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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Problem Set 3 MH2100 Solutions 1 Questions on understanding the lecture E±²³´µ¶² · Without doing any calculation, explain why any critical point of the function ± ( ²³´³µ ) = sin ² · arctan( ´ - ² ) + cos( ´ 2 - ² 2 ) is degenerate. S¸¹º»µ¸¼ Since the function is independent of µ , all its derivatives with respect to µ vanish and therefore the Hessian matrix looks like H ± = ± ²² ± ²´ 0 ± ²´ ± ´´ 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±²³´µ¶² ½ Find critical points and determine their type (degenerate/non-degenerate, Morse index if there is any) for the following functions: (a) ± ( ²³´ ) = ( ² - ´ + 1) 2 . (b) ± ( ) = ²´ + 50 ² + 20 ´ , where ²³´ > 0. (c) ± ( ) = ln( ² 2 + ´ 2 ). (d) ± ( ) = ² + ´ 2 4 ² + µ 2 ´ + 2 µ , where ²³´³µ > 0. 1
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
S±²³´µ±¶ (a) ± ( ²³´ ) = ( ² - ´ + 1) 2 . For critical points, we have ± ² = 2( ² - ´ + 1) = 0 ³ ± ´ = - 2( ² - ´ + 1) = 0 ³ so both equations are ² - ´ + 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) ± ( ) = ²´ + 50 ² + 20 ´ , where ²³´ > 0. For critical points, we have ± ² = ´ - 50 ² 2 = 0 ³ ± ´ = ² - 20 ´ 2 = 0 ³ so ´ = 50 ² 2 and ² = 20 ´ 2 = 20 (50 2 ) 2 = ² 4 125 . Now we get ² 3 = 125, that is, ² = 5 and ´ = 2. Further, H ± = · 100 ² 3 1 1 40 ´ 3 ¸ ³ H ± (5 ³ 2) = · 4 5 1 1 5 ¸ and det H ± = 3 > 0, tr H ± > 0. Hence (5 ³ 2) is a local minimum. (c) ± ( ) = ln( ² 2 + ´ 2 ). For critical points, we have ± ² = ´ ln( ² 2 + ´ 2 ) + 2 ² ² 2 + ´ 2 = 0 ³ ± ´ = ² ln( ² 2 + ´ 2 ) + 2 ´ ² 2 + ´ 2 = 0 µ From ± ² = 0, we see that either ´ = 0 or ln( ² 2 + ´ 2 )+ 2 ² 2 ² 2 + ´ 2 = 0. Let’s consider these two cases separately.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 6

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online