23.16 - 1 k Department of Mathematics The Ohio State...

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23.16. (With f as in the statement of the problem), we show that sup ( f, P ) = as P ranges over all partitions of [0 , 1]. By divergence of the harmonic series, it suffices to fix n N and produce a partition P of [0 , 1] such that ( f, P ) 2 π n X k =1 1 k . For k = 0 , . . . , n , put a k = 2 (2 k + 1) π ; then 0 < a n < · · · < a 0 < 1. Put P = { 0 , a n , . . . , a 0 , 1 } . Then: ( f, P ) = p a 2 n + f ( a n ) 2 + n X k =1 p ( a k - a k - 1 ) 2 + ( f ( a k ) - f ( a k - 1 )) 2 + p (1 - a 0 ) 2 + ( f (1) - f ( a 0 )) 2 n X k =1 p ( a k - a k - 1 ) 2 + ( f ( a k ) - f ( a k - 1 )) 2 n X k =1 | f ( a k ) - f ( a k - 1 ) | = n X k =1 ( a k + a k - 1 ) (high-school trig) = 2 π n X k =1 4 k 4 k
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Unformatted text preview: 1 k . Department of Mathematics, The Ohio State University, 231 West 18th Avenue, Colum-bus, OH 43210 E-mail address : [email protected] URL : http://www.math.ohio-state.edu/~miller Prepared by C. Miller for OSU Math H191, March 3, 2008. As always, if you spot typos or errors, please let me know. 1...
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