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# HW2Solns - HW2 Answers(Partial A4.d f(x = x1 x2 Applying...

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HW2 Answers (Partial) A4.d f ( x ) = x 1 x 2 Applying Jensen’s Inequality f ( θ x +(1 - θ ) y ) ? θf ( x ) + (1 - θ ) f ( y ) Let x = x 1 x 2 and y = y 1 y 2 where x , y R 2 , then f ±• θx 1 + (1 - θ ) y 1 θx 2 + (1 - θ ) y 2 ‚¶ ? θf ±• x 1 x 2 ‚¶ + (1 - θ ) f ±• y 1 y 2 ‚¶ θx 1 + (1 - θ ) y 1 θx 2 + (1 - θ ) y 2 ? θ x 1 x 2 + (1 - θ ) y 1 y 2 θx 1 + (1 - θ ) y 1 θx 2 + (1 - θ ) y 2 - θ x 1 x 2 - (1 - θ ) y 1 y 2 ? 0 ( θx 1 + (1 - θ ) y 1 ) x 2 y 2 - θx 1 y 2 ( θx 2 + (1 - θ ) y 2 ) - (1 - θ ) x 2 y 1 ( θx 2 + (1 - θ ) y 2 ) ? 0 (1 - θ ) ( x 1 x 2 y 2 + x 2 y 1 y 2 - x 1 y 2 2 - x 2 2 y 1 ) ? 0 (1 - θ )( x 1 y 2 - x 2 y 1 )( x 2 - y 2 ) ? 0 (1 - θ ) is always non-negative, moreover there is no guarantee that ( x 1 y 2 - x 2 y 1 )( x 2 - y 2 ) is negative deﬁnite, e.g. let x 1 = 1, x 2 = 1, y 1 =,3 y 2 = 2 , then ( x 1 y 2 - x 2 y 1 )( x 2 - y 2 ) = 1 , hence inequality test fails. f ( x ) = x 1 x 2 is not convex!. A4.e f ( x ) = x 2 1 x 2 Applying Jensen’s Inequality f ( θ x +(1 - θ ) y ) ? θf ( x ) + (1 - θ ) f ( y ) Let x = x 1 x 2 and y = y 1 y 2 where x , y R 2 , then f ±• θx 1 + (1 - θ ) y 1 θx 2 + (1 - θ ) y 2 ‚¶ ? θf ±• x 1 x 2 ‚¶ + (1 - θ ) f ±• y 1 y 2 ‚¶ 1

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( θx 1 + (1 - θ ) y 1 ) 2 θx 2 + (1 - θ ) y 2 ? θ x 2 1 x 2 + (1 - θ ) y 2 1 y 2 ( θx 1 + (1 - θ ) y 1 ) 2 x 2 y 2 - θx 2 1 y 2 ( θx 2 + (1 - θ ) y 2 ) - (1 - θ ) x 2 y 2 1 ( θx 2 + (1 - θ ) y 2 ) ?
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HW2Solns - HW2 Answers(Partial A4.d f(x = x1 x2 Applying...

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