ma017-2010-sol1

ma017-2010-sol1 - MA017 2010 HOMEWORK 1(1 Let a 1,a n be...

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Unformatted text preview: MA017 2010 HOMEWORK 1 (1) Let a 1 ,...,a n be real numbers, and define f ( x ) = ∑ n j =1 a j sin( jx ) . Suppose | f ( x ) | ≤ | sin x | for all real x . Prove, by induction , that | ∑ n j =1 ja j | ≤ 1 . (Considering f (0) gives a “slick” proof, but I am asking you to use induction.) Solution. We prove more generally that whenever a 1 ( x ) ,...,a n ( x ) are continuous and f ( x ) = ∑ n j =1 a j ( x )sin( jx ) satisfies | f ( x ) | ≤ | sin x | for all x ∈ R , we have | ∑ n j =1 ja j (0) | ≤ 1. For n = 1, if | a 1 ( x )sin x | ≤ | sin x | for all x ∈ R , then | a 1 ( x ) | ≤ 1 wherever sin x 6 = 0, so by continuity | a 1 (0) | ≤ 1. Assume the statement is true for n = k . Then for n = k + 1 we may rewrite f ( x ) as [ a 1 ( x ) + a k +1 ( x )cos( kx )]sin x + X 1 <j<k a j ( x )sin( jx ) + [ a k ( x ) + a k +1 ( x )cos x ]sin( kx ) , whence if | f ( x ) | ≤ | sin x | for all x ∈ R we conclude by our induction hypothesis that 1 ≥ [ a 1 (0) + a k +1 (0)] + X 1 <j<k ja j (0) + k [ a k (0) + a k +1 (0)] = k +1 X j =1 ja j (0) , as desired....
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ma017-2010-sol1 - MA017 2010 HOMEWORK 1(1 Let a 1,a n be...

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