%% Annotated Matlab transcript -- 4/17
%% CS 323 - Doug DeCarlo

% Numerical differentiation

n = 1:10;
h = 2.^(-n)

x = pi/6;
% approximate cos'(x) at x=pi/6 for different h's
d = (cos(x+h)-cos(x))./h;
% error (actualy answer is -1/2)
e = -0.5 - d;
plot(n,e,'.-')
% ratio approaches 2 as error is O(h)
e(1:end-1)./e(2:end)
% semilog plot of error shows a line
semilogy(n,e,'.-');

% we can also plot the upper/lower bounds on the error
plot(n,e,'.-',n,h/2*cos(pi/6),n,h/2.*cos(pi/6+h))
semilogy(n,e,'.-',n,h/2*cos(pi/6),n,h/2.*cos(pi/6+h))