Fourier Series Expansions of Even and Odd Functions. Half-Range Fourier Series Expansions Dr. Kamlesh Jangid Department of HEAS (Mathematics) Rajasthan Technical University, Kota-324010, India E-mail: [email protected] Dr. Kamlesh Jangid (RTU Kota) Fourier series 1 / 18 Outline Outline 1 Fourier Series Expansion of Even and Odd Functions 2 Fourier Half-Range Series Dr. Kamlesh Jangid (RTU Kota) Fourier series 2 / 18 Outline Recall (i) Let f (x) be periodic with period 2l and piecewise continuous in the interval [−l; l]. Then the Fourier series of f (x) is given by 1 X h nπ nπ i f (x) = a + a cos x + b sin x 0 n l n l n=1 where 1 R l 9 a0 = f (x) dx; > 2l −l > => 1 R l nπ an = f (x) cos x dx; n = 1; 2; 3; ::: l −l l > > 1 R l nπ ;> bn = l −l f (x) sin l x dx; n = 1; 2; 3; ::: Dr. Kamlesh Jangid (RTU Kota) Fourier series 3 / 18 Outline (ii) Let f (x) be periodic with period 2π and piecewise continuous in the interval [−π; π]. Then the Fourier series of f (x) is given by 1 X f (x) = a0 + (an cos nx + bn sin nx) n=1 with coefficients 9 a = 1 R π f (x) dx; > 0 2π −π > > => 1 R π an = π −π f (x) cos nx dx; n = 1; 2; 3; ::: > > 1 R π > bn = π −π f (x) sin nx dx; n = 1; 2; 3; ::: ; Dr. Kamlesh Jangid (RTU Kota) Fourier series 4 / 18 Fourier Series Expansion of Even and Odd Functions Fourier Series Expansion of Even and Odd Functions Let f (x) be a function defined in [−l; l]. Then, f (x) is an even function on [−l; l] if f (−x) = f (x); −l ≤ x ≤ l: (1) The function f (x) is odd if f (−x) = −f (x); −l ≤ x ≤ l: (2) Example (i) x2n, cos(nπx=l) are even functions on [−l; l]. (ii) x2n+1, sin(nπx=l) are odd functions on [−l; l]. Dr. Kamlesh Jangid (RTU Kota) Fourier series 5 / 18 Fourier Series Expansion of Even and Odd Functions Remark R l R l (i) −l f (x)dx = 2 0 f (x)dx, if f (x) is even. R l (ii) −l f (x)dx = 0, if f (x) is odd. If f (x) is an even function on [−l; l], then we have the following series 1 X nπx f (x) = a + a cos (3) 0 n l n=1 where, 9 1 R l 1 R l > a0 = 2l −l f (x)dx = l 0 f (x)dx; = (4) 1 R l nπx 2 R l nπx > an = l −l f (x) cos l dx = l 0 f (x) cos l dx: ; Dr. Kamlesh Jangid (RTU Kota) Fourier series 6 / 18 Fourier Series Expansion of Even and Odd Functions If f (x) is an odd function on [−l; l], then we have the following series 1 X nπx f (x) = b sin (5) n l n=1 where, 2 Z l nπx bn = f (x) sin dx: (6) l 0 l Dr. Kamlesh Jangid (RTU Kota) Fourier series 7 / 18 Fourier Series Expansion of Even and Odd Functions Example Find the Fourier series expansion of the periodic function f (x) = x, −π ≤ x ≤ π, f (x + 2π) = f (x). Solution Since, given function is an odd function, therefore a0 = an = 0. Hence the Fourier series is given by 1 1 X nπx X f (x) = b sin = b sin(nx) n π n n=1 n=1 where the Fourier coefficient bn is given as follows: 2 Z π bn = x sin(nx)dx π 0 Dr. Kamlesh Jangid (RTU Kota) Fourier series 8 / 18 Fourier Series Expansion of Even and Odd Functions 2 cos(nx) sin(nx)π 2 −π cos(nπ) 2 b = −x + = = (− )n+1: n 2 1 π n n 0 π n n Therefore, the Fourier series expansion of the given function is given by, sin 2x sin 3x sin 4x f (x) = x = 2 sin x − + − + ··· 2 3 4 Example Find the Fourier series expansion of f (x) = x2, −2 ≤ x ≤ 2. Dr. Kamlesh Jangid (RTU Kota) Fourier series 9 / 18 Fourier Series Expansion of Even and Odd Functions Solution 2 The given function f (x) = x is an even function, therefore bn = 0. Hence the Fourier series is given by 1 X nπx f (x) = a + a cos 0 n 2 n=1 where the Fourier coefficients a0 and an are given as follows: Z 2 1 2 4 a0 = x dx = 2 0 3 Z 2 2 2 an = x cos(nπx=2)dx 2 0 sin(nπx=2)2 Z 2 sin(nπx=2) = x2 − 2 x dx nπ=2 0 0 nπ=2 Dr. Kamlesh Jangid (RTU Kota) Fourier series 10 / 18 Fourier Series Expansion of Even and Odd Functions −4 cos(nπx=2) sin(nπx=2)2 a = −x + n 2 nπ (nπ=2) (nπ=2) 0 16 16(−1)n = cos(nπ) = n2π2 n2π2 Therefore, the Fourier series is given by 1 4 16 X (−1)n f (x) = + cos(nπx=2): 3 π2 n2 n=1 Dr. Kamlesh Jangid (RTU Kota) Fourier series 11 / 18 Fourier Series Expansion of Even and Odd Functions Example Find the Fourier series expansion of f (x) = x2, −2 ≤ x ≤ 2. Hence, show that 1 1 1 π2 (i) + + + ··· = . 12 22 32 6 1 1 1 1 π2 (ii) − + − + ··· = . 12 22 32 42 12 Solution The Fourier series expansion of f (x) is given by (see previous example) 1 4 16 X (−1)n f (x) = + cos(nπx=2): (7) 3 π2 n2 n=1 Dr. Kamlesh Jangid (RTU Kota) Fourier series 12 / 18 Fourier Series Expansion of Even and Odd Functions Note: At the end points of the interval [−l; l], the Fourier series 1 converges to [f (−l+) + f (l−)]. 2 Therefore at x = 2, the Fourier series converges to 1 1 [f (−2+) + (2−))] = [4 + 4] = 4: 2 2 (i) Substituting x = 2 in (7), we obtain 1 1 4 16 X (−1)n 4 16 X 1 4 = + cos(nπ) = + : 3 π2 n2 3 π2 n2 n=1 n=1 Therefore 1 1 1 π2 + + + ··· = : 12 22 32 6 Dr. Kamlesh Jangid (RTU Kota) Fourier series 13 / 18 Fourier Series Expansion of Even and Odd Functions (ii) At x = 0, the given function is continuous. The series converges to f (0) = 0. Therefore, 1 4 16 X (−1)n 0 = + 3 π2 n2 n=1 4 16 1 1 1 1 ) = − + − + ··· 3 π2 12 22 32 42 1 1 1 1 π2 ) − + − + ··· = 12 22 32 42 12 Dr. Kamlesh Jangid (RTU Kota) Fourier series 14 / 18 Fourier Half-Range Series Fourier Half-Range Series Suppose that a periodic function f (x) of period 2l is defined on a half-interval [0; l]. It is possible to extend the definition of f (x) to the other half [−l; 0] of the interval [−l; l] so that f (x) is either an even or an odd function. Hence, f (x) has either Fourier cosine or sine series, depends upon requirements of a particular problem. Dr. Kamlesh Jangid (RTU Kota) Fourier series 15 / 18 Fourier Half-Range Series Example Find the Fourier cosine series of the function 8 < x2; if 0 ≤ x ≤ 2 f (x) = : 4; if 2 ≤ x ≤ 4: Solution As per the requirement of the problem, we need to extend the function f (x) to the other half [−4; 0] of the interval [−4; 4] so that f (x) is an even function. Therefore, the Fourier cosine series of f (x) is given by 1 X nπx f (x) = a + a cos 0 n 4 n=1 Dr. Kamlesh Jangid (RTU Kota) Fourier series 16 / 18 Fourier Half-Range Series where the Fourier coefficient a0 and an are given as follows: 1 Z 4 1 Z 2 Z 4 8 a0 = f (x)dx = f (x)dx + f (x)dx = 4 0 4 0 2 3 and 2 Z 4 nπx an = f (x) cos dx 4 0 4 1 Z 2 nπx Z 4 nπx = x2 cos( )dx + 4 cos( )dx 2 0 4 2 4 32 nπ 2 nπ = cos( ) − sin( ) : n2π2 2 nπ 2 Therefore, the Fourier cosine series is 1 8 32 X 1 nπ 2 nπ nπx f (x) = + cos( ) − sin( ) cos( ): 3 π2 n2 2 nπ 2 4 n=1 Dr. Kamlesh Jangid (RTU Kota) Fourier series 17 / 18 Fourier Half-Range Series For the video lecture use the following link https://youtube.com/channel/ UCk9ICMqdkO0GREITx-2UaEw THANK YOU Dr. Kamlesh Jangid (RTU Kota) Fourier series 18 / 18.