Midterm Solutions

Midterm Solutions - SOLUTION FOR MATH417 MIDTERM Problem 1....

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Problem 1. (1). Since f ( x ) = 2 x for 0 < x < π , f ( x ) X n 1 A n sin( nx ) . Then by definition, we have A n = 2 π Z π 0 2 x sin( nx ) dx = - 4 cos( nx ) x | π 0 + Z π 0 4 cos( nx ) dx = - 4 n ( - 1) n + 4 n 2 π sin( nx ) | π 0 = ( - 1) n +1 4 n . (2). Let ˜ f be the odd and 2 π -periodic extension of f . By convergence theo- rem, the Fourier sine series S [ f ] of f equals to ˜ f at the point of continuity, and ˜ f ( x - )+ ˜ f ( x +) 2 at the point of discontinuity. In conclusion, we know that for S [ f ]( x ) = 0 x = - 3 π 2( x + 2 π ) - 3 π < x < - π 0 x = - π 2 x - π < x < π 0 x = π 2( x - 2 π ) π < x < 3 π 0 x = 3 π (3). If we use partial sum of eigenfunctions to approximate a function, we know that we have Gibbs phenomenon (roughly 9% overshoot) at the point of disconti- nuity. Here the only points of discontinuity are x k = π +2 , at x k , f ( x k +) - f ( x k - ) = 4 π , the overshoot is roughly 9% × 4 π = 0 . 36 π . (4). If we studied the Fourier cosine series instead, we need to consider ˆ f which is the even and 2 π -periodic extension of f . We can know that ˆ f is in fact a continuous function, then by convergence theorem, the Fourier cosine series C [ f ]( x ) of f equals to ˆ f . Problem 2. By the method of separation of variables, we assume first that u ( x, t ) = φ ( x ) h ( t ) . Then by the equation, we know that
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Midterm Solutions - SOLUTION FOR MATH417 MIDTERM Problem 1....

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