{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

solutions06

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

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

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 .

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}