📝 题目
解 $\displaystyle{a}_{0} = \frac{1}{\pi }{\int }_{0}^{2\pi }x\mathrm{\;d}x = {2\pi }}$ ,
$$ {a}_{n} = \frac{1}{\pi }{\int }_{0}^{2\pi }x\cos {nx}\mathrm{\;d}x = {\left. \frac{1}{n\pi }x\sin nx\right| }_{0}^{2\pi } - \frac{1}{n\pi }{\int }_{0}^{2\pi }\sin {nx}\mathrm{\;d}x $$
$$ = {\left. \frac{1}{{n}^{2}\pi }\cos nx\right| }_{0}^{2\pi } = 0, $$
$$ {b}_{n} = \frac{1}{\pi }{\int }_{0}^{2\pi }x\sin {nx}\mathrm{\;d}x = - {\left. \frac{1}{n\pi }x\cos nx\right| }_{0}^{2\pi } + \frac{1}{n\pi }{\int }_{0}^{2\pi }\cos {nx}\mathrm{\;d}x $$
$$ = - {\left. \frac{2}{n} + \frac{1}{{n}^{2}\pi }\sin nx\right| }_{0}^{2\pi } = - \frac{2}{n}. $$
因为 $f\left( x\right)$ 满足逐段单调条件,所以
$$ \pi - \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{2}{n}\sin {nx} = \left\{ \begin{array}{ll} x, & 0 < x < {2\pi }, \\ \pi , & x = 0,{2\pi }. \end{array}\right. $$
引申 由本题结果顺便可得到如下等式:
$$ \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{\sin {nx}}{n} = \frac{\pi - x}{2}\;\left( {0 < x < {2\pi }}\right) . $$
评注 前面两例中虽然函数表达式中都有 $x$ ,由于基本区间不同, 实际上是两个不同的周期函数.
💡 答案与解析
解 $\displaystyle{a}_{0} = \frac{1}{\pi }{\int }_{0}^{2\pi }x\mathrm{\;d}x = {2\pi }}$ ,
$$ {a}_{n} = \frac{1}{\pi }{\int }_{0}^{2\pi }x\cos {nx}\mathrm{\;d}x = {\left. \frac{1}{n\pi }x\sin nx\right| }_{0}^{2\pi } - \frac{1}{n\pi }{\int }_{0}^{2\pi }\sin {nx}\mathrm{\;d}x $$
$$ = {\left. \frac{1}{{n}^{2}\pi }\cos nx\right| }_{0}^{2\pi } = 0, $$
$$ {b}_{n} = \frac{1}{\pi }{\int }_{0}^{2\pi }x\sin {nx}\mathrm{\;d}x = - {\left. \frac{1}{n\pi }x\cos nx\right| }_{0}^{2\pi } + \frac{1}{n\pi }{\int }_{0}^{2\pi }\cos {nx}\mathrm{\;d}x $$
$$ = - {\left. \frac{2}{n} + \frac{1}{{n}^{2}\pi }\sin nx\right| }_{0}^{2\pi } = - \frac{2}{n}. $$
因为 $f\left( x\right)$ 满足逐段单调条件,所以
$$ \pi - \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{2}{n}\sin {nx} = \left\{ \begin{array}{ll} x, & 0 < x < {2\pi }, \\ \pi , & x = 0,{2\pi }. \end{array}\right. $$
引申 由本题结果顺便可得到如下等式:
$$ \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{\sin {nx}}{n} = \frac{\pi - x}{2}\;\left( {0 < x < {2\pi }}\right) . $$
评注 前面两例中虽然函数表达式中都有 $x$ ,由于基本区间不同, 实际上是两个不同的周期函数.