# Method of Least Squares Linear Fitting

by Defennder
Tags: fitting, linear, method, squares
 HW Helper P: 2,618 1. The problem statement, all variables and given/known data An experiment was conducted on a liquid at varying temperatures and the volume obtained at the differing temperatures are as follows: V/cm3 θ/oC 1.032 10 1.063 20 1.094 29.5 1.125 39.5 1.156 50 1.186 60.5 1.215 69.5 1.244 79.5 1.273 90 1.3 99 Assume that $$V = 1 + B\theta + D\theta^2$$ , where B and D are constants. Question: Linearize the above equation and plot the corresponding curve. 2. Relevant equations Microsoft Excel LINEST function, Least squares method statistical equations (too many to post) 3. The attempt at a solution Here's what I understand about linear least squares fitting. I attended a lab session where I was taught how to apply the least squares method in Excel to linearize a given equation and then use mbest and cbest to calculate the other unknown constants in the equation. My understanding is that firstly one takes the equation, and tries to express it, in any mathematical way possible in the form y=mx + c, where y is the dependent variable and x the independent variable. Hence while the y does not have to be the same independent variable as measured directly in the experimental setup, x has to be the same. Hence the resulting linearized equation cannot use expressions of x which are functions of x expressed in ways apart from simply x. eg. x^2 is not allowed, ln x is not allowed. Is this understanding correct? If so, then how is it possible for me to linearise the above given equation? I only got as far as: $$\theta + \frac{B}{2D} = \sqrt{ \frac{V-1}{D} + (\frac{B}{2D})^2$$ which should suggest (or at least it does to me) that mbest in Excel using the LINEST function should be 1, since the coefficient of θ is 1. But instead I get 0.003004423, which means I have somehow linearized the equation wrongly. How else could it have been linearized? EDIT: I didn't quite get what it means by "plot the corresponding curve". I assume this involves using Excel, but apart from using as source data the values of V and theta from the table, what else could it mean?
 HW Helper P: 4,125 Have a look at example 1 here: http://ceee.rice.edu/Books/CS/chapter3/data34.html Hope it helps.
 P: 1,295 I also got same value, 0.003004423*x+1.00 I instead used http://www.padowan.dk/graph/Download.php it much more better than excel! give it a try. A method to linearlize any equation dV/dt = 2Dt+B find dV/dt when t = 50 [almost middle point] and so y-1.156 = (dV/dt)[t-50]!!
HW Helper
P: 2,618

## Method of Least Squares Linear Fitting

 Quote by learningphysics Have a look at example 1 here: http://ceee.rice.edu/Books/CS/chapter3/data34.html Hope it helps.
Well, I've certainly haven't seen this approach before, but I don't see how relevant it is to my question. In the link you provided, there isn't any equation to linearize to start with, unlike in mine. The LINEST Excel function gives a straight line with R^2 = 0.9996, indicating a very close fit if the constant D is real close to 0. Without having to linearize V = 1 + B(theta) D(theta)^2, I could probably do the question simply by using MS Excel LINEST function. But the problem arises when I try to obtain a best fit line for the data from Excel as shown in my first post and obtain a gradient mbest which contradicts what I get from linearizing the equation directly.

 Quote by rootX I also got same value, 0.003004423*x+1.00 I instead used http://www.padowan.dk/graph/Download.php it much more better than excel! give it a try. A method to linearlize any equation dV/dt = 2Dt+B find dV/dt when t = 50 [almost middle point] and so y-1.156 = (dV/dt)[t-50]!!
I don't understand your method. Differentiating with respect to $$\theta$$ gives me a whole new equation. Although it's in a linear form, isn't it totally different from the original one? And why set t=50?
 P: 1,295 I am not really sure if you can use that equation, but it just gives an approximation. I chose 50, because this is the equation of the tangent line of this quadratic function at t = 50 (which is midpoint as 10 is min. and 99 is max)...

 Related Discussions Calculus & Beyond Homework 2 Introductory Physics Homework 3 Precalculus Mathematics Homework 0 Mathematics Learning Materials 0 Linear & Abstract Algebra 4