Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Unit VIII Fourier Series Unit VIII Fourier Series Fourier series are series of cosine and sine terms. Since sine and cosine are periodic functions we shall study Trigonometric series representation of what we call periodic functions. 7.1 Periodic Functions; Trigonometric Series. Definition 7.1 A function f is said to be periodic if there is some positive number p such that: f (x + p) = f (x) for all x dom. f. The number p is called the period of f and the smallest number p is called the primitive (or fundamental) period of f. Example 1 Let f (x) = sin x. Now dom. f = , f (x + p) = f (x) x sin (x + p) = sin x x sin x cos p + sin p cos x = sin x x sin x (1 cos p) = sin p cos x x If x = , then 1 cos p = 0. 2 Now 1 cos p = 0 p = 2n ,where n N. Therefore, 2n, where n N is a period of f and 2 is the primitive period of f. Example 2 Let g (x) = cos x. Now dom. g = , g (x + p) = g (x) x cos (x + p) = cos x x cos x cos p sin p sin x = cos x x cos x (cos p 1) = sin p sin x x If x = , then cos p 1= 0. Now cos p 1= 0 p = 2n, where n N. Therefore 2n, where n N is a period of g and 2 is the primitive period of g. Prepared by Tekleyohannes Negussie 143 Unit VIII Fourier Series Example 3. Let f (x) = tan x. Now dom. f =x x (2n + 1) , n Z. 2 f (x + p) = f (x) x dom. f. tan (x + p) = tan x x dom. f. tan x tan p = tan x x dom. f. 1 tan x tan p tan x + tan p = tan x tan p tan 2 x x dom. f. tan p (1 + tan 2 x) = 0 x dom. f. tan p = 0 x dom. f. p = n, where n N. Therefore n, where n N is a period of f and is the primitive period of f. Example 4. Let f (x) = sin (x), where 0. Now dom. f = . f (x + p) = f (x) x . sin (x + p) = sin x x . sin x cos p + sin p cos x = sin x x . sin x (1 cos p) = sin p cos x x . If x = , then 1 cos p = 0. 2 Now 1 cos p = 0 p = 2n, where n N. Therefore 2n , where n N is a period of f and 2 is the primitive period of f. Graphs of Periodic Functions. The graph of a periodic function of period p is obtained by periodic representation of its graph in any interval of length p. Example 1. Sketch the graph of f (x) = tan x. Solution. The primitive period of f is . Prepared by Tekleyohannes Negussie 144 Unit VIII Fourier Series Example 2 Sketch the graph of g (x) = sin x . Solution Dom. g. = . Now g (x + p) = g (x) x . sin (x + p) = sin x x . sin (x + p) = sin x x or sin (x + p) = sin x x . p = 2n, where n N or sin x (1 + cos p) = sin p cos x x . p = 2n, where n N or p = (2n 1), where n N p = n, where n N. Therefore is the primitive period of g. Prepared by Tekleyohannes Negussie 145 Unit VIII Fourier Series Example 3. Sketch the graph of f, where 1 f (x) = 1 if 1 x 0 if 0 x 1 and f (x + 2) = f (x) x . Properties of Periodic Functions. 1. If f is a periodic function with period p, then for any n N np is a period of f. Proof: (By the principle of mathematical induction) Since f (x + n p) = f (x) it is true for n = 1. Assume for n = k, i.e. f (x + k p) = f (x). We need to show for n = k + 1, i.e. f (x + (k + 1) p) = f (x). f (x + (k + 1) p) = f (x + k p + p) = f (x + p + k p) = f (x + p) = f (x). Therefore for any n N, np is a period of f. 2. If f and g are periodic functions with period p, then for any a, b h (x) = a f(x) + b g(x) x dom. h is a periodic function of period p. Prepared by Tekleyohannes Negussie 146 Unit VIII Fourier Series Proof: h (x + p) = a f(x + p) + b g(x + p) = a f(x) + b g(x) = h (x) x dom. h. Therefore h is a periodic function with period p. 3. If f (x) is a periodic function with period p, then for any b and b 0 a) f (b x) is periodic with period b) If ( p . b x ) is periodic with period b p. b Trigonometric Series. Definition 1.2 The system of functions: 1, cos x, sin x, cos 2x, sin 2x, cos 3x, sin 3x, … is called a trigonometric system. The series a0 + (an cos nx bn sin nx) n 1 where a0, an’s and bn’s are constants is called Trigonometric series, the constants are called coefficients. Definition 1.3 ( Orthogonality of the Trigonometric System) 1. Let f and g be functions defined on some interval a, b. f and g are said to be orthogonal on a, b if and only if b f ( x) g ( x) dx = 0. a 2. The set of functions n ( x) : n N is said to be orthogonal on a, b if and b only if n ( x)m ( x) dx = 0 for n m. a Prepared by Tekleyohannes Negussie 147 Unit VIII Fourier Series Example 1. Show that the set of functions n ( x) : n ( x) sin nx, where n N form an orthogonal set on , . Solution. Let n, m N and n m. n (x ) = sin nx. n ( x)m ( x) dx = sin nx sin mx dx 1 = cos (n m) x cos (n m) xdx 2 1 sin (n m) x sin (n m) x = 2 nm nm = 0. Therefore n ( x) : n ( x) sin nx, where n N form an orthogonal set on , . Example 2. Show that the trigonometric system 1, cos nx, sin nx, … , where n N, form an orthogonal set on , . Solution. We need to show that: a) sin nx dx = cos nx dx = sin nx cos mx dx = 0 for n, m N and m = n. b) sin nx cos mx dx = sin nx sin mx dx = cos nx cos mx dx = 0 where n, m N and m n. When m = n. 1 1 cos nx = sin nx dx = cos nx dx = sin nx = 0, n n and sin nx cos nx dx = 1 1 = = 0. sin 2 nx dx cos 2nx 2 4n Prepared by Tekleyohannes Negussie 148 Unit VIII Fourier Series When m n. 1 sin nx cos mx dx = 2 sin (n m) x sin (n m) xdx cos (n m) x cos (n m) x = 0. nm nm 1 = 2 1 sin nx sin mx dx = 2 cos (n m) x cos (n m) xdx 1 sin (n m) x sin (n m) x = = 0. 2 nm nm 1 cos nx cos mx dx = 2 cos (n m) x cos (n m) xdx = 1 sin (n m) x sin (n m) x = 0. 2 nm n m Therefore 1, cos nx, sin nx, … for n N form an orthogonal set on , . Definition 1.4 1. A function f (x) is said to be normal or normalized in a, b if and only if b f ( x) a 2 dx = 1 The norm of a function f (x) denoted by f (x) on a, b is defined by: b f (x) = f ( x) a 2 dx 2. An orthogonal set of functions n (x): n N is said to be an orthonormal set on a, b if and only if b a n ( x) 2 dx = 1 n N. Prepared by Tekleyohannes Negussie 149 Unit VIII Fourier Series Example 1. The set of functions n (x): n (x) = 1 sin nx, n Nis an orthonormal set on , . Solution. Let m, n N and m n. 1 cos (n m) x cos (n m) xdx sin nx sin mx dx = 2 1 1 sin (n m) x sin (n m) x = = 0. 2 nm nm Therefore n (x): n (x) = 1 sin nx, n Nis an orthogonal set on , . For m = n. 1 1 sin 2nx = = ( 1 cos 2 nx) dx sin nx dx x 1 2 2 2n 1 2 Therefore n (x): n (x) = 1 sin nx, n Nis an orthonormal set on , . Exercises. 1.1 1. Show that the following functions are non-periodic functions. i) f (x) = x ii) g ( x) = 2x2 iii) h (x) = e x 2. Sketch the graphs of the following functions, which are assumed to be periodic of period 2 and, for x , are given by the formulas: 1. f (x) = e x 2. f (x) = x2 3. f (x) = x x 4. f (x) = x if x 0 if 0 x if x 0 0 5. f (x) = sin x if 0 x 3. For the trigonometric system find the corresponding orthonormal system on , . Prepared by Tekleyohannes Negussie 150 Unit VIII Fourier Series 1.2 Fourier Series of Functions of Period 2 Let us assume that f (x) is a periodic function of period 2 that can be represented by a trigonometric series. f (x) = a0 + an cos nx bn sin nx (1) n 1 i.e. the series converges and has f (x) as its sum. Now we need to determine the coefficients a0, an and bn . i) To determine a0 integrate (1) from to . f ( x) dx = a0 dx + n1 an cos nx bn sin nx dx f ( x) dx = 2 a0 a0 = 1 2 f ( x) dx ii) To determine an, n N. Multiply (1) by cos mx and integrate from to where m N, fixed. f ( x) cos mx dx = a0 cos mx dx + n1 an cos mx cos nx bn sin nx cos mx dx If m n, then cos mx dx = cos mx cos nx dx = sin nx cos mx dx = 0. If m = n, then sin nx cos mx dx and = sin nx cos nx dx = 0. cos nx cos mx dx = cos 2 nx dx Prepared by Tekleyohannes Negussie 151 Unit VIII Fourier Series = 1 1 cos 2 nx dx 2 = 1 1 x + sin 2nx 2 4n = . Therefore an = 1 f ( x) cos nx dx n N. iii) To determine bn , n N. Multiply (1) by sin mx where m N and integrate from to . f ( x) sin mx dx = a0 sin mx dx + an sin mx cos nx bn sin nx sin mx dx n 1 If m n, then a0 sin mx dx = sin mx cos nx dx = sin nx sin mx dx = 0. If m = n, then a0 sin mx dx = 0, sin nx cos mx dx and = sin nx sin mx dx = Therefore bn = 1 sin nx cos nx dx = 0. sin 2 nx dx = . f ( x) sin nx dx n N. Prepared by Tekleyohannes Negussie 152 Unit VIII Fourier Series Therefore we get what we call Euler’s Formulas. a0 = an = bn = 1 2 1 1 f ( x) dx f ( x) cos nx dx f ( x) sin nx dx , where n N. Definition 1.5 The numbers a0, an and bn are called the Fourier coefficients of f (x). The trigonometric series a0 + an cos nx bn sin nx n 1 Where a0, an and bn are called the Fourier coefficients is called the Fourier Series of f (x) (Regardless of convergence). Remark: In the Euler’s Formulas the interval of integration may be replaced by any interval of length 2. Example 1 Find the Fourier coefficients of each of the following periodic functions. k i) f (x) = k ii) f (x) = if x 0 if 0 x and f (x + 2) = f (x) x2 for 0 x 2 and f (x + 2) = f (x). 4 Prepared by Tekleyohannes Negussie 153 Unit VIII Fourier Series 1 2 Solutions i) a0 = f ( x) dx = an = f ( x) cos nx dx = = 1 2 0 1 k dx k cos nxdx 0 0 k k + sin nx sin nx n n 0 f ( x) sin nx dx = = + + k cos nx dx = and bn = 2 0 1 1 k dx kx kx 0 + = 0, 2 2 0 = 1 0 1 0 k sin nx dx 1 + = 0, 1 k sin nxdx 0 0 k k + cos nx cos nx n n 0 2k ( 1 cos n) n 4k Thus bn = n 0 if n is odd . if n is even Therefore the Fourier series of the square wave function f is: 4k sin 2n 1x . n 1 2n 1 ii) a0 = an = 2 1 f ( x) dx = 2 2 2 2 x2 x3 2 = . dx = 2 4 3 0 4 0 2 1 x2 4 cos nx dx = 2 n 0 1 and bn = 0 1 1 2 2 x sin nx dx = 0 1 n2 2 1 x2 4 sin nx dx = n + 2 n x cos nx dx = n . 0 0 Therefore f (x) = 2 1 + ( 2 cos nx sin nx) . 3 n n 1 n Prepared by Tekleyohannes Negussie 154 Unit VIII Fourier Series Convergence and Sum of Fourier series Suppose f is any function of period 2 for which the integrals a0 = an = 1 2 1 f ( x) dx f ( x) cos nx dx n N. 1 and bn = f ( x) sin nx dx n N. exist. Then the Fourier series of f (x) is given by: a0 + (an cos nx bn sin nx) n 1 (1) i) If the series in (1) converges to f (x), then we write: f (x) = a0 + (an cos nx bn sin nx) . n 1 ii) If the series in (1) doesn’t converge to f (x), then we write: f (x) a0 + (an cos nx bn sin nx) . n 1 Question. When is the Fourier series in (1) converges to f (x)? To answer this question first we need to define piecewise continuity. Definition 1.6 Piecewise Continuity. A function f (x) is said to be piecewise continuous on an interval I if i) the interval can be divided into a finite number of subintervals in each of which f (x) is continuous. and ii) the limits of f (x) as x approaches the end points of each subinterval are finite. Remark: A piecewise continuous function is one that has only a finite number of jump discontinuities. Prepared by Tekleyohannes Negussie 155 Unit VIII Fourier Series Examples Consider the following functions. n i) f (x) = 0 4 for x n 1, n where n N and n 2. for x 0 for x 2 x ii) g (x) = x 1 x2 1 if 1 x 0 if 0 x 1 if 1 x 2 f and g are examples of piecewise continuous functions. Example. h (x) = tan x for x ( 3 , ). 2 4 lim h ( x) lim h ( x) and , h is not Since h (x) is discontinuous at x = and x x 2 2 2 piecewise continuous on ( 3 , ). 2 4 Theorem 1.1 (Representation by a Fourier series) If a periodic function with period 2 is piecewise continuous in the interval x and has a left hand derivative and a right hand derivative at each point of that interval, then the Fourier series (1) of f (x) is convergent. Its sum is f (x) except at a point x0 at which f (x) is discontinuous and the sum of the series is the average of the left and the right hand limits at x0. Example 1. Find the Fourier series of the function f where if x 0 k f (x) = if 0 x k and f (x + 2 ) = f (x). Solution. The Fourier coefficients are: 0 a0 = an = 0 and bn = 4k n if n is even. if n is odd . n N Prepared by Tekleyohannes Negussie 156 Unit VIII Fourier Series Therefore the Fourier series of f (x) is (an cos nx bn sin nx) = a0 + n 1 sin 2n 1 x . n 1 2n 1 4k Since f is piecewise continuous on ( , ) this series converges to f (x) for each x ( , ) \ 0. For x = 0, the Fourier series is given by lim f ( x) 1 lim f ( x) 1 k + k = 0 = x 0 x 0 2 2 sin 2n 1 x = 0. n 1 2n 1 4k and indeed at x = 0, Therefore f (x) = If x = sin 2n 1 x x ( , ) \ 0. n 1 2n 1 4k , then f ( ) = k. 2 2 Hence k = 1 1 n 2n 1 4k n0 1 1 n 2n 1 n0 = . 4 Therefore 1 1 n 2n 1 n0 = . 4 Example 2. For each of the following, find the Fourier series expansion of the function f (x) that has period 2. i) f (x) = x2 , x . ii) f (x) = x sin x , 0 x 2. iii) f (x) = x x2 , x . Prepared by Tekleyohannes Negussie 157 Unit VIII Fourier Series 1 x 2 dx 2 Solutions. i) a0 = x3 6 2 x cos nx dx 2 3 x 2 2x 2 sin nx 2 cos nx 3 sin nx n n n 1 an = 4 2 cos n . n bn = 1 2 2x 2 x cos nx 2 sin nx 3 cos nx n n 2 x sin nx dx = 0. Therefore f (x) = For x = , 2 3 4 n 1 n 2 cos n cos nx for x . lim f ( x) lim f ( x) 2 and 2 x x Hence 2 = 2 3 4 4 n 1 n 2 cos 2n Therefore 2 = 2 6. n 1 n 2 ii) a0 = an = 1 x sin x dx 2 0 1 1 2 1 x cos x sin x 1. 2 0 2 2 1 x sin x nx sin x nx dx 2 0 x sin x cos nx dx 0 If n 1, then an = bn = 1 2 2 2 n 1 x sin x sin nx dx 0 and if n = 1, then a1 = 1 . 2 2 1 x cos x nx cos x nx dx 2 0 Prepared by Tekleyohannes Negussie 158 Unit VIII Fourier Series If n 1, then bn = 0 and if n = 1, then b1 = Therefore f (x) = 1 1 2 cos x sin x 2 cos nx . 2 n2 n 1 If x = , then f () = 0. Thus n2 1 n 2 n 1 1 . 4 = Therefore 1 n n2 1 an = x x 2 cos nx dx 2 x3 2 x . 2 3 3 1 2 n 1 1 . 4 1 x cos nx dx 1 x 2 cos nx dx 1 x 1 4 4 2 cos n 2 cos n . sin nx 2 cos nx n n n n bn = 1 x x 2 dx 2 iii) a0 = = 2 1 x x 2 cos nx dx 2 cos n . n Therefore f (x) = 2 3 +2 1 1 n 1 x sin nx dx 1 x sin nx n 1 2 cos nx n n2 2 sin nx dx x , If x = 0, then 1 n 1 n 1 2 = . 12 n2 1 Therefore n 1 1 n 1 2 = . 12 n2 1 Prepared by Tekleyohannes Negussie 159 Unit VIII Fourier Series 1.3 Functions of any Period p = 2L Theorem 1.2 If a function f of period p = 2L has a Fourier series representation, then the series is: (an cos f (x) = a0 + n 1 n n x bn sin x) L L where the Fourier coefficients of f (x) are given by the Euler formulas: L 1 a0 = 2L an = and bn = L L 1 L 1 L f ( x) dx f ( x) cos L L f ( x) sin L n x dx n N, L n x dx n N. L Proof. (Refer page 578 Kreyszig) Remark: The interval of integration in the above formulas may be replaced by any interval of length p = 2L. Example 1. Find the Fourier series of the function f (x) with period p = 4, where 0 f (x) = 2 0 if 2 x 1 if 1 x 1 and f (x + 4) = f (x) if 1 x 2 Solution. p = 4 L = 2. 1 2 1 1 1 1 Now a0 = f ( x) dx = 2 dx = x = 1. 4 2 4 1 2 1 2 an = 1 1 n n f ( x) cos x dx cos x dx . 2 2 2 2 1 Prepared by Tekleyohannes Negussie 160 Unit VIII Fourier Series 2 4 n 1 n sin x sin , nN , n 2 1 n 2 0 = 1 n 1 n4 if n is even. i f n is odd . 2 bn 1 1 n n = f ( x) sin x dx sin x dx . 2 2 2 2 1 2 n 1 cos x 0, nN. n 2 1 Therefore f (x) = 1 + 4 1 n1 cos 2n 1 x . n 1 2n 1 2 If x = 0, then f (0) = 2. 1 n1 n 1 2n 1 = . 4 Example 2. Find the Fourier series representation of the periodic function f (x) with period p = 1, where f (x) = sin x for 0 x 1 and f (x + 1) = f (x). Solution. p = 1 L = 1 . 2 1 1 0 0 Thus a0 = sin x dx cos x 2. 1 an = 2 sin x cos 2nx dx 0 1 1 0 0 = sin 2n 1x dx sin 2n 1x dx Prepared by Tekleyohannes Negussie 161 Unit VIII Fourier Series = 1 1 1 1 cos 2n 1x cos 2 n 1 x + 2n 1 2n 1 0 0 2 2 2n 1 2n 1 = 4 = 1 4n 2 Thus an = 4 n N. 1 4n 2 1 bn = 2 sin x sin 2nx dx 0 1 1 0 0 = cos 2n 1x dx cos 2n 1x dx = 1 1 1 1 sin 2n 1x sin 2n 1x = 0. 2n 1 2n 1 0 0 Thus bn = 0 n N. Therefore f (x) = 2 + 4 cos 2nx 1 4 n2 n 1 . If x = 0, then 1 2 n 1 4 n 1 If x = 1 . 2 1 , then 2 1 n n 1 1 4 n 2 2 4 . Prepared by Tekleyohannes Negussie 162 Unit VIII Fourier Series 1.4 Even and Odd Functions. Definition 1.7 (Even and Odd Functions) 1. A function f (x) is even if f ( x) = f (x) x dom. f. 2. A function g (x) is odd if g ( x) = g (x) x dom. g. Example 1. f (x) = cos nx , x , where n N is even. g (x) = sin nx , x , where n N is odd. 2. Note that: 1. If f (x) is an even function, then L L f ( x) dx = 2 f ( x) dx L 2. 0 If g (x) is an odd function, then L g ( x) dx = 0 L 3. The product of i) even and odd functions is odd. ii) two odd functions is even. iii) two even functions is even. Prepared by Tekleyohannes Negussie 163 Unit VIII Fourier Series Theorem 1.3 (Fourier series of Odd and Even Functions) a. The Fourier series of an even function f of period 2L is a Fourier cosine series. f (x) = a0 + an cos n 1 L n x L with coefficients L 1 2 n a0 = f ( x) dx and an = f ( x) cos xdx n N L 0 L 0 L b. The Fourier series of an odd function f of period 2L is a Fourier sine series. f (x) = bn sin n 1 n x L with coefficients L 2 n bn = f ( x) sin xdx n N. L 0 L Theorem 1.4 (Sum of Functions) The Fourier coefficients of a sum of two function f1 and f2 are the sum of the corresponding Fourier coefficients of f1 and f2. The Fourier coefficients of cf are c times the corresponding Fourier coefficients of f. 2k Example 1. Let f (x) = 0 if 0 x if x 0 and f (x +2) = f (x) x dom. f. Find the Fourier series of f (x). k Solution. Let f1 (x) = k if 0 x if x 0 and f1 (x +2 ) = f1 (x) x dom. f and let f2 (x) = k x . Then (f1 + f2) (x) = f (x) x dom. f. But f1 (x) = 4k sin 2n 1x and f2 (x) = k. n 1 2n 1 4k sin 2n 1x Therefore f (x) = k + x dom. f. n 1 2n 1 Prepared by Tekleyohannes Negussie 164 Unit VIII Fourier Series Example 2. Find the Fourier series of the function f (x) where f (x) = x + if x and f (x + 2) = f (x) x dom. f. Solution. Let f1 (x) = x and f2 (x) = if x and be periodic functions with period 2. Now f1 (x) = 2 1 n 1 n 1 Therefore f (x) = + 2 sin nx and f2 (x) = . n 1 n 1 n 1 sin nx x dom. f. n Example 3. Find the Fourier series representation of f (x) = x + x2 on , . Solution. Let f1 (x) = x and f2 (x) = x2 for x , and f1 and f2 be periodic functions with period 2. Now f1 (x) is odd hence it has a Fourier sine series 1 x 1 and bn = x sin nx dx cos nx 2 sin nx n n 1 = 1 n 1 2 n nN Thus f1 (x) = 2 1 n 1 n 1 sin nx . n f2 (x) is even, hence it has a Fourier cosine series representation. 1 1 x 2 dx x 2 dx Now a0 = 2 0 and an = 0 2 3 2 1 2 cos nx dx x sin nx 2 x cos nx x cos nx dx 2 2 n n n 1 = 1 n Thus f2 (x) = x3 3 2 Therefore f (x) = 3 4 n2 nN +4 1 n 1 2 3 + 1 n 1 n cos nx n2 . 2 n 4 . cos nx sin nx n2 n Prepared by Tekleyohannes Negussie 165 Unit VIII Fourier Series Note that: If g (x) is defined x , then the function P (x) = 1 g ( x) g ( x) is even 2 and q (x) = 1 g ( x) g ( x) is odd. 2 Hence g (x) = p (x) + q (x) and g ( x) = p (x) q (x). Exercises 1.4 Represent the following functions as a sum of even and odd functions. i) ii) iii) 1 1 x 1 1 1 x 2 1 x 2 e 1 1 = 1 x kx e kx = 1 x2 1 x2 1 x2 2 + + x 1 x2 2x 1 x2 2 = cosh kx + sinh kx. 1.5 Half-range Expansion. Sometimes it is required to extend a function f (x) in the range (0, L) in a Fourier series of period 2L, and it is immaterial what the function may be outside the range 0 x L. We could extend f (x) with period 2L and then represent the extended function by a Fourier series. i) If we extend the function f (x) by reflecting it in the y-axis, then the Fourier series expansion contains only the cosine terms. ii) If we extend the function f (x) by reflecting it in the origin, then the Fourier series expansion contains only the sine terms. Prepared by Tekleyohannes Negussie 166 Unit VIII Fourier Series The cosine halt-range expansion is: f (x) = a0 + a n n 1 cos n x L L L 1 2 n where a0 = f ( x) dx and an = f ( x) cos x dx n N L0 L L0 The sine halt-range expansion is: L 2 n n f (x) = b sin x dx n N x , where bn = f ( x) cos n L0 L L n 1 Example. Find the two half-range expansions of the function 2k x L f (x) = 2k L x L if 0 x L 2 L if x L 2 where k 0. Solution. f (x) can be extended to (L, L) into two ways. I. Make f (x) even on (L, L). i.e. f (x) = f ( x) x (L, L), to get Fourier cosine expansion. II. Make f (x) odd on (L, L). i.e. f ( x) = f (x) x (L, L), to get Fourier sine expansion. Fourier cosine expansion L Now a0 = L 2 L 1 2k 1 2k 1 x dx + L x dx f ( x) dx = L0 L L0 L L L 2 L L k k k k 1 2k 1 = 2 x 2 2 + Lx x 2 = + = L 2 4 4 L L 2 L 0 2 Prepared by Tekleyohannes Negussie 167 Unit VIII Fourier Series L L 1 n 2 n f ( x) cos x dx = f ( x) cos x dx an = L L L L L 0 L 2 L n 4k n x dx + 2 ( L x) cos = 2 x cos x dx L L L 0 L L 4k 2 L xL n L2 n sin x cos x) 2 = 2 ( 2 L L L n (n ) 4k 0 L L2 n xL n L2 n 2 ( sin x sin x cos x) L 2 L n L L L n (n ) 4k 2 = n 4k cos n 1 2 cos 2 n 2 Thus an = 8k n cos cos 2 n 2 2 n n N. 2 Therefore f (x) = k 8k n n 2 n + cos cos cos x. 2 2 2 2 L n n 1 II. Fourier sine expansion. L Now an = L 1 n 2 n f ( x) sin x dx = f ( x) sin x dx L L L L L 0 L 2 L n 4k n x dx + 2 ( L x) sin = 2 x sin x dx L L L 0 L L 4k 2 L xL n L2 n cos x sin x) 2 = 2 ( 2 n L L L (n ) 4k 0 L L2 n xL n L2 n 2 ( cos x cos x sin x) L 2 n L n L L L (n ) 4k 2 Prepared by Tekleyohannes Negussie 168 Unit VIII = Fourier Series 8k n 2 sin 2 n Thus bn = 8k n 2 Therefore f (x) = sin n n N. 2 n n sin sin x. 2 L 2 n 1 n2 8k 1 lim f ( x) Since f (x) is piecewise continuous and x L lim f ( x ) L = k. = x 2 0 2 n 2 n = k, but sin sin 2 2 2 n 1 n2 1 8k 1 2 if n is even if n is odd 2 Thus = . 2 8 n 1 2n 1 1 Therefore n0 2 = . 8 2n 12 1 Exercises 1.5 1. Represent each of the following functions by a Fourier sine and cosine series extensions, where L + . i) f (x) = x, 0 x L. 0 x L. ii) f (x) = x2, iii) f (x) = sin x, L 0 x L. 2. Obtain a half-range sine and cosine series for i) ex in 0 x 1. ii) x2 x in 0 x . Prepared by Tekleyohannes Negussie 169