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April 19, 2005 Lecturer Dmitri Zaitsev Hilary Term 2005 Course 2E1 2004-05 (SF Engineers & MSISS & MEMS) S h e e t 20 Due: in the tutorial sessions first Wednesday/Thursday in the next term Exercise 1 Find the Fourier series of the function f (x): 1 if − π < x < 0 (i) f (x) = 0 if 0 < x < π; Solution. 1 a0 = 2π 1 an = π Z 1 bn = π Z π −π 1 f (x) cosnx dx = π −π π Z 1 , 2 f (x) dx = Z 1 f (x) sinnx dx = π −π π 0 cosnx dx = 0, −π Z 0 sinnx dx = −π 0 −1 + cosnπ 1 −cosnx −1 + (−1)n = = , π n −π nπ nπ so we have the Fourier Series expansion: ∞ 1 X −1 + (−1)n f (x) = + sinnx. 2 n=1 nπ (ii) f (x) = x, −π < x < π; Solution. The function is odd (f (−x) = −f (x)), so we only need to calculate bn : 1 bn = π Z π 1 x sinnx dx = π −π Z π u dv, −π where u = x and dv = sinnx dx, hence v = − cosnnx . Thus 1 1 π bn = (uv)|−π − π π Z π Z 1 −xcosnx 1 π cosnx vdu = − dx = −π π n n −π −π −π π − 2(−1)n+1 2 cosnπ +0= , n n so we have the Fourier Series expansion: ∞ X 2(−1)n+1 sinnx. f (x) = n n=1 (iii) f (x) = n x if − π < x < 0 −x if 0 < x < π. Solution. Now the function is even (f (−x) = f (x)), so we only need to calculate an . Furthermore, if g(x) is even, we can simplify our calculation by the formula Rπ Rπ g(x) dx = 2 0 g(x) dx. We have −π 1 a0 = 2π Z 1 f (x) dx = 2 2π −π 1 an = π π Z π Z π 0 1 f (x) dx = π 1 f (x) cosnx dx = 2 π −π Z Z π x dx = 0 π . 2 π x cosnx dx = ... 0 The end of calculation is similar to the previous case using integration by parts and is omitted.