i couldn't do the following question for hours
minimize $\sum_{i=1}^{n}x_{i}^{3}$
s.t. $\sum_{i=1}^{n}x_{i}=0$ and $\sum_{i=1}^{n}x_{i}^{2}=n$.
by Lagrange multiplier rule
?
i couldn't do the following question for hours
minimize $\sum_{i=1}^{n}x_{i}^{3}$
s.t. $\sum_{i=1}^{n}x_{i}=0$ and $\sum_{i=1}^{n}x_{i}^{2}=n$.
by Lagrange multiplier rule
?
This is a development of my comment, changing the restraints from two to one.
Since $x_{n}=-\sum_{i=1}^{n-1}x_{i}$, the question "find
$\min \sum_{i=1}^{n}x_{i}^{3}$
s.t.
$\sum_{i=1}^{n}x_{i}=0$
and
\sum_{i=1}^{n}x_{i}^{2}=n\ "
is equivalent to find
$\min f(x_{1},x_{2},\ldots ,x_{n-1}),$
where $f(x_{1},x_{2},\ldots ,x_{n-1})=\left( \sum_{i=1}^{n-1}x_{i}^{3}\right) -\left( \sum_{i=1}^{n-1}x_{i}\right) ^{3}$
s.t.
$c(x_{1},x_{2},\ldots ,x_{n-1})=0,$
where
$c(x_{1},x_{2},\ldots ,x_{n-1})=\left( \sum_{i=1}^{n-1}x_{i}\right) ^{2}-n+\sum_{i=1}^{n-1}x_{i}^{2}=0.$
The conditions regarding first derivatives are
$\nabla f(x_{1},x_{2},\ldots ,x_{n-1})-\nabla c(x_{1},x_{2},\ldots ,x_{n-1})\lambda =0\qquad (\ast ),$
where
$\nabla f(x)=\begin{pmatrix}\frac{\partial f}{\partial x_{1}} & \ldots & \frac{\partial f}{\partial x_{n-1}}\end{pmatrix}^{T}$
$=% \begin{pmatrix} 3x_{1}^{2}-3\left( \sum_{i=1}^{n-1}x_{i}\right) ^{2} & \ldots & 3x_{n-1}^{2}-3\left( \sum_{i=1}^{n-1}x_{i}\right) ^{2}% \end{pmatrix}% ^{T}$
$\nabla c(x)=% \begin{pmatrix} \frac{\partial c}{\partial x_{1}} & \ldots & \frac{\partial c}{\partial x_{n-1}}% \end{pmatrix}% ^{T}$
$=% \begin{pmatrix} 2\left( \sum_{i=1}^{n-1}x_{i}\right) +2x_{1} & \ldots & 2\left( \sum_{i=1}^{n-1}x_{i}\right) +2x_{n-1}% \end{pmatrix}% ^{T}.$
Condition $(\ast )$ is
$3x_{1}^{2}-3\left( \sum_{i=1}^{n-1}x_{i}\right) ^{2}-2\lambda \left( \sum_{i=1}^{n-1}x_{i}\right) -\lambda 2x_{1}=0$
$\ldots $
$3x_{n-1}^{2}-3\left( \sum_{i=1}^{n-1}x_{i}\right) ^{2}-2\lambda \left( \sum_{i=1}^{n-1}x_{i}\right) -\lambda 2x_{n-1}=0.$
All these equations have the same form (the coefficients of $x_k$, $1\lt k\lt n-1$, are independent of $k$). Thus the solution is atained when all the $n-1$ variables are equal
$x_{1}=\ldots =x_{n-1}=x^{\ast }$
Hence, we have to solve the new system of two equations
$3(x^{\ast })^{2}-3\left( \sum_{i=1}^{n-1}x^{\ast }\right) ^{2}-2\lambda \left( \sum_{i=1}^{n-1}x^{\ast }\right) -2\lambda x^{\ast }=0\qquad (\ast \ast )$
$\left( \sum_{i=1}^{n-1}x^{\ast }\right) ^{2}-n+\sum_{i=1}^{n-1}(x^{\ast })^{2}=0.$
We obtain, successively
$3(x^{\ast })^{2}-3n+3\sum_{i=1}^{n}(x^{\ast })^{2}-2\lambda \left( \sum_{i=1}^{n-1}x^{\ast }\right) -2\lambda x^{\ast }=0$
$3(x^{\ast })^{2}-3n+3(x^{\ast })^{2}(n-1)-2\lambda x^{\ast }(n-1)-2\lambda x^{\ast }=0$
$(x^{\ast })^{2}(n-1)-1=0$
$x^{\ast }=x_{1}=\ldots =x_{n-1}=\pm \frac{1}{\sqrt{n-1}}.$
Therefore
$x_{n}=-\sum_{i=1}^{n-1}x_{i}=\mp \frac{n-1}{\sqrt{n-1}}.$
Since for $n>2$
$\left( \frac{1}{\sqrt{n-1}}\right) ^{3}-\left( \frac{n-1}{\sqrt{n-1}}% \right) ^{3}<-\left( \frac{1}{\sqrt{n-1}}\right) ^{3}+\left( \frac{n-1}{% \sqrt{n-1}}\right) ^{3},$
the minimum is obtained at
$x^{\ast }=x_{1}=\ldots =x_{n-1}=\frac{1}{\sqrt{n-1}},x_{n}=-\frac{n-1}{% \sqrt{n-1}}$
and its value is equal to
$\left( \frac{1}{\sqrt{n-1}}\right) ^{3}-\left( \frac{n-1}{\sqrt{n-1}}% \right) ^{3}.$
Added: The original problem in $n$ variables is the solution of $n$ equations
$\nabla f(x_{1},x_{2},\ldots ,x_{n})-\nabla c(x_{1},x_{2},\ldots ,x_{n})\lambda =0\qquad (\ast\ast\ast )$
plus the two conditions
$c_{1}(x_{1},x_{2},\ldots ,x_{n})=0$
$c_{2}(x_{1},x_{2},\ldots ,x_{n})=0,$
where
$\nabla f=% \begin{pmatrix} \frac{\partial f}{\partial x_{1}} & \ldots & \frac{\partial f}{\partial x_{n}}% \end{pmatrix}% ^{T}$
$\nabla c=% \begin{pmatrix} \left( \nabla c_{1}\right) ^{T} & \left( \nabla c_{2}\right) ^{T}% \end{pmatrix}% $
$\lambda =\begin{pmatrix} \lambda _{1} & \lambda _{2}\end{pmatrix}^{T}.$
Note: The equations would be similar for $m$ constraints $c_1,\dots ,c_m.$
Remark: Since I don't know how to write here matrices with more than one row, I wrote them in the transposed form.
Using Lagrange multipliers we get $x_i^2 = \lambda x + \mu$, so that $x_i$ can take only two values, say X > 0 and $Y < 0$ (we can't have $X=Y=0$ for obvious reasons). Suppose that $X$ is taken $a$ times, and $Y$ is taken $b = n-a$ times. We have
$aX + bY = 0$
$aX^2 + bY^2 = n$
You can check that the solution is
$X^2 = b/a, Y^2 = a/b$
The objective function that we want to minimize is
$aX^3 + bY^3 = b\sqrt{b/a} - a\sqrt{a/b}$
We want to make the first summand small and the second big. Note that as we increase $a$, the first summand becomes smaller, and the second one becomes bigger, so we might as well choose $a=n-1$ and $b=1$. So $X = 1/\sqrt{(n-1)}$ and $Y = -\sqrt{n-1}$.
Concluding, the optimum is obtained with the sequence $-\sqrt{n-1}, 1/\sqrt{(n-1)} \times (n-1)$, and the optimal value is
$ -(n-1)\sqrt{n-1} + 1/\sqrt{n-1} $