The condition about some positive real numbers can be written as the sum the nearby two

Question

Given $n$ positive real numbers $x_1,...,x_n$. What is the condition that they can be written as $$x_1=y_1+y_2$$ $$x_2=y_2+y_3$$ $$\ldots$$ $$x_n=y_n+y_1$$ where $y_1,\ldots,y_n$ are also some positive real numbers?

Answer 1

If $n$ is odd, say $n=2m+1$, by assuming $x_{n+k}=x_k$ $$ x_{1+k}+x_{3+k}+\ldots+x_{2m+1+k} > x_{2+k}+x_{4+k}+\ldots x_{2m+k} $$ has to hold for every $k$. It is enough to consider the inverse of a circulant matrix, provided that it exists. If $n$ is even, $n=2m$, $$ x_1+x_3+\ldots+x_{2m-1} = x_2+x_4+\ldots+x_{2m} $$ has to hold, since $(1,-1,1,-1,\ldots)$ belongs to the kernel of the associated circulant matrix.

Answer 2

${\bf Edit}$ I totally overlooked the positivity condition. This answer is answering the question of finding all solutions when $x_i$ are any real numbers so it's not really addressing the given question. If we try to apply the positivity condition to the explicit formulas given here we just reproduces the conditions found by Jack so see his answer instead.

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When $n$ is odd this is always possible and the solution is unique. The equation system can be written as $A{\bf y} = {\bf x}$:

$$\pmatrix{ 1 & 1 & 0 & \ldots & 0 & 0\\ 0 & 1 & 1 & \ldots & 0 & 0\\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots\\ 0 & 0 & 0 & \ldots & 1 & 1\\ 1 & 0 & 0 & \ldots & 0 & 1 }\pmatrix{y_1\\y_2\\\ldots\\y_{n-1}\\y_n} = \pmatrix{x_1\\x_2\\\ldots\\x_{n-1}\\x_n}$$

Expanding the determinant $\det A$ along the first column we see that $\det A = 0$ when $n$ is even and $2$ when $n$ is odd (and larger than $1$). Thus when $n$ is odd, the matrix is invertible and we have the solution ${\bf y} = A^{-1}{\bf x}$ which gives

$$\matrix{ y_1 &=& \frac{s_n}{2} \\ y_2 &=& s_1-\frac{s_n}{2} \\ y_3 &=& -s_2+\frac{s_n}{2} \\ y_4 &=& s_3-\frac{s_n}{2} \\ \ldots &\ldots& \ldots\\ y_n &=& -s_{n-1}+\frac{s_{n}}{2} \\ }$$ where $s_k = x_1 - x_2 + x_3 - \ldots + (-1)^{k+1}x_k$.

When $n$ is even the determinant is zero so there are either no non-trivial solutions or infinitely many. The condition Jack puts up in his answer is sufficient to guarantee that there are infinitely many solutions as

$$ \matrix{ y_1 &=& \text{free}\\ y_2 &=& s_1-y_1\\ y_3 &=& -s_2+y_1\\ y_5 &=& s_3-y_1\\ \ldots &=& \ldots\\ y_n &=& s_{n-1} - y_1 } $$ where again $s_k = x_1 - x_2 + x_3 - \ldots + (-1)^{k+1}x_k$ is a solution provided $s_n=0$ which means that $x_1+x_3+\ldots = x_2+x_4+\ldots$.