How can I solve this system of equations??

Question

I have this system of equations and I need values for $c_1, c_2, c_3, x_1, x_2, x_3$. \begin{align*} c_1 + c_2 + c_3 &= 2\\ c_1x_1 + c_2x_2+ c_3x_3 &= 0\\ c_1x_1^2 + c_2x_2^2+c_3x_3^2 &= \frac{2}{3}\\ c_1x_1^3 + c_2x_2^3+c_3x_3^3 &= 0\\ c_1x_1^4 + c_2x_2^4 + c_3x_3^4 &= \frac{2}{5}\\ c_1x_1^5 + c_2x_2^5+c_3x_3^5 &= 0 \end{align*}

I have attempted using matrix but was unsuccessful. How can I obtain values for $x_i$ and $ c_i$?

Answer 1

Rewriting the system as \begin{align} \sum_{j=1}^3 c_j\,x_j^i&=v_i,\quad i=0,\dots,5 \tag{1}\label{1} , \end{align}

we can apply Prony's method as follows. Solve the linear system \begin{align} \left[\begin{matrix} v_0 & v_1 & v_2 \\ v_1 & v_2 & v_3 \\ v_2 & v_3 & v_4 \end{matrix}\right] \cdot \left[\begin{matrix} a_0 \\ a_1 \\ a_2 \end{matrix}\right] &= \left[\begin{matrix} v_3 \\ v_4 \\ v_5 \end{matrix}\right] \end{align}

for $a_0,a_1,a_2$.

The roots of polynomial

\begin{align} x^3-a_2\,x^2-a_1\,x-a_0 \end{align}

would be the triple $x_1,x_2,x_3$. Given that, the solution of another linear system

\begin{align} \left[\begin{matrix} 1&1&1 \\ x_1&x_2&x_3 \\ x_1^2&x_2^2&x_3^2 \end{matrix}\right] \cdot \left[\begin{matrix} c_1 \\ c_2 \\ c_3 \end{matrix}\right] &= \left[\begin{matrix} v_0 \\ v_1 \\ v_2 \end{matrix}\right] \end{align}

for $c_1,c_2,c_3$ completes the answer.

In numbers we have

\begin{align} \left[\begin{matrix} 2 & 0 & \tfrac23 \\ 0 & \tfrac23 & 0 \\ \tfrac23 & 0 & \tfrac25 \end{matrix}\right] \cdot \left[\begin{matrix} a_0 \\ a_1 \\ a_2 \end{matrix}\right] &= \left[\begin{matrix} 0 \\ \tfrac25 \\ 0 \end{matrix}\right] ,\\ \end{align}

\begin{align} a_0&=0,\quad a_1=\tfrac35.\quad a_2=0 ,\\ x^3-\tfrac35\,x&=0 ,\\ x_1&=0,\quad x_2=\tfrac{\sqrt{15}}5 ,\quad x_3=-\tfrac{\sqrt{15}}5 . \end{align}

The system for $c_j$:

\begin{align} \left[\begin{matrix} 1&1&\phantom{-}1 \\ 0& \tfrac{\sqrt{15}}5 & -\tfrac{\sqrt{15}}5 \\ 0& \tfrac35 & \phantom{-}\tfrac35 \end{matrix}\right] \cdot \left[\begin{matrix} c_1 \\ c_2 \\ c_3 \end{matrix}\right] &= \left[\begin{matrix} 2 \\ 0 \\ \tfrac23 \end{matrix}\right] ,\\ \end{align}

which results in \begin{align} c_1&=\tfrac89,\quad c_2= c_3=\tfrac59 . \end{align}