Log{convex} log(1+1/(1+x))

[Kirti_Kant_Sharma]

log(1+1/(1+x)) is a convex function over a<=x<=b, a,b>0, so how can i implement it in cvx?

[mcg (Michael C. Grant)]

Just because it’s convex doesn’t mean CVX can implement it. It does not claim to be able to handy any and all convex problems. Please read the FAQ.

[mcg (Michael C. Grant)]

You may be able to use a polynomial approximation; please see the function reference for the use of polyval.

[Kirti_Kant_Sharma]

Thank you for quick reply. I also find on the forum a way to do this

Minimize log(1+1/x) where 0<x<inf

Sorry for the late answer. May be you can rewrite your problem in the following form minimize y such that log(1 + 1/x) \leq y which is equivalent to minimize y such that exp(-y) + exp(-y-log(x)) \leq 1 The following code is working at least cvx_begin variables x y minimize( y ) exp(-y) + exp(-y - log(x) ) <= 1 cvx_end

[mcg (Michael C. Grant)]

Hey that’s a good find! Share here if you find a solution to your particular case.

It’s actually very useful to understand that you really never need to know how to “implement” a function f(x). Rather, you need a way to implement the inequality f(x) \leq y (or, f(x) \geq y if f is concave). Any set of equations that returns the same subset on (x,y) is a valid implementation of f(x) \leq y. That’s what the user is doing in that post.

[zhangqirui814]

Consisder another variable y satisfying 1/y=1+1/x. Then, we can minimize the convex objective function -log(y) s.t.a/(a+1) <=y<=b/(1+b)

[Aristruby (sun)]

log(1+1/(1+x))=rel_entr(x+1,x+2)+rel_entr(x+2,x+1)