Weighted average of arbitrary k points from a line

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Discussion Overview

The discussion revolves around the mathematical properties of a weighted barycenter of a set of arbitrary points selected from a straight line in R². Participants explore whether the resulting barycenter remains on the same line and the implications of weight choices on this property.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant proposes that if the points x_i are selected from a straight line, the weighted barycenter x_o can be shown to also lie on that line, depending on the choice of weights o_i.
  • Another participant suggests that the independence of the weights o_i allows for the selection of coefficients that match those of the original line, but notes that dividing by the sum of the weights introduces a loss of a degree of freedom.
  • A later reply simplifies the question by specifying that the weights o_i are non-negative and asks if this guarantees that x_o lies on the same line.
  • One participant questions whether the discussion relates to the triangle simplex in R².
  • Another participant emphasizes the need for a common gradient for the line and suggests using a specific line equation to demonstrate that the barycentric case leads to a common form in terms of the coefficients.
  • A participant states that a line in the plane can be expressed as a vector equation and asserts that showing the weighted average satisfies this equation is sufficient, provided the weights sum to 1.

Areas of Agreement / Disagreement

Participants express various viewpoints on the conditions under which the barycenter remains on the line, with no consensus reached on the necessity of specific weight conditions or the implications of the weights' sum.

Contextual Notes

Some participants mention constraints related to the weights and the implications of their sum, but these aspects remain unresolved and depend on the definitions used in the discussion.

onako
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Suppose a set of k arbitrary points, x_i, 1<=i<=k, x_i from R^2 are selected from a line. How can it be shown that a weighted barycenter x_o=(o_i*x_i)/(o_1+o_2+...+o_k) also belongs to that line (assume o_i are arbitrary weights)? Does the choice of weights restrict the solutions (ie, a particular choice to satisfy that x_o is within the 'convex hull' of other points)?
 
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Hey onako.

If this is a straight line then from this you have n-independent coeffecients (since the o_i's are independent) and since you can specify those then you can choose the coeffecients such that these match the ones of the original line.

If all the x_i's line on a straight-line, then provided that you can pick the o_i's, then you can solve for the particular values of o_i to provide equal coeffecients.

The only thing though about this is that you are dividing by the sum of the all the o_i's which means you lose a degree of freedom and the only way you can get a solution is if you have an extra constraint on the o_i's and this constraint is exactly what is imposed.

If your line is a unique line then the solution in terms of your o_i's should be unique as well since you have a constraint on the sum of the weights which means the scalar multiples should disappear as well (I'm consider scalar multiples of a linear equation).
 
Let's simplify the question. Suppose that non-negative arbitrary weights o_i are associated with each x_i chosen from the straight line. Does x_o lie on that same line? The point is: points x_i are chosen as arbitrary points from the line, and are associated coefficients o_i which are non-negative.
 
Are you looking at the triangle simplex in R^2?
 
I'm concerned with a 2D case, with a straight line.
 
If this line is in R^2 then they need a common gradient: try setting up a simple line based on y - y0 = m(x - x0) where m = (y1 - y0)/(x1 - x0) and showing that the gradient has a common form.

For the bary-centric case the sum of all the weights should be 1 so you can cancel that out, and see if calculating this line gives a common form in terms of the coeffecients.
 
A line in the plane satisfies a vector equation of the form
Nx=a
So if each xi satisfies that equation, you just have to show that a weighted average of them also satisfies that equation. As long as the weights add up to 1 it should work.
 

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