Tutorial03-solutions

# It then implies ln2 2 1 0 that is 2 2 and 1

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Unformatted text preview: 1 . Since here = ± , we get four more critical points √1 √1 2 2 2 √1 2 − √1 2 , − √1 2 , − √1 − √1 2 2 √1 2 −1 , . Now let’s check the type of these critical points. We have 2 = = ln( 2 2 2( 2 + + 2 +2 22 )+ 2 + + 2) − 2 ( 2 + 2 )2 3 − 2+ ( 2 2 2 2 +2 2 2 = 2 2 = + 2 ( 2 )2 2 2 + ( 2 + 2 )2 4 2 = ln( + 2 + )+2 2 (+ 4 2 )2 3 4 + 2 3 4 + + 2 )2 Now we get H (±1 0) = H (0 ±1) = det H = −4 &lt; 0 02 20 so (±1 0) and (0 ±1) are saddle points. Further, H ± 1 √ 2 1 √ 2 and therefore ± √1 2 2 (d) ( )= + √1 2 2 + H 20 02 = ± 1 √ 2 are local minima and ± 2 + , where 1 −√ 2 √1 2 = − √1 2 −2 0 0 −2 are local maxima. &gt; 0. For critical points, we have 4 2 =1− 4 2 =0 2 − = 2 2 2 =0 = − 2 2 =0 so 2 = 4 2 and since &gt; 0, we conclude that = 2 . Now substituting it to the second equation, we get = = 2 . Finally, from the third equation, we obtain 1 = 1, so the only critical point is 2 1 1 . Further, 2 3 2 H = −2 0 2 −2 2 2 1 + 23 2 − 22 0 − 22 2 + 43 3 H 1 11 2 4 −2 0 = −2 3 −2 0 −2 6 The characteristic polynomial is 4− −2 0 φ( ) := det(H − I ) = −2 3− −2 0 −2 6− =− 3 + 13 2 − 46 + 32 Now since det H = φ(0) &gt; 0, we have either 0 or 2 negative eigenval...
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## This document was uploaded on 02/10/2014.

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