Least squares fitting by a constant

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The discussion focuses on fitting a dataset to a constant using least squares fitting, specifically minimizing the sum of squared differences between the constant and the data points. The equation to minimize is expressed as the sum of (y(n) - c)^2, leading to a quadratic function in terms of c. By differentiating this function, the minimum can be found without the need for matrices, simplifying the process. An algorithm is mentioned that utilizes orthogonal polynomials to eliminate matrices, allowing for coefficients to be determined through finite summation series. This approach is applicable for both weighted and unweighted datasets, facilitating the fitting of constants effectively.
chuy52506
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say we have data set {y(t sub i), t sub i} Where i=1 2 3...m.
I know how to fit these into a line of the form ax+b, but how about fitting into a constant??
 
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chuy52506 said:
say we have data set {y(t sub i), t sub i} Where i=1 2 3...m.
I know how to fit these into a line of the form ax+b, but how about fitting into a constant??

Hey chuy52506 and welcome to the forums.

Think about the fact that you are minimizing the distances between y = c and the data points where we have the sum over (y(a)-c)^2 for all a belonging to the dataset being a minimum.

How much math have you taken? Have you taken any classes on optimization or linear algebra?
 
I only have taken an introductory course to linear algebra and no optimization...im sorry I am confused, so there is no need to use matrices? and why would (y(a),c) be squared?
 
chuy52506 said:
I only have taken an introductory course to linear algebra and no optimization...im sorry I am confused, so there is no need to use matrices? and why would (y(a),c) be squared?

You have to solve the following equation:

Minimize \sum (y(n) - c)^2

You can expand this out in terms of c and you will get a quadratic function of c in terms of f(c) = ac^2 + bc + d and then by differentiating this you need to find the minimum which is given by solving 2ac + b = 0.
 
chuy52506 said:
There is no need to use matrices?
For a polynomial fit, including y=c, the matrices can be eliminated using a polynomial that is the sum of orthognal (for the given data points) polynomials of increasing order. Link to description of algorithm, that includes a c code example at the end.

http://rcgldr.net/misc/opls.rtf

The algorithm uses a recursive definition for the set of polynomials, and then based on this recursive definition combined with the fact the generated polynomials will be orthogonal, it's able to elminate the need for matrices, allowing coefficients to be determined via finite summation series. The algoritm generates 3 sets of constants for the orthogonal polynomials, but the code example explains how generate standard coefficients for a single polynomial, which is what you'd really want.

Note that this algorithm assumes you enter a set of data points {x, y} or a weighted set {w, x, y}. For an unweighted set of data points, just use w = 1. For y = c, just use incrementing numbers for x values, with the y values representing the actual values to be fitted via least squares (in case you want to see if there is a slope using y = bx + c).
 
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Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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