Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider the following function $$\pi_{p}(x) = \left(\frac{1}{2}x_{1}-x_{2}\right)^{2}+\left(\frac{1}{4}x_{1}^{2}-\rho \min(0,x_{1}-1)\right)$$

For any $\rho$ we want to minimize this function. The minimum of $\pi$ is obtained at a point where $x_2 = \frac{1}{2}x_{1}$ and where $x_1$ minimizes the function defined by $$\begin{cases} \frac{1}{4}x_{1}^{2}-\rho(x_{1}-1),& \text{if} \ x_1 <1 \\\\ \frac{1}{4}x_{1}^2,&\text{if} \ x_{1} \geq 1 \end{cases}$$

Shouldn't it be $$\begin{cases} \frac{1}{4}x_{1}^{2}-\rho(x_{1}-1),&\text{if} \ x_1 \geq 1 \\\\ \frac{1}{4}x_{1}^2,&\text{if} \ x_{1} < 1 \end{cases}$$

share|improve this question

1 Answer 1

If $x_1<1$, then $x_1-1<0$, so $\min(0,x_1-1)=x_1-1$, and you want $\frac14x_1^2-\rho(x_1-1)$. If $x_1\ge 1$, then $x_1-1\ge 0$, so $\min(0,x_1-1)=0$, and you want simply $\frac14x_1^2$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.