sum((x/h)^0.25)sum((x^0.75)/h^-0.25)

given that x =[x1 x2…] and all positive such that sum(x)=1

and h=[h1,h2,…] also all positive and of equal dimension to x

sum((x/h)^0.25)sum((x^0.75)/h^-0.25)

given that x =[x1 x2…] and all positive such that sum(x)=1

and h=[h1,h2,…] also all positive and of equal dimension to x

If x has dimension 2 or greater, this is neither convex nor concave (as can be seen by looking at the eigenvalues of the Hessian), and therefore can not be entered into CVX. h being the vector of ones is sufficient to show this.

Edit (post left for context, but is withdrawn due to error).

yes x has dimension 2…but i derived the hessian of the whole objective function and it showed that it is positive semi-definite

what i have given above is just a portion of the whole objective function.

any ideas how to approach this problem

Sorry, I had a typo in my calculation and entered the wrong function. So I withdraw the non-convex claim.

It looks like it might be concave (definitely not convex, unless I have misread the function), but I haven’t proven it. If dimension is 2, you can reduce this to a 1 dimensional problem by substituting `x2 = 1- x1`

, and of course the constraint `0 <= x1 <= 1`

. I don’t see how to get it accepted by CVX. Can you build it up following CVX’s rules?

well, I tried the following

(sum(pow_p(x./h,0.25)*(sum(h.*(pow_p(x./h,0.75))))

but I get the following error

Error using cvx/times (line 173)

Disciplined convex programming error:

Cannot perform the operation: {concave} .* {concave}

I suggest you (re-)read the CVX Users’ Guide. Of course that operation is not allowed, as it does not necessarily preserve concavity.

thank you very much, will do that

Appreciate the help

It actually depends a lot on what you’re trying to do with this. Because in fact, it *is* possible to express

\left(\sum_i (x_i/h)^{1/4} \sum_i (x_i/h)^{3/4} \right)^{1/2}

in CVX as follows:

```
geo_mean([sum(pow_p(x/h,0.25)),sum(pow_p(x/h,0.75)])
```

But you can’t express it without the surrounding square root. The geometric mean here is concave, so you could maximize this, or you could put a lower bound on it.