📝 题目
例 4 设 $0 < a < 1$ ,将函数 $f\left( x\right) = \cos {ax}\left( {\left| x\right| < \pi }\right)$ 展开为傅氏级数.
💡 答案与解析
解 因为 $f\left( x\right)$ 是偶函数,所以 ${b}_{n} = 0$ ,且
$$ {a}_{0} = \frac{2}{\pi }{\int }_{0}^{\pi }\cos \left( {ax}\right) \mathrm{d}x = \frac{2}{\pi a}\sin \left( {a\pi }\right) , $$
$$ {a}_{n} = \frac{2}{\pi }{\int }_{0}^{\pi }\cos \left( {ax}\right) \cos \left( {nx}\right) \mathrm{d}x $$
$$ = \frac{1}{\pi }{\int }_{0}^{\pi }\left\lbrack {\cos \left( {a - n}\right) x + \cos \left( {a + n}\right) x}\right\rbrack \mathrm{d}x $$
$$ = \frac{1}{\pi }{\left\lbrack \frac{\sin \left( {a - n}\right) x}{a - n} + \frac{\sin \left( {a + n}\right) x}{a + n}\right\rbrack }_{0}^{\pi } $$
$$ = {\left( -1\right) }^{n}\frac{{2a}\sin \left( {a\pi }\right) }{\pi \left( {{a}^{2} - {n}^{2}}\right) }, $$
即得
$$ \frac{\sin {a\pi }}{\pi }\left\lbrack {\frac{1}{a} + \mathop{\sum }\limits_{{n = 1}}^{\infty }{\left( -1\right) }^{n}\frac{2a}{{a}^{2} - {n}^{2}}\cos {nx}}\right\rbrack = \cos {ax}\;\left( {\left| x\right| \leq \pi }\right) . $$