non-convex-optimizationnonlinear optimizationoptimization
Given $x_5 \geq x_4 \geq x_3 \geq x_2 \geq x_1 \geq 0$, solve the following optimization problem in $p_1, p_2,\dots, p_5$.
$$\max p_1p_2p_3p_4p_5$$
subject to:
$$p_1 x_1 + p_2 (x_2 – x_1) + p_3(x_3-x_2) + p_4(x_4-x_3)+p_5(x_5-x_4) = 1 $$
$$0\leq p_5 \leq p_4 \leq p_3 \leq p_2 \leq p_1$$
Is there a closed form solution to this problem? I am not sure how to go about approaching it.
The original minimization problem
\begin{align} \min z = 5y_1-10y_2+7y_3-3y_4 & \\ y_1+y_2+7y_3+2y_4 &= 3 \\ y_2+17y_3+7y_4 &= 8 \\ 6y_3+3y_4 &= 2 \\ y_i &\geq 0, i \in \{ 1, \dots, 4 \} \end{align}
We try to find a basic feasible solution to the original LPP by the two-phase method. Add the artificial variables $y_5,y_6 \ge 0$ into the LPP.
\begin{align} \min z = y_5+y_6 & \\ y_1+y_2+7y_3+2y_4 &= 3 \\ y_2+17y_3+7y_4+y_5 &= 8 \\ 6y_3+3y_4+y_6 &= 2 \\ y_i &\geq 0, i \in \{ 1, \dots, 6 \} \end{align}
We write the objective function as $z-y_5-y_6=0$. Since the coefficient of $z$ is always one, we omit it in the simplex tableaux to save ink.
\begin{equation*} \begin{array}{rrrrrrr|r} & y_1 & y_2 & y_3 & y_4 & y_5 & y_6 & \\ \hline y_1 & 1 & 1 & 7 & 2 & 0 & 0 & 3 \\ y_5 & 0 & 1 & 17 & 7 & 1 & 0 & 8 \\ y_6 & 0 & 0 & 6 & 3 & 0 & 1 & 2 \\ \hline z & 0 & 0 & 0 & 0 & -1 & -1 & 0 \end{array} \end{equation*}
Make it a simplex tableau.
\begin{equation*} \begin{array}{rrrrrrr|r} & y_1 & y_2 & y_3 & y_4 & y_5 & y_6 & \\ \hline y_1 & 1 & 1 & 7 & 2 & 0 & 0 & 3 \\ y_5 & 0 & 1 & 17 & 7 & 1 & 0 & 8 \\ y_6 & 0 & 0 & 6 & 3 & 0 & 1 & 2 \\ \hline z & 0 & 1 & 23 & 10 & 0 & 0 & 10 \end{array} \end{equation*}
Since it's minimization, we choose $y_j$ with the biggest $z_j - c_j$ as the entering variable.
\begin{equation*} \begin{array}{rrrrrrr|rr} & y_1 & y_2 & y_3 & y_4 & y_5 & y_6 & & \theta \\ \hline y_1 & 1 & 1 & 7 & 2 & 0 & 0 & 3 & \frac{3}{7} \\ y_5 & 0 & 1 & 17 & 7 & 1 & 0 & 8 & \frac{8}{17} \\ y_6 & 0 & 0 & 6^* & 3 & 0 & 1 & 2 & \frac{1}{3} \\ \hline z & 0 & 1 & 23 & 10 & 0 & 0 & 10 & \end{array} \tag{*} \label{min} \end{equation*}
\begin{equation*} \begin{array}{rrrrrrr|rr} & y_1 & y_2 & y_3 & y_4 & y_5 & y_6 & & \theta \\ \hline y_1 & 1 & 1^* & 0 & -\frac{3}{2} & 0 & -\frac{7}{6} & \frac{2}{3} & \frac{2}{3} \\ y_5 & 0 & 1 & 0 & -\frac{3}{2} & 1 & -\frac{17}{6} & \frac{7}{3} & \frac{7}{3} \\ y_3 & 0 & 0 & 1 & \frac{1}{2} & 0 & \frac{1}{6} & \frac{1}{3} & \\ \hline z & 0 & 1 & 0 & -\frac{3}{2} & 0 & -\frac{23}{6} & \frac{7}{3} & \end{array} \end{equation*}
Choose $y_2$ as the entering variable, $y_1$ as the leaving variable.
\begin{equation*} \begin{array}{rrrrrrr|r} & y_1 & y_2 & y_3 & y_4 & y_5 & y_6 & \\ \hline y_2 & 1 & 1 & 0 & -\frac{3}{2} & 0 & -\frac{7}{6} & \frac{2}{3} \\ y_5 & -1 & 0 & 0 & 0 & 1 & -\frac{5}{3} & \frac{5}{3} \\ y_3 & 0 & 0 & 1 & \frac{1}{2} & 0 & \frac{1}{6} & \frac{1}{3} \\ \hline z & -1 & 0 & 0 & 0 & 0 & -\frac{8}{3} & \frac{5}{3} \end{array} \end{equation*}
Since we still have the artificial variable $y_5$ in the basis in the optimal tableau, we conclude that this problem is not feasible.
I tried phase one in GNU Octave. You may run the following code online.
A=[1 1 7 2 0 0; 0 1 17 7 1 0; 0 0 6 3 0 1];
b = [3 8 2]'; c=[0 0 0 0 1 1]';
[x_min z_min] = glpk(c,A,b,zeros(6,1),[],"SSS","CCCCCC")
Results
x_min =
0.00000
1.66667
0.00000
0.66667
1.66667
0.00000
z_min = 1.6667
I also tried directly solving the original LPP, and the program returned NA.
A = [1 1 7 2; -2 -1 3 3; 2 2 8 1]; b = [3 2 4]'; c = [5 -10 7 -3]';
[x_min,z_min] = glpk(c,A,b,zeros(4,1),[],"SSS","CCCC")
Result
glp_simplex: unable to recover undefined or non-optimal solution
x_min =
NA
NA
NA
NA
z_min = NA
To get $z_j - c_j$, you may simply calculate $c_B^T B^{-1} a_j$. For the derivation, you may see my answer for another question about the regeneration of optimal simplex tableau from the optimal BFS. To actually calculate this number, compute $c_B^T y_j$, where $y_j$ is the column for $P_j$ in the current simplex tableau. For example, in the given tableau, $c_B = (-1,-1,-1)^T$.
\begin{align} z_1 - c_1 &= (-1,-1,-1)(1,-2,2)^T = -1 \\ z_2 - c_2 &= (-1,-1,-1)(1,-1,2)^T = -2 \\ z_3 - c_3 &= (-1,-1,-1)(7,3,8)^T = -18 \text{, etc} \end{align}
We want the artificial variables $x_5,x_6,x_7 = 0$, so we $\min x_5 + x_6 + x_7$, which is equivalent to $\mathbf{\max -x_5 - x_6 - x_7}$ (what you see in your book). That's why you see $-9$ in the $z$-row, $b$-column in the given simplex tableau.
Question: In simplex tableau $\eqref{min}$, the entering variable has $z_j - c_j > 0$. This seems contradictory to what you've learnt.
Answer: See my answer another question for detailed explanation.
\begin{array}{r|r|r|rrrrrrr|r|l} & & & 0 & 0 & 0 & 0 & -1 & -1 & -1 & & \\ \hline B & c_B & b & P_1 & P_2 & P_3 & P_4 & P_5 & P_6 & P_7 & \theta & \\ \hline \hline P_5 & -1 & 3 & 1 & 1 & 7 & 2 & 1 & 0 & 0 & & L_1 \\ P_6 & -1 & 2 & -2 & -1 & 3 & 3 & 0 & 1 & 0 & & L_2 \\ P_7 & -1 & 4 & 2 & 2 & 8 & 1 & 0 & 0 & 1 & & L_3 \\ \hline & z & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & & L_4 \\ \hline \hline P_5 & -1 & 3 & 1 & 1 & 7^* & 2 & 1 & 0 & 0 & 3/7^* & L_1' = L_1 \\ P_6 & -1 & 2 & -2 & -1 & 3 & 3 & 0 & 1 & 0 & 2/3 & L_2' = L_2 \\ P_7 & -1 & 4 & 2 & 2 & 8 & 1 & 0 & 0 & 1 & 4/8 & L_3' = L_3 \\ \hline & z & -9 & -1 & -2 & -18^* & -6 & 0 & 0 & 0 & & L_4' = -L_1 - L_2 - L_3 + L_4 \\ \hline \hline P_3 & 0 & 3/7 & 1/7 & 1/7 & 1 & 2/7 & 1/7 & 0 & 0 & 3/2 & L_1'' = \frac17 L_1' \\ P_6 & -1 & 5/7 & -17/7 & -10/7 & 0 & 15/7^* & -3/7 & 1 & 0 & 5/15^* & L_2'' = L_2' - \frac37 L_1' \\ P_7 & -1 & 4/7 & 6/7 & 6/7 & 0 & -9/7 & -8/7 & 0 & 1 & - & L_3'' = L_3' - \frac87 L_1' \\ \hline & z & -9/7 & 11/7 & 4/7 & 0 & -6/7^* & 18/7 & 0 & 0 & & L_4'' = L_4' + \frac{18}{7} L_1' \\ \hline \hline P_3 & 0 & 1/3 & 7/15 & 1/3 & 1 & 0 & 1/5 & -2/15 & 0 & & L_1''' = L_1'' - \frac{2}{15} L_2'' \\ P_4 & 0 & 1/3 & -17/15 & -2/3 & 0 & 1 & -1/5 & 7/15 & 0 & & L_2''' = \frac{7}{15} L_2'' \\ P_7 & -1 & 1 & -3/5 & 0 & 0 & 0 & -7/5 & 3/5 & 1 & & L_3''' = L_3'' + \frac{3}{5} L_2'' \\ \hline & z & -1 & 3/5 & 0 & 0 & 0 & 12/5 & 2/5 & 0 & & L_4''' = L_4'' + \frac{2}{5} L_2'' \end{array}
Since artificial variable $P_7$ is still in the basis, this LPP has no feasible solution.
Let $p_2=t$, $x_2-x_1=y\geq 0$ and $x_1=x$, then $$p_1 = {1-ty\over x}$$ so $$p_1p_2 = t{1-ty\over x} = -{y\over x}t^2+{1\over x}t$$
If $y\neq 0$ then this quadratic equation achieves maximum at $t= {1\over 2y}$ and that maximum is $${1\over 4xy}={1\over 4x_1(x_2-x_1)}$$
If $y=0$ we get $$p_1p_2 = {1\over x}t\leq {1\over x}$$
so in this case maximum is $1/x$
Best Answer
With the help of the slack variables $\epsilon_i$ and calling
$$ L(p,\mu,\epsilon,\lambda)= p_1 p_2 p_3 p_4 p_5+\mu _5 \left(p_1-p_2-\epsilon _5^2\right)+\mu _4 \left(p_2-p_3-\epsilon _4^2\right)+\mu _3 \left(p_3-p_4-\epsilon _3^2\right)+\mu _2 \left(p_4-p_5-\epsilon _2^2\right)+\mu _1 \left(p_5-\epsilon _1^2\right)+\lambda \left(p_1 x_1+p_2 \left(x_2-x_1\right)+p_3 \left(x_3-x_2\right)+p_4 \left(x_4-x_3\right)+p_5 \left(x_5-x_4\right)-1\right) $$
and solving the sationary conditions
$$ \nabla L = 0 $$
we obtain a set of solutions jointly with a set of conditions $\epsilon_i^2\ge 0$ to qualify those solutions. To be feasible the solution requires that $\epsilon_i^2\ge 0$ Also when $\epsilon_i^2 = 0$ it means that the corresponding restriction is active.
Due to the length of the symbolic response, we leave a script in MATHEMATICA that summarizes the results. There are sixteen non trivial solutions in res0 with structure $\{p_i,\epsilon_i^2,p_1p_2p_3p_4p_5\}$
Results for $n = 3$
$$ \left[ \begin{array}{ccccccc} p_1&p_2&p_3&\epsilon_1^2&\epsilon_2^2&\epsilon_3^2&p_1p_2p_3\\ \frac{1}{3 x_1} & \frac{1}{3 \left(x_2-x_1\right)} & \frac{1}{3 \left(x_3-x_2\right)} & \frac{1}{3 \left(x_3-x_2\right)} & \frac{1}{3 x_2-3 x_3}+\frac{1}{3 \left(x_2-x_1\right)} & \frac{1}{3} \left(\frac{1}{x_1-x_2}+\frac{1}{x_1}\right) & \frac{1}{27 \left(x_1^2-x_1 x_2\right) \left(x_2-x_3\right)} \\ \frac{1}{3 x_1} & \frac{2}{3 \left(x_3-x_1\right)} & \frac{2}{3 \left(x_3-x_1\right)} & \frac{2}{3 \left(x_3-x_1\right)} & 0 & \frac{1}{3} \left(\frac{2}{x_1-x_3}+\frac{1}{x_1}\right) & \frac{4}{27 x_1 \left(x_1-x_3\right){}^2} \\ \frac{2}{3 x_2} & \frac{2}{3 x_2} & \frac{1}{3 \left(x_3-x_2\right)} & \frac{1}{3 \left(x_3-x_2\right)} & \frac{1}{3 x_2-3 x_3}+\frac{2}{3 x_2} & 0 & \frac{4}{27 x_2^2 \left(x_3-x_2\right)} \\ \frac{1}{x_3} & \frac{1}{x_3} & \frac{1}{x_3} & \frac{1}{x_3} & 0 & 0 & \frac{1}{x_3^3} \\ \end{array} \right] $$
Results for $n = 4$
$$ \left[ \begin{array}{ccccccccc} p_1 & p_2 & p_3 & p_4 & \epsilon_1^2&\epsilon_2^2&\epsilon_3^2&\epsilon_4^2&p_1p_2p_3p_4\\ \frac{1}{4 x_1} & \frac{1}{4 \left(x_2-x_1\right)} & \frac{1}{4 \left(x_3-x_2\right)} & \frac{1}{4 \left(x_4-x_3\right)} & \frac{1}{4 \left(x_4-x_3\right)} & \frac{1}{4 x_3-4 x_4}+\frac{1}{4 \left(x_3-x_2\right)} & \frac{1}{4 x_2-4 x_3}+\frac{1}{4 \left(x_2-x_1\right)} & \frac{1}{4} \left(\frac{1}{x_1-x_2}+\frac{1}{x_1}\right) & -\frac{1}{256 \left(x_1^2-x_1 x_2\right) \left(x_2-x_3\right) \left(x_3-x_4\right)} \\ \frac{1}{4 x_1} & \frac{1}{4 \left(x_2-x_1\right)} & \frac{1}{2 \left(x_4-x_2\right)} & \frac{1}{2 \left(x_4-x_2\right)} & \frac{1}{2 \left(x_4-x_2\right)} & 0 & \frac{1}{2 x_2-2 x_4}+\frac{1}{4 \left(x_2-x_1\right)} & \frac{1}{4} \left(\frac{1}{x_1-x_2}+\frac{1}{x_1}\right) & -\frac{1}{64 x_1 \left(x_1-x_2\right) \left(x_2-x_4\right){}^2} \\ \frac{1}{4 x_1} & \frac{1}{2 \left(x_3-x_1\right)} & \frac{1}{2 \left(x_3-x_1\right)} & \frac{1}{4 \left(x_4-x_3\right)} & \frac{1}{4 \left(x_4-x_3\right)} & \frac{1}{4 x_3-4 x_4}+\frac{1}{2 \left(x_3-x_1\right)} & 0 & \frac{1}{4} \left(\frac{2}{x_1-x_3}+\frac{1}{x_1}\right) & \frac{1}{64 x_1 \left(x_1-x_3\right){}^2 \left(x_4-x_3\right)} \\ \frac{1}{4 x_1} & \frac{3}{4 \left(x_4-x_1\right)} & \frac{3}{4 \left(x_4-x_1\right)} & \frac{3}{4 \left(x_4-x_1\right)} & \frac{3}{4 \left(x_4-x_1\right)} & 0 & 0 & \frac{1}{4} \left(\frac{3}{x_1-x_4}+\frac{1}{x_1}\right) & \frac{27}{256 x_1 \left(x_4-x_1\right){}^3} \\ \frac{1}{2 x_2} & \frac{1}{2 x_2} & \frac{1}{4 \left(x_3-x_2\right)} & \frac{1}{4 \left(x_4-x_3\right)} & \frac{1}{4 \left(x_4-x_3\right)} & \frac{1}{4 x_3-4 x_4}+\frac{1}{4 \left(x_3-x_2\right)} & \frac{1}{4 x_2-4 x_3}+\frac{1}{2 x_2} & 0 & \frac{1}{64 x_2^2 \left(x_2-x_3\right) \left(x_3-x_4\right)} \\ \frac{1}{2 x_2} & \frac{1}{2 x_2} & \frac{1}{2 \left(x_4-x_2\right)} & \frac{1}{2 \left(x_4-x_2\right)} & \frac{1}{2 \left(x_4-x_2\right)} & 0 & \frac{1}{2} \left(\frac{1}{x_2-x_4}+\frac{1}{x_2}\right) & 0 & \frac{1}{16 x_2^2 \left(x_2-x_4\right){}^2} \\ \frac{3}{4 x_3} & \frac{3}{4 x_3} & \frac{3}{4 x_3} & \frac{1}{4 \left(x_4-x_3\right)} & \frac{1}{4 \left(x_4-x_3\right)} & \frac{1}{4 x_3-4 x_4}+\frac{3}{4 x_3} & 0 & 0 & \frac{27}{256 x_3^3 \left(x_4-x_3\right)} \\ \frac{1}{x_4} & \frac{1}{x_4} & \frac{1}{x_4} & \frac{1}{x_4} & \frac{1}{x_4} & 0 & 0 & 0 & \frac{1}{x_4^4} \\\end{array} \right] $$
and for $n = 5$
$$ \left[ \begin{array}{ccccccccccc} p_1&p_2&p_3&p_4&p_5&\epsilon_1^2&\epsilon_2^2&\epsilon_3^2&\epsilon_4^2&\epsilon_5^2&p_1p_2p_3p_4p_5\\ \frac{1}{5 x_1} & -\frac{1}{5 \left(x_1-x_2\right)} & -\frac{1}{5 \left(x_2-x_3\right)} & -\frac{1}{5 \left(x_3-x_4\right)} & -\frac{1}{5 \left(x_4-x_5\right)} & -\frac{1}{5 \left(x_4-x_5\right)} & \frac{x_3-2 x_4+x_5}{5 \left(x_3-x_4\right) \left(x_4-x_5\right)} & \frac{x_2-2 x_3+x_4}{5 \left(x_2-x_3\right) \left(x_3-x_4\right)} & \frac{x_1-2 x_2+x_3}{5 \left(x_1-x_2\right) \left(x_2-x_3\right)} & \frac{2 x_1-x_2}{5 x_1 \left(x_1-x_2\right)} & \frac{1}{3125 x_1 \left(x_1-x_2\right) \left(x_2-x_3\right) \left(x_3-x_4\right) \left(x_4-x_5\right)} \\ \frac{1}{5 x_1} & -\frac{1}{5 \left(x_1-x_2\right)} & -\frac{1}{5 \left(x_2-x_3\right)} & -\frac{2}{5 \left(x_3-x_5\right)} & -\frac{2}{5 \left(x_3-x_5\right)} & -\frac{2}{5 \left(x_3-x_5\right)} & 0 & \frac{2 x_2-3 x_3+x_5}{5 \left(x_2-x_3\right) \left(x_3-x_5\right)} & \frac{x_1-2 x_2+x_3}{5 \left(x_1-x_2\right) \left(x_2-x_3\right)} & \frac{2 x_1-x_2}{5 x_1 \left(x_1-x_2\right)} & \frac{4}{3125 x_1 \left(x_1-x_2\right) \left(x_2-x_3\right) \left(x_3-x_5\right){}^2} \\ \frac{1}{5 x_1} & -\frac{1}{5 \left(x_1-x_2\right)} & -\frac{2}{5 \left(x_2-x_4\right)} & -\frac{2}{5 \left(x_2-x_4\right)} & -\frac{1}{5 \left(x_4-x_5\right)} & -\frac{1}{5 \left(x_4-x_5\right)} & \frac{x_2-3 x_4+2 x_5}{5 \left(x_2-x_4\right) \left(x_4-x_5\right)} & 0 & \frac{2 x_1-3 x_2+x_4}{5 \left(x_1-x_2\right) \left(x_2-x_4\right)} & \frac{2 x_1-x_2}{5 x_1 \left(x_1-x_2\right)} & \frac{4}{3125 x_1 \left(x_1-x_2\right) \left(x_2-x_4\right){}^2 \left(x_4-x_5\right)} \\ \frac{1}{5 x_1} & -\frac{1}{5 \left(x_1-x_2\right)} & -\frac{3}{5 \left(x_2-x_5\right)} & -\frac{3}{5 \left(x_2-x_5\right)} & -\frac{3}{5 \left(x_2-x_5\right)} & -\frac{3}{5 \left(x_2-x_5\right)} & 0 & 0 & \frac{3 x_1-4 x_2+x_5}{5 \left(x_1-x_2\right) \left(x_2-x_5\right)} & \frac{2 x_1-x_2}{5 x_1 \left(x_1-x_2\right)} & \frac{27}{3125 x_1 \left(x_1-x_2\right) \left(x_2-x_5\right){}^3} \\ \frac{1}{5 x_1} & -\frac{2}{5 \left(x_1-x_3\right)} & -\frac{2}{5 \left(x_1-x_3\right)} & -\frac{1}{5 \left(x_3-x_4\right)} & -\frac{1}{5 \left(x_4-x_5\right)} & -\frac{1}{5 \left(x_4-x_5\right)} & \frac{x_3-2 x_4+x_5}{5 \left(x_3-x_4\right) \left(x_4-x_5\right)} & \frac{x_1-3 x_3+2 x_4}{5 \left(x_1-x_3\right) \left(x_3-x_4\right)} & 0 & \frac{3 x_1-x_3}{5 x_1 \left(x_1-x_3\right)} & \frac{4}{3125 x_1 \left(x_1-x_3\right){}^2 \left(x_3-x_4\right) \left(x_4-x_5\right)} \\ \frac{1}{5 x_1} & -\frac{2}{5 \left(x_1-x_3\right)} & -\frac{2}{5 \left(x_1-x_3\right)} & -\frac{2}{5 \left(x_3-x_5\right)} & -\frac{2}{5 \left(x_3-x_5\right)} & -\frac{2}{5 \left(x_3-x_5\right)} & 0 & \frac{2 \left(x_1-2 x_3+x_5\right)}{5 \left(x_1-x_3\right) \left(x_3-x_5\right)} & 0 & \frac{3 x_1-x_3}{5 x_1 \left(x_1-x_3\right)} & \frac{16}{3125 x_1 \left(x_1-x_3\right){}^2 \left(x_3-x_5\right){}^2} \\ \frac{1}{5 x_1} & -\frac{3}{5 \left(x_1-x_4\right)} & -\frac{3}{5 \left(x_1-x_4\right)} & -\frac{3}{5 \left(x_1-x_4\right)} & -\frac{1}{5 \left(x_4-x_5\right)} & -\frac{1}{5 \left(x_4-x_5\right)} & \frac{x_1-4 x_4+3 x_5}{5 \left(x_1-x_4\right) \left(x_4-x_5\right)} & 0 & 0 & \frac{4 x_1-x_4}{5 x_1 \left(x_1-x_4\right)} & \frac{27}{3125 x_1 \left(x_1-x_4\right){}^3 \left(x_4-x_5\right)} \\ \frac{1}{5 x_1} & -\frac{4}{5 \left(x_1-x_5\right)} & -\frac{4}{5 \left(x_1-x_5\right)} & -\frac{4}{5 \left(x_1-x_5\right)} & -\frac{4}{5 \left(x_1-x_5\right)} & -\frac{4}{5 \left(x_1-x_5\right)} & 0 & 0 & 0 & \frac{5 x_1-x_5}{5 x_1 \left(x_1-x_5\right)} & \frac{256}{3125 x_1 \left(x_1-x_5\right){}^4} \\ \frac{2}{5 x_2} & \frac{2}{5 x_2} & -\frac{1}{5 \left(x_2-x_3\right)} & -\frac{1}{5 \left(x_3-x_4\right)} & -\frac{1}{5 \left(x_4-x_5\right)} & -\frac{1}{5 \left(x_4-x_5\right)} & \frac{x_3-2 x_4+x_5}{5 \left(x_3-x_4\right) \left(x_4-x_5\right)} & \frac{x_2-2 x_3+x_4}{5 \left(x_2-x_3\right) \left(x_3-x_4\right)} & \frac{3 x_2-2 x_3}{5 x_2 \left(x_2-x_3\right)} & 0 & -\frac{4}{3125 x_2^2 \left(x_2-x_3\right) \left(x_3-x_4\right) \left(x_4-x_5\right)} \\ \frac{2}{5 x_2} & \frac{2}{5 x_2} & -\frac{1}{5 \left(x_2-x_3\right)} & -\frac{2}{5 \left(x_3-x_5\right)} & -\frac{2}{5 \left(x_3-x_5\right)} & -\frac{2}{5 \left(x_3-x_5\right)} & 0 & \frac{2 x_2-3 x_3+x_5}{5 \left(x_2-x_3\right) \left(x_3-x_5\right)} & \frac{3 x_2-2 x_3}{5 x_2 \left(x_2-x_3\right)} & 0 & -\frac{16}{3125 x_2^2 \left(x_2-x_3\right) \left(x_3-x_5\right){}^2} \\ \frac{2}{5 x_2} & \frac{2}{5 x_2} & -\frac{2}{5 \left(x_2-x_4\right)} & -\frac{2}{5 \left(x_2-x_4\right)} & -\frac{1}{5 \left(x_4-x_5\right)} & -\frac{1}{5 \left(x_4-x_5\right)} & \frac{x_2-3 x_4+2 x_5}{5 \left(x_2-x_4\right) \left(x_4-x_5\right)} & 0 & \frac{2 \left(2 x_2-x_4\right)}{5 x_2 \left(x_2-x_4\right)} & 0 & -\frac{16}{3125 x_2^2 \left(x_2-x_4\right){}^2 \left(x_4-x_5\right)} \\ \frac{2}{5 x_2} & \frac{2}{5 x_2} & -\frac{3}{5 \left(x_2-x_5\right)} & -\frac{3}{5 \left(x_2-x_5\right)} & -\frac{3}{5 \left(x_2-x_5\right)} & -\frac{3}{5 \left(x_2-x_5\right)} & 0 & 0 & \frac{5 x_2-2 x_5}{5 x_2 \left(x_2-x_5\right)} & 0 & -\frac{108}{3125 x_2^2 \left(x_2-x_5\right){}^3} \\ \frac{3}{5 x_3} & \frac{3}{5 x_3} & \frac{3}{5 x_3} & -\frac{1}{5 \left(x_3-x_4\right)} & -\frac{1}{5 \left(x_4-x_5\right)} & -\frac{1}{5 \left(x_4-x_5\right)} & \frac{x_3-2 x_4+x_5}{5 \left(x_3-x_4\right) \left(x_4-x_5\right)} & \frac{4 x_3-3 x_4}{5 x_3 \left(x_3-x_4\right)} & 0 & 0 & \frac{27}{3125 x_3^3 \left(x_3-x_4\right) \left(x_4-x_5\right)} \\ \frac{3}{5 x_3} & \frac{3}{5 x_3} & \frac{3}{5 x_3} & -\frac{2}{5 \left(x_3-x_5\right)} & -\frac{2}{5 \left(x_3-x_5\right)} & -\frac{2}{5 \left(x_3-x_5\right)} & 0 & \frac{5 x_3-3 x_5}{5 x_3 \left(x_3-x_5\right)} & 0 & 0 & \frac{108}{3125 x_3^3 \left(x_3-x_5\right){}^2} \\ \frac{4}{5 x_4} & \frac{4}{5 x_4} & \frac{4}{5 x_4} & \frac{4}{5 x_4} & -\frac{1}{5 \left(x_4-x_5\right)} & -\frac{1}{5 \left(x_4-x_5\right)} & \frac{5 x_4-4 x_5}{5 x_4 \left(x_4-x_5\right)} & 0 & 0 & 0 & -\frac{256}{3125 x_4^4 \left(x_4-x_5\right)} \\ \frac{1}{x_5} & \frac{1}{x_5} & \frac{1}{x_5} & \frac{1}{x_5} & \frac{1}{x_5} & \frac{1}{x_5} & 0 & 0 & 0 & 0 & \frac{1}{x_5^5} \\ \end{array} \right] $$