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Unformatted text preview: SOLUTION FOR MATH417 MIDTERM Problem 1. (40 Points) Let f ( x ) = π x for 0 ≤ x ≤ π . (1) Find the Fourier cosine series C [ f ] of f (with [0 ,π ] as the basic interval). (2) Over the interval ( 3 π, 3 π ), sketch the function to which the series con verges. (3) Find the Fourier sine series S [1] of 1 (0 ≤ x ≤ π ). (4) Find the Fourier sine series S [ g ] of g ( x ) = πx x 2 over [0 ,π ]. (1). Since f ( x ) = π x for 0 < x < π , f ( x ) ∼ C [ f ]( x ) = A + X n ≥ 1 A n cos( nx ) . Then by definition, we have A = 1 π Z π f ( x ) dx = 1 π ( πx x 2 2 )  π = 1 π ( π 2 π 2 2 ) = π 2 . For n ≥ 1, we have A n = 2 π Z π f ( x )cos( nx ) dx = 2 π Z π x cos( nx ) dx = 2 nπ sin( nx ) x  π + Z π 2 nπ sin( nx ) dx = 0 + 2 n 2 π ( cos( nx )  π = 2 n 2 π (1 cos( nπ )) = 2 n 2 π (1 ( 1) n ) = 4 n 2 π n odd n even Now we see that C [ f ] = π 2 + X n ≥ 1 , odd 4 n 2 π cos( nx ) . (2). Let ˜ f be the even and 2 πperiodic 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 ,π ] 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 nπ sin( nx ) ....
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 Spring '08
 GOLDBERG
 Math, Fourier Series, Boundary value problem, Partial differential equation, Joseph Fourier, NX

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