JIA-MING (FRANK) LIOU

1. Introduction A trigonometric is a finite sum n a0 X (1.1) t(x) = + {a cos kx + b sin kx}, 2 k k k=1 where a0, a1, ··· , an and b1, ··· , bn are real numbers. A trigonometric polynomial defines a continuous periodic of period 2π on R, i.e. t(x) is a on R with t(x + 2π) = t(x) for all x ∈ R. For each pair of integer (k, m), we set ( 1 if k = m, δk,m = 0 otherwise. Lemma 1.1. Let k, m be nonnegative integers. For all k, m Z 2π (1) cos kx cos mxdx = δk,mπ, 0 Z 2π (2) sin kx sin mxdx = δk,mπ, 0 Z 2π (3) cos kx sin mxdx = 0. 0 Proof. We leave it to the reader as an exercise.  Given a trigonometric polynomial t(x) of the form (1.1), using the fact that, Z 2π Z 2π cos kxdx = sin kxdx = 0, k ≥ 1, 0 0 we obtain Z 2π Z 2π n Z 2π Z 2π a0 X t(x)dx = dx + {a cos kxdx + b sin kxdx} 2 k k 0 0 k=1 0 0 = a0π. This implies that 1 Z 2π a0 = t(x)dx. π 0 We can also compute Z 2π Z 2π n Z 2π Z 2π a0 X t(x) cos mxdx = cos mxdx + {a cos kx cos mxdx + b sin kx cos mxdx} 2 k k 0 0 k=1 0 0 = amπ 1 2 JIA-MING (FRANK) LIOU and Z 2π Z 2π n Z 2π Z 2π a0 X t(x) sin mxdx = sin mxdx + {a cos kx sin mxdx + b sin kx sin mxdx} 2 k k 0 0 k=1 0 0 = bmπ. These give us 1 Z 2π 1 Z 2π am = t(x) cos mxdx, bm = t(x) sin mxdx. π 0 π 0 Fourier thought that all the (continuous) period functions are infinite sum of trigonometric functions: ∞ a0 X (1.2) + {a cos nx + b sin nx}. 2 n n n=1

Using the similar observation, he defined the Fourier coefficients a0, a1, ··· , and b1, b2, ··· of a y = f(x) by 1 Z 2π (1) a0 = f(x)dx, π 0 1 Z 2π (2) an = f(x) cos nxdx, π 0 1 Z 2π (3) bn = f(x) sin nxdx. π 0 The infinite series (1.2) is called the Fourier expansion of the function f(x); we also use the notation ∞ a0 X f(x) ∼ + {a cos nx + b sin nx} 2 n n n=1 to say that (1.2) is the Fourier series for y = f(x). Example 1.1. Let y = f(x) be a periodic function of period 2π so that f(x) = x on [0, 2π]. Then the Fourier series is given by ∞ 2π X  2  (1.3) x ∼ + − sin nx. 2 n n=1 Now, let us consider another trigonometric polynomial n α0 X s(x) = + {α cos jx + β sin jx}. 2 j j j=1 We compute the integral Z 2π Z 2π n Z 2π n Z 2π a0α0 X a0αj X a0βj t(x)s(x)dx = dx + cos jxdx + sin jxdx 4 2 2 0 0 j=1 0 j=1 0 n Z 2π n Z 2π X α0ak X α0bk + cos kxdx + sin kxdx 2 2 k=1 0 k=1 0 n n X X Z 2π + (ak cos kx + bk sin kx)(αj cos jx + βj sin jx)dx k=1 j=1 0 FOURIER SERIES 3

n n Z 2π n n Z 2π a0α0π X X X X = + a α cos kx cos jxdx + a β cos kx sin jxdx 2 k j k j k=1 j=1 0 k=1 j=1 0 n n n n X X Z 2π X X Z 2π + bkαj sin kx cos jxdx + bkβj sin kx sin jxdx k=1 j=1 0 k=1 j=1 0 n a0α0π X = + π {a α + b β }. 2 k k k k k=1 This shows that Z 2π n 1 a0α0 X t(x)s(x)dx = + {a α + b β }. π 2 k k k k 0 k=1 When t(x) = s(x), we obtain n 1 Z 2π a2 X t(x)2dx = 0 + {a2 + b2}. π 2 k k 0 k=1 These observation motivates the following theorem. Theorem 1.1. (Parseval Identity) Let us assume that f(x) and g(x) are periodic (continu- ous) functions of period 2π. Suppose that the Fourier expansions of f(x) and g(x) are given by ∞ ∞ a0 X α0 X f(x) ∼ + {a cos nx + b sin nx}, g(x) ∼ + {α cos nx + β sin nx}. 2 n n 2 n n n=1 n=1 Then Z 2π ∞ 1 a0α0 X f(x)g(x)dx = + {a α + b β }. π 2 n n n n 0 n=1 This also leads to Corollary 1.1. Let f(x) be as above. Then ∞ 1 Z 2π a2 X f(x)2dx = 0 + {a2 + b2 }. π 2 n n 0 n=1 Let s be a real number and define ∞ X 1 ζ(s) = . ns n=1 By p-test, the infinite series ζ(s) converges when s > 1. The function ζ(s) is called the Riemann zeta function. By (1.3) and the Parseval identity, we obtain ∞ 2 1 Z 2π (2π)2 X  2  x2dx = + − . π 2 n 0 n=1 Z 2π 8π3 The identity x2dx = implies that 0 3 ∞ 8π2 X 1 = 2π2 + 4 . 3 n2 n=1 4 JIA-MING (FRANK) LIOU

This shows that X 1 π2 = . n2 6 n=1 π2 In other words, ζ(2) = . 6 Example 1.2. The Fourier expansion of f(x) = x2, 0 ≤ x ≤ 2π, is given by 8π2 ∞ X  4   4π   x2 ∼ 3 + − cos nx + − sin nx . 2 n2 n n=1 Using the Parseval identity, we obtain

 2 2 2π 8π ∞ ∞ 1 Z 3 X 16 X 16π2 x4dx = + + . π 2 n4 n2 0 n=1 n=1 π2 Using the above identity and ζ(2) = , we obtain ζ(4) : 6 ∞ X 1 π4 = . n4 90 n=1 E-mail address: [email protected]