Math Group
Question:
pavelpimen
Student
High School - 10th Grade
Having the function fn(x)=x^n+n*x-1, demonstrate that fn is convex, for any n>2; n is a natural number.
Answers:
-
kjcdb8er Teacher
eNotes Editor
Posted by kjcdb8er on Tuesday June 9, 2009 at 1:55 AMA function is convex over a region R if every point on the function in R lies beneath the line connecting the boundaries of R. The second derivative of a function can tell us whether a function is convex over a region: if and only if f''(x) > 0, then the region is convex.
So for f(x) = x^n + nx - 1
f'(x) = nx^(n-1) + n
f''(x) = n(n-1)x^(n-2)
Note that the natural numbers are the set {0, 1, 2, 3, ...}
So for n = {0,1}, f''(x) = 0
For n = 2, f''(x) = 2, so f(x) is convex.
For n > 2, f''(x) = N*x^(n-2), where N > 0 and n-2 > 0
Hence, for n-2 even f(x) is strictly convex for all x, and for n-2 odd f(x) is convex for x > 0.
Sources:
