HW8_Solutions

HW8_Solutions - Problem 2. ikx ikx ik Let g(x) = f (x + )....

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Problem 2. Let g ( x ) = f ( x + α ). If f ( x ) = k e ikx f k then g ( x ) = k e ikx e ikα f k = k e ikx ˜ f k , where ˜ f k = e ikα f k . If f ( x ) = k ( a k sin kx + b k cos kx ) then by the above argument we get g ( x ) = k a k sin kx + ˜ b k cos kx ), where ˜ a k = a k cos - b k sin and ˜ b k = a k sin + b k cos . Problem 3. Consider function f ( x ) = x 2 . Then R 1 - 1 f ( x ) dx = 2 / 9, R 1 - 1 f 2 ( x ) dx = 4 / 5 and R 1 - 1 f ( x )cos nxdx = 4( - 1) n 2 n 2 . Then we apply Parseval equation and get the result. Problem 4. We know that s n ( f )( x ) = 1 / 2 π R π - π f ( x + t ) sin(2 n +1) / 2 t sin t/ 2 dt . Then s n ( f )( x 0 ) - f ( x 0 ) = 1 / 2 π Z π - π ( f ( x 0 + t ) - f ( x 0 )) sin(2 n + 1) / 2 t sin t/ 2 dt = 1 / 2 π Z π - π f ( x 0 + t ) - f ( x 0 ) t t sin t/ 2 sin(2 n + 1) / 2 tdt. Since the derivative exists, function f ( x 0 + t ) - f ( x 0 ) t is integrable. So, se can apply Riemann’s lemma to show that
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