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Unformatted text preview: SOLUTION FOR MATH417 MIDTERM Problem 1. (40 Points) Let f ( x ) =  x for 0 x . (1) Find the Fourier cosine series C [ f ] of f (with [0 , ] as the basic interval). (2) Over the interval ( 3 , 3 ), sketch the function to which the series con verges. (3) Find the Fourier sine series S [1] of 1 (0 x ). (4) Find the Fourier sine series S [ g ] of g ( x ) = x x 2 over [0 , ]. (1). Since f ( x ) =  x for 0 < x < , f ( x ) C [ f ]( x ) = A + X n 1 A n cos( nx ) . Then by definition, we have A = 1 Z f ( x ) dx = 1 ( x x 2 2 )  = 1 ( 2 2 2 ) = 2 . For n 1, we have A n = 2 Z f ( x )cos( nx ) dx = 2 Z x cos( nx ) dx = 2 n sin( nx ) x  + Z 2 n sin( nx ) dx = 0 + 2 n 2 ( cos( nx )  = 2 n 2 (1 cos( n )) = 2 n 2 (1 ( 1) n ) = 4 n 2 n odd n even Now we see that C [ f ] = 2 + X n 1 , odd 4 n 2 cos( nx ) . (2). Let f be the even and 2 periodic extension of f . By convergence theorem, the Fourier cosine series C [ f ] of f equals to f at the point of continuity, and 1 2 SOLUTION FOR MATH417 MIDTERM f ( x )+ f ( x +) 2 at the point of discontinuity. In this case, f is continuous on [0 , ] and so is the f . In conclusion, we know that for C [ f ]( x ) = f . (3). Notice that we have f continuous and f = 1 piecewise smooth, we can have term by term differentiation for C [ f ], and so S [ f ] = ( C [ f ]) = X n 1 ,odd A n (cos( nx )) = X n 1 ,odd nA n sin( nx ) Since f = 1, S [1] = S [ 1] = S [ f ] = X n 1 ,odd nA n sin( nx ) = X n 1 , odd 4 n sin( nx ) ....
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 Spring '08
 GOLDBERG
 Math

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