Red lines show the factors containing positive roots and blue lines show the factors containing negative roots as they are added. Dotted lines represent the factors accumulated from previous iterations. Light gray line represents the base factor f(x) = x.

Green line shows the approximant, notice the reasonable convergence in the interval [-pi, pi] as more factors are added.

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Let [x0, x1, x2, x3, x4, …] be an infinite sequence.

Let

suff(0, [x0, x1, x2, x3, x4, …] = [x0, x1, x2, x3, x4, …]

suff(1, [x0, x1, x2, x3, x4, …] = [x1, x2, x3, x4, …]

suff(2, [x0, x1, x2, x3, x4, …] = [x2, x3, x4, …]

suff(3, [x0, x1, x2, x3, x4, …] = [x3, x4, …]

etc.

Let “+” be the operator for sequence concatenation.

Let w^n be a word of length n.

Then for any equivalence class E, there is a fixed infinite sequence [s0, s1, s2, s3, s4, …] such that every sequence in E has the form w^n + suff(n, [s0, s1, s2, s3, s4, …])

With n=0, [s0, s1, s2, s3, s4, …] itself is in E, so take it as a representative.

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No. You need AC if you want a set of representatives of all equivalence classes. You can use this set to get a representative of a given equivalence class, I agree. But I do not agree that it is the only way.

1) Take the set of all equivalence classes E.

2) Take the desired equivalence class e in E (the one you are interested in)

3) Take a representative r in e.