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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 . ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online