📝 题目
例 6 将函数 $f\left( x\right) = {x}^{2}\left( {-\pi \leq x \leq \pi }\right)$ 展开为傅氏级数,并求级数 $\displaystyle{\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{{n}^{4}}}$ 的和.
💡 答案与解析
解 因为 $f\left( x\right)$ 是偶函数,所以 ${b}_{n} = 0$ ,且
$$ {a}_{0} = \frac{2}{\pi }{\int }_{0}^{\pi }{x}^{2}\mathrm{\;d}x = \frac{2}{3}{\pi }^{2}, $$
$$ {a}_{n} = \frac{2}{\pi }{\int }_{0}^{\pi }{x}^{2}\cos {nx}\mathrm{\;d}x = {\left( -1\right) }^{n}\frac{4}{{n}^{2}}, $$
即得
$$ \frac{{\pi }^{2}}{3} + 4\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( -1\right) }^{n}}{{n}^{2}}\cos {nx} = {x}^{2}\;\left( {-\pi \leq x \leq \pi }\right) . $$
由封闭性公式, 有
$$ \frac{1}{\pi }{\int }_{-\pi }^{\pi }{x}^{4}\mathrm{\;d}x = \frac{1}{2}{\left( \frac{2{\pi }^{2}}{3}\right) }^{2} + \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{16}{{n}^{4}}, $$
由此解得
$$ \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{{n}^{4}} = \frac{{\pi }^{4}}{90} $$