curve_fitting

##Problem 1

###Function least_squares.m

function [ a, fx, r2 ] = least_squares( Z, y )
%Function calculates the coefficients for a best fit function with least
%squares techniques
%   a = coeficients
%   fx = line of best fit
%   r2 = coefficient of determination

% evaluate coefficients for a polynomial of 3 degrees

a = Z\y; % evaluates constants for function

fx = Z*a; % evaluates function

Sr = sum((y-Z*a).^2); 

r2 = 1-Sr/sum((y-mean(y)).^2); % gives r^2 value

end

###Evaluated for case one yields

Z = [x.^0, x, 1./x];
>> [ a, fx, r2 ] = least_squares( Z, y )

a =

    0.3745
    0.9864
    0.8456


fx =

    2.2066
    2.7702
    3.6157
    4.5317
    5.4758


r2 =

    0.9996

###Evaluated for case two yields

>> Z = [x.^0, x, x.^2, x.^3]

>> [ a, fx, r2 ] = least_squares( Z, y )

a =

  -11.4887
    7.1438
   -1.0412
    0.0467

fx =

    1.8321
    3.4145
    4.0347
    3.5087
    2.9227
    2.4947
    3.2330
    4.9595

r2 =

    0.8290

###Evaluated for case three yields

>> Z = [exp(-1.5*x), exp(-.3*x),exp(-.05*x)];

>> [ a, fx, r2 ] = least_squares( Z, y )

a =

    4.0046
    2.9213
    1.5647

fx =

    5.9321
    4.5461
    3.2184
    2.5789
    2.1709
    1.8726
    1.6425
    1.4605
    1.1940

r2 =

    0.9971

##Problem 2 ###Function cost_logistic.m

function [J, grad] = cost_logistic(a, x, y)
% cost_logistic Compute cost and gradient for logistic regression
%   J = cost_logistic(theta, X, y) computes the cost of using theta as the
%   parameter for logistic regression and the gradient of the cost
%   w.r.t. to the parameters.

% Initialize some useful values
N = length(y); % number of training examples

% You need to return the following variables correctly 
J = 0;
grad = 0;

% ====================== YOUR CODE HERE ======================
% Instructions: Compute the cost of a particular choice of a.
%               Compute the partial derivatives and set grad to the partial
%               derivatives of the cost w.r.t. each parameter in theta
t = a(1)+a(2).*x;

sigm = 1./(1+exp(-t));


J = sum(-y.*log(sigm)- (1-y).*log(1-sigm));

costFun = @ (a) sum(-y.*log((1./(1+exp(-(a(1)+a(2).*x)))))- (1-y).*log(1-(1./(1+exp(-(a(1)+a(2).*x))))));

grad = (1/length(x))*sum((sigm - y).*t);

initial_a = [0 0];

% Set options for fminunc
options = optimset('GradObj', 'on', 'MaxIter', 400);
% Run fminunc to obtain the optimal theta
% This function will return theta and the cost
[theta, cost] = fminunc(costFun, initial_a);
tt = theta(1)+theta(2).*x;
sigmm = 1./(1+exp(-tt));
setdefaults
plot(x,y,'o', x, sigmm);
title('Best fit regression model')
xlabel('Temp (F)')
ylabel('Pass/Fail (0,1)')

% Note: grad should have the same dimensions as theta
% =============================================================

end

###Function evaluated

>> [J, grad] = cost_logistic(a, x, y)

J =

  115.5085


grad =

    5.0130

###Plot of best fit linear regression !

#Problem 3

The following function was created to evalaute problem 3

function sigma_z=boussinesq_lookup(q,a,b,z)
  % function that determines stress under corner of an a by b rectangular platform
  % z-meters below the platform. The calculated solutions are in the fmn data
  % m=fmn(:,1)
  % in column 2, fmn(:,2), n=1.2
  % in column 3, fmn(:,2), n=1.4
  % in column 4, fmn(:,2), n=1.6

  fmn= [0.1,0.02926,0.03007,0.03058
        0.2,0.05733,0.05894,0.05994
        0.3,0.08323,0.08561,0.08709
        0.4,0.10631,0.10941,0.11135
        0.5,0.12626,0.13003,0.13241
        0.6,0.14309,0.14749,0.15027
        0.7,0.15703,0.16199,0.16515
        0.8,0.16843,0.17389,0.17739];

  m=a/z;
  n=b/z;
  
  % find which n row to use
  
  nn = [ 1.2 1.4 1.6 ];
  r_n = round(n,1);
  n_loc = min(abs(nn - r_n));
  row_n = find( abs(nn - r_n) == n_loc)+1; % sets row for n 
  
 % find which m row to use
 
  mm = fmn(:,1)';
  r_m = round(m,1);
  m_loc = min(abs(mm - r_m));
  row_m = find( abs(mm - r_m) == m_loc); % sets row for m
  
  m_values = fmn(row_m-2:row_m+1, 1);
  n_values = fmn(row_m-2:row_m+1, row_n);
  
  pcon = polyfit(m_values', n_values', 3);
  x = m_values';
  func = @(x) pcon(4) + pcon(3)*x + pcon(2)*x.^2 + pcon(1)*x.^3;
  y = func(x);
  func(m);
  
  sigma_z = q*func(m);
  
end

##Spline Interpolation

function sigma_z=boussinesq_spline(q,a,b,z)
  % function that determines stress under corner of an a by b rectangular platform
  % z-meters below the platform. The calculated solutions are in the fmn data
  % m=fmn(:,1)
  % in column 2, fmn(:,2), n=1.2
  % in column 3, fmn(:,2), n=1.4
  % in column 4, fmn(:,2), n=1.6

  fmn= [0.1,0.02926,0.03007,0.03058
        0.2,0.05733,0.05894,0.05994
        0.3,0.08323,0.08561,0.08709
        0.4,0.10631,0.10941,0.11135
        0.5,0.12626,0.13003,0.13241
        0.6,0.14309,0.14749,0.15027
        0.7,0.15703,0.16199,0.16515
        0.8,0.16843,0.17389,0.17739];

  m=a/z;
  n=b/z;
  
  % find which n row to use
  
  nn = [ 1.2 1.4 1.6 ];
  r_n = round(n,1);
  n_loc = min(abs(nn - r_n));
  row_n = find( abs(nn - r_n) == n_loc)+1; % sets row for n 
  
 % find which m row to use
 
  mm = fmn(:,1)';
  r_m = round(m,1);
  m_loc = min(abs(mm - r_m));
  row_m = find( abs(mm - r_m) == m_loc); % sets row for m
  
  m_values = fmn(row_m-2:row_m+1, 1);
  n_values = fmn(row_m-2:row_m+1, row_n);
  
  pcon = interp3(m_values', n_values', 'cubic');
  x = m_values';
  func = @(x) pcon(4) + pcon(3)*x + pcon(2)*x.^2 + pcon(1)*x.^3;
  y = func(x);
  func(m);
  plot(x,y)
  
  sigma_z = q*func(m);
  
end

#END

akw13003/curve_fitting-1

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