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Unformatted text preview: In a similar way, we prove that c n = 1 κ n ² 1 f ( x ) X n ( x ) dx where κ n = ² 1 X 2 n ( x ) dx . This integral can be evaluated as we did in Exercise 15 or by straightforward computations, using the explicit formula for the X n ’s, as follows: ² 1 X 2 n ( x ) dx = ² 1 ( nπ cos nπxsin nπx ) 2 dx = ² 1 ( n 2 π 2 cos 2 nπx + sin 2 nπx2 nπ cos( nπx ) sin( nπx ) ) dx = =( n 2 π 2 ) / 2 ³ ´µ ¶ ² 1 n 2 π 2 cos 2 nπx dx + =1 / 2 ³ ´µ ¶ ² 1 sin 2 nπx dx2 nπ =0 ³ ´µ ¶ ² 1 cos( nπx ) sin( nπx ) dx = n 2 π 2 + 1 2 ....
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.
 Fall '11
 StuartChalk

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