m425b-ICfinsoln-s08

m425b-ICfinsoln-s08 - ± ± ( f ( x + h )-f ( x )) f ( x )...

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MATH 425b IN-CLASS FINAL EXAM SOLUTIONS SPRING 2008 Prof. Alexander (1)(a) We calculate = 2 z dx dy dz, all other terms being 0. By Stokes Theorem, Z ∂A ω = Z A dω. We can parametrize A by the identity so Z A = Z a - a Z a - a Z a - a z dx dy dz = Z a - a Z a - a Z a - a 2 z dz dx dy. Since z is an odd function, the innermost integral is 0. (b) It’s sufficient for the above argument that the innermost integral be from - a to a , so z 0 = 0 is sufficient. (2)(a) By the Cauchy-Schwarz inequality, | Ax · y | ≤ | Ax | | y | , so sup x,y 6 =0 | Ax · y | | x | | y | sup x,y 6 =0 | Ax | | y | | x | | y | = sup x 6 =0 | Ax | | x | = || A || . (b) Considering only x with Ax 6 = 0 and only y = Ax we get sup x,y 6 =0 | Ax · y | | x | | y | sup x : x 6 =0 and Ax 6 =0 | Ax · Ax | | x | | Ax | = sup x : x 6 =0 and Ax 6 =0 | Ax | | x | . But including also vectors x 6 = 0 with Ax = 0 does not change the last sup, since the value of | Ax | / | x | is 0 for these x anyway, and thus the last sup is the same as sup x 6 =0 | Ax | | x | = || A || . (3)(a) Basic algebra gives 1 f ( x + h ) - 1 f ( x ) + f 0 ( x ) h f ( x ) 2 = f ( x ) - f ( x + h ) + f 0 ( x ) h f ( x ) f ( x + h ) + ( f ( x + h ) - f ( x )) f 0 ( x ) h f ( x ) 2 f ( x + h ) . 1
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Then since f is differentiable we have | f ( x ) - f ( x + h ) + f 0 ( x ) h | = o ( | h | ), while | f 0 ( x ) h | ≤ || f 0 ( x ) || | h | , and so 1 | h | ± ± ± ± 1 f ( x + h ) - 1 f ( x ) + f 0 ( x ) h f ( x ) 2 ± ± ± ± = 1 | h | ± ± ± ± f ( x ) - f ( x + h ) + f 0 ( x ) h f ( x ) f ( x + h ) ± ± ± ± + 1 | h | ± ±
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Unformatted text preview: ± ± ( f ( x + h )-f ( x )) f ( x ) h f ( x ) 2 f ( x + h ) ± ± ± ± ≤ o ( | h | ) | h | 1 | f ( x ) f ( x + h ) | + | ( f ( x + h )-f ( x ) | || f ( x ) || | f ( x ) 2 f ( x + h ) | Since f is continuous, we have f ( x + h )-f ( x ) → 0, and by definition o ( | h | ) / | h | → 0, as | h | → 0, so we have 1 | h | ± ± ± ± 1 f ( x + h )-1 f ( x ) + f ( x ) h f ( x ) 2 ± ± ± ± → 0 as | h | → , which means 1 /f is differentiable at x with derivative ² 1 f ³ ( x ) =-f ( x ) f ( x ) 2 . (b) Let T : R n → R m be the 0 mapping. Then since g (0) must be 0, for h ∈ U we have 1 | h | | g (0 + h )-g (0)-T h | = | g ( h ) | | h | ≤ | h | α-1 → 0 as | h | → . This shows that g (0) = T = 0. 2...
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This note was uploaded on 06/23/2008 for the course MATH 425B taught by Professor Alexander during the Fall '07 term at USC.

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m425b-ICfinsoln-s08 - ± ± ( f ( x + h )-f ( x )) f ( x )...

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