18.100pset11

18.100pset11 - 18.100B/C: Fall 2008 Solutions to Homework...

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Unformatted text preview: 18.100B/C: Fall 2008 Solutions to Homework 11 1. (Rudin page 138 problem 7) (a) If f ∈ R on [0 , 1] then f is bounded, so say | f | ≤ M on [0 , 1]. Then for any fixed c ∈ [0 , 1] we have vextendsingle vextendsingle vextendsingle vextendsingle integraldisplay 1 f − integraldisplay 1 c f vextendsingle vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle integraldisplay c f vextendsingle vextendsingle vextendsingle vextendsingle ≤ cM → as c → 0, so the two definitions do agree. (b) Define the function f piecewise, as follows: on the interval (2 − ( n +1) , 2 − n ] for n ∈ Z ≥ , set f = ( − 1) n · 2 n +1 n +1 . For any c > 0, f is bounded with finitely many points of discontinuity on [ c, 1] and so is integrable. Define a n = 1 − 1 2 + 1 3 − . . . + ( − 1) n- 1 n . Then it’s easy to see that integraldisplay 1 2- n f = a n and that for c ∈ (2 − ( n +1) , 2 − n ), integraltext 1 c f lies between a n and a n +1 . Since a n → L for some L , this implies lim c → + integraldisplay 1 c f = L. However, integraldisplay 1 2- n | f | = 1 + 1 2 + . . . + 1 n → ∞ , so f has the properties we desire. (Once can design a similar function f based on any series that converges but not absolutely.) 2. (Rudin page 138 problem 8) Let s k = ∑ k i =1 f ( i ), t k = s k − f (1) = ∑ k i =2 f ( i ) and g ( t ) = integraltext t 1 f ( x ) dx . Because f is monotonically decreasing, s k is an upper estimate for g ( k + 1) and t k +1 is a lower estimate for g ( k + 1), i.e., t k +1 ≤ g ( k + 1) ≤ s k ....
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This note was uploaded on 12/07/2011 for the course MATH 18.100B taught by Professor Prof.katrinwehrheim during the Fall '10 term at MIT.

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18.100pset11 - 18.100B/C: Fall 2008 Solutions to Homework...

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