hello everyone!I met a problem about how to express sqrt(1/x+1) by CVX?I demonstrate mathematically that when 0 < x < 0.75, sqrt(1/x-1) is convex,but CVX rejects it.I konw 1/x is convex,sqrt(x) is concave,so CVX can’t judge that if sqrt(1/x+1) is convex.But everyone knows sqrt(1/x+1) is convex when 0<x<0.75.Although i define 0<x<0.75,it doesn’t work.So, How on should I express it?thank you for your answer,please help me
sqrt(1/(x+1)) is convex for x > -1.
Use
pow_p(x+1,-1/2)
You will benefit from re-reading (or reading for the first time) http://cvxr.com/cvx/doc/funcref.html#new-functions, as well as the rest of the CVX Users’ Guide.
i’m so sorry that i didn’t express clearly.the equation is sqrt((1/x)+1)
Let’s look at t\geq (\frac{1}{x}+1)^p for p>0.
The case p=\frac12 is a bit more elementary because the inequality is equivalent to x(t-1)(t+1)\geq 1, which you can do with a geometric mean of the three factors on the left.
For general p we can write
\log{t}\geq u\geq p\log(\frac{1}{x}+1),
where the left inequality is just the concativity of logarithm and the right one is convex and was discussed on the forum before. It is also in https://docs.mosek.com/cheatsheets/conic.pdf
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thank you,it works.
Hi,
Sir, I want to write D-sqrt(x^2-A^2) in cvx. Could you please help me writing it in cvx.
Thanks
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