  Taylor expansion, first order    y(x+h) = y(x) + dy/dx * h
%  As a side note, this is very similar to the "definition of derivative"
%                     dy/dx = ( y(x+h)-y(x) ) / h      as h gets small...

% so if we want to solve for when y is zero, we remove the assumption
% that "h" is small (and add the assumption that the function is linear)
% to get:

% y(x0+h) = y(x0) + y'(x0) * h

% 0 = y(x0) + y' * h

% h = -y(x0) / y'

% so the next guess for x is x0 + h.


function x = newton(x0)
%
% Special case coded for function y = x.^2 + 2x - 14
x = x0;
y = x.^2 + 2*x - 14;
% next class, when y = foo(x);
while abs(y) > 0.01
    % solve for value y = f(x0);
    y = x.^2 + 2*x - 14;
    % solve for value of y' = df(x0)/dx
    yprime = 2*x + 2;
    % solve for offset = -y / yprime
    offset = -y / yprime;
    x = x + offset
end



% Given some matlab function "foo" that calculates the value of the function.