方企勤 第五章 多元函数微分学 第1题
📝 题目
例 1 设 $f\left( x\right)$ 是周期为 ${2\pi }$ 的连续函数,令
$$ F\left( x\right) = \frac{1}{2h}{\int }_{x - h}^{x + h}f\left( t\right) \mathrm{d}t, $$
试求 $F\left( x\right)$ 的傅里叶系数.
💡 答案解析
解 由
$$ F\left( x\right) = \frac{1}{2h}{\int }_{x - h}^{x + h}f\left( t\right) \mathrm{d}t = \frac{1}{2h}{\int }_{-h}^{h}f\left( {x + y}\right) \mathrm{d}y $$
可看出 $F\left( x\right)$ 也是周期为 ${2\pi }$ 的连续函数. 记 $F\left( x\right)$ 的傅氏系数为 ${A}_{n}$ , ${B}_{n},f\left( x\right)$ 的傅氏系数为 ${a}_{n},{b}_{n}$ ,则
$$ {A}_{0} = \frac{1}{\pi }{\int }_{-\pi }^{\pi }F\left( x\right) \mathrm{d}x = \frac{1}{2\pi h}{\int }_{-\pi }^{\pi }\left\lbrack {{\int }_{-h}^{h}f\left( {x + y}\right) \mathrm{d}y}\right\rbrack \mathrm{d}x $$
$$ = \frac{1}{2\pi h}{\int }_{-h}^{h}{\int }_{-\pi }^{\pi }f\left( {x + y}\right) \mathrm{d}x\mathrm{\;d}y $$
$$ = \frac{1}{2\pi h}{\int }_{-h}^{h}{\int }_{y - \pi }^{y + \pi }f\left( t\right) \mathrm{d}t\mathrm{\;d}y = \frac{1}{2\pi h}{\int }_{-h}^{h}{\int }_{-\pi }^{\pi }f\left( t\right) \mathrm{d}t\mathrm{\;d}y $$
$$ = \frac{1}{2h}{\int }_{-h}^{h}{a}_{0}\mathrm{\;d}y = {a}_{0}, $$
$$ {A}_{n} = \frac{1}{\pi }{\int }_{-\pi }^{\pi }F\left( x\right) \cos {nx}\mathrm{\;d}x $$
$$ = \frac{1}{2\pi h}{\int }_{-\pi }^{\pi }{\int }_{-h}^{h}f\left( {x + y}\right) \cos {nx}\mathrm{\;d}y\mathrm{\;d}x $$
$$ = \frac{1}{2\pi h}{\int }_{-h}^{h}{\int }_{-\pi }^{\pi }f\left( {x + y}\right) \cos {nx}\mathrm{\;d}x\mathrm{\;d}y $$
$$ = \frac{1}{2\pi h}{\int }_{-h}^{h}{\int }_{-\pi }^{\pi }f\left( t\right) \cos n\left( {t - y}\right) \mathrm{d}t\mathrm{\;d}y $$
$$ = \frac{1}{2h}{\int }_{-h}^{h}\left( {{a}_{n}\cos {ny} + {b}_{n}\sin {ny}}\right) \mathrm{d}y = \frac{\sin {nh}}{nh}{a}_{n}. $$
同理可证 ${B}_{n} = \frac{\sin {nh}}{nh}{b}_{n}$ .
评注 在傅里叶级数习题中, 我们用傅氏级数一致收敛定理, 得出 $F\left( x\right)$ 的傅氏级数一致收敛于 $F\left( x\right)$ . 利用此题,我们又可得出 $F\left( x\right)$ 的傅氏级数一致收敛的另一证明.