
Real Analysis Real Analysis Math Forum 
 LinkBack  Thread Tools  Display Modes 
March 18th, 2016, 07:29 AM  #1 
Newbie Joined: Mar 2016 From: South Africa Posts: 1 Thanks: 0  log linear weighted least squares application
Hi, I am trying to fit a curve to a set of data using a weighted least squares approach. The reason I am using the weighted approach is to bias my solution to my more reliable data. I am however having a problem trying to derive the analytical solution to the problem. T curve I am trying to fit is a creep strain equation of the following form, where beta are the coefficients I am trying to solve for: $\displaystyle \epsilon(t,\sigma, T) = 10^{\beta_1}\sigma^{\beta_2}te^{\beta_3/T}$ I then log the equation in order to linearise it as follows: $\displaystyle log(\epsilon) = \beta_1 + \beta_2log(\sigma) + log(t)  \beta_3log(e^{1/T})$ I have experimental results for time t to reach strain $\displaystyle \epsilon$ at a constant test stress $\displaystyle \sigma$ and temperature T. My problem is when I try to implement the analytical solution for the weighted least squares problem which has the form, where W is a diagonal matrix of my chosen weightings: $\displaystyle \beta = (X^TWX)^{1}X^TWY $ As I understand I have the following for n data points: $\displaystyle \beta = [\beta_1,\ \beta_2,\ \beta_3]$ $\displaystyle Y^T = [log(\epsilon_1),\ log(\epsilon_2,\ ...\ ,log(\epsilon_n))]$ $\displaystyle X^T = [1,\ log(\sigma_i),\ log(e^{1/T_i})]$ with i=n rows My confusion is what do I do with the $\displaystyle \log(t)$ expression. As I see it, this term is an "x" data point with a coefficient of 1. However, I cannot see how to include this and solve for my $\displaystyle \beta$ coefficients. I know that I could go the optimisation route and solve it that way. I am however looking for an analytical solution to the least squares problem. Any assistance would be greatly appreciated. Thanks in advance. 
March 21st, 2016, 09:59 PM  #2 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,464 Thanks: 106 
The given equation says that strain is proportional to time and the constant of proportionality depends on stress $\displaystyle \sigma$ and temperature T. $\displaystyle \epsilon (t ,\sigma ,T)=f(\sigma ,T)t\\$ The objective is to determine f by measuring the slope of strain/time curves at different values of stress and temperature. If the form of f given in OP were exactly correct and repeatable, then two experiments would suffice to determine f: $\displaystyle f(\sigma _{1},T_{1})=K_{1}\\$ $\displaystyle f(\sigma _{2},T_{2})=K_{2}\\$ where $\displaystyle K_{1}$ and $\displaystyle K_2$ are the slopes of the strain\time curves under given conditions of stress and temperature. Solve for $\displaystyle \beta _{1}$ and $\displaystyle \beta_{2}$. In reality, f is an approximation and you have to use curve fitting for $\displaystyle \beta _{1}$ and $\displaystyle \beta_{2}$. t doesn't enter into it. 

Tags 
application, linear, log, squares, weighted 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Linear congruences application?  richardz03  Number Theory  0  January 14th, 2015 04:59 PM 
Matrix of a linear application  david940  Linear Algebra  3  June 29th, 2014 02:45 AM 
The Weighted Mean  bilano99  Algebra  2  March 14th, 2012 01:14 PM 
Numerical Analysis: Linear Least Squares Transformations  wattsup  Applied Math  0  February 9th, 2012 01:41 PM 
weighted averages  stillrussian1  Advanced Statistics  0  September 27th, 2010 08:35 PM 