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- An easy way to test for both is to connect two points on the curve with a straight line. If the line is above the curve, the graph is convex. If the line is below the curve, the graph is concave.
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Aug 16, 2019 · What you gave is the standard definition of a convex function. If $f$ is supposed to be continuous, it is enough to check that $$f\left(\frac{x+y}{2}\right) \le \frac{f(x)+f(y)}{2}$$ for all $x,y$. If $f$ is twice differentiable, it is enough to check that the second derivative is non negative.
The book "Convex Optimization" by Boyd, available free online here, describes methods to check. The standard definition is if f (θx + (1 − θ)y) ≤ θf (x) + (1 − θ)f (y) for 0≤θ≤1 and the domain of x,y is also convex. So if you could prove that for your function, you would know it's convex.
A set E \subseteq \mathbb {R} E ⊆ R is said to be convex if given x,y \in E x,y ∈ E such that x < y x <y then [x,y] \subseteq E [x,y] ⊆ E. If a < b a <b then [a,b], [a,b), (a,b], (a,b) [a,b],[a,b),(a,b],(a,b) are all convex sets. Another interesting example is \left \ { a \right \} {a}.
Answer. Prove that each of the following functions is convex on the given domain: f (x)=e^ {b x}, x \in \mathbb {R}, where b is a constant. f (x)=x^ {k}, x \in [0, \infty) and k \geq 1 is a constant. f (x)=-\ln (1-x), x \in (-\infty, 1). f (x)=-\ln \left (\frac {e^ {x}} {1+e^ {x}}\right), x \in \mathbb {R}.
Jun 3, 2022 · With functions of one variable, you would check for convexity by looking at the second derivative. Suppose you have $f(x)$: the function is convex on an interval $I$ if and only if $f''(x) \geq 0 \quad \forall x \in I$.
N X about x , we have f(x) f(x ) for all x 2 N. Suppose towards a contradiction that there exists ~x 2 X such that f(~x) < f(x ). Consider the line segment x(t) = tx + (1 t)~x; t 2 [0; 1], noting that x(t) 2 X by the convexity of X. Then by the convexity of f, f(x(t)) tf(x ) + (1. t)f(~x) < tf(x ) + (1.
2.1 That’s great, but how do I prove that a function is convex? 1. If you know calculus, take the second derivative. It is a well-known fact that if the second derivative f00(x) is 0 for all xin an interval I, then fis convex on I. On the other hand, if f(x) 0 for all x2I, then fis concave on I. 2.
