#Problem 1

The function 'least_squares' inputs a matrix Z and a scalar y and outputs best-dit constants, best-fit function as well as the coefficients of determination.

For each one of the proposed cases, the following set of values is obtained:

a) y=0.3745+0.98644x+0.84564/x

a	fx	r2
0.3745	2.2066	0.9996
0.9864	2.7702
0.8456	3.6157
	4.5317
	5.4758

b) y=-11.4887+7.143817x-1.04121 x^2+0.046676 x^3

a	fx	r2
-11.4887	1.8321	0.8290
7.1438	3.4145
-1.0412	4.0347
0.0467	2.9227
	2.4947
	3.2330
	4.9595

c) y=4.0046e^(-1.5x)+2.9213e^(-0.3x)+1.5647e^(-0.05x)

a	fx	r2
4.0046	5.9321	0.9971
2.9213	4.5461
1.5647	3.2184
	2.5789
	2.1709
	1.8726
	1.6425
	1.4605
	1.1940

#Problem 2 For this problem, we look at the case where the independent variable is temperature and the dependent variable is failure (1=fail, 0=pass). A function called cost_logistic.m takes the vector a, independent variable x and dependent variable y so that the output is [J,grad] or [cost, gradient]. Also, we solved for a0 and a1 on part b and plotted the data in part c.

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

J = 115.5085

grad = 5.0130

b) The result for a0 and a1 generated through cost_logistic

#Problem 3

The function boussinesq_lookup.m is writtent so that when you enter a force , q, dimensions of rectangular area a, b, and depth, z, it uses a third-order polynomial interpolation of the four closest values of m to determine the stress in the vertical direction as shown in the file in the repository.

b) The boussinesq_lookup.m code is copied to a file called boussinesq_spline.m using a cubic spline to interpolate in two dimensions, both m and n, that returns sigma_z.