📝 题目
例 3 将函数 $f\left( x\right) = {x}^{2}\left( {0 < x < \pi }\right)$ ,按如下要求展开为傅氏级数:
(1)按余弦展开； (2)按正弦展开.
💡 答案与解析
解 (1) 将 $f\left( x\right) = {x}^{2}\left( {0 < x < \pi }\right)$ 进行偶开拓,也就是考虑 $f\left( x\right)$ $= {x}^{2}\left( {-\pi < x < \pi }\right)$ 的傅氏展开. 这时 ${b}_{n} = 0\left( {n = 1,2,\cdots }\right)$ ,且
$$ {a}_{0} = \frac{2}{\pi }{\int }_{0}^{\pi }{x}^{2}\mathrm{\;d}x = \frac{2}{3}{\pi }^{2}, $$
$$ {a}_{n} = \frac{2}{\pi }{\int }_{0}^{\pi }{x}^{2}\cos {nx}\mathrm{\;d}x = {\left. \frac{2}{n\pi }{x}^{2}\sin nx\right| }_{0}^{\pi } - \frac{2}{n\pi }{\int }_{0}^{\pi }{2x} \cdot \sin {nx}\mathrm{\;d}x $$
$$ = \frac{4}{{n}^{2}\pi }\left\lbrack {{\left. x\cos nx\right| }_{0}^{\pi } - {\int }_{0}^{\pi }\cos {nx}\mathrm{\;d}x}\right\rbrack = {\left( -1\right) }^{n}\frac{4}{{n}^{2}}, $$
即得
$$ \frac{{\pi }^{2}}{3} + 4\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( -1\right) }^{n}}{{n}^{2}}\cos {nx} = {x}^{2}\;\left( {0 \leq x \leq \pi }\right) . $$
(2)将 $f\left( x\right) = {x}^{2}\left( {0 < x < \pi }\right)$ 进行奇开拓,也就是考虑 $f\left( x\right) =$ $x\left| x\right| \left( {-\pi < x < \pi }\right)$ 的傅氏展开. 这时 ${a}_{n} = 0\left( {n = 0,1,2,\cdots }\right)$ ,且
$$ {b}_{n} = \frac{2}{\pi }{\int }_{0}^{\pi }{x}^{2}\sin {nx}\mathrm{\;d}x $$
$$ = - {\left. \frac{2}{n\pi }{x}^{2}\cos nx\right| }_{0}^{\pi } + \frac{4}{n\pi }{\int }_{0}^{\pi }x\cos {nx}\mathrm{\;d}x $$
$$ = {\left( -1\right) }^{n - 1}\frac{2\pi }{n} + \frac{4}{{n}^{2}\pi }\left\lbrack {{\left. x\sin nx\right| }_{0}^{\pi } - {\int }_{0}^{\pi }\sin {nx}\mathrm{\;d}x}\right\rbrack $$
$$ = {\left( -1\right) }^{n - 1}\frac{2\pi }{n} + \frac{4}{{n}^{3}\pi }\left\lbrack {{\left( -1\right) }^{n} - 1}\right\rbrack , $$
即得
$$ {2\pi }\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( -1\right) }^{n - 1}}{n}\sin {nx} - \frac{8}{\pi }\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{\sin \left( {{2n} - 1}\right) x}{{\left( 2n - 1\right) }^{3}} $$
$$ = \left\{ \begin{matrix} {x}^{2}, & 0 \leq x < \pi , \\ 0, & x = \pi . \end{matrix}\right. $$
说明 利用函数的傅氏展开式,对于 $\left( {0,\pi }\right)$ 上的同一函数,我们可以用不同的三角级数来表示. 事实上,在 $\left( {-\pi ,0}\right)$ 上的任意开拓 $f\left( x\right)$ ,只要保证逐段单调,所得到的傅氏级数展式同样在 $\left( {0,\pi }\right)$ 上收敛到 $f\left( x\right)$ .