My Math Forum log linear weighted least squares application

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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 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,392 Thanks: 100 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.

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