Homework 5
#Problem 1 A least-squares function is created that accepts a Z-matrix and dependent variable y as input and returns the vector of best-fit constants, a, the best-fit function evaluated at each point, fx, and the coefficient of determniation, r2.
The following results of each variable are calculated for each case function:
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
c) The plot is located in the repository under Bestfitgraph.PNG
#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.