Fourier Series for Functions with Period 2L Suppose that we have a periodic function f (t) with arbitrary period P = 2L, generalizing the special case P = 2π which we have already seen. Then a simple re-scaling of the interval (−π, π ) to (− L, L) allows us to write down the general Fourier series and Fourier coefﬁcent formulas: f (t) = � π � � π � ∞ a0 + ∑ an cos n t + bn sin n t 2 L L n =1 (1) with Fourier coefﬁcients given by the general Fourier coefﬁcent formulas 1 L a0 = f (t) dt, L −L � 1 L π an = f (t) cos(n t) dt, L −L L � 1 L π bn = f (t) sin(n t) dt. L −L L � Note: The number L = P 2 (2) is called the half-period. Example. Let f (t) be the period 2 function, which is deﬁned on the window [−1, 1) by f (t) = |t|. Compute the Fourier series of f (t). The graph of f (t) below shows why this function is called either a tri angle wave or a continuous sawtooth function. −2 2 −1 1 Figure 1: The period 2 triangle wave. Solution. In this case the period is P = 2, so the half-period L = 1. This means n πL = nπ and we compute the coefﬁcients from the formulas in (2), using integration by parts, as follows. For n �= 0: an = 1 1 � 1 −1 � = 2 |t| cos(nπt) dt = 2 � 1 0 �1 � t cos(nπt) dt t sin(nπt) cos(nπt) � 2 + = 2 2 ((−1)n − 1) = nπ n2 π 2 �0 n π � − n24π2 0 for n odd for n even Fourier Series for Functions with Period 2L OCW 18.03SC and for n = 0: � � 1 1 1 a0 = |t| dt = 2 t dt = 1 1 −1 0 Since f (t) is an even function and sin(nπt) is odd, the sine coefﬁcients bn = 0. (We will justify this carefully in the next session. For now you can compute the integrals for bn as an exercise and verify it in this case.) Thus, the Fourier series for f (t) is � � 1 4 cos 3πt cos 5πt 1 4 f (t) = − 2 cos πt + + +··· = − 2 2 2 2 π 3 5 2 π cos(nπt) . n2 n odd ∑