pde1 - Solutions Math 21b, Spring 09 √ 1. The...

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Unformatted text preview: Solutions Math 21b, Spring 09 √ 1. The verification that cos(nx), sin(nx), 1/ 2 form an orthonormal family is a straightforward computation when using the identities provided. For example, cos(nx), sin(mx) = π 1π sin(nx) sin(mx) dx = 21 −π cos((n − m)x) − cos((n + m)x) dx which is equal to π −π π 1 if n = m and equal to 0 if n = m. The computations can be abbreviated by noting that integrating an odd 2π periodic function over [−π, π ] is zero. 2. To get the Fourier series of the function f (x) = |x|, note first that this is an even function so that it has a cos series. We compute √ √ π2 2π1 x √ dx = . a0 = f , 1/ 2 = π0 2 2 2 cos(nπ ) − 1 [ ]. π n2 0 √ √ 6 The Fourier coefficients of f (x) = 5 + |3x| are a0 = 3π2 2 + 5 2. and an = π [ cos(nπ)−1 ]. n2 an = f , cos(nx) = 2 π x cos(nx) dx = 3. The Fourier series of 4 cos2 (3x) + 5 sin2 (11x) + 90 is with cos2 (2x) = sin2 (2x) = 1 + cos(2x) 2 1 − cos(2x) 2 π given as 4/2 + 4 cos(6x)/2 − 5 cos(22x)/2 + 5/2 + 90 . All Fourier coefficients are zero except √ a0 = (189/2) · 2 and a6 = 2 and a22 = −5/2 . 4. To find the Fourier series of the function f (x) = | sin(x)|, we first note that this is an even function so that it has a cos-series. If we integrate from 0 to π and multiply the result by 2, we can take the function sin(x) instead of | sin(x)| so that an = 2 π π sin(x) cos(nx) dx . 0 We use one of the trigonometric identities provided in the text to solve this integral. f (x) = 2 4 − ππ cos(2x) cos(4x) cos(6x) +2 +2 + ... 22 − 1 4 −1 6 −1 5. The square of the length of the function f (x) is 1. Parceval’s identity shows that 1= a2 0 + √2 16 1 1 1 a2 = ( 2 )2 + 2 [ 2 +2 +2 + · · ·] n 2 2 π π (2 − 1) (4 − 1) (6 − 1)2 n=1 ∞ so that the sum is π 2 /16 − 1/2 . ...
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