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 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 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 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 Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post richardz03 Number Theory 0 January 14th, 2015 04:59 PM david940 Linear Algebra 3 June 29th, 2014 02:45 AM bilano99 Algebra 2 March 14th, 2012 01:14 PM wattsup Applied Math 0 February 9th, 2012 01:41 PM stillrussian1 Advanced Statistics 0 September 27th, 2010 08:35 PM

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