# How to address x^4/y^2 in cvx?

By using Hessian matrix for x^4/y^2, I confirm this matrix is semi-definite positve. But I don’t know which cvx expression can be used for x^4/y^2.

If we assume y>=0

x \leq (yyzt)^{1/4}, z=1, t \geq 0

which says x has to be less than geometric mean of some variables. Note that this implies

\frac{x^4}{y^2} \leq t, y \geq 0

In the reference guide you can find information about the geo_mean function.

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Thanks Erling. But in my formulation, y can not be assumed as y>0=0. But the term x^4/y^2 can still be convex. So, I don’t know how to address it in CVX.

Looking at

it seems the Hessian positive semi-definite. I have not figured the more general case out yet.

x^4/y^2 is convex separately on y positive and negative but not jointly because of the singularity at y=0 where it escapes to infinity. Consider the plot of 1/y^2 for example.

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Thanks Michal. I got it.

Necromancer here. If relevant, @Erling gave you a representation on y>=0 and you could very similarly obtain a representation on y<=0 using

x \leq ((-y)(-y)zt)^{1/4}.

So your nonconvex optimization problem has a spatial branching decision [y>=0] or [y<=0] that splits it into two convex optimization problems both representable in cvx. This is sometimes called backdoor branching. If you have many constraints of this form, however, the number of convex subproblems you need to solve grows exponentially.

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