题目

计算函数f（t）=t（0<=t<=2）的傅里叶系数

提问时间：2021-01-09

答案

a(0) = ∫[0,2] t dt = 2
a(n) = ∫[0,2] t * cos(nπt) dt = 1/(nπ) [ t * sin(nπt) | [0,2] - ∫[0,2] sin(nπt) dt ]
= 1/(nπ)² cos(nπt) | [0,2] = 0
b(n) = ∫[0,2] t * sin(nπt) dt = -1/(nπ) [ t * cos(nπt) | [0,2] - ∫[0,2] cos(nπt) dt ]
= -2/(nπ) + 1/(nπ)² sin(nπt) | [0,2] = -2 /(nπ)