lp}}} \right) = 0\\
{g_{{t_i}}} = \frac{{\partial {C_{\pi lp}}}}{{\partial {t_i}}} = \frac{1}{{{{\left( {1 + {{\bf{s}}^T}{\bf{s}}} \right)}^2}}}\frac{{\partial {{\bar C}_{\pi lp}}}}{{\partial {t_i}}} = 0,\;i = 1,2,3
\end{array}$$ Canceling the denominator of (\[equ:13\]) yields $$\label{equ:14}
\begin{array}{l}
{{\bar g}_{{s_i}}} = \left( {1 + {{\bf{s}}^T}{\bf{s}}} \right)\frac{{\partial {{\bar C}_{\pi lp}}}}{{\partial {s_i}}} - 4{s_i}{{\bar C}_{\pi lp}} = 0\\
{{\bar g}_{{t_i}}} = \frac{{\partial {{\bar C}_{\pi lp}}}}{{\partial {t_i}}} = 0,\;i = 1,2,3
\end{array}$$ $\frac{{\partial {{\bar C}_{\pi lp}}}}{{\partial {s_i}}}$ and $\frac{{\partial {{\bar C}_{\pi lp}}}}{{\partial {t_i}}}$ are of degree 5. Therefore, ${{\bar{g}}_{{{s}_{i}}}}$ and ${{\bar{g}}_{{{t}_{i}}}}$ are of degree 7 and 5, respectively. Although the Gröbner basis method gives a general way to solve the polynomial system, it is computationally demanding and numerically unstable to apply it to a high order polynomial system, as the experimental results in Fig. \[fig:vs\_DLSSol\]. The Newton-Raphson method provides an alternative way to find the roots of (\[equ:14\]). Denote an equation system as $\bar{G}\left({\bf{x}}\right)$. For the $k$th iteration, the Newton-Raphson method updates the solution as $$\label{equ:15}
{{\bf{x}}_k}  = {{\bf{x}}_{k-1}}  - J_{\bar G}^{ - 1}\left( {{\bf{x}}_{k-1}} \right)\bar G\left( {{\bf{x}}_{k-1}}  \right)$$ This iterative method requires an initial solution ${{\mathbf{s}}_{0}}$ and ${{\mathbf{t}}_{0}}$. In the next section, we will describe how to calculate an accurate ${{\mathbf{s}}_{0}}$ and ${{\mathbf{t}}_{0}}$.

Initial Estimation from Relaxation
