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Your constraint is non-convex, there's no getting around that (so you can't just replace it with some linear constraints, say). But what you might try is this: start by minimizing the objective witho …

It "can be" but usually is not. Think of it in reverse. Start with a problem of minimizing a convex function $f$ over a convex set $S$ in vector space $V$, where the minimum value happens to be posi …

$x_1 x_2 - x_3 x_4 = 0$ is inherently nonlinear, and maybe more importantly non-convex: e.g. the midpoint of two feasible solutions may not be (in fact, hardly ever is) a feasible solution. So in gen …

Are these really all the constraints? Let $u$ be the vector of all $1$'s. Consider solutions of the following form: $p$ is an arbitrary vector with $u^T p = 1$, $A = B + c u^T$ where $B p = 0$. …

You want a positive (I assume) number $1/s$ such that $\det(sA + B \pm iI) = 0$. For any given $A$ and choice of $\pm$, that is a polynomial in $s$ of degree at most $4$. So yes, it can be solved in …

Consider a problem $\min_x f(x)$ subject to $A x \le b$, for which you have an optimal solution $x^\star$ with Lagrange multipliers $\lambda^\star$ satisfying the Karush-Kuhn-Tucker conditions $$ \eqa …

Let $q - p = u$ and $\|u\| = d$. Let $x = p + t_1 u + v_1 \in B_{r_1}(p) \cap B_{s_1}(q)$ and $y = p + t_2 u + v_2 \in B_{r_2}(p) \cap B_{s_2}(q)$ where $t_i \in {\mathbb R}$ and $v_i \perp u$. The c …

In the case $n=3$,$p=2$, your 5 constraints for $v_1 = (1,0,0)^T$, $v_2 = (0,1,0)^T$, $v_3 = (0,0,1)^T$, $v_4 = (1,-2,0)^T$ and $v_5 = (1,-1,1)^T$ have solution $G = \pmatrix{1 & 1 & t\cr 1 & 1 & t\cr …

$t$ is in fact a tight bound. It's slightly tricky because the objective is not defined at what should be the optimal solution (due to zeros in numerators and denominators). What you want is first $p_ …

No, because this could easily have several local maxima.

Newton's is not really a "minimizing method". If you're using Newton's method to find a root of $f'$, the root you find might be a local minimum, local maximum or neither. To remove the root $x=0$ fr …

See the Michaelis-Menten equation in biochemistry.

The assumption of convexity can be weakened to quasiconvexity.

$\Phi(v,w) \le h$ if there is a path from $v$ to $w$ in the graph obtained from $G$ by removing all vertices of height $> h$. Moreover, the maximum height of any path in this graph from $v$ to $w$ gi …

For continuous problems, minimizing a convex function on a convex domain is considered an easy problem, because there is only ever one local minimum, and a local minimum is the global minimum. Findin …