##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