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Homework 1 § 11 . 1 1. If f ( x ) has period p , show that f ( ax ), a 6 = 0, is a periodic function of x of period p/a . Sol. Let g ( x ) = f ( ax ). Since p is a period of f , g ( x + p/a ) = f ( a ( x + p/a )) = f ( ax + p ) = f ( ax ) = g ( x ) . Thus g ( x ) = f ( ax ) is a periodic function of x of period p/a . 2. Find the Fourier series of the given periodic function f ( x ) of period 2 π : (a) (b) (c) f ( x ) = x 2 ( - π < x < π ) Sol. (a) a 0 = 1 2 π Z π - π f ( x ) dx = 1 π Z π/ 2 0 dx = 1 2 , a n = 1 π Z π - π f ( x ) cos nxdx = 2 π Z π/ 2 0 cos nxdx = 2 sin 2 , b n = 1 π Z π - π f ( x ) sin nxdx = 1 π Z π/ 2 - π/ 2 sin nxdx = 0 . Fourier series of f ( x ) f ( x ) = 1 2 + X n =1 2 sin 2 cos nx = 1 2 + 2 π £ cos x - 1 3 cos 3 x + 1 5 cos 5 x + · · · / . Computational Science & Engineering (CSE) C. K. Ko

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Homework 2 (b) a 0 = 1 2 π Z π - π f ( x ) dx = 1 π Z π 0 ( - x + π ) dx = π 2 , a n = 1 π Z π - π f ( x ) cos nxdx = 2 π Z π 0 ( - x + π ) cos nxdx = 2(1 - cos ) n 2 π , b n = 1 π Z π - π f ( x ) sin nxdx = 0 . Fourier series of f ( x ) f ( x ) = π 2 + X n =1 2(1 - cos ) n 2 π cos nx = π 2 + 4 π £ cos x + 1 3 2 cos 3 x + 1 5 2 cos 5 x + · · · / . (c) a 0 = 1 2 π Z π - π f ( x ) dx = 1 π Z π 0 x 2 dx = π 2 3 , a n = 1 π Z π - π f ( x ) cos nxdx = 2 π Z π 0 x 2 cos nxdx = 4 cos n 2 , b n = 1 π Z π - π f ( x ) sin nxdx = 1 π Z π - π x 2 sin nxdx = 0 . Fourier series of f ( x ) f ( x ) = π 2 3 + X n =1 4 cos n 2 cos nx = π 2 3 - 4 £ cos x - 1 2 2 cos 2 x + 1 3 2 cos 3 x + · · · / . 3. Using Problem 2, find the sum of the series. (a) X n =1 ( - 1) n - 1 1 2 n - 1 (b) X n =1 1 (2 n - 1) 2 (c) X n =1 ( - 1) n - 1 1 n 2 Sol. (a) Notice that f is continuous at x = 0. 1 = f (0) = 1 2 + 2 π £ 1 - 1 3 + 1 5 + · · · / Thus X n =1 ( - 1) n - 1 1 2 n - 1 = π 4 . Computational Science & Engineering (CSE) C. K. Ko
Homework 3 (b) Notice that f is continuous at x = 0. π = f (0) = π 2 + 4 π £ 1 + 1 3 2 + 1 5 2 + · · · / Thus X n =1 1 (2 n - 1) 2 = π 2 8 . (c) Notice that f is continuous at x = 0. 0 = f (0) = π 2 3 - 4 £ 1 - 1 2 2 + 1 3 2 + · · · / Thus X n =1 ( - 1) n - 1 1 n 2 = π 2 12 . § 11 . 2 1. Find the Fourier series of the given f ( x ) = x 2 ( - 1 < x < 1), of the period p = 2, and calculate n =1 ( - 1) n - 1 1 n 2 .

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