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Given the following problem \begin{align}\min&\quad x_1+2x_2+3x_3+4x_4+\sum_{i=1}^4x_i\ln(x_i)\\\text{s.t.}&\quad e^\top x=1\\&\quad x\geq0\end{align}

I was asked to solved the dual problem and I think I've managed to do so, I've got $$q(\lambda)=-e^{-\lambda}(e^{-2}+e^{-3}+e^{-4}+e^{-5})-\lambda.$$ I wanted to check if I'm right so I wanted to check with CVX via matlab but the following code doesn't work.

cvx_begin  
variable x(4)
minimize (x(1)+2*x(2)+3*x(3)+4*x(4)+x(1)*log(x(1))+x(2)*log(x(2))+x(3)*log(x(3))+x(4)*log(x(4)))
sum(x)==1
x>=0

How can I fix the objective function? (I know the problem is at $x_i\ln(x_i)$)

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Use CVX's entr function.

$\sum_{i=1}^ 4x_i\ln(x_i)$ can be entered as -sum(entr(x))

entr   Scalar entropy.
    entr(X) returns an array of the same size as X with the unnormalized
    entropy function applied to each element:
                 { -X.*LOG(X) if X > 0,
       entr(X) = { 0          if X == 0,
                 { -Inf       otherwise.
    If X is a vector representing a discrete probability distribution, then
    SUM(entr(X)) returns its entropy.
 
    Disciplined convex programming information:
        entr(X) is concave and nonmonotonic in X. Thus when used in CVX
        expressions, X must be real and affine. Its use will effectively 
        constrain X to be nonnegative: there is no need to add an
        additional X >= 0 to your model in order to enforce this.
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