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Answer by Riley for Weighted averaging of polynomial coefficients
Here is a counterexample. Assuming $x_{\rm avg} = M$ on the diagonal $x_1 = x_2 = x_3 = M$, then $y_{\rm avg} = d_0 + d_1 x_{\rm avg} = d_0 + d_1 M$, for some fixed $M$. For simplicity, fix $a_0 = 1,...
View ArticleWeighted averaging of polynomial coefficients
Suppose there are three polynomials $$y_1 = a_0 + a_1x_1, $$$$y_2 = b_0 + b_1x_2, $$$$y_3 = c_0 + c_1x_3,$$and define $$ x_{\rm avg} \triangleq...
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