Math 54 - Quiz 11 Ans, Spring 2006

Math 54 - Quiz 11 Ans, Spring 2006 - b m = 2 π ²-( x 2-π...

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Math 54, Quiz 11 Solutions Sections 202/203 GSI Carter 1. For each of the following statements, write the word “true” or “false.” (a) Let f ( x ) be an odd function and g ( x ) be an even function. Then f ( x ) g ( x ) is an odd function. True. (b) Let f ( x ) be an odd function and g ( x ) be an even function. Then f ( x ) + g ( x ) is an odd function. False. For a counterexample, let f ( x ) = x and g ( x ) = 1. 2. Find the Fourier series for the function f ( x ) defined by f ( x ) = ± - x 2 + π 2 if - π x 0 x 2 - π 2 if 0 < x < π and f ( x + 2 π ) = f ( x ) for any x . The function f ( x ) is odd, so each a m = 0. We only need to compute the coefficients b m , and b m = 1 π Z π - π f ( x ) sin( mx ) dx = 2 π Z π 0 ( x 2 - π 2 ) sin( mx ) dx. Using integration by parts twice, we obtain Z ( x 2 - π 2 ) sin( mx ) dx = - ( x 2 - π 2 ) cos( mx ) m + 2 m Z x cos( mx ) dx = - ( x 2 - π 2 ) cos( mx ) m + 2 x sin( mx ) m 2 - 2 m 2 Z sin( mx ) dx = - ( x 2 - π 2 ) cos( mx ) m + 2 x sin( mx ) m 2 + 2 cos( mx ) m 3 + C. Then we have that
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Unformatted text preview: b m = 2 π ²-( x 2-π 2 ) cos( mx ) m + 2 x sin( mx ) m 2 + 2 cos( mx ) m 3 ³´ ´ ´ ´ π x =0 = 2 π ² 2 cos( mπ ) m 3-π 2 m-2 m 3 ³ = 2 m 3 π ( 2(-1) m-m 2 π 2-2 ) . Therefore the Fourier series for f ( x ) is f ( x ) = ∞ X m =1 2 m 3 π ( 2(-1) m-2-m 2 π 2 ) sin( mx ) . This expression can be simplified (or made more complex, depending on how you look at it) by splitting the sum up into odd and even terms: f ( x ) = ∞ X m =1 b 2 m-1 sin((2 m-1) x ) + ∞ X m =1 b 2 m sin(2 mx ) =-2 π &quot; ∞ X m =1 4 + (2 m-1) 3 π (2 m-1) 3 sin((2 m-1) x ) + ∞ X m =1 π 2 m sin(2 mx ) # . 1...
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This note was uploaded on 02/20/2010 for the course AS a taught by Professor 11 during the Spring '10 term at École Normale Supérieure.

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