B note that when 0 0 2 from our expression in part

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Unformatted text preview: 0) = 0. (b) Note that when = 0, (0 ) = 2/ from our expression in part (b) and we also have (0 0) = 0. It readily follows that (0 0) doesn’t exist since (0 ) is not even continuous at = 0. :) E A function ( ) is called Morse if all its critical points are nondegenerate. A function ( ) is called harmonic if the equation + = 0 holds for all . Prove that a harmonic Morse function does not have local maxima or minima, so its critical points are always saddles. S Since = − , the determinant of the Hessian matrix is − 2 − 2 ≤ 0. Since all critical points are supposed to be non-degenerate, we have − 2 − 2 < 0 and therefore all of them are saddle points. :) E Let 2− 2 ( + 2 > 0; = =0 2 2+ 2 )= 0 (a) Find (0 0) and (0 0). (b) Explain why the fact that S For ( (0 0) = (0 0) does not contradict Clairaut’s Theorem. (a) As we have ( 0) = (0 ) = 0, it follows that (0 0) = (0 0) = 0. ) = (0 0), by the quotient rule we get (after some calculation) 4 ( )= +4 2 3− ( 2 + 2 )2 Hence (0 )= − 0 5 5 ( )= / 4 =0 = (− )| 0) = = 5 =− −4 3 2− ( 2 + 2 )2 ; ; =0 =− =0 and (0 0) = (0 )| =0 = −1| In the same manner, ( and (0 0) = 0)| ( 5 =0 = =1 =0 = −1 4 :) Hence (0 0) = ( (0 0) (b) After some calculation, you can verify, using usual differentiation rules, that for ) = (0 0), 6 +9 4 2−9 2 4− 6 ( )= ( )= ( 2 + 2 )3 Hence, approaching (0 0) along the -axis we have − lim (0 →0 ) = lim →0 6 6 = lim(−1) = −1 →0 while approaching along the -axis we have 6 lim →0 ( 0) = lim →0 6 = lim 1 = 1 →0 Since these are different, we conclude that ( ) (and also ) has no limit as ( )→ (0 0) and hence it is not continuous there. Continuity of the mixed derivatives was a condition for Clairaut’s theorem, so it is not contradicted by the result from (a). 6...
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