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Tutorial03-solutions - Problem Set 3 MH2100 Solutions 1...

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Problem Set 3 MH2100 Solutions 1 Questions on understanding the lecture E Without doing any calculation, explain why any critical point of the function ( ) = sin · arctan( - ) + cos( 2 - 2 ) is degenerate. S Since the function is independent of , all its derivatives with respect to vanish and therefore the Hessian matrix looks like H = 0 0 0 0 0 It follows that H has at least one eigenvalue being 0 so that any critical point of the function ( ) is degenerate. :) 2 Questions on calculation E Find critical points and determine their type (degenerate/non-degenerate, Morse index if there is any) for the following functions: (a) ( ) = ( - + 1) 2 . (b) ( ) = + 50 + 20 , where > 0 . (c) ( ) = ln( 2 + 2 ) . (d) ( ) = + 2 4 + 2 + 2 , where > 0 . 1
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S (a) ( ) = ( - + 1) 2 . For critical points, we have = 2( - + 1) = 0 = - 2( - + 1) = 0 so both equations are - + 1 = 0 , which is a whole straight line consisting of critical points. Since they are not isolated, they must be degenerate. Since the value of the function at this line is 0 and ( ) 0 , we see that this line consists of degenerate minima. (b) ( ) = + 50 + 20 , where > 0 . For critical points, we have = - 50 2 = 0 = - 20 2 = 0 so = 50 2 and = 20 2 = 20 (50 / 2 ) 2 = 4 125 . Now we get 3 = 125 , that is, = 5 and = 2 . Further, H = 100 3 1 1 40 3 H (5 2) = 4 5 1 1 5 and det H = 3 > 0 , tr H > 0 . Hence (5 2) is a local minimum. (c) ( ) = ln( 2 + 2 ) . For critical points, we have = ln( 2 + 2 ) + 2 2 + 2 = 0 = ln( 2 + 2 ) + 2 2 + 2 = 0 From = 0 , we see that either = 0 or ln( 2 + 2 )+ 2 2 2 + 2 = 0 . Let’s consider these two cases separately.
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