📝 题目
例 3 设 $\alpha$ 不是整数,试将函数
$$ f\left( t\right) = \cos {\alpha t},\;t \in \left\lbrack {-\pi ,\pi }\right\rbrack $$
展开成傅里叶级数.
💡 答案与解析
解 我们扩充这函数的定义,使它成为一个周期为 ${2\pi }$ 的函数. 扩充后的函数记为 $\widetilde{f}\left( t\right)$ . 因为 $\widetilde{f}\left( t\right)$ 是偶函数,所以它的傅里叶级数只含有余弦部分. 计算系数得:
$$ {a}_{0} = \frac{2}{\pi }{\int }_{0}^{\pi }\cos {\alpha t}\mathrm{\;d}t = \frac{2\sin {\alpha \pi }}{\alpha \pi }, $$
$$ {a}_{n} = \frac{2}{\pi }{\int }_{0}^{\pi }\cos {\alpha t} \cdot \cos {nt}\mathrm{\;d}t $$
$$ = \frac{1}{\pi }{\int }_{0}^{\pi }\left\lbrack {\cos \left( {\alpha - n}\right) t + \cos \left( {\alpha + n}\right) t}\right\rbrack \mathrm{d}t $$
$$ = \frac{1}{\pi }\left\lbrack {\frac{\sin \left( {\alpha - n}\right) \pi }{\alpha - n} + \frac{\sin \left( {\alpha + n}\right) \pi }{\alpha + n}}\right\rbrack $$
$$ = \frac{{\left( -1\right) }^{n}}{\pi }\sin {\alpha \pi }\left\lbrack {\frac{1}{\alpha - n} + \frac{1}{\alpha + n}}\right\rbrack $$
$$ = {\left( -1\right) }^{n}\frac{{2\alpha }\sin {\alpha \pi }}{\pi \left( {{\alpha }^{2} - {n}^{2}}\right) },\;n = 1,2,\cdots . $$
我们得到这函数的傅里叶展式:
$$ \widetilde{f}\left( t\right) = \frac{\sin {\alpha \pi }}{\pi }\left( {\frac{1}{\alpha } + \mathop{\sum }\limits_{{n = 1}}^{{+\infty }}{\left( -1\right) }^{n}\frac{{2\alpha }\cos {nt}}{{\alpha }^{2} - {n}^{2}}}\right) . $$
限制在 $\left\lbrack {-\pi ,\pi }\right\rbrack$ 上就得到:
$$ \cos {\alpha t} = \frac{\sin {\alpha \pi }}{\pi }\left( {\frac{1}{\alpha } + \mathop{\sum }\limits_{{n = 1}}^{{+\infty }}{\left( -1\right) }^{n}\frac{{2\alpha }\cos {nt}}{{\alpha }^{2} - {n}^{2}}}\right) , \tag{3.7} $$
$$ t \in \left\lbrack {-\pi ,\pi }\right\rbrack \text{ . } $$