% three4.m

% NEUR 1680, Spring 2008

% It is easy to show that in order to get equality
%    (rather than mere proportionality)
%    in the equation appearing in this exercise
%    we need to put a coefficient of 2/N in front of the summation sign,
%    N being the number of unit vectors, i.e., "neurons".

% This can be seen after a little algebra, and using the facts that
%   if theta_i, i=1,...,N are N independent uniformly distributed angles on [0,2*pi]
%   then

%   (1/N) * SUM (cos(theta_i))^2 is approximately .5, and
%   (1/N) * SUM (cos(theta_i))*(sin(theta_i)) is approximately 0.

% These two facts can be proven by realizing that the above sums are
%    discrete-sum approximations of the definite integrals of the functions cos(x)
%    and cos(x) * sin(x), respectively, over the interval [0,2*pi].


clear

V = [2.034 -5.45];              % pick any "stimulus"

N = 100000;                     % number of "neurons"
theta = 2*pi*rand(1,N);         % preferred orientations of the N neurons
C = [cos(theta); sin(theta)];   % these are the N unit vectors arranged as column vectors

responses = V*C;                % "responses" of the N neurons to stimulus V

population_vector = (2/N) * C*(responses');

error = sqrt(norm(V - population_vector'));
    % error, i.e. Euclidean distance between stimulus and population vector

display(['ERROR:  ', num2str(error)])