高等数学

设 $f\left( x\right)$ 是以 ${2\pi }$ 为周期的可积函数,其傅里叶系数为 ${a}_{0},{a}_{n},{b}_{n}$ ,记 ${S}_{0}\left( x\right) = \frac{{a}_{0}}{2}$ , ${S}_{n}\left( x\right) = \frac{{a}_{0}}{2} + \mathop{\sum }\limits_{k = 1}^{n}\left( {{a}_{k}\cos {kx} + {b}_{k}\sin {kx}}\right) ,{\sigma }_{n}\left( x\right) = \frac{1}{n}\mathop{\sum }\limits_{k = 0}^{n - 1}{S}_{k}\left( x\right)$ . 证明:
(1) ${S}_{n}\left( x\right) = \frac{1}{2\pi }{\int }_{-\pi }^{\pi }f\left( {x + t}\right) \frac{\sin \left( {n + 1/2}\right) t}{\sin \left( {t/2}\right) }\mathrm{d}t$
(2) ${\sigma }_{n}\left( x\right) = \frac{1}{2n\pi }{\int }_{-\pi }^{\pi }f\left( {x + t}\right) {\left[ \frac{\sin \left( {{nt}/2}\right) }{\sin \left( {t/2}\right) }\right] }^{2}\mathrm{\;d}t$ .

设 $f\left( x\right)$ 与 ${f}^{2}\left( x\right)$ 在 $\left[ {-\pi ,\pi }\right]$ 上可积, ${a}_{n},{b}_{n}$ 是 $f\left( x\right)$ 在 $\left[ {-\pi ,\pi }\right]$ 上的傅里叶系数.
(1) 记 ${S}_{n}\left( x\right) = \frac{{a}_{0}}{2} + \mathop{\sum }\limits_{k = 1}^{n}\left( {{a}_{k}\cos {kx} + {b}_{k}\sin {kx}}\right)$ ，证明 $\mathop{\max }\limits_{{x \in \left[ {-\pi ,\pi }\right] }}\left| {{S}_{n}^{\prime }\left( x\right) }\right| \leq n\sqrt{2n}\mathop{\max }\limits_{{x \in \left[ {-\pi ,\pi }\right] }}\left| {{S}_{n}\left( x\right) }\right|$
(2) 证明贝塞尔(Bessel)不等式: $\frac{1}{\pi }{\int }_{-\pi }^{\pi }{f}^{2}\left( x\right) \mathrm{d}x \geq \frac{{a}_{0}{}^{2}}{2} + \mathop{\sum }\limits_{n = 1}^{\infty }\left( {{a}_{n}^{2} + {b}_{n}^{2}}\right)$ .

设 $f\left( x\right)$ 是 $\left( {-\infty , + \infty }\right)$ 上以 ${2\pi }$ 为周期的具有二阶连续导数的函数. 记 ${b}_{n} = \frac{1}{\pi }{\int }_{-\pi }^{\pi }f\left( x\right) \sin {nx}\mathrm{\;d}x$ , ${b}_{n}^{\prime\prime} = \frac{1}{\pi }{\int }_{-\pi }^{\pi }{f}^{\prime \prime }\left( x\right) \sin {nx}\mathrm{\;d}x$ . 证明若 $\mathop{\sum }\limits_{n = 1}^{\infty }{b}_{n}^{\prime\prime}$ 绝对收敛,则 $\mathop{\sum }\limits_{n = 1}^{\infty }\sqrt{\left| {b}_{n}\right| } < \frac{1}{2}\left( {2 + \mathop{\sum }\limits_{n = 1}^{\infty }\left| {b}_{n}^{\prime\prime}\right| }\right)$ .

将函数 $f\left( x\right) = \sec x$ 在区间 $\left[ {-\frac{\pi }{4},\frac{\pi }{4}}\right]$ 上展开为傅里叶级数.