ps04_extra_soln

ps04_extra_soln - 18.03 PS4 Extra Credit Solution-Spring...

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18.03 PS4 Extra Credit Solution—Spring 09 (10 pts = 4 + 4 + 2) Xa. (4 pts) Start with y 0 < z 0 . The property | f y | ≤ A implies | f ( x 0 ,y 0 ) - f ( x 0 ,z 0 ) | = ± ± ± ± Z z 0 y 0 f y ( x 0 ,y ) dy ± ± ± ± Z z 0 y 0 | f y ( x 0 ,y ) | dy Z z 0 y 0 Ady = A ( z 0 - y 0 ) In particular, f ( x 0 ,z 0 ) - f ( x 0 ,y 0 ) ≥ - A ( z 0 - y 0 ) We have y 1 = y 0 + hf ( x 0 ,y 0 ) and z 1 = z 0 + hf ( x 0 ,z 0 ). Hence, since Ah < 1, z 1 - y 1 = ( z 0 - y 0 )+ h [ f ( x 0 ,z 0 ) - f ( x 0 ,y 0 )] ( z 0 - y 0 ) - hA ( z 0 - y 0 ) = (1 - Ah )( z 0 - y 0 ) > 0 In other words, we have shown that y 0 < z 0 = y 1 < z 1 . The same reasoning with indices shifted by 1 shows that y 1 < z 1 = y 2 < z 2 , and y 2 < z 2 = y 3 < z 3 , etc. In general, by mathematical induction, y n < z n for all positive integers n . Xb. (4 pts) The proof in part (a) really only depended on the lower bound f y ≥ - A , or equivalently f y > - 1 /h . We repeat the argument here. Suppose that f y ( x 0 ,y ) > - 1 /h for y 0 y z 0 . Then h Z z 0 y 0 f y ( x 0 ,y ) dy > h Z z 0 y 0 - 1 h dy = - ( z 0 - y 0 ) = z 1 - y 1
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This note was uploaded on 05/06/2010 for the course 18 18.03 taught by Professor Unknown during the Spring '09 term at MIT.

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