Convexity of a Function


by Izzhov
Tags: convexity, function
Izzhov
Izzhov is offline
#1
Apr12-07, 02:33 PM
P: 116
What does it mean for a function to be convex (or concave) on an interval [a,b]? I understand what a function is and what an interval is, but I don't get what "convexity" is.
Phys.Org News Partner Mathematics news on Phys.org
Math modeling handbook now available
Hyperbolic homogeneous polynomials, oh my!
Researchers help Boston Marathon organizers plan for 2014 race
Data
Data is offline
#2
Apr12-07, 02:47 PM
P: 998
A (twice-differentiable) function is concave at a point if its second derivative is negative at that point. Similarly a function is convex at a point if its second derivative is positive at that point.

You can extend the definition to functions that aren't differentiable also; see http://en.wikipedia.org/wiki/Concave_function.

Intuitively: A concave (or "concave down") function is one that is "cupped" downwards. For example, the parabola [itex]-x^2[/itex] is concave throughout its domain, and the parabola [itex]x^2[/itex] is convex throughout its domain.

There are functions which are "cupped" but don't actually have the cup shape. For example, [itex]1/x[/itex] is concave on the negative reals and convex on the positive reals, however it doesn't have any extrema at all.


Another way to present it is: A function [itex]f[/itex] is convex on an interval if the set of points above its graph on that interval is a convex set; that is, if[itex]p = (x_1, y_1)[/itex] and [itex]q = (x_2, y_2)[/itex] are points with [itex]x_1, x_2[/itex] on the interval of interest, [itex]y_1 \geq f(x_1)[/itex], and [itex] y_2 \geq f(x_2)[/itex], then the straight line joining [itex]p[/itex] to [itex]q[/itex] lies entirely above the graph of [itex]f[/itex]. Then you can define [itex]f[/itex] is concave whenever [itex]-f[/itex] is convex.
Eighty
Eighty is offline
#3
Apr12-07, 03:29 PM
P: 53
A convex set is a set where all points can be connected with a straight line inside the set (so every point can "see" every other). A function is convex if the set above it (ie the set {(x,y):y>f(x)}) is convex.

If the function is twice differentiable, this is equivalent with that the second derivative is everywhere non-negative.

Izzhov
Izzhov is offline
#4
Apr12-07, 03:32 PM
P: 116

Convexity of a Function


Quote Quote by Data View Post
A (twice-differentiable) function is concave at a point if its second derivative is negative at that point. Similarly a function is convex at a point if its second derivative is positive at that point.

You can extend the definition to functions that aren't differentiable also; see http://en.wikipedia.org/wiki/Concave_function.

Intuitively: A concave (or "concave down") function is one that is "cupped" downwards. For example, the parabola [itex]-x^2[/itex] is concave throughout its domain, and the parabola [itex]x^2[/itex] is convex throughout its domain.

There are functions which are "cupped" but don't actually have the cup shape. For example, [itex]1/x[/itex] is concave on the negative reals and convex on the positive reals, however it doesn't have any extrema at all.


Another way to present it is: A function [itex]f[/itex] is convex on an interval if the set of points above its graph on that interval is a convex set; that is, if[itex]p = (x_1, y_1)[/itex] and [itex]q = (x_2, y_2)[/itex] are points with [itex]x_1, x_2[/itex] on the interval of interest, [itex]y_1 \geq f(x_1)[/itex], and [itex] y_2 \geq f(x_2)[/itex], then the straight line joining [itex]p[/itex] to [itex]q[/itex] lies entirely above the graph of [itex]f[/itex]. Then you can define [itex]f[/itex] is concave whenever [itex]-f[/itex] is convex.
So this means the function [tex]x^x[/tex] is convex where x>0, correct?
Eighty
Eighty is offline
#5
Apr12-07, 05:19 PM
P: 53
Yes. (damn character limit)


Register to reply

Related Discussions
Convexity of set A = {(x,y) in R^2 | x^4+y^4 =< 1, x>=0 Calculus 1
convexity and concavity of a function Calculus 1
convexity Calculus & Beyond Homework 5
Convexity, convex hull, math software? General Math 0