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Extra resources for A Pictorial History of Jazz: People and Places from New Orleans to Modern Jazz
Using the theory of even and odd functions, and the theory of half-period symmetric and antisymmetric functions, which Fourier coefficients of tan θ have to be zero? Find the first non-zero coefficient. 3. Which Fourier coefficients vanish for a periodic function f (θ) of period 2π satisfying f (θ) = f (π − θ)? What about f (θ) = −f (π − θ)? 4. 1), then we recover the original function f (θ). The most obvious problem is that if two functions differ just at a single value of θ then the Fourier coefficients will be identical.
10. DAMPED HARMONIC MOTION 25 factors as (Y + 1)(Y + 3) = 0, so it has roots Y = −1 and Y = −3. 6) are given by y = Ae−t + Be−3t . y y = e−t + e−3t t 2. The equation y¨ + 2y˙ + 26y = 0 is underdamped. 7) Y 2 + 2Y + 26 = 0. Completing the square gives (Y + 1)2 + 25 = 0, so the solutions are Y = −1 ± 5i. 7) are given by y = e−t (Ae5it + Be−5it ), or y = e−t (A′ cos 5t + B ′ sin 5t). 8) y y = e−t sin 5t t 3. The equation y¨ + 4y˙ + 4y = 0 is critically damped. 9) Y 2 + 4Y + 4 = 0 factors as (Y + 2)2 = 0, so the only solution is Y = −2.
3 (1848), 198–201. 5. THE GIBBS PHENOMENON −φm (θ). 8). So the zeros of φ′m (θ) occur at θ = (2k+1)π m+1 m−1 8 2kπ θ = m , 0 ≤ k ≤ ⌊ 2 ⌋. Now sin 21 θ is positive throughout the interval 0 ≤ θ ≤ 2π. At θ = (2k+1)π 1 1 k k m+1 , sin 2 mθ has sign (−1) while cos 2 (m + 1)θ changes sign from (−1) to (−1)k+1 , so that φ′m (θ) changes from positive to negative. It follows that 2kπ 1 θ = (2k+1)π m+1 is a local maximum of φm . Similarly, at θ = m , cos 2 (m + 1)θ has sign (−1)k while sin 12 mθ changes sign from (−1)k−1 to (−1)k , so that φ′m (θ) changes from negative to positive.