# ps9 - version has the advantage of giving a more explicit...

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18.100B and 18.100C Fall 2011 Problem Set 9 Due December 1st at 4 pm in room 2-108. Hand in parts 1, 2 and 3 separately. Put your name and whether you are registered for 18.100B or 18.100C on each part. Part 1 1. Problem 10, parts a – c , from page 139. Hint: For part a , use the result of Problem 7 from Problem Set 8. 2. Let x > 0, let n N ∪ { 0 } , and let f : [0 ,x ] R be n + 1 times diﬀerentiable with f ( n +1) integrable. Use mathematical induction and integration by parts to prove that f ( x ) = f (0) + f 0 (0) x + f 00 (0) 2! x 2 + ··· + f ( n ) (0) n ! x n + I n ( x ) , where I n ( x ) = x n +1 n ! Z 1 0 (1 - t ) n f ( n +1) ( tx ) dt = 1 n ! Z x 0 ( x - x 0 ) n f ( n +1) ( x 0 ) dx 0 . Theorem 5.15 in Rudin uses the mean value theorem to prove another version of Taylor’s theorem under slightly weaker hypotheses, but this
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Unformatted text preview: version has the advantage of giving a more explicit remainder. Of course the case x &lt; 0 follows by applying the result to g ( x ) = f (-x ), and an expansion near a 6 = 0 follows by taking g ( x ) = f ( a + x ). Part 2 3. Problem 15 from page 141. 4. Problem 2 from page 165. Part 3 5. Problem 3 from page 165. 6. Let f : [0 , 1] [0 , 1] be continuously dierentiable with nonincreasing derivative and with f (0) = f (1) = 0. Prove that the arclength of the graph of f is at most 3. 1...
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## This note was uploaded on 02/15/2012 for the course MATH 18.100B taught by Professor Prof.katrinwehrheim during the Fall '10 term at MIT.

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