The title pretty much says it all. I have a complicated function $f:\mathbb{R}^n\rightarrow\mathbb{R}$ that is concave. I want to approximate this function using the first few terms of a Taylor series. Is the approximation also concave? Obviously, the first order approximation is concave. The second order should also be concave too since the Hessian is negative semi-definite. Is this correct or am I missing something?
2026-03-25 14:27:24.1774448844
Is the Taylor expansion of a concave function concave?
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The statement "$n$th degree Taylor expansion of a concave function is concave" is true for $n=1, 2$ as you observed.
It is false for $n\ge 3$ as gammatester noted in a comment. For one thing, there are no concave polynomials of degree 3 since the second derivative, being linear, changes its sign.