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Related to but different from Points on a circle with near-zero centroid I have the following puzzle:

Let's assume a set of points $\vec{a} = \{a_1, a_2, \ldots, a_n\}$, with $a_i$ being a real number between 0 and $2\pi$ (real number modulo $2\pi$).

We map to a new set of points $\vec{b} = \{b_1, b_2, \ldots, b_n\}$ via an alternating cumulative sum, $b_i = a_1 + \sum_{j=2}^{i}(-1)^{j-1}(a_j-a_{j-1})$. Each $b_i$ is also a real number modulo $2\pi$.

Notes:

  • The problem is set up like this so $a_i$ and $b_i$ correspond to phases on a unit circle.

  • The mapping is an involution, namely it is also true that $a_i = b_1 + \sum_{j=2}^{i}(-1)^{j-1}(b_j-b_{j-1})$. That is, in matrix/vector notation a matrix $\mathbf{M}$ exists where $\vec{b}=\mathbf{M}\vec{a}$ and $\mathbf{M}\mathbf{M}=\mathbf{1}$; $\mathbf{M}$ is triangular and self-inverse.

The question is then, if we fix $n$ and the centroid of the "$a$" phases on the circle, $c(\vec{a}) = \sum_{i=1}^n\exp(\mathrm{i}a_i)$, can we predict anything about the centroid of the "$b$" phases on the circle, $c(\vec{b}) = \sum_{i=1}^n\exp(\mathrm{i}b_i)$.

Suppose we want to find an $\vec{a}$ and $\vec{b}$ given the condition $c(\vec{a})=c(\vec{b})=0$? What non-brute-force recipe or theory would provide solutions?

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    $\begingroup$ I guess in the definition of $\bf M$ you mean $b_i:=a_1+\sum_{j=2}^{\bf i}$ etc, not $\sum_{j=2}^{\bf n}$ $\endgroup$
    – Pietro Majer
    Commented Oct 23 at 20:13
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    $\begingroup$ You are correct. I'm not sure how it got like that. Question edited so "i" is the highest index of the sum, not "n". $\endgroup$
    – MichaelT
    Commented Oct 24 at 7:37

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