# Express equality constraint involving exponentials cones

The exponential cone is define such that $$(x, y, z) \in \text{ExpCone: if } y \exp(x / y) \leq z \land y > 0.$$ The inequality $$\exp(a) \leq b$$ can be expressed as $$[a, 1, b] \in \text{ExpCone}$$.

How do i write $$\exp(a) = b$$ using cone programming?

• $y = 0$ is also possible -- you need to take the closure of this definition. May 2, 2022 at 23:02
• $y = 0$ is not possible in my definition includes $y > 0$ after an logic and ($\land$). May 2, 2022 at 23:12
• That's my point! $y=0$ must be included in the definition of Exponential Cone. The current definition is incomplete, and to complete it you must take the closure of the set of points that satisfy what you have written. May 2, 2022 at 23:40
• @independentvariable There are numerous sources with $y > 0$. I haven't seen any with $y \ge 0$. May 3, 2022 at 0:30
• @MarkL.Stone docs.mosek.com/modeling-cookbook/expo.html May 3, 2022 at 1:14

Q: "How do i write $$\text{exp}(a) = b$$ using cone programming?"
$$\text{exp}(a) = b$$ is a nonlinear equality constraint, and is therefore non-convex.
$$\text{exp}(a) \le b$$ is convex. But to produce $$\text{exp}(a) = b$$, it would need to be paired with $$\text{exp}(a) \ge b$$, which is going in the wrong direction to be convex.
• Me being the devil's advocate: $x^{2} = 0$ is a nonlinear equality constraint, and it is convex (as in, it defines a convex set). That's a horrible way of writing $x=0$, but it's convex nonetheless :) May 3, 2022 at 13:55
• @mtannea Yes, I have used that example myself. (if not on this forum, then on some optimization software forum) But I decided not to be pedantic this time. Actually, one way of looking at $x^2 = 0$ is as $x^2 \le 0$, which is convex, with the other direction, $x^2 \ge 0$ not being needed (which is similar to how complementarity constraint is changed from $.... = 0$ to$... \le 0$ for feeding to SQP solver). May 3, 2022 at 14:19