傅立叶级数的推导

傅立叶级数的核心思想是把一个周期函数分解成若干正余弦波的叠加。

推导目标

对于一个周期为 \(2\pi\) 的函数 \(f(t)\)，它可以表示成如下形式：

\[f(t) =a_0+ \sum_{k=1}^\infty a_k\cdot sin(kt) + \sum_{k=1}^\infty b_k \cdot cos(kt) \]

其中，

\[a_0 = \frac{1}{2\pi} \int_{-\pi}^\pi f(t)dt \\ a_k =\frac 1\pi \int_{-\pi}^\pi f(t)sin(kt) dt \\ b_k = \frac{1}{\pi} \int_{-\pi}^\pi f(t)cos(kt) dt \]

推导过程

几个预备公式

\(\int_{-\pi}^\pi sin(nt)dt = \int_{-\pi}^\pi cos(nt)dt = 0\)

\(\int_{-\pi}^\pi sin(mt)sin(nt)dt = \int_{-\pi}^\pi cos(mt)cos(nt)dt = 0,\ \forall m, n \in \{1,2,\cdots\} ,n\neq m\)

\(\int_{-\pi}^\pi sin(nt)sin(nt)dt = \int_{-\pi}^\pi cos(nt)cos(nt)dt = \pi,\ \forall n \in \{1,2,\cdots\}\)

\(\int_{-\pi}^\pi sin(mt)cos(nt)dt = 0,\ \forall m,n\in N\)

设 \(f(t) = a_0 + \sum_{k}a_k \cdot sin(kt) + \sum_k b_k \cdot cos(kt)\)

同时对两边求 \(-\pi \rightarrow \pi\) 的定积分：

\(\int_{-\pi}^\pi f(t) dt = \int_{-\pi}^\pi a_0 dt + \sum_k a_k \cdot \int_{-\pi}^\pi sin(kt) dt + \sum_k b_k \cdot \int_{-\pi}^\pi cos(kt) dt=\int_{-\pi}^\pi a_0 dt\)

\(a_0 = \frac{1}{2\pi} \int_{-\pi}^\pi f(t)dt\)

同时在两边乘以 \(sin(t)\)，并求 \(-\pi \rightarrow \pi\) 的定积分：

\(\int_{-\pi}^\pi f(t)sin(t) dt = \int_{-\pi}^\pi a_0sin(t)dt + \sum_{k}a_k \cdot \int_{-\pi}^\pi sin(kt)sin(t)dt + \sum_k b_k \cdot \int_{-\pi}^\pi cos(kt)sin(t)dt=a_1 \cdot \pi\)

\(a_1 = \frac 1\pi \int_{-\pi}^\pi f(t)sin(t) dt\)

同理，乘以 \(sin(2t), \ sin(3t), \cdots\) 可得：

\(a_k = \frac 1\pi \int_{-\pi}^\pi f(t)sin(kt) dt\)

同理，乘以 \(cos(t),\ cos(2t), \ cos(3t), \cdots\)，可得：

\(b_k = \frac{1}{\pi} \int_{-\pi}^\pi f(t)cos(kt) dt\)

推导完毕。