README.md

Homework #6

due 4/14 by 11:59pm

0. Create a new github repository called 'curve_fitting'.

   a. Add rcc02007 and pez16103 as collaborators.

   b. Clone the repository to your computer.

1. Create a least-squares function called least_squares.m 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 $f(x_{i})$, and the coefficient of determination, r2.

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

Test your function on the sets of data in script problem_1_data.m and show that the following functions are the best fit lines:

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

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

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

2. Use the Temperature and failure data from the Challenger O-rings in lecture_18 (challenger_oring.csv). Your independent variable is temerature and your dependent variable is failure (1=fail, 0=pass). Create a function called cost_logistic.m that takes the vector a, and independent variable x and dependent variable y. Use the function, $\sigma(t)=\frac{1}{1+e^{-t}}$ where $t=a_{0}+a_{1}x$. Use the cost function,

   $J(a_{0},a_{1})=1/m\sum_{i=1}^{n}\left[-y_{i}\log(\sigma(t_{i}))-(1-y_{i})\log((1-\sigma(t_{i})))\right]$

   and gradient

   $\frac{\partial J}{\partial a_{i}}= 1/m\sum_{k=1}^{N}\left(\sigma(t_{k})-y_{k}\right)x_{k}^{i}$

   where $x_{k}^{i} is the k-th value of temperature raised to the i-th power (0, and 1)

   a. edit cost_logistic.m so that the output is [J,grad] or [cost, gradient]

   b. use the following code to solve for a0 and a1

% 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(@(a)(costFunction(a, x, y)), initial_a, options);

c. plot the data and the best-fit logistic regression model

plot(x,y, x, sigma(a(1)+a(2)*x))

3. The vertical stress under a corner of a rectangular area subjected to a uniform load of intensity $q$ is given by the solution of the Boussinesq's equation:

   $\sigma_{z} = \frac{q}{4\pi}\left(\frac{2mn\sqrt{m^{2}+n^{2}+1}}{m^{2}+n^{2}+1+m^{2}n^{2}}\frac{m^{2}+n^{2}+2}{m^{2}+n^{2}+1}+sin^{-1}\left(\frac{2mn\sqrt{m^{2}+n^{2}+1}}{m^{2}+n^{2}+1+m^{2}n^{2}}\right)\right)$

   Typically, this equation is solved as a table of values where:

   $\sigma_{z}=q f(m,n)$

   where $f(m,n)$ is the influence value, q is the uniform load, m=a/z, n=b/z, a and b are width and length of the rectangular area and z is the depth below the area.

   a. Finish the function boussinesq_lookup.m 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, sigma_z=$\sigma_{z}$. Use a $0^{th}$-order, polynomial interpolation for the value of n (i.e. round to the closest value of n).

   b. Copy the boussinesq_lookup.m to a file called boussinesq_spline.m and use a cubic spline to interpolate in two dimensions, both m and n, that returns sigma_z.