Given the real, natural and binary successions $\{t_1,...,t_N\} \in \mathbb{R}$, $\{n_1,...,n_N\} \in \mathbb{N}$ and $\{E_1,...,E_N\} \in \{0,1\}$ we would like to find the $(x,y)$ that satisfies the following equations

$$\begin{cases} \displaystyle\sum_{i=1}^N \frac{E_i}{x-n_iy} = \displaystyle\sum_{i=1}^N n_i t_i \\ \displaystyle\sum_{i=1}^N \frac{E_in_i}{x-n_iy} = \displaystyle\sum_{i=1}^N n^2_i t_i \end{cases}$$

Both equations are quite similar, and my intuition says that there is an analytical solution, but I cannot find it.

Would you be able to find the explicit solution? (or prove that there is not solution?)

  • $\begingroup$ How large is $N$? $\endgroup$ – Rodrigo de Azevedo Jul 25 '16 at 15:17
  • $\begingroup$ It may be anything greater than 30 $\endgroup$ – Francisco Jul 26 '16 at 16:47


  • $b_1, b_2, \dots, b_n \in \{0,1\}$
  • $m_1, m_2, \dots, m_n \in \mathbb N$
  • $c_1, c_2 \in \mathbb R$


$$z_i := \dfrac{1}{x - m_i y}$$

We have a system of $2$ equations in $\mathrm z \in \mathbb R^n$

$$\begin{bmatrix} b_1 & b_2 & \dots & b_n\\ b_1 m_1 & b_2 m_2 & \dots & b_n m_n\end{bmatrix} \begin{bmatrix} z_1\\ z_2\\ \vdots\\ z_n\end{bmatrix} = \begin{bmatrix} c_1\\ c_2\end{bmatrix}$$

or, in a more succinct form, $\mathrm A \mathrm z = \mathrm c$. If $n > 2$, this is an underdetermined system whose least-norm solution is

$$\hat{\mathrm z} := \mathrm A^T (\mathrm A \mathrm A^T)^{-1} \mathrm c$$

If all the entries of $\hat{\mathrm z}$ are nonzero, then we have an overdetermined system of equations

$$\begin{array}{rl} x - m_1 y &= \hat z_1^{-1}\\ x - m_2 y &= \hat z_2^{-1}\\ &\vdots \\ x - m_n y &= \hat z_n^{-1}\end{array}$$

Lastly, we compute the least-squares solution $(\hat x, \hat y)$. If $\hat z_i = 0$, then the $i$-th equation, whose right-hand side is illegal, is simply discarded.

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  • $\begingroup$ Useful point of view. I am just wondering if there is some loss due to introducing $n$ variables while previously we had $2$. What I would like is to have an analytical solution of $x$ and $y$ as a function of data rather than a numerical approximation. It is better than what I had though. $\endgroup$ – Francisco Jul 26 '16 at 16:52
  • $\begingroup$ @Francisco The only alternative I see is to deal with two high-degree polynomial equations in $x$ and $y$, which isn't nice. $\endgroup$ – Rodrigo de Azevedo Jul 26 '16 at 16:58

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