问题：设函数 f ( x ) = ∑ n = 0 ∞ cos ( n x ) 2 n f(x)=\sum_{n = 0}^{\infty} \frac{\cos (nx)}{2^n} f(x)=∑n=0∞2ncos(nx). 求 A = ∫ 0 2 π f 2 ( x ) d x A=\int_0^{2\pi} f^2(x) dx A=∫02πf2(x)dx.
解：把 f f f看成是某个周期 2 π 2\pi 2π的偶函数的Fourier展开，用Parseval等式 1 2 π ∫ 0 2 π f 2 ( x ) d x = a 0 2 + 1 2 ∑ n = 1 ∞ ( a n 2 + b n 2 ) . \frac{1}{2\pi} \int_0^{2\pi} f^2(x) dx = a_0^2 + \frac{1}{2}\sum_{n=1}^{\infty}(a_n^2 + b_n^2). 2π1∫02πf2(x)dx=a02+21n=1∑∞(an2+bn2).
此处 a 0 = 1 , a n = 2 − n , b n = 0 a_0=1, a_n = 2^{-n}, b_n=0 a0=1,an=2−n,bn=0. 容易得到 ∑ n = 1 ∞ a n 2 = ∑ n = 1 ∞ 4 − n = 1 3 \sum_{n=1}^{\infty} a_n^2 = \sum_{n=1}^{\infty} 4^{-n} = \frac{1}{3} ∑n=1∞an2=∑n=1∞4−n=31. 所以 A = 2 π ( 1 + 1 2 ⋅ 1 3 ) = 7 3 π A=2\pi(1+\frac{1}{2}\cdot \frac{1}{3})=\frac{7}{3}\pi A=2π(1+21⋅31)=37π.
另解：由于 e i n x = cos n x + i sin n x e^{inx}=\cos n x + i \sin nx einx=cosnx+isinnx, 所以 cos n x = e i n x + e − i n x 2 \cos nx = \frac{e^{inx} + e^{-inx}}{2} cosnx=2einx+e−inx. 由此求出 f ( x ) = R e ∑ n = 0 ∞ e i n x 2 n = 4 − 2 cos x 5 − 4 cos x f(x)=\mathrm{Re} \sum_{n=0}^{\infty} \frac{e^{inx}}{2^n} = \frac{4-2\cos x}{5-4\cos x} f(x)=Re∑n=0∞2neinx=5−4cosx4−2cosx. 再直接计算定积分即可求出 A A A.
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