{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

solutions06 - 1 15.093J/2.098J Optimization Methods...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
1 15.093J/2.098J Optimization Methods Assignment 6 Solutions Exercise 6.1 These functions are twice differentiable; therefore, we can check the convexity using the Hessian matrices. A function f ( x ) is convex if its Hessian matrix is positive semidefinite for all x . The definiteness of matrices can be determined by using the definition, eigenvalues or principal minors. i) f ( x 1 , x 2 ) = x 2 + 2 x 1 x 2 10 x 1 + 5 x 2 1 2 2 H ( x 1 , x 2 ) = is indefinite (using principal minors), thus f is neither concave nor convex. 2 0 2 2 ii) f ( x 1 , x 2 , x 3 ) = x 1 3 x 2 2 x 2 + 4 x 1 x 2 + 2 x 1 x 3 + 4 x 2 x 3 3 2 4 2 H ( x 1 , x 2 , x 3 ) = 4 6 4 is indefinite (using principal minors again), thus f is neither convex 2 4 4 nor concave. ( x 1 + x 2 ) iii) f ( x 1 , x 2 ) = x 1 e H ( x 1 , x 2 ) = x 1 2 x 1 1 e ( x 1 + x 2 ) is not a definite matrix for all x , thus f is neither convex nor x 1 1 x 1 concave. 1 iv) f ( x 1 , x 2 , x 3 ) = 3 i =1 ln( x i ) + x 1 + e x 2 for x i > 0 We can prove this function to be convex by computing its Hessian matrix and showing that it is a positive 1 definite matrix. The second way is to show that f is the sum of convex functions, namely ln( x ), x , and x e .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}