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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$?

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    $\begingroup$ A solution to this system, indeed there is. Nonlinear, the system is. Use linear matrix methods, you cannot. To solve this system, methods exist. Newton is one. Secant another. Solving this problem well, Mathematica does. If pen and paper, you must use, then Gram-Schmidt integration may prove helpful. Though certain of this, I am not. $\endgroup$ – MasterYoda Feb 10 '18 at 3:36
  • $\begingroup$ @Corp.andLtd.: There are six solutions. An example is $$c_1 = \dfrac{5}{9}, c_2 = \dfrac{5}{9}, c_3 = \dfrac{8}{9}, x_1 =- \sqrt{\dfrac{3}{5}}, x_2 = \sqrt{\dfrac{3}{5}}, x_3 = 0$$ The other solutions are permutations of this solution. $\endgroup$ – Moo Feb 10 '18 at 7:39
  • $\begingroup$ I was eager to know solution $\endgroup$ – NEW SUN May 12 '18 at 12:50
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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}

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